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2900	https://www.cnblogs.com/samaritan-z/p/8432227.html	数学 - 线性代数导论 - #9 Ax=b的解：存在性、解法、解的结构、解的数量 - Samaritan_z - 博客园会员众包新闻博问闪存赞助商 HarmonyOS Chat2DB 所有博客当前博客我的博客我的园子账号设置会员中心简洁模式 ...退出登录注册登录 Samaritan_z 博客园首页新随笔联系订阅管理随笔 - 29 文章 - 0 评论 - 0 阅读 - 37415 数学 - 线性代数导论 - #9 Ax=b的解：存在性、解法、解的结构、解的数量终于，我们在b为参数的一般情况下，开始分析Ax=b的解，包括标题中的四个方面。线性代数导论 - #9 Ax=b的解：存在性、解法、解的结构、解的数量终于，我们在b为参数的一般情况下，开始分析Ax=b的解，包括标题中的四个方面。首先是解的存在性。从几何上说，当且仅当向量b位于列空间C(A)内时，Ax=b有解；从代数上说，不能出现类似于“非0数=0”的矛盾方程： 1.这为我们判定是否有解提供了一个简便的途径：根据Gauss消元法中对A和b进行行变换的同步性，行的相同线性组合的值一定相同。所以假如A中各行可以通过简单的线性组合得到零行，而b进行相同线性组合的结果非0，则该方程组一定无解。 2.这为我们面对b为参数的一般情况进行的分类讨论提供了依据：当我们使用Gauss消元法得到A中的零行时，回代前应该针对零行所对应的新b值是否为0进行分类讨论。其次是解法、解的结构和解的数量，这里要求我们运用之前解Ax=0时的知识。解法和解Ax=0大致相同。使用Guass消元法，确定主元，进一步确定主元变量和自由变量。 1.求出特解X p：置全部自由变量为0（简化运算），回代解出主元变量，得到Ax=b的一个解； 2.解出Ax=0的全部解X N：也即基向量的全部线性组合，含有1或2个常数c； 3.通解X=X p+X N：因为A(Xp+X N)=AX p+AX N=b+0=b，这也就是所谓“解的结构”，通解由一个特解和零空间内的全部向量组成。从几何上说，解空间由零空间平移得到。但是，这种方法存在缺陷，不通用。问题就出在第一步。如果没有自由变量怎么办？那后续的方法如何进行？解的结构还是那两个部分吗？还有，如果根本就没有解，怎么办？为了确定解的存在性；为了确定自由变量的个数，发掘其与解的数量及与之相对应的结构的关系，我们需要研究秩的概念。之前已经提及，秩r=主元数。如何利用r判定一个由mn矩阵A构成的方程Ax=b的解的数量呢？关键是： 1.自由变量的个数n-r（主元不同列），即r与n的相对关系； 2.零行（可能出现“非0数=0”的矛盾情况）的个数m-r（主元不同行以及主元非0），即r与m的相对关系；综合考虑，只可能出现以下四种情况（根据主元选取规则，r显然小于等于m和n）： 1.r=m=n（”满秩”），一定有唯一解：（1）没有零行，一定有解；（2）没有自由变量，解唯一（回代之后解出）。 2.r=m且r<n（“行满秩”），一定有无穷多个解：（1）没有零行，一定有解；（2）有自由变量，有无穷多个解； 3.r<m且r=n（“列满秩”），解的个数为0或1：（1）有零行，可能无解；（2）没有自由变量，如果有解，则解唯一； 4.r<m且r<n，解的个数为0或无穷大：（1）有零行，可能无解；（2）有自由变量，如果有解，则有无穷多个解。 r与m，n的相对关系可以作为判据，检查我们求出的解正确与否（是否存在以及是否完备）。分类: 数学标签: 数学, 线性代数导论好文要顶关注我收藏该文微信分享 Samaritan_z 粉丝 - 5关注 - 2 +加关注 0 0 升级成为会员 « 上一篇：数学 - 线性代数导论 - #8 求解Ax=0/构造零空间 » 下一篇：数学 - 线性代数导论 - #10 线性相关性、向量空间的基和维数 posted @ 2018-02-08 16:45Samaritan_z 阅读(2588) 评论(0)收藏举报刷新页面返回顶部登录后才能查看或发表评论，立即登录或者逛逛博客园首页【推荐】100%开源！大型工业跨平台软件C++源码提供，建模，组态！【推荐】HarmonyOS 专区 —— 闯入鸿蒙：浪漫、理想与「草台班子」【推荐】2025 HarmonyOS 鸿蒙创新赛正式启动，百万大奖等你挑战【推荐】天翼云爆款云主机2核2G限时秒杀，28.8元/年起！立即抢购编辑推荐： · 无意中在应用层瞥见了一个微内核的操作系统调度器 · Runtime Async - 步入高性能异步时代 · 基于 IOCP 的协程调度器——零基础深入浅出 C++20 协程 · 有有点意思！Java8后最有用新特性排行榜 · Ribbon LoadBalancer: 开源的客户端式负载均衡框架 HarmonyOS专区： · 闯入鸿蒙：浪漫、理想与「草台班子」 · 3天赚2万！开发者的梦想也可以掷地有声！ · Flutter 适配 HarmonyOS 5 开发知识地图 · HEIF：更高质量、更小体积，开启 HarmonyOS 图像新体验 · CodeGenie 的 AI 辅助调优让你问题定位效率大幅提升公告昵称： Samaritan_z 园龄： 7年8个月粉丝： 5 关注： 2 +加关注 <2025年9月> 日一二三四五六 31 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1 2 3 4 5 6 7 8 9 10 11 搜索常用链接我的随笔我的评论我的参与最新评论我的标签我的标签线性代数导论(12) 数学(12) Computer Science(11) 元学习(5) Learning How to Learn(5) Python(4) Tricks(2) Computer Vision(2) 随笔(1) Markdown(1) 更多积分与排名积分 - 29431 排名 - 56942 随笔分类计算机科学(11) 数学(12) 随笔(1) 元学习(5) 随笔档案 2018年2月(12) 2018年1月(17) 阅读排行榜 1. 数学 - 线性代数导论 - #4 矩阵分解之LU分解的意义、步骤和成立条件(6849) 2. 数学 - 线性代数导论 - #6 向量空间、列空间、R^n与子空间(4410) 3. 数学 - 线性代数导论 - #11 基于矩阵A生成的空间：列空间、行空间、零空间、左零空间(4351) 4. 数学 - 线性代数导论 - #5 矩阵变换之置换与转置(4168) 5. 数学 - 线性代数导论 - #2 用Gauss消元法解线性方程组(3677) 推荐排行榜 1. 数学 - 线性代数导论 - #2 用Gauss消元法解线性方程组(1) 博客园© 2004-2025 浙公网安备 33010602011771号浙ICP备2021040463号-3 点击右上角即可分享
2901	https://www.omnicalculator.com/statistics/hypergeometric-distribution	Board Last updated: Hypergeometric Distribution Calculator Use our hypergeometric distribution calculator whenever you need to find the probability (or cumulative probability) of a random variable following the hypergeometric distribution. If you want to learn what the hypergeometric distribution is and what the hypergeometric distribution formula looks like, keep reading! In addition to those essential facts, we also provide you with the properties of the hypergeometric distribution, an example of the hypergeometric distribution, and discuss when to use the hypergeometric probability distribution vs. the (more familiar) binomial distribution. What is hypergeometric distribution? The hypergeometric probability distribution describes the number of successes (objects with a specified feature, as opposed to objects without this feature) in a sample of fixed size when we know the total number of items and the number of success items (total number of objects with that feature). Importantly, we assume sampling is without replacement — when we choose an item from the population, we cannot select it again. The hypergeometric distribution turns out to be useful whenever an observed event cannot re-occur, e.g., in various card games, in which the fact that we drew a card implies we will not draw that card again. For example, the hypergeometric distribution appears in Fisher's exact test, which we use to test the difference between two proportions when the sample size is small (<=50). Check out our dedicated Fisher's exact test calculator to discover more. Note that, although the population's items are divided into two mutually exclusive categories (success/failure), the hypergeometric distribution is not the same as the binomial distribution. See the last section and the binomial distribution calculator for more details. But first, let's discuss the formula for the hypergeometric distribution. The properties of the hypergeometric distribution There are three properties of the hypergeometric distribution: The mean; The standard deviation; and The variance. Mean The mean of the hypergeometric distribution can be calculated by using the following formula: μ=nNk Where: n — the number of occurrences; k — is the number of successes; N — is the population size. Standard deviation The standard deviation is a property of the hypergeometric distribution; it can be calculated as: σ2=nNkNN−kN−1N−n Variance The variance of the hypergeometric distribution can be calculated by using the following formula: σ=nNkNN−kN−1N−n Hypergeometric distribution formula Three parameters define the hypergeometric probability distribution: N — Total number of items in the population; K — Number of success items in the population; and n — Number of drawn items (sample size). A random variable X follows the hypergeometric distribution if its probability mass function is given by: P(X=k)=(nN)(kK)(n−kN−K) where: k — Number of drawn success items. There are usually binomial coefficients in the hypergeometric distribution formula. With the use of the factorial operator !, we can rewrite the above equation as: P(X=k)=N!k!(K−k)!(n−k)!(N−K−n+k)!K!(N−K)!n!(N−n)! 🔎 See the factorial calculator if you're not sure what the exclamation mark ! means. The mean and variance of the hypergeometric distribution For a hypergeometric distribution with parameters N, K, n: The mean of hypergeometric distribution (expected value) is equal to: n × K / N The variance of hypergeometric distribution is equal to: n × K × (N - K) × (N - n) / [N² × (N - 1)] How to use this hypergeometric distribution calculator? As you can see, there are lots of formulae related to the hypergeometric distribution that are not so trivial to evaluate. Fortunately, there's our hypergeometric distribution calculator! 😁 Let's explain how to use it before we move on to an example of the hypergeometric distribution. Enter the parameters of the hypergeometric distribution you want to consider. Choose what to compute: P(X = k) or one of the four types of cumulative probabilities: P(X > k), P(X ≥ k), P(X < k), P(X ≤ k). Our hypergeometric distribution calculator returns the desired probability. At the very bottom of the calculator, you will find the variance and mean of your hypergeometric distribution shown. Example of the hypergeometric distribution As you now know what hypergeometric distribution is, let's have a look at an hypergeometric distribution example. Imagine a bag of chocolate bars with 12 dark and 36 white chocolate bars. You close your eyes and draw 10 bars without replacement. What is the probability that you have exactly 4 dark chocolate bars? The parameters are: N = 48, K = 12, n = 10, k = 4. So we apply the hypergeometric distribution formula and obtain: P(X = 4) = 12!×36!×10!×38! / (48!×4!×8!×6×30!) ≈ 0.1474 What is the probability that you have at least 4 dark chocolate bars? P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) ≈ 0.2023 What is the mean of hypergeometric distribution? 10 × 12 / 48 = 2.5 What is the variance of hypergeometric distribution? 10 × 12 × (48 - 12) × (48 - 10) / [48² × 47] ≈ 285 / 188 ≈ 1.5160 What is the standard deviation of this hypergeometric distribution? √1.5160 ≈ 1.2313 You can check these results with our hypergeometric distribution calculator! Hypergeometric distribution vs. binomial distribution The hypergeometric and binomial distributions both quantify the probability of k successes in n trials. However: For the hypergeometric distribution, there is sampling with no replacement, so each draw decreases the population. In consequence, after each trial, the probability of success in the next trial changes; and For the binomial distribution, there is sampling with replacement, so the probability of success remains the same for every trial. Tip: If the population size is large and the sample is small (relative to the population size), the hypergeometric distribution gives almost the same results as the binomial distribution. So we can summarize as follows: Use the hypergeometric distribution if you sample without replacement and the population has few enough elements that a trial changes the probability of the next trial significantly; and Use the binomial distribution for sampling with replacement or for sampling without replacement with a large population and a small sample size. Sum of Squares Calculator Midrange Calculator Coefficient of Variation Calculator 0 Parameters Results Share result Reload calculatorClear all changes Did we solve your problem today? Yes No Check out 32 similar distributions and plots calculators 🔔 Benford's law Beta distribution Binomial distribution Share Calculator Hypergeometric Distribution Calculator Share Calculator Hypergeometric Distribution Calculator Learn more about these settings
2902	https://math.stackexchange.com/questions/2235140/if-video-speed-increase-by-2x-total-time-is-reduced-by-50-how-do-i-understand	algebra precalculus - If video speed increase by 2x, total time is reduced by 50%. How do I understand this with math? Please recommend basic resources for my skill level.. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more If video speed increase by 2x, total time is reduced by 50%. How do I understand this with math? Please recommend basic resources for my skill level.. Ask Question Asked 8 years, 5 months ago Modified3 years, 1 month ago Viewed 12k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Problem: If you increase the speed of a video by 2x, you're reducing the total time by 50%. I have no understanding of how to calculate this, or how I got to this outcome. Request: Please recommend exactly what specific topics I should learn and understand. Continuation Of Problem: For example, if I increase playback speed by 2.5x, I have no idea what % the total time is reduced by. It's likely around ~66% or something, but I've no idea how to calculate this. Or more simply, I don't understand how to do this in math. Please recommend in comments or answers (doesn't matter; the important thing is being helpful): online textbooks or any other resources specifically on practical math for everyday life Other sources like Khan Academy has a lot of math that isn't useful or needed in everyday life. I wish there was a math curriculum that specifically listed the top 10 or so specific topics for practical math, and the math topics that tend to be more useful relative to other topics Whatever specific topic the question/problem I asked here should be on that top 10, 20 or whatever I'm highly knowledgeable and understand many concepts in many academic fields/areas outside of math (that don't require math), but I don't understand whatever basic math topic this is. I'm assuming the math-orientated had given this specific topic a specific label/word -- as to make communication easier as well as a host of many other benefits. It's generally said and understood that math is easier to learn via programming, but I do not know of any good sources, or if someone has made this yet as of 2017 Side note: I support all the people making progress in how math is being taught at all levels, besides the most abstract/theoretical. Please do not recommend any academic or theoretical math outside of the kind of practical resources that was asked for. algebra-precalculus self-learning education percentages Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Apr 15, 2017 at 21:14 ambwambw asked Apr 15, 2017 at 10:00 ambwambw 129 1 1 gold badge 1 1 silver badge 5 5 bronze badges 10 You should take a look at Proportionality. More precisely, Inverse Proportionality. Read this: en.m.wikipedia.org/wiki/…Filburt –Filburt 2017-04-15 10:15:58 +00:00 Commented Apr 15, 2017 at 10:15 Oh. And this link contains references... ;-)Filburt –Filburt 2017-04-15 10:20:11 +00:00 Commented Apr 15, 2017 at 10:20 Yep... super basic topic... classed under 'ratios', which is a subset of 'elementary mathematics' and 'algebra'.... these references are highly advanced and complicate/confuse things needlessly. I need a resource where I can learn/understand this and the other topics (that are useful to everyday life). This is an encyclopedia, which is largely historical. I read almost everything besides the math topics. I can't learn math like this. I need an actual learning resource.ambw –ambw 2017-04-15 10:40:54 +00:00 Commented Apr 15, 2017 at 10:40 2 What exactly do you classify as practical?I could list practical uses for about every precalculus topic,I also feel that Khan academy has one of the best easy to learn resources,and at least 70% of the lets say theoretical questions can be "translated" into practical problems/solutions.kingW3 –kingW3 2017-04-15 10:42:51 +00:00 Commented Apr 15, 2017 at 10:42 2 The Khan Academy knowledge map is not just videos, by the way, it's an actual learning system where you answer math problems to prove your proficiency and progress through the knowledge map. I recommend giving it a try -- actually doing the problems and earning the badges.littleO –littleO 2017-04-15 11:10:22 +00:00 Commented Apr 15, 2017 at 11:10 \|Show 5 more comments 4 Answers 4 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. I would actually recommend looking at this at a physics view point. The basic equation for constant-velocity motion is x=v t x=v t, where x x stands for the place, v v for the velocity, and t t for time. You can look at v v as the number of frames per second ("the speed of the video"), t t as the time it takes to watch the video, and x x as the total number of frames in the video. In the question you are asking, we have the same x x for any velocity (the number of frames don't change). So we are solving v t=c o n s t v t=c o n s t. In case the speed is say ×2.5×2.5, the time must be ×1 2.5×1 2.5 for us to have the same constant. The remaining part is to understand the relation between a number such as 1 2.5 1 2.5 and precentage. This is the definition of precentage - 1%=1 100 1%=1 100. So we have 1 2.5=40 100=40%1 2.5=40 100=40%, and the time would be 40%40% of the original time. To end my answer, I would recommend some generic middle school algebra book, these things are usually covered quite good there. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Apr 15, 2017 at 10:26 answered Apr 15, 2017 at 10:20 The way of lifeThe way of life 1,222 1 1 gold badge 8 8 silver badges 21 21 bronze badges 5 For the 3 letters, the only one that's unclear is x. I'm not sure what to put there or how I would understand what to put there. You of course did and put what's suppose to be there: 'total frames'. Looks like x is constant, but there's so many other things that are constant that I can put there. For example, a 'button on the UI' is constant but it doesn't seem like it would make sense to put that as v. But how do I figure out what to put in there for all cases and problems of this pattern? Is there a mathematical way to go about this? Or I don't understand.. What do I read or learn?ambw –ambw 2017-04-15 11:13:29 +00:00 Commented Apr 15, 2017 at 11:13 1 What chances in time? only the number of frames you have watched. Also, if you think about the relevant units - if we agree that time is in seconds, and the velocity is frames per second, we must get that the place is frames/sec sec which is frames.The way of life –The way of life 2017-04-15 11:15:09 +00:00 Commented Apr 15, 2017 at 11:15 I don't understand what you mean by 'chances in time'. Explaining is hard. Should read all the other comments.ambw –ambw 2017-04-15 22:20:38 +00:00 Commented Apr 15, 2017 at 22:20 1 changes in time, sorry The way of life –The way of life 2017-04-16 06:37:27 +00:00 Commented Apr 16, 2017 at 6:37 I meant x when I wrote v on that line: "would make sense to put that as v" -- where's a list of all 'changes in time' words so i know what to put in x?ambw –ambw 2017-04-17 01:08:41 +00:00 Commented Apr 17, 2017 at 1:08 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. This is what is called inversely proportional. Think about this example: it takes 1 1 man 24 24 hours to build a wall. However, if there were 2 2 men, they'd get it done twice as fast, so you half how long it takes --- it would take two of them 12 12 hours to build together. For 3 3 people, it would take a third of the time, so it would only take 8 8 hours. In general, if there are n n people, it would take 24/n 24/n hours to build this wall. Your video is the same --- you're playing it twice as fast, so it only plays for half the time. If you play it at 2.5 2.5 times speed, then it plays for 1/2.5 1/2.5 percent of the original time. EDIT: The law of proportionality is evident in a lot of things. You may conjecture that sunny skies improve ice cream sales; or that better tasing food will cost more; or that more time in the sun will give you a darker tan. These are each examples of relationships that are directly proportional, since an increase in one causes an increase in the other. Conversely, an increase in rainy weather may cause a decrease in ice cream sales, or an increase in burgers eaten per week may decrease how long you live, both of which are examples of inverse proportionality. Now, these are not strict relationships since there are more factors there, but I'm trying to convey the idea of proportional relationships. We will move to 'stricter' relationships. Say we can buy a box of five pens. If I buy one box, I get five pens. If I buy two boxes, I get ten pens. If I buy n n boxes, I get 5×n 5×n pens. The relationship is directly proportional and follows the formula y=5×x y=5×x where y y is the number of pens, and x x is the number of boxes bought. It is nonsense to talk of a 'proof' in this sense --- the relation is obvious (I hope!). In general, the formula for direct proportionality is y=k×x y=k×x where x x and y y are your variables (you have to think which is which --- there's not just some 'magic formula' for every possible situation, you have to apply some logic yourself), and k k is the proportionality factor. In the pens example, this is five. There are many more examples of this relationship, but you just need to think which was round it goes --- that is, if I buy more boxes, does it make sense that I get more or less pens? Of course, you get more! The other type of proportionality is inverse proportionality. This happens when an increase in one variable cause a decrease in the other. For my wall example, does it make sense that more workers means that it takes longer to build the wall? Of course not! (Unless the workers stand about chatting all day, but we're not considering that kind of situation.) The standard formula for inverse proportionality is y=k x y=k x where the variables are the same as before. You have to look at your situation and figure out which variable matches up with which letter in the formula. For your video example, you actually set k=1 k=1 because you are considering only one video, but we follow this thought process: "My video plays for some amount of time. If I speed up the video, do I expect it to play for a longer or for a shorter amount of time? Since it's playing faster, it plays more frames in each second. If it's playing more frames in each second, and it's playing the same amount of frames, logic dictates that it must play for a shorter amount of time (again, I stress that there is no formal proof for this in the sense that you are talking, you just need to think about what makes sense). If I play it at 2×2× the speed, then it can either double or half because of the 2 2. If it can't double since that doesn't make sense, it has to half (this is shown Mathematically in the other two answers). Now, we know that the relationship is (2×)(2×) speed ⟹⟹1 2 1 2 the play time, we can jump to the formula (s×)sped up⟹1 s times the original play time."(s×)sped up⟹1 s times the original play time." Again, this is 'proved' Mathematically in the other two answers, but it's such an obvious relationship that the above steps certainly suffice. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Apr 15, 2017 at 13:12 answered Apr 15, 2017 at 10:16 BilbottomBilbottom 2,738 2 2 gold badges 17 17 silver badges 37 37 bronze badges 10 As for further reading, Khan Academy is absolutely fantastic. You may not see the 'real life' applications of many of his videos directly, but a great deal of them are the foundations for bigger topics that do have applications in the real world. It seems that you need to learn a lot of the fundamentals of Mathematics --- I would suggest learning everything in the Mathematics GCSE schedule (or any foreign equivalent). There is very little Mathematics taught at the level that doesn't have applications, whether you can see it or not.Bilbottom –Bilbottom 2017-04-15 10:26:46 +00:00 Commented Apr 15, 2017 at 10:26 So looks like the formula generally at least is [total time taken] : [number of people, speed, or whatever the variable is] -- but of course this isn't helping me understand anything. I mean, do I always put the variable (the number that is being changed) in the second slot? Saying that something is 'twice as fast' equates to 'half the time' is a HUGE jump. How in the world did we get there? Why is it that 3 people is 1/3 the time? Every single step from the starting point (2x) to the end point (50% less) needs to be explained FULLY -- IN ABSOLUTE DETAIL. That's what understanding means,ambw –ambw 2017-04-15 10:48:53 +00:00 Commented Apr 15, 2017 at 10:48 3 @ambw Again, I'm going to stress the fact that there is no book that goes this in depth. If you're at school or college, talk to a teacher/lecturer about it. But, I would strongly suggest just sitting down and thinking about it yourself. Try writing out some of your own examples --- attempt to figure out these relationships, and what would actually make sense. Thinking independently and trying your own examples is so vastly important for coming to a full understanding of simple laws like these.Bilbottom –Bilbottom 2017-04-15 13:06:24 +00:00 Commented Apr 15, 2017 at 13:06 2 @ambw No they don't. There are some things that are objective truths. It is also common in the community to leave some work for the reader when it comes to self explanations. Being able to piece parts of a proof together to get an understanding of the whole proof is part pf the learning experience; giving someone all the information straight away prohibits free thinking and original thought. There are some things that you are expected to learn yourself, and this is totally fair.Bilbottom –Bilbottom 2017-04-15 22:54:53 +00:00 Commented Apr 15, 2017 at 22:54 1 Stack Exchange is that community. The reason no one develops on this topic is because people don't usually struggle with it --- irrespective of ones background, my explanation or the Physics explanation are two of the explanations that are easiest to understand for this topic. We're not 'set in our ways' --- it's like asking why dividing by two is the same and multiplying by a half, there's only so many ways it can be explained without overdoing it! I stand by the comment that some things should be left to you when they are simple steps like these as it helps you to develop yourself.Bilbottom –Bilbottom 2017-04-17 10:00:27 +00:00 Commented Apr 17, 2017 at 10:00 \|Show 5 more comments This answer is useful 1 Save this answer. Show activity on this post. Let's think about this in terms of something a little more concrete. Think of a sprinter running a 100-meter dash. If they run 5 5 meters every second, how long will it take them to complete the race? Well, they must cover a the distance D=100 m D=100 m at a speed (or rate) of r=5 m/s r=5 m/s. This will take them a time t=D v=100 m 5 m/s=20 s t=D v=100 m 5 m/s=20 s. I have derived this from the equation D=r×t D=r×t, which says that one travels a distance D D when moving at a speed r r for a length of time t t. Notice how the units (m m and s s) are treated algebraically, just as if they were numbers or variables. In particular, the unit of meters cancels as a common factor in the division, and the unit of seconds ends up in the numerator. In terms of your video example, the distance D D is the length of the video. So, say we have a video that is three minutes long: D=3 min=3 min×60 s 1 min=180 s D=3 min=3 min×60 s 1 min=180 s. We can multiply by the conversion factor of 60 s 1 min 60 s 1 min since it equals 1 1, and since multiplying by 1 1 leaves a number unchanged. If we play the video at normal speed (1 1 x), then we have playback rate r=1 s/s r=1 s/s. Of course, in this case, we have that the playback time t=D v=180 s 1 s/s=180 s=D t=D v=180 s 1 s/s=180 s=D is the same as the length of the video. What happens when we consider playback rate of 2.5x? In this case, we have r=2.5 s/s r=2.5 s/s, so that t=D r=180 s 2.5 s/s=72 s t=D r=180 s 2.5 s/s=72 s. So, at 2.5x playback, a 3 3-minute (D=180 s)D=180 s) video only takes 1 1 minute and 12 12 seconds (t=72 s t=72 s) to watch. This is a reduction of 1 1 minute and 48 48 seconds (D−t=108 s D−t=108 s), which is a reduction by D−t D=108 s 180 s=0.6=60%D−t D=108 s 180 s=0.6=60%. (Thus your guess of 66%66% was rather close.) Takeaway: The unit of the distance D D can be anything and the equation will still apply, assuming the rate r r remains constant for the duration t t. In the sprinter example, D D was traditional distance, length. In the video example, it was (video playback) time. But it could be pizzas. Final example: If I have to fill an order of 15 pizzas and can make one pizza in 7 minutes, how long will it take to finish all of the pizzas? Well, our distance is D=15 pizzas D=15 pizzas and our rate is r=1 pizza 7 min r=1 pizza 7 min. Thus it would take t=D r=15 pizzas 1 pizza 7 min=15×7 min=105 min t=D r=15 pizzas 1 pizza 7 min=15×7 min=105 min to complete this order. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 15, 2017 at 11:07 SylexzerSylexzer 53 4 4 bronze badges 1 1) How I do I know what equation to use? 2) How do I know what numbers to put into what letters? 3) I can put anything into D? - "distance D can be anything" - can I put the 'speed'? 4) In your example, [distance] = 100 m, so then... actually I have a billion other questions, but I'll check back tomorrow. What's a good source that would help give understanding for all these questions, and all the followup questions that would surely derive from the previous?ambw –ambw 2017-04-15 11:54:32 +00:00 Commented Apr 15, 2017 at 11:54 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. After reading these answers, I was left personally more confused, as the given examples used everything but video examples. So I made my own answer. I hope this helps you (and others) understand. Watch speed multiplier total time taken equation t=Y/nZ t=seconds Y=total frames Z=frames per second (fps) n=watch speed multiplier In this, we are assuming the video was filmed at 60fps and plays for 300 seconds at x1 speed. This gives us 18,000 frames. How many frames are in a given video won’t change, but the fps rate will change with your multiplier. So for the example video, the equation looks like this: 0.5x watch speed multiplier (half speed) t=18,000/0.560 t=18,000/30 t=600 seconds (10 minutes) 1x watch speed multiplier (normal speed) t=18,000/160 t=18,000/60 t=300 seconds (5 minutes) 2x watch speed multiplier (double speed) t=18,000/260 t=18,000/120 t=150 seconds (2 minutes 30 seconds) 2.5x watch speed multiplier t=18,000/2.560 t=18,000/150 t=120 seconds (2 minutes) 2.75x watch speed multiplier t=18,000/2.7560 t=18,000/165 t=109.09(repeating) seconds (1 minute 49.09(repeating) seconds) As you can see, the overall time and the fps both change, but the total frames never do. This is because video is made of frames. A typical frame rate is 60fps. So to determine how long your video will last given a specific watch speed multiplier, you need to know what fps your video has at normal (1x) watch speed. Then you can take the seconds it lasts and multiply the two together for total frames. Once you solve for total frames, you can use the simple equation above to determine how long any video will play for, for any given watch speed multiplier. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 26, 2022 at 15:29 Michael ChapmanMichael Chapman 1 2 1 Please use Mathjax Harish Chandra Rajpoot –Harish Chandra Rajpoot 2022-08-26 15:30:45 +00:00 Commented Aug 26, 2022 at 15:30 It seems the system likes to make a mess of line by line examples, so the equation and the examples got bunched up in a way I didn’t intend Michael Chapman –Michael Chapman 2022-08-26 15:32:15 +00:00 Commented Aug 26, 2022 at 15:32 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus self-learning education percentages See similar questions with these tags. 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2903	https://www.doubtnut.com/qna/643339402	[Assamese] Establish the relation T^(gamma) P^{(1-gamma)=constatnt for Download Allen App My Profile Logout Ncert Solutions English Medium Class 6 Maths Physics Chemistry Biology English Class 7 Maths Physics Chemistry Biology English Class 8 Maths Physics Chemistry Biology English Class 9 Maths Physics Chemistry Biology English Class 10 Maths Physics Chemistry Biology English Reasoning Class 11 Maths Physics Chemistry Biology English Class 12 Maths Physics Chemistry Biology English Courses IIT-JEE Class 11 Class 12 Class 12+ NEET Class 11 Class 12 Class 12+ UP Board Class 9 Class 10 Class 11 Class 12 Bihar Board Class 9 Class 10 Class 11 Class 12 CBSE Board Class 9 Class 10 Class 11 Class 12 Other Boards MP Board Class 9 Class 10 Class 11 Class 12 Jharkhand Board Class 9 Class 10 Class 11 Class 12 Haryana Board Class 9 Class 10 Class 11 Class 12 Himachal Board Class 9 Class 10 Class 11 Class 12 Chhattisgarh Board Class 9 Class 10 Class 11 Class 12 Uttarakhand Board Class 9 Class 10 Class 11 Class 12 Rajasthan Board Class 9 Class 10 Class 11 Class 12 Exam IIT-JEE NEET CBSE UP Board Bihar Board Study with ALLEN JEE NEET Class 6-10 Books Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Online Class Blog Select Theme Class 11 PHYSICS Establish the relation `T^(gamma) P^{(1-... Establish the relation T γ P(1−γ)=constatnt for adiabtic variation .Solution in Assamese Video Solution Know where you stand among peers with ALLEN's JEE Nurture Online Test Series P^{(1-gamma)=constatnt for adiabtic variation . by Physics experts to help you in doubts & scoring excellent marks in Class 11 exams. Updated on:21/07/2023 Class 11PHYSICSTHERMODYNAMICS Topper's Solved these Questions THERMODYNAMICS Book:BINA LIBRARY Chapter:THERMODYNAMICS Exercise:EXERCISE Explore 59 Videos THERMAL PROPERTIES OF MATTER Book:BINA LIBRARY Chapter:THERMAL PROPERTIES OF MATTER Exercise:EXERCISE Explore 119 Videos WAVES Book:BINA LIBRARY Chapter:WAVES Exercise:EXERCISE Explore 97 Videos Similar Questions State the first law of thermodynamics . View Solution Define co-efficient of linear expansion.Dose it depend on unit of length? View Solution What is a wave motion? View Solution In an isobaric process, Delta Q = (K gamma)/(gamma - 1) where gamma = C_(P)//C_(V) . What is K ? View Solution The dimensions of γ in the relation v=√γ p ρ (where v is velocity, p is pressure , ρ is density) View Solution If α,β,γ be the zeros of the polynomial p(x) such that (α+β+γ)=3,(α β+β γ+γ α)=−10 and α β γ=−24 then p(x) = ? View Solution Write the difference between alpha, beta and gamma variations? View Solution a:b=b:c হলে দেখাও যে (a+b)2:(b+c)2=a:c View Solution P=⎡⎢⎣√3 2 1 2−1 2√3 2⎤⎥⎦and A=[1 1 0 1] , Q=P A P T and P T Q 2020 P=[α γ β α] then α+β+γ View Solution a:b=b:c হলে দেখাও যে (a+b)2:(b+c)2=a:c View Solution Establish a relation between alpha and gamma of thermal expansion by calculus method. View Solution Establish the relation P V γ=constant, for adiabatic change of an ideal gas. View Solution Starting from PV^gamma= constant, establish the following relation for an adiabatic change of a perfect gas Tv^(gamma-1) =constant . View Solution Starting from P V γ= constant, establish the following relation for an adiabatic change of a perfect gas T γ P γ−1 =constant. View Solution obatain the adiabiatic equation PV^(gamma) =constatnt from the first law of thermodynamics . View Solution BINA LIBRARY-THERMODYNAMICS -EXERCISE how guasi-static achieved ? 02:43 what is reversible process ?under what condition can a process be reve... 03:46 Establish the relation T^(gamma) P^{(1-gamma)=constatnt for adiabtic v... 04:08 State and explain zeroth law of thermodynamics . 02:34 State the first law of thermodynamics . 03:48 what are limitations of first law of thermodynamics. 04:54 What is an adiabatic process ? 02:05 obatain the adiabiatic equation PV^(gamma)=constatnt from the first la... 08:08 From the first law of thermodynamics derive the relation Cp-Cv =R whwr... 12:04 Define effciency of a heat engine .can you design an engine which has... 02:48 Obtain an expression for work done in an isothermal expansion . 03:19 obtain the expression for work done in adiabatic expansion . 02:48 What is reversible and irreversible processes ? 03:37 Describe the various operation of a carnot's reversible engine and obt... 12:04 What is reversible and irreversible processes ? 03:37 state the second law of thermodynamics . 05:06 Assuming that the working substance of a carnot's engine is a perfect ... 10:58 On what does the efficiency of a heat engine depends ? 03:12 How can the efficiency of a reversible engine be 100%? 01:58 How does a gas do work when its expands adiabatically ? 02:49 Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics HC Verma Solutions for Physics Sunil Batra Solutions for Physics Pradeep Solutions for Physics Errorless Solutions for Physics Narendra Awasthi Solutions for Chemistry MS Chouhan Solutions for Chemistry Errorless Solutions for Biology Free Ncert Solutions English Medium NCERT Solutions NCERT Solutions for Class 12 English Medium NCERT Solutions for Class 11 English Medium NCERT Solutions for Class 10 English Medium NCERT Solutions for Class 9 English Medium NCERT Solutions for Class 8 English Medium NCERT Solutions for Class 7 English Medium NCERT Solutions for Class 6 English Medium Free Ncert Solutions Hindi Medium NCERT Solutions NCERT Solutions for Class 12 Hindi Medium NCERT Solutions for Class 11 Hindi Medium NCERT Solutions for Class 10 Hindi Medium NCERT Solutions for Class 9 Hindi Medium NCERT Solutions for Class 8 Hindi Medium NCERT Solutions for Class 7 Hindi Medium NCERT Solutions for Class 6 Hindi Medium Boards CBSE UP Board Bihar Board UP Board Resources Unit Converters Calculators Books Store Seminar Results Blogs Class 12 All Books Class 11 All Books Class 10 All Books Class 9 All Books Class 8 All Books Class 7 All Books Class 6 All Books Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. 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2904	https://periodictable.com/Properties/A/SpecificHeat.v.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| 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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| Specific Heat of the elements \| \| \| --- \| \| Text lists sorted by: \| Value \| Atomic Number \| Alphabetical \| \| Plots: \| Shaded \| Ball \| Crossed Line \| Scatter \| Sorted Scatter \| \| Log scale plots: \| Shaded \| Ball \| Crossed Line \| Scatter \| Sorted Scatter \| \| Good for this property: \| Scatter \| \| \| \| \| \| --- --- \| \| Technetium \| 63 J/(kg K)[note] \| Iodine \| 429 J/(kg K)[note] \| \| Radium \| 92 J/(kg K)[note] \| Nickel \| 445 J/(kg K)[note] \| \| Radon \| 93.65 J/(kg K)[note] \| Chromium \| 448 J/(kg K)[note] \| \| Protactinium \| 99.1 J/(kg K)[note] \| Iron \| 449 J/(kg K)[note] \| \| Uranium \| 116 J/(kg K)[note] \| Chlorine \| 478.2 J/(kg K)[note] \| \| Thorium \| 118 J/(kg K)[note] \| Manganese \| 479 J/(kg K)[note] \| \| Actinium \| 120 J/(kg K)[note] \| Vanadium \| 489 J/(kg K)[note] \| \| Bismuth \| 122 J/(kg K)[note] \| Titanium \| 520 J/(kg K)[note] \| \| Lead \| 127 J/(kg K)[note] \| Argon \| 520.33 J/(kg K)[note] \| \| Thallium \| 129 J/(kg K)[note] \| Scandium \| 567 J/(kg K)[note] \| \| Gold \| 129.1 J/(kg K)[note] \| Calcium \| 631 J/(kg K)[note] \| \| Osmium \| 130 J/(kg K)[note] \| Sulfur \| 705 J/(kg K)[note] \| \| Iridium \| 131 J/(kg K)[note] \| Carbon \| 710 J/(kg K)[note] \| \| Tungsten \| 132 J/(kg K)[note] \| Silicon \| 710 J/(kg K)[note] \| \| Platinum \| 133 J/(kg K)[note] \| Potassium \| 757 J/(kg K)[note] \| \| Rhenium \| 137 J/(kg K)[note] \| Phosphorus \| 769.7 J/(kg K)[note] \| \| Mercury \| 139.5 J/(kg K)[note] \| Fluorine \| 824 J/(kg K)[note] \| \| Tantalum \| 140 J/(kg K)[note] \| Aluminum \| 904 J/(kg K)[note] \| \| Hafnium \| 144 J/(kg K)[note] \| Oxygen \| 919 J/(kg K)[note] \| \| Ytterbium \| 154 J/(kg K)[note] \| Bromine \| 947.3 J/(kg K)[note] \| \| Lutetium \| 154 J/(kg K)[note] \| Magnesium \| 1020 J/(kg K)[note] \| \| Xenon \| 158.32 J/(kg K)[note] \| Boron \| 1030 J/(kg K)[note] \| \| Thulium \| 160 J/(kg K)[note] \| Neon \| 1030 J/(kg K)[note] \| \| Holmium \| 165 J/(kg K)[note] \| Nitrogen \| 1040 J/(kg K)[note] \| \| Dysprosium \| 167 J/(kg K)[note] \| Sodium \| 1230 J/(kg K)[note] \| \| Erbium \| 168 J/(kg K)[note] \| Beryllium \| 1820 J/(kg K)[note] \| \| Europium \| 182 J/(kg K)[note] \| Lithium \| 3570 J/(kg K)[note] \| \| Terbium \| 182 J/(kg K)[note] \| Helium \| 5193.1 J/(kg K)[note] \| \| Neodymium \| 190 J/(kg K)[note] \| Hydrogen \| 14300 J/(kg K)[note] \| \| Cerium \| 192 J/(kg K)[note] \| Promethium \| N/A \| \| Praseodymium \| 193 J/(kg K)[note] \| Polonium \| N/A \| \| Lanthanum \| 195 J/(kg K)[note] \| Astatine \| N/A \| \| Samarium \| 196 J/(kg K)[note] \| Francium \| N/A \| \| Tellurium \| 201 J/(kg K)[note] \| Neptunium \| N/A \| \| Barium \| 205 J/(kg K)[note] \| Plutonium \| N/A \| \| Antimony \| 207 J/(kg K)[note] \| Americium \| N/A \| \| Tin \| 217 J/(kg K)[note] \| Curium \| N/A \| \| Cadmium \| 230 J/(kg K)[note] \| Berkelium \| N/A \| \| Indium \| 233 J/(kg K)[note] \| Californium \| N/A \| \| Silver \| 235 J/(kg K)[note] \| Einsteinium \| N/A \| \| Ruthenium \| 238 J/(kg K)[note] \| Fermium \| N/A \| \| Rhodium \| 240 J/(kg K)[note] \| Mendelevium \| N/A \| \| Palladium \| 240 J/(kg K)[note] \| Nobelium \| N/A \| \| Gadolinium \| 240 J/(kg K)[note] \| Lawrencium \| N/A \| \| Cesium \| 242 J/(kg K)[note] \| Rutherfordium \| N/A \| \| Krypton \| 248.05 J/(kg K)[note] \| Dubnium \| N/A \| \| Molybdenum \| 251 J/(kg K)[note] \| Seaborgium \| N/A \| \| Niobium \| 265 J/(kg K)[note] \| Bohrium \| N/A \| \| Zirconium \| 278 J/(kg K)[note] \| Hassium \| N/A \| \| Yttrium \| 298 J/(kg K)[note] \| Meitnerium \| N/A \| \| Strontium \| 300 J/(kg K)[note] \| Darmstadtium \| N/A \| \| Selenium \| 321.2 J/(kg K)[note] \| Roentgenium \| N/A \| \| Germanium \| 321.4 J/(kg K)[note] \| Copernicium \| N/A \| \| Arsenic \| 328 J/(kg K)[note] \| Nihonium \| N/A \| \| Rubidium \| 364 J/(kg K)[note] \| Flerovium \| N/A \| \| Gallium \| 371 J/(kg K)[note] \| Moscovium \| N/A \| \| Copper \| 384.4 J/(kg K)[note] \| Livermorium \| N/A \| \| Zinc \| 388 J/(kg K)[note] \| Tennessine \| N/A \| \| Cobalt \| 421 J/(kg K)[note] \| Oganesson \| N/A \| Notes on the Specific Heat of particular elements: Hydrogen: Value given for gas phase of H . 2Helium: Value given for gas phase.Lithium: Value given for solid phase.Beryllium: Value given for solid phase.Boron: Value given for solid rhombic form.Carbon: Value given for solid graphite form.Nitrogen: Value given for gas phase of N . 2Oxygen: Value given for gas phase of O . 2Fluorine: Value given for gas phase of F . 2Neon: Value given for gas phase.Sodium: Value given for solid phase.Magnesium: Value given for solid phase.Aluminum: Value given for solid phase.Silicon: Value given for solid phase.Phosphorus: Value given for solid phase of P . 4Sulfur: Value given for solid rhombic form.Chlorine: gas (Cl ) 2Argon: Value given for gas phase.Potassium: Value given for solid phase.Calcium: Value given for solid phase.Scandium: Value given for solid phase.Titanium: Value given for solid phase.Vanadium: Value given for solid phase.Chromium: Value given for solid phase.Manganese: Value given for solid phase.Iron: Value given for solid phase.Cobalt: Value given for solid phase.Nickel: Value given for solid phase.Copper: Value given for solid phase.Zinc: Value given for solid phase.Gallium: Value given for solid phase.Germanium: Value given for solid phase.Arsenic: Value given for solid alpha form.Selenium: Value given for solid hexagonal form.Bromine: Value given for liquid phase.Krypton: Value given for gas phase.Rubidium: Value given for solid phase.Strontium: Value given for solid phase.Yttrium: Value given for solid phase.Zirconium: Value given for solid phase.Niobium: Value given for solid phase.Molybdenum: Value given for solid phase.Technetium: Value given for gas phase.Ruthenium: Value given for solid phase.Rhodium: Value given for solid phase.Palladium: Value given for solid phase.Silver: Value given for solid phase.Cadmium: Value given for solid gamma form.Indium: Value given for solid phase.Tin: Value given for solid gray form.Antimony: Value given for solid phase.Tellurium: Value given for solid phase.Iodine: Value given for solid phase.Xenon: Value given for gas phase.Cesium: Value given for solid phase.Barium: Value given for solid phase.Lanthanum: Value given for solid phase.Cerium: Value given for solid phase.Praseodymium: Value given for solid phase.Neodymium: Value given for solid phase.Samarium: Value given for solid phase.Europium: Value given for solid phase.Gadolinium: Value given for solid phase.Terbium: Value given for solid phase.Dysprosium: Value given for solid phase.Holmium: Value given for solid phase.Erbium: Value given for solid phase.Thulium: Value given for solid phase.Ytterbium: Value given for solid phase.Lutetium: Value given for solid phase.Hafnium: Value given for solid phase.Tantalum: Value given for solid phase.Tungsten: Value given for solid phase.Rhenium: Value given for solid phase.Osmium: Value given for solid phase.Iridium: Value given for solid phase.Platinum: Value given for solid phase.Gold: Value given for solid phase.Mercury: Value given for liquid phase.Thallium: Value given for solid phase.Lead: Value given for solid phase.Bismuth: Value given for solid phase.Radon: Value given for gas phase.Radium: Value given for gas phase.Actinium: Value given for solid phase.Thorium: Value given for solid phase.Protactinium: Value given for gas phase.Uranium: Value given for solid phase. Up to date, curated data provided by Mathematica'sElementData function from Wolfram Research, Inc. \| \| \| \| \| \| \| --- \| \| \| Click here to buy a book, photographic periodic table poster, card deck, or 3D print based on the images you see here! \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| Common Properties \| \| Abundance in Earth's Crust \| Discovery Year \| \| Abundance in Humans \| Electrical Conductivity \| \| Abundance in Meteorites \| Electron Affinity \| \| Abundance in the Ocean \| Electron Configuration \| \| Abundance in the Sun \| Electronegativity \| \| Abundance in the Universe \| Half Life \| \| Atomic Mass \| Heat of Fusion \| \| Atomic Number \| Heat of Vaporization \| \| Atomic Radius \| Ionization Energies \| \| Boiling Point \| Melting Point \| \| Covalent Radius \| Specific Heat \| \| Density \| \| \| \| Other Properties \| \| Adiabatic Index \| Mohs Hardness \| \| Allotrope Names \| Molar Magnetic Susceptibility \| \| Alternate Names \| Molar Volume \| \| Block \| Neutron Cross Section \| \| Brinell Hardness \| Neutron Mass Absorption \| \| Bulk Modulus \| NFPA Label \| \| CAS Number \| Period \| \| CID Number \| Phase \| \| Color \| Poisson Ratio \| \| Critical Pressure \| Quantum Numbers \| \| Critical Temperature \| Radioactive \| \| Crystal Structure \| Refractive Index \| \| Curie Point \| Resistivity \| \| Decay Mode \| RTECS Number \| \| DOT Hazard Class \| Series \| \| DOT Numbers \| Shear Modulus \| \| Electrical Type \| Space Group Name \| \| Gas Atomic Multiplicities \| Space Group Number \| \| Group \| Speed of Sound \| \| Isotope Abundances \| Superconducting Point \| \| Isotopes (All Known) \| Symbol \| \| Isotopes (Stable) \| Thermal Conductivity \| \| Lattice Angles \| Thermal Expansion \| \| Lattice Constants \| Valence \| \| Lifetime \| Van Der Waals Radius \| \| Liquid Density \| Vickers Hardness \| \| Magnetic Type \| Volume Magnetic Susceptibility \| \| Mass Magnetic Susceptibility \| Young Modulus \| \| Memberships \| \| \| \| \|
2905	https://flexbooks.ck12.org/cbook/ck-12-middle-school-physical-science-flexbook-2.0/section/9.8/primary/lesson/calculating-acceleration-from-velocity-and-time-ms-ps/	Calculating Acceleration from Velocity and Time \| CK-12 Foundation AI Teacher Tools – Save Hours on Planning & Prep. Try it out! Skip to content What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up Start Practice 9.8 Calculating Acceleration from Velocity and Time FlexBooks 2.0> CK-12 Physical Science for Middle School> Calculating Acceleration from Velocity and Time Written by:Jean Brainard, Ph.D. Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 02, 2025 Lesson Review Asked on Flexi Related Content Lesson [Figure 1] This cyclist is in constant motion as he competes in an off-road mountain bike race. Both his speed and his direction keep changing. Velocity is a measure that represents both speed and direction. Changes in velocity are measured by acceleration. Acceleration reflects how quickly velocity is changing. It may involve a change in speed, a change in direction, or both. Calculating Average Acceleration in One Direction Calculating acceleration is complicated if both speed and direction are changing or if you want to know acceleration at any given instant in time. However, it’s relatively easy to calculate average acceleration over a period of time when only speed is changing. Then acceleration is the change in velocity (represented by Δv) divided by the change in time (represented by Δt): acceleration=Δ v Δ t Accelerating on a Bike Look at the cyclist in the Figurebelow. With the help of gravity, he speeds up as he goes downhill on a straight part of the trail. His velocity changes from 1 meter per second at the top of the hill to 6 meters per second by the time he reaches the bottom. If it takes him 5 seconds to reach the bottom, what is his average acceleration as he races down the hill? acceleration=Δ v Δ t=6 m/s−1 m/s 5 s=5 m/s 5 s=1 m/s 1 s=1 m/s 2 In words, this means that for each second the cyclist travels downhill, his velocity (in this case, his speed) increases by 1 meter per second on average. Note that the answer to this problem is expressed in m/s 2, which is the SI unit for acceleration. [Figure 2] Q: The cyclist slows down at the end of the race. His velocity changes from 6 m/s to 2 m/s during a period of 4 seconds without any change in direction. What was his average acceleration during these 4 seconds? A: Use the equation given above for acceleration: acceleration=Δ v Δ t=2 m/s−6 m/s 4 s=−4 m/s 4 s=−1 m/s 1 s=−1 m/s 2 Notice that the change in velocity, or Δv, is negative because the cyclist is slowing down. This means that the acceleration is also negative, also known as a deceleration. Watch the following video to learn more about how to calculate acceleration: Summary To calculate average acceleration when direction is not changing, divide the change in velocity by the change in time using the formula:acceleration=Δ v Δ t The SI unit for acceleration is m/s 2. Review Write the equation for acceleration without a change in direction. What is the SI unit for acceleration? During the final 5 seconds of a race, a cyclist increased her velocity from 4 m/s to 7 m/s. What was her average acceleration during those last 5 seconds? Add Note CancelSave Discuss with Flexi NOTES / HIGHLIGHTS Please Sign In to create your own Highlights / Notes Add Note Edit Note Remove Highlight Image Attributions Asked by Students Here are the top questions that students are asking Flexi for this concept: How can one calculate distance traveled with acceleration? To calculate the distance traveled with acceleration, you can use the equation of motion: d = v i t + 0.5a t 2 where: - d is the distance traveled, - v i is the initial velocity, - t is the time, and - a is the acceleration. This equation assumes that the acceleration is constant over the time period. How can speed be determined from acceleration and distance? Speed can be determined from acceleration and distance using the equation v² = u² + 2as, where v is the final speed, u is the initial speed (which is zero if starting from rest), a is the acceleration, and s is the distance. If you're starting from rest, the equation simplifies to v = sqrt(2as). Kaya is riding her dirt bike eastward on a dirt road. She spots a pothole ahead. Kaya slows her car from 14.0 m/s to 5.5 m/s in 6.0 s. What is her acceleration? Acceleration is the rate of change of velocity. It can be calculated using the formula: a=v f−v i t where: a is acceleration, v f is the final velocity, v i is the initial velocity, and t is time. In this case, Kaya's initial velocity (v i) is 14.0 m/s, her final velocity (v f) is 5.5 m/s, and the time (t) is 6.0 s. Substituting these values into the formula gives: a=5.5 m/s−14.0 m/s 6.0 s=−1.42 m/s 2 The negative sign indicates that Kaya is decelerating, or slowing down. Calculate the acceleration of Josh riding his bicycle in a straight line that speeds up from 4 m/s to 6 m/s in 5 seconds. Acceleration is calculated by the formula: a=v f−v i t where: a = acceleration v f = final velocity v i = initial velocity t = time Substituting the given values: a=6 m/s−4 m/s 5 s=0.4 m/s 2 Velocity is speed. True/False Velocity and speed are related, but they are not the same. Speed is a scalar quantity that refers to "how fast an object is moving." Velocity is a vector quantity that refers to "the speed and direction of an object." So, velocity includes both the speed and the direction of the object's motion. Overview Velocity represents speed and direction, and changes are measured by acceleration. Average acceleration is calculated over time when only speed changes, as Δv divided by Δt. The SI unit for acceleration is m/s². Slowing down yields negative Δv, indicating negative acceleration or deceleration. Vocabulary speed velocity acceleration Test Your Knowledge Question 1 During the final 5 seconds of a race, a cyclist increased her velocity from 4 m/s to 7 m/s. What was her average acceleration during those last 5 seconds? a 0.16 m/s 2 b 0.60 m/s 2 c 1.6 m/s 2 d 0.12 m/s 2 Check It Q: The cyclist slows down at the end of the race. His velocity changes from 2 m/s to 6 m/s during a period of 4 seconds without any change in direction. What was his average acceleration during these 4 seconds? FlexCard™ Question 2 Choose the equation for acceleration without a change in direction. a @$\begin{align}\text{acceleration} = \frac{\Delta{t}}{\Delta{v}}\end{align}@$ b @$\begin{align}\text{acceleration} = \frac{\Delta{d}}{\Delta{t}}\end{align}@$ c @$\begin{align}\text{acceleration} = \frac{\Delta{g}}{\Delta{h}}\end{align}@$ d @$\begin{align}\text{acceleration} = \frac{\Delta{v}}{\Delta{t}}\end{align}@$ Check It To calculate average acceleration when direction is not changing, divide the change in velocity by the change in time using the formula: The SI unit for acceleration is m/s 2. FlexCard™ Asked by Students Ask your question Here are the top questions that students are asking Flexi for this concept: How can one calculate distance traveled with acceleration? To calculate the distance traveled with acceleration, you can use the equation of motion: d = v i t + 0.5a t 2 where: - d is the distance traveled, - v i is the initial velocity, - t is the time, and - a is the acceleration. This equation assumes that the acceleration is constant over the time period. How can speed be determined from acceleration and distance? Speed can be determined from acceleration and distance using the equation v² = u² + 2as, where v is the final speed, u is the initial speed (which is zero if starting from rest), a is the acceleration, and s is the distance. If you're starting from rest, the equation simplifies to v = sqrt(2as). Kaya is riding her dirt bike eastward on a dirt road. She spots a pothole ahead. Kaya slows her car from 14.0 m/s to 5.5 m/s in 6.0 s. What is her acceleration? Acceleration is the rate of change of velocity. It can be calculated using the formula: a=v f−v i t where: a is acceleration, v f is the final velocity, v i is the initial velocity, and t is time. In this case, Kaya's initial velocity (v i) is 14.0 m/s, her final velocity (v f) is 5.5 m/s, and the time (t) is 6.0 s. Substituting these values into the formula gives: a=5.5 m/s−14.0 m/s 6.0 s=−1.42 m/s 2 The negative sign indicates that Kaya is decelerating, or slowing down. Calculate the acceleration of Josh riding his bicycle in a straight line that speeds up from 4 m/s to 6 m/s in 5 seconds. Acceleration is calculated by the formula: a=v f−v i t where: a = acceleration v f = final velocity v i = initial velocity t = time Substituting the given values: a=6 m/s−4 m/s 5 s=0.4 m/s 2 Velocity is speed. True/False Velocity and speed are related, but they are not the same. Speed is a scalar quantity that refers to "how fast an object is moving." Velocity is a vector quantity that refers to "the speed and direction of an object." So, velocity includes both the speed and the direction of the object's motion. Related Content In a Flash! Cliff Diver How to Find Acceleration Position, Velocity and Acceleration How to calculate acceleration Back to Calculating Acceleration from Velocity and Time \| Image \| Reference \| Attributions \| --- \| \| [Figure 1] \| Credit:Flickr: Stig Nygaard Source: License:CC BY 2.0 \| \| \| [Figure 2] \| Credit:AndrzejRembowski Source: License:Pixabay License \| Ask me anything! CK-12 Foundation is a non-profit organization that provides free educational materials and resources. FLEXIAPPS ABOUT Our missionMeet the teamPartnersPressCareersSecurityBlogCK-12 usage mapTestimonials SUPPORT Certified Educator ProgramCK-12 trainersWebinarsCK-12 resourcesHelpContact us BYCK-12 Common Core MathK-12 FlexBooksCollege FlexBooksTools and apps CONNECT TikTokInstagramYouTubeTwitterMediumFacebookLinkedIn v2.11.10.20250923073248-4b84c670be © CK-12 Foundation 2025 \| FlexBook Platform®, FlexBook®, FlexLet® and FlexCard™ are registered trademarks of CK-12 Foundation. 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2906	https://askfilo.com/user-question-answers-smart-solutions/6-iv-why-the-second-ionization-potential-of-sodium-is-very-3237353631353637	Question asked by Filo student (6) (iv) Why the second ionization potential of sodium is very high as compared to first ionization potential ? Views: 5,568 students Updated on: Mar 3, 2025 Text SolutionText solutionverified iconVerified Concepts: Ionization potential, Sodium, Atomic structure Explanation: The first ionization potential of sodium is the energy required to remove the outermost electron from a neutral sodium atom. Sodium has the electronic configuration of [Ne] 3s1, meaning it has one electron in its outermost shell. When this electron is removed, sodium becomes a positively charged ion (Na+), which has a stable noble gas configuration ([Ne]). The second ionization potential, however, involves removing an electron from this positively charged ion (Na+). This process requires significantly more energy because the remaining electrons are held more tightly by the increased positive charge of the nucleus. The effective nuclear charge experienced by the remaining electrons increases, making it harder to remove an electron. Therefore, the second ionization potential of sodium is much higher than the first ionization potential. Step by Step Solution: Step 1 Understand that the first ionization potential is the energy required to remove the outermost electron from a neutral sodium atom. Step 2 Recognize that after the first ionization, sodium becomes Na+, which has a stable electronic configuration. Step 3 Realize that removing an electron from Na+ requires more energy due to the increased effective nuclear charge, resulting in a much higher second ionization potential. Final Answer: The second ionization potential of sodium is very high compared to the first because it involves removing an electron from a positively charged ion (Na+), where the remaining electrons are held more tightly due to the increased effective nuclear charge. Students who ask this question also asked Views: 5,286 Topic: Smart Solutions View solution Views: 5,658 Topic: Smart Solutions View solution Views: 5,727 Topic: Smart Solutions View solution Views: 5,757 Topic: Smart Solutions View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| (6) (iv) Why the second ionization potential of sodium is very high as compared to first ionization potential ? \| \| Updated On \| Mar 3, 2025 \| \| Topic \| All topics \| \| Subject \| Smart Solutions \| \| Class \| Class 9 \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
2907	https://www.cardinalspellman.org/ourpages/auto/2011/3/28/55151620/Roles%20of%20a_%20b_%20c.pdf	Roles of a, b, c 1 March 29, 2011 Roles of a, b, c in the graphs of Quadratic Functions The Standard Formula for Quadratic Functions a = Coefficient of 1st term b = Coefficient of 2nd term c = Constant Term ax2 + bx + c = 0 Roles of a, b, c 2 March 29, 2011 The Standard Formula for Quadratic Functions a - represents a change in orientation (increasing the value narrows the parabola) (decreasing the value widens the parabola) Negative will flip the graph (reflection) ax2 + bx + c = 0 Quadratic Function y = x2 Quadratic Function y = 3x2 Quadratic Function y = -x2 Quadratic Function y = .5x2 Roles of a, b, c 3 March 29, 2011 The Standard Formula for Quadratic Functions b - helps determine the axis of symmetry (and turning point) for a parabola ax2 + bx + c = 0 The Standard Formula for Quadratic Functions c - represents a vertical change of the graph (y-intercept) ax2 + bx + c = 0 Roles of a, b, c 4 March 29, 2011 Quadratic Function y = x2 + 3x - 2 Quadratic Function y = 3x2 + 1 y = x2 - 4x + 3 x y Roles of a, b, c 5 March 29, 2011 Homework Textbook Page 129 # 3 - 15 ALL PROBLEMS
2908	https://cjhb.site/Files.php/Books/Discrete%20Mathematics/(Discrete%20Mathematics%20and%20Its%20Applications)%20Gary%20L.%20Mullen%20-%20Handbook%20of%20Finite%20Fields-Chapman%20and%20Hall_CRC%20(2013).pdf	DISCRETE MATHEMATICS AND ITS APPLICATIONS Series Editor KENNETH H. ROSEN Gary L. Mullen Daniel Panario HANDBOOK OF FINITE FIELDS HANDBOOK OF FINITE FIELDS DISCRETE MATHEMATICS ITS APPLICATIONS Series Editor Kenneth H. Rosen, Ph.D. R. B. J. T. Allenby and Alan Slomson, How to Count: An Introduction to Combinatorics, Third Edition Craig P . Bauer, Secret History: The Story of Cryptology Juergen Bierbrauer, Introduction to Coding Theory Katalin Bimbó, Combinatory Logic: Pure, Applied and Typed Donald Bindner and Martin Erickson, A Student’s Guide to the Study, Practice, and Tools of Modern Mathematics Francine Blanchet-Sadri, Algorithmic Combinatorics on Partial Words Miklós Bóna, Combinatorics of Permutations, Second Edition Richard A. 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Gross, Combinatorial Methods with Computer Applications Jonathan L. Gross and Jay Yellen, Graph Theory and Its Applications, Second Edition Jonathan L. Gross and Jay Yellen, Handbook of Graph Theory David S. Gunderson, Handbook of Mathematical Induction: Theory and Applications Richard Hammack, Wilfried Imrich, and Sandi Klavžar, Handbook of Product Graphs, Second Edition Darrel R. Hankerson, Greg A. Harris, and Peter D. Johnson, Introduction to Information Theory and Data Compression, Second Edition Darel W. Hardy, Fred Richman, and Carol L. Walker, Applied Algebra: Codes, Ciphers, and Discrete Algorithms, Second Edition Daryl D. Harms, Miroslav Kraetzl, Charles J. Colbourn, and John S. Devitt, Network Reliability: Experiments with a Symbolic Algebra Environment Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words Leslie Hogben, Handbook of Linear Algebra Derek F. Holt with Bettina Eick and Eamonn A. O’Brien, Handbook of Computational Group Theory David M. Jackson and Terry I. Visentin, An Atlas of Smaller Maps in Orientable and Nonorientable Surfaces Richard E. Klima, Neil P . Sigmon, and Ernest L. Stitzinger, Applications of Abstract Algebra with Maple™ and MATLAB® , Second Edition Richard E. Klima and Neil P . Sigmon, Cryptology: Classical and Modern with Maplets Patrick Knupp and Kambiz Salari, Verification of Computer Codes in Computational Science and Engineering William Kocay and Donald L. Kreher, Graphs, Algorithms, and Optimization Donald L. Kreher and Douglas R. Stinson, Combinatorial Algorithms: Generation Enumeration and Search Hang T. Lau, A Java Library of Graph Algorithms and Optimization C. C. Lindner and C. A. Rodger, Design Theory, Second Edition San Ling, Huaxiong Wang, and Chaoping Xing, Algebraic Curves in Cryptography Nicholas A. Loehr, Bijective Combinatorics Toufik Mansour, Combinatorics of Set Partitions Alasdair McAndrew, Introduction to Cryptography with Open-Source Software Elliott Mendelson, Introduction to Mathematical Logic, Fifth Edition Alfred J. Menezes, Paul C. van Oorschot, and Scott A. Vanstone, Handbook of Applied Cryptography Stig F. Mjølsnes, A Multidisciplinary Introduction to Information Security Jason J. Molitierno, Applications of Combinatorial Matrix Theory to Laplacian Matrices of Graphs Titles (continued) Richard A. Mollin, Advanced Number Theory with Applications Richard A. Mollin, Algebraic Number Theory, Second Edition Richard A. Mollin, Codes: The Guide to Secrecy from Ancient to Modern Times Richard A. Mollin, Fundamental Number Theory with Applications, Second Edition Richard A. Mollin, An Introduction to Cryptography, Second Edition Richard A. Mollin, Quadratics Richard A. Mollin, RSA and Public-Key Cryptography Carlos J. Moreno and Samuel S. Wagstaff, Jr., Sums of Squares of Integers Gary L. Mullen and Daniel Panario, Handbook of Finite Fields Goutam Paul and Subhamoy Maitra, RC4 Stream Cipher and Its Variants Dingyi Pei, Authentication Codes and Combinatorial Designs Kenneth H. Rosen, Handbook of Discrete and Combinatorial Mathematics Douglas R. Shier and K.T. Wallenius, Applied Mathematical Modeling: A Multidisciplinary Approach Alexander Stanoyevitch, Introduction to Cryptography with Mathematical Foundations and Computer Implementations Jörn Steuding, Diophantine Analysis Douglas R. Stinson, Cryptography: Theory and Practice, Third Edition Roberto Togneri and Christopher J. deSilva, Fundamentals of Information Theory and Coding Design W. D. Wallis, Introduction to Combinatorial Designs, Second Edition W. D. Wallis and J. C. George, Introduction to Combinatorics Jiacun Wang, Handbook of Finite State Based Models and Applications Lawrence C. Washington, Elliptic Curves: Number Theory and Cryptography, Second Edition DISCRETE MATHEMATICS AND ITS APPLICATIONS Series Editor KENNETH H. ROSEN Gary L. Mullen Daniel Panario HANDBOOK OF FINITE FIELDS CRC Press is an imprint of the Taylor &" Francis Group, an infcrma business A CHAPMAN &; HAll BOOK CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2013 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: 20130515 International Standard Book Number-13: 978-1-4398-7382-3 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the valid-ity of all materials or the consequences of their use. 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Daniel Panario This page intentionally left blank This page intentionally left blank Contents Part I: Introduction 1 History of ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1 Finite ﬁelds in the 18-th and 19-th centuries Roderick Gow . . . . . . . . 3 1.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1.2 Early anticipations of ﬁnite ﬁelds . . . . . . . . . . . . . . . . . 4 1.1.3 Gauss’s Disquisitiones Arithmeticae . . . . . . . . . . . . . . . 4 1.1.4 Gauss’s Disquisitiones Generales de Congruentiis . . . . . . . . 5 1.1.5 Galois’s Sur la th´ eorie des nombres . . . . . . . . . . . . . . . . 6 1.1.6 Serret’s Cours d’algebre sup´ erieure . . . . . . . . . . . . . . . . 8 1.1.7 Contributions of Sch¨ onemann and Dedekind . . . . . . . . . . . 9 1.1.8 Moore’s characterization of abstract ﬁnite ﬁelds . . . . . . . . . 10 1.1.9 Later developments . . . . . . . . . . . . . . . . . . . . . . . . 10 2 Introduction to ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.1 Basic properties of ﬁnite ﬁelds Gary L. Mullen and Daniel Panario . . . . 13 2.1.1 Basic deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.1.2 Fundamental properties of ﬁnite ﬁelds . . . . . . . . . . . . . . 14 2.1.3 Extension ﬁelds . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.1.4 Trace and norm functions . . . . . . . . . . . . . . . . . . . . . 20 2.1.5 Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.1.6 Linearized polynomials . . . . . . . . . . . . . . . . . . . . . . 23 2.1.7 Miscellaneous results . . . . . . . . . . . . . . . . . . . . . . . 23 2.1.7.1 The ﬁnite ﬁeld polynomial Φ function . . . . . . . . 24 2.1.7.2 Cyclotomic polynomials . . . . . . . . . . . . . . . 24 2.1.7.3 Lagrange interpolation . . . . . . . . . . . . . . . . 26 2.1.7.4 Discriminants . . . . . . . . . . . . . . . . . . . . . 26 2.1.7.5 Jacobi logarithms . . . . . . . . . . . . . . . . . . . 27 2.1.7.6 Field-like structures . . . . . . . . . . . . . . . . . 27 2.1.7.7 Galois rings . . . . . . . . . . . . . . . . . . . . . . 28 2.1.8 Finite ﬁeld related books . . . . . . . . . . . . . . . . . . . . . 31 2.1.8.1 Textbooks . . . . . . . . . . . . . . . . . . . . . . . 31 2.1.8.2 Finite ﬁeld theory . . . . . . . . . . . . . . . . . . 31 2.1.8.3 Applications . . . . . . . . . . . . . . . . . . . . . 31 2.1.8.4 Algorithms . . . . . . . . . . . . . . . . . . . . . . 31 2.1.8.5 Conference proceedings . . . . . . . . . . . . . . . . 31 2.2 Tables David Thomson . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.2.1 Low-weight irreducible and primitive polynomials . . . . . . . . 32 2.2.2 Low-complexity normal bases . . . . . . . . . . . . . . . . . . . 37 2.2.2.1 Exhaustive search for low complexity normal bases . 38 2.2.2.2 Minimum type of a Gauss period admitting a normal basis of F2n over F2 . . . . . . . . . . . . . . . . . . . . 40 2.2.2.3 Minimum-known complexity of a normal basis of F2n over F2, n ≥40 . . . . . . . . . . . . . . . . . . . . . . 41 2.2.3 Resources and standards . . . . . . . . . . . . . . . . . . . . . 46 ix x Contents Part II: Theoretical Properties 3 Irreducible polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3.1 Counting irreducible polynomials Joseph L.Yucas . . . . . . . . . . . . . 53 3.1.1 Prescribed trace or norm . . . . . . . . . . . . . . . . . . . . . 54 3.1.2 Prescribed coeﬃcients over the binary ﬁeld . . . . . . . . . . . 55 3.1.3 Self-reciprocal polynomials . . . . . . . . . . . . . . . . . . . . 56 3.1.4 Compositions of powers . . . . . . . . . . . . . . . . . . . . . . 57 3.1.5 Translation invariant polynomials . . . . . . . . . . . . . . . . 58 3.1.6 Normal replicators . . . . . . . . . . . . . . . . . . . . . . . . 58 3.2 Construction of irreducibles Melsik Kyuregyan . . . . . . . . . . . . . . . 60 3.2.1 Construction by composition . . . . . . . . . . . . . . . . . . . 60 3.2.2 Recursive constructions . . . . . . . . . . . . . . . . . . . . . . 63 3.3 Conditions for reducible polynomials Daniel Panario . . . . . . . . . . . 66 3.3.1 Composite polynomials . . . . . . . . . . . . . . . . . . . . . . 66 3.3.2 Swan-type theorems . . . . . . . . . . . . . . . . . . . . . . . . 67 3.4 Weights of irreducible polynomials Omran Ahmadi . . . . . . . . . . . . 70 3.4.1 Basic deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . 70 3.4.2 Existence results . . . . . . . . . . . . . . . . . . . . . . . . . 70 3.4.3 Conjectures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 3.5 Prescribed coeﬃcients Stephen D. Cohen . . . . . . . . . . . . . . . . . 73 3.5.1 One prescribed coeﬃcient . . . . . . . . . . . . . . . . . . . . . 74 3.5.2 Prescribed trace and norm . . . . . . . . . . . . . . . . . . . . 75 3.5.3 More prescribed coeﬃcients . . . . . . . . . . . . . . . . . . . . 76 3.5.4 Further exact expressions . . . . . . . . . . . . . . . . . . . . . 78 3.6 Multivariate polynomials Xiang-dong Hou . . . . . . . . . . . . . . . . . 80 3.6.1 Counting formulas . . . . . . . . . . . . . . . . . . . . . . . . . 80 3.6.2 Asymptotic formulas . . . . . . . . . . . . . . . . . . . . . . . 81 3.6.3 Results for the vector degree . . . . . . . . . . . . . . . . . . . 81 3.6.4 Indecomposable polynomials and irreducible polynomials . . . . 83 3.6.5 Algorithms for the gcd of multivariate polynomials . . . . . . . 84 4 Primitive polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 4.1 Introduction to primitive polynomials Gary L. Mullen and Daniel Panario 87 4.2 Prescribed coeﬃcients Stephen D. Cohen . . . . . . . . . . . . . . . . . 90 4.2.1 Approaches to results on prescribed coeﬃcients . . . . . . . . . 91 4.2.2 Existence theorems for primitive polynomials . . . . . . . . . . 92 4.2.3 Existence theorems for primitive normal polynomials . . . . . . 93 4.3 Weights of primitive polynomials Stephen D. Cohen . . . . . . . . . . . . 95 4.4 Elements of high order Jos´ e Felipe Voloch . . . . . . . . . . . . . . . . . 98 4.4.1 Elements of high order from elements of small orders . . . . . . 98 4.4.2 Gao’s construction and a generalization . . . . . . . . . . . . . 98 4.4.3 Iterative constructions . . . . . . . . . . . . . . . . . . . . . . 99 5 Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 5.1 Duality theory of bases Dieter Jungnickel . . . . . . . . . . . . . . . . . 101 5.1.1 Dual bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 5.1.2 Self-dual bases . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 5.1.3 Weakly self-dual bases . . . . . . . . . . . . . . . . . . . . . . 104 5.1.4 Binary bases with small excess . . . . . . . . . . . . . . . . . . 106 5.1.5 Almost weakly self-dual bases . . . . . . . . . . . . . . . . . . 107 5.1.6 Connections to hardware design . . . . . . . . . . . . . . . . . 109 Contents xi 5.2 Normal bases Shuhong Gao and Qunying Liao . . . . . . . . . . . . . . . 109 5.2.1 Basics on normal bases . . . . . . . . . . . . . . . . . . . . . . 110 5.2.2 Self-dual normal bases . . . . . . . . . . . . . . . . . . . . . . 114 5.2.3 Primitive normal bases . . . . . . . . . . . . . . . . . . . . . . 115 5.3 Complexity of normal bases Shuhong Gao and David Thomson . . . . . . 117 5.3.1 Optimal and low complexity normal bases . . . . . . . . . . . . 117 5.3.2 Gauss periods . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 5.3.3 Normal bases from elliptic periods . . . . . . . . . . . . . . . . 121 5.3.4 Complexities of dual and self-dual normal bases . . . . . . . . . 123 5.3.4.1 Duals of Gauss periods . . . . . . . . . . . . . . . . 125 5.3.5 Fast arithmetic using normal bases . . . . . . . . . . . . . . . . 125 5.4 Completely normal bases Dirk Hachenberger . . . . . . . . . . . . . . . . 128 5.4.1 The complete normal basis theorem . . . . . . . . . . . . . . . 128 5.4.2 The class of completely basic extensions . . . . . . . . . . . . . 130 5.4.3 Cyclotomic modules and complete generators . . . . . . . . . . 131 5.4.4 A decomposition theory for complete generators . . . . . . . . . 133 5.4.5 The class of regular extensions . . . . . . . . . . . . . . . . . . 134 5.4.6 Complete generators for regular cyclotomic modules . . . . . . . 135 5.4.7 Towards a primitive complete normal basis theorem . . . . . . . 137 6 Exponential and character sums . . . . . . . . . . . . . . . . . . . . . . 139 6.1 Gauss, Jacobi, and Kloosterman sums Ronald J. Evans . . . . . . . . . . 139 6.1.1 Properties of Gauss and Jacobi sums of general order . . . . . . 139 6.1.2 Evaluations of Jacobi and Gauss sums of small orders . . . . . . 148 6.1.3 Prime ideal divisors of Gauss and Jacobi sums . . . . . . . . . . 151 6.1.4 Kloosterman sums . . . . . . . . . . . . . . . . . . . . . . . . . 154 6.1.5 Gauss and Kloosterman sums over ﬁnite rings . . . . . . . . . . 159 6.2 More general exponential and character sums Antonio Rojas-Le´ on . . . . 161 6.2.1 One variable character sums . . . . . . . . . . . . . . . . . . . 161 6.2.2 Additive character sums . . . . . . . . . . . . . . . . . . . . . 162 6.2.3 Multiplicative character sums . . . . . . . . . . . . . . . . . . . 166 6.2.4 Generic estimates . . . . . . . . . . . . . . . . . . . . . . . . . 167 6.2.5 More general types of character sums . . . . . . . . . . . . . . 168 6.3 Some applications of character sums Alina Ostafe and Arne Winterhof . . 170 6.3.1 Applications of a simple character sum identity . . . . . . . . . 170 6.3.1.1 Hadamard matrices . . . . . . . . . . . . . . . . . . 170 6.3.1.2 Cyclotomic complete mappings and check digit systems 171 6.3.1.3 Periodic autocorrelation of cyclotomic generators . . 172 6.3.2 Applications of Gauss and Jacobi sums . . . . . . . . . . . . . 172 6.3.2.1 Reciprocity laws . . . . . . . . . . . . . . . . . . . 173 6.3.2.2 Distribution of linear congruential pseudorandom num-bers . . . . . . . . . . . . . . . . . . . . . . . . . . 174 6.3.2.3 Diagonal equations, Waring’s problem in ﬁnite ﬁelds, and covering radius of certain cyclic codes . . . . . . . . 175 6.3.2.4 Hidden number problem and noisy interpolation . . 176 6.3.3 Applications of the Weil bound . . . . . . . . . . . . . . . . . . 176 6.3.3.1 Superelliptic and Artin-Schreier equations . . . . . . 177 6.3.3.2 Stable quadratic polynomials . . . . . . . . . . . . . 177 6.3.3.3 Hamming distance of dual BCH codes . . . . . . . . 178 6.3.4 Applications of Kloosterman sums . . . . . . . . . . . . . . . . 179 6.3.4.1 Kloosterman equations and Kloosterman codes . . . 179 xii Contents 6.3.4.2 Distribution of inversive congruential pseudorandom num-bers . . . . . . . . . . . . . . . . . . . . . . . . . . 180 6.3.4.3 Nonlinearity of Boolean functions . . . . . . . . . . 180 6.3.5 Incomplete character sums . . . . . . . . . . . . . . . . . . . . 181 6.3.5.1 Finding deterministically linear factors of polynomials 181 6.3.5.2 Measures of pseudorandomness . . . . . . . . . . . 182 6.3.6 Other character sums . . . . . . . . . . . . . . . . . . . . . . . 183 6.3.6.1 Distribution of primitive elements and powers . . . . 183 6.3.6.2 Distribution of Diﬃe-Hellman triples . . . . . . . . 183 6.3.6.3 Thin sets with small discrete Fourier transform . . . 184 6.3.6.4 Character sums over arbitrary sets . . . . . . . . . . 184 6.4 Sum-product theorems and applications Moubariz Z. Garaev . . . . . . . 185 6.4.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 6.4.2 The sum-product estimate and its variants . . . . . . . . . . . . 186 6.4.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 7 Equations over ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . . . . . . . . 193 7.1 General forms Daqing Wan . . . . . . . . . . . . . . . . . . . . . . . . . 193 7.1.1 Aﬃne hypersurfaces . . . . . . . . . . . . . . . . . . . . . . . . 193 7.1.2 Projective hypersurfaces . . . . . . . . . . . . . . . . . . . . . 195 7.1.3 Toric hypersurfaces . . . . . . . . . . . . . . . . . . . . . . . . 196 7.1.4 Artin-Schreier hypersurfaces . . . . . . . . . . . . . . . . . . . 197 7.1.5 Kummer hypersurfaces . . . . . . . . . . . . . . . . . . . . . . 198 7.1.6 p-Adic estimates . . . . . . . . . . . . . . . . . . . . . . . . . . 199 7.2 Quadratic forms Robert Fitzgerald . . . . . . . . . . . . . . . . . . . . . 201 7.2.1 Basic deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . 201 7.2.2 Quadratic forms over ﬁnite ﬁelds . . . . . . . . . . . . . . . . . 202 7.2.3 Trace forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 7.2.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 7.3 Diagonal equations Francis Castro and Ivelisse Rubio . . . . . . . . . . . 206 7.3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 7.3.2 Solutions of diagonal equations . . . . . . . . . . . . . . . . . . 207 7.3.3 Generalizations of diagonal equations . . . . . . . . . . . . . . 210 7.3.4 Waring’s problem in ﬁnite ﬁelds . . . . . . . . . . . . . . . . . 211 8 Permutation polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 215 8.1 One variable Gary L. Mullen and Qiang Wang . . . . . . . . . . . . . . . 215 8.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 8.1.2 Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 8.1.3 Enumeration and distribution of PPs . . . . . . . . . . . . . . . 217 8.1.4 Constructions of PPs . . . . . . . . . . . . . . . . . . . . . . . 220 8.1.5 PPs from permutations of multiplicative groups . . . . . . . . . 221 8.1.6 PPs from permutations of additive groups . . . . . . . . . . . . 224 8.1.7 Other types of PPs from the AGW criterion . . . . . . . . . . . 224 8.1.8 Dickson and reversed Dickson PPs . . . . . . . . . . . . . . . . 226 8.1.9 Miscellaneous PPs . . . . . . . . . . . . . . . . . . . . . . . . . 228 8.2 Several variables Rudolf Lidl and Gary L. Mullen . . . . . . . . . . . . . 230 8.3 Value sets of polynomials Gary L. Mullen and Michael E. Zieve . . . . . 232 8.3.1 Large value sets . . . . . . . . . . . . . . . . . . . . . . . . . . 233 8.3.2 Small value sets . . . . . . . . . . . . . . . . . . . . . . . . . . 233 8.3.3 General polynomials . . . . . . . . . . . . . . . . . . . . . . . 234 8.3.4 Lower bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 8.3.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 Contents xiii 8.3.6 Further value set papers . . . . . . . . . . . . . . . . . . . . . 235 8.4 Exceptional polynomials Michael E. Zieve . . . . . . . . . . . . . . . . . 236 8.4.1 Fundamental properties . . . . . . . . . . . . . . . . . . . . . . 236 8.4.2 Indecomposable exceptional polynomials . . . . . . . . . . . . . 237 8.4.3 Exceptional polynomials and permutation polynomials . . . . . 238 8.4.4 Miscellany . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 8.4.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 9 Special functions over ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . . . . 241 9.1 Boolean functions Claude Carlet . . . . . . . . . . . . . . . . . . . . . . 241 9.1.1 Representation of Boolean functions . . . . . . . . . . . . . . . 242 9.1.1.1 Algebraic normal form . . . . . . . . . . . . . . . . 242 9.1.1.2 Trace representation . . . . . . . . . . . . . . . . . 243 9.1.2 The Walsh transform . . . . . . . . . . . . . . . . . . . . . . . 244 9.1.3 Parameters of Boolean functions . . . . . . . . . . . . . . . . . 244 9.1.4 Equivalence of Boolean functions . . . . . . . . . . . . . . . . . 246 9.1.5 Boolean functions and cryptography . . . . . . . . . . . . . . . 246 9.1.6 Constructions of cryptographic Boolean functions . . . . . . . . 249 9.1.6.1 Primary constructions of resilient functions . . . . . 249 9.1.6.2 Secondary constructions of resilient functions . . . . 249 9.1.6.3 Constructions of highly nonlinear functions with optimal algebraic immunity . . . . . . . . . . . . . . . . . . 250 9.1.7 Boolean functions and error correcting codes . . . . . . . . . . 251 9.1.7.1 Reed-Muller codes . . . . . . . . . . . . . . . . . . 251 9.1.7.2 Kerdock codes . . . . . . . . . . . . . . . . . . . . 251 9.1.8 Boolean functions and sequences . . . . . . . . . . . . . . . . . 251 9.1.8.1 Boolean functions and cross correlation of m-sequences 252 9.2 PN and APN functions Pascale Charpin . . . . . . . . . . . . . . . . . . 253 9.2.1 Functions from F2n into F2m . . . . . . . . . . . . . . . . . . . 253 9.2.2 Perfect Nonlinear (PN) functions . . . . . . . . . . . . . . . . . 254 9.2.3 Almost Perfect Nonlinear (APN) and Almost Bent (AB) functions 255 9.2.4 APN permutations . . . . . . . . . . . . . . . . . . . . . . . . 256 9.2.5 Properties of stability . . . . . . . . . . . . . . . . . . . . . . . 257 9.2.6 Coding theory point of view . . . . . . . . . . . . . . . . . . . 258 9.2.7 Quadratic APN functions . . . . . . . . . . . . . . . . . . . . . 258 9.2.8 APN monomials . . . . . . . . . . . . . . . . . . . . . . . . . . 260 9.3 Bent and related functions Alexander Kholosha and Alexander Pott . . . 262 9.3.1 Deﬁnitions and examples . . . . . . . . . . . . . . . . . . . . . 262 9.3.2 Basic properties of bent functions . . . . . . . . . . . . . . . . 264 9.3.3 Bent functions and other combinatorial objects . . . . . . . . . 265 9.3.4 Fundamental classes of bent functions . . . . . . . . . . . . . . 266 9.3.5 Boolean monomial and Niho bent functions . . . . . . . . . . . 268 9.3.6 p-ary bent functions in univariate form . . . . . . . . . . . . . . 270 9.3.7 Constructions using planar and s-plateaued functions . . . . . . 271 9.3.8 Vectorial bent functions and Kerdock codes . . . . . . . . . . . 272 9.4 κ-polynomials and related algebraic objects Robert Coulter . . . . . . . . 273 9.4.1 Deﬁnitions and preliminaries . . . . . . . . . . . . . . . . . . . 273 9.4.2 Pre-semiﬁelds, semiﬁelds, and isotopy . . . . . . . . . . . . . . 275 9.4.3 Semiﬁeld constructions . . . . . . . . . . . . . . . . . . . . . . 275 9.4.4 Semiﬁelds and nuclei . . . . . . . . . . . . . . . . . . . . . . . 276 9.5 Planar functions and commutative semiﬁelds Robert Coulter . . . . . . . 278 9.5.1 Deﬁnitions and preliminaries . . . . . . . . . . . . . . . . . . . 278 xiv Contents 9.5.2 Constructing aﬃne planes using planar functions . . . . . . . . 279 9.5.3 Examples, constructions, and equivalence . . . . . . . . . . . . 279 9.5.4 Classiﬁcation results, necessary conditions, and the Dembowski-Ostrom Conjecture . . . . . . . . . . . . . . . . . . . . . . . . 280 9.5.5 Planar DO polynomials and commutative semiﬁelds of odd order 281 9.6 Dickson polynomials Qiang Wang and Joseph L. Yucas . . . . . . . . . . 282 9.6.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 9.6.2 Factorization . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 9.6.2.1 a-reciprocals of polynomials . . . . . . . . . . . . . 285 9.6.2.2 The maps Φa and Ψa . . . . . . . . . . . . . . . . 286 9.6.2.3 Factors of Dickson polynomials . . . . . . . . . . . 286 9.6.2.4 a-cyclotomic polynomials . . . . . . . . . . . . . . . 287 9.6.3 Dickson polynomials of the (k + 1)-th kind . . . . . . . . . . . 287 9.6.4 Multivariate Dickson polynomials . . . . . . . . . . . . . . . . 289 9.7 Schur’s conjecture and exceptional covers Michael D. Fried . . . . . . . . 290 9.7.1 Rational function deﬁnitions . . . . . . . . . . . . . . . . . . . 290 9.7.2 MacCluer’s Theorem and Schur’s Conjecture . . . . . . . . . . 292 9.7.3 Fiber product of covers . . . . . . . . . . . . . . . . . . . . . . 295 9.7.4 Combining exceptional covers; the (Fq, Z) exceptional tower . . 296 9.7.5 Exceptional rational functions; Serre’s Open Image Theorem . . 298 9.7.6 Davenport pairs and Poincar´ e series . . . . . . . . . . . . . . . 300 10 Sequences over ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . . . . . . . . 303 10.1 Finite ﬁeld transforms Gary McGuire . . . . . . . . . . . . . . . . . . . 303 10.1.1 Basic deﬁnitions and important examples . . . . . . . . . . . . 303 10.1.2 Functions between two groups . . . . . . . . . . . . . . . . . . 306 10.1.3 Discrete Fourier Transform . . . . . . . . . . . . . . . . . . . . 307 10.1.4 Further topics . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 10.1.4.1 Fourier spectrum . . . . . . . . . . . . . . . . . . . 309 10.1.4.2 Nonlinearity . . . . . . . . . . . . . . . . . . . . . 309 10.1.4.3 Characteristic functions . . . . . . . . . . . . . . . 309 10.1.4.4 Gauss sums . . . . . . . . . . . . . . . . . . . . . . 310 10.1.4.5 Uncertainty principle . . . . . . . . . . . . . . . . . 310 10.2 LFSR sequences and maximal period sequences Harald Niederreiter . . . 311 10.2.1 General properties of LFSR sequences . . . . . . . . . . . . . . 311 10.2.2 Operations with LFSR sequences and characterizations . . . . . 313 10.2.3 Maximal period sequences . . . . . . . . . . . . . . . . . . . . 315 10.2.4 Distribution properties of LFSR sequences . . . . . . . . . . . . 315 10.2.5 Applications of LFSR sequences . . . . . . . . . . . . . . . . . 316 10.3 Correlation and autocorrelation of sequences Tor Helleseth . . . . . . . . 317 10.3.1 Basic deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . 317 10.3.2 Autocorrelation of sequences . . . . . . . . . . . . . . . . . . . 318 10.3.3 Sequence families with low correlation . . . . . . . . . . . . . . 319 10.3.4 Quaternary sequences . . . . . . . . . . . . . . . . . . . . . . . 321 10.3.5 Other correlation measures . . . . . . . . . . . . . . . . . . . . 322 10.4 Linear complexity of sequences and multisequences Wilfried Meidl and Arne Winterhof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 10.4.1 Linear complexity measures . . . . . . . . . . . . . . . . . . . . 324 10.4.2 Analysis of the linear complexity . . . . . . . . . . . . . . . . . 327 10.4.3 Average behavior of the linear complexity . . . . . . . . . . . . 329 10.4.4 Some sequences with large n-th linear complexity . . . . . . . . 331 10.4.4.1 Explicit sequences . . . . . . . . . . . . . . . . . . 331 Contents xv 10.4.4.2 Recursive nonlinear sequences . . . . . . . . . . . . 332 10.4.4.3 Legendre sequence and related bit sequences . . . . 333 10.4.4.4 Elliptic curve sequences . . . . . . . . . . . . . . . 334 10.4.5 Related measures . . . . . . . . . . . . . . . . . . . . . . . . . 334 10.4.5.1 Kolmogorov complexity . . . . . . . . . . . . . . . 334 10.4.5.2 Lattice test . . . . . . . . . . . . . . . . . . . . . . 335 10.4.5.3 Correlation measure of order k . . . . . . . . . . . . 335 10.4.5.4 FCSR and p-adic span . . . . . . . . . . . . . . . . 335 10.4.5.5 Discrepancy . . . . . . . . . . . . . . . . . . . . . . 336 10.5 Algebraic dynamical systems over ﬁnite ﬁelds Igor Shparlinski . . . . . . 337 10.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 10.5.2 Background and main deﬁnitions . . . . . . . . . . . . . . . . . 337 10.5.3 Degree growth . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 10.5.4 Linear independence and other algebraic properties of iterates . 340 10.5.5 Multiplicative independence of iterates . . . . . . . . . . . . . . 341 10.5.6 Trajectory length . . . . . . . . . . . . . . . . . . . . . . . . . 341 10.5.7 Irreducibility of iterates . . . . . . . . . . . . . . . . . . . . . . 342 10.5.8 Diameter of partial trajectories . . . . . . . . . . . . . . . . . . 343 11 Algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 11.1 Computational techniques Christophe Doche . . . . . . . . . . . . . . . 345 11.1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 11.1.1.1 Prime ﬁeld generation . . . . . . . . . . . . . . . . 346 11.1.1.2 Extension ﬁeld generation . . . . . . . . . . . . . . 347 11.1.1.3 Primitive elements . . . . . . . . . . . . . . . . . . 349 11.1.1.4 Order of an irreducible polynomial and primitive polyno-mials . . . . . . . . . . . . . . . . . . . . . . . . . 349 11.1.1.5 Minimal polynomial of an element . . . . . . . . . . 350 11.1.2 Representation of ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . . 350 11.1.3 Modular reduction . . . . . . . . . . . . . . . . . . . . . . . . 351 11.1.3.1 Prime ﬁelds . . . . . . . . . . . . . . . . . . . . . . 351 11.1.3.2 Extension ﬁelds . . . . . . . . . . . . . . . . . . . . 353 11.1.4 Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 11.1.5 Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 11.1.5.1 Prime ﬁelds . . . . . . . . . . . . . . . . . . . . . . 354 11.1.5.2 Extension ﬁelds . . . . . . . . . . . . . . . . . . . . 355 11.1.6 Squaring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356 11.1.6.1 Finite ﬁelds of odd characteristic . . . . . . . . . . . 356 11.1.6.2 Finite ﬁelds of characteristic two . . . . . . . . . . . 356 11.1.7 Exponentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 356 11.1.7.1 Prime ﬁelds . . . . . . . . . . . . . . . . . . . . . . 356 11.1.7.2 Extension ﬁelds . . . . . . . . . . . . . . . . . . . . 357 11.1.8 Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358 11.1.8.1 Prime ﬁelds . . . . . . . . . . . . . . . . . . . . . . 359 11.1.8.2 Extension ﬁelds . . . . . . . . . . . . . . . . . . . . 360 11.1.9 Squares and square roots . . . . . . . . . . . . . . . . . . . . . 360 11.1.9.1 Finite ﬁelds of odd characteristic . . . . . . . . . . . 361 11.1.9.2 Finite ﬁelds of even characteristic . . . . . . . . . . 363 11.2 Univariate polynomial counting and algorithms Daniel Panario . . . . . . 364 11.2.1 Classical counting results . . . . . . . . . . . . . . . . . . . . . 364 11.2.2 Analytic combinatorics approach . . . . . . . . . . . . . . . . . 365 11.2.3 Some illustrations of polynomial counting . . . . . . . . . . . . 367 xvi Contents 11.2.3.1 Number of irreducible factors of a polynomial . . . . 367 11.2.3.2 Factorization patterns . . . . . . . . . . . . . . . . 368 11.2.3.3 Largest and smallest degree irreducibles . . . . . . . 369 11.2.3.4 Greatest common divisor of polynomials . . . . . . 371 11.2.3.5 Relations to permutations and integers . . . . . . . 372 11.3 Algorithms for irreducibility testing and for constructing irreducible polynomi-als Mark Giesbrecht . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374 11.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374 11.3.2 Early irreducibility tests of univariate polynomials . . . . . . . 375 11.3.3 Rabin’s irreducibility test . . . . . . . . . . . . . . . . . . . . . 376 11.3.4 Constructing irreducible polynomials: randomized algorithms . . 377 11.3.5 Ben-Or’s algorithm for construction of irreducible polynomials . 377 11.3.6 Shoup’s algorithm for construction of irreducible polynomials . . 378 11.3.7 Constructing irreducible polynomials: deterministic algorithms . 378 11.3.8 Construction of irreducible polynomials of approximate degree . 379 11.4 Factorization of univariate polynomials Joachim von zur Gathen . . . . . 380 11.5 Factorization of multivariate polynomials Erich Kaltofen and Gr´ egoire Lecerf 382 11.5.1 Factoring dense multivariate polynomials . . . . . . . . . . . . 382 11.5.1.1 Separable factorization . . . . . . . . . . . . . . . . 382 11.5.1.2 Squarefree factorization . . . . . . . . . . . . . . . 384 11.5.1.3 Bivariate irreducible factorization . . . . . . . . . . 384 11.5.1.4 Reduction from any number to two variables . . . . 386 11.5.2 Factoring sparse multivariate polynomials . . . . . . . . . . . . 387 11.5.2.1 Ostrowski’s theorem . . . . . . . . . . . . . . . . . 388 11.5.2.2 Irreducibility tests based on indecomposability of poly-topes . . . . . . . . . . . . . . . . . . . . . . . . . 388 11.5.2.3 Sparse bivariate Hensel lifting driven by polytopes . 388 11.5.2.4 Convex-dense bivariate factorization . . . . . . . . . 389 11.5.3 Factoring straight-line programs and black boxes . . . . . . . . 390 11.6 Discrete logarithms over ﬁnite ﬁelds Andrew Odlyzko . . . . . . . . . . . 393 11.6.1 Basic deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . 393 11.6.2 Modern computer implementations . . . . . . . . . . . . . . . . 394 11.6.3 Historical remarks . . . . . . . . . . . . . . . . . . . . . . . . . 394 11.6.4 Basic properties of discrete logarithms . . . . . . . . . . . . . . 395 11.6.5 Chinese Remainder Theorem reduction: The Silver–Pohlig–Hellman algorithm . . . . . . . . . . . . . . 395 11.6.6 Baby steps–giant steps algorithm . . . . . . . . . . . . . . . . . 396 11.6.7 Pollard rho and kangaroo methods for discrete logarithms . . . 397 11.6.8 Index calculus algorithms for discrete logarithms in ﬁnite ﬁelds . 397 11.6.9 Smooth integers and smooth polynomials . . . . . . . . . . . . 399 11.6.10 Sparse linear systems of equations . . . . . . . . . . . . . . . . 399 11.6.11 Current discrete logarithm records . . . . . . . . . . . . . . . . 400 11.7 Standard models for ﬁnite ﬁelds Bart de Smit and Hendrik Lenstra . . . 401 12 Curves over ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 12.1 Introduction to function ﬁelds and curves Arnaldo Garcia and Henning Stichtenoth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406 12.1.1 Valuations and places . . . . . . . . . . . . . . . . . . . . . . . 406 12.1.2 Divisors and Riemann–Roch theorem . . . . . . . . . . . . . . 409 12.1.3 Extensions of function ﬁelds . . . . . . . . . . . . . . . . . . . 413 12.1.4 Diﬀerentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 12.1.5 Function ﬁelds and curves . . . . . . . . . . . . . . . . . . . . 421 Contents xvii 12.2 Elliptic curves Joseph Silverman . . . . . . . . . . . . . . . . . . . . . . 422 12.2.1 Weierstrass equations . . . . . . . . . . . . . . . . . . . . . . . 423 12.2.2 The group law . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 12.2.3 Isogenies and endomorphisms . . . . . . . . . . . . . . . . . . . 427 12.2.4 The number of points in E(Fq) . . . . . . . . . . . . . . . . . . 430 12.2.5 Twists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431 12.2.6 The torsion subgroup and the Tate module . . . . . . . . . . . 432 12.2.7 The Weil pairing and the Tate pairing . . . . . . . . . . . . . . 433 12.2.8 The endomorphism ring and automorphism group . . . . . . . . 435 12.2.9 Ordinary and supersingular elliptic curves . . . . . . . . . . . . 436 12.2.10 The zeta function of an elliptic curve . . . . . . . . . . . . . . . 438 12.2.11 The elliptic curve discrete logarithm problem . . . . . . . . . . 439 12.3 Addition formulas for elliptic curves Daniel J. Bernstein and Tanja Lange 440 12.3.1 Curve shapes . . . . . . . . . . . . . . . . . . . . . . . . . . . 440 12.3.2 Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 12.3.3 Coordinate systems . . . . . . . . . . . . . . . . . . . . . . . . 442 12.3.4 Explicit formulas . . . . . . . . . . . . . . . . . . . . . . . . . 443 12.3.5 Short Weierstrass curves, large characteristic: y2 = x3 −3x + b 444 12.3.6 Short Weierstrass curves, characteristic 2, ordinary case: y2 + xy = x3 + a2x2 + a6 . . . . . . . . . . . . . . . . . . . . . . . . . . 444 12.3.7 Montgomery curves: by2 = x3 + ax2 + x . . . . . . . . . . . . 445 12.3.8 Twisted Edwards curves: ax2 + y2 = 1 + dx2y2 . . . . . . . . . 446 12.4 Hyperelliptic curves Michael John Jacobson, Jr. and Renate Scheidler . . 447 12.4.1 Hyperelliptic equations . . . . . . . . . . . . . . . . . . . . . . 447 12.4.2 The degree zero divisor class group . . . . . . . . . . . . . . . . 449 12.4.3 Divisor class arithmetic over ﬁnite ﬁelds . . . . . . . . . . . . . 450 12.4.4 Endomorphisms and supersingularity . . . . . . . . . . . . . . . 453 12.4.5 Class number computation . . . . . . . . . . . . . . . . . . . . 453 12.4.6 The Tate-Lichtenbaum pairing . . . . . . . . . . . . . . . . . . 454 12.4.7 The hyperelliptic curve discrete logarithm problem . . . . . . . 455 12.5 Rational points on curves Arnaldo Garcia and Henning Stichtenoth . . . 456 12.5.1 Rational places . . . . . . . . . . . . . . . . . . . . . . . . . . 457 12.5.2 The Zeta function of a function ﬁeld . . . . . . . . . . . . . . . 458 12.5.3 Bounds for the number of rational places . . . . . . . . . . . . 459 12.5.4 Maximal function ﬁelds . . . . . . . . . . . . . . . . . . . . . . 461 12.5.5 Asymptotic bounds . . . . . . . . . . . . . . . . . . . . . . . . 462 12.6 Towers Arnaldo Garcia and Henning Stichtenoth . . . . . . . . . . . . . 464 12.6.1 Introduction to towers . . . . . . . . . . . . . . . . . . . . . . 464 12.6.2 Examples of towers . . . . . . . . . . . . . . . . . . . . . . . . 466 12.7 Zeta functions and L-functions Lei Fu . . . . . . . . . . . . . . . . . . . 469 12.7.1 Zeta functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 12.7.2 L-functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474 12.7.3 The case of curves . . . . . . . . . . . . . . . . . . . . . . . . . 477 12.8 p-adic estimates of zeta functions and L-functions R´ egis Blache . . . . . 479 12.8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 12.8.2 Lower bounds for the ﬁrst slope . . . . . . . . . . . . . . . . . 480 12.8.3 Uniform lower bounds for Newton polygons . . . . . . . . . . . 481 12.8.4 Variation of Newton polygons in a family . . . . . . . . . . . . 483 12.8.5 The case of curves and abelian varieties . . . . . . . . . . . . . 485 12.9 Computing the number of rational points and zeta functions Daqing Wan 488 12.9.1 Point counting: sparse input . . . . . . . . . . . . . . . . . . . 488 xviii Contents 12.9.2 Point counting: dense input . . . . . . . . . . . . . . . . . . . . 489 12.9.3 Computing zeta functions: general case . . . . . . . . . . . . . 490 12.9.4 Computing zeta functions: curve case . . . . . . . . . . . . . . 491 13 Miscellaneous theoretical topics . . . . . . . . . . . . . . . . . . . . . . 493 13.1 Relations between integers and polynomials over ﬁnite ﬁelds Gove Eﬃnger 493 13.1.1 The density of primes and irreducibles . . . . . . . . . . . . . . 494 13.1.2 Primes and irreducibles in arithmetic progression . . . . . . . . 495 13.1.3 Twin primes and irreducibles . . . . . . . . . . . . . . . . . . . 495 13.1.4 The generalized Riemann hypothesis . . . . . . . . . . . . . . . 496 13.1.5 The Goldbach problem over ﬁnite ﬁelds . . . . . . . . . . . . . 497 13.1.6 The Waring problem over ﬁnite ﬁelds . . . . . . . . . . . . . . 498 13.2 Matrices over ﬁnite ﬁelds Dieter Jungnickel . . . . . . . . . . . . . . . . 500 13.2.1 Matrices of speciﬁed rank . . . . . . . . . . . . . . . . . . . . . 500 13.2.2 Matrices of speciﬁed order . . . . . . . . . . . . . . . . . . . . 501 13.2.3 Matrix representations of ﬁnite ﬁelds . . . . . . . . . . . . . . . 503 13.2.4 Circulant and orthogonal matrices . . . . . . . . . . . . . . . . 504 13.2.5 Symmetric and skew-symmetric matrices . . . . . . . . . . . . . 506 13.2.6 Hankel and Toeplitz matrices . . . . . . . . . . . . . . . . . . . 507 13.2.7 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 13.3 Classical groups over ﬁnite ﬁelds Zhe-Xian Wan . . . . . . . . . . . . . . 510 13.3.1 Linear groups over ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . 510 13.3.2 Symplectic groups over ﬁnite ﬁelds . . . . . . . . . . . . . . . . 512 13.3.3 Unitary groups over ﬁnite ﬁelds . . . . . . . . . . . . . . . . . 514 13.3.4 Orthogonal groups over ﬁnite ﬁelds of characteristic not two . . 516 13.3.5 Orthogonal groups over ﬁnite ﬁelds of characteristic two . . . . 519 13.4 Computational linear algebra over ﬁnite ﬁelds Jean-Guillaume Dumas and Cl´ ement Pernet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 520 13.4.1 Dense matrix multiplication . . . . . . . . . . . . . . . . . . . 521 13.4.1.1 Tiny ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . . 521 13.4.1.2 Word size prime ﬁelds . . . . . . . . . . . . . . . . 523 13.4.1.3 Large ﬁnite ﬁelds . . . . . . . . . . . . . . . . . . . 524 13.4.1.4 Large matrices: subcubic time complexity . . . . . . 524 13.4.2 Dense Gaussian elimination and echelon forms . . . . . . . . . . 525 13.4.2.1 Building blocks . . . . . . . . . . . . . . . . . . . . 525 13.4.2.2 PLE decomposition . . . . . . . . . . . . . . . . . . 526 13.4.2.3 Echelon forms . . . . . . . . . . . . . . . . . . . . . 527 13.4.3 Minimal and characteristic polynomial of a dense matrix . . . . 528 13.4.4 Blackbox iterative methods . . . . . . . . . . . . . . . . . . . . 530 13.4.4.1 Minimal polynomial and the Wiedemann algorithm . 530 13.4.4.2 Rank, determinant, and characteristic polynomial . . 531 13.4.4.3 System solving and the Lanczos algorithm . . . . . 531 13.4.5 Sparse and structured methods . . . . . . . . . . . . . . . . . . 532 13.4.5.1 Reordering . . . . . . . . . . . . . . . . . . . . . . 532 13.4.5.2 Structured matrices and displacement rank . . . . . 532 13.4.6 Hybrid methods . . . . . . . . . . . . . . . . . . . . . . . . . . 534 13.4.6.1 Hybrid sparse-dense methods . . . . . . . . . . . . 534 13.4.6.2 Block-iterative methods . . . . . . . . . . . . . . . 534 13.5 Carlitz and Drinfeld modules David Goss . . . . . . . . . . . . . . . . . 535 13.5.1 Quick review . . . . . . . . . . . . . . . . . . . . . . . . . . . 536 13.5.2 Drinfeld modules: deﬁnition and analytic theory . . . . . . . . . 537 13.5.3 Drinfeld modules over ﬁnite ﬁelds . . . . . . . . . . . . . . . . 539 Contents xix 13.5.4 The reduction theory of Drinfeld modules . . . . . . . . . . . . 539 13.5.5 The A-module of rational points . . . . . . . . . . . . . . . . . 540 13.5.6 The invariants of a Drinfeld module . . . . . . . . . . . . . . . 540 13.5.7 The L-series of a Drinfeld module . . . . . . . . . . . . . . . . 541 13.5.8 Special values . . . . . . . . . . . . . . . . . . . . . . . . . . . 542 13.5.9 Measures and symmetries . . . . . . . . . . . . . . . . . . . . . 542 13.5.10 Multizeta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544 13.5.11 Modular theory . . . . . . . . . . . . . . . . . . . . . . . . . . 544 13.5.12 Transcendency results . . . . . . . . . . . . . . . . . . . . . . . 545 Part III: Applications 14 Combinatorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 14.1 Latin squares Gary L. Mullen . . . . . . . . . . . . . . . . . . . . . . . . 550 14.1.1 Prime powers . . . . . . . . . . . . . . . . . . . . . . . . . . . 551 14.1.2 Non-prime powers . . . . . . . . . . . . . . . . . . . . . . . . . 552 14.1.3 Frequency squares . . . . . . . . . . . . . . . . . . . . . . . . . 553 14.1.4 Hypercubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553 14.1.5 Connections to aﬃne and projective planes . . . . . . . . . . . 554 14.1.6 Other ﬁnite ﬁeld constructions for MOLS . . . . . . . . . . . . 555 14.2 Lacunary polynomials over ﬁnite ﬁelds Simeon Ball and Aart Blokhuis . . 556 14.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 556 14.2.2 Lacunary polynomials . . . . . . . . . . . . . . . . . . . . . . . 556 14.2.3 Directions and R´ edei polynomials . . . . . . . . . . . . . . . . 557 14.2.4 Sets of points determining few directions . . . . . . . . . . . . . 558 14.2.5 Lacunary polynomials and blocking sets . . . . . . . . . . . . . 559 14.2.6 Lacunary polynomials and blocking sets in planes of prime order 561 14.2.7 Lacunary polynomials and multiple blocking sets . . . . . . . . 562 14.3 Aﬃne and projective planes Gary Ebert and Leo Storme . . . . . . . . . 563 14.3.1 Projective planes . . . . . . . . . . . . . . . . . . . . . . . . . 563 14.3.2 Aﬃne planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564 14.3.3 Translation planes and spreads . . . . . . . . . . . . . . . . . . 565 14.3.4 Nest planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567 14.3.5 Flag-transitive aﬃne planes . . . . . . . . . . . . . . . . . . . . 568 14.3.6 Subplanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569 14.3.7 Embedded unitals . . . . . . . . . . . . . . . . . . . . . . . . . 571 14.3.8 Maximal arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . 572 14.3.9 Other results . . . . . . . . . . . . . . . . . . . . . . . . . . . 573 14.4 Projective spaces James W.P. Hirschfeld and Joseph A. Thas . . . . . . . 574 14.4.1 Projective and aﬃne spaces . . . . . . . . . . . . . . . . . . . . 574 14.4.2 Collineations, correlations, and coordinate frames . . . . . . . . 576 14.4.3 Polarities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 578 14.4.4 Partitions and cyclic projectivities . . . . . . . . . . . . . . . . 582 14.4.5 k-Arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583 14.4.6 k-Arcs and linear MDS codes . . . . . . . . . . . . . . . . . . . 586 14.4.7 k-Caps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587 14.5 Block designs Charles J. Colbourn and Jeﬀrey H. Dinitz . . . . . . . . . 589 14.5.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 14.5.2 Triple systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 590 14.5.3 Diﬀerence families and balanced incomplete block designs . . . . 592 xx Contents 14.5.4 Nested designs . . . . . . . . . . . . . . . . . . . . . . . . . . . 594 14.5.5 Pairwise balanced designs . . . . . . . . . . . . . . . . . . . . . 596 14.5.6 Group divisible designs . . . . . . . . . . . . . . . . . . . . . . 596 14.5.7 t-designs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597 14.5.8 Packing and covering . . . . . . . . . . . . . . . . . . . . . . . 598 14.6 Diﬀerence sets Alexander Pott . . . . . . . . . . . . . . . . . . . . . . . 599 14.6.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 599 14.6.2 Diﬀerence sets in cyclic groups . . . . . . . . . . . . . . . . . . 601 14.6.3 Diﬀerence sets in the additive groups of ﬁnite ﬁelds . . . . . . . 603 14.6.4 Diﬀerence sets and Hadamard matrices . . . . . . . . . . . . . 604 14.6.5 Further families of diﬀerence sets . . . . . . . . . . . . . . . . . 605 14.6.6 Diﬀerence sets and character sums . . . . . . . . . . . . . . . . 606 14.6.7 Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606 14.7 Other combinatorial structures Jeﬀrey H. Dinitz and Charles J. Colbourn 607 14.7.1 Association schemes . . . . . . . . . . . . . . . . . . . . . . . . 607 14.7.2 Costas arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . 608 14.7.3 Conference matrices . . . . . . . . . . . . . . . . . . . . . . . . 609 14.7.4 Covering arrays . . . . . . . . . . . . . . . . . . . . . . . . . . 610 14.7.5 Hall triple systems . . . . . . . . . . . . . . . . . . . . . . . . 611 14.7.6 Ordered designs and perpendicular arrays . . . . . . . . . . . . 611 14.7.7 Perfect hash families . . . . . . . . . . . . . . . . . . . . . . . 612 14.7.8 Room squares and starters . . . . . . . . . . . . . . . . . . . . 614 14.7.9 Strongly regular graphs . . . . . . . . . . . . . . . . . . . . . . 617 14.7.10 Whist tournaments . . . . . . . . . . . . . . . . . . . . . . . . 617 14.8 (t, m, s)-nets and (t, s)-sequences Harald Niederreiter . . . . . . . . . . . 619 14.8.1 (t, m, s)-nets . . . . . . . . . . . . . . . . . . . . . . . . . . . 619 14.8.2 Digital (t, m, s)-nets . . . . . . . . . . . . . . . . . . . . . . . 621 14.8.3 Constructions of (t, m, s)-nets . . . . . . . . . . . . . . . . . . 623 14.8.4 (t, s)-sequences and (T, s)-sequences . . . . . . . . . . . . . . . 625 14.8.5 Digital (t, s)-sequences and digital (T, s)-sequences . . . . . . . 626 14.8.6 Constructions of (t, s)-sequences and (T, s)-sequences . . . . . 628 14.9 Applications and weights of multiples of primitive and other polynomials Brett Stevens . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 630 14.9.1 Applications where weights of multiples of a base polynomial are rel-evant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 630 14.9.1.1 Applications from other Handbook sections . . . . . 630 14.9.1.2 Application of polynomials to the construction of orthog-onal arrays . . . . . . . . . . . . . . . . . . . . . . 631 14.9.1.3 Application of polynomials to a card trick . . . . . . 632 14.9.2 Weights of multiples of polynomials . . . . . . . . . . . . . . . 633 14.9.2.1 General bounds on d((Cf n)⊥) . . . . . . . . . . . . 633 14.9.2.2 Bounds on d((Cf n)⊥) for polynomials of speciﬁc degree 635 14.9.2.3 Bounds on d((Cf n)⊥) for polynomials of speciﬁc weight 638 14.10 Ramanujan and expander graphs M. Ram Murty and Sebastian M. Cioab˘ a 642 14.10.1 Graphs, adjacency matrices, and eigenvalues . . . . . . . . . . . 643 14.10.2 Ramanujan graphs . . . . . . . . . . . . . . . . . . . . . . . . 646 14.10.3 Expander graphs . . . . . . . . . . . . . . . . . . . . . . . . . 648 14.10.4 Cayley graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 649 14.10.5 Explicit constructions of Ramanujan graphs . . . . . . . . . . . 652 14.10.6 Combinatorial constructions of expanders . . . . . . . . . . . . 655 14.10.7 Zeta functions of graphs . . . . . . . . . . . . . . . . . . . . . 657 Contents xxi 15 Algebraic coding theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 659 15.1 Basic coding properties and bounds Ian Blake and W. Cary Huﬀman . . 659 15.1.1 Channel models and error correction . . . . . . . . . . . . . . . 659 15.1.2 Linear codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 661 15.1.2.1 Standard array decoding of linear codes . . . . . . . 665 15.1.2.2 Hamming codes . . . . . . . . . . . . . . . . . . . . 666 15.1.2.3 Reed-Muller codes . . . . . . . . . . . . . . . . . . 667 15.1.2.4 Subﬁeld and trace codes . . . . . . . . . . . . . . . 668 15.1.2.5 Modifying linear codes . . . . . . . . . . . . . . . . 669 15.1.2.6 Bounds on codes . . . . . . . . . . . . . . . . . . . 670 15.1.2.7 Asymptotic bounds . . . . . . . . . . . . . . . . . . 673 15.1.3 Cyclic codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674 15.1.3.1 Algebraic prerequisites . . . . . . . . . . . . . . . . 675 15.1.3.2 Properties of cyclic codes . . . . . . . . . . . . . . . 676 15.1.3.3 Classes of cyclic codes . . . . . . . . . . . . . . . . 677 15.1.4 A spectral approach to coding . . . . . . . . . . . . . . . . . . 689 15.1.5 Codes and combinatorics . . . . . . . . . . . . . . . . . . . . . 690 15.1.6 Decoding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 692 15.1.6.1 Decoding BCH codes . . . . . . . . . . . . . . . . . 692 15.1.6.2 The Peterson-Gorenstein-Zierler decoder . . . . . . 692 15.1.6.3 Berlekamp-Massey decoding . . . . . . . . . . . . . 693 15.1.6.4 Extended Euclidean algorithm decoding . . . . . . . 694 15.1.6.5 Welch-Berlekamp decoding of GRS codes . . . . . . 695 15.1.6.6 Majority logic decoding . . . . . . . . . . . . . . . 696 15.1.6.7 Generalized minimum distance decoding . . . . . . . 696 15.1.6.8 List decoding - decoding beyond the minimum distance bound . . . . . . . . . . . . . . . . . . . . . . . . . 698 15.1.7 Codes over Z4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 699 15.1.8 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702 15.2 Algebraic-geometry codes Harald Niederreiter . . . . . . . . . . . . . . . 703 15.2.1 Classical algebraic-geometry codes . . . . . . . . . . . . . . . . 703 15.2.2 Generalized algebraic-geometry codes . . . . . . . . . . . . . . 705 15.2.3 Function-ﬁeld codes . . . . . . . . . . . . . . . . . . . . . . . . 708 15.2.4 Asymptotic bounds . . . . . . . . . . . . . . . . . . . . . . . . 710 15.3 LDPC and Gallager codes over ﬁnite ﬁelds Ian Blake and W. Cary Huﬀman 713 15.4 Turbo codes over ﬁnite ﬁelds Oscar Takeshita . . . . . . . . . . . . . . . 719 15.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 719 15.4.1.1 Historical background . . . . . . . . . . . . . . . . 719 15.4.1.2 Terminology . . . . . . . . . . . . . . . . . . . . . 719 15.4.2 Convolutional codes . . . . . . . . . . . . . . . . . . . . . . . . 721 15.4.2.1 Non-recursive convolutional codes . . . . . . . . . . 721 15.4.2.2 Distance properties of non-recursive convolutional codes 722 15.4.2.3 Recursive convolutional codes . . . . . . . . . . . . 723 15.4.2.4 Distance properties of recursive convolutional codes 723 15.4.3 Permutations and interleavers . . . . . . . . . . . . . . . . . . 724 15.4.4 Encoding and decoding . . . . . . . . . . . . . . . . . . . . . . 725 15.4.5 Design of turbo codes . . . . . . . . . . . . . . . . . . . . . . . 725 15.4.5.1 Design of the recursive convolutional code . . . . . . 726 15.4.5.2 Design of interleavers . . . . . . . . . . . . . . . . . 726 15.5 Raptor codes Ian Blake and W. Cary Huﬀman . . . . . . . . . . . . . . 727 15.5.1 Tornado codes . . . . . . . . . . . . . . . . . . . . . . . . . . . 728 xxii Contents 15.5.2 LT and fountain codes . . . . . . . . . . . . . . . . . . . . . . 730 15.5.3 Raptor codes . . . . . . . . . . . . . . . . . . . . . . . . . . . 733 15.6 Polar codes Simon Litsyn . . . . . . . . . . . . . . . . . . . . . . . . . . 735 15.6.1 Space decomposition . . . . . . . . . . . . . . . . . . . . . . . 735 15.6.2 Vector transformation . . . . . . . . . . . . . . . . . . . . . . . 736 15.6.3 Decoding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 737 15.6.4 Historical notes and other results . . . . . . . . . . . . . . . . . 739 16 Cryptography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 741 16.1 Introduction to cryptography Alfred Menezes . . . . . . . . . . . . . . . 741 16.1.1 Goals of cryptography . . . . . . . . . . . . . . . . . . . . . . . 742 16.1.2 Symmetric-key cryptography . . . . . . . . . . . . . . . . . . . 742 16.1.2.1 Stream ciphers . . . . . . . . . . . . . . . . . . . . 742 16.1.2.2 Block ciphers . . . . . . . . . . . . . . . . . . . . . 743 16.1.3 Public-key cryptography . . . . . . . . . . . . . . . . . . . . . 744 16.1.3.1 RSA . . . . . . . . . . . . . . . . . . . . . . . . . . 744 16.1.3.2 Discrete logarithm cryptosystems . . . . . . . . . . 746 16.1.3.3 DSA . . . . . . . . . . . . . . . . . . . . . . . . . . 746 16.1.4 Pairing-based cryptography . . . . . . . . . . . . . . . . . . . . 747 16.1.5 Post-quantum cryptography . . . . . . . . . . . . . . . . . . . 749 16.2 Stream and block ciphers Guang Gong and Kishan Chand Gupta . . . . 750 16.2.1 Basic concepts of stream ciphers . . . . . . . . . . . . . . . . . 751 16.2.2 (Alleged) RC4 algorithm . . . . . . . . . . . . . . . . . . . . . 753 16.2.3 WG stream cipher . . . . . . . . . . . . . . . . . . . . . . . . . 754 16.2.4 Basic structures of block ciphers . . . . . . . . . . . . . . . . . 758 16.2.5 RC6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 759 16.2.6 Advanced Encryption Standard (AES) RIJNDAEL . . . . . . . 760 16.3 Multivariate cryptographic systems Jintai Ding . . . . . . . . . . . . . . 764 16.3.1 The basics of multivariate PKCs . . . . . . . . . . . . . . . . . 765 16.3.1.1 The standard (bipolar) construction of MPKCs . . . 765 16.3.1.2 Implicit form MPKCs . . . . . . . . . . . . . . . . 766 16.3.1.3 Isomorphism of polynomials . . . . . . . . . . . . . 767 16.3.2 Main constructions and variations . . . . . . . . . . . . . . . . 767 16.3.2.1 Historical constructions . . . . . . . . . . . . . . . 767 16.3.2.2 Triangular constructions . . . . . . . . . . . . . . . 768 16.3.2.3 Big-ﬁeld families: Matsumoto-Imai (C∗) and HFE . 769 16.3.2.4 Oil and vinegar (unbalanced and balanced) and varia-tions . . . . . . . . . . . . . . . . . . . . . . . . . . 770 16.3.2.5 UOV as a booster stage . . . . . . . . . . . . . . . 772 16.3.2.6 Plus-Minus variations . . . . . . . . . . . . . . . . . 772 16.3.2.7 Internal perturbation . . . . . . . . . . . . . . . . . 773 16.3.2.8 Vinegar as an external perturbation and projection . 774 16.3.2.9 TTM and related schemes: “lock” or repeated triangular 774 16.3.2.10 Intermediate ﬁelds: MFE and ℓIC . . . . . . . . . . 775 16.3.2.11 Odd characteristic . . . . . . . . . . . . . . . . . . 776 16.3.2.12 Other constructions . . . . . . . . . . . . . . . . . . 776 16.3.3 Standard attacks . . . . . . . . . . . . . . . . . . . . . . . . . 776 16.3.3.1 Linearization equations . . . . . . . . . . . . . . . . 776 16.3.3.2 Critical bilinear relations . . . . . . . . . . . . . . 776 16.3.3.3 HOLEs (higher-order linearization equations) . . . . 777 16.3.3.4 Diﬀerential attacks . . . . . . . . . . . . . . . . . . 777 16.3.3.5 Attacking internal perturbations . . . . . . . . . . . 777 Contents xxiii 16.3.3.6 The skew symmetric transformation . . . . . . . . . 778 16.3.3.7 Multiplicative symmetry . . . . . . . . . . . . . . . 779 16.3.3.8 Rank attacks . . . . . . . . . . . . . . . . . . . . . 779 16.3.3.9 MinRank attacks on big-ﬁeld schemes . . . . . . . . 780 16.3.3.10 Distilling oil from vinegar and other attacks on UOV 780 16.3.3.11 Reconciliation . . . . . . . . . . . . . . . . . . . . . 781 16.3.3.12 Direct attacks using polynomial solvers . . . . . . . 781 16.3.4 The future . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 782 16.4 Elliptic curve cryptographic systems Andreas Enge . . . . . . . . . . . . 784 16.4.1 Cryptosystems based on elliptic curve discrete logarithms . . . . 784 16.4.1.1 Key sizes . . . . . . . . . . . . . . . . . . . . . . . 784 16.4.1.2 Cryptographic primitives . . . . . . . . . . . . . . . 784 16.4.1.3 Special curves . . . . . . . . . . . . . . . . . . . . . 785 16.4.1.4 Random curves: point counting . . . . . . . . . . . 787 16.4.2 Pairing based cryptosystems . . . . . . . . . . . . . . . . . . . 788 16.4.2.1 Cryptographic pairings . . . . . . . . . . . . . . . . 789 16.4.2.2 Pairings and twists . . . . . . . . . . . . . . . . . . 791 16.4.2.3 Explicit isomorphisms . . . . . . . . . . . . . . . . 792 16.4.2.4 Curve constructions . . . . . . . . . . . . . . . . . 793 16.4.2.5 Hashing into elliptic curves . . . . . . . . . . . . . . 795 16.5 Hyperelliptic curve cryptographic systems Nicolas Th´ eriault . . . . . . . 797 16.5.1 Cryptosystems based on hyperelliptic curve discrete logarithms . 797 16.5.2 Curves of genus 2 . . . . . . . . . . . . . . . . . . . . . . . . . 797 16.5.3 Curves of genus 3 . . . . . . . . . . . . . . . . . . . . . . . . . 798 16.5.4 Curves of higher genus . . . . . . . . . . . . . . . . . . . . . . 799 16.5.5 Key sizes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 799 16.5.6 Special curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 801 16.5.7 Random curves: point counting . . . . . . . . . . . . . . . . . . 802 16.5.8 Pairings in hyperelliptic curves . . . . . . . . . . . . . . . . . . 803 16.6 Cryptosystems arising from Abelian varieties Kumar Murty . . . . . . . 803 16.6.1 Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804 16.6.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 804 16.6.3 Jacobians of curves . . . . . . . . . . . . . . . . . . . . . . . . 804 16.6.4 Restriction of scalars . . . . . . . . . . . . . . . . . . . . . . . 804 16.6.5 Endomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . 805 16.6.6 The characteristic polynomial of an endomorphism . . . . . . . 805 16.6.7 Zeta functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 805 16.6.8 Arithmetic on an Abelian variety . . . . . . . . . . . . . . . . . 807 16.6.9 The group order . . . . . . . . . . . . . . . . . . . . . . . . . . 808 16.6.10 The discrete logarithm problem . . . . . . . . . . . . . . . . . . 808 16.6.11 Weil descent attack . . . . . . . . . . . . . . . . . . . . . . . . 809 16.6.12 Pairings based cryptosystems . . . . . . . . . . . . . . . . . . . 810 16.7 Binary extension ﬁeld arithmetic for hardware implementations M. Anwarul Hasan and Haining Fan . . . . . . . . . . . . . . . . . . . . . . . . . . . . 811 16.7.1 Preamble and basic terminologies . . . . . . . . . . . . . . . . 811 16.7.2 Arithmetic using polynomial operations . . . . . . . . . . . . . 812 16.7.3 Arithmetic using matrix operations . . . . . . . . . . . . . . . 816 16.7.4 Arithmetic using normal bases . . . . . . . . . . . . . . . . . . 817 16.7.5 Multiplication using optimal normal bases . . . . . . . . . . . . 819 16.7.6 Additional notes . . . . . . . . . . . . . . . . . . . . . . . . . . 822 xxiv Contents 17 Miscellaneous applications . . . . . . . . . . . . . . . . . . . . . . . . . . 825 17.1 Finite ﬁelds in biology Franziska Hinkelmann and Reinhard Laubenbacher 825 17.1.1 Polynomial dynamical systems as framework for discrete models in systems biology . . . . . . . . . . . . . . . . . . . . . . . . . . 825 17.1.2 Polynomial dynamical systems . . . . . . . . . . . . . . . . . . 826 17.1.3 Discrete model types and their translation into PDS . . . . . . 827 17.1.3.1 Boolean network models . . . . . . . . . . . . . . . 829 17.1.3.2 Logical models . . . . . . . . . . . . . . . . . . . . 829 17.1.3.3 Petri nets and agent-based models . . . . . . . . . . 831 17.1.4 Reverse engineering and parameter estimation . . . . . . . . . . 831 17.1.4.1 The minimal-sets algorithm . . . . . . . . . . . . . 831 17.1.4.2 Parameter estimation using the Gr¨ obner fan of an ideal 831 17.1.5 Software for biologists and computer algebra software . . . . . . 832 17.1.6 Speciﬁc polynomial dynamical systems . . . . . . . . . . . . . . 832 17.1.6.1 Nested canalyzing functions . . . . . . . . . . . . . 832 17.1.6.2 Parameter estimation resulting in nested canalyzing func-tions . . . . . . . . . . . . . . . . . . . . . . . . . . 834 17.1.6.3 Linear polynomial dynamical systems . . . . . . . . 834 17.1.6.4 Conjunctive/disjunctive networks . . . . . . . . . . 834 17.2 Finite ﬁelds in quantum information theory Martin Roetteler and Arne Win-terhof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834 17.2.1 Mutually unbiased bases . . . . . . . . . . . . . . . . . . . . . 835 17.2.2 Positive operator-valued measures . . . . . . . . . . . . . . . . 836 17.2.3 Quantum error-correcting codes . . . . . . . . . . . . . . . . . 837 17.2.4 Period ﬁnding . . . . . . . . . . . . . . . . . . . . . . . . . . . 839 17.2.5 Quantum function reconstruction . . . . . . . . . . . . . . . . . 840 17.2.6 Further connections . . . . . . . . . . . . . . . . . . . . . . . . 840 17.3 Finite ﬁelds in engineering Jonathan Jedwab and Kai-Uwe Schmidt . . . 841 17.3.1 Binary sequences with small aperiodic autocorrelation . . . . . 841 17.3.2 Sequence sets with small aperiodic auto- and crosscorrelation . . 842 17.3.3 Binary Golay sequence pairs . . . . . . . . . . . . . . . . . . . 843 17.3.4 Optical orthogonal codes . . . . . . . . . . . . . . . . . . . . . 844 17.3.5 Sequences with small Hamming correlation . . . . . . . . . . . 845 17.3.6 Rank distance codes . . . . . . . . . . . . . . . . . . . . . . . . 846 17.3.7 Space-time coding . . . . . . . . . . . . . . . . . . . . . . . . . 847 17.3.8 Coding over networks . . . . . . . . . . . . . . . . . . . . . . . 848 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 851 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1011 Preface The CRC Handbook of Finite Fields (hereafter referred to as the Handbook) is a reference book for the theory and applications of ﬁnite ﬁelds. It is not intended to be an introductory textbook. Our goal is to compile in one volume the state of the art in research in ﬁnite ﬁelds and their applications. Hence, our aim is a comprehensive book, with easy-to-access references for up-to-date facts and results regarding ﬁnite ﬁelds. The Handbook is organized into three parts. Part I contains just one chapter which is devoted to the history of ﬁnite ﬁelds through the 18-th and 19-th centuries. Part II contains theoretical properties of ﬁnite ﬁelds. This part of the Handbook contains 12 chapters. Chapter 2 deals with basic properties of ﬁnite ﬁelds; properties that are used in various places throughout the entire Handbook. Near the end of Section 2.1 is a rather ex-tensive list of recent ﬁnite ﬁeld-related books; these books include textbooks, books dealing with theoretical topics as well as books dealing with various applications to such topics as combinatorics, algebraic coding theory for the error-free transmission of information, and cryptography for the secure transmission of information. Also included is a list of recent ﬁnite ﬁeld-related conference proceedings volumes. Chapter 2 also provides rather extensive tables of polynomials useful when dealing with ﬁnite ﬁeld computational issues. The website 9781439873786 provides larger and more extensive versions of the tables presented in Sec-tion 2.2. The next two chapters deal with polynomials such as irreducible and primitive poly-nomials over ﬁnite ﬁelds. Chapter 5 discusses various kinds of bases over ﬁnite ﬁelds, and Chapter 6 discusses character and exponential sums over ﬁnite ﬁelds. In Chapter 7, results on solutions of equations over ﬁnite ﬁelds are discussed. Chapter 8 covers permutation polynomials in one and several variables, as well as a discussion of value sets of polynomials, and exceptional polynomials over ﬁnite ﬁelds. Chapter 9 discusses special functions over ﬁnite ﬁelds. This discussion includes Boolean, APN, PN, bent, kappa polynomials, planar functions and Dickson polynomials, and ﬁnishes with a discussion of Schur’s conjecture. Sequences over ﬁnite ﬁelds are considered in Chapter 10. This chapter includes material on ﬁnite ﬁeld transforms, LFSRs and maximal length sequences, correlation and autocorre-lation and linear complexity of sequences as well as algebraic dynamical systems over ﬁnite ﬁelds. Chapter 11 deals with various kinds of ﬁnite ﬁeld algorithms including basic ﬁnite ﬁeld computational techniques, formulas for polynomial counting, irreducible techniques, fac-torizations of polynomials in one and several variables, discrete logarithms, and standard models for ﬁnite ﬁelds. In Chapter 12, curves over ﬁnite ﬁelds are discussed in great detail. This discussion includes elliptic and hyperelliptic curves. Rational points on curves are considered as well as towers and zeta functions over ﬁnite ﬁelds. In addition, there is a discussion of p-adic estimates of zeta and L-functions over ﬁnite ﬁelds. Chapter 13 discusses a variety of topics over ﬁnite ﬁelds. These topics include relations between the integers and polynomials over ﬁnite ﬁelds, matrices over ﬁnite ﬁelds, linear algebra and related computational topics, as well as classical groups over ﬁnite ﬁelds, and Carlitz and Drinfeld modules. Part III of the Handbook, containing four chapters, discusses various important appli-cations, including mathematical as well as very practical applications of ﬁnite ﬁelds. Latin xxv xxvi Preface squares and the polynomial method, useful in various areas of combinatorics, are considered. In addition, aﬃne and projective planes, projective spaces, block designs, and diﬀerence sets are discussed in detail. In each of these areas, since these topics contain an immense num-ber of papers, we discuss only those techniques and topics related to ﬁnite ﬁelds. Other topics included in Chaper 14 are (t, m, s)-nets useful in numerical integration, applications of primitive polynomials over ﬁnite ﬁelds, and Ramanujan and expander graphs. Chapter 15 is another important chapter in the Handbook. It discusses algebraic coding theory and includes a long introductory section dealing with basics properties of codes. This is followed by sections on special kinds of codes including LDPC codes, turbo codes, algebraic geometry codes, raptor codes, and polar codes. Chapter 16 deals with cryptographic systems over ﬁnite ﬁelds. In the ﬁrst section various basic issues dealing with cryptography are discussed. Next to be discussed are stream and block ciphers, multivariate cryptographic systems, elliptic and hyperelliptic curve crypto-graphic systems as well as systems arising from Abelian varieties over ﬁnite ﬁelds. Finally, in Chapter 17 we discuss several additional applications of ﬁnite ﬁelds including ﬁnite ﬁelds in biology, quantum information theory, and various applications in engineer-ing including topics like optimal orthogonal codes, binary sequences with small aperiodic autocorrelation, and sequences with small Hamming correlation. In the bibliography, we have included for each reference, the pages where that reference is discussed in the Handbook. There is also a large index to help readers quickly locate various topics in the Handbook. The Handbook is not meant to be read in a sequential way. Instead, each section is meant to be self-contained. Basic properties of ﬁnite ﬁelds are included in Chapter 2. Proofs are not included in the Handbook; instead authors have given references where proofs of the important results can be found. In an eﬀort to help the reader locate proofs and important results for each section, at the end of each section we have provided a list of references used in that section. Those reference numbers refer to the main bibliography at the end of the Handbook which contains over 3,000 references. A short “See also” section is included for most sections; these are intended to provide the reader with references to other related sections and references of the Handbook. The following numbering system is in eﬀect in the Handbook. Within a given section, all results, theorems, corollaries, deﬁnitions, examples, etc., are numbered consecutively (with the exception of tables and ﬁgures). For example, the result numbered 2.1.5 happens to be a theorem which is the ﬁfth listing in Section 2.1 of Chapter 2. We have also included many remarks in each section. These are also numbered as part of the same system so that for example, Remark 2.1.4 is the fourth listing in Section 2.1. Readers are encouraged to make us aware of corrections to the material presented here. Readers should contact the author(s) of the section involved, as well as both of the Editors-in-Chief. We would of course like to ﬁrst thank the authors of the various sections for their time and eﬀort. Without their help, the Handbook would, quite simply, not exist. We also greatly appreciate the authors’ willingness to use our style and format so that the entire Handbook has a consistent and uniform style and format. While we appreciate the help of all of the authors, we would especially like to thank Ian Blake, Steve Cohen, Cary Huﬀman, Alfred Menezes, Harald Niederreiter, Henning Stichtenoth, and Arne Winterhof who not only wrote several sections, but who also provided the Editors-in-Chief with valuable input in numerous aspects of the Handbook. Every section was reviewed by at least two external reviewers, in addition to the Editors-in-Chief. We also would like to thank the many reviewers who took the time to read and send us comments on the various sections and drafts. Without their help, we would of course have ended up with a volume of considerably diminished quality. We would like to thank Brett Stevens and David Thomson for their help with various Preface xxvii L AT EX and ﬁle issues. Finally, we would like to thank Shashi Kumar for his invaluable help in setting up, reworking, and running the style ﬁles that deﬁne the overall look of the entire Handbook. His eﬀorts were of tremendous help to us. We would also like to thank Bob Stern for his continued support. Needless to say, this project has involved many, many hours. We thank Bevie Sue Mullen and Lucia Moura for their encouragement, support, love, and patience during the entire process. Gary L. Mullen Daniel Panario This page intentionally left blank This page intentionally left blank Editors-in-Chief Gary L. Mullen Department of Mathematics The Pennsylvania State University University Park, PA 16802 U.S.A. Email: mullen@math.psu.edu Daniel Panario School of Mathematics and Statistics Carleton University Ottawa ON K1S 5B6 Canada Email: daniel@math.carleton.ca Contributors Omran Ahmadi School of Mathematics Institute for Research in Fundamental Sci-ences (IPM) P. O. Box: 19395-5746, Tehran Iran Email: oahmadid@ipm.ir Simeon Ball Departament Matematica Aplicada IV Universitat Politecnica Catalunya c/Jordi Girona 1-3 08034 Barcelona Spain Email: simeon@ma4.upc.edu Daniel J. Bernstein Department of Computer Science (MC 152) University of Illinois at Chicago 851 S. Morgan Street Chicago, IL 60607-7053 U.S.A Email: djb@cr.yp.to R´ egis Blache IUFM de Guadeloupe Morne Ferret 97139 Les Abymes French West Indies Email: rblache@iufm.univ-ag.fr Ian F. Blake Department of Electrical and Computer Engineering University of British Columbia Vancouver, BC V6T 1Z4 Canada Email: ifblake@ece.ubc.ca Aart Blokhuis Department of Mathematics and Computer Science Eindhoven University of Technology P.O. Box 513 5600 MB The Netherlands Email: aartb@win.tue.nl Claude Carlet University of Paris 8 Department of Mathematics 2 rue de la Libert´ e 93526 Saint-Denis, Cedex France Email: claude.carlet@univ-paris8.fr Francis Castro Department of Mathematics University of Puerto Rico Rio Piedras Campus San Juan, 00931 Puerto Rico Email: franciscastr@gmail.com Pascale Charpin INRIA-Rocquencourt Domaine de Voluceau, B.P. 105 F-78153 Le Chesnay Cedex France Email: Pascale.Charpin@inria.fr Sebastian M. Cioab˘ a Department of Mathematical Sciences University of Delaware Newark, DE 19716-2553 U.S.A. Email: cioaba@math.udel.edu xxix xxx Contributors Stephen D. Cohen School of Mathematics & Statistics University of Glasgow Glasgow G12 8QW Scotland Email: Stephen.Cohen@glasgow.ac.uk Charles J. Colbourn School of CIDSE Arizona State University Tempe, AZ 85287-8809 U.S.A. Email: Charles.Colbourn@asu.edu Robert Coulter Department of Mathematical Sciences University of Delaware 520 Ewing Hall Newark, DE 19716 U.S.A. Email: coulter@math.udel.edu Jintai Ding Department of Mathematical Sciences University of Cincinnati 2815 Commons Way Cincinnati, OH 45221-0025 U.S.A. Email: jintai.ding@gmail.com JeﬀDinitz Department of Mathematics and Statistics University of Vermont Burlington, VT 05405 U.S.A. Email: Jeff.Dinitz@uvm.edu Christophe Doche Department of Computing Macquarie University North Ryde, NSW 2109 Australia Email: christophe.doche@mq.edu.au Jean-Guillaume Dumas Universit´ e Joseph Fourier, Grenoble I Laboratoire Jean Kuntzmann Math´ ematiques Appliqu´ ees et Informatique 38041 Grenoble France Email: Jean-Guillaume.Dumas@imag.fr Gary Ebert N3691 Sylvan Isle Drive Watersmeet, MI 49969 U.S.A. Email: ebert@math.udel.edu Gove Eﬃnger Department of Mathematics and Computer Science Skidmore College Saratoga Springs, NY 12866 U.S.A. Email: effinger@skidmore.edu Andreas Enge INRIA Bordeaux - Sud-Ouest Universit´ e Bordeaux 1 351 cours de la Lib´ eration 33405 Talence Cedex France Email: andreas.enge@inria.fr Ron Evans Department of Mathematics University of California at San Diego La Jolla, CA 92093-0112 U.S.A. Email: revans@ucsd.edu Haining Fan School of Software Tsinghua University Beijing China Email: fhn@tsinghua.edu.cn Robert Fitzgerald Department of Mathematics Southern Illinois University Carbondale, IL 62901-4408 U.S.A. Email: rfitzg@siu.edu Michael D. Fried 3548 Prestwick Rd. Billings, MT 59101 U.S.A. Email: mfried@math.uci.edu Contributors xxxi Lei Fu Chern Institute of Mathematics Nankai University Tianjin 300071 P. R. China Email: leifu@nankai.edu.cn Shuhong Gao Department of Mathematical Sciences Clemson University Clemson, SC 29634-0975 U.S.A. Email: sgao@clemson.edu Moubariz Z. Garaev Centro de Ciencias Matem´ aticas Universidad Nacional Aut´ onoma de M´ exico Morelia 58089, Michoac´ an M´ exico Email: garaev@matmor.unam.mx Arnaldo Garcia Instituto Nacional de Matem´ atica Pura e Aplicada, IMPA Estrada Dona Castorina 110 22460-320, Rio de Janeiro, RJ Brazil Email: garcia@impa.br Joachim von zur Gathen B-IT, Universit¨ at Bonn Dahlmannstr. 2 53179 Bonn Germany Email: gathen@bit.uni-bonn.de Mark Giesbrecht Cheriton School of Computer Science University of Waterloo Waterloo, Ontario, N2L 3G1 Canada Email: mwg@uwaterloo.ca Guang Gong Department of Electrical and Computer Engineering University of Waterloo Waterloo, Ontario N2L 3G1 Canada Email: ggong@uwaterloo.ca David Goss Department of Mathematics Ohio State University 231 West 18th Avenue Columbus, OH 43210 U.S.A. Email: goss@math.ohio-state.edu Roderick Gow School of Mathematical Sciences University College Dublin Belﬁeld, Dublin 4 Ireland Email: rod.gow@ucd.ie Kishan Chand Gupta Applied Statistics Unit Indian Statistical Institute 203 B. T. Road Kolkata 700108 India Email: kishan@isical.ac.in Dirk Hachenberger Institut f¨ ur Mathematik Universit¨ at Augsburg 86135 Augsburg Germany Email: hachenberger@math.uni-augsburg.de M. Anwarul Hasan Department of Electrical and Computer En-gineering University of Waterloo Waterloo, ON, N2L 3G1 Canada Email: ahasan@uwaterloo.ca Tor Helleseth Institutt for informatikk Universitetet i Bergen PB. 7803, N-5020 Bergen Norway Email: Tor.Helleseth@ii.uib.no Franziska Hinkelmann Mathematical Biosciences Institute Ohio State University 1735 Neil Ave Columbus, OH 43210 U.S.A. Email: hinkelmann.1@mbi.osu.edu xxxii Contributors James W.P. Hirschfeld Department of Mathematics University of Sussex Brighton BN1 9QH United Kingdom Email: jwph@sussex.ac.uk Xiang-dong Hou Department of Mathematics and Statistics University of South Florida 4202 E. Fowler Ave Tampa, FL 33620 U.S.A. Email: xhou@usf.edu W. Cary Huﬀman Department of Mathematics and Statistics Loyola University Chicago 1032 W. Sheridan Road Chicago, IL 60660 U.S.A. Email: whuffma@luc.edu Michael Jacobson, Jr. Department of Computer Science University of Calgary Calgary, Alberta, T2N 1N4 Canada Email: jacobs@ucalgary.ca Jonathan Jedwab Department of Mathematics Simon Fraser University Burnaby, British Columbia V5A 1S6 Canada Email: jed@sfu.ca Dieter Jungnickel Mathematical Institute University of Augsburg D-86135 Augsburg Germany Email: jungnickel@math.uni-augsburg.de Erich Kaltofen Department of Mathematics Campus Box 8205 North Carolina State University Raleigh, NC 27695-8205 U.S.A. Email: kaltofen@math.ncsu.edu Alexander Kholosha Department of Informatics University of Bergen P.O. Box 7800 N-5020 Bergen Norway Email: Alexander.Kholosha@uib.no Melsik Kyuregyan Institute for Informatics and Automation Problems National Academy of Sciences of Armenia 1, P. Sevak str., Yerevan 0014 Armenia Email: melsik@ipia.sci.am Tanja Lange Coding Theory and Cryptology, MF6.104 B Department of Mathematics and Computer Science Technische Universiteit Eindhoven P.O. Box 513 5600 MB Eindhoven Netherlands Email: tanja@hyperelliptic.org Reinhard Laubenbacher Virginia Bioinformatics Institute Blacksburg, VA 24061 U.S.A. Email: reinhard@vbi.vt.edu Gr´ egoire Lecerf Laboratoire d’informatique (LIX, UMR 7161 CNRS) Campus de l’´ Ecole polytechnique B´ atiment Alan Turing 91128 Palaiseau Cedex France Email: gregoire.lecerf@math.cnrs.fr Hendrik Lenstra Mathematisch Institut Universiteit Leiden Postbus 9512 2300 RA Leiden The Netherlands Email: hwl@math.leidenuniv.nl Contributors xxxiii Antonio Rojas-Le´ on Departamento de ´ Algebra - Facultad de Matem´ aticas Universidad de Sevilla Apdo. de correos 1160 41080 Sevilla Spain Email: arojas@us.es Qunying Liao Institute of Mathematics and Software Science Sichuan Normal University Chengdu, Sichuan Province, 610066 P. R. China Email: qunyingliao@sicnu.edu.cn Rudolf Lidl 7 Hill Street West Launceston Tasmania 7250 Australia Email: rudi@rlidl.com.au Simon Litsyn School of Electrical Engineering Tel Aviv University Ramat Aviv 69978 Israel Email: litsyn@eng.tau.ac.il Gary McGuire School of Mathematical Sciences University College Dublin Dublin 4 Ireland Email: gary.mcguire@ucd.ie Wilfried Meidl Faculty of Engineering and Natural Sciences Sabanci University Orhanli 34956, Tuzla-Istanbul Turkey Email: wmeidl@sabanciuniv.edu Alfred Menezes Department of Combinatorics & Optimization University of Waterloo Waterloo, Ontario N2L 3G1 Canada Email: ajmeneze@uwaterloo.ca Gary L. Mullen Department of Mathematics The Pennsylvania State University University Park, PA 16802 U.S.A. Email: mullen@math.psu.edu Kumar Murty Department of Mathematics University of Toronto 40 St. George Street Toronto, ON M5S 2E4 Canada Email: murty@math.toronto.edu M. Ram Murty Department of Mathematics and Statistics Queen’s University Kingston, Ontario K7L 3N6 Canada Email: murty@mast.queensu.ca Harald Niederreiter Radon Institute for Computational and Ap-plied Mathematics Austrian Academy of Sciences Altenbergerstr. 69 A-4040 Linz Austria Email: ghnied@gmail.com Andrew Odlyzko School of Mathematics University of Minnesota Minneapolis, MN 55455 U.S.A. Email: odlyzko@umn.edu Alina Ostafe Department of Computing Faculty of Science Macquarie University North Ryde, NSW 2109 Australia Email: alina.ostafe@mq.edu.au xxxiv Contributors Daniel Panario School of Mathematics and Statistics Carleton University Ottawa ON K1S 5B6 Canada Email: daniel@math.carleton.ca Cl´ ement Pernet INRIA/LIG MOAIS 51, avenue Jean Kuntzmann F-38330 Montbonnot Saint-Martin France Email: clement.pernet@imag.fr Alexander Pott Otto-von-Guericke University Magdeburg 39106 Magdeburg Germany Email: alexander.pott@ovgu.de Martin Roetteler NEC Laboratories America, Inc. 4 Independence Way, Suite 200 Princeton, NJ 08540 U.S.A. Email: mroetteler@nec-labs.com Ivelisse Rubio Department of Computer Science University of Puerto Rico Rio Piedras Campus P.O. Box 70377 San Juan, PR 00936-8377 Email: iverubio@gmail.com Renate Scheidler Department of Mathematics and Statistics University of Calgary 2500 University Drive NW Calgary, Alberta, T2N 1N4 Canada Email: rscheidl@ucalgary.ca Kai-Uwe Schmidt Faculty of Mathematics Otto-von-Guericke University Universit¨ atsplatz 2 39106 Magdeburg Germany Email: kaiuwe.schmidt@ovgu.de Igor Shparlinski Department of Computing Macquarie University Sydney, NSW 2109 Australia Email: igor.shparlinski@mq.edu.au Joseph H. Silverman Mathematics Department, Box 1917 Brown University Providence, RI 02912 U.S.A. Email: jhs@math.brown.edu Bart de Smit Mathematisch Institut Universiteit Leiden Postbus 9512 2300 RA Leiden The Netherlands Email: desmit@math.leidenuniv.nl Brett Stevens School of Mathematics and Statistics Carleton University Ottawa ON K1S 5B6 Canada Email: brett@math.carleton.ca Henning Stichtenoth Faculty of Engineering and Natural Sciences Sabanci University Orhanli 34956, Tuzla-Istanbul Turkey Email: henning@sabanciuniv.edu Leo Storme Department of Mathematics Ghent University Krijgslaan 281, Building S22 B-9000 Ghent, Belgium Email: ls@cage.ugent.be Oscar Takeshita Email: oscar_takeshita@hotmail.com Contributors xxxv Joseph A. Thas Department of Mathematics Ghent University Krijgslaan 281 - S22 9000 Gent Belgium Email: jat@cage.ugent.be Nicolas Th´ eriault Departamento de Matem´ atica Universidad del Bio-Bio Avda. Collao 1202 Casilla 5-C, Concepcion, 4051381 Chile Email: ntheriau@ubiobio.cl David Thomson School of Mathematics and Statistics Carleton University Ottawa ON K1S 5B6 Canada Email: dthomson@math.carleton.ca Jos´ e Felipe Voloch The University of Texas at Austin Mathematics Dept, RLM 8.100 2515 Speedway Stop C1200 Austin, Texas 78712-1202 U.S.A. Email: voloch@math.utexas.edu Daqing Wan Department of Mathematics University of California Irvine, CA 92697-3875 U.S.A. Email: dwan@math.uci.edu Zhe-Xian Wan Academy of Mathematics and Systems Science Chinese Academy of Sciences No 55, Zhongguancun East Road Zhongguancun, Beijing 100190 P. R. China Email: wan@amss.ac.cn Qiang Wang School of Mathematics and Statistics Carleton University Ottawa ON K1S 5B6 Canada Email: wang@math.carleton.ca Arne Winterhof Johann Radon Institute for Computational and Applied Mathematics Austrian Academy of Sciences Altenbergerstr. 69 4040 Linz, Austria Email: arne.winterhof@oeaw.ac.at Joseph L. Yucas 68 Rock Springs Rd. Makanda, IL 62958 U.S.A. Email: joeyucas@yahoo.com Michael E. Zieve Department of Mathematics University of Michigan Ann Arbor, MI 48109-1043 U.S.A. Email: zieve@umich.edu This page intentionally left blank This page intentionally left blank I Introduction 1 History of ﬁnite ﬁelds ............................................ 3 Finite ﬁelds in the 18-th and 19-th centuries 2 Introduction to ﬁnite ﬁelds ..................................... 13 Basic properties of ﬁnite ﬁelds • Tables 1 This page intentionally left blank This page intentionally left blank 1 History of ﬁnite ﬁelds 1.1 Finite ﬁelds in the 18-th and 19-th centuries .... 3 Introduction • Early anticipations of ﬁnite ﬁelds • Gauss’s Disquisitiones Arithmeticae • Gauss’s Disquisitiones Generales de Congruentiis • Galois’s Sur la th´ eorie des nombres • Serret’s Cours d’alg ebre sup´ erieure • Contributions of Sch¨ onemann and Dedekind • Moore’s characterization of abstract ﬁnite ﬁelds • Later developments 1.1 Finite ﬁelds in the 18-th and 19-th centuries Roderick Gow, University College Dublin 1.1.1 Introduction While the theory of ﬁnite ﬁelds emerged as an independent discipline at the end of the 19-th century, aspects of the subject can be traced back at least to the middle of the 17-th century. It is our intention to present here a survey of highlights of ﬁnite ﬁeld theory as they emerged in the 18-th and 19-th centuries, culminating in a description of Eliakim Hastings Moore’s , which began the study of abstract ﬁnite ﬁelds. Leonard Eugene Dickson (1874-1954), in the ﬁrst volume of his History of the Theory of Numbers gives many references to works that can be interpreted as dealing with ﬁnite ﬁelds, although not always described explicitly as such. Chapters VII and VIII are especially relevant, and give remarkably complete listings of what had been achieved before 1918. Chapter VIII, entitled Higher Congruences, occasionally uses the language of ﬁnite ﬁelds, although the emphasis is largely number theoretic. Dickson had already written a textbook, entitled Linear Groups with an Exposition of the Galois Field Theory which is probably the ﬁrst work devoted exclusively to ﬁnite ﬁelds. This book remained without any serious rival until the emergence in the 1950s of more geometric, less computational, methods, such as those pioneered by Artin in his Geometric Algebra . The ﬁrst 71 pages of Dickson’s work constitute a very full account of ﬁnite ﬁelds, and its exercises, partly based on the work of earlier researchers, are still a valuable source of problems and ideas. Of course, ﬁnite ﬁelds are mentioned in general histories of algebra, such as that of van der Waerden . Furthermore, Finite Fields by Lidl and Niederreiter contains much historical information and a very extensive bibliography, especially of the older litera-ture. Another brief but useful source of information is found in the historical notes scattered throughout Cox’s Galois Theory . 3 4 Handbook of Finite Fields The ﬁrst use of the English expressions ﬁeld of order s and Galois-ﬁeld of order s = qn occurs in a paper of E. H. Moore (1862-1932), which he presented in 1893. Moore states that the term ﬁeld was an equivalent to the German term endlicher K¨ orper, used by Heinrich Weber (1842-1913). We observe that Richard Dedekind (1831-1916) had already introduced such a term as Zahlenk¨ orper, which can be traced back to lectures he gave in 1858. The 1933 edition of the Oxford English Dictionary does not include a deﬁnition of the mathematical term ﬁeld, although it does deﬁne group in its mathematical meaning, but more recent editions of the dictionary include the mathematical use of ﬁeld, with an attribution to Moore. The name Galois ﬁeld is synonymous with ﬁnite ﬁeld, and it signiﬁes the importance to the subject of an innovatory paper by ´ Evariste Galois (1811-1832), published in 1830 , when the author was only 18. We will comment in greater detail on Galois’s work later in this article, but we will brieﬂy mention here that Galois lays the foundations of ﬁnite ﬁeld theory by showing that for each prime p and positive integer n, there is a ﬁnite ﬁeld of order pn, and its multiplicative group of non-zero elements is cyclic of order pn −1. Galois’s arguments are rather sketchy, but there is no doubt that he understood the fun-damental principles of the structure of a ﬁnite ﬁeld, including the role of the automorphism given by raising elements to the p-th power. As has proved to be the case on a number of occasions, it seems that most of Galois’s discoveries were already known to Gauss, in this case, in the late 1790s, but as Gauss never published an account of his work, Galois was unaware of Gauss’s priority. (Gauss is credited with the discovery of non-Euclidean geome-try before Bolyai and Lobachevsky, with the discovery of quaternions before Hamilton, and the discovery of the method of least squares before Legendre.) We will also give a sketch of Gauss’s approach to ﬁnite ﬁelds, which he called the theory of higher congruences, as it is described in Volume 2 of his Werke . 1.1.2 Early anticipations of ﬁnite ﬁelds Our approach to the early history of ﬁnite ﬁelds will be largely chronological. An early occurrence of a theorem that may be interpreted in the language of ﬁnite ﬁelds is Fermat’s Little Theorem, that xp−1 −1 is divisible by p when p is a prime and x an integer not divisible by p. Dickson states that the special case when x = 2 was already known to the ancient Chinese around 500 BCE. As was usual with Fermat (1601-1665), he did not give a formal proof, but he communicated his conjecture that the theorem holds true, in a letter to Bernard Fr´ enicle de Bessy (1605-1675), dated 18 October, 1640. Leonhard Euler (1707-1783) gave a complete proof of the theorem in 1736, but unpublished manuscripts of Gottfried Wilhelm Leibniz (1646-1716) show that he was in possession of a similar proof by 1680. In the 18-th century, further theorems, expressed in terms of congruences modulo a prime, that we can see as precursors of basic facts in ﬁnite ﬁeld theory were discovered by mathematicians such as Euler, Joseph-Louis Lagrange (1736-1813), and Adrien-Marie Legendre (1752-1833). However, the ﬁrst complete account of that body of knowledge that relates to the ﬁnite ﬁeld of prime order was presented by Carl Friedrich Gauss (1777-1855) in Sections I-IV of his Disquisitiones Arithmeticae , which we describe in the next section. 1.1.3 Gauss’s Disquisitiones Arithmeticae Gauss’s Disquisitiones Arithmeticae was an epoch making work in mathematics, introducing totally new ideas and demanding far higher standards of proof than had hitherto been required or expected. The book also serves as a commentary on the discoveries and shortcomings of his predecessors. For a very full account of the contents and inﬂuence of History of ﬁnite ﬁelds 5 Gauss’s magnum opus, we refer to the book The Shaping of Arithmetic . Gauss introduces the concept of congruence in Article (Art.) 1, and designates congru-ence by means of the now familiar symbol ≡. This is the ﬁrst published use of this symbol, which seems to have entered into conventional use quite rapidly. It occurs for instance in C. Kramp’s ´ El´ ements d’arithm´ etique universelle an elementary work much inﬂuenced by Gauss’s masterpiece (the use of the exclamation mark in n! makes its ﬁrst appearance here). In Section 2, Gauss proves in Art. 14 that if p is a prime integer and a, b are integers not divisible by p, then p does not divide the product ab. This basic result is fundamental for the proof that the integers modulo p form a ﬁeld. Gauss comments that the theorem was already in Euclid’s Elements. Oddly enough, for a person as notoriously meticulous as Gauss, he mistakenly says that it is Proposition 32 of Book VII, when it is in fact Propo-sition 30. Concerning this result, Gauss wrote magisterially: However we did not wish to omit it because many modern authors have employed vague computations in place of proof or have neglected the theorem entirely, and because by this very simple case we can more easily understand the nature of the method which will be used later for solving much more diﬃcult problems. He uses Art. 14 to prove Art. 16, a result often called the fundamental theorem of arithmetic: a composite number can be resolved into prime factors in only one way. This basic result is not in Euclid. Gauss describes Euler’s totient (or phi) function, which he denotes by the symbol φ (following Art. 38). (We recall that the totient function measures the number of totitives of a positive integer n, that is, the number of integers lying between 1 and n that are relatively prime to n.) This is again the ﬁrst occurrence of a now familiar symbol in mathematics. Euler himself, although introducing the idea of the function in 1760, did not use such notation. Art. 43 is a proof that an integer polynomial of degree m cannot have more than m incongruent roots modulo a prime. This basic theorem on polynomial arithmetic was ﬁrst published by Lagrange in 1768. Euler had shown that the congruence xn −1 ≡0 modulo a prime has at most n roots in 1774, and Gauss notes that Euler’s method is easily generalized. Section III, on residues of powers, contains Art. 49: if p is a prime number that does not divide a, and at is the lowest power of a that is congruent to unity to the modulus p, the exponent t will either = p −1 or be a factor of this number. Gauss notes that this implies Fermat’s Little Theorem, and he gives some of the history of this theorem that we described above. Art. 55 is the fundamental statement: There always exist numbers with the property that no power less than the p −1st is congruent to unity. This of course amounts to saying that the multiplicative group of the integers modulo a prime p is cyclic of order p −1. Again, it is interesting to observe the authority of Gauss’s language as he describes earlier approaches to Art. 55: This theorem furnishes an outstanding example of the need for circumspection in number theory so that we do not accept fallacies as certainties. . . . No one has attempted the demonstration except Euler . . . See especially his article 37 where he speaks at great length of the need for demonstration. But the demonstration which this shrewdest of men presents has two defects. . . . In Art. 57, Gauss adopts the nomenclature primitive roots, due originally to Euler, for the integers, or residues, described in Art. 55. 1.1.4 Gauss’s Disquisitiones Generales de Congruentiis Gauss had intended to include an eighth section of Disquisitiones Arithmeticae, and he even refers to this section at least twice in the published version. However, the section was omitted, possibly for reasons of saving space in an already long work. A manuscript of the missing section was found after Gauss’s death, and an edited version, with notes by Dedekind, was published in volume 2 of Gauss’s Werke in 1863, under the title Disquisitiones Generales de Congruentiis. A German translation followed in 1889. 6 Handbook of Finite Fields G¨ unther Frei, , has given a lengthy description of the genesis and contents of the unpublished Section Eight, and we will make use of some of his analysis here, since it has considerable bearing on the early theory of ﬁnite ﬁelds. Gauss’s work on ﬁnite ﬁelds can be traced back at least to 1796, as there are references to it in his Mathematical Diary . It is well known that Gauss was particularly fascinated by the law of quadratic reciprocity, and he gave several diﬀerent proofs of this fundamental theorem, the ﬁrst dating from 1796. The third and fourth of these proofs drew Gauss into the study of polynomials modulo a prime, and his surviving investigations enable us to discern much of the theory of ﬁnite extensions of a ﬁeld of prime order. In Frei’s translation, Gauss wrote But at the same time one sees that the solution of congruences constitutes only a part of a much higher investigation, namely the investigation of the decomposition of functions into factors. Accordingly, Gauss developed a theory of factorization of polynomials whose coeﬃcients are integers modulo a prime p, including the determination of greatest common divisors by Euclid’s algorithm. He introduced the concept of a prime polynomial, corresponding to irreducible polynomial in modern terminology, and showed that arbitrary polynomials can be factored into products of prime polynomials. Among the highlights of his discoveries, we may mention his proof that every irreducible polynomial modulo p, diﬀerent from x, and of degree m, is a divisor of xpm−1 −1. Further-more, xpm−1 −1 is the product of all monic irreducible polynomials of degree d dividing m, apart from x. From this fact, he obtained a formula for the number of irreducible monic polynomials of degree n with coeﬃcients integers modulo p. Frei also notes that Gauss ap-preciated the importance of the Frobenius automorphism, and came close to discovering a form of Hensel’s Lemma, signiﬁcant in p-adic analysis. The idea of using the imaginary roots of such irreducible polynomials to simplify some of his work had occurred to Gauss, and, in Frei’s translation, Gauss wrote Indeed, we could have shortened incomparably all our following investigations, had we wanted to introduce such imaginary quantities by taking the same liberty some more recent mathematicians have taken, but nevertheless, we have preferred to deduce everything from ﬁrst principles. It should be recalled that Gauss sometimes displayed a conservative approach to new concepts in mathematics, and his public aversion to using imaginary roots of congruences is akin to his disinclination to use complex numbers. Thus, for example, his thesis, published in 1799, states that every real polynomial is a product of real factors of degree one or two, rather than stating that every complex polynomial is a product of factors of degree 1. 1.1.5 Galois’s Sur la th´ eorie des nombres We turn now to presenting a synopsis of Galois’s 1830 paper Sur la th´ eorie des nombres on ﬁnite ﬁelds since it is a landmark in the subject. In Frei’s opinion, Galois establishes the additive and multiplicative structure of ﬁnite extensions of the ﬁeld of prime order. It certainly seems that the spirit of Galois’s paper is closer to the modern presentation of ﬁnite ﬁeld theory than Gauss’s version. It is worth noting that there are several misprints in the paper, which would have made it diﬃcult to follow for the uninitiated, and Galois’s attempts to illustrate the theory are hopelessly ﬂawed. Rather than translating the original French literally, we will instead try to convey some idea in modern terms of what Galois must have intended. For example, when Galois talks of a function, he means a polynomial in a single variable, with integer coeﬃcients. (This convention was common among mathematicians before the twentieth century.) He notes that we usually look for integer roots of the polynomial modulo a prime p, say. We call these real roots of the polynomial congruence. He proceeds to generalize the notion of real roots, and begins by introducing the concept of an integer polynomial F(x) being irreducible modulo p, meaning that it is impossible to ﬁnd three integer polynomials φ(x), ψ(x) and History of ﬁnite ﬁelds 7 χ(x) such that F(x) + pχ(x) = φ(x)ψ(x). (Galois does not use the term irreducible for this concept, but later in the paper speaks of an irreducible congruence.) Such an irreducible polynomial F(x) obviously has no integer roots modulo p, nor any irrational roots of degree less than that of F(x) (Galois does not explain these terms). He states that we must regard the roots of the congruence F(x) ≡0 (notation he attributes to Gauss) as a type of imaginary symbols, and opines that such imaginary roots will prove to be as useful as √−1 is in conventional analysis. These imaginary roots were subsequently called Galois imaginaries by later writers. Let i be a root of the congruence F(x) ≡0, where F has degree ν. (Galois does not justify why we may assume that F(x) ≡0 has roots, a point Serret attempted to rectify.) Galois then considers a general expression a + a1i + a2i2 + · · · + aν−1iν−1, where a, a1, . . . , aν−1 are integers modulo p. There are pν diﬀerent values for these expres-sions. Let α be an expression of the form above. If we raise α to the second, third, etc, powers, we obtain a sequence of expressions of the same form. Thus we must have αn = 1 for a certain positive integer n, which we choose to be as small as possible. We then have n diﬀerent expressions 1, α, . . . , αn−1. Galois shows that n divides pν −1, and thus αpν−1 = 1. Galois next aims to prove that there is some α for which the corresponding n is pν −1. He makes an analogy at this stage with existence of primitive roots modulo p in the theory of numbers. We did not ﬁnd that Galois provided a convincing argument for this key issue. Galois then draws the remarkable conclusion that all the algebraic quantities that arise in this theory are roots of equations of the form xpν = x. Furthermore, if F(x) is an integer polynomial of degree ν irreducible modulo p, there are integer polynomials f(x) and φ(x) such that f(x)F(x) = xpν −x + pφ(x). Galois also notes that if α is a root of the irreducible congruence F(x) ≡0, then the other roots are αp, αp2, . . . , αpν−1. This is a consequence of the fact that F(x)pn ≡F(xpn). We remark that this is an early indication of the role of the so-called Frobenius mapping as a generator of the associated Galois group. Galois later notes that all the roots of the congruence xpν ≡x depend only on the roots of a single irreducible polynomial of degree ν. To illustrate all this theory, Galois attempts to ﬁnd a primitive root of the congruence x73 ≡x (mod 7). He aims to do this by exhibiting elements having orders 9 and 19. In fact, he makes several noteworthy errors, which may have confused any readers of this exposition of the new theory. He begins by noting that x3 ≡2 (mod 7) is irreducible, and lets i be a root of the 8 Handbook of Finite Fields congruence. He claims that −1 −i has order 19, but this is false, as it has order 9 × 19. He then claims that α = i + i2 is primitive, but this is again false, as it has order 114, not 342 = 73 −1. Finally, Galois claims that his α satisﬁes α3 + 3α + 1 = 0, but this is again incorrect, as in truth it satisﬁes α3 + α + 1 = 0. Indeed, the polynomial x3 + 3x + 1 is not even irreducible modulo 7, as 4 is a root of it. As we mentioned earlier, the paper contains several misprints, possibly because the com-positor found the notation diﬃcult to handle, but Galois’s errors are not just typographical (although they are of course ultimately trivial and in no way invalidate his theory). Joseph-Alfred Serret gives a treatment of this problem of ﬁnding a primitive root in the second edition of Cours d’algebre sup´ erieure , pp. 367-370, following Galois’s methods. The required primitive element Galois might have had in mind was β = i −i2, not i + i2. This element β is a root of x3 −x+2, which is certainly an irreducible primitive polynomial. While i + i2 may have replaced i −i2 because of a typographical error, Galois nonetheless made further mistakes which are diﬃcult to explain. Serret himself made no comment on this strange aspect of Galois’s paper. As justiﬁcation for introducing this theory, Galois explains that it is required in the theory of permutations which arise in the study of primitive (rational) polynomials which are solvable by radicals. He alludes, in eﬀect, to what is the aﬃne group of the ﬁnite ﬁeld of order pν, which must be the Galois group of such a polynomial when the action on the roots is doubly transitive. He excludes degrees 9 and 25, where he must have known that there exist exceptional doubly transitive solvable permutation groups. There is another one in degree 49, which he did not mention. 1.1.6 Serret’s Cours d’alg ebre sup´ erieure The early editions of Serret’s textbook mentioned above provide us with a good opportunity to gauge the progress of the project to publicize Galois’s research, considered very advanced at the time. In the ﬁrst edition , Serret writes that his (Serret’s) work was a summary of lectures given at the Sorbonne, Paris, where he had been appointed to a chair in 1848. On p. 4, he notes that the diﬃcult problem of when an equation can be solved algebraically had been resolved, at least in the case of irreducible equations of prime degree, by ´ Evariste Gallois (sic), in a memoir of 1831. This memoir had been published in 1846 by Joseph Liouville in his Journal, and Liouville had wanted Serret to communicate part of Galois’s ﬁndings. Lesson 23 of this ﬁrst edition is devoted to the theory of congruence modulo a prime, but it includes nothing that was not already known at the time of Lagrange or Euler. The second edition of Serret’s work (1854) gives a fairly complete account of Galois’s theory of ﬁnite ﬁelds, as presented in his 1830 paper. Serret devotes almost 30 pages (the whole of Lesson 25) to the material that Galois had covered in six pages, with a view to making a diﬃcult part of the writings of this great mathematician more intelligible. Thus, for example, he proves a uniqueness theorem for factoring polynomials into irreducible factors, and also gives a lengthy discussion of why primitive elements exist. On the other hand, he does not attempt to give a rigorous explanation of why roots of irreducible congruences may be taken to exist. Lesson 25 seems to be the ﬁrst exposure of ﬁnite ﬁelds at the textbook level. Serret devotes 68 pages of the third edition of his textbook (1866) to the theory of ﬁnite ﬁelds. He considered his approach to be new, and based it on a memoir he had presented in History of ﬁnite ﬁelds 9 1865. In fact, it bears many similarities to Gauss’s unpublished Section 8 (itself published for the ﬁrst time in Latin in 1863), and to Dedekind’s 1857 paper (for details, see later in this section). Serret was presumably unaware of this material, as he made no mention of it. We can recognize several classical theorems of ﬁnite ﬁeld theory described clearly in Serret’s Chapter 3 of Volume 2 of the third edition. Thus for example, if the integer g is not divisible by the prime p, the polynomial (later said to be of Artin-Schreier type) xp −x −g is irreducible modulo p. Art. 372 presents six theorems summarizing Galois’s ﬁndings of 1830, Art. 349 gives a formula for the number of monic irrreducible polynomials modulo a prime p, and Art. 350 gives upper and lower bounds for their number. 1.1.7 Contributions of Sch¨ onemann and Dedekind Another early contribution to ﬁnite ﬁeld theory is a paper by Theodor Sch¨ onemann, Grundz¨ uge einer allgemeinen Theorie der H¨ oheren Congruenzen, deren Modul eine reele Primzahl ist published in 1845. Sch¨ onemann is described as an oberlehrer (head teacher) at the Gymnasium in Brandenburg. He begins his paper with an apology, acknowl-edging that Gauss’s unpublished Section 8 was to contain contributions to the theory of higher congruences, and that he (Sch¨ onemann) may have inadvertently rediscovered some of Gauss’s results. Sch¨ onemann starts with a monic integer polynomial f of degree n which is irreducible modulo a prime p. He then takes a complex root α of f and considers in eﬀect the quotient ring, Z[α]/pZ[α], which is a ﬁnite ﬁeld of order pn. In this way, he avoids the question of whether imaginary roots of irreducible congruences may be taken to exist. This is described well in Cox , p. 296. One of Sch¨ onemann’s main theorems is that if we allow α also to represent a root of f modulo p, then f ≡(x −α)(x −αp) · · · (x −αpn−1) (mod p). He also obtained a formula for the number of monic irreducible polynomials of degree n modulo p. Sch¨ onemann’s paper is long (56 pages) and not very clear. It is also written in a very formal style, each result being presented in the form of Erkl¨ arung and Lehrsatz, followed by Beweis, in imitation of the approach characteristic of Euclid’s Elements. Nonetheless, Sch¨ onemann did innovatory work, which, even if anticipated by Gauss, was quoted reason-ably frequently in the second half of the nineteenth century, for instance, by Kronecker. In his paper Abriss einer Theorie der H¨ oheren Congruenzen in Bezug auf einen reellen Primzahl-Modulus written in late 1856, and published in 1857, Dedekind covered much of the same ground pioneered by Gauss in Disquisitiones Generales de Congruentiis. While we pointed out above that Dedekind was responsible for editing Gauss’s manuscript for publication in 1863, Frei presents several strong reasons to suppose that, at the time he wrote, Dedekind was unacquainted with this key work, and did not see it until 1860. Frei suggests that Dedekind was more concerned to give a solid foundation to Kummer’s theory of ideal numbers. In any case, Dedekind notes that there is a strong analogy between the theory of polynomials modulo a prime and elements of number theory. By way of illustrating this analogy, let p be an odd prime and let P and Q be diﬀerent irreducible monic polynomials of degrees m and n, respectively. Then working modulo Q, P determines an element of the ﬁeld of order pn, and this element is either a square or a non-square. By analogy with the Legendre symbol, we set ( P Q) equal to 1 if P is a square modulo Q, and ( P Q) equal to −1 if it is a non-square. Working modulo P, we may likewise deﬁne ( Q P ). Then, in complete 10 Handbook of Finite Fields analogy with the law of quadratic reciprocity, we have P Q Q P = (−1)( p−1 2 )mn. In his proof, Dedekind uses a version of Gauss’s Lemma, employed in one of Gauss’s proofs of the quadratic reciprocity theorem. 1.1.8 Moore’s characterization of abstract ﬁnite ﬁelds Towards the end of the 19th century, ﬁnite ﬁelds began to assume a more prominent position in contemporary algebraic research, partly because of their importance in the construction of ﬁnite analogues of the classical linear groups. Camille Jordan, in his Trait´ e des substitutions had investigated classical groups, such as the general linear group (also called the linear homogeneous group), over ﬁnite ﬁelds of prime order. E. H. Moore observed that certain of these constructions could be extended to arbitrary ﬁnite ﬁelds and he discovered the simple groups usually denoted by PSL2(qn). (He became aware by 1895 that these groups were already known to ´ Emile Mathieu, who had come across them in 1860, , pp. 38-42.) In any case, Moore was led to the investigation of abstract ﬁnite ﬁelds. We quote from Moore’s paper A doubly inﬁnite system of simple groups, read on August 25, 1893, at the International Mathematical Congress held in Chicago . This paper gives the details of Moore’s discoveries. Suppose that we have a system of s distinct symbols or marks (s being some positive integer) and suppose that these marks may be combined by the four fundamental operations of algebra–addition, subtraction, multiplication, and division–these operations being subject to the ordinary abstract operational identities of algebra µi + µj = µj + µi; µiµj = µjµi; (µi + µj)µk = µiµk + µjµk; etc, and that when the marks are so combined the results of these operations are in every case uniquely determined and belong to the system of marks. Such a system of marks we shall call a ﬁeld of order s, using the notation F[s]. . . . We are led at once to seek [t]o determine all such ﬁelds of order s, F[s]. Moore notes that Galois had deﬁned a ﬁeld of order qn, for each prime q and each positive integer n. Moore denotes this ﬁeld by GF[qn], presumably in honor of Galois. This GF[qn] is deﬁned via an irreducible polynomial of degree n modulo q, and is unique, in the sense that such irreducible polynomials exist for all q and n, and the GF[qn] so constructed is independent of the particular irreducible polynomial chosen. Moore’s main theorem is then stated as: Every existent ﬁeld F[s] is the abstract form of a Galois ﬁeld GF[qn]; s = qn. Moore remarks: This interesting result I have not seen stated before. Moore’s proof occupies pages 212-220 of his paper, and he derives further properties of GF[qn] in the next few pages. We feel that Moore’s paper marks the beginning of the abstract theory of ﬁnite ﬁelds. In 1896, Dickson was awarded the ﬁrst doctorate in mathe-matics at the new University of Chicago, for a thesis written under Moore’s direction, the subject matter being permutation polynomials over ﬁnite ﬁelds. Dickson’s 1901 book gave a streamlined proof of Moore’s uniqueness theorem on pp. 13-14. 1.1.9 Later developments Following the work of his thesis, Dickson was to extend Jordan’s analysis of classical groups to their counterparts over arbitrary ﬁnite ﬁelds, and his research was the subject of his monograph of 1901. Dickson even generalized ideas of ´ Elie Cartan on continuous groups History of ﬁnite ﬁelds 11 and their Lie algebras, and published in 1901 (with later additions) details of his discovery of versions of the groups of type E6 and G2 over ﬁnite ﬁelds [842, 848, 843, 844]. It was not until later work of Chevalley in 1955 that further ﬁnite analogues of the exceptional continuous groups were constructed in a uniform way. The theory of ﬁnite ﬁelds may be said to have acquired a more conceptual form in the twentieth century after Emil Artin (1898-1962) introduced the notion of a zeta function for a quadratic extension of the rational function ﬁeld Fp(t), where p is a prime. Artin formulated a version of the Riemann hypothesis for these zeta functions, and veriﬁed the hypothesis for a number of curves in his dissertation, published in 1924. Helmut Hasse (1899-1979) subsequently proved the Riemann hypothesis for function ﬁelds of genus 1 in 1934, but the complete proof for arbitrary non-singular curves by Andr´ e Weil (1906-1998) in 1948 employed sophisticated methods of algebraic geometry. The analogy between counting rational points on algebraic varieties over ﬁnite ﬁelds and the cohomology theories of complex varieties has been a powerful motivating force in the more recent theory of ﬁnite ﬁelds. References Cited: [135, 749, 793, 842, 843, 844, 848, 850, 851, 1168, 1257, 1258, 1259, 1297, 1621, 1804, 1939, 2021, 2139, 2556, 2596, 2848] This page intentionally left blank This page intentionally left blank 2 Introduction to ﬁnite ﬁelds 2.1 Basic properties of ﬁnite ﬁelds .................... 13 Basic deﬁnitions • Fundamental properties of ﬁnite ﬁelds • Extension ﬁelds • Trace and norm functions • Bases • Linearized polynomials • Miscellaneous results • Finite ﬁeld related books 2.2 Tables ................................................. 32 Low-weight irreducible and primitive polynomials • Low-complexity normal bases • Resources and standards 2.1 Basic properties of ﬁnite ﬁelds Gary L. Mullen, The Pennsylvania State University Daniel Panario, Carleton University Proofs for most of the results in this chapter can be found in Chapters 2 and 3 of ; see also [1631, 1938, 2017, 2049, 2077, 2179, 2921]. We refer the reader to Section 2.1.8 for a comprehensive list of other ﬁnite ﬁeld related books. 2.1.1 Basic deﬁnitions 2.1.1 Deﬁnition A ring (R, +, ·) is a nonempty set R together with two operations, “+” and “·” such that: (1) (R, +) is an abelian group; (2) · is associative, that is for all a, b, c ∈R, a · (b · c) = (a · b) · c; (3) left and right distributive laws hold: for all a, b, c ∈R a · (b + c) = a · b + a · c and (b + c) · a = b · a + c · a. 2.1.2 Deﬁnition Let R be a ring. (1) R is a ring with identity if the ring has a multiplicative identity; (2) R is commutative if “·” is commutative; (3) R is an integral domain if it is commutative with identity and a·b = 0 implies a = 0 or b = 0, for any a, b ∈R; 13 14 Handbook of Finite Fields (4) R is a division ring (also called a skew ﬁeld) if the nonzero elements of R form a group under “·”; (5) R is a ﬁeld if it is a commutative division ring. 2.1.3 Deﬁnition The order of a ﬁnite ﬁeld F is the number of distinct elements in F. 2.1.4 Remark The following theorem is a famous result due to Wedderburn. 2.1.5 Theorem Every ﬁnite division ring is a ﬁeld. 2.1.6 Deﬁnition If R is a ring and there exists a positive integer n such that nr = 0 for all r ∈R, then the least such positive integer n is the characteristic of the ring, and R has positive characteristic. Otherwise, R has characteristic zero. 2.1.7 Theorem A ring R ̸= {0} of positive characteristic having an identity and no zero divisors must have prime characteristic. 2.1.8 Corollary A ﬁnite ﬁeld has prime characteristic. 2.1.9 Proposition For a commutative ring R of characteristic p, we have (a1 + · · · + as)pn = apn 1 + · · · + apn s for every n ≥1 and ai ∈R. 2.1.2 Fundamental properties of ﬁnite ﬁelds 2.1.10 Lemma Suppose F is a ﬁnite ﬁeld with a subﬁeld K containing q elements. Then F is a vector space over K and \|F\| = qm, where m is the dimension of F viewed as a vector space over K. 2.1.11 Deﬁnition A ﬁeld containing no proper subﬁeld is a prime ﬁeld. 2.1.12 Theorem Let F be a ﬁnite ﬁeld. The cardinality of F is pn, where p is the characteristic of F and n is the dimension of F over its prime subﬁeld. 2.1.13 Remark We denote by Fq a ﬁnite ﬁeld with q elements. We note that by Remark 2.1.34 there is only one ﬁnite ﬁeld (up to isomorphism) with q elements. 2.1.14 Remark Another common notation for a ﬁeld of order q is GF(q), where GF stands for Galois ﬁeld. This name is used in honor of ´ Evariste Galois (1811–1832), who in 1830 was the ﬁrst person to seriously study properties of general ﬁnite ﬁelds (ﬁelds with a prime power but not necessarily a prime number of elements). 2.1.15 Remark The recent publication of The Mathematical Writings of Evariste Galois by Neu-mann will make Galois’s own words available to readers. 2.1.16 Lemma If Fq is a ﬁnite ﬁeld with q elements and a ̸= 0 ∈Fq, then aq−1 = 1, and thus aq = a, for all a in Fq. 2.1.17 Remark An immediate consequence of the previous lemma is that the multiplicative inverse of any a ̸= 0 in a ﬁeld of order q is aq−2, because aq−2 · a = aq−1. 2.1.18 Theorem The sum of all elements of a ﬁnite ﬁeld is 0, except for the ﬁeld F2. Introduction to ﬁnite ﬁelds 15 2.1.19 Deﬁnition A polynomial f over Fq is an expression of the form f(x) = Pn i=0 aixi, where n is a nonnegative integer, and ai ∈Fq for i = 0, 1, . . . , n. A polynomial is monic if the coeﬃcient of the highest power of x is 1. The ring formed by the polynomials over Fq with sum and product of polynomials is the ring of polynomials over Fq and is denoted by Fq[x]. 2.1.20 Deﬁnition A polynomial f ∈Fq[x] is an irreducible polynomial over Fq if f has positive degree and f = gh with g, h ∈Fq[x] implies that either g or h is a constant polynomial. 2.1.21 Remark Both Fq[x] and the ring of polynomials in n ≥1 variables, Fq[x1, . . . , xn], have unique factorization into irreducibles. 2.1.22 Deﬁnition The M¨ obius µ function is deﬁned on the set of positive integers by µ(m) =      1 if m = 1, (−1)k if m = m1m2 · · · mk, where the mi are distinct primes, 0 otherwise, i.e., if p2 divides m for some prime p. 2.1.23 Deﬁnition The number of monic irreducible polynomials of degree n over Fq is denoted by Iq(n). 2.1.24 Theorem For all n ≥1 and any prime power q, we have Iq(n) = 1 n X d\|n µ(d)qn/d. 2.1.25 Remark We have that Iq(n) > 0 for all prime powers q and all integers n > 1: Iq(n) = 1 n X d\|n µ(d)qn/d ≥1 n qn −qn−1 −qn−2 −· · · −q > 0. 2.1.26 Remark For a polynomial f ∈Fq[x], we have (f(x))q = f(xq). This property is of great use in ﬁnite ﬁeld calculations. 2.1.27 Lemma If Fq is a ﬁnite ﬁeld with q elements then in Fq[x] we have xq −x = Y a∈Fq (x −a). 2.1.28 Remark The next theorem is crucial for fast polynomial irreducibility testing and factor-ization algorithms over ﬁnite ﬁelds; see Sections 11.3 and 11.4. 2.1.29 Theorem Let f be an irreducible polynomial of degree n over Fq. Then f(x)\|(xqr −x) if and only if n\|r. 16 Handbook of Finite Fields 2.1.30 Deﬁnition Let f ∈F[x] be of positive degree and E an extension of F. Then f splits in E if f can be written as a product of linear factors in E[x], that is, there exist α1, α2, . . . , αn ∈E such that f(x) = a(x −α1)(x −α2) · · · (x −αn), where a ∈F is the leading coeﬃcient of f and E is the smallest such ﬁeld. The ﬁeld E is a splitting ﬁeld of f over F if f splits in E. 2.1.31 Theorem If F is a ﬁeld, and f any polynomial of positive degree in F[x], then there exists a splitting ﬁeld of f over F. Any two splitting ﬁelds of f over F are isomorphic under an isomorphism which keeps the elements of F ﬁxed and maps the roots of f into each other. 2.1.32 Theorem For every prime p and positive integer n ≥1 there is a ﬁnite ﬁeld with pn elements. Any ﬁnite ﬁeld with pn elements is isomorphic to the splitting ﬁeld of xpn −x over Fp. 2.1.33 Remark The previous theorem shows that a ﬁnite ﬁeld of a given order is unique up to ﬁeld isomorphism because splitting ﬁelds are unique up to isomorphism. Thus we speak of “the” ﬁnite ﬁeld of a particular order q. 2.1.34 Remark We note that when p is a prime the ﬁeld Fp is the same as (isomorphic to) the ring Zp of integers modulo p. The ring Zp is also denoted by Z/pZ. When n > 1 the ﬁnite ﬁeld Fpn is not the same as the ring Zpn of integers modulo pn. Indeed, Zpn is not a ﬁeld if n > 1. 2.1.35 Theorem Let Fpn be the ﬁnite ﬁeld with pn elements. Every subﬁeld of Fpn has pm elements for some positive integer m dividing n. Conversely, for any positive integer m dividing n there is a unique subﬁeld of Fpn of order pm. 2.1.36 Remark The subﬁelds of Fq36 are illustrated in the following diagram: Fq18 Fq18 Fq2 Fq6 Fq12 Fq3 Fq4 Fq9 Fq36 Fq1 Figure 2.1.1 The subﬁelds of Fq36. 2.1.37 Theorem The multiplicative group F∗ q of all nonzero elements of the ﬁnite ﬁeld Fq is cyclic. 2.1.38 Deﬁnition An element α ∈Fq which multiplicatively generates the group F∗ q of all nonzero elements of the ﬁeld Fq is a primitive element, sometimes also a primitive root. 2.1.39 Remark Let θ be a primitive element of a ﬁnite ﬁeld Fq. Then every nonzero element of Fq can be written as a power of θ. This representation makes multiplication of ﬁeld elements Introduction to ﬁnite ﬁelds 17 very easy to compute. However, in general, it may not be easy to ﬁnd the power s of θ such that θt + θr = θs; see Subsection 2.1.7.5. Conversely, as we will see later in our discussion of bases for ﬁnite ﬁelds, representations which make exponentiation easy to compute often have a more complex multiplicative structure. 2.1.40 Deﬁnition Let α ∈F∗ q. The order of α is the smallest positive integer n such that αn = 1. 2.1.41 Remark We use the notation (a, b) or gcd(a, b) to represent the greatest common divisor (gcd) of a and b, where a and b belong to a Euclidean domain (usually integers or polyno-mials). 2.1.42 Lemma If g is a primitive element of Fq then gt is a primitive element of Fq if and only if (t, q −1) = 1. 2.1.43 Deﬁnition The number of positive integers e ≤n such that (n, e) = 1 is denoted by φ(n), and is the Euler function. 2.1.44 Remark The Euler function is multiplicative: if (m, n) = 1, then φ(mn) = φ(m)φ(n). 2.1.45 Remark It follows from Lemma 2.1.42 that there are exactly φ(q −1) primitive elements in Fq. 2.1.46 Deﬁnition A monic polynomial all of whose roots are primitive elements is a primitive polynomial. 2.1.47 Remark Primitive polynomials are treated in Chapter 4. 2.1.48 Deﬁnition The reciprocal f ∗of a monic polynomial f of degree n is deﬁned by f ∗(x) = xnf(1/x). The polynomial f is self-reciprocal if f ∗= f. 2.1.49 Remark The reciprocal polynomial of an irreducible polynomial f, f(x) ̸= x, over Fq is again irreducible over Fq. In addition, the monic reciprocal polynomial deﬁned by f(x)/f(0) of a primitive polynomial is also primitive. 2.1.50 Remark If f is a self-reciprocal irreducible polynomial of degree n > 1, in Fq[x], then n must be even. 2.1.51 Deﬁnition Let f ∈Fq[x] be a nonzero polynomial. If f(0) ̸= 0, the order of f is the least positive integer e such that f\|xe −1. If f(0) = 0, let f(x) = xrg(x) for some integer r ≥1 and g ∈Fq[x] with g(0) ̸= 0. In this case, the order of f is the order of g. 2.1.52 Remark We denote the order of f by ord(f). The order of a polynomial is also called the period or exponent of the polynomial. 2.1.53 Theorem Let f ∈Fq[x] be an irreducible polynomial over Fq of degree n with f(0) ̸= 0. Then ord(f) is equal to the order of any root of f in the multiplicative group of F∗ qn. 2.1.54 Corollary If f ∈Fq[x] is an irreducible polynomial over Fq of degree n, then ord(f)\|(qn−1). 2.1.55 Theorem Let Fq be a ﬁnite ﬁeld of characteristic p, and let f ∈Fq[x] be a polynomial of positive degree with f(0) ̸= 0. Let f = af b1 1 · · · f bk k be the canonical factorization of f into irreducibles in Fq[x], where a ∈Fq, b1, . . . , bk ∈N, and f1, . . . , fk are distinct monic irreducible polynomials in Fq[x]. Then ord(f) = ept, where e is the least common multiple of ord(f1), . . . , ord(fk) and t is the smallest integer with pt ≥max(b1, . . . , bk). 18 Handbook of Finite Fields 2.1.3 Extension ﬁelds 2.1.56 Deﬁnition Let K be a subﬁeld of F and let M be a subset of F. Then K(M) denotes the intersection of all subﬁelds of F containing K and M as subsets. This ﬁeld is K adjoin M. When M is ﬁnite, say M = {α1, . . . , αk}, we write K(α1, . . . , αk) for K(M). 2.1.57 Deﬁnition Let K ⊆F, α ∈F, and f(α) = 0 where f is a monic polynomial in K[x]. Then f is the minimal polynomial of α if α is not a root of any nonzero polynomial in K[x] of lower degree. 2.1.58 Proposition The minimal polynomial of any extension ﬁeld element is irreducible over the base ﬁeld. This result provides a method by which one can obtain irreducible polynomials. 2.1.59 Deﬁnition A ﬁeld F is a ﬁnite extension of K if K ⊆F and F is a ﬁnite dimensional vector space over K. In this case we refer to the dimension m of F over K as the degree of the extension, and we write [F : K] = m. 2.1.60 Theorem Let F be a ﬁnite extension of K and let E be a ﬁnite extension of F. Then E is a ﬁnite extension of K. Moreover, we have [E : K] = [E : F][F : K]. 2.1.61 Deﬁnition Let K ⊆F and let α ∈F. Then α is algebraic over K if there is a nonzero polynomial f ∈K[x] such that f(α) = 0 in F[x]. An extension ﬁeld is algebraic if every element of the extension ﬁeld is algebraic. 2.1.62 Theorem Every ﬁnite extension of a ﬁeld is algebraic. 2.1.63 Theorem Let K be a subﬁeld of F with α ∈F algebraic of degree n over K and let g be the minimal polynomial of α over K. Then: 1. The ﬁeld K(α) is isomorphic to the factor ring K[x]/(g). 2. The dimension of K(α) over K is n. 3. The set {1, α, α2, . . . , αn−1} is a basis for K(α) over K. 4. Every element of K(α) is algebraic over K with degree dividing n. 2.1.64 Remark An extension obtained by adjoining a single element is a simple extension. The next theorem gives an important property of ﬁnite ﬁelds which is not shared by inﬁnite ﬁelds (there are ﬁnite extensions of inﬁnite ﬁelds which are not simple). 2.1.65 Theorem Let Fq be a ﬁnite ﬁeld and let Fr be a ﬁnite extension of Fq. Then Fr is a simple algebraic extension of Fq, and for any primitive element α of Fr the relation Fr = Fq(α) holds. 2.1.66 Corollary For any prime power q and any integer n ≥1 there is an irreducible polynomial of degree n over Fq. 2.1.67 Example Consider q = 2100. We can identify the elements of Fq with polynomials of the form a0 +a1α+a2α2 +· · ·+a99α99, where 0 ≤ai < 2 for each i and where α is a root of an irreducible polynomial of degree 100 over the ﬁeld F2. Corollary 2.1.66 shows that such an irreducible polynomial always exists. Using Theorem 2.1.24 we have that there are exactly 1 100 2100 −250 −220 + 210 Introduction to ﬁnite ﬁelds 19 irreducible polynomials of degree 100 over F2. 2.1.68 Remark Corollary 2.1.66 can also be derived using the formula for the number of irreducible polynomials in Theorem 2.1.24. 2.1.69 Example Consider the polynomial f(x) = x2 +x+1 over the ﬁeld F2. Since f does not have a root in F2, f is irreducible over F2. Let α be a root of f so that α2 + α + 1 = 0, that is, α2 = −(α+1) = α+1. The ﬁeld F4 = F22 can be represented as the set {aα+b: a, b ∈F2}. We give the addition and multiplication tables for the ﬁeld F22. We note that α is a primitive element in the ﬁeld F4, so α1 = α, α2 = α + 1 and α3 = 1. + 0 1 α α + 1 0 0 1 α α + 1 1 1 0 α + 1 α α α α + 1 0 1 α + 1 α + 1 α 1 0 × 0 1 α α + 1 0 0 0 0 0 1 0 1 α α + 1 α 0 α α + 1 1 α + 1 0 α + 1 1 α 2.1.70 Example Consider the ﬁeld F9, which is a vector space of dimension 2 over F3. Consider f(x) = x2+x+2 in F3[x]. This polynomial has no roots in F3 so it is irreducible over F3. Let α be a root of f, so α2 +α+2 = 0. Hence α2 = −α−2 = 2α+1. The ﬁeld F32 is isomorphic to the set {aα+b \| a, b ∈F3} with its natural operations. We can compute the addition and multiplication tables by hand. For example, 2α(α+2) = 2α2 +4α = 2(2α+1)+α = 2α+2. The following addition and multiplication tables are obtained. We can use the multiplication table to check that the multiplicative order of α in F9 is 8, and thus α is a primitive element of F9. + 0 1 2 α α + 1 α + 2 2α 2α + 1 2α + 2 0 0 1 2 α α + 1 α + 2 2α 2α + 1 2α + 2 1 1 2 0 α + 1 α + 2 α 2α + 1 2α + 2 2α 2 2 0 1 α + 2 α α + 1 2α + 2 2α 2α + 1 α α α + 1 α + 2 2α 2α + 1 2α + 2 0 1 2 α + 1 α + 1 α + 2 α 2α + 1 2α + 2 2α 1 2 0 α + 2 α + 2 α α + 1 2α + 2 2α 2α + 1 2 0 1 2α 2α 2α + 1 2α + 2 0 1 2 α α + 1 α + 2 2α + 1 2α + 1 2α + 2 2α 1 2 0 α + 1 α + 2 α 2α + 2 2α + 2 2α 2α + 1 2 0 1 α + 2 α α + 1 × 0 1 2 α α + 1 α + 2 2α 2α + 1 2α + 2 0 0 0 0 0 0 0 0 0 0 1 0 1 2 α α + 1 α + 2 2α 2α + 1 2α + 2 2 0 2 1 2α 2α + 2 2α + 1 α α + 2 α + 1 α 0 α 2α 2α + 1 1 α + 1 α + 2 2α + 2 2 α + 1 0 α + 1 2α + 2 1 α + 2 2α 2 α 2α + 1 α + 2 0 α + 2 2α + 1 α + 1 2α 2 2α + 2 1 α 2α 0 2α α α + 2 2 2α + 2 2α + 1 α + 1 1 2α + 1 0 2α + 1 α + 2 2α + 2 α 1 α + 1 2 2α 2α + 2 0 2α + 2 α + 1 2 2α + 1 α 1 2α α + 2 2.1.71 Example Let f(x) = x2 + 1 ∈F3[x]. It is straightforward to check that f is irreducible over the ﬁeld F3. Let α be a root of f. We compute α2 = −1 and α4 = 1. Hence no root of f can have order 8, that is, no root of f can be a primitive element. Nevertheless, the splitting ﬁeld of f over F3 is F9. It can be seen that α+1 has order 8 and is thus a primitive element for F9 over F3. 20 Handbook of Finite Fields 2.1.72 Remark Tables of irreducible and primitive polynomials can be found in Section 2.2. In that section is a discussion of some computer algebra packages for implementing ﬁnite ﬁeld arithmetic. 2.1.73 Theorem If f is an irreducible polynomial of degree n over Fq then f has a root α in Fqn. Moreover all of the roots of f are simple and are given by α, αq, αq2, . . . , αqn−1. 2.1.74 Deﬁnition Let α ∈Fqn. Then α, αq, αq2, . . . , αqn−1 are the conjugates of α over Fq. 2.1.75 Lemma Let α ∈Fqn and let the minimal polynomial of α over Fq have degree d. Consider the set α, αq, αq2, . . . , αqn−1 of conjugates of α. The elements of this set are distinct if n = d; otherwise each distinct conjugate is repeated n/d times. 2.1.76 Theorem The distinct automorphisms of Fqn over Fq are given by the functions σ0, σ1, . . . , σn−1 where σj : Fqn →Fqn and is deﬁned by σj(α) = αqj for any α ∈Fqn. 2.1.77 Remark The set of automorphisms of Fq forms a group with the operation of functional composition. This group is called the Galois group of Fqn over Fq. It is a cyclic group with generator σ1 : Fqn →Fqn that maps α ∈Fqn to αq, and is called the Frobenius automorphism. The conjugates of α are thus the elements to which α is sent by iterated applications of the Frobenius automorphism. 2.1.78 Remark The subﬁelds of Fqn are exactly the ﬁelds of the form Fqm where m\|n. The sub-groups of the Galois group of Fqn over Fq are exactly the groups generated by σm 1 where m\|n. Moreover, σm 1 (α) = α if and only if α ∈Fqm. Thus there is a one-to-one correspondence between the subﬁelds of Fqn and the subgroups of its Galois group. 2.1.79 Remark In general, if F is an extension of a ﬁeld K then the set of automorphisms of F that leave K ﬁxed pointwise is the Galois group of F over K. The ﬁeld of Galois theory is the study of Galois groups. Thus, if K is ﬁnite and F is a ﬁnite extension of K then the Galois group is cyclic. When K is inﬁnite, the Galois group need not be cyclic, even if F is a ﬁnite extension of K. 2.1.4 Trace and norm functions 2.1.80 Deﬁnition Let K = Fq and F = Fqn. For α ∈F, we deﬁne the trace of α over K as TrF/K(α) = α + αq + · · · + αqn−1. Equivalently, TrF/K(α) is the sum of the conjugates of α. If K is the prime subﬁeld of F then the trace function is the absolute trace. 2.1.81 Example Let K = F2 and F = F24. Then TrF/K(α) = α + α2 + α4 + α8. For K = F4 and F = F16 we have TrF/K(α) = α + α4. 2.1.82 Remark Since TrF/K(α) q = TrF/K(α) the trace of an element always lies in the base ﬁeld K. 2.1.83 Theorem Let K = Fq and F = Fqn. The trace function has the following properties: 1. for any α ∈F, TrF/K(α) ∈K; 2. TrF/K(α + β) = TrF/K(α) + TrF/K(β) for α, β ∈F; 3. TrF/K(cα) = cTrF/K(α) for α ∈F and c ∈K; 4. the trace function is a K-linear map from F onto K; Introduction to ﬁnite ﬁelds 21 5. TrF/K(α) = nα for α ∈K; 6. TrF/K(αq) = TrF/K(α) for α ∈F; 7. for any α ∈K, we have \|{β ∈F \| TrF/K(β) = α}\| = qn−1; 8. Suppose that K ⊆F ⊆E are ﬁnite ﬁelds; then for any α ∈E TrE/K(α) = TrF/K(TrE/F (α)). 2.1.84 Theorem For β ∈F let Lβ be the map α 7→TrF/K(βα). Then Lβ ̸= Lγ if β ̸= γ. Moreover the K-linear transformations from F to K are exactly the maps of the form Lβ as β varies over the elements of the ﬁeld F. 2.1.85 Remark The result in Theorem 2.1.84 provides a method to generate all of the linear transformations from the extension ﬁeld F to the subﬁeld K. 2.1.86 Deﬁnition Let K = Fq and F = Fqn. The norm over K of an element α ∈F is deﬁned by NormF/K(α) = ααq · · · αqn−1 = n−1 Y i=0 αqi = α(qn−1)/(q−1). 2.1.87 Remark The norm of an element α is thus calculated by taking the product of all of the conjugates of α, just as the trace of α is obtained by taking the sum of all of the conjugates of α. 2.1.88 Theorem Let K = Fq and F = Fqn. The norm function has the following properties: 1. NormF/K(α) ∈K; 2. NormF/K(αβ) = NormF/K(α) NormF/K(β) for α, β ∈F; 3. the norm maps F onto K and F ∗onto K∗; 4. NormF/K(α) = αn if α ∈K; 5. NormF/K(αq) = NormF/K(α); 6. if K ⊆F ⊆E are ﬁnite ﬁelds then NormE/K(α) = NormF/K(NormE/F (α)) . 2.1.5 Bases 2.1.89 Remark Every ﬁnite ﬁeld F is a vector space over each of its subﬁelds, and thus has a vector space basis over each of its subﬁelds. There are several diﬀerent kinds of bases for ﬁnite ﬁelds. Each kind of basis facilitates certain computations. When doing computations in ﬁnite ﬁelds, there are some important operations like addition, multiplication, q-th powering and ﬁnding inverses. With some bases computing inverses and q-th powers are easy, while multiplication could be more involved. With other bases, one can calculate multiplications quickly at the cost of more complicated inverse computations or exponentiations. 2.1.90 Remark The vector space of all n × r matrices over a ﬁeld Fq is of dimension nr over Fq. Taking into account the order of the elements, the total number of distinct bases of Fqn over Fq is given by (qn −1)(qn −q) · · · (qn −qn−1), which is also equal to the number of elements in the general linear group GLn(Fq), the ring of nonsingular n × n matrices over Fq. 22 Handbook of Finite Fields 2.1.91 Remark Consider Fqn as a vector space over Fq of dimension n. We know there are many bases for this vector space. Given B = {α1, . . . , αn} ⊆Fqn, how can we tell if B is a basis for Fqn over Fq? We begin with a test which determines whether a set of elements of Fqn is independent over Fq. If this result is applied to a set containing n elements, it can thus be used to determine whether these elements form a basis of Fqn over Fq. We require the following notation. 2.1.92 Deﬁnition Let K = Fq and F = Fqn. Let {α1, . . . , αn} be a set of n elements of F viewed as a vector space over the subﬁeld K. We deﬁne the discriminant ∆F/K(α1, . . . , αn) with the following rule: ∆F/K(α1, . . . , αn) = TrF/K(α1α1) · · · TrF/K(α1αn) . . . ... . . . TrF/K(αnα1) · · · TrF/K(αnαn) . 2.1.93 Theorem If α1, . . . , αn ∈Fqn, then the set {α1, . . . , αn} is a basis for Fqn over Fq if and only if ∆Fqn/Fq(α1, . . . , αn) is nonzero. 2.1.94 Remark The following result provides an alternative method to determine if a given set of elements forms a basis. We note that the calculations for this method must be done in the extension ﬁeld, not in the base ﬁeld. Working in the extension ﬁeld may have a signiﬁcant computational cost. For example, if the base ﬁeld is F2 and the extension ﬁeld is F21000 then computations in the base ﬁeld are much faster than computations in the extension ﬁeld. 2.1.95 Corollary The set {α1, . . . , αn} is a basis for Fqn over Fq if and only if α1 · · · αn αq 1 · · · αq n . . . ... . . . αqn−1 1 · · · αqn−1 n ̸= 0. 2.1.96 Deﬁnition Let α be a root of an irreducible polynomial of degree n over Fq. The set {1, α, α2, . . . , αn−1} is a polynomial basis of the ﬁeld Fqn over Fq. 2.1.97 Remark When we use a polynomial basis for Fqn we can regard ﬁeld elements, which in reality are polynomials in α of degree at most n −1, as vectors. We can then easily add vectors in the usual way by adding the corresponding coeﬃcients. Field multiplication is more complicated since we must gather terms with like powers of the basis elements when we simplify a product. 2.1.98 Deﬁnition If α ∈Fqn and {α, αq, . . . , αqn−1} is a basis for Fqn over Fq, then the basis is a normal basis of Fqn over Fq, and α is a normal element. 2.1.99 Remark If β = a0α + a1αq + · · · + an−1αqn−1 so that β is represented by the vector (a0, . . . , an−1), then αq is simply represented by the shifted vector (an−1, a0, . . . , an−2). Thus if we have a normal basis, it is extremely easy to raise a ﬁeld element to the power q. Addition is of course also still easy to compute using a normal basis. We note that multiplication of ﬁeld elements is quite complicated using a normal basis. In Section 5.2 we Introduction to ﬁnite ﬁelds 23 give important properties of normal bases including their existence for any ﬁnite extension ﬁeld of Fq. 2.1.100 Deﬁnition Two ordered bases of Fqn over Fq {α1, . . . , αn} and {β1, . . . , βn} are comple-mentary (or dual) if TrFqn/Fq(αiβj) = δij, where δij = 0 if j ̸= i and δij = 1 if i = j. An ordered basis is self-dual if it is dual with itself. 2.1.101 Deﬁnition A primitive normal basis for an extension ﬁeld Fqn over Fq is a basis of the form {α, αq, αq2, . . . , αqn−1}, where α is a primitive element for Fqn over Fq. 2.1.102 Remark Further kinds of bases for ﬁnite ﬁelds and their properties are discussed in detail in Chapter 5. For example, we show that each basis of Fqn has a unique dual basis. We give fundamental properties of normal bases and primitive normal bases in Section 5.2. We give there, among other results, the fundamental theorem that for any prime power q and any integer n ≥2 there exists a primitive normal basis for Fqn over Fq. 2.1.6 Linearized polynomials 2.1.103 Deﬁnition Let L(x) = Pn−1 i=0 αixqi, where αi ∈Fqn. A polynomial of this form is a linearized polynomial over Fqn (also a q-polynomial because the exponents are all powers of q). 2.1.104 Remark These polynomials form an important class of polynomials over ﬁnite ﬁelds because they are Fq-linear functions from Fqn to Fqn. 2.1.105 Theorem Let L(x) be a linearized polynomial. Then for all α, β ∈Fqn and all c ∈Fq, we have 1. L(α + β) = L(α) + L(β), 2. L(cα) = cL(α). 2.1.106 Theorem Let L be a nonzero linearized polynomial over Fqn and assume that the roots of L lie in the ﬁeld Fqs, an extension ﬁeld of Fqn. Then each root of L has the same multiplicity, which is either 1, or a positive power of q. 2.1.107 Remark The Frobenius automorphism x 7→xq is one such example, and the trace function Tr(x) = Pn−1 i=0 xqi provides another important example of a linearized polynomial over Fq. 2.1.108 Deﬁnition Let L be a linearized polynomial over Fqn. A polynomial of the form A(x) = L(x) −α, for α ∈Fqn, is an aﬃne polynomial over Fqn. 2.1.109 Theorem Let A be a nonzero aﬃne polynomial over Fqn and assume that the roots of A lie in the ﬁeld Fqs, an extension ﬁeld of Fqn. Then each root of A has the same multiplicity, which is either 1, or a positive power of q. 2.1.7 Miscellaneous results 2.1.110 Remark We collect here some concepts and results needed in later sections of the handbook. 24 Handbook of Finite Fields 2.1.7.1 The ﬁnite ﬁeld polynomial Φ function 2.1.111 Deﬁnition For f ∈Fq[x], Φq(f) denotes the number of polynomials over Fq which are of smaller degree than the degree of f and which are relatively prime to f. This is also the number of units in the ring Fq[x]/(f(x)). 2.1.112 Remark Similarly to the corresponding properties for the Euler function from elementary number theory, we have the following result (see Lemma 3.69 of and Deﬁnition 2.1.43). 2.1.113 Lemma The function Φq has the following properties: 1. Φq(f) = 1 if the degree of f is 0; 2. Φq(fg) = Φq(f)Φq(g) if f and g are relatively prime; 3. if f has degree n ≥1 then Φq(f) = qn(1 −q−n1) · · · (1 −q−nr), where n1, . . . , nr are the degrees of the distinct monic irreducible polynomials appearing in the unique factorization of f in Fq[x]. 2.1.114 Remark One important consequence of Lemma 2.1.113 is that if f is irreducible of degree n over Fq, then Φq(f e) = qne −qn(e−1) for any positive integer e. 2.1.7.2 Cyclotomic polynomials 2.1.115 Remark The following is a synopsis of properties of roots of unity and cyclotomic polyno-mials, which can be found in [1939, Chapters 2 and 3]. 2.1.116 Remark Let n be a positive integer. The polynomial xn−1 has many special properties over any ﬁeld. For example, xn −1 is the minimal polynomial of the Frobenius automorphism which generates the Galois group of Fqn over Fq, which is useful when studying normal bases over ﬁnite ﬁelds, see Section 5.2. Many of the basic properties of cyclotomic polynomials (and their roots) hold over arbitrary ﬁelds, however in this section we restrict to the ﬁnite ﬁeld case. 2.1.117 Deﬁnition The roots α1, α2, . . . , αn ∈Fqn of the polynomial xn −1 ∈Fq[x] are the n-th roots of unity over Fq. 2.1.118 Remark The roots of any degree n polynomial over Fq must be in Fqn. Thus, the n-th roots of unity of Fq are all contained in Fqn. 2.1.119 Theorem Let n be a positive integer and let Fq be a ﬁnite ﬁeld of characteristic p. If p does not divide n, the roots of unity form a cyclic group of order n with respect to multiplication in F∗ q. Otherwise, let n = mpe, where e > 0 and gcd(m, p) = 1. Then xn −1 = (xm −1)pe and the n-th roots of unity are the m-th roots of unity with multiplicity pe. 2.1.120 Deﬁnition Let Fq be a ﬁnite ﬁeld of characteristic p which does not divide n. Denote the cyclic group of n-th roots of unity as Un. Suppose that Un is generated by α ∈Fqn, that is Un = ⟨α⟩. Then α is a primitive n-th root of unity over Fq. Introduction to ﬁnite ﬁelds 25 2.1.121 Deﬁnition Let Fq have characteristic p, not dividing n, and let ζ be a primitive n-th root of unity over Fq. Then the polynomial Qn(x) = n Y s=1 gcd(s,n)=1 (x −ζs) is the n-th cyclotomic polynomial over Fq. 2.1.122 Remark The n-th cyclotomic polynomial does not depend on the choice of primitive root of unity chosen, since ζs, gcd(s, n) = 1, runs over all primitive n-th roots of unity. 2.1.123 Theorem Let Fq be a ﬁnite ﬁeld with characteristic p which does not divide n. Then 1. deg(Qn) = φ(n); 2. xn −1 = Q d\|n Qd(x); 3. the coeﬃcients of Qn(x) lay within Fp. 2.1.124 Proposition Let r be a prime and let k be a positive integer. Then Qrk(x) = 1 + xrk−1 + x2rk−1 + · · · + x(r−1)rk−1. 2.1.125 Theorem Suppose gcd(n, q) = 1, then Qn factors into φ(n)/d distinct monic irreducible polynomials in Fq[x] of degree d, where d is the order of q modulo n. Furthermore, Fqd is the splitting ﬁeld of any such factor. 2.1.126 Corollary The cyclotomic polynomial Qn is irreducible over Fq if and only if n = 4, rk, 2rk, k ≥0, where r is an odd prime and q is a primitive root modulo n. 2.1.127 Proposition Let p be a prime and let m and k be positive integers. The following properties of cyclotomic polynomials hold over any ﬁeld for which they are deﬁned: 1. Qmp(x) = Qm(xp)/Qm(x), if p does not divide m; 2. Qmp(x) = Qm(xp), if p divides m; 3. Qmpk(x) = Qmp(xpk−1); 4. Q2n(x) = Qn(−x) if n ≥3 and n is odd; 5. Qn(0) = 1 if n ≥2; 6. Qn(x−1)xφ(n) = Qn(x) if n ≥2; 7. Qn(1) =      0 if n = 1, p if n = pe, 1 if n has at least two distinct prime factors; 8. Qn(−1) =          −2 if n = 1, 0 if n = 2, p if n = 2pe, 1 otherwise. 2.1.128 Theorem Let n be a positive integer not divisible by the characteristic of Fq. An explicit factorization of the Qn over Fq is given by Qn(x) = Y d\|n (xd −1)µ(n/d) = Y d\|n (xn/d −1)µ(d), 26 Handbook of Finite Fields where µ is the M¨ obius function, see Deﬁnition 2.1.22. 2.1.129 Proposition Suppose gcd(n, q) = 1, then Qn is irreducible over Fq if and only if the multi-plicative order of q modulo n is φ(n). 2.1.130 Theorem The product of all monic irreducible polynomials of degree n over Fq, denoted I(q, n; x), is given by I(q, n; x) = Y m Qm(x), where the product is taken over all positive divisors m of qn −1 such that n is the multi-plicative order of q modulo m. 2.1.7.3 Lagrange interpolation 2.1.131 Theorem If f : Fq →Fq, there is a unique polynomial Pf with coeﬃcients in Fq and of degree at most q −1 so that Pf represents the function f, that is, Pf(b) = f(b) for all b ∈Fq. In particular, Pf(x) = X a∈Fq f(a)[1 −(x −a)q−1]. 2.1.132 Remark The property that every function over a ﬁnite commutative ring with identity can be represented by a polynomial with coeﬃcients in that ring characterizes ﬁnite ﬁelds. In particular, if a ﬁnite commutative ring R with unity has the property that every function from the ring to itself can be represented by a polynomial with coeﬃcients in the ring, then R is a ﬁnite ﬁeld, and conversely. 2.1.133 Remark The Lagrange Interpolation Formula can also be stated in the following form: for n ≥0, let a0, . . . , an be n + 1 distinct elements of Fq, and let b0, . . . , bn be n + 1 arbitrary elements of Fq. Then, there exists exactly one polynomial f ∈Fq[x] of degree less than or equal to n such that f(ai) = bi, i = 0, . . . , n. This polynomial is given by f(x) = n X i=0 bi n Y k=0,k̸=i x −ak ai −ak . 2.1.134 Theorem Let f : Fn q → Fq. The polynomial Pf(x1, . . . , xn) represents f, that is, Pf(b1, . . . , bn) = f(b1, . . . , bn) for all (b1, . . . , bn) ∈Fn q , where Pf(x1, . . . , xn) = X (a1,...,an)∈Fn q f(a1, . . . , an)[1 −(x1 −a1)q−1] · · · [1 −(xn −an)q−1]. 2.1.7.4 Discriminants 2.1.135 Deﬁnition Let f be a polynomial of degree n in Fq[x] with leading coeﬃcient a, and with roots α1, α2, . . . , αn in its splitting ﬁeld, counted with multiplicity. The discriminant of f is given by D(f) = a2n−2 Y 1≤ia di Padova) for this section. 2.1.142 Deﬁnition A left (resp. right) prequasiﬁeld is a set Q together with two operations, “+” and “·” such that: (1) (Q, +) is an abelian group; (2) for all a, b, c ∈Q there exist unique x, y, z ∈Q such that a · x = b and y · a = b and a · z = b · z + c; (3) left (resp. right) distributive laws hold: for all a, b, c ∈Q a · (b + c) = a · b + a · c (resp.(b + c) · a = b · a + c · a). 2.1.143 Deﬁnition Let Q be a left prequasiﬁeld. (1) A left prequasiﬁeld is a left quasiﬁeld if it has a multiplicative identity. (2) A left prequasiﬁeld is a presemiﬁeld if it is also a right prequasiﬁeld. (3) A presemiﬁeld is a semiﬁeld if it has a multiplicative identity. (4) A semiﬁeld is commutative if “·” is commutative. (5) A left quasiﬁeld is a left nearﬁeld if “·” is associative. 28 Handbook of Finite Fields 2.1.144 Remark All left prequasiﬁelds have prime power order. Left prequasiﬁelds coordinatize translation planes. The smallest left prequasiﬁeld which is not a ﬁeld has order 9. The smallest semiﬁeld which is not a ﬁeld has order 16. For more on the above structures see or . 2.1.145 Remark The multiplicative structure of a left prequasiﬁeld is a quasigroup. The multi-plicative structure of a semiﬁeld is a loop. The multiplicative structure of a nearﬁeld is a group. 2.1.146 Deﬁnition Let Q be a set together with two operations, “+” and “·”, containing additive identity 0 and multiplicative identity 1, such that: (1) (Q/{0}, ·) is a group; (2) left and right distributive laws hold: for all a, b, c ∈Q a · (b + c) = a · b + a · c and (b + c) · a = b · a + c · a; (3) there exists a unique element 0 such that for all a ∈Q a + 0 = 0 + a = a. (4A) Q is a neoﬁeld if in addition to (1), (2) and (3) it satisﬁes for all a, b ∈Q there exist unique x, y ∈Q such that a + x = b and y + a = b. (4B) Q is a division semiring if in addition to (1), (2) and (3) above it also satisﬁes + is associative and commutative. 2.1.147 Remark The additive structure of a neoﬁeld is a loop. The additive structure of a division semiring is a commutative monoid. 2.1.148 Remark For more properties of semirings see . Note that a division semiring in which multiplication is commutative is sometimes also referred to as a semiﬁeld, but this deﬁnition does not coincide with the previously deﬁned structures. 2.1.149 Remark For more properties of neoﬁelds see . 2.1.7.7 Galois rings 2.1.150 Remark We brieﬂy describe Galois rings. We are indebted to Horacio Tapia-Recillas (Uni-versidad Aut´ onoma Metropolitana, Unidad Iztapalapa, M´ exico) for this subsection. 2.1.151 Remark Galois rings represent a natural (Galois) extension of the (local) modular ring of integers Z/pmZ where p is a prime and m a positive integer. Krull recognized their existence and later, Janusz and Raghavendran independently obtained additional properties of these rings. More details on Galois rings can be found in [283, 1409, 2045, 2921]. 2.1.152 Deﬁnition Let Z/pmZ be the ring of integers modulo pm, p a prime and m > 1 an integer. A monic irreducible (primitive) polynomial f ∈(Z/pmZ)[x] of degree n is a monic basic irreducible (primitive) if its reduction modulo p is irreducible (primitive) in (Z/pZ)[x]. Introduction to ﬁnite ﬁelds 29 2.1.153 Remark Monic basic irreducible (primitive) polynomials in (Z/pmZ)[x] can be determined by means of Hensel’s Lifting Lemma from monic irreducible (primitive) polynomials in (Z/pZ)[x]. 2.1.154 Deﬁnition Let f ∈(Z/pmZ)[x] be a monic basic irreducible polynomial of degree n. Then the Galois ring determined by f is GR(pm, n) = (Z/pmZ)[x]/⟨f(x)⟩, where ⟨f(x)⟩is the principal ideal of (Z/pmZ)[x] generated by f(x). 2.1.155 Remark With the above notation, an equivalent deﬁnition of a Galois ring is the following. 2.1.156 Deﬁnition Let Z be the ring of (rational) integers and let f ∈Z[x] be a monic polynomial of degree n such that its reduction modulo pZ is irreducible, then GR(pm, n) = Z[x]/⟨pm, f⟩. 2.1.157 Remark The Galois ring GR(pm, n) can also be deﬁned by means of the p-adic numbers in the following way. 2.1.158 Deﬁnition Let p be a prime, Qp be the ﬁeld of p-adic (rational) numbers and Zp be the ring of p-adic integers (for details see ). Let n be a positive integer and let ω be a (pn −1) root of unity. Then Qp(ω) is an unramiﬁed Galois extension of degree n of Qp. Let Zp[ω] be the ring of elements of Qp(ω) integral over Zp. Let pZp[ω] be the (unique) maximal ideal of Zp[ω] generated by p. Then the quotient Zp[ω]/pZp[ω] is a ﬁeld isomorphic to the Galois ﬁeld Fpn. 2.1.159 Deﬁnition With the notation as above let m be a positive integer and let pmZp[ω] be the principal ideal of Zp[ω] generated by pm. Then the Galois ring GR(pm, n) is deﬁned as: GR(pm, n) = Zp[ω]/pmZp[ω]. 2.1.160 Remark This ring contains as a subring the ring of integers modulo pm, Z/pmZ, and can be thought of as an extension of Z/pmZ by adjoining a (pn −1) root of unity ω: GR(pm, n) = (Z/pmZ) [ω]. 2.1.161 Theorem With the notation as above, the basic properties of the Galois ring GR(pm, n) are the following [283, 1409, 2045, 2921]: 1. GR(pm, n) contains Z/pmZ as a subring, it has characteristic pm and cardinality pmn. The integer m is the nilpotency index of the Galois ring. 2. The ring GR(pm, n) is local with maximal ideal M = ⟨p⟩= pGR(pm, n) gen-erated by p, and a principal ideal ring where any ideal is of the form ⟨pi⟩for i = 0, 1, 2, . . . , m. Furthermore, it is a ﬁnite chain ring: GR(pm, n) = ⟨p0⟩⊃⟨p⟩⊃· · · ⊃⟨pm−1⟩⊃⟨pm⟩= {0}. The ideal ⟨pi⟩has cardinality pn(m−i) for i = 0, 1, . . . , m. 30 Handbook of Finite Fields 3. Each non-zero element of the Galois ring GR(pm, n) can be written as upk, where u is a unit and 0 ≤k ≤m −1. In this representation the integer k is unique and the unit u is unique modulo the ideal ⟨pm−k⟩. 4. The canonical homomorphism φ : GR(pm, n) − →GR(pm, n)/M, between the Galois ring and its residue ﬁeld GR(pm, n)/M is such that φ(ξ) = ξ is a root of φ(f(x)). The residue ﬁeld is isomorphic to the Galois ﬁeld GF(pn) = Fpn with pn elements. Furthermore, GF(pn)∗= ⟨ξ⟩. 5. The Galois ring is a (Z/pmZ)-module: GR(pm, n) = (Z/pmZ)[ξ] = (Z/pmZ) + ξ(Z/pmZ) + · · · + ξn−1(Z/pmZ). 6. The group of units U of the Galois ring GR(pm, n) has the following structure: U = C × G, where C is a cyclic group of order pn −1 generated by ξ and G is an abelian group of order p(m−1)n. Furthermore, (a) if p is odd, or if p = 2 and m ≤2 then G is a direct product of n cyclic groups, each of order pm−1; (b) if p = 2 and m ≥3 the group G is the direct product of a cyclic group of order 2, a cyclic group of order 2m−2 and n −1 cyclic groups each of order 2m−1. 7. There is a subset T of GR(pm, n), the Teichm¨ uller set of representatives of the Galois ring, such that any element β ∈GR(pm, n) has a unique p-adic (multi-plicative) representation: β = ρ0(β) + ρ1(β)p + · · · + ρm−1(β)pm−1, where ρi(β) ∈T for 0 ≤i ≤m −1. The elements of the maximal ideal of the Galois ring correspond to ρ0(β) = 0. There is a bijection, induced by the canonical homomorphism φ, between T and the residue ﬁeld of the Galois ring GR(pm, n). 8. The Teichm¨ uller set of representatives of the Galois ring can be taken as T = {0, 1, ξ, ξ2, . . . , ξq−2} = {0} ∪C, where q = pn. 9. Given a prime p and an integer n > 1, for each divisor r of n there is a unique Galois ring GR(pm, r), and any subring of the Galois ring GR(pm, n) is of this form. 10. For each positive integer t, there is a natural injective ring homomorphism GR(pm, n) − →GR(pm, nt). 11. There is a natural surjective ring homomorphism GR(pm, n) − →GR(pm−1, n) with kernel ⟨pm−1⟩. 12. The group of automorphisms of the Galois ring GR(pm, n) is a cyclic group of order n. 13. The Galois ring GR(pm, n) is quasi-Frobenius. 2.1.162 Example GR(p, n) = GF(p, n) = Fpn, GR(pm, 1) = (Z/pmZ). 2.1.163 Example The polynomial f(x) = x3 + x + 1 ∈(Z/22Z)[x] is monic basic irreducible over (Z/22Z). Then GR(22, 3) = (Z/22Z)[x]/⟨f(x)⟩. Introduction to ﬁnite ﬁelds 31 2.1.164 Example The polynomial g(x) = x3 + 2x2 + x −1 ∈(Z/22Z)[x] is also monic basic irreducible over (Z/22Z). Then GR(22, 3) = (Z/22Z)[x]/⟨g(x)⟩. 2.1.165 Example The polynomial g(x) = x3 −2x2 −x −1 ∈(Z/23Z)[x] is monic basic irreducible over (Z/23Z). Then GR(23, 3) = (Z/23Z)[x]/⟨g(x)⟩. 2.1.8 Finite ﬁeld related books 2.1.166 Remark We give a list of ﬁnite ﬁeld related books, divided into categories and listed without duplication even though a number of these books could be listed in two or more categories. 2.1.8.1 Textbooks 2.1.167 Remark We begin by listing a number of books that could be used as textbooks. Ref-erence by Lidl and Niederreiter is, by far, the most comprehensive. Other text-books include Jungnickel , Lidl and Niederreiter , Masuda and Panario , McEliece , Menezes et al. , Mullen and Mummert , Small , and Wan [2921, 2923]. 2.1.8.2 Finite ﬁeld theory 2.1.168 Remark We list a number of books dealing with various theoretical topics related to ﬁnite ﬁelds: [240, 398, 557, 850, 961, 1121, 1122, 1333, 1389, 1511, 1631, 1701, 1756, 1773, 1843, 1845, 1922, 1936, 1938, 1939, 2017, 2049, 2054, 2077, 2107, 2548, 2637, 2641, 2667, 2670, 2672, 2681, 2711, 2714, 2793, 2920, 2921, 2923, 2949, 2950]. 2.1.8.3 Applications 2.1.169 Remark The use of ﬁnite ﬁelds in algebraic coding theory has been the focus for numerous books: [231, 270, 304, 311, 1558, 1943, 1945, 1991, 2252, 2281, 2404, 2405, 2819, 2820, 2849]. 2.1.170 Remark Theoretical and applied aspects of cryptography have been treated in: [245, 312, 313, 661, 759, 762, 922, 1105, 1303, 1413, 1521, 1563, 1694, 1774, 2076, 2080, 2644, 2720]. 2.1.171 Remark There have been several books on the applications of ﬁnite ﬁelds in combinatorics, especially in combinatorial design theory and ﬁnite geometries: [131, 141, 211, 260, 261, 262, 453, 484, 706, 785, 807, 819, 1509, 1510, 1515, 1560, 1875, 2445, 2719, 2781, 2851]. 2.1.8.4 Algorithms 2.1.172 Remark Several books contain results on algorithmic and computational ﬁnite ﬁeld topics. These include [761, 1227, 2632]. 2.1.8.5 Conference proceedings 2.1.173 Remark The Finite Fields and Applications Conferences (Fq n series) have been held every two years (except 2005) since 1991. The proceedings from these conferences are: [663, 1636, 1870, 2052, 2181, 2182, 2184, 2185, 2187, 2197]. Other conference proceedings volumes include: [533, 535, 596, 869, 1057, 1228, 1306, 1316, 1436, 1478, 1480, 1928]. References Cited: [131, 136, 141, 211, 231, 240, 245, 260, 261, 262, 270, 304, 311, 312, 313, 398, 453, 484, 533, 535, 557, 596, 661, 663, 706, 759, 761, 762, 785, 797, 807, 819, 850, 32 Handbook of Finite Fields 869, 922, 961, 1057, 1105, 1121, 1122, 1227, 1228, 1291, 1303, 1306, 1316, 1333, 1389, 1413, 1436, 1478, 1480, 1509, 1510, 1511, 1515, 1521, 1558, 1560, 1563, 1570, 1584, 1631, 1636, 1694, 1701, 1756, 1773, 1774, 1843, 1845, 1848, 1875, 1922, 1928, 1936, 1938, 1939, 1943, 1945, 1991, 2017, 2049, 2052, 2054, 2076, 2077, 2080, 2107, 2144, 2179, 2181, 2182, 2184, 2185, 2187, 2197, 2223, 2252, 2280, 2281, 2343, 2404, 2405, 2445, 2548, 2632, 2637, 2641, 2644, 2667, 2670, 2672, 2681, 2711, 2714, 2719, 2720, 2781, 2793, 2819, 2820, 2849, 2851, 2920, 2921, 2923, 2949, 2950] 2.2 Tables David Thomson, Carleton University 2.2.1 Remark Unless otherwise stated, all of the data given in this section was created by the author and, when possible, was veriﬁed with known results. Basic algorithms (for example, brute force) were preferred due to their reliability and ease of veriﬁcation. Unless stated, all simulations were done in C/C++ using the NTL version 5.5.2 library for modular computations. NTL was compiled using the GMP version 4.3.2 library for multi-precision arithmetic. Extended and machine-readable versions of the tables found in this section can be found on the book’s website . 2.2.2 Remark Since most computer algebra packages can readily handle basic ﬁnite ﬁeld compu-tations, our aim is not to repeat tables whose purpose is to improve hand-calculations. For reference, we brieﬂy recall the list of tables found in . Tables A and B are aids to perform fast arithmetic by hand over small ﬁnite ﬁelds. Table A is a list of all elements over small ﬁnite ﬁelds and their discrete logarithms with respect to a primitive element. Table B provides a list of Jacobi’s logarithms L(·) for F2n, 2 ≤n ≤6. These logarithms allow the computation of ﬁeld elements by the relationship ζα + ζβ = ζα+L(β−α). Table C provides a list of all monic irreducible polynomials of degree n over small prime ﬁelds. Particularly, these tables cover p = 2 and n ≤11, p = 3 and n ≤7, p = 5 and n ≤5, p = 7 and n ≤4. Tables D, E, and F deal with primitive polynomials. Table D lists one primitive poly-nomial over F2 for degrees n ≤100. Table E lists all quadratic primitive polynomials for 11 ≤p ≤31 and Table F lists one primitive polynomial of degree n over Fp for all n ≥2 with p < 50 and pn < 109. 2.2.1 Low-weight irreducible and primitive polynomials 2.2.3 Remark Low-weight irreducible polynomials are highly desired due to their eﬃciency in hardware and software implementations of ﬁnite ﬁelds. Irreducible polynomials of degree at least 2 over F2 must have an odd number of terms. The use of irreducible trinomials (having 3 terms) and, in their absence, irreducible pentanomials (having 5 terms) are useful; see, for example, [1413, Chapter 2]. For cryptographic use, the irreducible trinomial or pentanomial of lowest lexicographical order (for a ﬁxed n, prefer the trinomial xn+xk+1 over xn+xk1 +1 when k < k1, the analogue for pentanomials is obvious) is often preferred for transparency reasons. However, the irreducible with the optimal performance for a given implementation is not necessarily the lowest lex-order, see and Section 11.1. A list of the lowest-weight Introduction to ﬁnite ﬁelds 33 lowest-lex-order irreducible over F2 is given in for degree n ≤10000. Table 2.2.1 gives the lowest-weight, lowest-lex-order irreducible polynomial for n ≤1025. The output of the table follows the format n, k (for trinomials xn + xk + 1) or n, k1, k2, k3 (for pentanomials xn + xk1 + xk2 + xk3 + 1). We have extended these tables to larger n and to larger q for small values of n. Furthermore, the computer algebra package Magma contains similar tables, due to Steel (2004-2007), for the following values of q and n: q n ≤ q n ≤ q n ≤ q n ≤ 2 120, 000 3 50, 000 4, 5, 7 2000 9 ≤q ≤127 1000 (or more). Sections 3.4 and 4.3 give more information on weights of irreducible and primitive polyno-mials. 2,1 3,1 4,1 5,2 6,1 7,1 8,4,3,1 9,1 10,3 11,2 12,3 13,4,3,1 14,5 15,1 16,5,3,1 17,3 18,3 19,5,2,1 20,3 21,2 22,1 23,5 24,4,3,1 25,3 26,4,3,1 27,5,2,1 28,1 29,2 30,1 31,3 32,7,3,2 33,10 34,7 35,2 36,9 37,6,4,1 38,6,5,1 39,4 40,5,4,3 41,3 42,7 43,6,4,3 44,5 45,4,3,1 46,1 47,5 48,5,3,2 49,9 50,4,3,2 51,6,3,1 52,3 53,6,2,1 54,9 55,7 56,7,4,2 57,4 58,19 59,7,4,2 60,1 61,5,2,1 62,29 63,1 64,4,3,1 65,18 66,3 67,5,2,1 68,9 69,6,5,2 70,5,3,1 71,6 72,10,9,3 73,25 74,35 75,6,3,1 76,21 77,6,5,2 78,6,5,3 79,9 80,9,4,2 81,4 82,8,3,1 83,7,4,2 84,5 85,8,2,1 86,21 87,13 88,7,6,2 89,38 90,27 91,8,5,1 92,21 93,2 94,21 95,11 96,10,9,6 97,6 98,11 99,6,3,1 100,15 101,7,6,1 102,29 103,9 104,4,3,1 105,4 106,15 107,9,7,4 108,17 109,5,4,2 110,33 111,10 112,5,4,3 113,9 114,5,3,2 115,8,7,5 116,4,2,1 117,5,2,1 118,33 119,8 120,4,3,1 121,18 122,6,2,1 123,2 124,19 125,7,6,5 126,21 127,1 128,7,2,1 129,5 130,3 131,8,3,2 132,17 133,9,8,2 134,57 135,11 136,5,3,2 137,21 138,8,7,1 139,8,5,3 140,15 141,10,4,1 142,21 143,5,3,2 144,7,4,2 145,52 146,71 147,14 148,27 149,10,9,7 150,53 151,3 152,6,3,2 153,1 154,15 155,62 156,9 157,6,5,2 158,8,6,5 159,31 160,5,3,2 161,18 162,27 163,7,6,3 164,10,8,7 165,9,8,3 166,37 167,6 168,15,3,2 169,34 170,11 171,6,5,2 172,1 173,8,5,2 174,13 175,6 176,11,3,2 177,8 178,31 179,4,2,1 180,3 181,7,6,1 182,81 183,56 184,9,8,7 185,24 186,11 187,7,6,5 188,6,5,2 189,6,5,2 190,8,7,6 191,9 192,7,2,1 193,15 194,87 195,8,3,2 196,3 197,9,4,2 198,9 199,34 200,5,3,2 201,14 202,55 203,8,7,1 204,27 205,9,5,2 206,10,9,5 207,43 208,9,3,1 209,6 210,7 211,11,10,8 212,105 213,6,5,2 214,73 215,23 216,7,3,1 217,45 218,11 219,8,4,1 220,7 221,8,6,2 222,5,4,2 223,33 224,9,8,3 225,32 226,10,7,3 227,10,9,4 228,113 229,10,4,1 230,8,7,6 231,26 232,9,4,2 233,74 234,31 235,9,6,1 236,5 237,7,4,1 238,73 239,36 240,8,5,3 241,70 242,95 243,8,5,1 244,111 245,6,4,1 246,11,2,1 247,82 248,15,14,10 249,35 250,103 251,7,4,2 252,15 253,46 254,7,2,1 255,52 256,10,5,2 257,12 258,71 259,10,6,2 260,15 261,7,6,4 262,9,8,4 263,93 264,9,6,2 265,42 266,47 267,8,6,3 268,25 269,7,6,1 270,53 271,58 272,9,3,2 273,23 274,67 275,11,10,9 276,63 277,12,6,3 278,5 279,5 280,9,5,2 281,93 282,35 283,12,7,5 284,53 285,10,7,5 286,69 287,71 288,11,10,1 289,21 290,5,3,2 291,12,11,5 292,37 293,11,6,1 294,33 295,48 296,7,3,2 297,5 298,11,8,4 299,11,6,4 300,5 301,9,5,2 302,41 303,1 304,11,2,1 305,102 306,7,3,1 307,8,4,2 308,15 309,10,6,4 310,93 311,7,5,3 312,9,7,4 313,79 314,15 315,10,9,1 316,63 317,7,4,2 318,45 319,36 320,4,3,1 321,31 322,67 323,10,3,1 324,51 325,10,5,2 326,10,3,1 327,34 328,8,3,1 329,50 330,99 331,10,6,2 332,89 333,2 334,5,2,1 335,10,7,2 336,7,4,1 337,55 338,4,3,1 339,16,10,7 340,45 341,10,8,6 342,125 343,75 344,7,2,1 345,22 346,63 347,11,10,3 348,103 349,6,5,2 350,53 351,34 352,13,11,6 353,69 354,99 355,6,5,1 356,10,9,7 357,11,10,2 358,57 359,68 360,5,3,2 361,7,4,1 362,63 363,8,5,3 364,9 365,9,6,5 366,29 367,21 368,7,3,2 369,91 370,139 371,8,3,2 372,111 373,8,7,2 374,8,6,5 375,16 376,8,7,5 377,41 378,43 379,10,8,5 380,47 381,5,2,1 382,81 383,90 384,12,3,2 385,6 386,83 387,8,7,1 388,159 389,10,9,5 390,9 391,28 392,13,10,6 393,7 394,135 395,11,6,5 396,25 397,12,7,6 398,7,6,2 399,26 400,5,3,2 401,152 402,171 403,9,8,5 404,65 405,13,8,2 406,141 407,71 408,5,3,2 409,87 410,10,4,3 411,12,10,3 412,147 413,10,7,6 414,13 415,102 416,9,5,2 417,107 418,199 419,15,5,4 420,7 421,5,4,2 422,149 423,25 424,9,7,2 425,12 426,63 427,11,6,5 428,105 429,10,8,7 430,14,6,1 431,120 432,13,4,3 433,33 434,12,11,5 435,12,9,5 436,165 437,6,2,1 438,65 439,49 440,4,3,1 441,7 442,7,5,2 443,10,6,1 444,81 445,7,6,4 446,105 447,73 448,11,6,4 449,134 450,47 451,16,10,1 452,6,5,4 453,15,6,4 454,8,6,1 455,38 456,18,9,6 457,16 458,203 459,12,5,2 460,19 461,7,6,1 462,73 463,93 464,19,18,13 465,31 466,14,11,6 467,11,6,1 468,27 469,9,5,2 470,9 471,1 472,11,3,2 473,200 474,191 475,9,8,4 476,9 477,16,15,7 478,121 479,104 480,15,9,6 481,138 34 Handbook of Finite Fields 482,9,6,5 483,9,6,4 484,105 485,17,16,6 486,81 487,94 488,4,3,1 489,83 490,219 491,11,6,3 492,7 493,10,5,3 494,17 495,76 496,16,5,2 497,78 498,155 499,11,6,5 500,27 501,5,4,2 502,8,5,4 503,3 504,15,14,6 505,156 506,23 507,13,6,3 508,9 509,8,7,3 510,69 511,10 512,8,5,2 513,26 514,67 515,14,7,4 516,21 517,12,10,2 518,33 519,79 520,15,11,2 521,32 522,39 523,13,6,2 524,167 525,6,4,1 526,97 527,47 528,11,6,2 529,42 530,10,7,3 531,10,5,4 532,1 533,4,3,2 534,161 535,8,6,2 536,7,5,3 537,94 538,195 539,10,5,4 540,9 541,13,10,4 542,8,6,1 543,16 544,8,3,1 545,122 546,8,2,1 547,13,7,4 548,10,5,3 549,16,4,3 550,193 551,135 552,19,16,9 553,39 554,10,8,7 555,10,9,4 556,153 557,7,6,5 558,73 559,34 560,11,9,6 561,71 562,11,4,2 563,14,7,3 564,163 565,11,6,1 566,153 567,28 568,15,7,6 569,77 570,67 571,10,5,2 572,12,8,1 573,10,6,4 574,13 575,146 576,13,4,3 577,25 578,23,22,16 579,12,9,7 580,237 581,13,7,6 582,85 583,130 584,14,13,3 585,88 586,7,5,2 587,11,6,1 588,35 589,10,4,3 590,93 591,9,6,4 592,13,6,3 593,86 594,19 595,9,2,1 596,273 597,14,12,9 598,7,6,1 599,30 600,9,5,2 601,201 602,215 603,6,4,3 604,105 605,10,7,5 606,165 607,105 608,19,13,6 609,31 610,127 611,10,4,2 612,81 613,19,10,4 614,45 615,211 616,19,10,3 617,200 618,295 619,9,8,5 620,9 621,12,6,5 622,297 623,68 624,11,6,5 625,133 626,251 627,13,8,4 628,223 629,6,5,2 630,7,4,2 631,307 632,9,2,1 633,101 634,39 635,14,10,4 636,217 637,14,9,1 638,6,5,1 639,16 640,14,3,2 641,11 642,119 643,11,3,2 644,11,6,5 645,11,8,4 646,249 647,5 648,13,3,1 649,37 650,3 651,14 652,93 653,10,8,7 654,33 655,88 656,7,5,4 657,38 658,55 659,15,4,2 660,11 661,12,11,4 662,21 663,107 664,11,9,8 665,33 666,10,7,2 667,18,7,3 668,147 669,5,4,2 670,153 671,15 672,11,6,5 673,28 674,11,7,4 675,6,3,1 676,31 677,8,4,3 678,15,5,3 679,66 680,23,16,9 681,11,9,3 682,171 683,11,6,1 684,209 685,4,3,1 686,197 687,13 688,19,14,6 689,14 690,79 691,13,6,2 692,299 693,15,8,2 694,169 695,177 696,23,10,2 697,267 698,215 699,15,10,1 700,75 701,16,4,2 702,37 703,12,7,1 704,8,3,2 705,17 706,12,11,8 707,15,8,5 708,15 709,4,3,1 710,13,12,4 711,92 712,5,4,3 713,41 714,23 715,7,4,1 716,183 717,16,7,1 718,165 719,150 720,9,6,4 721,9 722,231 723,16,10,4 724,207 725,9,6,5 726,5 727,180 728,4,3,2 729,58 730,147 731,8,6,2 732,343 733,8,7,2 734,11,6,1 735,44 736,13,8,6 737,5 738,347 739,18,16,8 740,135 741,9,8,3 742,85 743,90 744,13,11,1 745,258 746,351 747,10,6,4 748,19 749,7,6,1 750,309 751,18 752,13,10,3 753,158 754,19 755,12,10,1 756,45 757,7,6,1 758,233 759,98 760,11,6,5 761,3 762,83 763,16,14,9 764,6,5,3 765,9,7,4 766,22,19,9 767,168 768,19,17,4 769,120 770,14,5,2 771,17,15,6 772,7 773,10,8,6 774,185 775,93 776,15,14,7 777,29 778,375 779,10,8,3 780,13 781,17,16,2 782,329 783,68 784,13,9,6 785,92 786,12,10,3 787,7,6,3 788,17,10,3 789,5,2,1 790,9,6,1 791,30 792,9,7,3 793,253 794,143 795,7,4,1 796,9,4,1 797,12,10,4 798,53 799,25 800,9,7,1 801,217 802,15,13,9 803,14,9,2 804,75 805,8,7,2 806,21 807,7 808,14,3,2 809,15 810,159 811,12,10,8 812,29 813,10,3,1 814,21 815,333 816,11,8,2 817,52 818,119 819,16,9,7 820,123 821,15,11,2 822,17 823,9 824,11,6,4 825,38 826,255 827,12,10,7 828,189 829,4,3,1 830,17,10,7 831,49 832,13,5,2 833,149 834,15 835,14,7,5 836,10,9,2 837,8,6,5 838,61 839,54 840,11,5,1 841,144 842,47 843,11,10,7 844,105 845,2 846,105 847,136 848,11,4,1 849,253 850,111 851,13,10,5 852,159 853,10,7,1 854,7,5,3 855,29 856,19,10,3 857,119 858,207 859,17,15,4 860,35 861,14 862,349 863,6,3,2 864,21,10,6 865,1 866,75 867,9,5,2 868,145 869,11,7,6 870,301 871,378 872,13,3,1 873,352 874,12,7,4 875,12,8,1 876,149 877,6,5,4 878,12,9,8 879,11 880,15,7,5 881,78 882,99 883,17,16,12 884,173 885,8,7,1 886,13,9,8 887,147 888,19,18,10 889,127 890,183 891,12,4,1 892,31 893,11,8,6 894,173 895,12 896,7,5,3 897,113 898,207 899,18,15,5 900,1 901,13,7,6 902,21 903,35 904,12,7,2 905,117 906,123 907,12,10,2 908,143 909,14,4,1 910,15,9,7 911,204 912,7,5,1 913,91 914,4,2,1 915,8,6,3 916,183 917,12,10,7 918,77 919,36 920,14,9,6 921,221 922,7,6,5 923,16,14,13 924,31 925,16,15,7 926,365 927,403 928,10,3,2 929,11,4,3 930,31 931,10,9,4 932,177 933,16,6,1 934,22,6,5 935,417 936,15,13,12 937,217 938,207 939,7,5,4 940,10,7,1 941,11,6,1 942,45 943,24 944,12,11,9 945,77 946,21,20,13 947,9,6,5 948,189 949,8,3,2 950,13,12,10 951,260 952,16,9,7 953,168 954,131 955,7,6,3 956,305 957,10,9,6 958,13,9,4 959,143 960,12,9,3 961,18 962,15,8,5 963,20,9,6 964,103 965,15,4,2 966,201 967,36 968,9,5,2 969,31 970,11,7,2 971,6,2,1 972,7 973,13,6,4 974,9,8,7 975,19 976,17,10,6 977,15 978,9,3,1 979,178 980,8,7,6 981,12,6,5 982,177 983,230 984,24,9,3 985,222 986,3 987,16,13,12 988,121 989,10,4,2 990,161 991,39 992,17,15,13 993,62 994,223 995,15,12,2 996,65 997,12,6,3 998,101 999,59 1000,5,4,3 1001,17 1002,5,3,2 1003,13,8,3 1004,10,9,7 1005,12,8,2 1006,5,4,3 1007,75 1008,19,17,8 1009,55 1010,99 1011,10,7,4 1012,115 1013,9,8,6 1014,385 1015,186 1016,15,6,3 1017,9,4,1 1018,12,10,5 1019,10,8,1 1020,135 1021,5,2,1 1022,317 1023,7 1024,19,6,1 1025,294 Table 2.2.1 Lowest weight lowest-lexicographical order irreducible polynomial of degree n over F2. Out-put: n, k (for trinomials xn + xk + 1) or n, k1, k2, k3 (for pentanomials xn + xk1 + xk2 + xk3 + 1). 2.2.4 Remark Constructions of irreducible low-weight polynomials are rare; see Sections 3.4 and 3.5. Instead, conditions for reducibility are often more tractable; see Section 3.3. Swan gives conditions for when a trinomial xn + xk + 1 ∈F2[x] is reducible. In particular, the trinomial is reducible when 8 divides n. Only partial results for Swan-like Introduction to ﬁnite ﬁelds 35 conditions on pentanomials over F2 exist in the literature; see, for example, and Section 3.3. 2.2.5 Conjecture For every n, there exists either an irreducible trinomial of degree n over F2 or, in the absence of an irreducible trinomial, there exists an irreducible pentanomial of degree n over F2. 2.2.6 Remark A polynomial over Fq is primitive if all of its roots are generators of the (cyclic) multiplicative group F∗ q. We give an analogous table to Table 2.2.1 but instead list the lowest-weight lowest-lexicographical order primitive polynomial of degree n ≤577 over F2. To compute primitivity, we used the Cunningham project to ﬁnd the factorization of 2n −1; see Section 2.2.3 for more details. 2,1 3,1 4,1 5,2 6,1 7,1 8,4,3,2 9,4 10,3 11,2 12,6,4,1 13,4,3,1 14,5,3,1 15,1 16,5,3,2 17,3 18,7 19,5,2,1 20,3 21,2 22,1 23,5 24,4,3,1 25,3 26,6,2,1 27,5,2,1 28,3 29,2 30,6,4,1 31,3 32,7,6,2 33,13 34,8,4,3 35,2 36,11 37,6,4,1 38,6,5,1 39,4 40,5,4,3 41,3 42,7,4,3 43,6,4,3 44,6,5,2 45,4,3,1 46,8,7,6 47,5 48,9,7,4 49,9 50,4,3,2 51,6,3,1 52,3 53,6,2,1 54,8,6,3 55,24 56,7,4,2 57,7 58,19 59,7,4,2 60,1 61,5,2,1 62,6,5,3 63,1 64,4,3,1 65,18 66,9,8,6 67,5,2,1 68,9 69,6,5,2 70,5,3,1 71,6 72,10,9,3 73,25 74,7,4,3 75,6,3,1 76,5,4,2 77,6,5,2 78,7,2,1 79,9 80,9,4,2 81,4 82,9,6,4 83,7,4,2 84,13 85,8,2,1 86,6,5,2 87,13 88,11,9,8 89,38 90,5,3,2 91,8,5,1 92,6,5,2 93,2 94,21 95,11 96,10,9,6 97,6 98,11 99,7,5,4 100,37 101,7,6,1 102,6,5,3 103,9 104,11,10,1 105,16 106,15 107,9,7,4 108,31 109,5,4,2 110,6,4,1 111,10 112,11,6,4 113,9 114,11,2,1 115,8,7,5 116,6,5,2 117,5,2,1 118,33 119,8 120,9,6,2 121,18 122,6,2,1 123,2 124,37 125,7,6,5 126,7,4,2 127,1 128,7,2,1 129,5 130,3 131,8,3,2 132,29 133,9,8,2 134,57 135,11 136,8,3,2 137,21 138,8,7,1 139,8,5,3 140,29 141,13,6,1 142,21 143,5,3,2 144,7,4,2 145,52 146,5,3,2 147,11,4,2 148,27 149,10,9,7 150,53 151,3 152,6,3,2 153,1 154,9,5,1 155,7,5,4 156,9,5,3 157,6,5,2 158,8,6,5 159,31 160,5,3,2 161,18 162,8,7,4 163,7,6,3 164,12,6,5 165,9,8,3 166,10,3,2 167,6 168,16,9,6 169,34 170,23 171,6,5,2 172,7 173,8,5,2 174,13 175,6 176,12,11,9 177,8 178,87 179,4,2,1 180,12,10,7 181,7,6,1 182,8,6,1 183,56 184,9,8,7 185,24 186,9,8,6 187,7,6,5 188,6,5,2 189,6,5,2 190,13,6,2 191,9 192,15,11,5 193,15 194,87 195,8,3,2 196,11,9,2 197,9,4,2 198,65 199,34 200,5,3,2 201,14 202,55 203,8,7,1 204,10,4,3 205,9,5,2 206,10,9,5 207,43 208,9,3,1 209,6 210,12,4,3 211,11,10,8 212,105 213,6,5,2 214,5,3,1 215,23 216,7,3,1 217,45 218,11 219,8,4,1 220,12,10,9 221,8,6,2 222,8,5,2 223,33 224,12,7,2 225,32 226,10,7,3 227,10,9,4 228,12,11,2 229,10,4,1 230,8,7,6 231,26 232,11,9,4 233,74 234,31 235,9,6,1 236,5 237,7,4,1 238,5,2,1 239,36 240,8,5,3 241,70 242,11,6,1 243,8,5,1 244,9,4,1 245,6,4,1 246,11,2,1 247,82 248,15,14,10 249,86 250,103 251,7,4,2 252,67 253,7,3,2 254,7,2,1 255,52 256,10,5,2 257,12 258,83 259,10,6,2 260,10,8,7 261,7,6,4 262,9,8,4 263,93 264,10,9,1 265,42 266,47 267,8,6,3 268,25 269,7,6,1 270,53 271,58 272,9,6,2 273,23 274,67 275,11,10,9 276,6,3,1 277,12,6,3 278,5 279,5 280,9,5,2 281,93 282,35 283,12,7,5 284,119 285,10,7,5 286,69 287,71 288,11,10,1 289,21 290,5,3,2 291,12,11,5 292,97 293,11,6,1 294,61 295,48 296,11,9,4 297,5 298,11,8,4 299,11,6,4 300,7 301,9,5,2 302,41 303,13,12,6 304,11,2,1 305,102 306,7,3,1 307,8,4,2 308,15,9,2 309,10,6,4 310,8,5,1 311,7,5,3 312,11,10,5 313,79 314,15 315,10,9,1 316,135 317,7,4,2 318,8,6,5 319,36 320,4,3,1 321,31 322,67 323,10,3,1 324,6,4,3 325,10,5,2 326,10,3,1 327,34 328,9,7,5 329,50 330,8,7,2 331,10,6,2 332,123 333,2 334,7,4,1 335,10,7,2 336,7,4,1 337,55 338,6,3,2 339,16,10,7 340,11,4,3 341,14,11,5 342,125 343,75 344,11,10,6 345,22 346,11,7,2 347,11,10,3 348,8,7,4 349,6,5,2 350,53 351,34 352,13,11,6 353,69 354,14,13,5 355,6,5,1 356,10,9,7 357,11,10,2 358,14,8,7 359,68 360,26,25,1 361,7,4,1 362,63 363,8,5,3 364,67 365,9,6,5 366,29 367,21 368,17,9,7 369,91 370,139 371,8,3,2 372,15,7,3 373,8,7,2 374,8,6,5 375,16 376,8,7,5 377,41 378,43 379,10,8,5 380,47 381,5,2,1 382,81 383,90 384,16,15,6 385,6 386,83 387,9,8,2 388,14,3,1 389,10,9,5 390,89 391,28 392,13,10,6 393,7 394,135 395,11,6,5 396,25 397,12,7,6 398,14,6,5 399,86 400,5,3,2 401,152 402,9,4,3 403,9,8,5 404,189 405,17,8,7 406,157 407,71 408,7,5,1 409,87 410,10,4,3 411,12,10,3 412,147 413,10,7,6 414,16,13,9 415,102 416,9,5,2 417,107 418,15,3,1 419,15,5,4 420,13,10,8 421,5,4,2 422,149 423,25 424,9,7,2 425,12 426,14,12,11 427,11,6,5 428,105 429,10,8,7 430,15,13,11 431,120 432,13,4,3 433,33 434,12,11,5 435,12,9,5 436,165 437,6,2,1 438,65 439,49 440,4,3,1 441,31 36 Handbook of Finite Fields 442,7,5,2 443,10,6,1 444,13,12,9 445,7,6,4 446,105 447,73 448,11,6,4 449,134 450,79 451,16,10,1 452,6,5,4 453,15,6,4 454,10,9,5 455,38 456,23,11,2 457,16 458,203 459,12,5,2 460,61 461,7,6,1 462,73 463,93 464,23,9,4 465,59 466,14,11,6 467,11,6,1 468,15,9,4 469,9,5,2 470,149 471,1 472,11,3,2 473,8,6,3 474,191 475,9,8,4 476,15 477,16,15,7 478,121 479,104 480,16,13,7 481,138 482,9,6,5 483,9,6,4 484,105 485,17,16,6 486,14,8,5 487,94 488,4,3,1 489,83 490,219 491,11,6,3 492,8,7,1 493,10,5,3 494,137 495,76 496,16,5,2 497,78 498,11,9,3 499,11,6,5 500,10,6,1 501,5,4,2 502,8,5,4 503,3 504,21,14,2 505,156 506,95 507,13,6,3 508,109 509,8,7,3 510,12,10,9 511,10 512,8,5,2 513,85 514,7,5,3 515,14,7,4 516,7,5,2 517,12,10,2 518,33 519,79 520,17,13,11 521,32 522,15,13,4 523,13,6,2 524,167 525,6,4,1 526,9,5,1 527,47 528,11,6,2 529,42 530,10,7,3 531,12,6,2 532,1 533,4,3,2 534,7,5,1 535,8,6,2 536,7,5,3 537,94 538,5,2,1 539,10,5,4 540,179 541,13,10,4 542,9,3,2 543,16 544,13,9,6 545,122 546,8,2,1 547,13,7,4 548,10,5,3 549,16,4,3 550,193 551,135 552,20,5,2 553,39 554,11,8,3 555,10,9,4 556,153 557,7,6,5 558,14,9,5 559,34 560,11,9,6 561,71 562,11,4,2 563,14,7,3 564,163 565,11,6,1 566,153 567,143 568,17,11,10 569,77 570,67 571,10,5,2 572,12,8,1 573,10,6,4 574,13 575,146 576,13,4,3 577,25 Table 2.2.2 Lowest weight lowest-lexicographical order primitive polynomial of degree n ≤577 over F2. Output: n, k (for trinomials xn +xk +1) or n, k1, k2, k3 (for pentanomials xn +xk1 +xk2 +xk3 +1). 2.2.7 Remark Table 2.2.3 is the analogous table to Table 2.2.1, giving the lowest-weight, lowest-lexicographical order irreducible polynomial of degree n ≤516 over F3. 2,(1) 3,1(2),(1) 4,1(1),(2) 5,1(2),(1) 6,1(1),(2) 7,2(1),(2) 8,2(1),(2) 9,4(1),(2) 10,2(2),(1) 11,2(1),(2) 12,2(1),(2) 13,1(2),(1) 14,1(1),(2) 15,2(1),(2) 16,4(1),(2) 17,1(2),(1) 18,7(1),(2) 19,2(1),(2) 20,5(1),(2) 21,5(2),(1) 22,4(2),(1) 23,3(2),(1) 24,4(1),(2) 25,3(2),(1) 26,2(2),(1) 27,7(2),(1) 28,2(1),(2) 29,4(1),(2) 30,1(1),(2) 31,5(2),(1) 32,5(1),(2) 33,5(2),(1) 34,2(2),(1) 35,2(1),(2) 36,14(1),(2) 37,6(1),(2) 38,4(2),(1) 39,7(2),(1) 40,1(1),(2) 41,1(2),(1) 42,7(1),(2) 43,17(2),(1) 44,3(1),(2) 45,17(2),(1) 46,5(1),(2) 47,15(2),(1) 48,8(1),(2) 49,3(2),2(1),(1) 50,6(2),(1) 51,1(2),(1) 52,7(1),(2) 53,13(2),(1) 54,1(1),(2) 55,11(2),(1) 56,3(1),(2) 57,7(1),2(1),(2) 58,8(2),(1) 59,17(2),(1) 60,2(1),(2) 61,7(2),(1) 62,10(2),(1) 63,26(1),(2) 64,3(1),(2) 65,5(1),3(1),(1) 66,10(2),(1) 67,2(1),(2) 68,3(1),2(1),(1) 69,17(2),(1) 70,4(2),(1) 71,20(1),(2) 72,28(1),(2) 73,1(2),(1) 74,12(2),(1) 75,5(2),4(1),(1) 76,9(1),(2) 77,16(1),(2) 78,13(1),(2) 79,26(1),(2) 80,2(1),(2) 81,40(1),(2) 82,2(2),(1) 83,27(2),(1) 84,14(1),(2) 85,16(1),(2) 86,13(1),(2) 87,26(1),(2) 88,6(1),(2) 89,13(2),(1) 90,19(1),(2) 91,17(2),(1) 92,10(1),(2) 93,23(2),(1) 94,30(2),(1) 95,47(2),(1) 96,16(1),(2) 97,12(1),(2) 98,4(1),3(1),(1) 99,19(2),(1) 100,25(1),(2) 101,31(2),(1) 102,2(2),(1) 103,47(2),(1) 104,5(1),(2) 105,6(1),2(1),(1) 106,26(2),(1) 107,3(2),(1) 108,2(1),(2) 109,9(2),(1) 110,22(2),(1) 111,2(1),(2) 112,6(1),(2) 113,19(2),(1) 114,7(1),(2) 115,32(1),(2) 116,15(1),(2) 117,52(1),(2) 118,34(2),(1) 119,2(1),(2) 120,4(1),(2) 121,1(2),(1) 122,14(2),(1) 123,7(1),4(1),(2) 124,25(1),(2) 125,52(1),(2) 126,49(1),(2) 127,8(1),(2) 128,6(1),(2) 129,3(2),2(1),(1) 130,10(1),6(1),(1) 131,27(2),(1) 132,19(1),14(1),(1) 133,15(2),(1) 134,4(2),(1) 135,44(1),(2) 136,57(1),(2) 137,1(2),(1) 138,34(2),(1) 139,59(2),(1) 140,59(1),(2) 141,5(2),(1) 142,40(2),(1) 143,35(2),(1) 144,56(1),(2) 145,24(1),(2) 146,2(2),(1) 147,8(1),(2) 148,3(1),(2) 149,11(2),10(1),(1) 150,73(1),(2) 151,2(1),(2) 152,18(1),(2) 153,59(2),(1) 154,32(2),(1) 155,12(1),(2) 156,26(1),(2) 157,22(1),(2) 158,52(2),(1) 159,32(1),(2) 160,4(1),(2) 161,9(1),5(1),(1) 162,19(1),(2) 163,59(2),(1) 164,15(1),(2) 165,22(1),(2) 166,54(2),(1) 167,71(2),(1) 168,28(1),(2) 169,24(1),(2) 170,32(2),(1) 171,20(1),(2) 172,19(1),(2) 173,7(2),(1) 174,52(2),(1) 175,10(1),8(1),(1) 176,12(1),(2) 177,52(1),(2) 178,11(1),(2) 179,59(2),(1) 180,38(1),(2) 181,37(2),(1) 182,25(1),(2) 183,2(1),(2) 184,20(1),(2) 185,64(1),(2) 186,46(2),(1) 187,8(1),(2) 188,11(1),(2) 189,9(1),7(1),(1) 190,94(2),(1) 191,71(2),(1) 192,32(1),(2) 193,12(1),(2) 194,24(2),(1) 195,26(1),(2) 196,79(1),(2) 197,9(1),7(1),(1) 198,29(1),(2) 199,35(2),(1) 200,3(1),(2) 201,88(1),(2) 202,62(2),(1) 203,3(2),(1) 204,50(1),(2) 205,9(2),(1) 206,61(1),(2) 207,11(2),8(1),(1) 208,10(1),(2) 209,40(1),(2) 210,7(1),(2) 211,89(2),(1) 212,14(1),3(1),(1) 213,17(2),4(1),(1) 214,6(2),(1) 215,36(1),(2) 216,4(1),(2) 217,85(2),(1) 218,18(2),(1) 219,25(2),(1) 220,15(1),(2) 221,12(1),2(1),(1) 222,4(2),(1) 223,8(1),5(2),(1) 224,12(1),(2) 225,16(1),(2) 226,38(2),(1) 227,11(2),(1) 228,14(1),(2) 229,72(1),(2) 230,64(2),(1) 231,8(1),7(1),(2) 232,30(1),(2) 233,6(1),2(1),(1) 234,91(1),(2) 235,26(1),(2) 236,9(1),(2) 237,70(1),(2) 238,4(2),(1) 239,5(2),(1) 240,8(1),(2) 241,88(1),(2) 242,2(2),(1) 243,121(2),(1) 244,31(1),(2) 245,97(2),(1) 246,13(1),(2) 247,122(1),(2) 248,50(1),(2) 249,59(2),(1) 250,104(2),(1) 251,9(2),(1) 252,98(1),(2) 253,7(2),(1) 254,16(2),(1) 255,26(1),(2) 256,12(1),(2) Introduction to ﬁnite ﬁelds 37 257,22(1),(2) 258,7(1),(2) 259,65(2),(1) 260,35(1),(2) 261,119(2),(1) 262,54(2),(1) 263,69(2),(1) 264,23(1),16(1),(1) 265,61(2),(1) 266,30(2),(1) 267,9(1),2(1),(2) 268,15(1),(2) 269,7(2),(1) 270,88(2),(1) 271,50(1),(2) 272,114(1),(2) 273,46(1),(2) 274,2(2),(1) 275,12(1),(2) 276,10(1),2(1),(2) 277,24(1),(2) 278,118(2),(1) 279,7(2),(1) 280,15(1),(2) 281,10(1),7(1),(2) 282,10(2),(1) 283,23(2),(1) 284,5(1),(2) 285,89(2),(1) 286,70(2),(1) 287,101(2),(1) 288,112(1),(2) 289,73(2),(1) 290,43(1),(2) 291,25(2),(1) 292,13(1),12(1),(1) 293,7(2),(1) 294,16(2),(1) 295,83(2),(1) 296,6(1),(2) 297,3(2),2(1),(1) 298,13(1),3(1),(2) 299,51(2),(1) 300,146(1),(2) 301,30(1),(2) 302,4(2),(1) 303,8(1),2(1),(1) 304,36(1),(2) 305,46(1),(2) 306,118(2),(1) 307,17(2),(1) 308,53(1),(2) 309,3(2),2(1),(1) 310,24(2),(1) 311,13(2),12(1),(1) 312,52(1),(2) 313,93(2),(1) 314,44(2),(1) 315,127(2),(1) 316,87(1),(2) 317,7(2),(1) 318,64(2),(1) 319,10(1),9(1),(2) 320,3(1),(2) 321,83(2),(1) 322,71(1),(2) 323,9(2),(1) 324,38(1),(2) 325,157(2),(1) 326,118(2),(1) 327,7(2),(1) 328,3(1),(2) 329,52(1),(2) 330,11(1),(2) 331,2(1),(2) 332,13(1),6(1),(1) 333,94(1),(2) 334,142(2),(1) 335,8(1),(2) 336,56(1),(2) 337,3(2),(1) 338,48(2),(1) 339,49(2),(1) 340,86(1),(2) 341,25(2),(1) 342,40(2),(1) 343,12(1),10(1),(1) 344,38(1),(2) 345,101(2),(1) 346,14(2),(1) 347,18(1),(2) 348,146(1),(2) 349,54(1),(2) 350,157(1),(2) 351,20(1),(2) 352,7(1),(2) 353,142(1),(2) 354,104(2),(1) 355,41(2),(1) 356,15(1),(2) 357,71(2),(1) 358,77(1),(2) 359,15(2),(1) 360,76(1),(2) 361,157(2),(1) 362,74(2),(1) 363,26(1),(2) 364,1(1),(2) 365,88(1),(2) 366,4(2),(1) 367,107(2),(1) 368,27(1),(2) 369,11(2),(1) 370,11(1),(2) 371,27(2),(1) 372,94(1),(2) 373,25(2),(1) 374,16(2),(1) 375,67(2),(1) 376,9(1),(2) 377,160(1),(2) 378,7(1),(2) 379,44(1),(2) 380,9(1),(2) 381,143(2),(1) 382,137(1),(2) 383,80(1),(2) 384,64(1),(2) 385,22(1),(2) 386,24(2),(1) 387,152(1),(2) 388,87(1),(2) 389,76(1),(2) 390,13(1),(2) 391,22(1),21(2),(1) 392,158(1),(2) 393,185(2),(1) 394,14(1),9(1),(1) 395,23(2),(1) 396,58(1),(2) 397,12(1),5(1),(2) 398,70(2),(1) 399,181(2),(1) 400,3(1),(2) 401,11(2),10(1),(1) 402,176(2),(1) 403,161(2),(1) 404,9(1),2(1),(1) 405,25(1),18(1),(2) 406,6(2),(1) 407,48(1),(2) 408,100(1),(2) 409,99(2),(1) 410,18(2),(1) 411,8(1),2(1),(1) 412,79(1),(2) 413,22(1),(2) 414,37(1),(2) 415,13(1),3(1),(1) 416,20(1),(2) 417,40(1),(2) 418,80(2),(1) 419,26(1),(2) 420,14(1),(2) 421,13(2),(1) 422,178(2),(1) 423,68(1),(2) 424,45(1),(2) 425,61(2),(1) 426,9(1),7(1),(2) 427,167(2),(1) 428,71(1),(2) 429,65(2),(1) 430,72(2),(1) 431,66(1),(2) 432,8(1),(2) 433,120(1),(2) 434,67(1),(2) 435,8(1),2(1),(1) 436,13(1),2(1),(1) 437,14(1),3(2),(1) 438,17(1),(2) 439,16(1),3(2),(1) 440,11(1),(2) 441,7(1),6(1),(2) 442,11(1),3(1),(2) 443,188(1),(2) 444,178(1),(2) 445,141(2),(1) 446,1(1),(2) 447,157(2),(1) 448,24(1),(2) 449,52(1),(2) 450,32(2),(1) 451,17(2),(1) 452,17(1),(2) 453,17(2),4(1),(1) 454,22(2),(1) 455,32(1),(2) 456,28(1),(2) 457,67(2),(1) 458,144(2),(1) 459,13(2),6(1),(1) 460,57(1),(2) 461,13(2),(1) 462,73(1),(2) 463,15(1),13(1),(1) 464,60(1),(2) 465,41(2),(1) 466,167(1),(2) 467,48(1),(2) 468,182(1),(2) 469,166(1),(2) 470,52(2),(1) 471,8(1),(2) 472,18(1),(2) 473,73(2),(1) 474,83(1),(2) 475,17(2),(1) 476,10(1),(2) 477,101(2),(1) 478,10(2),(1) 479,221(2),(1) 480,16(1),(2) 481,22(1),(2) 482,127(1),(2) 483,26(1),(2) 484,39(1),(2) 485,1(2),(1) 486,125(1),(2) 487,29(2),(1) 488,62(1),(2) 489,7(1),5(1),(1) 490,194(2),(1) 491,11(2),(1) 492,26(1),(2) 493,4(1),(2) 494,244(2),(1) 495,7(2),(1) 496,85(1),(2) 497,7(2),6(1),(1) 498,118(2),(1) 499,20(1),(2) 500,39(1),(2) 501,88(1),(2) 502,18(2),(1) 503,35(2),(1) 504,196(1),(2) 505,61(2),(1) 506,14(2),(1) 507,80(1),(2) 508,91(1),(2) 509,151(2),(1) 510,52(2),(1) 511,215(2),(1) 512,24(1),(2) 513,14(1),10(1),(1) 514,44(2),(1) 515,8(1),(2) 516,14(1),(2) Table 2.2.3 Lowest weight lowest lexicographical order irreducible polynomial of degree n over F3. Output: n, {degrees, (coeﬃcients)}, (constant term). 2.2.8 Remark Necessary and suﬃcient conditions for the existence of an irreducible binomial of degree n over ﬁnite ﬁelds of odd characteristic are given in [1939, Theorem 3.75]. A constructive derivation of the degrees for which there exists an irreducible binomial over Fq, q odd, is given in . The following conjecture summarizes empirical observations of extending Tables 2.2.1 and 2.2.3 to higher characteristics. 2.2.9 Conjecture Let q > 2. For every n, there is an irreducible polynomial of degree n over Fq of weight at most 4. 2.2.2 Low-complexity normal bases 2.2.10 Remark Normal bases are often required in hardware implementations of ﬁnite ﬁelds due to the eﬃciency of exponentiation when the ﬁnite ﬁeld is represented using a normal basis. 38 Handbook of Finite Fields The complexity of a normal basis N, CN, is deﬁned in Deﬁnition 5.3.1. Normal bases with low complexity are highly preferred. An optimal normal basis of Fqn over Fq is a normal basis attaining the minimum complexity CN = 2n −1. See Sections 5.2 and 5.3 for more details on normal bases and their complexities. 2.2.2.1 Exhaustive search for low complexity normal bases 2.2.11 Remark Table 2.2.4 is due to an exhaustive search for normal bases of F2n over F2, n ≤39, originally given in . The table gives the number of normal bases, the smallest and largest complexities (mCN , MCN ), the average and variance (AvgCN , V arCN ) of complexities and the smallest and largest complexities for self-dual normal elements. In Table 2.2.4, we ﬁx a typo on the minimum complexity of n = 37, originally noted in , and make some minor corrections to the calculations of the averages and variances. In the “Notes” column, “Optimal” indicates that the basis with minimal complexity is an optimal normal basis (Theorem 5.3.6), and “sd” indicates that the minimal complexity basis is self-dual. Self-dual n # Normal bases mCN MCN AvgCN V arCN mCN MCN Notes 2 1 3 3 3.00 0 3 3 Optimal, sd 3 1 5 5 5.00 0 5 5 Optimal, sd 4 2 7 9 8.00 1.00 --5 3 9 15 11.67 6.22 9 9 Optimal, sd 6 4 11 17 15.00 6.00 11 15 Optimal, sd 7 7 19 27 23.00 9.14 21 21 mCN = 3n −2 8 16 21 35 29.00 11 --mCN = 3n −3 9 21 17 45 35.57 41.57 17 29 Optimal, sd 10 48 19 61 44.83 61.31 27 51 11 93 21 71 55.82 57.65 21 57 Optimal, sd 12 128 23 83 64.13 107.23 --13 315 45 101 78.38 71.07 45 81 sd 14 448 27 135 91.07 108.42 27 135 Optimal, sd 15 675 45 137 105.89 127.36 45 105 sd 16 2048 85 157 115.82 114.59 --17 3825 81 177 136.83 136.67 81 171 sd 18 5376 35 243 153.51 185.12 35 243 Optimal, sd 19 13797 117 229 172.00 171.91 117 201 sd 20 24576 63 257 190.81 205.81 --21 27783 95 277 210.97 216.43 105 237 22 95232 63 363 231.93 238.56 63 363 mCN = 3n −3 23 182183 45 325 254.02 254.60 45 309 Optimal, sd 24 262144 105 375 276.89 281.01 --25 629145 93 383 301.01 300.37 93 357 sd 26 1290240 51 555 325.96 328.59 51 555 Optimal, sd 27 1835001 141 443 351.99 351.38 141 413 28 3670016 55 517 378.98 379.12 --Optimal 29 9256395 57 521 407.00 406.21 57 465 Optimal, sd 30 11059200 59 759 435.95 438.52 59 759 Optimal, sd 31 28629151 237 587 466.00 465.21 237 537 sd 32 67108864 361 621 497.00 496.07 --33 97327197 65 693 529.00 528.44 65 693 Optimal, sd 34 250675200 243 819 562.00 561.52 243 819 sd 35 352149515 69 779 596.00 595.08 69 693 Optimal, sd 36 704643060 71 1017 630.99 630.51 --Optimal 37 1857283155 141 823 667 666.04 141 sd 38 3616800703 207 1131 704.00 703.18 207 39 5282242828 77 933 742.00 741.09 77 Optimal, sd Table 2.2.4 Statistics for normal bases of F2n over F2 obtained by exhaustive search, n ≤39. 2.2.12 Remark Our ﬁrst conjecture based on Table 2.2.4 appears in and elsewhere. We also summarize the conjectures found in . 2.2.13 Conjecture When no optimal normal basis of F2n over F2 exists, the minimum complexity of a normal basis of F2n over F2 is 3n −3. Introduction to ﬁnite ﬁelds 39 2.2.14 Remark Normal bases of F2n over F2 achieving a complexity of 3n −3 are given in Propo-sition 5.3.46 and this complexity is the minimal found when n = 8 and n = 22. 2.2.15 Conjecture The number of normal bases of F2n over F2 are normally distributed with respect to their complexities. Furthermore, the average complexity of a normal basis of F2n over F2 is (n2 −n + 3)/2 and the variance is also n2/2 −cn, for a small positive constant c. 2.2.16 Remark We remark that the conspicuous wording in Conjecture 2.2.15, that normal bases are normally distributed, is mostly coincidental. Indeed, as n grows, the number of normal bases grow like 2n/ log(n), see Theorem 5.2.13, so the Central Limit Theorem supports this conjecture. The precise distribution of the complexities is still an open and interesting problem. 2.2.17 Remark Self-dual normal bases are often preferred in normal basis implementations due to their highly symmetric properties; see Sections 5.1, 5.2, 5.3, 16.7 as well as [1264, 2925], for more information on self-dual normal bases and their implementations. Exhaustive searches of self-dual normal bases of F2n over F2 appear in [130, 1263, 1631, 2015] and gives an exhastive search of self-dual normal bases of Fqn over Fq for larger q and odd n. Ta-bles 2.2.5, 2.2.6, and 2.2.7 are directly from ; we note that we did not implement their algorithm. Table 2.2.5 gives the minimum complexity Cn of a self-dual normal basis of F2n over F2 for odd n ≤45, Table 2.2.6 for q a power of 2 and small n, and Table 2.2.7 for Fqn over Fq for odd q ≤19 and small n. n 3 5 7 9 11 13 15 17 19 21 23 Cn 5 9 21 17 21 45 45 81 117 105 45 n 25 27 29 31 33 35 37 39 41 43 45 Cn 93 141 57 237 65 69 141 77 81 165 153 Table 2.2.5 The lowest complexity for self-dual normal bases of F2n over F2 for odd n, n ≤45. q/n 3 5 7 9 11 13 15 17 19 21 23 25 2 5 9 21 17 21 45 45 81 117 105 45 93 4 5 9 21 17 21 45 45 81 117 105 45 93 8 9 9 21 45 21 45 81 81 16 5 9 21 17 21 45 32 5 19 21 17 21 64 9 9 21 45 128 5 9 37 256 5 9 Table 2.2.6 Lowest complexity for self-dual normal bases of Fqn over Fq where q is a power of 2 for small odd values of n. q/n 3 5 7 9 11 13 15 17 19 21 23 25 3 7 13 25 37 55 67 – 91 172 – 127 135 5 6 13 25 46 64 85 – 157 153 150 7 6 16 19 41 61 96 87 – 11 6 13 25 52 31 100 78 13 6 13 25 51 64 37 17 8 13 25 51 64 100 – 19 8 13 31 51 67 – Table 2.2.7 Lowest complexity for self-dual normal bases of Fqn over Fq for odd primes q ≤19 and small odd values of n. 40 Handbook of Finite Fields 2.2.2.2 Minimum type of a Gauss period admitting a normal basis of F2n over F2 2.2.18 Remark We brieﬂy recall the deﬁnition of a Gauss period (Deﬁnition 5.3.16). Let r = nk+1 be a prime not dividing q and let γ be a primitive r-th root of unity in Fqnk. Furthermore, let K be the unique subgroup of order k in Z∗ r and Ki = {a · qi : a ∈K} ⊆Z∗ r be cosets of K, 0 ≤i ≤n −1. The elements αi = X a∈Ki γa ∈Fqn, 0 ≤i ≤n −1, are Gauss periods of type (n, k) over Fq. Gauss periods over ﬁnite ﬁelds are highly desirable as normal bases since, when they exist, they have low complexity; see Theorem 5.3.23. Normal bases due to Gauss periods of type (n, 1), for all q, and of type (n, 2), for q = 2, characterize the optimal normal bases (Theorem 5.3.6) and have complexity 2n −1. Gauss periods also often have high order, see Remark 5.3.49. For conditions on when Gauss periods of type (n, k) admit normal bases of Fqn over Fq, see Theorem 5.3.17. In particular, we note that there is no Gauss period of F2n over F2 which admits a normal basis when 8 divides n. Table 2.2.8 gives the lowest k for which a Gauss period of type (n, k) admits a normal basis of F2n pver F2 for n ≤577. We give a similar table over F3 in Table 2.2.9. This range was chosen to cover degrees for common implementations of ﬁnite ﬁeld arithmetic. The output of the table is in the format “n, k′′ where k is the minimum number admitting a type (n, k) Gauss period over Fqn, where q = 2, 3. 2,1 3,2 4,1 5,2 6,2 7,4 9,2 10,1 11,2 12,1 13,4 14,2 15,4 17,6 18,1 19,10 20,3 21,10 22,3 23,2 25,4 26,2 27,6 28,1 29,2 30,2 31,10 33,2 34,9 35,2 36,1 37,4 38,6 39,2 41,2 42,5 43,4 44,9 45,4 46,3 47,6 49,4 50,2 51,2 52,1 53,2 54,3 55,12 57,10 58,1 59,12 60,1 61,6 62,6 63,6 65,2 66,1 67,4 68,9 69,2 70,3 71,8 73,4 74,2 75,10 76,3 77,6 78,7 79,4 81,2 82,1 83,2 84,5 85,12 86,2 87,4 89,2 90,2 91,6 92,3 93,4 94,3 95,2 97,4 98,2 99,2 100,1 101,6 102,6 103,6 105,2 106,1 107,6 108,5 109,10 110,6 111,20 113,2 114,5 115,4 116,3 117,8 118,6 119,2 121,6 122,6 123,10 124,3 125,6 126,3 127,4 129,8 130,1 131,2 132,5 133,12 134,2 135,2 137,6 138,1 139,4 140,3 141,8 142,6 143,6 145,10 146,2 147,6 148,1 149,8 150,19 151,6 153,4 154,25 155,2 156,13 157,10 158,2 159,22 161,6 162,1 163,4 164,5 165,4 166,3 167,14 169,4 170,6 171,12 172,1 173,2 174,2 175,4 177,4 178,1 179,2 180,1 181,6 182,3 183,2 185,8 186,2 187,6 188,5 189,2 190,10 191,2 193,4 194,2 195,6 196,1 197,18 198,22 199,4 201,8 202,6 203,12 204,3 205,4 206,3 207,4 209,2 210,1 211,10 212,5 213,4 214,3 215,6 217,6 218,5 219,4 220,3 221,2 222,10 223,12 225,22 226,1 227,24 228,9 229,12 230,2 231,2 233,2 234,5 235,4 236,3 237,10 238,7 239,2 241,6 242,6 243,2 244,3 245,2 246,11 247,6 249,8 250,9 251,2 252,3 253,10 254,2 255,6 257,6 258,5 259,10 260,5 261,2 262,3 263,6 265,4 266,6 267,8 268,1 269,8 270,2 271,6 273,2 274,9 275,14 276,3 277,4 278,2 279,4 281,2 282,6 283,6 284,3 285,10 286,3 287,6 289,12 290,5 291,6 292,1 293,2 294,3 295,16 297,6 298,6 299,2 300,19 301,10 302,3 303,2 305,6 306,2 307,4 308,15 309,2 310,6 311,6 313,6 314,5 315,8 316,1 317,26 318,11 319,4 321,12 322,6 323,2 324,5 325,4 326,2 327,8 329,2 330,2 331,6 332,3 333,24 334,7 335,12 337,10 338,2 339,8 340,3 341,8 342,6 343,4 345,4 346,1 347,6 348,1 349,10 350,2 351,10 353,14 354,2 355,6 356,3 357,10 358,10 359,2 361,30 362,5 363,4 364,3 365,24 366,22 367,6 369,10 370,6 371,2 372,1 373,4 374,3 375,2 377,14 378,1 379,12 380,5 381,8 382,6 383,12 385,6 386,2 387,4 388,1 389,24 390,3 391,6 393,2 394,9 395,6 396,11 397,6 398,2 399,12 401,8 402,5 403,16 404,3 405,4 406,6 407,8 409,4 410,2 411,2 412,3 413,2 414,2 415,28 417,4 418,1 419,2 420,1 421,10 422,11 423,4 425,6 426,2 427,16 428,5 429,2 430,3 431,2 433,4 434,9 435,4 436,13 437,18 438,2 439,10 441,2 442,1 443,2 444,5 445,6 446,6 447,6 449,8 450,13 451,6 452,11 453,2 454,19 455,26 457,30 458,6 459,8 460,1 461,6 462,10 463,12 465,4 466,1 467,6 468,21 469,4 470,2 471,8 473,2 474,5 475,4 476,5 477,46 478,7 479,8 481,6 482,5 483,2 484,3 485,18 486,10 487,4 489,12 490,1 491,2 492,13 493,4 494,3 495,2 497,20 498,9 499,4 500,11 501,10 502,10 503,6 505,10 506,5 507,4 508,1 509,2 510,3 511,6 513,4 514,33 515,2 516,3 517,4 518,14 519,2 521,32 522,1 523,10 524,5 525,8 526,3 527,6 529,24 530,2 531,2 532,3 533,12 534,7 535,4 537,8 538,6 539,12 540,1 541,18 542,3 543,2 545,2 546,1 547,10 548,5 549,14 550,7 551,6 553,4 554,2 555,4 556,1 557,6 558,2 559,4 561,2 562,1 563,14 564,3 565,10 566,3 567,4 569,12 570,5 571,10 572,5 573,4 574,3 575,2 577,4 Table 2.2.8 Lowest type of a Gauss period forming a normal basis for q = 2 and n ≤577. Introduction to ﬁnite ﬁelds 41 2,2 3,2 4,1 5,2 6,1 7,4 8,2 9,2 10,3 11,2 13,4 14,2 15,2 16,1 17,6 18,1 19,10 20,5 21,2 22,3 23,2 25,4 26,2 27,4 28,1 29,2 30,1 31,10 32,8 33,6 34,3 35,2 37,4 38,15 39,2 40,7 41,2 42,1 43,4 44,2 45,4 46,3 47,6 49,4 50,2 51,8 52,1 53,2 54,3 55,6 56,2 57,4 58,4 59,12 61,6 62,21 63,2 64,4 65,2 66,3 67,4 68,2 69,2 70,3 71,8 73,4 74,2 75,8 76,10 77,6 78,1 79,4 80,5 81,2 82,9 83,2 85,16 86,2 87,4 88,1 89,2 90,7 91,10 92,5 93,4 94,3 95,2 97,4 98,2 99,2 100,1 101,6 102,11 103,6 104,5 105,2 106,10 107,6 109,10 110,3 111,2 112,1 113,2 114,5 115,4 116,2 117,8 118,9 119,2 121,6 122,3 123,6 124,13 125,2 126,1 127,4 128,2 129,8 130,4 131,2 133,16 134,2 135,4 136,1 137,6 138,1 139,4 140,2 141,2 142,4 143,6 145,10 146,2 147,10 148,1 149,8 150,5 151,6 152,5 153,14 154,3 155,2 157,10 158,2 159,34 160,4 161,6 162,1 163,4 164,5 165,2 166,3 167,14 169,4 170,8 171,12 172,1 173,2 174,9 175,4 176,2 177,4 178,15 179,2 181,6 182,14 183,4 184,7 185,8 186,15 187,6 188,5 189,2 190,3 191,2 193,4 194,2 195,10 196,1 197,18 198,1 199,4 200,2 201,10 202,3 203,12 205,4 206,3 207,4 208,10 209,2 210,1 211,10 212,5 213,6 214,3 215,6 217,6 218,15 219,4 220,4 221,2 222,1 223,12 224,2 225,8 226,15 227,24 229,12 230,2 231,2 232,1 233,2 234,5 235,4 236,8 237,6 238,4 239,2 241,6 242,3 243,2 244,4 245,24 246,3 247,6 248,11 249,8 250,3 251,2 253,4 254,2 255,12 256,1 257,6 258,5 259,10 260,2 261,6 262,3 263,6 265,4 266,8 267,4 268,1 269,8 270,3 271,6 272,5 273,10 274,3 275,12 277,4 278,2 279,10 280,1 281,2 282,1 283,6 284,2 285,2 286,3 287,6 289,12 290,20 291,6 292,1 293,2 294,5 295,12 296,2 297,8 298,4 299,2 301,10 302,3 303,2 304,4 305,6 306,7 307,4 308,2 309,4 310,15 311,6 313,6 314,14 315,2 316,1 317,26 318,17 319,4 320,2 321,18 322,3 323,2 325,4 326,2 327,10 328,7 329,2 330,1 331,6 332,8 333,6 334,15 335,6 337,10 338,2 339,10 340,4 341,8 342,13 343,4 344,5 345,2 346,3 347,6 349,10 350,2 351,22 352,1 353,14 354,3 355,12 356,11 357,4 358,4 359,2 361,30 362,3 363,4 364,7 365,18 366,5 367,6 368,11 369,2 370,4 371,2 373,4 374,3 375,2 376,7 377,14 378,1 379,12 380,5 381,20 382,10 383,12 385,6 386,2 387,14 388,1 389,24 390,5 391,6 392,8 393,10 394,9 395,6 397,6 398,2 399,12 400,1 401,8 402,5 403,4 404,2 405,2 406,21 407,8 409,4 410,2 411,2 412,19 413,2 414,9 415,30 416,5 417,6 418,15 419,2 421,10 422,21 423,4 424,4 425,12 426,3 427,4 428,2 429,2 430,3 431,2 433,4 434,3 435,4 436,13 437,18 438,9 439,10 440,2 441,6 442,3 443,2 445,6 446,15 447,4 448,1 449,8 450,33 451,6 452,8 453,2 454,28 455,2 457,30 458,15 459,8 460,1 461,6 462,1 463,12 464,2 465,10 466,3 467,6 469,10 470,2 471,8 472,4 473,2 474,3 475,4 476,2 477,14 478,4 479,8 481,28 482,3 483,10 484,7 485,2 486,1 487,4 488,2 489,12 490,15 491,2 493,4 494,3 495,30 496,13 497,14 498,11 499,4 500,8 501,16 502,9 503,6 505,10 506,2 507,18 508,1 509,2 510,7 511,6 512,23 513,4 514,3 515,2 517,4 518,9 519,2 520,1 521,32 522,3 523,10 524,2 525,20 526,3 527,6 529,24 530,2 531,2 532,4 533,12 534,27 535,4 536,8 537,8 538,4 539,12 541,18 542,3 543,2 544,10 545,6 546,5 547,10 548,2 549,18 550,21 551,2 553,4 554,2 555,6 556,1 557,6 558,5 559,4 560,5 561,2 562,9 563,14 565,6 566,3 567,28 568,1 569,12 570,1 571,10 572,5 573,4 574,3 575,2 577,4 Table 2.2.9 Lowest type of a Gauss period forming a normal basis for q = 3 and n ≤577. 2.2.2.3 Minimum-known complexity of a normal basis of F2n over F2, n ≥40 2.2.19 Remark Table 2.2.10 gives the minimum complexity of a normal basis of F2n over F2 for 40 ≤n ≤721 by using a combination of the exhaustive search data of Table 2.2.4 and theo-rems from Section 5.3. In each row, we give the degree n, the minimum complexity Cn of a normal basis of F2n over F2, the method by which the normal basis was obtained and what property or parameters were used. In the “Method” column, “Optimal” indicates existence of an optimal normal basis, “GNB” indicates the basis arises as a Gauss period and their type is given in the “Property” column. Proposition 5.3.38 constructs normal bases of Fqn using normal bases of subﬁelds of coprime degree. When this method wins, the values of these coprime factors are indicated in the “Property” column. Corollary 5.3.15 requires an optimal normal basis of F2kn and the type of the optimal normal basis and the value of k are indicated in the “Property” column. Finally, “sd” indicates that the basis is self-dual. When n is a power of 2, the best result, when available, is by random search since known methods do not apply. Gauss periods cannot form normal bases when 8 divides n, see Propo-sition 5.3.20, and n contains no coprime factors with which to apply Proposition 5.3.38. By Conjecture 2.2.15, the complexity of these bases is likely to approach n2/2. 2.2.20 Problem Find constructions of low complexity normal bases of F2n over F2 when n is a prime power, speciﬁcally a power of 2. 42 Handbook of Finite Fields n Cn Method Property 40 189 Prop. 5.3.38 5, 8 41 81 Optimal Type 2, sd 42 135 Prop. 5.3.38 3, 14 43 165 GNB k = 4, sd 44 147 Prop. 5.3.38 4, 11 45 153 Search , sd 46 135 Prop. 5.3.38 2, 23 47 261 GNB k = 6, sd 48 425 Prop. 5.3.38 3, 16 49 189 GNB k = 4, sd 50 99 Optimal Type 2, sd 51 101 Optimal Type 2, sd 52 103 Optimal Type 1 53 105 Optimal Type 2, sd 54 209 GNB k = 3 55 189 Prop. 5.3.38 5, 11 56 399 Prop. 5.3.38 7, 8 57 497 Search , sd 58 115 Optimal Type 1 59 597 Search , sd 60 119 Optimal Type 1 61 345 GNB k = 6, sd 62 351 GNB k = 6, sd 63 323 Prop. 5.3.38 7, 9 64 1829 Random 65 129 Optimal Type 2, sd 66 131 Optimal Type 1 67 261 GNB k = 4, sd 68 567 Prop. 5.3.38 4, 17 69 137 Optimal Type 2, sd 70 207 Prop. 5.3.38 2, 35 71 567 GNB k = 8, sd 72 357 Prop. 5.3.38 8, 9 73 285 GNB k = 4, sd 74 147 Optimal Type 2, sd 75 465 Prop. 5.3.38 3, 25 76 297 GNB k = 3 77 399 Prop. 5.3.38 7, 11 78 231 Prop. 5.3.38 2, 39 79 309 GNB k = 4, sd 80 765 Prop. 5.3.38 5, 16 81 161 Optimal Type 2, sd 82 163 Optimal Type 1 83 165 Optimal Type 2, sd 84 275 Prop. 5.3.38 3, 28 85 729 Prop. 5.3.38 5, 17 86 171 Optimal Type 2, sd 87 285 Prop. 5.3.38 3, 29 88 441 Prop. 5.3.38 8, 11 89 177 Optimal Type 2, sd 90 179 Optimal Type 2, sd 91 525 GNB k = 6, sd 92 315 Prop. 5.3.38 4, 23 93 365 GNB k = 4, sd 94 369 GNB k = 3 95 189 Optimal Type 2, sd 96 1805 Prop. 5.3.38 3, 32 97 381 GNB k = 4, sd 98 195 Optimal Type 2, sd 99 197 Optimal Type 2, sd 100 199 Optimal Type 1 101 585 GNB k = 6, sd 102 303 Prop. 5.3.38 2, 51 103 597 GNB k = 6, sd 104 945 Prop. 5.3.38 8, 13 105 209 Optimal Type 2, sd 106 211 Optimal Type 1 107 621 GNB k = 6, sd 108 627 GNB k = 5 109 1081 Cor. 5.3.15 Type 2, k = 5, sd 110 399 Prop. 5.3.38 10, 11 111 705 Prop. 5.3.38 3, 37 112 1615 Prop. 5.3.38 7, 16 113 225 Optimal Type 2, sd 114 663 GNB k = 5 115 405 Prop. 5.3.38 5, 23 116 399 Prop. 5.3.38 4, 29 117 765 Prop. 5.3.38 9, 13 118 687 GNB k = 6, sd 119 237 Optimal Type 2, sd 120 945 Prop. 5.3.38 3, 40 121 705 GNB k = 6, sd 122 711 GNB k = 6, sd 123 405 Prop. 5.3.38 3, 41 124 489 GNB k = 3 125 729 GNB k = 6, sd 126 459 Prop. 5.3.38 9, 14 127 501 GNB k = 4, sd n Cn Method Property 128 7821 Random 129 825 Prop. 5.3.38 3, 43 130 259 Optimal Type 1 131 261 Optimal Type 2, sd 132 455 Prop. 5.3.38 4, 33 133 1595 GNB k = 12, sd 134 267 Optimal Type 2, sd 135 269 Optimal Type 2, sd 136 1701 Prop. 5.3.38 8, 17 137 801 GNB k = 6, sd 138 275 Optimal Type 1 139 549 GNB k = 4, sd 140 483 Prop. 5.3.38 4, 35 141 1127 GNB k = 8, sd 142 831 GNB k = 6, sd 143 837 GNB k = 6, sd 144 1445 Prop. 5.3.38 9, 16 145 513 Prop. 5.3.38 5, 29 146 291 Optimal Type 2, sd 147 861 GNB k = 6, sd 148 295 Optimal Type 1 149 1191 GNB k = 8, sd 150 495 Prop. 5.3.38 3, 50 151 885 GNB k = 6, sd 152 2457 Prop. 5.3.38 8, 19 153 605 GNB k = 4, sd 154 567 Prop. 5.3.38 11, 14 155 309 Optimal Type 2, sd 156 515 Prop. 5.3.38 3, 52 157 1561 Cor. 5.3.15 Type 2, k = 5, sd 158 315 Optimal Type 2, sd 159 525 Prop. 5.3.38 3, 53 160 3249 Prop. 5.3.38 5, 32 161 855 Prop. 5.3.38 7, 23 162 323 Optimal Type 1 163 645 GNB k = 4, sd 164 567 Prop. 5.3.38 4, 41 165 585 Prop. 5.3.38 5, 33 166 495 Prop. 5.3.38 2, 83 167 2325 Cor. 5.3.15 Type 2, k = 7, sd 168 1995 Prop. 5.3.38 3, 56 169 669 GNB k = 4, sd 170 999 GNB k = 6, sd 171 1989 Prop. 5.3.38 9, 19 172 343 Optimal Type 1 173 345 Optimal Type 2, sd 174 347 Optimal Type 2, sd 175 693 GNB k = 4, sd 176 1785 Prop. 5.3.38 11, 16 177 701 GNB k = 4, sd 178 355 Optimal Type 1 179 357 Optimal Type 2, sd 180 359 Optimal Type 1 181 1065 GNB k = 6, sd 182 721 GNB k = 3 183 365 Optimal Type 2, sd 184 945 Prop. 5.3.38 8, 23 185 1269 Prop. 5.3.38 5, 37 186 371 Optimal Type 2, sd 187 1101 GNB k = 6, sd 188 1107 GNB k = 5 189 377 Optimal Type 2, sd 190 567 Prop. 5.3.38 2, 95 191 381 Optimal Type 2, sd 192 9145 Prop. 5.3.38 3, 64 193 765 GNB k = 4, sd 194 387 Optimal Type 2, sd 195 645 Prop. 5.3.38 3, 65 196 391 Optimal Type 1 197 3529 Cor. 5.3.15 Type 2, k = 9, sd 198 591 Prop. 5.3.38 2, 99 199 789 GNB k = 4, sd 200 1953 Prop. 5.3.38 8, 25 201 1305 Prop. 5.3.38 3, 67 202 1191 GNB k = 6, sd 203 1083 Prop. 5.3.38 7, 29 204 707 Prop. 5.3.38 4, 51 205 729 Prop. 5.3.38 5, 41 206 817 GNB k = 3 207 765 Prop. 5.3.38 9, 23 208 3825 Prop. 5.3.38 13, 16 209 417 Optimal Type 2, sd 210 419 Optimal Type 1 211 2101 Cor. 5.3.15 Type 2, k = 5, sd 212 735 Prop. 5.3.38 4, 53 213 845 GNB k = 4, sd 214 849 GNB k = 3 215 1269 GNB k = 6, sd Introduction to ﬁnite ﬁelds 43 n Cn Method Property 216 2961 Prop. 5.3.38 8, 27 217 1281 GNB k = 6, sd 218 1287 GNB k = 5 219 869 GNB k = 4, sd 220 873 GNB k = 3 221 441 Optimal Type 2, sd 222 735 Prop. 5.3.38 3, 74 223 2665 Cor. 5.3.15 Type 2, k = 6, sd 224 6859 Prop. 5.3.38 7, 32 225 1581 Prop. 5.3.38 9, 25 226 451 Optimal Type 1 227 5447 GNB k = 24, sd 228 1485 Prop. 5.3.38 3, 76 229 2747 GNB k = 12, sd 230 459 Optimal Type 2, sd 231 461 Optimal Type 2, sd 232 1197 Prop. 5.3.38 8, 29 233 465 Optimal Type 2, sd 234 867 Prop. 5.3.38 9, 26 235 933 GNB k = 4, sd 236 937 GNB k = 3 237 1545 Prop. 5.3.38 3, 79 238 711 Prop. 5.3.38 2, 119 239 477 Optimal Type 2, sd 240 3825 Prop. 5.3.38 3, 80 241 1425 GNB k = 6, sd 242 1431 GNB k = 6, sd 243 485 Optimal Type 2, sd 244 969 GNB k = 3 245 489 Optimal Type 2, sd 246 815 Prop. 5.3.38 3, 82 247 1461 GNB k = 6, sd 248 4977 Prop. 5.3.38 8, 31 249 825 Prop. 5.3.38 3, 83 250 2187 Prop. 5.3.38 2, 125 251 501 Optimal Type 2, sd 252 935 Prop. 5.3.38 9, 28 253 945 Prop. 5.3.38 11, 23 254 507 Optimal Type 2, sd 255 909 Prop. 5.3.38 5, 51 256 No data Prime power 257 1521 GNB k = 6, sd 258 855 Prop. 5.3.38 3, 86 259 2581 Cor. 5.3.15 Type 2, k = 5, sd 260 903 Prop. 5.3.38 4, 65 261 521 Optimal Type 2, sd 262 783 Prop. 5.3.38 2, 131 263 1557 GNB k = 6, sd 264 1365 Prop. 5.3.38 8, 33 265 945 Prop. 5.3.38 5, 53 266 1575 GNB k = 6, sd 267 885 Prop. 5.3.38 3, 89 268 535 Optimal Type 1 269 2151 GNB k = 8, sd 270 539 Optimal Type 2, sd 271 1605 GNB k = 6, sd 272 6885 Prop. 5.3.38 16, 17 273 545 Optimal Type 2, sd 274 2403 Prop. 5.3.38 2, 137 275 1953 Prop. 5.3.38 11, 25 276 959 Prop. 5.3.38 4, 69 277 1101 GNB k = 4, sd 278 555 Optimal Type 2, sd 279 1109 GNB k = 4, sd 280 1449 Prop. 5.3.38 8, 35 281 561 Optimal Type 2, sd 282 1671 GNB k = 6, sd 283 1677 GNB k = 6, sd 284 1129 GNB k = 3 285 945 Prop. 5.3.38 3, 95 286 1071 Prop. 5.3.38 11, 26 287 1539 Prop. 5.3.38 7, 41 288 6137 Prop. 5.3.38 9, 32 289 3457 Cor. 5.3.15 Type 2, k = 6, sd 290 1035 Prop. 5.3.38 5, 58 291 1725 GNB k = 6, sd 292 583 Optimal Type 1 293 585 Optimal Type 2, sd 294 975 Prop. 5.3.38 3, 98 295 4719 GNB k = 16, sd 296 2961 Prop. 5.3.38 8, 37 297 1761 GNB k = 6, sd 298 1767 GNB k = 6, sd 299 597 Optimal Type 2, sd 300 995 Prop. 5.3.38 3, 100 301 3001 Cor. 5.3.15 Type 2, k = 5, sd 302 1201 GNB k = 3 303 605 Optimal Type 2, sd n Cn Method Property 304 9945 Prop. 5.3.38 16, 19 305 1809 GNB k = 6, sd 306 611 Optimal Type 2, sd 307 1221 GNB k = 4, sd 308 1155 Prop. 5.3.38 11, 28 309 617 Optimal Type 2, sd 310 927 Prop. 5.3.38 2, 155 311 1845 GNB k = 6, sd 312 1617 Prop. 5.3.38 8, 39 313 1857 GNB k = 6, sd 314 1863 GNB k = 5 315 1173 Prop. 5.3.38 9, 35 316 631 Optimal Type 1 317 8217 Cor. 5.3.15 Type 2, k = 13, sd 318 1055 Prop. 5.3.38 3, 106 319 1197 Prop. 5.3.38 11, 29 320 16461 Prop. 5.3.38 5, 64 321 3105 Prop. 5.3.38 3, 107 322 1215 Prop. 5.3.38 14, 23 323 645 Optimal Type 2, sd 324 1127 Prop. 5.3.38 4, 81 325 1293 GNB k = 4, sd 326 651 Optimal Type 2, sd 327 2615 GNB k = 8, sd 328 1701 Prop. 5.3.38 8, 41 329 657 Optimal Type 2, sd 330 659 Optimal Type 2, sd 331 1965 GNB k = 6, sd 332 1155 Prop. 5.3.38 4, 83 333 2397 Prop. 5.3.38 9, 37 334 2629 GNB k = 7 335 2349 Prop. 5.3.38 5, 67 336 8075 Prop. 5.3.38 3, 112 337 3361 Cor. 5.3.15 Type 2, k = 5, sd 338 675 Optimal Type 2, sd 339 1125 Prop. 5.3.38 3, 113 340 1353 GNB k = 3 341 2727 GNB k = 8, sd 342 2031 GNB k = 6, sd 343 1365 GNB k = 4, sd 344 3465 Prop. 5.3.38 8, 43 345 1233 Prop. 5.3.38 5, 69 346 691 Optimal Type 1 347 2061 GNB k = 6, sd 348 695 Optimal Type 1 349 3481 Cor. 5.3.15 Type 2, k = 5, sd 350 699 Optimal Type 2, sd 351 3501 Cor. 5.3.15 Type 2, k = 5, sd 352 7581 Prop. 5.3.38 11, 32 353 4929 Cor. 5.3.15 Type 2, k = 7, sd 354 707 Optimal Type 2, sd 355 2109 GNB k = 6, sd 356 1239 Prop. 5.3.38 4, 89 357 1185 Prop. 5.3.38 3, 119 358 1071 Prop. 5.3.38 2, 179 359 717 Optimal Type 2, sd 360 3213 Prop. 5.3.38 5, 72 361 10801 Cor. 5.3.15 Type 2, k = 15, sd 362 2151 GNB k = 5 363 1445 GNB k = 4, sd 364 1449 GNB k = 3 365 2565 Prop. 5.3.38 5, 73 366 1095 Prop. 5.3.38 2, 183 367 2181 GNB k = 6, sd 368 3825 Prop. 5.3.38 16, 23 369 1377 Prop. 5.3.38 9, 41 370 1323 Prop. 5.3.38 5, 74 371 741 Optimal Type 2, sd 372 743 Optimal Type 1 373 1485 GNB k = 4, sd 374 1489 GNB k = 3 375 749 Optimal Type 2, sd 376 5481 Prop. 5.3.38 8, 47 377 2565 Prop. 5.3.38 13, 29 378 755 Optimal Type 1 379 4547 GNB k = 12, sd 380 1323 Prop. 5.3.38 4, 95 381 2505 Prop. 5.3.38 3, 127 382 1143 Prop. 5.3.38 2, 191 383 4595 GNB k = 12, sd 384 39105 Prop. 5.3.38 3, 128 385 1449 Prop. 5.3.38 11, 35 386 771 Optimal Type 2, sd 387 1541 GNB k = 4, sd 388 775 Optimal Type 1 389 9335 GNB k = 24, sd 390 1295 Prop. 5.3.38 3, 130 391 2325 GNB k = 6, sd 44 Handbook of Finite Fields n Cn Method Property 392 3969 Prop. 5.3.38 8, 49 393 785 Optimal Type 2, sd 394 3915 Cor. 5.3.15 Type 1, k = 9 395 2349 GNB k = 6, sd 396 1379 Prop. 5.3.38 4, 99 397 2361 GNB k = 6, sd 398 795 Optimal Type 2, sd 399 4777 Cor. 5.3.15 Type 2, k = 6, sd 400 7905 Prop. 5.3.38 16, 25 401 3207 GNB k = 8, sd 402 1335 Prop. 5.3.38 3, 134 403 6447 GNB k = 16, sd 404 1609 GNB k = 3 405 1449 Prop. 5.3.38 5, 81 406 1539 Prop. 5.3.38 14, 29 407 2961 Prop. 5.3.38 11, 37 408 2121 Prop. 5.3.38 8, 51 409 1629 GNB k = 4, sd 410 819 Optimal Type 2, sd 411 821 Optimal Type 2, sd 412 1641 GNB k = 3 413 825 Optimal Type 2, sd 414 827 Optimal Type 2, sd 415 1485 Prop. 5.3.38 5, 83 416 16245 Prop. 5.3.38 13, 32 417 1661 GNB k = 4, sd 418 835 Optimal Type 1 419 837 Optimal Type 2, sd 420 839 Optimal Type 1 421 4209 GNB k = 10, sd 422 5053 GNB k = 11 423 1685 GNB k = 4, sd 424 2205 Prop. 5.3.38 8, 53 425 2529 GNB k = 6, sd 426 851 Optimal Type 2, sd 427 6555 Prop. 5.3.38 7, 61 428 2547 GNB k = 5 429 857 Optimal Type 2, sd 430 1539 Prop. 5.3.38 5, 86 431 861 Optimal Type 2, sd 432 11985 Prop. 5.3.38 16, 27 433 1725 GNB k = 4, sd 434 3843 Prop. 5.3.38 2, 217 435 1733 GNB k = 4, sd 436 6091 GNB k = 13 437 5265 Prop. 5.3.38 19, 23 438 875 Optimal Type 2, sd 439 4381 Cor. 5.3.15 Type 2, k = 5, sd 440 3969 Prop. 5.3.38 5, 88 441 881 Optimal Type 2, sd 442 883 Optimal Type 1 443 885 Optimal Type 2, sd 444 1475 Prop. 5.3.38 3, 148 445 1593 Prop. 5.3.38 5, 89 446 2655 GNB k = 6, sd 447 2661 GNB k = 6, sd 448 34751 Prop. 5.3.38 7, 64 449 3591 GNB k = 8, sd 450 1683 Prop. 5.3.38 9, 50 451 1701 Prop. 5.3.38 11, 41 452 1575 Prop. 5.3.38 4, 113 453 905 Optimal Type 2, sd 454 9025 Cor. 5.3.15 Type 1, k = 19 455 2451 Prop. 5.3.38 7, 65 456 10437 Prop. 5.3.38 8, 57 457 13681 Cor. 5.3.15 Type 2, k = 15, sd 458 2727 GNB k = 6, sd 459 3671 GNB k = 8, sd 460 919 Optimal Type 1 461 2745 GNB k = 6, sd 462 1383 Prop. 5.3.38 2, 231 463 5545 Cor. 5.3.15 Type 2, k = 6, sd 464 4845 Prop. 5.3.38 16, 29 465 1545 Prop. 5.3.38 3, 155 466 931 Optimal Type 1 467 2781 GNB k = 6, sd 468 1751 Prop. 5.3.38 9, 52 469 1869 GNB k = 4, sd 470 939 Optimal Type 2, sd 471 3767 GNB k = 8, sd 472 12537 Prop. 5.3.38 8, 59 473 945 Optimal Type 2, sd 474 1575 Prop. 5.3.38 3, 158 475 1893 GNB k = 4, sd 476 1659 Prop. 5.3.38 4, 119 477 1785 Prop. 5.3.38 9, 53 478 1431 Prop. 5.3.38 2, 239 479 3831 GNB k = 8, sd n Cn Method Property 480 16245 Prop. 5.3.38 3, 160 481 2865 GNB k = 6, sd 482 2871 GNB k = 5 483 965 Optimal Type 2, sd 484 1929 GNB k = 3 485 3429 Prop. 5.3.38 5, 97 486 1455 Prop. 5.3.38 2, 243 487 1941 GNB k = 4, sd 488 7245 Prop. 5.3.38 8, 61 489 3225 Prop. 5.3.38 3, 163 490 979 Optimal Type 1 491 981 Optimal Type 2, sd 492 1863 Prop. 5.3.38 12, 41 493 1965 GNB k = 4, sd 494 1969 GNB k = 3 495 989 Optimal Type 2, sd 496 20145 Prop. 5.3.38 16, 31 497 9921 Cor. 5.3.15 Type 2, k = 10, sd 498 1815 Prop. 5.3.38 6, 83 499 1989 GNB k = 4, sd 500 5103 Prop. 5.3.38 4, 125 501 5001 Cor. 5.3.15 Type 2, k = 5, sd 502 1503 Prop. 5.3.38 2, 251 503 2997 GNB k = 6, sd 504 6783 Prop. 5.3.38 7, 72 505 5041 Cor. 5.3.15 Type 2, k = 5, sd 506 2835 Prop. 5.3.38 2, 253 507 2021 GNB k = 4, sd 508 1015 Optimal Type 1 509 1017 Optimal Type 2, sd 510 1919 Prop. 5.3.38 10, 51 511 3045 GNB k = 6, sd 512 No data Prime power 513 2045 GNB k = 4, sd 514 4563 Prop. 5.3.38 2, 257 515 1029 Optimal Type 2, sd 516 1715 Prop. 5.3.38 3, 172 517 2061 GNB k = 4, sd 518 2793 Prop. 5.3.38 7, 74 519 1037 Optimal Type 2, sd 520 2709 Prop. 5.3.38 8, 65 521 16671 GNB k = 32, sd 522 1043 Optimal Type 1 523 5221 Cor. 5.3.15 Type 2, k = 5, sd 524 1827 Prop. 5.3.38 4, 131 525 3465 Prop. 5.3.38 3, 175 526 2097 GNB k = 3 527 3141 GNB k = 6, sd 528 5525 Prop. 5.3.38 16, 33 529 12695 GNB k = 24, sd 530 1059 Optimal Type 2, sd 531 1061 Optimal Type 2, sd 532 2121 GNB k = 3 533 3645 Prop. 5.3.38 13, 41 534 1775 Prop. 5.3.38 3, 178 535 2133 GNB k = 4, sd 536 5481 Prop. 5.3.38 8, 67 537 1785 Prop. 5.3.38 3, 179 538 3207 GNB k = 6, sd 539 3969 Prop. 5.3.38 11, 49 540 1079 Optimal Type 1 541 9737 GNB k = 18, sd 542 2161 GNB k = 3 543 1085 Optimal Type 2, sd 544 29241 Prop. 5.3.38 17, 32 545 1089 Optimal Type 2, sd 546 1091 Optimal Type 1 547 5469 GNB k = 10, sd 548 3267 GNB k = 5 549 5865 Prop. 5.3.38 9, 61 550 2079 Prop. 5.3.38 11, 50 551 3285 GNB k = 6, sd 552 2877 Prop. 5.3.38 8, 69 553 2205 GNB k = 4, sd 554 1107 Optimal Type 2, sd 555 2213 GNB k = 4, sd 556 1111 Optimal Type 1 557 3321 GNB k = 6, sd 558 1115 Optimal Type 2, sd 559 2229 GNB k = 4, sd 560 5865 Prop. 5.3.38 16, 35 561 1121 Optimal Type 2, sd 562 1123 Optimal Type 1 563 7869 Cor. 5.3.15 Type 2, k = 7, sd 564 2249 GNB k = 3 565 2025 Prop. 5.3.38 5, 113 566 2257 GNB k = 3 567 2261 GNB k = 4, sd Introduction to ﬁnite ﬁelds 45 n Cn Method Property 568 11907 Prop. 5.3.38 8, 71 569 6817 Cor. 5.3.15 Type 2, k = 6, sd 570 2079 Prop. 5.3.38 6, 95 571 5709 GNB k = 10, sd 572 2163 Prop. 5.3.38 11, 52 573 1905 Prop. 5.3.38 3, 191 574 2187 Prop. 5.3.38 14, 41 575 1149 Optimal Type 2, sd 576 31093 Prop. 5.3.38 9, 64 577 2301 GNB k = 4, sd 578 3447 GNB k = 6, sd 579 3825 Prop. 5.3.38 3, 193 580 2313 GNB k = 3 581 3135 Prop. 5.3.38 7, 83 582 1935 Prop. 5.3.38 3, 194 583 2205 Prop. 5.3.38 11, 53 584 5985 Prop. 5.3.38 8, 73 585 1169 Optimal Type 2, sd 586 1171 Optimal Type 1 587 8205 Cor. 5.3.15 Type 2, k = 7, sd 588 1955 Prop. 5.3.38 3, 196 589 2349 GNB k = 4, sd 590 6183 Prop. 5.3.38 5, 118 591 3525 GNB k = 6, sd 592 11985 Prop. 5.3.38 16, 37 593 1185 Optimal Type 2, sd 594 4389 Prop. 5.3.38 11, 54 595 2133 Prop. 5.3.38 5, 119 596 2377 GNB k = 3 597 2381 GNB k = 4, sd 598 1791 Prop. 5.3.38 2, 299 599 4791 GNB k = 8, sd 600 9765 Prop. 5.3.38 3, 200 601 3585 GNB k = 6, sd 602 3249 Prop. 5.3.38 7, 86 603 4437 Prop. 5.3.38 9, 67 604 4789 GNB k = 7 605 3609 GNB k = 6, sd 606 1211 Optimal Type 2, sd 607 3621 GNB k = 6, sd 608 42237 Prop. 5.3.38 19, 32 609 2429 GNB k = 4, sd 610 5427 Prop. 5.3.38 2, 305 611 1221 Optimal Type 2, sd 612 1223 Optimal Type 1 613 6121 Cor. 5.3.15 Type 2, k = 5, sd 614 1227 Optimal Type 2, sd 615 1229 Optimal Type 2, sd 616 8379 Prop. 5.3.38 7, 88 617 4935 GNB k = 8, sd 618 1235 Optimal Type 1 619 2469 GNB k = 4, sd 620 2163 Prop. 5.3.38 4, 155 621 3705 GNB k = 6, sd 622 2481 GNB k = 3 623 3363 Prop. 5.3.38 7, 89 624 6545 Prop. 5.3.38 16, 39 625 22465 Cor. 5.3.15 Type 2, k = 18, sd 626 5571 Prop. 5.3.38 2, 313 627 2085 Prop. 5.3.38 3, 209 628 4981 GNB k = 7 629 1257 Optimal Type 2, sd 630 2415 Prop. 5.3.38 18, 35 631 6301 Cor. 5.3.15 Type 2, k = 5, sd 632 6489 Prop. 5.3.38 8, 79 633 10505 Prop. 5.3.38 3, 211 634 8839 Cor. 5.3.15 Type 1, k = 13 635 4509 Prop. 5.3.38 5, 127 636 2415 Prop. 5.3.38 12, 53 637 2541 GNB k = 4, sd 638 1275 Optimal Type 2, sd 639 1277 Optimal Type 2, sd 640 70389 Prop. 5.3.38 5, 128 641 1281 Optimal Type 2, sd 642 3831 GNB k = 6, sd 643 7705 Cor. 5.3.15 Type 2, k = 6, sd 644 2475 Prop. 5.3.38 23, 28 n Cn Method Property 645 1289 Optimal Type 2, sd 646 1935 Prop. 5.3.38 2, 323 647 9045 Cor. 5.3.15 Type 2, k = 7, sd 648 3381 Prop. 5.3.38 8, 81 649 6481 Cor. 5.3.15 Type 2, k = 5, sd 650 1299 Optimal Type 2, sd 651 1301 Optimal Type 2, sd 652 1303 Optimal Type 1 653 1305 Optimal Type 2, sd 654 6435 Prop. 5.3.38 3, 218 655 2349 Prop. 5.3.38 5, 131 656 6885 Prop. 5.3.38 16, 41 657 4845 Prop. 5.3.38 9, 73 658 1315 Optimal Type 1 659 1317 Optimal Type 2, sd 660 1319 Optimal Type 1 661 3945 GNB k = 6, sd 662 2641 GNB k = 3 663 2205 Prop. 5.3.38 3, 221 664 3465 Prop. 5.3.38 8, 83 665 3591 Prop. 5.3.38 7, 95 666 2499 Prop. 5.3.38 9, 74 667 2565 Prop. 5.3.38 23, 29 668 7985 Cor. 5.3.15 Type 1, k = 11 669 2669 GNB k = 4, sd 670 2403 Prop. 5.3.38 5, 134 671 4005 GNB k = 6, sd 672 34295 Prop. 5.3.38 3, 224 673 2685 GNB k = 4, sd 674 4023 GNB k = 5 675 13113 Prop. 5.3.38 25, 27 676 1351 Optimal Type 1 677 5415 GNB k = 8, sd 678 2255 Prop. 5.3.38 3, 226 679 6789 GNB k = 10, sd 680 15309 Prop. 5.3.38 5, 136 681 14961 Cor. 5.3.15 Type 2, k = 11, sd 682 4071 GNB k = 6, sd 683 1365 Optimal Type 2, sd 684 2729 GNB k = 3 685 2733 GNB k = 4, sd 686 1371 Optimal Type 2, sd 687 6869 GNB k = 10, sd 688 14025 Prop. 5.3.38 16, 43 689 4725 Prop. 5.3.38 13, 53 690 1379 Optimal Type 2, sd 691 6901 Cor. 5.3.15 Type 2, k = 5, sd 692 2415 Prop. 5.3.38 4, 173 693 3743 Prop. 5.3.38 7, 99 694 2769 GNB k = 3 695 4941 Prop. 5.3.38 5, 139 696 5985 Prop. 5.3.38 3, 232 697 2781 GNB k = 4, sd 698 4167 GNB k = 5 699 2325 Prop. 5.3.38 3, 233 700 1399 Optimal Type 1 701 12601 Cor. 5.3.15 Type 2, k = 9, sd 702 7191 Prop. 5.3.38 26, 27 703 4197 GNB k = 6, sd 704 38409 Prop. 5.3.38 11, 64 705 4209 GNB k = 6, sd 706 14787 Prop. 5.3.38 2, 353 707 4221 GNB k = 6, sd 708 1415 Optimal Type 1 709 2829 GNB k = 4, sd 710 2833 GNB k = 3 711 5253 Prop. 5.3.38 9, 79 712 3717 Prop. 5.3.38 8, 89 713 1425 Optimal Type 2, sd 714 2607 Prop. 5.3.38 6, 119 715 2709 Prop. 5.3.38 11, 65 716 2499 Prop. 5.3.38 4, 179 717 2385 Prop. 5.3.38 3, 239 718 2151 Prop. 5.3.38 2, 359 719 1437 Optimal Type 2, sd 720 13005 Prop. 5.3.38 5, 144 721 4305 GNB k = 6, sd Table 2.2.10 Minimum found complexity of a normal basis of F2n over F2, 40 ≤n ≤721. 46 Handbook of Finite Fields 2.2.3 Resources and standards 2.2.21 Remark The Combinatorial Object Server (COS) allows the user to specify a type of combinatorial object with speciﬁc parameter values and COS will return a list of the objects having the desired parameters. In many cases, the format of the output can be chosen to be more machine-readable or human-readable. COS does not rely on a list, rather it generates the objects requested on-the-ﬂy; for this reason, the output is restricted to 200 objects. Examples of the objects generated are permutations, subsets and combinations, set and integer partitions, irreducible and primitive polynomials over small ﬁnite ﬁelds and spanning trees of a graph. 2.2.22 Remark The Cunningham project produces a set of tables to factor the numbers bn ± 1 for b = 2, 3, 5, 6, 7, 10, 11, 12 for n as large as possible. The current factorization methods employed are the elliptic curve method, the multiple polynomial quadratic sieve and the number ﬁeld sieve. For more information on factorization methods, see [2080, Chapter 3]. The Cunningham tables appear in published form and as an electronic resource . 2.2.23 Remark The Great Internet Mersenne Prime Search (GIMPS) is a distributed com-puting eﬀort dedicated to ﬁnding and verifying Mersenne primes (that is, primes of the form 2p −1, where p is also a prime). GIMPS uses a combination of trial factoring using the Sieve of Eratosthenes, followed by the Pollard P −1 method and ending with the Lucas-Lehmer primality test. For more information on primality testing, see [724, Chapter 31], for exam-ple. GIMPS provides the Prime95 software, which automates all factoring and distributed computing processes. The (currently) largest known Mersenne prime is 243112609 −1 con-taining 12978189 decimal digits . The search for Mersenne primes is of particular interest in searching for primitive tri-nomials of large degrees. Primitive polynomials of low-weight are useful in cryptographic applications and pseudo-random number generation; see Sections 14.9 and 16.2. If p is a Mersenne prime, then any irreducible polynomial of degree p over F2 is primitive. Since bi-nomials of degree at least 2 cannot be irreducible over F2, we consider trinomials xp+xr +1, for some 0 ≤r ≤p −1. Sieving trinomials for reducibles is possible by Swan’s Theorem; see Section 3.3. For more details on the algorithms and methods used in the search for primitive trinomials, see . An implementation of polynomial arithmetic over F2 which was motivated by the GIMPS project, entitled gf2x, is necessarily highly optimized and is preferred in some ﬁnite ﬁeld software implementations; see Table 2.2.11 for more details. 2.2.24 Remark The On-Line Encyclopedia of Integer SequencesTM (OEISTM) is a constantly-updated, searchable database of integer sequences. Examples of famous sequences in the OEISTM are the Catalan numbers (A000108), prime numbers (A000040), and the Fibonnacci numbers (A000045). Users can search by sequence, “word” (for example, “num-ber of irreducible polynomials” yields sequence A001037) or sequence number. Sequences are sorted lexicographically, so the sequence references may have changed since the date of publication. 2.2.25 Remark In Table 2.2.11, we present a number of software packages which are useful for ﬁnite ﬁeld implementations. We distinguish between packages which are open-source and commercial. We refer the reader to the citation, which provides a current (as of the date of publication) Web URL to the most recent build of the software. We note that this is not an exhaustive list of software packages, simply a useful list of packages used or researched by the author. Introduction to ﬁnite ﬁelds 47 Open-source software packages for computations in discrete mathematics Name Ver. Description Fast Library for Number Theory (FLINT) 2.3 A C library for performing computations in number theory. Routines include fast algorithms, on par with the other most eﬃcient packages listed, for arbitrary precision integers, rational numbers, modular arith-metic, and p-adic numbers. Most libraries also con-tain vector, polynomial and matrix methods. Multi-core support to come in future versions. Groups, Algo-rithms, Pro-gramming (GAP) 4.4.x A system for computational discrete algebra em-phasizing computational group theory. Can do ba-sic computations with arbitrary integers, rationals, ﬁnite ﬁelds, p-adic numbers, polynomials, rational functions, and more. Contains a coding theory pack-age, combinatorial functions and prime factorization routines. Provides its own programming language, li-braries of algebraic algorithms written in the GAP language as well as data libraries of algebraic objects, particularly various types of groups. gf2x 1.0 Library for eﬃcient arithmetic of single-variable polynomials over F2. Primarily introduces fast-fourier transform (FFT) for large-degree polynomial multiplication. The GNU Mul-tiple Precision Arithmetic Library (GMP) 5.0.x C/C++ library providing fast arbitrary precision arithmetic on integers, rational numbers, and ﬂoat-ing point numbers. Macaulay2 1.4 Software system focusing on algebraic geometry and commutative algebra. Contains core algorithms com-bined with a high-level interpreted language and de-bugger to support package creation. Uses elements of PARI, NTL, and others in its routines. Number Theory Library (NTL) 5.5.x C++ library providing data structures and routines for arbitrary length integer arithmetic, arbitrary pre-cision ﬂoating-point arithmetic, and ﬁnite ﬁeld arith-metic. Also contains lattice basis reduction algo-rithms and basic linear algebra packages. Interfaces with gf2x and GMP libraries for additional speed-ups. PARI/GP 2.5.x C library designed for fast computations in num-ber theory including integer factorization and elliptic curve compututations. Also contains useful function for use with matrices, polynomials, power series, and others. GP is a scripting language used by the gp interactive shell, which accesses the PARI functions. A subset of the GP language can be compiled as C code, resulting in a substantial speed-up. 48 Handbook of Finite Fields Open-source software packages for computations in discrete mathematics Name Ver. Description Singular 3.1.x A computer algebra system focusing on polynomial computations. Specializes on commutative and non-commutative algebra, algebraic geometry, and sin-gularity theory. Provides a C-like programming lan-guage, extendable using libraries. Its core algorithms handle Gr¨ obner bases, polynomial factorization, re-sultants, and root ﬁnding. Advanced libraries and third-party software provide further functionality. SAGE 5.0 Comprehensive Python-based open-source computer algebra package. Natively contains a ﬁnite ﬁeld im-plementation as well as wrappers for other useful packages including Flint, GAP, NTL, PARI, and Sin-gular. Interpreted but contains the ability to compile, using Cython, as C code for a drastic improvement in speed. Commercial stand-alone packages containing ﬁnite ﬁeld implementations Name Ver. Description Magma 2.18-x Computational algebra system focusing on alge-bra, algebraic combinatorics, algebraic geometry and number theory. Language built to closely approxi-mate the user’s mode of thought and usual notation. Major algorithms are designed to give comparable performance to specialized programs. Also contains a number of large databases of elliptic curves, lin-ear codes, irreducible polynomials over ﬁnite ﬁelds, graphs, Cunningham factorizations, and others. Maple 16 Comprehensive computer algebra suite, contains a full featured programming language to create scripts or full applications. A “smart” document environ-ment allows embedding equations, visualizations, or components in the document. Can take advantage of parallelism, multi-threading and multi-process pro-graming. Finite ﬁeld arithmetic natively given by the “GF” package. Mathematica 8 Development platform concetrating on integrating computation into workﬂows. Finite ﬁeld computa-tions are performed using the “FiniteFields” package and “GF” class. Matlab R2012a Programming environment for algorithm develop-ment and data analysis. Contains arithmetic over ﬁ-nite ﬁelds F2n over F2 for n ≤16 within the “Com-munications System Toolbox.” Table 2.2.11 Software packages useful for ﬁnite ﬁeld implementations. Introduction to ﬁnite ﬁelds 49 See Also §3.2, §3.3, §3.4 For reducibility and irreducibility of low-weight polynomials. §5.2, §5.3 For normal bases and their complexities. §11.1 For computational techniques over ﬁnite ﬁelds. , For patents and standards of elliptic curve cryptography, most of which contain guidelines for ﬁnite ﬁeld implementations. References Cited: [130, 399, 408, 415, 712, 724, 792, 1263, 1264, 1355, 1413, 1426, 1631, 1777, 1939, 2002, 2015, 2080, 2084, 2180, 2356, 2507, 2573, 2582, 2633, 2709, 2753, 2796, 2797, 2799, 2800, 2801, 2890, 2925, 3004, 3036] This page intentionally left blank This page intentionally left blank II Theoretical Properties 3 Irreducible polynomials .......................................... 53 Counting irreducible polynomials • Construction of irreducibles • Con-ditions for reducible polynomials • Weights of irreducible polynomials • Prescribed coeﬃcients • Multivariate polynomials 4 Primitive polynomials ............................................ 87 Introduction to primitive polynomials • Prescribed coeﬃcients • Weights of primitive polynomials • Elements of high order 5 Bases ................................................................. 101 Duality theory of bases • Normal bases • Complexity of normal bases • Completely normal bases 6 Exponential and character sums ............................... 139 Gauss, Jacobi, and Kloosterman sums • More general exponential and character sums • Some applications of character sums • Sum-product theorems and applications 7 Equations over ﬁnite ﬁelds ...................................... 193 General forms • Quadratic forms • Diagonal equations 8 Permutation polynomials ........................................ 215 One variable • Several variables • Value sets of polynomials • Exceptional polynomials 9 Special functions over ﬁnite ﬁelds ............................. 241 Boolean functions • PN and APN functions • Bent and related functions • κ-polynomials and related algebraic objects • Planar functions and commutative semiﬁelds • Dickson polynomials • Schur’s conjecture and exceptional covers 10 Sequences over ﬁnite ﬁelds ...................................... 303 Finite ﬁeld transforms • LFSR sequences and maximal period sequences • Correlation and autocorrelation of sequences • Linear complexity of sequences and multisequences • Algebraic dynamical systems over ﬁnite ﬁelds 11 Algorithms .......................................................... 345 Computational techniques • Univariate polynomial counting and algo-rithms • Algorithms for irreducibility testing and for constructing ir-reducible polynomials • Factorization of univariate polynomials • Fac-51 52 Theoretical Properties torization of multivariate polynomials • Discrete logarithms over ﬁnite ﬁelds • Standard models for ﬁnite ﬁelds 12 Curves over ﬁnite ﬁelds .......................................... 405 Introduction to function ﬁelds and curves • Elliptic curves • Addition formulas for elliptic curves • Hyperelliptic curves • Rational points on curves • Towers • Zeta functions and L-functions • p-adic estimates of zeta functions and L-functions • Computing the number of rational points and zeta functions 13 Miscellaneous theoretical topics ............................... 493 Relations between integers and polynomials over ﬁnite ﬁelds • Matrices over ﬁnite ﬁelds • Classical groups over ﬁnite ﬁelds • Computational linear algebra over ﬁnite ﬁelds • Carlitz and Drinfeld modules 3 Irreducible polynomials 3.1 Counting irreducible polynomials ................. 53 Prescribed trace or norm • Prescribed coeﬃcients over the binary ﬁeld • Self-reciprocal polynomials • Compositions of powers • Translation invariant polynomials • Normal replicators 3.2 Construction of irreducibles ....................... 60 Construction by composition • Recursive constructions 3.3 Conditions for reducible polynomials ............. 66 Composite polynomials • Swan-type theorems 3.4 Weights of irreducible polynomials ............... 70 Basic deﬁnitions • Existence results • Conjectures 3.5 Prescribed coeﬃcients .............................. 73 One prescribed coeﬃcient • Prescribed trace and norm • More prescribed coeﬃcients • Further exact expressions 3.6 Multivariate polynomials ........................... 80 Counting formulas • Asymptotic formulas • Results for the vector degree • Indecomposable polynomials and irreducible polynomials • Algorithms for the gcd of multivariate polynomials 3.1 Counting irreducible polynomials Joseph L.Yucas, Southern Illinois University 3.1.1 Remark In this section we∗are concerned with exact formulae for the number of (univariate) irreducible polynomials over ﬁnite ﬁelds possessing various properties. There is some overlap with Section 3.5 where speciﬁcally polynomials with prescribed coeﬃcients are discussed. Formulae and asymptotic expressions for mutivariate polynomials are given in Section 3.6. 3.1.2 Theorem (Theorem 2.1.24). Denote the number of monic irreducible polynomials of degree n over Fq by Iq(n). Then Iq(n) = 1 n X d\|n µ(d)qn/d. ∗The author wishes to thank Stephen Cohen for a number of helpful improvements in this section. 53 54 Handbook of Finite Fields 3.1.3 Deﬁnition For a positive integer n set Dn = {r : r\|qn −1 but r does not divide qm −1 for m < n}. 3.1.4 Theorem We have Iq(n) = 1 n X r∈Dn φ(r), where φ denotes Euler’s function. 3.1.1 Prescribed trace or norm 3.1.5 Deﬁnition The trace of a monic polynomial f of degree n over Fq is −a1, where a1 is the ﬁrst coeﬃcient of f, i.e., the coeﬃcient of xn−1 in f. The norm of a monic polynomial f of degree n over Fq is (−1)nan, where an is the last coeﬃcient of f, i.e., the constant term in f. The trace and norm of a monic irreducible polynomial are, respectively, the trace and norm of any of its roots in Fqn over Fq. 3.1.6 Remark In [2508, 3048] (cited below) and sometimes in the literature, the trace of a poly-nomial f is taken to be the ﬁrst coeﬃcient a1 itself. 3.1.7 Theorem [541, 2508] For a non-zero a ∈Fq the number Iq(n, a) of monic irreducible poly-nomials of degree n over Fq with trace a is Iq(n, a) = 1 qn X d\|n (d,q)=1 µ(d)qn/d. 3.1.8 Theorem Let q be a power of the prime p. Write n = pkm with m being p-free (i.e., p does not divide m). The number Iq(n, 0) of monic irreducible polynomials of degree n over Fq with trace 0 is Iq(n, 0) = 1 qn X d\|m µ(d)qn/d −ϵ n X d\|m µ(d)qn/dp, where ϵ = 1 if k > 0 and ϵ = 0 if k = 0. 3.1.9 Deﬁnition For r ∈Dn, write r = drmr where dr = r, qn−1 q−1 . 3.1.10 Theorem 1. Let r ∈Dn and suppose a ∈Fq has order mr. Further, let Iq(n, r, a) denote the number of monic irreducible polynomials over Fq of degree n, order r and (non-zero) norm a. Then Iq(n, r, a) = φ(r) nφ(mr). 2. Suppose a ∈F∗ q has order m and let Iq(n, a) denote the number of monic irre-ducible polynomials over Fq of degree n with (non-zero) norm a. Then Iq(n, a) = 1 nφ(m) X r∈Dn mr=m φ(r). Irreducible polynomials 55 3.1.11 Remark In , Carlitz obtained formulae for the number of monic irreducible polynomials over Fq, q odd, with prescribed trace and whose norm is a (non-zero) square or a non-square, respectively. These involve the quadratic character λ on Fq; thus for 0 ̸= b ∈Fq, λ(b) = 1 or −1 according as b is a square or non-square in Fq, respectively. (See also Remark 3.5.49.) 3.1.12 Theorem [541, 1783] Let q = pm be an odd prime power. For a ∈Fq denote by Iq(n, a, h), h = 1, −1, respectively, the number of monic irreducible polynomials of degree n over Fq with trace a whose norm is a (non-zero) square or a non-square. Then, 1. if a = 0 Iq(n, 0, h) = 1 2p(qp−1 −q) for n = p, 1 2n(qn−1 −1) for n ̸= p; 2. if a ̸= 0 Iq(n, a, h) = 1 2p(qp−1 + S) for n = p, 1 2n(qn−1 + S −(−1)hλ(na) −1) for n ̸= p, where S = (−1)hq n−1 2 λ((−1) n−1 2 a), and λ is the quadratic character in Fq. 3.1.2 Prescribed coeﬃcients over the binary ﬁeld 3.1.13 Remark Subsection 3.5.4 contains various formulae for the numbers of irreducible polyno-mials with some prescribed coeﬃcients in a general ﬁnite ﬁeld Fq. We list here a specialized result over the binary ﬁeld F2 not included there. 3.1.14 Deﬁnition For β ∈F2n let Tj(β) = X 0≤i1<i2<···<ij≤n−1 β2i1 β2i2 . . . β2ij ; Tj maps F2n to F2 and T1 is the usual trace function. For n = 2 and j = 3, we deﬁne T3(β) = 0 for all β ∈F4. For an integer r with 1 ≤r ≤n, deﬁne F(n, t1, t2, . . . , tr) to be the number of elements β ∈F2n with Tj(β) = tj for j = 1, . . . , r and let (I2(n, t1, t2, . . . , tr) =) I(n, t1, t2, . . . , tr) be the number of monic irreducible polyno-mials f(x) over F2 of degree n with coeﬃcient of xn−j = tj for j = 1, . . . , r. 3.1.15 Theorem We have nI(n, 0, 0, 0) = X d\|n,d odd µ(d)F(n/d, 0, 0, 0) − X d\|n,d odd,n/d even µ(d)2 n 2d −1; nI(n, 0, 0, 1) = X d\|n,d odd µ(d)F(n/d, 0, 0, 1); nI(n, 0, 1, 0) = X d\|n,d odd µ(d)F(n/d, 0, 1, 0) − X d\|n,d odd,n/d even µ(d)2 n 2d −1; nI(n, 0, 1, 1) = X d\|n,d odd µ(d)F(n/d, 0, 1, 1); nI(n, 1, 0, 0) = X d\|n,d≡1 µ(d)F(n/d, 1, 0, 0) + X d\|n,d≡3 µ(d)F(n/d, 1, 1, 1); nI(n, 1, 0, 1) = X d\|n,d≡1 µ(d)F(n/d, 1, 0, 1) + X d\|n,d≡3 µ(d)F(n/d, 1, 1, 0); 56 Handbook of Finite Fields nI(n, 1, 1, 0) = X d\|n,d≡1 µ(d)F(n/d, 1, 1, 0) + X d\|n,d≡3 µ(d)F(n/d, 1, 0, 1); nI(n, 1, 1, 1) = X d\|n,d≡1 µ(d)F(n/d, 1, 1, 1) + X d\|n,d≡3 µ(d)F(n/d, 1, 0, 0). 3.1.16 Remark Explicit formulae for I(n, t1, t2, t3) (the number of irreducible polynomials over F2 whose ﬁrst three coeﬃcients are prescribed) can be recovered from Theorem 3.1.15 via the next theorem. 3.1.17 Theorem [1076, 3049] For n ≥3, F(n, t1, t2, t3) = 2n−3 + G(n, t1, t2, t3), where the values of G(n, t1, t2, t3) are displayed in the following tables. 1. Case n = 2m + 1 (in the ﬁrst column m is calculated modulo 12): m 000 001 010 011 100 101 110 111 0 −3 · 2m−2 3 · 2m−2 2m−2 2m−2 0 0 0 0 1 or 5 2m−2 −2m−2 2m−2 −2m−2 −2m−2 −2m−2 3 · 2m−2 −2m−2 2 or 10 0 2m−1 0 −2m−1 2m−1 −2m−1 −2m−1 2m−1 3 2m−2 −2m−2 2m−2 −2m−2 2m−1 0 2m−1 −2m 4 or 8 −2m−1 0 −2m−1 2m 0 0 0 0 6 3 · 2m−2 −2m−2 −3 · 2m−2 2m−2 2m−1 −2m−1 −2m−1 2m−1 7 or 11 2m−2 −2m−2 2m−2 −2m−2 −2m−2 3 · 2m−2 −2m−2 −2m−2 9 2m−2 −2m−2 2m−2 −2m−2 −2m 2m−1 0 2m−1 2. Case n = 2m with 3 not dividing n (in the ﬁrst column m is calculated modulo 4): m 000 001 010 011 100 101 110 111 0 3 · 2m−2 −2m−2 −2m−2 −2m−2 3 · 2m−2 −2m−2 −2m−2 −2m−2 1 −3 · 2m−2 2m−2 2m−2 2m−2 2m−2 2m−2 2m−2 −3 · 2m−2 2 −3 · 2m−2 2m−2 2m−2 2m−2 −3 · 2m−2 2m−2 2m−2 2m−2 3 3 · 2m−2 −2m−2 −2m−2 −2m−2 −2m−2 −2m−2 −2m−2 3 · 2m−2 3. Case n = 2m with 3 dividing n (in the ﬁrst column m is calculated modulo 4): m 000 001 010 011 100 101 110 111 0 0 2m−1 2m−1 −2m 0 2m−1 −2m 2m−1 1 0 −2m−1 −2m−1 2m −2m−1 2m −2m−1 0 2 0 −2m−1 −2m−1 2m 0 −2m−1 2m −2m−1 3 0 2m−1 2m−1 −2m 2m−1 −2m −2m−1 0 3.1.18 Remark Formulae for I(n, t1, t2) and F(n, t1, t2) can be obtained by adding appropriate terms from above . For the binary ﬁeld these will agree with the earlier general expres-sions of Kuz’min (Theorems 3.5.43 and 3.5.45) from . 3.1.3 Self-reciprocal polynomials 3.1.19 Remark Self-reciprocal polynomials were deﬁned in Remark 2.1.48. For results on these polynomials see . Any monic self-reciprocal polynomial R of (even) degree 2n has the form xnf x2 + 1 x , where f is a monic polynomial of degree n in Fq[x]. (We observe that, Irreducible polynomials 57 if R is irreducible, then necessarily f is irreducible.) Let SRMIq(2n) denote number of monic irreducible self-reciprocal polynomials of degree 2n over Fq. 3.1.20 Theorem [547, 666, 2091, 2200] If q is odd, then SRMIq(2n) = 1 2n X d\|n d odd µ(d)(qn/d −1), and, if q is even, then SRMIq(2n) = 1 2n X d\|n d odd µ(d)qn/d. 3.1.21 Remark The notion of self-reciprocal polynomial has recently been generalized and Theorem 3.1.20 extended correspondingly . 3.1.22 Deﬁnition Let g(x) = a1x2 + b1x + c1 and h(x) = a2x2 + b2x + c2 be relatively prime polynomials over Fq with max(deg f, deg g) = 2 (so that a1 and a2 are not both zero). Let Iq(2n, g, h) denote the number of irreducible polynomials (not necessarily monic) of degree 2n that can be expressed in the form h(x)nf g(x) h(x) , where f is a monic polynomial (necessarily irreducible) of degree n. 3.1.23 Theorem Suppose n > 1 and g, h are polynomials over Fq as in Deﬁnition 3.1.22. Then Iq(2n, g, h) =              0 if q is even and b1 = b2 = 0, 1 2n(qn −1) if q is odd and n = 2m, 1 2n X d\|n d odd µ(d)qn/d otherwise. 3.1.24 Remark Theorem 3.1.20 is recovered from Theorem 3.1.23 on setting g(x) = x2 +1, h(x) = x (using X d\|n µ(d) = 0 whenever n > 1). 3.1.4 Compositions of powers 3.1.25 Deﬁnition The radical of an integer m (> 1) (denoted here by m∗) is the product of the distinct primes dividing m. 3.1.26 Deﬁnition For t > 1, a t-polynomial T over Fq of degree tn is one that has the form T(x) = f(xt) for some monic polynomial f of degree n. 3.1.27 Remark If T is irreducible, then f is also irreducible. Further (see Theorem 3.2.5), (i) t∗\|(qn −1), and (ii) if 4\|t, then 4\|(qn −1). In fact, if (i) holds then n can be expressed as n = klm, where k is the order of q (mod t∗), l∗\|t, and m and t are relatively prime . 58 Handbook of Finite Fields 3.1.28 Theorem Suppose t > 1 and that (i) and (ii) of Remark 3.1.27 hold with n = klm. Let TMIq(tn) be the number of monic irreducible t-polynomials of degree tn. Then TMIq(tn) =        φ(t) tn (qn −1) for m = 1, mφ(t) tn Iqn/m(m) for m > 1. 3.1.29 Remark For t > 1, a t-reciprocal polynomial T over Fq of degree 2tn is one that is both a t-polynomial and a self-reciprocal polynomial. Thus it has the form T(x) = xtnf x2t+1 xt , where f is a monic polynomial of degree n. If T is irreducible then f is irreducible. Moreover, from we have (i) t is odd and (ii) t∗\|(qn + 1). Also 2n = klm, where l∗\|2t and m and 2t are relatively prime (so that m\|n). 3.1.30 Theorem [666, 2200] Suppose t > 1 and that (i) and (ii) of Remark 3.1.29 hold with 2n = klm. Let TSRMIq(2tn) be the number of monic irreducible t-reciprocal polynomials of degree 2tn. Then TSRMIq(2tn) =        φ(t) 2tn (qn + 1) for m = 1, mφ(t) 2tn Iqn/m(m) for m > 1. 3.1.5 Translation invariant polynomials 3.1.31 Deﬁnition A polynomial f over Fq is translation invariant if f(x+a) = f(x) for all a ∈Fq. 3.1.32 Theorem Let TIMIq(qn) denote the number of translation invariant monic irre-ducible polynomials of degree qn over Fq. Then TIMIq(qn) = q −1 qn X d\|n (q,d)=1 µ(d)qn/d. 3.1.6 Normal replicators 3.1.33 Remark Many of the above formulae and other similar ones can be obtained using the general counting technique which follows. 3.1.34 Deﬁnition A rational function r = f/g ∈Fq(x) is a replicator over Fq if for every n ≥1, xqn−1 −1 divides f(x)qn −f(x)g(x)qn−1. In this case, we write f(x)qn −f(x)g(x)qn−1 = (xqn−1 −1)b r(n, x) for some polynomial b r(n, x) ∈Fq[x]. The polynomial b r(n, x) is the n-th order transform of r. 3.1.35 Deﬁnition Let k be a positive integer. A rational function r = f/g ∈Fq(x) is k-normal if for every n ≥1 and for each λ ∈Fqn, the degrees of the factors of the associated polynomial f −λg divide k, and for some λ, at least one factor has degree equal to k. Irreducible polynomials 59 3.1.36 Remark For a polynomial f(x) = xm + am−1xm−1 + · · · + a1x + a0 over Fqn we deﬁne the following sequence of polynomials: f (j)(x) = xm + aqj m−1xm−1 + · · · + aqj 1 x + aqj 0 . We observe that f (s) = f where s is the least common multiple of the degrees of the minimal polynomials of the coeﬃcients of f. 3.1.37 Deﬁnition Deﬁne the spin Sf(x) of the polynomial f(x) by Sf(x) = s−1 Y j=0 f (j)(x). 3.1.38 Deﬁnition Let r = f/g be a k-normal replicator over Fq and suppose h is an irreducible factor of f −λg for λ ∈Fqn of degree d. Then, Sh is hard for r if the degree of Sh does not divide n. 3.1.39 Remark Write f qn(x) −f(x)gqn−1(x) = (xqn−1 −1)G(n, x)¯ r(n, x), where ¯ r(n, x) is the factor of f qn −fgqn−1 of largest degree which is square-free and satisﬁes (G(n, x), ¯ r(n, x)) = 1. Further let g(n) = deg(G(n, x)). Let HMIq(n, r(x)) denote the number of monic irreducible polynomials of degree n which are hard for r. 3.1.40 Theorem We have HMIq(kn, r(x)) = 1 kn X d\|n d ∤(n/k) µ(n/d)[(m −1)qn −g(d) + 1] = 1 kn X d\|n k ∤d µ(d)[(m −1)qn/d −g(n/d) + 1]. See Also §3.5 For further formulae and estimates for irreducible polynomials with prescribed coeﬃcients. §3.6 For formulae and asymptotic expressions for irreducible multivariate polynomials. References Cited: [48, 541, 547, 565, 666, 1076, 2091, 2200, 2508, 3048, 3049, 3050] 60 Handbook of Finite Fields 3.2 Construction of irreducibles Melsik Kyuregyan, Armenian National Academy of Sciences 3.2.1 Construction by composition 3.2.1 Remark Known constructions of irreducible polynomials depend on the composition of an initial irreducible polynomial with a further polynomial or rational function. Often this process can be iterated or continued recursively to produce an inﬁnite sequence of irreducible polynomials of increasing degrees. 3.2.2 Theorem [505, 666] Let f, g ∈Fq[x] be relatively prime polynomials and let P ∈Fq[x] be an irreducible polynomial of degree n. Then the composition F(x) = (g(x))n P (f(x)/g(x)) is irreducible over Fq if and only if f −αg is irreducible over Fqn for any zero α ∈Fqn of P. 3.2.3 Remark Theorem 3.2.2 was employed by several authors [589, 678, 685, 1172, 1820, 1819, 1821, 1822, 1823, 1824, 1939, 2091] to give iterative constructions of irreducible polynomials over ﬁnite ﬁelds. A further extension of the theorem is produced in , which is also instrumental in the construction of irreducible polynomials of relatively higher degree from given ones. 3.2.4 Theorem Let P ∈Fq[x] be irreducible of degree n. Then for any a, b, c, d ∈Fq such that ad −bc ̸= 0, F(x) = (cx + d)n P ax + b cx + d is also irreducible over Fq. 3.2.5 Theorem Let t be a positive integer and P ∈Fq[x] be irreducible of degree n and exponent e (equal to the order of any root of P). Then P(xt) is irreducible over Fq if and only if 1. (t, (qn −1)/e) = 1, 2. each prime factor of t divides e, and 3. if 4\|t then 4\|(qn −1). 3.2.6 Theorem Let f1, f2, . . . , fN be all the distinct monic irreducible polynomials in Fq[x] of degree m and order e, and let t ≥2 be an integer whose prime factors di-vide e but not (qm −1) /e. Assume also that qm ≡1(mod 4) if t ≡0(mod 4). Then f1(xt), f2(xt), . . . , fN(xt) are all the distinct monic irreducible polynomials in Fq[x] of degree mt and order et. 3.2.7 Remark Agou has established a criterion for f(g(x)) to be irreducible over Fq, where f, g ∈Fq[x] are monic and f is irreducible over Fq. This criterion was used in Agou [36, 38, 39, 40] to characterize irreducible polynomials of special types such as f xpr −ax , f(xp −x −b), and others. Such irreducible compositions of polynomials are also studied in Cohen [666, 671], Long [1954, 1955], and Ore . Irreducibility criteria for compositions of polynomials of the form f(xt) have been estab-lished by Agou [34, 35, 36], Butler , Cohen , Pellet , Petterson , and Serret [2597, 2600]. Berlekamp [231, Chapter 6] and Varshamov and Ananiashvilii discussed the relationship between the orders of f(xt) and that of f(x). Irreducible polynomials 61 3.2.8 Theorem Let q = 2m and let P(x) = Pn i=0 cixi ∈Fq[x] be irreducible over Fq of degree n and P ∗(x) = xnP 1 x . Denote F2m = F and F2 = K. Then 1. xnP x + x−1 is irreducible over F if and only if TrF/K (c1/c0) = 1; 2. xnP ∗x + x−1 is irreducible over F if and only if TrF/K (cn−1/cn) = 1. 3.2.9 Remark Part 1 of Theorem 3.2.8 was obtained by Meyn and by Kyuregyan in the present general form; for the case q = 2 it was earlier obtained by Varshamov and Garakov . 3.2.10 Theorem Let q be an odd prime power. If P is an irreducible polynomial of degree n over Fq, then xnP x + x−1 is irreducible over Fq if and only if the element P(2)P(−2) is a non-square in Fq. 3.2.11 Theorem Let q be odd, P(x) ̸= x be an irreducible polynomial of degree n ≥1 over Fq, and ax2 + bx + c and dx2 + rx + h be relatively prime polynomials in Fq[x] with a or d being non-zero and r2 ̸= 4dh. Suppose (ah)2 + (cd)2 + acr2 + b2dh −bcdr −abhr −2acdh = δ2 for some δ ̸= 0 from Fq. Then the polynomial F(x) = dx2 + rx + h n P ax2 + bx + c dx2 + rx + h is irreducible over Fq if and only if the element r2 −4dh n P br −2 (cd + ah −δ) r2 −4hd P br −2 (cd + ah + δ) r2 −4hd is a non-square in Fq. 3.2.12 Remark The case a = c = r = 1 and b = d = h = 0 of Theorem 3.2.11 reduces to Theorem 3.2.10. 3.2.13 Remark We brieﬂy describe some constructive aspects of irreducibility of certain types of polynomials, particularly binomials and trinomials. 3.2.14 Deﬁnition A binomial is a polynomial with two nonzero terms, one of them being the constant term. 3.2.15 Remark Irreducible binomials can be characterized explicitly. For this purpose it suﬃces to consider nonlinear, monic binomials. 3.2.16 Theorem Let t ≥2 be an integer and a ∈F∗ q. Then the binomial xt −a is irreducible in Fq[x] if and only if the following two conditions are satisﬁed: 1. each prime factor of t divides the order e of a in F∗ q, but not (q −1)/e; 2. q ≡1(mod 4) if t ≡0(mod 4). 3.2.17 Remark Theorem 3.2.16 was essentially shown by Serret for ﬁnite prime ﬁelds. Fur-ther characterizations of irreducible binomials can be found in Albert [70, Chapter 5], Cap-peli [505, 506, 507], Dickson [850, Part I, Chapter 3]. Lowe and Zelinsky , R´ edei [2443, Chapter 11], and Schwarz . 3.2.18 Theorem Let a be a nonzero element in an extension ﬁeld of Fq, q = 2Au −1, with A ≥2 and u odd. Suppose e is the order of the subgroup of F∗ q generated by a and the condition 62 Handbook of Finite Fields of Part 1 in Theorem 3.2.16 is satisﬁed for some natural number t divisible by 2A = 2B. Then the binomial xt −a factors as a product of B monic irreducible polynomials in Fq[x] of degrees υ = t/B, that is in Fq[x] we have the canonical factorization xt −a = B Y j=1 xυ −bcjxυ/2 −b2 , where b = ar, 2Br = e/2 + 1(mod (q −1)), and the elements c1, . . . , cB are the roots of the polynomial F(x) = B/2 X i=0 (B −i −1)!B i!(B −2i)! xB−2i ∈Fq[x], and all cj, 1 ≤j ≤B are in Fq. 3.2.19 Remark The factorization in Theorem 3.2.18 is due to Serret , see also Albert [70, Chapter 5] and Dickson [850, Part I, Chapter 3]. Shiva and Allard discuss a method for factoring x2k−1 + 1 over F2. The factorization of xq−1 −a over Fq is considered in Dickson , see also Agou . Schwarz has a formula for the number of monic irreducible factors of ﬁxed degree for a given binomial and R´ edei gives a short proof of it; see also Agou , Butler , and Schwarz . Gay and V´ elez prove a formula for the degree of the splitting ﬁeld of an irreducible binomial over an arbitrary ﬁeld that was shown by Darbi for ﬁelds of characteristic 0. Agou studied the factor-ization of an irreducible binomial over Fq in an extension ﬁeld of Fq. Beard and West and McEliece tabulate factorizations of the binomials xn −1. The factorization of more general polynomials g(x)t −a over ﬁnite prime ﬁelds is considered in Ore and Petterson . Applications of factorizations of binomials are contained in Agou , Berlekamp , and Vaughan . 3.2.20 Deﬁnition A trinomial is a polynomial with three nonzero terms, one of them being the constant term. 3.2.21 Remark The trinomials that we consider are also aﬃne polynomials. 3.2.22 Theorem Let a ∈Fq and let p be the characteristic of Fq. Then the trinomial xp−x−a is irreducible in Fq[x] if and only if it has no root in Fq. 3.2.23 Corollary With the notation of Theorem 3.2.22 the trinomial xp −x−a is irreducible in Fq[x] if and only if the absolute trace TrF/K(a) ̸= 0, where F = Fq and K = Fp. 3.2.24 Remark Theorem 3.2.22 and Corollary 3.2.23 were ﬁrst shown by Pellet . The fact that xp−x−a is irreducible over Fp if a ∈F∗ p was already established by Serret [2597, 2600]. See also Dickson , [850, Part I, Chapter 3] and Albert [70, Chapter 5] for these results. 3.2.25 Remark Since for b ∈F∗ p the polynomial f(x) is irreducible over Fq if and only if f(bx) is irreducible over Fq, the criteria above hold also for trinomials of the form bpxp −bx −a. 3.2.26 Remark If we consider more general trinomials of the above type for which the degree is a higher power of the characteristic, then these criteria need not be valid any longer. In fact, the following decomposition formula can be established. 3.2.27 Theorem For xq −x −a with a being an element of the subﬁeld K = Fr of F = Fq we have the decomposition xq −x −a = q/r Y j=1 (xr −x −βj) Irreducible polynomials 63 in Fq[x], where the βj are the distinct elements of Fq with TrF/K(βj) = a. 3.2.28 Remark Theorem 3.2.27 is due to Dickson , [850, Part I, Chapter 3], but in the special case a = 0 it was already noted by Mathieu . 3.2.29 Theorem [2857, 2858] Let P(x) = xn+an−1xn−1+· · ·+a1x+a0 be an irreducible polynomial over the ﬁnite ﬁeld Fq of characteristic p and let b ∈Fq. Then the polynomial P(xp −x −b) is irreducible over Fq if and only if the absolute trace TrF/K(nb −an−1) ̸= 0, where F = Fq and K = Fp. 3.2.30 Remark Theorem 3.2.29 was shown in this general form by Varshamov [2857, 2858]; see also Agou . The case b = 0 received considerable attention much earlier. The corresponding result for b = 0 and ﬁnite prime ﬁelds was stated by Pellet and proved in Pellet . Polynomials f(xp −x) over Fp with deg(f) a power of p were treated by Serret [2598, 2599]. The case b = 0 for arbitrary ﬁnite ﬁelds was considered in Dickson [850, Part I, Chapter 3] and Albert [70, Chapter 5]. More general types of polynomials such as f(xpr −ax), f(xp2r −axpr −bx) and others have also been studied, see Agou [36, 37, 38, 39, 40, 41, 42], Cohen , Long [1953, 1954, 1955, 1956], Long and Vaughan [1957, 1958], and Ore . 3.2.31 Theorem Let f(x) = xr −ax −b ∈Fq[x], where r > 2 is a power of the characteristic of Fq, and suppose that the binomial xr−1 −a is irreducible over Fq. Then f(x) is the product of a linear polynomial an an irreducible polynomial over Fq of degree r −1. 3.2.32 Remark Theorem 3.2.31 generalizes results of Dickson and Albert [70, Chapter 5]. See Schwarz for further results in this direction. 3.2.2 Recursive constructions 3.2.33 Theorem Let q = ps be a prime power and let f(x) = Pn u=0 cuxu be a monic irreducible polynomial over Fq. Denote Fq = F and Fp = K. Suppose that there exists an element δ0 ∈Fq such that f (δ0) = a with a ∈F∗ p, and TrF/K (nδ0 + cn−1) · TrF/K (f ′(δ0)) ̸= 0, where f ′ is the formal derivative of f. Let g0(x) = xp−x+δ0 and gk(x) = xp−x+δk, where δk ∈F∗ p, k ≥1. Deﬁne f0(x) = f (g0(x)), and fk(x) = f ∗ k−1 (gk(x)) for k ≥1, where f ∗ k−1(x) is the monic reciprocal polynomial of fk−1(x), i.e., f ∗ k−1(x) = 1 fk−1(0)xnk−1fk−1 1 x . Then for each k ≥0 the polynomial fk(x) is irreducible over Fq of degree nk = n · pk+1. 3.2.34 Remark The case s = 1 and the sequence (δk)k≥0 is constant, i.e., δk = δ ∈F∗ p of Theo-rem 3.2.33, has been studied by Varshamov in , where no proof is given. For a proof see [1172, 2077]. 3.2.35 Theorem [1820, 1821, 1823] Let δ ∈F∗ 2s and f1(x) = Pn u=0 cuxu be a monic irreducible polynomial over F2s whose coeﬃcients satisfy the conditions TrF/K c1δ c0 = 1 and TrF/K cn−1 δ = 1, where F2s = F and F2 = K. Then all the terms in the sequence (fk(x))k≥1 deﬁned as fk+1(x) = x2k−1nfk x + δ2x−1 , k ≥1, are irreducible polynomials over F2s. 64 Handbook of Finite Fields 3.2.36 Theorem [1820, 1821, 2077] Let f(x) = Pn i=0 cixi be irreducible over F2m of degree n. Denote F2m = F and F2 = K. Suppose that TrF/K (c1/c0) ̸= 0 and TrF/K (cn−1/cn) ̸= 0. Deﬁne the polynomials ak(x) and bk(x) recursively by a0(x) = x, b0(x) = 1 and for k ≥1 ak+1(x) = ak(x)bk(x), bk+1(x) = a2 k(x) + b2 k(x). Then fk(x) = (bk(x))n f (ak(x)/bk(x)) is irreducible over F2m of degree n2k for all k ≥0. 3.2.37 Remark 1. For the case q = 2 in Theorem 3.2.36 the trace function is the identity map on Fq. 2. Let f(x) = Pn i=0 cixi be a monic irreducible polynomial over F2 of degree n with c1cn−1 ̸= 0. Then fk(x) = Pn i=0 ciai k(x)bn−i k (x) is irreducible over F2 of degree n2k for all k ≥0. 3. The irreducibility of fk over F2 has been studied by several authors, includ-ing Varshamov , Wiedemann , Meyn , Gao , Menezes et al. . 4. The irreducibility of fk over F2s has been studied by Kyuregyan [1820, 1819, 1821] and Menezes et al. [1172, 2077]. 3.2.38 Theorem [678, 2077] Let f be a monic irreducible polynomial of degree n ≥1 over Fq, q odd, where n is even if q ≡3(mod 4). Suppose that f(1)f(−1) is a non-square in Fq. Deﬁne f0(x) = f(x), fk(x) = (2x)tk−1fk−1 x + x−1 /2 , k ≥1, where tk = n2k denotes the degree of fk(x). Then fk(x) is an irreducible polynomial over Fq of degree n2k for every k ≥1. 3.2.39 Remark Further constructions similar to the one from Theorem 3.2.38 can be found in [1822, 1824]. 3.2.40 Theorem Let P(x) ̸= x be an irreducible polynomial of degree n ≥1 over Fq, where n is even if q ≡3(mod 4), with r, h, δ ∈Fq and r ̸= 0, δ ̸= 0. Suppose that P 2δ−rh r2 P −2δ+rh r2 is a non-square in Fq. Deﬁne F0(x) = P(x), Fk(x) = 2x + 2h r tk−1 Fk−1 x2 + 4δ2 −(hr)2 r4 . 2x + 2h r , k ≥1, where tk = n2k denotes the degree of Fk(x). Then Fk(x) is an irreducible polynomial over Fq of degree n2k for every k ≥1. 3.2.41 Remark For r = δ = 2 and h = 0 Theorem 3.2.40 coincides with Theorem 3.2.38 due to Cohen [678, 685]; see also [2077, Theorem 3.24]. 3.2.42 Theorem Let P(x) ̸= x be an irreducible polynomial of degree n ≥1 over Fq. Suppose that the elements P(0), hn and (2r)n are squares in Fq and the element P 2h r is Irreducible polynomials 65 a non-square in Fq. Deﬁne F0(x) = P(x), Fk(x) = (Fk−1(0))−1 (rx + 2h)2 4h !tk−1 Fk (4h)2 x (rx + 2h)2 ! , k ≥1, where tk = n2k denotes the degree of Fk(x). Then Fk(x) is an irreducible polynomial over Fq of degree n2k for every k ≥1. 3.2.43 Theorem Let P be an irreducible polynomial of degree n ≥1 over Fq, where n is even if q ≡3(mod 4) and b ∈Fq. Suppose that the element P −b 2 is a non-square in Fq. Deﬁne F0(x) = P(x), Fk(x) = Fk−1 x2 + bx + b2 4 −b 2 , k ≥1. Then for every k ≥1, Fk(x) is an irreducible polynomial over Fq of degree n2k. 3.2.44 Deﬁnition For a given polynomial P(x) = Pn u=0 auxu ∈Fq[x] deﬁne the polynomial gP as gP (x) = (−1)n n X j=0 2j X u=0 (−1)uaua2j−uxj. 3.2.45 Theorem Let q be an odd prime power and P(x) = Pn u=0 auxu be an irreducible polynomial of degree n > 1 over Fq with at least one coeﬃcient a2i+1 ̸= 0 0 ≤i ≤⌊n 2 ⌋ . Let ax2 + 2hx + ahd−1 and dx2 + 2ax + h be relatively prime, where a, d, h, ∈F∗ q and a2 ̸= hd. Suppose that the element hd−1n is a non-zero square in Fq and the element hd −a2n gF0 h d is non-square in Fq (see Deﬁnition 3.2.44 for gP ). Deﬁne F0(x) = P(x), Fk(x) = Hk−1(a, d)−1 dx2 + 2ax + h tk−1 Fk−1 ax2+2hx+ahd−1 dx2+2ax+h , for k ≥1, where Hk−1(a, d) = dtk−1Fk−1 a d , and tk is the degree of Fk(x). Then Fk(x) is an irre-ducible polynomial over Fq of degree tk = n2k for every k ≥1. 3.2.46 Theorem Let q and P satisfy the hypothesis of Theorem 3.2.45. Suppose a, c ∈F∗ q, (ac)n is a square in Fq and the element (−1)n gF0 c a is a non square in Fq (see Deﬁni-tion 3.2.44 for gP ). Deﬁne F0(x) = P(x), Fk(x) = (2x)tk−1Fk−1 ax2 + c 2ax , k ≥1, where tk is the degree of Fk(x). Then Fk(x) is an irreducible polynomial over Fq of degree tk = n2k for every k ≥1. 3.2.47 Remark In particular, the case q ≡3(mod 4)) and F0(x) = x2 + 2x + c with a = 1 of Theorem 3.2.46 was considered by McNay . The case a = c = 1 was derived by Cohen; see [678, 685, 2077]. 66 Handbook of Finite Fields See Also §3.3 For composite polynomials. §3.4 For weights of irreducible polynomials. §3.5 For irreducible polynomials with prescribed coeﬃcients. References Cited: [33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 70, 213, 229, 231, 469, 505, 506, 507, 589, 666, 671, 678, 685, 769, 841, 849, 850, 1172, 1261, 1819, 1820, 1821, 1822, 1823, 1824, 1825, 1939, 1953, 1954, 1955, 1956, 1957, 1958, 1962, 2022, 2046, 2077, 2091, 2323, 2324, 2376, 2377, 2378, 2392, 2442, 2443, 2568, 2569, 2571, 2597, 2598, 2599, 2600, 2616, 2857, 2858, 2859, 2860, 2861, 2863, 2977] 3.3 Conditions for reducible polynomials Daniel Panario, Carleton University We present qualitative results on the reducibility of univariate polynomials over ﬁnite ﬁelds. First we provide some classical work; see the comments at the end of Chapters 3 and 4 of Lidl and Niederreiter for other results published before 1983. Then we cover several reducibility results that follow from a theorem of Pellet and Stickelberger [2379, 2716]. 3.3.1 Composite polynomials 3.3.1 Remark There has been substantial work showing that some classes of polynomials are irreducible using several types of composition of polynomials. Here we are interested in “if and only if” irreducibility statements that as a consequence provide reducibility results. 3.3.2 Remark Let f be a polynomial of degree m over Fq, q = pk, and let L be the linearized polynomial L(x) = Pn i=0 aixpi. Ore considers the irreducibility of f(L). Agou in several articles [37, 39, 41] considers special types of linearized polynomials including xpr−ax and xp2r −axpr −bx. 3.3.3 Theorem Let f(x) = xm + bm−1xm−1 + · · · + b0 be an irreducible polynomial over Fq, q = pk, with root β. Then, for any nonzero a in Fq, f(xp −ax) is irreducible over Fq if and only if a(q−1) gcd(m,p−1)/(p−1) = 1 and Trkm(β/Ap) ̸= 0, where A ∈Fqm satisﬁes Ap−1 = a and Trkm(x) = x + xp + · · · + xpkm−1. In particular, if A is in Fq, then f(xp −Ap−1x) is irreducible over Fq if and only if Trk(bm−1/Ap) ̸= 0. 3.3.4 Remark Similar results can be found in the papers by Agou cited above. 3.3.5 Remark We are also interested in results that guarantee classes of polynomials that are reducible. The concluding reducibility result for compositions of the type f(L), where f and L are as above, is given next. 3.3.6 Theorem Let f be an irreducible polynomial of degree m over Fq, q = pk, and let L be the linearized polynomial L(x) = Pn i=0 aixpi. If n ≥3, f(L) is reducible. 3.3.7 Remark The cases when n ≤2 in the previous theorem were studied by Agou [39, 40]. 3.3.8 Remark Cohen [671, 672] gives alternative proofs for Agou’s results. Irreducible polynomials 67 3.3.9 Remark Moreno considers the irreducibility of a related composition of functions. 3.3.10 Theorem Let f and g be polynomials over Fq, q = pk, and let f be irreducible of degree m. The polynomial f(g(x)) is irreducible over Fq if and only if g(x)+β is irreducible over Fqm for any root β of f. 3.3.11 Remark Brawley and Carlitz deﬁne root-based polynomial compositions called com-posed products. 3.3.12 Deﬁnition Let f and g be monic polynomials in F∗ q with factorizations in the algebraic closure of Fq, given by f(x) = Y α (x −α) and g(x) = Y β (x −β). The composed products f ◦g and f ⋆g are deﬁned, respectively, by (f ◦g)(x) = Y α Y β (x −αβ), and (f ⋆g)(x) = Y α Y β (x −(α + β)). 3.3.13 Theorem The composed products of f and g, f ◦g and f ⋆g, are irreducible if and only if f and g are irreducible with coprime degrees. 3.3.14 Remark Related results to composed products can be found in [394, 395, 2103]. 3.3.2 Swan-type theorems 3.3.15 Remark Pellet and Stickelberger relate the parity of the number of irreducible factors of a squarefree polynomial with its discriminant. When the parity of the number of irreducible factors of a polynomial is even, the polynomial is reducible. 3.3.16 Remark We recall, from Section 2.1, the deﬁnition of discriminant (Deﬁnition 2.1.135). 3.3.17 Deﬁnition Let f be a polynomial of degree n in Fq[x] with leading coeﬃcient a, and with roots α1, α2, . . . , αn in its splitting ﬁeld, counted with multiplicity. The discriminant of f is given by D(f) = a2n−2 Y 1≤i<j≤n (αi −αj)2. 3.3.18 Remark The discriminant of f is zero if and only if f has multiple roots. 3.3.19 Remark Next is a result given by Stickelberger although the theorem was originally shown by Pellet ; see also . 3.3.20 Theorem [2379, 2716] Let p be an odd prime and suppose that f is a monic polynomial of degree n with integral coeﬃcients in a p-adic ﬁeld F. Let ¯ f be the result of reducing the coeﬃcients of f modulo p. Assume further that ¯ f has no repeated roots. If ¯ f has r irreducible factors over the residue class ﬁeld, then r ≡n (mod 2) if and only if D(f) is a square in F. 3.3.21 Proposition Let q be a power of an odd prime p and let Fq be the ﬁnite ﬁeld with q elements. Let g be a polynomial over Fq of degree n with no repeated roots. Furthermore, 68 Handbook of Finite Fields let r be the number of irreducible factors of g over Fq. Then r ≡n (mod 2) if and only if D(f) is a square in Fq. 3.3.22 Remark Swan extends the previous result to the case p = 2 by noting that a p-adic integer a coprime to p, is a p-adic square if and only if a is a square modulo 4p. 3.3.23 Corollary Let g be a polynomial of degree n over F2 with D(g) ̸= 0 and let f be a monic polynomial over the 2-adic integers such that g is the reduction of f modulo 2. Furthermore, let r be the number of irreducible factors of g over F2. Then r ≡n (mod 2) if and only if D(f) ≡1 (mod 8). 3.3.24 Remark Swan also characterizes the parity of the number of irreducible factors of a trino-mial over F2. These polynomials are of practical importance when implementing ﬁnite ﬁeld extensions; for example, see and Section 3.4. 3.3.25 Theorem Let n > k > 0. Assume precisely one of n, k is odd. If r is the number of irreducible factors of f(x) = xn + xk + 1 ∈F2[x], then r is even (and hence f is not irreducible) in the following cases: 1. n even, k odd, n ̸= 2k and nk/2 ≡0, 1 (mod 4); 2. n odd, k even, k ∤2n and n ≡3, 5 (mod 8); 3. n odd, k even, k \| 2n and n ≡1, 7 (mod 8). In other cases f has an odd number of factors. 3.3.26 Remark The case when n and k are both odd can be covered by making use of the fact that the reciprocal polynomial of f has the same number of irreducible factors as f; for reciprocal polynomials see Deﬁnition 2.1.48. If both n and k are even the trinomial is a square and has an even number of irreducible factors. 3.3.27 Corollary Let n be a positive integer divisible by 8. Then every trinomial over F2 of degree n has an even number of irreducible factors in F2[x], and hence it is not irreducible. 3.3.28 Remark Many results have been given following Swan’s technique for other types of poly-nomials and ﬁnite ﬁeld extensions and characteristics. We state several of them starting from characteristic two results and then focusing on odd characteristic. 3.3.29 Remark Vishne considers trinomials in ﬁnite extensions of F2; see also Theorems 6.69 and 6.695 in . Evaluating the discriminant of a trinomial (modulo 8R, where R is the valuation ring of the corresponding extension of 2-adic numbers), Vishne’s studies are a direct analogue of Swan’s proof over F2. 3.3.30 Corollary Let F2s be an even degree extension of F2 and n an even number. Then, g(x) = xn + axk + b ∈F2s[x] has an odd number of irreducible factors only when g(x) = x2d + axd + b and t2 + at + b has no root in F2s. 3.3.31 Remark Similar results to the one in Corollary 3.3.30 are given in . Special cases of trinomials of low degrees over extensions of F2 are given in . 3.3.32 Remark Hales and Newhart give a Swan-like result for binary tetranomials; see Theorem 2 in . 3.3.33 Remark It is convenient to use irreducible trinomials over F2 when constructing extension ﬁelds. The usage of pentanomials (polynomials with 5 nonzero coeﬃcients) when trinomials do not exist is in the IEEE standard speciﬁcations for public-key cryptography . However, Scott shows that some of the recommended irreducible polynomials are not optimal. Irreducible polynomials 69 3.3.34 Remark Next we show some partial results attempting to characterize the reducibility of binary pentanomials. 3.3.35 Remark Koepf and Kim give a Swan-like result for a class of binary pentanomials of the form xm + xn+1 + xn + x + 1, where m is even and 1 ≤n ≤⌊m/2⌋−1. Ahmadi shows that there are no self-reciprocal irreducible pentanomials of degree n over F2 if n is a multiple of 12. 3.3.36 Problem Characterize completely the reducible pentanomials over F2. 3.3.37 Remark The reducibility of some classes of binary polynomials with more than ﬁve nonzero coeﬃcients has also been studied. In all cases, the polynomials have a special form. 3.3.38 Remark Fredricksen, Hales, and Sweet give a Swan-like result for polynomials of the form xnf(x) + g(x) where f, g ∈F2[x]. 3.3.39 Remark Let n and v ≥2 be coprime positive integers, let r be the least positive residue of n modulo v, and set d = (n −r)/v. A polynomial of the form xrg(xv) + h(xv) with g monic of degree d and h of degree ≤d is an (n, v)-windmill polynomial. These polynomials are related to a method to generate pseudorandom bit sequences by combining linear feedback shift registers in a “windmill” conﬁguration . 3.3.40 Theorem Suppose that n and v are as above and f is a squarefree (n, v)-windmill polynomial over F2m. Let nm(f) denote the number of irreducible factors of f over F2m. 1. If n ≡±1 (mod 8) or m is even, then nm(f) is odd. 2. If n ≡±3 (mod 8) and m is odd, then nm(f) is even. 3.3.41 Theorem Let f(x) = xn + P iS xi + 1 ∈F2[x], where S ⊂{i: i odd, 0 < i < n/3} ∪{i : i ≡n (mod 4), 0 < i < n}. Then f has no repeated roots. If n ≡±1 (mod 8), then f has an odd number of irreducible factors. If n ≡±3 (mod 8), then f has an even number of irreducible factors. 3.3.42 Remark A short proof of Theorem 3.3.41 is given in . 3.3.43 Remark Swan-type results for composite, linearized, and aﬃne polynomials over F2 are given in [1735, 3063] 3.3.44 Problem The complete characterization of reducible polynomials over F2 is a hard open problem. Provide new reducibility characterizations for classes of polynomials over F2. 3.3.45 Remark There have also been reducibility results over ﬁnite ﬁelds of odd characteristic. Binomials over ﬁnite ﬁelds of odd characteristic can be treated easily with a Swan-type approach . 3.3.46 Remark For trinomials over ﬁnite ﬁelds of odd characteristic only partial results are known [46, 1223, 1417]. 3.3.47 Remark Over F3, Loidreau gives the parity for the number of irreducible factors for any trinomial over F3 by examining the discriminant using all possible congruences of n and k modulo 12; see also . This type of analysis holds for higher characteristic, but the number of cases grows quickly with the characteristic, making a complete analysis for large q hard to achieve. 3.3.48 Problem Completely characterize the reducibility of trinomials over ﬁnite ﬁelds of charac-teristic diﬀerent from 2 and 3. 70 Handbook of Finite Fields See Also §3.2 For results on composition of polynomials. §3.4 For results on low weight irreducible polynomials. §4.3 For results on low weight primitive polynomials. §11.2 For counting polynomials with given factorization patterns. §11.3 For irreducibility tests. Chapter 6, for extensions of Swan’s theorem. For results on quadratic and cubic polynomials in extension ﬁelds of characteristic two. Chapter 5, for factorizations of binary trinomials. For a family of reducible polynomials related to word-oriented feedback shift registers; see [827, 828, 2818]. References Cited: [37, 39, 40, 41, 45, 46, 52, 231, 236, 333, 394, 395, 570, 671, 672, 673, 827, 828, 846, 1096, 1223, 1300, 1399, 1417, 1567, 1735, 1777, 1939, 1952, 2103, 2145, 2324, 2379, 2573, 2686, 2716, 2753, 2818, 2878, 3054, 3063] 3.4 Weights of irreducible polynomials Omran Ahmadi, Institute for Research in Fundamental Sciences (IPM) 3.4.1 Basic deﬁnitions 3.4.1 Deﬁnition The weight of a polynomial f(x) = Pn i=0 aixi in Fq[x] is the number of its nonzero coeﬃcients. 3.4.2 Deﬁnition A polynomial in Fq[x] is a binomial, trinomial, tetranomial, or a pentanomial if it has two, three, four, or ﬁve nonzero coeﬃcients, respectively. 3.4.3 Deﬁnition A polynomial in Fq[x] is a fewnomial or a sparse polynomial if it has a small number of nonzero coeﬃcients. 3.4.2 Existence results 3.4.4 Theorem Let the order of a ∈F∗ q be e. Then the binomial xn −a ∈Fq[x], n ≥2, is irreducible over Fq if and only if the following conditions are satisﬁed: 1. if r is a prime number and r\|n, then r \| e and r ∤(q −1)/e; 2. q ≡1 (mod 4) if n ≡0 (mod 4). 3.4.5 Remark The study of irreducible binomials over ﬁnite ﬁelds can be traced back to the works of researchers who were trying to generalize Fermat’s little theorem – see for example the Irreducible polynomials 71 expositions of Poinsot and Serret . Theorem 3.4.4 in the format given above appears in [850, 1938, 1939, 2077]. 3.4.6 Corollary Let n be an odd number, and let a ∈Fq. Then xn −a is irreducible over Fq if and only if for every prime divisor r of n, a is not an r-th power of an element of Fq. 3.4.7 Corollary [1938, 1939, 2077] Let the order of a ∈F∗ q be e, and let r be a prime factor of q −1 which does not divide (q −1)/e. Assume that q ≡1 (mod 4) if r = 2 and k ≥2. Then for any non-negative integer k, xrk −a is irreducible over Fq. 3.4.8 Corollary Let Fq be a ﬁnite ﬁeld of odd characteristic p, p ≥5. There exists an irreducible binomial over Fq of degree m, m ̸≡0 (mod 4), if and only if every prime factor of m is also a prime factor of q −1. For m ≡0 (mod 4) then there exists an irreducible binomial over Fq of degree m if and only if q ≡1 (mod 4) and every prime factor of m is also a prime factor of q −1. 3.4.9 Example 1. x6 −a is irreducible over Fq if and only if a is neither a quadratic nor a cubic residue in Fq. 2. x2k ± 2 are irreducible over F5 . 3. x3k ± 2 and x3k ± 3 are irreducible over F7 . 4. x3k + ω is irreducible over F4 where F4 = F2(ω) . 3.4.10 Remark It is clear that there is no irreducible binomial of degree > 1 over F2. From results above, it follows that x2 +1 is the only irreducible binomial of degree > 1 over F3 and there are inﬁnitely many irreducible binomials over Fq for q ≥4. Also, it is clear that for every q there are inﬁnitely many positive integers m, for example p the characteristic of Fq, such that there is no irreducible binomial of degree m over Fq. This fact motivates the following results. 3.4.11 Theorem Let Fq be of characteristic p. Then the trinomial f(x) = xp −x −b is irreducible over Fq if and only if one of the following equivalent conditions are satisﬁed 1. f has no root in Fq; 2. TrFq(b) ̸= 0; 3. p does not divide [Fq : Fp]. 3.4.12 Corollary Let Fq be of characteristic p. For a, b ∈F∗ q, the trinomial xp −ax −b is irreducible over Fq if and only if a = cp−1for some c ∈Fq and TrFq(a/cp) ̸= 0. 3.4.13 Corollary Let xp −x −a ∈Fq[x] be an irreducible trinomial over Fq of characteristic p, and let α be a root of this polynomial in an extension ﬁeld of Fqp. Then xp −x −aαp−1 is irreducible over Fq(α). 3.4.14 Theorem Let p ≡3 (mod 4) be a prime and let p + 1 = 2rs with s odd. Then, for any integer k ≥1, x2k −2ax2k−1 −1 is irreducible over Fp, and hence irreducible over any odd degree extension Fq, where a = ar is obtained recursively as follows: 1. a1 = 0; 2. for j from 2 to r −1, set aj = aj−1+1 2 (p+1)/4 ; 3. ar = ar−1−1 2 (p+1)/4 . 3.4.15 Remark The following constitutes a partial result towards proving Conjecture 3.4.31. 72 Handbook of Finite Fields 3.4.16 Theorem [667, 2446] Let Fq be a ﬁnite ﬁeld of characteristic p, and let n ≥2 be such that p does not divide 2n(n −1). Let Tn(q) denote the number of a ∈Fq for which the trinomial xn + x + a is irreducible over Fq. Then Tn(q) = q n + O(q1/2), (3.4.1) where the implied constant depends only on n. 3.4.17 Remark There is not any substantial result proving the existence of irreducible fewnomials of weight at least four over ﬁnite ﬁelds. 3.4.18 Remark There are many conjectures concerning the existence of irreducible fewnomials over ﬁnite ﬁelds (see the next section) whose resolution currently seems to be out of reach. The following is a partial result towards proving the existence of irreducible fewnomials over binary ﬁelds. 3.4.19 Theorem There exists a primitive polynomial of degree n over F2 whose weight is n/4 + o(n). 3.4.20 Remark The weight of a monic polynomial of degree n over a ﬁnite ﬁeld is between 1 and n + 1 inclusive. On one end of the weight spectrum we have the monomial x and binomials, whose irreducibility is well understood. We have some partial results concerning the irreducibility of polynomials corresponding to the other end of the weight spectrum. 3.4.21 Theorem The all one polynomial xn + xn−1 + · · · + x2 + x + 1 ∈Fq[x], which is of weight n+1, is irreducible over Fq if and only if n+1 is a prime number and q is a primitive root modulo n + 1. 3.4.22 Example 1. If n = 2, 4, 10, 12, 18, 28, 36, 52, 58, 60, 66, then xn + xn−1 + · · · + x2 + x + 1 is irreducible over F2. 2. Let m > 1 be an integer. If m is even with m ≡0 or 6 (mod 8), then the number of irreducible factors of f(x) = xm + xm−1 + · · · + x3 + x2 + x + 1 is an even number and hence f is reducible over F2 . 3.4.23 Conjecture (Artin) Let a be a non-square integer diﬀerent from 1 and −1. Then there are inﬁnitely many primes p so that a is a primitive element in Fp. 3.4.24 Remark Using Theorem 3.4.21 and Conjecture 3.4.23, naturally we have the following conjecture. 3.4.25 Conjecture There are inﬁnitely many n for which xn +xn−1 +· · ·+x2 +x+1 is irreducible over Fq. 3.4.26 Theorem Artin’s conjecture holds provided that the generalized Riemann hypothesis is true. 3.4.27 Remark The following is an immediate corollary of Theorem 3.4.21 and Theorem 3.4.26. 3.4.28 Theorem There are inﬁnitely many n for which xn + xn−1 + · · · + x2 + x + 1 is irreducible over Fq provided that the generalized Riemann hypothesis is true. 3.4.3 Conjectures 3.4.29 Remark The following conjectures are supported by extensive computations, but it seems that resolving them will be very hard. The ﬁrst three conjectures have become part of Irreducible polynomials 73 folklore and it is very diﬃcult to trace back their origin. Here we provide a reference for the interested readers who wish to have more information about these conjectures. Note that the following is by no means an exhaustive list of conjectures related to the weight distribution of irreducible polynomials over ﬁnite ﬁelds. 3.4.30 Conjecture For every n, there exists a polynomial of degree n and of weight at most ﬁve which is irreducible over F2. 3.4.31 Conjecture Let q ≥3 be a prime power. For every n, there exists a polynomial of degree n and of weight at most four which is irreducible over Fq. 3.4.32 Conjecture The set of positive integers n for which there exists an irreducible trino-mial of degree n over F2 has a positive density in the set of positive integers. 3.4.33 Conjecture The number of irreducible trinomials over F2 of degree at most n is 3n + o(n). 3.4.34 Conjecture Let i, j be two positive integers such that j < i, and let Fi,j(x) = xi+1 + 1 x + 1 + xj. Then the number of irreducible polynomials Fi,j of degree at most n over F2 is 2n + o(n). 3.4.35 Conjecture For every positive integer n, there exists a polynomial g of degree at most logq n + 3 such that f(x) = xn + g(x), called a sedimentary polynomial, is irreducible over Fq. See Also §3.3 For results on reducibility of low weight polynomials. §3.5 For results on polynomials with prescribed coeﬃcients. §4.2 For results on primitive polynomials with prescribed coeﬃcients. §4.3 For results on low weight primitive polynomials. References Cited: [52, 55, 307, 667, 850, 1233, 1531, 1938, 1939, 2077, 2186, 2324, 2356, 2378, 2407, 2446, 2600, 2634] 3.5 Prescribed coeﬃcients Stephen D. Cohen, University of Glasgow 3.5.1 Deﬁnition Given a monic polynomial f(x) = xn + Pn i=1 aixn−i of degree n in Fq[x] then ai is the i-th coeﬃcient. For 0 ≤k, m ≤n, {a1, . . . , ak} are the ﬁrst k coeﬃcients and {an−m+1, . . . , an} are the last m coeﬃcients. In particular −a1 is the trace of f and (−1)nan is the norm of f. 74 Handbook of Finite Fields 3.5.2 Remark Results on the distribution of irreducible polynomials of degree n with certain coeﬃcients prescribed generally fall into the categories (i) existence, (ii) asymptotic esti-mates, (iii) explicit estimates, (iv) exact formulae or expressions. For those (few) in category (iv) further classiﬁcation under (i)–(iii) may be desirable. Historically, the distribution of irreducible polynomials with prescribed ﬁrst and/or last coeﬃcients is approachable by number-theoretic methods [134, 541, 1447, 2834] yielding asymptotic behaviour. Recently, more specialized ﬁnite ﬁeld techniques have reﬁned these into explicit estimates, whilst re-taining asymptotic features. As for existence, in some cases there are theorems which yield the existence of polynomials of degree n with the stronger property of being primitive for all but a few pairs (q, n); see Section 4.2. 3.5.3 Remark To keep notation simple, the number of irreducible polynomials of any particular type will be denoted at that speciﬁc juncture by I (or, for example, I(a, b)). All polynomials will be monic polynomials in Fq[x] (with q a power of the characteristic p) of degree n unless mentioned otherwise. 3.5.4 Remark See Section 3.1 for further formulae for irreducible polynomials. 3.5.1 One prescribed coeﬃcient 3.5.5 Remark For convenience, the results of Theorems 3.1.7, 3.1.8, and 3.1.10 are summarized again here. 3.5.6 Theorem Write n = pjn0, where j ≥0 and p ∤n0. Then the number of irreducible polyno-mials with trace a ∈Fq is I = 1 nq X d\|n0 µ(d)qn/d −ε n X d\|n0 qn/dp, where ε = 1 if a = 0 and j > 0; otherwise ε = 0. 3.5.7 Theorem As in Deﬁnition 3.1.3 set Dn = {r : r\|qn −1; r ∤qi −1, i < n}. For r ∈Dn, write mr = r/ gcd(r, qn−1 q−1 ). Then an irreducible polynomial with norm a ∈F∗ q is such that a has order mr for some r ∈Dn and the number of irreducible polynomials of order r and norm a is I = φ(r) nφ(mr). Thus, the total number of irreducible polynomials of norm a, having order m, is I = 1 nφ(m) X r∈Dn mr=m φ(r). 3.5.8 Remark An asymptotic description of the number in Theorem 3.5.7 occurs in Theorem III.5 of . 3.5.9 Theorem The number I of irreducible polynomials with prescribed non-zero constant term satisﬁes 1 n qn −1 q −1 −2qn/2 ≤I ≤ qn −1 n(q −1). 3.5.10 Remark The next theorem, like Theorem 3.5.9, eﬀectively concerns polynomials with one ﬁxed coeﬃcient, although it involves three coeﬃcients a1, an−1 and an. Irreducible polynomials 75 3.5.11 Theorem Let q be a power of p, where p = 2 or 3 and n = pjn0, p ∤n0. Then the number I(b) of irreducible polynomials, with a1 + an−1/an having prescribed value b, satisﬁes I(b) −qn−1 n ≤3q n 2 n . More generally, this estimate holds for I(0) for any prime power q whenever p ∤n. 3.5.12 Remark The general existence theorem for irreducible polynomials with a prescribed coeﬃ-cient which follows had been conjectured by Hansen and Mullen . The key theoretical work (a dual approach eﬀective for small values of m, n −m, respectively) on weighted (as opposed to exact) numbers of irreducibles is in . 3.5.13 Theorem [1405, 2893] Given integers m, n with 0 ≤m < n and a ∈Fq (non-zero if m = 0) there exists an irreducible polynomial with the coeﬃcient of xm equal to a, except when n = 2, m = 1 and a = 0. 3.5.2 Prescribed trace and norm 3.5.14 Remark By considering each monic reciprocal polynomial f ∗(x) = a−1 n xnf(1/x) (see Def-inition 2.1.48), provided an ̸= 0, one sees that the number of irreducible polynomials with prescribed ﬁrst coeﬃcient a1 and last coeﬃcient an is the same as that with last two coeﬃ-cients a1/an and 1/an, respectively. Hence it suﬃces to consider the ﬁrst of these situations. Equivalently, examine the set of irreducible polynomials with prescribed trace and norm. 3.5.15 Theorem [512, 2893] The number of irreducible polynomials I(a, b) with prescribed trace a and prescribed non-zero norm b satisﬁes I(a, b) − qn−1 n(q −1) ≤3 nqn/2. 3.5.16 Theorem The number I(a, b) of irreducible polynomials of degree n with trace a and non-zero norm b satisﬁes I(0, b) −qn−1 −1 n(q −1) ≤s −1 n q n−2 2 + q n 2 −1 q −1 < 2 q −1q n 2 , where s = gcd(n, q −1), and, when a ̸= 0, I(a, b) − qn −1 nq(q −1) ≤q n−2 2 + q n 2 −1 q(q −1) + n 2 q n−4 4 < 2 q −1q n 2 . 3.5.17 Remark Theorem 3.5.16 is an improvement on Theorem 3.5.15 whenever n ≤3 2qn/2. 3.5.18 Deﬁnition Given a pair (q, n) let P be the largest prime factor of qn −1. Then (q, n) is an lps (largest prime survives) pair if P ∈Dn as in Theorem 3.5.7. 3.5.19 Remark According to , empirical evidence indicates that many pairs (q, n) are lps pairs although there are some “sporadic” pairs that are not. 3.5.20 Theorem Suppose n ≥3 and (q, n) is an lps pair. Then the number I(a, b) of irreducible polynomials with non-zero trace a and non-zero norm b satisﬁes I(a, b) ≥(1 −1 P )(qn −1) −qn−1 −2nq n 2 + 1 n(q −1)2 . 76 Handbook of Finite Fields 3.5.21 Example Take (q, n) = (9, 7). Since 97 −1 = 23 × 547 × 1093, then (9, 7) is an lps pair — though (3, 14) which also relates to the factorization of 97 −1 is not an lps pair! Theorem 3.5.15 yields a lower bound for I(a, b) of 8552.73; Theorem 3.5.16 yields lower bounds of 9216.75 and 9198.47 when a is zero and non-zero, respectively, and, when a ̸= 0, Theorem 3.5.20 yields an improved lower bound of 9411.91. 3.5.22 Remark Even when (q, n) is not an lps pair a classical theorem of Zsigmondy (see e.g., or Wikipedia) implies there is always a prime l ∈Dn except when q is a Mersenne prime and n = 2 and when (q, n) = (2, 6). Given (q, n) with these exceptions, such a prime is a Zsigmondy prime and the largest Zsigmondy prime is the largest of such primes. The authors of Theorem 3.5.20 were unaware of Zsigmondy’s theorem and it is evident that the same argument yields the following modiﬁcation which is generally applicable. It appears here for the ﬁrst time. 3.5.23 Theorem Suppose n ≥3 and (q, n) ̸= (2, 6). Let PZ be the corresponding largest Zsigmondy prime. Then the number I(a, b) in Theorem 3.5.20 satisﬁes I(a, b) ≥ (1 − 1 PZ )(qn −1) −qn−1 −2nq n 2 + 1 n(q −1)2 . 3.5.24 Example Take (q, n) = (47, 4), not an lps pair since q2 −1 = 25 × 3 × 23 and q2 + 1 = 2 × 5 × 13 × 17. Here 5, 13, and 17 are Zsigmondy primes with PZ = 17. Then Theorem 3.5.15 provides no information, whereas Theorem 3.5.16 yields lower bounds for I(a, b) of 504.4 and 515.23, respectively, when a is zero and non-zero. Now, when a ̸= 0, Theorem 3.5.23 delivers an improved lower bound for I(a, b) of 528.25. 3.5.25 Remark also provides an expression for upper bounds for I(a, b). These are always better than those of Theorem 3.5.16 when n is a multiple of q −1. 3.5.26 Remark Existence results for (merely) irreducible polynomials with prescribed trace and norm follow from those on the existence of primitive polynomials (Section 4.2). 3.5.3 More prescribed coeﬃcients 3.5.27 Remark The question of estimating the number of irreducible polynomials f with the ﬁrst k coeﬃcients a1, . . . , ak and the last m coeﬃcients an−m+1, . . . , an of f prescribed (where 2 ≤k + m < n) can be generalized as follows and tackled by number-theoretical methods. Given a monic polynomial M ∈Fq[x] of degree m, where 1 ≤m < n, let R ∈Fq[x] be a (not necessarily monic) polynomial prime to M. Then consider polynomials f of degree n with the ﬁrst k coeﬃcients prescribed and such that f ≡R (mod M). The original question of choosing k ﬁrst and m last coeﬃcients is recovered by selecting M(x) = xm and R(x) = an−m+1xm−1 + · · · + an, an ̸= 0. In this connection, note from Lemma 2.1.113 that Φq(xm) = qm−1(q −1). 3.5.28 Theorem [512, 1551, 2454] Let 2 ≤k + m < n. Suppose a1, . . . , ak ∈Fq, M is a monic polynomial in Fq[x] of degree m, and R ∈Fq[x] is a ﬁxed (not necessarily monic) polynomial of degree < m prime to M. Then the number I of irreducible polynomials f of degree n with prescribed ﬁrst k coeﬃcients and such that f ≡R (mod M) satisﬁes 1 n qn−k Φq(M) −(k + m + 1)q n 2 ≤I ≤1 n qn−k Φq(M) + (1 −δ)(k + m −1)q n 2 , where 0 < δ = 1 − 1 qkΦq(M) < 1. Irreducible polynomials 77 3.5.29 Remark Theorem 3.5.28 leads to explicit existence results such as those which follow. 3.5.30 Corollary Under the conditions of Theorem 3.5.28, suppose that k + m < n/2 and that q > n 2 (n even); q > n + 1 2 2 (n odd). Then there exists an irreducible polynomial f ≡R (mod M) of degree n with its ﬁrst k coeﬃcients prescribed. In particular, there exists an irreducible polynomial with its ﬁrst k coeﬃcients and its last m coeﬃcients prescribed (and non-zero constant term). 3.5.31 Corollary There exists an irreducible polynomial of degree n with its ﬁrst k and last m coeﬃcients prescribed whenever 2 ≤k + m ≤n/3. 3.5.32 Remark In contrast to Corollaries 3.5.30 and 3.5.31 giving existence conditions for ir-reducible polynomials with prescribed ﬁrst and last coeﬃcients, investigates those whose middle coeﬃcients are ﬁxed providing these are all zero. 3.5.33 Theorem Suppose 1 < k ≤m < n and q2k−m−2 ≥q(n −k + 1)4. Then there exists an irreducible polynomial of degree n with ak = · · · = am = 0. 3.5.34 Corollary For any c with 0 < c < 1 and any positive integer n such that (1 −3c)n ≥2 + 8 logq n, there exists an irreducible polynomial of degree n over Fq with any ⌊cn⌋consecutive coeﬃcients (other than the ﬁrst or last) equal to 0. 3.5.35 Example With c = 1/4 in Corollary 3.5.34, there exists an irreducible polynomial of degree n with ⌊n/4⌋consecutive coeﬃcients equal to 0 whenever q ≥61 and n ≥37 and for smaller prime powers for n ≥nq where nq ≤n2 = 266. 3.5.36 Remark When the prescribed coeﬃcients do not comprise ﬁrst and last, or middle coeﬃ-cients, the only general estimates are asymptotic. In the following two theorems n−m ≤n−2 coeﬃcients aj of f are prescribed and given their assigned values. The remaining m coeﬃ-cients A = {aj1, . . . , ajm}, say, are allowed to take any values in Fq. 3.5.37 Theorem Let the n−m coeﬃcients of f not in A (as in Remark 3.5.36) be prescribed and given their assigned values in Fq. Regard the members of A as algebraically independent indeterminates (or transcendentals). Suppose that f is absolutely irreducible in Fq[x, A]. Then, for suﬃciently large q, the number I of irreducible polynomials f ∈Fq[x] of degree n with the coeﬃcients not in A as prescribed satisﬁes I = cqm + O(qm/2), where c satisﬁes 1/n ≤c < 1. 3.5.38 Theorem Suppose m ≤n −2 and the n −m coeﬃcients of f not in A are prescribed and given their assigned values (with an ̸= 0, if prescribed). Suppose that there is not a d > 1 such that f ∈Fq(A)[xd]. Assume also that p > n. Then, for suﬃciently large q, the number I as in Theorem 3.5.37 satisﬁes I = 1 nqm + O qm−1 2 , where the implied constant depends only on n. 3.5.39 Remark From , alternative versions of Theorem 3.5.38 apply even if p ≤n. Indeed, f need not be monic, in the sense that the coeﬃcient of xn may be prescribed in F∗ q or allowed to vary in Fq so long as the total number of prescribed coeﬃcients does not exceed n −1. 78 Handbook of Finite Fields 3.5.40 Remark The ﬁnal theorem in this section relates to irreducible polynomials of even degree of a speciﬁc type, namely self-reciprocal polynomials of degree 2n (see Deﬁnition 2.1.48). Note that, if F is a self-reciprocal polynomial of degree 2n, then its last coeﬃcient a2n = 1 and its ﬁrst n coeﬃcients are the same as its last n non-constant coeﬃcients in the sense that ai = a2n−i, i = 1, . . . , n. 3.5.41 Theorem The number I of irreducible self-reciprocal polynomial of degree 2n with the ﬁrst m (< n/2) coeﬃcients prescribed satisﬁes I −qn−m 2n ≤m + 5 n q n 2 +1. See also . 3.5.4 Further exact expressions 3.5.42 Remark Expressions for the number of irreducible polynomials with the ﬁrst two coeﬃcients prescribed are given in . These are described in terms of the function H(a, n), a ∈Fq. Here, if p ∤n, then H(a, n) = qn−2 −λ((−1)l−1la)ql−1 for n = 2l, qn−2 + δ(a)λ((−1)ln)ql−1 for n = 2l + 1, where λ denotes the quadratic character on Fq, δ(a) = −1 (a ̸= 0) and δ(0) = q −1. If p\|n, then H(a, n) = qn−2 −δ(a)λ((−1)l)ql−1 for n = 2l, qn−2 + λ((−1)l2a)ql for n = 2l + 1. Given a1, a2 ∈Fq, in Theorems 3.5.43 and 3.5.45 (which relate, respectively, to odd and even q), I(a1, a2) denotes the number of irreducible polynomials of degree n over Fq with the indicated ﬁrst two coeﬃcients. 3.5.43 Theorem Suppose q is odd. Then, if p ∤n and a ∈Fq, I(0, −a) = 1 n X d\|n µ(d)H(a/d, n/d). Further, if p\|n with n = pjn0 (j ≥1, p ∤n0), then I(0, −a) = 1 n X d\|n0 µ(d)H(a/d, n/d), a ̸= 0, I(0, 0) = 1 n X d\|n0 µ(d)[H(0, n/d) −qn/pd], I(1, 0) = 1 nq2 X d\|n0 µ(d)qn/d. 3.5.44 Remark The general value of I(a1, a2) can be recovered from Theorem 3.5.43 using, if p ∤n, I(a1, a2) = I(0, a2 −(n−1) 2n a2 1). If p\|n and a1 ̸= 0, then I(a1, a2) = I(1, 0). 3.5.45 Theorem Suppose q is even. Then for a ∈Fq, I(0, a) =          1 n X d\|n µ(d)H(a, n/d) for n odd, 1 n X d\|n, d odd µ(d)[H(a, n/d) −q n 2d −1] for n even; Irreducible polynomials 79 I(1, a) = 1 n X d\|n, d odd µ(d)H∗ a + d −1 2 , n d , where, with q = 2r and χ the canonical additive character on Fq, H∗(a, n) =        qn−2 for n = 4l, qn−2 + (−1)lrδ(a)q n−3 2 for n = 4l + 1, qn−2 + (−1)lrδ(1 + a)q n−3 2 for n = 4l −1, qn−2 −(−1)lrχ(a)q n−2 2 for n = 4l + 2. 3.5.46 Remark The general value of I(a1, a2) can be recovered from Theorem 3.5.45 since, for a1 ̸= 0, I(a1, a2) = I(1, a2/a2 1) and I(0, a) = I(0, 1), a ̸= 0. 3.5.47 Remark For the binary ﬁeld F2, contains alternative expressions for the number I(a1, a2) in Theorem 3.5.45. For expressions for the number of irreducible polynomials over F2 with the ﬁrst three coeﬃcients prescribed, see [1076, 3049]. The ﬂavor of these results is caught by the following conjecture which holds for k ≤3. 3.5.48 Conjecture Let n = 2l be even and I(a1, . . . , as) denote the number of irreducible polynomials of degree n (over F2) whose ﬁrst s coeﬃcients are as shown. Then I(a1, . . . , as) = X 2\|n, d odd µ(d)J(n/d, a1, . . . , as), where J(n, a1, . . . , as) = 2n−s + cl−r+12l−r+1 + · · · + cl2l, for some r with 1 ≤r ≤l and ci ∈{−1, 0, 1}. 3.5.49 Remark Let γ be a primitive element of Fq and s\|q −1. A coset C of the subgroup of F∗ q generated by γs takes the form {γis+h : 0 ≤i < (q −1)/s}, for some h with 0 ≤h < s. In general expressions are obtained for the number of irreducible polynomials of degree n with prescribed trace and norm in a speciﬁed coset C. These involve quantities like Gauss and Jacobi sums. Such expressions are made explicit (i.e., the trigonometric sums are precisely determined) in the following cases: 1. s = 2 (compare with ), s = 3 and s = 4; 2. q = p2er and s (> 1) a factor of pe + 1. 3.5.50 Remark For q powers of 2 or 3, contains completely explicit expressions for the number of irreducible polynomials of degrees n in the indicated ranges for polynomials with two prescribed coeﬃcients as follows: 1. a1 any prescribed value, an−1 = 0, for n ≤10; 2. a1 = 0, a3 any prescribed value, for n ≤30. See Also §4.2 For discussion of primitive polynomials with prescribed coeﬃcients. §6.1 For evaluation of Gauss and Jacobi sums appearing in estimates. §13.1 For arithmetical comparisons of irreducibles with primes. References Cited: [134, 310, 512, 541, 565, 669, 685, 1076, 1209, 1210, 1211, 1405, 1416, 1447, 1551, 1783, 1815, 2121, 2123, 2318, 2454, 2467, 2710, 2834, 2893, 3049] 80 Handbook of Finite Fields 3.6 Multivariate polynomials Xiang-dong Hou, University of South Florida 3.6.1 Theorem The polynomial ring Fq[x1, . . . , xk] is a unique factorization domain. Let f(x1, . . . , xk) = a0 + a1xk + · · · + anxn k ∈Fq[x1, . . . , xk], where n > 0, ai ∈Fq[x1, . . . , xk−1], 0 ≤i ≤n, an ̸= 0. Then f is irreducible in Fq[x1, . . . , xk] if and only if it is irreducible in (Fq(x1 . . . , xk−1))[xk] and gcdFqx1,...,xk−1 = 1, where Fq(x1 . . . , xk−1) is the ﬁeld of rational functions in x1, . . . , xk−1 over Fq and gcdFqx1,...,xk−1 denotes the gcd of a0, . . . , an in Fq[x1, . . . , xk−1]. 3.6.2 Deﬁnition For a k-tuple of indeterminates z = (z1, . . . , zk) and for n = (n1, . . . , nk) ∈Nk, where N = {0, 1, 2, · · · }, deﬁne zn = zn1 1 · · · znk k . Let [zn]P(z) denote the coeﬃcient of zn in a formal power series P(z). 3.6.1 Counting formulas 3.6.3 Deﬁnition Let Nk = Fq[x1, . . . , xk]/∼, where for f, g ∈Fq[x1, . . . , xk], f ∼g means that f = cg for some c ∈F∗ q. Elements of Nk are normalized polynomials in Fq[x1, . . . , xk]. Let Nk(m) = {f ∈Nk : deg f = m}, where deg f denotes the total degree of f. Let Nk(m) = \|Nk(m)\| = 1 q−1 h q( m+k k ) −q( m+k−1 k )i , Ik(m) = \|{f ∈Nk(m) : f is irreducible}\|, Pk(m; n) = \|{(f, g) ∈Nk(m) × Nk(n) : gcd(f, g) = 1}\|. 3.6.4 Remark We have that Nk(m) is the number of normalized polynomials of (total) degree m in Fq[x1, . . . , xk], Ik(m) is the number of normalized irreducible polynomials of degree m, and Pk(m; n) is the number of relatively prime pairs of normalized polynomials of degrees m and n, respectively. 3.6.5 Theorem We have Ik(0) = 0, and for m > 0, Ik(m) is given by the recursive formula Ik(m) = Nk(m) − X 1a1+2a2+···+(m−1)am−1=m Ik(1) + a1−1 a1 · · · Ik(m−1)+am−1−1 am−1 . 3.6.6 Theorem Let k ≥1 be a ﬁxed integer. Deﬁne N(z) = P n≥0 Nk(n)zn and I(z) = P n≥1 Ik(n)zn. Then 1. I(z) = P m≥1 µ(m) m log N(zm). 2. Ik(n) = P m\|n µ(m) m [zn/m] log N(z). 3.6.7 Theorem We have Pk(m; n) = X 0≤d≤min{m,n} Nk(m −d)Nk(n −d)Ak(d), where Ak(d) = X 1a1+2a2+···+dad=d (−1)a1+···+ad Ik(1) a1 · · · Ik(d) ad . 3.6.8 Remark In Theorem 3.6.7, when k ≥2, no closed formula for Ak(d) is known. When k = 1, it is known that A1(0) = 1, A1(1) = −q, and A1(d) = 0 for d ≥2 [537, 668]. Irreducible polynomials 81 3.6.2 Asymptotic formulas 3.6.9 Theorem Let k ≥2 and t ≥0 be integers. Then as m →∞, Ik(m) = t X i=0 αiNk(m −i) + O q( m−t−1+k k ) , where α0 = 1, and for i > 0 αi = X 1a1+···+iai=i (a1 + · · · + ai)! a1! · · · ai! (−1)a1+···+aiNk(1)a1 · · · Nk(i)ai, where the constant in the O-term depends only on q, k, t. 3.6.10 Remark There is a fundamental diﬀerence between the distribution of irreducible polyno-mials over Fq in the univariate case and in the multivariate case. For the univariate case, limm→∞ I1(m) N1(m) = 0. For k ≥2, limm→∞ Ik(m) Nk(m) = 1. 3.6.11 Theorem Let k ≥2. Then limm+n→∞ Pk(m;n) Nk(m)Nk(n) = 1. 3.6.12 Remark We have for the univariate case P1(m;n) N1(m)N1(n) = 1−1 q; see Section 11.2 for more results on counting univariate polynomials. 3.6.13 Theorem Let k ≥2 and t ≥0 be ﬁxed integers. Then Pk(m; n) = t X d=0 Nk(m −d)Nk(n −d)Ak(d) + O Nk(m −t −1)Nk(n −t −1) , where Ak(d) is deﬁned in Theorem 3.6.7. The constant in the O-term depends only on q, k, t. 3.6.3 Results for the vector degree 3.6.14 Deﬁnition Let f ∈Fq[x1, . . . , xk]. The vector degree of f, denoted by Def f, is the k-tuple (degx1 f, . . . , degxk f). 3.6.15 Deﬁnition For m = (m1, . . . , mk) and n = (n1, . . . , nk) ∈Nk, let Nk(m) = {f ∈Nk : Deg f = m}, Nk(m) = \|Nk(m)\|, Ik(m) = \|{f ∈Nk(m) : f is irreducible}\| and Pk(m; n) = \|{(f, g) ∈Nk(m) × Nk(n) : gcd(f, g) = 1}\|. 3.6.16 Remark We have that Nk(m) is the number of normalized polynomials of vector degree m in Fq[x1, . . . , xk], Ik(m) is the number of normalized irreducible polynomials of vector degree m, and Pk(m; n) is the number of relative prime pairs of normalized polynomials of vector degrees m and n, respectively. 3.6.17 Remark [664, 1546] For m = (m1, . . . , mk) ∈Nk, Nk(m) = 1 q −1 X (δ1,...,δk)∈{0,1}k (−1)k+δ1+···+δkq(m1+δ1)···(mk+δk). 82 Handbook of Finite Fields 3.6.18 Deﬁnition Let o = (0, . . . , 0) ∈Nk. For i = (i1, . . . , ik), m = (m1, . . . , mk) ∈Nk, deﬁne i ≤m if ij ≤mj for all 1 ≤j ≤k. 3.6.19 Theorem We have Ik(o) = 0, and for m > o, Ik(m) is given by the recursive formula Ik(m) = Nk(m) − X (ai)o<i<m P i aii=m Y i Ik(i) + ai −1 ai . 3.6.20 Theorem Let k ≥1 be a ﬁxed integer and let z = (z1, . . . , zk) be a k-tuple of indeterminates. Deﬁne N(z) = P n∈Nk Nk(n)zn, I(z) = P o̸=n∈Nk Ik(n)zn. Then 1. I(z) = P m≥1 µ(m) m log N(zk). 2. Ik(z) = P m\|gcd(n) µ(m) m [z 1 m n] log N(z). 3.6.21 Theorem Let k ≥2 and (m1, . . . , mk) ∈(Z+)k with m1 = max1≤i≤k−1 mi. Further assume that m1 ≥3 if k = 2 and m1 ≥2 if k = 3. Then Ik(m1, . . . , mk) = Nk(m1, . . . , mk) −qNk(m1, . . . , mk−1, mk −1) + O qm1(m2+1)···(mk+1) . 3.6.22 Remark The asymptotic formula in Theorem 3.6.21 is interesting only when mk > m1 since otherwise the O-term is bigger than or comparable to the term qNk(m1, . . . , mk−1, mk −1). When mk > m1, Theorem 3.6.21 indicates that most of the reducible polynomials in Nk(m1, . . . , mk) are of the form (xk + α)f for some α ∈Fq, f ∈Nk(m1, . . . , mk−1, mk −1). 3.6.23 Corollary Under the assumptions of Theorem 3.6.21, we have Ik(m1, . . . , mk) = (1 −q−Mk)N(m1, . . . , mk) + O qm1(m2+1)···(mk+1) , where Mk = (m1 + 1) · · · (mk−1 + 1) −1. The main term here was given by Cohen [664, Theorem 1]. For ﬁxed m1, . . . , mk−1, it is not the case that almost all polynomials in Nk(m1, . . . , mk) are irreducible: we have Ik(m1,...,mk) Nk(m1,...,nk) →1 −q−Mk as mk →∞. 3.6.24 Remark For k ≥2, Ik(m1,...,mk) Nk(m1,...,mk) →1 as both m1, mk →∞. 3.6.25 Theorem For m = (m1, . . . , mk), n = (n1, . . . , nk) ∈Nk, deﬁne min{m, n} = (min{m1, n1}, . . . , min{mk, nk}) ∈Nk. We have Pk(m; n) = X 0≤d≤min{m,n} Nk(m −d)Nk(n −d)Ak(d), where Ak(d) = X (ai)0<i≤d P 0<i≤d aii=d (−1) P 0<i≤d ai Y 0<i≤d Ik(i) ai . 3.6.26 Theorem Let k ≥2 and m = (m1, . . . , mk), n = (n1, . . . , nk) ∈Nk. Then limmk−1,mk→∞ Pk(m;n) Nk(m)Nk(n) = 1. 3.6.27 Theorem Let k ≥2, m = (m1, . . . , mk), n = (n1, . . . , nk), t = (t1, . . . , tk) ∈Nk such that t ≤min{m, n} and max{mi, ni} > 2ti + 1 for all 1 ≤i ≤k. Then Pk(m; n) = X 0≤d≤t Nk(m −d)Nk(n −d)Ak(d) + O q max1≤j≤k mj −tj mj +1 Qk i=1(mi+1)+ nj −tj nj +1 Qk i=1(ni+1) . Irreducible polynomials 83 3.6.28 Theorem Let k ≥2, m = (m1, . . . , mk), n = (n1, . . . , nk) ∈Nk such that mk > 0, nk > 0 and max{mi, ni} > 1 for all 1 ≤i ≤k −1. Then Pk(m; n) = Nk(m)Nk(n) −qNk(m1, . . . , mk−1, mk −1)Nk(n1, . . . , nk−1, nk −1) + O q max1≤j≤k−1 mj mj +1 Qk i=1(mi+1)+ nj nj +1 Qk i=1(ni+1) . 3.6.29 Theorem Let k ≥2 and m = (m1, . . . , mk) ∈Nk with max1≤i≤k−1 mi ≥1. Then as mk →∞(with m1, . . . , mk−1 ﬁxed), Pk(m;m) Nk(m)2 →1 −q1−2(m1+1)···(mk−1+1). 3.6.4 Indecomposable polynomials and irreducible polynomials 3.6.30 Deﬁnition Let F be a ﬁeld and k ≥2. A nonconstant polynomial f ∈F[x1, . . . , xk] is indecomposable over F if there do not exist h ∈F[x1, . . . , xk] and u ∈F[t] with deg u ≥2 such that f = u(h(x1, . . . , xk)). 3.6.31 Theorem Let k ≥2. Assume that f ∈Fq[x1, . . . , xk] is indecomposable over Fq, the algebraic closure of Fq. For each λ ∈Fq, let Iλ denote the set of all distinct irreducible factors of f −λ over Fq. Then X λ∈Fq (\|Iλ\| −1) ≤min λ∈Fq X g∈Iλ deg g −1. In particular, f −λ is reducible over Fq for at most deg f −1 values of λ. 3.6.32 Remark [338, Theorem 4.2] f ∈Fq[x1, . . . , xk] is indecomposable over Fq if and only if it is indecomposable over Fq. 3.6.33 Deﬁnition Let F be a ﬁeld and ﬁx a term order in F[x1, . . . , xk] that respects the total degree. A monic polynomial f ∈F[x1, . . . , xk] with f(0, . . . , 0) = 0 is monic original. 3.6.34 Remark [338, 1224] Let F be a ﬁeld and let k ≥2. Every monic original polynomial f ∈F[x1, . . . , xk] has a unique decomposition f = u ◦h, where u ∈F[t], h ∈F[x1, . . . , xk] are both monic original and h is indecomposable. 3.6.35 Theorem Let F be an algebraically closed ﬁeld and let k ≥2 and n ≥2 be integers. Denote by l the smallest prime divisor of n. Then the set of all decomposable monic original polynomials in F[x1, . . . , xk] of degree n is an aﬃne algebraic set of dimension k+ n m k +m−3, where m = ( n if k = 2, n l is a prime and n l ≤2l −5, l otherwise. 3.6.36 Deﬁnition Let k ≥2 and n ≥1. Denote the number of monic original (respectively indecomposable monic original, decomposable monic original) polynomials of (total) degree n in Fq[x1, . . . , xk] by Pk,n (respectively Ik,n, Dk,n). 3.6.37 Theorem We have Ik,1 = qk−1, and for n > 1, Ik,n is given by the recursive formula Ik,n = Pk,n − X m\|n, m<n q n m −1 Ik,m. 84 Handbook of Finite Fields 3.6.38 Remark Let k ≥2. We have Ik,n Pk,n →1 when n →∞with q ﬁxed or when q →∞with n ﬁxed. 3.6.39 Theorem Assume k ≥2 and n ≥2. Let l be the smallest prime divisor of n and let m deﬁned in Theorem 3.6.35. Then the following hold. 1. \|Dk,n −αk,n\| ≤αk,nβ∗ k,n where αk,n = q( k+ n m k )+m−3 · 1 −q−( k−1+ n m k−1 ) 1 −q−1 , β∗ k,n = 2 1 −q−1 q−1 2( k−1+ n l k−1 )+1. 2. Ik,n ≥Pk,n −2αk,n. 3.6.40 Remark For a reﬁnement of the bound in Theorem 3.6.39, see [1224, Theorem 4.1], where β∗ k,n is replaced by an expression βk,n which is of smaller magnitude but more complicated; also see for related results. 3.6.5 Algorithms for the gcd of multivariate polynomials 3.6.41 Remark Computation of the gcd of multivariate polynomials over ﬁnite ﬁelds is more diﬃcult than that of univariate polynomials; the Extended Euclidean Algorithm alone is not suﬃcient to produce the gcd due to the fact that Fq[x1, . . . , xk] with k > 1 is no longer a Euclidean domain. In this subsection we gather several algorithms for computing the multivariate gcd based on diﬀerent approaches. 3.6.42 Deﬁnition Let R be a unique factorization domain. For f ∈R[x], lc(f) denotes the leading coeﬃcient of f; pp(f) denotes the primitive part of f, i.e., f divided by the gcd of its coeﬃcients. The resultant of f, g ∈R[x] is denoted by res(f, g). 3.6.43 Algorithm Input: Primitive f, g ∈(Fq[y])[x], where y = (y1, . . . , yk). Output: gcdFqx,y. 1. Use the Extended Euclidean Algorithm to compute the monic v= gcd(Fq(y))x. 2. Compute b = gcdFqy, lc(g)). 3. gcdFqx,y = pp(bv). 3.6.44 Algorithm (Modular bivariate gcd, small primes version) Input: Primitive f, g ∈(Fq[y])[x], degx f = n ≥degx g ≥1, degy f, degy g ≤d and q ≥(4n + 2)d + 2. Output: gcdFqx,y. 1. Compute b = gcdFqy, lcx(g)). Let l = d + 1 + degy b. 2. Repeat steps 3 – 7 until the conditions in step 7 are satisﬁed. 3. Choose S ⊂Fq with \|S\| = 2l. 4. Delete the roots of b from S. For each u ∈S, use the Extended Euclidean Algo-rithm to compute the monic vu = gcdFqx, g(x, u)). 5. Determine e = min{degx vu : u ∈S}. Delete {u ∈S : degx vu > e} from S. If \|S\| ≥l, delete \|S\| −l elements from S; otherwise, go to step 3. Irreducible polynomials 85 6. Compute by interpolation the coeﬃcients in Fq[y] of w, f ∗, g∗∈(Fq[y])[x] of degrees in y less than l such that w(x, u) = b(u)vu, f ∗(x, u)w(x, u) = b(u)f(x, u), g∗(x, u)w(x, u) = b(u)g(x, u) for all u ∈S. 7. Check if degy(f ∗w) = degy(bf) and degy(g∗w) = degy(bg). 8. gcdFqx,y = ppx(w). 3.6.45 Remark Let h = gcdFqx,y. The conditions in Step 7 of Algorithm 3.6.44 are satisﬁed if and only if S does not contain any root of resx(f/h, g/h). 3.6.46 Algorithm (Gr¨ obner basis) Input: f, g ∈Fq[x1, . . . , xk], f, g ̸= 0. Output: gcd(f, g). 1. Compute the reduced Gr¨ obner basis G for the ideal ⟨wf, (1 −w)f⟩of Fq[x1, . . . , xk, w] with respect to an elimination order with x1, . . . , xk smaller than w. 2. lcm(f, g) is the polynomial in G which does not involve w. 3. gcd(f, g) = fg lcm(f,g). 3.6.47 Remark For the correctness of Algorithm 3.6.43, see [1227, Theorem 6.12]. The cost of the Extended Euclidean Algorithm is given in [1227, Theorem 3.11]. For the correctness and the cost of Algorithm 3.6.44, see [1227, Theorem 6.37]. The correctness of Algorithm 3.6.46 is given in [13, p.72]. For the complexity of computing reduced Gr¨ obner bases, see [1227, Section 21.7]. See Also §3.1 For counting univariate irreducible polynomials. §11.4 For factorization algorithms of univariate polynomials. §11.5 For factorization algorithms of multivariate polynomials. References Cited: [13, 226, 336, 337, 338, 537, 664, 668, 1224, 1227, 1244, 1546, 1561, 2214] This page intentionally left blank This page intentionally left blank 4 Primitive polynomials 4.1 Introduction to primitive polynomials............ 87 4.2 Prescribed coeﬃcients .............................. 90 Approaches to results on prescribed coeﬃcients • Existence theorems for primitive polynomials • Existence theorems for primitive normal polynomials 4.3 Weights of primitive polynomials ................. 95 4.4 Elements of high order ............................. 98 Elements of high order from elements of small orders • Gao’s construction and a generalization • Iterative constructions 4.1 Introduction to primitive polynomials Gary L. Mullen, The Pennsylvania State University Daniel Panario, Carleton University 4.1.1 Deﬁnition An element α ∈Fq is a primitive element if α generates the multiplicative group F∗ q of nonzero elements in Fq. 4.1.2 Deﬁnition A polynomial f ∈Fq[x] of degree n ≥1 is a primitive polynomial if it is the minimal polynomial of a primitive element of Fqn. 4.1.3 Theorem The number of primitive polynomials of degree n over Fq is φ(qn −1)/n, where φ denotes Euler’s function. 4.1.4 Remark [1939, Section 3.1] The reciprocal polynomial f ∗(with leading coeﬃcient diﬀerent from 0) of a primitive polynomial f of degree n is deﬁned by f ∗(x) = xnf(1/x); see Deﬁni-tion 2.1.48. The monic reciprocal polynomial of a primitive polynomial is again primitive. In general, for any polynomial f, the order of f ∗is the same as the order of f. For the deﬁnition of order of a polynomial see Deﬁnition 2.1.51. 4.1.5 Theorem [1939, Theorem 3.16] A polynomial f ∈Fq[x] of degree n is primitive if and only if f is monic, f(0) ̸= 0, and the order of f is qn −1. 4.1.6 Theorem [1939, Theorem 3.18] A monic polynomial f ∈Fq[x] of degree n ≥1 is a primitive polynomial if and only if the smallest positive integer r for which xr is congruent modulo f to some element of Fq is r = (qn −1)/(q −1) and (−1)nf(0) is a primitive element in Fq. In case f is primitive over Fq, xr ≡(−1)nf(0) (mod f(x)). 87 88 Handbook of Finite Fields 4.1.7 Theorem Let f be an irreducible polynomial of degree k over Fq. Set m = qk −1 and deﬁne g(x) = (xm −1)/((x −1)f(x)). Then f is primitive if and only if g has exactly (q −1)qk−1 −1 nonzero terms. 4.1.8 Theorem Let f(x) = fk + fk−1x + · · · + f0xk be irreducible over Fq of degree k ≥2. Set m = qk −1, t = m/(q −1), y = xq−1 and g(y) = yt −1 (y −1)f(y) = h(y) + r0 + r1y + · · · + rk−1yk−1 f(y) , where h(y) = ϵt−k+ϵt−k−1y+· · ·+ϵ1yt−k−1. Then f is primitive if and only if the number of nonzero terms in h(y), considered as a polynomial in y over Fq, is equal to qk−1(q−1)−1−N, where N is the number of nonzero terms in the ﬁnite sequence ϵt−k+1, ϵt−k+2, . . . , ϵm deﬁned by ϵt−n = rn −Pk−n−1 i=1 fiϵt−n−i, n = 0, 1, . . . , k −1 where the empty sum is interpreted as 0, and ϵt+n = 1 −Pk i=1 fiϵt+n−i, n = 1, 2 . . . , m −t. 4.1.9 Corollary An irreducible polynomial is primitive if and only if the ﬁnite sequence ϵ1, . . . , ϵm deﬁned in Theorem 4.1.8 contains no two identical periodic subsequences. 4.1.10 Deﬁnition Deﬁne wn(q) and Wn(q) as the minimal weight (the number of nonzero coeﬃcients) among all monic irreducible and primitive, respectively, polynomials of de-gree n over Fq. 4.1.11 Remark It is stated in that wn(p) ≤Wn(p) ≤(n + 1)/2 for any suﬃciently large prime p. For p = 2, wn(2) ≤Wn(2) ≤n/4 + o(n). 4.1.12 Problem Extend these results to Fq when q is a power of a prime. 4.1.13 Conjecture Wn(2) = 3 inﬁnitely often. 4.1.14 Problem Find examples of ﬁelds Fqn with Wn(q) = o(n) or at least with wn(q) = o(n) for inﬁnitely many n. 4.1.15 Remark The following conjectures require the notions of primitive normal polynomials and completely normal primitive polynomials; see Sections 5.2 and 5.4. 4.1.16 Conjecture [1416, 2156] 1. For each prime p and n ≥2, wn(p) ≤5 and if p ̸= 2, 3, then wn(p) ≤4. 2. For each prime p and n ≥2 there is a primitive normal polynomial of degree n over Fp with weight at most 5. 3. If p ≥11 weight 5 can be replaced by weight 4. 4.1.17 Deﬁnition For a prime p ≥3, assume that the ﬁeld Fp consists of the elements 0, ±1, . . . , ±(p −1)/2. The height of a polynomial is the maximum absolute value of its coeﬃcients. We deﬁne hn(p) and Hn(p) as the minimal height of all monic irreducible and primitive, respectively, polynomials of degree n over Fp. 4.1.18 Remark It is stated in that hn(p) = O(p2/3) and Hn(p) = O(pn/(n+1)+ε). The bound hn(p) = O(p2/3) has been improved to hn(p) ≤p1/2+o(1) in . 4.1.19 Problem Improve the above bounds for hn(p) and Hn(p). 4.1.20 Problem Extend the bounds for Fq when q is a power of a prime. Primitive polynomials 89 4.1.21 Theorem For n ≥2 and a ∈F∗ q, there is a primitive normal polynomial of degree n over Fq with trace a. 4.1.22 Theorem There is a primitive normal polynomial of degree n ≥3 with given norm and nonzero trace; see for n ≥5, for n = 4, and for n = 3. 4.1.23 Conjecture For each n ≥2 there is a completely normal primitive polynomial of degree n over Fq. (This is true if n is a prime, or n = 4, or if qn ≤231 with q ≤97.) 4.1.24 Remark For q ≥n log n there is a completely normal primitive polynomial of Fqn over Fq. 4.1.25 Conjecture For any k there is a trinomial f so that f(x), x2k−1 + 1 is a primitive polynomial of degree k over F2. (This conjecture has been proved for k ≤500 .) 4.1.26 Deﬁnition Let f(x) = xn + Pt k=0 akxnk, ak ̸= 0, and 0 = n0 < n1 < · · · < nt < n. We deﬁne the excess of f by E(f) = X t≥j≥(t+1)/2 nj − X t/2≥j≥1 nj. 4.1.27 Remark The excess of a polynomial is related to self-dual, weakly self-dual and almost weakly self-dual bases; see Section 5.1. 4.1.28 Remark In it is proved that there are binary primitive polynomials of degree n with E(f) ≤ 3 32n2 + o(n). 4.1.29 Problem Let E(f) denote the excess of the polynomial f. Prove or disprove that for p a prime, there is a primitive polynomial f for each degree n ≥2 with excess E(f) at most as follows: 1. E(f) ≤1, if p > 5; 2. E(f) ≤2, if p = 5; 3. E(f) ≤3, if p = 3; 4. E(f) ≤6, if p = 2. 4.1.30 Remark See for more results related to the excess of a polynomial. 4.1.31 Remark Several classes of irreducible and primitive polynomials seem to have asymptotic density δirr (as q →∞) and δprim given by δirr = 1 n and δprim = φ(qn −1) nqn . 4.1.32 Problem Find natural examples of families of polynomials over Fq having densities of irreducible and/or primitive polynomials diﬀerent from δirr and δprim. Some such examples are given in Table 7 of and in in connection to windmill polynomials. 90 Handbook of Finite Fields See Also §2.2 For tables of primitives of various kinds and weights. §4.2 For discussion of primitive polynomials with prescribed coeﬃcients. §4.3 For discussion of primitive polynomials with prescribed weights. §5.2 For information on primitive normal polynomials. §5.4 For discussion of completely normal primitive polynomials. §10.2 For connections to linear feedback shift registers. §11.3 For algorithms related to primitive polynomials. §14.9 For combinatorial applications of primitive polynomials; see also . Considers formulas for primitive polynomials of special forms. Develops the notion of “typical primitive polynomials” over Zn. References Cited: [307, 588, 673, 682, 690, 692, 1071, 1181, 1416, 1557, 1857, 1939, 2156, 2157, 2158, 2186, 2204, 2654, 2807, 2816] 4.2 Prescribed coeﬃcients Stephen D. Cohen, University of Glasgow 4.2.1 Deﬁnition Given a positive integer r deﬁne τ(r) = φ(r)/r, where φ is Euler’s function. 4.2.2 Remark For a pair (q, n) with q (as always) a prime power, τ(qn −1) is the proportion of non-zero elements of Fqn that are primitive (see Theorem 4.1.3). It also signiﬁes the proportion of powers of irreducible polynomials of degree n that are primitive. 4.2.3 Remark For a primitive polynomial f ∈Fq and non-zero a ∈Fq the polynomial F(x + a) need not be primitive. The arithmetical structure of the set of primitive polynomials is less marked than that of the set of irreducible polynomials. 4.2.4 Remark The monic reciprocal f ∗(x) = a−1 n xnf(1/x) (see Remark 3.5.14) of a primitive polynomial is also primitive (Remark 4.1.4). 4.2.5 Remark Except as mentioned, all polynomials (in particular in statements of theorems, etc) will be assumed to be monic polynomials of degree n over Fq taken to have the form f(x) = xn + Pn i=1 aixn−i. Here ai is the i-th coeﬃcient. The meaning of the terms ﬁrst (or last) m coeﬃcients will be as described in Deﬁnition 3.5.1. In particular −a1 is the trace of f and (−1)nan is the norm of f. 4.2.6 Remark The norm of a primitive polynomial has to be a primitive element of Fq (Theorem 4.1.6). 4.2.7 Remark Asymptotic estimates (for large n) of the number of primitive polynomials of degree n with prescribed coeﬃcients (e.g., with prescribed ﬁrst or last coeﬃcients) in themselves do not lead to strong existence results. All the theorems in this section (except Theorem 4.2.14) are existence results that are unaccompanied by useful lower bounds on the number of primitive polynomials over a range of pairs (q, n). Primitive polynomials 91 4.2.8 Problem Supply non-trivial bounds for the number of primitive polynomials with prescribed coeﬃcients in the problems in the following subsections. 4.2.9 Deﬁnition A normal polynomial of degree n over Fq is an irreducible polynomial whose roots comprise a normal basis of Fqn over Fq (Deﬁnition 2.1.98, Section 5.2). 4.2.10 Remark This section will also feature results on the distribution of polynomials with pre-scribed coeﬃcients that are simultaneously primitive and normal. 4.2.1 Approaches to results on prescribed coeﬃcients 4.2.11 Remark Character sum techniques and estimates constitute the principal underlying mech-anism. Speciﬁcally, one can characterize both primitivity and the prescribed coeﬃcient re-quirement in terms of such sums. The primitivity condition is in terms of multiplicative character sums over Fqn, whereas the coeﬃcient conditions involve lifted (multiplicative and additive) characters over Fq. 4.2.12 Remark Theorem 4.2.14 is an illustration of a typical asymptotic type of lower bound estimate that can be attained. All existence results described derive from such estimates (perhaps “weighted”). Further examples are not given here because the best estimates for absolute existence purposes are not optimal asymptotically (see Remark 4.2.17). 4.2.13 Deﬁnition A square-free divisor of a positive integer r is any factor of the radical of r, i.e., the product of the distinct primes dividing r. Thus the number of square-free divisors or r (denoted by W(r)) is given by W(r) = 2ω(r), where ω(r) is the number of distinct primes dividing r (with ω(1) = 0). 4.2.14 Theorem The number N of primitive polynomials whose ﬁrst m coeﬃcients are arbitrarily prescribed satisﬁes \|N −τ(qn −1)(qn −1)\| ≤mτ(qn −1)W(qn −1)(qm −1)qn/2, (4.2.1) where τ is given in Deﬁnition 4.2.1. 4.2.15 Remark Necessarily, Theorem 4.2.14 can give a positive result on the existence of a primitive polynomial with prescribed ﬁrst m coeﬃcients only if m < n/2, and indeed for m close to n/2 only asymptotically as n →∞. An improved asymptotic result on the existence of primitive normal polynomials with prescribed ﬁrst m coeﬃcients occurs in . 4.2.16 Remark If the last m coeﬃcients of a primitive polynomial are prescribed, one can obtain estimates by considering instead the number of monic primitive reciprocal polynomials f ∗(x) = a−1 n xnf(1/x) whose ﬁrst m −1 coeﬃcients and trace are prescribed. 4.2.17 Remark The eﬀect of the factor W(qn −1) in estimates such as (4.2.1) can be signiﬁcantly reduced through a sieving technique described in many of the papers cited below. In this way W(qn −1) is reduced to W(k) where k is an essential “core,” a factor of qn −1, and the sieving process generally proceeds over the remaining primes in qn −1 not in the core. This produces sharper existence results (while simultaneously blunting asymptotic quality). 4.2.18 Remark When the characteristic p of Fq is less than m, an equivalent process to prescribing directly the ﬁrst m coeﬃcients is to consider (irreducible) polynomials f with the ﬁrst m values σj prescribed, where σj is the trace over Fq of γj for a root γ of f. By these means it suﬃces to prescribe m alternative parameters (or conditions). 92 Handbook of Finite Fields 4.2.19 Remark One can focus on polynomials with a speciﬁc coeﬃcient prescribed (the m-th, say), . For m ≤n/2 this can be achieved with around m/2 constraints (rather than the m conditions needed to ﬁx the ﬁrst m coeﬃcients, noted in Remark 4.2.18). For m exceeding n/2 the reciprocal polynomial can be considered but then an extra condition (relating to ﬁxing the norm as a selected primitive element of Fq) is needed. 4.2.20 Remark Dealing with the situation when one (or more) of the ﬁrst m coeﬃcients are prescribed via an alternative set of prescribed values as in Remark 4.2.18 breaks down if p ≥m. To overcome this, a p-adic method, devised initially by Fan and Han, for example in [1030, 1031, 1032], and improved by Cohen, for example in , is used in many of the papers cited below. This eliminates from the situation diﬃculties caused by the characteristic. 4.2.21 Remark To establish a speciﬁc existence result, it is shown that it is valid for pairs (q, n) satisfying an arithmetical condition that holds for almost all pairs (q, n). Computation is required; ﬁrstly numerical checks to show that some of the ﬁnite number of remaining pairs satisfy the condition and then, for pairs which fail, direct working in the ﬁeld to exhibit a polynomial with the required property. 4.2.2 Existence theorems for primitive polynomials 4.2.22 Remark A conjecture of Hansen and Mullen, , has been the driver for the key theorem on the existence of a primitive polynomial with an arbitrary prescribed coeﬃcient. Prior to its formulation the only known result was the following existence theorem for primitive polynomials with arbitrary trace [675, 700, 1637]. Paper establishes a self-contained proof of the full theorem. 4.2.23 Theorem [686, 701, 702] Given m with 1 ≤m < n and a ∈Fq, there exists a primitive polynomial with m-th coeﬃcient a, with (genuine) exceptions only when (q, n, m, a) = (q, 2, 1, 0), (4, 3, 1, 0), (4, 3, 2, 0), (2, 4, 2, 1). 4.2.24 Remark An existence result on primitive polynomials with prescribed norm and trace can be derived by combining a theorem on primitive normal polynomials with prescribed norm and trace (Theorem 4.2.46) with one on the last two coeﬃcients prescribed (Theorem 4.2.50), since these two coeﬃcient prescriptions are equivalent for primitive polynomials, as for irreducible polynomials (Remark 3.5.14). (Note, however, that the monic reciprocal of a normal polynomial need not be normal and therefore the equivalence breaks down for primitive normal polynomials.) For the question of prescribing the norm and trace it is sensible to assume n ≥3 and that the prescribed norm is a primitive element of Fq. 4.2.25 Theorem [682, 691, 692, 1037, 1557] Let a, b ∈Fq with b a primitive element of Fq. Suppose n ≥3 if a ̸= 0 and n ≥5 if a = 0. Then there exists a primitive polynomial with trace a and norm b. 4.2.26 Remark Theorem 4.2.25 is complete except for the cases in which a = 0 and n = 3, 4; these remaining cases have recently been resolved in : the only exceptions occur when n = 3, a = 0, and q = 4 or 7. 4.2.27 Problem Similarly, other existence results on primitive polynomials may be derived as a consequence from those on primitive normal polynomials (Section 4.2.3). Enunciate these and ﬁll gaps that arise because the trace of a normal polynomial must be non-zero. 4.2.28 Theorem Suppose n ≥5. Then there exists a primitive polynomial with a1 = an−1 = 0, except when (q, n) = (4, 5), (2, 6), (3, 6). Primitive polynomials 93 4.2.29 Remark As an existence result, Theorem 4.2.28 is complete: the question must have a negative answer when n ≤4. It can be rephrased as asserting that the existence of a primitive f for which both f and its monic reciprocal (Remark 3.5.14) have trace 0. 4.2.30 Theorem Suppose n ≥5 and a, b ∈Fq. Then there exists a primitive polynomial f such that f has trace a and its monic reciprocal has trace b. 4.2.31 Problem Extend Theorem 4.2.30 to cover degrees n = 3, 4, perhaps with some listed ex-ceptions. 4.2.32 Theorem [198, 695, 698, 1411, 1412, 2658] Suppose n ≥5 if q is odd, and n ≥7 if q is even. Then there exists a primitive polynomial with its ﬁrst two coeﬃcients arbitrarily prescribed. If n = 4 and q is odd, then the same conclusion holds for suﬃciently large q. 4.2.33 Remark In it is claimed that when q is even, then the conclusion of Theorem 4.2.32 holds (with some exceptions) when n ≥4, although some of the given detail assumes n ≥7. 4.2.34 Theorem [695, 1030, 1033, 2104] Suppose n ≥7. Then there exists a primitive polynomial with its ﬁrst three coeﬃcients arbitrarily prescribed. 4.2.35 Remark An asymptotic result Theorem 4.2.52 on the existence on a primitive normal polynomial with its ﬁrst ⌊n−1 2 ⌋coeﬃcients prescribed yields a corresponding one for a primitive polynomial [695, 1032] (see Problem 4.2.27). What follows now is an unconditional result on the existence of a primitive polynomial with up to one-third of its ﬁrst coeﬃcients prescribed. 4.2.36 Theorem Suppose m ≤ n 3 (except that m ≤ n 4 when q = 2). Then there exists a primitive polynomial with its ﬁrst m coeﬃcients arbitrarily prescribed, with the exception that there is no primitive cubic over F4 with zero ﬁrst coeﬃcient. 4.2.3 Existence theorems for primitive normal polynomials 4.2.37 Remark Underlying these results is the fundamental existence theorem of Lenstra and Schoof (see also ). 4.2.38 Theorem [693, 1899] For every pair (q, n) there exists a primitive normal polynomial of degree n over Fq. 4.2.39 Remark The original proof of Theorem 4.2.38 in requires signiﬁcant numerical com-putation to verify plus direct examination of a few ﬁelds. For the modiﬁed approach of a computer is not required. 4.2.40 Remark If n ≤2, then a primitive polynomial is automatically normal (theorem statements may or may not include these values). 4.2.41 Remark For normal polynomials the character sums referred to in Remark 4.2.11 involve additive characters over Fqn. The resulting estimates (corresponding to Theorem 4.2.14, for example) now feature such quantities as W(xn −1), deﬁned as the number of square-free polynomial divisors of the polynomial xn −1 ∈Fq[x]. Furthermore, the sieve (referred to in Remark 4.2.17) now can additionally relate to the factorization of the polynomial xn −1, wherein the sieving process proceeds over the irreducible factors in xn −1 not in the “core.” Indeed, because the factorization of xn −1 can be examined more systematically than the corresponding numerical factorization of qn−1, theoretical existence can be treated initially more eﬀectively by means of an additive rather than a multiplicative sieve. Naturally, the resulting arithmetic conditions are more demanding than in the case of (merely) primitive polynomials and the consequent computations more substantial. 94 Handbook of Finite Fields 4.2.42 Remark In the study of polynomials which are both primitive and normal, prescribed ﬁrst or last coeﬃcients cannot be so easily interchanged via use of reciprocal polynomials since the monic reciprocal of a normal polynomial is not necessarily normal. Similarly, the requirements for specifying one coeﬃcient (see Remark 4.2.19) are more diﬃcult. There are, however, few new diﬃculties associated with the characteristic (see Remark 4.2.20). 4.2.43 Theorem Suppose n ≥15 and 1 ≤m < n and that a ∈Fq (with a ̸= 0 if m = 1). Then there exists a primitive normal polynomial with m-th coeﬃcient am = a. 4.2.44 Remark Paper relies on some substantial computations which are only brieﬂy sum-marized. Its authors note that the theory extends unchanged to polynomials of smaller degree and indeed there is a manuscript that claims an extension of Theorem 4.2.43 to degrees n ≥9, though there are some problems with the details. The problem is sensible for n ≥3 but the present method cannot currently be applied eﬀectively to small degrees. 4.2.45 Conjecture Suppose n ≥2 and 1 ≤m < n and that a ∈Fq (with a ̸= 0 if m = 1). Then there exists a primitive normal polynomial with am = a, except when (q, n, m, a) takes any of the values (2, 3, 2, 1), (2, 4, 2, 1), (2, 4, 3, 1), (2, 6, 3, 1), (3, 4, 2, 2), (5, 3, 4, 3), (4, 3, 2, 1 + γ), where F4 = F2(γ) with γ2 + γ + 1 = 0. 4.2.46 Theorem [682, 691, 692, 1557] (Compare with Theorem 4.1.22) Assume n ≥3. Suppose a ̸= 0 ∈Fq and b is a primitive element in Fq. Then there exists a primitive normal polynomial with trace a and norm b. 4.2.47 Remark Theorem 4.2.46 is a striking complete existence result. Thus, in the case of cubics, there is a primitive polynomial x3 + ax2 + cx + b with a, b ﬁxed appropriately and only c allowed to vary. Further results listed below are incomplete. 4.2.48 Problem The existence of primitive polynomials of the lowest degrees remains to be estab-lished. 4.2.49 Theorem Let a ̸= 0, b ∈Fq. Suppose n ≥7. Then there exists a primitive normal polynomial with ﬁrst coeﬃcient a1 = a and second coeﬃcient a2 = b. 4.2.50 Theorem Let a ∈Fq and b be a primitive element in Fq. Suppose n ≥5. Then there exists a primitive normal polynomial with penultimate coeﬃcient an−1 = a and last coeﬃcient an = (−1)nb. 4.2.51 Remark The next results (on prescribed ﬁrst and last coeﬃcients) are asymptotic, i.e., apply for suﬃciently large values of q. 4.2.52 Theorem Suppose n ≥2. Then there exists a constant C(n) such that, for q > C(n), there exists a primitive normal polynomial with its ﬁrst ⌊n 2 ⌋coeﬃcients arbitrarily prescribed, subject to the constraint that the ﬁrst coeﬃcient is non-zero. 4.2.53 Theorem Suppose n ≥2. Then there exists a constant C(n) such that, for q > C(n), there exists a primitive normal polynomial with its last ⌊n 2 ⌋coeﬃcients arbitrarily prescribed subject to the constraint that the last coeﬃcient is (−1)nb, where b is a given primitive element of Fq. Moreover, if the prescribed coeﬃcients (other than the last) are all zero, the total number of prescribed coeﬃcients may be taken to be ⌊n+1 2 ⌋. 4.2.54 Remark Further constraints on primitive normal polynomials with prescribed coeﬃcients could lead to new areas of research as in the illustrations which follow. Primitive polynomials 95 4.2.55 Deﬁnition An element α of Fqn is completely normal if it generates over any intermediate ﬁeld Fqd (where d\|n) a normal basis of Fqn over Fqd. The minimal polynomial of α over Fq is a completely normal polynomial. 4.2.56 Conjecture For every n there exists a primitive completely normal polynomial. 4.2.57 Remark Major contributions towards establishing this conjecture have been made by Hachenberger [1391, 1394] (see Section 5.4). Part of the construction involves trace-compatible sequences. Also the methods employed emphasize algebraic structure rather than character sums and lead to useful lower bounds on the number of constructed polyno-mials. 4.2.58 Problem For which pairs (q, n) does there exist a primitive completely normal polynomial with prescribed (non-zero) trace and/or (primitive) norm? 4.2.59 Remark Although the (monic) reciprocal of a primitive polynomial is primitive, the recip-rocal of a normal polynomial need not be normal. 4.2.60 Deﬁnition A strong primitive normal polynomial f is such that both f and its monic reciprocal are primitive normal polynomials. 4.2.61 Theorem For every pair (q, n) there exists a strong primitive normal polynomial, except when (q, n) = (2, 3), (2, 4), (3, 4), (4, 3), (5, 4). 4.2.62 Problem For which pairs (q, n) does there exist a strong primitive normal polynomial with prescribed (non-zero) trace and/or (primitive) norm? See Also §3.5 For discussion of irreducible polynomials with prescribed coeﬃcients. §5.2 For information on primitive normal polynomials. §5.4 For discussion of completely normal primitive polynomials. References Cited: [198, 627, 675, 682, 683, 684, 686, 691, 692, 693, 694, 695, 698, 700, 701, 702, 1029, 1030, 1031, 1032, 1033, 1034, 1035, 1036, 1037, 1391, 1394, 1411, 1412, 1416, 1557, 1637, 1899, 2104, 2157, 2658] 4.3 Weights of primitive polynomials Stephen D. Cohen, University of Glasgow 4.3.1 Deﬁnition As in Deﬁnition 4.1.10, denote by Wn(q) the minimal weight (or number of nonzero coeﬃcients) of a primitive polynomial of degree n over Fq. 4.3.2 Remark Hansen and Mullen list for each prime p < 100 and each degree n, with pn < 1050, a primitive polynomial over Fp of weight Wn(p). Earlier, Stahnke had listed a primitive polynomial over F2 for each degree n ≤168 of weight Wn(2). In every 96 Handbook of Finite Fields case Wn(p) = 3 or 5. Though Deﬁnition 4.3.1 makes sense for any prime power q, most relevant literature relates to the binary ﬁeld F2. Because of numerous applications there is particular interest in primitive polynomials of low or minimal weight. When n ≥2, Wn(2) ≥3: thus primitive trinomials (failing which pentanomials) are especially sought. In this research area there are few “theorems,” most work being empirical, heuristic or conjectural. References are scattered in journals in diverse ﬁelds. The citations given here are selective and incomplete. 4.3.3 Theorem (See Remark 4.1.11) For any suﬃciently large prime p, Wn(p) ≤n + 1 2 . 4.3.4 Theorem (See Remark 4.1.11) Wn(2) ≤n 4 + o(1). 4.3.5 Conjecture [1302, 2186] For all n, Wn(2) ≤5. 4.3.6 Conjecture [1302, 2186] For inﬁnitely many values of n, Wn(2) ≤3. 4.3.7 Remark Progress on Conjectures 4.3.5 and 4.3.6 may be diﬃcult. The next conjecture has a less ambitious goal. 4.3.8 Conjecture There is a positive integer m such that for inﬁnitely many values of n, Wn(2) ≤m. 4.3.9 Remark When 2n −1 is a (Mersenne) prime any irreducible polynomial of degree n over F2 is primitive. This means that polynomials of these degrees and small weight are valuable, because primitivity can be tested (or at least ruled out) with the aid of theorems such as Swan’s theorem for trinomials from . See also Theorem 3.3.25. 4.3.10 Theorem [408, 2753] Let n > s > 0 and assume n+s is odd. Then the trinomial xn+xs+1 ∈ F2[x] has an even number of irreducible factors if and only if one of the following holds: 1. n even, n ̸= 2s, ns/2 ≡0 or 1 (mod 4); 2. n odd, s ∤2n, n ≡±3 (mod 8); 3. n odd s\|2n, n ≡1 (mod 8). 4.3.11 Remark A trinomial of degree n over F2 where 2n−1 is a prime is a Mersenne trinomial (of degree n). To date 47 Mersenne primes are known (numbered in increasing order) as M1 = 2, . . . , M47 = 43112609. A total of 30 of these yield primitive trinomials. Zierler lists Mersenne trinomials of degrees up to 11213 (corresponding to M23) and additional primitive trinomials have been listed in [406, 1489, 1811]. The “Great Trinomial Hunt,” driven by Brent and Zimmermann , parallels GIMPS, the “Great Internet Prime Search” (see www.mersenne.org). Note that M47 is the largest known prime (as of November 2011), having 12978189 digits. Table 4.3 gives a list of all known Mersenne trinomials xn + xs + 1 with Mr = 2n −1 and s < n/2. Their reciprocals (wherein s > n/2) are also primitive. 4.3.12 Remark Lists of primitive Mersenne pentanomials have also been produced, see for exam-ple [1811, 3011]. Those in have the form xn + xn−1 + xm + xm−1 + 1. In particular, it has been suggested that, for random number generation, the use of pentanomials might be preferable to trinomials. Lists of primitive polynomials over F2 of weights 5, 7, and 9 and all degrees between 9 and 660 occur in . These possess the quality that the dif-ference between each pair of consecutive indices is almost the same. This property promotes the implementation of the generation of linear recurring sequences based on highly mod-ular devices (ring generators) leading to enhanced performance. On the other hand, there are cryptographic reasons why primitive polynomials of high weight might be important especially if all multiples of moderate degree also have high weight . Primitive polynomials 97 r n s 1 2 1 2 3 1 3 5 2 4 7 1, 3 6 17 3, 5, 6 8 31 3, 6, 7, 13 10 89 38 12 127 1, 7, 15, 30, 63 13 521 32, 48, 158, 168 14 607 105, 147, 273 15 1 279 216, 418 17 2 281 715, 915, 1029 18 3 217 67, 576 20 4 423 271, 369, 370, 649, 1393, 1419, 2098 21 9 689 84, 471, 1836, 2444, 4187 24 19 937 881, 7083, 9842 26 23 209 1530, 6619, 9739 27 44 497 8575, 21034 29 110 503 25230, 53719 30 132 049 7000, 33912, 41469, 52549, 54454 32 756 839 215747, 267428, 279695 33 859 433 170340, 288477 37 3 021 377 361604, 1010202 38 6 972 593 3037958 41 24 036 583 8412642, 8785528 42 25 964 951 880890, 4627670, 4830131, 6383880 43 30 402 457 2162059 44 32 582 657 5110722, 5552421, 7545455 46 42 643 801 55981, 3706066, 3896488, 12899278, 20150445 47 43 112 609 3569337, 4463337, 17212521, 21078848 Table 4.3.1 Mersenne trinomials. 4.3.13 Example There is no Mersenne trinomial of degree 61 (M9) but x61+x60+x46+x45+1 is a primitive pentanomial. See Also §3.3 For Swan’s theorem type results. §3.4 For discussion of the weights of irreducible polynomials. §14.9 For applications of primitive polynomials. References Cited: [406, 408, 1302, 1416, 1489, 1811, 1994, 2186, 2437, 2634, 2699, 2753, 3011, 3071] 98 Handbook of Finite Fields 4.4 Elements of high order Jos´ e Felipe Voloch, University of Texas at Austin 4.4.1 Remark There is no known explicit construction of primitive elements (Deﬁnition 2.1.38). For some applications, it suﬃces to construct elements of suﬃciently large order. The current state of the art on explicit constructions of elements of large order usually proves lower bounds for their orders which are much smaller than the expected actual order. 4.4.2 Deﬁnition The (absolute) degree of an element α of a ﬁnite ﬁeld of characteristic p is deg α = [Fp(α) : Fp]. The order of α ̸= 0 is deﬁned in Deﬁnition 2.1.40 and is denoted by ord α. 4.4.3 Remark If α ̸= 0, 1 is an element of a ﬁnite ﬁeld of characteristic p, then deg α < ord α ≤ pdeg α −1. 4.4.1 Elements of high order from elements of small orders 4.4.4 Theorem There exists an absolute constant c > 0 and, for every ϵ > 0, there exists a δ > 0, such that whenever α ̸= 0, 1 is an element of a ﬁnite ﬁeld of characteristic p with deg α = n. 1. [1241, 1243] If ord α = n + 1 then ord(1 −α) ≥exp(c√n). 2. If ord α < n2−ϵ then ord(1 −α) ≥exp(cnδ). 4.4.5 Remark If α is as in Theorem 4.4.4 Part 1, then n + 1 is prime and p is a primitive root modulo n + 1. Likewise, the possible values of n in Theorem 4.4.4 Part 2 are restricted. 4.4.6 Remark Similar, but weaker, results can be proved about the order of R(α), R ∈Fp(x) or even β, F(α, β) = 0, F(x, y) ∈Fp[x, y], with α as in the previous theorem . 4.4.7 Remark Poonen conjectured (as part of a more general conjecture, see ) that, with notation as in the previous theorem, max{ord α, ord(1 −α)} ≥exp(cn). A special case was also conjectured by Cheng . 4.4.8 Theorem Let α satisfy αm = g where m\|(q −1) and let g be a primitive element in Fq. Then, deg α = m deg g and ord(1 −α) ≥exp(cm). 4.4.2 Gao’s construction and a generalization 4.4.9 Theorem Given an integer n and a prime p, let m = ⌈logp n⌉. If g ∈Fp[x], deg(g) ≤ 2m is such that xpm −g(x) has an irreducible factor of degree n then any root of this factor in Fpn has order at least exp((log n)2/ log log n). 4.4.10 Remark A search for the polynomial g satisfying the hypotheses in the above theorem can be done in time polynomial in n log p. It is conjectured that such a polynomial exists for all p, n. 4.4.11 Remark An improvement on the bounds of was given in . Primitive polynomials 99 4.4.3 Iterative constructions 4.4.12 Theorem Let α0 = 1 ∈F2, αk a root of x2 + αk−1x + 1, k > 0. Then F22k = F2(αk), n = deg αk = 2k and ordαk ≥exp(nδ), for some absolute δ > 0. 4.4.13 Theorem Deﬁne f(x, y) := y2 + (6 −8x2)y + (9 −8x2). If q = pm is an odd prime power such that q ≡1 (mod 4), α0 ∈Fq is such that α2 0 −1 is not a square in Fq, deﬁne αk by f(αk−1, αk) = 0, k > 0. Let δk = α2 k −1. Then n = deg δk = m2k and ord(δk) ≥ exp(c(log n)2) for some constant c > 0. See Also For multiplicative orders of Gauss periods. For high order elements using subspace polynomials. For high order elements using Gauss periods. References Cited: [54, 465, 611, 612, 713, 1173, 1241, 1242, 1243, 2884, 2886] This page intentionally left blank This page intentionally left blank 5 Bases 5.1 Duality theory of bases ............................. 101 Dual bases • Self-dual bases • Weakly self-dual bases • Binary bases with small excess • Almost weakly self-dual bases • Connections to hardware design 5.2 Normal bases......................................... 109 Basics on normal bases • Self-dual normal bases • Primitive normal bases 5.3 Complexity of normal bases ....................... 117 Optimal and low complexity normal bases • Gauss periods • Normal bases from elliptic periods • Complexities of dual and self-dual normal bases • Fast arithmetic using normal bases 5.4 Completely normal bases........................... 128 The complete normal basis theorem • The class of completely basic extensions • Cyclotomic modules and complete generators • A decomposition theory for complete generators • The class of regular extensions • Complete generators for regular cyclotomic modules • Towards a primitive complete normal basis theorem 5.1 Duality theory of bases Dieter Jungnickel, University of Augsburg As noted in Section 2.1, Fqm may be viewed as a vector space of dimension m over Fq, and therefore has a basis (in fact, many bases) over Fq. Some fundamental deﬁnitions and results on bases were already given there. In particular, Theorem 2.1.93 and Corollary 2.1.95 provide criteria when a set {α1, . . . , αm} of elements in Fqm forms a basis over Fq. The present section provides a more detailed treatment of the general theory of bases, the unifying theme being the notion of duality introduced in Deﬁnition 2.1.100. Proofs for most of the results in this section can be found in ; however, a somewhat diﬀerent notation is used there. 5.1.1 Dual bases We restate and expand the basic Deﬁnition 2.1.100 as follows. 5.1.1 Deﬁnition Two ordered bases {α1, . . . , αm} and {β1, . . . , βm} of F = Fqm over K = Fq are dual (or complementary) if TrF/K(αiβj) = δij, where δij = 0 if i ̸= j and δij = 1 if i = j. An ordered basis is trace-orthogonal if it satisﬁes TrF/K(αiαj) = 0 whenever j ̸= i; and 101 102 Handbook of Finite Fields it is self-dual if it is dual with itself, that is, if it additionally satisﬁes TrF/K(α2 i ) = 1 for all i. 5.1.2 Remark The following simple but fundamental result guarantees the existence of dual bases. Remark 5.1.4 provides a proof by giving an explicit construction using the well-known concept of dual bases in linear algebra. 5.1.3 Theorem For every ordered basis B = {α1, . . . , αm} of F = Fqm over K = Fq, there is a uniquely determined dual ordered basis B∗= {β1, . . . , βm}. 5.1.4 Remark As F is a ﬁnite-dimensional vector space over K, it is isomorphic to the dual vector space F ∗consisting of all linear transformations from F to K. In Theorem 2.1.84, these transformations were given in terms of the trace function TrF/K in the form Lβ with β ∈F. Then the map L : F →F ∗with β 7→Lβ is an isomorphism. This allows one to obtain the following explicit description of the dual basis B∗: βj = L−1(α∗ j), (5.1.1) where α∗ j is the linear transformation deﬁned by the requirement α∗ j(αi) = δij for i = 1, . . . , m. (5.1.2) 5.1.5 Remark One reason for the importance of dual bases is the fact that they provide an easy way for determining the coordinate representation of arbitrary elements of F, using the concept of primal and dual coordinates. 5.1.6 Deﬁnition Given a dual pair of ordered bases B, B∗as above, we use the following notation. Let ξ = x1α1 + · · · + xmαm = (x)1β1 + · · · + (x)mβm. Then rB(ξ) = (x1, . . . , xm) and rB∗(ξ) = ((x)1, . . . , (x)m) are, respectively, the primal coordinates and the dual coordinates of ξ (with respect to B). 5.1.7 Lemma Let B = {α1, . . . , αm} and B∗= {β1, . . . , βm} be a dual pair of ordered bases of F = Fqm over K = Fq, let ξ ∈F, and let rB(ξ) and rB∗(ξ) be as in Deﬁnition 5.1.6. Then xi = TrF/K(ξβi) and (xi) = TrF/K(ξαi). (5.1.3) 5.1.8 Remark The following two results deal with the dual basis of a basis B in the important special cases where B is either a polynomial basis or a normal basis; see Deﬁnitions 2.1.96 and 2.1.98. 5.1.9 Theorem Let θ ∈F = Fqm, and assume that B = {α1 = θ, α2 = θq, . . . , αm = θqm−1} is a normal basis for F over K = Fq. Then the dual basis B∗= {β1, . . . , βm} is likewise normal: B∗= {ζ, ζq, . . . , ζqm−1}, where ζ = β1. 5.1.10 Remark The dual normal basis B∗can be described explicitly as follows. For i = 0, . . . , m −1, put ti := TrF/K(θθqi). Let t be the polynomial t(x) = tm−1xm−1 + · · · + t1x + t0 ∈K[x], Bases 103 and let d(x) = dm−1xm−1 + · · · + d1x + d0 be the unique monic polynomial of degree < m in K[x] satisfying d(x)t(x) ≡1 (mod xm −1). Then B∗is the normal basis generated by the element ζ = d0θ + d1θq + · · · + dm−1θqm−1. Normal bases will be studied in detail in Section 5.2. 5.1.11 Remark In contrast to the case of normal bases considered in Theorem 5.1.9, the dual basis of a polynomial basis is usually not a polynomial basis. Nevertheless, an explicit description is possible. 5.1.12 Theorem Let θ be a root of a monic irreducible polynomial f of degree m over K = Fq, and let B = {1, θ, θ2, . . . , θm−1} be the corresponding polynomial basis of F = Fqm over K. Assume that f splits over F as f(x) = (x −θ) γm−1xm−1 + · · · + γ1x + γ0 . Then the dual basis B∗= {β1, . . . , βm} can be computed as follows: βi = γi−1/f ′(θ) for i = 1, . . . , m, where f ′ denotes the formal derivative of f. 5.1.13 Theorem [1265, 1298] Let θ be a root of a monic irreducible polynomial f of degree m over K = Fq, and let B = {1, θ, θ2, . . . , θm−1} be the corresponding polynomial basis of F = Fqm over K. Then the dual basis B∗of B is likewise a polynomial basis if and only if f is a binomial and m ≡1 (mod p), where q is a power of the prime p. 5.1.14 Corollary There exists a dual pair of polynomial bases of Fqm over Fq if and only if the following three conditions are satisﬁed: 1. m ≡1 (mod p); 2. every prime r dividing m also divides q −1; 3. m ≡0 (mod 4) implies q ≡1 (mod 4). 5.1.15 Corollary Let B = {1, θ, θ2, . . . , θm−1} be a polynomial basis of of F2m over F2, where m ≥2. Then the dual basis of B cannot be a polynomial basis. 5.1.16 Example There is no dual pair of polynomial bases of F3m over F3. There exists a dual pair of polynomial bases of F4m over F4 if and only if m is a power of 3. There exists a dual pair of polynomial bases of F5m over F5 if and only if m is a power of 16. 5.1.2 Self-dual bases 5.1.17 Remark Lemma 5.1.7 shows that the computation of coordinates is particularly simple when the basis used is self-dual. This also has important applications in the design of hardware implementations for the multiplication in ﬁnite (extension) ﬁelds; see, for instance, Sections 4.1, 4.4, and 5.5 of . Unfortunately, a self-dual basis does not always exist, which motivates considering a slightly weaker notion. 5.1.18 Theorem There exists a self-dual basis of Fqm over Fq if and only if either q is even or both q and n are odd. 104 Handbook of Finite Fields 5.1.19 Deﬁnition A basis {α1, . . . , αm} of F = Fqm over K = Fq is almost self-dual if it is trace-orthogonal and if it additionally satisﬁes TrF/K(α2 i ) = 1 for i = 1, . . . , m−1, with possibly one exception. 5.1.20 Theorem There always exists an almost self-dual basis of Fqm over Fq. 5.1.21 Remark If there exists a self-dual basis for F over K, there are many such bases. All these bases are related by suitable transformations, namely via orthogonal matrices; see Deﬁnition 13.2.35. The number of such matrices is given in Theorem 13.2.37, which implies an explicit formula for the number of self-dual bases. 5.1.22 Lemma Let B = {α1, . . . , αm} be a self-dual ordered basis of F = Fqm over K = Fq, and let A = (aij) be an invertible m × m matrix over K. Then the ordered basis B′ = {β1, . . . , βm} with βi = m X i=1 aijαj for i = 1, . . . , m is likewise self-dual if and only if A is an orthogonal matrix. In particular, the number of ordered self-dual bases of F over K equals the number of orthogonal m × m matrices over K, provided that one such basis exists. 5.1.23 Theorem The number of (unordered) self-dual bases of Fqm over Fq is equal to sd(m, q) = γ m! m−1 Y i=1 (qi −ϵi), (5.1.4) where ϵi = ( 1 if i is even, 0 if i is odd, and γ =      2 if q is even, 1 if q and m are odd, 0 otherwise. 5.1.3 Weakly self-dual bases 5.1.24 Theorem For m ≥2, there does not exist a self-dual polynomial basis of Fqm over Fq. 5.1.25 Remark In contrast to Theorem 5.1.24, self-dual normal bases often exist; see Section 5.2. 5.1.26 Remark For computational purposes, it would be helpful to have a self-dual polynomial basis. However, Theorem 5.1.24 excludes this possibility. Fortunately, there are weaker no-tions which are still useful for hardware implementations; one such notion is discussed in the present subsection. 5.1.27 Deﬁnition A basis {α1, . . . , αm} of F = Fqm over K = Fq with associated dual basis B∗= {β1, . . . , βm} is weakly self-dual if there exist an element δ ∈F ∗and a permutation σ of {1, . . . , m} such that the following condition holds: βi = δασ(i) for i = 1, . . . , m. (5.1.5) 5.1.28 Remark For computational purposes, weakly self-dual polynomial bases are quite attractive, as they lead to rather simple transformations between dual and primal coordinates. Consider Bases 105 some element ξ ∈F, with primal and dual coordinates as in Deﬁnition 5.1.6. Using Equation (5.1.5), one obtains ξδ = m X j=1 xσ(j)βj. (5.1.6) Thus the dual coordinates of the product ξδ arise by simply permuting the primal coordi-nates of ξ according to σ. 5.1.29 Remark The observation in Remark 5.1.28 is of particular interest for hardware implemen-tations of the multiplication in F if a polynomial basis B is used. Consider a further element η ∈F, given in primal coordinates (y1, . . . , ym), and write π = ξη. Then one may design a simple hardware device – called a dual basis multiplier with respect to B – which computes the product πδ = (ξη)δ = (ξδ)η in dual coordinates from ξδ in dual coordinates (which, ac-cording to Equation (5.1.6), are obtained from the primal coordinates of ξ by just applying the permutation σ) and η in primal coordinates. Using Equation (5.1.6) for π instead of ξ, the primal coordinates of the product π can then be obtained by simply permuting the dual coordinates of πδ according to σ−1, so that all required coordinate transformations reduce to permutations. Moreover, the permutations arising are very simple ones; see Theorem 5.1.30 below. In the particularly important binary case, this allows the hardware design of eﬃcient dual basis multipliers in many instances of practical interest; see for more details and for examples of dual basis multipliers. 5.1.30 Theorem [1298, 2936] A polynomial basis B = {1, θ, θ2, . . . , θm−1} of Fqm over Fq is weakly self-dual if and only if the minimal polynomial f of θ is either a trinomial with constant term −1 or a binomial. Moreover, for i = 1, . . . , m, δ = βk = 1 θkf ′(θ) and σ(i) := k −i + 1 (mod m) if f(x) = xm + axk −1, and δ = 1 f ′(θ) = 1 mθm−1 and σ(i) := 1 −i (mod m) if f(x) = xm −a. 5.1.31 Corollary Let B = {1, θ, θ2, . . . , θm−1} be a polynomial basis of F = Fqm over Fq. Then the transformation from the primal coordinates of an arbitrary element ξ ∈F to the dual coordinates of the element ξδ (for a suitable constant element δ of F) is just a permutation if and only if the minimal polynomial of θ is either a trinomial with constant term −1 or a binomial. 5.1.32 Remark As the binary case is of particular practical importance, we state some results for this special case explicitly. 5.1.33 Theorem Let f(x) = xm + xk + 1 be an irreducible trinomial over F2, let θ be a root of f, and B∗= {β1, . . . , βm} the dual basis of the polynomial basis B = {1, θ, θ2, . . . , θm−1} generated by θ. Then B is weakly self-dual, and Equation (5.1.5) is satisﬁed with δ = βk = 1 θkf ′(θ) and σ(i) := k −i + 1 (mod m) for i = 1, . . . , m. 5.1.34 Example For m = 6, we may use a root θ of the cyclotomic polynomial Φ9(x) = x6 +x3 +1 over F2 to generate F = F26. Then the polynomial basis B = {1, θ, θ2, θ3, θ4, θ5} is weakly self-dual with δ = 1 θ3θ2 = θ4 and σ = 1 2 3 4 5 6 3 2 1 6 5 4 . 106 Handbook of Finite Fields Explicitly, the dual basis of B is given by B∗= {θ6 = θ3 + 1, θ5, θ4, θ9 = 1, θ8 = θ5 + θ2, θ7 = θ4 + θ}. It is easily checked that B and B∗indeed form a pair of dual bases. 5.1.35 Corollary Let θ be a root of a monic irreducible polynomial f of degree m over K = F2, and let B = {1, θ, θ2, . . . , θm−1} be the corresponding polynomial basis of F2m over K. Then the dual basis B∗of B is given by a permutation of B if and only if m is odd and f(x) = xm + x + 1. 5.1.36 Remark As noted before, the weakly self-dual polynomial bases are those polynomial bases for which the transformation between primal and dual coordinates can be realized with a minimum of complexity. In view of Theorem 5.1.33, the irreducible polynomials one should use for designing dual basis multipliers are therefore the irreducible trinomials. Fortunately, these exist in many – though by no means all – cases; see Sections 2.2 and 3.4. For n ≤ 10, 000, this holds in about half of the cases. 5.1.4 Binary bases with small excess 5.1.37 Remark In cases where no weakly self-dual polynomial basis can exist, further general-izations of the notion of weak self-duality are useful. We begin with some results for the binary case; proofs for these results can be found in . Consider a polynomial basis B = {1, θ, θ2, . . . , θm−1} of F2m over F2 and a scalar multiple C = B∗/δ of its dual basis B∗. According to Deﬁnition 5.1.27, B is a weakly self-dual basis (with respect to the given value of δ) if and only if the coordinate transformation from C to B is just a permutation; equivalently, the matrix S associated with this change of basis has to be a permutation matrix. If no weakly self-dual basis exists, one wants to ﬁnd a polynomial basis B and a suitable element δ for which the associated transformation matrix S is as simple as possible, meaning that S should have the smallest possible number of non-zero entries. This leads to the following deﬁnition. 5.1.38 Deﬁnition The weight w(S) of an invertible m × m matrix S is the number of non-zero entries of S. As the minimum weight is always at least m, one deﬁnes the excess of S as e(S) = w(S) −m. 5.1.39 Remark In the hardware design of dual basis multipliers over F2, one wants to use an irreducible polynomial for which the matrix S associated with this change of basis has the smallest possible excess, as this turns out to be the number of XOR-gates required for computing the primal coordinates from the generalized dual coordinates, that is, from the coordinates with respect to C = B∗/δ; see, for instance, [1631, Section 4.5]. Weakly self-dual bases correspond to the smallest possible case, namely e(S) = 0. 5.1.40 Example Consider the irreducible polynomial f(x) = x8 + x4 + x3 + x2 + 1 over F2, and let θ be a root of f. Then f = (x −θ)g, where g(x) = θ−1 + θ−2x + (1 + θ + θ5)x2 + (1 + θ4)x3 + θ3x4 + θ2x5 + θx6 + x7. By Theorem 5.1.12, the dual basis B∗= {β1, . . . , βm} of the polynomial basis B deﬁned by θ is obtained by dividing the coeﬃcients of g by f ′(θ) = θ2. Choosing δ = (θ3f ′(θ))−1 = 1/θ5 gives B∗/δ = {γ1, . . . , γm}, where γ1 = θ2, γ2 = θ, γ3 = 1 + θ2, γ4 = θ3 + θ7, γ5 = θ6, γ6 = θ5, γ7 = θ4, γ8 = θ3. Bases 107 Thus the transformation matrix S from generalized dual to primal coordinates has excess 2 in this case, giving indeed a quite simple coordinate transformation. This heavily depends on the choice of δ; for instance, the choice δ = f ′(θ)−1 = 1/θ2 would lead to a transformation matrix with excess 16. 5.1.41 Theorem Let θ be a root of an irreducible polynomial f of degree m over K = F2, and let B = {1, θ, θ2, . . . , θm−1} be the corresponding polynomial basis of F2m over K, and B∗the dual basis of B. Write f(x) = xm + xmt + · · · + xm1 + xm0, where 0 = m0 < m1 < · · · < mt < m, and put δ = 1 f ′(θ)θms , where s = ⌈t/2⌉. Then the transformation matrix S from C = B∗/δ to B has excess e(S) = t X j=s+1 mj − s−1 X j=1 mj. 5.1.42 Corollary Let θ be a root of an irreducible pentanomial f of degree m over F2, say f(x) = xm + xm3 + xm2 + xm1 + 1, and let B be a corresponding polynomial basis of F2m over F2, and B∗the dual basis of B. Put δ = (f ′(θ)θm2)−1. Then the transformation matrix S from C = B∗/δ to B has excess e(S) = m3 −m1. 5.1.43 Corollary The binary irreducible polynomials leading to a transformation matrix of excess 2 are precisely the irreducible pentanomials of the special form xm + xk+1 + xk + xk−1 + 1. 5.1.44 Theorem A binary irreducible polynomial leads to a transformation matrix of excess 1 if and only if it has the form xm + x + 1, where m is even. 5.1.45 Remark It is also of interest to investigate the possible spectra of the number of elements of trace 1 in a polynomial basis for F2m over F2. This problem was ﬁrst considered in where it was noted that using a polynomial basis with a small number of elements of trace 1 is desirable, since it allows a particularly eﬃcient implementation of the trace function. For example, this is important for halving a point on an elliptic curve over F2m, and for generating pseudo random sequences using elliptic curves. In particular, it is of interest to ﬁnd trinomials and pentanomials associated with bases with a small number of elements of trace 1. We mention one striking result in this direction; more results on the trace spectra of polynomial bases can be found in [47, 51, 2646]. 5.1.46 Theorem Suppose that there exists an irreducible trinomial of degree m over F2. Then there also exists an irreducible trinomial such that the corresponding polynomial basis for F2m over F2 contains exactly one element with trace 1. 5.1.5 Almost weakly self-dual bases 5.1.47 Remark In this subsection, we present some results on polynomial bases corresponding to matrices with small excess over a general ﬁeld Fq. These results are taken from . For q ̸= 2, a matrix of excess 0 is not necessarily a permutation matrix. This leads to the following generalization of weakly self-dual bases. 108 Handbook of Finite Fields 5.1.48 Deﬁnition A basis {α1, . . . , αm} of F = Fqm over K = Fq with associated dual basis B∗= {β1, . . . , βm} is almost weakly self-dual if there exist an element δ ∈F ∗, elements c1, . . . , cn ∈K∗, and a permutation σ of {1, . . . , m} such that the following condition holds: βi = ciδασ(i) for i = 1, . . . , m. (5.1.7) 5.1.49 Remark The almost weakly self-dual bases are precisely those bases corresponding to a transformation matrix S with e(S) = 0. An almost weakly self-dual basis is actually weakly self-dual if and only if c1 = · · · = cn. 5.1.50 Theorem A polynomial basis B = {1, θ, θ2, . . . , θm−1} of Fqm over Fq is almost weakly self-dual if and only if the minimal polynomial f of θ is either a trinomial or a binomial. 5.1.51 Remark There are no proper almost weakly self-dual bases which belong to irreducible binomials: in this case, the basis is actually weakly self-dual by Theorem 5.1.30. Hence it suﬃces to consider the case of irreducible trinomials in Theorem 5.1.52; in the special case where d = 1, one recovers Theorem 5.1.30. 5.1.52 Theorem Let f(x) = xm −axk −d be an irreducible trinomial over F2, let θ be a root of f, and B∗= {β1, . . . , βm} the dual basis of the polynomial basis B = {1, θ, θ2, . . . , θm−1} generated by θ. Then B is almost weakly self-dual, and Equation (5.1.7) is satisﬁed with δ = βk = d θkf ′(θ), ci = ( 1 if i ≤k, d−1 if i > k and σ(i) := k −i + 1 (mod m) for i = 1, . . . , m. 5.1.53 Theorem Let θ be a root of a monic irreducible polynomial f of degree m over K = Fq, and let B = {1, θ, θ2, . . . , θm−1} be the corresponding polynomial basis of Fqm over K, and B∗the dual basis of B. Write f(x) = xm + fmtxmt + · · · + fm1xm1 + fm0xm0, where 0 = m0 < m1 < · · · < mt < m, and put δ = 1 f ′(θ)θms , where s = ⌈t/2⌉. Then the transformation matrix S from C = B∗/δ to B has excess e(S) =    Pt j=s+1 mj −Ps−1 j=1 mj if t is odd, Pt j=s+1 mj −Ps j=1 mj if t is even. 5.1.54 Corollary Let θ be a root of an irreducible polynomial f of degree m over Fq, let B = {1, θ, θ2, . . . , θm−1} be the corresponding polynomial basis of Fqm over K, and B∗the dual basis of B. Then the transformation matrix S from C = B∗/δ to B has excess e(S) =          0 if f is either a trinomial or a binomial, 1 if f is of the form f(x) = xm + fkxk + fk−1xk1 + f0, 2 if f is of the form f(x) = xm + fk+1xk+1 + fkxk + fk−1xk−1 + f0 or f(x) = xm + fk+1xk+1 + fk−1xk−1 + f0. Bases 109 5.1.6 Connections to hardware design 5.1.55 Remark Many aspects of the study of various types of bases for Fqm over Fq are to a large extent motivated by the hardware design of eﬃcient multipliers for Fqm. The seminal paper in this area is due to Berlekamp . Another seminal idea – which motivated the study of optimal and low complexity normal bases, see Section 5.3 – was contained in a 1981 US patent application by Massey and Omura; see “Computational Method and Apparatus for Finite Field Arithmetic,” US Patent No. 4,587,627, 1986. As already mentioned, intro-ductory examples and some references can be found in Sections 4.1, 4.4, and 5.5 of . There is an abundance of papers in this area, largely due to its importance for hardware architectures for public key cryptography; the interested reader should consult the relevant sections of the extensive survey . A more recent survey concerning the special topic of polynomial basis multipliers in the binary case is given in . See Also §3.4, §3.5 For irreducible trinomials. §5.2 For normal bases. §5.3 For complexities of normal bases. §13.2 For matrices over ﬁnite ﬁelds. §16.7 For hardware implementations. References Cited: [47, 51, 208, 234, 982, 1172, 1265, 1298, 1573, 1631, 1635, 2158, 2583, 2646, 2718, 2936] 5.2 Normal bases Shuhong Gao, Clemson University Qunying Liao, Sichuan Normal University We present basic results on normal and self-dual normal elements, and we give a uniﬁed approach following the PhD thesis . Let p be a prime and let q be a power of p. Denote by σ the Frobenius map of Fqn: σ(α) = αq, for α ∈Fqn. Then σi(α) = αqi, for all i ≥0, and σn = 1 as a map on Fqn (since σn(α) = αqn = α for all α ∈Fqn). The Galois group of Fqn over Fq consists of the n maps σi, 0 ≤i ≤n −1. Recall from Deﬁnition 2.1.98 that an element α ∈Fqn is a normal element over Fq if the conjugates σi(α), 0 ≤i ≤n −1, are linearly independent over Fq. Hence a normal basis for Fqn over Fq is of the form {α, σ(α), . . . , σn−1(α)}, where α ∈Fqn is normal over Fq. Also, an irreducible polynomial f ∈Fq[x] of degree n is an N-polynomial (or normal polynomial) if its roots are linearly independent over Fq, that is, its roots form a normal basis of Fqn over Fq. 110 Handbook of Finite Fields 5.2.1 Basics on normal bases 5.2.1 Theorem (Normal basis theorem) For every prime power q and every integer n ≥1, Fqn has a normal basis over Fq. 5.2.2 Remark The normal basis theorem was proved ﬁrst by Hensel . A more general normal basis theorem for any ﬁnite Galois extension of an arbitrary ﬁeld was proved by Noether and Deuring . 5.2.3 Proposition 1. For any two integers m, n ≥1 and for any normal element α ∈Fqmn over Fq, the element β = TrFqmn/Fqn (α), the trace of α from Fqmn to Fqn, is a normal element of Fqn over Fq. 2. If gcd(m, n) = 1, then any normal element α ∈Fqn over Fq is still normal in Fqmn over Fqm. 3. If gcd(m, n) = 1, then for any normal elements α ∈Fqn and β ∈Fqm over Fq, the product αβ is a normal element of Fqmn over Fq. 5.2.4 Deﬁnition Let Fq[σ] denote the set of all polynomials in σ with coeﬃcients in Fq. The ring Fq[σ] acts on Fqn in a natural way: for any f(σ) = Pm i=0 ciσi ∈Fq[σ] and α ∈Fqn, f ◦α = m X i=0 ciσi(α) ∈Fqn. 5.2.5 Remark 1. An element α ∈Fqn is normal over Fq if and only if every β ∈Fqn is equal to f(σ) ◦α for some f ∈Fq[x]. 2. For any f, g ∈Fq[σ], we have (fg) ◦α = f ◦(g ◦α). 3. For any nonzero polynomial f(σ) = Pm i=0 ciσi ∈Fq[σ] with m < n, f is not equal to the zero map on Fqn. In fact, if f ◦α = 0 for all α ∈Fqn, then the polynomial Pm i=0 aixqi has qn zeros in Fqn, which is more than the degree qm. Hence xn −1 is the minimal polynomial of σ and Fq[σ] ∼ = Fq[x]/(xn −1) as rings over Fq. This implies that Fq[σ] = (n−1 X i=0 ciσi : ci ∈Fq, 0 ≤i ≤n −1 ) . There is an obvious correspondence between polynomials of degree at most n−1 in Fq[x] and elements in Fq[σ], so we use the two notations interchangeably. 4. An element f(σ) ∈Fq[σ] is invertible if there is g(σ) ∈Fq[σ] so that f(σ)g(σ) = 1. By the ring isomorphism above, f(σ) is invertible in Fq[σ] if and only if gcd(f(x), xn −1) = 1 in Fq[x]. Bases 111 5. Each element f(σ) = a0 + a1σ + · · · + an−1σn−1 ∈Fq[σ] corresponds to an n × n circulant matrix (see Deﬁnition 13.2.29) C(f) = (aj−i) whose rows are indexed by i and columns by j, where 0 ≤i, j ≤n −1, with j −i computed modulo n. For any f, g ∈Fq[σ], we have C(fg) = C(f)C(g), hence f is invertible in Fq[σ] if and only if the matrix C(f) is nonsingular. 5.2.6 Remark An Fq-linear subspace V of Fqn is σ-invariant (or σ-stable) if σ(α) ∈V for every α ∈V . We can characterize normal elements in Fqn by invariant subspaces. Let n = em, e = pv, where p is the characteristic of Fq and gcd(p, m) = 1. Suppose xn −1 factors in Fq[x] as xn −1 = (g1(x)g2(x) · · · gr(x))e , (5.2.1) where the gi(x)’s are distinct monic irreducible factors of xm −1 in Fq[x]. Let Ei(x) ∈Fq[x] for 1 ≤i ≤r so that, with 1 ≤j ≤r, Ei(x) ≡ 1 (mod ge j(x)) if j = i, 0 (mod ge j(x)) if j ̸= i. This means that, for 1 ≤i, j ≤r, Ei(σ)Ej(σ) = Ei(σ) if j = i, 0 if j ̸= i. Let Ri = Ei(σ)Fq[σ] for 1 ≤i ≤r. Then Ri ∼ = Fq[x]/(gi(x)e), 1 ≤i ≤r, and Fq[σ] = R1 + R2 + · · · + Rr, (5.2.2) is a direct sum of subrings. For 1 ≤i ≤r, let Vi = Ri ◦Fqn = {f ◦α : f ∈Ri and α ∈Fqn} ⊆Fqn. Then Vi is a σ-invariant subspace of Fqn annihilated by gi(x)e, that is gi(σ)e ◦α = 0, for all α ∈Vi. Also, Vi has dimension e · deg(gi(x)) over Fq. Let Wi be the subspace of Vi annihilated by gi(x)e−1, 1 ≤i ≤r, that is Wi = {α ∈Vi : gi(σ)e−1 ◦α = 0}. Then Wi is a σ-invariant subspace of Vi with dimension (e −1) deg(gi(x)) over Fq. 5.2.7 Theorem [2395, 2580] With the notation as in the above remark, we have that Fqn = V1 + V2 + · · · + Vr 112 Handbook of Finite Fields is a direct sum of Fq-vector spaces. Furthermore, an arbitrary element α ∈Fqn, written as α = α1 + α2 + · · · + αr, where αi ∈Vi, is normal over Fq if and only if αi ̸∈Wi, 1 ≤i ≤r. 5.2.8 Corollary [1486, 2324] Let di = deg(gi(x)) for 1 ≤i ≤r. Then the number of normal elements in Fqn over Fq is r Y i=1 qedi −q(e−1)di = Φq(xn −1), where, Φq is the Euler Phi function for polynomials, see Deﬁnition 2.1.111. 5.2.9 Corollary Let n be a power of p, the characteristic of Fq. Then 1. an element α ∈Fqn is normal over Fq if and only if TrFqn/Fq(α) ̸= 0; 2. an irreducible polynomial f(x) ∈Fq[x] of degree n is an N-polynomial if and only if the coeﬃcient of xn−1 in f(x) is nonzero. 5.2.10 Corollary Let n be a prime such that q is primitive modulo n. Then 1. an element α ∈Fqn is normal over Fq if and only if α ̸∈Fq and TrFqn/Fq(α) ̸= 0; 2. an irreducible polynomial f(x) ∈Fq[x] of degree n is an N-polynomial if and only if the coeﬃcient of xn−1 in f(x) is nonzero. 5.2.11 Theorem For any α ∈Fqn, deﬁne Tα(x) = n−1 X i=0 σi(α)xi ∈Fqn[x], tα(x) = n−1 X i=0 tixi ∈Fq[x], where ti = TrFqn/Fq(α σi(α)) ∈Fq, 0 ≤i ≤n −1. We have the following characterizations of normal elements. 1. α ∈Fqn is normal over Fq if and only if gcd(Tα(x), xn −1) = 1 in Fqn[x]. 2. α ∈Fqn is normal over Fq if and only if gcd(tα(x), xn −1) = 1 in Fq[x], that is, if tα(σ) is invertible in Fq[σ]. 3. Suppose that α ∈Fqn is normal over Fq. Then, for any g(σ) ∈Fq[σ], the element β = g(σ) ◦α is normal over Fq if and only if g(σ) is invertible in Fq[σ], that is, gcd(g(x), xn −1) = 1 in Fq[x]. 5.2.12 Remark Part 3 above shows again that Φq(xn−1) is equal to the number of normal elements in Fqn over Fq. There is a nice formula for Φq(xn −1) due to Ore . Suppose n = pvm where m is not divisible by p. For a positive integer d, let τm(d) denote the multiplicative order of d modulo m, and φ(d) be the Euler φ-function (equal to the number of integers between 1 and d that are relatively prime to d). Then Φq(xn −1) = qn Y d\|m 1 − 1 qτm(d) φ(d)/τm(d) . Also, for a general polynomial f ∈Fq[x] with r distinct irreducible factors in Fq[x] with degrees d1, d2, . . . , dr (the degrees need not be distinct), we have Φq(f(x)) = qn r Y i=1 1 −1 qdi . Bases 113 5.2.13 Theorem For any f(x) ∈Fq[x] of degree n ≥1 with f(0) ̸= 0, we have Φq(f) ≥    qn e if n < q, qn e γ+ 1 2(1+logq n) (1+logq n) > qn e0.83(1+logq n) if n ≥q, where γ ≈0.577216 is Euler’s constant and e ≈2.71828 is Euler’s number. 5.2.14 Remark In , it is also proved that, for f(x) = xn −1, the lower bound above is almost tight for an inﬁnite sequence of values of n. 5.2.15 Proposition Let α be a normal basis generator for Fqn over Fq and let C = (cij) be an invertible matrix over Fq. Then the basis {β0, β1, . . . , βn−1} deﬁned by βi = n−1 X j=0 cijαqi, i = 0, 1, . . . , n −1, is a normal basis for Fqn over Fq if and only if C is circulant. 5.2.16 Corollary The number of normal basis generators of Fqn over Fq is the order of the group C(n, q) of invertible circulant n × n matrices over Fq. 5.2.17 Proposition [2077, 2859] Deﬁne two sequences of polynomials ak(x), bk(x) ∈F2[x] as fol-lows: a0(x) = x, b0(x) = 1 ak+1(x) = ak(x)bk(x), bk+1(x) = a2 k(x) + b2 k(x), k ≥0. Then, for every k ≥1, the polynomial ak(x) + bk(x) is an N-polynomial over F2 of degree 2k. 5.2.18 Remark The above sequences of polynomials are used by Gao and Mateer to perform fast additive Fourier transforms over ﬁelds of characteristic two. 5.2.19 Proposition Let p be any prime. Deﬁne a sequence of polynomials fk(x) ∈Fp[x] as follows: f1(x) = xp + xp−1 + · · · + x −1, f2(x) = f1(xp −x −1), fk+1(x) = f ∗ k(xp −x −1), k ≥2, where f ∗(x) denotes the reciprocal polynomial of f(x), that is, f ∗(x) = xdf(1/x) where d is the degree of f(x); see Deﬁnition 2.1.48. Then, for every k ≥1, f ∗ k(x) is an N-polynomial over Fp of degree pk. 5.2.20 Proposition 1. Suppose q ≡1 (mod 4) and let a ∈Fq be a non-square. Then the polynomial x2k −a(x −1)2k is an N-polynomial over Fq for all k ≥1. 2. Suppose q ≡3 (mod 4) and x2 −bx −c2 ∈Fq[x] is irreducible with b ̸= 2. Then the polynomial (x −1)2k+1 −b(x2 −x)2k −cx2k is an N-polynomial over Fq for all k ≥0. 114 Handbook of Finite Fields 3. Let q be any prime power, r1, . . . , rt be the distinct prime factors of q −1 and n = ri1 1 · · · rit t , i1, . . . , it ≥0. Assume that q ≡1 (mod 4) if some ri is equal to 2. Let a ∈Fq be any element of order q −1. Then the polynomial xn −a(x −1)n is an N-polynomial. 5.2.2 Self-dual normal bases 5.2.21 Deﬁnition For any α, β ∈Fqn, deﬁne the trace polynomial of α and β over Fq to be tα,β(σ) = n−1 X i=0 Tr(α σi(β)) σi ∈Fq[σ]. When α = β, tα,α(σ) is denoted by tα(σ), which agrees with the deﬁnition of tα in Theorem 5.2.11, and is the trace polynomial of α over Fq. An element α is dual to β in Fqn over Fq if tα,β(σ) = 1, and α is self-dual if it is dual to itself. 5.2.22 Remark Note that α ∈Fqn is normal over Fq if and only if tα(σ) is invertible, while α ∈Fqn is self-dual over Fq if and only if tα(σ) = 1. Hence an element α ∈Fqn is self-dual over Fq if and only if it generates a self-dual normal basis for Fqn over Fq. 5.2.23 Theorem There is a self-dual normal basis for Fqn over Fq if and only if q and n are odd, or q is even and n is not divisible by 4. 5.2.24 Remark Imamura and Morii proved the “only if” part of the theorem, and Lempel and Weinberger proved the “if” part. For self-dual normal bases in Galois extensions of arbitrary ﬁelds, see Bayer-Fluckier and Lenstra . 5.2.25 Proposition Suppose gcd(m, n) = 1. Then 1. for any self-dual element α ∈Fqn over Fq, α remains self-dual in Fqmn over Fqm; 2. for any self-dual elements α ∈Fqn over Fq and β ∈Fqm over Fq, the product αβ is a self-dual element of Fqmn over Fq. 5.2.26 Lemma Let α ∈Fqn be any normal element over Fq. For any β = f(σ) ◦α and γ = g(σ) ◦α where f(σ), g(σ) ∈Fq[σ], we have the following equation in Fq[σ]: tβ,γ(σ) = f(σ)g(σ−1)tα(σ). 5.2.27 Proposition Let α ∈Fqn be any normal element over Fq, and let hα(σ) be the inverse of tα(σ), that is, hα(x)tα(x) ≡1 (mod xn −1). 1. The element β = hα(σ)◦α is dual to α; hence, the dual basis of the normal basis generated by α is a normal basis and is generated by β. 2. An element β = f(σ) ◦α in Fqn has a dual if and only if f(σ) is invertible, and its dual is equal to γ = g(σ) ◦α where g(σ) = (f(σ−1)tα(σ−1))−1. 3. An element β = f(σ) ◦α is self-dual if and only if f(σ)f(σ−1) = hα(σ). 4. Suppose Fqn has a self-dual element α over Fq (so tα(σ) = 1). Then β = f(σ) ◦α is self-dual if and only if f(σ)f(σ−1) = 1. (5.2.3) Bases 115 5.2.28 Remark In terms of circulant matrices, Equation (5.2.3) is equivalent to C(f)C(f)t = I, that is, C(f) is orthogonal. Let oc(n, q) denote the number of orthogonal n × n circulant matrices over Fq. When Fqn has a self-dual normal basis for Fq, oc(n, q) is the number of self-dual elements in Fqn over Fq, and oc(n, q)/n is the number of self-dual normal bases in Fqn over Fq. Suppose xn −1 factors as in Equation (5.2.1) where e = pv. We classify the irreducible factors gi(x) into three types: 1. x −1 and possibly x + 1 (if n is even and q is odd); 2. self-reciprocal factors gi(x) with roots consisting of pairs {ξ, ξ−1} with ξ ̸= ξ−1, hence gi(x) = xdgi(1/x)/gi(0) where d is the degree of gi(x). Suppose there are s such irreducible factors in Equation (5.2.1) and their degrees are 2d1, . . . , 2ds; 3. the remaining irreducible factors, which come in pairs gi(x) and ˜ gi(x) of the same degree so that, for each root ξ of gi(x), ξ−1 is a root of ˜ gi(x), hence ˜ gi(x) = xdgi(1/x)/gi(0). Suppose there are t such pairs of irreducible factors in Equation (5.2.1) and their degrees are e1, . . . , et. We order the components of R in Equation (5.2.2) accordingly: R = R0 + R1 + · · · Rs + (Rs+1 + ˜ Rs+1) + · · · + (Rs+t + ˜ Rs+t), where R0 represents the component for x −1 plus that for x + 1 if it is a factor. Then the space of solutions for Equation (5.2.3) is the direct sum of solutions from each component. This leads to the following theorem. 5.2.29 Theorem Let n = pvm, di’s and ej’s be as deﬁned in the above remark. Then oc(n, q) = ( ϵ1 Qs i=1(qdi + 1) Qt i=1(qei −1) if v = 0, ϵ2 q(pv−1)m/2 · oc(m, q) if v ≥1, where ϵ1 =    1 if q is even, 2 if q and m are odd, 4 if q is odd and m is even, ϵ2 =    1 if p ̸= 2, q1/2 if p = 2 and v = 1, 2q1/2 if p = 2 and v ≥2. 5.2.30 Corollary [130, 258, 1635] If the condition in Theorem 5.2.23 is satisﬁed, then there are oc(n, q) self-dual normal elements in Fqn over Fq. 5.2.3 Primitive normal bases 5.2.31 Deﬁnition An element α ∈Fqn is primitive normal over Fq if it is normal over Fq and has multiplicative order qn −1. A normal basis generated by a primitive normal element is a primitive normal basis. A polynomial of degree n in Fq[x] is primitive normal if its roots are primitive normal in Fqn over Fq. 5.2.32 Theorem (Primitive normal basis theorem) For any prime power q and any integer n ≥1, Fqn has a primitive normal element over Fq. 5.2.33 Remark The primitive normal basis theorem was proved by Carlitz for qn suﬃciently large, by Davenport when q is a prime and by Lenstra and Schoof in the general 116 Handbook of Finite Fields case. For a theoretical proof which does not require any machine calculation, see Cohen and Huczynska . The next few theorems are further strengthenings of the primitive normal basis theorem. 5.2.34 Theorem For any prime power q, any integer n ≥2 and any nonzero a ∈Fq, there is a primitive normal polynomial f(x) = xn + c1xn−1 + c2xn−2 + · · · + cn−1x + cn ∈Fq[x] with c1 = a. 5.2.35 Theorem For any prime power q, any integer n ≥15, any integer m with 1 ≤ m < n, and any a ∈Fq (with a ̸= 0 if m = 1), there is a primitive normal polynomial f(x) = xn + c1xn−1 + c2xn−2 + · · · + cn−1x + cn ∈Fq[x] with cm = a. 5.2.36 Theorem [694, 2806] For any prime power q and any integer n ≥2, there is an element α ∈Fqn such that both α and α−1 are primitive normal over Fq except when (q, n) is one of the pairs (2, 3), (2, 4), (3, 4), (4, 3) and (5, 4). 5.2.37 Theorem For any prime power q, any integer n ≥7 and any a, b ∈Fq with a ̸= 0, there is a primitive normal polynomial f(x) = xn+c1xn−1+c2xn−2+· · ·+cn−1x+cn ∈Fq[x] with c1 = a and c2 = b. 5.2.38 Theorem For any integer n ≥2 and suﬃciently large prime power q, there is a primitive normal polynomial f(x) = xn + c1xn−1 + c2xn−2 + · · · + cn−1x + cn ∈Fq[x] with its ﬁrst ⌊n/2⌋coeﬃcients arbitrarily prescribed except that c1 ̸= 0. 5.2.39 Theorem For any integer n ≥2 and suﬃciently large prime power q, there is a primitive normal polynomial f(x) = xn + c1xn−1 + c2xn−2 + · · · + cn−1x + cn ∈Fq[x] with its last ⌊n/2⌋coeﬃcients arbitrarily prescribed except that (−1)ncn must be a primitive element of Fq. See Also §2.2 For tables of primitive polynomials of various kinds and standards requiring normal basis arithmetic. §5.3 For normal bases of low complexity. §5.4 For completely normal bases. §16.7 For hardware implementations of ﬁnite ﬁeld arithmetic. References Cited: [130, 212, 258, 309, 540, 690, 692, 694, 775, 824, 1029, 1034, 1035, 1036, 1172, 1185, 1186, 1486, 1574, 1631, 1635, 1890, 1899, 1990, 2077, 2297, 2324, 2385, 2395, 2580, 2806, 2859]. Bases 117 5.3 Complexity of normal bases Shuhong Gao, Clemson University David Thomson, Carleton University 5.3.1 Optimal and low complexity normal bases 5.3.1 Deﬁnition Let α ∈Fqn be normal over Fq and let N = (α0, α1, . . . , αn−1) be the normal basis of Fqn over Fq generated by α, where αi = αqi, 0 ≤i ≤n −1. Denote by T = (tij) the n × n matrix given by ααi = n−1 X j=0 tijαj, 0 ≤i ≤n −1, where tij ∈Fq. The matrix T is the multiplication table of the basis N. Furthermore, the number of non-zero entries of T, denoted by CN, is the complexity (also called the density) of the basis N. 5.3.2 Remark An exhaustive search for normal bases of F2n over F2 for n < 40 is given in , extending previous tables such as those found in . Using data from the search, the authors in indicate that normal bases of F2n over F2 follow a normal distribution (with respect to their complexities) which is tightly compacted about a mean of roughly n2/2. We deﬁne low complexity normal bases loosely to mean normal bases known to have sub-quadratic bounds, with respect to n, on their complexity. 5.3.3 Remark In addition, gives the minimum-known complexity of a normal basis of F2n over F2 for many values of n using a variety of constructions that appear in this section. Further tables on normal bases are provided in Section 2.2. 5.3.4 Proposition The complexity CN of a normal basis N of Fqn over Fq is bounded by 2n −1 ≤CN ≤n2 −n + 1. 5.3.5 Deﬁnition A normal basis is optimal normal if it achieves the lower bound in Proposi-tion 5.3.4. 5.3.6 Theorem [139, 2199] 1. (Type I optimal normal basis) Suppose n + 1 is a prime and q is a primitive element in Zn+1. Let α be a primitive (n + 1)-st root of unity. Then α generates an optimal normal basis of Fqn over Fq. 2. (Type II optimal normal basis) Suppose 2n+1 is a prime and let γ be a primitive (2n+1)-st root of unity. Assume that the multiplicative group of Z2n+1 is gener-ated by 2 and −1 (that is, either 2 is a primitive element in Z2n+1, or 2n + 1 ≡3 (mod 4) and 2 generates the quadratic residues in Z2n+1). Then α = γ + γ−1 generates an optimal normal basis of F2n over F2. 118 Handbook of Finite Fields 5.3.7 Theorem (Optimal normal basis theorem) Every optimal normal basis is equivalent to either a Type I or a Type II optimal normal basis. More precisely, suppose Fqn has an optimal normal basis over Fq generated by α and let b = Tr(α) ∈Fq. Then one of the following must hold: 1. n + 1 is a prime, q is primitive modulo n + 1 and −α/b is a primitive (n + 1)-st root of unity; 2. q = 2v with gcd(v, n) = 1, 2n + 1 is a prime such that 2 and −1 generate the multiplicative group of Z2n+1, and α/b = γ + γ−1 for some primitive (2n + 1)-st root of unity γ. 5.3.8 Remark Gao and Lenstra prove a more general version of the optimal normal basis theorem. They show that if a ﬁnite Galois extension L/K, where K is an arbitrary ﬁeld, has an optimal normal basis, say generated by α, then there is a prime number r, an r-th root of unity γ in some algebraic extension of L and a nonzero constant c ∈K so that one of the following holds: 1. α = cγ and L has degree r −1 over K (so the polynomial xr−1 + xr−2 + x + 1 is irreducible over K); 2. α = c(γ + γ−1) and L has degree (r −1)/2 over K (so the minimal polynomial of γ + γ−1 over K has degree (r −1)/2). 5.3.9 Theorem Let F(x) = xq+1 + dxq −(ax + b) with a, b, d ∈Fq and b ̸= ad. Let f be an irreducible factor of F of degree n > 1 and let α be a root of f. Then all the roots of f are αi = αqi = ϕi(α), i = 0, 1, . . . , n −1, where ϕ(x) = (ax + b)/(x + d). If τ = TrFqn/Fq(α) ̸= 0, then (α0, α1, . . . , αn−1) is a normal basis of Fqn over Fq such that α        α0 α1 α2 . . . αn−1        =        τ ∗ −en−1 −en−2 · · · −e1 e1 en−1 e2 en−2 . . . ... en−1 e1               α0 α1 α2 . . . αn−1        +       b∗ b b b b       , (5.3.1) where e1 = a, ei+1 = ϕ(ei) (i ≥1), b∗= −b(n −1) and τ ∗= τ −ϵ with ϵ = n−1 X i=0 ei =          (n −1)(a −d)/2 if p ̸= 2, a = d if p = n = 2, a −d if p = 2 and n ≡3 (mod 4), 0 if p = 2 and n ≡1 (mod 4). 5.3.10 Corollary The following are two special cases of the above theorem. 1. For every a, β ∈F∗ q with TrFq/Fp(β) = 1, xp −1 β axp−1 −1 β ap is irreducible over Fq and its roots form a normal basis of Fqp over Fq of complexity at most 3p−2. This corresponds to the case of Theorem 5.3.9 with n = p, e1 = a, ϕ(x) = ax/(x + a), b = b∗= 0, and τ ∗= a/β if p ̸= 2 and τ ∗= a/β −a if p = 2. Bases 119 2. Let n be any factor of q −1. Let β ∈Fq have multiplicative order t such that gcd(n, (q −1)/t) = 1 and let a = β(q−1)/n. Then xn −β(x −a + 1)n is irreducible over Fq and its roots form a normal basis of Fqn over Fq with complexity at most 3n −2. This corresponds to the case of Theorem 5.3.9 with e1 = a, ϕ(x) = ax/(x + 1), b = b∗= 0, and τ ∗= −n(a −1)β/(1 −β) −ϵ, with ϵ given as in Theorem 5.3.9 with d = 1. 5.3.11 Conjecture If there does not exist an optimal normal basis of Fqn over Fq, then the complexity of a normal basis of Fqn over Fq is at least 3n −3. 5.3.12 Remark Explicit constructions of low complexity normal bases beyond the optimal normal bases and the constructions given in Theorem 5.3.10 are rare. In Section 5.3.2 we give a generalization of optimal normal bases arising from Gauss periods. Below, we illustrate how to construct new normal bases of low complexity arising from previously known normal bases. 5.3.13 Proposition [1172, 2578, 2580] Suppose gcd(m, n) = 1 and α and β generate normal bases A and B for Fqm and Fqn over Fq, respectively. By Proposition 5.2.3, αβ generates a normal basis N for Fqmn over Fq. Furthermore, we have CN = CACB and if α and β both generate optimal normal bases, then CN = 4mn −2m −2n + 1. 5.3.14 Proposition Let n = mk and suppose α ∈ Fqn generates a normal basis (α0, α1, . . . , αn−1) over Fq with multiplication table T = (tij) for 0 ≤i, j ≤n −1. Then β = TrFqn/Fqm(α) = α0 + αm + α2m + · · · + α(k−1)m generates a normal basis (β0, β1, . . . , βm−1) for Fqm over Fq with ββi = m−1 X j=0 sijβj, 0 ≤i ≤m −1, where sij = X 0≤u,v≤k−1 tum+i,vm+j, 0 ≤i, j ≤m −1. 5.3.15 Corollary [633, 634, 1931] Let n = mk. Upper-bounds on the complexity obtained from traces of optimal normal bases of Fqn over Fq are given in Table 5.3.1. Type I (q odd): Type I (q = 2): Type II (q = 2): m even, k odd, p \| k km −(k + 1)/2 – – m even, k odd, k ≡1 (mod p) (k + 1)m −(3k + 1)/2 (k + 1)m −3k + 2∗ 2km −2k + 1 m even, k odd, all other k (k + 1)m −(3k + 1)/2 – – k even, p \| k km −k/2† km −k + 1 2km −2k + 1 k even, k ≡2 (mod p) (k + 1)m −3k/2 + 1† km −k + 1 2km −2k + 1 k even, all other k (k + 1)m −k – – Table 5.3.1 Upper-bounds on the complexity obtained from of traces of optimal normal bases of Fqn over Fq, where n = mk. ∗Tight when k = 3; †tight when k = 2, 3. 120 Handbook of Finite Fields 5.3.2 Gauss periods 5.3.16 Deﬁnition Let r = nk + 1 be a prime not dividing q and let γ be a primitive r-th root of unity in Fqnk. Furthermore, let K be the unique subgroup of order k in Z∗ r and Ki = {a · qi : a ∈K} ⊆Z∗ r be cosets of K, 0 ≤i ≤n −1. The elements αi = X a∈Ki γa ∈Fqn, 0 ≤i ≤n −1, are Gauss periods of type (n, k) over Fq. 5.3.17 Theorem [1180, 2951] Let αi ∈Fqn be Gauss periods of type (n, k) as deﬁned in Deﬁni-tion 5.3.16. The following are equivalent: 1. N = (α0, α1, . . . , αn−1) is a normal basis of Fqn over Fq; 2. gcd(nk/e, n) = 1, where e is the order of q modulo r; 3. the union of K0, K1, . . . , Kn−1 is Z∗ r; equivalently, Z∗ r = ⟨q, K⟩. 5.3.18 Remark Gauss periods of type (n, 1) deﬁne Type I optimal normal bases and Gauss periods of type (n, 2) deﬁne Type II optimal normal bases when q = 2. 5.3.19 Remark For the remainder of this section, we are concerned with Gauss periods which are admissible as normal bases, that is, where the properties in Theorem 5.3.17 hold. When the characteristic p does not divide n, the existence of admissible Gauss periods of type (n, k) is shown assuming the ERH in [14, 159] for any n with k ≤(cn)3(log(np))2. For any k and prime power q, assuming the GRH, there are inﬁnitely many n such that there is an admissible Gauss period of Fqn over Fq . In contrast, when p divides n, contains necessary and suﬃcient conditions for admissible Gauss periods, thus showing the non-existence of admissible Gauss periods in certain cases. 5.3.20 Proposition There is no admissible Gauss period of type (n, k) over F2 if 8 divides nk. 5.3.21 Deﬁnition Let Ki be deﬁned as in Deﬁnition 5.3.16 for i = 0, 1, . . . , n −1. The cyclotomic numbers are given by cij = \|(1 + Ki) ∩Kj\|. 5.3.22 Proposition [259, 1180] Let N = (α0, α1, . . . , αn−1) be the normal basis arising from Gauss periods of type (n, k) for Fqn over Fq. Let j0 < n be the unique index such that −1 ∈κj0, and let δj = 1 if j = j0 and 0 if j ̸= j0. Then ααi = δik + n−1 X j=0 cijαj, 0 ≤i ≤n −1, hence CN ≤(n −1)k + n. 5.3.23 Proposition [139, 259] Let p be the characteristic of Fq and let N = (α0, α1, . . . , αn−1) be the normal basis of Fqn over Fq arising from Gauss periods of type (n, k). 1. If p divides k, then CN ≤nk −1. 2. If p = 2, then ( kn −(k2 −3k + 3) ≤CN ≤(n −1)k + 1 if k even, (k + 1)n −(k2 −k + 1) ≤CN ≤(n −2)k + n + 1 if k odd. Bases 121 3. If q = 2 and k = 2vr, where either r = 1 or both r is an odd prime and v ≤2, then the lower bounds above are tight for suﬃciently large n. 5.3.24 Problem Find the complexity of Gauss periods of type (n, k) over F2 for all n when k is not a prime, twice an odd prime, or four times an odd prime. 5.3.25 Remark [139, 634] The complexities of normal bases arising from Gauss periods of type (n, k), 2 ≤k ≤6, are given in Table 5.3.2 for all characteristics p when n > p. Type (n, 1) Type (n, 2) Type (n, 3) Type (n, 4) all p p = 2 p > 2 p = 2 p = 3 p > 3 p = 2 p = 3 p > 3 2n −1 2n −1 3n −2 4n −7 3n −2 4n −4 4n −7 5n −7 5n −6 Type (n, 5) Type (n, 6) p = 2 p = 3 p = 5 p > 5 p = 2 p = 3 p = 5 p > 5 6n −21 6n −11 5n −7 6n −11 6n −21 6n −11 7n −15 7n −14 Table 5.3.2 Complexities of normal bases from Gauss periods of Type (n, k), 2 ≤k ≤6, n > p. 5.3.26 Remark Let q = 2. Proposition 5.3.13 can be used to create normal bases of large extension degree by combining normal bases of subﬁelds with coprime degree. By Proposition 5.3.20 Gauss periods of type (n, k) do not exist when 8 divides nk. Hence, Proposition 5.3.13 cannot be used to construct low complexity normal bases when the degree is a prime power. Thus, when n is a prime power (speciﬁcally a power of two), there are no constructions of low-complexity normal bases arising from the above propositions. 5.3.27 Problem Find explicit constructions of low-complexity normal bases of F2n over F2 when n is a power of two. 5.3.28 Remark Normal bases of low complexity are useful in fast encoding and decoding of network codes, see for more details. 5.3.3 Normal bases from elliptic periods 5.3.29 Remark Proposition 5.2.20, Theorem 5.3.9, and its corollaries show how the multiplicative group of Fq or Fq2 can be used to construct irreducible polynomials and normal bases for those degrees n whose prime factors divide q −1 or q + 1. Also, Gauss periods use the multiplicative group of Fqr−1 for some prime r. Couveignes and Lercier show how these methods can be generalized by using elliptic curve groups. The normal bases from their construction may not have low complexity, but these bases still allow a fast algorithm for multiplication. We outline their construction below; for more details on how to perform fast multiplication using elliptic periods, we refer the reader to . For properties of elliptic curves, see Section 12.2. 5.3.30 Remark Let E be an elliptic curve over Fq deﬁned by a Weierstrass equation Y 2 + a1XY + a3Y = X3 + a2X2 + a4X + a6, where ai ∈Fq. The points of E over every extension of Fq form an additive group with the point O at inﬁnity as the identity. The order of the group E(Fq) is q +1−t for some integer t with \|t\| ≤2√q. Let n > 1 be an integer such that E(Fq) has a cyclic subgroup F of order n. The quotient E′ = E/F is also an elliptic curve over Fq and there is an isogeny φ : E →E′, 122 Handbook of Finite Fields that has F as its kernel, and φ is deﬁned by rational functions in Fq[X, Y ]. For any point P ∈E, let x(P) denote the x-coordinate of P and similarly denote y(P), thus P = (x(P), y(P)). V´ elu gives a formula for E′ and φ. In fact, for P ∈E, φ(P) =  x(P) + X T ∈F \{O} (x(P + T) −x(T)) , y(P) + X T ∈F \{O} (y(P + T) −y(T))  . 5.3.31 Remark We describe here an explicit formula due to Kohel for E′ and φ when E is of the form E : Y 2 = X3 + aX + b. We denote by D the kernel polynomial given by D(X) = Y Q∈F \{O} (X −x(Q)) = Xn −c1Xn−1 + c2Xn−2 −c3Xn−3 + · · · + (−1)ncn ∈Fq[X]. Then, for P = (x, y) ∈E, φ(P) = N(x) D(x), y · N(x) D(x) ′! , where N(x) is determined by the equation: N(x) D(x) = nx −c1 −(3x2 + a)D′(x) D(x) −2(x3 + ax + b) D′(x) D(x) ′ . Furthermore, E′ is deﬁned by Y 2 = X3 + (a −5v)X + (b −7w), where v = a(n −1) + 3(c2 1 −2c2), w = 3ac1 + 2b(n −1) + 5(c3 1 −3c1c2 + 3c3). 5.3.32 Deﬁnition Let T ∈E(Fq) be a point of order n and φ be the corresponding isogeny with its kernel generated by T. For any point P ∈E(Fqn) with φ(P) ∈E′(Fq), let θ(P, T) denote the slope of the line passing through the two points T and P + T, that is θ(P, T) = y(P + T) −y(T) x(P + T) −x(T) ∈Fqn. The element θ(P, T) is an elliptic period over Fq. 5.3.33 Theorem Let T ∈E(Fq) be a point of order n ≥3 and φ be the corresponding isogeny with its kernel generated by T. Suppose there is a point P ∈E(Fqn) so that n P ̸= O in E and φ(P) ∈E′(Fq). Then either 1. the elliptic period θ(P, T) is a normal element of Fqn over Fq if the trace of θ(P, T) from Fqn to Fq is nonzero, or Bases 123 2. the element 1 + θ(P, T) is a normal element of Fqn over Fq if the trace of θ(P, T) is zero. 5.3.34 Example Consider the following curve over F7 E : y2 + xy −2y = x3 + 3x2 + 3x + 2. The point T = (3, 1) ∈E(F7) has order n = 5, so the subgroup F = ⟨T⟩has order 5. By V´ elu’s formula, the equation for E′ = E/F is E′ : y2 + xy −2y = x3 + 3x2 −3x −1, and the corresponding isogeny is φ(x, y) = x5 + 2x2 −2x −1 x4 + 3x2 −3 , x6 −3x4 + 3x3 −x2 + 3x −3 y + 3x5 + x4 + x3 + 3x2 −3x + 1 x6 + x4 −2x2 −1 ! . Take A = (4, 2) ∈E′(F7). We note that the polynomial f(X) = (X5 + 2X2 −2X −1) + 3(X4 + 3X2 −3) = X5 + 3X4 −3X2 −2X −3 is irreducible over F7. Hence F75 = F7[α], where α is a root of f. Compute β so that φ(α, β) = (4, 2). We ﬁnd that β = α4756 = −α3 −α2 + 3α + 2, and P = (α, β) ∈E(F75). We check that 5 P ̸= O in E and we note that P + T = (−3α4 + 3α2 + 2α −1, −α4 + α3 + α2 + 1). Hence θ(P, T) = Y (P + T) −Y (T) X(P + T) −X(T) = −α4 + α3 + 3α2 −3α −3 is a normal element in F75 over F7. 5.3.4 Complexities of dual and self-dual normal bases 5.3.35 Remark For the deﬁnition of dual and self-dual bases, see Deﬁnition 2.1.100. Self-dual nor-mal bases have been well studied due to their eﬃciency in implementation, see Section 16.7. A complete treatment of dual bases over ﬁnite ﬁelds can be found in [1631, Chapter 4], see also Sections 5.1 and 5.2. 5.3.36 Remark It is computationally easier to restrict an exhaustive search to self-dual normal bases. Geiselmann in [1263, 1631] computes the minimum complexity for a self-dual normal basis of F2n over F2 for all n ≤47. These computations are repeated for odd degrees n ≤45 in and the authors also give tables of minimum complexity self-dual normal bases over ﬁnite ﬁelds of odd characteristic and for extensions of F2ℓ, ℓ> 1. Some additional searches for self-dual normal bases can be found in . 5.3.37 Proposition Let N be a normal basis with multiplication table T. Then N is self-dual if and only if T is symmetric. 124 Handbook of Finite Fields 5.3.38 Proposition Let gcd(m, n) = 1. Suppose α and β generate normal bases A and B for Fqm and Fqn over Fq, respectively. Then γ = αβ generates a self-dual normal basis N for Fqmn over Fq if and only if both A and B are self-dual, as in Proposition 5.2.3. The complexity of the basis N is CN = CACB, as in Proposition 5.3.13. 5.3.39 Proposition [1632, 2422] Let n be even, α ∈F2n and γ = 1 + α. Then, 1. the element α generates a self-dual normal basis for F2n over F2 if and only if γ does; 2. if α and γ = 1+α generate self-dual normal bases B and ¯ B, respectively, for F2n over F2, then the complexities of B and ¯ B are related by C ¯ B = n2 −3n + 8 −CB. 5.3.40 Corollary Suppose n ≡2 (mod 4), then the following hold 1. the average complexity of a self-dual normal basis of F2n over F2 is 1 2(n2−3n+8); 2. if B is a self-dual normal basis for F2n over F2 , we have 2n −1 ≤CB ≤n2 −5n + 9, and one of the equalities holds if and only if either B or its complement ¯ B is optimal. 5.3.41 Proposition Let q be a power of a prime p. For any β ∈F∗ q with TrFq/Fp(β) = 1, xp −xp−1 −βp−1 is irreducible over Fq and its roots form a self-dual normal basis of Fqp over Fq with com-plexity at most 3p −2. The multiplication table is as in Theorem 5.3.10 with e1 = β, ei+1 = ϕ(ei) for i ≥1, ϕ(x) = βx/(x + β), τ ∗= 1 if p ̸= 2 and τ ∗= 1 −β if p = 2. 5.3.42 Proposition Let n be an odd factor of q −1 and let ξ ∈Fq have multiplicative order n. Then there exists u ∈Fq such that (u2)(q−1)/n = ξ. Let x0 = (1 + u)/n and x1 = (1 + u)/(nu). Then the monic polynomial 1 1 + u2 (x −x0)n −u2(x −x1)n is irreducible over Fq and its roots form a self-dual normal basis of Fqn over Fq. The multiplication table is as in Theorem 5.3.9 with a = (x0 −ξx1)/(1 −ξ), b = −x0x1, d = a −(x0 −x1) and τ = 1. 5.3.43 Proposition Let n be an odd factor of q + 1 and let ξ ∈Fq2 be a root of xq+1 −1 with multiplicative order n. Then there is a root u of xq+1 −1 such that (u2)q+1/n = ξ. Let x0 = (1 + u)/n and x1 = (1 + u)/(nu). Then 1 1 −u2 (x −x0)n −u2(x −x1)n is irreducible over Fq and its roots form a self-dual normal basis of Fqn over Fq. The multiplication table is as in Theorem 5.3.9 with a = (x1 −ξx0)/(1 −ξ), b = −x0x1, d = a −(x0 + x1) and τ = 1. Bases 125 5.3.4.1 Duals of Gauss periods 5.3.44 Proposition [1180, 1930] Let α be a type (n, k) Gauss period generating a normal basis N and let j0 = 0 if k is even and j0 = n/2 if k is odd. Then the element γ = αqj0 −k nk + 1 is dual to α, and hence γ generates the dual basis ˜ N of N. Furthermore, the complexity of the dual basis ˜ N is C ˜ N ≤ ( (k + 1)n −k if p ∤k, kn −1 if p \| k. 5.3.45 Corollary For n > 2, a normal basis of Fqn over Fq arising from Gauss periods of type (n, k) is self-dual if and only if k is even and divisible by the characteristic of Fq. In particular, Type II optimal normal bases are self-dual. 5.3.46 Proposition The complexity of the dual of a Type I optimal normal basis is 3n −2 if q is odd and 3n −3 if q is even. 5.3.47 Remark Upper bounds on the complexities of the dual basis of the Fqm-trace of optimal normal bases of Fqn over Fq, where n = mk, are given in Table 5.3.3. Type I (q odd) Type I (q even) Type II (q even) m odd (k + 2)m −2 (k + 2)(m −1) + 1 2k(m −1) + 1 m even (k + 3)m −k −4 (k + 3)m −2k −3 2k(m −1) + 1 Table 5.3.3 Upper bounds on complexities of the dual bases of the trace of optimal normal bases. 5.3.5 Fast arithmetic using normal bases 5.3.48 Remark In practical applications it is important to know how to do fast arithmetic in ﬁnite ﬁelds, for example addition, multiplication, and division; and for cryptographic applications it is also desirable to have elements of high orders and a fast algorithm for exponentiation. Details for the basic operations discussed in this section can be found in Section 11.1, see also . In hardware implementations, normal bases are often preferred, see Section 16.7 for details on hardware implementations. This subsection presents some theoretical results related to fast multiplication and exponentiation under normal bases generated by Gauss periods. 5.3.49 Remark Gao and Vanstone ﬁrst observed that a Type II optimal normal basis gen-erator has high order, which was proved later by von zur Gathen and Shparlinski ; for more details see Section 4.4. Computer experiments by Gao, von zur Gathen, and Pa-nario indicate that Gauss periods of type (n, k) with k > 2 also have high orders; however, it is still open whether one can prove a subexponential lower bound on their orders. 5.3.50 Problem Give tight bounds on the orders of Gauss periods of type (n, k), k > 2. 5.3.51 Proposition [1179, 1188] Suppose α ∈Fqn is a Gauss period of type (n, k) over Fq. Then for any integer 1 ≤t < qn −1, αt can be computed using at most n2k operations in Fq. 5.3.52 Theorem Suppose γ is an element of order r (not necessarily a prime) and α = X i∈K γi 126 Handbook of Finite Fields generates a normal basis N for Fqn over Fq, where K is a subgroup of Z× r . With Fqn represented under the normal basis N, we have 1. addition and subtraction can be computed in O(n) operations in Fq; 2. multiplication can be computed in O(r log r log log r) operations in Fq; 3. division can be computed in O(r log2 r log log r) operations in Fq; 4. exponentiation of an arbitrary element in Fqn can be computed in O(nr log r log log r) operations in Fq. 5.3.53 Remark Theorem 1.5 in tells us when and how to ﬁnd such a subgroup K in the above theorem. The element α can be a Gauss period or a generalized Gauss period, see Example 5.3.54 for more information. We outline the algorithm from for fast mul-tiplication and division. The basic idea is to convert the normal basis representation to a polynomial basis in the ring R = Fq[x]/(xr −1), do fast multiplication of polynomials in the ring, then convert the result back to the normal basis. More precisely, let γ and α be as in Theorem 5.3.52. The condition of the theorem implies that that K, qK, . . . , qn−1K are disjoint subsets of Zr and qnK = K. For 0 ≤j ≤n −1, let Kj = qjK ⊆Zr, and αj = X i∈Kj βi. Then (α0, α1, . . . , αn−1) is the normal basis generated by α, with the following property: α0 + α1 + · · · + αn−1 = −1. For each element A = a0α0 + a1α1 + · · · + an−1αn−1 ∈Fqn, where ai ∈Fq, we associate a polynomial A(x) = r−1 X i=0 uixi, where ui = aj if i ∈Kj for some j, and ui = 0 if i is not in any Kj. This can be viewed as a map from Fqn to R = Fq[x]/(xr −1). The map is in fact a ring homomorphism. Suppose we have two arbitrary elements A, B ∈Fqn. To compute AB, we ﬁrst write them as polynomials A(x), B(x) ∈Fq[x] of degree at most r −1 as above. Then we use a fast algorithm to compute the product polynomial C1(x) = A(x)B(x) of degree at most 2r −2. This step needs O(r log r log log r) operations in Fq, see . Next, we reduce C1 modulo xr −1 (just reduce the exponents of x modulo r) to get a polynomial C2(x) = c0 + c1x + · · · + cr−1xr−1. The coeﬃcients satisfy the property that ci = cj whenever i, j ∈Kℓfor some 0 ≤ℓ≤n−1. Since Pn−1 j=0 αj = −1, we conclude that AB = d0α0 + d1α1 + · · · + dn−1αn−1, where di = cj −c0 for any j ∈Ki. To compute A−1 (assuming A ̸= 0), we apply a fast gcd algorithm to the two polynomials A(x) and xr −1 to get a polynomial U(x) of degree at most r −1 so that A(x)U(x) ≡1 (mod xr −1). The element in Fqn corresponding to the polynomial U(x) is the desired inverse of A. The fast gcd step needs O(r log2 r log log r) operations in Fq, see . 5.3.54 Example (Generalized Gauss Periods [1047, 1174]) For any normal basis from Gauss periods of type (n, k), we can apply Theorem 5.3.52 to perform fast arithmetic in Fqn. To obtain an admissible Gauss period of type (n, k), r = nk + 1 must be a prime. Here we give an Bases 127 example of generalized Gauss periods where r is not prime. Suppose we want to perform fast arithmetic in F2954. Let n = 954 and note that the smallest k so that there is an admissible Gauss period of type (n, k) over F2 is k = 49. The corresponding r = nk + 1 = 46747 is a little big in this case. We observe that 954 = 106 · 9 and that there is an admissible Gauss period α1 of type (106, 1) over F2, and an admissible Gauss period α2 of type (9, 2). Then α = α1α2 is a normal element of F2n over F2. We construct this α as follows. Let r = (106 · 1 + 1)(9 · 2 + 1) = 2033, K = {1, 322}. Then K is a subgroup of Z× r satisfying the condition in Theorem 5.3.52. Let γ be any primitive r-th root of unity in an extension ﬁeld of F2. Then α = γ + γ322 is a generalized Gauss period that is normal for Fqn over Fq. Now we can apply Theorem 5.3.52 to perform fast arithmetic in Fqn with a much smaller r. In , it is shown how to ﬁnd generalized Gauss periods with minimum r and the related subgroups K; see [1174, Tables 2-4] for many more examples for which generalized Gauss periods are better than Gauss periods. 5.3.55 Example (Fast arithmetic under type II optimal normal bases) For type II optimal normal bases over F2, we describe below a slightly faster algorithm from [248, 1238]. Suppose 2n+1 is a prime and the multiplicative group of Z2n+1 is generated by −1 and 2. Let γ ∈F22n be an element of order 2n + 1. For any i ≥0, deﬁne γi = γi + γ−i. Then N = (γ1, γ2, . . . , γn) is a permutation of the normal basis for F2n over F2 generated by α = γ1 = γ + γ−1. We note that γ0 = 2 and γ1 + γ2 + · · · + γn = 1. To do fast multiplication and division in F2n, we ﬁrst perform a basis transition from N to the polynomial basis P = (α, α2, . . . , αn), then perform a fast multiplication of polynomials and ﬁnally transform the result back to the basis N. To do the basis transitions, we need the following properties: γi+j = γiγj + γj−i, for all i, j. To see how to go from the basis N to the basis P, suppose we have an expression A = a1γ1 + a2γ2 + · · · + aℓγℓ, where ai ∈F2 and ℓ≥1 is arbitrary. We want to express A as a combination of α, α2, . . . , αℓ over F2. Let m be a power of 2 so that ℓ/2 ≤m < ℓ. Then γm = αm. We observe that a1γ1 + a2γ2 + · · · + aℓγℓ = a1γ1 + · · · + amγm + am+1(γmγ1 + γm−1) + · · · + aℓ(γmγℓ−m + γm−(ℓ−m)), = (a1γ1 + · · · + amγm + am+1γm−1 + · · · + aℓγm−(ℓ−m)) +αm (am+1γ1 + am+2γ2 + · · · + aℓγℓ−m) . 128 Handbook of Finite Fields We note that the ﬁrst part is of the form U = u1γ1 + · · · + umγm, where ui ∈F2, which can be computed using m bit operations. Let V = am+1γ1 + am+2γ2 + · · · + aℓγℓ−m, where ℓ−m ≤m. Apply the method recursively to convert U and V into the power basis, say U = b1α + · · · + bmαm, V = bm+1α + · · · + bℓαℓ−m, where bi ∈F2. Then A = U + αmV = b1α + · · · + bmαm + bm+1αm+1 + · · · + bℓαℓ. This gives an algorithm for going from N to P using at most 1 2n log2(n) operations in F2, where log2(n) is the logarithm of n in base 2. By reversing the above procedure, we get an algorithm for going from P to N using at most 1 2n log2(n) operations in F2. This shows that multiplication in F2n under N can be computed by (a) two transforma-tions from N to P, (b) one multiplication of polynomials of degree at most n in F2[x], and (c) one transformation from (α, α2, . . . , α2n) to (γ1, γ2, . . . , γ2n) which is easily converted to N, as γn+1+i = γn−i. The number of bit operations used is the cost for one multiplication of polynomials of degree at most n, plus 2n log2 n bit operations for the basis transformations. Finally, for division in F2n, we need to precompute the minimal polynomial of α and apply a fast gcd algorithm for polynomials of degree at most n in F2[x]. See Also §2.2 For standards requiring normal basis arithmetic. §5.2 For general results on normal bases. §11.1 For basic operations over ﬁnite ﬁelds. §16.7 For hardware implementarions of ﬁnite ﬁelds arithmetic. [1179, 1188, 1240] For orders and cryptographic applications of Gauss Periods. References cited: [14, 130, 139, 159, 248, 259, 308, 633, 634, 746, 1047, 1172, 1174, 1179, 1180, 1184, 1188, 1227, 1236, 1238, 1240, 1263, 1631, 1632, 1779, 1930, 1931, 2015, 2199, 2422, 2578, 2580, 2665, 2865, 2924, 2951, 2952, 3036]. 5.4 Completely normal bases Dirk Hachenberger, University of Augsburg We present some theoretical results concerning algebraic extensions of ﬁnite ﬁelds. The starting point is the Complete Normal Basis Theorem, which is a strengthening of the classical Normal Basis Theorem. The search for completely normal elements leads to an interesting structure theory for ﬁnite ﬁelds comprising a generalization of the class of ﬁnite Galois ﬁeld extensions to the class of cyclotomic modules. 5.4.1 The complete normal basis theorem 5.4.1 Remark Let Fq denote an algebraic closure of the ﬁnite ﬁeld Fq. The Frobenius automor-phism of Fq/Fq (throughout denoted by σ) is the ﬁeld automorphism mapping each θ ∈Fq Bases 129 to its q-th power θq. For every integer m ≥1 there is a unique subﬁeld Em of Fq such that Fq ⊆Em and \|Em\| = qm. As usual we write Fqm for Em. Given integers d, m ≥1, one has Fqd ⊆Fqm if and only if d\|m. Moreover, if d is a divisor of m, then Fqm/Fqd is a Galois ex-tension of degree m d ; its Galois group is cyclic and generated by σd (when restricted to Fqm). The (Fqm, Fqd)-trace of w ∈Fqm is Pm/d−1 i=0 wqdi, while Qm/d−1 i=0 wqdi is the (Fqm, Fqd)-norm of w. 5.4.2 Remark Recall from Deﬁnition 2.1.98 that θ ∈Fqm is a normal element of Fqm over Fq provided its conjugates θ, θq, . . . , θqm−1 under the Galois group of Fqm/Fq form an Fq-basis of Fqm. 5.4.3 Deﬁnition An element θ ∈Fqm is a completely normal element over Fq, provided θ (simultaneously) is a normal element of Fqm over Fqd for every intermediate ﬁeld Fqd of Fqm over Fq, i.e., for every divisor d ≥1 of m. A monic irreducible polynomial g ∈Fq[x] of degree m is completely normal over Fq if one (and hence all) of its roots is a completely normal element of Fqm/Fq. 5.4.4 Example (Based on [1387, 1388]) Let q = 7 and m = 3c, where c ≥1, and let η ∈F7 be a primitive 3c+1-th root of unity. Then F73c is obtained by adjoining η to F7. For every i = 1, . . . , c, let τ(i) := 3⌊(i−1)/2⌋. Then θ := 1 + c X i=1 2·τ(i) X j=1, gcd(3,j)=1 ηj·3c−i is a completely normal element of F73c over F7. 5.4.5 Theorem (Complete Normal Basis Theorem) Given any ﬁnite ﬁeld Fq and any integer m ≥1 there exists a completely normal element of Fqm over Fq. 5.4.6 Remark The Complete Normal Basis Theorem was ﬁrst proved by Blessenohl and Johnsen in 1986. A simpliﬁcation of the proof is given in . The analogous result for (ﬁnite dimensional) Galois extensions over inﬁnite ﬁelds was already settled by Faith in 1957. For an outline of the general proof, see [1389, Chapter I]. Often, an element that is completely normal is also called completely free (or vollst¨ andig frei or vollst¨ andig regul¨ ar) [315, 317, 1386, 1387, 1388, 1389, 2090]. 5.4.7 Remark Let m = Qk i=1 rνi i be the prime power decomposition of m. For every i = 1, . . . , k assume that θi is completely normal in Fqrνi i over Fq. Then θ := Qk i=1 θi is completely normal in Fqm/Fq. This is an application of Hilfssatz 4.4 of (see also the Reduction Theorem in Section 4 of ); it reduces the existence problem for completely normal elements to extensions of prime power degree. 5.4.8 Remark Concerning the existence of completely normal elements for prime power extensions Fqrn /Fq for some prime r, the cases where r = p (the characteristic of Fq) and r ̸= p have to be dealt with separately. The case of equal characteristic is covered by Subsection 5.4.2 on completely basic extensions. The case r ̸= p is much harder to establish; it is covered by the structure theory on completely normal elements to be outlined in Subsections 5.4.3-5.4.6. For explicit constructions (in the spirit of Example 5.4.4) of completely normal elements in Fqrn over Fq with r ̸= p, see [1387, 1388]. 130 Handbook of Finite Fields 5.4.9 Deﬁnition Let M be a nonempty set of positive integers such that m ∈M and d\|m imply d ∈M, and k, n ∈M imply lcm(k, n) ∈M (where lcm denotes the least common multiple). Then M is a Steinitz number. 5.4.10 Remark The intermediate ﬁelds of Fq/Fq are in one-to-one correspondence with the Steinitz numbers (see Brawley and Schnibben ). The ﬁeld corresponding to the Steinitz number M is the union FqM := S m∈M Fqm. This algebraic extension of Fq is inﬁnite if and only if M is inﬁnite. Any ﬁnite Steinitz number is of the form {d : d ∈Z, d ≥1, d\|m} for some positive integer m; the corresponding ﬁeld is Fqm. 5.4.11 Deﬁnition Let M be a Steinitz number. Then a sequence (wm)m∈M of elements of Fq is trace-compatible, if for all d, m ∈M with d\|m the (Fqm, Fqd)-trace of wm is equal to wd. Similarly, the sequence is norm-compatible, if the (Fqm, Fqd)-norm of wm is equal to wd for every d, m ∈M such that d\|m. 5.4.12 Remark For an inﬁnite Steinitz number M a trace-compatible sequence (wm)m∈M such that every wm is a normal element of Fqm/Fq can be interpreted as an (inﬁnite) normal ba-sis for FqM over Fq. The existence of sequences of that kind in inﬁnite Galois extensions over an arbitrary ﬁeld is proved by Lenstra . Representations of ﬁnite ﬁelds within com-puter algebra systems which rely on subﬁeld embeddings and trace-compatible sequences of normal elements are studied by Scheerhorn [2537, 2538]. For ﬁnite ﬁelds, provides an elementary proof of the existence of normal trace-compatible sequences for the entire algebraic closure Fq/Fq (i.e., for the Steinitz number N). 5.4.13 Theorem [1389, Section 26] Consider the Steinitz number N and let Fq be any ﬁnite ﬁeld. Then there exists a trace-compatible sequence (wm)m∈N in Fq such that, for every m, the element wm is completely normal in Fqm over Fq. 5.4.2 The class of completely basic extensions 5.4.14 Deﬁnition A Galois ﬁeld extension Fqm/Fq is completely basic, if every normal element of Fqm/Fq is completely normal in Fqm/Fq. 5.4.15 Remark The notion of a completely basic extension in the context of a general ﬁnite di-mensional Galois extension goes back to Faith . Blessenohl and Johnsen have characterized the completely basic extensions among the abelian extensions. Previously, Blessenohl considered cyclic extensions of prime power degree. Meyer has ex-tended to a certain class of not necessarily abelian Galois extensions. For ﬁnite ﬁelds an elementary proof of Theorem 5.4.18 below is given in Section 15 of . 5.4.16 Deﬁnition For relatively prime integers q, ℓ≥1, the order of q modulo ℓis the least integer n ≥1 such that qn ≡1 (mod ℓ); it is denoted by ordℓ(q). 5.4.17 Deﬁnition When considering a ﬁnite ﬁeld with characteristic p, for an integer t ≥1 the p-free-part t′ of t is the largest divisor of t which is relatively prime to p. 5.4.18 Theorem [1389, Section 15] The following are equivalent: 1. Fqm is completely basic over Fq. Bases 131 2. For every prime divisor r of m, every normal element of Fqm/Fq is normal in Fqm/Fqr. 3. For every prime divisor r of m, the number ord(m/r)′(q) is not divisible by r. 5.4.19 Example Let Fq be any ﬁnite ﬁeld. Then Fqm is completely basic over Fq in all the following cases: 1. m = r or m = r2, where r is a prime; 2. m divides q −1; 3. m = pb, where p is the characteristic of Fq and b ≥0 is any integer. 5.4.20 Remark Actually, in the latter case of Example 5.4.19, θ is a normal element for Fqpb /Fq if and only if the (Fqpb , Fq)-trace of θ is non-zero (see [1389, Section 5]). For this class of ﬁeld extensions, Blake, Gao, and Mullin provide an iterative construction of completely normal elements. 5.4.3 Cyclotomic modules and complete generators 5.4.21 Remark In the present subsection the class of ﬁnite extensions of a ﬁnite ﬁeld is generalized to the class of cyclotomic modules; thereby, completely normal elements are generalized to complete generators of cyclotomic modules. This generalization is necessary in order to understand how completely normal elements are additively composed. The main references are [1389, 1390]. 5.4.22 Remark Consider again an algebraic closure Fq of Fq. For every integer d ≥1 the additive group of Fq is equipped with a module structure over the algebra Fqd[x] of polynomials in the variable x. The corresponding scalar multiplication (of ﬁeld elements by polynomials) is carried out by ﬁrst evaluating a polynomial f ∈Fqd[x] at σd (the Frobenius automorphism over Fqd) and afterwards by applying the Fqd-endomorphism f(σd) to an element ω ∈Fq, resulting in f(σd)(ω). In this context, Fq is an Fqd[x]-module (with respect to σd). For θ ∈Fq and an integer d ≥1 let Fqd[x]θ := {h(σd)(θ) : h ∈Fqd[x]} denote the Fqd[x]-submodule generated by θ. 5.4.23 Deﬁnition For θ ∈Fq and an integer d ≥1, the qd-order of θ is the monic polynomial g ∈Fqd[x] of least degree such that g(σd)(θ) = 0. The qd-order of θ is denoted by Ordqd(θ). 5.4.24 Theorem [1389, Section 8] Let d ≥1 be an integer. 1. The ﬁnite Fqd[x]-submodules of Fq correspond bijectively to the monic polynomi-als of Fqd[x] that are not divisible by x: If f ∈Fqd[x] is monic and f(0) ̸= 0, the corresponding Fqd[x]-submodule is the kernel of the Fqd-linear mapping f(σd). 2. Every ﬁnite Fqd[x]-submodule of Fq is cyclic. If U is the ﬁnite Fqd[x]-submodule of Fq corresponding to the monic polynomial f ∈Fqd[x] (not divisible by x), then ω ∈U if and only if Ordqd(ω) divides f, and Fqd[x]ω = U if and only if Ordqd(ω) = f. 5.4.25 Remark For integers m, d ≥1 with d\|m, the ﬁeld Fqm is the Fqd[x]-submodule of Fq corresponding to the polynomial xm/d −1 ∈Fqd[x]. The generators of Fqm as Fqd[x]-module are precisely the normal elements of Fqm over Fqd. 132 Handbook of Finite Fields 5.4.26 Deﬁnition For an integer k ≥1 which is relatively prime to the characteristic p of the underlying ﬁeld Fq, let Φk(x) ∈Fq[x] denote the k-th cyclotomic polynomial (see Deﬁnition 2.1.121). Given a further integer t ≥1, the generalized cyclotomic polynomial corresponding to the pair (k, t) is the polynomial Φk(xt) ∈Fq[x]. 5.4.27 Deﬁnition For an integer m ≥1, its square-free part ν(m) is the product of the distinct prime divisors of m. We let ν(1) := 1. 5.4.28 Proposition [1389, Section 18] Consider Fq[x] and let (k, t) be as in Deﬁnition 5.4.26. Write t = pbt′, where p is the characteristic of Fq and t′ the p-free part of t. Then 1. Φk(xt) = Φk(xt′)pb. 2. If d is a common divisor of k and t, then Φk(xt) = Φkd(xt/d). In particular, Φk(xt) = Φν(k)(xkt/ν(k)). 3. If k and t are relatively prime, then Φk(xt) = Q e\|t′ Φke(x)pb. The latter is the canonical decomposition of Φk(xt) over Fq. 5.4.29 Remark Observe that diﬀerent pairs (k, t) and (ℓ, s) may lead to the same generalized cyclotomic polynomial. However, Φk(xt) = Φℓ(xs) if and only if ν(k) = ν(ℓ) and kt ν(k) = ℓs ν(ℓ). 5.4.30 Deﬁnition Consider an algebraic closure Fq of Fq. Given a pair (k, t) as in Deﬁnition 5.4.26, the cyclotomic module corresponding to (k, t) is the kernel of Φk(σt), where σ is the Frobenius automorphism of Fq/Fq. Throughout, this Fq[x]-submodule is denoted by Ck,t, hence Ck,t = {θ ∈Fq : Φk(σt)(θ) = 0}. The module character of Ck,t is the number kt ν(k). 5.4.31 Remark In the setting of Deﬁnition 5.4.30, one has λω ∈Ck,t for all ω ∈Ck,t if and only if λ ∈Fqn where n = kt ν(k) [1389, Section 18]. Therefore, Ck,t carries the structure of an Fqd[x]-module if and only if d is a divisor of kt ν(k). This motivates the notion of the module character in Deﬁnition 5.4.30. 5.4.32 Deﬁnition Let Ck,t be a cyclotomic module over Fq. An element θ is a complete generator for Ck,t over Fq, provided θ (simultaneously) generates Ck,t as an Fqd[x]-module for every divisor d of its module character kt ν(k). 5.4.33 Theorem (Complete Cyclotomic Generator Theorem; [1389, Section 18]) Given any ﬁnite ﬁeld Fq and any cyclotomic module Ck,t over Fq, there exists a complete generator for Ck,t over Fq. 5.4.34 Remark Theorem 5.4.33 generalizes the Complete Normal Basis Theorem 5.4.5 from the class of ﬁnite ﬁeld extensions of Fq to the class of cyclotomic modules over Fq: If (k, t) = (1, m), then the cyclotomic module Ck,t is equal to the extension ﬁeld Fqm; over Fq, the module character of Fqm is equal to m and the complete generators of Fqm/Fq are exactly the completely normal elements of Fqm over Fq. The following theorem generalizes Remark 5.4.7. 5.4.35 Theorem (Cyclotomic Reduction Theorem; [1389, Section 25] and [1390, Section 3]) Con-sider two cyclotomic modules Ck,s and Cℓ,t over Fq. Assume that ks and ℓt are relatively prime. Then, for θ ∈Ck,s and ω ∈Cℓ,t the following two assertions are equivalent: Bases 133 1. θ and ω are complete generators for Ck,s and Cℓ,t over Fq, respectively. 2. θ · ω is a complete generator for the cyclotomic module Ckℓ,st over Fq. 5.4.4 A decomposition theory for complete generators 5.4.36 Deﬁnition Consider a generalized cyclotomic polynomial Φk(xt) ∈Fq[x]. A set ∆⊆Fq[x] of generalized cyclotomic polynomials is a cyclotomic decomposition of Φk(xt), if the members of ∆are pairwise relatively prime and Φk(xt) = Y Ψ(x)∈∆ Ψ(x). In that case, i(∆) denotes a corresponding set of pairs (ℓ, s) with Φℓ(xs) ∈∆. 5.4.37 Proposition [1389, Section 19] Let ∆be a cyclotomic decomposition of Φk(xt) ∈Fq[x]. Then 1. Ck,t = L (ℓ,s)∈i(∆) Cℓ,s is a decomposition into a direct sum of cyclotomic mod-ules. 2. For θ ∈Ck,t let θ = P (ℓ,s)∈i(∆) θℓ,s be the corresponding decomposition into its ∆-components. If θ is a complete generator of Ck,t over Fq, then (necessarily) for every (ℓ, s) ∈i(∆) the component θℓ,s is a complete generator of Cℓ,s over Fq. 5.4.38 Deﬁnition Let ∆be a cyclotomic decomposition of Φk(xt) considered over Fq. Then ∆ is an agreeable decomposition, provided the following holds: if for every (ℓ, s) ∈i(∆), one chooses any element θℓ,s that is a complete generator of Cℓ,s over Fq, then the sum θ = P (ℓ,s)∈i(∆) θℓ,s is a complete generator of Ck,t over Fq. 5.4.39 Theorem (Complete Decomposition Theorem; [1389, Section 19] and [1390, Section 5]) Consider a generalized cyclotomic polynomial Φk(xt) over the ﬁnite ﬁeld Fq with charac-teristic p. Let r be a prime divisor of t. Assume that r ̸= p and that r does not divide k. Then ∆r := {Φk(xt/r), Φkr(xt/r)} is a cyclotomic decomposition of Φk(xt). Moreover, the following two statements are equiv-alent: 1. ∆r is an agreeable decomposition over Fq. 2. ordν(kt′)(q) is not divisible by ra, where a ≥1 is maximal such that ra divides t (recall that t′ is the p-free part of t and ν(kt′) is the square-free part of kt′). 5.4.40 Remark Consider Φk(xt) and ∆r as in Theorem 5.4.39. Then the module character of each cyclotomic module corresponding to a member of ∆r is equal to kt rν(k) and therefore a proper divisor of the module character kt ν(k) of Ck,t over Fq. 5.4.41 Example [1389, Section 19] Consider an extension Fqm/Fq, where m > 1 is not a power of the characteristic p. Let r\|m be the largest prime divisor of m that is diﬀerent from p. Then {xm/r −1, Φr(xm/r)} is an agreeable decomposition of xm −1 over Fq. If in particular m = rn, this decomposition is equal to {xrn−1 −1, Φrn(x)}. The Complete Decomposition Theorem 5.4.39 may then be applied to xrn−1 −1 if n ≥2, and an induction argument 134 Handbook of Finite Fields shows that the canonical decomposition {x −1, Φr(x), Φr2(x), . . . , Φrn(x)} is an agreeable decomposition of xrn −1 over Fq. 5.4.42 Remark In Section 6 of a Uniqueness Theorem is proved: Starting from a generalized cyclotomic polynomial Φk(xt) over Fq, one obtains a unique (ﬁnest) agreeable decomposi-tion of Φk(xt) by a recursive application of the Complete Decomposition Theorem 5.4.39 independent of the order the various primes r are chosen. 5.4.43 Example [1389, Section 19] The set {x −1, Φ2(x), Φ4(x), Φ3(x4), Φ9(x4), Φ7(x36)} is an agreeable decomposition of x252 −1 over Fq, where q is any prime power relatively prime to 252. When specializing q to be equal to 5, the Complete Decomposition Theorem may be applied several further times to reach the following agreeable decomposition of x252 −1 over F5: {x −1, Φ2(x), Φ4(x), Φ3(x2), Φ12(x), Φ9(x2), Φ36(x), Φ7(x6), Φ28(x3), Φ63(x2), Φ252(x)}. 5.4.5 The class of regular extensions 5.4.44 Remark Consider a cyclotomic module Ck,t over Fq, where k and t are relatively prime. Let t = pbt′, where p is the characteristic of Fq and t′ is the p-free part of t. The canonical de-composition {Φke(x)pb : e\|t′} is (by deﬁnition) the ﬁnest possible cyclotomic decomposition of Φk(xt). By Theorem 19.10 of , the canonical decomposition is agreeable provided that t′ and ordν(kt′)(q) are relatively prime. (Recall that ν(kt′) is the square-free part of kt′.) 5.4.45 Theorem [1389, Section 19] Let Fq be a ﬁnite ﬁeld and p its characteristic. Write m = m′pb (with m′ being the p-free part of m). Then the canonical decomposition {Φd(x)pb : d\|m′} of xm −1 is agreeable over Fq if and only if m′ and ordν(m′)(q) are relatively prime. 5.4.46 Deﬁnition Consider a cyclotomic module Ck,t over Fq, with k and t being relatively prime. Then Ck,t is regular over Fq, provided that ordν(kt′)(q) and kt are relatively prime (t′ is the p-free part of t and ν(kt′) is the square-free part of kt′). In particular, in the case where (k, t) = (1, m), the extension Fqm is regular over Fq, provided that m and ordν(m′)(q) are relatively prime. 5.4.47 Example These examples (taken from [1391, Section 1]), indicate that the class of regular extensions is quite large. Given any ﬁnite ﬁeld Fq, we have that Fqm is regular over Fq in all the following cases: 1. m is a power of a prime r; 2. ν(m) divides ν(q −1); 3. m is a power of a Carmichael number (a Carmichael number is an odd composite integer n ≥1 such that r −1 divides n −1 for every prime divisor r of n; by Alford, Granville, and Pomerance there are inﬁnitely many Carmichael numbers, examples being 561, 1105, 1729, and 2465); 4. m has all its prime divisors from {7, 11, 13, 17, 19, 31, 41, 47, 49, 61, 73, 97, 101, 107, 109, 139, 151, 163, 167, 173, 179, 181, 193}, without any restriction on their multiplicity. 5.4.48 Remark Any completely basic extension (Deﬁnition 5.4.14) is regular. In order to compare the completely basic extensions with the regular ones, consider a ﬁnite set π of prime num-bers, and let ν be the product of all r ∈π. Let further N(π) := {m ∈Z : m ≥1, ν(m)\|ν}. If Fqν is completely basic over Fq, then Fqm is regular over Fq for every m ∈N(π). In Bases 135 contrast, the subset of those m ∈N(π) for which Fqm is completely basic over Fq is only ﬁnite (see [1391, Proposition 3.3]). 5.4.6 Complete generators for regular cyclotomic modules 5.4.49 Remark When considering the complete generation of a regular cyclotomic module Ck,t over Fq, due to Remark 5.4.44, one can independently work on the components (which are also regular) arising from the canonical decomposition of Φk(xt). It is therefore suﬃcient to consider the subclass of (regular) cyclotomic modules of the form Cn,pb (corresponding to the generalized cyclotomic polynomial Φn(x)pb), where p is the characteristic of Fq and n is relatively prime to q. 5.4.50 Deﬁnition Let p be the characteristic of Fq and n be relatively prime to q. Assume that Cn,pb is a regular cyclotomic module over Fq. Write n = 2c · n with n odd. Then Cn,pb is exceptional over Fq, provided the following conditions are satisﬁed: q ≡3 (mod 4) and c ≥3 and the order of q modulo 2c is equal to 2. In all other cases, Cn,pb is non-exceptional over Fq. 5.4.51 Remark For an integer n ≥1 let π(n) denote the set of prime divisors of n. If n and q are relatively prime, the order of q modulo n is of the form ordn(q) = ordν(n)(q) · Y r∈π(n) rα(r), where α(r) ≥0 for all r ∈π(n). Assume next that Cn,pb is a regular cyclotomic module over Fq, hence ordν(n)(q) is relatively prime to npb. Let further τ = τ(q, n) := Y r∈π(n) r⌊α(r)/2⌋. 5.4.52 Theorem [1389, Section 20] Let p be the characteristic of Fq and n be relatively prime to q. Assume that Cn,pb is a regular cyclotomic module over Fq. Let τ = τ(q, n) be as in Remark 5.4.51. Then τ divides n ν(n) in general, and 2τ divides n ν(n) if Cn,pb is exceptional. Moreover the following hold: 1. If Cn,pb is non-exceptional, then θ is a complete generator of Cn,pb over Fq if and only if Ordqτ (θ) = Φn/τ(x)pb. 2. If Cn,pb is exceptional, then θ is a complete generator of Cn,pb over Fq if and only if Ordqτ (θ) = Φn/τ(x)pb and Ordq2τ (θ) = Φn/(2τ)(x)pb. 5.4.53 Theorem [1389, Section 21] Let p be the characteristic of Fq and n be relatively prime to q. Assume that Cn,pb is a regular cyclotomic module over Fq. Let τ = τ(q, n) be as in Remark 5.4.51 and let ϕ denote Euler’s function (see Deﬁnition 2.1.43). Then the number of complete generators of Cn,pb over Fq is equal to 1. qordn(q)/τ −1 τϕ(n)/ordn(q) · q(pb−1)ϕ(n), if Cn,pb is non-exceptional; 2. q2ordn(q)/τ −4qordn(q)/τ + 3 τϕ(n)/(2ordn(q)) · q(pb−1)·ϕ(n), if Cn,pb is exceptional. 5.4.54 Theorem [1389, Section 21] Let Ck,t be a regular cyclotomic module over Fq and write t = pbt′, where p is the characteristic of Fq and t′ is the p-free part of t. Then the number of complete generators of Ck,t over Fq is at least (q −1)ϕ(k)t′ · qϕ(k)t′(pb−1), 136 Handbook of Finite Fields where ϕ denotes Euler’s function. Moreover, equality holds if and only if kt′ divides q −1, in which case θ is a complete generator of Ck,t if and only if the q-order of θ is Φk(xt). 5.4.55 Conjecture If Fqm/Fq is a regular extension, then the number of completely normal elements of Fqm/Fq is at least (q −1)m′ · qm′(pb−1) by Theorem 5.4.54 (where m = pbm′ with m′ being the p-free part of m and p the characteristic of Fq); moreover, equality holds if and only if m′ divides q −1, in which case Fqm is completely basic over Fq. We conjecture that, for any extension Fqm/Fq, the number of completely normal elements of Fqm/Fq is at least (q −1)m′ · qm′(pb−1). 5.4.56 Remark The aim of Theorem 5.4.57 below and its subsequent remarks is a construction (in the spirit of Example 5.4.4) of complete generators for regular cyclotomic modules over Fq that are of the form Cn,1. It is based on Theorem 5.4.52. A suitable application of the Complete Decomposition Theorem 5.4.39 in combination with the Cyclotomic Reduction Theorem 5.4.35 gives rise to a complete generator for a general cyclotomic module, in particular a completely normal element for an arbitrary extension Fqm over Fq. Thereby, Remark 5.4.20 on extensions of the form Fqpb may be used to cover the cases Ck,t and Fqm, where the characteristic p of Fq divides t and m, respectively. Throughout, gcd denotes the greatest common divisor. 5.4.57 Theorem [1389, Chapter VI] Consider a ﬁnite ﬁeld Fq and an integer n ≥1 with gcd(n, q) = 1. Let s := ordν(n)(q). Assume that Cn,1 is a regular cyclotomic module over Fq (hence n and s are relatively prime). If n is odd, or if q ≡1 (mod 4), or if n ≡2 (mod 4) and q ≡3 (mod 4), let (Q, N) := (qs, n). If Cn,1 is exceptional over Fq, let (Q, N) := (q2s, n 2 ). Write Q−1 = ρ·ρ, where ν(ρ) = ν(N) and gcd(N, ρ) = 1 (hence ρ is the largest divisor of Q−1 composed from prime divisors of N), and let a := gcd(ρ, N) and I := {i ∈Z : 1 ≤i ≤a, gcd(i, a) = 1}. On I there is deﬁned an equivalence relation by i ∼j if and only if i ≡qℓj (mod a) for some ℓ∈Z. Let R be a set of representatives of ∼on I. Finally, let y ∈Fq be a primitive (Nρ)-th root of unity, and w := X i∈R yi and u := X i∈R (yi + yiq). With Tr denoting the (Fqns, Fqn)-trace mapping, the following hold: 1. Ordq(Tr(w)) = Φn(x) in the ﬁrst case, that is, n odd, or q ≡1 (mod 4), or n ≡2 (mod 4) and q ≡3 (mod 4). 2. Ordq(Tr(u)) = Φn(x) and Ordq2(Tr(u)) = Φ n 2 (x) in the second case, that is, where Cn,1 is exceptional over Fq. 5.4.58 Remark The case where 4\|n and q ≡3 (mod 4) and Cn,1 is regular but non-exceptional over Fq is missed in the formulation of Theorem 5.4.57. It can be covered, however, by applying Theorem 5.4.57 to the pair (q2, n 2 ) in order to determine an element having q2-order Φn/2(x). Any such element has q-order Φn(x) (Section 24 of ). 5.4.59 Remark When searching for a complete generator of the regular cyclotomic module Cn,1 over Fq, where gcd(n, q) = 1, deﬁne the parameter τ = τ(q, n) as in Theorem 5.4.52. Then (q, n) is exceptional if and only if (qτ, n τ ) is exceptional. Consider Cn,1 = {θ ∈Fq : Φn(σ)(θ) = 0} as the cyclotomic module C n τ ,1 over Fqτ . Apply the construction of Theorem 5.4.57 (and Remark 5.4.58) to the pair (qτ, n τ ) instead of (q, n). Then the resulting elements constitute complete generators of Cn,1 over Fq by Theorem 5.4.52. Bases 137 5.4.7 Towards a primitive complete normal basis theorem 5.4.60 Remark Completing previous work of Carlitz in 1952 and Davenport in 1968, Lenstra and Schoof proved the Primitive Normal Basis Theorem (Theorem 5.2.32) in 1987. It states that for any ﬁnite ﬁeld Fq and any integer m ≥1 there exists a primitive element of Fqm that is normal over Fq. Recall that a primitive element of Fqm is a generator of the (cyclic) multiplicative group of Fqm; see Theorem 2.1.37 and Deﬁnition 2.1.38. In this subsection we consider the existence of primitive elements that are completely normal. 5.4.61 Remark By means of a computer search, Morgan and Mullen calculated for every pair (p, m), with p ≤97 a prime and with pm < 1050, a monic irreducible polynomial of degree m over Fp whose roots are primitive and completely normal elements for Fpm over Fp. It is conjectured in that for every extension Fqm/Fq there exists a primitive element of Fqm that is completely normal over Fq. 5.4.62 Theorem Let q be a prime power and assume that Fqm is a regular extension over Fq (see Deﬁnition 5.4.46). Assume further that q −1 is divisible by 4 if q is odd and m is even. Then there exists a primitive element of Fqm that is completely normal over Fq. 5.4.63 Remark Blessenohl settles the existence of primitive completely normal elements for extensions Fqm/Fq, where m = 2ℓis a divisor of q2 −1 and ℓ≥3 and q ≡3 (mod 4). In , the case where q ≡3 (mod 4) and where m is a suﬃciently large power of 2 is handled, giving rise to the ﬁrst bound in Theorem 5.4.64 below. In a not yet published work by the author , the existence of primitive completely normal elements is proved for all regular extensions, i.e., the assertion of Theorem 5.4.62 also holds without the additional assumption that q −1 is divisible by 4 if q is odd and m is even. 5.4.64 Theorem Let Fq be a ﬁnite ﬁeld with characteristic p. For an integer m ≥1 let PCN(q, m) denote the number of primitive elements of Fqm that are completely normal over Fq. Then (with ϕ denoting Euler’s function): 1. PCN(q, 2ℓ) ≥4(q −1)2ℓ−2, if q ≡3 (mod 4) and ℓ≥e + 3 (where e is maximal such that 2e\|q2 −1), or if q ≡1 (mod 4) and ℓ≥5. 2. PCN(q, rℓ) ≥r2(q −1)rℓ−2, if r ̸= p is an odd prime and ℓ≥2. 3. PCN(q, rℓ) ≥r(q −1)rℓ−1 · ϕ(qrℓ−1 −1), if r ≥7 and r ̸= p is a prime and ℓ≥2. 4. PCN(q, pℓ) ≥pqpℓ−1−1(q −1), if ℓ≥2. 5. PCN(q, pℓ) ≥pqpℓ−1−1(q −1) · ϕ(qpℓ−1 −1), if p ≥7 and ℓ≥2. 5.4.65 Deﬁnition Consider a ﬁnite ﬁeld Fq and let M be a Steinitz number. Assume that (wm)m∈M is a sequence in FqM that is both, norm-compatible and trace-compatible over Fq (see Deﬁnition 5.4.11). Then (wm)m∈M is a complete universal generator of FqM over Fq if, for every m ∈M, wm is a primitive element of Fqm that is completely normal over Fq. 5.4.66 Theorem Let q > 1 be any prime power and r ≥7 be any prime. Furthermore, let M := {rn : n ∈N}. Then there exists a complete universal generator for FqM over Fq. 5.4.67 Remark The conclusion of Theorem 5.4.66 could also be proved for r = 5 when Fq has characteristic 5, or when q mod 25 is not equal to 1, 7, 18 or 24, or when q is suﬃciently large [1392, 1393]. For the cases r = 2 and r = 3 similar results are available only when the assumption that M is a Steinitz number is weakened, e.g., M has to consist of powers of 9 or powers of 8, respectively. 138 Handbook of Finite Fields See Also §5.1, §5.2 For self-dual, weakly self-dual, and primitive normal bases. §5.3 For information on low-complexity normal bases. , , For explicit constructions of completely normal polynomials. , Section 27 discusses aspects of completely normal polynomials and iterative constructions. , For connections to Dickson polynomials; see also Section 9.6. References Cited: [75, 309, 315, 316, 317, 318, 398, 539, 589, 775, 1023, 1386, 1387, 1388, 1389, 1390, 1391, 1392, 1393, 1394, 1395, 1896, 1899, 2090, 2157, 2537, 2538, 2539, 2540] 6 Exponential and character sums 6.1 Gauss, Jacobi, and Kloosterman sums ........... 139 Properties of Gauss and Jacobi sums of general order • Evaluations of Jacobi and Gauss sums of small orders • Prime ideal divisors of Gauss and Jacobi sums • Kloosterman sums • Gauss and Kloosterman sums over ﬁnite rings 6.2 More general exponential and character sums .. 161 One variable character sums • Additive character sums • Multiplicative character sums • Generic estimates • More general types of character sums 6.3 Some applications of character sums ............. 170 Applications of a simple character sum identity • Applications of Gauss and Jacobi sums • Applications of the Weil bound • Applications of Kloosterman sums • Incomplete character sums • Other character sums 6.4 Sum-product theorems and applications ......... 185 Notation • The sum-product estimate and its variants • Applications 6.1 Gauss, Jacobi, and Kloosterman sums Ronald J. Evans, University of California at San Diego In this section, we focus mainly on Gauss, Jacobi, and Kloosterman sums over ﬁnite ﬁelds, with brief mention of Eisenstein and Jacobsthal sums. Throughout, Fq is a ﬁnite ﬁeld of characteristic p with q = pr elements. (In Subsection 6.1.3, the exponent r will be taken to be the order of p (mod k) for a ﬁxed integer k.) We refer to for proofs of many of the results in this section. In some cases, the proofs need modiﬁcation because of diﬀering deﬁnitions of the trivial character χ0: in Deﬁnition 6.1.1 below, χ0(0) = 0, while in , χ0(0) = 1. 6.1.1 Properties of Gauss and Jacobi sums of general order 6.1.1 Deﬁnition A multiplicative character χ on F∗ q is a map from the cyclic group F∗ q into the group of complex roots of unity such that χ(αβ) = χ(α)χ(β) for all α, β ∈F∗ q. We extend χ to a function on Fq by setting χ(0) = 0. The trivial character χ0 satisﬁes χ0(α) = 1 for every α ∈F∗ q. The order of χ is the smallest positive integer n for which χn = χ0. The unique character ρ of order 2 is the quadratic character. 139 140 Handbook of Finite Fields 6.1.2 Deﬁnition Write ζp = e2πi/p. For a character χ of order k on Fq and for β ∈Fq, the Gauss sum G(β, χ) of order k over Fq is deﬁned by G(β, χ) = X α∈Fq χ(α)ζTr(αβ) p , where Tr(α) denotes the trace of α from Fq to the prime ﬁeld Fp. When β = 1, we abbreviate G(χ) = G(β, χ). 6.1.3 Remark The next theorem shows that G(β, χ) can be evaluated in terms of G(χ). 6.1.4 Theorem [240, p. 9]. For a character χ on Fq and β ∈Fq, G(β, χ) = ( q −1 if β = 0, χ = χ0, χ(β)G(χ) otherwise. In particular, G(χ) = χ(−1)G(χ). 6.1.5 Remark For proofs of the following two theorems, see [240, p. 10]. 6.1.6 Theorem For a character χ on Fq, \|G(χ)\| = √q if χ ̸= χ0, and G(χ) = −1 if χ = χ0. 6.1.7 Theorem For a character χ on Fq and β ∈Fq, G(β, χp) = G(βp, χ). 6.1.8 Remark We next present two theorems on uniform distribution of Gauss sums. The ﬁrst, due to Katz and Zheng , was subsequently extended by Shparlinski . For the second and some generalizations thereof, see Iwaniec and Kowalski [1581, Theorem 21.6], Katz [1701, Chapter 9], and Fu and Liu . 6.1.9 Theorem Consider the collection of (q −1)(q −2) normalized Gauss sums G(β, χ)/√q, β ∈F∗ q, χ ̸= χ0. As q tends to inﬁnity, this collection is asymptotically equidistributed on the complex unit circle. 6.1.10 Theorem Consider the collection of q −2 normalized Gauss sums G(χ)/√q, χ ̸= χ0. As q tends to inﬁnity, this collection is asymptotically equidistributed on the complex unit circle. 6.1.11 Deﬁnition Let β ∈F∗ q and suppose that q = kf + 1 for some positive integer k. The (reduced) f-nomial Gaussian periods g(β, k) of order k are deﬁned by g(β, k) = X α∈Fq ζTr(βαk) p . Exponential and character sums 141 When β = 1, we abbreviate g(k) = g(β, k). For example, if q = p, then g(k) = p X m=1 ζmk p . The periods g(β, k) are also known as Gauss sums. 6.1.12 Remark Gauss sums and periods have been used for counting solutions to diagonal equa-tions over Fq [240, Chapters 10, 12] and for counting points on more general varieties [1343, 1344, 2167]. Thaine has given an application to class groups of cyclotomic ﬁelds. For some applications to coding theory, see [1078, 2225]. 6.1.13 Theorem [240, p. 11]. Let β ∈F∗ q. If χ is a character on Fq of order k, then g(β, k) = k−1 X j=1 G(β, χj). In particular, \|g(β, k)\| ≤(k −1)√q. 6.1.14 Remark The inequality above has been strengthened in several diﬀerent ways, depending on the relationship between p and k; see Heath-Brown and Konyagin . For further estimates for Gauss sums, see [374, 1790, 2645]. 6.1.15 Deﬁnition Let q = pr = kf + 1 and let γ be a generator of the cyclic group F∗ q. The polynomial Rk(x) = k−1 Y s=0 (x −g(γs, k)), whose zeros are (reduced) f-nomial Gaussian periods, is the (reduced) period polynomial of degree k. 6.1.16 Example R2(x) = x2 −q(−1)f, by Theorem 6.1.86. 6.1.17 Remark The period polynomial Rk(x) deﬁned above has integral coeﬃcients and is inde-pendent of the choice of generator γ. It is irreducible over Q when r = 1, but not necessarily irreducible when r > 1 [240, p. 426]. See for examples of factorizations of Rk(x), par-ticularly in the case k = 5. Determinations of period polynomials over Fp have applications to residuacity criteria mod p; see for example . 6.1.18 Deﬁnition For any positive integer n, write ζn = e2πi/n. 6.1.19 Remark The following theorem is a restatement of Theorem 6.1.13. 6.1.20 Theorem (Finite Fourier expansion of Gaussian periods.) Let γ be a generator of the group F∗ q and let χ be a character of order k on Fq such that χ(γ) = ζk. Then for each period g(γs, k), we have the Fourier expansion g(γs, k) = k−1 X v=1 G(χv)ζ−sv k . 142 Handbook of Finite Fields 6.1.21 Deﬁnition For a character χ on Fq, deﬁne the Eisenstein sum E(χ) by E(χ) = X α∈Fq Tr(α)=1 χ(α). 6.1.22 Remark Eisenstein sums can be applied to obtain congruences for binomial coeﬃcients [240, Section 12.9] and to evaluate Brewer sums [240, Chapter 13]. They are also useful for evaluating Gauss sums g(k), via the following theorem. 6.1.23 Theorem [240, p. 421] Let χ be a character of order k on Fq, and let χ∗denote the restriction of χ to the prime ﬁeld Fp, so that χ∗is a character on Fp of order k∗= k gcd(k, (q −1)/(p −1)). Then g(k) = k−1 X j=1 k∗∤j G(χ∗j)E(χj) −p k−1 X j=1 k∗\| j E(χj). 6.1.24 Theorem [240, p. 391]. Let χ be a nontrivial character on Fq, and let χ∗denote the restric-tion of χ to Fp. Then the Eisenstein sum E(χ) can be expressed in terms of the Gauss sum G(χ) over Fq and the Gauss sum G(χ∗) over Fp as follows: E(χ) = ( G(χ)/G(χ∗) if χ∗is nontrivial, −G(χ)/p if χ∗is trivial. As a consequence, \|E(χ)\| = ( p(r−1)/2 if χ∗is nontrivial, p(r−2)/2 if χ∗is trivial. 6.1.25 Deﬁnition Let χ, ψ be multiplicative characters on Fq. The Jacobi sum J(χ, ψ) over Fq is deﬁned by J(χ, ψ) = X α∈Fq χ(α)ψ(1 −α). We say that the Jacobi sum has order k if k is the least common multiple of the orders of its arguments. 6.1.26 Remark Clearly J(χ, ψ) = J(ψ, χ) ∈Q(ζq−1). The next four theorems follow easily from the results in [240, Section 2.1]. 6.1.27 Theorem (Trivial Jacobi sums.) For characters χ, ψ on Fq, J(χ, ψ) =      q −2 if ψ, χ are both trivial, −1 if exactly one of ψ, χ is trivial, −χ(−1) if χψ is trivial with χ nontrivial. 6.1.28 Theorem If χ and ψ are characters on Fq with χψ nontrivial, then J(χ, ψ) = G(χ)G(ψ)/G(χψ). Exponential and character sums 143 Thus if χ, ψ, and χψ are all nontrivial, then \|J(χ, ψ)\| = √q. 6.1.29 Theorem If χ and ψ are characters on Fq with χ nontrivial, then J(χ, ψ) = ψ(−1)G(ψ)G(χψ)/G(χ). 6.1.30 Theorem If χ is a character on Fq of order k > 1, then G(χ)k = qχ(−1) k−2 Y j=1 J(χ, χj). Thus if f denotes the order of p (mod k), then G(χ)k lies in a subﬁeld of Q(ζk) of index f. 6.1.31 Remark Louboutin used Theorem 6.1.30 for power residue characters attached to the simplest real cyclic cubic, quartic, quintic, and sextic number ﬁelds, to eﬃciently compute class numbers of these ﬁelds. 6.1.32 Deﬁnition Let χ1, . . . , χt be characters on Fq. Deﬁne the multiple Jacobi sum J(χ1, . . . , χt) by J(χ1, . . . , χt) = X α1,...,αt∈Fq α1+···+αt=1 χ1(α1) · · · χt(αt). Similarly deﬁne J0(χ1, . . . , χt) = X α1,...,αt∈Fq α1+···+αt=0 χ1(α1) · · · χt(αt). 6.1.33 Remark Jacobi sums have applications to solving diagonal equations over ﬁnite ﬁelds and to discrete log cryptosystems [240, Chapter 10]. They have been used to determine the cardinality of certain classes of irreducible polynomials over Fq with prescribed trace and restricted norm [1365, 1783]. See [2948, Chapter 16] for an application to primality testing, and for an application to coding theory. 6.1.34 Remark The next four theorems can be proved by a straightforward modiﬁcation of the proofs in [240, Sections 10.1–10.3]. 6.1.35 Theorem If χ1, . . . , χt are all trivial characters on Fq, then J(χ1, . . . , χt) = (q −1)t −(−1)t q , J0(χ1, . . . , χt) = J(χ1, . . . , χt) + (−1)t. 6.1.36 Theorem Suppose that χ1, . . . , χt are characters on Fq which are not all trivial. Then J0(χ1, . . . , χt) = ( (1 −q)J(χ1, . . . , χt) if χ1 · · · χt is trivial, 0 otherwise. 6.1.37 Theorem (Reduction formula) Suppose that χ1, . . . , χt are characters on Fq such that χt is nontrivial. Then J(χ1, . . . , χt) =      −(−1)t if χ1, . . . , χt−1 are all trivial, J(χt, χ1 · · · χt−1)J(χ1, . . . , χt−1) if χ1 · · · χt−1 is nontrivial, −qJ(χ1, . . . , χt−1) otherwise. 144 Handbook of Finite Fields 6.1.38 Theorem Suppose that the characters χ1, . . . , χt on Fq are not all trivial. Then J(χ1, . . . , χt) = ( G(χ1) · · · G(χt)/G(χ1 · · · χt) if χ1 · · · χt is nontrivial, −G(χ1) · · · G(χt)/q otherwise. Thus if χ1, . . . , χt are all nontrivial, \|J(χ1, . . . , χt)\| = ( q(t−1)/2 if χ1 · · · χt is nontrivial, q(t−2)/2 otherwise. 6.1.39 Remark We next present two theorems on the uniform distribution of Jacobi sums. The ﬁrst is due to Katz and Zheng . For the second and generalizations thereof, see Katz [1711, Corollary 20.3]. (Katz’s Corollary 20.3 for r = 1 is equivalent to his Theorem 17.5 with n = 1.) 6.1.40 Theorem Consider the collection of (q −2)(q −3) normalized Jacobi sums J(χ, ψ)/√q, χ, ψ, χψ all nontrivial on Fq. As q tends to inﬁnity, this collection is asymptotically equidistributed on the complex unit circle. 6.1.41 Theorem For a ﬁxed nontrivial character ψ on Fq, consider the collection of q−3 normalized Jacobi sums J(χ, ψ)/√q, χ ̸= χ0, χ ̸= ψ. As q tends to inﬁnity, this collection is asymptotically equidistributed on the complex unit circle. 6.1.42 Remark Suppose that in place of the collection above, one considers the more general collection of normalized multiple Jacobi sums J(χ1, . . . , χm, ψ1, . . . , ψn)/q(n+m−1)/2, where the ψj are ﬁxed nontrivial characters and where the χi run through all nontrivial characters for which χ1 · · · χmψ1 · · · ψn is nontrivial. Katz [email communication, 2011] has shown that as q tends to inﬁnity, this collection is asymptotically equidistributed on the complex unit circle, except in the “degenerate” case where both m = 1 and ψ1 · · · ψn is trivial. 6.1.43 Deﬁnition (Lifted Gauss sums) Let χ be a character on Fq, and let m be a positive integer. Recall that the Gauss sum G(χ) on Fq is deﬁned by G(χ) = X α∈Fq χ(α)ζTr(α) p . The lift of G(χ) to the extension ﬁeld Fqm is the Gauss sum Gm(χ′) = X δ∈Fqm χ′(δ)ζTr(δ) p , where Tr is the trace from Fqm to Fp and χ′ is the character on Fqm deﬁned by χ′(δ) = χ Norm Fqm/Fq(δ) , δ ∈Fqm. We call χ′ the lift of the character χ from Fq to Fqm. Exponential and character sums 145 6.1.44 Deﬁnition (Lifted Jacobi sums) Suppose that χ1, . . . , χt are characters on Fq, and let m be a positive integer. The lift of the Jacobi sum J(χ1, . . . , χt) = X α1,...,αt∈Fq α1+···+αt=1 χ1(α1) · · · χt(αt) on Fq to the extension ﬁeld Fqm is the Jacobi sum Jm(χ′ 1, . . . , χ′ t) = X δ1,...,δt∈Fqm δ1+···+δt=1 χ′ 1(δ1) · · · χ′ t(δt), where χ′ i is the the lift of χi from Fq to Fqm. 6.1.45 Theorem (Hasse-Davenport theorem on lifted Gauss sums [240, p. 360]) Let χ be a character on Fq, and let m be a positive integer. Then in the notation of Deﬁnition 6.1.43, Gm(χ′) = (−1)m−1G(χ)m. 6.1.46 Remark The next corollary follows with the aid of Theorem 6.1.38. 6.1.47 Corollary (Hasse-Davenport theorem on lifted Jacobi sums.) Suppose that χ1, . . . , χt are characters on Fq which are not all trivial, and let m be a positive integer. Then in the notation of Deﬁnition 6.1.44, Jm(χ′ 1, . . . , χ′ t) = (−1)(m−1)(t−1)J(χ1, . . . , χt)m. 6.1.48 Deﬁnition A Gauss or Jacobi sum is pure if some positive integral power of it is real. 6.1.49 Example Quadratic Gauss sums are pure, by Theorem 6.1.86. Another example is given by the following theorem of Stickelberger, proved in [240, Section 11.6]. 6.1.50 Theorem Let χ be a character of order k > 2 on Fq, where q = pr. Suppose that there is a positive integer t such that pt ≡−1 (mod k), with t chosen minimal. Then r = 2ts for some positive integer s, and q−1/2G(χ) = ( (−1)s−1 if p = 2, (−1)s−1+(pt+1)s/k if p > 2. 6.1.51 Theorem Let χ be a character of order k on Fq. Then the Gauss sums G(χj) are pure for all integers j if and only if −1 is a power of p (mod k). In the special case that k is a prime power, G(χ) is pure if and only if −1 is a power of p (mod k). 6.1.52 Theorem Let χ be a character of order k > 1 on Fq, where q = pr. If G(χ) is pure, then 2(q −1)/(k(p −1)) is an integer with the same parity as r. In particular, if r = 1 and k > 2, then G(χ) is not pure, i.e., the normalized Gauss sum G(χ)/√p on Fp cannot equal a root of unity when χ is a character on Fp of order > 2. Also, if r = 2, then G(χ) is pure if and only if k \| (p + 1). 6.1.53 Remark The next theorem, due to Aoki , gives further examples of pure Gauss sums. See also Aoki . 146 Handbook of Finite Fields 6.1.54 Theorem Let χ be a character of order k > 2 on Fq, where q = pr. For r = 3, if G(χ) is pure, then either (a) k = 14 and p ≡2 or 4 (mod 7); (b) k = 42 and p ≡4 or 16 (mod 21); or (c) k = 78 and p ≡16 or 32 (mod 39). For r = 4, G(χ) is pure if and only if k \| (p2 + 1), except in the following four cases: (i) k = 20 and p ≡13 or 17 (mod 20); (ii) k = 30 and p ≡17 or 23 (mod 30); (iii) k = 60 and p ≡17 or 53 (mod 60); (iv) k = 120 and p ≡83 or 107 (mod 120). 6.1.55 Remark The next theorem gives examples of pure Jacobi sums over Fp2. It is due to Shiratani and Yamada, who point out an application to algebraic combinatorics . It is also due independently to Akiyama, who gave a diﬀerent proof . Other examples have been given by Aoki . 6.1.56 Theorem Let r = 2, i.e., q = p2, and consider the Jacobi sum J(ρ, χ) on Fq, where ρ, χ are characters on Fq of orders 2, k, respectively, with k > 2. Then J(ρ, χ) = ±p (and is thus pure) if either (a) k \| (p + 1); (b) 2(p −1)/k is an odd integer; (c) k = 24 and p ≡17 or 19 (mod 24); or (d) k = 60 and p ≡41 or 49 (mod 60). Moreover, if none of (a)–(d) hold, then J(ρ, χ) is not pure. 6.1.57 Remark The Jacobi sums satisfying one of (a)–(d) above of course generate the subﬁeld Q of Q(ζk). Aoki has determined the subﬁeld of Q(ζk) generated by more general Jacobi sums in Q(ζk). 6.1.58 Remark Pure multiple Jacobi sums are connected with supersingularity and ranks of certain elliptic curves; see Aoki and Ulmer [2835, Section 5]. 6.1.59 Theorem (Hasse-Davenport product formula for Gauss sums [240, p. 351]) Let ψ be a character on Fq of order ℓ> 1. For every character χ on Fq, ℓ−1 Y i=1 G(χψi)/G(ψi) = χℓ(ℓ)G(χℓ)/G(χ). 6.1.60 Remark The Hasse-Davenport formula for products of Gauss sums (Theorem 6.1.59) is the ﬁnite ﬁeld analogue of the Gauss multiplication formula for gamma functions. Work of Kubert and Lichtenbaum on Jacobi sum Hecke characters led to identities involving products of Gauss sums which extended the Hasse-Davenport product formula. (For calcula-tion of conductors of Jacobi sum Hecke characters, see .) Other examples of identities involving products of Gauss sums may be found in [813, 1010, 1013, 1014, 1015, 2854]. For evaluations of special hypergeometric character sums over Fq in terms of products of Gauss sums, see papers of Evans and Greene [1007, 1008]. 6.1.61 Example If ℓ= 2 and ψ is the quadratic character ρ on Fq, then the Hasse-Davenport product formula becomes G(χρ)/G(ρ) = χ(4)G(χ2)/G(χ), which is equivalent to the following theorem. 6.1.62 Theorem [240, p. 59] Let χ, ρ be characters on Fq with χ nontrivial and ρ quadratic. Then J(χ, ρ) = χ(4)J(χ, χ). 6.1.63 Remark The next theorem gives a variant of Theorem 6.1.62. 6.1.64 Theorem [240, p. 60] Let χ, ρ be characters on Fq such that χ has order > 2 and ρ is quadratic. Then ρ(−1)χρ(4)J(χρ, χρ) = χ(4)J(χ, χ) = χ(−1)J(χ, χρ). Exponential and character sums 147 6.1.65 Remark We next give two congruences for Jacobi sums; see [240, pp. 60, 97]. More general congruences are given in . 6.1.66 Theorem Let χ, ψ be nontrivial characters on Fq of orders a, b, respectively. Then J(χ, ψ) ≡−q (mod (1 −ζa)(1 −ζb)) . If moreover a = b > 2, then the right member −q may be replaced by −1. 6.1.67 Theorem Let χ be a character on Fq of order 2ℓ, where ℓ> 1 is odd, and let n denote an even integer not divisible by ℓ. Then J(χ, χn) ≡−χn(4) mod (1 −ζℓ)2 . 6.1.68 Deﬁnition Let γ be a generator of the cyclic group F∗ q, and let χ be a character of order k on Fq for which χ(γ) = ζk. For any pair of integers a, b (mod k), deﬁne the cyclotomic number C(a, b) = C(γ, a, b) of order k over Fq to be the number of α ∈F∗ q for which χ(α/γa) = χ((α + 1)/γb) = 1. 6.1.69 Remark Cyclotomic numbers are useful for obtaining residuacity criteria. For example, the cyclotomic numbers of order 12 can be used to prove that for a prime p ≡1 (mod 12), 3 is a quartic residue (mod p) if and only if a3 ≡−1 (mod 4), where a3 is as in Deﬁnition 6.1.74. See [240, p. 231]. 6.1.70 Theorem [240, p. 365] In the notation of Deﬁnition 6.1.68, the cyclotomic numbers C(a, b) are related to the Jacobi sums J(χu, χv) by the following ﬁnite Fourier series expansions: k2C(a, b) = k−1 X u=0 k−1 X v=0 χu(−1)J(χu, χv)ζ−au−bv k and χu(−1)J(χu, χv) = k−1 X a=0 k−1 X b=0 C(a, b)ζau+bv k . 6.1.71 Deﬁnition Let p be an odd prime. For a positive integer k and an integer a not divisible by p, deﬁne the Jacobsthal sums Uk(a), Vk(a) over Fp by Uk(a) = p−1 X m=0 m p mk + a p , Vk(a) = p−1 X m=0 mk + a p , where the symbols in the summands are Legendre symbols. 6.1.72 Remark Jacobsthal sums have applications to the distribution of quadratic residues [240, Chapter 6], to evaluations of Brewer sums [240, Chapter 13], and to the evaluation of certain hypergeometric character sums [240, Equation (13.3.2)]. The next theorem expresses Jacobsthal sums in terms of Jacobi sums. 6.1.73 Theorem [240, pp. 188–189]. For a prime p ≡1 (mod 2k), let χ be a character on Fp of order 2k. Let a be an integer not divisible by p. Then Uk(a) = χ(−1) a p k−1 X j=0 χ2j+1(4a)J(χ2j+1, χ2j+1) 148 Handbook of Finite Fields and Vk(a) = a p k−1 X j=1 χ2j(4a)J(χ2j, χ2j). 6.1.2 Evaluations of Jacobi and Gauss sums of small orders 6.1.74 Deﬁnition For a prime p = 6f + 1, deﬁne integer parameters a3, b3, r3, s3, u3, v3 as follows: p = a2 3 + 3b2 3, a3 ≡−1 (mod 3), 4p = r2 3 + 3s2 3, r3 ≡1 (mod 3), s3 ≡0 (mod 3), 4p = u2 3 + 3v2 3, u3 ≡1 (mod 3), where ( v3 ≡±1 (mod 6) if 2 is a cubic nonresidue (mod p), v3 = s3 ≡0 (mod 3) if 2 is a cubic residue (mod p). 6.1.75 Remark Given p, the parameters a3, r3, and u3 are uniquely determined, but b3, s3, and v3 are determined only up to sign. These parameters appear below in the evaluations of cubic and sextic Gauss and Jacobi sums over Fp. In the case that 2 is a cubic nonresidue (mod p), we have 3 ∤b3 and the parameters r3, s3, u3, v3 are odd. In the case that 2 is a cubic residue (mod p), we have 3 \| b3 and r3, s3, u3, v3 are even; see [240, Section 3.1]. 6.1.76 Theorem (Cubic Jacobi sums [240, Section 3.1]) For q = p = 6f + 1, let χ be a character of order 3 on Fp. Then J(χ, χ) = (r3 + is3 √ 3)/2. 6.1.77 Theorem (Sextic Jacobi sums [240, Section 3.1]) For q = p = 6f + 1, let χ be a character of order 6 on Fp. Then J(χ, χ) = (−1)f(u3 + iv3 √ 3)/2 = (−1)fJ(χ, χ4), J(χ, χ2) = J(χ3, χ2) = a3 + ib3 √ 3 = (−1)fJ(χ, χ3). 6.1.78 Deﬁnition For a prime p = 4f + 1, deﬁne integer parameters a4, b4 by p = a2 4 + b2 4, a4 ≡−(−1)f (mod 4). 6.1.79 Theorem (Quartic Jacobi sums [240, Section 3.2]) For q = p = 4f + 1, let χ be a character of order 4 on Fp. Then J(χ, χ) = (−1)f(a4 + ib4) = (−1)fJ(χ, χ2). 6.1.80 Deﬁnition For a prime p = 8f + 1, deﬁne integer parameters a8, b8 by p = a2 8 + 2b2 8, a8 ≡−1 (mod 4). Exponential and character sums 149 6.1.81 Theorem (Octic Jacobi sums [240, Section 3.3]) For q = p = 8f + 1, let χ be a character of order 8 on Fp. Then J(χ, χ) = χ(4)(a8 + ib8 √ 2) = χ(−4)J(χ, χ3), J(χ, χ2) = χ(−4)(a4 + ib4). 6.1.82 Theorem (Duodecic Jacobi sums [240, Section 3.5]) For q = p = 12f +1, let χ be a character of order 12 on Fp so that J(χ3, χ3) = (−1)f(a4 + ib4) as in Theorem 6.1.79. Then χ(−1)J(χ, χ5) = χ(4)J(χ, χ) = χ(4)5J(χ5, χ5) = ±(a4 + ib4), where the plus sign is chosen if 3 \| b4 and the minus sign is chosen if 3 \| a4. The three additional duodecic Jacobi sums below are expressed in terms of previously evaluated Jacobi sums of orders 6, 4, 3, respectively: J(χ, χ2) = χ(4)2J(χ2, χ2)/c12, J(χ, χ3) = J(χ3, χ3)/c12, J(χ, χ4) = J(χ4, χ4), where ( c12 = ±1 with c12 ≡−a4 (mod 3) if 3 \| b4, c12 = ±i with c12 ≡−ib4 (mod 3) if 3 \| a4. 6.1.83 Remark Values of all duodecic Jacobi sums may be deduced from Theorem 6.1.82. For example, J(χ3, χ5) = σ5J(χ, χ3) = J(χ, χ3), where σ5 is as in Deﬁnition 6.1.99 with k = 12. Niitsuma applied duodecic Jacobi sum evaluations to count rational points on certain hyperelliptic curves over Fp. 6.1.84 Remark Jacobi sums of various other small orders are explicitly evaluated in ; e.g., quintic Jacobi sums are computed in [240, Section 3.7]. For the quintic case, see also Hoshi for an analysis of Gauss sums, Jacobi sums, and period polynomials. Values of Jacobi sums of order 16 have been applied to construct regular Hadamard matrices . 6.1.85 Theorem (Quadratic multiple Jacobi sums [240, p. 299]) Suppose that χ1, . . . , χt are all equal to the quadratic character ρ on Fq. Then J(χ1, . . . , χt) = ( ρ(−1)(t−1)/2q(t−1)/2 if t is odd, −ρ(−1)t/2q(t−2)/2 if t is even. 6.1.86 Theorem (Quadratic Gauss sums [240, p. 362]) Let q = pr for an odd prime p, and let ρ be the quadratic character on Fq. Then g(2) = G(ρ) = ( (−1)r−1√q if p ≡1 (mod 4), (−1)r−1ir√q if p ≡3 (mod 4). In particular, if q = p, we obtain the evaluation of Gauss: p−1 X n=0 ζn2 p = p−1 X n=0 n p ζn p = (√p if p ≡1 (mod 4), i√p if p ≡3 (mod 4), where (n/p) is the Legendre symbol. 150 Handbook of Finite Fields 6.1.87 Remark The Gauss sums in Theorem 6.1.86 lie in a quadratic extension of Q. For evalu-ations of general Gauss sums over Fq lying in quadratic and multi-quadratic extensions of Q, see [115, 2042, 3026, 3027]. 6.1.88 Theorem (Cubic periods [240, Section 4.1]) Let q = p ≡1 (mod 6), so that 4p = r2 3 + 3s2 3 with r3 ≡1 (mod 3), s3 ≡0 (mod 3). Let b be a primitive root of p. Then the cubic irreducible polynomial x3 −3px−pr3 has the three real zeros g(3), g(b, 3), and g(b2, 3), with one zero in each of the three intervals (−2√p, −√p), (−√p, √p), (√p, 2√p). 6.1.89 Remark No simple criterion is known for determining which of the three intervals above contains the cubic Gauss sum g(3). However, g(3) and the Gauss sums G(χ) for cubic characters χ have been evaluated in terms of products of values of Weierstrass ℘-functions; see [240, p. 158]. The next theorem, due to Heath-Brown and Patterson , gives an equidistribution result for cubic Gauss character sums over Fp. 6.1.90 Problem Determining the distribution of n-th order Gauss sums over Fp for a general ﬁxed n is an open problem. 6.1.91 Theorem Consider the collection of all normalized cubic Gauss sums G(χ)/√p, χ cubic on Fp, q = p ≡1 (mod 3), p < x. As x tends to inﬁnity, this collection is asymptotically equidistributed on the complex unit circle. 6.1.92 Remark The next theorem determines the sextic Gaussian period g(6) unambiguously, once g(3) is known. 6.1.93 Theorem (Sextic periods ) Let q = p ≡1 (mod 6), so that 4p = r2 3 + 3s2 3 with r3 ≡1 (mod 3), s3 ≡0 (mod 3). In the case that 2 is a cubic residue (mod p), g(6) = g(3) + i(p−1)2/4 g(3)2 −p /√p. In the case that 2 is a cubic nonresidue (mod p), then with the sign of s3 speciﬁed by s3 ≡−r3 (mod 4), g(6) = g(3) + i(p−1)2/4{4p −g(3)2 + s−1 3 2pg(3) + 2pr3 −r3g(3)2 }/(2√p). 6.1.94 Remark For the history behind the quartic Gauss sum evaluations in the two theorems below, see [240, p. 162]. 6.1.95 Theorem (Quartic Gauss sums [240, Section 4.2]) Let q = p ≡1 (mod 4), and let χ be a quartic character on Fp. As in Theorem 6.1.79, write J(χ, χ) = a + bi, where p = a2 + b2 with a ≡−1 (mod 4). (Note that the sign of b depends on the choice of the quartic character χ.) Deﬁne C = ±1 by C ≡(−1)(p−1)/4 \|b\| a p −1 2 ! (mod p). Then G(χ) = \|b\| \|a\| C(−1)(b2+2b)/8 r p + a√p 2 + i\|b\| b r p −a√p 2 ! , where the ﬁrst symbol on the right is the Jacobi symbol. Exponential and character sums 151 6.1.96 Theorem (Quartic periods [240, Section 4.2]) In the notation of the previous theorem, if p ≡1 (mod 8), g(4) = √p + C \|b\| \|a\| (−1)(b2+2\|b\|)/8q 2p + 2a√p, while if p ≡5 (mod 8), g(4) = √p + iC \|b\| \|a\| (−1)(b2+2\|b\|)/8q 2p −2a√p. 6.1.97 Remark The next theorem determines the Gaussian period g(12) unambiguously, once g(3) is known. For an extension to Fq, see Gurak . 6.1.98 Theorem (Duodecic periods ) Let q = p = 12f + 1, so that as in Theorem 6.1.78, p = a2 + b2 with a ≡−(−1)f (mod 4). In the case that −3 is a quartic residue (mod p), g(12) = g(6) + (g(4) −√p) 1 + −a 3 g(3)/√p . In the case that −3 is a quartic nonresidue (mod p) (which is equivalent to 3 ∤b by [240, Section 7.2]) then with the sign of b speciﬁed by b ≡−1 (mod 3), g(12) = g(6) + g(4) −√p + 2b 2 p g(3)/ (g(4) −√p) . 6.1.3 Prime ideal divisors of Gauss and Jacobi sums 6.1.99 Deﬁnition Fix an integer k > 1. In this subsection, q = pf, where f is the order of p (mod k). For the group R = (Z/kZ)∗, let T denote a complete set of φ(k)/f coset representatives of the quotient group R/⟨p⟩. For an integer a, let ℓ(a) denote the least nonnegative integer congruent to a (mod k). Write s(a) = f−1 X i=0 ai , t(a) = f−1 Y i=0 ai! , where the ai are the digits in the base p expansion ℓ(a)(q −1)/k = f−1 X i=0 aipi, 0 ≤ai ≤p −1. Consider the cyclotomic ﬁelds K = Q(ζk), M = Q(ζk, ζp) with rings of integers OK, OM, respectively. Let P be a prime ideal of OK above p. Since Norm(P) = q, the ﬁnite ﬁeld OK/P is isomorphic to Fq. Since P is totally ramiﬁed in OM, we have OMP = Pp−1 for a prime ideal P ⊂OM. For j ∈R, let σj denote the element in Gal(Q(ζk, ζp)/Q(ζp)) for which σj(ζk) = ζj k, and write Pj = σj(P), Pj = σj(P), Pj−1 = σj −1(P). 152 Handbook of Finite Fields 6.1.100 Theorem [240, p. 343] Let π = ζp −1. We have the prime ideal factorizations πOM = Y j∈T Pj, pOM = Y j∈T Pp−1 j , pOK = Y j∈T Pj. 6.1.101 Deﬁnition (Power residue symbol χP ) Deﬁne the character χP of order k on the ﬁnite ﬁeld OK/P by setting χP (α + P) equal to the unique power of ζk which is congruent to α(q−1)/k (mod P), for every α ∈OK with α / ∈P. If α ∈P, set χP (α + P) = 0. 6.1.102 Theorem (Stickelberger’s congruence for Gauss sums [240, p. 344]) For any integer a, the Gauss sum G(χ−a P ) over the ﬁnite ﬁeld OK/P is an element of OM satisfying the congruence G(χ−a P ) ≡−πs(a) t(a) mod Ps(a)+1 . 6.1.103 Remark For the following two corollaries, see Conrad . 6.1.104 Corollary If 0 ≤a, b < k with a, b not both 0, then with u := (q −1)/k, J(χ−a P , χ−b P ) ≡−(−1)au q −1 −bu au (mod P). 6.1.105 Corollary Let k = q −1 (so that χP has order q −1), and let 0 ≤bi < q −1, i = 1, 2, . . . , t, where not all bi equal 0. Then J(χ−b1 P , χ−b2 P , . . . , χ−bt P ) ≡(−1)t+1 (b1 + b2 + · · · + bt)! b1!b2! · · · bt! (mod P). 6.1.106 Remark A consequence of Stickelberger’s congruence is the prime ideal factorization of Gauss sums (Theorem 6.1.107). An important application of this factorization is the deter-mination of the annihilator in Z[Gal(L/Q)] of the ideal class group of an abelian number ﬁeld L; see Washington [2948, Theorem 6.10]. For applications of Theorem 6.1.107 to the theory of diﬀerence sets and coding theory, see [144, 580, 1009]. 6.1.107 Theorem (Prime ideal factorization of Gauss sums [240, p. 346]) For any integer a, G(χ−a P )OM = Y j∈T Ps(aj) j−1 . 6.1.108 Theorem [240, p. 347] For any integer a, we have G(χ−a P )k ∈OK and G(χ−a P )kOK = Y j∈T P ks(aj)/(p−1) j−1 = Y j∈R P ℓ(aj) j−1 . 6.1.109 Theorem (Prime ideal factorization of Jacobi sums [240, p. 346]) Suppose that a, b are integers such that a + b is not divisible by k. Then J(χ−a P , χ−b P )OK = Y j∈T P v(a,b,j) j−1 , with v(a, b, j) = s(aj) + s(bj) −s(aj + bj) p −1 . Exponential and character sums 153 In the special case q = p, this reduces to J(χ−a P , χ−b P )OK = Y j∈T ℓ(aj)+ℓ(bj)>k Pj−1. In particular, for q = p, J(χP , χP )OK = Y 1≤j<k/2 (j,k)=1 Pj−1. 6.1.110 Deﬁnition Let Qp denote the ﬁeld of p-adic rationals, and let Zp be its ring of p-adic integers. Consider the extension ﬁeld Qp(ζ), where ζ is a primitive p-th root of the element 1 ∈Zp. For π = ζ −1, let λ denote the prime element in Qp(ζ) satisfying λp−1 = −p, λ ≡π (mod π2). 6.1.111 Deﬁnition (Morita’s p-adic gamma function Γp) Deﬁne Γp : Zp →Z∗ p by Γp(z) = lim N→z(−1)N Y 0<j<N p ∤j j, where N runs through any sequence of positive integers p-adically approaching z. 6.1.112 Remark The following theorem of Gross-Koblitz expresses the p-adic Gauss sum in terms of Morita’s p-adic gamma functions. For a relatively elementary proof, see Robert . 6.1.113 Theorem (Gross-Koblitz formula for p-adic Gauss sums) Let a be any integer. Viewing the Gauss sum G(χ−a P ) ∈OM as embedded in the subﬁeld Qp(ζ) of the P-adic completion of M, we have the following equality in Zp[ζ]: G(χ−a P ) = −λs(a) f−1 Y i=0 Γp ℓ(api) k . 6.1.114 Remark The next corollary follows with the aid of Theorem 6.1.38. 6.1.115 Corollary (Gross-Koblitz formula for p-adic Jacobi sums) Let b1, . . . , bt be integers such that c := b1 +· · ·+bt is not divisible by k. Viewing the Jacobi sum J(χ−b1 P , . . . , χ−bt P ) ∈OK as embedded in the subﬁeld Qp of the P-adic completion of K, we have the following equality in Zp: J(χ−b1 P , . . . , χ−bt P ) = (−1)t−1(−p)u f−1 Y i=0 Γp ℓ(b1pi) k · · · Γp ℓ(btpi) k Γp ℓ(cpi) k , with u = {(s(b1) + · · · + s(bt)) −s(c)}/(p −1). 6.1.116 Remark The exponent u in Theorem 6.1.115 is an integer, since for any integer x, we have s(x) ≡x(q −1)/k (mod p −1). 154 Handbook of Finite Fields 6.1.4 Kloosterman sums 6.1.117 Remark In this subsection, we revert back to the notation q = pr. 6.1.118 Deﬁnition For u ∈Fq and a multiplicative character χ on Fq, deﬁne the (twisted) Kloost-erman sum K(u, χ) over Fq by K(u, χ) = X α∈F∗ q χ(α)ζTr(α+u/α) p . Note that K(0, χ) = G(χ). When χ is trivial, we abbreviate K(u) = K(u, χ). 6.1.119 Remark Kloosterman sums occur frequently in the theory of modular forms, and they have many applications in analytic number theory [1453, 1581, 2524]. For some applications to coding theory, see [620, 1562, 1732, 2124]. For further applications, see the references in [503, p. 448]. Some congruences for Kloosterman sums may be found in [592, 1299, 1946, 2128]. In , it is proved that if K(u) is an integer and p > 3, then K(u) is even. This proves in particular the nonvanishing of 1 + K(u) for p > 3. For analysis of the cases p = 2, 3, see . The vanishing of 1 + K(u) has applications to bent functions, deﬁned in Section 9.3. (See for recent work on bent and hyper-bent functions.) 6.1.120 Theorem For odd q and u ∈Fq, K(u) = X y∈Fq ρ(y2 −4u)ζTr(y) p , where ρ is the quadratic character on Fq. 6.1.121 Remark The theorem above follows easily from the deﬁnition of K(u), by counting, for each y ∈Fq, the number of α ∈F∗ q for which α + u/α = y. For the case where q is even, see Conrad . 6.1.122 Deﬁnition For u ∈Fq and characters χ1, . . . , χm on Fq, deﬁne the (twisted) multiple Kloosterman sum K(u, χ1, . . . , χm) over Fq by K(u, χ1, . . . , χm) = X α0,...,αm∈Fq α0···αm=u χ1(α1) · · · χm(αm)ζTr(α0+···+αm) p . This is also known as a (twisted) hyper-Kloosterman sum. It reduces to the sum given in Deﬁnition 6.1.118 when m = 1. 6.1.123 Remark Wan studied the algebraic degree of multiple Kloosterman sums. For further results on the degrees of Kloosterman sums, see . The next theorem gives an easily proved expression for multiple Kloosterman sums in terms of Gauss sums. 6.1.124 Theorem [1701, p. 47] For u ∈F∗ q and characters χ1, . . . , χm on Fq, K(u, χ1, . . . , χm) = 1 q −1 X χ χ(u) m Y i=0 G(χχi), where χ0 is the trivial character and where the sum is over all characters χ on Fq. Exponential and character sums 155 6.1.125 Theorem For a positive integer m, let ψ be a character on Fq of order m + 1. Then for any u ∈Fq, K(u, ψ, ψ2, . . . , ψm) = ϵ(q, m)qm/2 X α∈Fq αm+1=u ζTr(α(m+1)) p , where ϵ(q, m) = ( 1 if 2 \| m, (−1)r−1+(q−1)(m−1)/8i(p−1)2r/4 if 2 ∤m. 6.1.126 Remark Theorem 6.1.125 is due to Duke , who showed it to be equivalent to the Hasse-Davenport product formula for Gauss sums (Theorem 6.1.59). For a related identity, see Ye . The Kloosterman sum K(u, ψ, ψ2, . . . , ψm) is connected to Fourier expansions of certain Poincar´ e series . In the case m = 1, Theorem 6.1.125 reduces to the following well-known evaluation of the Sali´ e sum K(u, ρ). 6.1.127 Theorem (Sali´ e sum) Let u ∈Fq and let ρ be the quadratic character on Fq. Then K(u, ρ) =      (−1)r−1i(p−1)2r/4√q if u = 0, 0 if u ∈F∗ q is not a square, 2(−1)r−1i(p−1)2r/4√q cos(4πTr(b)/p) if u = b2 for some b ∈F∗ q. 6.1.128 Remark Theorem 6.1.129 below reduces to Theorem 6.1.127 when χ = ρ and reduces to Theorem 6.1.120 when χ is trivial. 6.1.129 Theorem [715, Equation (4)] Let u ∈Fq and let ρ denote the quadratic character on Fq. Then for any character χ on Fq, K(u, χ) = G(ρ)χ(4) G(χρ) X y∈Fq χρ(y2 −4u)ζTr(y) p . 6.1.130 Remark Sums closely related to the above sum on y form an orthogonal set of eigenfunctions for a collection of adjacency matrices of “ﬁnite upper half plane” Cayley graphs . 6.1.131 Theorem (Upper bound for multiple Kloosterman sums) For u ∈Fq and characters χ1, . . . , χm on Fq, \|K(u, χ1, . . . , χm)\| ≤(m + 1)qm/2. 6.1.132 Remark The upper bound above is due to Deligne for trivial characters, and in full general-ity to Katz [1701, p. 49]. See Conrad for a nice proof of the special case \|K(u, χ)\| ≤2√q, patterned on Weil’s original proof for trivial χ. Equality in Theorem 6.1.131 can occur; for example, let u = 1, (m + 1) \| (p −1), p \| r in Theorem 6.1.125. However, equality cannot occur in Theorem 6.1.131 in the case that all m characters are trivial, since then K(u, χ1, . . . , χm) ≡(−1)m (mod (1 −ζp)). Nor can equality occur in the case m = 1, q = p ; see [1004, p. 446]. 6.1.133 Deﬁnition The Weil bound \|K(u)\| < 2√q is a consequence of the formula expressing −K(u) as a sum of conjugate Frobenius eigenvalues: −K(u) = g(u) + g(u), u ∈F∗ q, 156 Handbook of Finite Fields where g(u) = √q exp(iθ(q, u)), θ(q, u) ∈(0, π). We call θ(q, u) the Kloosterman angle, noting that −K(u) = 2√q cos(θ(q, u)), u ∈F∗ q. 6.1.134 Deﬁnition Let u ∈F∗ q. For a positive integer n, let Kn(u) denote the Kloosterman sum over Fqn (so that K1(u) = K(u)). We call Kn(u) the lift of the Kloosterman sum K(u) from Fq to Fqn. 6.1.135 Remark It is shown in that Kn(u) is an integer if and only if K(u) is an integer. 6.1.136 Remark The following theorem, analogous to the Hasse-Davenport lifting formula for Gauss sums (Theorem 6.1.45), oﬀers a formula of Carlitz for the lift Kn(u) in terms of a Dickson polynomial in K(u). For an extension to K(u, χ), see [1581, p. 281]. (Dickson polynomials over Fq are discussed in Section 9.6.) 6.1.137 Theorem We have Kn(u) = (−1)n−1Dn(K(u), q), where Dn(x, a) is the Dickson polynomial of degree n. 6.1.138 Remark In the notation of Deﬁnition 6.1.133, −Kn(u) = g(u)n + g(u)n = 2qn/2 cos(nθ(q, u)), u ∈F∗ q. Since g(u) = −K(u) ± p K(u)2 −4q /2, we have −2nKn(u) = −K(u) + p K(u)2 −4q n + −K(u) − p K(u)2 −4q n . This is equivalent to Theorem 6.1.137; see [240, pp. 440–441]. 6.1.139 Theorem (Equidistribution of Kloosterman angles) The set of Kloosterman angles {θ(q, u) : u ∈F∗ q} is asymptotically equidistributed with respect to the Sato-Tate measure on (0, π), as q tends to inﬁnity. In other words, as q tends to inﬁnity, the proportion of these q −1 Kloosterman angles that lie in a ﬁxed subinterval [x, y] ⊂(0, π) approaches 2 π Z y x sin2 t dt. 6.1.140 Conjecture Fix a positive integer u and consider the collection of Kloosterman angles {θ(p, u) : u < p < x}. As x tends to inﬁnity, this collection is asymptotically equidistributed with respect to the Sato-Tate measure on (0, π). 6.1.141 Remark Theorem 6.1.139 is due to Katz [1701, p. 240]. For a quantitative reﬁnement and related results, see [2092, 2245, 2650]. For an extension to Kloosterman sums over rings, see . Statements of the conjecture above may be found in Katz’s books [1700, Conjecture 1.2.5] and [1701, p. 5]. See also the references in Shparlinski [2650, p. 420]. Exponential and character sums 157 6.1.142 Deﬁnition Fix an integer u which is not a perfect square. Motivated by Theorem 6.1.127 with q = p > 2, we deﬁne the Sali´ e angle ϑ(p, u) ∈(0, π) for each prime p with u p = 1 by ϑ(p, u) = 2πb(u, p)/p, where b = b(u, p) is the smallest positive integer for which b2 ≡u (mod p). 6.1.143 Remark The next theorem gives an equidistribution result for angles of Sali´ e sums K(u, ρ); see [1581, p. 496] and the references in Shparlinski , where one ﬁnds further results of this type. 6.1.144 Theorem (Equidistribution of Sali´ e angles) Fix an integer u which is not a perfect square. The set of Sali´ e angles {ϑ(p, u) : u p = 1, p < x} is asymptotically equidistributed in the interval (0, π), as x tends to inﬁnity. 6.1.145 Remark The following equidistribution result for normalized Kloosterman sums K(−1, χ)/√q was conjectured by Evans and proved by Katz . Note that each such sum is real and lies in the interval [−2, 2] by Theorem 6.1.131. 6.1.146 Theorem (Equidistribution of K(−1, χ)) Consider the collection of q−1 normalized Kloost-erman sums K(−1, χ)/√q, where χ runs through the characters on Fq. As q tends to inﬁnity, the “angles” of this collection are asymptotically equidistributed with respect to the Sato-Tate measure on [−2, 2]. In other words, as q tends to inﬁnity, the proportion of the members of this collection that lie in a ﬁxed subinterval [v, w] ⊂[−2, 2] approaches 1 2π Z w v p 4 −x2 dx. 6.1.147 Deﬁnition Let n be a positive integer. Deﬁne the n-th power moment Sn of the Klooster-man sums K(u) by Sn = X u∈Fq K(u)n. 6.1.148 Remark Moisio [2119, 2122] gave evaluations of Sn for n ≤10 when q is a power of 2 or 3, and he related them to cyclic codes; see also . In the remainder of this subsection, we restrict our attention to the case q = p > n. 6.1.149 Remark Various congruences have been given for Sn, but explicit evaluations of Sn for all q = p > n are known only for n = 1, 2, 3, 4, 5, 6 . The values for n ≤4 below, due to Sali´ e, may be found in Iwaniec’s book [1580, Section 4.4] (where −p should be replaced by −3p in Equation (4.25)). 6.1.150 Theorem (Power moments of Kloosterman sums ) Let q = p > n. The power moments Sn are integer multiples of p. We have S1 = 0, S2 = p2 −p, S3 = p 3 p2 + 2p, S4 = 2p3 −3p2 −3p. For p > 5, S5 = 4 p 3 p3 + (ap + 5)p2 + 4p, 158 Handbook of Finite Fields where ap is the integer of absolute value < 2p satisfying ap =      2p −12u2 if p = 3u2 + 5v2, 4x2 −2p if p = x2 + 15y2, 0 if p ≡7, 11, 13, or 14 (mod 15). Also for p > 5, S6 = 5p4 −10p3 −(bp + 9)p2 −5p, where bp is the integer of absolute value < 2p3/2 deﬁned to be the coeﬃcient of qp in the q-expansion of the weight 4, level 6 newform {η(6z)η(3z)η(2z)η(z)}2. Here η(z) is the Dedekind eta function. 6.1.151 Remark Evans has conjectured an explicit formula for S7 in terms of the coeﬃcient of qp in the q-expansion of a weight 3, level 525 newform. 6.1.152 Deﬁnition Let q = p > n ≥1. In the notation of Deﬁnition 6.1.133, the Kloosterman power moments Sn can be expressed as Sn = (−1)n + (−1)n p−1 X u=1 (g(u) + g(u))n. Closely related to Sn is the sum Tn = p−1 X u=1 n X j=0 g(u)n−jg(u)j = p−1 X u=1 pn/2Un(2 cos(θ(p, u))), where Un is the n-th monic Chebychev polynomial of the second kind. We normalize the sum Tn by deﬁning Yn := (−1 −Tn)/p2. 6.1.153 Remark For some twists of Sn and Tn, see Liu and Evans . 6.1.154 Theorem If q = p > n, then Yn is an integer. In particular, Y1 = Y2 = 0, Y3 = p 3 , Y4 = 1, Y5 = ap, Y6 = bp, where ap, bp are deﬁned in Theorem 6.1.150. 6.1.155 Remark Theorem 6.1.154 can be found in Evans . There it is moreover conjectured (for p > 7) that Y7 = p 105 (\|A(p)\|2 −p2), where A(p) is the p-th Fourier coeﬃcient of a weight 3, level 525 eigenform with quartic nebentypus of conductor 105. Furthermore, Evans [1005, p. 523] conjectured (for p > 7) that Y8 = B(p) + p2, Exponential and character sums 159 where B(p) is the p-th Fourier coeﬃcient of a weight 6, level 6 newform with trivial neben-typus. These conjectured values for Yn satisfy the following upper estimate due to Katz [1701, Theorem 0.2]. 6.1.156 Theorem For q = p > n, \|Yn\| ≤ n −1 2 p(n−3)/2. 6.1.5 Gauss and Kloosterman sums over ﬁnite rings 6.1.157 Remark Let k be a positive integer. We brieﬂy discuss some basic Gauss and Kloosterman sums over Z/kZ, as they are natural extensions of sums over the ﬁnite ﬁeld Z/pZ. 6.1.158 Deﬁnition For integers m and k > 0, deﬁne the quadratic Gauss sum qk(m) over Z/kZ by qk(m) = k−1 X n=0 ζmn2 k . 6.1.159 Remark The special case qp(m) is the quadratic Gaussian period g(m, 2) over Fp given in Deﬁnition 6.1.11. 6.1.160 Deﬁnition For integers a, b, c with ac ̸= 0, deﬁne a generalized quadratic Gauss sum S(a, b, c) by S(a, b, c) = \|c\|−1 X n=0 exp(πi(an2 + bn)/c). 6.1.161 Remark The special case S(2m, 0, k) is the quadratic Gauss sum qk(m) given in Deﬁni-tion 6.1.158. 6.1.162 Theorem (Reciprocity theorem for Gauss sums [240, p. 13]) For integers a, b, c with ac ̸= 0 and ac + b even, S(a, b, c) = \|c/a\|1/2 exp πi 4 sgn(ac) −b2 ac S(−c, −b, a). 6.1.163 Theorem [240, Section 1.5] If (m, k) = 1 and k > 1, then qk(m) =          k m (1 + im) √ k if k ≡0 (mod 4), m k √ k if k ≡1 (mod 4), 0 if k ≡2 (mod 4), m k i √ k if k ≡3 (mod 4). In particular (cf. Theorem 6.1.86), qk(1) = 1 2(1 + i)(1 + i−k) √ k. 160 Handbook of Finite Fields 6.1.164 Theorem [240, p. 47] If (m, k)=1 with k > 1 odd and squarefree, then (cf. Theorem 6.1.86) qk(m) = k−1 X n=1 n k ζmn k . 6.1.165 Deﬁnition (Gauss character sums over Z/kZ.) Let χ be a Dirichlet character (mod k), where k > 1. For any integer m, deﬁne the Gauss sum τk(m, χ) over Z/kZ by τk(m, χ) = k−1 X n=1 χ(n)ζmn k . 6.1.166 Remark When χ is trivial, τk(m, χ) is a Ramanujan sum. When k = p, τk(m, χ) is the Gauss sum G(m, χ) over Fp. The next two theorems deal with primitive Gauss sums; for proofs and generalizations, see [1581, pp. 48–49]. 6.1.167 Theorem For k > 1, let χ (mod k) be a primitive Dirichlet character [240, pp. 28–29]. Then \|τk(1, χ)\| = √ k. If further χ is quadratic, then τk(1, χ) = (√ k if χ(−1) = 1, i √ k if χ(−1) = −1. 6.1.168 Theorem If χ (mod k) is primitive, then for any integer m, τk(m, χ) = χ(m)τk(1, χ). 6.1.169 Remark There is a reduction formula which reduces the problem of evaluating τk(m, χ) to the case where χ (mod k) is primitive, m = 1, and k is a prime power ps; see [240, p. 29]. When s > 1, such Gauss sums have known closed form evaluations; see [240, Section 1.6] and . There are similar reduction formulae for Kloosterman sums. Evaluations and bounds for Kloosterman sums over Z/psZ with s > 1 are given in [655, 1004]. For Gauss and Kloosterman sums over rings of algebraic or p-adic integers, see [1002, 1364, 1366]. Extensions of Gaussian periods to ﬁnite rings are discussed in . For Hecke Gauss sums in quadratic number ﬁelds, see . Gauss sums connected with Hecke L-functions are discussed in [1581, p. 60]. 6.1.170 Remark Let k > 1 be an odd integer. We close with an evaluation of a Sali´ e sum over the ring Z/kZ, which for prime k = p reduces to the evaluation of the Sali´ e sum K(u, ρ) over Fp given in Theorem 6.1.127. For a short proof, see . 6.1.171 Deﬁnition Fix an odd integer k > 1. For an integer a with (a, k) = 1, deﬁne the Sali´ e sum S(a) over Z/kZ by S(a) = X x∈(Z/kZ)∗ x k ζx+a/x k , where (x/k) is the Jacobi symbol. Exponential and character sums 161 6.1.172 Theorem (Sali´ e sums over Z/kZ) Fix an odd integer k > 1 and let (a, k) = 1. If a is not congruent to a square mod k, then S(a) = 0. If a ≡b2 (mod k) for some integer b, then S(a) = i(k−1)2/4√ k X x∈Z/kZ x2=1 ζ2xb k . See Also §3.1, §3.5 For counting irreducible polynomials with prescribed norm or trace. §6.2 For more general exponential and character sums. §6.3 For further applications of character sums. §7.3 For solutions to diagonal equations over ﬁnite ﬁelds. §10.1 For the discrete Fourier transform and Gauss sums. §12.2, §12.4, §12.5, §12.7, §12.8, §12.9 For curves and varieties, counting rational points, zeta and L-functions. §14.6 For applications to diﬀerence sets. §15.1 For cyclic codes. References Cited: [49, 66, 111, 112, 113, 114, 115, 144, 240, 374, 382, 503, 548, 580, 592, 620, 622, 655, 659, 714, 715, 813, 928, 999, 1000, 1001, 1002, 1003, 1004, 1005, 1006, 1007, 1008, 1009, 1010, 1011, 1012, 1013, 1014, 1015, 1078, 1139, 1146, 1299, 1343, 1344, 1363, 1364, 1365, 1366, 1453, 1454, 1455, 1538, 1562, 1580, 1581, 1700, 1701, 1711, 1713, 1725, 1732, 1733, 1783, 1784, 1785, 1790, 1809, 1908, 1946, 1947, 1960, 2042, 2086, 2092, 2119, 2122, 2124, 2128, 2167, 2225, 2245, 2288, 2463, 2524, 2610, 2615, 2645, 2649, 2650, 2652, 2791, 2815, 2835, 2854, 2915, 2948, 3026, 3027, 3034] 6.2 More general exponential and character sums Antonio Rojas-Le´ on, University of Sevilla 6.2.1 One variable character sums 6.2.1 Theorem Let f ∈Fq[x] be a polynomial of degree d > 0 and ψ : Fq →C∗a non-trivial additive character. If d is prime to p, then X x∈Fq ψ(f(x)) ≤(d −1)√q. If d is divisible by p and ψ(t) = ψ(tp) for every t ∈Fq then X x∈Fq ψ(f(x)) ≤(d′ −1)√q 162 Handbook of Finite Fields if d′ ̸= 0, where d′ is the lowest degree of a polynomial in Fq[x] Artin-Schreier equivalent to f (i.e., of the form f + gp −g with g ∈Fq[x]). 6.2.2 Theorem Let f ∈Fq[x] be a polynomial of degree d > 0 and χ : F∗ q →C∗a non-trivial multiplicative character of order m (extended by zero to Fq). Then, if f is not an m-th power in Fq[x] (where Fq is the algebraic closure of Fq), X x∈Fq χ(f(x)) ≤(d −1)√q. 6.2.3 Theorem Let f, g ∈Fq[x] be polynomials of degrees d > 0 and e > 0 respectively, ψ : Fq →C∗a non-trivial additive character and χ : F∗ q →C∗a non-trivial multiplica-tive character of order m (extended by zero to Fq), such that either f is not of the form ¯ f p −¯ f with ¯ f ∈Fq[x] or g is not an m-th power in Fq[x]. Then X x∈Fq ψ(f(x))χ(g(x)) ≤(d + e −1)√q. 6.2.4 Remark The previous three results are a consequence of Weil’s conjectures for curves, as pointed out by Hasse and Weil , so they follow from Weil’s proof of the conjectures . They are also particular cases of the more general higher dimensional results stated in this section. 6.2.2 Additive character sums 6.2.5 Deﬁnition Let V ⊆AN Fq be an aﬃne variety, f : V →A1 Fq a regular map and ψ : Fq →C∗ a non-trivial additive character. We denote by S(V, f, ψ) the additive character sum S(V, f, ψ) = X x∈V (Fq) ψ(f(x)). 6.2.6 Deﬁnition The L-function associated to V , f and ψ is the power series L(V, f, ψ; T) = exp ∞ X m=1 Sm(V, f, ψ)T m m ! ∈1 + TC, where Sm(V, f, ψ) = S(V ×Fq Fqm, f, ψ ◦TrFqm/Fq). 6.2.7 Remark Character sums can be used to count the number of rational points on a variety. In particular, zeta functions of aﬃne varieties are special cases of L-functions of additive character sums: it is easy to check that, if V is the aﬃne variety deﬁned by the vanishing of f1, . . . , fr ∈Fq[x1, . . . , xn], then for any non-trivial character ψ, and variables y1, . . . , yr, Sm(An+r, y1f1 + · · · + yrfr, ψ) = qmr · #V (Fqm). Exponential and character sums 163 6.2.8 Deﬁnition A number z ∈C is a Weil integer if it is an algebraic integer and all its conjugates over Q have the same absolute value. If A ∈R, A > 0 is ﬁxed, z has weight w ∈R (relative to A) if all its conjugates have absolute value Aw/2. 6.2.9 Theorem For every aﬃne variety V ⊆AN Fq of dimension n, every regular map f : V →A1 Fq and every non-trivial additive character ψ : Fq →C∗, the L-function L(V, f, ψ; T) is rational, and all its reciprocal roots and poles are Weil integers of weight ≤2n relative to q. 6.2.10 Remark This result is just a particular case of the more general theory of L-functions of ℓ-adic sheaves on varieties over ﬁnite ﬁelds; see Section 12.7 for the general statements and some references. For additive exponential sums, rationality was ﬁrst proven by Bombieri following Dwork’s ideas . 6.2.11 Remark The L-function encodes the sequence of exponential sums Sm(V, f, ψ) for m ≥1. More precisely, if L(V, f, ψ; T) = Q i(1 −αiT) Q j(1 −βjT) is the decomposition into linear factors, then Sm(V, f, ψ) = X j βm j − X i αm i . In this way, estimates about character sums can be derived from statements about the number of roots and poles of the corresponding L-function and their absolute values. 6.2.12 Remark The type of statements that one tries to prove about these sums are estimates of the form \|Sm(V, f, ψ)\| ≤Cqm(n+i)/2, where i is as small as possible and C is a constant that depends only on certain quantities attached to the polynomials that deﬁne the variety V and the regular map f. In some particularly nice cases, one can give geometric conditions that imply the optimal bound (i = 0). 6.2.13 Theorem [795, Th´ eor eme 8.4] Let f ∈Fq[x1, . . . , xn] be a polynomial of degree d > 0. Suppose that 1. d is prime to p. 2. The projective hypersurface deﬁned by the highest degree homogeneous form fd of f is smooth, that is, the polynomials fd, ∂fd ∂x1 , . . . , ∂fd ∂xn do not have any common zero in Pn−1(Fq). Then for every non-trivial ψ, L(An Fq, f, ψ; T)(−1)n is a polynomial of degree (d −1)n, all whose reciprocal roots have weight n relative to q. In particular, for every m ≥1 the following estimate holds: \|Sm(An, f, ψ)\| ≤(d −1)nqmn/2. 6.2.14 Theorem [28, Theorems 1.4, 1.11][2469, Corollary 3] Let f ∈Fq[x1, . . . , xn] be a polynomial of degree d. Suppose that 1. d is divisible by p. 2. The projective hypersurface deﬁned by the highest degree homogeneous form fd of f is smooth. 3. The projective hypersurface deﬁned by the homogeneous form fd−1 of degree d −1 of f does not contain any of the common roots of the partial derivatives of fd. 164 Handbook of Finite Fields Then for every non-trivial ψ, L(An Fq, f, ψ; T)(−1)n is a polynomial of degree ((d −1)n+1 − (−1)n+1)/d, all whose reciprocal roots have weight n relative to q. In particular, for every m ≥1 the following estimate holds: \|Sm(An, f, ψ)\| ≤(d −1)n+1 −(−1)n+1 d qmn/2. 6.2.15 Theorem [341, Theorem 1] Let f ∈Fq[x1, . . . , xn] be a polynomial of degree d. Then for every non-trivial ψ, the total degree of the L-function L(An Fq, f, ψ; T) does not exceed (4d + 5)n. 6.2.16 Theorem [341, Theorem 2] Let V ⊆AN Fq be an aﬃne variety of dimension n and degree e and f ∈Fq[x1, . . . , xN] a polynomial of degree d. Then for every non-trivial ψ, the total degree of the L-function L(V, f, ψ; T) does not exceed (4 max(e + 1, d) + 5)2N+1. 6.2.17 Theorem [1532, Theorem 5] Let V ⊆A3 Fq be the surface deﬁned by g = 0, where g ∈Fq[x1, x2, x3], and let f ∈Fq[x1, x2, x3]. Suppose that 1. All geometric ﬁbres of f : V →A1 Fq have dimension ≤1. 2. The generic ﬁbre of f : V →A1 Fq is a geometrically irreducible curve. Then for every non-trivial ψ : Fq →C∗and every m ≥1 the following estimate holds: \|Sm(V, f, ψ)\| ≤C(f, g)qm, where C(f, g) depends only on the degrees of f and g. 6.2.18 Theorem [342, Theorem 1] Let V ⊆AN Fq be a geometrically irreducible variety of dimension n ≥3 and f : V →A1 Fq a regular map. Suppose that 1. Every geometric ﬁbre Vt of f has a unique irreducible component V d−1 t of di-mension d −1, and possibly other irreducible components of lower dimensions. 2. For all but ﬁnitely many t ∈Fq, Vt = V d−1 t and the Albanese varieties of V and Vt [1844, Chapter II, Section 3] have the same dimension. Then, if Fq has suﬃciently large characteristic, for every non-trivial ψ all reciprocal roots and poles of L(V, f, ψ; T) have weight ≤2n−3 relative to q. Morover, its degree is bounded by a constant C depending only on the number and the degrees of the polynomials that deﬁne V and f. In particular, for every m ≥1 the following estimate holds: \|Sm(V, f, ψ)\| ≤C(qm)n−3 2 . 6.2.19 Theorem [1700, Th´ eoreme 5.1.1] Let X ⊆PN Fq be a geometrically connected smooth pro-jective variety of dimension n, V = X ∩AN Fq its aﬃne part, f ∈Fq[x1, . . . , xN] a polynomial of degree d and F ∈Fq[x0, x1, . . . , xn] its homogenized with respect to x0. Suppose that 1. d is prime to p. 2. The hyperplane at inﬁnity L = PN Fq\AN Fq intersects X transversally (i.e., the scheme-theoretic intersection X ∩L is smooth of dimension n −1). 3. The hypersurface H deﬁned by F = 0 intersects X ∩L transversally (i.e., the scheme-theoretic intersection X ∩H ∩L is smooth of dimension n −2). Then for every non-trivial ψ, L(V, f, ψ; T)(−1)n is a polynomial of degree C(X, d) = Z X c(X) (1 + L)(1 + dL) Exponential and character sums 165 where c(X) is the total Chern class of X [799, Expos´ e XVII, 5.2] and L the class of a hyperplane section, all whose reciprocal roots have weight n relative to q. In particular, for every m ≥1 the following estimate holds: \|Sm(V, f, ψ)\| ≤C(X, d)qmn/2. 6.2.20 Theorem [1704, Theorem 4] Let X ⊆PN Fq be a geometrically irreducible integral projective variety of dimension n, V = X ∩AN Fq its aﬃne part, f ∈Fq[x1, . . . , xN] a polynomial of degree d and F ∈Fq[x0, x1, . . . , xn] its homogenized with respect to x0. Suppose that 1. d is prime to p. 2. The scheme-theoretic intersection X ∩H ∩L has dimension n −2, where L = PN Fq\AN Fq is the hyperplane at inﬁnity and H is the hypersurface deﬁned by F = 0. Let δ ≥−1 be the dimension of its singular locus. 3. The dimension of the singular locus of the scheme-theoretic intersection X ∩L is smaller than or equal to δ. Then for every non-trivial ψ, all reciprocal roots and poles of L(V, f, ψ; T) have weight smaller than or equal to n+δ +1 relative to q, and its total degree is bounded by a constant C(X, d) depending only on number and the degrees of the forms deﬁning X and on d. In particular, for every m ≥1 the following estimate holds: \|Sm(V, f, ψ)\| ≤C(X, d)qm(n+δ+1)/2. 6.2.21 Deﬁnition Let f = P i∈Zn aixi ∈Fq[x±1 1 , . . . , x±1 n ] be a Laurent polynomial. The Newton polyhedron ∆∞(f) of f at inﬁnity is the convex hull in Rn of the set {0}∪{i ∈Zn\|ai ̸= 0}. A polynomial f is non-degenerate with respect to ∆∞(f) if, for every face δ of ∆∞(f) that does not contain the origin, the equations ∂fδ ∂x1 = · · · = ∂fδ ∂xn = 0 do not have any common solution in (F ∗ q)n, where fδ = P i∈Zn∩δ aixi is the restriction of f to the face δ. 6.2.22 Theorem [23, Theorem 4.2][812, Theorem 1.3] Let Tn be the n-dimensional split torus over Fq, f ∈Fq[x±1 1 , . . . , x±1 n ] a Laurent polynomial and ∆∞(f) its Newton polyhedron at inﬁnity. Suppose that 1. dim ∆∞(f) = n. 2. f is non-degenerate with respect to ∆∞(f). Then for every non-trivial ψ, L(Tn(Fq), f, ψ; T)(−1)n is a polynomial of degree n!Vol(∆∞(f)), all whose reciprocal roots have weight smaller than or equal to n relative to q. If, in addition, the origin is an interior point of ∆∞(f), then all its reciprocal roots have weight n relative to q. In particular, for every m ≥1 the following estimate holds: \|Sm(Tn, f, ψ)\| ≤n!Vol(∆∞(f))qmn/2. 166 Handbook of Finite Fields 6.2.3 Multiplicative character sums 6.2.23 Deﬁnition Let V ⊆AN Fq be an aﬃne variety, f : V →A1 Fq a regular map and χ : F∗ q →C∗ a non-trivial multiplicative character, extended by zero to Fq. We denote by S(V, f, χ) the multiplicative character sum S(V, f, χ) = X x∈V (Fq) χ(f(x)). 6.2.24 Deﬁnition The L-function associated to V , f and χ is the power series L(V, f, χ; T) = exp ∞ X m=1 Sm(V, f, χ)T m m ! ∈1 + TC, where Sm(V, f, χ) = S(V ⊗Fqm, f, χ ◦NormFqm/Fq). 6.2.25 Theorem For every aﬃne variety V ⊆AN Fq of dimension n, every regular map f : V →A1 Fq and every non-trivial multiplicative character χ : F∗ q →C∗, the L-function L(V, f, χ; T) is rational, and all its reciprocal roots and poles are Weil integers of weight at most 2n relative to q. 6.2.26 Remark As in the additive case, this result is a particular case of the more general theory of L-functions of ℓ-adic sheaves on varieties over ﬁnite ﬁelds, see Section 12.7 for the general statements and some references. 6.2.27 Remark More generally, one can construct “mixed” character sums of the form X x∈V (Fq) ψ(f(x))χ(g(x)) where ψ : Fq →C∗is a non-trivial additive character, χ : F∗ q →C∗a non-trivial multiplica-tive character and f, g : V →A1 Fq are regular functions. The corresponding L-function is again rational, and all its reciprocal roots and poles are Weil integers of weight at most 2n relative to q. 6.2.28 Theorem [1707, Theorems 2.1, 2.2][2468, Theorem 4.4] Let f ∈Fq[x1, . . . , xn] be a polyno-mial of degree d > 0. Suppose that 1. The aﬃne scheme deﬁned by f is smooth. 2. The projective scheme deﬁned by the highest degree homogeneous form fd of f is smooth. Then for every non-trivial χ, L(An(Fq), f, χ; T)(−1)n is a polynomial of degree (d −1)n, all whose reciprocal roots have weight smaller than or equal to n relative to q. In particular, for every m ≥1 the following estimate holds: \|Sm(An, f, χ)\| ≤(d −1)nqmn/2. 6.2.29 Theorem [1707, Theorem 3.1, 3.2][2468, Corollary 4.3] Let X ⊆AN Fq be a geometri-cally connected smooth projective variety of dimension n, V = X ∩AN Fq its aﬃne part, f ∈Fq[x1, . . . , xN] a polynomial of degree d and F ∈Fq[x0, x1, . . . , xn] its homogenized with respect to x0. Suppose that Exponential and character sums 167 1. The hypersurface H deﬁned by F = 0 intersects X transversally (i.e., the scheme-theoretic intersection X ∩H is smooth of dimension n −1). 2. The hyperplane at inﬁnity L = PN Fq\AN Fq intersects X ∩H transversally (i.e., the scheme-theoretic intersection X ∩H ∩L is smooth of dimension n −2). Then for every non-trivial χ, L(V, f, χ; T)(−1)n is a polynomial of degree C(X, d) = Z X c(X) (1 + L)(1 + dL) where c(X) is the total Chern class of X [799, Expos´ e XVII, 5.2] and L the class of a hyperplane section, all whose reciprocal roots have weight smaller than or equal to n relative to q. In particular, for every m ≥1 the following estimate holds: \|Sm(V, f, χ)\| ≤C(X, d)qmn/2. 6.2.30 Theorem [2468, Theorem 1.1] Let X ⊆PN Fq be a geometrically irreducible Cohen-Macaulay projective variety of dimension n, V = X ∩AN Fq its aﬃne part, f ∈Fq[x1, . . . , xN] a poly-nomial of degree d and F ∈Fq[x0, x1, . . . , xn] its homogenized with respect to x0. Let χ : (Fq)⋆→C∗be a non-trivial multiplicative character. Suppose that 1. The scheme-theoretic intersection X ∩H ∩L has dimension n −2 and singular locus of dimension δ ≥−1, where L = PN Fq\AN Fq is the hyperplane at inﬁnity and H is the hypersurface deﬁned by F = 0. 2. Either the singular locus of the scheme-theoretic intersection X∩H has dimension at most δ, or else d is prime to p, χd is non-trivial and the singular locus of the scheme-theoretic intersection X ∩L has dimension at most δ. Then all reciprocal roots and poles of L(V, f, χ; T) have weights at most n+δ +1 relative to q, and its total degree is bounded by a constant C(X, d) depending only on the number and degrees of the forms that deﬁne X and on d. In particular, for every m ≥1 the following estimate holds: \|Sm(V, f, χ)\| ≤C(X, d)qm(n+δ+1)/2. 6.2.4 Generic estimates 6.2.31 Theorem [1712, Theorem 5.5.1] Let V ⊆AN Fq be a geometrically connected smooth variety of dimension n, f1, . . . , fr ∈Fq[x1, . . . , xN] such that the map (f1, . . . , fr) : V →Ar Fq is ﬁnite and ψ : Fq →C∗a non-trivial additive character. Then there exists a non-zero polynomial g ∈Fq[x1, . . . , xr] and a constant C ≥0 such that for every m ≥1 and every (a1, . . . , ar) ∈Fr qm with g(a1, . . . , ar) ̸= 0 the following estimate holds: \|Sm(V, a1f1 + · · · + arfr, ψ)\| ≤Cqmn/2. 6.2.32 Theorem [1703, Corollary 4.1.3] Let V be an aﬃne variety of dimension n over Fq, f : V →A1 Fq a regular map, π = (π1, . . . , πN) : V →AN Fq a quasi-ﬁnite map and ψ : Fq →C∗a non-trivial additive character. Then there exists a non-zero polyno-mial g ∈Fq[x1, . . . , xN+1] and a constant C ≥0 such that for every m ≥1 and every (a1, . . . , aN, b) ∈FN+1 qm with g(a1, . . . , aN, b) ̸= 0 the following estimate holds: \|Sm(V, f + a1π1 + · · · + aNπN + b, ψ)\| ≤Cqmn/2. 6.2.33 Theorem [1703, Corollary 4.1.3] Let V be an aﬃne variety of dimension n over Fq, f : V →A1 Fq a regular map, π = (π1, . . . , πN) : V →AN Fq a quasi-ﬁnite map and 168 Handbook of Finite Fields χ : Fq →C∗a non-trivial multiplicative character. Then there exists a non-zero poly-nomial g ∈Fq[x1, . . . , xN+1] and a constant C ≥0 such that for every m ≥1 and every (a1, . . . , aN, b) ∈FN+1 qm with g(a1, . . . , aN, b) ̸= 0 the following estimate holds: \|Sm(V, f + a1π1 + · · · + aNπN + b, χ)\| ≤Cqmn/2. 6.2.5 More general types of character sums 6.2.34 Theorem [560, Theorem 13] Let X be a smooth projective curve of genus g over Fq, f, h ∈Fq(X) rational functions and ψ : Fq →C∗(respectively χ : F∗ q →C∗) a non-trivial additive (resp. multiplicative) character. Suppose that f is not of the form ¯ f p −¯ f with ¯ f ∈Fq(X), and g is not an ord(χ)-th power in Fq(X). Then for every m ≥1, X x∈V (Fqm) ψ(TrFqm/Fqf(x))χ(NormFqm/Fq h(x)) ≤(2g −2 + s + l + d)qm/2, where V ⊆X is the open set where f and h are deﬁned, l is the number of poles of f, s is the number of zeroes and poles of h and d is the degree of the polar part of the divisor (f). 6.2.35 Theorem [1709, Theorem 1.1] Let f, g ∈Fq[x1, . . . , xn] be polynomials of degrees d and e, respectively, prime to p. Suppose that the projective hypersurfaces deﬁned by the highest degree forms of f and g are smooth and intersect transversally. Then for every m ≥1 the following estimate holds: X x∈Fqm ψ(TrFqm/Fqf(x))χ(NormFqm/Fq g(x)) ≤Cn,d,eqmn/2, where Cn,d,e = X a+b=n (d −1)a(e −1)b + X a+b=n−1 (d −1)a(e −1)b. 6.2.36 Theorem [1138, Proposition 0.1] Let Tn be the n-dimensional split torus over Fq, f ∈Fq[x±1 1 , . . . , x±1 n ] a Laurent polynomial, ∆∞(f) its Newton polyhedron at inﬁnity and χ : Tn(Fq) →C∗a character. Suppose that 1. dim ∆∞(f) = n. 2. f is non-degenerate with respect to ∆∞(f). Then for every non-trivial ψ and every m ≥1 the following estimate holds: X x∈(F∗ qm)n ψ(TrFqm/Fqf(x))χ(NormFqm/Fq x) ≤n!Vol(∆∞(f))qmn/2. 6.2.37 Theorem [1702, Theorem 1][2893, Corollary 2.2] Let a ∈Fqm such that Fqm = Fq(a) and χ : F∗ qm →C∗a non-trivial multiplicative character. Then the following estimate holds: X x∈Fq χ(x −a) ≤(m −1)√q. Exponential and character sums 169 6.2.38 Theorem [2893, Corollary 2.8] Let f ∈Fq[x] be a monic polynomial, and χ : (Fq[x]/(f))∗→ C∗a non-trivial character. Then X χ(g mod f) ≤1 d(deg(f) + 1)qd/2, where the sum is taken over the set of monic irreducible polynomials of degree d in Fq[x] which are coprime to f. 6.2.39 Theorem [1140, Theorem 3.7] Let f ∈Fq[x1, . . . , xn, y1, . . . , yn′] be a polynomial of degree d, r ≥1 an integer and let g be the polynomial r X j=1 f(x1,j, . . . , xn,j, y1, . . . , yn′) in Fq[x1,1, . . . , xn,r, y1, . . . , yn′]. Suppose that 1. d is prime to p. 2. The degree d homogeneous part of g deﬁnes a smooth hypersurface in Prn+n′−1. Then for every non-trivial additive character ψ and every m ≥1 the following estimate holds: X xi∈Fqmr ,yj∈Fqm ψ(TrFqmr /Fqf(x1, . . . , xn, y1, . . . , yn′)) ≤(d −1)nr+n′qm(nr+n′)/2. The constant (d−1)nr+n′ can be replaced by C(p, f)r3(d+1)n−1, where C(p, f) depends only on p and f. See Also §6.1 For speciﬁc results about Gauss, Jacobi, and Kloosterman sums. §12.7 For the general theory of ℓ-adic sheaves and their L-functions. , For a study of exponential sums and their L-functions using Dwork’s p-adic methods. For applications of p-adic cohomology to the study of exponential sums. , For the general theory of exponential sums using ℓ-adic cohomology. , For a study of the total degree of the L-function associated to an exponential sum and its change with the characteristic of the base ﬁeld. For a comprehensive survey of the main estimates for exponential sums obtained using ℓ-adic cohomology. References Cited: [22, 23, 28, 253, 339, 341, 342, 560, 795, 796, 799, 812, 940, 1138, 1140, 1532, 1700, 1702, 1703, 1704, 1707, 1709, 1712, 1844, 1867, 1869, 2384, 2468, 2469, 2589, 2698, 2893, 2961, 2962] 170 Handbook of Finite Fields 6.3 Some applications of character sums Alina Ostafe, Macquarie University Arne Winterhof, Austrian Academy of Sciences The main goal of this chapter is to show that character sums are very useful and friendly tools for a variety of problems in many areas such as coding theory, cryptography, and algorithms. There are so many applications of character sums that any survey will be incomplete. Here we chose a combination of some classical applications and newer, less known applications using diﬀerent types of character sums. 6.3.1 Applications of a simple character sum identity 6.3.1 Proposition [1631, Lemma 7.3.7] Let χ denote a nontrivial multiplicative character of Fq. Then we have X x∈Fq χ(x + a)χ(x + b) = −1, a, b ∈Fq, a ̸= b. 6.3.1.1 Hadamard matrices 6.3.2 Deﬁnition A Hadamard matrix of order n is an n×n matrix H with entries from {−1, +1} satisfying HHT = nI. 6.3.3 Construction (Paley) Let q be the power of an odd prime, η the quadratic character of Fq and Fq = {ξ1, . . . , ξq} any ﬁxed ordering of Fq. For q ≡3 (mod 4) there exists a Hadamard matrix H = (hij) of order n = q + 1 deﬁned by hi,n = hn,i = 1, i = 1, . . . , n, hj,j = −1, j = 1, . . . , n −1, hi,j = η(ξj −ξi), i, j = 1, . . . , n −1, i ̸= j. 6.3.4 Remark 1. The inner product of two diﬀerent rows of H is zero by Proposition 6.3.1. 2. Paley also presented a similar, but slightly more complicated construction for Hadamard matrices of order n = 2(q + 1) if q ≡1 (mod 4). 3. The Hadamard conjecture proposes that a Hadamard matrix of order n = 4k exists for every positive integer k. The smallest order n for which no Hadamard matrix of order n = 4k has been constructed is n = 668. The last progress known by the authors was the construction of a Hadamard matrix of order n = 428 by . 4. For a monograph on Hadamard matrices see ; see also Subsection 14.6.4. 5. Hadamard matrices can be used to construct good error correcting codes . 6. For relations between Hadamard matrices and designs see . Exponential and character sums 171 6.3.1.2 Cyclotomic complete mappings and check digit systems 6.3.5 Deﬁnition Let n be a positive divisor of q −1 and γ a primitive element of Fq. Then the sets Ci = γjn+i : j = 0, 1, . . . , (q −1)/n −1 , i = 0, 1, . . . , n −1, are cyclotomic cosets of order n. For a0, a1, . . . , an−1 ∈F∗ q we deﬁne a cyclotomic mapping fa0,a1,...,an−1 (of index n) by fa0,a1,...,an−1(0) = 0 and fa0,a1,...,an−1(ξ) = aiξ if ξ ∈Ci, i = 0, 1, . . . , n −1. 6.3.6 Proposition [998, Theorem 3.7] The mapping fa0,a1,...,an−1 is a permutation of Fq if and only if aiCi ̸= ajCj for all 0 ≤i < j ≤n −1. 6.3.7 Corollary For n ≥2 let j be an integer with 0 ≤j < n, χ a multiplicative character of Fq of order n, and a, b ∈Fq with a ̸= b. If aj = a and ai = b for i ̸= j, then ga,b = fa0,a1,...,an−1 is a permutation if and only if χ(a) = χ(b). 6.3.8 Deﬁnition A permutation f of Fq is a complete mapping of Fq if f(x) + x is also a permu-tation of Fq. 6.3.9 Remark Complete mappings are pertinent to the problem of constructing orthogonal Latin squares [1939, Section 9.4]. 6.3.10 Remark Substituting b = ac we see that ga,b is a complete mapping if and only if χ(c) = 1 and χ(a + 1) = χ(ac + 1) and the number N of complete mappings ga,ac with c ̸= 1 is N = X c∈F∗ q, χ(c)=1, c̸=1 1 n n−1 X i=0 X a∈F∗ q\{1,c−1} χi(a + 1)χi(a −c−1). Proposition 6.3.1 implies the following theorem. 6.3.11 Theorem [2278, Theorem 3] Let n ≥2 be a divisor of q −1 and j ∈{0, 1, . . . , n −1}. Then the number N of ordered pairs (a, b) ∈F∗ q ×F∗ q with a ̸= b such that the cyclotomic mapping ga,b = fa0,a1,...,an−1 with aj = a and ai = b for i ̸= j is a complete mapping of Fq equals N = (q −n −1)(q −2n −1)/n2. 6.3.12 Deﬁnition A check digit system over Fq consists of s permutations p1, . . . , ps of Fq and a symbol c ∈Fq such that each word a1 . . . as−1 ∈Fs−1 q is extended by a check digit as such that p1(a1) + · · · + ps(as) = c. 6.3.13 Remark All single errors are detected since all pi are permutations. Another frequent family of errors are adjacent transpositions . . . ab . . . →. . . ba . . . which are all detected if pi+1(x)p−1 i (x) −x are also permutations for i = 1, . . . , s −1. A permutation f such that f(x)−x is also a permutation is an orthomorphism. Since f is an orthomorphism whenever −f is a complete mapping, the number of orthomorphisms and complete mappings of the form ga,b is the same and the probability that a random choice of the parameters (a, b) gives an orthomorphism is asymptotically n−2 by Theorem 6.3.11. 6.3.14 Example [International Standard Book Number (ISBN-10)] An ISBN-10 consists of 10 digits x1 −x2x3x4x5x6 −x7x8x9 −x10. The ﬁrst digit x1 characterizes the language group, x2x3x4x5x6 is the actual book number, x7x8x9 is the number of the publisher, and x10 is a 172 Handbook of Finite Fields check digit. A correct ISBN satisﬁes x1+2x2+3x3+4x4+5x5+6x6+7x7+8x8+9x9+10x10 = 0 ∈F11, i.e., pi(x) = ix, i = 1, . . . , 10, and pi+1p−1 i (x) = (i−1 + 1)x, i = 1, . . . , 9, which are all orthomorphisms. 6.3.1.3 Periodic autocorrelation of cyclotomic generators 6.3.15 Deﬁnition Put εn = exp(2πi/n). Let (sk) be a T-periodic sequence over Zn. The (periodic) autocorrelation of (sk) is the complex-valued function deﬁned by A(t) = 1 T T −1 X k=0 εsk+t−sk n , 1 ≤t < T. 6.3.16 Remark Sequences with low autocorrelation have several applications in wireless commu-nication, cryptography, and radar, see the monograph . 6.3.17 Deﬁnition The p-periodic sequence (sk) over Zn deﬁned by s0 = 0 and sk = j if k ∈Cj for 0 ≤j < n, 1 ≤k < p, where Cj denotes the jth cyclotomic coset of order n deﬁned by Deﬁnition 6.3.5, is a cyclotomic generator of order n. 6.3.18 Remark Proposition 6.3.1 implies the exact values of the autocorrelation function of the cyclotomic generator of order n; see for the proof of a generalization to arbitrary ﬁnite ﬁelds. 6.3.19 Theorem The autocorrelation function f(t) of the cyclotomic generator of order n is given by A(t) = (−1 + εj n + ε−j−k n )/p if t ∈Cj and −1 ∈Ck. 6.3.2 Applications of Gauss and Jacobi sums 6.3.20 Deﬁnition Let χ be a multiplicative and ψ be an additive character of Fq and a ∈Fq of order T. The sums G(χ, ψ) = X c∈F∗ q χ(c)ψ(c) and Ga(ψ) = T −1 X n=0 ψ(an) are Gauss sums of ﬁrst kind and second kind, respectively. Let χ1, . . . , χk be k ≥2 multiplicative characters of Fq. The sum J(χ1, . . . , χk) = X c1+···+ck=1 χ1(c1) . . . χk(ck), is a Jacobi sum, where the summation is extended over all (c1, . . . , ck) ∈Fk q such that c1 + · · · + ck = 1. 6.3.21 Remark Gauss sums of the ﬁrst and second kinds are closely related by Ga(ψ) = T q −1 (q−1)/T −1 X j=0 G(χj, ψ), Exponential and character sums 173 where χ is a multiplicative character of order (q −1)/T, and Gauss sums of ﬁrst kind and Jacobi sums by J(χ1, . . . , χk) = G(χ1, ψ) · · · G(χk, ψ) G(χ1 · · · χk, ψ) if all involved characters are nontrivial. For background on Gauss and Jacobi sums see Section 6.1, or [1939, Chapter 5]. In particular, we have (if all characters are nontrivial) \|G(χ, ψ)\| = q1/2, \|Ga(ψ)\| ≤q1/2, and \|J(χ1, . . . , χk)\| = q(k−1)/2. (6.3.1) We note that the bound on \|Ga(ψ)\| is only nontrivial if T > q1/2. If q = p is a prime, bounds which are nontrivial for T ≥p1/3+ε and T ≥pε are given in [380, 1454] (see also for an improvement), respectively. 6.3.2.1 Reciprocity laws 6.3.22 Remark Gauss and Jacobi sums are involved in the proofs and statements of reciprocity laws. For example, let p and q be two distinct odd primes, let r be the order of p modulo q, ξ be a primitive q-th root of unity in Fpr, and G = P x∈Fq x q ξx be a Gauss sum of ﬁrst kind over Fq in Fpr. Then from G2 = (−1)(q−1)/2q we get Gp = (−1)(p−1)(q−1)/4 q p G. On the other hand, simple calculations give Gp = p q G (mod p) and we get the following result. 6.3.23 Theorem (Gauss law of quadratic reciprocity) [1939, Theorem 5.17] For any two distinct odd primes p and q we have p q q p = (−1)(p−1)(q−1)/4. 6.3.24 Deﬁnition For an integer n ≥2 and two distinct primes p, q ≡1 (mod n) we denote by χp,n(a) = a p n the character of order n with values in Fq deﬁned by γj p n = ξj, j = 0, . . . , n −1, where γ is a ﬁxed primitive element of Fp and ξ a primitive n-th root of unity in Fq. 6.3.25 Remark The sums Kp,n = (−1)(p−1)(q−1)/4 P x∈Fp χp,n(x)χp,n(1 −x) are up to signs Jacobi sums in Fq and appear in reciprocity laws of higher order. 6.3.26 Theorem (Cubic and biquadratic reciprocity laws) For two distinct primes p, q ≡1 (mod 3) we have p q 3 q p 3 Kp,3 q 3 = 1. For two distinct primes p, q ≡1 (mod 4) we have p q 4 p q 4 Kp,4 q 2 = 1. 174 Handbook of Finite Fields 6.3.27 Remark In terms of the decompositions of p and q in the rings of Eisenstein and Gaussian integers we have Kp,3 = ad −bc 3 if p = a2 −ab + b2; q = c2 −cd + d2; a, c ≡2 (mod 3); b, d ≡0 (mod 3) and Kp,4 = ad −bc 2 if p = a2 + b2; q = c2 + d2; a, c ≡1 (mod 4); b, d ≡0 (mod 2). 6.3.2.2 Distribution of linear congruential pseudorandom numbers 6.3.28 Deﬁnition The sequences xn+1 = axn + b, n ≥0, where x0, a, b ∈Fp with x0, a ̸= 0, and a ̸= 1, are linear congruential pseudorandom number generators. 6.3.29 Remark If a ̸= 1, they can also be given explicitly by xn = anx0 + an −1 a −1 b, n ≥0. (6.3.2) The sequence (xn) is T-periodic if b ̸= (1 −a)x0, where T is the order of a. 6.3.30 Deﬁnition Let Γ be a sequence of N elements (γn)N n=1 in the unit interval [0, 1). The discrepancy DN(Γ) is deﬁned by DN(Γ) = sup B⊆[0,1) TΓ(B) N −\|B\| , where the supremum is taken over all subintervals B = [α, β) ⊆[0, 1), and TΓ(B) is the number of elements of Γ inside B. 6.3.31 Remark The discrepancy is a measure for the uniform distribution of Γ and a small dis-crepancy is a desirable feature for (quasi-)Monte Carlo integration . The problem of estimating the discrepancy can be reduced to the problem of estimating certain exponential sums. 6.3.32 Proposition (Erd˝ os-Turan inequality) [922, Theorem 1.2.1] Let Γ be a sequence (γn)N n=1 in [0, 1). We have for any integer H ≥1, DN(Γ) ≪1 H + 1 N H X h=1 1 h\|SN(h)\|, where SN(h) = N−1 X n=0 exp (2πihγn) . 6.3.33 Remark For the sequence (xn/p), n = 0, 1, . . . , T −1, in [0, 1) derived from a linear pseu-dorandom number generator (xn) (where we identify Fp with the integers {0, 1, . . . , p −1}), the absolute value of the sums SN(h) equals the absolute value of Gauss sums of second kind provided that b ̸= (1 −a)x0 and a ̸= 1. A discrepancy bound can be easily obtained by combining Proposition 6.3.32 with the bound \|Ga(ψ)\| ≤p1/2. 6.3.34 Theorem [2230, Theorem 1] For the sequence Γ = (xn/p : n = 0, . . . , N −1), where xn is deﬁned by (6.3.2), N < T and T is the order of a, we have DN(Γ) ≪N −1p1/2(log p)2. Exponential and character sums 175 6.3.2.3 Diagonal equations, Waring’s problem in ﬁnite ﬁelds, and covering radius of certain cyclic codes 6.3.35 Deﬁnition A diagonal equation over Fq is an equation of the type c1xk1 1 + · · · + csxks s = b (6.3.3) for any positive integers k1, . . . , ks, c1, . . . , cs ∈F∗ q, and b ∈Fq. We denote by Nb the number of solutions in Fs q of (6.3.3); see Section 7.3. 6.3.36 Theorem [1939, Theorem 6.34] The number Nb of solutions of (6.3.3) for b ∈F∗ q is Nb = qs−1 + d1−1 X j1=1 . . . ds−1 X js=1 χj1 1 (bc−1 1 ) . . . χjs s (bc−1 s )J(χj1 1 , . . . , χjs s ), where χi denotes a multiplicative character of order di = gcd(ki, q −1). 6.3.37 Remark Put Tj = (q −1)/ gcd(kj, q −1) and let aj ∈F∗ q be an element of order Tj for j = 1, . . . , s. Since {xkj : x ∈Fq} = {an j : 0 ≤n < Tj} ∪{0} =: Gj, we can also express Nb in terms of Gauss sums of the second kind, Nb = 1 q X (y1,...,xs)∈G1×...Gs X ψ ψ(c1y1 + · · · + csys −b) = 1 q X ψ ψ(−b) s Y j=1 1 + Gaj(ψcj) , where the sum over ψ runs over all additive characters of Fq and ψcj(x) = ψ(cjx). 6.3.38 Deﬁnition Let g(k, q) be the smallest s such that every element b ∈Fq can be written as a sum of at most s summands of k-th powers in Fq. The problem of determining or estimating g(k, q) is Waring’s problem in Fq. 6.3.39 Remark We note that g(k, q) = g(d, q) if d = gcd(k, q −1) and we may restrict ourselves to the case that k \| (q −1). Combining Theorem 6.3.36 (with k1 = · · · = ks = k \| q −1 and c1 = · · · = cs) with the result on the absolute value of Jacobi sums (6.3.1) we get immediately Nb ≥qs−1 −(k −1)sq(s−1)/2 which implies the following bound. 6.3.40 Theorem For any divisor k of q −1 we have g(k, q) ≤s if qs−1 > (k −1)2s. 6.3.41 Remark Theorem 6.3.40 applies only to k < q1/2−ε and for q3/7 +1 ≤k < q1/2 we have the improvement g(k, q) ≤8 of [650, Corollary 7]. Moreover, from [1283, Theorem 6] it follows that for any ε > 0, k ≤q1−ε and if Fq = Fp(xk) for some x ∈Fq, there is a constant c(ε) such that g(k, q) ≤c(ε). However, if q = p is a prime, a very moderate but nontrivial bound on Gauss sums of the second kind from leads to the nontrivial bound on g(k, p) ≪(ln k)2+ε if k < p(log log p)1−ε/ log p. 6.3.42 Deﬁnition The covering radius ρ(C) of a code C ⊆Fn q is ρ(C) = max x∈Fn q min c∈C d(c, x), where d is the Hamming distance. 176 Handbook of Finite Fields 6.3.43 Proposition [1468, Lemma 1.1] Let H be the parity check matrix of a linear [n, k]-code C over Fq, i.e., C = {c ∈Fn q : HcT = 0}. The covering radius is the least integer ρ such that every x ∈Fn−k q is a linear combination of at most ρ columns of H. 6.3.44 Remark Let g ∈Fq[X] be the minimal polynomial of an element α ∈F∗ q of order n and r be the order of q modulo n, i.e., Fq(α) = Fqr. Then the cyclic code C = (g) is the [n, n−r]-code with parity check matrix H = (1, α, α2, . . . , αn−1), where the elements of Fqr are identiﬁed with r-dimensional column vectors. Put N = (qr −1)/n. Then α = γN for some primitive element γ of Fqr and the columns of H consist of the nonzero N-th powers in Fqr. By Proposition 6.3.43, ρ(C) is the least integer ρ such that any x ∈Fqr can be written as a linear combination of at most ρ N-th powers in Fqr. Hence, we have ρ(C) ≤g(N, q), where we have equality for q = 2. 6.3.2.4 Hidden number problem and noisy interpolation 6.3.45 Deﬁnition ((Extended) Hidden number problem) [347, 348] Let T ⊆Fp. Recover a number a ∈Fp if for many known t ∈T the l most signiﬁcant bits of at are given. 6.3.46 Remark If l is of order log1/2 p and T has some uniform distribution property, a lattice reduction technique solves the hidden number problem in polynomial time. The uniform distribution property is fulﬁlled if the maximum over all nontrivial additive character sums of Fp over T is small, max ψ X t∈T ψ(t) = O(#T 1−ε). If T is a subgroup of F∗ p, the sums are Gauss sums of the second kind and the desired uniform distribution property is fulﬁlled by the bounds of and [380, 371] if #T ≥p1/3+ε and #T ≥pε, respectively. The bound of and ideas reminiscent to Waring’s problem solve the problem for smaller #T ≥log p/(log log p)1−ε using more bits, that is, l of order log4 p. 6.3.47 Deﬁnition The sparse polynomial noisy interpolation problem consists of ﬁnding an unknown polynomial f ∈Fp[X] of small weight from approximate values of f(t) at polynomially many points t ∈Fp selected uniformly at random. 6.3.48 Remark 1. The case f(X) = aX corresponds to the hidden number problem. 2. For more details we refer to the survey and [2644, Chapter 30]. 6.3.3 Applications of the Weil bound 6.3.49 Theorem [2548, Theorem 2G] Let g ∈Fq[X] be of degree n and f ∈Fq[X] have d distinct roots in its splitting ﬁeld over Fq. Let χ be a multiplicative character of Fq of order s and let ψ be an additive character of Fq. If either s > 1 and f is not, up to a multiplicative constant, an s-th power, or ψ is nontrivial and gcd(n, q) = 1, we have X c∈Fq χ(f(c))ψ(g(c)) ≤(d + n −1)q1/2. Exponential and character sums 177 6.3.50 Remark The condition gcd(n, q) = 1 can be replaced by the weaker condition that Y q −Y −g(X) is absolutely irreducible. The condition on f is fulﬁlled if gcd(deg(f), s) = 1. 6.3.3.1 Superelliptic and Artin-Schreier equations 6.3.51 Deﬁnition For polynomials f, g ∈Fq[X] and a divisor s of q −1 Y s = f(X) and Y q −Y = g(X) are superelliptic and Artin-Schreier equations deﬁned over Fq, respectively. For s = 2 these equations are hyperelliptic equations and, if additionally deg(f) = 3, elliptic equa-tions. 6.3.52 Proposition [2711, Lemma 1 and Lemma 2] The number Nf,s,qr of solutions (x, y) ∈F2 qr of a superelliptic equation is Nf,s,qr = X ord(χ)\|s X x∈Fqr χ(f(x)), where the outer sum runs over all multiplicative characters χ of Fqr such that χs is trivial. The number Ng,qr of an Artin-Schreier equation is Ng,qr = X ψr X x∈Fqr ψr(g(x)), where the outer sum runs over all additive characters ψ of Fq and ψr(x) = ψ(Tr(x)), and Tr denotes the trace from Fqr to Fq. 6.3.53 Remark After isolating the trivial characters, the Weil bound implies immediately bounds on Nf,s,qr and Ng,qr. 6.3.54 Theorem [2711, Chapter 1.4] If f ∈Fq[X] with gcd(deg(f), s) = 1 and d > 0 diﬀerent zeros (in the algebraic closure of Fq), then the number of solutions Nf,s,qr over Fqr of the superelliptic equation Y s = f(X) satisﬁes \|Nf,s,qr −qr\| ≤(s −1)(d −1)qr/2. If g ∈Fq[X] has degree n with gcd(n, q) = 1, then the number of solutions Ng,qr over Fqr of the Artin-Schreier equation Y q −Y = g(X) satisﬁes \|Ng,qr −qr\| ≤(n −1)(q −1)qr/2. 6.3.3.2 Stable quadratic polynomials 6.3.55 Deﬁnition For a polynomial f ∈Fq[X], q odd, we deﬁne f (0)(X) = X, f (n)(X) = f f (n−1)(X) , n = 1, 2, . . . . A polynomial f is stable if all polynomials f (n) are irreducible over Fq. 178 Handbook of Finite Fields 6.3.56 Deﬁnition For a quadratic polynomial f(X) = aX2 + bX + c ∈Fq[X], a ̸= 0, we deﬁne α = −b/2a as the unique critical point of f (that is, the zero of the derivative f ′) and the critical orbit of f, Orb = {f (n)(α) : n = 2, 3, . . .} = {f (n)(α) : n = 2, . . . , tf}, where tf is the smallest value of t with f (t)(α) = f (s)(α) for some positive integer s < t, i.e., the orbit length. 6.3.57 Proposition [1620, Proposition 3] Let η denote the quadratic character of Fq. A quadratic polynomial f ∈Fq[X] is stable if and only if η(x) = −1 for all elements x of the adjusted orbit Orb(f) = {−f(α)} S Orb(f). 6.3.58 Remark If f is stable, then the elements f (n)(α), n = 2, . . . , tf −1, are diﬀerent and Proposition 6.3.57 implies for any positive integer K, tf −2 = 1 2K tf −1 X n=2 K Y k=1 (1 −η( f (k)(f (n)(α)) )) ≤1 2K X x∈Fq K Y k=1 1 −η f (k)(x) . Expanding the product on the right hand side, we obtain one trivial sum equal to q and 2K −1 sums which can be bounded by the Weil bound giving tf = q 2K + O(2Kq1/2). The optimal choice of K gives the following result. 6.3.59 Theorem [2331, Theorem 1] For any odd q and any stable quadratic polynomial f ∈Fq[X] we have tf = O q3/4 . 6.3.60 Remark Similarly, Gomez and Nicol´ as estimated in the number of stable quadratic polynomials over Fq for an odd prime power q. As for Theorem 6.3.59, this reduces to estimating the character sum X (a,b,c)∈F∗ q×Fq×Fq K Y k=1 (1 −η (Fk(a, b, c))) , where Fk(a, b, c) is the k-th element of the critical orbit of f and K any positive integer parameter. Gomez and Nicol´ as have proved that there are O(q5/2(log q)1/2) stable quadratic polynomials over Fq for the power q of an odd prime. 6.3.3.3 Hamming distance of dual BCH codes 6.3.61 Deﬁnition Let γ be a primitive element of F2r, t be a positive integer with 1 ≤2t −1 ≤ 2⌈r/2⌉+ 1, and Gt denote the set of polynomials Gt = ( t X i=1 giX2i−1 ∈F2r[X] ) . For each g ∈Gt we deﬁne a binary word cg of length 2r −1, cg = Tr(g(1)), Tr(g(γ)), . . . , Tr(g(γ2r−2)) , Exponential and character sums 179 where Tr denotes the absolute trace from F2r to F2, and deﬁne a code Ct = {cg : g ∈Gt}. 6.3.62 Remark The code Ct is linear and is the dual of the primitive, binary BCH code with designed distance 2t + 1 . 6.3.63 Remark Following , one can use the Weil bound to estimate the Hamming weight for Ct. To do this, we recall that any nonzero codeword cg ∈Ct comes from a nonzero polynomial g(X) = Pt i=1 giX2i−1. Hence, we have 2r −2wtH(cg) = X x∈F2r (−1)Tr(g(x)) = X x∈F2r ψ(g(x)), where ψ(x) = (−1)Tr(x) is the additive canonical character [2372, Equation (4)]. Applying now the Weil bound, we get the following result. 6.3.64 Theorem [2372, Theorem 4] For t ≥1 the minimum Hamming weight of Ct is at least 2r−1 −(t −1)2r/2. 6.3.4 Applications of Kloosterman sums 6.3.65 Deﬁnition Let ψ be a nontrivial additive character of Fq and let a, b ∈Fq. We deﬁne the Kloosterman sum (see Section 6.1) by K(ψ; a, b) = X x∈F∗ q ψ(ax + bx−1). 6.3.66 Proposition [1939, Theorem 5.45] If ab ̸= 0, we have \|K(ψ; a, b)\| ≤2q1/2. 6.3.4.1 Kloosterman equations and Kloosterman codes 6.3.67 Deﬁnition For a, b, c ∈Fq with ab ̸= 0 Y q −Y = aX + bX−1 + c is a Kloosterman equation. 6.3.68 Theorem The number Na,b,c of solutions (x, y) ∈F2 qr of a Kloosterman equation satisﬁes \|Na,b,c −qr\| ≤2(q −1)qr/2. 6.3.69 Remark We note that this estimate is obtained in a similar way as the number of solutions to an Artin-Schreier equation in Theorem 6.3.54, but using the estimate in Proposition 6.3.66 instead of the Weil bound. 6.3.70 Deﬁnition Let γ be a primitive element of F2r. The codewords of the Kloosterman code C = {ca,b : a, b ∈F2r} are deﬁned by ca,b = (Tr(a + b), Tr(aγ + bγ−1) + · · · + Tr(aγ2r−2 + bγ2−2r)), a, b ∈F2r. 180 Handbook of Finite Fields 6.3.71 Theorem The minimum weight of the Kloosterman code is at least 2r−1 −2r/2. 6.3.4.2 Distribution of inversive congruential pseudorandom numbers 6.3.72 Deﬁnition For a ∈F∗ p, b ∈Fp we deﬁne, with the convention 0−1 = 0, the sequence (un) by the recurrence relation un+1 = au−1 n + b, n = 0, 1, . . . , (6.3.4) with u0 the initial value. Then the numbers un/p, n = 0, 1, . . ., in the interval [0, 1) form a sequence of inversive congruential pseudorandom numbers. 6.3.73 Remark The sequence (un) deﬁned by (6.3.4) is purely periodic with some period T ≤p. Using the Erd˝ os-Turan inequality given by Theorem 6.3.32, one reduces the problem of estimating the discrepancy of the elements in the sequence (un/p), n = 0, . . . , T −1, to estimating character sums given by ST (h) = T −1 X n=0 ψ(hun) with a ﬁxed integer h ̸≡0 (mod p) and ψ is the additive canonical character of Fp. Niederre-iter and Shparlinski developed a method to reduce the problem of estimating \|ST (h)\| to estimating Kloosterman sums. 6.3.74 Theorem [2270, Theorem 2] For the sequence Γ = (un/p : n = 0, . . . , T −1), where (un) is deﬁned by (6.3.4) and T is the period of the sequence (un), we have DT (Γ) ≪T −1/2p1/4 log p. 6.3.4.3 Nonlinearity of Boolean functions 6.3.75 Deﬁnition Let Br = {0, 1}r . The Fourier coeﬃcients (or Walsh-Hadamard coeﬃcients) b B(a) of B(U1, . . . , Ur), where a ∈Br, are deﬁned as b B(a) = X u∈Br (−1)B(u)+<a,u> , where < a, u >= a1u1 + a2u2 + · · · + arur denotes the standard inner product. The nonlinearity of the Boolean function B(U1, . . . , Ur) is deﬁned by N(B) = 2r−1 −1 2 max a∈Br b B(a) . 6.3.76 Remark Boolean functions used in cryptography must have high nonlinearity, see for ex-ample and Section 9.1. 6.3.77 Remark Each Boolean function B : Fr 2 →F2 can be represented with a polynomial f ∈F2r and the absolute trace function as B(x1, . . . , xr) = Tr(f(x1β1 + · · · + xrβr)) with some ﬁxed ordered basis {β1, . . . , βr} of F2r. Moreover, if we identify x ∈F2r with its coordinate Exponential and character sums 181 vector, we have that x →(−1)<a,x> is an additive character of F2r. Hence, we get N(B) = 2r−1 −1 2 max b∈F2r X u∈F2r ψ(B(u) + bu) , where ψ is the additive canonical character of F2r. For example, if f(x) = x−1 (with the convention 0−1 = 0), we get Kloosterman sums on the right hand side. 6.3.5 Incomplete character sums 6.3.78 Theorem (P olya-Vinogradov-Weil bound) Let χ be a nontrivial multiplicative character of order s of Fp and f ∈Fp[X] with d > 0 diﬀerent zeros such that f is not, up to a constant multiple, an s-th power. Then we have N−1 X n=0 χ(f(n)) ≤dp1/2 log p, 1 ≤N < p. 6.3.79 Remark The bound of Theorem 6.3.78 can be obtained using the standard method for reducing incomplete character sums to complete ones, see for example [1581, Section 12.2], and the Weil bound. This reducing method can be traced back to Polya and Vinogradov; for references see . Generalizations to certain incomplete character sums over arbitrary ﬁnite ﬁelds are given in [776, 2989, 2991, 2993]. 6.3.5.1 Finding deterministically linear factors of polynomials 6.3.80 Algorithm Let f ∈Fp[X], p an odd prime, be a squarefree polynomial with f(0) ̸= 0 which splits over Fp. For t = 0, 1, . . . , N we compute Lt(X) = gcd((X + t)(p−1)/2 −1, f(X)) = gcd(gt(X) −1, f(X)) via the Euclidean algorithm, where N is the main parameter of the algorithm, hoping that at least one polynomial Lt is nontrivial, that is, is equal to neither 1 nor f. For each t, the polynomial gt(X) ≡(X + t)(p−1)/2 (mod f(X)), deg(gt) < deg(f) is calculated eﬃciently using repeated squaring. 6.3.81 Remark Since x(p−1)/2 = 1 if and only if x is a quadratic residue modulo p, Lt is trivial, i.e., either 1 or f, only if a + t p = b + t p for any two distinct roots of f. Therefore, the algorithm does not determine a nontrivial factor of f only if N + 1 = N X t=0 (a + t)(b + t) p ≪p1/2 log p by Theorem 6.3.78. 6.3.82 Remark Any polynomial f can be factorized into squarefree polynomials by calculating gcd(f, f ′). The part of a squarefree polynomial f which splits over Fp and is not divisible 182 Handbook of Finite Fields by X is gcd(Xp−1 −1, f). More details about the factorization of univariate polynomials over ﬁnite ﬁelds can be found in Section 11.4. 6.3.83 Remark Shoup extended Legendre’s idea to design a deterministic factoring algo-rithm for all squarefree polynomials. 6.3.5.2 Measures of pseudorandomness 6.3.84 Deﬁnition [Mauduit and S´ ark¨ ozy ] For a ﬁnite binary sequence EN = (e1, . . . , eN) ∈ {−1, +1}N the well-distribution measure of EN is deﬁned by W(EN) = max a,b,M M X j=1 ea+bj , where the maximum is taken over all a, b, M ∈Z and b, M > 0 such that 1 ≤a + b ≤ a + bM ≤N, and the correlation measure of order ℓof EN is deﬁned as Cℓ(EN) = max M,D M X n=1 en+d1en+d2 · · · en+dℓ , where the maximum is taken over all D = (d1, . . . , dℓ) and M is such that 0 ≤d1 < . . . < dℓ≤N −M. 6.3.85 Deﬁnition Let p be an odd prime. The binary sequence Ep = (1, e1, . . . , ep−1) deﬁned by en = n p , 0 ≤n < p, is the Legendre sequence. 6.3.86 Remark The sums in the deﬁnitions of well-distribution measure and correlation measure of order ℓfor the Legendre sequence are essentially sums of products of Legendre symbols which can be estimated by Theorem 6.3.78. 6.3.87 Remark We note that W(EN) and Cℓ(EN) of a “truly random” sequence are of the order of magnitude N 1/2 log N and N 1/2(log N)c(ℓ), respectively, see [82, 554]. 6.3.88 Theorem For the Legendre sequence Ep we have W(Ep) ≪p1/2 log p and Cℓ(Ep) ≪ℓp1/2 log p. 6.3.89 Remark The linear complexity is a measure for the unpredictability and thus suitability of a sequence in cryptography. For sequences (u0, . . . , uN−1) ∈FN 2 it is closely related to the correlation measure of order ℓof the sequence (e0, . . . , eN−1) ∈{−1, +1}N deﬁned by en = (−1)un, n = 0, . . . , N −1. Hence, from a suitable upper bound on Cℓ(EN) up to a suﬃciently large ℓwe can derive a lower bound on the linear complexity of (un), see and Subsection 10.4.5. Exponential and character sums 183 6.3.6 Other character sums 6.3.6.1 Distribution of primitive elements and powers 6.3.90 Theorem (Vinogradov’s Formula) [1631, Lemma 7.5.3] For a subset S ⊆F∗ q, the number Q of primitive elements in S is Q = ϕ(q −1) q −1 X d\|(q−1) µ(d) ϕ(d) X χ ord(χ)=d X x∈S χ(x), where µ and ϕ denote M¨ obius’ and Euler’s totient function, respectively, and χ is a nontrivial multiplicative character of Fq. 6.3.91 Theorem [1631, Lemma 7.5.3] Let S be a subset of F∗ q. Then the number R of s-th powers in S is R = 1 s X ord(χ)\|s X x∈S χ(x). 6.3.92 Remark [1939, Theorem 5.4] For a set S of 1 ≤N ≤p consecutive elements in Fp, the Burgess bound M+N X n=M+1 χ(n) ≪N 1−1/rp(r+1)/4r2+ε, which is nontrivial for any N ≥p1/4+ε, implies Q = ϕ(p−1) p−1 (N + O(N 1−1/rp(r+1)/4r2+ε)) and R = N s + O(N 1−1/rp(r+1)/4r2+ε). For generalizations of the Burgess bound see [375, 583, 584, 1791]. 6.3.93 Remark For Fpr with 2 ≤r < p1/2 and a deﬁning element α of Fpr, the bound X x∈Fp χ(α + x) ≤(r −1)p1/2 of guarantees the existence of primitive elements in rather small subsets of Fpr. 6.3.6.2 Distribution of Diﬃe-Hellman triples 6.3.94 Remark Let g ∈Fp be an element of order T \| (p −1). The Diﬃe-Hellman key ex-change is a way for two parties to establish a common secret key over an insecure channel. The question of studying the distribution of the Diﬃe-Hellman triples (gx, gy, gxy), x, y = 0, . . . , T −1, is motivated by the assumption that these triples cannot be distinguished from totally random triples in feasible computational time, see [487, 488]. In the series of papers [197, 368, 487, 488, 587, 1191] it has been shown that such triples are uniformly distributed via estimating double exponential sums with linear combinations of the entries in such triples. 6.3.95 Deﬁnition Let ψ be the additive canonical character of Fp. For integers a, b, c, we deﬁne the exponential sum Sa,b,c(T) = T X x,y=1 ψ (agx + bgy + cgxy) . 184 Handbook of Finite Fields 6.3.96 Remark Bounds on the sums Sa,b,c(T) were obtained by actually estimating diﬀerent sums Wa,c(T) = T X y=1 T X x=1 ψ(agx + cgxy) since \|Sa,b,c(T)\| ≤Wa,c(T). In the bound Wa,c(T) ≪T 5/3p1/4 is obtained, which is nontrivial for T > p3/4+ε. Moreover, Bourgain gave a nontrivial estimate for T > pε for any ε > 0 and Garaev improved not only the bound in obtain-ing Wa,c(T) ≪T 7/4p1/8+ε, but also the range of nontriviality T > p1/2+ε. Furthermore, Chang and Yao gave an estimate that works in a range that was not covered by any explicit bound in any other work. 6.3.6.3 Thin sets with small discrete Fourier transform 6.3.97 Deﬁnition Let ψ denote the additive canonical character of the prime ﬁeld Fp. Given a set T = {t1, t2, . . . , tn} of n elements in Fp, the sequence fT (k) = n X j=1 ψ(tjk), k = 0, 1, . . . , p −1, is the discrete Fourier transform of T. 6.3.98 Remark It is well known that the discrete Fourier transform (see Section 10.1) captures a lot of information about the “random-like” behavior of sets, that is, “good” sets have a small discrete Fourier transform for all 1 ≤k < p. In [59, 1702], several constructions of thin sets T were given, that is, sets of size \|T\| = O((log p)2+ε) with small maximum discrete Fourier transform max 1≤k≤p−1 \|fT (k)\| = O(\|T\|(log p)−ε). 6.3.99 Remark Constructions of thin sets have applications in graph theory, computer sci-ence , and in combinatorial number theory [2510, 2996]. For example, in additive number theory, for an inﬁnite set A of natural numbers, one deﬁnes the lower density of A as d(A) = lim inf x→∞ N(x) x , where N(x) = #{a ∈A \| a ≤x}. One problem of interest is to construct essential com-ponents, that are sets H with the property that d(A + H) > d(A) for all such A with 0 < d(A) < 1, and investigate how thin an essential component is. For such estimates, see [59, 2510, 2996]. 6.3.6.4 Character sums over arbitrary sets 6.3.100 Theorem [1384, Corollaries 1 and 5] Let A, B ⊆Fq and ψ and χ be a nontrivial additive and multiplicative character, respectively. Then we have X a∈A,b∈B ψ(ab) ≤(\|A\|\|B\|q)1/2 and X a∈A,b∈B χ(ab + 1) ≤(\|A\|\|B\|q)1/2. 6.3.101 Remark Based on Theorem 6.3.100, Gyarmati and S´ ark¨ ozy showed in [1385, Corollaries 1 and 2] that for all subsets A, B, C, D ⊆Fq with \|A\|\|B\|\|C\|\|D\| > q3, the equation a + b = cd, a ∈A, b ∈B, c ∈C, d ∈D Exponential and character sums 185 has a solution, and with \|A\|\|B\|\|C\|\|D\| > 100q3, the equation ab + 1 = cd, a ∈A, b ∈B, c ∈C, d ∈D has a solution. 6.3.102 Remark When q = p and ψ is the additive canonical character of Fp, Bourgain and Garaev proved in [378, Theorem 1.2] the following estimate for any subsets A, B, C of F∗ p, X a∈A,b∈B,c∈C ψ(abc) < (\|A\|\|B\|\|C\|)13/16p5/18+o(1). See Also §3.1, §3.5, §4.2 For estimating the number of polynomials with certain features. §6.1, §6.2 For basics on character sums. §6.4, §7.3 For diagonal equations and Waring’s problem. §7.1 For Kummer and Artin-Schreier curves. §9.1, §9.3 For Boolean functions and nonlinearity. §10.3, §10.4, §17.3 For correlation and related measures. §10.5 For nonlinear recurrence sequences. §11.4 For univariate factorization. §11.6, §17.2 For discrete logarithm based cryptosystems. §14.1 For Latin squares. §14.5, §14.6 For Hadamard matrices and related combinatorial structures. §15.1 For basics on coding theory. §17.2 For applications of character sums in quantum information theory. References Cited: [59, 82, 197, 240, 260, 347, 348, 368, 371, 375, 378, 380, 393, 463, 464, 487, 488, 523, 554, 583, 584, 587, 650, 776, 859, 922, 998, 1161, 1191, 1283, 1303, 1310, 1384, 1385, 1454, 1468, 1536, 1581, 1620, 1631, 1702, 1730, 1789, 1791, 1887, 1911, 1939, 1991, 2037, 2068, 2230, 2248, 2270, 2278, 2331, 2344, 2372, 2510, 2548, 2626, 2644, 2655, 2647, 2711, 2989, 2990, 2991, 2993, 2996] 6.4 Sum-product theorems and applications Moubariz Z. Garaev, Universidad Nacional Aut´ onoma de M´ exico 6.4.1 Notation 6.4.1 Remark The order q = pn of the ﬁeld Fq is assumed to be suﬃciently large. The elements of the prime residue ﬁeld Fp are occasionally associated with their concrete representatives. The cardinality of a ﬁnite set X is denoted by \|X\|. 186 Handbook of Finite Fields 6.4.2 Deﬁnition Given nonempty sets A and B the sum set A + B is deﬁned by A + B := {a + b : a ∈A, b ∈B} and the product set AB is deﬁned by AB = {ab : a ∈A, b ∈A}. The number of solutions of the equation a1b1 = a2b2, (a1, a2) ∈A × A, (b1, b2) ∈B × B is denoted by E×(A, B) and is the multiplicative energy between the sets A and B. 6.4.3 Remark There is a simple and important connection between the cardinality of the product set AB and the multiplicative energy E×(A, B): \|AB\| ≥\|A\|2\|B\|2 E×(A, B). 6.4.4 Remark All subsets in this section are assumed to be nonempty. For quantities U and V , the notations U = O(V ), U ≪V and V ≫U are all equivalent to the statement that the inequality \|U\| ≤cV holds with some absolute constant c > 0. We use the abbreviation ep(x) to denote e2πix/p. 6.4.2 The sum-product estimate and its variants 6.4.5 Remark Bourgain, Katz, and Tao , with subsequent reﬁnement by Bourgain, Glibichuk and Konyagin , proved the following theorem, which is called the sum-product estimate in prime ﬁelds. 6.4.6 Theorem For any ε > 0 there exists δ = δ(ε) > 0 such that, if A ⊂Fp and \|A\| < p1−ε, then max{\|A + A\|, \|AA\|} ≥\|A\|1+δ. 6.4.7 Remark The proof of Theorem 6.4.6 uses results from additive combinatorics and ideas of Edgar and Miller . The condition \|A\| < p1−ε is essential, because if the cardinality of \|A\| is close to p, then \|A + A\| and \|AA\| have cardinalities close to \|A\|. 6.4.8 Theorem There is a positive constant c such that for any nonempty set A ⊂Fp the following bound holds: max{\|A + A\|, \|AA\|} > c min n\|A\|2 p1/2 , p1/2\|A\|1/2o . 6.4.9 Remark Theorem 6.4.8 is meaningful for sets A of cardinality larger than p1/2. Explicit sum-product estimates for large subsets for the ﬁrst time were given by Hart, Iosevich, and Solymosi . Theorem 6.4.8 is due to Garaev . It follows that if \|A\| > p2/3, then max{\|A + A\|, \|AA\|} > cp1/2\|A\|1/2. This bound is optimal in the sense that for any positive integer N < p there exists A ⊂Fp with \|A\| = N such that max{\|A + A\|, \|AA\|} < c1p1/2\|A\|1/2, Exponential and character sums 187 where c1 is some positive constant. 6.4.10 Theorem For any subset A ⊂Fp there exists a subset A′ ⊂A with \|A′\| ≥0.1\|A\| such that E×(A′, A′)4 ≪ \|A −A\| + \|A\|3 p · \|A\|3 · \|A −A\|7 · log4(e\|A\|). 6.4.11 Remark Theorem 6.4.10 in this formulation is convenient for applications to explicit expo-nential sum estimates. It also implies that if \|A\| < p1/2 then one has max{\|A −A\|, \|AA\|} > \|A\|13/12+o(1). An explicit sum-product estimate for subsets of cardinality \|A\| < p1/2 was given for the ﬁrst time by Garaev in the form max{\|A + A\|, \|AA\|} > \|A\|15/14+o(1). This estimate subsequently was improved by several authors; the most recent one is due to Rudnev : if \|A\| < p1/2, then max{\|A + A\|, \|AA\|} > \|A\|12/11+o(1). 6.4.12 Theorem Let A, B ⊂F∗ p, L = min n \|B\|, p\|A\|−1o . Then \|A −A\|2 · \|A\|2\|B\|2 E×(A, B) ≫\|A\|3 L1/9 (log L)−1. 6.4.13 Remark Theorem 6.4.12 is an explicit version of Bourgain’s sum product estimate for subsets of incomparable sizes. The presence of the multiplicative energy is important in applications to multilinear exponential sum estimates. The proof can be found in and is based on ideas from [371, 1192, 1193]. 6.4.14 Theorem Let A, B ⊂F∗ p. Then \|8AB −8AB\| ≥1 2 min{\|A\|\|B\|, p −1}. 6.4.15 Remark Here 8AB −8AB = {P8 i=1 aibi −P16 i=9 aibi : ai ∈A, bi ∈B}. From the result of Glibichuk and Konyagin it was known that \|3AA −3AA\| ≥1 2 min{\|A\|2, p −1}. 6.4.16 Remark Theorem 6.4.14 is proved by Bourgain with subsequent application to mul-tilinear exponential sum estimates with nearly optimal entropy conditions. 6.4.17 Theorem Let A, B ⊂Fq such that \|A\| > 1 and B is not contained in any proper subﬁeld of Fq. Then max{\|A + AB\|, \|A −AB\|} ≥1 2\|A\|6/7 min{\|A\|\|B\|, q}1/7. 6.4.18 Remark Theorem 6.4.17 is proved by Bourgain and Glibichuk with subsequent appli-cations to exponential sum estimates over small subgroups of F∗ q. Because of the presence of subﬁelds an additional condition on subsets is needed. An explicit sum-product estimate in Fq for the ﬁrst time was obtained by Katz and Shen in the following form: suppose that A is a subset of Fq such that for any A′ ⊂A with \|A′\| ≥\|A\|18/19 and for any G ⊂Fq 188 Handbook of Finite Fields a subﬁeld (not necessarily proper) and for any elements c, d ∈Fq if A′ ⊂cG + d then \|A′\| ≤\|G\|1/2. Then it must be that max{\|A + A\|, \|AA\|} ≫\|A\|20/19−ε, where the implied constant depends only on a small positive quantity ε. This estimate subsequently has been improved in several works, the most recent one is due to Li and Rocher-Newton . 6.4.19 Remark There are many interesting versions of sum-product estimates, see the works of Ahmadi and Shparlinski , Shparlinski , Hart, Li, and Shen among others. The following useful result is due to Bukh and Tsimerman . 6.4.20 Theorem Let f be a polynomial over Fp of degree d ≥2. Then, for every set A ⊂Fp of size \|A\| ≤p1/2, we have \|A + A\| + \|f(A) + f(A)\| ≫\|A\|1+1/(16·6d), where the implied constant depends only on d. 6.4.21 Theorem Let A, B be ﬁnite subsets of an additive group. Assume that E ⊂A × B is such that \|E\| ≥\|A\|\|B\|/K. Then there exists a subset A′ ⊂A such that \|A′\| ≥0.1\|A\|/K and \|{a −b : (a, b) ∈E\|4 ≥\|A′ −A′\|\|A\|\|B\|2 104K5 . 6.4.22 Remark Theorem 6.4.21 is a version of the Balog-Szemer´ edi-Gowers estimate, which is an important tool in additive combinatorics. It takes its origin from the works by Balog and Sze-mer´ edi and Gowers . Balog-Szemer´ edi-Gowers type estimates are also important in multilinear exponential sum estimates over small subsets. The proof of Theorem 6.4.21 can be found in and is based on ideas and results from [193, 2738, 2781]. 6.4.3 Applications 6.4.23 Remark The sum-product estimate and its variants have found many spectacular applica-tions in various areas of mathematics. Using sum-product estimates, Bourgain, Glibichuk, and Konyagin obtained a new estimate of multilinear rational trigonometric sums, which has important applications to classical Gauss trigonometric sums. 6.4.24 Theorem For any ε > 0 there exists δ = δ(ε) > 0 and a positive integer k = k(ε) such that if X ⊂Fp with \|X\| > pε, then max (a,p)=1 X x1∈X . . . X xk∈X ep(ax1 . . . xk) < \|X\|kp−δ. 6.4.25 Corollary For any ε > 0 there exists δ = δ(ε) > 0 such that if H is a subgroup of the multiplicative group F∗ p with \|H\| > pε, then max (a,p)=1 X h∈H ep(ah) < \|H\|p−δ. 6.4.26 Theorem Let 0 < δ < 1/4 and r ≥2 be an integer. There is a δ′ > (δ/r)Cr such that if p is a suﬃciently large prime and X1, X2, . . . , Xr ⊂Fp satisfy \|Xi\| ≥pδ for all 1 ≤i ≤r and if r Y i=1 \|Xi\| > p1+δ, Exponential and character sums 189 then there is the exponential sum bound X x1∈X1 . . . X xr∈Xr ep(x1 . . . xr) < \|X1\| · · · \|Xr\|p−δ′. 6.4.27 Remark Theorem 6.4.26 is due to Bourgain . It gives a nontrivial estimate for multi-linear exponential sums under nearly optimal conditions on the sizes of the sets Xi. 6.4.28 Theorem Let 3 ≤r ≤1.44 log log p be an integer, ε be a ﬁxed positive constant. Let X1, X2, . . . , Xr be subsets of F∗ p such that \|X1\| · \|X2\| · \|X3\| · · · \|Xr\| 1/81 > p1+ε. Then X x1∈X1 . . . X xr∈Xr ep(x1 . . . xr) < \|X1\| · · · \|Xr\|p−0.45 ε/2r for all suﬃciently large primes p > p0(ε). 6.4.29 Remark Theorem 6.4.28 is an explicit version of Bourgain’s exponential sum estimate from . 6.4.30 Corollary Let H be a subgroup of F∗ p with \|H\| > e57 log p/ log log p. Then, as p →∞, we have max (a,p)=1 X x∈H ep(ax) = o(\|H\|). 6.4.31 Corollary Let g be a generator of F∗ p and let N > e57 log p/ log log p. Then, as p →∞, we have max (a,p)=1 X x≤N ep(agx) = o(N). 6.4.32 Theorem Let X, Y, Z ⊂F∗ p be such that \|X\|\|Y \| > δ p for some constant δ > 0. Then, for any ε > 0 one has the bound X x∈X X y∈Y X z∈Z ep(xyz) ≪\|X\| · \|Y \| · \|Z\|539/540+ε, where the implied constant depends only on ε and δ. 6.4.33 Theorem For any δ > 0 there exists δ′ > 0 such that if θ ∈F∗ p is of multiplicative order t and t ≥t1 > pδ, t ≥t2 > pδ, then max (a,b,p)=1 t1 X x=1 t2 X y=1 ep(aθy + bθxy) < t1t2p−δ′. 6.4.34 Remark Theorem 6.4.33 is due to Bourgain . Previously known results only applied for large values of t, t1, t2. The exponential sum that appears in Theorem 6.4.33 for the ﬁrst time was investigated by Canetti, Friedlander, and Shparlinski motivated by cryptographic applications. Subsequently Banks, Conﬂitti, Friedlander, and Shparlinski found ap-plications to estimate exponential sums with Mersenne numbers. For further details, see Bibak , Chang and the references therein. 190 Handbook of Finite Fields 6.4.35 Theorem Given a positive integer r and ε > 0, there exists δ = δ(r, ε) > 0 satisfying the following property: if f(x) = r X i=1 aixki ∈Z[x], (ai, p) = 1, where the exponents ki, 1 ≤ki < p1−ε, satisfy (ki, p −1) < p1−ε for all i, 1 ≤i ≤r, (ki −kj, p −1) < p1−ε for all i, j, 1 ≤i < j ≤r, then there is an exponential sum estimate p−1 X x=1 ep(f(x)) < p1−δ. 6.4.36 Remark Theorem 6.4.35 is due to Bourgain, its proof is based on sum-product estimates for subsets of Fp×Fp. It gives a nontrivial bound for the exponential sums under essentially op-timal conditions on the exponents ki. The exponential sum that appears in Theorem 6.4.33 was estimated by Mordell in 1932, see Cochrane, Coﬀelt, and Pinner for more details. 6.4.37 Deﬁnition Denote Tr(x) = 1 + xp + · · · + xpn−1 the trace of x ∈Fq. Let ψ(x) = ep(aTr(x)), a ∈F∗ q, be a nontrivial additive character of Fq. 6.4.38 Theorem Let 0 < δ, δ2 < 1 and r ≥2 be integer. Let A1, . . . , Ar ⊂Fq satisfy \|Ai\| > qδ for 1 ≤i ≤r, \|Ai ∩(aG + b)\| < q−δ2 for 1 ≤i ≤r, whenever a, b ∈Fq and G a proper subﬁeld, and let \|A1\|\|A2\| r Y i=3 \|Ai\|1/2 > q1+δ. Then X x1∈A1 . . . X xr∈Ar ψ(x1 . . . xr) < \|A1\| · · · \|Ar\|q−δ′, where we may take δ′ = C−r/δ2(δ/r)Cr for some positive constant C. 6.4.39 Theorem Let 0 < η ≤1 be ﬁxed and H be a multiplicative subgroup of F∗ q with \|H\| ≥q max(135/η,180000) log2 log2 q . Suppose that for any subﬁeld G we have \|H ∩G\| ≤\|H\|1−η. Then for suﬃciently large q there is an exponential sum estimate X h∈H ψ(h) < 2−0.0045(log2 q)0.1\|H\|. Exponential and character sums 191 6.4.40 Remark Using the sum-product estimate in Fp, Bourgain, Katz, and Tao obtained the following Fp - analogy of the Szemer´ edi-Trotter theorem. 6.4.41 Theorem Let P (correspondingly L) be a set of points (correspondingly a set of lines) in the plane Fp × Fp. Assume that for some α < 2 we have max{\|P\|, \|L\|} ≤M < pα. Then for some constant γ = γ(α) > 0 one has W(P, L) := # n (x, y), ℓ ∈P × L : (x, y) ∈ℓ o ≪M 3/2−γ. 6.4.42 Remark An explicit bound for W(P, L) has been obtained by Helfgott and Rudnev . When the cardinalities of P and L are large, a sharp bound for W(P, L) has been obtained by Vinh ; see also Cilleruelo . 6.4.43 Remark The following two results are due to Bourgain . They solve one of the questions of Wigderson on expander maps in two variables. Their proofs are based on the Fp - analogy of Szemer´ edi-Trotter theorem. 6.4.44 Theorem Let A, B be subsets of Fp with \|A\| = \|B\| = N < pα, where α < 1 is a positive constant. Then for some constant β = β(α) > 0 one has \|{a2 + ab : a ∈A, b ∈B}\| ≥N 1+β. 6.4.45 Theorem Let A, B be subsets of Fp with \|A\| < p1/2, \|B\| < p1/2. Then max (a,p)=1 X x∈A X y∈B ep(a(xy + x2y2)) < p1−γ, where γ > 0 is an absolute constant. 6.4.46 Remark From Theorem 6.4.45 one can deduce that there are absolute constants ρ > 0 and γ > 0 such that for any subsets A, B ⊂Fp of cardinalities \|A\| > p1/2−ρ, \|B\| > p1/2−ρ, the following bound holds: X x∈A X y∈B sign sin 2π p (xy + x2y2) < p−γ\|A\|\|B\|. Thus, at the same time Bourgain constructed 2-source extractors with entropies less than 1/2, breaking the barrier 1/2. Regarding the problem of extractors and applications of sum-product results to problems of computer science, see [200, 268, 370]. 6.4.47 Deﬁnition Let G be a ﬁnite group, A ⊂G be a set of generators of G (that is, every g ∈G can be expressed as a product of elements of A ∪A−1). The Cayley graph Γ(G, A) is the graph (V, E) with vertex set V = G and edge set E = {(ag, g) : g ∈G, a ∈A}. The diameter of a graph X = (V, E) is maxv1,v2∈V d(v1, v2), where d(v1, v2) is the length of the shortest path between v1 and v2 in X. 6.4.48 Theorem Let p be a prime. Let A be a set of generators of G = SL2(Z/pZ). Then the Cayley graph Γ(G, A) has diameter O((log p)c), where c and the implied constant are absolute. 6.4.49 Remark Theorem 6.4.48 is due to Helfgott , the sum-product estimate played a crucial role in the proof of this result. It initiated a series of very important works in the diameter problem and expansion theory of Cayley graphs of ﬁnite groups. See also [376, 377, 1464, 1466] for further references. 192 Handbook of Finite Fields 6.4.50 Remark The sum product estimates have found numerous applications in many other prob-lems. For instance, in the work of Chang the sum-product estimate and its versions have found a number of applications to multiplicative character sum estimates. Cochrane and Pinner applied the sum-product estimates to ﬁnite ﬁeld versions of the Waring problem. Ostafe and Shparlinski applied the result of Glibichuk and Rudnev to a version of the Waring problem with Dickson’s function. References Cited: [53, 193, 194, 196, 200, 268, 368, 370, 371, 373, 376, 377, 378, 379, 380, 381, 456, 488, 587, 643, 654, 656, 955, 1192, 1193, 1194, 1195, 1282, 1284, 1347, 1424, 1425, 1463, 1464, 1465, 1466, 1696, 1697, 2143, 2333, 2500, 2651, 2738, 2781, 2876] 7 Equations over ﬁnite ﬁelds 7.1 General forms ........................................ 193 Aﬃne hypersurfaces • Projective hypersurfaces • Toric hypersurfaces • Artin-Schreier hypersurfaces • Kummer hypersurfaces • p-Adic estimates 7.2 Quadratic forms ..................................... 201 Basic deﬁnitions • Quadratic forms over ﬁnite ﬁelds • Trace forms • Applications 7.3 Diagonal equations .................................. 206 Preliminaries • Solutions of diagonal equations • Generalizations of diagonal equations • Waring’s problem in ﬁnite ﬁelds 7.1 General forms Daqing Wan, University of California Irvine 7.1.1 Remark There are lots of results on equations over ﬁnite ﬁelds. In this section, we give a collection of sample results and examples focusing on hypersurfaces. Additional results can be found in [1708, 2548, 2902] and in the references of this section. 7.1.1 Aﬃne hypersurfaces 7.1.2 Deﬁnition For a polynomial f(x1, . . . , xn) ∈Fq[x1, . . . , xn], let Af denote the aﬃne hyper-surface in the aﬃne n-space An deﬁned by the equation f = 0. Then, Af(Fq) denotes the set of Fq-rational points (x1, . . . , xn) ∈Fn q such that f(x1, . . . , xn) = 0. The notation #Af(Fq) denotes the cardinality of the set Af(Fq), namely, the number of Fq-rational points on the aﬃne hypersurface Af. 7.1.3 Remark More generally, for a system of polynomials f1, . . . , fm ∈Fq[x1, . . . , xn], let Af1,··· ,fm(Fq) denote the set of Fq-rational points (x1, . . . , xn) ∈Fn q such that f1(x1, . . . , xn) = · · · = fm(x1, . . . , xn) = 0. It is straightforward to check the simple relation #Af1,...,fm(Fq) = 1 qm −qm−1 (#Ay1f1+···+ymfm(Fq) −qm+n−1), 193 194 Handbook of Finite Fields where y1, . . . , ym are variables. Thus, for simplicity and without too much loss of generality, we shall focus on the hypersurface case. 7.1.4 Theorem For a non-zero polynomial f of (total) degree d in Fq[x1, . . . , xn], we have 1. 0 ≤#Af(Fq) ≤dqn−1. 2. If #Af(Fq) ≥1, then #Af(Fq) ≥qn−d. 3. If f is homogenous of degree d, then qn−d ≤#Af(Fq) ≤d(qn−1 −1) + 1. 7.1.5 Deﬁnition Let Ωd,n(q) denote the Fq-vector space of polynomials in Fq[x1, ..., xn] with degree at most d. 7.1.6 Theorem For positive integers d and n, we have 1 \|Ωd,n(q)\| X f∈Ωd,n(q) #Af(Fq) = qn−1, 1 \|Ωd,n(q)\| X f∈Ωd,n(q) (#Af(Fq) −qn−1)2 = qn−1 −qn−2. 7.1.7 Remark The above average or probabilistic result suggests that for most polynomials f ∈Ωd,n(q), one should expect that #Af(Fq) is approximately qn−1 for large q. This is indeed the case, as the next few theorems show. The ﬁrst one is an eﬀective version of the Lang-Weil theorem. 7.1.8 Deﬁnition A polynomial f ∈F[x1, . . . , xn] over a ﬁeld F is absolutely irreducible if it is irreducible over an algebraic closure of F. 7.1.9 Theorem [473, 1849] If f ∈Fq[x1, . . . , xn] is an absolutely irreducible polynomial over Fq of degree d > 0, then \|#Af(Fq) −qn−1\| ≤(d −1)(d −2)qn−3 2 + 5d13/3qn−2. If in addition, q > 15d13/3, one has \|#Af(Fq) −qn−1\| ≤(d −1)(d −2)qn−3 2 + (5d2 + d + 1)qn−2. 7.1.10 Deﬁnition A polynomial f ∈Fq[x1, . . . , xn] (or the aﬃne hypersurface it deﬁnes) is smooth if the following system of equations ∂f ∂x1 = · · · = ∂f ∂xn = f = 0 has no solutions in the algebraic closure of Fq. A homogenous polynomial f (or the projective hypersurface it deﬁnes) is smooth if the above system of equations has no solutions other than possibly the trivial one (0, . . . , 0). 7.1.11 Theorem Let f ∈Fq[x1, . . . , xn] be a smooth non-homogenous polynomial of degree d such that its homogenous leading form is also smooth. Then, \|#Af(Fq) −qn−1\| ≤(d −1)nq(n−1)/2. 7.1.12 Remark This result follows from Deligne’s well known theorem for projective hypersurfaces in the next subsection. One simply applies Deligne’s theorem twice, to the projective closure of Af (which is itself smooth) and to the inﬁnite part of Af. Equations over ﬁnite ﬁelds 195 7.1.13 Deﬁnition The singular locus Sing(f) of a polynomial f ∈Fq[x1, . . . , xn] (or the aﬃne hypersurface it deﬁnes) is the aﬃne algebraic set in An deﬁned by the following system of equations ∂f ∂x1 = · · · = ∂f ∂xn = f = 0. Similarly, the singular locus Sing(f) of a homogeneous polynomial f (or the projective hypersurface it deﬁnes) is the projective algebraic set in the projective space Pn−1 deﬁned by the same system of equations. 7.1.14 Theorem Let f ∈Fq[x1, . . . , xn] be a polynomial of degree d > 0 such that the singular loci of both f and its homogenous leading form have dimension at most s for some integer s ≥−1. Then, \|#Af(Fq) −qn−1\| ≤Cd,nq(n+s)/2, where Cd,n is an explicit constant depending only on d and n. 7.1.15 Remark This result is the aﬃne version of the Hooley-Katz theorem which uniﬁes the previous two theorems, at least in a qualitative way. If f is absolutely irreducible, then we can take s ≤n −2 and this recovers the Lang-Weil theorem. If f and its leading form are both smooth, then we can take s = −1 and this recovers Deligne’s theorem. 7.1.2 Projective hypersurfaces 7.1.16 Deﬁnition For a homogeneous polynomial f ∈Fq[x0, x1, . . . , xn] of degree d, let Pf denote the projective hypersurface in Pn deﬁned by f = 0. Thus, Pf(Fq) denotes the set of projective solutions (x0, . . . , xn) ∈Fn+1 q of the equation f(x0, . . . , xn) = 0, where two solutions are identiﬁed if they diﬀer by a non-zero scalar multiple. 7.1.17 Theorem For a homogeneous polynomial f ∈Fq[x0, x1, . . . , xn] of degree d > 0, we have qn+1−d −1 q −1 ≤#Pf(Fq) ≤dqn −1 q −1 . 7.1.18 Remark This is simply a consequence of the corresponding aﬃne theorem in the previous subsection. 7.1.19 Theorem If f ∈Fq[x0, x1, . . . , xn] is a smooth homogeneous polynomial of degree d > 0, then #Pf(Fq) −qn −1 q −1 ≤d −1 d ((d −1)n −(−1)n)q(n−1)/2. 7.1.20 Theorem [1270, 1533] If f ∈Fq[x0, x1, . . . , xn] is a homogeneous polynomial of degree d > 0 such that the projective singular locus Sing(f) is of dimension at most s for some integer s ≥−1, then #Pf(Fq) −qn −1 q −1 ≤Cd,nq(n+s)/2, where Cd,n is an explicit constant depending only on d and n. 7.1.21 Remark Similar results hold for more general complete intersections, see . 196 Handbook of Finite Fields 7.1.3 Toric hypersurfaces 7.1.22 Deﬁnition Let f ∈Fq[x±1 1 , . . . , x±1 n ]. The Newton!polytope ∆(f) of f is the convex closure in Rn of the exponents of the non-zero monomials in f. We shall assume that ∆(f) = ∆ is a ﬁxed n-dimensional integral polytope in Rn. 7.1.23 Deﬁnition A Laurent polynomial f ∈Fq[x±1 1 , . . . , x±1 n ] is ∆-regular if ∆(f) = ∆and for every face δ (of any dimension) of ∆, the system f δ = x1 ∂f δ ∂x1 = · · · = xn ∂f δ ∂xn = 0 has no solutions in (¯ F∗ q)n, where f δ = X u∈δ∩Zn auxu is the restriction of f to the face δ. 7.1.24 Deﬁnition For a Laurent polynomial f ∈Fq[x±1 1 , . . . , x±1 n ], let Tf denote the toric aﬃne hypersurface in the n-torus Gn m deﬁned by f = 0. Thus, Tf(Fq) is the set of n-tuples (x1, . . . , xn) ∈(F∗ q)n such that f(x1, . . . , xn) = 0. 7.1.25 Theorem [23, 812, 2902] Assume that f ∈Fq[x±1 1 , . . . , x±1 n ] is ∆-regular. Then, #Tf(Fq) −(q −1)n −(−1)n q ≤(n!Vol(∆) −1)q(n−1)/2. 7.1.26 Example Let f = x1 + · · · + xn + 1 x1···xn −λ, where λ ∈Fq and n ≥2. Assume that the characteristic p does not divide n+ 1. If (λ/(n+ 1))n+1 ̸= 1, then f is ∆(f)-regular and thus #Tf(Fq) −(q −1)n −(−1)n q ≤nq(n−1)/2. If (λ/(n + 1))n+1 = 1, then f is singular and we have the improved bound #Tf(Fq) −(q −1)n −(−1)n q ≤(n −1)q(n−1)/2. If, in addition, n is even, then we have the even better bound #Tf(Fq) −(q −1)n −(−1)n q ≤(n −2)q(n−1)/2 + q(n−2)/2. These results are used to obtain improved bounds for the number of elements with given trace and norm, see . The toric hypersurface Tf in this example is a toric Calabi-Yau hypersurface. It is the most important example in arithmetic mirror symmetry, see . 7.1.27 Remark For results on more general toric complete intersections, see . Equations over ﬁnite ﬁelds 197 7.1.4 Artin-Schreier hypersurfaces 7.1.28 Deﬁnition Let f ∈Fq[x1, . . . , xn] be a polynomial of degree d > 0. Let ASf denote the Artin-Schreier hypersurface in An+1 deﬁned by the equation yq −y = f(x1, . . . , xn). For any positive integer k, let #ASf(Fqk) denote the number of Fqk-rational points on ASf. 7.1.29 Remark We note that for k = 1, one has the relation #ASf(Fq) = q#Af(Fq), where Af is the aﬃne hypersurface in An deﬁned by f = 0. 7.1.30 Theorem Assume that d is not divisible by p and the leading form of f is smooth. Then, \|#ASf(Fqk) −qkn\| ≤(d −1)nq(kn+2)/2. In particular, taking k = 1, we deduce the estimate \|#Af(Fq) −qn−1\| ≤(d −1)nqn/2. 7.1.31 Remark This is simply a consequence of Deligne’s estimate for exponential sums, see Section 6.2 for more details. It can be improved and generalized in various ways as the next few theorems show. The next theorem is a consequence of Katz’s estimate for singular exponential sums. 7.1.32 Theorem Assume that d is not divisible by p and the singular locus of the leading form of f has dimension at most s for some integer s ≥−1. Then, \|#ASf(Fqk) −qkn\| ≤Cd,nq(kn+s+3)/2, where Cd,n is an explicit constant depending only on d and n. In particular, taking k = 1, we deduce the estimate \|#Af(Fq) −qn−1\| ≤Cd,nq(n+s+1)/2. 7.1.33 Remark In the case s = −1, that is, the leading form of f is smooth, the above result reduces to Deligne’s estimate. In this case, the subscheme deﬁned by the Jacobian ideal < ∂f/∂x1, . . . , ∂f/∂xn > has at most a ﬁnite number of points and the above Deligne estimate can be improved further in some cases. 7.1.34 Theorem Assume that d is not divisible by p > 2 and the leading form of f is smooth. Assume further that the Jacobian subscheme in An deﬁned by the ideal < ∂f/∂x1, . . . , ∂f/∂xn > is a zero-dimensional smooth variety and the image of its ¯ Fq-points under f are distinct (a Morse condition). If nk is even, suppose additionally that the hypersurface deﬁned by f(x11, . . . , x1n) + · · · + f(xk1, . . . , xkn) = 0 is also smooth in Akn. Then, \|#ASf(Fqk) −qkn\| ≤Cd,k,nq(kn+1)/2, for some explicit constant Cd,k,n. 198 Handbook of Finite Fields 7.1.35 Corollary Assume d is not divisible by p > 2, the leading form of f is smooth, and the Ja-cobian subscheme in An deﬁned by the ideal < ∂f/∂x1, . . . , ∂f/∂xn > is a zero-dimensional smooth variety and the images of its ¯ Fq-points under f are distinct. Assume that n is odd, then \|#Af(Fq) −qn−1\| ≤Cd,1,nq(n−1)/2. 7.1.36 Deﬁnition Let f(x, y) ∈Fq[x1, . . . , xn, y1, . . . , yn′] be a polynomial with two sets of vari-ables, where n, n′ ≥1. For a positive integer k, let Nk(f) denote the number of solutions of the equation xp 0 −x0 = f(x1, . . . , xn, y1, . . . , yn′) such that (x0, x1, . . . , xn) ∈Fn+1 qk and (y1, . . . , yn′) ∈Fn′ q . 7.1.37 Remark We note that the two sets of coordinates in the previous deﬁnition run over diﬀerent extension ﬁelds of Fq. 7.1.38 Deﬁnition Let f(x, y) be a polynomial as in the previous deﬁnition. For a positive integer k, we deﬁne the k-th ﬁbred sum of f along y to be the new polynomial ⊕k yf := f(x11, . . . , x1n, y1, . . . , yn′) + · · · + f(xk1, . . . , xkn, y1, . . . , yn′). 7.1.39 Theorem Given f of degree d > 0 as above. Let fd be the homogeneous leading form of f. Assume that the k-th ﬁbred sum ⊕k yfd is smooth in Pkn+n′−1 and assume that d is not divisible by p. Then, we have the following two estimates \|Nk(f) −qkn+n′\| ≤(p −1)(d −1)kn+n′q(kn+n′)/2, \|Nk(f) −qkn+n′\| ≤c(p, n, n′)k3(d+1)n−1q(kn+n′)/2, where the constant c(p, n, n′), depending only on p, n, n′, is not known to be eﬀective if n′ ≥2. 7.1.40 Remark In the case k = 1, the ﬁrst estimate reduces to Deligne’s theorem as above. More generally, for a degree d polynomial f ∈Fq[x1, . . . , xn] and positive integers k1, . . . , kn, let Nk1,...,kn(f) denote the number of solutions of the equation f(x1, . . . , xn) = 0 with xi ∈Fqki for all 1 ≤i ≤n. Under suitable nice conditions, one expects the estimate of the type \|Nk1,...,kn(f) −qk1+···+kn−k\| ≤Cd,n,kq(k1+···+kn)/2, where k is the least common multiple of the integers ki. The above theorem is one example of this kind. For two additional examples, see [1705, 2471]. 7.1.5 Kummer hypersurfaces 7.1.41 Deﬁnition Let f ∈Fq[x1, . . . , xn] be a polynomial of degree d > 0. For a positive integer m not divisible by p, let Kf,m denote the Kummer hypersurface in An+1 deﬁned by ym = f(x1, . . . , xn). Equations over ﬁnite ﬁelds 199 7.1.42 Theorem [1707, 2468] Assume that f is smooth in An and its leading form fd is also smooth in Pn−1. Then, \|#Kf,m(Fq) −qn\| ≤(m −1)(d −1)nqn/2. 7.1.43 Remark Since #Kf,m(Fq) = #Kf,(m,q−1)(Fq), we may assume that m divides q −1. Write m = (q −1)/e, then for ﬁxed e, the above bound reduces to \|#Kf,m(Fq) −qn\| ≤1 e(d −1)nq(n+2)/2. Similar to the Artin-Schreier hypersurface case, we expect that for ﬁxed e, under suitable hypotheses, an improved estimate of the form \|#Kf,m(Fq) −qn\| ≤c(n, k, d)q(n+1)/2, see Rojas-Leon for new results in this direction. 7.1.6 p-Adic estimates 7.1.44 Remark In this subsection, we review several results on p-divisibility of the solution number for an aﬃne algebraic set, where p is the characteristic of the ﬁnite ﬁeld Fq. We begin with the well known Chevalley-Warning theorem. 7.1.45 Theorem Let f1, . . . fm ∈Fq[x1, . . . , xn] be polynomials of degrees d1, . . . , dm respec-tively. Assume that n > d1 + · · · + dm. Then, #Af1,...,fm(Fq) ≡0 (mod p). 7.1.46 Theorem Let f1, . . . fm ∈Fq[x1, . . . , xn] be polynomials of degrees d1, . . . , dm respectively. Let µ be the smallest integer such that µ ≥n −(d1 + · · · + dm) maxi di . Then, #Af1,...,fm(Fq) ≡0 (mod qµ). 7.1.47 Remark This theorem is due to Ax in the case m = 1 and to Katz for gen-eral m. Ax’s proof is elementary and based on the Stickelberger relation for Gauss sums (for Gauss sums, see Section 6.1). Katz’s proof uses more advanced methods from Dwork’s p-adic theory. In , it was noted that Katz’s result can be proved by Ax’s more ele-mentary method. Later, a short elementary reduction of Katz’s theorem for general m to Ax’s theorem for m = 1 was given in . This is consistent with Remark 7.1.3 that a system of equations can often be reduced to the one equation case. Additional simpler proofs to more general results are given in . If one takes into account the actual terms (the polytope) of the polynomials f1, . . . , fm, the Ax-Katz theorem can be generalized or improved in certain cases, see Adolphson-Sperber for p-divisibility of exponential sums. We next describe its consequence for equations over ﬁnite ﬁelds. 7.1.48 Theorem Let ∆be an integral convex polytope in Rm+n which contains all the exponents of the monomials in the polynomial y1f1(x1, . . . , xn)+· · ·+ymfm(x1, . . . , xn) with coeﬃcients in Fq. Let ω be the smallest positive integer such that the dilation ω∆contains a lattice point in Zm+n with all coordinates positive. Then, #Af1,...,fm(Fq) ≡0 (mod qω−m). 200 Handbook of Finite Fields 7.1.49 Remark For non-prime ﬁelds (i.e., q is not a prime), the Ax-Katz theorem can be improved in some cases by considering the p-weights of the exponents of the polynomials fi and the Weil descent, see Moreno-Moreno . Let q = pr, and let {α1, . . . , αr} be an Fp-basis of Fq. Then, any element xi ∈Fq can be written uniquely as xi = r X j=1 xijαj, xij ∈Fp. Let e = e0 + e1p + · · · + er−1pr−1 be the p-digit expansion of an integer e ∈[0, q −1]. Then, one has the relation xe i = r−1 Y j=0 (xi1αpj 1 + · · · + xirαpj r )ej. In this way, one ﬁnds a system of mr polynomials gij ∈Fp[x11, . . . , xmr] such that #Af1,...,fm(Fq) = #Ag11,...,gmr(Fp). One can then apply the Ax-Katz theorem to the right side over the prime ﬁeld Fp. 7.1.50 Remark For a homogeneous polynomial f ∈Fq[x0, . . . , xn] of degree d > 0, the Ax-Katz theorem implies that the projective hypersurface Pf in Pn satisﬁes the congruence #Pf(Fq) ≡#Pn−1(Fq) (mod qu), where µ is the smallest positive integer greater than or equal to (n + 1 −d)/d. This gives a non-trivial congruence only in the case n + 1 > d (Fano varieties). In the case n + 1 = d (Calabi-Yau variety) or n + 1 < d (varieties of general type), one cannot expect such general p-divisibility results. However, the following example suggests that one may still expect some congruences for some pair of varieties. For λ ∈Fq, let Xλ denote the Dwork family of Calabi-Yau hypersurfaces in Pn deﬁned by the equation xn+1 0 + · · · + xn+1 n + λx0x1 · · · xn = 0. Let Yλ be the projective closure in the projective toric variety P∆of the toric aﬃne hyper-surface deﬁned by x1 + · · · + xn + 1 x1 · · · xn + λ = 0, where ∆is the simplex in Rn with vertices {(1, 0, . . . , 0), . . . , (0, . . . , 0, 1), (−1, −1, . . . , −1)}. Then, we have the mirror congruence #Xλ(Fq) ≡#Yλ(Fq) (mod q). In this example, the mirror variety Yλ is a quotient of Xλ by a ﬁnite group G, see [254, 1141] for extensions of such congruence to a pair of more general quotient varieties. 7.1.51 Remark From a zeta function point of view, see Section 12.7 for more on zeta functions, the p-adic estimate in this subsection corresponds to an estimate for the ﬁrst non-trivial slope. The study of all slopes for the zeta function is signiﬁcantly deeper, and the reader is refered to Section 12.8. Equations over ﬁnite ﬁelds 201 See Also §6.1 For Gauss, Jacobi, and Kloosterman sums. §6.2 For other related results to character sums. §7.3 For diagonal equations. §11.6 For discrete logarithms. §12.7 For zeta functions. §12.8 For p-adic estimates of zeta functions. References Cited: [21, 23, 27, 150, 254, 473, 795, 812, 1140, 1141, 1270, 1533, 1540, 1698, 1704, 1705, 1707, 1708, 1849, 2125, 2151, 2468, 2470, 2471, 2472, 2548, 2898, 2900, 2902, 2907, 2914] 7.2 Quadratic forms Robert Fitzgerald, Southern Illinois University 7.2.1 Basic deﬁnitions 7.2.1 Deﬁnition A quadratic form f over a ﬁeld F is a homogeneous polynomial over F of degree two: f(X) = X i≤j aijxixj aij ∈F, where X is the (column) vector of the variables xi. Let CMf be the upper triangular matrix of coeﬃcients. Two quadratic forms, f and g are equivalent, denoted f ≃g, if there is an invertible matrix P over F such that f(PX) = g(X). 7.2.2 Remark The theory is diﬀerent in the characteristic 2 and odd characteristic cases. 7.2.3 Deﬁnition A quadratic space is a pair (Q, V ) where V is a ﬁnite dimensional vector space over F and Q : V →F satisﬁes 1. Q(λv) = λ2v, for all λ ∈F and v ∈V , and 2. for char(F) = 2, we require BQ(v, w) = Q(v + w) + Q(v) + Q(w) to be a sym-metric bilinear form; for char(F) ̸= 2, we require that BQ(v, w) = 1 2(Q(v +w)− Q(v) −Q(w)) to be a symmetric bilinear form. 7.2.4 Remark Each choice of a basis {e1, . . . , en} of V yields a quadratic form f(X) = Q(P i xiei); a diﬀerent choice of basis yields an equivalent form. 7.2.5 Remark Suppose char(F) = 2. The associated matrix of f is CMf. Then f(X) = XT CMfX. There are many matrices M with f(X) = XT MX but CMf is the only up-per triangular one. The matrix for the associated symmetric bilinear form bf(X, Y ) = 202 Handbook of Finite Fields BQ(P i xiei, P j yjej) is CMf + CM T f . Inequivalent f may have the same associated sym-metric bilinear form. 7.2.6 Deﬁnition The radical of a quadratic space is: rad(Q) = {v ∈V : BQ(v, w) = 0 for all w ∈V }. If rad(Q) = 0, Q is non-degenerate. 7.2.7 Remark Continue to assume char(F) = 2. Suppose Q is non-degenerate. Then there is a symplectic basis e1, f1, e2, f2, . . . , en, fn with associated matrix          α1 1 0 0 . . . 0 0 0 β1 1 0 . . . 0 0 0 0 α2 1 . . . 0 0 . . . . . . . . . . . . . . . . . . . . . 0 0 0 0 . . . αn 1 0 0 0 0 . . . 0 βn          . Let P(F) be the additive subgroup {a2 + a : a ∈F}. The Arf invariant ∆(Q) is P αiβi ∈ F/P(F). ∆(Q) does not depend on the choice of the symplectic basis. If Q is degenerate then write V = rad(Q) ⊕W. Then Q\|W is non-degenerate and we deﬁne ∆(Q) = ∆(Q\|W ). Now suppose char(F) ̸= 2. The associated matrix of f is Mf = 1 2(CMf + CM T f ). Then f(X) = XT MfX. There are many matrices M with f(X) = XT MX but Mf is the only symmetric one. The matrix for the associated symmetric bilinear form bf is also Mf. The function BQ determines Q (and bf determines f) via Q(v) = BQ(v, v). Suppose Q is non-degenerate. Then there is an orthogonal basis with the associated matrix diagonal. Let F ∗2 be the multiplicative subgroup {a2 : a ∈F}. The determinant of f is det Mf ∈F ∗/F ∗2 (equivalently, the product of the entries in the diagonalization) and it does not depend on the choice of basis. Again, if Q is degenerate then write V = rad(Q) ⊕W and set det(Q) = det(Q\|W ). 7.2.2 Quadratic forms over ﬁnite ﬁelds 7.2.8 Remark Details on the results of this section can be found in [1939, Section 6.2]. 7.2.9 Theorem Let q be even and let (Q, V ) be a quadratic space over Fq. Let n = dim V and r = dim rad(Q). Then: 1. n −r = 2s is even. 2. \|Fq/P(Fq)\| = 2. 3. Fix 0 ̸= d ∈Fq/P(Fq). Then V has a symplectic basis such that the resulting quadratic form is one of the following: (a) E1 : x1x2 + x3x4 + · · · + x2s−1x2s, (b) E2 : x2 1 + x1x2 + dx2 2 + x3x4 + · · · + x2s−1x2s, (c) E3 : x2 0 + x1x2 + x3x4 + · · · + x2s−1x2s. 4. E1 occurs if and only if Q(rad(Q)) = 0 and ∆(Q) = 0; E2 occurs if and only if Q(rad(Q)) = 0 and ∆(Q) = d ̸= 0; E3 occurs if and only if Q(rad(Q)) ̸= 0. Equations over ﬁnite ﬁelds 203 7.2.10 Deﬁnition The rank of Q is the minimal number of variables in a quadratic form induced from Q. 7.2.11 Remark The rank of Q is dim V −dim rad(Q) except for the case E3 when it is dim V − dim rad(Q) + 1. 7.2.12 Theorem Let q be odd and let (Q, V ) be a quadratic space over Fq. Let n = dim V and r = dim rad(Q). Write n −r = 2s or 2s + 1. Then 1. \|F∗ q/F∗2 q \| = 2. 2. Fix 1 ̸= d ∈F∗ q/F∗2 q . Then V has a basis such the resulting quadratic form is one of the following: (a) O1 : x1x2 + x3x4 + · · · + x2s−1x2s, (b) O2 : x2 1 −dx2 2 + x3x4 + · · · + x2s−1x2s, (c) O3 : x2 0 + x1x2 + x3x4 + · · · + x2s−1x2s, (d) O4 : dx2 0 + x1x2 + x3x4 + · · · + x2s−1x2s. 3. In cases O1, O3, det Q = (−1)s, and in cases O2, O4, det Q = (−1)sd. 7.2.13 Remark When char(F) ̸= 2: 1. we always have Q(rad(Q)) = 0; 2. the rank of Q is always dim V −dim rad(Q); and 3. xy ≃x2 −y2 so that each form in Theorem 7.2.12 can be written as a diagonal form. 7.2.14 Deﬁnition For a quadratic space (Q, V ) over a ﬁnite ﬁeld, let N(Q = 0) denote the number of v ∈V such that Q(v) = 0. 7.2.15 Theorem Let (Q, V ) be a quadratic space over Fq, where q is even. Let n = dim V and r = dim rad(Q). Then N(Q = 0) = 1 q h qn + (q −1)Λ(Q) p qn+r i , where Λ(Q) =      +1, if Q is type E1, −1, if Q is type E2, 0, if Q is type E3. 7.2.16 Theorem Let (Q, V ) be a quadratic space over Fq, where q is odd. Let n = dim V and r = dim rad(Q). Then N(Q = 0) = 1 q h qn + (q −1)Λ(Q) p qn+r i , where Λ(Q) =      +1, if Q is type O1, −1, if Q is type O2, 0, if Q is type O3 or O4. 7.2.17 Remark With a little more work [1745, 1746], one can give the number of solutions to Q(x) = c, for any scalar c, and even to Q(x) + F(x) = c, where F(x) is an arbitrary linear function. 204 Handbook of Finite Fields 7.2.3 Trace forms 7.2.18 Remark In almost all applications the quadratic spaces arise as follows: Let F = Fq, K = Fqn and let L(x) = Pm i=0 αixqi be a linearized polynomial over K, see Deﬁnition 2.1.103. 7.2.19 Deﬁnition A trace form over F = Fq is the quadratic space (QK L , K) where K = Fqn and QK L : K →F is given by QK L (x) = TrK/F (x · L(x)). 7.2.20 Remark If (Q, V ) is a quadratic space with dim V = n then, replacing V by K, Q is a QK L for some linearized L(x). In even characteristic, for m = ⌊n/2⌋, L is uniquely determined except that if n = 2m then αm is only determined modulo Fqm . We set L∗(x) = m X i=1 αi(xqm+i −xqm−i). 7.2.21 Proposition For F, K and L as above: 1. \|rad(QK L )\| is equal to the number of roots of L∗(x) in K. 2. If all αi ∈F then dim rad(QK L ) = deg(L∗(x)dn, xk −1). Here L∗ dn is the polynomial associated to L∗, namely L∗(x)dn = m X i=1 αi(xm+i −xm−i). 7.2.22 Remark The invariants dim rad(Q) = d and Λ(Q) are completely determined only in the case of q even and L having one term. Let v2(k) demote the highest power of 2 dividing k. 7.2.23 Theorem Let F = Fq with q even. Let Q(x) = TrK/F (γx·xqa), where K = Fqn and γ ∈K. Set d = (n, a). 1. If v2(n) ≤v2(a) then dim rad(Q) = d and Λ(Q) = 0. 2. If v2(n) = v2(a) + 1 then (dim rad(Q), Λ(Q)) = ( (2d, +1) if γ is a (qa + 1)-th power in K, (0, −1) if γ is not a (qa + 1)-th power in K. 3. If v2(n) > v2(a) + 1 then (dim rad(Q), Λ(Q)) = ( (2d, −1) if γ is a (qa + 1)-th power in K, (0, +1) if γ is not a (qa + 1)-th power in K. 7.2.24 Theorem Let F = Fq with q odd. Let Q(x) = TrK/F (γx · xqa), where K = Fqn and γ ∈K. Set d = (n, a). Let ω be a primitive element of K and write γ = ωg for some 0 ≤g < qn −1. 1. If v2(n) ≤v2(a) then dim rad(Q) = 0. 2. If v2(n) = v2(a) + 1 then (dim rad(Q), Λ(Q)) = ( (2d, +1), if g ≡1 2(qd + 1) (mod qd + 1), (0, −1), if g ̸≡1 2(qd + 1) (mod qd + 1). Equations over ﬁnite ﬁelds 205 3. If v2(n) > v2(a) + 1 then (dim rad(Q), Λ(Q)) = ( (2d, −1), if g ≡0 (mod qd + 1), (0, +1), if g ̸≡0 (mod qd + 1). 7.2.25 Problem Find the value of Λ(Q) in Case 1 above. 7.2.26 Remark In the following we use the Jacobi symbol ( a n). 7.2.27 Theorem Let F = F2, K = F2n and Q(x) = TrK/F (x(x2a + x2b)) with a < b. Set d = (b −a, b + a) and M = max{v2(b −a), v2(b + a)}. Then dim rad(Q) = ( (b −a, n) + (b + a, n) −(d, n) if v2(n) ≤M, (b −a, n) + (b + a, n) if v2(n) > M. We have Λ(Q) = 0 if and only if v2(b −a) = v2(b + a) = v2(n) −1. For an intermediate ﬁeld E = F2m write Λ(m) for the invariant of TrE/F (x(x2a + x2b)). Let k = v2(n). 1. If n is odd then Λ(n) = Q( 2 p), where the product is over odd prime divisors p of n with vp(n) + min{vp(n), max{vp(b −a), vp(b + a)}} odd. 2. If n is even and b ± a is odd then Λ(n) = Q( 2 p)Λ(2k), where the product is over odd prime divisors p of n with min{vp(n), vp(b−a)}+min{vp(n), vp(b+a)} odd. 3. If n is even and b ± a is even then Λ(n) = Λ(2k). 4. For n = 2k: (a) If k ≤M then Λ(n) = +1. (b) If k = 1 + M and v2(b −a) ̸= v2(b + a) then Λ(n) = −1. (c) If k = 1 + M and v2(b −a) = v2(b + a) then Λ(n) = 0. (d) If k ≥2 + M and v2(b −a) ̸= v2(b + a) then Λ(n) = −1. (e) If k ≥2 + M and v2(b −a) = v2(b + a) then i. If k = 2 and one of a, b is odd and the other is equivalent to 2 (mod 4) then Λ(n) = +1. ii. If k ≥3 and one of a, b is odd and the other is divisible by 4 then Λ(n) = +1. iii. Otherwise, Λ(n) = −1. 7.2.4 Applications 7.2.28 Remark The ﬁrst appearance of trace forms seems to be Welch’s Theorem which is Theorem 7.2.23 for γ = 1 and 2a dividing n. It was used to compute weight enumerators of double-error correcting BCH codes. Other particular trace forms have been used to compute weight enumerators of second order Reed-Muller codes and minimal codes . Weights of irreducible codes have been found via counting the number of polynomials L with ﬁxed invariants by Feng and Luo [1056, 1981]. 7.2.29 Remark Quadratic forms have been used to construct Artin-Schreier curves yq +y = xL(x) with many rational points [1072, 1075, 2844]. Quadratic forms over F2, partitioned by their bilinear forms, were used to construct systems of linked symmetric designs in . Maximal rank quadratic forms give rise to the ﬁrst examples of bent functions which have cryptographic importance, see Section 9.3. Piecewise functions, with each piece a trace form, 206 Handbook of Finite Fields yield other bent functions and a presentation of the Kerdock code in . Families of trace forms have been used to construct Gold-like sequences [1077, 1731]. Correlations of maximal and other sequences have been computed with trace forms [1475, 1736, 1745, 1746]. See Also §9.3, §10.3, §12.6, §15.2 For more information on applications of quadratic forms. References Cited: [231, 483, 536, 1056, 1072, 1073, 1074, 1075, 1077, 1475, 1731, 1736, 1745, 1746, 1747, 1939, 1981, 1991, 2844, 3000] 7.3 Diagonal equations Francis Castro and Ivelisse Rubio, University of Puerto Rico 7.3.1 Preliminaries We present a summary of results on diagonal equations. The selection gives an overview of the area as well as some of its recent developments. General references for diagonal equations are: Chapter 10 in , Chapter 8 in , Chapters 3-6 in , Chapter 6, Section 3 in , and Chapter 6 in . 7.3.1 Deﬁnition A diagonal equation over Fq is an equation of the type c1xk1 1 + · · · + csxks s = b (7.3.1) for any positive integers k1, . . . , ks; c1, . . . , cs ∈F∗ q, and b ∈Fq. We denote by Nb the number of solutions in Fs q of (7.3.1). A deformed diagonal equation is a diagonal equation with b = g (x1, . . . , xs) ∈Fq [x1, . . . , xs]. 7.3.2 Remark The number of solutions Nb can be expressed in terms of Jacobi and Gauss sums. The relation between Jacobi and Gauss sums and other results on these sums are included in Section 6.3. 7.3.3 Theorem [1939, Theorem 6.33] The number N0 of solutions of (7.3.1) for b = 0 is N0 = qs−1 + X (j1,...,js)∈T χj1 1 (c1) · · · χjs s (cs) J0 χj1 1 , . . . , χjs s , (7.3.2) where T is the set of all (j1, . . . , js) ∈Zs such that 1 ≤ji ≤di −1 for 1 ≤i ≤s, χj1 1 . . . χjs s is trivial, χi is a multiplicative character of order di = gcd(ki, q −1), and J0 is a Jacobi sum. 7.3.4 Theorem [1939, Theorem 6.34] The number Nb of solutions of (7.3.1) for b ∈F∗ q is Nb = qs−1 + d1−1 X j1=1 · · · ds−1 X js=1 χj1 1 bc−1 1 · · · χjs s bc−1 s J χj1 1 , . . . , χjs s , (7.3.3) Equations over ﬁnite ﬁelds 207 where χi is a multiplicative character of order di = gcd(ki, q −1), and J is a Jacobi sum. 7.3.5 Remark Results on N0 and Nb can be related to the s-tuples (j1, . . . , js) ∈Zs, ji ≥1, such that j1 k1 + · · · + js ks (7.3.4) is an integer. 7.3.6 Deﬁnition Let I(k1, . . . , ks) be the number of s-tuples (j1, . . . , js) ∈Zs such that 1 ≤ji ≤ ki −1 and expression (7.3.4) is an integer. 7.3.7 Remark Note that I(k1, . . . , ks) ≤(k1 −1) · · · (ks −1). The number I(k1, . . . , ks) can be interpreted as the degree of the numerator of the zeta-function of s P i=1 cixki i ; see . This number appears in the estimates (7.3.5) and (7.3.6) for the number of solutions N0 and Nb, b ̸= 0, of Equation (7.3.1). 7.3.8 Theorem [1939, Theorems 6.36, 6.37] Let di = gcd(ki, q −1) for i = 1, . . . , s. Then, 1. For b = 0, \|N0 −qs−1\| ≤I(d1, . . . , ds)(q −1)q s−2 2 . (7.3.5) 2. For b ̸= 0, \|Nb −qs−1\| ≤ h (d1 −1) · · · (ds −1) − 1 −q−1/2 I(d1, . . . , ds) i q s−1 2 . (7.3.6) 7.3.9 Remark Note that if q is suﬃciently large with respect to d1, . . . , ds then (7.3.5) and (7.3.6) imply that Equation (7.3.1) is solvable. One can obtain an improvement of (7.3.5) and (7.3.6) if the p-weights of all the di’s are small and the solutions are in Fq2 . 7.3.2 Solutions of diagonal equations 7.3.10 Remark When dealing with any type of equation, the ﬁrst question that one might ask is whether or not the equation has solutions over a given ﬁeld. A classical result is Chevalley’s theorem that guarantees a non-trivial solution whenever the number of variables is larger than the degree of the polynomial and there is no constant term. In the authors improve Chevalley’s result by considering the p-weight degree of the polynomial instead of its degree. The following results determine solvability of some families of diagonal equations. 7.3.11 Theorem Let q be any prime power and k a positive integer such that k ̸= p −1 in the case q = p. The equation Ps i=1 cixk i = 0 has a nontrivial solution for s ≥k+3 2 . 7.3.12 Theorem [2679, Theorems 1, 2] Let Fq′ ⊆Fq be ﬁnite ﬁelds, and suppose that c1, . . . , cs ∈ Fq′. 1. If s ≥2 and q > Qs i=1(ki −1) 2 s−1 , then Equation (7.3.1) is solvable in Fq for any b ∈Fq. 2. If s ≥3 and qs−1 −1 (q −1)q s 2 −1 > 1 D D−1 X l=0  Y ki \| l (1 −ki)   , 208 Handbook of Finite Fields where D = Q i ki, then Equation (7.3.1) has a nontrivial solution over Fq for b = 0. 7.3.13 Theorem If Ps i=1 1 ki > 1, then Equation (7.3.1) is solvable over Fq for any b. Fur-thermore, for b = 0 there is a nontrivial solution. 7.3.14 Remark It is not easy to ﬁnd general formulas for the exact number of solutions of the diagonal Equation (7.3.1) and, in most cases, the formulas are rather complex. Results on the exact number of solutions of particular families of diagonal equations can be found in [199, 1548, 2120]. The case b = 0 has been studied separately and some general results are included below. 7.3.15 Theorem [3001, Theorem 1] Let q = p2t, s ≥2, nk = q −1, and consider Ps i=1 cixk i = b. If there exists a divisor r of t such that pr ≡−1 (mod k), then N0 = qs−1 + ϵsq s 2 −1(q −1) k k−1 X j=0 (1 −k)τ(j), and, for b ̸= 0, Nb = qs−1 −ϵs+1q s 2 −1  (1 −k)θ(b)q 1 2 −(q 1 2 −ϵ) k k−1 X j=0 (1 −k)τ(j)  , where ϵ = (−1) t r , θ(b) = \|{i \| cn i = (−b)n}\| , τ(j) = i \| cn i = (αj)n , 1 ≤i ≤s, and α is a primitive root in Fq. 7.3.16 Theorem Let kj = 2mj for j = 1, . . . , s −2, ks−1 = kms−1, ks = krms, where r ≥1, (2k, m1 · · · ms) = 1 and m1, . . . , ms are pairwise coprime. 1. If k = 1 or if s and k are odd, then N0 = qs−1. 2. If s is even, then N0 = qs−1 + (−1) (s−2)(q−1) 4 (q −1)q s−2 2   1 + (−1) q−1 k k 2 −1  . 7.3.17 Remark Theorem 7.3.16 holds when the number of variables s is even. Explicit formulas for the case when s is odd are given in . 7.3.18 Theorem Let s > 2. Then I(k1, . . . , ks) = 0 (and hence N0 = qs−1) if and only if one of the following holds: 1. For some i, ki, k1···ks ki = 1. 2. If {ki1, . . . , kir \| 1 ≤i1 < · · · < ir ≤s} is the set of all even integers among {k1, . . . , ks}, then 2 ∤r, ki1 2 , . . . , kir 2 are pairwise coprime, and kij is coprime to any odd number in {k1, . . . , ks} for j = 1, . . . , r and r < s. 7.3.19 Remark Diagonal equations with few variables have received special attention [2144, 2714]. Hermitian curves, a particular case of “Fermat like” equations, have been studied extensively because they provide good examples of curves with maximal number of rational points [1199, 1278, 1514, 2499]. We present a recent result on Fermat equations and refer the reader to [110, 1717] for results on other speciﬁc families of “Fermat like” equations. Equations over ﬁnite ﬁelds 209 7.3.20 Theorem [1798, Theorem 1.1] The number of solutions of Xk + Y k + Zk = 0 over Fqm is N0 = 3k + k2(q −2) + (d −1)(d −2) provided that p >   2 t+1 q sin( kπ 2N ) + 1   N(t−1) k−d , where N = qm−1 q−1 , k \| N, d = ( N k , t + 1), and q ≡t (mod N k ) with 0 < t < N k . 7.3.21 Deﬁnition Let L(k1, . . . , ks) be the least positive integer represented by (7.3.4) if there is such an integer, or, otherwise, let L(k1, . . . , ks) = s −1. 7.3.22 Deﬁnition For k an integer, let vq(k) denote the highest power of q dividing k. 7.3.23 Remark Many results on diagonal equations give bounds for the number of solutions Nb of (7.3.1), while others give bounds for vp (Nb). Most of these bounds depend on the numbers I and L of Deﬁnitions 7.3.6 and 7.3.21. Exact values for I and L are hard to compute in general; includes their exact values for certain equations, and provides sharp general lower bounds for I. 7.3.24 Theorem [1939, 2679, 2744] I(k1, . . . , ks) = (−1)s + s X r=1 (−1)s−r X 1≤i1<i2<···<ir≤s ki1 · · · kir lcm [ki1, . . . , kir]. 7.3.25 Theorem [2906, Theorem 1] If there is a positive integer µ such that 1 k1 + · · · + 1 ks > µ ≥1, then vq (Nb) ≥µ. 7.3.26 Theorem Let q = pn, ki\|(q −1), ri be the least integer such that ki\| (pri −1), Kiki = pri −1 for i = 1, . . . , s, and µ1 be the least integer such that µ1 ≥ s X i=1 n (Ki, p −1) ri(p −1) . If µ = µ1 n −1, then vq (Nb) ≥µ, where [a] denotes the integer part of a. 7.3.27 Theorem Suppose ki\|(q −1), I(k1, . . . , ks) > 0 and let wi = gcd (ki, lcm [kj\|i ̸= j]). Then L(k1, . . . , ks) = lPs i=1 1 wi m if one of the following conditions holds: 1. Ps i=1 1 wi ≡0 (mod 1), 2. lcm [w1, . . . , ws] \|ws, 3. I(k1, . . . , ks) ≤10, 4. s ≤3. 7.3.28 Theorem [2745, Theorems 2,4] For each i, deﬁne ui = gcd ki, k1k2···ks ki . Then, 1. I(k1, . . . , ks) = I(u1, . . . , us) and L(k1, . . . , ks) = L(u1, . . . , us). 2. For all 1 ≤j ≤s, I(k1, . . . , ks) ≤Q i̸=j(ui −1). 210 Handbook of Finite Fields 3. The following are equivalent: (a) I(k1, . . . , ks) = 1, (b) s is even and ui = 2 for all i except uj = 2m, m > 0 for one j. 7.3.29 Corollary Assume that each ki satisﬁes one of the conditions of Part 3 of Theorem 7.3.28. Then L(k1, . . . , ks) = s 2. 7.3.30 Theorem [2906, Theorem 3] vq (N0) ≥L(k1, . . . , ks) −1. 7.3.31 Deﬁnition Let k = a0 + a1p + · · · + arpr, where 0 ≤ai ≤p −1. The p-weight of k is deﬁned and denoted as σp(k) = Pr i=0 ai. The p-weight degree of a monomial xk1 1 · · · xks s is deﬁned and denoted as wp(xk1 1 · · · xks s ) = σp(k1) + · · · + σp(ks). The p-weight degree wp(g) of a polynomial g is the largest p-weight degree of the terms in g. 7.3.32 Remark Some of the bounds for the powers of p dividing the number of solutions of diagonal equations can be improved if one considers the p-weight σp(ki) of the degrees of the terms in the equation instead of their degrees ki. 7.3.33 Theorem [2149, Theorem 10] Let q = pn. If µ is the least integer satisfying µ ≥n 1 σp(k1) + · · · + 1 σp(ks) −1 , then vp (Nb) ≥µ. 7.3.34 Remark If Ps i=1 1 σp(ki) > 1 then the equation Ps i=1 cixki i = 0 has a non-trivial solution. 7.3.35 Theorem [2149, Theorem 2] Let γ = min(j1,...,js) n Ps i=1 σp(ji(q−1)/ki) p−1 o −1, where (j1, . . . , js) satisﬁes (7.3.4). If N0 is the number of solutions of Equation (7.3.1) over Fqm, then vp (N0) ≥mγ. Also, mγ is best possible, i.e., there exists an equation c1xk1 1 +· · ·+csxks 1 = 0 such that vp (N0) = mγ. 7.3.36 Remark We note that Theorem 7.3.35 uses a calculation on Fq to give information on vp (N0) for any extension of Fq. 7.3.3 Generalizations of diagonal equations 7.3.37 Remark Generalizations of diagonal equations have been considered by several authors. Some examples are systems of diagonal equations [21, 1698, 2151, 2808], deformed diagonal equations [546, 2698], and equations with terms of disjoint support [501, 502, 563]. 7.3.38 Theorem [1051, Theorem 5] Let F = c1xk 1 +· · ·+csxk s +g(x1, . . . , xs) be a polynomial over Fp, where c1 · · · cs ̸= 0 and deg(g) < k. Then F = 0 is solvable in Fs p if s ≥ p−1 ⌊p−1 k ⌋ . 7.3.39 Remark Theorem 7.3.38 was proved using the combinatorial nullstellensatz and it general-izes Theorem 1 in to include the case k ∤(p−1). Prior to this result, Newton polyhedra were used in to prove results that allows one to estimate the number of zeros of poly-nomials of the type in Theorem 7.3.38. Note that, even if s > k, Chevalley’s theorem does not guarantee the solvability of the equations in Theorem 7.3.38. Equations over ﬁnite ﬁelds 211 7.3.40 Theorem [2746, Theorem 1.1] Let q = pn and F = xk 1 + · · · + xk s + g(x1, . . . , xs) be a polynomial over Fq where deg(g) < k. Then, for nonempty subsets A1, . . . , As ⊂Fq, \|{F(a1, . . . , as) \| ai ∈Ai}\| ≥min ( p, s X i=1 \|Ai\| −1 k + 1 ) . 7.3.41 Remark If q = p and A1 = · · · = As = F∗ p, s > k, one obtains Theorem 1.3 in . 7.3.42 Theorem Let q = pn and ki > 1 be positive integers satisfying ki \|(p −1). Let F = xk1 1 + · · · + xks s + g(x1, . . . , xs) be a polynomial over Fq, where wp(g) < mini {ki}, and N0 be the number of solutions of F = 0. Then, vp (N0) = n Ps i=1 1 ki −1 whenever Ps i=1 1 ki is an integer. In particular, F is solvable over Fq. 7.3.43 Remark The result in Theorem 7.3.42 is a special case of Theorem 15 of . 7.3.4 Waring’s problem in ﬁnite ﬁelds 7.3.44 Remark Waring’s problem is to ﬁnd the smallest s = g(k, q) such that every element b ∈Fq can be written as a sum of s summands of k-th powers in Fq. This is an active area of research and the introduction of new techniques from arithmetic combinatorics has allowed the improvement of the bounds on g(k, q). Some results on Waring’s problem and their application to coding theory were presented in Section 6.3. 7.3.45 Deﬁnition The smallest s = g(k, q) such that the equation xk 1 +· · ·+xk s = b has a solution for every b ∈Fq is Waring’s number for Fq with respect to k. 7.3.46 Theorem Waring’s number g(k, pn) exists if and only if pn−1 pd−1 ∤k for all d\|n, d ̸= n. Also, if d = gcd(k, q −1), then g(k, q) = g(d, q). 7.3.47 Remark Because of the previous theorem it is enough to consider k\| (q −1). From now on we assume that g(k, q) exists and k\| (q −1). 7.3.48 Theorem We have g(k, p) ≤k and equality holds if k = 1, 2, p−1 2 , p −1. 7.3.49 Theorem If 2 ≤k < q 1 4 + 1, then g(k, q) = 2. 7.3.50 Remark To ﬁnd the exact value for g(k, q) is a diﬃcult problem and, given Theorem 7.3.49, one might ask, for each k, which is the largest q such that g(k, q) ̸= 2 . The next table contains some of the exact values known for g(k, p). These and other exact values can be found in [12, 2148, 2676, 2678]. k k ̸= p −1, p−1 2 5 6 6 7 7 8 31, 41 37 ≤p ≤67 71, 113 p p ≤29 61 31 109, 139, 223 43 127 41 g(k, p) k 2 + 1 3 4 3 4 3 4 k 8 9 10 10 10 11 11 11 73, 109 71 ≤p ≤491 p 73 ≤p ≤137 127, 163 31 41 521 ≤p ≤631 89 67 199, 331 233, 257 181, 199 61 661 ≤p ≤881 353, 419 337, 761 271, 307 641, 911 463, 617 g(k, p) 3 3 5 4 3 4 5 3 7.3.51 Remark If the value of g(k, p) for any 3 ≤k ≤11, k ̸= p −1, p−1 2 is not included in the table above, then g(k, p) = 2. Hence, the table and Theorem 7.3.48 provide all the exact values of Waring’s number for k = 1, . . . , 11. 212 Handbook of Finite Fields 7.3.52 Theorem [2149, Theorem 18] Let l ̸= 1. If k\|(pn + 1), k ̸= pn + 1, then g(k, p2nl) = 2. 7.3.53 Theorem [651, Theorem 2] Let t = p−1 k . If a, b are the unique positive integers with a > b and a2+b2+ab = p, then, for t = 3, g(k, p) = a+b−1, and for t = 6, g(k, p) = 2 3a + 1 3b . If t = 4 and a, b are the unique positive integers with a > b and a2+b2 = p, then g(k, p) = a−1. 7.3.54 Theorem [1782, Theorems 1,2] Let φ denote Euler’s function, m be a positive integer and p,r be primes such that p is a primitive root modulo rm. Then, g pφ(rm) −1 rm , pφ(rm) = (p −1)φ(rm) 2 . If, in addition, p and r are odd primes, then g pφ(rm) −1 2rm , pφ(rm) =      rm−1 pr 4 −p 4r if r < p, rm−1 j pr 4 −r 4p k if r ≥p. 7.3.55 Remark Theorem 7.3.54 generalizes the results in . A good survey on bounds for g(k, q) can be found in . 7.3.56 Theorem [1282, Theorem 5] If k < √q, then g(k, q) ≤8. 7.3.57 Theorem [2990, Theorem 2] If d = p−1 gcd( pn−1 k ,p−1), then g(k, pn) ≤ng(d, p). 7.3.58 Theorem [1206, Theorem 4] If k ≥2 is a proper divisor of p −1 and k ≥(p −1)4/7, then g(k, p) ≤170 k7/3 (p−1)4/3 log p. 7.3.59 Remark By using a result in an improvement to Theorem 7.3.58 can be obtained. 7.3.60 Theorem [1789, Theorem 2] For any ϵ > 0 there exists cϵ > 0 such that for any k ≥2, p ≥ k ln k (ln(ln k+1))1−ϵ , we have g(k, p) ≤cϵ (ln k)2+ϵ. 7.3.61 Theorem [651, Theorem 1] Let t = p−1 k and l be a positive integer. If φ(t) ≥l, then g(k, p) ≤C(l)k 1 l for some constant C(l), where φ is Euler’s function. 7.3.62 Remark Theorems 7.3.60 and 7.3.61 prove conjectures made by Heilbronn in . The next theorem gives an explicit value for the constant C(l) in Theorem 7.3.61 when l = 2. Other estimates for C(l) are also given in . 7.3.63 Theorem [656, Theorem 1.1] For t = p−1 k > 2 we have the uniform upper bound g(k, p) ≤ 83k1/2. 7.3.64 Deﬁnition Let Ak := xk : x ∈Fq deﬁne the set of k-th powers of the elements in the ﬁeld Fq, and A′ k := Ak ∩Fp. 7.3.65 Theorem [650, Theorems 1, 3, 4], [2550, Theorem 4.1] 1. g(k, pn) ≤8n l (k+1)1/n−1 \|A′ k\|−1 m . If \|A′ k\| ≤3, then g(k, pn) ≤4n l (k+1)1/n−1 \|A′ k\|−1 + 2 m . 2. For any ϵ > 0, if \|A′ k\| ≥42/ϵn, then g(k, pn) ≤C(ϵ)kϵ, for some constant C(ϵ). 3. g(k, p2) ≤16 √ k + 1. If n ≥3, then g(k, pn) ≤10 √ k + 1. 4. If \|A′ k\| ≥pϵ for ϵ > 41 83, then, for all suﬃciently large p, we have g(k, p) ≤6. Equations over ﬁnite ﬁelds 213 7.3.66 Remark Part 1 of Theorem 7.3.65 improves Theorem 1 in . Parts 2 and 3 prove the extensions of Heilbronn’s conjectures to arbitrary ﬁelds. See also . 7.3.67 Remark Waring’s problem has been generalized to systems of diagonal equations [561, 2808], to general polynomials [372, 549, 658], to Dickson polynomials , to factorials , and to reciprocals [754, 2643]. 7.3.68 Theorem [1196, Theorem 1] Any residue class λ modulo p can be represented in the form P5 i=1 mi!ni! ≡λ (mod p) for some positive integers m1, n1, . . . , m5, n5 with max1≤i≤5{mi, ni} ≤cp27/28, where c is an absolute constant. 7.3.69 Theorem [1196, Theorem 2] Let l(p) be the smallest integer such that for every integer λ the congruence n1! + · · · + nl! ≡λ (mod p) has a solution in positive integers. Then l(p) ≤C log3(p), for some constant C. 7.3.70 Deﬁnition Let Dk(x, a) be the Dickson polynomial of degree k and parameter a ∈Fq (see Section 8.3). The Waring problem for Dickson polynomials over Fq is to ﬁnd the smallest positive integer s = ga(k, q) such that the equation Dk(x1, a) + · · · + Dk(xs, a) = b, x1, . . . , xs ∈Fq is solvable for any b ∈Fq. 7.3.71 Theorem [2333, Theorem 1] Let ga(k, q) be deﬁned as in Deﬁnition 7.3.70. The inequality ga(k, q) ≤16 holds 1. For any a ∈F∗ q and (k, q −1) ≤2−3/2(q −1)1/2. 2. For any a that it is a square in F∗ q and (k, q + 1) ≤2−3/2(q −1)1/2. See Also , , , References on solvability/divisibility of diagonal equations. , , , , , , , , , For solvability/divisibility of general systems of equations. , , , , , For studies of diagonal equations over function ﬁelds. , For results on diagonal equations over p-adic ﬁelds. , , , For applications of diagonal equations to coding theory. , References Cited: [12, 21, 23, 24, 30, 110, 150, 168, 199, 240, 372, 374, 416, 494, 501, 502, 511, 515, 538, 546, 549, 561, 562, 563, 566, 615, 650, 651, 656, 658, 754, 1051, 1196, 1197, 1199, 1206, 1274, 1278, 1282, 1314, 1350, 1460, 1514, 1540, 1548, 1575, 1617, 1698, 1717, 1757, 1782, 1789, 1798, 1838, 1917, 1918, 1939, 2030, 2120, 2137, 2144, 2146, 2147, 2148, 2149, 2150, 2151, 2165, 2333, 2432, 2449, 2499, 2550, 2643, 2676, 2677, 2678, 2679, 2681, 2697, 2698, 2714, 2743, 2744, 2745, 2746, 2808, 2809, 2883, 2906, 2956, 2990, 2992, 2995, 3001, 3002, 3003, 3066] This page intentionally left blank This page intentionally left blank 8 Permutation polynomials 8.1 One variable ......................................... 215 Introduction • Criteria • Enumeration and distribution of PPs • Constructions of PPs • PPs from permutations of multiplicative groups • PPs from permutations of additive groups • Other types of PPs from the AGW criterion • Dickson and reversed Dickson PPs • Miscellaneous PPs 8.2 Several variables ..................................... 230 8.3 Value sets of polynomials .......................... 232 Large value sets • Small value sets • General polynomials • Lower bounds • Examples • Further value set papers 8.4 Exceptional polynomials ........................... 236 Fundamental properties • Indecomposable exceptional polynomials • Exceptional polynomials and permutation polynomials • Miscellany • Applications 8.1 One variable Gary L. Mullen, The Pennsylvania State University Qiang Wang, Carleton University 8.1.1 Introduction 8.1.1 Deﬁnition For q a prime power, let Fq denote the ﬁnite ﬁeld containing q elements. A polynomial f ∈Fq[x] is a permutation polynomial (PP) of Fq if the function f : c →f(c) from Fq into itself induces a permutation. Alternatively, f is a PP of Fq if the equation f(x) = a has a unique solution for each a ∈Fq. 8.1.2 Remark The set of all PPs on Fq forms a group under composition modulo xq−x, isomorphic to the symmetric group Sq of order q!. For q > 2, the group Sq is generated by xq−2 and all linear polynomials ax + b, and if c is a primitive element in Fq, Sq is generated by cx, x + 1, and xq−2. 8.1.3 Remark Given a permutation g of Fq, the unique permutation polynomial Pg(x) of Fq of degree at most q−1 representing the function g can be found by the Lagrange Interpolation Formula (see Theorem 1.71 in ). In particular Pg(x) = P a∈Fq g(a)(1 −(x −a)q−1); see also Theorem 2.1.131. 215 216 Handbook of Finite Fields 8.1.4 Remark If f is a PP and a ̸= 0, b ̸= 0, c ∈Fq, then f1 = af(bx+c) is also a PP. By suitably choosing a, b, c we can arrange to have f1 in normalized form so that f1 is monic, f1(0) = 0, and when the degree n of f1 is not divisible by the characteristic of Fq, the coeﬃcient of xn−1 is 0. 8.1.5 Remark A few well known classes of PPs from : Monomials: The monomial xn is a PP of Fq if and only if (n, q −1) = 1. Dickson: For a ̸= 0 ∈Fq, the polynomial Dn(x, a) = P⌊n/2⌋ i=0 n n−i n−i i (−a)ixn−2i is a PP of Fq if and only if (n, q2 −1) = 1; see Section 9.6. Linearized: The polynomial L(x) = Pn−1 s=0 asxqs ∈Fqn[x] is a PP of Fqn if and only if det(aqj i−j) ̸= 0, 0 ≤i, j ≤n −1. The set of linearized PPs forms the Betti-Mathieu group isomorphic to the group GL(n, Fq), the general linear group of all non-singular n×n matrices over Fq under matrix multiplication. Recently, there have been several papers devoted to explicit constructions of linearized PPs; see for example, [504, 3047, 3065]. For odd q, f(x) = x(q+1)/2 + ax is a PP of Fq if and only if a2 −1 is a nonzero square in Fq. Moreover, the polynomial f(x) + cx is a PP of Fq for (q −3)/2 values of c ∈Fq . The polynomial xr(f(xd))(q−1)/d is a PP of Fq if (r, q −1) = 1, d \| q −1, and f(xd) has no nonzero root in Fq. 8.1.6 Remark For more information, we refer to Chapter 7 of , and survey papers [681, 1933, 1935, 2176, 2178]. 8.1.2 Criteria 8.1.7 Theorem (Hermite) Let p be the characteristic of Fq. A polynomial f ∈Fq[x] is a PP if and only if 1. the polynomial f has exactly one root in Fq; 2. for each integer t with 1 ≤t ≤q −2 and t ̸≡0 (mod p), the reduction of (f(x))t (mod xq −x) has degree at most q −2. 8.1.8 Remark Hermite’s criterion was used by Dickson to obtain all normalized PPs of degree at most 5 in the list below. Normalized PPs of Fq q x any q x2 q ≡0 (mod 2) x3 q ̸≡1 (mod 3) x3 −ax (a not a square) q ≡0 (mod 3) x4 ± 3x q = 7 x4 + a1x2 + a2x (if its only root in Fq is 0) q ≡0 (mod 2) x5 q ̸≡1 (mod 5) x5 −ax (a not a fourth power) q ≡0 (mod 5) x5 + ax (a2 = 2) q = 9 x5 ± 2x2 q = 7 x5 + ax3 ± x2 + 3a2x (a not a square) q = 7 x5 + ax3 + 5−1a2x (a arbitrary) q ≡±2 (mod 5) x5 + ax3 + 3a2x (a not a square) q = 13 x5 −2ax3 + a2x (a not a square) q ≡0 (mod 5) A list of PPs of degree 6 over ﬁnite ﬁelds with odd characteristic can be found in . A list of PPs of degree 6 and 7 over ﬁnite ﬁelds with characteristic two can be found in . Permutation polynomials 217 A recent preprint tabulates all monic PPs of degree 6 in the normalized form. 8.1.9 Theorem The polynomial f is a PP of Fq if and only if P c∈Fq χ(f(c)) = 0 for all nontrivial additive characters χ of Fq. 8.1.10 Remark Another criterion for PPs conjectured by Mullen in terms of the size \|Vf\| of the value set Vf = {f(a) \| a ∈Fq} of a polynomial f of degree n was proved by Wan . Namely, if \|Vf\| > q −q−1 n then f is a PP of Fq . We refer to Section 8.3 for more information on value sets. A variation of Hermite’s criterion in terms of combinatorial identities is given in . Hermite’s criterion can be rewritten in terms of the invariant, up(f), the smallest positive integer k such that P x∈Fq f(x)k ̸= 0. That is, f is a PP of Fq if and only if up(f) = q −1. In the case q = p, this criterion was improved in and only requires k > p−1 2 . Using Teichm¨ uller liftings, Wan et al. obtained an upper bound for \|Vf\| and improved Hermite’s criterion. Several other criteria were obtained by Turnwald in terms of invariants associated with elementary symmetric polynomials, without using Teichm¨ uller liftings. 8.1.11 Theorem Let f ∈Fq[x] be a polynomial of degree n with 1 ≤n < q. Let u be the smallest positive integer k with sk ̸= 0 if such k exists and otherwise set u = ∞, where sk denotes the k-th elementary symmetric polynomial of the values f(a). Let v be the size of the value set Vf = {f(a) \| a ∈Fq}. Let w be the smallest positive integer k with pk = P a∈Fq f(a)k ̸= 0 if such k exists and otherwise set w = ∞. The following are equivalent: (1) f is a PP of Fq; (2) u = q −1; (3) u > q −q n; (4) u > q −v; (5) v > q −q−1 n ; (6) w = q −1; (7) 2q 3 −1 < w < ∞; (8) q −q+1 n < w < ∞; (9) q −u ≤w < ∞; (10) u > q−1 2 and w < ∞. 8.1.12 Remark A criterion in terms of resultants was given by von zur Gathen . Using the Euclidean algorithm to compute the resultant, von zur Gathen provided a probabilistic test to determine whether a given polynomial is a PP or not. The number of operations in Fq has a softly linear running time O(n log q(log(n log q))k) for some k. Furthermore, Ma and von zur Gathan showed that this decision problem has a zero-error probabilistic polynomial time in and provided a random polynomial time test for rational functions over ﬁnite ﬁelds, along with several related problems in . Earlier, Shparlinski had given a deterministic superpolynomial time algorithm for testing PP . In 2005 Kayal provided a deterministic polynomial-time algorithm for testing PP . 8.1.3 Enumeration and distribution of PPs 8.1.13 Problem Let Nn(q) denote the number of PPs of Fq which have degree n. We have the trivial boundary conditions: N1(q) = q(q −1), Nn(q) = 0 if n is a divisor of q −1 larger than 1, and P Nn(q) = q! where the sum is over all 1 ≤n < q −1 such that n is either 1 or is not a divisor of q −1. Find Nn(q). 8.1.14 Remark In an invited address before the MAA in 1966, Carlitz conjectured that for each even integer n, there is a constant Cn so that for each ﬁnite ﬁeld of odd order q > Cn, there does not exist a PP of degree n over Fq. A polynomial f over Fq is exceptional if the only absolutely irreducible factors of f(x) −f(y) in Fq[x, y] are scalar multiples of x −y. One can also characterize an exceptional polynomial as a polynomial which induces a permutation of inﬁnitely many ﬁnite extension ﬁelds of Fq. As ﬁrst proved by Cohen in , any exceptional polynomial is a PP, and the converse holds if q is large compared to 218 Handbook of Finite Fields the degree of f. Cohen’s equivalent statement of Carlitz’s conjecture says that there is no exceptional polynomial of even degree in odd characteristic. This was proved by Fried, Guralnick, and Saxl in ; in fact an even stronger result was obtained through the use of powerful group theoretic methods, including the classiﬁcation of ﬁnite simple groups. 8.1.15 Remark For the next theorem we require the concept of an exceptional cover; see Section 9.7. 8.1.16 Theorem There is no exceptional cover of nonsingular absolutely irreducible curves over Fq of degree 2p where q is a power of p and p is prime. 8.1.17 Remark Several partial results on Carlitz’s conjecture can be found in [677, 2908]. Moreover, Wan generalized Carlitz’s conjecture in proving that if q > n4 and (n, q −1) = 1 then there is no PP of degree n over Fq. Later Cohen and Fried gave an elementary proof of Wan’s conjecture following an argument of Lenstra and this result was stated in terms of exceptional polynomials; see Section 8.4 for more information on exceptional polynomials. 8.1.18 Theorem [688, 2909] There is no exceptional polynomial of degree n over Fq if (n, q−1) > 1. 8.1.19 Theorem 1. Let ℓ> 1. For q suﬃciently large, there exists a ∈Fq such that the polynomial x(x(q−1)/ℓ+ a) is a PP of Fq. 2. Let ℓ> 1, (r, q −1) = 1, and k be a positive integer. For q suﬃciently large, there exists a ∈Fq such that the polynomial xr(x(q−1)/ℓ+ a)k is a PP of Fq. 8.1.20 Remark Any non-constant polynomial h(x) ∈Fq[x] of degree ≤q −1 can be written uniquely as axrf(x(q−1)/ℓ) + b with index ℓ. Namely, write h(x) = a(xn + an−i1xn−i1 + · · · + an−ikxn−ik) + b, where a, an−ij ̸= 0, j = 1, . . . , k. Here we suppose that j ≥1 and n −ik = r. Then h(x) = axrf(x(q−1)/ℓ) + b, where f(x) = xe0 + an−i1xe1 + · · · + an−ik−1xek−1 + ar, ℓ= q −1 (n −r, n −r −i1, . . . , n −r −ik−1, q −1), and (e0, e1, . . . , ek−1, ℓ) = 1. Clearly, h is a PP of Fq if and only if g(x) = xrf(x(q−1)/ℓ) is a PP of Fq. Then ℓis the index of h. 8.1.21 Remark If ℓ= 1 then f(x) = 1 so that g(x) = xr. In this case g(x) is a PP of Fq if and only if (r, q −1) = 1. We can assume ℓ> 1. 8.1.22 Remark More existence and enumerative results for binomials can be found in [61, 64, 1833, 2018, 2019, 2020, 2825, 2905, 2913]. In , Laigle-Chapuy proved the ﬁrst assertion of Theorem 8.1.19 assuming q > ℓ2ℓ+2 1 + ℓ+1 ℓℓ+2 2 . In , Masuda and Zieve obtained a stronger result for more general binomials of the form xr(xe1(q−1)/ℓ+ a). More precisely they showed the truth of Part 1 of Theorem 8.1.19 for q > ℓ2ℓ+2. Here we present a general result of Akbary-Ghioca-Wang (Theorem 8.1.25) which shows that there exist permutation polynomials of index ℓfor any prescribed exponents satisfying conditions (8.1.1). This result generalizes all the existence results from [551, 1833, 2020]. 8.1.23 Deﬁnition Let q be a prime power, and ℓ≥2 be a divisor of q −1. Let m, r be positive integers, and ¯ e = (e1, . . . , em) be an m-tuple of integers that satisfy the following conditions: 0 < e1 < e2 < · · · < em ≤ℓ−1, (e1, . . . , em, ℓ) = 1 and r + ems ≤q −1, (8.1.1) Permutation polynomials 219 where s := (q −1)/ℓ. For a tuple ¯ a := (a1, . . . , am) ∈ F∗ q m, we let g¯ a r,¯ e(x) := xr (xems + a1xem−1s + · · · + am−1xe1s + am) . We deﬁne N m r,¯ e(ℓ, q) as the number of all tuples ¯ a ∈ F∗ q m such that g¯ a r,¯ e(x) is a PP of Fq. In other words N m r,¯ e(ℓ, q) is the number of all monic permutation (m + 1)-nomials g¯ a r,¯ e(x) = xrf(x(q−1)/ℓ) over Fq with vanishing order at zero equal to r, set of exponents ¯ e for f(x), and index ℓ. Note that if r and ¯ e satisfy (8.1.1) then g¯ a r,¯ e(x) has index ℓ. 8.1.24 Theorem With the above notation, we have N m r,¯ e(ℓ, q) −ℓ! ℓℓqm < ℓ!ℓqm−1/2. 8.1.25 Theorem For any q, r, ¯ e, m, ℓthat satisfy (8.1.1), (r, s) = 1, and q > ℓ2ℓ+2, there exists an ¯ a ∈(F∗ q)m such that the (m + 1)-nomial g¯ a r,¯ e(x) is a permutation polynomial of Fq. 8.1.26 Remark For q ≥7 we have ℓ2ℓ+2 < q if ℓ< log q 2 log log q. 8.1.27 Theorem The value Np−2(p) ∼(ϕ(p)/p)p! as p →∞, where ϕ is the Euler function. More precisely, Np−2(p) −ϕ(p) p p! ≤ q pp+1(p−2)+p2 p−1 . 8.1.28 Theorem Let q be a prime power. Then \|Nq−2(q) −(q −1)!\| ≤ q 2e π q q 2 . 8.1.29 Theorem Fix j integers k1, . . . , kj with the property that 0 < k1 < · · · < kj < q −1 and deﬁne N(k1, . . . , kj; q) as the number of PPs of Fq of degree less than q −1 such that the coeﬃcient of xki equals 0, for i = 1, . . . , j. Then N(k1, . . . , kj; q) −q! qj < 1 + r 1 e !q ((q −k1 −1)q)q/2. In particular, Nq−2(q) = q! −N(q −2; q). 8.1.30 Remark We note that for 1 ≤t ≤q −2 the number of PPs of degree at least q −t −1 is q! −N(q −t −1, q −t, . . . , q −2; q). In Konyagin and Pappalardi proved that N(q −t −1, q −t, . . . , q −2; q) ∼ q! qt holds for q →∞and t ≤0.03983 q. This result guarantees the existence of PPs of degree at least q −t −1 for t ≤0.03983 q (as long as q is suﬃciently large). However, the following theorem establishes the existence of PPs with exact degree q −t −1. 8.1.31 Theorem Let m ≥1. Let q be a prime power such that q −1 has a divisor ℓwith m < ℓ and ℓ2ℓ+2 < q. Then for every 1 ≤t < (ℓ−m) ℓ (q −1) coprime with (q −1)/ℓthere exists an (m + 1)-nomial g¯ a r,¯ e(x) of degree q −t −1 which is a PP of Fq. 8.1.32 Corollary Let m ≥1 be an integer, and let q be a prime power such that (m+1) \| (q−1). Then for all n ≥2m + 4, there exists a permutation (m + 1)-nomial of Fqn of degree q −2. 8.1.33 Deﬁnition Let mk be the number of permutations of Fq which are k-cycles and are represented by polynomials of degree q −k. 8.1.34 Theorem Every transposition of Fq is represented by a unique polynomial of degree q −2. Moreover, m3 =    2 3q(q −1) if q ≡1 (mod 3), 0 if q ≡2 (mod 3), 1 3q(q −1) if q ≡0 (mod 3). 220 Handbook of Finite Fields 8.1.35 Theorem 1. If q ≡1 (mod k), then mk ≥ϕ(k) k q(q −1). 2. If char(Fq) > e(k−3)/e, then mk ≤(k−1)! k q(q −1). 8.1.36 Remark It is conjectured in that the above upper bound for mk holds for any k < q. Small k-cycles such as k = 4, 5 are also studied in [1995, 1996, 1997]. 8.1.37 Theorem Let r and s be ﬁxed positive integers, and k2, . . . , ks be non-negative integers such that Ps i=2 iki = r. Let P(k2, . . . , ks) be the set of permutations of Fq which are the disjoint products of k2 transpositions, k3 3-cycles, etc. Then the number of permutations in P(k2, . . . , ks) represented by a polynomial of degree q −2 is asymptotic to the number of all permutations in P(k2, . . . , ks) as q goes to ∞. 8.1.4 Constructions of PPs 8.1.38 Remark For the purpose of introducing the construction of PPs in the next few sections, we present the following recent result by Akbary-Ghioca-Wang (AGW). 8.1.39 Theorem (AGW’s criterion, ) Let A, S, and ¯ S be ﬁnite sets with #S = # ¯ S, and let f : A →A, ¯ f : S →¯ S, λ : A →S, and ¯ λ : A →¯ S be maps such that ¯ λ ◦f = ¯ f ◦λ. If both λ and ¯ λ are surjective, then the following statements are equivalent: 1. f is a bijection (a permutation of A); 2. ¯ f is a bijection from S to ¯ S and f is injective on λ−1(s) for each s ∈S. 8.1.40 Remark We note that this criterion does not require any restriction on the structures of the sets S and ¯ S in ﬁnding new classes of PPs of a set A. In particular, if we take A as a group and S and ¯ S as homomorphic images of A, then we obtain the following general result for ﬁnding permutations of a group. 8.1.41 Theorem Let (G, +) be a ﬁnite group, and let ϕ, ψ, ¯ ψ ∈End(G) be group endomor-phisms such that ¯ ψ ◦ϕ = ϕ ◦ψ and #im(ψ) = #im( ¯ ψ). Let g : G − →G be any mapping, and let f : G − →G be deﬁned by f(x) = ϕ(x) + g(ψ(x)). Then, 1. f permutes G if and only if the following two conditions hold: a. ker(ϕ)∩ker(ψ) = {0} (or equivalently, ϕ induces a bijection between ker(ψ) and ker( ¯ ψ)); and b. the function ¯ f(x) := ϕ(x) + ¯ ψ(g(x)) restricts to a bijection from im(ψ) to im( ¯ ψ). 2. For any ﬁxed endomorphisms ϕ, ψ and ¯ ψ satisfying Part 1.a, there are (#im(ψ))! · # ker( ¯ ψ) #im(ψ) such permutation functions f (when g varies). 3. Let g : G − →G be such that ¯ ψ ◦g \|im(ψ) = 0. Then f = ϕ + g ◦ψ permutes G if and only if ϕ is a permutation of G. 4. Assume ϕ ◦ψ = 0 and g : G − →G is a mapping such that g(x) restricted to im(ψ) is a permutation of im(ψ). Then f(x) = ϕ(x)+g(ψ(x)) permutes G if and only if ϕ and ψ satisfy Part 1.a, and ¯ ψ restricted to im(ψ) is a bijection from im(ψ) to im( ¯ ψ). 8.1.42 Remark One can apply this result to a multiplicative group of a ﬁnite ﬁeld, an additive group of a ﬁnite ﬁeld, or the group of rational points of an elliptic curve over ﬁnite ﬁelds . This reduces a problem of determining whether a given polynomial over a ﬁnite ﬁeld Permutation polynomials 221 Fq is a permutation polynomial to a problem of determining whether another polynomial permutes a smaller set. 8.1.43 Corollary [2359, 2940, 3076] Let q −1 = ℓs for some positive integers ℓand s. Then P(x) = xrf(xs) is a PP of Fq if and only if (r, s) = 1 and xrf(x)s permutes the set µℓof all distinct ℓ-th roots of unity. 8.1.44 Remark The above corollary is a consequence of Theorem 8.1.41 when taking multiplicative group endomorphisms xr and xs. There are several other equivalent descriptions of PPs of the form xrf(xs); see for example, [65, 2359, 2916, 2940, 3076]. All of the classes of PPs in Subsection 8.1.5 are of this type. There are also many recent results on new classes of PPs when taking additive group endomorphisms in Theorem 8.1.41, see [62, 2003, 3046, 3077]. We give some classes of PPs in Subsection 8.1.6. We note that AGW’s criterion does not require any restriction on the structures of subsets S and ¯ S, which has even broader applications in ﬁnding new classes of PPs. Many classes of PPs in Subsection 8.1.7 can be obtained through this general construction method. 8.1.5 PPs from permutations of multiplicative groups 8.1.45 Deﬁnition [2278, 2940] Let γ be a primitive element of Fq, q −1 = ℓs for some positive integers ℓand s, and the set of all nonzero ℓ-th powers of Fq be C0 = {γℓj : j = 0, 1, . . . , s−1}. Then C0 is a subgroup of F∗ q of index ℓ. The elements of the factor group F∗ q/C0 are the cyclotomic cosets Ci := γiC0, i = 0, 1, . . . , ℓ−1. For any integer r > 0 and any A0, A1, . . . , Aℓ−1 ∈Fq, we deﬁne an r-th order cyclotomic mapping f r A0,A1,...,Aℓ−1 of index ℓfrom Fq to itself by f r A0,A1,...,Aℓ−1(0) = 0 and f r A0,A1,...,Aℓ−1(x) = Aixr if x ∈Ci, i = 0, 1, . . . , ℓ−1. Moreover, f r A0,A1,...,Aℓ−1 is an r-th order cyclotomic mapping of the least index ℓif the mapping cannot be written as a cyclotomic mapping of any smaller index. 8.1.46 Remark Cyclotomic mapping permutations were introduced in when r = 1 and in for any positive r. Let ζ = γs be a primitive ℓ-th root of unity in Fq and P(x) = xrf(xs) be a polynomial of index ℓover Fq with positive integer r. Then P(x) = xrf(xs) = f r A0,A1,...,Aℓ−1(x) where Ai = f(ζi) for 0 ≤i ≤ℓ−1. We note that the least index of a cyclotomic mapping is equal to the index of the corresponding polyno-mial. If P is a PP of Fq then (r, s) = 1 and Ai = f(ζi) ̸= 0 for 0 ≤i ≤ℓ−1. Under these two necessary conditions, P(x) = xrf(xs) is a PP of Fq if and only if f r A0,A1,...,Aℓ−1 is a PP of Fq . The concept of cyclotomic mapping permutations have recently been general-ized in allowing each branch to take a diﬀerent ri value so that P(x) has the form Pn i=0 xrifi(xs). More results can be found in and related piecewise constructions in [1061, 3058]. 8.1.47 Remark There are several other equivalent descriptions of PPs of the form xrf(xs), see [65, 2359, 2916, 2940, 3076] for example. In particular, in , it is shown that P(x) = xrf(xs) is a PP of Fq if and only if (r, s) = 1, Ai = f(ζi) ̸= 0 for 0 ≤i ≤ℓ−1, and ℓ−1 X i=0 ζcriAcs i = 0 for all c = 1, . . . , ℓ−1. This criterion is equivalent to Hermite’s criterion when the index 222 Handbook of Finite Fields ℓequals q −1. However, if ℓ< q −1, this criterion in fact improves Hermite’s criterion because we only need to verify that the coeﬃcient of xq−1 is 0 for ℓ−1 diﬀerent powers of P instead of all of the q −2 powers of P. We note that ℓ= 2 implies that q must be odd. 8.1.48 Remark For odd q, the polynomial P(x) = xrf(x(q−1)/2) is a PP of Fq if and only if (r, (q −1)/2) = 1 and η(f(−1)f(1)) = (−1)r+1. Here η is the quadratic character of Fq with the standard convention η(0) = 0. 8.1.49 Remark Let P(x) = xk+axr be a binomial of index ℓand s = q−1 ℓ. Then P(x) = xr(xes+a) for some e such that (e, ℓ) = 1. If a = bs for some b ∈Fq, then xr(xes + a) is a PP of Fq if and only if xr(xes + 1) is a PP of Fq. 8.1.50 Remark Necessary conditions for P(x) = xr(xes + 1) of index ℓto be a PP are as follows: (r, s) = 1, (2e, ℓ) = 1, (2r + es, ℓ) = 1, and 2s = 1. (8.1.2) 8.1.51 Remark For ℓ= 3, the conditions in (8.1.2) are suﬃcient to determine whether P is a PP of Fq . However, for ℓ> 3, it turns out not to be the case (for example, see [63, 64, 2927]). For general ℓ, a characterization of PPs of the form xr(xes+1) in terms of generalized Lucas sequences of order k := ℓ−1 2 is given in [2940, 2942]. 8.1.52 Deﬁnition For any integer k ≥1 and η a ﬁxed primitive (4k + 2)-th root of unity, the generalized Lucas sequence (or unsigned generalized Lucas sequence) of order k is deﬁned as {an}∞ n=0 such that an = 2k X t=1 t odd (ηt + η−t)n = k X t=1 ((−1)t+1(ηt + η−t))n. The characteristic polynomial of the generalized Lucas sequence is deﬁned by g0(x) = 1, g1(x) = x −1, gk(x) = xgk−1(x) −gk−2(x) for k ≥2. 8.1.53 Theorem [2940, 2942] Let q = pm be an odd prime power and q −1 = ℓs with odd ℓ≥3 and (e, ℓ) = 1. Let k = ℓ−1 2 . Then P(x) = xr(xes + 1) is a PP of Fq if and only if (r, s) = 1, (2r + es, ℓ) = 1, 2s = 1, and Rjc,k(L)(acs) = −1, for all c = 1, . . . , ℓ−1, (8.1.3) where acs is the (cs)-th term of the generalized Lucas sequence {ai}∞ i=0 of order k over Fp, jc = c(2eφ(ℓ)−1r +s) (mod 2ℓ), Rjc,k(x) is the remainder of the Dickson polynomial Djc(x) of the ﬁrst kind of degree jc divided by the characteristic polynomial gk(x), and L is a left shift operator on sequences. In particular, all jc are distinct even numbers between 2 and 2ℓ−2. 8.1.54 Remark We note that the degree of any remainder Rn,k is at most k−1 and Rn,k is either a Dickson polynomial of degree ≤k −1 or the degree k −1 characteristic polynomial gk−1(x) of the generalized Lucas sequence of order k, or a negation of the above polynomials . We can extend the deﬁnition of {an} to negative subscripts n using the same recurrence relations. We also remark that, for ℓ≤7, the sequences used in the descriptions of per-mutation binomials have simple structures and are fully described [63, 2927]. We note that signed generalized Lucas sequences are deﬁned in and they are used to compute the coeﬃcients of the compositional inverse of permutation binomials xr(xes + 1). Finally we note that Equation (8.1.3) always holds if the sequence {an} is s-periodic over Fp, which means that an ≡an+ks (mod p) for integers k and n. We remark that these sequences are Permutation polynomials 223 deﬁned over prime ﬁelds and checking the s-periodicity of these sequences is a much simpler task than checking whether the polynomial is a PP over the extension ﬁeld directly. 8.1.55 Theorem Assume the conditions (8.1.2) on ℓ, r, e, and s hold. If {an} is s-periodic over Fp, then the binomial P(x) = xr(xes + 1) is a permutation binomial of Fq. 8.1.56 Theorem Let p be an odd prime and q = pm and let ℓ, r, s be positive integers satisfying that q −1 = ℓs, (r, s) = 1, (e, ℓ) = 1, and ℓodd. Let p ≡−1 (mod ℓ) or p ≡1 (mod ℓ) and ℓ\| m. Then the binomial P(x) = xr(xes + 1) is a permutation binomial of Fq if and only if (2r + es, ℓ) = 1. In particular, if p ≡1 (mod ℓ) and ℓ\| m, then the conditions (r, s) = 1, (e, ℓ) = 1, and ℓodd imply that (2r + es, ℓ) = 1 . 8.1.57 Remark For a = 1 (equivalent to a = bs for some b), under the assumptions on q, s, ℓ, r, e, it is shown in that the s-periodicity of the generalized Lucas sequence implies that (η +η−1)s = 1 for every (2ℓ)-th root of unity η. However, we note that these two conditions are in fact equivalent for a = 1. The following result extends Theorem 8.1.55 as it also deals with even characteristic. 8.1.58 Theorem For q, s, ℓ, e, r, a satisfying q −1 = ℓs, (r, s) = 1, (e, ℓ) = 1, r, e > 0 and a ∈F∗ q, suppose (−a)ℓ̸= 1 and (z + a/z)s = 1 for every (2ℓ)-th root of unity z. Then P(x) = xr(xes + a) is a permutation binomial of Fq if and only if (2r + es, 2ℓ) ≤2. 8.1.59 Theorem Suppose xr(xes + a) permutes Fp, where a ∈F∗ p and r, e, s > 0 such that p −1 = ℓs and (ℓ, e) = 1. Then s ≥ p p −3/4 −1/2 > √p −1. 8.1.60 Remark For earlier results on permutation binomials, we refer to [569, 2018, 2020, 2127, 2680, 2681, 2825, 2905, 2913]. 8.1.61 Theorem Let q −1 = ℓs. Assume that f(ζt)s = 1 for any t = 0, . . . , ℓ−1. Then P(x) = xrf(xs) is a PP of Fq if and only if (r, q −1) = 1. 8.1.62 Corollary Let q −1 = ℓs and g be any polynomial over Fq. Then P(x) = xrg(xs)ℓis a PP of Fq if and only if (r, q −1) = 1 and g(ζt) ̸= 0 for all 0 ≤t ≤ℓ−1. 8.1.63 Corollary [65, 1833] Let p be a prime, ℓbe a positive integer and v be the order of p in Z/ℓZ. For any positive integer n, let q = pm = pℓvn and ℓs = q−1. Assume f is a polynomial in Fpvn[x]. Then the polynomial P(x) = xrf(xs) is a PP of Fq if and only if (r, q −1) = 1 and f(ζt) ̸= 0 for all 0 ≤t ≤ℓ−1. 8.1.64 Remark Corollary 8.1.63 is reformulated as Theorem 2.3 in . Namely, let ℓ, r > 0 satisfy ℓs = q −1. Suppose q = qm 0 where q0 ≡1 (mod ℓ) and ℓ\| m, and f ∈Fq0[x]. Then P(x) = xrf(xs) permutes Fq if and only if (r, s) = 1 and f has no roots in the µℓ, the set of ℓ-th roots of unity. 8.1.65 Theorem Let q −1 = ℓs, and suppose that Fq (the algebraic closure of Fq) contains a primitive (jℓ)-th root of unity η. Assume that η−utf(ηjt) s = 1 for any t = 0, . . . , ℓ−1 and a ﬁxed u. Moreover assume that j \| us. Then P(x) = xrf(xs) is a PP of Fq if and only if (r, s) = 1 and (r + us j , ℓ) = 1. 8.1.66 Remark Some concrete classes of PPs satisfying the above assumptions can be found in [60, 65, 2033, 3075, 3076]. For example, let hk(x) := xk + · · · + x + 1. Then the permutation behavior of the polynomials xrhk(xs) = xr(xks + · · · + xs + 1) and xrhk(xes)t has been studied in detail. Moreover, for certain choices of indices ℓand ﬁnite ﬁelds Fq (for example, p ≡−1 (mod 2ℓ) where ℓ> 1 is either odd or 2ℓ1 with ℓ1 odd), several concrete classes of PPs can be obtained [65, 3075, 3076]. 224 Handbook of Finite Fields 8.1.67 Remark When ℓ≤5, much simpler descriptions involving congruences and gcd conditions can be found in [60, 2033]. A reformulation of these results in terms of roots of unity can be found in which also covers the case of ℓ= 7. For larger index ℓ, one can also construct PPs of this form when ℓis an odd prime such that ℓ< 2p + 1. 8.1.68 Theorem Let ℓbe an odd prime such that ℓ< 2p+1, then P(x) = xr(xks+· · ·+xs+1) is a PP of Fq if and only if (r, s) = 1, (ℓ, k + 1) = 1, (2r + ks, ℓ) = 1, and (k + 1)s ≡1 (mod p). 8.1.6 PPs from permutations of additive groups 8.1.69 Remark There are several results on new classes of PPs when using additive group endo-morphisms ψ, ¯ ψ, and ϕ in Theorem 8.1.39 [62, 2003, 3046, 3077]. 8.1.70 Theorem Consider any polynomial g ∈Fqn[x], any additive polynomials ϕ, ψ ∈Fqn[x], any Fq-linear polynomial ¯ ψ ∈Fqn[x] satisfying ϕ ◦ψ = ¯ ψ ◦ϕ and #ψ(Fqn) = # ¯ ψ(Fqn), and any polynomial h ∈Fqn[x] such that h(ψ(Fqn)) ⊆Fq \ {0}. Then 1. f(x) := h(ψ(x))ϕ(x) + g(ψ(x)) permutes Fqn if and only if a. ker(ϕ) ∩ker(ψ) = {0}; and b. ¯ f(x) := h(x)ϕ(x) + ¯ ψ(g(x)) is a bijection between ψ(Fqn) and ¯ ψ(Fqn). 2. For any ﬁxed h, ϕ, ψ and ¯ ψ satisfying the above hypothesis and Part 1.a, there are (#im(ψ))! · # ker( ¯ ψ) #im(ψ) such permutation functions f (when g varies) (where ψ and ¯ ψ are viewed as endomorphisms of (Fqn, +)). 3. Assume in addition that ¯ ψ ◦g \|im(ψ) = 0. Then f(x) = h(ψ(x))ϕ(x) + g(ψ(x)) permutes Fqn if and only if ker(ϕ)∩ker(ψ) = {0} and h(x)ϕ(x) induces a bijection from ψ(Fqn) to ¯ ψ(Fqn). 4. Assume in addition that ϕ ◦ψ = 0, and that g(x) restricted to im(ψ) is a permu-tation of im(ψ). Then f(x) = h(ψ(x))ϕ(x) + g(ψ(x)) permutes Fqn if and only if ker(ϕ) ∩ker(ψ) = {0} and ¯ ψ restricted to im(ψ) is a bijection between im(ψ) and im( ¯ ψ). 8.1.71 Theorem [62, 3046] Let q = pe for some positive integer e. 1. If k is an even integer or k is odd and q is even, then fa,b,k(x) = axq + bx + (xq −x)k, a, b ∈Fq2, permutes Fq2 if and only if b −aq ∈F∗ q and a + b ̸= 0. 2. If k and q are odd positive integers, then fa,k(x) = axq + aqx + (xq −x)k, a ∈F∗ q2 and a + aq ̸= 0, permutes Fq2 if and only if (k, q −1) = 1. 8.1.72 Remark The classes fa,b,k with a, b ∈Fq and k even and fa,k for a ∈Fq and p and k odd were ﬁrst constructed in . The remaining classes were obtained in . For other concrete classes of PPs of additive groups, we refer to [62, 325, 730, 1832, 2003, 3046, 3077]. 8.1.7 Other types of PPs from the AGW criterion 8.1.73 Remark In this subsection, we give several other constructions of PPs that can be obtained by using arbitrary surjective maps λ and ¯ λ in Theorem 8.1.39, instead of using multiplicative or additive group homomorphism. 8.1.74 Theorem Let q be a prime power, let n be a positive integer, and let L1, L2, L3 be Fq-linear polynomials over Fq seen as endomorphisms of (Fqn, +). Let g ∈Fqn[x] be such that g(L3(Fqn)) ⊆Fq. Then f(x) = L1(x) + L2(x)g(L3(x)) is a PP of Fqn if and only if Permutation polynomials 225 1. ker(Fy) ∩ker(L3) = {0}, for any y ∈im(L3), where Fy(x) := L1(x) + L2(x)g(y); and 2. ¯ f(x) := L1(x) + L2(x)g(x) permutes L3(Fqn). 8.1.75 Remark The above result extends some constructions in [325, 730]. 8.1.76 Theorem Let q be any power of the prime number p, let n be any positive integer, and let S be any subset of Fqn containing 0. Let h, k ∈Fqn[x] be polynomials such that h(0) ̸= 0 and k(0) = 0, and let B ∈Fqn[x] be any polynomial satisfying h(B(Fqn)) ⊆S and B(aα) = k(a)B(α) for all a ∈S and all α ∈Fqn. Then the polynomial f(x) = xh(B(x)) is a PP of Fqn if and only if ¯ f(x) = xk(h(x)) induces a permutation of the value set B(Fqn). 8.1.77 Remark The case that S = Fq and k(x) = x2 was considered in . Some examples of B are given in . It is remarked in that Theorem 8.1.76 can be generalized for f(x) = A(x)h(B(x)) and ¯ f(x) = C(x)k(h(x)) where A, C ∈Fqn[x] are polynomials such that B(A(x)) = C(B(x)) with C(0) = 0 and A is injective on B−1(s) for each s ∈B(Fqn), under the similar assumptions h(B(Fqn)) ⊆S \ {0} and B(aα) = k(a)B(α) for all a ∈S and all α ∈Fqn. 8.1.78 Deﬁnition Let S ⊆Fq and let γ, b ∈Fq. Then γ is a b-linear translator with respect to S for the mapping F : Fq − →Fq, if F(x + uγ) = F(x) + ub for all x ∈Fqn and for all u ∈S. 8.1.79 Remark The above deﬁnition is a generalization of the concept of b-linear translator studied in [593, 595, 1817], which deals with the case q = pmn, and S = Fpm. The relaxation on the condition for S to be any subset of Fq provides a much richer class of functions (see examples in ). Using the original deﬁnition of linear translators, several classes of PPs of the form G(x)+γTr(H(x)) are constructed in [593, 595, 1817]. In the case that G is also a PP, it is equivalent to constructing polynomials of the form x + γTr(H′(x)). 8.1.80 Theorem Let S ⊆Fq and F : Fq − →S be a surjective map. Let γ ∈Fq be a b-linear translator with respect to S for the map F. Then for any G ∈Fq[x] which maps S into S, we have that x + γG(F(x)) is a PP of Fq if and only if x + bG(x) permutes S. 8.1.81 Deﬁnition A complete mapping f of Fq is a permutation polynomial f(x) of Fq such that f(x) + x is also a permutation polynomial of Fq. 8.1.82 Corollary [62, 593] Under the conditions of Theorem 8.1.80, we have 1. If G(x) = x then x + γF(x) is a PP of Fq if and only if b ̸= −1. 2. If q is odd and 2S = S, then x + γF(x) is a complete mapping of Fq if and only if b ̸∈{−1, −2}. 8.1.83 Theorem Let L : Fqn − →Fqn be an Fq-linear mapping of Fqn with kernel αFq, α ̸= 0. Suppose α is a b-linear translator with respect to Fq for the mapping f : Fqn − →Fq and h : Fq − →Fq is a permutation of Fq. Then the mapping G(x) = L(x) + γh(f(x)) permutes Fqn if and only if b ̸= 0 and γ does not belong to the image set of L. 8.1.84 Corollary Let t be a positive integer with (t, q −1) = 1, H ∈Fqn[x] and γ, β ∈Fqn. Then the mapping G(x) = xq −x + γ (Tr(H(xq −x)) + βx)t is a PP of Fqn if and only if Tr(γ) ̸= 0 and Tr(β) ̸= 0. 226 Handbook of Finite Fields 8.1.85 Theorem Let A be a ﬁnite ﬁeld, S and ¯ S be ﬁnite sets with #S = # ¯ S such that the maps ψ : A →S and ¯ ψ : A →¯ S are surjective and ψ is additive, i.e., ¯ ψ(x+y) = ¯ ψ(x)+ ¯ ψ(y), for all x, y ∈A. Let g : S →A and f : A →A be maps such that ¯ ψ(f + g ◦ψ) = f ◦ψ and ¯ ψ(g(ψ(x))) = 0 for every x ∈A. Then the map f(x) + g(ψ(x)) permutes A if and only if f permutes A. 8.1.86 Corollary Let n and k be positive integers such that (n, k) = d > 1, and let s be any positive integer with s(qk−1) ≡0 (mod qn−1). Let L1(x) = a0x+a1xqd +· · ·+an/d−1xqn−d be a polynomial with L1(1) = 0 and let L2 ∈Fq[x] be a linearized polynomial and g ∈Fqn[x]. Then f(x) = (g(L1(x)))s + L2(x) permutes Fqn if and only if L2 permutes Fqn. 8.1.87 Corollary Let n and k be positive integers such that (n, k) = d > 1, let s be any positive integer with s(qk −1) ≡0 (mod qn −1). Then h(x) = (xqk −x + δ)s + x permutes Fqn for any δ ∈Fqn. 8.1.88 Remark More classes of PPs of the form (xqk −x + δ)s + L(x) and their generalization can be found in [1061, 3058]. See [1481, 3043, 3044, 3057] for more classes of PPs of the form (xpk +x+δ)s +L(x). One can also ﬁnd several classes of PPs of the form xd +L(x) over F2n in [1929, 2363, 2362]. It is proven in that, under the assumption gcd(d, 2n −1) > 1, if xd + L(x) is a PP of F2n then L must be a PP of F2n. Hence some of these classes of PPs of the form xd + L(x) are compositional inverses of PPs of the form L1(x)d + x. 8.1.89 Remark See [872, 3045] for some explicit classes of PPs over F3m. 8.1.8 Dickson and reversed Dickson PPs 8.1.90 Remark The permutational behavior of Dickson polynomials of the ﬁrst kind is simple and classical; see Remark 8.1.5. For more information on Dickson polynomials, see Section 9.6. 8.1.91 Deﬁnition For any positive integer n, let En(x, a) be the Dickson polynomial of the second kind (DPSK) deﬁned by En(x, a) = ⌊n/2⌋ X i=0 n −i i (−a)ixn−2i. 8.1.92 Remark Matthews observed in his Ph.D. thesis that if q is a power of an odd prime p and n satisﬁes the system of congruences n + 1 ≡±2 (mod p), n + 1 ≡±2 (mod 1 2(q −1)), n + 1 ≡±2 (mod 1 2(q + 1)), (8.1.4) then En is a PP of Fq. However, the above is not a necessary condition in general. When p = 3 or 5 and q is composite there are examples of DPSK En known which are PP for which (8.1.4) does not hold, see [1592, 2175]. Moreover, there are several papers concentrating on the permutational behavior of DPSK over ﬁnite ﬁelds with small characteristics including characteristic two and general a which is not necessarily ±1 [733, 1482, 1483, 1484]. On the other hand, when q = p or p2 and a = 1, Cipu and Cohen proved that the condition (8.1.4) is also necessary [652, 653, 679, 680]. Permutation polynomials 227 8.1.93 Deﬁnition For any positive integer n, the reversed Dickson polynomial, Dn(a, x), is deﬁned by Dn(a, x) = ⌊n/2⌋ X i=0 n n −i n −i i (−x)ian−2i. 8.1.94 Remark It is easy to check that Dn(0, x) = 0 if n is odd and Dn(0, x) = 2(−x)k if n = 2k. Hence Dn(0, x) is a PP of Fq if and only if q is odd, n = 2k, and (k, q −1) = 1. For a ̸= 0, we can also check that Dn(a, x) = anDn(1, x/a2). Hence Dn(a, x) is a PP of Fq where a ̸= 0 if and only if Dn(1, x) is a PP of Fq. If Dn(1, x) is a PP of Fq, then (q, n) is a desirable pair. 8.1.95 Deﬁnition A mapping f from Fq to Fq is almost perfect nonlinear (APN) if the diﬀerence equation f(x + a) −f(x) = b has at most two solutions for any ﬁxed a ̸= 0, b ∈Fq. 8.1.96 Remark We refer to Section 9.2 for more information on APN functions and their applica-tions. 8.1.97 Theorem Let q = pe with p a prime and e > 0. If p = 2 or p > 3 and n is odd, then xn is an APN on Fq2 implies that Dn(1, x) is a PP of Fq, which also implies that xn is an APN over Fq. 8.1.98 Theorem The pair (pe, n) is a desirable pair in each of the following cases: p e n 2 2k + 1, (k, 2e) = 1 [1295, 1547] 2 22k −2k + 1, (k, 2e) = 1 [1547, 1690] 2 even 2e + 2k + 1, k > 0, (k −1, e) = 1 2 5k 28k + x6k + 24k + 22k −1 [906, 1547] 3 (3k + 1)/2, (k, 2e) = 1 3 even 3e + 5 5 (5k + 1)/2, (k, 2e) = 1 ≥3 pk + 1, k ≥0, pk ≡1 (mod 4), v2(e) ≤v2(k) ≥3 pe + 2, k ≥0, pe ≡1 (mod 3) [1479, 1547] ≥5 3 8.1.99 Remark Two pairs (q, n1) and (q, n2), where n1 and n2 are positive integers, are equivalent if n1 and n2 are in the same p-cyclotomic coset modulo q2 −1, i.e., Dn1(1, x) ≡Dn2(1, x) (mod xq −x). No desirable pairs outside the ten families (up to equivalence) given in Theo-rem 8.1.98 are known. There are several papers on necessary conditions for a reverse Dickson polynomial to be a PP [1544, 1545]. In particular, it is proved in that (pe, n) is a desirable pair if and only if fn(x) = P j≥0 n 2j xj is a PP of Fpe. However, it is not known whether the above classes are the only non-equivalent desirable pairs. Several new classes of reversed Dickson polynomials can be found in [1542, 1544]. 8.1.100 Remark Dickson polynomials are used to prove the following class of PPs. 8.1.101 Theorem Let m ≥1 and 1 ≤k, r ≤m −1 be positive integers satisfying that kr ≡1 (mod m). Let q = 2m, σ = 2k, and Tr(x) := x + x2 + · · · + x2m−1. For α, γ in {0, 1}, we deﬁne Hα,γ(x) := γTr(x) + (αTr(x) + Pr−1 i=0 xσi)σ+1 x2 . Then Hα,γ(x) is a PP of F2m if and only if r + (α + γ)m ≡1 (mod 2). 228 Handbook of Finite Fields 8.1.9 Miscellaneous PPs 8.1.102 Remark Cyclic and Dickson PPs play a vital role in the Schur Conjecture from 1922 which postulated that if f is a polynomial with integer coeﬃcients which is a PP of Fp (when considered modulo p) for inﬁnitely many primes p, then f must be a composition of binomials axn +b and Dickson polynomials. This has been shown to be true; see , the notes to Chapter 7 of and . A proof without the use of complex analysis can be found in . More generally, a matrix analogue of the Schur conjecture was studied in . Let Fm×m q denote the ring of m × m matrices over the ﬁnite ﬁeld Fq. A polynomial f ∈Fq[x] is a permutation polynomial (PP) on Fm×m q if it gives rise to a permutation of Fm×m q . Using a characterization of PPs of the matrix ring over ﬁnite ﬁelds Fq in , it is shown by Mullen in that any polynomial f with integral coeﬃcients which permutes the matrices of ﬁxed size over a ﬁeld of p elements for inﬁnitely many p is a composition of linear polynomials and Dickson polynomials Dn(x, a) with n ̸= 3 an odd prime and a ̸= 0 an integer. Several related questions are also addressed in ; see Section 9.7 for more information on Schur’s conjecture. 8.1.103 Remark Let f be an integral polynomial of degree n ≥2. Cohen proved one of the Chowla and Zassenhaus conjectures (concerning irreducible polynomials), which postulated that if f is a PP over Fp of degree n modulo p for any p > (n2 −3n + 4)2, then f(x) + cx is not a PP of Fp unless c = 0. This shows that if both f and g are integral PPs of degree n ≥2 over a large prime ﬁeld, then their diﬀerence h = f −q cannot be a linear polynomial cx where c ̸= 0. Suppose that t ≥1 denotes the degree of h. More generally, it is proved in that t ≥3n/5. Moreover, if n ≥5 and t ≤n −3 then (t, n) > 1. Roughly speaking, two PPs over a large prime ﬁeld Fp cannot diﬀer by a polynomial with degree less than 3n/5. 8.1.104 Remark Evans considers orthomorphisms, mappings θ with θ(0) = 0 so that θ and θ(x)−x are both PPs of Fq. He studies connections between orthomorphisms, latin squares, and aﬃne planes. A map θ is an orthomorphism if and only if θ(x) −x is a complete map-ping. Complete mappings of small degrees and existence of complete mappings (in partic-ular, binomials) are studied in . Enumeration results for certain types of cyclotomic orthomorphisms are provided in . It is proved in for odd q and in for even q that the degree of a complete mapping is at most q −3. It is known that families of permutation polynomials of the form f(x) + cx can be used in the construction of maximal sets of mutually orthogonal Latin squares [997, 2918]. Let C(f) be the number of c in Fq such that f(x) + cx is a permutation polynomial over Fq. Cohen’s theorem on the Chowla-Zassenhaus conjecture shows that C(f) ≤1 if the degree n of f is not divisible by p and q is suﬃciently large compared to n. Chou showed that C(f) ≤q −1 −n in his thesis . Then Evans, Greene, and Niederreiter proved C(f) ≤q −q−1 n−1 in , which also proves a conjecture of Stothers when q is prime. In the case that q is an odd prime, it gives the best possible result C(f) ≤(q −3)/2 for polynomials of the form x(q+1)/2 + cx. A general bound for C(f) which implies Chou’s bound was obtained by Wan, Mullen, and Shiue in , as well as a signiﬁcant bound C(f) ≤r where r is the least nonnegative residue of q −1 modulo n under certain mild conditions. It is conjectured in that f(x) −f(0) is a linearized p-polynomial over Fq if C(f) ≥⌊q/2⌋; this was proved to be true for q = p or any monomial f(x) = xe in . Wan observed that this conjecture holds also for q = p2 from the results in [624, 1016]. Several other related results on the function C(f) can be found in [997, 2826, 2912]. 8.1.105 Remark Results on the cycle structure of monomials and of Dickson polynomials can be found in and , respectively. Cycle decomposition, in particular, decomposition of Permutation polynomials 229 them into cycles of the same length (which are motivated by Turbo codes [2742, 2769]), are studied in [44, 2152, 2296, 2495, 2496, 2519]. Moreover, we refer readers to [68, 575, 625] for cycle structure of permutation polynomials with small Carlitz rank or with full cycles. 8.1.106 Remark Finding the compositional inverse of a PP is a hard problem except for the triv-ial well known classes such as the inverses of linear polynomial, monomials, and Dickson polynomials. There are several papers on the explicit format of the inverses of some special classes of permutation polynomials, for example, [727, 1817, 2205, 2206, 2942]. Because the problem is equivalent to ﬁnding the inverses of PPs of the form xrf(xs), the most general result can be found in . 8.1.107 Remark PPs are related to special functions. For example, Dobbertin constructed several classes of PPs [903, 904] over ﬁnite ﬁelds of even characteristic and used them to prove several conjectures on APN monomials. The existence of APN permutations on F22n is a long-term open problem in the study of vectorial Boolean functions. Hou proved that there are no APN permutations over F24 and there are no APN permutations on F22n with coeﬃcients in F2n. Only recently the authors in found the ﬁrst APN permutation over F26. However, the existence of APN permutations on F22n for n ≥4 remains open. Over ﬁnite ﬁelds of odd characteristics, a function f is a planar function if f(x + a) −f(x) is a PP for each nonzero a; see Sections 9.2 and 9.5. In , Ding and Yuan constructed a new family of planar functions over F3m, where m is odd, and then obtained the ﬁrst examples of skew Hadamard diﬀerence sets, which are inequivalent to classical Paley diﬀerence sets. Permutation polynomials of F32h+1 obtained from the Ree-Tits slice symplectic spreads in PG(3, 32h+1) were studied in . Later on, they were used in to construct a family of skew Hadamard diﬀerence sets in the additive group of this ﬁeld. For more information on these special functions and their applications, we refer the readers to Sections 9.2, 9.5, and 14.6. 8.1.108 Remark Golomb and Moreno show that PPs are useful in the construction of circular Costas arrays, which are useful in sonar and radar communications. They gave an equivalent conjecture for circular Costas arrays in terms of permutation polynomials and provided some partial results. The connection between Costas arrays and APN permutations of integer rings Zn was studied in . Composed with discrete logarithms, permutation polynomials of ﬁnite ﬁelds are used to produce permutations of integer rings Zn with optimum ambiguity and deﬁciency [2353, 2355], which generate APN permutations in many cases. Earlier results on PPs of Zn can be found in Section 5.6 of and [2171, 2460, 3062]. See Also §8.2 For discussion of PPs in several variables. §8.3 For value sets of polynomials. §8.4 For exceptional polynomials over ﬁnite ﬁelds. §9.2 For discussion of PN and APN functions. §9.5 For studies of planar functions. §9.6 For Dickson polynomials over ﬁnite ﬁelds. §9.7 For connections to Schur’s Conjecture. For connections with monomial graphs. For connections with check digit systems. References Cited: [44, 60, 61, 62, 63, 64, 65, 68, 192, 325, 396, 424, 504, 551, 569, 575, 593, 230 Handbook of Finite Fields 595, 624, 625, 632, 652, 653, 667, 676, 677, 679, 680, 681, 688, 699, 727, 730, 733, 771, 840, 870, 871, 872, 874, 902, 903, 904, 906, 917, 997, 998, 1016, 1061, 1109, 1120, 1221, 1295, 1305, 1479, 1481, 1482, 1483, 1484, 1527, 1541, 1542, 1544, 1545, 1547, 1592, 1690, 1716, 1787, 1788, 1813, 1817, 1832, 1833, 1916, 1929, 1933, 1934, 1935, 1936, 1939, 1983, 1984, 1995, 1996, 1997, 2003, 2018, 2019, 2020, 2033, 2034, 2127, 2152, 2171, 2175, 2176, 2178, 2192, 2205, 2206, 2268, 2278, 2296, 2353, 2355, 2359, 2362, 2363, 2460, 2495, 2496, 2519, 2603, 2605, 2638, 2680, 2681, 2729, 2742, 2769, 2825, 2826, 2827, 2904, 2905, 2908, 2909, 2911, 2912, 2913, 2916, 2918, 2919, 2927, 2940, 2941, 2942, 2943, 2967, 3043, 3044, 3045, 3046, 3047, 3057, 3058, 3062, 3065, 3075, 3076, 3077] 8.2 Several variables Rudolf Lidl, University of Tasmania Gary L. Mullen, The Pennsylvania State University 8.2.1 Deﬁnition A polynomial f ∈Fq[x1, . . . , xn] is a permutation polynomial in n variables over Fq if the equation f(x1, . . . , xn) = α has exactly qn−1 solutions in Fn q for each α ∈Fq. 8.2.2 Remark A permutation polynomial f(x1, . . . , xn) induces a mapping from Fqn to Fq but does not induce a 1 −1 mapping unless n = 1. 8.2.3 Remark A combinatorial computation shows that there are (qn)!/((qn−1)!)q permu-tation polynomials in n variables over Fq. 8.2.4 Deﬁnition A set of polynomials fi(x1, . . . , xn) ∈Fq[x1, . . . , xn], 1 ≤i ≤r, forms an orthogonal system in n variables over Fq if the system of equations fi(x1, . . . , xn) = αi, 1 ≤i ≤r, has exactly qn−r solutions in Fn q for each (α1, . . . , αr) ∈Fr q. 8.2.5 Theorem For every orthogonal system f1, . . . , fm ∈Fq[x1, . . . , xn], 1 ≤m < n, over Fq and every r, 1 ≤r ≤n −m, there exist fm+1, . . . , fm+r ∈Fq[x1, . . . , xn] so that f1, . . . , fm+r forms an orthogonal system over Fq. 8.2.6 Theorem The system f1, . . . , fm, 1 ≤m ≤n, is orthogonal over Fq if and only if X (c1,...,cn)∈Fn q χb1(f1(c1, . . . , cn)) · · · χbm(fm(c1, . . . , cn)) = 0 for all additive characters χb1, . . . , χbm of Fq with (b1, . . . , bm) ̸= (0, . . . , 0). 8.2.7 Corollary A polynomial f ∈Fq[x1, . . . , xn] is a permutation polynomial over Fq if and only if X (c1,...,cn)∈Fn q χ(f(c1, . . . , cn)) = 0 for all nontrivial additive characters χ of Fq. 8.2.8 Theorem A set of polynomials f1, . . . , fr in n variables over Fq forms an orthogonal system over Fq if and only if the polynomial b1f1 + · · · + brfr is a permutation polynomial for each (b1, . . . , br) ̸= (0, . . . , 0) ∈Fr q. Permutation polynomials 231 8.2.9 Theorem Suppose f ∈Fq[x1, . . . , xn] is of the form f(x1, . . . , xn) = g(x1, . . . , xm) + h(xm+1, . . . , xn), 1 ≤m < n. If at least one of g and h is a permutation polynomial over Fq then f is a permutation polynomial over Fq. If q is prime, then the converse holds. 8.2.10 Theorem Let f1, . . . , ft be polynomials in disjoint sets of variables, where fi is a polynomial in vi variables, i = 1, . . . , t. Then f1 + · · · + ft is a permutation polynomial over Fq, q = pe, if and only if, for every subgroup H of the additive group G of Fq of order pe−1, there is an fi which distributes Fvi q uniformly over the cosets of H in G. 8.2.11 Theorem If q is not prime then for 1 ≤m < n there exist polynomials g(x1, . . . , xm) and h(xm+1, . . . , xn) over Fq such that g(x1, . . . , xm) + h(xm+1, . . . , xn) is a permutation polynomial over Fq but neither g(x1, . . . , xm) nor h(xm+1, . . . , xn) is a permutation poly-nomial over Fq. 8.2.12 Theorem If n = mk for positive integers n, m, k, there is a 1-1 correspondence between orthogonal systems over Fq consisting of m polynomials over Fq of degree less than q in each of their n variables and permutation polynomials over Fqm of degree less than qm in each of their k variables. 8.2.13 Corollary There is a 1-1 correspondence between orthogonal systems over Fq con-sisting of n polynomials over Fq of degree less than q in each of their n variables and permutation polynomials in one variable over Fqn of degree less than qn. 8.2.14 Remark Dickson permutation polynomials are studied in Section 8.1 and in Section 9.6 where results related to them in both one as well as several variables are provided. 8.2.15 Theorem 1. For a ∈F∗ q the system gk(a) (see Deﬁnition 9.6.51) is orthogonal over Fq if and only if (k, qs −1) = 1 for s = 1, . . . , n + 1. 2. The system gk(0) is orthogonal if and only if (k, qs −1) = 1 for s = 1, . . . , n. 8.2.16 Deﬁnition Two polynomials are equivalent if one can be transformed into the other by a transformation of the form xi = Pn j=1 aijyj + bi, 1 ≤i ≤n, where aij, bi ∈Fq and the matrix (aij) is nonsingular. 8.2.17 Theorem Let f ∈Fq[x1, . . . , xn] with the degree of f at most two and n ≥2. For q odd, f is a permutation polynomial over Fq if and only if f is equivalent to a polynomial of the form g(x1, . . . , xn−1) + xn for some g. For q even, f is a permutation polynomial over Fq if and only if f is equivalent to a polynomial of the form g(x1, . . . , xn−1) + xn or g(x1, . . . , xn−1) + x2 n. 8.2.18 Theorem If q is odd and n ≥2, then f(x1, . . . , xn) of degree at most two is a non-singular feedback function if and only if f(x1, . . . , xn) = cx1 + f0(x2, . . . , xn), c ∈Fq where f0(x2, . . . xn) is any polynomial in the variables x2, . . . , xn of degree at most two over Fq. 8.2.19 Remark Niederreiter provides criteria for quadratic polynomials over Z to be per-mutation polynomials modulo p, i.e., permutation polynomials over Fp, that involve the rank of a matrix of coeﬃcients. A result similar to Theorem 8.2.18 is obtained in for q even; see the corresponding result for quadratic forms in Theorem 8.2.17. These results provide an application of several variable permutation polynomials. 8.2.20 Remark Results are given on permutation polynomials in several variables over a ﬁnite ﬁeld and orthogonal systems of polynomials whose image spaces are allowed to be 232 Handbook of Finite Fields arbitrary subﬁelds of the ﬁnite ﬁeld. The results developed in this paper are used to con-struct additional complete sets of frequency squares, rectangles and hyperrectangles, and orthogonal arrays. Some of the theorems present a relationship between permutation poly-nomials of a ﬁnite ﬁeld and ﬁeld permutation functions, an implicit bound on the possible number of functions in an orthogonal ﬁeld system, and results related to the generation of orthogonal ﬁeld systems. 8.2.21 Remark For ﬁnite rings there are two concepts to distinguish: permutation polynomials (as above) and strong permutation polynomials. The latter are deﬁned via the cardinality of the inverse image. Polynomials are strong permutation polynomials (or strong orthogonal systems) in n variables if they can be completed to an orthogonal system of n polynomials in n variables. 8.2.22 Theorem If R is a local ring whose maximal ideal has a minimal number m of generators, then for every n > m there exists a permutation polynomial in n variables that is not a strong permutation polynomial. 8.2.23 Corollary Every permutation polynomial in any number of variables over a local ring R is strong if and only if R is a ﬁnite ﬁeld. 8.2.24 Remark Wei and Zhang showed that if n ≤m in the setting of Theorem 8.2.22, then every orthogonal system of k polynomials in n variables can be completed to an orthogonal system of n polynomials (and in particular, every permutation polynomial is strong). See Also §9.4 Considers κ-polynomials used for constructions of semiﬁelds. §14.1 Discusses orthogonal latin squares and hypercubes. Section 7.5 discusses permutations and orthogonal systems in several variables. Considers bounds for value sets of polynomial vectors in several variables. Considers an application of permutation polynomials and orthogonal systems to pseudorandom number generation. Considers permutation polynomials over ﬁnite commutative rings. References Cited: [542, 1132, 1937, 1939, 1940, 2032, 2172, 2174, 2227, 2228, 2229, 2326, 2736, 2957, 3062] 8.3 Value sets of polynomials Gary L. Mullen, The Pennsylvania State University Michael E. Zieve, University of Michigan 8.3.1 Deﬁnition For f ∈Fq[x], the value set of f is the set Vf = {f(a)\|a ∈Fq}; the cardinality of Vf is denoted by #Vf. Permutation polynomials 233 8.3.2 Remark Every subset of Fq occurs as Vf for some f ∈Fq[x] of degree at most q −1 (by the Lagrange Interpolation Formula); see Theorem 2.1.131. 8.3.1 Large value sets 8.3.3 Remark Any f ∈Fq[x] satisﬁes #Vf ≤q; equality occurs precisely when f is a permutation polynomial; see Section 8.1. 8.3.4 Theorem Suppose f ∈Fq[x] of degree n is not a permutation polynomial. Then: 1. [2826, 2911, 2982] #Vf ≤q −⌈(q −1)/n⌉. 2. [1086, 1367, 1368] If #Vf ̸= (1−1/n)q and n > 5 then #Vf ≤(1−2/n)q+On(√q). 3. If gcd(n, q) = 1 then #Vf ≤(5/6)q + On(√q). 8.3.5 Example Let q = rk where r is a prime power and k is a positive integer. Then f(x) := xr + xr−1 satisﬁes #Vf = q −q/r, and hence achieves equality in (1). 8.3.2 Small value sets 8.3.6 Remark If f ∈Fq[x] has degree n, then #Vf ≥⌈q/n⌉(since each α ∈Fq has at most n preimages under f). 8.3.7 Deﬁnition A polynomial f ∈Fq[x] of degree n is a minimal value set polynomial if #Vf = ⌈q/n⌉. 8.3.8 Theorem Let f ∈Fp[x] have degree n, where p is prime. If n < p and #Vf = ⌈p/n⌉≥3, then n divides p −1 and f(x) = a(x + b)n + c with a, b, c ∈Fp. 8.3.9 Theorem Let f ∈Fq[x] be monic of degree n, where gcd(n, q) = 1 and n ≤√q. If #Vf = ⌈q/n⌉, then n divides q −1 and f(x) = (x + b)n + c with b, c ∈Fq. 8.3.10 Problem Determine all minimal value set polynomials over Fpk. This is done for k ≤2 in . 8.3.11 Remark Minimal value set polynomials whose values form a subﬁeld are characterized in . A connection between minimal value set polynomials and Frobenius non-classical curves is given in . 8.3.12 Theorem [628, 1308] If f(x) ∈Fq[x] is monic of degree n > 15, where n4 < q and #Vf < 2q/n, then f(x) has one of the forms: 1. (x + a)n + b, where n \| (q −1); 2. ((x + a)n/2 + b)2 + c, where n \| (q2 −1); 3. ((x + a)2 + b)n/2 + c, where n \| (q2 −1). 8.3.13 Theorem Let f ∈Fp[x] have degree less than 3 4(p−1), where p is prime. If #f(F∗ p) = 2 then f is a polynomial in x(p−1)/d for some d ∈{2, 3}. 8.3.14 Remark This result indicates that some phenomena become apparent only when one con-siders #f(F∗ p) rather than #Vf. 234 Handbook of Finite Fields 8.3.3 General polynomials 8.3.15 Theorem [285, 2981] Fix n, and let en := Pn j=1(−1)j−1/j!. There is a constant an such that, for each q, there are qn +On(qn−1) monic polynomials f ∈Fq[x] of degree n satisfying \|#Vf −enq\| ≤an√q. 8.3.16 Remark The previous result says that if q is large compared to n then most polynomials over Fq of degree n take approximately enq values. Note that en →1 −1/e as n →∞. 8.3.17 Theorem For ﬁxed n, there is a ﬁnite set Tn of rational numbers such that: for any q, and any f ∈Fq[x] of degree n, there is an element cf ∈Tn such that #Vf = cfq +On(√q). 8.3.18 Remark The set Tn may be chosen to be {a/n!: (n −1)! ≤a ≤n!}. 8.3.19 Remark For ﬁxed n, if q is large and f ∈Fq[x] has degree n, then #Vf/q lies in a tiny interval around a member of a ﬁnite set; crucially, this ﬁnite set depends only on n, and not on q. 8.3.20 Theorem Let f ∈Fq[x] have degree n, and write f = g(xpj) where j ≥0 and g ∈Fq[x] \ Fq[xp]; here p is the characteristic of Fq. Let t be transcendental over Fq, and let A and G be the Galois groups of g(x) −t over Fq(t) and ¯ Fq(t), respectively, where ¯ Fq denotes an algebraic closure of Fq. Then G is a normal subgroup of A, and A/G is cyclic. The quantity cf in Theorem 8.3.17 may be taken to be the proportion of elements in a generating coset of A/G which ﬁx at least one of the roots of g(x) −t. 8.3.21 Example Let f ∈Fq[x] have degree n. 1. If n = 2 then #Vf ∈{q/2, (q + 1)/2, q}. 2. If n = 3 then #Vf ∈{q/3, (q + 2)/3, (2q −1)/3, 2q/3, (2q + 1)/3, q}. 3. If n = 4 and q is an odd prime then #Vf is either (q + 3)/4, (q + 1)/2, (3q + 4 + i)/8 with ±i ∈{1, 3, 5}, or 5q/8 + O(√q). 8.3.4 Lower bounds 8.3.22 Theorem If µq(f) is the smallest positive integer i so that P a∈Fq(f(a))i ̸= 0, then #Vf ≥µq(f) + 1. 8.3.23 Remark Assume that for a polynomial f the degree n of f satisﬁes n < q −1. Write (f(x))i = Pq−1 j=0 aij mod (xq −x). Let Af be the matrix Af = (aq−1 ij ), for 1 ≤i, j ≤q −1 so that the (i, j)-th entry of Af is 1 if the coeﬃcient of xj in (f(x))i mod (xq −x) is nonzero. If f is not the zero polynomial, then Af has at least one nonzero column. If the j-th column of Af consists entirely of 0s or entirely of 1s, set lj = 0. Otherwise, for a nonzero j-th column of Af, arrange the entries in a circle and deﬁne lj to be the maximum number of consecutive zeros appearing in this circular arrangement. Let Lf be the maximum of the values of lj, where the maximum is taken over all of the q −1 columns of the matrix Af. 8.3.24 Theorem With notation as above, \|Vf\| ≥Lf + 2. 8.3.25 Remark If f is a polynomial over Fq and A ′ f is the matrix from Remark 8.3.23 without the (q −1)-st powers, i.e., the matrix A ′ f = (aij), then #Vf = 1 + rank(A ′ f). 8.3.26 Corollary Since \|Vf\| ≥lq−1 + 2, we have the result of Theorem 2.1 of . 8.3.27 Remark The Hermite/Dickson criterion from Section 8.1 is essentially equivalent to the ﬁrst q −2 consecutive elements of the last column of the matrix Af being 0, with the last Permutation polynomials 235 element being 1. Thus f is a permutation polynomial if and only if Lf = q −2. 8.3.28 Remark See for further inequalities. 8.3.5 Examples 8.3.29 Theorem #Vxn = 1 + (q −1)/(n, q −1). 8.3.30 Deﬁnition The Dickson polynomial of degree n and parameter a ∈Fq is deﬁned by Dn(x, a) = ⌊n/2⌋ X i=0 n n −i n −i i (−a)ixn−2i. 8.3.31 Remark See Section 9.6 for a discussion of Dickson polynomials over Fq. 8.3.32 Theorem Suppose q is odd with 2r \|\| (q2 −1). Then for each n ≥1, and each a ∈F∗ q, #VDn(x,a) = q −1 2(n, q −1) + q + 1 2(n, q + 1) + α, where α = 1 if 2r−1 \|\| n; α = 1/2 if 2t \|\| n and 1 ≤t ≤r −2; α = 0 otherwise. Here η denotes the quadratic character deﬁned by η(0) = 0, η(a) = 1 if a is a square in Fq and η(a) = −1 if a is a nonsquare in Fq. 8.3.33 Remark If (n1, q2 −1) = (n2, q2 −1), then #VDn1(x,a) = #VDn2(x,a). 8.3.34 Theorem Suppose q is even. Then for each n ≥1, and each a ∈F∗ q, #VDn(x,a) = q −1 2(n, q −1) + q + 1 2(n, q + 1) 8.3.35 Remark For further examples, see for instance [757, 758]. 8.3.6 Further value set papers 8.3.36 Remark There are many other papers describing results concerning value sets of polynomi-als over ﬁnite ﬁelds; however, for lack of space, we are unable to precisely state them. Pages 379-381 of provide a wealth of descriptions of older papers dealing with value sets; pages 388-389 of provide summaries of value set results for polynomials in several variables. The paper presents the state of knowledge about value sets as of 1995. 8.3.37 Remark Since the publication of , in are given formulas for the number of polynomials of degree q −1 with a value set of cardinality k. Paper describes values sets of diagonal equations over ﬁnite ﬁelds by giving a new proof of the Cauchy-Davenport theorem. See and for a discussion of polynomials over F2n which take on each nonzero value only a small number of times (at most six). 8.3.38 Remark Paper shows that if f is not a permutation polynomial over Fq and q ≥n4, then #Vf < q −q/(2n), while shows that by using the polynomial (x + 1)xq−1, Wan’s bound is sharp for every extension of the base ﬁeld. The paper discusses results concerning the size of the intersection of the value sets of two nonconstant polynomials and discusses lower bounds for the size of the value set for the polynomial (xm + b)n improving the bound given in . Paper discusses cardinalities of value sets for 236 Handbook of Finite Fields diagonal kinds of polynomials in several variables where the preimage points come from subsets of the ﬁeld rather than the entire ﬁeld. See Also §8.1 Discusses permutation polynomials in one variable. §8.2 Discusses permutation polynomials in several variables. §8.4 Considers exceptional polynomials. Considers polynomials whose value sets lie in a subﬁeld. Studies value sets as they relate to Dembowski-Ostrom and planar polynomials. Considers bounds for value sets of polynomial vectors in several variables. References Cited: [57, 285, 289, 350, 351, 549, 628, 629, 630, 667, 734, 757, 758, 760, 772, 773, 774, 1086, 1222, 1307, 1308, 1367, 1368, 1458, 1939, 2043, 2105, 2188, 2746, 2826, 2911, 2919, 2981, 2982] 8.4 Exceptional polynomials Michael E. Zieve, University of Michigan 8.4.1 Fundamental properties 8.4.1 Deﬁnition An exceptional polynomial over Fq is a polynomial f ∈Fq[x] which is a permu-tation polynomial on Fqm for inﬁnitely many m. 8.4.2 Remark If f ∈Fq[x] is exceptional over Fqk for some k, then f is exceptional over Fq. 8.4.3 Deﬁnition A polynomial F(x, y) ∈Fq[x, y] is absolutely irreducible if it is irreducible in ¯ Fq[x, y], where ¯ Fq is an algebraic closure of Fq. 8.4.4 Theorem A polynomial f ∈Fq[x] is exceptional over Fq if and only if every absolutely irreducible factor of f(x) −f(y) in Fq[x, y] is a constant times x −y. 8.4.5 Corollary If f ∈Fq[x] is exceptional, then there are integers 1 < e1 < e2 < · · · < ek such that: f is exceptional over Fqn if and only if n is not divisible by any ei. 8.4.6 Corollary If f ∈Fq[x] is exceptional, then there is an integer M > 1 such that f permutes each ﬁeld Fqm for which m is coprime to M. 8.4.7 Corollary For g, h ∈Fq[x], the composition g ◦h is exceptional if and only if both g and h are exceptional. Permutation polynomials 237 8.4.8 Deﬁnition A polynomial f ∈Fq[x] is indecomposable if it cannot be written as the com-position f = g ◦h of two nonlinear polynomials g, h ∈Fq[x]. 8.4.9 Corollary A polynomial f ∈Fq[x] is exceptional if and only if it is the composition of indecomposable exceptional polynomials. 8.4.2 Indecomposable exceptional polynomials 8.4.10 Theorem Let f be an indecomposable exceptional polynomial over Fq of degree n, and let p be the characteristic of Fq. Then either 1. n is coprime to p, or 2. n is a power of p, or 3. n = pr(pr−1) 2 where r > 1 is odd and p ∈{2, 3}. 8.4.11 Theorem [1755, 2192] The indecomposable exceptional polynomials over Fq of degree co-prime to q are precisely the polynomials of the form ℓ1 ◦f ◦ℓ2 where ℓ1, ℓ2 ∈Fq[x] are linear and either 1. f(x) = ax + b with a ∈F∗ q and b ∈Fq, or 2. f(x) = xn where n is a prime which does not divide q −1, or 3. f(x) = Dn(x, a) (a Dickson polynomial) where a ∈F∗ q and n is a prime which does not divide q2 −1. 8.4.12 Theorem [1372, 1374] The indecomposable exceptional polynomials over Fq of degree s(s −1)/2, where s = pr > 3 and q = pm with p prime, are precisely the polynomials of the form ℓ1 ◦f ◦ℓ2 where ℓ1, ℓ2 ∈Fq[x] are linear, r > 1 is coprime to 2m, and f is one of the following polynomials: 1. x−s T(axe)(s+1)/e where p = 2, T(x) = xs/2 + xs/4 + · · · + x, e \| (s + 1), and a ∈F∗ q, 2. T(x) + a x s · T(x) + T(x) + a a + 1 · T x(a2 + a) (T(x) + a)2 where p = 2, a ∈Fq \ F2, and T(x) = xs/2 + xs/4 + · · · + x, 3. x(x2e −a)(s+1)/(4e) (x2e −a)(s−1)/2 + a(s−1)/2 x2e (s+1)/(2e) where p = 3, e s+1 4 , and a ∈F∗ q is an element whose image in F∗ q/(F∗ q)2e has even order. 8.4.13 Remark The proofs of Theorems 8.4.10 and 8.4.12 rely on the classiﬁcation of ﬁnite simple groups. 8.4.14 Theorem [1120, 1755] For prime p, the degree-p exceptional polynomials over Fpm are pre-cisely the polynomials ℓ1◦f ◦ℓ2 where ℓ1, ℓ2 ∈Fpm[x] are linear and f(x) = x(x(p−1)/r −a)r with r \| (p −1) and a ∈Fpm such that ar(pm−1)/(p−1) ̸= 1. 8.4.15 Proposition [674, 840] Let L be a linearized polynomial (i.e., L(x) = Pd i=0 aixpi with ai ∈Fpm), and let S(x) = xjH(x)k where H ∈Fpm[x] satisﬁes L(x) = xjH(xk). Then S is exceptional over Fpm if and only if S has no nonzero roots in Fpm. 238 Handbook of Finite Fields 8.4.16 Proposition Let s = pr where p is an odd prime. If a ∈Fpm is not an (s −1)-th power, then (xs −ax −a) · (xs −ax + a)s + (xs −ax + a)2 + 4a2x (s+1)/2 2xs is an indecomposable exceptional polynomial over Fpm. 8.4.17 Proposition Let s = 2r. If a ∈F2m is not an (s −1)-th power, then (xs + ax + a)s+1 xs · xs + ax xs + ax + a + T a2x (xs + ax + a)2 is an indecomposable exceptional polynomial over F2m, where T(x) = xs/2 + xs/4 + · · · + x. 8.4.18 Remark The previous three Propositions describe all known indecomposable exceptional polynomials over Fpm of degree pr with r > 0, up to composing on both sides with linear polynomials. It is expected that there are no further examples. Theorem 8.4.14 shows this when r = 1, and [1371, 2126] show it under diﬀerent hypotheses. 8.4.3 Exceptional polynomials and permutation polynomials 8.4.19 Theorem A permutation polynomial over Fq of degree at most q1/4 is exceptional over Fq. 8.4.20 Remark A weaker version of Theorem 8.4.19 was proved in ; the stated result is obtained from the same proof by using the fact that an absolutely irreducible degree-d bivariate polynomial over Fq has at least q + 1 −(d −1)(d −2)√q roots in Fq × Fq. For proofs of this estimate, see [145, 1122, 1886]. A stronger (but false) version of this estimate was stated in , and deduced Theorem 8.4.19 from this false estimate. Finally, states a stronger version of Theorem 8.4.19, but the proof is ﬂawed and when ﬁxed it yields Theorem 8.4.19. 8.4.21 Remark Up to composing with linears on both sides, the only known non-exceptional permutation polynomials over Fq of degree less than √q are x10+3x over F343 and (x+1)N+1 x over F24r−1, where r ≥3 and N = (4r + 2)/3. 8.4.22 Remark Heuristics predict that “at random” there would be no permutation polynomials over Fq of degree less than q 2 log q. 8.4.23 Remark There are no known examples of non-exceptional permutation polynomials over Fq of degree less than q 2 log q when q is prime. 8.4.24 Remark Nearly all known examples of permutation polynomials over Fq of degree less than q 2 log q can be written as the restriction to Fq of a permutation π of an inﬁnite algebraic extension K of Fq, where π is induced by a rational function in the symbols σi(x), with σ being a ﬁxed automorphism of K. Such a permutation π may be viewed as an exceptional rational function over the diﬀerence ﬁeld (K, σ); see [703, 1912, 1913]. 8.4.4 Miscellany 8.4.25 Theorem [543, 3074] Every permutation of Fq is induced by an exceptional polynomial. 8.4.26 Theorem [688, 1369, 1898] Exceptional polynomials over Fq have degree coprime to q −1. Permutation polynomials 239 8.4.27 Remark Theorem 8.4.26 is called the Carlitz–Wan conjecture. It follows from Theo-rems 8.4.10, 8.4.11, and 8.4.12. However, the known proofs of Theorems 8.4.10 and 8.4.12 rely on the classiﬁcation of ﬁnite simple groups, whereas [688, 1369, 1898] present short self-contained proofs of Theorem 8.4.26. 8.4.28 Theorem If f ∈Z[x] is a permutation polynomial over Fp for inﬁnitely many primes p, then f is the composition of linear and Dickson polynomials. 8.4.29 Remark Theorem 8.4.28 was proved in when f has prime degree. It was shown in (conﬁrming an assertion in ) that the full Theorem 8.4.28 follows quickly from the main lemma in together with a group-theoretic result from . A diﬀerent proof of Theorem 8.4.28 appears in [1109, 1936, 2827], which combines this group-theoretic result with Weil’s bound on the number of Fq-rational points on a genus-g curve over Fq. 8.4.30 Remark Theorem 8.4.28 is called the Schur conjecture, although Schur did not pose this conjecture. The paper made the incorrect assertion that Schur had conjectured The-orem 8.4.28 in , and this assertion has become widely accepted despite its falsehood. 8.4.31 Remark The concept of exceptionality can be extended to rational functions or more general maps between varieties . In particular, many exceptional rational functions arise as coordinate projections of isogenies of elliptic curves [1115, 1370, 2193]. 8.4.5 Applications 8.4.32 Remark Exceptional polynomials were used in to produce families of hyperelliptic curves whose Jacobians have an unusually large endomorphism ring. These curves were used in to realize certain groups PSL2(q) as Galois groups of extensions of certain cyclotomic ﬁelds. 8.4.33 Remark Exceptional polynomials were used in [516, 2341] to produce curves whose Jacobian is isogenous to a power of an elliptic curve, and in particular to produce maximal curves (see Section 12.5). 8.4.34 Lemma We have (x + 1)N + xN + 1 = f(x2 + x) in F2[x], where N = 4r −2r + 1 and f(x) = T(x)2r+1/x2r with T(x) = x2r−1 +x2r−2 +· · ·+x. This polynomial f(x) is obtained from case 1 of Theorem 8.4.12 by putting a = 1 and e = 1. 8.4.35 Remark This result (together with exceptionality of f) has been used to produce new examples of binary sequences with ideal autocorrelation , cyclic diﬀerence sets with Singer parameters , almost perfect nonlinear functions , and bent functions [864, 3012]. See Sections 10.3, 14.6, 9.2, and 9.3, respectively. 8.4.36 Remark For further results about the polynomials f from Lemma 8.4.34, including formulas for a polynomial inducing the inverse of the permutation induced by f on F2m, see . These polynomials are shown to be exceptional in [696, 697, 864, 905, 3073]. 8.4.37 Remark The polynomials in cases 1 and 3 of Theorem 8.4.12 have been used to produce branched coverings of the projective line in positive characteristic whose Galois group is either symplectic or orthogonal . 240 Handbook of Finite Fields See Also §8.1 For discussion of permutation polynomials in one variable. §8.3 For value sets of polynomials. , For Davenport pairs, which are pairs (f, g) of polynomials in Fq[x] such that f(Fqm) = g(Fqm) for inﬁnitely many m. This notion generalizes exceptionality, since f ∈Fq[x] is exceptional if and only if (f, x) is a Davenport pair. , , For the factorization of f(x) −f(y) where f(x) is a polynomial from case 1 or 3 of Theorem 8.4.12. , , For the discovery of some of the polynomials in Theorem 8.4.12. For a thorough study of exceptional polynomials using only the Hermite–Dickson criterion, and the discovery of the polynomials in Theorem 8.4.11 and Proposition 8.4.15. References Cited: [8, 9, 58, 145, 516, 543, 667, 674, 688, 696, 697, 703, 770, 777, 840, 862, 863, 864, 905, 1109, 1115, 1120, 1122, 1222, 1368, 1369, 1370, 1371, 1372, 1373, 1374, 1755, 1886, 1898, 1900, 1912, 1913, 1936, 1939, 2126, 2190, 2192, 2193, 2341, 2563, 2564, 2785, 2827, 3012, 3073, 3074] 9 Special functions over ﬁnite ﬁelds 9.1 Boolean functions ................................... 241 Representation of Boolean functions • The Walsh transform • Parameters of Boolean functions • Equivalence of Boolean functions • Boolean functions and cryptography • Constructions of cryptographic Boolean functions • Boolean functions and error correcting codes • Boolean functions and sequences 9.2 PN and APN functions ............................. 253 Functions from F2n into F2m • Perfect Nonlinear (PN) functions • Almost Perfect Nonlinear (APN) and Almost Bent (AB) functions • APN permutations • Properties of stability • Coding theory point of view • Quadratic APN functions • APN monomials 9.3 Bent and related functions ......................... 262 Deﬁnitions and examples • Basic properties of bent functions • Bent functions and other combinatorial objects • Fundamental classes of bent functions • Boolean monomial and Niho bent functions • p-ary bent functions in univariate form • Constructions using planar and s-plateaued functions • Vectorial bent functions and Kerdock codes 9.4 κ-polynomials and related algebraic objects ..... 273 Deﬁnitions and preliminaries • Pre-semiﬁelds, semiﬁelds, and isotopy • Semiﬁeld constructions • Semiﬁelds and nuclei 9.5 Planar functions and commutative semiﬁelds ... 278 Deﬁnitions and preliminaries • Constructing aﬃne planes using planar functions • Examples, constructions, and equivalence • Classiﬁcation results, necessary conditions, and the Dembowski-Ostrom Conjecture • Planar DO polynomials and commutative semiﬁelds of odd order 9.6 Dickson polynomials ................................ 282 Basics • Factorization • Dickson polynomials of the (k + 1)-th kind • Multivariate Dickson polynomials 9.7 Schur’s conjecture and exceptional covers ....... 290 Rational function deﬁnitions • MacCluer’s Theorem and Schur’s Conjecture • Fiber product of covers • Combining exceptional covers; the (Fq, Z) exceptional tower • Exceptional rational functions; Serre’s Open Image Theorem • Davenport pairs and Poincar´ e series 9.1 Boolean functions Claude Carlet, Universit´ e Paris 8, LAGA 241 242 Handbook of Finite Fields 9.1.1 Deﬁnition Let n be a positive integer. 1. An n-variable Boolean function (or Boolean function in dimension n) is a func-tion deﬁned over the vector space Fn 2 and valued in F2. Notation: f(x), where x = (x1, . . . , xn) ∈Fn 2. 2. The domain Fn 2 can be endowed with the structure of the ﬁeld F2n. Same no-tation: f(x), where x ∈F2n. 9.1.2 Remark In this section, we study single-output Boolean functions. Multi-output (or vecto-rial) Boolean functions are studied in Section 9.2. 9.1.3 Remark Endowing Fn 2 with the structure of F2n allows taking advantage of the ﬁeld struc-ture for designing Boolean functions (this is more true for multi-output Boolean functions, however, see Section 9.2), despite the fact that the important parameters in applications are more related to the vector space structure of Fn 2 than to the ﬁeld structure of F2n. 9.1.4 Remark We emphasize that in all the deﬁnitions and propositions of this section, “x ∈Fn 2” can be replaced by “x ∈F2n,” except when the coordinates of x are speciﬁcally involved or when the Hamming weight of x plays a role. 9.1.1 Representation of Boolean functions 9.1.5 Remark The simplest way of representing a Boolean function is its truth table. But this representation gives little insight on the function. Another well-known way is with the disjunctive and conjunctive normal forms. But these representations, which do not allow uniqueness, are not well adapted to coding, cryptography, and sequences for communica-tions, which are main applications of Boolean functions. 9.1.1.1 Algebraic normal form 9.1.6 Proposition Every n-variable Boolean function f can be represented by a multivariate polynomial (i.e., is a polynomial mapping) over F2 of the form f(x) = X I⊆{1,...,n} aI Y i∈I xi ! ∈F2[x1, . . . , xn]/(x2 1 + x1, . . . , x2 n + xn). (9.1.1) This representation is unique. 9.1.7 Remark The sum in (9.1.1) is in F2. Every coordinate xi appears in this polynomial with exponents at most 1 (more precisely, we consider the polynomial f modulo the ideal gener-ated by x2 1 + x1, . . . , x2 n + xn) because xi represents a bit and every bit in F2 equals its own square. 9.1.8 Deﬁnition Representation (9.1.1) is the algebraic normal form of f (in brief, ANF). The terms Q i∈I xi are monomials. 9.1.9 Remark The ANF can also be represented in the form f(x) = X u∈Fn 2 au   n Y j=1 xj uj  = X u∈Fn 2 auxu, (9.1.2) Special functions over ﬁnite ﬁelds 243 with the convention 00 = 1. 9.1.10 Example The 3-variable function f(x) = x1x2x3 +x1x3 +x2 +1 = x111 +x101 +x010 +x000 takes at input (1, 0, 1), for instance, the value 0 + 1 + 0 + 1 = 0. 9.1.11 Proposition Let the Boolean function f be given by (9.1.1) or equivalently by (9.1.2). We have f(x) = X I⊆supp(x) aI = X u⪯x au, (9.1.3) where supp(x) denotes the support {i ∈{1, . . . , n} \| xi = 1} of x and u ⪯x means that supp(u) ⊆supp(x). Conversely, for every I ⊆{1, . . . , n}, respectively, u ∈Fn 2 aI = X x∈Fn 2 \| supp(x)⊆I f(x), respectively, au = X x∈Fn 2 \| x⪯u f(x). (9.1.4) 9.1.12 Remark The transform giving the expression of the coeﬃcients of the ANF by means of the values of f, and vice versa, is the binary M¨ obius transform. 9.1.13 Remark Proposition 9.1.11 results in an algorithm for calculating the ANF from the truth table of the function and vice versa, with complexity O(n2n) (so, a little higher than linear, since the size of the input f is 2n), see for example . 9.1.14 Remark A multivariate representation similar to the ANF but over Z also exists, called numerical normal form, with similar formulas involving the M¨ obius transform over Z ,[523, Section 8.2.1]. 9.1.1.2 Trace representation 9.1.15 Proposition Let Fn 2 be identiﬁed with F2n and let f be an n-variable Boolean function of even weight (i.e., of algebraic degree at most n−1). There exists a unique representation of f as a univariate polynomial mapping of the form f(x) = X j∈Γn TrF2o(j)/F2(Ajxj), x ∈F2n, (9.1.5) where Γn is the set of integers obtained by choosing one element in each cyclotomic coset of 2 (mod 2n −1), o(j) is the size of the cyclotomic coset containing j, Aj ∈F2o(j) and TrF2o(j)/F2 is the trace function from F2o(j) to F2. 9.1.16 Remark The Aj’s can be calculated by using the Mattson-Solomon polynomial [523, 1992]. 9.1.17 Deﬁnition Representation (9.1.5) is the trace representation (or univariate representation) of f. 9.1.18 Example Let n be even. Then we can deﬁne the function f(x) = TrF2n/F2(x3) + TrF2n/2/F2(x2n/2+1). 9.1.19 Remark Any Boolean function f can be simply represented in the form TrF2n/F2(P(x)) where P is a polynomial over F2n, but there is no uniqueness of such a representation, unless o(j) = n for every j such that Aj ̸= 0. 244 Handbook of Finite Fields 9.1.2 The Walsh transform 9.1.20 Deﬁnition The Walsh transform Wf of an n-variable Boolean function f is the discrete Fourier transform (or Hadamard transform) of the sign function (−1)f(x). Given an inner product x · y in Fn 2 (for instance x · y = Pn i=1 xiyi over Fn 2, or x · y = TrF2n/F2(xy) over F2n), the value of the Walsh transform of f at u ∈Fn 2 is given by: Wf(u) = X x∈Fn 2 (−1)f(x)+x·u, where the sum is over the integers. The set {u ∈Fn 2 \| Wf(u) ̸= 0} is the Walsh support of f. 9.1.21 Remark There exists also a normalized version, in which the value Wf(u) above is divided by √ 2n, which simpliﬁes the inverse Walsh transform formula (see below) but gives a non-integer value. 9.1.22 Remark There exists an algorithm for calculating the Walsh transform whose complexity is O(n2n), see for example . A more general deﬁnition and an example are given in Section 9.3. 9.1.23 Proposition (Inverse Walsh transform) , [523, Section 8.2.2] For every x ∈Fn 2 and every n-variable Boolean function f, we have X u∈Fn 2 Wf(u)(−1)u·x = 2n(−1)f(x). 9.1.24 Proposition (Parseval’s relation) , [523, Section 8.2.2] For every n-variable Boolean function f, we have X u∈Fn 2 W 2 f (u) = 22n. 9.1.25 Proposition (Poisson summation formula) [523, Section 8.2.2] For every n-variable Boolean function f, for every vector subspace E of Fn 2, and for all elements a and b of Fn 2, we have X u∈a+E⊥ (−1)b·u Wf(u) = \|E⊥\| (−1)a·b X x∈b+E (−1)f(x)+a·x, where E⊥is the orthogonal subspace of E (that is, E⊥= {u ∈Fn 2 \| u·x = 0, for all x ∈E}) and \|E⊥\| denotes the cardinality of E⊥. 9.1.26 Proposition Let E and E′ be two supplementary vector subspaces of Fn 2. Then, for every element a of Fn 2, we have X u∈a+E⊥ W 2 f (u) = \|E⊥\| X b∈E′ X x∈b+E (−1)f(x)+a·x !2 . 9.1.3 Parameters of Boolean functions 9.1.27 Deﬁnition The degree d◦f of the ANF is the algebraic degree of the function: if f is given by (9.1.1), respectively, by (9.1.2), then d◦f = max{\|I\| \| aI ̸= 0} = max{wt(u) \| au ̸= 0}. Special functions over ﬁnite ﬁelds 245 9.1.28 Remark A function is aﬃne (respectively, quadratic) if it has algebraic degree at most 1 (respectively, 2). Aﬃne functions are those functions of the form x 7→u · x + ϵ, u ∈Fn 2, ϵ ∈F2. 9.1.29 Example The function of Example 9.1.10 has algebraic degree 3. 9.1.30 Proposition [523, Section 8.2.1] Let f be represented by its trace representation (9.1.5). Then f has algebraic degree max j∈Γn \| Aj̸=0 w2(j), where w2(j) is the Hamming weight of the binary expansion of j (called the 2-weight of j). 9.1.31 Example The function of Example 9.1.18 has algebraic degree 2 (i.e., is quadratic). 9.1.32 Proposition [523, Section 8.2.1] The algebraic degree of any n-variable Boolean function f equals the maximum dimension of the subspaces {x ∈Fn 2 \| supp(x) ⊆I}, where I is any subset of {1, . . . , n} (equivalently, of all aﬃne subspaces of Fn 2), on which f takes value 1 an odd number of times. 9.1.33 Deﬁnition The Hamming weight of an n-variable Boolean function f is the integer wt(f) = \|{x ∈Fn 2 \| f(x) = 1}\|. The function is balanced if it has Hamming weight 2n−1 (that is, if its output is uniformly distributed over F2). 9.1.34 Proposition An n-variable Boolean function has algebraic degree at most n −1 if and only if it has even Hamming weight. 9.1.35 Proposition (McEliece’s theorem) Let f be an n-variable Boolean function of algebraic degree at most r with 0 < r < n. Then the Hamming weight of f is divisible by 2⌈n r ⌉−1 = 2⌊n−1 r ⌋(equivalently, the Walsh transform of f takes values divisible by 2⌈n r ⌉= 2⌊n−1 r ⌋+1). 9.1.36 Proposition , [523, Section 8.2.2] For n ≥2 and 1 ≤k ≤n, if the Walsh transform of f takes values divisible by 2k, then f has algebraic degree at most n −k + 1. 9.1.37 Proposition A quadratic Boolean function f is balanced if and only if its restriction to the vector subspace Ef = {a ∈Fn 2 \| Daf(x) := f(x) + f(x + a) ≡cst} = {a ∈Fn 2 \| Daf(x) = Daf(0), for all x ∈Fn 2} (the linear kernel of f) is not constant. If it is not balanced, then its Hamming weight equals 2n−1 ± 2 n+k 2 −1 where k is the dimension of Ef. 9.1.38 Deﬁnition The Hamming distance between two n-variable Boolean functions f and g equals the size of the set {x ∈Fn 2 \| f(x) ̸= g(x)}, that is, equals wt(f + g). The nonlinearity NL(f) of a Boolean function f is its minimal distance to aﬃne functions. 9.1.39 Proposition , [523, Section 8.4.1] For every n-variable Boolean function, we have NL(f) = 2n−1 −1 2 max u∈Fn 2 \|Wf(u)\|. (9.1.6) 9.1.40 Proposition , [523, Section 8.4.1] Parseval’s relation implies the covering radius bound NL(f) ≤2n−1 −2n/2−1. 9.1.41 Deﬁnition The Boolean functions achieving the covering radius bound with equality are bent (see Section 9.3 and the references therein). 9.1.42 Proposition An n-variable Boolean function is bent if and only if its Walsh transform takes values ±2n/2 only (n even). 246 Handbook of Finite Fields 9.1.43 Deﬁnition For every 0 ≤r ≤n, the r-th order nonlinearity NLr(f) of a Boolean function f equals its minimum Hamming distance to Boolean functions of algebraic degrees at most r. 9.1.44 Remark This notion is related to the Gowers norm . 9.1.45 Proposition Let f be any n-variable function and r a positive integer smaller than n. Denoting (again) by Daf(x) = f(x) + f(x + a) the ﬁrst-order derivatives of f, we have NLr(f) ≥max  1 2 max a∈Fn 2 NLr−1(Daf); 2n−1 −1 2 s 22n −2 X a∈Fn 2 NLr−1(Daf)  . 9.1.46 Remark Proposition 9.1.45, iteratively applied, allows deducing lower bounds on the higher order nonlinearity of a function from lower bounds on the nonlinearities (i.e., the ﬁrst-order nonlinearities) of its higher-order derivatives. 9.1.4 Equivalence of Boolean functions 9.1.47 Remark The parameters above (algebraic degree, Hamming weight, nonlinearity) are in-variant under composition of the Boolean function on the right by any aﬃne automorphism x 7→L(x) + a (where L is linear bijective). The algebraic degree and the nonlinearity are invariant under addition of any aﬃne Boolean function (in the case of the algebraic degree, though, the invariance needs the function to be non-aﬃne) . 9.1.48 Deﬁnition Two n-variable Boolean functions f and g are extended-aﬃne equivalent (in brief, EA-equivalent) if there exists a linear automorphism L, an aﬃne Boolean function ℓand a vector a such that g(x) = f(L(x) + a) + ℓ(x). A parameter is EA-invariant if it is preserved by EA-equivalence. 9.1.49 Remark If x is in Fn 2 (and is viewed as a 1×n matrix over F2), then L(x) = x×A where A is a non-singular n×n matrix over F2. If x lives in F2n, then L(x) = a1x+a2x2+· · ·+an−1x2n−1 where a1, a2, . . . , an−1 are chosen in F2n such that the kernel of L is reduced to {0}. 9.1.50 Remark Little is known on the number of n-variable Boolean functions up to EA-equivalence, except of course that it is larger than the number 22n of Boolean functions divided by the number 2n(2n −1)(2n −2)(2n −22) · · · (2n −2n−1) of aﬃne automorphisms of Fn 2 and by the number 2n+1 of aﬃne Boolean functions. 9.1.51 Remark There exists another notion of equivalence: the CCZ-equivalence; for vectorial functions, it is more general than EA-equivalence, but for Boolean functions, it coincides with EA-equivalence . 9.1.5 Boolean functions and cryptography 9.1.52 Remark Boolean functions are used in pseudo-random generators, in stream ciphers (in conventional cryptography), to ensure a suﬃcient “nonlinearity” (since linear ciphers are weak ), see Section 16.2. They must then be balanced to avoid distinguishing attacks. 9.1.53 Remark A high algebraic degree of a Boolean function f ensures a good resistance of the stream ciphers (using it as the nonlinear part) against the Berlekamp-Massey attack and the Rønjom-Helleseth attack and its variants. Special functions over ﬁnite ﬁelds 247 9.1.54 Remark A large nonlinearity (that is, a nonlinearity near the covering radius bound) is an important cryptographic characteristic which ensures resistance to the fast correlation attack . 9.1.55 Remark Vectorial Boolean functions are also used in cryptography (in block ciphers); they are addressed in Section 9.2. 9.1.56 Deﬁnition Let n be a positive integer and m < n a non-negative integer. A Boolean function over Fn 2 is m-resilient (respectively, m-th order correlation immune) if any of its restrictions obtained by ﬁxing at most m of its input coordinates xi is balanced (respectively, has same output distribution as f itself). The resiliency order of f is the largest value of m such that f is m-resilient. 9.1.57 Remark The notions of resilient and correlation immune functions have a sense for x ∈Fn 2 only (that is, not for x ∈F2n). 9.1.58 Remark A function is m-resilient if and only if it is balanced and m-th order correlation immune. Since Boolean functions used in stream ciphers must be balanced, we shall address only resiliency in the sequel. 9.1.59 Remark The resiliency order quantiﬁes the resistance to the Siegenthaler correlation attack of a stream cipher using f as a combiner (that is, combining the outputs to linear feedback registers by applying f, see Section 16.2). 9.1.60 Remark The resiliency order is not EA-invariant. It is invariant under permutations of the coordinates of x ∈Fn 2. 9.1.61 Proposition (Siegenthaler’s bound) , [523, Section 8.7] Any m-resilient n-variable Boolean function has algebraic degree smaller than or equal to n −m −1 if 0 ≤m < n −1 and is aﬃne if m = n −1. 9.1.62 Proposition (Xiao-Massey’s characterization) Any n-variable Boolean function f is m-resilient if and only if Wf(u) = 0 for all u ∈Fn 2 such that 0 ≤wt(u) ≤m. 9.1.63 Proposition (Improved Sarkar-Maitra’s bound) [520, 534, 2522], [523, Section 8.7] Let f be any n-variable m-resilient function (m ≤n−2) and let d be its algebraic degree. The values of the Walsh transform of f are divisible by 2m+2+⌊n−m−2 d ⌋. Hence, according to Equation (9.1.6), the nonlinearity of f is divisible by 2m+1+⌊n−m−2 d ⌋and is therefore bounded above by the largest number divisible by 2m+1+⌊n−m−2 d ⌋and smaller than 2n−1 −2n/2−1. In particular, if m + 1 + n−m−2 d < n 2 and n is even, we have NL(f) ≤2n−1 −2n/2−1 − 2m+1+⌊n−m−2 d ⌋. 9.1.64 Remark The divisibility property in Proposition 9.1.63, combined with known bounds on the nonlinearity, implies improvements of these bounds (as shown in Proposition 9.1.63 for the covering radius bound). 9.1.65 Deﬁnition A function g such that fg = 0 is an annihilator of f. The minimum algebraic degree of the nonzero functions g such that fg = 0 or (f + 1)g = 0 is the algebraic immunity of f. It is denoted by AI(f). 9.1.66 Example The function f(x1, x2, x3) = x1x2x3 + x2 + 1 has algebraic immunity 1 since, being non-constant, it does not have algebraic immunity 0, and since f + 1 has annihilator x2 + 1. 248 Handbook of Finite Fields 9.1.67 Remark The algebraic immunity quantiﬁes the resistance to the algebraic attack of the ciphers using f as a combiner, or as a ﬁlter (taking as input n bits at ﬁxed positions in an LFSR, see Section 16.2). 9.1.68 Remark The set of annihilators of f equals the ideal of all multiples of f + 1. 9.1.69 Remark Let g be any function of algebraic degree at most d, then expressing that g is an annihilator of f (that is, f(x) = 1 implies g(x) = 0, for every x ∈Fn 2) by means of the (unknown) coeﬃcients of the ANF of g results in a system of homogeneous linear equations. In this system, we have Pd i=0 n i number of variables (the coeﬃcients of the monomials of degrees at most d) and wt(f) many equations. 9.1.70 Proposition For every n-variable Boolean function, we have: AI(f) ≤ n 2 . 9.1.71 Remark Let f ﬁlter an LFSR. Let us assume for instance that the LFSR has length 256 (it can then be initialized for instance with a key of length 128 and an IV of length 128 as well). Then Proposition 9.1.70 requires n ≥16 to ensure a complexity of the algebraic attack larger than exhaustive search, see e.g., . A minimal security margin would be n ≥18. Algebraic attacks have therefore changed the number of variables of the Boolean functions used in practice (before them, for reasons of eﬃciency, the number of variables was rarely more than 10). 9.1.72 Proposition (Bounds on the weight, Lobanov’s bound on the nonlinearity) [523, Section 8.9] For every n and every n-variable Boolean function f, we have PAI(f)−1 i=0 n i ≤wt(f) ≤ Pn−AI(f) i=0 n i (in particular, if n is odd and f has optimal algebraic immunity n+1 2 , then f is balanced) and NL(f) ≥2 PAI(f)−2 i=0 n−1 i . 9.1.73 Remark Bounds also exist for the higher order nonlinearities [523, Section 8.9], . 9.1.74 Proposition If an n-variable balanced Boolean function f, with n odd, has no non-zero annihilator of algebraic degree at most n−1 2 , then it has optimal algebraic immunity. 9.1.75 Remark 1. Stream ciphers must also resist fast algebraic attacks, which work if one can ﬁnd g ̸= 0 of low degree and h of algebraic degree not much larger than ⌈n/2⌉, such that fg = h [735, 1446]. These attacks need more data (of a particular shape) than standard algebraic attacks but can be faster if g has lower degree due to the relaxation of the condition on h. 2. Stream ciphers must also resist algebraic attacks on the augmented function . 3. Finally, they must resist the already mentioned attack by Rønjom and Helleseth (designed for the ﬁlter generator and later generalized) which requires f to have an algebraic degree close to n. 9.1.76 Deﬁnition Let 1 ≤l ≤n. An n-variable Boolean function f satisﬁes the propagation criterion of order l, denoted by PC(l), if for every a ∈Fn 2 such that 1 ≤wt(a) ≤l, the derivative Daf is balanced. Property PC(1) is the strict avalanche criterion. 9.1.77 Remark This characteristic is related to the diﬀusion of the cipher in which f is involved. It is less important than the characteristics seen above, but it has attracted some attention. 9.1.78 Remark Other cryptographic characteristics (less essential than the algebraic degree, the nonlinearity and the algebraic immunity) exist : the non-existence of nonzero linear Special functions over ﬁnite ﬁelds 249 structure, the global avalanche criterion, the maximum correlation to subsets, the algebraic thickness, the nonhomomorphicity. 9.1.6 Constructions of cryptographic Boolean functions 9.1.79 Remark Boolean functions being tools for the design of cryptosystems, an important as-pect of the research in this domain is to design constructions of Boolean functions having the necessary cryptographic features (contrary to other domains of the study of Boolean functions, which mainly study their properties). 9.1.80 Remark A Boolean function obtained by some construction and satisfying a given crypto-graphic criterion, or several criteria, will be considered as new if it is EA-inequivalent to all previously found functions satisfying the same criteria. 9.1.81 Remark We call secondary the constructions which use already deﬁned functions satisfying a given property, to build a new one satisfying the same property. A construction from ﬁrst principles will be primary. 9.1.6.1 Primary constructions of resilient functions 9.1.82 Proposition (Maiorana-McFarland’s construction) [485, 519], [523, Section 8.7] Let r and s be positive integers; let g be any s-variable Boolean function and let φ be a mapping from Fs 2 to Fr 2. Then the function fφ,g(x, y) = x · φ(y) + g(y), x ∈Fr 2, y ∈Fs 2, (9.1.7) where “·” is an inner product in Fr 2, is (at least) k-resilient where k = min{wt(φ(y)), y ∈ Fs 2} −1, and has nonlinearity satisfying: 2n−1 −2r−1 max a∈Fr 2 \|φ−1(a)\| ≤NL(fφ,g) ≤2n−1 −2r−1r max a∈Fr 2 \|φ−1(a)\|. 9.1.83 Remark This construction was originally developed for bent functions (see Section 9.3) and has been later adapted to resilient functions. 9.1.84 Remark The resiliency order of fφ,g can be larger, for some well-chosen functions g . 9.1.85 Remark The Maiorana-McFarland construction has been generalized in several ways . 9.1.6.2 Secondary constructions of resilient functions 9.1.86 Proposition (Indirect sum) , [523, Section 8.7] Let r and s be positive integers and let 0 ≤t < r and 0 ≤m < s. Let f1 and f2 be two r-variable t-resilient functions. Let g1 and g2 be two s-variable m-resilient functions. Then the function h(x, y) = f1(x) + g1(y) + (f1 + f2)(x) (g1 + g2)(y); x ∈Fr 2, y ∈Fs 2 is an (r+s)-variable (t+m+1)-resilient function. The Walsh transform of h takes the value Wh(a, b) = 1 2Wf1(a) [Wg1(b) + Wg2(b)] + 1 2Wf2(a) [Wg1(b) −Wg2(b)] . (9.1.8) If the Walsh transforms of f1 and f2 have disjoint supports and if the Walsh transforms of g1 and g2 have disjoint supports, then NL(h) = min i,j∈{1,2} 2r+s−2 + 2r−1NL(gj) + 2s−1NL(fi) −NL(fi)NL(gj) . (9.1.9) 250 Handbook of Finite Fields In particular, if f1 and f2 have nonlinearity 2r−1 −2t+1 and disjoint Walsh supports, if g1 and g2 have nonlinearity 2s−1 −2m+1 and disjoint Walsh supports, and if f1 + f2 has algebraic degree r −t −1 and g1 + g2 has algebraic degree s −m −1, then h achieves Siegenthaler’s and (improved) Sarkar-Maitra’s bounds with equality. 9.1.87 Remark Some particular choices of functions f1, f2, g1, g2 give secondary constructions pre-viously introduced by several authors (Siegenthaler, Tarannikov) [523, Section 8.7]; in par-ticular the well-known direct sum h(x, y) = f(x) + g(y) is obtained by putting g1 = g2 and/or f1 = f2. 9.1.88 Remark The indirect sum gives also a secondary construction of bent functions . 9.1.89 Proposition (Secondary construction without extension of the number of variables) [523, Section 8.7] Let 0 ≤k ≤n. Let f1, f2 and f3 be three k-resilient n-variable functions. Then the function s1 = f1 +f2 +f3 is k-resilient if and only if the function s2 = f1f2 +f1f3 +f2f3 is k-resilient. Moreover NL(s2) ≥1 2 NL(s1) + 3 X i=1 NL(fi) ! −2n−1, (9.1.10) and if the Walsh supports of f1, f2 and f3 are pairwise disjoint, then NL(s2) ≥1 2 NL(s1) + min 1≤i≤3 NL(fi) . (9.1.11) 9.1.90 Remark It has been impossible until now to obtain resilient functions of suﬃcient orders with good algebraic immunity. For this reason, the ﬁlter model is preferred to the combiner model (the correlation attack works on the latter). 9.1.6.3 Constructions of highly nonlinear functions with optimal algebraic immunity 9.1.91 Proposition Let n be any positive integer and α a primitive element of the ﬁeld F2n. Let f be the balanced Boolean function on F2n whose support equals {0, αs, . . . , αs+2n−1−2} for some s. Then f has optimum algebraic immunity ⌈n/2⌉. Moreover, f has algebraic degree n −1 and nonlinearity NL(f) ≥2n−1 −n · ln 2 · 2 n 2 −1. 9.1.92 Remark This bound on the nonlinearity, which has been recently slightly improved, is not enough for ensuring that the function allows resisting the fast correlation attacks, but it has been checked, for n ≤26, that the exact value of NL(f) is much better than this lower bound. Improving signiﬁcantly the bound of Proposition 9.1.91 is an open problem. 9.1.93 Remark The function in Proposition 9.1.91 shows also good immunity against fast algebraic attacks as shown by Liu et al. . 9.1.94 Remark Despite the fact that complexity for computing the output is roughly the same as for computing the discrete logarithm, the function can be eﬃciently computed because n is small; the Pohlig-Hellman method is eﬃcient, at least for some values (n = 18 or 20). 9.1.95 Remark We note that a similar function had been previously studied by Brandst¨ atter, Lange, and Winterhof in but the algebraic immunity was not addressed by these authors. 9.1.96 Remark Other inﬁnite classes of balanced functions with optimal algebraic immunity have been found [2777, 3056] which have good nonlinearity and good resistance to fast algebraic attacks (checked by computer), at least for small n. More classes exist but, either their Special functions over ﬁnite ﬁelds 251 optimal algebraic immunity is not completely proved, or they are closely related to the classes mentioned above, or they do not have good nonlinearity, or they have bad resistance to fast algebraic attacks. 9.1.7 Boolean functions and error correcting codes 9.1.97 Remark Boolean functions are used to design error correcting codes whose lengths are powers of 2, in particular Reed-Muller codes (see Section 15.1). 9.1.7.1 Reed-Muller codes 9.1.98 Deﬁnition Let n be any positive integer and 0 ≤r ≤n. The Reed-Muller code of length 2n and order r is the set of binary words of length 2n corresponding to the output columns of the truth-tables of all the n-variable Boolean functions of algebraic degrees at most r. 9.1.99 Proposition The Reed-Muller code of length 2n and order r is a linear code over F2 (i.e., is a vector subspace of F2n 2 ) of dimension 1+n+ n 2 +· · ·+ n r and minimum distance 2n−r. 9.1.100 Remark The Reed-Muller code of order 1 is optimal according to the Griesmer bound . 9.1.7.2 Kerdock codes 9.1.101 Remark The Reed-Muller code of length 2n and of order 2 contains a nonlinear optimal code: the Kerdock code of the same length, introduced in (not as in the deﬁnition below, though). The Kerdock code of length 2n is the union of cosets of the ﬁrst order Reed-Muller code, chosen such that the sum of two elements from diﬀerent cosets is a bent function. 9.1.102 Deﬁnition The Kerdock code of length 2n is the set of functions of the form (x, ϵ) ∈F2n−1× F2 7→f(ux, ϵ) + TrF2n−1/F2(ax) + ηϵ + τ where f(x, ϵ) = TrF2n−1/F2 P n 2 −1 i=1 x2i+1 + ϵ TrF2n−1/F2(x), with u, a ∈F2n−1, and η, τ ∈F2. 9.1.103 Proposition The Kerdock code of length 2n has size 22n and minimum distance 2n−1 −2n/2−1. 9.1.104 Remark The Kerdock code of length 2n is optimal, as proved by Delsarte . 9.1.105 Remark Other codes with the same parameters exist, called generalized Kerdock codes . 9.1.106 Remark The Kerdock code of length 2n is not linear but it is the image of a linear code over Z/4Z by a distance preserving mapping called the Gray map . 9.1.8 Boolean functions and sequences 9.1.107 Remark Boolean functions are related to sequences for communications: see Chapter 10. 252 Handbook of Finite Fields 9.1.8.1 Boolean functions and cross correlation of m-sequences 9.1.108 Deﬁnition A sequence (si)i≥0 over F2 satisfying an order n linear homogeneous recurrence relation (with constant coeﬃcients), is an m-sequence if it has (optimal) period 2n −1. 9.1.109 Proposition The m-sequences of period 2n −1 are the sequences of the form si = TrF2n/F2(λαi), where λ ∈F∗ 2n and α is a primitive element of F2n. Given such an m-sequence, any other m-sequence of the same period diﬀers with (si)i≥0, up to a cyclic shift, by a decimation d such that gcd(d, 2n−1) = 1 (that is, the second sequence equals (sdi+t)i≥0, where t is some integer). 9.1.110 Remark Since sequences are viewed up to cyclic shifts, we shall, in the sequel, take λ equal to 1 and the integer t equal to 0. 9.1.111 Remark According to Proposition 9.1.109, the crosscorrelation Cd(τ) = Pn−1 t=0 (−1)st+τ +sdt between two m-sequences equals X x∈F∗ 2n (−1)TrF2n /F2(cx+xd) where c = ατ; hence, 1 + Cd(τ) is the value at c of the Walsh transform of the monomial function f(x) = TrF2n/F2(xd). The nonlinearity of f(x) = TrF2n/F2(xd) is therefore NL(f) = 2n−1 −1 2 max τ \|1 + Cd(τ)\|. (9.1.12) 9.1.112 Remark A family of balanced sequences of period 2n −1 with good correlation properties and large linear complexity can be constructed from a bent function. These sequences are bent sequences . Let n = 0 (mod 4), k = n/2 and h : F2k →F2 be a bent function. Let (e1, e2, . . . , ek) be a basis of the vector space F2k over F2 and let σ ∈F2n \ F2k. Let hi(y1, y2, . . . , yk) = h(y1, y2, . . . , yk) + a1y1 + a2y2 + · · · + akyk for some ordering of the 2k possible choices of (a1, . . . , ak) in Fk 2. Let fi(x) = hi(TrF2n/F2(e1x), TrF2n/F2(e2x), . . . , TrF2n/F2(ekx)) + TrF2n/F2(σx) for i = 1, 2, . . . , 2k. The family is composed of the sequences s(i) t 0≤t≤2n−2 for 1 ≤i ≤2k, where s(i) t = fi(αt) for some ﬁxed primitive element α in F2n. See Also §15.1 For properties of Reed-Muller codes and their sub-codes; see also . For a survey on Boolean functions for coding and cryptography (the chapter which follows it in the same monograph deals with vectorial functions, which are the subject of Section 9.2 in the present handbook). This survey includes binary bent functions (see also Section 9.3 in the present handbook). For a recent survey on sequences. References Cited: [267, 391, 444, 485, 490, 491, 519, 520, 521, 522, 523, 528, 530, 534, 735, 741, 800, 1069, 1303, 1409, 1446, 1673, 1728, 1855, 1951, 1992, 2011, 2073, 2075, 2085, 2317, 2474, 2522, 2663, 2777, 3015, 3056] Special functions over ﬁnite ﬁelds 253 9.2 PN and APN functions Pascale Charpin, INRIA 9.2.1 Remark Around 1992, two cryptanalysis methods had been introduced in the literature devoted to symmetric cryptosystems, the diﬀerential cryptanalysis , and the linear cryptanalysis . It was shown later that these methods are basically linked . In order to resist these attacks the round function used in an iterated block cipher must satisfy some mathematical properties. These properties are mainly covered by the concepts of nonlinearity and of diﬀerential uniformity for functions on extension ﬁelds. 9.2.1 Functions from F2n into F2m 9.2.2 Deﬁnition Any function from F2n into F2m is an (n, m)-function. It is a Boolean function when m = 1 and a function on F2n when m = n. Here we assume that n ≥m. Basic properties on Boolean functions can be found in Section 9.1. 9.2.3 Deﬁnition Let F be an (n, m)-function. The component functions of F are the Boolean functions fλ : x ∈F2n 7→Tr(λF(x)), λ ∈F∗ 2m, where Tr is the absolute trace on F2m (see Deﬁnition 2.1.80). 9.2.4 Deﬁnition An (n, m)-function F is balanced when it is uniformly distributed, i.e., F takes every value of F2m each 2n−m times. 9.2.5 Proposition An (n, m)-function is balanced if and only if all its component functions are balanced. 9.2.6 Remark When m = n, a balanced function is a permutation of F2n, as it is shown in a more general context in [1939, Theorem 7.7]. 9.2.7 Remark The nonlinearity of a Boolean function f is usually computed by means of the highest magnitude of its Walsh spectrum Wf. These quantities are denoted by NL(f) and L(f) respectively (see the deﬁnitions in Section 9.1). 9.2.8 Deﬁnition A Boolean function f is plateaued if its Walsh coeﬃcients take at most three values, namely 0, ±L(f). Then, L(f) = 2s with s ≥n/2. If s = n/2 (and n even) then f is bent and its Walsh coeﬃcients take two values only, namely ±2 n 2 . Also, f is semi-bent if s = (n + 1)/2 for odd n and s = (n + 2)/2 for even n. An (n, m)-function is plateaued when its components are plateaued. 9.2.9 Deﬁnition Let F be an (n, m)-function with components fλ. Let NL(fλ) be the nonlin-earity of fλ. The nonlinearity of F, say NL(F), is the lowest nonlinearity achieved by one of its components: NL(F) = min λ∈F2m (NL(fλ)) = 2n−1 −L(F) 2 , where L(F) is the highest magnitude appearing in the Walsh spectrum of all fλ. 254 Handbook of Finite Fields 9.2.10 Remark By deﬁnition the nonlinearity of an (n, m)-function F satisﬁes NL(F) ≤2n−1 −2n/2−1. Indeed, this holds for any fλ (see the covering radius bound in Deﬁnition 9.1.41). 9.2.11 Deﬁnition Let an (n, m)-function F be expressed as a univariate polynomial of degree less than 2n. The algebraic degree of F is the maximal Hamming weight of its exponents, considering the 2-ary expansion of exponents. 9.2.12 Deﬁnition Let F be an (n, m)-function. For any a ∈F2n, the derivative of F with respect to a is the (n, m)-function DaF deﬁned for all x ∈F2n by DaF(x) = F(x + a) + F(x). 9.2.2 Perfect Nonlinear (PN) functions 9.2.13 Deﬁnition An (n, m)-function F is a bent function, also called a perfect nonlinear (PN) function, if and only if NL(F) = 2n−1 −2(n/2−1), i.e., all its components are Boolean bent functions. 9.2.14 Theorem An (n, m)-function can be PN only if n is even and m ≤n/2. 9.2.15 Remark To construct a PN function is exactly to characterize a subspace of Boolean bent functions. This PN function is an (n,m)-function if this subspace is of dimension m composed of bent functions on F2n. Primary constructions, by means of the main classes of bent functions are explained in [524, Section 3.1.1]. To ﬁnd subspaces of bent functions is an important research problem (see recent results in [264, 495]). 9.2.16 Proposition An (n, m)-function F is PN if and only if all of its derivatives are balanced. In other terms, let δa,b(F) = \|{x ∈F2n\|DaF(x) = b}\|, a ∈F∗ 2n, b ∈F2m. (9.2.1) Then F is PN if and only δa,b(F) = 2n−m for any a and any b. 9.2.17 Remark From Theorem 9.2.14, PN functions do not exist for m = n. However they do exist for odd characteristics. They are planar functions and were introduced in . They are functions on Fpn, p odd, such that for every non-zero α ∈F the diﬀerence mapping x 7→F(x + α) −F(x) is a permutation of Fpn (see Section 9.5). 9.2.18 Remark In cryptology, PN functions were introduced in 1990-95 as functions which provide the optimal resistance to linear attacks and to diﬀerential attacks (see and ). They have the best nonlinearity, by Deﬁnition 9.2.13, and the δa,b are as small as possible, by Proposition 9.2.16. A major drawback is that these optimal functions are not balanced; also, they presuppose the use of non-invertible round functions (see , [496, Chapter 3]). Special functions over ﬁnite ﬁelds 255 9.2.3 Almost Perfect Nonlinear (APN) and Almost Bent (AB) functions 9.2.19 Remark In cryptography, most works focused on optimal functions with respect to diﬀe-rential attacks. The aim is to exhibit such functions that, moreover, are bijective and oppose a good resistance to linear attacks. According to Remark 9.2.18, functions on F2n, i.e., (n, n)-functions) are generally considered. Algebraic properties of almost perfect nonlinear (resp. almost bent) functions and their links with error-correcting codes are introduced in . 9.2.20 Deﬁnition A function F on F2n is almost perfect nonlinear (APN) if and only if all the equations F(x) + F(x + a) = b, a ∈F2n, a ̸= 0, b ∈F2n, (9.2.2) have at most two solutions. The function F is almost bent (AB) if and only if the value of WF (β, λ) = X x∈F2n (−1)Tr(λF (x)+βx) (9.2.3) is equal either to 0 or to ±2 n+1 2 , for any β and λ in F2n, β ̸= 0. 9.2.21 Theorem AB functions exist for n odd only. Any AB function is APN. 9.2.22 Remark A function F on F2n is APN if and only if all its derivatives are 2-to-1. This can be derived from the deﬁnition. 9.2.23 Proposition [525, Theorem 1] Let F be an AB function on F2n. Then the algebraic degree of F is less than or equal to (n + 1)/2. 9.2.24 Remark According to (9.2.3), a function F on F2n is an AB function if and only if its components fλ are semi-bent. Thus NL(fλ) = 2n−1 −2(n−1)/2, i.e., L(fλ) = 2(n+1)/2, for all λ ∈F∗ 2n. Consequently, L(F) = 2(n+1)/2. 9.2.25 Proposition Let n be odd. Then F is an AB function on F2n if and only if F is APN and all Walsh coeﬃcients of all components fλ are divisible by 2(n+1)/2. 9.2.26 Remark APN functions are optimal concerning the resistance of a cipher to diﬀerential at-tacks and to its variants. More generally, this resistance is related to the following quantities, introduced in . 9.2.27 Deﬁnition Let F be a function on F2n. For any a ∈F∗ 2n and b ∈F2n, we denote δ(a, b) = \|{x ∈F2n\| DaF(x) = b}\|. Then, the diﬀerential uniformity of F is δ(F) = max a̸=0, b∈F2n δ(a, b). 9.2.28 Remark Those functions for which δ(F) = 2 are APN. It is worth noticing that such functions are rare and hard to ﬁnd. From a recent result of Voloch , it follows that these functions asymptotically have density zero in the set of all functions. Few inﬁnite 256 Handbook of Finite Fields families are known. The monomial APN functions are related with exceptional objects (see Subsection 9.2.8). The known inﬁnites families of non-monomial APN functions are listed in . They are all quadratic. Edel and Pott propose in an original construction providing a sporadic non-quadratic non-monomial APN function. 9.2.29 Problem AB functions on F2n, with n odd, provide an optimal resistance to both diﬀerential attacks and linear attacks. There are several classes of AB permutations. The situation is diﬀerent for even n: there are APN functions F such that L(F) = 2(n+2)/2 and it is conjectured that this value is the minimum; also the existence of APN permutations, for n > 6, is not resolved. 9.2.30 Example The inverse function∗, F(x) = x−1, is an APN (not AB) permutation on F2n when n is odd. For even n, it is a permutation too, but δ(a, b) takes three values, namely 0, 2 and 4 so that δ(F) = 4. This function has the highest degree and satisﬁes L(F) = 2(n+2)/2 for n even. The inverse function on F28 is used in the AES S-boxes (see Section 16.2). 9.2.31 Remark Regarding the PN property, the APN property can be extended on ﬁelds of odd characteristic (see [908, 1479], for instance). 9.2.4 APN permutations 9.2.32 Remark The ﬁrst APN permutation of F26 (n = 6) was presented by Dillon at the conference Finite Fields and their Applications (Fq 9) in 2009. Thus, a long-standing (and famous) conjecture stating that there is no APN permutation of F2n when n is even was disproved. The method, explained in , uses mostly the representation of APN functions by codes (see Theorem 9.2.47). However the existence of APN permutations of an even number n (with n ≥8) of variables remains a research problem of great interest. The next theorem, is due to several authors [228, 1541, 2303]. 9.2.33 Conjecture APN permutations on F2n exist for any n ≥6. 9.2.34 Theorem [228, Theorem 3] Let F be any permutation polynomial on F2n, n = 2t. Then F is not APN when one of the statements below is satisﬁed: 1. n = 4; 2. F ∈F4[x]; 3. F ∈F2t[x]; 4. F is plateaued, i.e., all its components fλ are plateaued. 9.2.35 Remark When n is odd, any APN monomial is bijective; little is known about the other APN functions. Clearly, they are not generally bijective. For example, in the case where n is even the APN monomials are 3-to-1. 9.2.36 Proposition [228, Proposition 3] Let r be a divisor of n. Let F be any function on F2n such that F ∈F2r[x]. If F satisﬁes for some a ∈F2r F(y) + F(y + a) = β, β ∈F2r, for some y ̸∈F2r such that y2r + y + a ̸= 0, then F is not APN. Consequently, if F is an APN monomial, F(x) = xd, then gcd(d, 2n −1) = 1 for odd n and gcd(d, 2n −1) = 3 for even n. ∗Note that, as a function on F2n, F(x) = x2n−2 so that F(0) = 0. Special functions over ﬁnite ﬁelds 257 9.2.37 Remark It was conjectured in that for any AB function F, there exists a linear function L such that F + L is a permutation. A counterexample for this conjecture is given in [449, Remark 4]. 9.2.38 Example [447, Proposition 1] Let n = 3k where k is odd and not divisible by 3. Let u be any element in F∗ 2n of order 22k + 2k + 1. Then any function F(x) = x2s+1 + ux2ik+2tk+s with gcd(s, 3k) = 1, i ≡sk (mod 3), t = 3 −i, is an AB permutation. 9.2.5 Properties of stability 9.2.39 Deﬁnition Let F and F ′ be two functions on F2n. They are equivalent if F ′ is obtained from F by compositions of 1 and/or 2: 1. F 7→A1 ◦F ◦A2 + A, where A1 and A2 are aﬃne permutations and A is any function which is aﬃne or constant; 2. F 7→F −1, the inverse function of F when F is a permutation. They are extended aﬃne equivalent (EA-equivalent) when F ′ is obtained from F by transformations of type 1. 9.2.40 Proposition Let F be a function on F2n which is APN (resp. AB). Then any function which is equivalent to F is an APN (resp. AB) function too. 9.2.41 Remark The deﬁnition of Carlet-Charpin-Zinoviev (CCZ)-equivalence, naturally derived from [525, Proposition 3], was proposed in . It necessitates to introduce the graph of any function F on F2n: GF = { (x, F(x)) \| x ∈F2n}. 9.2.42 Deﬁnition Let F and F ′ be two functions on F2n. They are CCZ-equivalent if their graphs are aﬃne equivalent, i.e., if there exists an aﬃne automorphism A of F2n × F2n such that A(GF ) = GF ′. 9.2.43 Theorem [449, Proposition 2] Let F, F ′ be CCZ-equivalent functions. Then F is APN (resp. AB) if and only if F ′ is APN (resp. AB). In this case, F and F ′ have the same nonlinearity but may have diﬀerent algebraic degrees. 9.2.44 Remark In , Budaghyan, Carlet, and Pott proved that CCZ-equivalence is more general than equivalence. They are then able to present APN functions which are EA-inequivalent to all known APN functions. 9.2.45 Remark The important question whether aﬃnely inequivalent functions are CZZ-equivalent was ﬁrst investigated by Edel, Kyureghyan, and Pott . They exhibited an APN bi-nomial, which is new since it cannot be obtained from another known APN function (see example below). A number of constructions of APN functions followed the deﬁnition of CCZ-equivalence (see, for instance: [386, 423, 443, 447]). A classiﬁcation up to CCZ-equivalence, for small dimensions, is proposed in . 9.2.46 Example [953, Theorem 2] Let the function F on F210, F(x) = x3+ux36. Let ω ∈F4\{0, 1}. Then F is APN for any u = ωiy, i ∈{1, 2} and y ∈F∗ 25. Moreover, these APN functions are not CCZ-equivalent to any APN monomial on F210. 258 Handbook of Finite Fields 9.2.6 Coding theory point of view 9.2.47 Theorem Let F be a function on F2n such that F(0) = 0. Let α be a primitive element of F2n. Let us denote by CF the linear binary code of length 2n −1 deﬁned by its parity check matrix HF = 1 α α2 . . . α2n−2 F(1) F(α) F(α2) . . . F(α2n−2) where each entry is viewed as a binary vector. The dual code is denoted by (CF )⊥. Then we have 1. The function F is APN if and only if the code CF has minimum distance ﬁve. 2. The function F is AB if and only if the weights of the non zero codewords of the code (CF )⊥form the following set: {2n−1, 2n−1 ± 2(n−1)/2}. 9.2.48 Corollary Let F be a function on F2n such that F(0) = 0. Then we have 1. If F is APN then the dimension of CF is equal to 2n −2n −1. 2. If F is APN then (CF )⊥does not contain the all-one vector. 3. If F is AB then the weight distribution of (CF )⊥is unique and the same as the weight distribution of the dual of the 2-error-correcting BCH code. 9.2.49 Remark The weight distribution of (CF )⊥exactly corresponds to the Walsh spectrum of F, that is the multiset of the values WF (β, λ) given by (9.2.3). Thus, if this weight distribution is known, the nonlinearity of F is also known. 9.2.50 Deﬁnition A binary code C is 2ℓ-divisible if the weight of any of its codewords is divisible by 2ℓ. Moreover C is exactly 2ℓ-divisible if, additionally, it contains at least one codeword whose weight is not divisible by 2ℓ+1. 9.2.51 Theorem Let F be a function on F2n, with n = 2ℓ+ 1. Then F is AB if and only if F is APN and the code (CF )⊥, deﬁned in Theorem 9.2.47, is 2ℓ-divisible. 9.2.52 Remark The determination of the weight distributions of codes of type CF remains an open problem except when the code is optimal in a certain sense. The work of Kasami remains fundamental, proving notably the uniqueness of the weight enumerator of these optimal codes ([1688, 1689]; see also an extensive study in [590, Section 3.4.2]). 9.2.53 Example The functions F(x) = xd, d = 22i −2i + 1 with gcd(i, n) = 1 are APN for any n. Such exponents d are called the Kasami exponents. When n is odd, the code CF is equivalent to the cyclic code with two zeros α2i+1 and α23i+1 (α being a primitive root of F2n) and F is AB. The proof is due to Kasami for odd n (see also [590, Theorem 3.32]) and to Janwa and Wilson for even n. 9.2.54 Remark The code CF always contains a subcode which is a cyclic code. This is of most interest when F is a monomial as we will see later (see for more details). 9.2.7 Quadratic APN functions 9.2.55 Deﬁnition A function F on F2n is quadratic when it has algebraic degree 2. 9.2.56 Proposition [525, Section 3.4] Let F be a quadratic function on F2n with n odd. Then F is AB if and only if F is APN. Special functions over ﬁnite ﬁelds 259 9.2.57 Proposition Let F be a quadratic function. Then F is APN if and only if the code CF does not contain any codeword of weight three. 9.2.58 Remark When F is quadratic, C⊥ F is contained in the punctured Reed-Muller code of order 2 (see [1991, Chapter 15]). 9.2.59 Example The functions F(x) = x2i+1 with gcd(i, n) = 1 are APN for any n (see [1689, 2303]). Such quadratic exponents are called Gold exponents. 9.2.60 Remark There are several constructions of quadratic non-monomial APN functions; no-tably, there are inﬁnite families of such functions (see [272, 386, 388, 443, 447, 449] and Remark 9.2.28). There is only one inﬁnite family of binomials, which was introduced and extensively studied in [447, Section II] by Budaghyan, Carlet, and Leander. Their Walsh spectrum was computed in . Some such binomials are bijective (see Example 9.2.38). 9.2.61 Problem The Walsh spectrum of a non-monomial quadratic function is generally not known. Concerning APN such functions, the problem is discussed in . For the computation of some spectrum, see [384, 385, 387]. 9.2.62 Example The function F(x) = x3 + Tr(x9), was recently discovered . It is APN for any n and inequivalent (in any sense) to the Gold functions, while it has the same Walsh spectrum . 9.2.63 Theorem [389, Theorem 3] Let F be a quadratic APN function and G be a Gold function, i.e., G(x) = x2k+1 with gcd(k, n) = 1, on F2n. If F and G are CCZ-equivalent then they are EA-equivalent. 9.2.64 Theorem [228, Theorem 6] There are no APN quadratic mappings on F2n of the form F(x) = n−1 X i=1 cix2i+1 , ci ∈F2n (9.2.4) unless F is a monomial: F(x) = cx2k+1 with gcd(k, n) = 1, c ∈F2n. 9.2.65 Deﬁnition The APN function F, on F2n, is crooked if for every a ∈F∗ 2n the image of its derivative DaF is a hyperplane or the complement of a hyperplane. 9.2.66 Proposition Any APN quadratic function is crooked. 9.2.67 Remark Deﬁnition 9.2.65 was introduced in , as an extension of the original deﬁnition where crooked functions are implicitly bijective (see [224, 2838] for crooked AB functions). The next conjecture is strengthened by the work of Bierbrauer and Kyureghyan (Theo-rem 9.2.69). 9.2.68 Conjecture Any APN function, which is crooked, is quadratic. 9.2.69 Theorem [277, 1816] Let F(x) = xd + βxe be an APN function on F2n, where β ∈F2n and d, e are integers modulo 2n −1. If F is crooked then F is quadratic. 9.2.70 Proposition Let F be an APN permutation on F2n which is crooked. Then n is odd and F is an AB function. 9.2.71 Proposition [1816, 2838] Any APN function, which is is crooked, is plateaued. 260 Handbook of Finite Fields 9.2.8 APN monomials 9.2.72 Remark In cryptography, monomials are usually called power functions. They were inten-sively studied, since they have a lower implementation cost in hardware. Moreover, their properties regarding diﬀerential attacks can be studied more easily, since they are related to the weight enumerators of some cyclic codes with two zeros [590, Section 3.4.2]. 9.2.73 Proposition Let F(x) = xd be a function on F2n. Then CF , deﬁned as in Theorem 9.2.47, is the binary cyclic code of length 2n −1 whose zeros are α, αd and their conjugates; CF is a cyclic code with two zeros. 9.2.74 Remark According to Corollary 9.2.48, if F is APN then the cyclotomic coset of d modulo 2n −1 has size n (with F(x) = xd). In this case, CF has minimum distance 5 and dimension 2n −2n −1. 9.2.75 Example The Melas code is the cyclic code of length 2n −1 with two zeros α and α−1. Its minimum distance is 5 for odd n and 3 for even n. It is the code CF corresponding to the inverse function F(x) = x−1 (see Example 9.2.30). Lachaud and Wolfmann described in the set of weights of C⊥ F . 9.2.76 Proposition [494, 1467] Let F(x) = xd with gcd(d, 2n −1) = 1. Then C⊥ F is exactly 2-divisible if and only if CF is the Melas code, i.e., d = −1. 9.2.77 Remark Let F(x) = xd with gcd(d, 2n −1) = 1. The study of the weights of the code C⊥ F corresponds to the study of the crosscorrelation of a pair of maximal length linear sequences, called m-sequences (see Section 10.3). 9.2.78 Remark Janwa, McGuire, and Wilson characterized several classes of cyclic codes with two zeros whose minimum distance is at most four. More precisely, by applying a form of Weil’s theorem, they showed that, for a large class of such codes, only a ﬁnite number could be good, i.e., have minimum distance ﬁve [1594, 1595]. 9.2.79 Theorem For any ﬁxed d satisfying d ≡3 (mod 4) and d > 3, the cyclic codes CF , F(x) = xd, of length 2n −1, with two zeros α and αd, have codewords of weight 4 for all but ﬁnitely many values of n. 9.2.80 Remark By their work , the authors strengthened the conjecture that APN (a fortiori AB) functions are exceptional. Their main conjecture was solved recently by Hernando and McGuire (see Theorem 9.2.83 below). 9.2.81 Deﬁnition Let F(x) = xd. The exponent d is exceptional if F is APN on inﬁnitely many extension ﬁelds of F2. 9.2.82 Remark The epithet exceptional is due to Dillon who observed that Gold and Kasami exponents can be deﬁned by means of exceptional polynomials (see Section 8.4). Dillon extensively describes the links between these exceptional codes, maps, diﬀerence sets, and polynomials in , a review of great interest. More can be found in [862, 864]. Moreover, reversed permutation Dickson polynomials are closely related with APN monomials . 9.2.83 Theorem Only the Gold and the Kasami numbers are exceptional exponents. 9.2.84 Problem Gold functions and Kasami functions, presented in Examples 9.2.59 and 9.2.53, form the exceptional classes of APN power functions. The other classes, known as sporadic, are listed in Example 9.2.85. It is currently conjectured that any APN power function, which is not exceptional, belongs to one of the known “sporadic” classes (up to equivalence). Special functions over ﬁnite ﬁelds 261 9.2.85 Example The “sporadic” classes, which are known, are the following classes of APN power functions. Such a function F on F2n, F(x) = xd, is designed by its exponent d. 1. The Welch functions: d = 2t + 3 with n = 2t + 1. Any Welch function is AB, proved by Canteaut, Charpin, and Dobbertin (see also [492, 494]). This was conjectured by Welch around 1968, concerning the crosscorrelation function of binary m-sequences. Dobbertin previously proved that F is APN . 2. The Niho functions: for n = 2t + 1, d = 2t + 2t/2 −1 if t is even, 2t + 2(3t+1)/2 −1 otherwise (proposed by Niho , also concerning properties of sequences). The APN property was proved by Dobbertin . Hollmann and Xiang proved that Niho functions are AB in , where they also give another proof of Welch’s conjec-ture. 3. The Dobbertin functions: d = 24k + 23k + 22k + 2k −1 with n = 5k. Dobbertin introduced these functions as potential APN and in proved that these func-tions are APN. For odd n, the Dobbertin functions are not AB [494, Proposition 7.13]. 4. The inverse function: d = 2n−1 −1 for odd n (see Examples 9.2.30 and 9.2.75). 9.2.86 Remark The nonlinearity of APN functions is generally not known. Our last general result concerns monomials: Let n = 2t and F(x) = xd with s = gcd(d, 2n −1), s > 1. Then L(F) ≥2t+1 and, if equality holds, then s = 3 (see and [489, Theorem 8.14]). We note that if F is APN then s = 3 (Proposition 9.2.36). 9.2.87 Conjecture If gcd(d, 2n −1) = 1, n even, then L(xd) ≥2(n+2)/2. 9.2.88 Remark APN monomials appear as speciﬁc objects with very particular properties. Al-though these are rare they are not completely classiﬁed, as explained in Problem 9.2.84. Very little is known about diﬀerential properties of other power functions. It is important to explore these properties, in view of future possible applications . 9.2.89 Remark Monomial APN functions in odd characteristic are investigated in [908, 1479]. We note that there exist planar monomial PN functions (see Section 9.5). See Also §8.4 For exceptional polynomials. §9.1 For basic properties on Boolean functions. §9.5 For planar functions, that is, PN functions in odd characteristics. §10.3 For connections to m-sequences. §16.2 For the design of block ciphers. References Cited: [224, 228, 264, 272, 277, 278, 331, 384, 385, 386, 387, 388, 389, 417, 423, 424, 443, 447, 448, 449, 489, 492, 493, 494, 495, 496, 524, 525, 577, 590, 808, 862, 863, 864, 903, 904, 906, 908, 953, 954, 1467, 1479, 1490, 1526, 1541, 1547, 1594, 1595, 1688, 1689, 1816, 1828, 1939, 1991, 2026, 2287, 2302, 2303, 2304, 2838, 2885] 262 Handbook of Finite Fields 9.3 Bent and related functions Alexander Kholosha, University of Bergen Alexander Pott, Otto-von-Guericke-Universit¨ at Magdeburg 9.3.1 Remark In this section, we give basics and mainly focus on recent results on Boolean bent functions and also provide a rather comprehensive survey of generalized bent functions. For other results on Boolean bent functions, together with most of the proofs, the reader is referred to . 9.3.1 Deﬁnitions and examples 9.3.2 Deﬁnition (Walsh transform) Let f : Fn p →C be a complex-valued function, where p is a prime. Let ζp = e 2πi p be a primitive complex p-th root of unity. Then the Walsh transform of f is the function ˆ f : Fn p →C such that ˆ f(y) = P x∈Fn p f(x) · ζy·x p , where “·” denotes some non-degenerate symmetric bilinear form on Fn p, sometimes also called an inner product. If f : Fn p →Fp, we may deﬁne a corresponding complex-valued function fc by fc(x) = ζf(x) p . The Walsh transform of f is by deﬁnition the Walsh transform of fc. The normalized Walsh coeﬃcients are the numbers p−n/2 ˆ f(y). 9.3.3 Remark 1. When we compute ζy·x p , we interpret the exponent as an integer in {0, . . . , p−1}. 2. The Walsh transform is a special case of the discrete Fourier transform of Abelian groups. 3. More information about the Walsh transform of Boolean functions is contained in Section 9.1, where ˆ f(y) is denoted Wf(y), instead. 4. If we identify the vector space Fn p with the additive group of Fpn, we may use the trace function to deﬁne the bilinear form x · y that occurs in Deﬁnition 9.3.2. We take the bilinear form x · y := Trn(xy), where x, y ∈Fpn. 5. The standard inner product on Fn p is (x1, . . . , xn) · (y1, . . . , yn) = Pn i=1 xiyi. 9.3.4 Remark 1. If we identify Fn p with Fpn, all p-ary functions can be described by Trn(F(x)) for some function F : Fpn →Fpn of degree at most pn −1. This is the univari-ate representation. If we do not identify Fn p with Fpn, the p-ary function has a representation as a multinomial in x1, . . . , xn, where the variables xi occur with exponent at most p −1. This is the multivariate representation. 2. The multivariate representation is unique. 3. The univariate representation is not unique. Unique univariate form of a Boolean function, trace representation, is discussed in Section 9.1. The case of odd p is similar. 9.3.5 Deﬁnition The algebraic degree of a p-ary function is the degree of the polynomial giving its multivariate representation. Special functions over ﬁnite ﬁelds 263 9.3.6 Remark The algebraic degree of a p-ary function is not the degree of the polynomial F in the univariate representation, see Section 9.1. Throughout this section, degree always means the algebraic degree. 9.3.7 Remark The Walsh transform can also be deﬁned for functions F : Fn p →Fm p . In this case, one considers all functions Fa : Fpn →Fp of the form Fa(x) = a · F(x), where a · x denotes an inner product on Fm p . In this section, we mostly consider functions Fn p →Fp, where p is prime. Interesting classes of functions Fn p →Fn p include planar functions, APN functions and permutation polynomials, see Chapter 8 and Sections 9.2, 9.5. 9.3.8 Deﬁnition A function f : Fn p →Fp is a p-ary bent function (or generalized bent function) if all its Walsh coeﬃcients satisfy \| ˆ f(y)\|2 = pn. A bent function f is regular if for every y ∈Fn p, the normalized Walsh coeﬃcient p−n/2 ˆ f(y) is equal to a complex p-th root of unity, i.e., p−n/2 ˆ f(y) = ζf ∗(y) p for some function f ∗mapping Fn p into Fp. A bent function f is weakly regular if there exists a complex number u having unit magnitude such that up−n/2 ˆ f(y) = ζf ∗(y) p for all y ∈Fn p. Hereafter, pn/2 with odd n stands for the positive square root of pn. If f is a weakly regular p-ary bent function (in particular, a Boolean bent function) then f ∗is the dual of f. 9.3.9 Deﬁnition A Boolean function f : Fn 2 →F2 is bent if \| ˆ f(y)\| = \| X x∈Fn 2 (−1)f(x)+y·x\| = 2n/2 for all y ∈Fn 2. 9.3.10 Remark Deﬁnition 9.3.9 is a special case of Deﬁnition 9.3.8. We include it here since most papers on bent functions just deal with the Boolean case. In the Boolean case, the dimension n must obviously be even. Boolean bent functions are regular. 9.3.11 Remark Boolean bent functions that are equal to their dual are self-dual and those whose dual is the complement of the function are anti self-dual [526, 1543]. The class of self-dual bent functions can be extended by deﬁning formally self-dual Boolean functions . 9.3.12 Example 1. The function f(x1, . . . , x2m) := x1x2 + x3x4 + · · · + x2m−1x2m is bent on F2m 2 . 2. If we identify Fn p with the additive group of Fpn, the function f(x) = Trn(x2) is p-ary bent for all odd primes p. This example shows that in the odd characteristic case, the dimension n need not be even. 9.3.13 Deﬁnition (EA-equivalence) Functions f, g : Fn p →Fp are extended-aﬃne equivalent (in brief, EA-equivalent) if there exists an aﬃne permutation L of Fn p, an aﬃne function l : Fn p →Fp and a ∈F∗ p such that g(x) = a(f ◦L)(x) + l(x). 264 Handbook of Finite Fields 9.3.14 Deﬁnition A class of functions is complete if it is a union of EA-equivalence classes. The completed class is the smallest possible complete class that contains the original one. 9.3.15 Remark 1. EA-equivalence of Boolean functions from Deﬁnition 9.1.48 is the case p = 2 in Deﬁnition 9.3.13. 2. If a function f is not aﬃne then any function EA-equivalent to it has the same algebraic degree as f. On the other hand, the function f and its dual f ∗do not necessarily have the same degree (this can be seen in many examples below). 3. Any function which is EA-equivalent to a bent function is bent. 4. For bent functions, CCZ equivalence coincides with EA equivalence (see [451, Theorem 3]). 5. Each Boolean quadratic bent function over F2m 2 is EA-equivalent to the function in Example 9.3.12 (1). Thus, the class of Boolean quadratic bent functions is complete and consists of a single equivalence class. For the case of odd p, see Remark 9.3.36. 9.3.16 Deﬁnition A Boolean function f over F2n is hyper bent if f(xk) is bent for any k coprime with 2n −1 [529, 591]. 9.3.2 Basic properties of bent functions 9.3.17 Theorem [1539, 2486] If f is a p-ary bent function on Fn p then 2 ≤deg f ≤(p −1)n 2 + 1. Moreover, if f is a weakly regular bent function (that includes the case of Boolean bent functions) with (p −1)n ≥4 then 2 ≤deg f ≤(p −1)n 2 . 9.3.18 Remark The Maiorana-McFarland and Dillon classes of bent functions (see Construc-tion 9.3.37, Remark 9.3.39, and Theorem 9.3.59) contain examples of regular functions where the upper bound on the degree is achieved when n is even. Other examples of ternary (weakly) regular bent functions in even dimension that achieve the maximal degree come from Theorem 9.3.62 and are given by Coulter-Matthews exponents with k = n −1 (see Remark 9.3.65). Inﬁnite classes of ternary bent functions in odd dimension (both weakly and non weakly regular) attaining the bound are obtained from plateaued functions using Construction 9.3.69. This construction also gives examples of weakly regular functions in odd dimension with p = 5 that have maximal degree. 9.3.19 Problem It is not known whether the bound for the degree of non weakly regular bent functions is sharp for general p. The same question remains open for weakly regular bent functions with odd n and p > 5. 9.3.20 Theorem [1812, 2486] The dual of a weakly regular bent function is again a weakly regular bent function. Special functions over ﬁnite ﬁelds 265 9.3.21 Theorem [1812, Property 8] The Walsh transform coeﬃcients of a p-ary bent function f with odd p satisfy p−n/2 ˆ f(y) = ( ± ζf ∗(y) p if n is even or n is odd and p ≡1 (mod 4), ± i ζf ∗(y) p if n is odd and p ≡3 (mod 4), where i is a complex primitive 4-th root of unity. Regular bent functions can only be found for even n and for odd n with p ≡1 (mod 4). Also, for a weakly regular bent function, the constant u in Deﬁnition 9.3.8 can only be equal to ±1 and ±i. 9.3.22 Theorem [1812, 2486] (Propagation criterion) A function f : Fn p →Fp is bent if and only if the mappings Da(f) : Fn p →Fp with Da(f)(x) = f(x + a) −f(x) are balanced for all a ̸= 0, i.e., the number of solutions f(x + a) −f(x) = b is pn−1 for all a ̸= 0 and all b ∈Fp. 9.3.23 Example If f(x) = Trn(xp+1), deﬁned on Fpn with p odd and n odd, then Da(f)(x) = f(x + a) −f(x) = Trn(xpa + apx + ap+1). The polynomial xpa + apx is linearized and xpa + apx = 0 if and only if x = 0 or (x/a)p−1 + 1 = 0. But the latter expression has no solution if n is odd, hence xpa+apx is a permutation polynomial and Trn(xpa+apx+ap+1) is balanced, hence f is bent. 9.3.24 Theorem Let n = 2m and f : Fpn →Fp be a p-ary bent function. Then N(0) = pn−1 −µ(pm −pm−1), N(1) = · · · = N(p −1) = pn−1 + µpm−1 , where µ = ±1 and N(j) = #{x ∈Fpn : f(x) = j}. In particular, ˆ f(0) = ±pm. 9.3.3 Bent functions and other combinatorial objects 9.3.25 Remark Weakly regular bent functions are useful for constructing certain combinatorial objects such as partial diﬀerence sets, strongly regular graphs, and association schemes (see [598, 1058, 2424, 2775]). 9.3.26 Theorem A function f : Fn 2 →F2 is bent if and only if the set {x ∈Fn 2 : f(x) = 1} is a diﬀerence set in Fn 2 with parameters (2n, 2n−1 ± 2(n/2)−1, 2n−2 ± 2(n/2)−1; 2n−2) (see Section 14.6.4). Equivalent diﬀerence sets give rise to EA-equivalent bent functions, but EA-equivalent bent functions need not necessarily give rise to equivalent diﬀerence sets, see Example 9.3.28. 9.3.27 Remark Theorem 9.3.26 does not hold for p-ary bent functions with p odd. For general p, the set Rf := {(x, f(x)) : x ∈Fn p} ⊂Fn+1 p is a relative diﬀerence set . 9.3.28 Example If f and g are EA-equivalent Boolean bent functions, the corresponding diﬀer-ence sets need not be equivalent. One reason is that the complemented function f, i.e., f + 1 describes the complementary diﬀerence set. But, more seriously, adding a linear map-ping l to f may not preserve equivalence of the corresponding diﬀerence sets. Using the bivariate notation introduced in Remark 9.3.52, the functions f(x, y) = Tr4(x · y7) and g(x, y) = f(x, y) + Tr4(x) with x, y ∈F24, are EA equivalent. Both functions describe dif-ference sets with parameters (256, 120, 56; 64). These diﬀerence sets are inequivalent since the corresponding designs are not isomorphic. 9.3.29 Theorem The function f : Fn 2 →F2 is bent if and only if #{x ∈Fn 2 : f(x) ̸= l(x) + ϵ} = 2n−1 ± 2(n/2)−1 266 Handbook of Finite Fields for all linear mappings l : Fn 2 →F2 and all ϵ ∈{0, 1}. 9.3.30 Remark Theorem 9.3.29 shows that f has the largest distance to all aﬃne functions, hence bent functions solve the covering radius problem for ﬁrst order Reed-Muller codes of length 2n with n even. In other words, bent functions are maximum nonlinear functions Fn 2 →F2 if n is even. 9.3.4 Fundamental classes of bent functions 9.3.31 Example The following is a complete list of Boolean bent functions on F2m 2 for 1 ≤m ≤3 (up to equivalence). For the functions F3, F4, F5, and F6, we have m = 3. 1. x1x2 for m = 1, 2. x1x2 + x3x4 for m = 2, 3. x1x4 + x2x5 + x3x6 = F3, 4. F3 + x1x2x3 = F4, 5. F4 + x2x4x6 + x1x2 + x4x6 = F5, 6. F5 + x3x4x5 + x1x2 + x3x5 + x4x5 = F6. 9.3.32 Remark The corresponding diﬀerence sets are inequivalent, but the design corresponding to the diﬀerence set from F3 is equivalent to the design corresponding to F4. 9.3.33 Proposition [1991, Chapter 15] Let f be a quadratic function P i,j ai,jxixj in 2m variables over F2. Then f is bent if and only if one of the following equivalent conditions is satisﬁed: 1. The matrix (ai,j + aj,i)i,j=1,...,2m ∈F(2m,2m) 2 is invertible. 2. The symplectic form B(x, y) := f(x + y) + f(x) + f(y) is nondegenerate. 9.3.34 Example The matrix corresponding to the Boolean function f(x) = x1x2 + x2x3 + x3x4 is     0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0    and it is invertible, hence f is bent. Similarly, the function x1x2 + x2x3 + x3x4 + x1x4 is not bent, since the corresponding matrix     0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0    is singular. 9.3.35 Proposition Any quadratic p-ary function f mapping Fpn to Fp is bent if and only if the bilinear form B(x, y) := f(x+y)−f(x)−f(y)+f(0) associated with f is nondegenerate. Moreover, all quadratic p-ary bent functions are (weakly) regular. 9.3.36 Remark There are exactly two inequivalent nondegenerate quadratic forms on Fpn with p odd: f1 := x2 1 + x2 2 + · · · + x2 n, f2 := x2 1 + x2 2 + · · · + g · x2 n, where g is a nonsquare in Fp. Here “equivalence” of quadratic forms is the usual linear algebra equivalence. Both functions are bent. If n is odd, then g · f2 is EA-equivalent to f1, hence there is only one quadratic bent function if n is odd. If n is even, the two quadratic bent functions are EA-inequivalent. Special functions over ﬁnite ﬁelds 267 9.3.37 Construction [861, 1812, 2051] (Maiorana-McFarland) Take any permutation π of Fm p and any function σ : Fm p →Fp. Then f : Fm p × Fm p →Fp with f(x, y) := x · π(y) + σ(y) is a bent function. Moreover, the bijectiveness of π is necessary and suﬃcient for f being bent. Such bent functions are regular and the dual function is equal to f ∗(x, y) = y·π−1(x)+σ(π−1(x)). Boolean quadratic bent functions all belong to the completed Maiorana-McFarland class. On the other hand, for odd p and even n, quadratic bent function f2 in Remark 9.3.36 does not belong to the completed Maiorana-McFarland class . 9.3.38 Example The Boolean function x1x4 +x2x5 +x3x6 +x4x5x6 is a Maiorana-McFarland bent function on 6 variables. 9.3.39 Remark The Maiorana-McFarland construction can be used to construct bent functions of degree (p −1)m by choosing a function σ of degree (p −1)m. 9.3.40 Construction (Partial spread) Let V = Fn 2 with n = 2m. Let Ui, i ∈I, be a collection of subspaces of dimension m with Ui ∩Uj = {0} for all i ̸= j. If \|I\| = 2m−1, the set D− I = S i∈I Ui \{0} is a diﬀerence set with parameters (22m, 22m−1 −2m−1, 22m−2 −2m−1; 22m−2). Similarly, if \|I\| = 2m−1 + 1, the union D+ I = S i∈I Ui is a diﬀerence set with parameters (22m, 22m−1 + 2m−1, 22m−2 + 2m−1; 22m−2). The functions f −: V →F2 (or f +) with f −(x) = 1 if and only if x ∈D− I (or f +(x) = 1 if and only if x ∈D+ I ) are bent functions (see Theorem 9.3.26). For more information on diﬀerence sets see Section 14.6. 9.3.41 Remark These functions are bent functions of partial spread type, since a collection of subspaces of dimension m with pairwise trivial intersection is a partial spread. The functions f + are of type PS+, the others of type PS−. 9.3.42 Example Let q = 2m, and view Fq2 as a 2-dimensional space over Fq, but also as a 2m-dimensional vector space over F2. Then the 1-dimensional subspaces of Fq2 (viewed as a 2-dimensional vector space) are m-dimensional subspaces over F2. The q + 1 subspaces of dimension 1 over Fq are Uα := {α · x : x ∈Fq} where αq+1 = 1, i.e., α is in the multiplicative group of order q + 1 in Fq2. These subspaces intersect pairwise trivially, hence we may take any 2m−1 or 2m−1 + 1 of these subspaces to construct f −or f +. 9.3.43 Remark 1. The parameters of D+ I are complementary to the parameters of D− I , but the diﬀerence sets are, in general, not complements of each other. 2. All functions in PS−have algebraic degree n/2. On the contrary, if n/2 is even then class PS+ contains all quadratic bent functions . 3. A partial spread where the union of the subspaces covers the entire vector space is called a spread. 4. The spread constructed in Example 9.3.42 is the regular spread. 5. Spreads of subspaces of dimension m in 2m-dimensional subspaces can be used to describe translation planes. 6. There are numerous spreads, hence many partial spreads. Many partial spreads are not contained in spreads. For partial spreads contained in spreads, the two constructions in Construction 9.3.40 are complements of each other, i.e., for any f −, there is another partial spread such that f −= f + + 1. 268 Handbook of Finite Fields 9.3.44 Deﬁnition The class PSap is a subclass of PS−where a subspread of the regular spread is used. 9.3.45 Remark All of the functions in the PSap class are hyper bent (see Deﬁnition 9.3.16). 9.3.46 Remark Some of the known constructions of bent functions are direct, that is, do not use as building blocks previously constructed bent functions. We call these primary constructions. The others, sometimes leading to recursive constructions, will be called secondary construc-tions. Most of the secondary constructions of Boolean bent functions are not explained here and can be found in [523, Section 6.4.2]. 9.3.47 Construction (Recursive construction) Let f1 : Fm1 p →Fp and f2 : Fm2 p →Fp be p-ary bent functions, then f1 ∗f2 : Fm1+m2 p with (f1 ∗f2)(x1, x2) = f1(x1) + f2(x2) is p-ary bent. 9.3.48 Remark This construction was formulated originally in a much more general form using relative diﬀerence sets . Since the function f : Fp →Fp with f(x) = x2 is a p-ary bent function if p is odd, Construction 9.3.47 yields bent functions Fn p →Fp for all n (p odd). 9.3.5 Boolean monomial and Niho bent functions 9.3.49 Example The following monomial functions f(x) = Trn(αxd) are bent on F2n with n = 2m: 1. d = 2k + 1 with n/ gcd(k, n) being even and α / ∈{yd : y ∈F2n} ; 2. d = r(2m −1) with gcd(r, 2m +1) = 1 and α ∈F2m being −1 of the Kloosterman sum (see Remark 9.3.61) [591, 861]; 3. d = 22k −2k + 1 with gcd(k, n) = 1 and α / ∈{y3 : y ∈F2n} [864, 1856]; 4. d = (2k + 1)2 with n = 4k and k odd, α ∈ωF2k with ω ∈F4 \ F2 [594, 1879]; 5. d = 22k + 2k + 1 with n = 6k and k > 1, α ∈F23k with TrF23k /F2k (α) = 0 . 9.3.50 Remark These functions are monomial bent functions. Functions in Part 1 are quadratic, and those in Parts 4 and 5 belong to the Maiorana-McFarland class. Functions in Part 2 are in the class PSap. An exhaustive search shows that there are no other monomial bent functions for n ≤20. 9.3.51 Example [531, 532, 907, 1474, 1878] (Niho bent functions) A positive integer d (always understood modulo 2n −1 with n = 2m) is a Niho exponent if d ≡2j (mod 2m −1) for some j < n. As we consider Trn(axd) with a ∈F2n, without loss of generality, we can assume that d is in normalized form, i.e., with j = 0. Then we have a unique representation d = (2m −1)s + 1 with 2 ≤s ≤2m. Following are examples of bent functions consisting of one or more Niho exponents: 1. Quadratic function Trm(ax2m+1) with a ∈F∗ 2m (here s = 2m−1 + 1). 2. Binomials of the form f(x) = Trn(α1xd1+α2xd2), where 2d1 ≡2m + 1 (mod 2n −1) and α1, α2 ∈F∗ 2n are such that (α1 + α2m 1 )2 = α2m+1 2 . Equivalently, denoting a = (α1 + α2m 1 )2 and b = α2 we have a = b2m+1 ∈F∗ 2m and f(x) = Trm(ax2m+1) + Trn(bxd2). We note that if b = 0 and a ̸= 0 then f is a bent function listed under number 1. Special functions over ﬁnite ﬁelds 269 The possible values of d2 are: d2 = (2m −1)3 + 1, 6d2 = (2m −1) + 6 (with the condition that m is even). These functions have algebraic degree m and do not belong to the completed Maiorana-McFarland class . 3. Take r > 1 with gcd(r, m) = 1 and deﬁne f(x) = Trn ax2m+1 + 2r−1−1 X i=1 xdi ! , where 2rdi = (2m −1)i + 2r and a ∈F2n is such that a + a2m = 1. The dual of f, calculated using Proposition 9.3.56, is equal to f ∗(y) = Trm u(1 + y + y2m) + u2n−r + y2m (1 + y + y2m)1/(2r−1) , where u ∈F2n is arbitrary with u + u2m = 1. Moreover, if d < m is a positive integer deﬁned uniquely by dr ≡1 (mod m) then the algebraic degree of f ∗is equal to d + 1. Both functions f and its dual belong to the completed Maiorana-McFarland class. On the other hand, f ∗is not a Niho bent function. 4. Bent functions in a bivariate representation obtained from the known o-polynomials (see Remarks 9.3.52 and 9.3.54). 9.3.52 Remark The bivariate representation of a Boolean function is deﬁned as follows: we identify Fn 2 with F2m × F2m and consider the argument of f as an ordered pair (x, y) of elements in F2m. There exists a unique bivariate polynomial P 0≤i,j≤2m−1 ai,jxiyj over F2m that represents f. The algebraic degree of f is equal to max(i,j) \| ai,j̸=0(w2(i) + w2(j)) (w2(j) is the Hamming weight of the binary expansion of j). Since f is Boolean the bivariate representation can be written in the form of f(x, y) = Trm(P(x, y)), where P(x, y) is some polynomial of two variables over F2m. 9.3.53 Construction [532, 861] Deﬁne class H of functions in their bivariate representation as follows g(x, y) = Trm xH( y x) if x ̸= 0, Trm(µy) if x = 0, (9.3.1) where µ ∈F2m and H is a mapping from F2m to itself with G(z) := H(z) + µz satisfying z 7→G(z) + βz is 2-to-1 on F2m for any β ∈F∗ 2m. (9.3.2) Condition (9.3.2) is necessary and suﬃcient for g to be bent. 9.3.54 Remark 1. Any mapping G on F2m that satisﬁes (9.3.2) is an o-polynomial. 2. Any o-polynomial deﬁnes a hyperoval in PG(2, 2m). Using Construction 9.3.53, every o-polynomial results in a bent function in class H. For a list of known, up to equivalence, o-polynomials, see . 9.3.55 Remark A bent function F2n →F2 with n = 2m belongs to class H if and only if its restriction to each coset uF2m with u ∈F∗ 2n is linear. Thus, Niho bent functions from 270 Handbook of Finite Fields Example 9.3.51 are just functions of class H viewed in their univariate representation. In particular, binomial Niho bent functions with d2 = (2m −1)3 + 1 correspond to Subiaco hyperovals , functions with 6d2 = (2m −1)+6 correspond to Adelaide hyperovals and functions listed under number 3 (consisting of 2r terms) are obtained from Frobenius map G(z) = z2m−r (i.e., translation hyperovals) . 9.3.56 Proposition Let g be a bent function having the form of (9.3.1). Then its dual function g∗, represented in its bivariate form, satisﬁes g∗(α, β) = 1 if and only if the equation H(z) + βz = α has no solutions in F2m. 9.3.6 p-ary bent functions in univariate form 9.3.57 Theorem Consider an odd prime p and nonzero a ∈Fpn. Then for any j ∈{1, . . . , n}, the quadratic p-ary function f mapping Fpn to Fp and given by f(x) = Trn axpj+1 is bent if and only if p gcd(2j,n) −1 ̸ pn −1 2 −ind(a)(pj −1) , (9.3.3) where ind(a) denotes the unique integer t ∈{0, . . . , pn −1} with a = ξt and ξ being a primitive element of Fpn. 9.3.58 Remark The functions in Theorem 9.3.57 are quadratic, hence they are (weakly) regular. In , the dual function has been determined explicitly. 9.3.59 Theorem Let n = 2m and t be an arbitrary positive integer with gcd(t, pm + 1) = 1 for an odd prime p. For any nonzero a ∈Fpn, deﬁne the following p-ary function mapping Fpn to Fp f(x) = Trn axt(pm−1) . (9.3.4) Let K denote the Kloosterman sum (see Deﬁnition 6.1.118). Then for any y ∈F∗ pn, the corresponding Walsh transform coeﬃcient of f is equal to ˆ f(y) = 1 + K apm+1 + pmζ−Trn(apmyt(pm−1)) p and ˆ f(0) = 1 −(pm −1)K apm+1 . Assuming pm > 3, then f is bent if and only if K apm+1 = −1. Moreover, if the latter holds then f is a regular bent function of degree (p −1)m. 9.3.60 Remark According to the result of Katz and Livn´ e (see also [2121, Theorem 6.4]), with c running over F∗ 3m, the Kloosterman sum K(c) takes on all the integer values in the range (−2 √ 3m, 2 √ 3m) that are equal to −1 modulo 3. In particular, there exists at least one a ∈F3n such that K a3m+1 = −1. This means that in the ternary case (i.e., when p = 3) and given the conditions of Theorem 9.3.59, there exists at least one a ∈F3n such that the function (9.3.4) is bent. Moreover, there are no bent functions having the form of (9.3.4) when p > 3 since in this case, the Kloosterman sum never takes on the value −1 as shown in . More on Kloosterman sums is contained in Section 6.1. 9.3.61 Remark Let p = 2 and without loss of generality assume a ∈F2m. Then exactly the same result as in Theorem 9.3.59 holds for any m in the binary case giving the Dillon class of Special functions over ﬁnite ﬁelds 271 bent functions [591, 861, 1879] (see Example 9.3.49 Part 2). Moreover, the Kloosterman sum over F2m takes on all the integer values in the closed range [−2 √ 2m, 2 √ 2m] that are equal to −1 modulo 4 . This means that binary Dillon bent functions exist. 9.3.62 Theorem [1315, 1469, 1470] Let n = 2m with m odd and a = ξ 3m+1 4 , where ξ is a primitive element of F3n. Then the ternary function f mapping F3n to F3 and given by f(x) = Trn ax 3n−1 4 +3m+1 is a weakly regular bent function of degree n. For any y ∈F3n the Walsh transform coeﬃcient of f is equal to ˆ f(y) = −3mζg(y) 3 with g(y) = −Trn a−1y 3n−1 4 +3m+1 X t∈Im Tro(t) (ay−2)t(3m+1) , where o(t) is the size of the cyclotomic coset modulo 3m −1 that contains t and the set Im is deﬁned as follows. Select all such integers in the range {0, . . . , 3m −1} that do not contain 2-digits in the ternary expansion and none of 1-digits are adjacent (the least signiﬁcant digit is cyclically linked with the most signiﬁcant). Split this set into cyclotomic cosets modulo 3m −1, take coset leaders and denote this subset Im. 9.3.63 Theorem Let n = 4k. Then the p-ary function f mapping Fpn to Fp and given by f(x) = Trn xp3k+p2k−pk+1 + x2 is a weakly regular bent function of degree (p −1)k + 2. Moreover, for any y ∈Fpn the corresponding Walsh transform coeﬃcient of f is equal to ˆ f(y) = −p2kζTrk(x0)/4 p , where x0 is a unique root in Fpk of the polynomial yp2k+1 + (y2 + X)(p2k+1)/2 + ypk(p2k+1) + (y2 + X)pk(p2k+1)/2. In particular, if y2 ∈Fp2k then x0 = −TrFp2k /Fpk y2 . 9.3.64 Remark Some sporadic examples of ternary (non) weakly regular bent functions consisting of one or two terms in the univariate representation can be found in . 9.3.7 Constructions using planar and s-plateaued functions 9.3.65 Remark Deﬁnition and other information on planar functions can be found in Section 9.5. In particular, except for one class, all known planar functions are quadratic which means that they can be represented by so called Dembowski-Ostrom polynomials (see Def-inition 9.5.17 and ). The only known example of a nonquadratic planar function is F(x) = x 3k+1 2 over F3n with gcd(k, n) = 1 and odd k, known as Coulter-Matthews function . The following theorem shows that every planar function gives a family of generalized bent functions. 9.3.66 Theorem A function F mapping Fpn to itself is planar if and only if for every nonzero a ∈Fpn the function Trn(aF) is generalized bent. 272 Handbook of Finite Fields 9.3.67 Remark The generalized bent functions Trn(aF) obtained from Dembowski-Ostrom poly-nomials are quadratic, hence they are weakly regular (see Proposition 9.3.35). It was shown in [1055, 3042] that the bent functions coming from the Coulter-Matthews planar functions are also weakly regular. 9.3.68 Deﬁnition A function f mapping Fpn to Fp is s-plateaued if its Walsh coeﬃcients either equal zero or satisfy \| ˆ f(y)\|2 = pn+s. The case s = 0 corresponds to bent functions. 9.3.69 Construction [571, 572] For every a = (a1, a2, . . . , as) ∈Fs p let fa be an s-plateaued function from Fpn to Fp such that ˆ fa(t)· ˆ fb(t) ≡0 for any a, b ∈Fs p with a ̸= b and t ∈Fpn. Then function f(x, y1, y2, . . . , ys) from Fpn × Fs p to Fp deﬁned by f(x, y1, y2, . . . , ys) = (−1)s X a∈Fs p Qs i=1 yi(yi −1) · · · (yi −(p −1)) (y1 −a1) · · · (ys −as) fa(x) is bent. Moreover, for any t ∈Fpn and a ∈Fs p, the corresponding Walsh transform coeﬃcient of f is equal to ˆ f(t, a) = ζ−a·y p ˆ fy(t), where y ∈Fs p is unique with ˆ fy(t) ̸= 0. 9.3.70 Remark Quadratic functions are always plateaued. Then one can construct s-plateaued functions with the prescribed properties by adding suitably chosen linear functions to such quadratic s-plateaued functions. In this case, the constructed bent function f has degree (p −1)s + 2 (resp. (p −1)s + 1) if P a∈Fs p fa is quadratic (resp. aﬃne). It is possible to construct speciﬁc families of quadratic s-plateaued functions that lead to weakly regular bent functions when n −s is even. Similar constructions with n −s odd can lead both to weakly and non weakly regular bent functions. On the other hand, one can take a suitable family of (n −1)-plateaued quadratic functions (maximal order achievable by a non aﬃne function). Then in the case when the degree of the corresponding bent function exceeds (p −1)n/2 being the maximum for a weakly regular function (see Theorem 9.3.17), the obtained functions are non weakly regular. 9.3.8 Vectorial bent functions and Kerdock codes 9.3.71 Deﬁnition A mapping f : Fn p →Fm p is a vectorial bent function if all its component functions fa : Fn p →Fp deﬁned by fa(x) = a · x are bent for all a ̸= 0. Here “·” denotes a symmetric bilinear form (see Deﬁnition 9.3.2). 9.3.72 Example We identify the vector spaces Fn p and Fm p with the ﬁnite ﬁelds Fpn and Fpm. 1. Any planar function is a vectorial bent function Fpn →Fpn. 2. If n = 2m, then f(x, y) = xy is a vectorial bent function Fpm × Fpm →Fpm. All component functions are quadratic. 3. If n = 2m, then f(x, y) = xypm−2 is a vectorial bent function Fpm × Fpm →Fpm. 9.3.73 Theorem If f : F2m 2 →Fn 2 is vectorial bent, then n ≤m and equality can be attained (see Example 9.3.72). 9.3.74 Remark A collection of 22m−1 −1 quadratic Boolean bent functions in 2m variables such that the sum of any two of them is bent again, gives rise to a Kerdock code of length 22m. The Boolean functions in Example 9.3.72 Part 2 only give rise to 2m−1 such bent functions. Special functions over ﬁnite ﬁelds 273 See Also §6.1 Kloosterman sums are related to some classes of bent functions. §7.2 The classical bent functions are related to quadratic forms. §9.1 Boolean bent functions are special types of Boolean functions. §9.2 Via the trace mapping, PN and some APN functions give rise to bent functions. §9.5 Any planar function is a vectorial bent function. §14.6 Boolean bent functions are special types of diﬀerence sets. A very good source for vectorial functions. This paper contains a lot of information about the symmetries of bent functions. A description of Kerdock codes which initiated work on codes over rings. Lander’s book includes information about bent functions and diﬀerence sets. A thesis about bent functions has been written by Timo Neumann. References Cited: [445, 446, 451, 495, 523, 524, 526, 527, 529, 531, 532, 571, 572, 591, 594, 598, 732, 809, 861, 864, 907, 1055, 1058, 1295, 1315, 1409, 1469, 1470, 1471, 1472, 1473, 1474, 1475, 1539, 1543, 1565, 1695, 1785, 1812, 1828, 1842, 1856, 1878, 1879, 1991, 2051, 2121, 2224, 2302, 2423, 2424, 2486, 2775, 3038, 3042] 9.4 κ-polynomials and related algebraic objects Robert Coulter, The University of Delaware 9.4.1 Deﬁnitions and preliminaries 9.4.1 Deﬁnition A polynomial f(x1, . . . , xn) is a κ-polynomial over Fq if ka = #{(x1, . . . , xn) ∈Fn q : f(x1, . . . , xn) = a}, is independent of a for a ∈F∗ q. 9.4.2 Theorem The polynomial f(x1, . . . , xn) is a κ-polynomial over Fq if and only if X a∈Fn q χ(f(a)) = 0 for all non-trivial multiplicative characters χ of Fq. 9.4.3 Remark This should be compared to Corollary 8.2.7 of Section 8.2. There are almost no results in the literature directly discussing κ-polynomials. This seems altogether surprising since the speciﬁed regularity on preimages of all non-zero elements of the ﬁeld suggests such polynomials must almost certainly appear in many guises other than those outlined here. 9.4.4 Lemma Let f(x1, . . . , xn) be a κ-polynomial over Fq and g be any polynomial in variables disjoint from x1, . . . , xn. Then fg is a κ-polynomial. 274 Handbook of Finite Fields 9.4.5 Deﬁnition Let S = ⟨S, +, ⋆⟩be a set with two binary operations, addition + and multi-plication ⋆and where ⟨S, +⟩forms a group. 1. S is a Cartesian group if it has no zero divisors and ⟨S∗, ⋆⟩forms a loop. 2. S is a left (resp. right) quasiﬁeld if, in addition to being a Cartesian group, S also has a left (resp. right) distributive law. 3. S is a semiﬁeld if it is a non-associative division ring – that is, a Cartesian group with both distributive laws. If we do not insist on a multiplicative identity, so that ⟨S∗, ⋆⟩forms a quasigroup instead of a loop, then we may speak of a pre-Cartesian group, prequasiﬁeld, or pre-semiﬁeld. 9.4.6 Remark Each algebraic object corresponds with a type of projective plane under the Lenz-Barlotti classiﬁcation. Speciﬁcally, Cartesian groups correspond to type II, quasiﬁelds to type IV, and semiﬁelds to type V. For more information along these lines see Section 14.3, as well as the classical texts of Dembowski and Hughes and Piper . See also Subsection 2.1.7.6. 9.4.7 Theorem 1. The additive group of a quasiﬁeld is necessarily an elementary abelian p-group. 2. Any semiﬁeld of prime or prime squared order is necessarily a ﬁnite ﬁeld. 3. Proper semiﬁelds, i.e., semiﬁelds which are not ﬁelds, exist for all prime power orders pe ≥16 with e ≥3. 9.4.8 Theorem Let M(x, y) be a κ-polynomial over Fq satisfying M(x, y) = 0 if and only if xy = 0. Deﬁne the sets L and R as follows: L = {M(x, a) ∈Fq[x] : a ∈F∗ q}; R = {M(a, x) ∈Fq[x] : a ∈F∗ q}. Consider the algebraic object S = ⟨Fq, +, ⋆⟩where x ⋆y = M(x, y) for all x, y ∈Fq. 1. If L∪R contains only permutation polynomials over Fq, then S is a pre-Cartesian group. 2. If in addition to Part 1, precisely one of L or R contains only linearized polyno-mials, then S is a prequasiﬁeld. 3. If in addition to Part 1, L ∪R contains only linearized polynomials, then S is a presemiﬁeld. 9.4.9 Remark Theorem 9.4.8 shows how κ-polynomials underpin all of these algebraic structures. Indeed, it follows from Theorem 9.4.7 Part 1 that there is a one-to-one correspondence between either quasiﬁelds or semiﬁelds and κ-polynomials satisfying the corresponding part of Theorem 9.4.8. Hence, when we talk of a quasiﬁeld or semiﬁeld S = ⟨Fq, +, ⋆⟩, we may refer to the κ-polynomial corresponding to S, by which we mean the reduced bivariate polynomial M ∈Fq[x, y] satisfying M(x, y) = x ⋆y for all x, y ∈Fq. It is this point that underlines the philosophy behind the presentation of material in this section, which is essentially to provide a polynomial-based or, if one prefers, algebraic approach to this area, thematically linking the material with the chapter in which it resides. That said, it needs to be underlined that the literature on quasiﬁelds and especially semiﬁelds is vast, and that κ-polynomials represent only one approach to this subject. An excellent and detailed survey of what might be termed the geometric approach to semiﬁelds was recently given by Lavrauw and Polverino ; there one will ﬁnd many more results, along with discussions on several topics reluctantly omitted here, such as the Knuth operations. Special functions over ﬁnite ﬁelds 275 9.4.10 Remark There are many examples of quasiﬁeld planes – for example, the classiﬁcation of translation planes of order 49 by Mathon and Royle reveals there are 1347 non-isomorphic such planes. However, there does not appear to have been any attempt to study the corresponding κ-polynomials, nor the non-linearized permutation polynomials in L or R that arise from quasiﬁelds (through the known examples or by attempting to prove restrictions on the permissible sets of permutation polynomials which can represent the non-distributive side of a quasiﬁeld). 9.4.11 Remark Unsurprisingly, given the signiﬁcant amount of extra structure, much more is known about semiﬁelds than quasiﬁelds or Cartesian groups. Consequently, the remainder of this section focuses on the semiﬁeld case. 9.4.2 Pre-semiﬁelds, semiﬁelds, and isotopy 9.4.12 Deﬁnition Let S1 = ⟨Fq, +, ⋆⟩and S2 = ⟨Fq, +, ∗⟩be two presemiﬁelds. Then S1 and S2 are isotopic if there exists three non-singular linear transformations L, M, N ∈Fq[x] of Fq over Fp such that for all x, y ∈Fq : M(x) ⋆N(y) = L(x ∗y). The triple (M, N, L) is an isotopism between S1 and S2. An isotopism of the form (N, N, L) is a strong isotopism and two presemiﬁelds are strongly isotopic if there exists a strong isotopism between them. 9.4.13 Remark This deﬁnition of equivalence, which is clearly much weaker than the standard ring isomorphism, arises from projective geometry, where its importance is underlined by the following result of Albert. 9.4.14 Theorem Two presemiﬁelds coordinatize isomorphic planes if and only if they are isotopic. 9.4.15 Remark Any presemiﬁeld S = ⟨Fq, +, ◦⟩is isotopic to a semiﬁeld via the following trans-formation: choose any α ∈F∗ q and deﬁne a new multiplication ⋆by (x ◦α) ⋆(α ◦y) = x ◦y for all x, y ∈Fq. Then S′ = ⟨Fq, +, ⋆⟩is a semiﬁeld with identity α ◦α, isotopic to S. Any such S′ ia a semiﬁeld corresponding to S. If S is commutative, then S′ is strongly isotopic to S. 9.4.16 Remark There is no general classiﬁcation result for semiﬁelds, in direct contrast to ﬁnite ﬁelds, which were classiﬁed in the 1890s. However, some computational results on the clas-siﬁcation problem do exist. In light of Menichetti’s results – see Theorem 9.4.24 below – the problem remains open for semiﬁelds of order pe with e ≥4. Semiﬁelds of order N have been classiﬁed up to isotopism for the following N: N = 16 , N = 32 , N = 64 , N = 81 , and N = 243 . 9.4.3 Semiﬁeld constructions 9.4.17 Proposition (Dickson’s commutative semiﬁelds) Let q = pe with p an odd prime, e > 1 and let {1, λ} be basis for Fq2 over Fq. For any j a non-square of Fq and any non-trivial automorphism σ of Fq, we deﬁne a binary operation ⋆by 276 Handbook of Finite Fields for all a, b, c, d ∈Fq, (a + λb) ⋆(c + λd) = ac + j(bd)σ + λ(ad + bc). 1. Dj,σ = ⟨Fq2, +, ⋆⟩is a commutative semiﬁeld. 2. For j, k distinct non-squares of Fq, Dj,σ and Dk,σ are isotopic. 3. [466, 2521] Dj,σ and Dj,τ are isotopic if and only if σ = τ or σ = τ −1. In such cases, they are strongly isotopic. 9.4.18 Proposition (Albert’s generalized twisted ﬁelds) [72, 73] For any prime power q, select σ, τ ∈Aut(Fq) and let j ∈Fq be any element satisfying (xy)−1xσyτ = j has no solution for x, y ∈F∗ q. Deﬁne a binary operation ⋆by for all x, y ∈Fq, x ⋆y = xy −jxσyτ. 1. A(σ, τ, j) = ⟨Fq, +, ⋆⟩is a presemiﬁeld. 2. A(σ, τ, j) is isotopic to a commutative semiﬁeld if and only if σ = τ −1 and j = −1. 3. A(σ, τ, j) is isomorphic to a ﬁnite ﬁeld if and only if σ = τ. 9.4.19 Remark The Dickson semiﬁelds of Proposition 9.4.17 were the ﬁrst published examples of non-associative ﬁnite division rings, while Albert’s construction is both historically im-portant and fundamental, as it has played a signiﬁcant role in subsequent classiﬁcation results, see Theorem 9.4.24 below. These two constructions are the only ones needed for our discussion. To highlight the variety of construction techniques developed, even for isotopic semiﬁelds, we direct the interested reader to the following not exhaustable list: 1. the paper of Knuth which contains several constructions; 2. the Cohen-Ganley and Ganley commutative semiﬁelds [689, 1170]; 3. the Jha-Johnson semiﬁelds [1608, 1680]; 4. the Hiramine-Matsumoto-Oyama quasiﬁeld construction [1508, 1679]; 5. the Kantor-Williams semiﬁelds ; 6. the semiﬁeld construction using spread sets viewed as linear maps [518, 952]. 9.4.4 Semiﬁelds and nuclei 9.4.20 Deﬁnition Let S = ⟨Fq, +, ⋆⟩be a ﬁnite semiﬁeld. We deﬁne the left, middle, and right nucleus of S, denoted by Nl, Nm and Nr, respectively, as follows: Nl(S) = {α ∈S \| (α ⋆x) ⋆y = α ⋆(x ⋆y) for all x, y ∈S}, Nm(S) = {α ∈S \| (x ⋆α) ⋆y = x ⋆(α ⋆y) for all x, y ∈S}, Nr(S) = {α ∈S \| (x ⋆y) ⋆α = x ⋆(y ⋆α) for all x, y ∈S}. The set N(S) = Nl ∩Nm ∩Nr is the nucleus of S. 9.4.21 Remark It is easy to show that all nuclei are ﬁnite ﬁelds. The nuclei measure how far S is from being associative. Moreover, the orders of the nuclei are invariants of S under isotopy and so act as a coarse signature of the semiﬁeld. We note that if S is commutative, then Nl = Nr ⊂Nm. 9.4.22 Theorem (Restricting the corresponding κ-polynomial) Let n and e be natural num-bers. Set q = pe for some odd prime p and t(x) = xq −x. Let S be a semiﬁeld of order qn with middle nucleus containing Fq. Then there exists a semiﬁeld S′ = ⟨Fqn, +, ⋆⟩, isotopic to Special functions over ﬁnite ﬁelds 277 S, with corresponding κ-polynomial M ∈Fqn[x, y] satisfying M(x, y) = K(t(x), t(y))+ 1 2xy and where K(x, y) is of the shape K(x, y) = (n−1)e−1 X i,j=0 aijxpiypj. Furthermore, if S′ is commutative and N = Fpk, then K is symmetric in x and y and the only non-zero terms of K are of the form xpkiypkj. 9.4.23 Remark Any semiﬁeld S can be represented as a right vector space over Nl, a left vector space over Nr and both a left or right vector space over Nm. The concept of dimension therefore naturally arises in the study of semiﬁelds. 9.4.24 Theorem (Menichetti’s classiﬁcation results based on dimension) 1. Any three dimensional semiﬁeld over N is necessarily either a ﬁnite ﬁeld or a twisted ﬁeld. 2. Fix d to be a prime. For suﬃciently large q, any semiﬁeld of dimension d over N = Fq is necessarily either a ﬁnite ﬁeld or a twisted ﬁeld. 9.4.25 Theorem (Dimension two results for commutative semiﬁelds) Let S be a commutative semiﬁeld of order q2n with [S : Nm] = 2. 1. If q is even, then S is a ﬁnite ﬁeld. 2. If q is odd and q ≥4n2 −8n + 2, then S is necessarily a ﬁnite ﬁeld or a Dickson semiﬁeld. 9.4.26 Theorem (Strong isotopy for commutative semiﬁelds) Let S1 = ⟨Fq, +, ⋆⟩and S2 = ⟨Fq, +, ∗⟩be isotopic commutative presemiﬁelds and let S′ 1 be any commutative semi-ﬁeld corresponding to S1. Set d = [Nm(S′ 1) : N(S′ 1)]. 1. If d is odd, then S1 and S2 are strongly isotopic. 2. If d is even, then either S1 and S2 are strongly isotopic or the only isotopisms between any two corresponding commutative semiﬁelds S′ 1 and S′ 2 are of the form (α ⋆N, N, L) where α is a non-square element of Nm(S′ 1). In particular, if q = pe with p = 2 or e odd, then S1 and S2 are strongly isotopic. 278 Handbook of Finite Fields See Also For a recent attempt to resolve the problem of establishing inequivalence between projective planes. Whether dealing with Cartesian groups, quasiﬁelds or semiﬁelds, present methods for establishing the inequivalence of examples are technical and generally unwieldy. Of major interest would be a new and eﬃcient method for doing so. Determines many ovals in semiﬁelds, as well as in the planes generated by the planar functions of Proposition 9.5.11 Part 2 of the next section. Gives a good discussion on the coordinatization method for projective planes. Gives a conjecture concerning the asymptotic number of pairwise non-isotopic semiﬁelds of ﬁxed characteristic. The conjecture is proved for characteristic two by Kantor ; see also [1679, 1682]. For a recent survey emphasizing the geometric approach to semiﬁelds by two of the leading authors in that ﬁeld. Conjectures the existence of left or right primitive elements (suitably deﬁned) in ﬁnite semiﬁelds. This was proved by Gow and Sheekey for semiﬁelds of suﬃciently large order relative to the characteristic. See [1487, 2490] for counterexamples of small order in characteristic two. References Cited: [71, 138, 327, 466, 518, 689, 728, 729, 787, 807, 810, 845, 847, 952, 1170, 1346, 1487, 1508, 1560, 1608, 1677, 1678, 1679, 1680, 1682, 1751, 1764, 1871, 2025, 2034, 2081, 2082, 2490, 2491, 2492, 2521, 2891, 2969] 9.5 Planar functions and commutative semiﬁelds Robert Coulter, The University of Delaware 9.5.1 Deﬁnitions and preliminaries 9.5.1 Deﬁnition Let G and H be arbitrary ﬁnite groups, written additively, but not necessarily abelian. A function f : G →H is a planar function if for every non-identity a ∈G the functions ∆f,a : x 7→f(a + x) −f(x) and ∇f,a : x 7→−f(x) + f(x + a) are bijections. A polynomial f ∈Fq[x] is planar over Fq if the function induced by f on Fq is a planar function (on ⟨Fq, +⟩). 9.5.2 Remark A few points should be made clear from the outset. 1. Planar functions do not exist over groups of even order. 2. If f : G →H is planar with G and H abelian, then so is f + φ + c where φ is any homomorphism from G to H and c ∈H a constant. In particular, if f ∈Fq[x] is planar, then so is f + L for any linearized polynomial L ∈Fq[x]. 3. At the time of writing, all known planar functions can be deﬁned over ﬁnite ﬁelds; that is, when G = H = ⟨Fq, +⟩with q odd. Special functions over ﬁnite ﬁelds 279 4. The polynomial x2 is planar over any ﬁnite ﬁeld of odd characteristic. 5. A planar function is a perfect non-linear function in one variable, see Section 9.2. 9.5.3 Theorem Suppose f : G →H is a planar function with G and H abelian. Then G×H is a p-group of order p2b for some natural number b and the minimum number of generators of G × H is at least b + 1. 9.5.4 Remark In a letter to the author, Jill C.D.S. Yaqub claimed to have a proof of “about 8 hand-written pages” in length which showed that if a planar function existed mapping G to H, then G and H were necessarily abelian. Along with the above result, this would go a long way to proving that G and H are necessarily elementary abelian. Unfortunately, Yaqub passed away before she received my return correspondence urging her to publish. In her memory, we record this as a conjecture. 9.5.5 Conjecture (Yaqub’s Conjecture) If f : G →H is a planar function, then G and H are abelian. 9.5.2 Constructing aﬃne planes using planar functions 9.5.6 Remark Given groups G and H as in Deﬁnition 9.5.1 and a function f : G →H, an incidence structure I(G, H; f) may be deﬁned as follows: “Points” are the elements of G×H; “Lines” are the symbols L(a, b) with (a, b) ∈G × H, together with the symbols L(c) with c ∈G; incidence is deﬁned by (x, y) ∈L(a, b) if and only if y = f(x −a) + b; and (x, y) ∈L(c) if and only if x = c. 9.5.7 Theorem Let G and H be groups and f : G →H be a function. 1. The function f is planar if and only if I(G, H; f) is an aﬃne plane. 2. Suppose G × H is abelian and f is planar. Then I(G, H; f) allows a collineation group, isomorphic to G × H, which acts transitively on the aﬃne points as well as on the set of lines {L(a, b) : (a, b) ∈G × H}. 9.5.8 Remark Theorem 9.5.7 constitutes Dembowski and Ostrom’s original motivation for study-ing planar functions. In the case where f is planar over Fq, we use I(f) to denote the corresponding aﬃne plane and P(f) its projective closure. 9.5.9 Lemma If f ∈Fq[x] is a planar polynomial, then P(f) can only be of Lenz-Barlotti type II.1, V.1, or VII.2. 9.5.10 Lemma Let Γ = ΓL(1, q) denote the group of all semilinear automorphisms of Fq, viewed as a vector space over its prime ﬁeld, and set G = ⟨Fq, +⟩. If xn is planar over Fq and P(xn) is Lenz-Barlotti type II.1, then Aut(P(xn)) ≈Γ · (G × G). 9.5.3 Examples, constructions, and equivalence 9.5.11 Proposition 1. xpα+1 is planar over Fpe if and only if e/ gcd(α, e) is odd. 2. x(3α+1)/2 is planar over F3e if and only if gcd(α, 2e) = 1. 3. [732, 874] For any a ∈F3e, x10 + ax6 −a2x2 is planar over F3e if and only if e is odd or e = 2. 280 Handbook of Finite Fields 9.5.12 Remark Many more examples can be generated from commutative semiﬁelds of odd order, see Section 9.5.5. The examples of Proposition 9.5.11 Part 2 do not generalize to larger characteristic – apply Proposition 9.5.24 Part 2. When e is odd, these examples generate the only Lenz-Barlotti type II planes of non-square order known ; see [263, 787, 1758] for more concerning these planes. 9.5.13 Deﬁnition Let f, g ∈Fq[x] be two planar polynomials over Fq. Set f1 (respectively g1) to be the polynomial which results when f (respectively g) is stripped of all linearized and constant terms. Then f and g are planar equivalent if there exist two linearized permutation polynomials L, M ∈Fq[x] satisfying L(f1(x)) ≡g1(M(x)) mod (xq −x). 9.5.14 Theorem Let f ∈Fq[x] and let L ∈Fq[x] be a linearized polynomial. Then the following are equivalent. 1. f(L) is a planar polynomial. 2. L(f) is a planar polynomial. 3. f is a planar polynomial and L is a permutation polynomial. 9.5.15 Theorem (Isomorphic planes and planar equivalence) 1. Let f, g ∈Fq[x] be planar polynomials. If f and g are planar equivalent, then I(f) and I(g) are isomorphic. 2. Let xm and xn be planar over Fpe. Then I(xm) and I(xn) are isomorphic if and only if m ≡npi mod (pe −1) for some suitable choice of integer i. 9.5.16 Remark The converse of Theorem 9.5.15 Part 1 is false as both situations of Theorem 9.4.26 Part 2 do actually occur; see Theorem 9.5.25 below. 9.5.4 Classiﬁcation results, necessary conditions, and the Dembowski-Ostrom Conjecture 9.5.17 Deﬁnition A Dembowski-Ostrom (or DO) polynomial over a ﬁeld of characteristic p is any polynomial of the shape X i,j aijxpi+pj. 9.5.18 Theorem Let f ∈Fq[x] with deg(f) < q. The following are equivalent. 1. f = D + L + c, where D is a Dembowski-Ostrom polynomial, L is a linearized polynomial and c ∈Fq is a constant. 2. For each a ∈F∗ q, ∆f,a = La + ca where La is a linearized polynomial and ca ∈Fq is a constant (both depending on a). 9.5.19 Conjecture (The Dembowski-Ostrom Conjecture) Let q be a power of a prime p ≥5. If f ∈Fq[x] is a planar polynomial, then f(x) = D(x)+L(x)+c, where D is a DO polynomial, L is a linearized polynomial and c is a constant. 9.5.20 Remark Though it was only formally made a conjecture by R´ onyai and Sz¨ onyi , the conjecture stems from a question posed by Dembowski and Ostrom in , hence the attribution. Originally stated for all odd characteristics, Proposition 9.5.11 Part 2 shows the conjecture is false in characteristic 3. However, the class of all polynomials planar equivalent Special functions over ﬁnite ﬁelds 281 to those examples remain the only known counterexamples in any characteristic and it is eminently possible that no other non-DO examples exist. 9.5.21 Theorem (Classiﬁcation results) 1. [1286, 1506, 2481] The polynomial f ∈Fp[x] is planar over Fp if and only if the reduced form of f is a quadratic. 2. The polynomial xn is planar over Fp2 if and only if n ≡2pi mod (p2 −1) for some integer i ∈{0, 1}. 3. For prime p ≥5, the polynomial xn is planar over Fp4 if and only if n ≡ 2pi mod (p4 −1) for some integer i ∈{0, 1, 2, 3}. 9.5.22 Remark These results represent the only general classiﬁcation results on planar polynomials so far obtained. 9.5.23 Proposition (General necessary conditions) [734, 871, 1818] Let f ∈Fq[x] and let V (f) = {f(a) : a ∈Fq}. A necessary condition for f to be planar over Fq is #V (f) ≥(q + 1)/2. If f is a DO polynomial, then this condition is also suﬃcient. 9.5.24 Proposition (Necessary conditions for monomials) Let xn be planar over Fpe. The following statements hold. 1. gcd(n, pe −1) = 2. 2. n ≡2 mod (p −1). 3. If 2\|e, then n ≡2pi mod (p2 −1). 4. If 4\|e, then n ≡2pi mod (p4 −1). 9.5.5 Planar DO polynomials and commutative semiﬁelds of odd order 9.5.25 Theorem (Correspondence) If S = ⟨Fq, +, ⋆⟩is a commutative presemiﬁeld of odd order, then f(x) = 1 2(x⋆x) describes a planar DO polynomial. Conversely, let f ∈Fq[x] be a planar DO polynomial and deﬁne the κ-polynomial M ∈Fq[x, y] by M(x, y) = f(x + y) −f(x) − f(y). Then Sf = ⟨Fq, +, ⋆⟩, where x ⋆y = M(x, y) for all x, y ∈Fq, deﬁnes a commutative presemiﬁeld, the presemiﬁeld corresponding to f. 9.5.26 Remark More information on κ-polynomials can be found in Section 9.4. 9.5.27 Remark By Theorem 9.5.25, classifying commutative semiﬁelds of odd order is equivalent to determining the “isotopism classes” of planar DO polynomials. Moreover, Theorem 9.4.26 shows that any strong isotopism class of commutative semiﬁelds can split into at most two isotopism classes, so that there are sound reasons for considering the strong isotopism problem on planar DO polynomials instead of the more diﬃcult isotopism problem. 9.5.28 Theorem (Strong isotopy and planar equivalence) Let f, g ∈Fq[x] be planar DO polynomials with corresponding commutative presemiﬁelds Sf and Sg. There is a strong isotopism (N, N, L) between Sf and Sg if and only if f(N(x)) ≡L(g(x)) mod (xq −x). 282 Handbook of Finite Fields See Also §9.2 For APN functions that are closely related to planar functions. §9.3 For bent functions that are closely related to planar functions. §14.3 For aﬃne and projective planes; the seminal paper clearly outlines the main properties of the planes constructed via planar functions. §14.6 Discusses diﬀerence sets. Ding and Yuan used the examples of Proposition 9.5.11 Part 3 to disprove a long-standing conjecture on skew Hadamard diﬀerence sets; see also [871, 2972]. Construct further classes of planar DO polynomials; see also , , , , , . The problem of planar (in)equivalence between these constructions is not completely resolved at the time of writing. An incredible new class, which combines Albert’s twisted ﬁelds with Dickson’s semiﬁelds, was very recently discovered by Pott and Zhou . Classiﬁes planar DO polynomials over ﬁelds of order p2 and p3. This does not constitute a classiﬁcation of planar polynomials over ﬁelds of these orders. Applies Theorem 9.4.22 to commutative presemiﬁelds of odd order to restrict both the form of the DO polynomials and the isotopisms that need to be considered; see also . A promising alternative approach (which applies also to APN functions, see Section 9.2) is outlined in , while a third approach was given recently in . For results on possible forms of planar functions not deﬁned over ﬁnite ﬁelds. Gives speciﬁc forms for planar DO polynomials corresponding to the Dickson semiﬁelds , the Cohen-Ganley semiﬁelds , the Ganley semiﬁelds , and the Penttila-Williams semiﬁeld . References Cited: [263, 273, 274, 326, 450, 451, 689, 726, 728, 729, 731, 732, 734, 787, 808, 811, 847, 871, 874, 1170, 1286, 1506, 1507, 1611, 1758, 1799, 1818, 2383, 2393, 2425, 2481, 2972, 2973, 3059, 3060] 9.6 Dickson polynomials Qiang Wang, Carleton University Joseph L. Yucas, Southern Illinois University 9.6.1 Basics 9.6.1 Deﬁnition Let n be a positive integer. For a ∈Fq, we deﬁne the n-th Dickson polynomial of the ﬁrst kind Dn(x, a) over Fq by Dn(x, a) = ⌊n/2⌋ X i=0 n n −i n −i i (−a)ixn−2i. Special functions over ﬁnite ﬁelds 283 9.6.2 Theorem (Waring’s formula, [1939, Theorem 1.76]) Let σ1, . . . , σk be elementary symmetric polynomials in the variables x1, . . . , xk over a ring R and sn = sn(x1, . . . , xk) = xn 1 + · · · + xn k ∈R[x1, . . . , xk] for n ≥1. Then we have sn = X (−1)i2+i4+i6+··· (i1 + i2 + · · · + ik −1)!n i1!i2! · · · ik! σi1 1 σi2 2 · · · σik k , for n ≥1, where the summation is extended over all tuples (i1, i2, . . . , in) of nonnegative integers with i1 + 2i2 + · · · + kik = n. The coeﬃcients of the σi1 1 σi2 2 · · · σik k are integers. 9.6.3 Theorem Dickson polynomials of the ﬁrst kind are the unique monic polynomials satisfying the functional equation Dn y + a y , a = yn + an yn , where a ∈Fq and y ∈Fq2. Moreover, they satisfy the recurrence relation Dn(x, a) = xDn−1(x, a) −aDn−2(x, a), for n ≥2 with initial values D0(x, a) = 2 and D1(x, a) = x. 9.6.4 Remark The Dickson polynomial Dn(x, a) of the ﬁrst kind satisﬁes a commutative type of relation under composition. That is, Dmn(x, a) = Dm(Dn(x, a), an). Hence the set of all Dickson polynomials Dn(x, a) of even degree over Fq are closed under composition if and only if a = 0 or a = 1. In particular, if a = 0 or 1 then Dm(x, a) ◦Dn(x, a) = Dn(x, a) ◦Dm(x, a). Moreover, the set of all Dickson polynomials Dn(x, a) of odd degree over Fq is closed under composition if and only if a = 0, a = 1 or a = −1. See [1936, 1939] for more details. 9.6.5 Deﬁnition For a ∈Fq, we deﬁne the n-th Dickson polynomial of the second kind En(x, a) over Fq by En(x, a) = ⌊n/2⌋ X i=0 n −i i (−a)ixn−2i. 9.6.6 Theorem Dickson polynomials of the second kind have a functional equation En(x, a) = En y + a y , a = yn+1 −an+1/yn+1 y −a/y , for y ̸= ±√a; for y = ±√a, we have En(2√a, a) = (n + 1)(√a)n and En(−2√a, a) = (n + 1)(−√a)n; here a ∈Fq and y ∈Fq2. Moreover, they satisfy the recurrence relation En(x, a) = xEn−1(x, a) −aEn−2(x, a), for n ≥2 with initial values E0(x, a) = 1 and E1(x, a) = x. 9.6.7 Remark In the case a = 1, denote Dickson polynomials of degree n of the ﬁrst and the second kinds by Dn(x) and En(x), respectively. These Dickson polynomials are closely related over the complex numbers to the Chebychev polynomials through the connections Dn(2x) = 2Tn(x) and En(2x) = Un(x), where Tn(x) and Un(x) are Chebychev polynomials of degree n of the ﬁrst and the second kind, respectively. In recent years these polynomials have received an extensive examination. The book is devoted to a survey of the algebraic and number theoretic properties of Dickson polynomials. 284 Handbook of Finite Fields 9.6.8 Remark Suppose q is odd and a is a nonsquare in Fq. Then (x + √a)n = rn(x) + sn(x)√a, where rn(x) = ⌊n/2⌋ X i=0 n 2i aixn−2i, sn(x) = ⌊n/2⌋ X i=0 n 2i + 1 aixn−2i−1. The R´ edei function is the rational function Rn(x) = rn(x) sn(x). It is shown in that 2rn(x) = Dn(2x, x2 −a). 9.6.9 Remark Permutation properties of Dickson polynomials are important; see Section 8.1. The famous Schur conjecture postulating that every integral polynomial that is a permutation polynomial for inﬁnitely many primes is a composition of linear polynomials and Dickson polynomials was proved by Fried . We refer readers to Section 9.7. 9.6.2 Factorization 9.6.10 Remark The factorization of the Dickson polynomials of the ﬁrst kind over Fq was given and simpliﬁed in . 9.6.11 Theorem [266, 626] If q is even and a ∈F∗ q then Dn(x, a) is the product of squares of irreducible polynomials over Fq which occur in cliques corresponding to the divisors d of n, d > 1. Let kd be the least positive integer such that qkd ≡±1 (mod d). To each such d there corresponds φ(d)/(2kd) irreducible factors of degree kd, each of which has the form kd−1 Y i=0 (x −√a(ζqi + ζ−qi)) where ζ is a d-th root of unity and φ is Euler’s totient function. 9.6.12 Theorem [266, 626] If q is odd and a ∈F∗ q then Dn(x, a) is the product of irreducible polynomials over Fq which occur in cliques corresponding to the divisors d of n for which n/d is odd. Let md is the least positive integer satisfying qmd ≡±1 (mod 4d). To each such d there corresponds φ(4d)/(2Nd) irreducible factors of degree Nd, each of which has the form Nd−1 Y i=0 (x − p aqi(ζqi + ζ−qi)) where ζ is a 4d-th root of unity and Nd =    md/2 if √a / ∈Fq, md ≡2 (mod 4) and qmd/2 ≡2d ± 1 (mod 4d), 2md if √a / ∈Fq and md is odd, md otherwise. . 9.6.13 Example Let (q, n) = (5, 12). Then D12(x, 2) = x12 +x10 +x8 +4x6 +3x2 +3 is the product of irreducible polynomials over F5 which occur in cliques corresponding to the divisors d = 4 and d = 12 of n = 12. By direct computation, m4 = N4 = 4 and m12 = N12 = 4. For d = 4, there corresponds one irreducible factor of degree 4, while there are two irreducible factors of degree 4 for d = 12, each of which has the form QNd−1 i=0 (x − √ aqi(ζqi + ζ−qi)), where ζ is a 4d-th root of unity. 9.6.14 Remark Similar results hold for Dickson polynomials of the second kind and they can be found in and . Dickson polynomials of other kinds are deﬁned in and the Special functions over ﬁnite ﬁelds 285 factorization of the Dickson polynomial of the third kind is obtained similarly in . We note that the factors appearing in the above results are over Fq, although their description uses elements in an extension ﬁeld of Fq. In Fitzgerald and Yucas showed that these factors can be obtained from the factors of certain cyclotomic polynomials. This in turn gives a relationship between a-self-reciprocal polynomials and these Dickson factors. In the subsequent subsections we explain how this works. These results come mainly from . 9.6.2.1 a-reciprocals of polynomials 9.6.15 Deﬁnition Let q be an odd prime power and ﬁx a ∈F∗ q. For a monic polynomial f over Fq of degree n, with f(0) ̸= 0, deﬁne the a-reciprocal of f by ˆ fa(x) = xn f(0)f(a/x). The polynomial f is a-self-reciprocal if f(x) = ˆ fa(x). 9.6.16 Remark We note that the notion of a 1-self-reciprocal is the usual notion of a self-reciprocal. 9.6.17 Lemma 1. If α is a root of f then a/α is a root of ˆ fa. 2. The polynomial f is irreducible over Fq if and only if ˆ fa is irreducible over Fq. 9.6.18 Remark The a-reciprocal of an irreducible polynomial f may not have the same order as f. For example, consider f(x) = x3 + 3 when q = 7. Then f has order 9 while ˆ f3(x) = x3 + 2 has order 18. 9.6.19 Theorem Suppose f is a polynomial of even degree n = 2m over Fq. The following statements are equivalent: 1. f is a-self-reciprocal; 2. n = 2m and f has the form f(x) = bmxm + m−1 X i=0 b2m−i(x2m−i + am−ixi) for some bj ∈Fq. 9.6.20 Deﬁnition Let n be an even integer. Deﬁne Dn = {r : r divides qn −1 but r does not divide qs −1 for s < n}. For r ∈Dn and even n, we write r = drtr where dr = (r, qm + 1) and m = n/2. 9.6.21 Theorem Suppose f is an irreducible polynomial of degree n = 2m over Fq. The following statements are equivalent: 1. f is a-self-reciprocal for some a ∈F∗ q with ord(a) = t, 2. f has order r ∈Dn and tr = t. 9.6.22 Theorem Let r ∈Dn and suppose tr divides q −1. Then the cyclotomic polynomial Q(r, x) factors into all a-self-reciprocal monic irreducible polynomials of degree n and order r where a ranges over all elements of Fq of order tr. 286 Handbook of Finite Fields 9.6.2.2 The maps Φa and Ψa 9.6.23 Deﬁnition Deﬁne the mapping Φa : Pm →Sn from the polynomials over Fq of degree m to the a-self-reciprocal polynomials over Fq of degree n = 2m by Φa(f(x)) = xmf(x + a/x). 9.6.24 Remark In the case a = 1 this transformation has appeared often in the literature. The ﬁrst occurrence is Carlitz . Other authors writing about Φ are Chapman , Cohen , Fitzgerald-Yucas , Kyuregyan , Miller , Meyn , and Scheerhorn . 9.6.25 Deﬁnition Deﬁne Ψa : Sn →Pm by Ψa bmxm + m−1 X i=0 b2m−i(x2m−i + am−ixi) ! = bm + m−1 X i=0 b2m−iDm−i(x, a). 9.6.26 Theorem Maps Φa and Ψa are multiplicative and are inverses of each other. 9.6.2.3 Factors of Dickson polynomials 9.6.27 Theorem The polynomial Dn(x, a) is mapped to x2n + an by the above deﬁned Φa, namely, Φa(Dn(x, a)) = x2n + an. 9.6.28 Theorem The polynomial x2n + an factors over Fq as x2n + an = Y f(x), where each f is either an irreducible a-self-reciprocal polynomial or a product of an irre-ducible polynomial and its a-reciprocal over Fq. 9.6.29 Theorem The polynomial Dn(x, a) factors over Fq as Dn(x, a) = Y Ψa(f(x)), where x2n + an = Q f(x) such that f is either an irreducible a-self-reciprocal polynomial or a product of an irreducible polynomial and its a-reciprocal. 9.6.30 Theorem The following is an algorithm for factoring Dn(x, a) over Fq. 1. Factor x2n + an. 2. For each factor f of x2n + an which is not a-self-reciprocal, multiply f with ˆ fa. 3. Apply Ψa. 9.6.31 Example We factor D12(x, 2) = x12 + x10 + x8 + 4x6 + 3x2 + 3 when q = 5. x24 + 212 = [(x4 + x2 + 2)(x4 + 2x2 + 3)][(x4 + 3)(x4 + 2)][(x4 + 4x2 + 2)(x4 + 3x2 + 3)] = (x8 + 3x6 + 2x4 + 2x2 + 1)(x8 + 1)(x8 + 2x6 + 2x4 + 3x2 + 1). Special functions over ﬁnite ﬁelds 287 Then apply Ψ2 to obtain D12(x, 2) = (D4(x, 2) + 3D2(x, 2) + 2)D4(x, 2)(D4(x, 2) + 2D2(x, 2) + 2) = (x4 + 3)(x4 + 2x2 + 3)(x4 + 4x2 + 2). 9.6.32 Deﬁnition For a ∈F∗ q, deﬁne η(n, a) by η(n, a) = n · ord(an) if n is odd and a is a non-square, 4n · ord(an) otherwise. 9.6.33 Theorem For a monic irreducible polynomial f over Fq and a ∈F∗ q, the following statements are equivalent: 1. f divides Dn(x, a). 2. There exists a divisor d of n with n/d odd and ord(Φa(f)) = η(d, a), where Φa is deﬁned in Deﬁnition 9.6.23. 9.6.2.4 a-cyclotomic polynomials 9.6.34 Deﬁnition For a ∈F∗ q, deﬁne the a-cyclotomic polynomial Qa(n, x) over Fq by Qa(n, x) = Y d\|n d even (xd −ad/2)µ(n/d). 9.6.35 Remark When n ≡0 (mod 4), we have Q1(n, x) = Q(n, x), the n-th cyclotomic poly-nomial over Fq. When n ≡2 (mod 4), we have Q1(n, x) = Q(n/2, −x2). Similar to the factorization of xn −1 = Q d\|n Q(d, x) , we can reduce the factorization of x2n ± an to the factorization of a-cyclotomic polynomials. 9.6.36 Theorem We have 1. x2n −an = Y d\|n Qa(2d, x); 2. x2n + an = Y d\|n d odd Qa(4d, x). 9.6.37 Remark A factorization of these a-cyclotomic polynomials Qa(m, x) is also given in . 9.6.3 Dickson polynomials of the (k + 1)-th kind 9.6.38 Deﬁnition For a ∈Fq, any integers n ≥0 and 0 ≤k < p, we deﬁne the n-th Dickson polynomial of the (k + 1)-th kind Dn,k(x, a) over Fq by D0,k(x, a) = 2 −k and Dn,k(x, a) = ⌊n/2⌋ X i=0 n −ki n −i n −i i (−a)ixn−2i. 288 Handbook of Finite Fields 9.6.39 Deﬁnition For a ∈Fq, any integers n ≥0 and 0 ≤k < p, we deﬁne the n-th reversed Dickson polynomial of the (k + 1)s-th kind Dn,k(a, x) over Fq by D0,k(a, x) = 2 −k and Dn,k(a, x) = ⌊n/2⌋ X i=0 n −ki n −i n −i i (−1)ian−2ixi. 9.6.40 Remark It is easy to see that Dn,0(x, a) = Dn(x, a) and Dn,1(x, a) = En(x, a). Moreover, if char(Fq) = 2, then Dn,k(x, a) = Dn(x, a) if k is even and Dn,k(x, a) = En(x, a) if k is odd. 9.6.41 Theorem For any integer k ≥1, we have Dn,k(x, a) = kDn,1(x, a) −(k −1)Dn,0(x, a) = kEn(x, a) −(k −1)Dn(x, a). 9.6.42 Theorem The fundamental functional equation for Dn,k is Dn,k(y + ay−1, a) = y2n + kay2n−2 + · · · + kan−1y2 + an yn = y2n + an yn + ka yn y2n −an−1y2 y2 −a , for y ̸= 0, ±√a. 9.6.43 Theorem The Dickson polynomial of the (k + 1)-th kind satisﬁes the following re-currence relation Dn,k(x, a) = xDn−1,k(x, a) −aDn−2,k(x, a), for n ≥2 with initial values D0,k(x, a) = 2 −k and D1,k(x, a) = x. 9.6.44 Theorem The generating function of Dn,k(x, a) is ∞ X n=0 Dn,k(x, a)zn = 2 −k + (k −1)xz 1 −xz + az2 . 9.6.45 Remark The Dickson polynomial Dn,k(x, a) of the (k + 1)-th kind satisﬁes a second order diﬀerential equation; see [2723, 2945] for more details. 9.6.46 Theorem Suppose ab is a square in F∗ q. Then Dn,k(x, a) is a PP of Fq if and only if Dn,k(x, b) is a PP of Fq. Furthermore, Dn,k(α, a) = ( p a/b)nDn,k(( p b/a)α, b). 9.6.47 Deﬁnition Deﬁne Sq−1, Sq+1, and Sp by Sq−1 = {α ∈Fq : uq−1 α = 1}, Sq+1 = {α ∈Fq : uq+1 α = 1}, Sp = {±2}, where uα ∈Fq2 satisﬁes α = uα + 1 uα ∈Fq. 9.6.48 Theorem As functions on Fq, we have Dn,k(α) =    D(n)2p,k(α) if α ∈Sp, D(n)q−1,k(α) if α ∈Sq−1, D(n)q+1,k(α) if α ∈Sq+1, Special functions over ﬁnite ﬁelds 289 where for positive integers n and r we use the notation (n)r to denote n (mod r), the smallest positive integer congruent to n modulo r. 9.6.49 Theorem Let α = uα + 1 uα where uα ∈Fq2 and α ∈Fq. Let ϵα = uc α ∈{±1} where c = p(q2−1) 4 . As functions on Fq we have Dc+n,k(α) = ϵαDn,k(α). Moreover, Dn,k(x) is a PP of Fq if and only if Dc+n,k(x) is a PP of Fq. 9.6.50 Theorem For k ̸= 1, let k′ = k k−1 (mod p) and ϵα = uc α ∈{±1} where c = p(q2−1) 4 . For n < c, as functions on Fq we have Dc−n,k′(α) = −ϵα k −1Dn,k(α). Moreover, Dn,k(x) is a PP of Fq if and only if Dc−n,k′(x) is a PP of Fq. 9.6.4 Multivariate Dickson polynomials 9.6.51 Deﬁnition The Dickson polynomial of the ﬁrst kind D(i) n (x1, . . . , xt, a), 1 ≤i ≤t, is given by the functional equations D(i) n (x1, . . . , xt, a) = si(un 1, . . . , un t+1), 1 ≤i ≤t, where xi = si(u1, . . . , ut+1) are elementary symmetric functions and u1 · · · ut+1 = a. The vector D(t, n, a) = (D(1) n , . . . , D(t) n ) of the t Dickson polynomials is a Dickson polynomial vector. 9.6.52 Remark Let r(c1, . . . , ct, z) = zt+1 −c1zt + c2zt−1 + · · · + (−1)tctz + (−1)t+1a be a poly-nomial over Fq and β1, . . . , βt+1 be the roots (not necessarily distinct) in a suitable ex-tension of Fq. For any positive integer n, we let rn(c1, . . . , ct, z) = (z −βn 1 ) . . . (z −βn t+1). Then rn(c1, . . . , ct, z) = zt+1 −D(1) n (c1, . . . , ct, a)zt + D(2) n (c1, . . . , ct, a)zt−1 + · · · + (−1)tD(t) n (c1, . . . , ct, a)z + (−1)t+1at. 9.6.53 Remark For the Dickson polynomial D(1) n (x1, . . . , xt, a), an explicit expression, a generating function, a recurrence relation, and a diﬀerential equation satisﬁed by D(1) n (x1, . . . , xt, a) can be found in . Here we only give the generating function and recurrence relation. 9.6.54 Theorem The Dickson polynomial of the ﬁrst kind D(1) n (x1, . . . , xt, a) satisﬁes the generat-ing function ∞ X n=0 D(1) n (x1, . . . , xt, a)zn = Pt i=0(t + 1 −i)(−1)ixizi Pt+1 i=0(−1)ixizi , for n ≥0, and the recurrence relation D(1) n+t+1 −x1D(1) n+t + · · · + (−1)txkD(1) n+1 + (−1)t+1aD(1) n = 0, with the t + 1 initial values D(1) 0 = t + 1, D(1) j = j X r=1 (−1)r−1xrD(1) j−r + (−1)j(t + 1 −j)xj, for 0 < j ≤t. 290 Handbook of Finite Fields 9.6.55 Remark Much less is known for the multivariate Dickson polynomials of the second kind. The same recurrence relation of D(1) n (x1, . . . , xt, a) is used to deﬁne the multivariate Dickson polynomials of the second kind E(1) n (x1, . . . , xt, a) with the initial condition E0 = 1, Ej = Pj r=1(−1)r−1xrEj−r for 1 ≤j ≤t. The generating function is P∞ n=0 Enzn = 1 Pt+1 i=0(−1)ixizi ; see for more details. See Also §8.1 For permutation polynomials with one variable. §8.3 For value sets of polynomials over fnite ﬁelds. §9.7 For Schur’s conjecture and exceptional covers. For a comprehensive book on Dickson polynomials. References Cited: [266, 547, 589, 626, 678, 1079, 1109, 1116, 1821, 1936, 1939, 2091, 2100, 2538, 2723, 2945] 9.7 Schur’s conjecture and exceptional covers . Michael D. Fried, Irvine campus of the University of California 9.7.1 Rational function deﬁnitions 9.7.1 Remark (Extend values) The historical functions of this section are polynomials and ratio-nal functions: f(x) = Nf(x)/Df(x) with Nf and Df relatively prime (nonzero) polynomials, denoted f ∈F(x), F a ﬁeld (almost always Fq or a number ﬁeld). The subject takes oﬀ by including functions f – covers – where the domain and range are varieties of the same dimension. Still, we emphasize functions between projective algebraic curves (nonsingular), often where the target and domain are projective 1-space. 9.7.2 Deﬁnition The degree of f ∈F(x), deg(f), is the maximum of deg(Nf) and deg(Df). Add a point at ∞to F, F ∪{∞} = P1 x(F), to get the F points of projective 1-space. 9.7.3 Remark (Plug in ∞) Using Deﬁnition 9.7.2 requires plugging in and getting out ∞. We sometimes use the notion of value sets Vf and their cardinality #Vf (Section 8.3). 1. The value of f(x′) for x′ ∈F is ∞if x′ is a zero of Df(x). 2. The value of f(∞) is respectively ∞, 0, or the ratio of the Nf and Df leading coeﬃcients, if the degree of Nf is greater, less than, or equal to the degree of Df. If z is a variable indicating the range, this gives f as a function from P1 x(F) to P1 z(F). We abbreviate this as f : P1 x →P1 z. Special functions over ﬁnite ﬁelds 291 9.7.4 Deﬁnition (M¨ obius equivalence) Denote the group – under composition – of M¨ obius trans-formations x 7→ax+b cx+d with ad −bc ̸= 0, a, b, c, d ∈F by PGL(F). Refer to f1, f2 ∈F(x) as M¨ obius equivalent if f2 = α ◦f1 ◦β for α, β ∈PGL(F). 9.7.5 Example If f(x) = xn, with gcd(n, q −1) = 1, then #Vf = qk + 1 on P1 x(Fqk) exactly for those inﬁnitely many k with gcd(n, qk −1) = 1. 9.7.6 Remark Initial motivation came from Schur’s Conjecture Theorem 9.7.32, which starts over a number ﬁeld K – a ﬁnite extension of Q, the rational numbers – with its ring of integers OK. That asks about Vf over residue class ﬁelds, OK/p p p of prime ideals p p p, denoting this Vf(O/p p p) (Vf(Fp) if O = Z). Assume Nf and Df have coeﬃcients in OK. Avoid p p p – it is a bad prime – if it contains the leading coeﬃcient of either Nf or Df. 9.7.7 Deﬁnition For f ∈F(x), if f = f1 ◦f2 with f1, f2 ∈F(x), deg(fi) > 1, i = 1, 2, we say f decomposes over F. Then, the fi s are composition factors of f. 9.7.8 Deﬁnition (Coﬁnite) For B a subset of A, we say B is coﬁnite in A if A \ B is ﬁnite. 9.7.9 Proposition [2203, p. 390] Consider X′ h = {(x, y) \| h(x, y) = 0}, an algebraic curve, deﬁned by h ∈K[x, y]. Then, there is a unique nonsingular curve Xh – the normalization of X′ h – and a morphism µh : X′ h →Xh that is an isomorphism on the complement of a ﬁnite subset of points in Xh. Indeed, every variety X′ h has such a unique normalization, but in higher dimensions it may be singular, and µh is an isomorphism oﬀa codimension 1 set. 9.7.10 Deﬁnition (Components) A deﬁnition ﬁeld for an algebraic set W is a ﬁeld containing all coeﬃcients of all polynomials deﬁning W. Components of W over F are algebraic subsets which are not the union of two closed non-empty proper algebraic subsets over F [1427, p. 3]. We say W is a variety if it has just one component. It is absolutely irreducible if it has just one component over ¯ F, an algebraic closure of F. 9.7.11 Remark (Points on varieties) [1427, Chapters 1 and 2] and [2203, Section 2] introduce aﬃne and projective algebraic sets, and their components (Deﬁnition 9.7.10), except they are over an algebraically closed ﬁeld. For perfect ﬁelds F (including ﬁnite ﬁelds and number ﬁelds) this extends for normal varieties. Since their components do not meet, taking any disjoint union of distinct varieties under the action of the absolute Galois group of F deﬁnes components in general. Points on an algebraic set X over F refers here to geometric points: points with coordinates in ¯ F. It is an F point if its coordinates are in F. 9.7.12 Deﬁnition A general f : X →Z is a cover means it is a ﬁnite, ﬂat morphism (see Deﬁnition 9.7.25) of quasi-projective varieties [2203, p. 432, Proposition 2]. 9.7.13 Lemma Deﬁnition 9.7.12 simpliﬁes for curves, because all our varieties will be normal, and so for curves, nonsingular. Then, any nonconstant morphism is a cover: That includes any nonconstant rational function f : P1 x →P1 z. 9.7.14 Example If f : X →Z is ﬁnite and X and Z are nonsingular, generalizing what happens for curves, and no matter their dimension, then f is automatically ﬂat [1427, p. 266, 9.3a)]. This does not extend to weakening nonsingular to normal varieties. [2203, p. 434] has a ﬁnite morphism, where X is nonsingular (it is aﬃne 2-space), and Z is normal. But, the ﬁber degree is 2 over each z ∈Z, excluding one point where it is 3. 292 Handbook of Finite Fields 9.7.15 Remark (Assuming normality) Starting with Subsection 9.7.2 all results assume that the algebraic sets are normal. Some constructions (especially Deﬁnition 9.7.45) momentarily produce nonnormal sets, that we immediately replace with their normalizations. 9.7.2 MacCluer’s Theorem and Schur’s Conjecture 9.7.16 Deﬁnition An f ∈Fq(x) is exceptional if it maps one-one on P1(Fqk) for inﬁnitely many k. Similarly, with K a number ﬁeld, f ∈K(x) is exceptional if it is exceptional mod p p p for inﬁnitely many primes p p p. 9.7.17 Remark We use K, allowing decoration, for a number ﬁeld. Section 8.3 refers to the split-ting ﬁeld, Ωf (respectively Ωf, ¯ F ), of f(x) −z over F(z) (respectively over ¯ F(z)). The automorphism group of the extension Ωf/F(z) (respectively Ωf, ¯ F / ¯ F(z)) is the arithmetic (respectivelygeometric) monodromy group A (respectively G) of a separable function (Def-inition 9.7.25) f ∈F(x). When there are several functions, we denote these Af and Gf. They act on the zeros, {x1, . . . , xn} (often denoted {1, . . . , n}), of f(x)−z, giving a natural permutation representation on n symbols. 9.7.18 Deﬁnition Every cover f : X →Z over a ﬁeld F with X irreducible has an associated extension of function ﬁelds that determines the cover up to birational morphisms (see Lemma 9.7.43). 9.7.19 Remark Essentially all the Galois theory of ﬁelds translates to useful statements about a cover f : X →Z (over F) of an irreducible variety Z. It does this by corresponding to f the composite of the function ﬁeld extensions F(X′)/F(Z) where X′ runs over the components of X [2203, p. 396]. Several papers in our references (say, [1114, Section 0.C]) give oft-used examples, with Lemma 9.7.20 a simple archetype. 9.7.20 Lemma (See Remark 9.7.21) Any separable cover f : X →Y over F has a Galois closure cover ˆ f : ˆ X →Z over F. Then, Af is the group of ˆ f with its natural permutation repre-sentation TAf (of degree the degree of f). Do this over ¯ F to get the geometric monodromy Gf. Then, X is irreducible (resp. absolutely irreducible) if and only if TAf (resp. TGf ) is transitive. For f a rational function it is automatic that TGf (and so TAf ) is transitive. 9.7.21 Remark [1118, Section 2.1] explains how to form the Galois closure cover of a cover using ﬁber products (see Remark 9.7.54). This shows how to form the Galois closure cover of any collection of covers as in Lemma 9.7.50. 9.7.22 Remark Normalization gives a nearly invertible process to Remark 9.7.19: going from ﬁeld extensions of F(Z) to covers of Z. While this does not translate all arithmetic cover prob-lems to Galois theory, we apply the phrase “monodromy precision” (Remark 9.7.26) to when it does. Example: It does in the topic of exceptional covers, as in Proposition 9.7.28. 9.7.23 Deﬁnition Denote the elements of a group G, under a representation TG, that ﬁx 1 by G(1). When TG is transitive, refer to TG as primitive (respectively doubly transitive) if there is no group properly between G(1) and G (respectively G(1) is transitive on {2, . . . , n}). 9.7.24 Theorem [1112, Theorem 1]: An f ∈Fq(x) is exceptional if and only if the following holds for each orbit O of Af(1) on {2, . . . , n}: O breaks into strictly smaller orbits under Gf(1). (9.7.1) Special functions over ﬁnite ﬁelds 293 Denote the projective normalization of {(x, y) \| f(x)−f(y) x−y = 0} by Xf,f \ ∆. Also equivalent to (9.7.1): Each Fq component of Xf,f \ ∆has at least 2 components over ¯ Fq. Similarly, an f ∈K(x), K a number ﬁeld, is exceptional if and only if (9.7.1) holds for f mod p p p for inﬁnitely many primes p p p. 9.7.25 Deﬁnition (Covers) Let f ∈Fq(x) be nonconstant and separable: not g(xp) for some g ∈Fq(x). Then, f : P1 x( ¯ Fq) →P1 z( ¯ Fq) by x 7→f(x) has these cover properties. 1. Excluding a ﬁnite set {z1, . . . , zr} ⊂P1 z( ¯ Fq), branch points of f, there are exactly n = deg(f) points over z′. 2. For z′ a branch point, counting zeros, x′ of f(x) −z′ with multiplicity, the sum at all x′ s over z′ is still n. An x′ ∈P1 x with multiplicity > 1 is a ramiﬁed point. For K a number ﬁeld, the same properties hold, without any separable condition. 9.7.26 Remark (MacCluer’s Theorem) Theorem 9.7.24 has a surprise: (9.7.1) implies exceptionality over Fq. An error term in applying Chebotarev’s density theorem with branch points (as in Section 8.3, in Section 8.3.3) vanishes. A ramiﬁed point with p not dividing its multiplicity is tame. Macluer’s thesis responded to a Davenport-Lewis conjecture by showing The-orem 9.7.24 for a polynomial tame at every point. We say: MacCluer’s Theorem shows tame polynomial exceptional covers exhibit monodromy precision [1119, Section 3.2.1]. Proposi-tion 9.7.28 shows monodromy precision holds for general exceptional covers. 9.7.27 Example A polynomial f over Fq for which p\| deg(f) is not tame at ∞. 9.7.28 Proposition combined with [1118, Principle 3.1]: Let f : X →Z be any cover (Deﬁnition 9.7.12) over Fq with X absolutely irreducible. Then [1118, Corollary 2.5]: 1. the extended meaning of (9.7.1) is that the 2-fold ﬁber product (Section 9.7.3) of f minus the diagonal has no absolutely irreducible Fq components; and 2. (9.7.1) is equivalent to f being exceptional: X(Fqk) →Y (Fqk) is one-one (and onto) for inﬁnitely many k. 9.7.29 Remark As noted in [1118, Comments on Principle 3.1], the proof of applies without change to give Proposition 9.7.28 Part 2 when X and Z are non-singular; indeed, it applies to pr-exceptionality (Deﬁnition 9.7.93). Without, however, this nonsingularity assumption, there are complications considered in [1119, Section A.4.1] (see Example 9.7.14). 9.7.30 Deﬁnition Let f in Proposition 9.7.28 over Fq be an exceptional cover. Denote values k where (9.7.1) holds with Fqk replacing Fq, by Ef,q: the exceptionality set of f. Similarly, for f satisfying the hypotheses of Proposition 9.7.28 over a number ﬁeld K, denote those primes p p p where f mod p p p has Ef,O/p p p inﬁnite, by Ef,K. 9.7.31 Deﬁnition The equation Tu(cos(θ)) = cos(uθ) deﬁnes the u-th Chebychev polynomial, Tu. From it deﬁne a Chebychev conjugate: α ◦Tu ◦α−1 with α(x) = αz′(x) = z′x and either z′ = 1, or z′ and −z′ are conjugate in a quadratic extension of K. 9.7.32 Theorem (Schur’s Conjecture) [1109, Theorem 2]: With K a number ﬁeld, the f ∈O[x] for which Ef,K is inﬁnite are compositions with maps a 7→ax + b (aﬃne) over K with polynomials of the following form for some odd prime u: xu (cyclic) or, α ◦Tu ◦α−1, u > 3, a Chebychev conjugate. (9.7.2) 294 Handbook of Finite Fields 9.7.33 Remark Many still refer to Theorem 9.7.32 as Schur’s Conjecture, though Schur conjectured it only over Q. Paper refers to all Chebychev conjugates as Chebychev polynomials, rather than Dickson as in Remark 9.7.34. Reference assiduously distinguishes Dickson polynomials. Here is a simple branch point Chebychev Conjugate characterization: f has two ﬁnite (̸= ∞) branch points, ±z′ ∈P1 z(¯ Q), which identify with the unique unramiﬁed points (in P1 x(¯ Q)) over the branch points, as in [1109, Proof of Lemma 9]. 1. A corollary of [1124, Theorem 3.5] is that any cover with a unique totally and tamely ramiﬁed point decomposes over F if and only if it decomposes over ¯ F. This applies if f ∈F[x] has deg(f) prime to the characteristic of F. 2. If f from Part 1 is indecomposable, then Gf is primitive (see Deﬁnition 9.7.23) and it contains an n-cycle. 3. If f ∈K[x] is exceptional, since (9.7.1) says Gf cannot be doubly transitive, up to composing with K aﬃne maps, f from Part 2 is in (9.7.2). 9.7.34 Remark (Dickson doppelgangers, see Section 9.6) Each Chebychev conjugate is a constant times a Dickson polynomial [1118, Proposition 5.3]. The Remark 9.7.33 characterization – by locating their branch points – avoids using equations. That is the distinction at the last step between the proof of Theorem 9.7.32 and [1936, Chapter 6]. 9.7.35 Remark Use the notation in Theorem 9.7.32. Suppose f ∈OK[x] is an exceptional polyno-mial. Deﬁne nf,c (resp. nf,C) to be the product of distinct primes s for which f has a degree s cyclic (resp. Chebychev conjugate) composition factor. One can check that Corollary 9.7.36 follows from 9.7.28 combined with 9.7.32. 9.7.36 Corollary For f ∈OK[x] an exceptional polynomial, one can determine Ef,K (excluding bad primes, Remark 9.7.6) from nc,f and nf,C by congruences. When OK = Z, then p ∈Ef,Q if and only if gcd(p −1, s) = 1 for each s\|nf,c and gcd(p2 −1, s) = 1 for each s\|nf,C. 9.7.37 Example (Inﬁnite Ef,Q) It is necessary that gcd(2, nc) = 1 and gcd(6, nC) = 1 for there to be inﬁnitely many p that satisfy the conclusion of Corollary 9.7.36. But it is suﬃcient, too. Without loss, assume gcd(nc, nC) = 1. If 3̸ \| nc, then Dirichlet’s Theorem on primes in arithmetic progressions gives an inﬁnite set of p ≡3 (mod ncnC). They are in Ef,Q. If 3\|nc, the Chinese remainder theorem gives an arithmetic progression of p satisfying p ≡3 (mod nC) and p ≡−1 (mod nc). So, Ef,Q is inﬁnite whenever it has a chance to be. 9.7.38 Remark Combine [1113, Lemma 1] with monodromy precision in Proposition 9.7.39. This shows, the Proposition 9.7.28 ﬁber product statement is equivalent to f ∈K(x) being ex-ceptional, and therefore permutation, mod p p p. If OK/p p p is suﬃciently large, the ﬁber product statement is also necessary for f to be permutation (well-known, for example [1109, proof of Theorem 2, last paragraph]). 9.7.39 Proposition (Permutation functions) From Remark 9.7.38, for f ∈Fq(x), those k where f permutes P1(Fqk) contains Ef,Fq as a coﬁnite subset. Similarly, for K a number ﬁeld, those p p p where f functionally permutes P1(O/p p p) contains Ef,K as a coﬁnite subset. 9.7.40 Remark Section 8.1 shows permutation polynomials are abundant. Exceptional polynomials satisfy a much stronger property, but Corollary 9.7.36 shows they are abundant, too. One diﬀerence: Section 9.7.3 combines them in ways with no analog for permutation polynomials. 9.7.41 Corollary An analog of Theorem 9.7.32 holds over Fq to characterize exceptional poly-nomials of degree prime to p ([1120, Introduction to Section 5] or [1118, Proposition 5.1]). There, z′ in αz′ is either 1 or in the unique quadratic extension of Fq. Consider a Chebychev Special functions over ﬁnite ﬁelds 295 conjugate αz′ ◦Tn ◦α−1 z′ as a permutation polynomial on Fqk with gcd(q2k −1, n) = 1. Then, when n · m ≡1 (mod q2k −1), αz′ ◦Tm ◦α−1 z′ is its functional inverse. 9.7.3 Fiber product of covers 9.7.42 Deﬁnition For any ﬁeld extension F1/F2 containing Fp, there is the notion of being sep-arable [1121, p. 111]. For f ∈Fq(x), the extension ¯ Fq(x)/ ¯ Fq(f(x)) being separable is equivalent to f is separable (Deﬁnition 9.7.25). Many of our examples inherit separa-bleness from this special case. 9.7.43 Lemma (Curve covering maps [1427, Chapter I, Section 6]) Any nonsingular projective al-gebraic curve X over a perfect ﬁeld F has a ﬁeld of functions F(X) that uniquely determines X up to isomorphism over F. Each non-constant element f ∈F(X) determines a ﬁnite map X →P1 z over F [1427, Chapter I, Exercise 6.4]. If F(X)/F(f) is separable, then f has the covering properties of (9.7.25): ﬁnite number of branch points, and uniform count of points in a ﬁber over ¯ F (including multiplicity in the ﬁber) [1427, Chapter IV, Proposition 2.2]. 9.7.44 Deﬁnition Refer to any f in the conclusion of Lemma 9.7.43 as a nonsingular cover of P1 z. 9.7.45 Deﬁnition (Fiber product) Let fi : Xi →P1 z, i = 1, 2, be two nonsingular covers of P1 z. The set theoretic ﬁber product consists of the algebraic curve {(x1, x2) ∈X1 × X2\|f1(x1) = f2(x2)}. Denote this X1 ×set P1 z X2. Its normalization (Proposition 9.7.9), X1 ×P1 z X2, is the ﬁber product of f1 and f2. 9.7.46 Remark Deﬁnition 9.7.45 works equally for any covers Xi →Z, i = 1, 2, with Z a normal projective variety. Then, X1 ×Z X2 is normal and projective (possibly with several compo-nents) with natural maps pri : X1 ×Z X2 →Xi, i = 1, 2, given by its projection on each factor. The functions fi ◦pri, i = 1, 2 are identical, giving a well-deﬁned map: (f1, f2) : X1 ×Z X2 →Z. (9.7.3) 9.7.47 Remark (Fiber equations) Consider x′ ∈X1 ×Z X2 that is simultaneously over x′ i ∈Xi, i = 1, 2, where both x′ i s ramify over pr1(x′ 1) = pr2(x′ 2). Then, f1(x1) = f2(x2), with xi in a neighborhood of x′ i, is not a correct local description around x′. There is another complication when Z is not a curve (dimension 1). The ﬁber product might be singular even when the Xi s are not. So (f1, f2) in (9.7.3) may not be a cover because it is not ﬂat (Remark 9.7.67). 9.7.48 Example Consider two polynomials, f1, f2 ∈K[x], of the same degree n. They deﬁne fj : P1 xj →P1 z, j = 1, 2. Then, there are n points over z = ∞on P1 y1 ×P1 z P1 y2, but only one point on the set theoretic ﬁber product over ∞. Proposition 1 of gives the generalization of this, showing – when the covers are tame – how to compute the genus of the ﬁber product components from the covers fj, j = 1, 2. 9.7.49 Deﬁnition The ﬁber product Xf,f = X ×Z X for a cover f : X →Z of degree exceeding 1 has at least two components. One is the diagonal: the set ∆(X) = {(x, x) \| x ∈X}. The normal variety X ×Z X \ ∆(X) generalizes the set in Theorem 9.7.24. 296 Handbook of Finite Fields 9.7.50 Lemma (Fiber product monodromy [1118, Section 2.1.3]) Consider the covers in Deﬁni-tion (9.7.45). To each fj there is an arithmetic (resp. geometric) monodromy group Afj (resp. Gfj), j = 1, 2. Similarly, for (f1, f2) in (9.7.3). Then, A(f1,f2) maps naturally, surjec-tively, to Afj by homomorphisms pr∗ j, j = 1, 2. There is a largest simultaneous quotient, H, of both Afj s given by homomorphisms mi : Afj →H, j = 1, 2, so that A(f1,f2) = {(σ1, σ2) ∈Af1 × Af2 \| m1(σ1) = m2(σ2)}. Similarly with geometric replacing arithmetic monodromy. 9.7.51 Corollary (Components) With the hypotheses of Lemma 9.7.50, let {1j, . . . , nj}, be integers on which Aj acts, j = 1, 2. Then, A(f1,f2) acts on the pairs (i1, i2) and on each of the sets {1j, . . . , nj} separately. If X1 is absolutely irreducible, then the components of X1 ×Z X2 over F (resp. ¯ F) correspond to the orbits of A(f1,f2)(11) (resp. G(f1,f2)(11); see Deﬁnition 9.7.23) on {12, . . . , n2}. We note that the degrees n1 and n2 may be diﬀerent. 9.7.52 Deﬁnition (Absolute components) Given X1 ×Z X2 in Corollary 9.7.51, denote the union of its absolutely irreducible F components by X1 ×abs Z X2. Denote the complementary set, X1 ×Z X2 \ X1 ×abs Z X2, of components by X1 ×cp Z X2. 9.7.53 Theorem (Explicit Ef,q – see Remark 9.7.54) Let f : X →Z (as in Proposition 9.7.28) be an exceptional cover over Fq. For X′ i, an Fq component of X ×cp Z X, denote the number of components in its breakup over ¯ Fq by si, i = 1, . . . , u. With sexc = lcm(s1, . . . , su), Ef,q = {k (mod sexc) \| gcd(k, si) < si, i = 1, . . . , u}. The group G(Fqsexc /Fq) is naturally a quotient of Af/Gf. We can interpret all quantities using Af and Gf. 9.7.54 Remark All but the last sentence of Theorem 9.7.53 is [1118, Corollary 2.8]. The last sen-tence is from [1118, Lemma 2.6], using that the Galois closure cover of f is a(ny) component (over Fq) of the deg(f) = n-fold ﬁber product of f with itself. Project that ﬁber product onto the 2-fold ﬁber product of f over Fq to ﬁnish. Corollary 9.7.51 shows the orbit lengths of Af(1) on {2, . . . , n} divided by the corresponding orbit lengths of Gf(1), give the si s. 9.7.55 Theorem (Explicit Ef,K – see Remark 9.7.56) Now change Fq to K (number ﬁeld) in the ﬁrst sentence of Theorem 9.7.53. For each cyclic subgroup C ≤Af/Gf denote those σ ∈Af that map to C by AC. As previously, denote the stabilizers of 1 in the representation by AC(1) and GC(1). Consider the set, Cf,K, of cyclic C (as in (9.7.1)): {C \| each orbit of AC(1) on {2, . . . , n} breaks into strictly smaller orbits under GC(1)}. Then, f is exceptional over K if and only if Cf,K is nonempty. Further, Ef,K consists of those primes p p p for which the Frobenius attached to p p p is a generator of some C ∈Cf,K. 9.7.56 Remark Theorem 9.7.55 comes from applying [1113, Section 2] exactly as in Remark 9.7.28. If Ef,K is inﬁnite, then X ×Z X \ ∆(X) has no absolutely irreducible component. The converse, however, does not hold. 9.7.4 Combining exceptional covers; the (Fq, Z) exceptional tower 9.7.57 Deﬁnition (Category of exceptional covers) For Z absolutely irreducible over Fq, denote the collection of exceptional covers of Z over Fq by TZ,Fq. Special functions over ﬁnite ﬁelds 297 9.7.58 Theorem [1118, Section 4.1] Given (fi, Xi) ∈TZ,Fq, i = 1, 2, X1 ×abs Z X2 (Deﬁnition 9.7.52) has one component. We conclude that: (f1 ◦pr1, X1 ×abs Z X2) ∈TZ,Fq. Also, there is at most one morphism between any two objects in TZ,Fq. 9.7.59 Remark (When f1 = f2 in Theorem 9.7.58) We deﬁnitely include the ﬁber product of a cover in TZ,Fq with itself. Then, the only absolutely irreducible component of the ﬁber product is the diagonal (Deﬁnition 9.7.49), which is equivalent to the original cover. 9.7.60 Deﬁnition We call X1 ×abs Z X2 the ﬁber product of f1 and f2 in TZ,Fq, and continue to denote its morphism to Z by (f1, f2). This deﬁnes TZ,Fq as a category with ﬁber products. Theorem 9.7.53 shows Ef1,q ∩Ef2,q = E(f1,f2),Fq is inﬁnite. 9.7.61 Remark Consider (fi, Xi) ∈TZ,Fq, i = 1, 2, for which there exists ψ : X1 →X2 over Fq that factors through f2: f2 ◦ψ = f1. Then, Theorem 9.7.58 says ψ is unique. 9.7.62 Corollary For (f, X) ∈TZ,Fq, denote the group of the Galois closure cover of f over X by Af(1). Then, Af has the representation Tf by acting on cosets of Af(1). If (fi, Xi) ∈TZ,Fq, i = 1, 2, we write (f1, X1) > (f2, X2) if f1 factors through X2. [1118, Proposition 4.3] pro-duces from these pairs a canonical group AZ,Fq with a proﬁnite permutation representation TZ,Fq. 9.7.63 Remark (A projective limit) Given (fi, Xi) ∈TZ,Fq, i = 1, 2, there is a 3rd (f, X) ∈TZ,Fq, given by the ﬁber product, that factors through both. This is the condition deﬁning a projective sequence. So, AZ,Fq in Corollary 9.7.62 is a projective limit. 9.7.64 Deﬁnition (AZ,Fq, TZ,Fq) is the (arithmetic) monodromy group, in its natural permutation representation, of the exceptional tower TZ,Fq. 9.7.65 Theorem Let fi : Xi →Z, i = 1, 2, be exceptional covers over K: Efi,K is inﬁnite, i = 1, 2. Then, X1 ×Z X2 is exceptional in the sense that X1 mod p p p ×abs Z mod p p p X2 mod p p p is exceptional for inﬁnitely p p p if and only if Ef1,K ∩Ef2,K is inﬁnite. 9.7.66 Remark Theorem 9.7.65 forces considering if there is an inﬁnite intersection of two excep-tionality sets over K. As Theorem 9.7.53 shows, this is automatic over Fq. Example 9.7.37 shows it is not automatic over a number ﬁeld. Subsections 9.7.5 and 9.7.6 have examples along these lines: If both fi s, i = 1, 2, are exceptional rational functions, then their com-position is again exceptional over K if and only if Ef1,K ∩Ef2,K is inﬁnite. Beyond cyclic and Chebychev situations, it is very diﬃcult to decide when this intersection is inﬁnite. 9.7.67 Remark The same deﬁnition for exceptional works for any ﬁnite, surjective, map of normal varieties over Fq. Such maps may not be ﬂat (say, when Remark 9.7.14 does not apply), so they may not be covers. Normalization of any projective variety is projective: Segre’s Embedding [2203, Theorem 4, p. 400]. For irreducible X, ﬂatness says the multiplicity sum of points in the ﬁber over z is constant in z: the function ﬁeld extension degree, [K(X) : K(Z)] [2203, Proposition 2, p. 432]. That is, Deﬁnition 9.7.25, Part 2, holds. For ﬁnite morphisms that characterizes ﬂatness [2203, Corollary p. 432]. With normality, but not ﬂatness, this may hold only outside a codimension 2 set in the target. Appendix A.4 of has a liesurely discussion; see Example 9.7.14. 298 Handbook of Finite Fields 9.7.5 Exceptional rational functions; Serre’s Open Image Theorem 9.7.68 Deﬁnition Deﬁnition 9.7.31 explains Chebychev conjugates. Consider lz′ : x 7→ x−z′ x+z′ , mapping ±z′ to 0, ∞, with a = (z′)2 ∈K, z′ ̸∈K. Then, for n odd, characterize Rn,a = (lz′)−1 ◦(lz′(x))n, a cyclic conjugate, by these conditions: ±z′ are its sole ramiﬁed points, Rn,a(±z′) = ±z′ and it maps ∞7→∞. (9.7.4) 9.7.69 Remark According to [1936, Chapter 2, Section 5]), the function Rn,a in Deﬁnition 9.7.68 is a Redei function. From [1936, Theorem 3.11], under the hypotheses on z′, the exceptionality set ERn,a,K is {p p p \| (\|OK/p p p\| −1, n) = 1} if z′ is a quadratic residue modulo p p p, and {p p p \| (\|OK/p p p\| + 1, n) = 1} if not. 9.7.70 Remark (Addendum Remark 9.7.69) Quadratic reciprocity determines nonempty arith-metic progressions for which z′ is a quadratic residue and those for which it is not. If z′ in Deﬁnition 9.7.68 were in K, then – of course – the exceptional set is the same as for xn. Whether or not z′ ∈K, we refer to Rn,a as a cyclic conjugate. 9.7.71 Deﬁnition [1118, Section 4.2] Suppose a collection C of covers from an exceptional tower TY,Fq is closed under the categorical ﬁber product. We say C is a subtower. We also speak of the (minimal) subtower any collection generates under ﬁber product. 9.7.72 Remark Section 4.3 of uses that the ﬁber product of two unramiﬁed covers is un-ramiﬁed to create cryptographic exceptional subtowers. Section 5.2.3 of computes the arithmetic monodromy attached to the Dickson subtower generated by all the exceptional Chebychev conjugates over Fq. The analog of Remark 9.7.69 over Fq gives a similar – Redei – subtower of TP1 z,Fq generated by exceptional cyclic conjugates. 9.7.73 Remark Theorem 9.7.65 requires common exceptional intersection (Remark 9.7.66) to form ﬁber products in TZ,K, Z absolutely irreducible over a number ﬁeld K. For ﬁber products (or composites) of Chebychev and cyclic conjugates, we easily decide if exceptional sets have inﬁnite intersection. Exceptional rational functions from Serre’s O(pen) I(mage) T(heorem) give much harder versions of such problems. 9.7.74 Deﬁnition (j-line P1 j) A special copy of projective 1-space, the j-line, occurs in the study of modular curves (see Theorem 9.7.76). Each j ∈P1 j \{∞}(¯ Q) = A1 j(¯ Q) has an attached isomorphism class of elliptic curves Ej. For each integer n > 0, consider a special case of a modular curve, µ0(n) : X0(n) →P1 j, with its cover of P1 j. Denote the points of X0(n) not lying over j = ∞by Y0(n). 9.7.75 Deﬁnition For E an elliptic curve, denote by E →E/C an isogeny from quotienting E by a (ﬁnite) torsion subgroup C of E. When C is a cyclic, generated by e′ ∈E (resp. all torsion points killed by multiplication by n), write C = ⟨e′⟩(resp. Cn). 9.7.76 Theorem There are two approaches to giving “meaning” to each algebraic point y ∈Y0(n), whose image in P1 j is jy Special functions over ﬁnite ﬁelds 299 1. [2311, p. 108] or [1115, p. 158]: y 7→[Ejy →Ejy/⟨e′ y⟩] with e′ y ∈Ejy of order n where brackets, [ ] , indicate an isomorphism class of isogenies. 2. [1115, Lemma 2.1]: y 7→fy ∈¯ Q(x) (up to M¨ obius equivalence) of degree n. 9.7.77 Theorem [1115, Theorem 2.1] Suppose f ∈K(x) is exceptional and of prime degree u. Then, f is M¨ obius equivalent over K to either: 1. a cyclic (Remark 9.7.70) or a Chebychev (Remark 9.7.34) conjugate; or 2. to some fy (u = n) in Theorem 9.7.76, Part 2. 9.7.78 Deﬁnition For a dense set of j′ ∈A1 j, the corresponding Ej′ is of CM -type if its ring of isogenies, tensored by Q, has dimension 2 over Q. Such isogenies form a complex quadratic extension of Q (containing j′, which is an algebraic integer; [2586, II-28] or [2614, Chapter 2, Section 5.2]). Otherwise, j′ is of GL2-type. 9.7.79 Theorem [1115, (2.10)] Continue the notation of Theorem 9.7.77. Except for the two cases where jy is one of the two ﬁnite branch points of µ0(u), the geometric monodromy Gfy is the order 2u dihedral group Du, and fy has four branch points (Deﬁnition 9.7.25). For u in Theorem 9.7.77, Part 2, for which Ej′ has good reduction, the coordinates of e′ y generate a constant extension of K with group Afy/Gfy (explained in Theorem 9.7.85). 9.7.80 Theorem [1115, Section 2.B] For j′ of CM-type, complex multiplication theory gives (an inﬁnite) Efy,K. Computing this would use [1370, Sections 6.3.1-6.3.2]. 9.7.81 Remark (Addendum to Theorem 9.7.80) Using adelic (modular) arithmetic gives analogs of Corollary 9.7.36; and Corollary 9.7.41 for explicitly ﬁnding the functional inverse of a CM-type reduced modulo a prime in the exceptional set Efy,K. If K = Q(j′), then Efy,K depends on the congruence deﬁning the Frobenius in the (cyclic of degree u−1 over K) constant ﬁeld. Only ﬁnitely many j′ in Q have CM-type, corresponding to class number 1 for complex quadratic extensions. 9.7.82 Problem Take one of the CM-type j s in Q. Then, consider two allowed values of u, ui, i = 1, 2, denoting the corresponding fy s by fi, i = 1, 2. Test for explicitness in Remark 9.7.81 as to whether Ef1,Q ∩Ef2,Q is inﬁnite. 9.7.83 Deﬁnition (Composition factor deﬁnition ﬁeld) For f ∈F(x) consider a minimal ﬁeld Ff(ind) over which f decomposes into composition factors indecomposable over ¯ F. Similarly, denote the minimal ﬁeld over which Xf,f \ ∆in Theorem 9.7.24 breaks into absolutely irreducible components by ˆ Ff(2). 9.7.84 Proposition [1118, Proposition 6.5] If f :X →Z is a cover over F, then Ff(ind) ⊂ˆ Ff(2). 9.7.85 Theorem (See Remark 9.7.87) Assume j′ ∈A1 j is of GL2-type. For K = Q(j′), consider C = Cu in Deﬁnition 9.7.75 with u a prime. The corresponding fy ∈K(x), y over j′, has degree u2. Use the monodromy groups of Deﬁnition 9.7.17. There is a constant M1,j′ so that if u > M1,j′, then the arithmetic/geometric monodromy quotient Afy/Gfy is GL2(Z/u)/{±1}. Further, fy decomposes into two degree u rational functions over Kf(ind), but it is indecomposable over K. 9.7.86 Theorem [1118, Proposition 6.6] Continue Theorem 9.7.85 hypotheses. For a second con-stant M2,j′, and for any prime p p p of OK with \|OK/p p p\| > M2,j′ assume Ap p p ∈GL2(Z/u)/⟨±1⟩ represents the conjugacy class of the Frobenius for p p p. Then, fy mod p p p is an exceptional indecomposable rational function, and it decomposes over the algebraic closure of OK/p p p, 300 Handbook of Finite Fields precisely when ⟨Ap p p⟩acts irreducibly on (Z/u)2 = Vu. This holds for inﬁnitely many primes p p p. In particular, fy is exceptional over K (Deﬁnition 9.7.30). 9.7.87 Remark (Using Serre’s OIT) lays the groundwork for . The latter has the existence of the constant M1,j′. Appendix A.1 and Section 3.2 of proves that it exists when j′ ∈A1 j(¯ Q) is not an algebraic integer. Then, the computation of Mi,j′, i = 1, 2, in Theorems 9.7.85 and 9.7.86 is eﬀective. Even after all these years, there is no eﬀective computation of these constants when j is not CM-type, but is an algebraic integer. Section 2 of gets Theorem 9.7.85 from the OIT using the relation between Parts 1 and 2 in Theorem 9.7.76. 9.7.88 Remark (More elementary, but less precise than Theorem 9.7.86) Theorem 2.2 of shows, for every K and any prime u > 3, the j′ ∈K, with fy satisfying the exceptionality and decomposability conclusions of Theorem 9.7.86, are dense. Applying the [1113, Theorem 3] (or [1121, Theorem 12.7]) version of Hilbert’s Irreducibility Theorem to X0(u) gives the corresponding M2,j′ explicitly. 9.7.89 Example (M1,j′ eﬀectiveness?) Appendix A.1 and Section 3.3 of gives Ogg’s example with j′ ∈Q. Section 6.2.2 of reviews this case, where M2,j′ = 6, to show how to pick an Ap p p acting irreducibly on Vu as in Theorem 9.7.86 (for inﬁnitely many p p p), assuring that Efy,Q is inﬁnite for u > M2,j′. Section 6.3.2 of – still Ogg’s case – aims at ﬁnding an automorphic function, a la Langland’s Program, that would characterize the primes in Efy,Q. This is akin to the unrelated examples of , but uses results on automorphic functions in [2590, Theorem 22]. Primes of Efy,Q do not lie in arithmetic progressions. So, Problem 9.7.90 is much harder than Problem 9.7.82. 9.7.90 Problem (Analog of Problem 9.7.82) For the Ogg curve in Example 9.7.89, consider two allowed values of u, ui, i = 1, 2, denoting the corresponding fy s by fi, i = 1, 2. Test for explicitness in Remark 9.7.81 as to whether Ef1,Q ∩Ef2,Q is inﬁnite. 9.7.91 Remark Paper connects “variables separated factors” of Xf,f and composition factors of f. Reference used this to eﬀectively test for composition factors (and primitivity) of covers. 9.7.92 Theorem [1370, Chapter 3] Excluding ﬁnitely many degrees, all indecomposable exceptional f ∈K(x) (K a number ﬁeld) are M¨ obius equivalent to a cyclic or Chebychev conjugate, or to a CM function from Theorem 9.7.77 of prime degree; or they are from Theorem 9.7.86 and of prime degree squared. 9.7.6 Davenport pairs and Poincar´ e series 9.7.93 Deﬁnition [1118, Deﬁnition 2.2] Consider f : X →Z, a cover of normal varieties over Fq, with Z absolutely irreducible, but X possibly reducible. Then f is pr-exceptional if it is surjective on Fqk points for inﬁnitely many k. There is a similar deﬁnition extending Deﬁnition 9.7.30 over a number ﬁeld, and for both a notation for exceptional sets. 9.7.94 Deﬁnition Use the value set notation of Remark 9.7.3. We say fi ∈Fq(x), i = 1, 2, is a Davenport pair over Fq if Vf1(P1(Fqk)) = Vf2(P1(Fqk)) for inﬁnitely many k. So, take f2(x) = x to see Davenport pairs generalize exceptional functions. The notion applies to any pair of covers fi : Xi →Z, i = 1, 2. For K a number ﬁeld, this similarly generalizes Deﬁnition 9.7.30: f1, f2 ∈K(x) are a Davenport pair if they are a Davenport pair for inﬁnitely many residue class ﬁelds. Special functions over ﬁnite ﬁelds 301 9.7.95 Theorem [1118, Corollary 3.6] Monodromy precision (Deﬁnition 9.7.26) applies to pr-exceptional covers and so to Davenport pairs. That is, generalizing Theorems 9.7.53 and 9.7.55, a precise monodromy statement generalizes MacCluer’s Theorem (Proposition 9.7.28) to pr-exceptional covers and to Davenport pairs. 9.7.96 Theorem [1118, Section 3.1.2] With the notation of Deﬁnition 9.7.93, a pr-exceptional cover over Fq is exceptional if and only if X is absolutely irreducible. 9.7.97 Remark The proof of Schur’s Conjecture began the solution of Davenport’s problem for polynomial pairs (f1, f2) over a number ﬁeld, the main result of . Section 3.2 in shows the exceptional set characterization for Davenport pairs in general is given by the intersection of exceptionality sets for pr-exceptionality correspondences. A full description of many authors’ results that came from the solution of Davenport’s problem – especially the study of general zeta functions attached to diophantine problems – is in [1119, Section 7.3]. 9.7.98 Remark (The Genus 0 Problem) Geometric monodromy groups of rational functions are severely limited. The mildest statement for f ∈¯ Q(x) is that excluding cyclic and alternating groups the composition factors of Gf fall among a ﬁnite set of simple groups. That is the original genus 0 problem. There is a large literature distinguishing between geometric monodromy of f ∈¯ Q(x) and those in ¯ Fq(x), because of wild (not tame; Remark 9.7.26) ramiﬁcation. The contrast starts from the [2191, Section 8.1.2, Guralnick’s Optimistic Conjecture] list of all primitive monodromy groups of indecomposable f ∈¯ Q[x]. 9.7.99 Example (Davenport pairs) A signiﬁcant part of the exceptional primitive monodromy groups (Remark 9.7.98), without cyclic or alternating group composition factors, came from the ﬁnitely many possible degrees of Davenport pairs f1, f2 ∈K[x] (polynomials) over number ﬁelds, with f1 indecomposable and Vf1(OK/p p p) = Vf2(OK/p p p). Important hints about what to expect for primitive monodromy groups of f ∈¯ Fq(x) came also from Davenport pairs. Section 3.3.3 in (explicitly in ): Over every Fq, there are inﬁnitely many degrees of Davenport pairs, where (deg(f1), p) = 1, f1 is indecomposable, and Vf1(Fqk) = Vf2(Fqk) for all k. 9.7.100 Example Theorem 14.1 in described the geometric monodromy (PSL2(pa), p = 2, 3, a odd) of the only possible exceptional polynomials over Fp whose degrees were neither prime to p or a power of p. Then, produced these: the ﬁrst exceptional polynomials over ﬁnite ﬁelds with nonsolvable monodromy. 9.7.101 Remark (Zeta functions attached to problems) Chapters 25 and 26 in details how Davenport pairs led to attaching Poincar´ e series – based on the Galois stratiﬁcation pro-cedure of – to counting the values of parameters for any diophantine problem inter-pretable over all extensions of Fq, or for inﬁnitely many primes p p p of K. 9.7.102 Example Denote w1, . . . , wu by w w w. Suppose f(w w w, x), g(w w w, y) ∈Fq[w w w, x, y]. Denote the car-dinality of w w w′ ∈Au(Fqk) with V (f(w w w′, x))(P1(Fqk)) = V (g(w w w′, x))(P1(Fqk)) (9.7.5) by Nf,g,k. Deﬁne Pf,g,Fq(t) to be the Poincar´ e series P∞ i=1 Nf,g,ktk. 9.7.103 Example With notation over Z, as in Example 9.7.102, suppose f(w w w, x), g(w w w, y) ∈Z[w w w, x, y]. Denote the cardinality of w w w′ ∈Au(Fpk) with (9.7.5) holding over Fpk by Nf,g,Z/p,k. Deﬁne Pf,g,Z/p(t) to be P∞ i=1 Nf,g,Z/p,ktk. 9.7.104 Theorem [1121, Chapter 25], [1119, Section 7.3.3] For any diophantine problem over Fq ex-pressed in a ﬁrst order language, the attached Poincar´ e series is a rational function. Further, 302 Handbook of Finite Fields there is an eﬀective computation of the coeﬃcients of its numerator and denominator based on expressing those coeﬃcients in p-adic Dwork cohomology. 9.7.105 Theorem [Theorem 9.7.104 continued] Given a diophantine problem D over Z (or OK) expressed in a ﬁrst order language, there is an eﬀective split of the primes of Q (or over K) into two sets: LD,1 and LD,2, with LD,2 ﬁnite. Further, there is a set of varieties V1, . . . , Vs over Z, from which we produce linear equations in variables Y1, . . . , Ys′ that serve as the coeﬃcients of the numerator and denominator of a rational function PD(t). To each (p, Yi), p ∈LD,1 there is a universal attachment of a p-adic Dwork cohomology group, H(p, Yi), computed in the category of such Dwork cohomology attached to V1, . . . , Vs. The corresponding Poincar´ e series PD,p at p ∈LD,1 comes by substituting H(p, Yi) for each Y1, . . . , Ys′ in PD(t). Then apply the Frobenius operator at p to these coeﬃcients. 9.7.106 Remark In Theorem 9.7.105 it is possible to take V1, . . . , Vs to be nonsingular projec-tive varieties with Yi representing a Chow motive (over Q). Applying the Frobenius operator at p is meaningful as Chow motives are formed from ´ etale cohomology groups of V1, . . . , Vs. 9.7.107 Remark The eﬀectiveness of Theorem 9.7.104 is based on Dwork cohomology , and the explicit calculations of . Theorem 9.7.105 and Remark 9.7.106 both rest on the Galois stratiﬁcation procedure of or [1121, Chapter 24]. On the plus side, the uniform use of ´ etale cohomology from characteristic 0 produces wonderful invariants – like, Euler characteristics – attached to diophantine problems. On the negative, all the eﬀectiveness disappears. In particular, the relation between the sets denoted LD,1 in the two results is a mystery. 9.7.108 Remark Relating exceptional covers (and Davenport pairs) and other problems about al-gebraic equations is a running theme in and . Detecting these relations comes from pr-exceptional correspondences [1118, Section 3.2]. We catch the possible appearance of such correspondences when two Poincar´ e series have inﬁnitely many identical coeﬃcients. 9.7.109 Example An exceptional cover, X →P1 z, over Q, will be a curve whose Poincar´ e series is the same as that of P1 z at inﬁnitely many primes. The systematic use of such characterizations combines monodromy precision (where it applies) and Theorem 9.7.110. 9.7.110 Theorem ([1119, Proposition 7.17], based on ) The zero support of the diﬀerence of two Poincar´ e series consists of the union of arithmetic progressions. See Also §8.1 Discusses the large literature on permutation polynomials (as in Proposition 9.7.39). This contrasts with the use of a cover given by an exceptional polynomial, where one ﬁxed polynomial works for inﬁnitely many ﬁnite ﬁelds. §8.3 Mentions several explicit Chebotarev density theorem error terms. Such error terms have improved over time, but, like Proposition 9.7.28, this sections’ results exhibit monodromy precision: the error term vanishes. §9.6 Discusses Dickson polynomials in detail, including their various combinatorial formulas. This contrasts with Remark 9.7.34 which provides a formula free characterization. References Cited: [84, 332, 341, 696, 777, 814, 943, 1021, 1109, 1110, 1111, 1112, 1113, 1114, 1115, 1117, 1118, 1119, 1120, 1121, 1123, 1124, 1370, 1427, 1936, 1986, 2031, 2191, 2203, 2310, 2311, 2586, 2587, 2590, 2595, 2614] 10 Sequences over ﬁnite ﬁelds 10.1 Finite ﬁeld transforms .............................. 303 Basic deﬁnitions and important examples • Functions between two groups • Discrete Fourier Transform • Further topics 10.2 LFSR sequences and maximal period sequences 311 General properties of LFSR sequences • Operations with LFSR sequences and characterizations • Maximal period sequences • Distribution properties of LFSR sequences • Applications of LFSR sequences 10.3 Correlation and autocorrelation of sequences ... 317 Basic deﬁnitions • Autocorrelation of sequences • Sequence families with low correlation • Quaternary sequences • Other correlation measures 10.4 Linear complexity of sequences and multisequences ....................................... 324 Linear complexity measures • Analysis of the linear complexity • Average behavior of the linear complexity • Some sequences with large n-th linear complexity • Related measures 10.5 Algebraic dynamical systems over ﬁnite ﬁelds .. 337 Introduction • Background and main deﬁnitions • Degree growth • Linear independence and other algebraic properties of iterates • Multiplicative independence of iterates • Trajectory length • Irreducibility of iterates • Diameter of partial trajectories 10.1 Finite ﬁeld transforms Gary McGuire, University College Dublin 10.1.1 Basic deﬁnitions and important examples A ﬁnite ﬁeld contains two important ﬁnite abelian groups, the additive group and the multiplicative group. We shall ﬁrst deﬁne the Fourier transform for an arbitrary ﬁnite abelian group, and then we shall focus on the two groups in a ﬁnite ﬁeld. The deﬁnition involves the characters of a ﬁnite abelian group. 10.1.1 Deﬁnition Let G be a ﬁnite abelian group. A character of G is a group homomorphism G − →C×, where C× is the multiplicative group of nonzero complex numbers. 303 304 Handbook of Finite Fields 10.1.2 Remark Let b G denote the group of characters of G, which is a group under the operation (χχ′)(x) = χ(x)χ′(x). Sometimes b G is called the dual group of G. The dual group b G is isomorphic to G, although this isomorphism is not canonical. 10.1.3 Deﬁnition Let G be a ﬁnite abelian group. The Fourier transform of any function f : G − →C is the function b f : b G − →C deﬁned by b f(χ) = X x∈G f(x)χ(x) where χ is a character of G. 10.1.4 Remark The Fourier transform can also be deﬁned without the complex conjugation of χ(x). 10.1.5 Remark It is common to choose an identiﬁcation of G and b G. If χα denotes the image of α ∈G under some isomorphism from G to b G, we write the Fourier transform as b f(α) = X x∈G f(x)χα(x) and as a result we may consider b f to be deﬁned on G. 10.1.6 Example The characters of the additive group of Fq have the form µα(x) = ζ⟨α,x⟩ p where ζp is a ﬁxed primitive p-th root of unity, and ⟨, ⟩is any Fp-valued inner product on Fq. If we take ⟨α, x⟩to be Tr(αx) (absolute trace) then the Fourier transform of a function f : Fq − →C becomes b f(α) = X x∈Fq f(x)ζ−Tr(αx) p . 10.1.7 Example The characters of F∗ q (the multiplicative group of nonzero elements of Fq) have the form χj(γk) = ζjk q−1 where 0 ≤j ≤q −2, ζq−1 is a ﬁxed primitive complex (q −1)-th root of unity, and γ is a ﬁxed generator of F∗ q. For example, if q is odd and j = (q −1)/2 then χj is the quadratic character. The Fourier transform of a function f : F∗ q − →C can then be written b f(j) = q−2 X k=0 f(γk)ζ−jk q−1 10.1.8 Example Let n > 1 be a positive integer. If G = Z/nZ, the characters are the functions χj : Z/nZ →C, for j ∈Z/nZ, where χj(k) = ζjk n where ζn is a complex primitive n-th root of unity. The Fourier transform of a function f : Z/nZ − →C can then be written b f(j) = n−1 X k=0 f(k)ζ−jk n . See Subsection 10.1.3 for a discussion of the Discrete Fourier Transform. Sequences over ﬁnite ﬁelds 305 10.1.9 Remark The space CG of all complex-valued functions deﬁned on G is a Hermitian inner product space via ⟨f, g⟩= 1 \|G\| X x∈G f(x)g(x). This vector space has dimension \|G\|. 10.1.10 Theorem The characters form an orthonormal basis of CG. 10.1.11 Remark We note that b f(χ) = ⟨f, χ⟩. 10.1.12 Remark The Fourier transform can be inverted, in the sense that f(x) = 1 \|G\| X χ∈b G ⟨f, χ⟩χ(x) for all x ∈G. This expresses f as a linear combination of the basis of characters. Note that it also expresses each value of f as a linear combination of roots of unity. This expression explains why the values b f(χ) are sometimes called the Fourier coeﬃcients of f. 10.1.13 Remark Next we present some of the fundamental theorems in Fourier analysis. Proofs can be found in . 10.1.14 Theorem Plancherel’s Theorem states that ⟨f, g⟩= 1 \|G\|⟨b f, b g⟩. 10.1.15 Theorem Parseval’s Theorem states that ⟨f, f⟩= 1 \|G\|⟨b f, b f⟩. 10.1.16 Theorem The Poisson Summation Formula states that, for any subgroup H of G, 1 \|H\| X x∈H f(x) = 1 \|G\| X χ∈H⊥ b f(χ) where H⊥consists of all the characters of G that are trivial on H. 10.1.17 Remark This is a discrete version of the Poisson Summation Formula. There are many other versions, and many applications, see for further details. 10.1.18 Deﬁnition The convolution of f and g is deﬁned by (f ∗g)(a) = X x∈G f(a −x)g(x) for all a ∈G. 10.1.19 Theorem The convolution theorem of Fourier analysis states that the Fourier transform of a convolution of two functions is equal to the ordinary product of their Fourier transforms: [ f ∗g = b f · b g. 306 Handbook of Finite Fields 10.1.2 Functions between two groups 10.1.20 Remark We have deﬁned above the Fourier transform of a function G − →C. The deﬁnition can be extended to functions deﬁned on G taking values in another abelian group, B. The deﬁnition involves characters of B as well as characters of G. 10.1.21 Deﬁnition Let f : G − →B be a function between ﬁnite abelian groups. The Fourier transform of f is the function b f : b G × b B − →C deﬁned by b f(χ, ψ) = X x∈G ψ(f(x)) χ(x). (10.1.1) 10.1.22 Remark We observe that if ψ ∈b B is the principal character, the Fourier transform evaluates to 0 if χ is not principal, and evaluates to \|G\| if χ is principal. 10.1.23 Remark This Fourier transform of f : G − →B can also be viewed as the ordinary Fourier transform (Deﬁnition 10.1.3) on the group G × B of the characteristic function of the set {(x, f(x)) : x ∈G}. 10.1.24 Remark If we use isomorphisms α 7→χα from G to b G and β 7→ψβ from B to b B, we may consider b f to be deﬁned on G×B and we write the Fourier transform of f at (α, β) ∈G×B as b f(α, β) = X x∈G ψβ(f(x)) χα(x). (10.1.2) 10.1.25 Example The Fourier transform of a function f : Fq − →Fq is b f(α, β) = X x∈Fq ζTr(βf(x)−αx) because the additive characters (see Example 10.1.6) have the form Tr(αx). 10.1.26 Remark We note that in the important special case where f(x) = xd and gcd(d, q −1) = 1, we may write any β ∈Fq as cd, and then b f(α, β) = X x∈Fq ζTr(cdxd−αx) = X x∈Fq ζTr(xd−αc−1x) = b f(αc−1, 1). It follows that, when f(x) = xd and gcd(d, q −1) = 1, we may often assume without loss of generality that β = 1. 10.1.27 Example For a Boolean function f : Fn 2 − →F2, the characters of Fn 2 have the form χα(x) = (−1)⟨α,x⟩ where ⟨, ⟩is any inner product on Fn 2. Also, the only nonzero β in F2 is β = 1, so we may drop the dependence on β and the Fourier transform is written b f(α) = X x∈Fn 2 (−1)f(x)+⟨α,x⟩. This is the Walsh transform of the Boolean function f, or the Hadamard transform of the ±1 valued function (−1)f. Sequences over ﬁnite ﬁelds 307 10.1.28 Remark One often uses the ﬁnite ﬁeld F2n for the vector space Fn 2, and the trace inner product ⟨α, x⟩= Tr(αx). In this case the Walsh transform of a Boolean function f is b f(α) = X x∈Fn 2 (−1)f(x)+Tr(αx). See Section 9.1 for further details on Boolean functions. 10.1.29 Example Let n > 1 be a positive integer. If G = B = Z/nZ, the characters are the functions χj(k) = ζjk n where ζn is a complex primitive n-th root of unity (see Example 10.1.8). For a function f : Z/nZ − →Z/nZ it follows that b f(α, β) = X x∈Z/nZ ζβf(x)−αx n . (10.1.3) 10.1.3 Discrete Fourier Transform 10.1.30 Remark A function f : G − →B can also be considered as a vector or sequence (f(g1), . . . , f(gn)) ∈Bn when G = {g1, . . . , gn}. Through this correspondence, transforms of functions are sometimes considered as transforms of vectors and sequences. 10.1.31 Deﬁnition If G = {g1, . . . , gn} and b G = {χ1, . . . , χn}, the character table of G is the matrix with (i, j) entry χi(gj). 10.1.32 Remark If we identify a function f : G − →C with its vector (f(g1), . . . , f(gn)), the Fourier Transform of f is the vector obtained by multiplying the vector f by the character table of G: b f = Xf where X is the character table of G. 10.1.33 Remark We have already presented a case of the Discrete Fourier Transform in Example 10.1.8, however we shall give the matrix formulation here, which is more common. In fact, the ﬁeld K below does not need to be a ﬁnite ﬁeld. Further information can be found in many places, see for example. 10.1.34 Deﬁnition Let K be a ﬁeld containing all the n-th roots of unity. Let ζn be a primitive n-th root of unity in K. Let Fn be the n × n matrix whose (i, j) entry is ζij n , where 0 ≤i, j ≤n −1. The matrix Fn is the n-th Fourier matrix. 10.1.35 Remark When K = C the Fourier matrix Fn is an important example of a Butson-Hadamard matrix. We note that Fn is also a Vandermonde matrix. 10.1.36 Remark We modify the matrix Fn to the matrix Dn whose (i, j) entry is ζ−ij n (we complex conjugate each entry of Fn). We remark that it does not matter whether we use Fn or Dn in the deﬁnition of Discrete Fourier Transform, but we shall use Dn to be consistent with our earlier deﬁnitions. 308 Handbook of Finite Fields 10.1.37 Deﬁnition Let f = (f0, f1, . . . , fn−1) be in Kn. The Discrete Fourier Transform of f is the vector b f = Dnf = ( b f0, b f1, . . . , b fn−1) where b fj = Pn−1 k=0 fkζ−jk n . 10.1.38 Remark The Fourier matrix Fn is the character table of G = Z/nZ, and this deﬁnition is a case of Remark 10.1.32. 10.1.39 Remark If K = C the Discrete Fourier Transform (DFT) is the same as Example 10.1.8. If K = Fq and n is a divisor of q −1, the DFT is known as the Discrete Fourier Transform over a ﬁnite ﬁeld. 10.1.40 Remark The inverse of Fn has (i, j) entry 1 nζ−ij n , i.e., DnFn = nIn. Therefore the Discrete Fourier Transform has an inverse, which is almost a Discrete Fourier Transform itself, except for the factor of 1/n. Sometimes authors include a factor of 1/√n in the deﬁnition, which then also appears in the inverse, so with this deﬁnition the inverse DFT is also a DFT. If K is a ﬁeld of characteristic p, we assume here that n is relatively prime to p so that 1/n exists in K. If p divides n, a generalized DFT has been deﬁned in using the values of the Hasse derivatives. 10.1.41 Remark The DFT as deﬁned here may also be viewed as the DFT on the group ring K[G] where G is a cyclic group. This deﬁnition can be generalized to a DFT on other group rings. Rings which support Fourier transforms are characterized in . 10.1.42 Remark Sometimes we identify the vector f = (f0, f1, . . . , fn−1) with the polynomial f(x) = f0 + f1x + · · · + fn−1xn−1, and then b f = (f(1), f(ζn), f(ζ2 n), . . . , f(ζn−1 n )). Thus the Discrete Fourier Transform of f is the vector of values of the polynomial f at the n-th roots of unity. 10.1.43 Remark The Fast Fourier Transform is a computationally eﬃcient way of computing the Discrete Fourier Transform, and has many applications. Traditionally, n is a power of 2 and one uses polynomial evaluations as in Remark 10.1.42. There is a huge literature on this topic, see for example, so we do not go into this here. 10.1.44 Remark Given a vector f = (f0, f1, . . . , fn−1) in Kn, we construct a circulant matrix F with f as its top row. Similarly we construct a circulant matrix b F with b f = ( b f0, b f1, . . . , b fn−1) as its top row. There is a result sometimes known as Blahut’s theorem stating that the weight of f is equal to the rank of this circulant b F, and the weight of b f is equal to the rank of F. This is true because if we let D be the diagonal matrix diag(f0, f1, . . . , fn−1), we observe that FnDFn = b F, and because Fn is invertible, the rank of D (which is the weight of f) is equal to the rank of b F. Because the linear complexity of f is equal to the rank of F, this result can be useful for linear complexities of sequences. 10.1.4 Further topics 10.1.45 Remark In this subsection, as before, G and B denote ﬁnite abelian groups. Sequences over ﬁnite ﬁelds 309 10.1.4.1 Fourier spectrum 10.1.46 Remark The Fourier spectrum is the set of values of the Fourier transform. Most authors take the Fourier spectrum to be the multi-set of values, i.e., the values including their multiplicities. The Fourier spectrum is an important invariant in many applications. 10.1.47 Example The Fourier spectrum of the function x3 on F2n is {0, ±2(n+1)/2} if n is odd, and {0, ±2n/2, ±2(n+2)/2} if n is even. 10.1.48 Remark There are many papers calculating the Fourier spectrum of speciﬁc functions that are of particular interest, see or for example. There are also general papers on the structure of the Fourier spectrum of arbitrary functions. The Weil bound gives an upper bound on the absolute value of any Fourier coeﬃcient, a result that we state here, and that has many variations. 10.1.49 Deﬁnition The Weil sum associated to an additive character µ of Fq and a polynomial f ∈Fq[x] is X x∈Fq µ(f(x)). 10.1.50 Theorem (Weil bound) If the Weil sum is non-degenerate then X x∈Fq µ(f(x)) ≤(deg(f) −1)√q. 10.1.4.2 Nonlinearity 10.1.51 Deﬁnition The linearity of a function f : G − →B is deﬁned to be the maximum of the absolute values of the numbers in the Fourier spectrum, i.e., we deﬁne the linearity of f by L(f) = max α∈G,β∈B∗\| b f(α, β)\|. 10.1.52 Deﬁnition The nonlinearity of f : G − →B is deﬁned by NL(f) = (\|G\| −L(f))/\|B\|. 10.1.53 Remark For a Boolean function f : Fn 2 − →F2, the nonlinearity of f is (2n −L(f))/2. The nonlinearity of a Boolean function is often a useful measure in cryptography; see Section 9.1. 10.1.4.3 Characteristic functions 10.1.54 Deﬁnition If A is a subset of Fq, the characteristic function of A is χA(x) = ( 1 if x ∈A 0 if x / ∈A. 310 Handbook of Finite Fields 10.1.55 Remark The Fourier Transform of χA can be used to obtain information about A. There are some general principles, such as: if all the Fourier coeﬃcients of χA are small, relatively speaking, then A is usually a fairly “random” subset; see for further details. In a diﬀerent direction, the paper is an example of utilizing the Fourier Transform of the characteristic function of the support of a speciﬁc function of interest. 10.1.4.4 Gauss sums 10.1.56 Deﬁnition The Gauss sum associated to a multiplicative character χ (i.e., a character of F∗ q) and an additive character µ (i.e., a character of Fq) is S(χ, µ) = X x∈F∗ q χ(x)µ(x). (10.1.4) See Section 6.1 for further details. 10.1.57 Remark Here is one interpretation of Gauss sums. Working in the space of functions F∗ q − → C we consider µ as a function on F∗ q by restriction. Using Fourier inversion on (10.1.4), we can express an additive character µ in terms of the basis of multiplicative characters; and the coeﬃcients in this expansion are Gauss sums. 10.1.4.5 Uncertainty principle 10.1.58 Remark If G is any ﬁnite abelian group, the uncertainty principle states that \|supp (f)\| · \|supp ( b f)\| ≥\|G\| for any nonzero complex-valued function f deﬁned on G. In particular, if G = Fq we have \|supp (f)\| · \|supp ( b f)\| ≥q. Thus, a function and its Fourier transform cannot both have “small” support. Tao recently showed that, in the case q = p, the uncertainty principle can be improved as follows: \|supp (f)\| + \|supp ( b f)\| ≥p + 1. See Also §6.2 For discussion of character sums. §6.3 For applications of character sums. §9.1 For discussion of Boolean functions. §9.2 For discussion of PN and APN functions. §9.3 For a study of bent functions. References Cited: [141, 155, 384, 387, 418, 865, 2013, 2780, 2789] Sequences over ﬁnite ﬁelds 311 10.2 LFSR sequences and maximal period sequences Harald Niederreiter, KFUPM 10.2.1 General properties of LFSR sequences 10.2.1 Deﬁnition Let k be a positive integer and let a0, a1, . . . , ak−1 be ﬁxed elements of the ﬁnite ﬁeld Fq. A sequence s0, s1, . . . of elements of Fq satisfying the linear recurrence relation sn+k = k−1 X i=0 aisn+i for n = 0, 1, . . . is an LFSR sequence (or a linear feedback shift register sequence, also a linear recurring sequence) in Fq. The integer k is the order of the LFSR sequence or of the linear recurrence relation. 10.2.2 Remark We usually abbreviate the sequence s0, s1, . . . by (sn). The sequence (sn) in Def-inition 10.2.1 is uniquely determined by the linear recurrence relation and by the initial values s0, s1, . . . , sk−1. 10.2.3 Remark In electrical engineering, LFSR sequences in Fq are generated by special switching circuits called linear feedback shift registers. A linear feedback shift register consists of adders and multipliers for arithmetic in Fq as well as delay elements. In the binary case q = 2, multipliers are not needed. 10.2.4 Theorem Any LFSR sequence (sn) in Fq of order k is ultimately periodic with least period at most qk −1. A suﬃcient condition for (sn) to be (purely) periodic is that the coeﬃcient a0 in the linear recurrence relation in Deﬁnition 10.2.1 is nonzero. 10.2.5 Deﬁnition Let (sn) be an LFSR sequence in Fq of order k satisfying the linear recurrence relation in Deﬁnition 10.2.1. Then the polynomial f(x) = xk − k−1 X i=0 aixi ∈Fq[x] is a characteristic polynomial of (sn) and also a characteristic polynomial of the linear recurrence relation. The reciprocal polynomial f ∗(x) = xkf(1/x) of f is a connection polynomial of (sn) and also a connection polynomial of the linear recurrence relation. 10.2.6 Deﬁnition Let (sn) be an LFSR sequence in Fq of order k. Then for n = 0, 1, . . ., the vector sn = (sn, sn+1, . . . , sn+k−1) ∈Fk q is the n-th state vector of (sn). 10.2.7 Remark Let (sn) be an LFSR sequence in Fq with characteristic polynomial f ∈Fq[x] and let A be the companion matrix of f. Then the linear recurrence relation for (sn) in Deﬁnition 10.2.1 can be written as the identity sn+1 = snA, n = 0, 1, . . ., for the state vectors. Consequently, we get sn = s0An for n = 0, 1, . . . . Since An can be calculated by O(log n) matrix multiplications using the standard square-and-multiply technique, this 312 Handbook of Finite Fields identity leads to an eﬃcient algorithm for computing remote terms of the LFSR sequence (sn). 10.2.8 Remark Let F∞ q be the sequence space over Fq, viewed as a vector space over Fq under termwise operations for sequences. Let T be the shift operator Tω = (wn+1) for all ω = (wn) ∈F∞ q . Then for a characteristic polynomial f ∈Fq[x] of an LFSR sequence σ = (sn) in Fq we have f(T)(σ) = (0), where (0) denotes the zero sequence. The set I(σ) = {g ∈Fq[x] : g(T)(σ) = (0)} of annihilating polynomials is a nonzero ideal in Fq[x]. 10.2.9 Deﬁnition The uniquely determined monic polynomial over Fq generating the ideal I(σ) in Remark 10.2.8 is the minimal polynomial of the LFSR sequence σ in Fq. 10.2.10 Remark If σ is the zero sequence, then its minimal polynomial is the constant polynomial 1. If σ is a nonzero LFSR sequence, then its minimal polynomial has positive degree and is the characteristic polynomial of the linear recurrence relation of least possible order satisﬁed by σ. 10.2.11 Theorem The minimal polynomial of an LFSR sequence (sn) in Fq divides any characteristic polynomial of (sn). A characteristic polynomial of (sn) of degree k ≥1 is the minimal polynomial of (sn) if and only if the corresponding state vectors s0, s1, . . . , sk−1 are linearly independent over Fq. 10.2.12 Example Let (sn) be an LFSR sequence in Fq of order k with initial values s0 = s1 = · · · = sk−2 = 0, sk−1 = 1 (s0 = 1 if k = 1). Then the linear independence property in the second part of Theorem 10.2.11 is clearly satisﬁed, and so the characteristic polynomial of (sn) of degree k is also the minimal polynomial of (sn). An LFSR sequence with these special initial values is an impulse response sequence. 10.2.13 Theorem If m ∈Fq[x] is the minimal polynomial of the LFSR sequence σ in Fq, then the least period of σ is equal to the order of m and the least preperiod of σ is equal to the multiplicity of 0 as a root of m. 10.2.14 Corollary An LFSR sequence in Fq is periodic if and only if its minimal polynomial m ∈ Fq[x] satisﬁes m(0) ̸= 0. 10.2.15 Remark The basic theory of LFSR sequences in ﬁnite ﬁelds, as presented above, has quite a long history. An important early paper is Zierler . Other milestones in the history of LFSR sequences in ﬁnite ﬁelds are the lecture notes of Selmer and the book of Golomb . A treatment of LFSR sequences in the wider context of general feedback shift register sequences in ﬁnite ﬁelds, i.e., those including also nonlinear feedback functions, is given in the monograph of Ronse . The proofs of many results in this section can be found in Chapters 6 and 7 of the book . 10.2.16 Theorem Let (sn) be an LFSR sequence in Fq with characteristic polynomial f ∈Fq[x]. Let e0 be the multiplicity of 0 as a root of f, where we can have e0 = 0, and let α1, . . . , αh be the distinct nonzero roots of f (in its splitting ﬁeld F over Fq) with multiplicities e1, . . . , eh, respectively. Then sn = tn + h X i=1 ei−1 X j=0 n + j j βijαn i for n = 0, 1, . . . , Sequences over ﬁnite ﬁelds 313 where all tn ∈Fq, tn = 0 for n ≥e0, and all βij ∈F. 10.2.17 Corollary If a characteristic polynomial f of (sn) is irreducible over Fq and α is a root of f in the splitting ﬁeld F of f over Fq, then there exists a uniquely determined β ∈F such that sn = TrF/Fq(βαn) for n = 0, 1, . . . . 10.2.2 Operations with LFSR sequences and characterizations 10.2.18 Theorem For each i = 1, . . . , h, let σi be an LFSR sequence in Fq with minimal polynomial mi ∈Fq[x] and least period ri. If m1, . . . , mh are pairwise coprime, then the minimal polynomial of the (termwise) sum sequence σ1 + · · · + σh is equal to the product m1 · · · mh and the least period of σ1 + · · · + σh is equal to the least common multiple of r1, . . . , rh. 10.2.19 Remark In general, operations with LFSR sequences are treated in terms of the spaces S(f), where for a monic f ∈Fq[x] we let S(f) be the kernel of the linear operator f(T) on F∞ q (compare with Remark 10.2.8). Any S(f) is a linear subspace of F∞ q of dimension deg(f). The following result characterizes the spaces S(f). 10.2.20 Theorem A subset E of F∞ q is equal to S(f) for some monic f ∈Fq[x] if and only if E is a ﬁnite-dimensional subspace of F∞ q which is closed under the shift operator T. 10.2.21 Theorem For any monic f1, . . . , fh ∈Fq[x], we have S(f1) ∩· · · ∩S(fh) = S(gcd(f1, . . . , fh)), S(f1) + · · · + S(fh) = S(lcm(f1, . . . , fh)). 10.2.22 Remark The (termwise) product sequence σ1 · · · σh of LFSR sequences σ1, . . . , σh in Fq is more diﬃcult to analyze. For monic polynomials f1, . . . , fh ∈Fq[x], let S(f1) · · · S(fh) be the subspace of F∞ q spanned by all product sequences σ1 · · · σh with σi ∈S(fi) for 1 ≤i ≤h. It follows from Theorem 10.2.20 that S(f1) · · · S(fh) = S(g) for some monic g ∈Fq[x]. If each fi, 1 ≤i ≤h, is nonconstant and has only simple roots, then g is the monic polynomial whose roots are the distinct elements of the form α1 · · · αh, where each αi is a root of fi in the splitting ﬁeld of f1 · · · fh over Fq. For the general case, a procedure to determine g can be found in Zierler and Mills . 10.2.23 Deﬁnition If σ = (sn) is a sequence of elements of Fq and d is a positive integer, then the operation of decimation produces the decimated sequence σ(d) = (snd). Thus, σ(d) is obtained by taking every d-th term of σ, starting from s0. 10.2.24 Theorem [939, 2237] Let σ be an LFSR sequence in Fq. Then so is σ(d) for any positive integer d. If f is a characteristic polynomial of σ and f(x) = Qk j=1(x−βj) is the factorization of f in its splitting ﬁeld over Fq, then gd(x) = Qk j=1(x−βd j ) is a characteristic polynomial of σ(d). Furthermore, if f is the minimal polynomial of σ and d is coprime to the least period of σ, then gd is the minimal polynomial of σ(d). 10.2.25 Deﬁnition An LFSR sequence σ in Fq which has a characteristic polynomial f ∈Fq[x] and satisﬁes σ(q) = σ is a characteristic sequence for f. 314 Handbook of Finite Fields 10.2.26 Remark Characteristic sequences play an important role in the Niederreiter algorithm for factoring polynomials over ﬁnite ﬁelds [2249, 2260]. Characteristic sequences can be described explicitly in terms of their generating functions (see Theorem 10.2.35 below). 10.2.27 Remark In the case q = 2, an interesting operation on sequences is that of binary comple-mentation. If σ is a sequence of elements of F2, then its binary complement σ is obtained by replacing each term 0 in σ by 1 and each term 1 in σ by 0. 10.2.28 Theorem Let σ be an LFSR sequence in F2 with minimal polynomial m ∈F2[x]. Write m in the form m(x) = (x+1)hm1(x) with an integer h ≥0 and m1 ∈F2[x] satisfying m1(1) = 1. Then the minimal polynomial m of the binary complement σ is given by m(x) = (x+1)m(x) if h = 0, m(x) = m1(x) if h = 1, and m(x) = m(x) if h ≥2. 10.2.29 Remark LFSR sequences in Fq can be characterized in terms of Hankel determinants. For an arbitrary sequence (sn) of elements of Fq and for integers n ≥0 and b ≥1, deﬁne the Hankel determinant D(b) n = det((sn+i+j)0≤i,j≤b−1). 10.2.30 Theorem The sequence (sn) of elements of Fq is an LFSR sequence in Fq if and only if there exists an integer b ≥1 such that D(b) n = 0 for all suﬃciently large n. Furthermore, (sn) is an LFSR sequence in Fq with minimal polynomial of degree k if and only if D(b) 0 = 0 for all b ≥k + 1 and k + 1 is the least positive integer for which this holds. 10.2.31 Remark If an LFSR sequence in Fq is known to have a minimal polynomial of degree at most k for some integer k ≥1, then the Berlekamp-Massey algorithm produces the minimal polynomial from the ﬁrst 2k terms of the sequence [231, 2011]. 10.2.32 Remark LFSR sequences in Fq can also be characterized in terms of their generating functions. There are two diﬀerent characterizations, depending on whether the generating function is a formal power series in x or in x−1. 10.2.33 Theorem The sequence (sn) of elements of Fq is an LFSR sequence in Fq of order k with connection polynomial c ∈Fq[x] if and only if ∞ X n=0 snxn = g(x) c(x) with g ∈Fq[x] and deg(g) < k. 10.2.34 Theorem The sequence (sn) of elements of Fq is an LFSR sequence in Fq with minimal polynomial m ∈Fq[x] if and only if ∞ X n=0 snx−n−1 = g(x) m(x) with g ∈Fq[x] and gcd(g, m) = 1. 10.2.35 Theorem Let f ∈Fq[x] be a monic polynomial of positive degree. Then the sequence (sn) of elements of Fq is a characteristic sequence for f if and only if ∞ X n=0 snx−n−1 = h X i=1 bi p′ i(x) pi(x) with b1, . . . , bh ∈Fq, where p1, . . . , ph ∈Fq[x] are the distinct monic irreducible factors of f and p′ i is the ﬁrst derivative of pi. Sequences over ﬁnite ﬁelds 315 10.2.3 Maximal period sequences 10.2.36 Deﬁnition An LFSR sequence in Fq whose minimal polynomial is a primitive polynomial over Fq is a maximal period sequence (or an m-sequence) in Fq. 10.2.37 Theorem Any maximal period sequence σ in Fq is periodic with least period qk −1, where k is the degree of the minimal polynomial of σ. 10.2.38 Remark The terminology “maximal period sequence” stems from the fact that, by Theo-rems 10.2.4 and 10.2.37, qk −1 is the largest value that can be achieved by the least period of an LFSR sequence in Fq of order k. 10.2.39 Theorem Let (sn) be a maximal period sequence in Fq with minimal polynomial of degree k. Then the state vectors s0, s1, . . . , sqk−2 of (sn) run exactly through all nonzero vectors in Fk q. 10.2.40 Theorem A nonzero periodic sequence σ of elements of Fq is a maximal period sequence in Fq if and only if all its shifted sequences T hσ, h = 0, 1, . . ., together with the zero sequence form an Fq-linear subspace of F∞ q . 10.2.41 Theorem Let σ be a maximal period sequence in Fq with minimal polynomial of degree k. Then every LFSR sequence in Fq having an irreducible minimal polynomial g with g(0) ̸= 0 and deg(g) dividing k can be obtained from σ by applying a shift and then a decimation. 10.2.42 Remark For maximal period sequences, the autocorrelation function has a simple form. Let (sn) be a maximal period sequence in Fq with least period r. Then its autocorrelation function Cr is deﬁned by Cr(h) = r−1 X n=0 χ(sn −sn+h) for all positive integers h, where χ is a ﬁxed nontrivial additive character of Fq. 10.2.43 Theorem For any maximal period sequence in Fq with least period r, its autocorrelation function Cr satisﬁes Cr(h) = r if h ≡0 (mod r) and Cr(h) = −1 if h ̸≡0 (mod r). 10.2.4 Distribution properties of LFSR sequences 10.2.44 Remark For LFSR sequences in Fq for which the least period is suﬃciently large compared to the order, the terms in the full period are almost evenly distributed over Fq. Without loss of generality, it suﬃces to consider periodic LFSR sequences in Fq. For such a sequence σ = (sn) with least period r and for b ∈Fq, let Z(b; σ) be the number of integers n with 0 ≤n ≤r −1 such that sn = b. In other words, Z(b; σ) is the number of occurrences of b in a full period of σ. 10.2.45 Theorem Let σ be a periodic LFSR sequence in Fq of order k and with least period r. Then for any b ∈Fq we have Z(b; σ) −r q ≤ 1 −1 q qk/2. 10.2.46 Remark If σ is a maximal period sequence in Fq with minimal polynomial of degree k, then it follows from Theorem 10.2.39 that Z(b; σ) = qk−1 for b ̸= 0 and Z(0; σ) = qk−1 −1. 10.2.47 Remark Let the sequence σ = (sn) be as in Remark 10.2.44. For b ∈Fq and a positive integer N, let Z(b; N; σ) be the number of integers n with 0 ≤n ≤N −1 such that sn = b. 316 Handbook of Finite Fields 10.2.48 Theorem Let σ be a periodic LFSR sequence in Fq of order k and with least period r. Then for any b ∈Fq and any integer N with 1 ≤N ≤r we have Z(b; N; σ) −N q ≤ 1 −1 q qk/2 2 π log r + 7 5 . 10.2.49 Remark The distribution of blocks of elements in a periodic LFSR sequence σ = (sn) in Fq has also been investigated. Let r be the least period of σ. For b = (b1, . . . , bt) ∈Ft q with a positive integer t, let Z(b; σ) be the number of integers n with 0 ≤n ≤r −1 such that sn+i−1 = bi for 1 ≤i ≤t. The simplest case is that of a maximal period sequence σ in Fq. If k is the degree of the minimal polynomial of σ and 1 ≤t ≤k, then Z(b; σ) = qk−t for b ∈Ft q with b ̸= 0, whereas Z(0; σ) = qk−t −1. 10.2.50 Theorem Let σ be a periodic LFSR sequence in Fq with least period r. Let m ∈Fq[x] be the minimal polynomial of σ and put k = deg(m). Suppose that the positive integer t is less than or equal to the degree of any irreducible factor of m in Fq[x]. Then for any b ∈Ft q we have Z(b; σ) −r qt ≤ 1 −1 qt qk/2. 10.2.51 Remark More general results on distribution properties of LFSR sequences are available in . For instance, there is an analog of Theorem 10.2.48 for the distribution of blocks of elements in parts of the full period. Furthermore, one can consider not only blocks of successive terms of an LFSR sequence as in Remark 10.2.49, but also blocks of terms with arbitrary lags. Reﬁned results for various special cases can be found in [2635, 2662]. 10.2.5 Applications of LFSR sequences 10.2.52 Remark LFSR sequences in Fq have numerous applications. In this subsection, we mention some typical applications. We start with an application to combinatorics. 10.2.53 Deﬁnition An (h, k) de Bruijn sequence is a ﬁnite sequence d0, d1, . . . , dN−1 with N = hk terms from a nonempty ﬁnite set of h elements such that the k-tuples (dn, dn+1, . . . , dn+k−1), n = 0, 1, . . . , N −1, with subscripts considered modulo N are all diﬀerent. 10.2.54 Remark Let s0, s1, . . . be an impulse response sequence of order k (see Example 10.2.12) which is also a maximal period sequence in Fq with minimal polynomial of degree k. Then 0, s0, s1, . . . , sqk−2 is a (q, k) de Bruijn sequence. This follows immediately from Theorem 10.2.39. 10.2.55 Remark Any periodic sequence σ = (sn) of elements of Fq, say with period r, satisﬁes the linear recurrence relation sn+r = sn for n = 0, 1, . . . and is thus an LFSR sequence in Fq. The linear complexity (or the linear span) of σ is deﬁned to be the degree of the minimal polynomial of σ. The linear complexity is an important complexity measure in the theory of stream ciphers in cryptology. For details on the linear complexity, the reader is referred to Section 10.4. 10.2.56 Remark There is a family of cryptosystems which are based on LFSR sequences in Fq and the operation of decimation (see Deﬁnition 10.2.23). These cryptosystems were introduced in and are FSR cryptosystems. 10.2.57 Remark LFSR sequences in Fq can be used for the encoding of cyclic codes. We refer to Section 8.7 in the book of Peterson and Weldon for an account of this application. Sequences over ﬁnite ﬁelds 317 This connection between LFSR sequences and cyclic codes, when combined with results of the type stated in Theorem 10.2.45, yields information on the weight distribution of cyclic codes . 10.2.58 Remark Maximal period sequences in ﬁnite prime ﬁelds are used in methods for generating uniform pseudorandom numbers in the interval [0, 1]. Well-known methods of this type are the digital multistep method and the generalized feedback shift register (GFSR) method. We refer to Chapter 9 in the book for a detailed discussion of these methods. 10.2.59 Remark LFSR sequences in Fq have important applications in digital communication sys-tems. A celebrated example is code division multiple access (CDMA) in wireless communi-cation. The book of Viterbi is the standard reference for CDMA. See Also §10.3 For applications to correlation of sequences. §10.4 For connections with linear complexity and cryptology. References Cited: [231, 939, 1063, 1294, 1300, 1938, 2011, 2231, 2232, 2233, 2237, 2241, 2248, 2249, 2260, 2390, 2475, 2579, 2635, 2662, 2880, 3070, 3072] 10.3 Correlation and autocorrelation of sequences Tor Helleseth, University of Bergen 10.3.1 Basic deﬁnitions 10.3.1 Deﬁnition Let {u(t)} and {v(t)} be two complex-valued sequences of period n. The periodic correlation of {u(t)} and {v(t)} at shift τ is the inner product θu,v(τ) = n−1 X t=0 u(t + τ)v(t), 0 ≤τ < n, where a denotes the complex conjugation of a and t + τ is calculated modulo n. If the sequences {u(t)} and {v(t)} are the same, the correlation θu,u(τ) is denoted by θu(τ) and is the autocorrelation. When they are distinct θu,v(τ) is the crosscorrelation. 10.3.2 Remark Sequences with good correlation properties have numerous applications in commu-nication systems and lead to many challenging problems in ﬁnite ﬁelds. The main problems from an application point of view are to ﬁnd single sequences with low autocorrelation for all nonzero shifts and families of sequences where the maximum nontrivial auto- and crosscorrelation values between any two sequences in the family is low. For a more detailed survey on the design and analysis of sequences with low correlation the reader is referred to . Other related references are [1215, 1303, 1476, 2526, 2673] and Chapter V (Section 7) in . 318 Handbook of Finite Fields 10.3.3 Remark The crosscorrelation between two sequences {a(t)} and {b(t)} that take on values in Zq = {0, 1, . . . , q −1} is deﬁned using Deﬁnition 10.3.1 where u(t) = ωa(t), v(t) = ωb(t) and ω is a complex q-th root of unity, i.e., θa,b(τ) = n−1 X t=0 ωa(t+τ)−b(t), 0 ≤τ < n. 10.3.2 Autocorrelation of sequences 10.3.4 Deﬁnition A sequence {s(t)} has ideal autocorrelation if θs(τ) = 0 for all τ ̸≡0 (mod n). 10.3.5 Remark Sequences with ideal autocorrelation do not always exist so a sequence has optimal autocorrelation if the maximal value of its autocorrelation is as small as it can be for a sequence of the given period and symbol alphabet. Many optimal sequences have constant autocorrelation of −1 for all out-of-phase shifts, i.e., when τ ̸≡0 (mod n). 10.3.6 Deﬁnition A q-ary maximal-length linear sequence {s(t)} (or m-sequence) is a sequence of elements with symbols from Fq of period qm −1 generated from a nonzero initial state (s(0), s(1), . . . , s(m −1)) and a linear recursion of degree m given by m X i=0 fis(t + i) = 0, where the characteristic polynomial of the recursion, deﬁned by f(x) = Pm i=0 fixi, is a primitive polynomial in Fq[x] of degree m. 10.3.7 Theorem [1939, Theorem 8.24] An m-sequence, after a suitable cyclic shift, can be described using the trace function from F = Fqm to K = Fq as s(t) = TrF/K(αt), where α is a zero of the primitive characteristic polynomial of the m-sequence. 10.3.8 Remark Maximal-length linear sequences are balanced (with each nonzero element occur-ring qm−1 times and 0 occurring qm−1 −1 times). Furthermore, for any τ ̸= 0 (mod qm −1) there is a δ such that s(t + τ) −s(t) = s(t + δ). In combination with Remark 10.3.3 this leads to the following well-known theorem. 10.3.9 Theorem Let q be a prime. The autocorrelation of the q-ary m-sequence {s(t)} is two-valued, θs(τ) = qm −1 if τ = 0 (mod qm −1), −1 if τ ̸= 0 (mod qm −1). 10.3.10 Construction (GMW sequences) [1324, 2555] Let k, m be integers with k \| m, k ≥1 and let q be a prime power. Let also gcd(r, qk −1) = 1, 1 ≤r ≤qk −2. Let F = Fqm, M = Fqk, and K = Fq. Let α be a primitive element of F and c ∈F. A GMW sequence of period qm −1 is deﬁned by s(t) = TrM/K TrF/M cαtr . 10.3.11 Theorem A GMW sequence is balanced and, if q is a prime, has a two-valued autocorrelation with value -1 for all out-of-phase shifts. Sequences over ﬁnite ﬁelds 319 10.3.12 Remark There is a close connection between a balanced binary sequence {s(t)} of period 2m −1 with autocorrelation −1 for all shifts τ ̸= 0 (mod 2m −1) and diﬀerence sets with Singer parameters (2m −1, 2m−1 −1, 2m−2 −1). The connection is that {s(t)} gives a Singer diﬀerence set (mod 2m −1) deﬁned by D = {t \| s(t) = 0} (see Section 14.6). 10.3.13 Theorem Let 1 ≤k < m 2 and gcd(k, m) = 1 and let α be a primitive element in F2m. Deﬁne the subset of F2m by Uk = n (x + 1)22k−2k+1 + x22k−2k+1 + 1 \| x ∈F2m o . The binary sequence deﬁned by sk(t) = 1 if αt ∈F2m \ Uk and sk(t) = 0 otherwise, is balanced with 2m−1 −1 zeros and 2m−1 ones, and has two-valued autocorrelation with out-of-phase correlation value −1. 10.3.14 Construction (Legendre sequence) The Legendre sequence is a binary sequence {s(t)} of a prime period p, and deﬁned by s(t) = 1 if t = 0 or t is a nonsquare (mod p), 0 otherwise. 10.3.15 Theorem If p ≡3 (mod 4) then the Legendre sequence has a two-valued autocor-relation with values θs(0) = p and θs(τ) = −1 for τ ̸= 0 (mod p). If p ≡1 (mod 4) the autocorrelation is inferior and takes on values 1 and −3 when τ ̸= 0 (mod p) in addition to θs(0) = p. 10.3.16 Construction (Binary Sidelnikov sequences) Let α be a primitive element in Fpm. The binary Sidelnikov sequence has period pm −1 and is deﬁned by s(t) = 1 if αt + 1 is a nonsquare in Fpm, 0 otherwise. 10.3.17 Theorem Binary Sidelnikov sequences are balanced with three-valued optimal au-tocorrelation with out-of-phase values 0 or −4 when n ≡0 (mod 4) and 2 and −2 when n ≡2 (mod 4). 10.3.18 Theorem Let h be a function from Zq to Zq. The sequence {h(t)} of period q with symbols from Zq has ideal autocorrelation if and only if h is a bent function. 10.3.19 Remark Information on bent functions can be found in Section 9.3. 10.3.3 Sequence families with low correlation 10.3.20 Remark In code-division multiple-access (CDMA) systems it is important to ﬁnd large families of sequences where the (nontrivial) auto- and crosscorrelation between all pairs of sequences in the family are low. 10.3.21 Deﬁnition Let F = {{si(t)} \| i = 1, 2, . . . , M} be a family of M sequences of period n with symbols from the alphabet Zq = {0, 1, . . . , q −1}. Let θi,j(τ) denote the crosscorrelation between the sequences {si(t)} and {sj(t)} at shift τ. The parameters of family F are (n, M, θmax) where θmax = max{\|θi,j(τ)\| : either i ̸= j or τ ̸= 0}. 320 Handbook of Finite Fields 10.3.22 Remark There are three well-known bounds on θmax for a family of sequences with given period n and family size M. These bounds are due to Welch , and Sidelnikov and Levenshtein . The Welch and Sidelnikov bounds are based on bounds on the inner products between complex vectors. In the Welch (resp. Sidelnikov) bound the sequences are considered as complex vectors of norm √n (resp. complex q-th roots of unity). 10.3.23 Theorem (The Welch bound) Let k ≥1 be an integer and F a family of M cyclically distinct sequences of period n. Then (θmax)2k ≥ 1 Mn −1 Mn2k+1 k+n−1 n−1 −n2k ! . 10.3.24 Corollary (The Welch bound for k = 1) θmax ≥n r M −1 Mn −1. 10.3.25 Remark For even moderate values of M this implies that θmax ≥√n. 10.3.26 Theorem (The Sidelnikov bound) 1. In the case q = 2, then (θmax)2 > (2k + 1)(n −k) + k(k + 1) 2 − 2kn2k+1 M(2k)! n k , 0 ≤k < 2n 5 . 2. In the case q > 2, then (θmax)2 > k + 1 2 (2n −k) − 2kn2k+1 M(k!)22n k , k ≥0. 10.3.27 Remark The crosscorrelation between two m-sequences of the same period and with sym-bols from K = Fp, p prime, is equivalent to calculating the following exponential sum for all nonzero c ∈F = Fpm, θa,b(τ) = −1 + X x∈F ωTrF/K(cx−xd), where two zeros α and β of the two characteristic polynomials are related by β = αd where gcd(d, pm −1) = 1 and c = ατ. General results and open problems can be found in [1467, 1475]. 10.3.28 Remark Many binary families of sequences with excellent correlation properties are con-structed from m-sequences. The most well-known is the family of Gold sequences. 10.3.29 Construction (Gold sequences) Let m be odd, d = 2k + 1 and gcd(k, m) = 1. Let {s(t)} be an m-sequence of period n = 2m −1. The family of Gold sequences is deﬁned by G = {s(t)} ∪{s(dt)} ∪{{s(t + τ) + s(dt)} \| 0 ≤τ ≤n −1}. 10.3.30 Theorem The parameters of the Gold family are n = 2m −1, M = 2m + 1, and θmax = 2 m+1 2 + 1. 10.3.31 Remark The Gold sequence family is optimal with respect to the Sidelnikov bound and has the smallest possible θmax for the given period, alphabet, and family size. Sequences over ﬁnite ﬁelds 321 10.3.32 Construction (The small family of Kasami sequences) Let m = 2k, k ≥2, and α be a primitive element of F = F2m. Let also M = F2k, K = F2, and sa(t) = TrF/K(αt) + TrM/K aα(2k+1)t where a ∈M. The small family of Kasami sequences is K = {{sa(t)} \| a ∈M}. 10.3.33 Theorem The parameters of the small Kasami family are n = 2m −1, M = 2k, and θmax = 2k + 1. 10.3.34 Remark The small Kasami family is optimal with respect to the Welch bound and has the smallest possible θmax for the given period, alphabet, and family size. 10.3.35 Construction (No sequences) Let m = 2k, k ≥2, and α be a primitive element of F = F2m. Let also 1 ≤r ≤2k −1, r ̸= 2i for any i and gcd(r, 2k −1) = 1. Let M = F2k, K = F2, and deﬁne sa(t) = TrM/K TrF/M αt + aα(2k+1)tr where a ∈F. The No sequence family is deﬁned by N = {{sa(t)} \| a ∈F}. 10.3.36 Theorem The parameters of the No sequence family are the same as for the small Kasami family. No sequences have higher linear complexity than the Kasami sequences. 10.3.37 Theorem (Sidelnikov sequence family) Let p be a prime, 0 < d < p and α be an element of order pm −1. Let also F = Fpm and K = Fp. Let {s(t)} be a sequence over Fp of the form s(t) = TrF/K d X k=1 akαkt ! , where ak ∈F for 1 ≤k ≤d. The parameters of the Sidelnikov sequence family are n = pm −1, M ≥pm(d−1) and θmax ≤(d −1)p m 2 + 1. 10.3.38 Remark The condition on d can be relaxed but leads to a more complicated bound on the parameters. The main idea is that θmax is bounded by the Carlitz-Uchiyama bound by the maximal degree of the polynomials involved in the correlation computations. 10.3.4 Quaternary sequences 10.3.39 Remark Families of sequences with symbols from Z4 can have correlation properties that are superior to binary sequences. Family A in Construction 10.3.42 is an important example. 10.3.40 Deﬁnition (Lifting of f(x)) The lifting of a binary polynomial f(x) is the quaternary polynomial g(x) = Pm i=0 gixi where g(x2) ≡(−1)mf(x)f(−x) (mod 4). 10.3.41 Example The lifting of the primitive binary polynomial f(x) = x3 + x + 1 is g(x) ≡ x3 + 2x2 + x + 3 (mod 4) since g(x2) = (−1)3f(x)f(−x) ≡x6 + 2x4 + x2 + 3 (mod 4). 10.3.42 Construction (Family A) Let g(x) be a lifting of a binary primitive polynomial of degree m. The recursion Pm i=0 gis(t + i) ≡0 (mod 4) generates 4m −1 quaternary nonzero sequences corresponding to all nonzero initial states (s(0), s(1), . . . , s(m −1)). These sequences are known to have period 2m −1. Family A is constructed by selecting 2m +1 cyclically distinct sequences from this set. 322 Handbook of Finite Fields 10.3.43 Theorem [383, 2692] Family A has parameters n = 2m−1, M = 2m+1 and θmax ≤2 m 2 +1. 10.3.44 Remark Compared with the family of binary Gold sequences, family A has the same period n = 2m −1 and family size M = 2m + 1 but θmax is a factor of √ 2 lower than for the Gold sequence family. 10.3.45 Deﬁnition Let s = a + 2b ∈Z4 where a, b ∈Z2. The most signiﬁcant bit map π is deﬁned by π(s) = b. 10.3.46 Remark Any quaternary sequence {s(t)} of odd period n deﬁnes a quaternary sequence {(−1)ts(t)} of period 2n. Thus a binary sequence {b(t)} of period 2n is obtained by b(t) = π((−1)ts(t)). 10.3.47 Construction (Family of binary Kerdock sequences) Let g(x) be the lifting of a primitive binary polynomial of odd degree m and deﬁne h(x) = −g(−x). The recursion with char-acteristic polynomial h(x) is applied to all initial states that are nonzero modulo 2. This generates 4m −2m quaternary sequences of period 2(2m −1). Selecting cyclically distinct sequences from this set and using the most signiﬁcant bit map π to these sequences leads to the Kerdock family K of 2m−1 binary sequences of period 2(2m −1). 10.3.48 Theorem The parameters of family K are n = 2(2m −1), M = 2m−1 and θmax ≤ 2 m+1 2 + 2. 10.3.49 Remark The sequence family K is superior to the small Kasami set. The size of family K can be increased by a factor of 2 without increasing θmax. For further details see . 10.3.5 Other correlation measures 10.3.50 Deﬁnition The aperiodic correlation of two complex-valued sequences {u(t)} and {v(t)} for t = 0, 1, · · · , n −1 is deﬁned by ρu,v(τ) = min{n−1,n−1−τ} X t=max{0,−τ} u(t + τ)v(t), −(n −1) ≤τ ≤n −1. 10.3.51 Remark If {s(t)} is a binary {+1, −1} sequence then ρs,s(τ) = ρs.s(−τ) and the aperiodic autocorrelation is determined by the values ρs(τ) = n−1−τ X t=0 s(t + τ)s(t), 0 ≤τ ≤n −1. 10.3.52 Deﬁnition A Barker sequence is a binary {−1, +1} sequence of length n if the aperiodic values ρs(τ) satisfy \|ρs(τ)\| ≤1 for all τ, 1 ≤τ ≤n −1. 10.3.53 Remark Barker sequences are only known for the following lengths n = 2, 3, 4, 5, 7, 11, 13. For example the sequence (+1 + 1 + 1 + 1 + 1 −1 −1 + 1 + 1 −1 + 1 −1 + 1) of length n = 13 is the longest known Barker sequence; see also Section 17.3. 10.3.54 Remark It has been shown by Turyn and Storer that there are no Barker sequences of odd length n > 13 and if they exist then n = 0 (mod 4). There is an overwhelming evidence that no Barker sequence of length n > 13 exists. Sequences over ﬁnite ﬁelds 323 10.3.55 Deﬁnition The merit factor of a binary {−1, +1} sequence {s(t)} of length n is deﬁned by F = n2 2 Pn−1 τ=1 ρ2 s(τ). 10.3.56 Remark The highest known merit factor for a sequence is F = 14.08 coming from a Barker sequence of length 13. For a long time the largest proven asymptotical value of the merit factor for a family of arbitrarily long sequences was 6 . In and new con-structions were presented of families of sequences with asymptotic merit factor believed to be greater than 6.34. Recently, this claim has been proved . 10.3.57 Remark Low correlation zone sequences (LCZ) are designed with small auto- and crosscor-relation values for small values of their relative time shifts. The parameters of a family of LCZ sequences are the period n of the sequences, the number M of sequences, the length L of the low correlation zone, and the upper bound δ on the correlation value in the low correlation zone. For further details see . 10.3.58 Deﬁnition A family of low correlation zone sequences is deﬁned by the parameters (n, M, L, δ), where \|θi,j(τ)\| ≤δ when 0 ≤\|τ\| < L, i ̸= j and for 1 ≤\|τ\| < L when i = j. 10.3.59 Theorem For an (n, M, L, δ) LCZ sequence family it holds that ML −1 ≤n −1 1 −δ2 n . 10.3.60 Deﬁnition The periodic Hamming correlation between a pair of binary sequences {s1(t)} and {s2(t)} of period n is the integer θ1,2(τ) = n−1 X t=0 s1(t + τ)s2(t), 0 ≤τ < n. 10.3.61 Remark The periodic Hamming correlation is important in evaluating optical communica-tion systems where 0s and 1s indicate presence or absence of pulses of transmitted light. 10.3.62 Deﬁnition An (n, w, λ) optical orthogonal code (OOC) is a family F = {{si(t)} \| i = 1, 2, . . . , M} of M binary sequences of period n and constant Hamming weight w where 1 ≤w ≤n−1 and θi,j(τ) ≤λ when either i ̸= j or τ ̸= 0. 10.3.63 Remark The close relations between (n, w, λ) OOCs and constant weight codes provides good bounds on OOC from known bounds on constant weight codes. For further information see . 10.3.64 Remark Other correlation measures include the partial-period correlation between two very long sequences where the correlation is calculated over a partial period. In practice, there 324 Handbook of Finite Fields is also some interest in the mean-square correlation of a sequence family rather than in θmax. For the evaluation of these correlation measures coding theory sometimes plays an important role. For more information the reader is referred to . See Also §6.1 Exponential sums are crucial in calculating the correlation of sequences. §9.1 Boolean functions are closely related to sequences. §9.3 Bent functions can be used in sequence constructions and vice versa. §10.2 Linear Feedback Shift Registers (LFSRs) are important in constructing sequences with low correlation. §14.6 Cyclic diﬀerence sets are related to sequences with two-level autocorrelation. §17.3 Describes some applications of sequences and results on aperiodic correlation. An overview of some recent advances of low correlation sequences. This textbook by Golomb and Gong gives important information on sequences and their applications to signal design and cryptography. Provides an extensive survey of sequences with low correlation and their connections to coding theory. An elementary introduction to pseudonoise sequences. A classical paper on the crosscorrelation of pseudorandom sequences. References Cited: [352, 383, 550, 706, 864, 1215, 1295, 1303, 1324, 1467, 1475, 1476, 1523, 1605, 1688, 1806, 1939, 2295, 2388, 2526, 2555, 2659, 2660, 2661, 2673, 2692, 2778, 2828, 2964] 10.4 Linear complexity of sequences and multisequences Wilfried Meidl, Sabanci University Arne Winterhof, Austrian Academy of Sciences 10.4.1 Linear complexity measures 10.4.1 Deﬁnition A sequence S = s0, s1, . . . over the ﬁnite ﬁeld Fq is called a (homogeneous) linear recurring sequence over Fq with characteristic polynomial f(x) = l X i=0 cixi ∈Fq[x] of degree l, if S satisﬁes the linear recurrence relation l X i=0 cisn+i = 0 for n = 0, 1, . . . . (10.4.1) Sequences over ﬁnite ﬁelds 325 10.4.2 Deﬁnition The minimal polynomial of a linear recurring sequence S is the uniquely deﬁned monic polynomial M ∈Fq[x] of smallest degree for which S is a linear recurring sequence with characteristic polynomial M. The linear complexity L(S) of S is the degree of the minimal polynomial M. 10.4.3 Remark Without loss of generality one can assume that f is monic, i.e., cl = 1. A sequence S over Fq is a linear recurring sequence if and only if S is ultimately periodic, if c0 in (10.4.1) is nonzero then S is purely periodic, see [1939, Chapter 8]. Consequently Deﬁnition 10.4.1 is only meaningful for (ultimately) periodic sequences. Using the notation of [1134, 2064], we let M(1) q (f) be the set of sequences over Fq with characteristic polynomial f. The set of sequences with a ﬁxed period N is then M(1) q (f) with f(x) = xN −1. The minimal polynomial M of a sequence S ∈M(1) q (f) is always a divisor of f. For an N-periodic sequence S we have L(S) ≤N; see Section 10.2. 10.4.4 Remark The linear complexity of a sequence S can alternatively be deﬁned as the length of the shortest linear recurrence relation satisﬁed by S. In engineering terms, L(S) is 0 if S is the zero sequence and otherwise it is the length of the shortest linear feedback shift register (Section 10.2) that can generate S [1631, 1939, 2502, 2503]. 10.4.5 Deﬁnition For n ≥1 the n-th linear complexity L(S, n) of a sequence S over Fq is the length L of a shortest linear recurrence relation sj+L = cL−1sj+L−1 + · · · + c0sj, 0 ≤j ≤n −L −1, over Fq satisﬁed by the ﬁrst n terms of the sequence. The polynomial PL i=0 cixi ∈Fq[x] is an n-th minimal polynomial of S. The linear complexity L(S) of a periodic sequence can then be deﬁned by L(S) := sup n≥1 L(S, n). 10.4.6 Remark Again one may assume that the n-th minimal polynomial is monic. Then it is unique whenever L ≤n/2. Deﬁnition 10.4.5 is also applicable for ﬁnite sequences, i.e., strings of elements of Fq of length n. 10.4.7 Deﬁnition For an inﬁnite sequence S, the non-decreasing integer sequence L(S, 1), L(S, 2), . . . is the linear complexity proﬁle of S. 10.4.8 Remark Linear complexity and linear complexity proﬁle of a given sequence (as well as the linear recurrence deﬁning it) can be determined by using the Berlekamp-Massey algorithm; see Section 15.1 or [1631, Section 6.7], and . The algorithm is eﬃcient for sequences with low linear complexity and hence such sequences can easily be predicted. 10.4.9 Remark A sequence used as a keystream in stream ciphers must consequently have a large linear complexity, but also altering a few terms of the sequence should not cause a signiﬁcant decrease of the linear complexity. An introduction to the stability theory of stream ciphers is the monograph . For a general comprehensive survey on the theory of stream ciphers we refer to [2502, 2503]. 326 Handbook of Finite Fields 10.4.10 Deﬁnition The k-error linear complexity Lk(S, n) of a sequence S of length n is deﬁned by Lk(S, n) = min T L(T, n), where the minimum is taken over all sequences T of length n with Hamming distance d(T, S) from S at most k. For an N-periodic sequence S over Fq the k-error linear complexity is deﬁned by Lk(S) = min T L(T), where the minimum is taken over all N-periodic sequences T over Fq for which the ﬁrst N terms diﬀer in at most k positions from the corresponding terms of S. 10.4.11 Remark The concept of the k-error linear complexity is based on the sphere complexity introduced in . 10.4.12 Remark Recent developments in stream ciphers point toward an increasing interest in word-based or vectorized stream ciphers (see for example [784, 1445]), which requires the study of multisequences. 10.4.13 Deﬁnition For an arbitrary positive integer m, an m-fold multisequence S = (S1, . . . , Sm) over Fq (of ﬁnite or inﬁnite length) is a string of m parallel sequences S1, . . . , Sm over Fq (of ﬁnite or inﬁnite length, respectively). Let f1, . . . , fm ∈Fq[x] be arbitrary monic polynomials with deg(fi) ≥1, 1 ≤i ≤m. The set Mq(f1, . . . , fm) is deﬁned to be the set of m-fold multisequences (S1, . . . , Sm) over Fq such that for each 1 ≤i ≤m, Si is a linear recurring sequence with characteristic polynomial fi. 10.4.14 Deﬁnition The joint minimal polynomial of an m-fold multisequence S ∈Mq(f1, . . . , fm) is the (uniquely determined) monic polynomial M ∈Fq[x] of smallest degree which is a characteristic polynomial of Si for all 1 ≤i ≤m. The joint linear complexity of S is the degree of the joint minimal polynomial M. 10.4.15 Remark The set of N-periodic m-fold multisequences is Mq(f1, . . . , fm) with f1 = · · · = fm = xN −1, alternatively denoted by M(m) q (f) with f(x) = xN −1. The joint linear complexity of an m-fold multisequence can also be deﬁned as the length of the shortest linear recurrence relation the m parallel sequences satisfy simultaneously. The joint minimal polynomial M of S ∈Mq(f1, . . . , fm) is always a divisor of lcm(f1, . . . , fm). 10.4.16 Deﬁnition For an integer n ≥1 the n-th joint linear complexity L(S, n) of an m-fold multisequence S = (S1, . . . , Sm) is the length of the shortest linear recurrence relation the ﬁrst n terms of the m parallel sequences S1, . . . , Sm satisfy simultaneously. The joint linear complexity proﬁle of S is the non-decreasing integer sequence L(S, 1), L(S, 2), . . .. 10.4.17 Remark As the Fq-linear spaces Fm q and Fqm are isomorphic, an m-fold multisequence S can also be identiﬁed with a single sequence S having its terms in the extension ﬁeld Fqm. If s(i) j denotes the j-th term of the i-th sequence Si, 1 ≤i ≤m, and {β1, . . . , βm} is a basis of Fqm over Fq, then the j-th term of S is σj = Pm i=1 βis(i) j . The (n-th) joint linear complexity of S coincides then with the Fq-linear complexity of S, which is the length of the shortest linear recurrence relation with coeﬃcients exclusively in Fq (the ﬁrst n terms of) S satisﬁes (see [759, pp. 83–85]). Sequences over ﬁnite ﬁelds 327 10.4.18 Deﬁnition We identify an m-fold multisequence S of length n (or period n) with an m × n matrix and write S ∈Fm×n q (S ∈(Fm×n q )∞). For two m-fold multisequences S = (S1, . . . , Sm), T = (T1, . . . , Tm) ∈Fm×n q the term distance dT (S, T) between S and T is the number of terms in the matrix for S that are diﬀerent from the corresponding terms in the matrix for T. The column distance dC(S, T) between S and T is the number of columns in which the matrices of S and T diﬀer. The individual distances vector for S, T is deﬁned by dV (S, T) = (dH(S1, T1), . . . , dH(Sm, Tm)), where dH denotes the Hamming distance. 10.4.19 Example For q = 2, m = 2, n = 5, and S = 1 1 0 0 1 0 1 0 1 1 , T = 1 1 0 1 1 1 1 0 0 1 , we have dT (S, T) = 3, dC(S, T) = 2 and dV (S, T) = (1, 2). 10.4.20 Deﬁnition For an integer k with 0 ≤k ≤mn, the (n-th) k-error joint linear complexity Lk(S, n) of an m-fold multisequence S over Fq is deﬁned by Lk(S, n) = min T∈Fm×n q ,dT (S,T)≤k L(T, n). For an integer 0 ≤k ≤n the (n-th) k-error Fq-linear complexity Lq k(S, n) of S is deﬁned by Lq k(S, n) = min T∈Fm×n q ,dC(S,T)≤k L(T, n). We deﬁne a partial order on Zm by k = (k1, . . . , km) ≤k′ = (k′ 1, . . . , k′ m) if ki ≤k′ i, 1 ≤i ≤m. For k = (k1, . . . , km) ∈Zm such that 0 ≤ki ≤n for 1 ≤i ≤m, the (n-th) k-error joint linear complexity Lk(S, n) of S is Lk(S, n) = min T∈Fm×n q ,dV (S,T)≤k L(T, n), i.e., the minimum is taken over all m-fold length n multisequences T = (T1, . . . , Tm) over Fq with Hamming distances dH(Si, Ti) ≤ki, 1 ≤i ≤m. The deﬁnitions for periodic multisequences are analogous. 10.4.2 Analysis of the linear complexity 10.4.21 Proposition Let f ∈Fq[x] be a nonconstant monic polynomial. 1. [1631, Theorem 6.1.2], [1939, Chapter 8] For a sequence S = s0, s1, . . . over Fq consider the element P∞ i=0 sixi in the ring Fq of formal power series over Fq. Then S is a linear recurring sequence with characteristic polynomial f if and only if P∞ i=0 sixi = g(x)/f ∗(x) with g ∈Fq[x], deg(g) < deg(f) and f ∗(x) = xdeg(f)f(1/x) is the reciprocal polynomial of f(x). 2. [2240, Lemma 1] For a sequence S = s1, s2, . . . over Fq consider the element P∞ i=1 six−i in the ﬁeld Fq((x−1)) of formal Laurent series in x−1 over Fq. Then S is a linear recurring sequence with characteristic polynomial f if and only if P∞ i=1 six−i = g(x)/f(x) with g ∈Fq[x] and deg(g) < deg(f). 328 Handbook of Finite Fields 10.4.22 Remark For more information and discussion of linear recurring sequences, we refer to Section 10.2. 10.4.23 Remark The reciprocal of a characteristic polynomial of a sequence S is also called a feedback polynomial of S. 10.4.24 Remark Proposition 10.4.21 implies a one-to-one correspondence between sequences in M(1) q (f) and rational functions g/f with deg(g) < deg(f) (when the approach via Laurent series is used), and more generally between m-fold multisequences in Mq(f1, . . . , fm) and m-tuples of rational functions (g1/f1, . . . , gm/fm) with deg(gi) < deg(fi), 1 ≤i ≤m. We note that in Proposition 10.4.21 Part 2 it is more convenient to start the indices for the sequence elements si with i = 1. 10.4.25 Proposition Let (g1/f1, . . . , gm/fm) be the m-tuple of rational functions corre-sponding to S ∈Mq(f1, . . . , fm). The joint minimal polynomial of S is the unique monic polynomial M ∈Fq[x] such that g1 f1 = h1 M , . . . , gm fm = hm M for some (unique) polynomials h1, . . . , hm ∈Fq[x] with gcd(M, h1, . . . , hm) = 1. 10.4.26 Remark For an N-periodic sequence S = s0, s1, . . ., let SN(x) be the polynomial SN(x) = s0 + s1x + · · · + sN−1xN−1 of degree at most N −1. Then P∞ i=0 sixi = SN(x)/(1 −xN), which gives rise to the following theorem. 10.4.27 Theorem [759, Lemma 8.2.1], The joint linear complexity of an N-periodic m-fold multisequence S = (S1, . . . , Sm) is given by L(S) = N −deg(gcd(xN −1, SN 1 (x), . . . , SN m(x))). 10.4.28 Remark Theorem 10.4.27 implies the famous Blahut theorem [303, 2503], [1631, Theorem 6.8.2] for the linear complexity of N-periodic sequences over Fq, gcd(N, q) = 1, which we state in 3 commonly used diﬀerent versions. 10.4.29 Theorem (Blahut’s Theorem) Let S be an N-periodic sequence over Fq, let gcd(N, q) = 1, and let α be a primitive N-th root of unity in an extension ﬁeld of Fq. Then L(S) = N −\|{j : SN(αj) = 0, 0 ≤j ≤N −1}\|. 10.4.30 Theorem (Blahut’s Theorem) Let gcd(N, q) = 1, α be a primitive N-th root of unity in an extension ﬁeld of Fq and let A = (aij) be the N × N Vandermonde matrix with aij = αij, 0 ≤i, j ≤N −1. Let s = (s0, s1, . . . , sN−1) be the vector corresponding to one period of an N-periodic sequence S over Fq. The linear complexity L(S) of S is the Hamming weight of the vector AsT . 10.4.31 Remark The vector a = AsT is called the discrete Fourier transform of s. Several gener-alizations of the discrete Fourier transform have been suggested in the literature that can be used to determine the linear complexity of periodic sequences and multisequences with period not relatively prime to the characteristic of the ﬁeld. We refer to [297, 2013, 2061]. 10.4.32 Theorem (Blahut’s Theorem) Let S = s0, s1, . . . be a sequence over Fq with period N dividing q−1, and let g ∈Fq[x] be the unique polynomial of degree at most N −1 satisfying g(αj) = sj, j = 0, 1, . . ., where α is a ﬁxed element of Fq of order N. Then L(S) = w(g), where w(g) denotes the weight of g, i.e., the number of nonzero coeﬃcients of g. 10.4.33 Theorem [298, Theorem 8] Let f be a polynomial over a prime ﬁeld Fp with degree of f at most p −1 and let S = s0, s1, . . . be the p-periodic sequence over Fp deﬁned by sj = f(j), j = 0, 1, . . .. Then L(S) = deg(f) + 1. Sequences over ﬁnite ﬁelds 329 10.4.34 Remark Theorem 8 in more generally describes the linear complexity of pr-periodic sequences over Fp. A generalization of Theorem 10.4.33 to arbitrary ﬁnite ﬁelds is given in Theorem 1 of . 10.4.35 Remark The linear complexity of an N-periodic sequence over Fq can be determined by the Berlekamp-Massey algorithm in O(N 2) elementary operations. For some classes of period lengths, faster algorithms (of complexity O(N)) are known, the earliest being the Games-Chan algorithm for binary sequences with period N = 2v. A collection of algorithms for several period lengths can be found in (see also [2958, 2959, 2960, 3014]). Some techniques to establish fast algorithms for arbitrary periods are presented in [85, 599, 600, 2057]. Stamp and Martin established a fast algorithm for the k-error linear complexity for binary sequences with period N = 2v. Generalizations are presented in [1642, 1864, 2520], and for odd characteristic in [2056, 3013]. 10.4.36 Remark In contrast to the faster algorithms introduced in the literature for certain period lengths, the Berlekamp-Massey algorithm also can determine the linear complexity proﬁle of a (single) sequence. As an application, the general behavior of linear complexity proﬁles can be analyzed. 10.4.37 Theorem [1631, Theorem 6.7.4], Let S = s1, s2, . . . be a sequence over Fq. If L(S, n) > n/2 then L(S, n + 1) = L(S, n). If L(S, n) ≤n/2, then L(S, n + 1) = L(S, n) for exactly one choice of sn+1 ∈Fq and L(S, n+1) = n+1−L(S, n) for the remaining q −1 choices of sn+1 ∈Fq. 10.4.38 Remark The linear complexity proﬁle is uniquely described by the increment sequence of S, i.e., by the sequence of the positive integers among L(S, 1), L(S, 2) −L(S, 1), L(S, 3) −L(S, 2), . . . [2246, 2935, 2937]. Another tool for the analysis of the linear com-plexity proﬁle arises from a connection to the continued fraction expansion of Laurent series [2239, 2240]. 10.4.39 Theorem Let S = s1, s2, . . . be a sequence over Fq, let S(x) = P∞ i=1 six−i ∈ Fq((x−1)) be the corresponding formal Laurent series, and let A1, A2, . . . be the polyno-mials in the continued fraction expansion of S(x), i.e., S(x) = 1/(A1 + 1/(A2 + · · · )) where Aj ∈Fq[x], deg(Aj) ≥1, j ≥1. Let Q−1 = 0, Q0 = 1 and Qj = AjQj−1 + Qj−2 for j ≥1. Then L(S, n) = deg(Qj) where j is determined by deg(Qj−1) + deg(Qj) ≤n < deg(Qj) + deg(Qj+1). The n-th minimal polynomials are all (monic) polynomials of the form M = aQj + gQj−1, a ∈F∗ q, g ∈Fq[x] with deg(g) ≤2 deg(Qj) −n −1. In particular, the increment sequence of S is deg(A1), deg(A2), . . .. 10.4.40 Remark Generalizations of the Berlekamp-Massey algorithm and of continued fraction anal-ysis for the linear complexity of multisequences can be found in [127, 763, 764, 765, 766, 873, 1053, 1671, 2516, 2517, 2944]. 10.4.3 Average behavior of the linear complexity 10.4.41 Remark We use the notation N (m) n (L) and E(m) n for the number of m-fold multisequences over Fq with length n and joint linear complexity L and the expected value for the joint linear complexity of a random m-fold multisequence over Fq of length n. 10.4.42 Theorem [1378, 2502, 2685] For 1 ≤L ≤n N (1) n (L) = (q −1)qmin(2L−1,2n−2L). 330 Handbook of Finite Fields The expected value for L(S, n) for a random sequence S over Fq is E(1) n = 1 qn X S∈Fn q L(S, n) = ( n 2 + q (q+1)2 −q−n n(q+1)+q (q+1)2 for even n, n 2 + q2+1 2(q+1)2 −q−n n(q+1)+q (q+1)2 for odd n. 10.4.43 Remark Theorem 10.4.42 was obtained by an analysis of the Berlekamp-Massey algorithm. Rueppel and Smeets [2502, 2685] provide closed formulas for the variance, showing that the variance is small. A detailed analysis of the linear complexity proﬁle of sequences over Fq is given by Niederreiter in the series of papers [2235, 2239, 2240, 2243, 2246]. As a main tool, the continued fraction expansion of formal Laurent series is used. For a more elementary combinatorial approach, see . 10.4.44 Theorem The linear complexity proﬁle of a random sequence follows closely but irregularly the n/2-line, deviations from n/2 of the order of magnitude log n must appear for inﬁnitely many n. 10.4.45 Remark The asymptotic behavior of the joint linear complexity is investigated by Niederre-iter and Wang in the series of papers [2275, 2276, 2933] using a sophisticated multisequence linear feedback shift-register synthesis algorithm based on a lattice basis reduction algorithm in function ﬁelds [2549, 2928, 2934]. 10.4.46 Theorem [2253, 2275, 2276] N (m) n (L) = (qm −1)q(m+1)L−m, 1 ≤L ≤n/2, N (m) n (L) ≤ C(q, m)Lmq2mn−(m+1)L, 1 ≤L ≤n, where C(q, m) is a constant only depending on q and m. We have N (m) n (L) ≤q(m+1)L. 10.4.47 Remark In a method to determine N (m) n (L) is presented and a closed formula for N (2) n (L) is given. A closed formula for N (3) n (L) is presented in . In [2275, 2276] it is shown that the joint linear complexity proﬁle of a random m-fold multisequence follows closely the mn/(m + 1)-line, generalizing Theorem 10.4.44 for m = 1. 10.4.48 Theorem [2275, 2276] E(m) n = mn m + 1 + o(n) as n →∞. For m = 2, 3 [1059, 2276, 2933] E(m) n = mn m + 1 + O(1), as n →∞. 10.4.49 Remark Feng and Dai obtained their result with diﬀerent methods, namely with multi-dimensional continued fractions. 10.4.50 Conjecture E(m) n = mn m + 1 + O(1) as n →∞. 10.4.51 Remark For a detailed survey on recent developments in the theory of the n-th joint linear complexity of m-fold multisequences we refer to . 10.4.52 Theorem [1134, 1136] For a monic polynomial f ∈Fq[x] with deg(f) ≥1, let f = re1 1 re2 2 · · · rek k Sequences over ﬁnite ﬁelds 331 be the canonical factorization of f into monic irreducible polynomials over Fq. For 1 ≤i ≤k, let αi = qm deg(ri). Then for an arbitrary positive integer m the expected value E(m)(f) of the joint linear complexity of a random m-fold multisequence from M(m) q (f) is E(m)(f) = deg(f) − k X i=1 1 −α−ei i αi −1 deg(ri). 10.4.53 Remark In [1134, 1136] an explicit formula for the variance Var(m)(f) of the joint linear complexity of random multisequences of M(m) q (f) is given. In [1135, 1136] it is shown how to obtain from Theorem 10.4.52 closed formulas for the more general case of m-fold multisequences in Mq(f1, . . . , fm). 10.4.54 Remark Since for f(x) = xN −1 the set M(m)(f) is the set of N-periodic sequences, earlier formulas on expectation (and variance) of the (joint) linear complexity of periodic (multi)sequences can be obtained as a corollary of Theorem 10.4.52: [2059, Theorem 3.2], [2060, Theorem 1], [3025, Theorem 1] on E(1)(xN −1), and [1137, Theorem 1], [2061, Theorem 1] on E(m)(xN −1) for arbitrary m. 10.4.55 Remark In [1137, 2061] lower bounds on the expected joint linear complexity for periodic multisequences are presented, estimating the magnitude of the formula for E(m)(xN −1) in Theorem 10.4.52. In it is also noted that the variance Var(m)(xN −1) is small, showing that for random N-periodic multisequences over Fq the joint linear complexity is close to N (the trivial upper bound), with a small variance. 10.4.56 Remark Lower bounds for the expected n-th k-error joint linear complexity, the expected n-th k-error Fq-linear complexity and the expected n-th k-error joint linear complexity for an integer vector k = (k1, . . . , km) for a random m-fold multisequence over Fq are established in . These results generalize earlier bounds for the case m = 1 presented in . 10.4.57 Remark For periodic sequences, lower bounds on the expected k-error linear complexity have been established in [2059, 2060]. For periodic multisequences (with prime period N diﬀerent from the characteristic), lower bounds for the expected error linear complexity are presented in for all 3 multisequence error linear complexity measures. 10.4.58 Remark In the papers [2062, 2254, 2273, 2274, 2866] the question is addressed if linear complexity and k-error linear complexity can be large simultaneously. Among others, the existence of N-periodic sequences attaining the upper bounds N and N −1 for linear and k-error linear complexity is shown for inﬁnitely many period lengths (and a certain range for k depending on the period length), and it is shown that for several classes of period length a large number of N-periodic (multi)sequences with (joint) linear complexity N also exhibits a large k-error linear complexity. 10.4.59 Remark In methods from function ﬁelds are used to construct periodic multise-quences with large linear complexity and k-error linear complexity simultaneously for vari-ous period lengths. 10.4.4 Some sequences with large n-th linear complexity 10.4.4.1 Explicit sequences 10.4.60 Deﬁnition For a, b ∈Fp with a ̸= 0 the explicit inversive congruential sequence Z = z0, z1, . . . is zj = (aj + b)p−2, j ≥0. (10.4.2) 332 Handbook of Finite Fields 10.4.61 Theorem We have L(Z, n) ≥    (n −1)/3 for 1 ≤n ≤(3p −7)/2, n −p + 2 for (3p −5)/2 ≤n ≤2p −3, p −1 for n ≥2p −2. 10.4.62 Remark We note that jp−2 = j−1 for j ∈F∗ p. Since inversion is a fast operation this sequence is, despite its high n-th linear complexity, still highly predictable. 10.4.63 Remark Analogous sequences of (10.4.2) over arbitrary ﬁnite ﬁelds Fq are studied in . Multisequences of this form are investigated in . Explicit inversive sequences and multisequences can also be deﬁned using the multiplicative structure of Fq. 10.4.64 Deﬁnition For m ≥1, αi, βi ∈F∗ q, 1 ≤i ≤m, and an element γ ∈Fq of order N, the explicit inversive congruential sequence of period N, Z = (Z1, . . . , Zm), with Zi = σ(i) 0 , σ(i) 1 , . . . is σ(i) j = (αiγj + βi)q−2, j ≥0. (10.4.3) 10.4.65 Remark Sequences of the form (10.4.3) are analyzed in [2069, 2070]. With an appropriate choice of the parameters one can obtain (multi)sequences with perfect linear complexity proﬁle, i.e., L(Z, n) ≥mn/(m + 1). 10.4.66 Theorem Let m < (q −1)/N and let C1, . . . , Cm be diﬀerent cosets of the group ⟨γ⟩ generated by γ, such that none of them contains the element −1. For 1 ≤i ≤m choose αi, βi such that αiβ−1 i ∈Ci, then L(Z, n) ≥min mn m + 1, N , n ≥1. 10.4.67 Deﬁnition Given an element ϑ ∈F∗ q, the quadratic exponential sequence Q = q0, q1, . . . is qj = ϑj2, j ≥0. 10.4.68 Theorem We have L(Q, n) ≥min {n, N} 2 , n ≥1. 10.4.69 Remark The period N of Q is at least half of the multiplicative order of ϑ. 10.4.4.2 Recursive nonlinear sequences 10.4.70 Deﬁnition Given a polynomial f ∈Fp[x] of degree d ≥2, the nonlinear congruential sequence U = u0, u1, . . . is deﬁned by the recurrence relation uj+1 = f(uj), j ≥0, (10.4.4) with some initial value u0 ∈Fp such that U is purely periodic with some period N ≤p. Sequences over ﬁnite ﬁelds 333 10.4.71 Theorem Let U be as in (10.4.4), where f ∈Fp[x] is of degree d ≥2, then L(U, n) ≥min {logd(n −⌊logd n⌋), logd N} , n ≥1. 10.4.72 Remark For some special classes of polynomials much better results are available, see [1359, 1382, 2642]. For instance, in case of the largest possible period N = p we have L(U, n) ≥min{n −p + 1, p/d}, n ≥1. 10.4.73 Theorem The inversive (congruential) sequence Y = y0, y1, . . . deﬁned by yj+1 = ayp−2 j + b, j ≥0, with a, b, y0 ∈Fp, a ̸= 0, has linear complexity proﬁle L(Y, n) ≥min n −1 3 , N −1 2 , n ≥1. 10.4.74 Theorem [1359, 2642] The power sequence P = p0, p1, . . ., deﬁned as pj+1 = pe j, j ≥0, with some integer e ≥2 and initial value 0 ̸= p0 ∈Fp satisﬁes L(P, n) ≥min n2 4(p −1), N 2 p −1 , n ≥1. 10.4.75 Remark Two more classes of nonlinear sequences provide much better results than in the general case, nonlinear sequences with Dickson polynomials and R´ edei functions . See Section 9.6 and for the deﬁnitions. 10.4.4.3 Legendre sequence and related bit sequences 10.4.76 Deﬁnition Let p > 2 be a prime. The Legendre sequence Λ = l0, l1, . . ., for j ≥0, is lj = ( 1 if j p = −1, 0 otherwise, where · p is the Legendre symbol. 10.4.77 Theorem [759, 2829] The linear complexity of the Legendre sequence is L(Λ) =        (p −1)/2 if p ≡1 (mod 8), p if p ≡3 (mod 8), p −1 if p ≡5 (mod 8), (p + 1)/2 if p ≡7 (mod 8). 10.4.78 Theorem [2644, Theorem 9.2] The linear complexity proﬁle of the Legendre sequence sat-isﬁes L(Λ, n) > min{n, p} 1 + p1/2(1 + log p) −1, n ≥1. 10.4.79 Remark For similar sequences, that are deﬁned by the use of the quadratic character of arbitrary ﬁnite ﬁelds and the study of their linear complexity proﬁles, see [1786, 2065, 2994]. 334 Handbook of Finite Fields 10.4.80 Deﬁnition Let γ be a primitive element and η be the quadratic character of the ﬁnite ﬁeld Fq of odd characteristic. The Sidelnikov sequence σ = σ0, σ1, . . . for j ≥0, is σj = 1 if η(γj + 1) = −1, 0 otherwise. 10.4.81 Remark In many cases one is able to determine the linear complexity L(σ) over F2 exactly, see Meidl and Winterhof . For example, if (q −1)/2 is an odd prime such that 2 is a primitive root modulo (q −1)/2, then σ attains the largest possible linear complexity L(σ) = q −1. Moreover we have the lower bound L(σ, n) ≫min{n, q} q1/2 log q , n ≥1. The k-error linear complexity of the Sidelnikov sequence seen as a sequence over Fp has been estimated in [86, 641, 1198]. For results on similar sequences with composite modulus see and [759, Chapter 8.2]. 10.4.4.4 Elliptic curve sequences 10.4.82 Deﬁnition Let p > 3 be a prime and E be an elliptic curve over Fp of the form Y 2 = X3 + aX + b with coeﬃcients a, b ∈Fp such that 4a3+27b2 ̸= 0. For a given initial point W0 ∈E(Fp), a ﬁxed point G ∈E(Fp) of order N and a rational function f ∈Fp(E) the elliptic curve congruential sequence W = w0, w1, . . . (with respect to f) is wj = f(Wj), j ≥0, where Wj = G ⊕Wj−1 = jG ⊕W0, j ≥1. 10.4.83 Remark Obviously, W is N-periodic. 10.4.84 Remark For example, choosing the function f(x, y) = x, the work of Hess and Shparlin-ski gives the lower bound L(W, n) ≥min{n/3, N/2}, n ≥2. 10.4.5 Related measures 10.4.5.1 Kolmogorov complexity 10.4.85 Remark The Kolmogorov complexity is a central topic in algorithmic information theory. The Kolmogorov complexity of a binary sequence is, roughly speaking, the length of the shortest computer program that generates the sequence. The relationship between linear complexity and Kolmogorov complexity was studied in [257, 2946]. The Kolmogorov com-plexity is twice the linear complexity for almost all sequences over F2 of suﬃciently (but only moderately) large length. In contrast to the linear complexity the Kolmogorov complexity is in general not computable and so of no practical signiﬁcance. Sequences over ﬁnite ﬁelds 335 10.4.5.2 Lattice test 10.4.86 Deﬁnition Let S = s0, s1, . . . be a sequence over Fq, and for s ≥1 let V (S, s) be the subspace of Fs q spanned by the vectors sj −s0, j = 1, 2, . . ., where sj = (sj, sj+1, . . . , sj+s−1), j ≥0. The sequence S passes the s-dimensional lattice test for some s ≥1, if V (S, s) = Fs q. For given s ≥1 and n ≥2 we say that S passes the s-dimensional n-lattice test if the subspace spanned by the vectors sj −s0, 1 ≤j ≤n −s, is Fs q. The largest s for which S passes the s-dimensional n-lattice test is the lattice proﬁle at n and is denoted by S(S, n). 10.4.87 Theorem We have either S(S, n) = min{L(S, n), n + 1 −L(S, n)} or S(S, n) = min{L(S, n), n + 1 −L(S, n)} −1. 10.4.88 Remark The results of on the expected value of the lattice proﬁle show that a “ran-dom” sequence should have S(S, n) close to min{n/2, N}. 10.4.5.3 Correlation measure of order k 10.4.89 Deﬁnition The correlation measure of order k of a binary sequence S is Ck(S) = max M,D M−1 X n=0 (−1)sn+d1 · · · (−1)sn+dk , k ≥1, where the maximum is taken over all D = (d1, d2, . . . , dk) with non-negative integers d1 < d2 < · · · < dk and M such that M −1 + dk ≤T −1. Obviously, C2(S) is bounded by the maximal absolute value of the aperiodic autocorrelation of S. 10.4.90 Remark The correlation measure of order k was introduced by Mauduit and S´ ark¨ ozy in . The linear complexity proﬁle of a given N-periodic sequence can be estimated in terms of its correlation measure and a lower bound on L(S, n) can be obtained whenever an appropriate bound on max Ck(S) is known. 10.4.91 Theorem We have L(S, n) ≥n − max 1≤k≤L(S,n)+1 Ck(S), 1 ≤n ≤N −1. 10.4.5.4 FCSR and p-adic span 10.4.92 Remark In an alternative feedback shift register architecture was presented, feedback with carry shift registers (FCSR). For binary sequences the procedure is as follows: Diﬀer-ently to linear recurring sequences the bits are added as integers (again following a linear recurrence relation). The result is added to the content of a memory, which is a nonnegative integer m, to obtain an integer σ. The parity bit σ (mod 2), of σ is then the next term of the sequence, and the higher order bits ⌊σ/2⌋are the new content of the memory. FCSR-sequences share many properties with linear recurring sequences, but for their analysis instead of arithmetics in ﬁnite ﬁelds, arithmetics in the 2-adic numbers is used - or in the more general case of sequences modulo p in the p-adic numbers. 336 Handbook of Finite Fields An FCSR-equivalent to the linear complexity is the 2-adic span, respectively the p-adic span of a sequence, which measures the size of the smallest FCSR that generates the sequence. Since their introduction, FCSR-sequences attracted a lot of attention. We refer to [129, 1330, 1331, 1749, 2774] and the references therein. 10.4.5.5 Discrepancy 10.4.93 Deﬁnition Let X = x0, x1, . . . be a sequence in the unit interval [0, 1). For 0 ≤d1 < · · · < dk < n we put xj = xj(d1, . . . , dk) = (xj+d1, . . . , xj+dk), 1 ≤j ≤n −dk. The discrepancy of the vectors x1(d1, . . . , dk), . . . , xn−dk(d1, . . . , dk) is sup I A(I, x1, . . . , xn−dk) n −dk −V (I) , where the supremum is taken over all subintervals of [0, 1)k, V (I) is the volume of I and A(I, x1, . . . , xn−dk) is the number of points xj, j = 1, . . . , n −dk, in the interval I. 10.4.94 Remark We can derive a binary sequence B = e0, e1, . . . from X by ej = 1 if 0 ≤xj < 1/2 and ej = 0 otherwise. 10.4.95 Remark In [2036, Theorem 1] the correlation measure of order k of B is estimated in terms of the above discrepancy of vectors derived from the sequence X. Hence, using the relation between linear complexity proﬁle and correlation measure of B we can obtain (weak) linear complexity proﬁle lower bounds for B from discrepancy upper bounds for X. See Also §6.3, §10.2, §17.3 For related measures. §9.1, §9.3 For Boolean functions and nonlinearity. §10.2 For LFSR. §10.5, §17.1 For nonlinear recurrence sequences. §12.2, §12.3, §16.4 For elliptic curves. §15.1 For basics on coding theory and the Berlekamp-Massey algorithm. §16.2 For stream ciphers. References Cited: [85, 86, 87, 127, 129, 257, 297, 298, 303, 392, 393, 599, 600, 641, 759, 763, 764, 765, 766, 784, 873, 913, 914, 1053, 1059, 1134, 1135, 1136, 1137, 1169, 1198, 1330, 1331, 1359, 1378, 1382, 1445, 1493, 1631, 1642, 1671, 1748, 1749, 1786, 1864, 1936, 1939, 2011, 2013, 2036, 2037, 2056, 2057, 2058, 2059, 2060, 2061, 2062, 2063, 2064, 2065, 2066, 2067, 2069, 2070, 2071, 2072, 2235, 2239, 2240, 2242, 2243, 2246, 2253, 2254, 2257, 2273, 2274, 2275, 2276, 2502, 2503, 2516, 2517, 2520, 2549, 2642, 2644, 2685, 2700, 2774, 2829, 2866, 2928, 2933, 2934, 2935, 2937, 2944, 2946, 2958, 2959, 2960, 2994, 3013, 3014, 3016, 3025] Sequences over ﬁnite ﬁelds 337 10.5 Algebraic dynamical systems over ﬁnite ﬁelds Igor Shparlinski, Macquarie University 10.5.1 Introduction 10.5.1 Deﬁnition Let F1, . . . , Fm ∈Fq(X1, . . . , Xm) be m rational functions in m variables over the ﬁnite ﬁeld Fq of q elements. The algebraic dynamical system (ADS) generated by F = {F1, . . . , Fm} is the dynamical system formed by the iterations F (k) i = Fi(F (k−1) 1 , . . . , F (k−1) m ), k = 1, 2, . . . , i = 1, . . . , m, where F (0) i = Xi. 10.5.2 Remark ADSs have proved to be exciting and challenging mathematical objects. They have very interesting algebraic and number theoretic properties and also exhibit a very complex behavior; see [91, 964, 1020, 1087, 1313, 1618, 1619, 2421, 2547, 2668] for the foundations of the theory. This makes them an invaluable building block for various applications including pseudorandom number generators (PRNGs), which are of crucial value in quasi-Monte Carlo methods and cryptography; see [2271, 2648, 2814]. Recently very surprisingly links with other natural sciences such as biology [1596, 1860, 2717] and physics [106, 154, 216, 222, 1348, 1498, 2464, 2870, 2871] have emerged. 10.5.2 Background and main deﬁnitions 10.5.3 Deﬁnition Given an initial vector u0 = (u0,1, . . . , u0,m) ∈Fm q , the trajectory of an ADS (see Deﬁnition 10.5.1) originating at this vector, is the sequence of vectors un = (un,1, . . . , un,m) ∈Fm p deﬁned by the recurrence relation un+1,i = Fi(un,1, . . . , un,m), n = 0, 1, . . . , i = 1, . . . , m. 10.5.4 Remark Using the following vector notation F(X1, . . . , Xm) = (F1(X1, . . . , Xm), . . . , Fm(X1, . . . , Xm)), we have the recurrence relation un+1 = F(un), n = 0, 1, . . . . (10.5.1) In particular, for any n, k ≥0 and i = 1, . . . , m we have un+k,i = F (k) i (un) = F (k) i (un,1, . . . , un,m) or un+k = F(k)(un). 10.5.5 Remark If F1, . . . , Fm are rational functions, then one has to decide what to do if un is a pole of some of them. A canonical way to resolve this is to deﬁne these functions on the 338 Handbook of Finite Fields set of their poles separately (for example, deﬁne 0−1 = 0). Certainly this problem does not occur if all functions F1, . . . , Fm are polynomials. 10.5.6 Remark Recently, new applications have emerged to cryptography and quasi-Monte Carlo methods, where ADSs have been shown to provide a very attractive alternative to the classical linear congruential PRNG. 10.5.7 Deﬁnition An attack on a PRNG is an algorithm that observes several outputs of a PRNG and then is able to continue to generate the same sequence with a nontrivial probability. 10.5.8 Remark The interest in nonlinear dynamical systems as sources of pseudorandom numbers [2271, 2814] has been driven by a series of devastating attacks on traditional linear con-structions which have made them useless for cryptographic purposes; see [716, 1831] and references therein. 10.5.9 Remark Although nonlinear PRNGs are believed to be cryptographically stronger; see, however [210, 299, 300, 301, 1309, 1380] for some attacks. Yet, obtaining concrete eﬃcient constructions with good rigorously proven estimates on their statistical and other properties, has also been a challenging task [1358, 1379, 2329]. 10.5.3 Degree growth 10.5.10 Deﬁnition The algebraic entropy of the ADS generated by F = {F1, . . . , Fm} is δ(F) = lim n→∞ log Dn(F) n , where Dk(F) is the degree of F(k), deﬁned as the largest degree of the components F (k) 1 , . . . , F (k) m . 10.5.11 Remark The existence of the limit follows immediately from the inequality Dk+m(F) ≤ Dk(F)Dm(F). 10.5.12 Remark Studying how the iterates of an ADS grow is a classical research direction with a rich history and a variety of results. In the case of ADSs over the complex and p-adic numbers there is a well-studied measure of “size” called the height, [91, 964, 1020, 2668]. Unfortunately this measure does not apply to ADSs over ﬁnite ﬁelds. However, in this case the degree Dk(F) provides a very natural and adequate substitute for the notion of height. Thus, the algebraic entropy plays a very essential role in the theory of ADSs over ﬁnite ﬁelds [215, 216, 222, 1498, 2817, 2870, 2871]. 10.5.13 Remark The degree growth is important for many applications of ADSs. In the univariate case, it is obvious that the n-th iterate of a polynomial of degree d is a polynomial of degree dn. This however is not true anymore in the multidimensional case. Yet, the exponential growth is expected for a “typical” ADS. For example in [1358, 1379] the exponential growth is shown for some very special classes of polynomial systems, corresponding to nonlinear recurrence sequences of order m, that is, sequences satisfying a recurrence relation of the form wn+m = F(wn+m−1, . . . , wn), n = 0, 1, . . . , with F ∈Fq(X1, . . . , Xm). An alternative approach with a combinatorial ﬂavor to studying such sequences has been suggested in . Sequences over ﬁnite ﬁelds 339 10.5.14 Remark Recently, a family of multivariate ADSs with polynomial degree growth of their iterates has been constructed in [1312, 2328, 2330, 2332, 2334]. These ADSs are formed by systems of rational functions of the form: F1(X1, . . . , Xm) = Xe1 1 G1(X2, . . . , Xm) + H1(X2, . . . , Xm), . . . (10.5.2) Fm−1(X1, . . . , Xm) = Xem−1 m−1 Gm−1(Xm) + Hm−1(Xm), Fm(X1, . . . , Xm) = gmXem m + hm, where ei ∈{−1, 1}, i = 1, . . . , m, gm, hm ∈Fq, gm ̸= 0, and Gi has a unique leading monomial: Gi(Xi+1, . . . , Xm) = giXsi,i+1 i+1 · · · Xsi,m m + f Gi(Xi+1, . . . , Xm), which “dominates” all other terms: gi ̸= 0, degXj f Gi < si,j, degXj Hi ≤si,j, for 1 ≤i < j ≤m. 10.5.15 Remark The structure and the degree of iterates of ADSs in 10.5.14 is given by Theo-rem 10.5.16 which is essentially [2332, Lemma 1]. 10.5.16 Theorem [2332, Lemma 1] In the case of ei = 1, i = 1, . . . , m, for any ADS of the form (10.5.2), we have F (k) i = XiGi,k(Xi+1, . . . , Xm) + Hi,k(Xi+1, . . . , Xm), i = 1, . . . , m, k = 0, 1, . . . , where Gi,k, Hi,k ∈Fq[Xi+1, . . . , Xm], i = 1, . . . , m −1 and deg Gi,k = 1 (m −i)!km−isi,i+1 . . . sm−1,m + ψi(k), with ψi(T) ∈Q[T], deg ψi < m −i, i = 1, . . . , m −1, and Gm,k = gk m ∈F∗ q. 10.5.17 Remark In [2328, 2334] a more general (but also more technically involved) form of Theo-rem 10.5.16 is given which applies to exponents ei = ±1, i = 1, . . . , m. 10.5.18 Problem Find new constructions of ADSs with polynomial degree grows of the iterates. 10.5.19 Remark In yet another new class of ADSs is given that also leads to good pseudo-random number generators, namely, given ADSs that are formed by systems of multivariate polynomials F1, . . . , Fm ∈Fp[X1, . . . , Xm] of the form: F1 = (X1 −h1)e1 G1 + h1, . . . (10.5.3) Fm−1 = (Xm−1 −hm−1)em−1 Gm−1 + hm−1, Fm = gm (Xm −hm)em + hm, 340 Handbook of Finite Fields where Gi ∈Fp[Xi+1, . . . , Xm], i = 1, . . . , m −1, ej are positive integers and gm, hj ∈Fp, j = 1, . . . , m. The corresponding pseudorandom number generator generalizes the classical power generator (see Section 10.4) but is free of some of its undesirable features such as homogeneity. Although the degree of the iterates of (10.5.3) grows exponentially, under some additional condition using a recent result of Cochrane and Pinner , one can get a reasonably good estimate of the discrepancy of the corresponding sequence; see [2335, Theorem 8] for details. This direction has been further developed in . 10.5.20 Problem Find new constructions of ADSs with sparse iterates (that is, having a constant or slowly growing with the number of iterations number of monomials). 10.5.4 Linear independence and other algebraic properties of iterates 10.5.21 Remark Surprisingly enough, investigating the distribution and other number theoretic properties of PRNGs ultimately requires to study additive character sums with their ele-ments. In turn, this leads to investigation of their algebraic properties such as the degree growth or absolute irreducibility of certain linear combinations of the iterates. Deeper alge-braic properties such as the dimension of the singularity locus of certain associated algebraic varieties are also of interest. Below we explain how these properties become ultimately re-lated to the rather analytic question of the uniformity of distribution. 10.5.22 Remark An application of the celebrated Koksma–Sz¨ usz inequality (see the original works [1780, 2759] and also Section 6.3) reduces the question of studying the distribu-tion of the vectors (10.5.1) to the question of estimating the following additive character sums Sa(N) = N−1 X n=0 ψ m X i=1 aiun,i ! , where ψ is a ﬁxed additive character of Fq and a = (a1, . . . , am) ∈Fm q . The standard methodology, suggested in [2269, 2270] and recently improved in , after a certain chain of standard transformations used in estimating character sums, leads to additive character sums with La,k,l(X) = m X i=1 ai F (k) i (X) −F (l) i (X) , where X = (X1, . . . , Xm). So a natural next step is to use the Weil bound, see Section 6.3. However, for this the rational function La,k,l(X) has to be “exponential sums friendly,” that is, 1. nontrivial (that is, with a nontrivial trace); 2. of small degree (so that the Weil bound is nontrivial for this function); 3, if possible, linear in some of the variables (thus to avoid using the Weil bound at all); 4. if possible, have a low dimensional locus of singularity (to apply the Deligne bound, see Section 6.3 or Deligne-like bounds by Katz ). 10.5.23 Problem Find general (necessary and suﬃcient) conditions under which, the linear forms La,k,l(X) are not constant for any non-zero vector a ∈Fm q and k ̸= l. 10.5.24 Remark Over Fq, for a special class of iterations, some suﬃcient conditions are given in [1358, 1379]. In , in some special case, the above condition of on La,k,l(X) has been replaced by some combinatorial argument, which, however, does not seem to generalize any further. Sequences over ﬁnite ﬁelds 341 10.5.25 Remark For ADSs of the shape (10.5.2) the degree of the iterates grows polynomially. Furthermore the iterates are linear in one of the variables. Both properties together, have led to rather strong results about the distribution of the vectors (10.5.1); see [2330, 2332]. This idea and construction has been further developed in [2326, 2327, 2336, 2337]. Furthermore, in a new construction of hash functions is suggested that is based on the above ADSs. 10.5.26 Remark It is clear that the systems of the shape (10.5.2) with ei = 1, i = 1, . . . , m (or for arbitrary ei = ±1, if one deﬁnes 0−1 = 0) deﬁne a permutation of Fm q if and only if the “coeﬃcients” Gi, i = 1, . . . , m −1, have no zeros over Fq. For such permutation systems, Ostafe has established rather strong results about the distribution of elements in trajectories on average over all initial values; see also [2334, 2653]. 10.5.27 Remark At the same time it has become clear that in order to improve the results of [2330, 2332] one needs to obtain more detailed information about the algebraic structure of polynomial iterates, in particular about the number of absolutely irreducible components of the polynomials Gi,k1 −Gi,k2 where 0 ≤k2 < k1 and Gi,k is as in Theorem 10.5.16. 10.5.28 Remark Sums of multiplicative characters along trajectories of ADSs have also been con-sidered in the literature [2272, 2277, 2337]. Such sums can be estimated within the same lines that have been used for additive character sums, however instead of linear combi-nations La,k,l(X) one has to study some other algebraic expressions including the iter-ates. For instance, the argument of is based on studying the bilinear combinations Gi,kHi,l −Gi,lHi,k (in the notation of Theorem 10.5.16) and showing that they are not constant. 10.5.5 Multiplicative independence of iterates 10.5.29 Theorem [1173, Theorem 1.4] Suppose that a polynomial f ∈Fq[X] is not a monomial or a binomial of the form axpℓ+ b where p is the characteristic of Fq. Then for any integer N ≥1, the polynomials f, f (1), . . . , f (N) are multiplicative independent. 10.5.30 Remark Theorem 10.5.29 is used in an construction of elements of large multiplicative order over ﬁnite ﬁelds; see [1173, Theorem 1.1]. 10.5.31 Remark There are several possible interpretations of what a multivariate analogue of The-orem 10.5.29 may look like. All of them are interesting, however no results in this direction have been obtained so far. 10.5.6 Trajectory length 10.5.32 Remark Clearly the sequence of vectors {un}, given by (10.5.1) is eventually periodic with some period τ. That is, for some integer s ≥0 we have un+τ = un for n ≥s. 10.5.33 Deﬁnition If s is the smallest integer with this above property of Remark 10.5.32, then T = s + τ is the trajectory length. 10.5.34 Remark If we work in a ﬁnite ﬁeld of q elements, then the trajectory length T satisﬁes T ≤qm, where m is the number of variables. 10.5.35 Remark No general lower bounds on the trajectory length of sequences generated by ADSs are known. Furthermore, assuming that the map generated by F behaves as a random map (and for a generic polynomial system this seems to be a natural and well tested assumption), one should expect that in fact T is of order qm/2. On the other hand, most of the results 342 Handbook of Finite Fields about the distribution of the sequence (10.5.1) are nontrivial only if the trajectory length T is close to its largest possible value [2326, 2327, 2330, 2332, 2336, 2337]. Hence we see that “generic” ADSs are not likely to satisfy this property. Thus one needs constructions of special ADSs tailored for these applications. 10.5.36 Remark Very little is known about constructing ADSs with guaranteed large trajectory length. The only known rigorous results are those of Ostafe and their generalizations in , that gives a complete characterization (which in turn leads to explicit construc-tions) of ADSs of the type (10.5.2) which achieve the largest possible value of the trajectory length T = qm. 10.5.37 Remark A result of gives a nontrivial (albeit very weak) lower bound on the length of a reduction of a trajectory of an ADS over the rationals modulo a prime p that holds for almost all p (in fact the result applies to more general settings). 10.5.7 Irreducibility of iterates 10.5.38 Remark As we have seen in Section 10.5.4, the algebraic structure of iterates becomes very important for studying the distribution of trajectories. Questions of this type are also of great intrinsic interest for the theory of ADSs. Unfortunately, they are notoriously hard, and most of the few known results apply only to iterates of univariate quadratic polynomials; see [76, 152, 153, 1310, 1313, 1618, 1619, 1620, 2331] and references therein. 10.5.39 Deﬁnition A polynomial f over a ﬁeld K is stable if all its iterations f (n) are irreducible over K. 10.5.40 Deﬁnition For a quadratic polynomial f(X) = aX2 + bX + c ∈K[X], over a ﬁeld K of characteristic p ̸= 2, we deﬁne the critical orbit of f as the set of consecutive iterations Orb(f) = {f (n)(γ) : n = 2, 3, . . .}, where γ = −b/2a. 10.5.41 Remark Clearly γ in Deﬁnition 10.5.40 is the unique critical point of f (that is, the zero of the derivative f ′). 10.5.42 Remark It is shown in [1618, 1619, 1620] that critical orbits play a very important role in the dynamics of polynomial iterations. 10.5.43 Remark Capelli’s Lemma, describing the conditions on the irreducibility of polynomial compositions, plays a prominent role in this area. 10.5.44 Theorem [1620, Proposition 3] A quadratic polynomial f ∈K[X] is stable if the set {−f(γ)} ∪Orb(f) contains no squares. If K = Fq is a ﬁnite ﬁeld of odd characteristic, this property is also necessary. 10.5.45 Remark If K = Fq is a ﬁnite ﬁeld, there is some integer t such that f (t)(γ) = f (s)(γ) for some positive integer s < t. Then f (n+t)(γ) = f (n+s)(γ) for any n ≥0. Accordingly, for the smallest value of t with the above condition denoted by tf, we have Orb(f) = {f (n)(γ) : n = 2, . . . , tf} and #Orb(f) = tf −1 or #Orb(f) = tf −2 (depending whether s = 1 or s ≥2 in the above). Sequences over ﬁnite ﬁelds 343 10.5.46 Remark Remark 10.5.45 immediately implies that a quadratic polynomial f ∈Fq[X] can be tested for stability in q1+o(1) arithmetic operations over Fq. Using bounds of character sums, it has been shown in that this can be improved. 10.5.47 Theorem A quadratic polynomial over Fq can be tested for stability in q3/4+o(1) arithmetic operations over Fq. 10.5.48 Remark In , a generalization of Theorem 10.5.47 is given to polynomials g(f) where f is quadratic and g is an arbitrary polynomial over Fq. 10.5.49 Remark Since a random polynomial of degree d is irreducible over Fq with probability about 1/d and the degree of the iterations f (n) grows exponentially with n, it is natural to expect that there are only very few stable polynomials over Fq. For quadratic polynomials this has been conﬁrmed in a quantitative form by Gomez and Nicol´ as . With several ingenious extensions of the method of , Gomez, Nicol´ as, Ostafe, and Sadornil have shown this for polynomials of arbitrary degree d ≥2. 10.5.50 Theorem For any odd prime power q, there are at most q5/2+o(1) stable quadratic polynomials over Fq. 10.5.51 Remark The number of irreducible divisors and other arithmetic properties of iterations of polynomials over large ﬁnite ﬁelds have been studied in . 10.5.52 Problem Give explicit constructions of polynomial systems, over some “interesting” ﬁelds such as Q and Fq, such that all polynomials F (k) i , i = 1, . . . , m, k = 1, 2, . . ., are 1. irreducible over Fq; 2. absolutely irreducible over Fq. 10.5.53 Remark Over the ﬁeld of rational numbers Q we deﬁne the class Ep,m (where p is an arbitrary prime) of m-variate analogues of the Eisenstein polynomials: We say that F ∈ Z[X1, . . . , Xm] belongs to Ep,m if F(X1, . . . , Xm) = A0Xd1 1 · · · Xdm m + pf(X1, . . . , Xm) for some f ∈Z[X1, . . . , Xm], where A0 ̸≡0 (mod p), d1 + · · · + dm > deg f and such that F(0) ̸≡0 (mod p2), where 0 = (0, . . . , 0). Clearly if F = GH, where G, H ∈Z[X1, . . . , Xm], is reducible then F ≡GH (mod p). Since if F is a monomial modulo p then so are G and H. As deg F = d1 +· · ·+dm, we conclude that G and H are nonconstant monomials modulo p. Therefore G(0) ≡H(0) ≡0 (mod p) which implies that F(0) = G(0)H(0) ≡0 (mod p2), contradicting F ∈Em,p. If F, G1, . . . , Gm ∈Ep,m then because F(G1(0), . . . , Gm(0)) ≡F(0) ̸≡0 (mod p2), we also have F(G1, . . . , Gm) ∈Ep,m. Therefore if F1, . . . , Fm ∈Em,p for some prime p then their iterations F (k) 1 , . . . , F (k) m , k = 1, 2, . . . are all irreducible over Q. This however does not lead to a construction of absolutely irreducible iterates. There is no obvious ﬁnite ﬁeld analogue of this construction. 10.5.8 Diameter of partial trajectories 10.5.54 Deﬁnition The diameter DF,u0(N) of the sequence of the ﬁrst N vectors (10.5.1) over Fp is deﬁned as DF,u0(N) = max 0≤k,n≤N−1 ∥uk −un∥, where ∥u∥is the Euclidean norm of a vector u and we assume that the elements of Fp are represented by the set {0, . . . , p −1}. 344 Handbook of Finite Fields 10.5.55 Remark Certainly results about the asymptotically uniform distribution of the ﬁrst N vectors (10.5.1) (mentioned in Section 10.5.4) immediately imply that DF,u0(N) is close to the largest possible value n1/2p in this case. However such results are known only for very long segments of the trajectories. 10.5.56 Remark The notion of the diameter (under a slightly diﬀerent name) is introduced in and then has also been studied in [585, 586, 644] where a wide variety of methods has been used. However, the only known results about the diameter are in the univariate case, that is, when F = {f} ⊆Fp[X] and u0 = u0 ∈Fp. 10.5.57 Theorem [1381, Theorem 6] For any ﬁxed ε > 0 and Tf,u0 ≥N ≥p1/2+ε where Tf,u0 is the trajectory length corresponding to iterations of f ∈Fp[X] originating at u0, we have Df,u0(N) = p1+o(1) as p →∞. 10.5.58 Remark Theorem 10.5.57, when it applies, provides the asymptotically best possible bound. For smaller values of N, one can use a variety of the results from [585, 586, 644] that are nontrivial in essentially the best possible range Tf,u0 ≥N ≥pε for any ﬁxed ε > 0. 10.5.59 Problem Let Fqs be an extension of degree s ≥2 of Fq. Obtain a lower bound on the smallest dimension of an aﬃne space over Fq containing the ﬁrst N elements of the trajectory of iterations of f ∈Fqs[X] originating at u0 ∈Fqs. 10.5.60 Problem In the multidimensional case, besides estimating DF,u0(N), one can also study other geometric characteristics, such as 1. the volume and the number of vertices of the convex hull of the set {u0, . . . , uN−1}; 2. the number of directions and the number of distances deﬁned by pairs of vectors (uk, un), 0 ≤k, n ≤N −1; 3. the number of directions and the number of distances deﬁned by pairs of consec-utive vectors (un, un+1), 0 ≤n ≤N −1. References Cited: [50, 76, 91, 106, 152, 153, 154, 210, 215, 216, 222, 299, 300, 301, 585, 586, 644, 657, 716, 964, 1020, 1087, 1309, 1310, 1311, 1312, 1313, 1348, 1358, 1379, 1380, 1381, 1498, 1596, 1618, 1619, 1620, 1704, 1780, 1831, 1860, 2269, 2270, 2271, 2272, 2277, 2279, 2326, 2327, 2328, 2329, 2330, 2331, 2332, 2334, 2335, 2336, 2337, 2421, 2464, 2547, 2648, 2653, 2668, 2669, 2717, 2759, 2814, 2817, 2870, 2871] 11 Algorithms 11.1 Computational techniques ......................... 345 Preliminaries • Representation of ﬁnite ﬁelds • Modular reduction • Addition • Multiplication • Squaring • Exponentiation • Inversion • Squares and square roots 11.2 Univariate polynomial counting and algorithms 364 Classical counting results • Analytic combinatorics approach • Some illustrations of polynomial counting 11.3 Algorithms for irreducibility testing and for constructing irreducible polynomials ............. 374 Introduction • Early irreducibility tests of univariate polynomials • Rabin’s irreducibility test • Constructing irreducible polynomials: randomized algorithms • Ben-Or’s algorithm for construction of irreducible polynomials • Shoup’s algorithm for construction of irreducible polynomials • Constructing irreducible polynomials: deterministic algorithms • Construction of irreducible polynomials of approximate degree 11.4 Factorization of univariate polynomials .......... 380 11.5 Factorization of multivariate polynomials ....... 382 Factoring dense multivariate polynomials • Factoring sparse multivariate polynomials • Factoring straight-line programs and black boxes 11.6 Discrete logarithms over ﬁnite ﬁelds.............. 393 Basic deﬁnitions • Modern computer implementations • Historical remarks • Basic properties of discrete logarithms • Chinese Remainder Theorem reduction: The Silver–Pohlig–Hellman algorithm • Baby steps–giant steps algorithm • Pollard rho and kangaroo methods for discrete logarithms • Index calculus algorithms for discrete logarithms in ﬁnite ﬁelds • Smooth integers and smooth polynomials • Sparse linear systems of equations • Current discrete logarithm records 11.7 Standard models for ﬁnite ﬁelds .................. 401 11.1 Computational techniques Christophe Doche, Macquarie University This section presents algorithmic methods for ﬁnite ﬁelds. It deals with concrete imple-mentations and describes how some fundamental operations on the elements are performed. 345 346 Handbook of Finite Fields Finite ﬁelds play a central role in many areas, such as cryptography, coding theory, and ran-dom number generation, where the speed of the computations is paramount. Therefore we discuss algorithms with a particular attention to eﬃciency and we address implementation techniques and complexity aspects as often as possible. In many instances, the complexity of the algorithms that we describe depends on the ﬁeld multiplication method that is imple-mented. In the following, M(log p) denotes the complexity to multiply two positive integers less than p. Similarly, Mq(n) represents the complexity to multiply two polynomials in Fq[x] of degree less than n. We note that many software tools or libraries implement some of the al-gorithms that are presented next. A non-exhaustive list includes Magma , Pari , Sage , Mathematica , Maple , NTL , GMP , MPFQ , FLINT , and ZEN . Detailed algorithms and complexity analyses can be found in [660, 661, 751, 1227, 1413, 1768, 2080, 2632]. 11.1.1 Preliminaries 11.1.1.1 Prime ﬁeld generation 11.1.1 Remark For many applications, a random prime ﬁeld of a given size is needed. This implies ﬁnding a prime number of a given bit length that is random in the following sense: the probability for any particular prime of that size to be selected is suﬃciently small so that it is impossible for anyone to take advantage of this event. There are two diﬀerent ways to ﬁnd such a random prime number. The ﬁrst technique is to identify a prime number among integers of a prescribed size successively picked at random. The second technique is to construct integers with special properties so that it is easy to prove that they are prime and whose distribution over the set of all the primes of the desired size is close to uniform. 11.1.2 Algorithm (Prime number random search) Input: An integer ℓ≥2. Output: An ℓ-bit prime number. 1. repeat 2. Pick an ℓ-bit random integer n 3. until n is not composite 4. return n 11.1.3 Remark The prime number theorem [2080, Section 4.1] ensures that it takes O(ℓ) random draws among ℓ-bit integers before picking an integer that is actually prime. 11.1.4 Remark The approach used at Line 3 of Algorithm 11.1.2 to test if n is composite or not determines the quality of the prime number that is returned. In practice, Algorithm 11.1.2 relies on trial division, to quickly eliminate most composite numbers, and on the Miller– Rabin compositeness test, properly set up depending on the size of the desired prime. See Re-mark 11.1.5 and [2080, Section 4.4] for implementation details. In the end, Algorithm 11.1.2 returns a probable prime, i.e., not certiﬁed but prime with very high probability. 11.1.5 Remark The ﬁrst compositeness test of real practical signiﬁcance is due to Solovay and Strassen . It was superseded by the Miller–Rabin test [2099, 2435], which is faster and easier to implement. The error probability in the Miller–Rabin test can be adjusted using a security parameter t that also drives the complexity of the algorithm. The Miller–Rabin test is always correct when it declares that an integer n is composite, but it may be wrong with probability at most 4−t when it declares that n is prime. Taking into account the distribution of prime numbers and using advanced probabilistic analysis techniques, it can be shown that Algorithm 11.1.2 coupled with the Miller–Rabin test requires a very small t in order to return a high quality probable prime. For instance, t = 5 is enough to produce Algorithms 347 a 500-bit integer that is prime with probability greater than 1 −2−85; see [2080, Note 4.47 and Fact 4.48] and [2632, Section 10.3] for more examples. 11.1.6 Remark Given the parameter t and the factorization of n−1 as 2sm with m odd, the Miller– Rabin test computes at most t modular exponentiations to determine if n is composite. Each exponentiation is of the form ak, where a is a O(n) random value and k = 2rm for some r < s. Its complexity is therefore O(t log3 n). As noted in , the algorithm is signiﬁcantly faster when single precision integers a are used instead of random values in [0, n −1]. Although the error bound may no longer hold in that case. 11.1.7 Remark An interesting variant of Algorithm 11.1.2 consists in testing integers in an arith-metic progression. For instance, considering the odd numbers in some interval may reduce the complexity of the process, see [2080, Section 4.4]. 11.1.8 Remark When a provable prime is required, we can run a more expensive primality test on the integer n returned by Algorithm 11.1.2. There are mainly two primality tests used in practice. APRCL [660, 662], named after its inventors, relies on Jacobi sums and is a simpliﬁed version of the test presented in . Its complexity is O(logc log log log n n), for some eﬀective constant c. The Elliptic Curve Primality Proving test (ECPP) uses elliptic curves and its complexity is ˜ O(log5 n). The fast version of ECPP, called fastECPP runs in ˜ O(log4 n), see for implementation details. Both ECPP and fastECPP oﬀer the advantage of producing a certiﬁcate that can be used to prove that n is prime much quicker than running the test again. In 2002, Agrawal, Kayal, and Saxena announced the ﬁrst deterministic polynomial time primality testing algorithm. For an input n, the complexity of the so-called AKS algorithm is O(log10.5 n). One variant of AKS runs in time ˜ O(log4 n). Although this algorithm has the same complexity as fastECPP, the constants are much larger and as a result fastECPP is still the method of choice to prove the primality of general integers. 11.1.9 Remark As hinted in Remark 11.1.1, a radically diﬀerent approach to ﬁnd a prime number is to construct it from scratch. There are two popular provable prime generation methods in the literature, respectively due to Mih˘ ailescu and Maurer [2038, 2080]. They both use Pocklington’s lemma recursively, but Maurer’s method generates random provable primes whose distribution is close to uniform over the set of all primes of a given size, whereas Mih˘ ailescu’s approach is more eﬃcient but slightly reduces the set of primes that may be produced. 11.1.1.2 Extension ﬁeld generation 11.1.10 Remark To deﬁne Fqn, it is enough to construct a basis of the vector space Fqn over Fq. A normal basis oﬀers several advantages, such as a cheap evaluation of the Frobenius automorphism, see Section 5.2. Alternatively, any irreducible polynomial of degree n with coeﬃcients in Fq can be used to represent Fqn. This gives rise to a polynomial basis; see Deﬁnition 2.1.96. 11.1.11 Remark Quite similarly to the prime ﬁeld case, there are two diﬀerent ways to ﬁnd an irreducible polynomial of a given degree. We can identify an irreducible polynomial among many polynomials generated at random. We may also construct an irreducible polynomial directly. This is well illustrated in . 11.1.12 Remark It follows from Theorem 2.1.24 that a random monic polynomial of degree n with coeﬃcients in Fq is irreducible with probability close to 1 n· So it takes an expected O(n) attempts to ﬁnd an irreducible polynomial of degree n at random; see Subsection 11.3.2, for a description of some irreducibility testing algorithms. 348 Handbook of Finite Fields 11.1.13 Remark It is well known that diﬀerent irreducible polynomials of degree n generate iso-morphic ﬁnite ﬁelds. The isomorphism can even be computed explicitly [77, 1895]. So it may seem that the choice of the irreducible polynomial used to generate Fqn is irrelevant. However, certain polynomials are more suitable than others when it comes to the eﬃciency of the operations in Fqn. In particular, polynomials with a low number of nonzero terms have a clear advantage, as they provide a faster modular reduction, see Subsection 11.1.3.2. 11.1.14 Deﬁnition Let f ∈Fq[x]. The polynomial f is s-sparse, if f has s nonzero terms. The terms binomial, trinomial, quadrinomial, and pentanomial are frequently used to refer to 2-sparse, 3-sparse, 4-sparse, and 5-sparse polynomials, respectively. Furthermore, f is t-sedimentary if f = xn + h where deg h = t. 11.1.15 Remark According to Deﬁnition 11.1.14, any polynomial is s-sparse. Similarly any poly-nomial is t-sedimentary. However only those with suﬃciently small parameters s or t are considered in practice. The relevance of s-sparse polynomials is known for a long time, whereas the interest for t-sedimentary polynomials is more recent [717, 2306]. 11.1.16 Deﬁnition We denote by σq(n) the minimal s such that there exists an irreducible s-sparse polynomial of degree n in Fq[x]. Similarly, τq(n) denotes the minimal t such that there exists an irreducible polynomial of degree n in Fq[x] of the form xn + h with deg h = t. 11.1.17 Conjecture Let n be a positive integer and let q be a power of a prime greater than 2, then we have σq(n) ≤4. If q = 2, then σq(n) ≤5 . 11.1.18 Remark Conjecture 11.1.17 states that except for certain extensions of degree n of F2 where a pentanomial is required because there is no irreducible trinomial of degree n, it is always possible to ﬁnd an irreducible binomial, trinomial, or quadrinomial to deﬁne Fqn over Fq. A similar conjecture exists for primitive polynomials, see Section 4.1. 11.1.19 Example We have σ2(8) = 5. An exhaustive search shows that there is no irreducible trinomial of degree 8 over F2 but it is easy to see that x8 + x4 + x3 + x + 1 is irreducible over F2. The ﬁeld F28 is the smallest extension of F2 that requires a pentanomial. Similarly, σ3(49) = 4. Again, an exhaustive search shows that there is no irreducible binomial or trinomial of degree 49 over F3 but x49 + 2x3 + x2 + 1 is irreducible over F3. The ﬁeld F349 is the smallest extension of F3 that requires a quadrinomial. 11.1.20 Conjecture Let n be a positive integer and let q ≥2 be a prime power, then we have τq(n) ≤3 + logq n . 11.1.21 Remark Conjectures 11.1.17 and 11.1.20 are supported by extensive computations [1181, 1187, 1233, 1327]. 11.1.22 Remark We refer to for a table of irreducible trinomials and pentanomials in F2[x] of degree ranging from 2 to 10, 000; see also Section 2.2. 11.1.23 Remark See for a dedicated algorithm with reduced space complexity designed to test the irreducibility of trinomials of large degree in F2[x]. See for additional algorithms speciﬁcally designed for trinomials. 11.1.24 Remark Finally, we refer to Subsection 11.3.2 for methods to construct an irreducible polynomial of degree n over Fq. We note that there exist inﬁnite families of irreducible polynomials. For instance, the polynomials x2.3k + x3k + 1 with k ≥0 are all irreducible in F2[x] . Algorithms 349 11.1.1.3 Primitive elements 11.1.25 Remark Finding a primitive element of a ﬁnite ﬁeld is an interesting problem both from a theoretical and a practical point of view [439, 927, 2627, 2628, 2640]. Given the complete factorization of q −1, Algorithm 11.1.26 returns a generator of F∗ q in polynomial time in log q. No eﬃcient method is available when the factorization of q −1 is not known. 11.1.26 Algorithm (Primitive element random search) Input: A prime power q and the complete factorization of q −1 as pd1 1 · · · pdk k . Output: A generator γ of F∗ q. 1. for i = 1 to k do 2. repeat 3. Choose α ∈F∗ q at random and compute β = α(q−1)/pi 4. until β ̸= 1 5. γi = α(q−1)/p di i 6. end for 7. return γ = Qk i=1 γi 11.1.27 Remark Algorithm 11.1.26 is more eﬃcient than the na¨ ıve approach, which consists in searching for an element γ such that γ(q−1)/pi ̸= 1, for all i. This is especially true as the number of prime factors of q −1 grows. The complexity of ﬁnding a generator of the multiplicative group of a prime ﬁeld F∗ p with Algorithm 11.1.26 is O(log4 p) . 11.1.28 Remark To generate a random prime ﬁeld Fp together with a primitive element, simply modify Algorithm 11.1.2 so that the random integer n produced at Line 2 is of the form n = m+ 1, where all the prime factors of m are known. An algorithm to generate a random factored number is given in [2632, Section 9.6]. Then use Algorithm 11.1.26 to return a generator of F∗ p. 11.1.1.4 Order of an irreducible polynomial and primitive polynomials 11.1.29 Remark The order of a polynomial is introduced in Deﬁnition 2.1.51. The order of an irreducible polynomial f is equal to the order of any of its roots. It can be found with Algorithm 11.1.30 . 11.1.30 Algorithm (Order of an irreducible polynomial) Input: An irreducible polynomial f of degree n with coeﬃcients in Fq and the complete factorization of qn −1 as qe1 1 · · · qeℓ ℓ. Output: The order of f. 1. for i = 1 to ℓdo 2. Find the smallest nonnegative integer fi such that f \| xq1 e1···qi fi···qℓ eℓ−1 3. end for 4. return q1f1 · · · qℓfℓ 11.1.31 Remark If f is monic of order qn −1 such that f(0) ̸= 0 then f is a primitive polynomial. In fact, the conditions are equivalent [1939, Theorem 3.16]; see also Section 4.1 for more details on primitive polynomials. Algorithm 11.1.32 is speciﬁcally designed to quickly test if a random polynomial is primitive or not. 11.1.32 Algorithm (Primitive polynomial testing) Input: An irreducible polynomial f of degree n with coeﬃcients in Fq and the list of all the prime factors of qn −1. Output: true if f is primitive and false otherwise. 350 Handbook of Finite Fields 1. if (−1)nf(0) is not a primitive element of Fq then 2. return false 3. end if 4. if x(qn−1)/(q−1) ̸≡(−1)nf(0) (mod f) then 5. return false 6. end if 7. for each prime factor ri of qn−1 q−1 that does not divide q −1 do 8. if x(qn−1)/((q−1)ri) (mod f) ∈Fq then 9. return false 10. end if 11. end for 12. return true 11.1.33 Remark The prime factors of q −1 are needed at Line 1 to test if (−1)nf(0) is a primitive element or not. If q is small enough, it is worth evaluating f(α) for all α ∈F∗ q in order to detect linear factors. Another useful test, if n is not too large, is to form the Berlekamp matrix Q (see Subsection 11.3.2) and compute the rank of Q −I. If the rank is strictly less than n −1 then the polynomial is not irreducible and hence not primitive. Those two optional tests should be implemented after the initial one that ends at Line 3. 11.1.34 Remark Given a single primitive polynomial f of degree n over Fq, it is quite easy to ﬁnd them all. Simply consider γ, the primitive element whose minimal polynomial is f and form the minimal polynomial of γk, for each k coprime with q −1. A more eﬃcient algorithm is presented in . 11.1.35 Remark If n = 2k −1 is a Mersenne prime, then an irreducible polynomial of degree k is necessarily primitive in F2[x]. We refer to [403, 404, 406, 408] for a search of binary primitive trinomials by Brent and Zimmermann. 11.1.1.5 Minimal polynomial of an element 11.1.36 Remark The simplest approach to ﬁnd the minimal polynomial of α ∈Fqn is to compute the diﬀerent conjugates αqi of α until we have αqk = α. The minimal polynomial of α is then equal to the product (x −α)(x −αq) · · · (x −αqk−1). The worst case complexity of this algorithm is O(Mq(n)n log q) in time and O(n) elements in space. This technique is reﬁned in for the special case Fq = F2. 11.1.37 Remark In , Shoup proposes a more general algorithm with a better time complexity but also with an increased space complexity. Indeed, given the ring Fq[α][β] of dimension n over Fq, Shoup’s algorithm ﬁnds the minimal polynomial of an element in that ring with complexity O(Mq(n)n1/2 + n2) in time and O(n3/2) elements in space. It relies on several subroutines, including Wiedemann’s projection method , a polynomial evalu-ation technique by Brent–Kung , and ﬁnally, the Berlekamp/Massey algorithm applied to recover the minimal polynomial from a sequence of projected points. Kedlaya and Umans give an algorithm for constructing a minimal polynomial in n1+o(1) log1+o(1) q bit operations. It relies on a fast modular composition method via multivariate multipoint evaluation, see Remark 11.1.86. 11.1.2 Representation of ﬁnite ﬁelds 11.1.38 Remark There are essentially three diﬀerent ways to represent the elements of a ﬁnite ﬁeld. Small ﬁnite ﬁelds can be represented with Jacobi logarithms, also known as Zech’s Algorithms 351 logarithms, see Subsection 2.1.7.5. Extension ﬁelds Fqn can always be represented with a normal basis over Fq, see Section 5.2. Finally, prime ﬁelds and extension ﬁelds can be seen as a quotient set, respectively modulo a prime number and an irreducible polynomial. 11.1.39 Remark Elements in Fp are usually represented as integers in [0, p−1] or in [−⌊p/2⌋, ⌊p/2⌋]. We can also use alternative systems, such as the Montgomery representation, see Re-mark 11.1.46, redundant systems , or even ﬂoating point numbers . 11.1.40 Remark When Fqn is deﬁned as Fq[x]/(f), where f is an irreducible polynomial of degree n in Fq[x], elements are usually represented as polynomials in Fq[x] of degree strictly less than n. However, a generalization of Montgomery representation, see Remark 11.1.57, and various redundant systems [405, 909, 3009] exist for extension ﬁelds as well. 11.1.41 Remark The Kronecker substitution allows to represent a polynomial in Fq[x] as an integer by formally replacing x by a suitable integer, usually a suﬃciently large power of 2. Polynomial multiplication can be done faster with this system [930, 1430]. This technique is implemented in the FLINT library . 11.1.3 Modular reduction 11.1.42 Remark The reduction of an integer u modulo a prime number p and the reduction of a polynomial g modulo an irreducible polynomial f are denoted by u (mod p) and g (mod f), respectively. In any case, an eﬃcient reduction method is crucial for fast ﬁnite ﬁeld arith-metic. Indeed, a reduction is usually performed after each operation to ensure that the size of the operands, i.e., the bit length or the degree, remains under control. Assuming that the operands are always reduced, an addition requires at worst one straightforward reduction. On the contrary, reducing the result of a multiplication is more involved even if we know that the size of the product is at most twice the size of the modulus. 11.1.3.1 Prime ﬁelds 11.1.43 Remark We assume that integers are represented using radix b. For most architectures, we have b = 2w, for a ﬁxed w ≥2. A word then corresponds to w bits and we assume that multiplications and divisions by b, which correspond respectively to word left shift and word right shift instructions, are free. The following is presented in [2080, Algorithm 14.42]. 11.1.44 Algorithm (Barrett reduction) Input: A 2n-word integer u, the n-word integer p, and the value µ = b2n/p . Output: The n-word integer r such that u ≡r (mod p). 1. ˆ q ← ⌊(u/bn−1)⌋µ/bn+1 2. r1 ←u (mod bn+1) 3. r2 ←(ˆ qp) (mod bn+1) 4. r ←r1 −r2 5. if r < 0 then 6. r ←r + bn+1 7. end if 8. while r ≥p do 9. r ←r −p 10. end while 11. return r 352 Handbook of Finite Fields 11.1.45 Remark The quantity ˆ q computed at Line 1 of Algorithm 11.1.44 is an approximation of q = ⌊u/p⌋. It satisﬁes q −2 ≤ˆ q ≤q but it is obtained with reduced eﬀorts thanks to the precomputed value µ. When b is large enough with respect to k, it is possible to implement Algorithm 11.1.44 so that it requires at most n(n + 4) + 1 single-precision multiplications; see [2080, Note 14.45] for a precise analysis. 11.1.46 Remark Given R some ﬁxed power of the radix b larger than p, Montgomery developed in a very interesting arithmetic system modulo p where an integer x ∈[0, p −1] is represented by xR (mod p). The Montgomery representation of x is denoted by [x]. The advantage of this system lies in a notion of reduction that can be evaluated very eﬃciently. Namely, the Montgomery reduction of u ∈[0, Rp−1] denoted by Redc(u) is deﬁned as uR−1 (mod p). Assuming that p′ = −p−1 (mod R) has been precomputed, simply determine k ≡up′ (mod R) with k ∈[0, p −1] and let t be the result of the exact division of (u + kp) by R. Then it is easy to see that t = Redc(u) or t = Redc(u) + p. Also, since only the bottom half of up′ and the upper half of (u+kp) are relevant, the overall complexity of Redc is close to one multiprecision multiplication of size n. Algorithm 11.1.47 is a multiprecision variant with a reduction at each step [2080, Subsection 14.3.2]. 11.1.47 Algorithm (Montgomery reduction) Input: An n-word integer p, R = bn, p′ = −p−1 (mod b) and a 2n-word integer u = (u2n−1 . . . u0)b < Rp. Output: The n-word integer t = uR−1 (mod p). 1. (t2n−1 . . . t0)b ←(u2n−1 . . . u0)b 2. for i = 0 to n −1 do 3. ki ←tip′ (mod b) 4. t ←t + kipbi 5. end for 6. t ←t/R 7. if t ⩾p then 8. t ←t −p 9. end if 10. return t 11.1.48 Remark The computation of Redc(u) does not require any division, only n word right shifts at Line 6. A precise analysis shows that Algorithm 11.1.47 needs n(n + 1) single-precision multiplications [2080, Note 14.34]. A classical reduction modulo p of a 2n-word integer rely-ing on a Euclidean division also requires approximately n2 single-precision multiplications but it also needs n single-precision divisions on top. 11.1.49 Remark The conversion to the Montgomery representation is not particularly cheap. How-ever, once the conversion is done, it is possible to eﬃciently add, multiply, and even invert values within this system, see Remarks 11.1.67 and 11.1.101. At the end of the whole com-putation, for instance a modular exponentiation, the result is then converted back to its normal representation using the relation Redc([z]) = z. 11.1.50 Remark Other representation systems, such as the residue number system [163, 2754], the modular number system , and the polynomial modular number system oﬀer interesting features for prime ﬁeld arithmetic. All the computations can easily be parallelized for increased performance. 11.1.51 Remark If working with a random prime is not crucial, choosing a modulus of a special form can lead to substantial savings. An extreme example is illustrated by a Mersenne prime p = 2k −1, for which a reduction modulo p is extremely cheap. Indeed, since a shift by a full word is free, reducing x < p2 modulo p only requires to shift x by exactly r = k Algorithms 353 (mod w) bits, i.e., less than w bits. Unfortunately, Mersenne primes are too scarce to be of any use for practical applications. This explains the introduction of generalizations of the form p = 2k +c with c small and later p = 2nkw ±2nk−1w ±· · ·±2n1w ±1 where w = 16, 32, or 64 . For instance, p = 2192−264−1 and p = 2256−2224+2192+296−1 are prime and their low Hamming weight allows a fast reduction. Those integers are part of a list of recommended primes of various sizes known as NIST primes . A further generalization is to consider prime numbers of the form p = g(t), where g is a sparse polynomial and t is an integer not necessarily equal to 2 . 11.1.3.2 Extension ﬁelds 11.1.52 Remark The remainder g (mod f) can always be computed with the Euclidean division algorithm, but if f is a sparse polynomial, the following algorithm is particularly well suited and more eﬃcient. 11.1.53 Algorithm (Polynomial modular reduction) Input: Two polynomials f and g with coeﬃcients in Fq, where f = xn + Ps−1 i=1 aixbi with 0 = b1 < b2 < · · · < bs−1 < n. Output: The polynomial r such that g ≡r (mod f) with deg r < n. 1. r ←g 2. while deg(r) ≥n do 3. k ←max{n, deg r −n + bs−1 + 1} 4. Write r as r1xk + r2 with deg r2 < k 5. r ←r2 −r1 f −xn xk−n 6. end while 7. return r 11.1.54 Remark For an s-sparse polynomial f of degree n, the reduction of g modulo f requires at most 2(s −1)(deg g −n + 1) operations in Fq. The impact of s on the overall complexity is obvious in this complexity analysis and justiﬁes the interest for low weight irreducible polynomials; see Subsection 11.1.1.2. 11.1.55 Remark The concept of an almost irreducible trinomial, i.e., a trinomial f deﬁned over F2 and having an irreducible factor of degree n, is introduced in and . The arithmetic is then performed in the ring F2[x]/(f) containing the ﬁeld F2n, using a redundant set of representatives. In , Scott observes that for a given architecture, some well chosen irreducible pentanomials over F2 may provide a faster arithmetic than trinomials, including irreducible trinomials. 11.1.56 Remark Dedicated reduction methods have been developed for speciﬁc polynomials. For instance, [1413, Algorithms 2.41 and 2.42] gives highly optimized reduction methods modulo x163 + x7 + x6 + x3 + 1 and x233 + x74 + 1. 11.1.57 Remark A generalization of the Montgomery representation for polynomials over ﬁnite ﬁelds of characteristic 2 is described in . 11.1.58 Remark For cryptographic applications, especially for applications running on embedded devices, a new type of ﬁnite ﬁeld, called an optimal extension ﬁeld, has been recently intro-duced [161, 2096]. 11.1.59 Deﬁnition An optimal extension ﬁeld is one of the form Fpn where p is a generalized Mersenne prime of the form 2k + c that ﬁts in a word and such that the irreducible 354 Handbook of Finite Fields polynomial f used to deﬁne Fpn is the binomial xn −w, where w ∈Fp. The ﬁeld is of Type I if c = ±1 and it is of Type II when w = 2. 11.1.60 Remark Considering ﬁnite ﬁelds of similar cardinality, optimal extension ﬁelds compare favorably against prime ﬁelds thanks to an inversion that is usually faster. Also, due to the lack of dedicated instructions in some old processors to multiply polynomials in F2[x], a multiplication in an optimal extension ﬁeld is usually faster than in characteristic 2 on those platforms. 11.1.4 Addition 11.1.61 Remark Adding, or subtracting, elements in a ﬁnite ﬁeld is in general pretty straightfor-ward; see for instance [1413, Algorithms 2.7 and 2.32]. This is not the case when nonzero elements are expressed as a power of a generator of the multiplicative group. The notion of Jacobi logarithm, see Subsection 2.1.7.5, gives a way to add elements easily. However, Ja-cobi logarithms must be precomputed and stored for each nonzero element [661, Subsection 2.3.3] and this explains why they are only used for relatively small ﬁelds. 11.1.62 Remark For prime ﬁelds represented as Z/pZ, a reduction is sometimes needed after adding two integers modulo a prime number p and this might lead to branch mispredictions . In , Zimmermann discuss speciﬁcally designed algorithms, which do not need any adjustment step to perform additions, subtractions, and multiplications under certain con-ditions. 11.1.5 Multiplication 11.1.63 Remark Multiplying two elements is a task signiﬁcantly more complicated than adding them. There is a wide range of multiplication methods whose eﬃciency and level of sophis-tication increase with the size of the operands. 11.1.5.1 Prime ﬁelds 11.1.64 Remark One approach to perform a modular multiplication is to compute the product ﬁrst and then reduce it independently. This is especially eﬀective for large values where it is worth using advanced multiplication techniques, such as Karatsuba , Toom-Cook , or fast Fourier transform . For smaller values, Algorithm 11.1.65, which is based on the schoolbook method, reduces the result while it is computed for increased performance. 11.1.65 Algorithm (Interleaved multiplication-reduction) Input: The n-word prime p and two n-word integers u = (un−1 . . . u0)b and v. Output: An integer r such that r ≡uv (mod p). 1. r ←0 2. for i = 0 to n −1 do 3. r ←rb + un−i−1v 4. Approximate q = ⌊r/p⌋by ˆ q 5. r ←r −ˆ qp 6. end for 7. while r ≥p do 8. r ←r −p 9. end while 10. return r Algorithms 355 11.1.66 Remark Although r is relatively small, diﬀerent techniques, some of them quite involved, exist to determine ˆ q at Line 4 of Algorithm 11.1.65; see [829, Section 2.2] and [661, Subsec-tion 11.1.2]. 11.1.67 Remark Remark 11.1.46 gives a presentation of the notions of Montgomery representation and of Montgomery reduction. Montgomery multiplication consists in multiplying elements in Montgomery representation, before applying Montgomery reduction in order to have the result of the product, again, in Montgomery representation. Formally, we have the relation Redc([x][y]) = [xy]. 11.1.68 Remark Karatsuba’s method relies on a clever use of the divide and conquer strategy. While the schoolbook multiplication has complexity O(n2) to compute the product of two n-word integers, Karatsuba multiplication [1683, 1684] has asymptotic complexity O(nlog2 3) = O(n1.585). Karatsuba multiplication is thus faster than the na¨ ıve approach, but due to a certain overhead this occurs only for integers of size larger than some threshold n0, which depends on the platform. It is reported in that this threshold can vary from 10 up to more than 100 words. Note that GMP uses speciﬁcally optimized values for a wide range of architectures. 11.1.69 Algorithm (Karatsuba multiplication) Input: Two n-word integers u = (un−1 . . . u0)b, v = (vn−1 . . . v0)b, and n0 ≥2. Output: The 2n-word integer uv. 1. if n ≤n0 then 2. return uv computed with the schoolbook method 3. else 4. k ←⌈n/2⌉ 5. Split u and v in two parts: 6. U1 ←(un−1 . . . uk)b and U0 ←(uk−1 . . . u0)b 7. V1 ←(vn−1 . . . vk)b and V0 ←(vk−1 . . . v0)b 8. Us ←U0 + U1 and Vs ←V0 + V1 9. Compute recursively U0V0, U1V1, and UsVs 10. return U1V1b2k + (UsVs −U1V1 −U0V0)bk + U0V0 11. end if 11.1.70 Remark Since multiplications by bq are free, multiplying two n-word integers with Algo-rithm 11.1.69 requires only three multiplications of size n/2 and a few additions of size n. This observation applied recursively justiﬁes the complexity O(nlog2 3) of Karatsuba’s method. 11.1.71 Remark Using polynomial evaluation and interpolation techniques, Toom 3-ways reduces one multiplication of size n to ﬁve multiplications of size n/3 and thus runs in O(nlog3 5) = O(n1.465) [1768, Subsection 4.3.3.A]. Generalizing this idea, Sch¨ onhage–Strassen multiplica-tion using the fast Fourier transform runs in O(n log n log log n) [751, Section 9.5]. The overhead is such that this method is only worth using for integers having several thousand digits . 11.1.5.2 Extension ﬁelds 11.1.72 Remark For ﬁnite ﬁelds Fqn deﬁned via an irreducible polynomial f ∈Fq[x] of degree n, all the techniques presented for prime ﬁelds apply in this context as well. In particular, there is a generalization of Algorithm 11.1.65 where polynomial multiplications and reductions modulo f are interleaved. Also, more advanced multiplication methods, such as Karatsuba or based on the fast Fourier transform are available as well, with similar complexities as in the integer case. 356 Handbook of Finite Fields 11.1.73 Remark Many practical applications use ﬁnite ﬁelds of characteristic two and speciﬁc mul-tiplication methods, such as the right-to-left comb, left-to-right comb, or even window meth-ods involving some precomputations, have been developed in this context; see for some explicit algorithms and for a discussion focused on the implementation of Karatsuba, Toom-Cook, Sch¨ onhage, and Cantor methods. 11.1.74 Remark We refer to Section 5.3 for multiplication in a normal basis, where the complexity is thoroughly discussed and the concept of optimal normal basis is introduced. 11.1.6 Squaring 11.1.6.1 Finite ﬁelds of odd characteristic 11.1.75 Remark The formula n−1 X i=0 uibi !2 = n−1 X i=0 u2 i b2i + 2 X i<j uiujbi+j suggests that it takes less eﬀort to compute u2 than to compute uv, where u and v are distinct integers of the same size. In practice, a dedicated modular squaring algorithm can be up to 20% faster than its general counterpart . 11.1.76 Remark For each multiplication algorithm, there exists a speciﬁc variant exclusively de-signed to square elements; see [661, Subsection 10.3.3] for a description of the schoolbook and Karatsuba squaring methods. 11.1.6.2 Finite ﬁelds of characteristic two 11.1.77 Remark For the binary ﬁeld F2n deﬁned over F2 with a normal basis, a squaring corresponds to a circular shift of the coordinates. With a polynomial basis modulo f, the squaring of an element requires slightly more work, as we have n−1 X i=0 aixi !2 = n−1 X i=0 aix2i. Thus, only a reduction modulo f is necessary to ﬁnd the result. 11.1.7 Exponentiation 11.1.78 Remark The exponentiation operation consists in computing αm, for a nonnegative integer m and α ∈Fq. It should be optimized as much as possible as it is a crucial subroutine in many algorithms. For instance, exponentiation is key for ﬁnding a generator of the multi-plicative group or a primitive polynomial, for computing an inverse or the square root of an element, or for factoring a polynomial; see for a very complete survey dedicated to exponentiation methods in ﬁnite ﬁelds. 11.1.7.1 Prime ﬁelds 11.1.79 Remark When the exponent m is ﬁxed, it may be worth searching for an addition chain with low complexity computing m; see [661, Section 9.2]. If the element α ∈Fp is ﬁxed while the exponent varies, precomputing some powers of α can lead to considerable speedups. See Algorithms 357 in particular Yao’s method [1767, 3032] also known as BGMW and ﬁxed-base comb algorithm [242, 2396] presented in as well. More details are available in [661, Section 9.3]. In the general case, where both the element and the exponent vary, Algorithm 11.1.80 is the most eﬃcient exponentiation method that is known. 11.1.80 Algorithm (Sliding window exponentiation) Input: An element α ∈Fp, a nonnegative exponent m = Pℓ−1 i=0 mi2i, a parameter k ≥1, and the stored values α, α3, . . . , α2k−1. Output: The element αm ∈Fp. 1. β ←1 and i ←ℓ−1 2. while i ≥0 do 3. if mi = 0 then 4. β ←β2 (mod p) and i ←i −1 5. else 6. s ←max{i −k + 1, 0} 7. while ms = 0 do 8. s ←s + 1 9. end while 10. for j = 1 to i −s + 1 do 11. β ←β2 (mod p) 12. end for 13. Form t = (mi . . . ms)2 14. β ←β × αt (mod p) 15. i ←s −1 16. end if 17. end while 18. return β 11.1.81 Remark The parameter k controls the size of the window used to scan the bits of n. For k = 1, Algorithm 11.1.80 coincides with the well-known square and multiply method. For k > 1, Algorithm 11.1.80 relies on 2k−1 −1 precomputed values obtained at a cost of 2k−1 multiplications. Given an ℓ-bit exponent m, it requires ℓ/(k + 1) extra ﬁeld multiplications, on average . The size of the window k should therefore be selected to minimize the quantity 2k−1 + ℓ/(k + 1). The number of squarings is independent of k and is equal to ℓin any case. Algorithm 11.1.80 is implemented in NTL . 11.1.7.2 Extension ﬁelds 11.1.82 Remark The main diﬀerence with the prime ﬁeld case is the existence of the Frobenius automorphism that can be used in an extension ﬁeld to speed up the computation of an exponentiation. 11.1.83 Algorithm (Fast exponentiation in polynomial basis) Input: Two polynomials f and g with coeﬃcients in Fq such that deg f = n and deg g < n, an exponent 0 < m < qn, and a positive integer r. Output: The polynomial gm (mod f). 1. Write m in base qr as m = (mℓ−1 . . . m0)qr 2. for i = 0 to ℓ−1 do 3. Compute and store gmi (mod f) 4. end for 5. h ←xqr (mod f) and t ←1 6. for i = ℓ−1 down to 0 do 358 Handbook of Finite Fields 7. t ←t(h) (mod f) 8. t ←tgmi (mod f) 9. end for 10. return t 11.1.84 Remark Algorithm 11.1.83 relies on a generalization of the square and multiply method in base qr. It is discussed in where r is set to be ⌈n/ logq n⌉and where the computation of the residues gmi (mod f) at Line 3 is done with the BGMW method . The computation t(h) (mod f) at Line 7 is a modular composition; see Remarks 11.1.85 and 11.1.86. 11.1.85 Remark The ﬁrst nontrivial method to compute t(h) (mod f) is due to Brent and Kung . Assuming that deg h, deg t < deg f = n and that square matrices of di-mension n can be multiplied with O(nw) ﬁeld multiplications, the Brent–Kung approach takes O(Mq(n)n1/2 + n(w+1)/2) ﬁeld operations. With Mq(n) = O(nlog3 2) correspond-ing to Karatsuba’s multiplication (Subsection 11.1.5), the overall complexity becomes O(nlog3 2+1/2) = O(n2.085). If instead fast integer multiplication methods a la Sch¨ onhage– Strassen are used, the complexity is O(n(w+1)/2). The natural bound w = 3 was improved by Strassen who introduced a method with w = log2 7 < 2.8074 . The best known upper bound is w < 2.3727 [2984, 2985]. 11.1.86 Remark Umans proposed a completely diﬀerent approach to perform a modular composi-tion, based on multivariate multipoint evaluation . Initially, this work only addressed characteristic at most no(1) but was later generalized to any characteristic by Umans and Kedlaya [1721, 1722]. In any case, the asymptotic complexity is n1+o(1) log1+o(1) q bit oper-ations, which is optimal up to lower order terms. 11.1.87 Remark For the particular case Fq = F2n and for r = ⌈n/ log2 n⌉, the complexity of Algorithm 11.1.83 becomes O(M2(n)n/ log n). This complexity does not depend on the method, either or , used to perform the modular composition. 11.1.88 Remark To compute αm ∈Fqn, where Fqn is represented using a normal basis over Fq, the exponent m is again expressed in base qk, for some ﬁxed k, in order to take advantage of the Frobenius automorphism that allows one to compute αq with just a cyclic shift of the coordinates of α. In this case, only Line 7 of Algorithm 11.1.83 needs to be modiﬁed and replaced by t ←tqk; see also Subsection 5.3.5. 11.1.89 Remark When Fq is an optimal extension ﬁeld (Deﬁnition 11.1.59), the action of the Frobe-nius automorphism can also be computed extremely eﬃciently, leading to very fast expo-nentiation techniques; see and [661, Subsection 11.3.3]. 11.1.8 Inversion 11.1.90 Remark There are two ways to compute the inverse of a ﬁeld element α ∈Fqn. If the ﬁeld is deﬁned as a quotient set, we can use the extended Euclidean algorithm, see Algo-rithm 11.1.92. Alternatively, Lagrange’s theorem implies that we have αqn−2 = 1/α. This last method is totally general but it is particularly adapted to ﬁnite ﬁelds deﬁned with a normal basis or when the action of the Frobenius automorphism α 7→αq can be computed eﬃciently. 11.1.91 Remark Let R be a Euclidean ring with Euclidean function ϕ. For elements a, b ∈R, we can always write a = bq + r with ϕ(r) < ϕ(b). In practice, R = Z with natural order < or R = Fq[x] with the degree function. Assuming that b is a prime number or an irreducible polynomial and a is an element such that ϕ(a) < ϕ(b), we then have gcd(a, b) = 1. The B´ ezout identity au + bv = 1, whose coeﬃcients u and v are returned by Algorithm 11.1.92, implies that au ≡1 (mod b), i.e., u is the inverse of a modulo b. Algorithms 359 11.1.92 Algorithm (Extended Euclidean Algorithm) Input: Two elements a, b in a Euclidean ring R such that ϕ(a) < ϕ(b). Output: Elements (u, v, d) in R such that au + bv = d with d = gcd(a, b). 1. t ←0, u ←1, c ←b, and d ←a 2. while c ̸= 0 do 3. Write d = cq + r with ϕ(r) < ϕ(c) 4. s ←u −tq, u ←t, t ←s, d ←c, and c ←r 5. end while 6. v ←(d −au)/b 7. return (u, v, d) 11.1.93 Remark The division at Line 6 of Algorithm 11.1.92 is exact. Dedicated methods to compute the quotient of an exact division are given in [1601, 1805]. 11.1.94 Remark In a ﬁnite ﬁeld, an inversion is in general considerably more expensive than a multi-plication. When k elements a1, . . . , ak need to be inverted, a trick due to Montgomery allows one to replace k inversions by one inversion and 3k−3 ﬁeld multiplications. The prin-ciple is to compute the inverse (a1 · · · ak)−1 and then multiply it by suitable precomputed terms in order to recover successively each inverse individually. It was introduced initially to speed up the elliptic curve factorization method and is therefore described for integers deﬁned modulo a composite number. However, this trick can be applied to any structure where the notion of inverse exists. It is also presented in and . 11.1.95 Remark In order to compute the division e/a, where e, a ∈Fqn, one could invert a and multiply the result by e. However, e/a can be obtained directly with Algorithm 11.1.92. Simply replace u ←1 by u ←e at Line 1. 11.1.8.1 Prime ﬁelds 11.1.96 Remark Algorithm 11.1.92 is usually preferred over Lagrange’s method to compute the inverse of α modulo p. There are at least two reasons for that. Computing the exponentia-tion αp−2 (mod p) requires on average twice as many arithmetic operations as the extended Euclidean method . Also, there are a number of improvements and variants of Algo-rithm 11.1.92 that can be implemented to speed up its execution on diﬀerent platforms. 11.1.97 Remark Algorithm 11.1.92 returns the inverse modulo p after O(log p) steps. When suitably implemented, in particular if the precision of the Euclidean division is adjusted and decreases with the size of its arguments d and c, then the overall time complexity is O(log2 p) [660, Section 1.3]. In , Lehmer suggests to replace the exact computation of the quotient q at Line 3 of Algorithm 11.1.92 by an approximation obtained by dividing the most signiﬁcant digits of d and c. This remark has been explored further by many authors [710, 1602, 1903]. 11.1.98 Remark In a variant introduced by Brent and Kung [402, 661], divisions are eliminated and replaced by shifts, additions, and subtractions. This approach, known as the binary method or the plus-minus method, is well-suited for architectures where a division is expensive. Not surprisingly, the number of steps needed is still O(log p). Indeed, the quotient q at each step of Algorithm 11.1.92 is small most of the time. The probability that q = 1 is close to 0.415 and q ≤5 in more than 77% of the cases . 11.1.99 Remark Sch¨ onhage , improving on Knuth’s work , showed how the ex-tended Euclidean algorithm, and hence inversion, can be done asymptotically in time O(M(log p) log log p); see also and for a description of the method. Stehl´ e and Zimmermann developed a binary recursive gcd method. Although it does not im-prove on the O(M(log p) log log p) asymptotic complexity, its description, implementation, and proof of correctness are simpler than Sch¨ onhage’s method. 360 Handbook of Finite Fields 11.1.100 Remark A very simple approach presented in allows one to ﬁnd an inverse modulo p. Interestingly, it is not related to the extended Euclidean gcd method nor Lagrange’s method. It is particularly eﬃcient for certain types of primes, such as Mersenne primes. The method is recalled in [661, Subsection 11.1.3]. 11.1.101 Remark There is a notion of Montgomery inverse , which completes the other op-erations already existing in Montgomery representation; see Remarks 11.1.46 and 11.1.67, and [1644, 2536] for additional improvements regarding the Montgomery inverse. 11.1.8.2 Extension ﬁelds 11.1.102 Remark As in the integer case, there is a binary version of Algorithm 11.1.92 that does not require any division; see for an eﬃcient version in even characteristic. 11.1.103 Remark A notion of reduction in a ﬁnite ﬁeld deﬁned by a normal basis is discussed in . It is then possible to compute the inverse of an element with a variant of Al-gorithm 11.1.92. However, since the Frobenius automorphism can be evaluated for free in a normal basis, Lagrange’s method, which computes α−1 as αqn−2 is preferred. Algo-rithm 11.1.104 follows this idea with an additional improvement, i.e., the use of addition chains computing q −2 and n −1 . See [661, Section 9.2] for a deﬁnition of addition chains and related techniques to ﬁnd short chains. 11.1.104 Algorithm (Inversion using Lagrange’s theorem) Input: An element α ∈F∗ qn, two addition chains, namely (a0, a1, . . . , as1) computing q −2 and (b0, b1, . . . , bs2) computing n −1. Output: The inverse of α, i.e., α−1 = αqn−2. 1. Compute β ←αq−2 using the addition chain (a0, a1, . . . , as1) 2. T ←α × β 3. for i = 1 to s2 do 4. γ ←T[k]qbj where bi = bk + bj 5. T[i] ←γ × T[j] 6. end for 7. γ ←T[s2] 8. return β × γq 11.1.105 Remark Algorithm 11.1.104 needs s1 + s2 + 2 multiplications in Fqn and 1 + b1 + · · · + bs2 applications of the Frobenius automorphism to compute α−1. 11.1.106 Remark Algorithm 11.1.104 is particularly well suited for q = 2. For q > 2, set r = (qn−1)/(q−1) and decompose α−1 as α−rαr−1. Since r = qn−1+· · ·+q+1, it follows that αr is the norm of α. Thus, it is in Fq and can be easily inverted. Also, αr−1 can be obtained with at most n−1 evaluations of the Frobenius automorphism and at most n−2 multiplications in Fq. This is the standard way to compute an inversion in an optimal extension ﬁeld. Further optimizations exist for speciﬁc extension degrees; see [661, Subsections 11.3.4 and 11.3.6] for more details and concrete examples. 11.1.9 Squares and square roots 11.1.107 Remark Computing square roots in Fpn eﬃciently is important in many applications and is the subject of active research; see for instance [204, 1410, 1781, 2194]. In a ﬁnite ﬁeld of characteristic 2, every element is a square and it is straightforward to compute a square root. In odd characteristic, not every element is a square. When it exists, a square root can be computed in most cases with a closed formula. If that is not the case, and in fact in any case, there are algorithms returning the result fairly eﬃciently. Algorithms 361 11.1.9.1 Finite ﬁelds of odd characteristic 11.1.108 Remark Let us assume that p is an odd prime number and that q is some power of p. We know that F∗ q is a cyclic group, generated by, say, γ. Because the cardinality of F∗ q is even, all the square elements in F∗ q must be even powers of γ and all the nonsquare elements correspond to the odd powers of γ. Thus, there are (q −1)/2 squares and just as many nonsquare elements in F∗ q. 11.1.109 Remark To check if a nonzero element α is a square or not, it is enough to perform an exponentiation. Indeed, α is a square if and only if α(q−1)/2 = 1. The outcome is −1 when α is not a square. 11.1.110 Remark For the prime number p, we introduce the Legendre symbol denoted by α p such that 0 p = 0 and α p = α(p−1)/2 (mod p) otherwise; see Algorithm 11.1.111 for an eﬃcient way to compute α p and thus decide if α is a square or not in Fp. 11.1.111 Algorithm (Legendre symbol) Input: An integer α and an odd prime number p. Output: The Legendre symbol α p · 1. k ←1 2. while p ̸= 1 do 3. if α = 0 then 4. return 0 5. end if 6. v ←0 7. while α ≡0 (mod 2) do 8. v ←v + 1 and α ←α/2 9. end while 10. if v ≡1 (mod 2) and p ≡±3 (mod 8) then 11. k ←−k 12. end if 13. if α ≡3 (mod 4) and p ≡3 (mod 4) then 14. k ←−k 15. end if 16. r ←α, α ←p (mod r), and p ←r 17. end while 18. return k 11.1.112 Remark Algorithm 11.1.111 relies on a quadratic reciprocity law that allows one to reduce the size of the operands in a way that is similar to the computation of the gcd with Euclid’s algorithm. 11.1.113 Remark The evaluation of α p via the exponentiation α(p−1)/2 (mod p) has complexity O(log3 p) with the square and multiply method. By contrast, Algorithm 11.1.111 has com-plexity O(log2 p). In , Brent and Zimmermann describe a new algorithm to compute α p with complexity O(M(log p) log log p). 11.1.114 Remark There is a generalization of the Legendre symbol for the elements of Fpn. Let f ∈Fp[x] be an irreducible polynomial of degree n and let g ∈Fp[x]. The Legendre symbol for polynomials, again denoted by g f satisﬁes 0 f = 0 and g f = g(q−1)/2 (mod f) for nonzero g. Algorithm 11.1.115 relies also on a quadratic reciprocity law, quite similarly to Algorithm 11.1.111, and allows to eﬃciently determine if g is a square or not modulo f. 362 Handbook of Finite Fields 11.1.115 Algorithm (Legendre symbol for polynomials) Input: A polynomial g ∈Fp[x] and an irreducible polynomial f ∈Fp[x]. Output: The Legendre symbol for polynomials g f · 1. k ←1 2. repeat 3. if g = 0 then 4. return 0 5. end if 6. a ←the leading coeﬃcient of g 7. g ←g/a 8. if deg f ≡1 (mod 2) then 9. k ← a p k 10. end if 11. if pdeg f ≡3 (mod 4) and deg f deg g ≡1 (mod 2) then 12. k ←−k 13. end if 14. r ←g, g ←f (mod r), and f ←r 15. until deg f = 0 16. return k 11.1.116 Remark Once we know that a nonzero element α ∈Fp is a square, it may be necessary to compute a square root of α, i.e., an element ρ ∈Fp, satisfying ρ2 = α. If ρ is one square root, then −ρ is the other one and there exist closed formulas to compute ρ in most cases [660, Section 1.5]. Indeed, we have 1. ρ = ±α(q+1)/4, if q ≡3 (mod 4); 2. ρ = ±α(q+3)/8, if q ≡5 (mod 8) and α(q−1)/4 = 1; 3. ρ = ±2α(4α)(q−5)/8 (mod p), if q ≡5 (mod 8), α(q−1)/4 = −1, and q is an odd power of p. 11.1.117 Remark For the other cases, i.e., when q ≡1 (mod 8) or when q ≡5 (mod 8), α(q−1)/4 = −1, and q is an even power of p, there exist several polynomial time methods to compute a square root. The most eﬃcient factorization methods, applied to the polynomial x2 −α, return a square root of α in Fq using O(log q) ﬁeld operations; see Remark 11.4.3. There are also dedicated methods, such as Tonelli and Shanks algorithm , which returns a square root in the prime ﬁeld Fp in time O(log4 p). Another example is Algorithm 11.1.118, which is a generalization of Cipolla’s method [751, Subsection 2.3.9]. Algorithm 11.1.118 is remarkably simple and easy to implement. It works in the quadratic extension Fq2 and also requires O(log q) ﬁeld operations. 11.1.118 Algorithm (Square root computation) Input: A square α in Fq. Output: A square root ρ ∈Fq of α. 1. repeat 2. Choose β ∈Fq at random 3. until β2 −4α is not a square in Fq 4. T ←x2 −βx + α 5. ρ ←x(q+1)/2 (mod T) 6. return ρ Algorithms 363 11.1.9.2 Finite ﬁelds of even characteristic 11.1.119 Remark Every element α ∈F2n is a square and the square root ρ of α can be easily obtained thanks to the multiplicative structure of F∗ 2n, which implies that ρ = α2n−1. With a normal basis, the computation is immediate. Using a polynomial representation, modulo an irreducible polynomial f, there is a diﬀerent approach. If α is represented by Pn−1 i=0 gixi, then observe that √α = X i even gixi/2 + √x X i odd gix i−1 2 where √x has been precomputed modulo f. We note that √x can be obtained very easily when f is a trinomial of odd degree; see [661, Subsection 11.2.6]. 11.1.120 Remark Unlike for ﬁnite ﬁelds of odd characteristic, solving a quadratic equation in F2n is more involved than just extracting square roots. Considering the polynomial x2 + x + β in F2n[x], we can show that it has a root in F2n if and only if the trace of β is zero. In that case, a solution τ is given by τ = (n−3)/2 X i=0 β22i+1 when n is odd. When n is even, τ satisﬁes τ = n−1 X i=0 i X j=0 β2j ! ω2i where ω ∈F2n is any element of trace 1. In any case, the other solution is τ + 1. Reference gives techniques to speed up the computation of a square root in characteristic two at the expense of extra storage. See Also §3.4 For discussion on the weights of irreducible polynomials. §5.3 For discussion on optimal and low complexity normal bases. §11.6 For discussion on discrete logarithms. §16.4 For discussion on elliptic cryptographic systems. §16.5 For discussion on hyperelliptic cryptographic systems. References Cited: [19, 43, 56, 77, 143, 161, 163, 165, 166, 204, 209, 242, 400, 401, 402, 403, 404, 405, 406, 407, 408, 409, 414, 436, 439, 576, 610, 640, 660, 661, 662, 709, 710, 717, 750, 751, 829, 909, 927, 930, 933, 1066, 1088, 1102, 1180, 1181, 1187, 1223, 1227, 1232, 1233, 1255, 1327, 1328, 1410, 1413, 1416, 1426, 1428, 1430, 1576, 1601, 1602, 1644, 1683, 1684, 1721, 1722, 1766, 1767, 1768, 1775, 1781, 1805, 1888, 1895, 1903, 1939, 1941, 2002, 2011, 2038, 2080, 2095, 2096, 2097, 2099, 2111, 2132, 2133, 2141, 2194, 2305, 2306, 2396, 2420, 2435, 2473, 2536, 2557, 2559, 2573, 2582, 2627, 2628, 2631, 2632, 2633, 2640, 2693, 2694, 2705, 2709, 2731, 2748, 2754, 2798, 2803, 2836, 2976, 2984, 2985, 3004, 3009, 3030, 3032, 3033, 3078] 364 Handbook of Finite Fields 11.2 Univariate polynomial counting and algorithms Daniel Panario, Carleton University We∗give basic counting estimates for univariate polynomials over ﬁnite ﬁelds. First, we provide some classical counting results. Then we focus on a methodology based on analytic combinatorics that allows the derivation of many nontrivial counting results. A series of counting results for univariate polynomials over ﬁnite ﬁelds, some of them linked to the analysis of algorithms, is then provided. 11.2.1 Classical counting results 11.2.1 Remark The most classical counting estimate is for the number In of monic irreducible polynomials of degree n over Fq; see Theorem 2.1.24. (We observe that in the results of this section there is only one ﬁnite ﬁeld involved, and so we drop the notation Iq(n) for the simpler one In. Sometimes, however, we let q go to inﬁnity.) 11.2.2 Theorem For all n ≥1 and any prime power q, we have In = 1 n X d\|n µ(d)qn/d = qn n + O qn/2 n . 11.2.3 Remark Since In > 0 for any prime power q and integer n > 1 and the number of monic polynomials of degree n over Fq is qn, then the probability of a polynomial being irreducible is close to 1/n. This probability tends to zero with n →∞. 11.2.4 Remark A polynomial is squarefree if it has no repeated factors. The number of squarefree polynomials over Fq was ﬁrst given by Carlitz . 11.2.5 Theorem Let Qn be the number of squarefree polynomials of degree n over Fq. Then, Qn = qn −qn−1 for n ≥2, qn for n = 0, 1. 11.2.6 Remark Theorem 11.2.5 implies, for n ≥2, that the proportion of squarefree polynomials is 1 −1/q. As a consequence, for large ﬁnite ﬁelds Fq most polynomials are squarefree. 11.2.7 Remark Let us consider the number of irreducible factors of ﬁxed degree d in a random polynomial of degree n over Fq. Zsigmondy concentrates on the prime ﬁeld case Fp and gives results for the number of monic polynomials having no irreducible factors of degree ∗Originally, this section was to be written by Philippe Flajolet and the author, but sadly, Philippe passed away before we started working on it. This section is dedicated to the memory of my friend Philippe Flajolet, for all his many lessons and guidance. Algorithms 365 d, 1 ≤d ≤n, and for the number of monic polynomials having a given number of distinct roots. We will return to this problem in Subsection 11.2.3.2. 11.2.8 Remark The notes at the end of Chapter 4 of Lidl and Niederreiter’s book point to several classical counting references published before 1983. 11.2.9 Remark Other classical results related to polynomials with prescribed trace or norm and to self-reciprocal polynomials are given in Sections 3.1 and 3.5, respectively. 11.2.10 Remark As a ﬁnal classical estimate we consider the probability that two polynomials are coprime. It has been known since at least the 1960s (see Berlekamp and Knuth ) but most likely for a long time before then, that with probability 1 −1/q, the greatest common divisor of two polynomials over a ﬁnite ﬁeld is 1, independently of the degrees of the polynomials. This result has been reinvented a large number of times. In Subsection 11.2.3.4, a much more precise reﬁnement of this classical result is given. 11.2.2 Analytic combinatorics approach 11.2.11 Remark Flajolet has made major methodological contributions to the research area known as analytic combinatorics. Among other things, analytic combinatorics provides a general methodology that can be successfully applied to the analysis of algorithms from many di-verse areas. Its main and classical reference is the book by Flajolet and Sedgewick . In this section this general framework is presented only in relation to polynomials over a ﬁnite ﬁeld Fq although it can be used in much more general settings. For a longer introduction to this methodology and its application to the analysis of polynomial factorization algorithms see Flajolet, Gourdon, and Panario . 11.2.12 Remark This framework has two basic components: generating functions to express, com-binatorially, properties of interest, and asymptotic analysis for the derivation of estimates when exact extraction of coeﬃcients is not possible. The studied properties could be for pure mathematical interest or for their application to the analysis of algorithms for polynomials over ﬁnite ﬁelds [1080, 2347]. 11.2.13 Remark Generating functions for counting some properties of polynomials over ﬁnite ﬁelds have been previously used in some speciﬁc cases by Berlekamp [231, Chapter 3], Knuth [1765, Subsection 4.6.2], and Odlyzko . The global usage of this technique to count many interesting expressions, and its usage in the analysis of polynomial over ﬁnite ﬁelds algorithms, only became possible after the establishment of analytic combinatorics . 11.2.14 Deﬁnition Let In be the number of monic irreducible polynomials of degree n in Fq. The generating functions of monic irreducible polynomials and monic polynomials are, respectively, I(z) = X j≥1 Ijzj, and P(z) = X j≥0 Pjzj. We denote by [zn]T(z) the coeﬃcient of zn in the generating function T(z). 366 Handbook of Finite Fields 11.2.15 Proposition We have P(z) = Y k≥1 (1 + zk + z2k + · · · )Ik = Y k≥1 (1 −zk)−Ik. 11.2.16 Remark Since [zn]P(z) is qn, it follows that P(z) = (1 −qz)−1. This expression and the one in Proposition 11.2.15 implicitly determine that In satisﬁes In = 1 n X k\|n µ(k)qn/k. 11.2.17 Remark Carlitz’s result for squarefree polynomials (Theorem 11.2.5) can be easily recovered under this framework. Indeed, the generating function for squarefree polynomials is Q(z) = Y k≥1 1 + zkIk . Moreover, considering the multiplicity of its irreducible factors, each polynomial f factors as f = st2, where s is squarefree and t is an arbitrary polynomial. We thus have P(z) = Q(z)P(z2), and therefore, Q(z) = P(z) P(z2) = 1 −qz2 1 −qz . Carlitz’s estimate, given in Theorem 11.2.5, can be recovered after extracting coeﬃcients from Q(z). 11.2.18 Remark Generating functions encode exact counting information in their coeﬃcients. How-ever, their extraction from a given generating function is in general a diﬃcult task. Neverthe-less, when considering generating functions as analytic functions, their behavior near their dominant singularities (those with smallest modulus) is an important source of information to extract coeﬃcient asymptotics as their index tends to inﬁnity. 11.2.19 Remark Most of the generating functions f(z) of interest in this section are singular at z = 1/q with an isolated singularity of the algebraic-logarithmic type. In that case, we can apply the following important result due to Flajolet and Odlyzko . 11.2.20 Theorem Let f(z) be a function analytic in a domain D = {z : \|z\| ≤z1, \|Arg(z −1/q)\| > π/2 −ε}, where z1 > 1/q and ε are positive real numbers. Let k ≥0 be any integer, and α a real number with α ̸= 0, −1, −2, . . .. If in a neighborhood of z = 1/q, f(z) has an expansion of the form f(z) = 1 (1 −qz)α log 1 1 −qz k (1 + o(1)), then the coeﬃcients satisfy, asymptotically as n →∞, [zn]f(z) = qn nα−1 Γ(α) (log n)k (1 + o(1)). Algorithms 367 11.2.21 Remark This theorem requires analytic continuation of f(z) outside its circle of conver-gence. However, there are some situations in which generating functions do not satisfy this hypothesis. For instance in some of the generating functions related to smooth polynomials (Subsection 11.2.3.3) analytic continuation is not possible. Saddle point methods are used in these cases. All asymptotic enumeration methods required in this section are explained in detail in the excellent presentations by Flajolet and Sedgewick and Odlyzko . 11.2.3 Some illustrations of polynomial counting 11.2.22 Remark We list several results that are mostly derived in the framework of analytic com-binatorics. 11.2.23 Remark Bivariate generating functions are used to study important parameters of inter-est. The exact counting problem is now reﬁned with two parameters, namely, the degree of the polynomial and an additional property to be studied (for example, the number of its irreducible factors). With an appropriate normalization, successive diﬀerentiation of the bivariate generating function with respect to the additional parameter (evaluated at 1) gives the factorial moments of interest. The classical book by Flajolet and Sedgewick presents a comprehensive explanation of this methodology. The complete study of the num-ber of irreducible factors of a random polynomial is presented below as an example. 11.2.3.1 Number of irreducible factors of a polynomial 11.2.24 Remark Let us consider the bivariate generating function P(u, z) = Y j≥1 (1 + uzj + u2z2j + · · · )Ij = Y j≥1 (1 −uzj)−Ij, where [ukzn]P(u, z) is the number of polynomials of degree n with k irreducible factors. Successive diﬀerentiation of P(u, z) with respect to u (evaluated at u = 1) give univariate generating functions for the factorial moments of this parameter. Asymptotic analysis of these univariate generating functions gives an expectation of log n and standard deviation √log n. More can be said for this problem as the next theorem states. Flajolet and So-ria prove that the number of irreducible factors in a random polynomial over a ﬁnite ﬁeld satisﬁes a central limit theorem with mean and variance asymptotic to log n. We note that this result is equivalent to the Erd¨ os-Kac theorem stating that the number of prime factors in a random integer at most n satisﬁes a central limit theorem with mean and vari-ance asymptotic to log log n. We refer to Subsection 11.2.3.5 for analogies among irreducible decompositions of polynomials over ﬁnite ﬁelds, prime decompositions of integers, and cycle decompositions of permutations. 11.2.25 Theorem Let Ωn be a random variable counting the number of irreducible factors of a random polynomial of degree n over Fq, where each factor is counted with its order of multiplicity. 1. The mean value of Ωn is asymptotic to log n [231, 1765]. 2. The variance of Ωn is asymptotic to log n [1084, 1765]. 368 Handbook of Finite Fields 3. For any two real constants λ < µ, we have Pr n log n + λ p log n < Ωn < log n + µ p log n o → 1 √ 2π Z µ λ e−t2/2dt. 4. The distribution of Ωn admits exponential tails . 5. A local limit theorem holds . 6. The behavior of Pr{Ωn = m} for all m is known [509, 665, 1564]. 11.2.3.2 Factorization patterns 11.2.26 Remark As a ﬁrst example of a factorization pattern, let us consider the number of irre-ducible factors in a random polynomial of a given ﬁxed degree. As it is indicated in Remark 11.2.7, the number of roots was studied by Zsigmondy for prime ﬁelds. Knopfmacher and Knopfmacher present a detailed analysis, for any ﬁnite ﬁeld, including variance. 11.2.27 Remark The case of polynomials with no roots is interesting when studying the distinct values that a polynomial can take. This is related to permutation polynomials (Section 8.1) and was studied by Uchiyama ; see also . 11.2.28 Remark The generating function of polynomials with no linear factors is Y k≥2 1 1 −zk Ik = 1 1 −qz (1 −z)I1 = 1 1 −qz (1 −z)q. As a consequence, the number of polynomials of degree n with no irreducible factors of degree 1 is asymptotic, as n →∞, to qn(1 −1/q)q. That is, the probability of a polynomial with no linear factor tends to 1/e = 0.3678 . . . when q grows. This implies that “most” polynomials are reducible and have at least one irreducible linear factor. 11.2.29 Remark The number of irreducible factors of a given degree d in a polynomial of degree n was studied by Williams . In can be found a detailed analysis of this problem as well as a determination of the variance, in both the cases where repetitions are allowed, and where they are not allowed. 11.2.30 Remark Knopfmacher, Knopfmacher, and Warlimont provide the mean and variance of what they call the “length” of a general factorization pattern by studying the number of polynomial factorizations into exactly k factors. 11.2.31 Remark When factoring univariate polynomials (Section 11.4) using the method based on the squarefree, distinct-degree, and equal-degree factorizations, it is relevant to determine if a polynomial has all its irreducible factors of diﬀerent degrees. In this case, the third stage, the “equal-degree factorization,” is not required. 11.2.32 Theorem [1080, 1763] The probability that all irreducible factors of a random polynomial of degree n over Fq have diﬀerent degrees (but with single factors possibly repeated) is Algorithms 369 asymptotic to cq = Y k≥1 1 + Ik qk −1 (1 −q−k)Ik, where c2 = 0.6656 . . . , c257 = 0.5618 . . . , c∞= e−γ = 0.5614 . . ., where γ is Euler’s constant. 11.2.33 Remark Several related results used in the analysis of polynomial factorization algorithms can be found in . Recent fast factorization algorithms like [1226, 1239, 1667] require a study of the distribution of irreducible factors in parts of interval partitions of [1, n]. See von zur Gathen, Panario, and Richmond for ﬁrst steps towards an understanding of these advanced algorithms. 11.2.3.3 Largest and smallest degree irreducibles 11.2.34 Remark Information on the degrees of the largest irreducible factors is crucial to measure stopping rules for factorization algorithms . Information on the degrees of the smallest irreducible factors helps in the analysis of irreducible test algorithms [2349, 2350]. 11.2.35 Remark A random polynomial of degree n has with high probability several irreducible factors whose degrees sum to near n; see Theorem 11.2.37 and Remark 11.2.38. Let D[j] n be the j-th largest degree of the factors of a random polynomial of degree n in Fq. Car ob-tained an asymptotic expression for the cumulative distribution function of D n in terms of the Dickman function. This number-theoretic function was originally introduced to describe the distribution of the largest prime divisor of a random integer . 11.2.36 Deﬁnition The Dickman function is deﬁned as the unique continuous solution of the diﬀerence-diﬀerential equation ρ(u) = 1 (0 ≤u ≤1), uρ ′(u) = −ρ(u −1) (u > 1). 11.2.37 Theorem The distribution of the largest degree D n satisﬁes for all x ∈(0, 1): lim n→∞Pr{D n ≤x} = F1(x), where F1(x) = ρ(1/x) and ρ denotes the Dickman function. In particular, one has E(D n ) ∼ gn, where g = 0.62432 . . . is known as the Golomb-Dickman constant. 11.2.38 Remark The most complete results about D[j] n , for any ﬁxed positive integer j, are due to Gourdon ; see also . For instance, we also have E(D n ) ∼0.20958 . . . n, and E(D n ) ∼0.08831 . . . n. 11.2.39 Remark Information on the relation between the ﬁrst and second largest degree irreducible factors is used to compute the average-case analysis of the classical factorization method based on squarefree, distinct-degree, and equal-degree factorization under the “early-abort” stopping strategy . This also requires information on the joint distribution of the ﬁrst two largest degree irreducible factors. 370 Handbook of Finite Fields 11.2.40 Remark Estimates for the largest degree of the irreducible factors are related to the study of smooth polynomials that play an important role in the discrete logarithm problem in ﬁnite ﬁelds, especially in the index calculus method; see Section 11.6. 11.2.41 Deﬁnition A polynomial of degree n over Fq is m-smooth if all its irreducible factors have degree at most m. 11.2.42 Remark In the index calculus method a search is repeated until an m-smooth polynomial is found. Hence, the analysis of the index calculus method requires information on the number of polynomials that are m-smooth. The generating function for the number Nq(m; n) of monic polynomials over Fq of degree n which are m-smooth is Sm(z) = X n≥0 Nq(m; n) zn = m Y k=1 1 1 −zk Ik . For the cryptographic applications, m tends to inﬁnity with n. More precisely, we have m = √n log n/√2 log 2; see . Hence, singularity analysis does not apply since analytic continuation is not possible. Odlyzko uses the saddle point method for deriving an asymptotic estimate for the numbers Nq(m; n) as n →∞, uniformly for m in the range n1/100 ≤m ≤n99/100. (Actually, his results hold for nδ ≤m ≤n1−δ, where δ > 0.) 11.2.43 Remark A variant of the index calculus method over F2n (the Waterloo algorithm) was introduced in ; see also . It improves the running time of the method but it does not improve its asymptotic order. The running time was proven rigorously by Drmota and Panario using a bivariate saddle point analysis that follows closely Odlyzko’s estimates in . 11.2.44 Remark For related estimates for the index calculus method without using smooth poly-nomials see [1212, 1213]. The fastest variant of the index calculus method for F2n is still due to Coppersmith ; see Section 11.6 for more details and references. 11.2.45 Remark We focus now on the smallest degrees of the irreducible factors of a polynomial. These estimates are useful, for example, to analyze Ben-Or’s irreducible test; see Section 11.3. This analysis requires the study of the probability that a random polynomial of degree n contains no irreducible factors of degree up to a certain value m (such polynomials are sometimes called m-rough) and are related to the Buchstab function. 11.2.46 Deﬁnition The Buchstab function is the unique continuous solution of the diﬀerence-diﬀerential equation uω(u) = 1 1 ≤u ≤2, (uω(u)) ′ = ω(u −1) u > 2. 11.2.47 Remark This function was introduced by Buchstab when studying the analogous problem for integer numbers, that is, numbers with no small prime factors. This function Algorithms 371 has been largely studied . It is known that it tends quickly to e−γ = 0.56416 . . ., where γ is Euler’s constant. 11.2.48 Remark Car gives estimates for m-roughness that depend on the Buchstab function for m large with respect to n, say m > c1 n log log n/ log n. Gao and Panario show that for m small with respect to n, say m < c2 log n, the estimate e−γ/m holds; see also . 11.2.49 Remark The study of the probability that a random polynomial is m-rough for the complete range 1 ≤m ≤n, is given by Panario and Richmond . The estimates are in terms of the Buchstab function when m →∞. When m is ﬁxed Flajolet and Odlyzko singularity analysis is applied. 11.2.50 Theorem The smallest degree Sn among the irreducible factors of a random poly-nomial of degree n over Fq satisﬁes Pr(Sn ≥m) = 1 m ω n m + O max 1 m2 , log n mn , when m tends to inﬁnity with n. 11.2.51 Remark Using Theorem 11.2.50 it is not diﬃcult to prove that the expected smallest degree among the irreducible factors of a random polynomial is asymptotic to e−γ log n . More generally, the expected r-th smallest degree among the irreducible factors of a random polynomial is asymptotic to e−γ logr n/r!. 11.2.3.4 Greatest common divisor of polynomials 11.2.52 Remark As it is pointed out in Remark 11.2.10 two polynomials over Fq are coprime with probability 1 −1/q. Much more can be said about the distribution of the degrees of the irreducible factors in the gcd of several polynomials. Indeed, the limiting distribution of a random variable counting the total degree of the greatest common divisor of two or more random univariate polynomials over the ﬁnite ﬁeld Fq is geometric, and the distributions of random variables counting the number of common factors (with and without repetitions) are very close to Poisson distributions when q is large. The main reference for these results, from where we extract a couple of main theorems, is . For simplicity we state the results for two polynomials but they immediately generalize to several polynomials. 11.2.53 Theorem Let us consider two polynomials over Fq of degrees n1 and n2, respectively, and the random variables Zd(n1, n2) for the number of distinct irreducible factors in the gcd, Zr(n1, n2) for the number of irreducible factors in the gcd counting repetitions, and Zt(n1, n2) for the total degree of the gcd of the two polynomials. Then, as n1 →∞and n2 →∞and for I(z) the generating function of irreducible polynomials, we have 1. the probability generating function for Zd is PD(u) = exp  − X m≥1 (1 −u)m m I q−2m  ; 372 Handbook of Finite Fields 2. the probability generating function for Zr is PR(u) = exp  X m≥1 um −1 m I(q−2m)  ; 3. the probability generating function for Zt is PT(u) = q −1 q −u. 11.2.54 Remark As a corollary we obtain, for example, the probability that the gcd has zero (coprime polynomials), one or two irreducible factors: P(Zr = 0) = 1 −1/q, P(Zr = 1) = (1 −1/q)I(1/q2), and P(Zr = 2) = (1/2)(1 −1/q) I2(1/q2) + I(1/q4) . We also obtain, from the total degree results, that the probability that the gcd has degree k is asymptotic to q−k(1 −q−1) as the degrees grow. 11.2.55 Remark Exact values of these probabilities can be computed for small values of the degrees n1 and n2 and of ﬁeld size q. For tables of probabilities for a few common irreducible factors (counted with, or without repetitions), and for the mean and variance of Zd and Zr for small values of q, see . 11.2.3.5 Relations to permutations and integers 11.2.56 Remark There are several analogies among the irreducible decomposition of polynomi-als over ﬁnite ﬁelds, the prime decomposition of integers, and the cycle decomposition of permutations. We exemplify below with several speciﬁc results and then give a heuristic argument to justify these analogies. 11.2.57 Remark We focus ﬁrst on the splitting degree of a random polynomial of degree n over Fq. Let λ be a partition of n (denoted λ ⊢n) and write λ in the form 1k12k2 · · · nkn where λ has ks parts of size s. We say that a polynomial is of shape λ if it has ks irreducible factors of degree s for each s, 1 ≤s ≤n. Let w(λ, q) be the proportion of polynomials of degree n over Fq which have shape λ. Let m(λ) be the least common multiple of the sizes of the parts of λ. Then the degree of the splitting ﬁeld over Fq of a polynomial of shape λ is m(λ). The average degree of a splitting ﬁeld is then given by En(q) := X λ⊢n w(λ, q)m(λ). 11.2.58 Remark Consider the following classes of polynomials: 1. M1(q): the class of all monic polynomials over Fq. In this class the number of polynomials of degree n is qn. 2. M2(q): the class of all monic square-free polynomials over Fq. In this class the number of polynomials of degree n is (1 −q−1)qn. Algorithms 373 3. M3(q): the class of all monic square-free polynomials over Fq whose irreducible factors have distinct degrees. In this class the number of polynomials of degree n is a(n, q)qn where, a(n, q) →a(q) := Q k≥1(1 + Ikq−k) exp(−1/k) as n →∞. Let us deﬁne, for x > 0, Φn(x) := λ ⊢n: log m(λ) −1 2(log n)2 > x √ 3(log n)3/2 . 11.2.59 Theorem Fix one of the classes Mi(q) described above. For each λ ⊢n, let w(λ, q) denote the proportion of polynomials in this class whose factorizations have shape λ. Then there exists a constant c0 > 0 (independent of the class) such that for each x ≥1 there exists n0(x) such that X λ∈Φn(x) wi(λ, q) ≤c0e−x/4 for all q and all n ≥n0(x). In particular, almost all polynomials of degree n over Fq in Mi(q) have splitting ﬁelds of degree exp(( 1 2 + o(1))(log n)2), as n →∞. 11.2.60 Theorem In each of the classes described above the average degree En(q) of a splitting ﬁeld of a polynomial of degree n in that class satisﬁes log En(q) = C r n log n + O √n log log n log n , uniformly in q, and for C = 2.99047... an explicitly deﬁned constant. 11.2.61 Remark The constant C in the above theorem was obtained by Goh and Schmutz and Stong in their study of the analogous problem in the symmetric group Sn. A permutation in Sn is of type λ = 1k12k2 · · · nkn if it has exactly ks cycles of length s for each s, and its order is then equal to m(λ). If w(λ) denotes the proportion of permutations in Sn which are of type λ, then the average order of a permutation in Sn is equal to En := P λ⊢n w(λ)m(λ). We can think of m(λ) as a random variable where λ ranges over the partitions of n and the probability of λ is w(λ). Properties of the random variable m(λ) (and related random variables) under the distribution w(λ) have been studied by Erd¨ os and Tur´ an [983, 984, 985, 986]. In particular, the distribution of log m(λ) is approximated by a normal distribution with mean 1 2(log n)2 and variance 1 3(log n)3 in a precise sense . 11.2.62 Remark There is a general heuristic that Flajolet, Gourdon, and Panario call the permutation model. Probabilistic properties of the decomposition of polynomials into irre-ducible factors are expected to have a shape resembling (as q →∞) that of the correspond-ing properties of the cycle decomposition of permutations. As the cardinality q of the ﬁnite ﬁeld Fq goes to inﬁnity (with n staying ﬁxed!), the joint distribution of the degrees of the irreducible factors in a random polynomial of degree n converges to the joint distribution of the lengths of cycles in a random permutation of size n. As stated in : This prop-erty is visible at the generating functions level when any generating function of polynomials taken at z/q converges (as q →∞) to the corresponding exponential generating function of permutations. For example, the generating function of monic polynomials, when normalized with the change of variable z 7→z/q, is the exponential generating function of permutations: P z q = 1 1 −z = ∞ X n=1 n!zn n! . 374 Handbook of Finite Fields Similarly, we have I z q → (q→∞) log 1 1 −z = ∞ X n=1 (n −1)!zn n! , the exponential generating function for the number of cycles in permutations. 11.2.63 Remark Several of the results for polynomials and permutations have been generalized to problems in the exp-log combinatorial class; see for instance [1084, 1085, 1341, 1342, 2351]. The exp-log class includes problems such as 2-regular graphs, several types of permutations, random mappings (functional digraphs), polynomials over ﬁnite ﬁelds, random mappings patterns for unlabelled objects, and arithmetical semigroups. 11.2.64 Remark Relations between the irreducible decomposition of polynomials over ﬁnite ﬁelds and the prime decomposition of integers are discussed in detail in Section 13.1. See Also §3.1 For formulas for self-reciprocal polynomials. §3.5 For formulas for irreducible polynomials with prescribed coeﬃcients. §11.3 For irreducible test algorithms. §11.4 For polynomial factorization algorithms. §11.6 For the index calculus method for discrete logarithms. §13.1 For relations between polynomials over ﬁnite ﬁelds and integers. References Cited: [231, 306, 442, 509, 510, 537, 670, 665, 717, 901, 921, 983, 984, 985, 986, 1080, 1081, 1083, 1084, 1085, 1187, 1189, 1190, 1212, 1213, 1226, 1235, 1239, 1290, 1341, 1342, 1564, 1667, 1760, 1761, 1762, 1763, 1765, 1939, 2306, 2307, 2347, 2349, 2350, 2351, 2352, 2724, 2788, 2833, 2983, 3083] 11.3 Algorithms for irreducibility testing and for construct-ing irreducible polynomials Mark Giesbrecht, University of Waterloo 11.3.1 Introduction We consider the problem of testing polynomials for irreducibility and constructing irre-ducible polynomials of prescribed degree. Such constructions are essential in many compu-tations with ﬁnite ﬁelds, including cryptosystems, error-correcting codes, random number generators, combinatorial designs, complexity theory, and many other mathematical com-putations. In particular, they allow us to construct ﬁnite ﬁelds of speciﬁed order. We conﬁne ourselves to univariate polynomials in this section. Factorization algorithms and irreducibil-ity tests for multivariate polynomials are discussed in Section 11.5. Algorithms 375 11.3.1 Deﬁnition The following notation will be used in the analyses of algorithms. M(n): N →N is deﬁned such that two polynomials over a ﬁeld F of degree at most n can be multiplied with O(M(n)) operations in F. M(n) = n2 using the “school” method, while M(n) = n log n log log n for FFT-based methods . MM(n): N →N is deﬁned such that two n × n matrices over a ﬁeld F can be multiplied with O(MM(n)) operations in F. Using the standard algorithm, MM(n) = n3 and MM(n) = n2.3727 for the best currently known algorithm . For f, g: R →R, f = O˜(g) if f = O(g(log \|g\|)c) for some absolute constant c ≥0. 11.3.2 Early irreducibility tests of univariate polynomials 11.3.2 Remark An essential early construction for certifying irreducibility, as well as for factoring, was established by Petr in 1937 , though it was not presented in algorithmic terms. 11.3.3 Deﬁnition Let f ∈Fq[x] be a squarefree polynomial of degree n over a ﬁnite ﬁeld Fq. The Petr/Berklekamp matrix Q ∈Fn×n q of f is deﬁned such that xiq ≡ X 0≤j<n Qijxj (mod f). This is the matrix representation of the Frobenius map (a 7→aq (mod f)) on the basis ⟨1, x, x2, . . . , xn−1⟩for Fq[x]/(f). 11.3.4 Theorem [2391, 2567, 2570] Let f = f e1 1 · · · f ek k ∈Fq[x], for irreducible f1, . . . , fk ∈Fq[x], have Petr/Berlekamp matrix Q ∈Fn×n q . The characteristic polynomial det(Q −λI) ∈Fq[x] of Q satisﬁes det(Q −λI) = (−1)n  Y 1≤i≤k (λei −1)  · λ(e1−1)···(ek−1). 11.3.5 Remark Theorem 11.3.4 could, in principle, have been cast as an algorithm for testing irre-ducibility with the technology of the day, using the method for computing the characteristic polynomial of a matrix by Danilevsky from 1937. The characteristic polynomial of Q has the form xn −1 for an irreducible polynomial. 11.3.6 Remark In 1954 Butler presented an explicit method for determining the number of irreducible factors of a polynomial based on the following theorem. 11.3.7 Theorem Let f ∈Fq[x] have degree n, with Petr/Berlekamp matrix Q ∈Fn×n q . Then rank(Q−I) = n−k, where k is the number of distinct irreducible factors of f. A squarefree f is irreducible if and only if rank(Q −I) = n −1. 11.3.8 Remark The Petr/Butler approach is developed into a complete algorithm for polynomial factorization by Berlekamp in 1967. While the cost of Butler’s method was not analyzed in the modern sense, it is straightforward that constructing Q and taking the rank of Q −I would cost O(M(n)(n + log q) + MM(n)) or O˜(n log q + MM(n)) operations in Fq. 376 Handbook of Finite Fields 11.3.3 Rabin’s irreducibility test 11.3.9 Remark In 1980, Rabin based a more eﬃcient test around the following theorem. It was originally presented for polynomials over prime ﬁelds, but works for polynomials over any ﬁnite ﬁeld Fq. 11.3.10 Theorem A polynomial f ∈Fq[x] of degree n ≥1 is irreducible if and only if 1. f \| xqn −x, and 2. gcd(xqm −x, f) = 1 for all divisors m of n. 11.3.11 Algorithm: Rabin’s Irreducibility Test Input: f ∈Fq[x] of degree n Output: “Irreducible” if f is irreducible in Fq[x]; “Reducible” otherwise 1. If (xqn rem f) ̸= x then return “Reducible” 2. For all prime factors d of n do 3. h ←xqn/d rem f 4. If gcd(h −x, f) ̸= 1 return “Reducible” 5. Return “Irreducible” 11.3.12 Remark For h ∈Fq[x], h rem f is deﬁned as the unique polynomial of degree less than that of f which is equivalent to h modulo f. 11.3.13 Remark Using repeated squaring to compute powers (see Section 11.1), and the fact that n has at most log2 n prime factors, Rabin provides a cost estimate of O(n2 log2(n) log log(n) log(q)) operations in Fq to test irreducibility of a polynomial. A similar, alternative algorithm based on the Petr/Berlekamp matrix is presented by Calmet and Loos . 11.3.14 Remark Panario and Gao show that a slight variant of Rabin’s algorithm requires O(nM(n) log q + M(n) log2 n) operations in Fq, deterministically. 11.3.15 Remark The fast modular composition algorithm of von zur Gathen and Shoup allows us to compute xqm from xq using O((MM(n1/2)n1/2 + n1/2M(n)) log m) operations in Fq. Employing this in Rabin’s algorithm gives an algorithm for testing irreducibility which requires O(M(n) log q + (MM(n1/2)n1/2 + n1/2M(n) log2 n)) or O˜(MM(n1/2)n1/2 + n log q) operations in Fq. 11.3.16 Remark The algorithm for modular composition of Kedlaya and Umans requires n1+o(1) · (log q)1+o(1) bit operations, for prime q. This yields an algorithm for testing irre-ducibility which requires n1+o(1) · (log q)1+o(1) bit operations. Note that this algorithm is not in the algebraic model of all operations in Fq, and counts bit operations instead. 11.3.17 Remark A more precise average-case analysis of the cost of Rabin’s algorithm is given by Panario et al. [2357, 2349]. This involves the study of the probability that a polynomial has an irreducible factor of degree dividing a maximal divisor of n. They give an exact expression for this probability when n is prime or a product of two primes, as well as an asymptotic analysis. They also present analyses of variants of Rabin’s algorithm suggested in [1187, 1227]. 11.3.18 Remark Any modern factoring algorithm can be used to test irreducibility, possibly at the cost of randomization. Asymptotically, the fastest currently known algorithms are those of von zur Gathen and Shoup and Kaltofen and Shoup , with improvements in Algorithms 377 the bit complexity model by Kedlaya and Umans . These are slower than the above irreducibility tests; see Section 11.4. 11.3.4 Constructing irreducible polynomials: randomized algorithms 11.3.19 Remark Density estimates for irreducible polynomials easily reduce the problem of ﬁnding irreducible polynomials to that of certifying a polynomial is irreducible, assuming we have a way to generate random ﬁeld elements. Such an approach was even suggested by Galois in 1830 . However, one can do considerably better than the most na¨ ıve reduction, and both the asymptotic complexity of this problem and more practical concerns hold considerable interest. 11.3.20 Remark Gauss gives an explicit formula for the number Iq(n) of irreducible polynomials of degree n, from which it is easily derived that qn/(2n) < I(n, q) < qn/n (see , Exercises 3.26 and 3.27). Thus, given a way to randomly generate elements of Fq, any algorithm for testing irreducibility also yields a probabilistic algorithm for ﬁnding irreducible polynomials of any speciﬁed degree. The expected number of operations is n times the cost of the chosen irreducibility test. 11.3.5 Ben-Or’s algorithm for construction of irreducible polynomials 11.3.21 Remark In 1981, Ben-Or described a simple probabilistic algorithm for constructing an irreducible polynomial with a better expected cost than the straightforward approach suggested in Remark 11.3.20. 11.3.22 Algorithm: Ben-Or’s irreducible polynomial constructor Input: n ∈Z>0 Output: A uniformly random monic irreducible polynomial of degree n in Fq[x] 1. Randomly choose a monic f ∈Fq[x] of degree n 2. For i from 1 to ⌊n/2⌋do 3. If gcd(xqi −x, f) ̸= 1 goto Step 1. 4. Return f 11.3.23 Theorem . The expected value of the degree of the smallest factor of a randomly and uniformly chosen polynomial of degree n in Fq[x] is O(log n). 11.3.24 Theorem . Using Ben-Or’s algorithm we can construct an irreducible polynomial of degree n over Fq with an expected number of O(nM(n) log n log(nq)) or O˜(n2 log q) oper-ations in Fq. 11.3.25 Remark A more precise analysis is provided by Panario and Gao , who consider the probability that a polynomial is rough, i.e., has all irreducible factors of degree greater than some bound. Reducible polynomials with no small degree factors cause Ben-Or’s algorithm to perform more operations. 11.3.26 Theorem Let P I q (n, m) be the probability that a polynomial in Fq[x] of degree n is irreducible if it has no factors less than or equal to m = O(log n). Then P I q (n, m) ∼eγm/n as n, m, and q approach inﬁnity, where γ is Euler’s constant. 11.3.27 Remark Panario and Richmond give a much more precise average-case analysis of Ben-Or’s algorithm. They study the probability that the smallest degree irreducible factor of a degree n polynomial is greater than some m. As well, they study the expectation and 378 Handbook of Finite Fields variance of the smallest degree among the irreducible factors of a random polynomial of degree n. Results are stated in terms of the Buchstab function, a classical number-theoretic function used in the study of numbers with no prime factors smaller than some bound; see for deﬁnitions and details. They also show that the expected average-case cost of an irreducibility test in Ben-Or’s algorithm is O(M(n) log(n)(log n + log q)) operations in Fq, and provide an asymptotic analysis with explicit constants for this complexity. The expected cost of actually ﬁnding an irreducible polynomial is n times this cost. 11.3.28 Remark Von zur Gathen and Gerhard , Section 14.9 suggest that a reasonable ap-proach to ﬁnding irreducible polynomials might be a hybrid of Ben-Or’s algorithm, with a switch to the irreducibility test of Rabin when testing for factors of degrees over some prescribed bound such as log2 n. 11.3.6 Shoup’s algorithm for construction of irreducible polynomials 11.3.29 Remark Shoup gives an asymptotically faster probabilistic algorithm for constructing an irreducible polynomial of degree n, which requires O((n2 log n + n log q) log n log log n) operations in Fq. It also has the beneﬁt of reducing use of randomness to O(n) random elements from Fq (the algorithms described above all require an expected O(n2) random elements). Shoup’s algorithm is very diﬀerent from the Rabin’s and Ben-Or’s, and more closely resembles the deterministic constructions discussed in Section 11.3.7 below. The algorithm proceeds by looking at each prime power re dividing n (where r is prime). There are a number of special cases. In particular, if r is the characteristic of Fq, then Artin-Schreier type polynomials can be employed (see Remark 11.3.33 below), and when r = 2 a simple construction suﬃces. In the remaining cases, for each prime-power-factor re of n, it ﬁrst factors the r-th cyclotomic polynomial Φr ∈Fq[x]. This can be done quickly. It then ﬁnds an r-th power non-residue in ξ ∈Fq[θ], where θ is an adjoined root of Φr; xre −ξ is irreducible of degree re in Fq[θ][x], from which an irreducible polynomial of degree re over Fq[x] is constructed via traces. The good asymptotic cost is maintained through a fast algorithm for ﬁnding minimal polynomials in algebraic extensions, employing a very elegant use of the “Transposition Principle” (also known as Tellegen’s theorem). Despite the intricate construction, the polynomial produced is uniformly selected from the set of all irreducible polynomials of degree n in Fq[x]. 11.3.30 Remark An algorithm for ﬁnding irreducible polynomials of degree n, along the lines of Shoup’s algorithm above, is exhibited by Couveignes and Lercier and has an expected cost of n1+o(1)(log q)5+o(1). The key innovation is the fast construction of irreducible polyno-mials (for all but some exceptional degrees) using isogenies of a random elliptic curve. This is combined with the recent fast modular composition algorithm of Kedlaya and Umans , along with the fast algorithm of Bostan et al. to compute “composed sums” of polynomials. 11.3.7 Constructing irreducible polynomials: deterministic algorithms 11.3.31 Remark Deterministic algorithms for constructing univariate polynomials of speciﬁed de-gree over a ﬁnite ﬁeld of characteristic p are considerably more diﬃcult. The goal is al-gorithms which require time polynomial in n and log p. Given such an algorithm it is straightforward to construct an irreducible polynomial over an extension ﬁeld Fq of Fp, at least up to polynomial time in n and log q. We thus concern ourselves with constructing irreducible polynomials over prime ﬁelds Fp. A more eﬃcient algorithm for non-prime ﬁelds is established by Shoup . Algorithms 379 11.3.32 Remark The algorithms of Chistov from 1984 and Semaev are the ﬁrst to require time (np)O(1), and hence are eﬀective over small ﬁnite ﬁelds; see also . 11.3.33 Remark In 1986 Adleman and Lenstra exhibited an algorithm for constructing irre-ducible polynomials which runs in polynomial time in n and log p, assuming that the Ex-tended Riemann Hypothesis is true. Their algorithm proceeds by ﬁrst factoring n = pen0, where gcd(n0, p) = 1. An extension of degree n0 of Fp is built, and then an extension of degree pe is constructed on top of that. A minimal polynomial of a generating element is irreducible of degree n. The degree n0 extension requires ﬁnding a prime r ≡1 (mod n0) such that p is inert in the unique subﬁeld K of the r-th cyclotomic ﬁeld such that [K : Q] = n. From this an extension of Fp of degree n0 is constructed via Gauss periods; see Section 5.3. To ﬁnd the inert primes, they simply test numbers of the form n0k + 1 for k = 1, 2, . . . in sequence for primality and inertness. The Extended Riemann Hypothesis guarantees that a k = O(n4 0(log np)2) exists, but without this assumption (or randomization) there is no way known to ﬁnd such a p. To ﬁnd an extension of degree pe, Artin-Schreier polynomials of the form xp −x −α ∈Fq[x] are employed, which are irreducible when α ̸= βp −β for some β ∈K, where K is any extension of Fp. A tower of ﬁelds is constructed, starting with Fpn0 , and ending with Fpn = Fpn0pe . 11.3.34 Remark Evdokimov , in 1986, independently presented a similar result to Adelman and Lenstra’s. In this construction n is factored as a product re1 1 · · · reℓ ℓ for distinct primes ri, and an irreducible polynomial of each degree rei i is constructed. When ri = p the Artin-Schreier polynomials are again employed. When ri ̸= p he employs a reducibility theorem for the polynomial xr ei i −a (where a lies in a small cyclotomic extension) which requires ﬁnding ri-th non-residues, small examples of which are ensured by the Extended Riemann Hypothesis. Irreducible polynomials of diﬀerent prime power degrees are combined by com-puting the minimal polynomial of the sum of the roots of the two polynomials. This is easily accomplished with basic linear algebra, though the “composed sum” algorithm of can now accomplish this in quasi-linear time; see Remark 11.3.30 above. 11.3.35 Remark In 1989 Shoup [2624, 2625] gave the currently fastest completely unconditional deterministic algorithm for ﬁnding irreducible polynomials, along similar lines to . He demonstrates a polynomial-time (in n and log p) reduction from the problem of ﬁnding irreducible polynomials of degree n to the problem of factoring polynomials in Fp[x]. Shoup shows that if, for every prime r \| n, we are given a representation of a splitting ﬁeld K of xr −1 (i.e., a non-trivial irreducible factor of xr −1), and an r-th nonresidue in K, then we can construct an irreducible polynomial of degree n. Both ﬁnding the factor of xr −1 and the non-residue (by repeatedly taking r-th roots in K) can be accomplished through factorization. Shoup [2624, 2625] also provides the fastest (unconditional) deterministic algorithm for factoring polynomials over Fp[x]. Employing this in his irreducible polynomial construction method yields a deterministic algorithm requiring O(p1/2(log p)3n3(log n)c + (log p)2n4(log n)c) or O˜(p1/2n3 + n4) operations in Fp, for some absolute constant c > 0. 11.3.8 Construction of irreducible polynomials of approximate degree 11.3.36 Remark If the requirement that the degree of the constructed polynomial in Fp[x] be exactly n is relaxed, and only an approximation of this degree is required, some unconditional algorithms are known. 11.3.37 Remark In 1986, von zur Gathen gave a polynomial-time algorithm (in n and log p) which ﬁnds an irreducible polynomial of degree at least n, assuming we can do preprocessing only with respect to p (and requiring time polynomial in p). 380 Handbook of Finite Fields 11.3.38 Remark Adleman and Lenstra present an algorithm which outputs an irreducible polynomial f ∈Fp[x] with n/(c log p) < deg f ≤n, for some absolute constant c, and requires time polynomial in n and log p. 11.3.39 Remark For a ﬁxed prime p and ˜ n ∈N, Shparlinski [2639, Theorem 1], gives a deter-ministic, unconditional algorithm that ﬁnds an irreducible polynomial of degree ˜ n · (1 + exp(−(log log ˜ n)1/2−o(1))), and requires time polynomial in p and log ˜ n. 11.3.40 Remark For a suﬃciently large ˜ q, Shparlinski [2639, Theorem 2], provides a deterministic, unconditional algorithm that constructs a prime p, an integer n ≥0, and an irreducible polynomial f ∈Fp[x] of degree n, such that pn = ˜ q + o(˜ q). The algorithm requires time polynomial in log ˜ q; see also [2636, Section 2]. See Also [1227, Section 14.9] For an exposition of Rabin’s algorithm for testing irreducible polynomials, and Ben-Or’s algorithm for generating irreducible polynomials, and their complexity analyses. [2636, Section 2] For early deterministic algorithms for constructing irreducible polynomials; see also the introduction of . References Cited: [14, 223, 230, 363, 468, 482, 498, 621, 747, 768, 1017, 1168, 1187, 1219, 1227, 1239, 1667, 1722, 1939, 2349, 2350, 2357, 2391, 2434, 2567, 2570, 2580, 2624, 2625, 2629, 2636, 2639, 2856, 2985] 11.4 Factorization of univariate polynomials Joachim von zur Gathen, Universit¨ at Bonn 11.4.1 Remark The pioneering algorithms for the factorization of univariate polynomials over a ﬁnite ﬁeld are due to Berlekamp [230, 232]. Subsequent improvements of the asymptotic running time were presented by Cantor and Zassenhaus , von zur Gathen and Shoup , Huang and Pan , and Kaltofen and Shoup . The currently fastest method yields the following main result of this section. 11.4.2 Theorem The factorization of a univariate polynomial of degree n over Fq can be computed with an expected number (n log q)1+o(1) · (n1/2 + log q) of bit operations. 11.4.3 Remark Ignoring asymptotically small factors, this running time corresponds to n3/2 + n log q ﬁeld operations. These methods rely on a long line of algorithmic developments. The most basic nontrivial task is that of multiplying polynomials. At degree n, its cost M(n) is discussed in Deﬁnition 11.3.1. The current record of F¨ urer for integer multiplica-tion stands at n log n 2O(log∗n) bit operations, where log∗n is the smallest k for which the k-fold iterated logarithm log log · · · log n is less than 1. F¨ urer’s algorithm can be adapted to multiplication of polynomials. For a k-bit prime p and two polynomials in Fp[x] of degree at most n it takes O(M(n(k + log n))) bit operations. Next come division with remainder at O(M(n)) (Sieveking , Brent and Kung ), and univariate gcd, multipoint eval-Algorithms 381 uation, and interpolation at O(M(n) log n) ﬁeld operations (Strassen ). Further tasks include squarefree factorization, with O(M(n) log n) operations (Yun ). Furthermore, the matrix multiplication exponent ω is such that MM(n) ≤nω; see Deﬁnition 11.3.1. It appears in an algorithm of Brent and Kung for modular composition, which computes f ◦g (mod h) from f, g, and h with O(n1.687) operations. Multivariate multipoint evaluation asks for the evaluation of a multivariate polynomial, of degree at most d in each of its r variables, at n points in Fr q. One can achieve this with O(d(ω+1)(r−1)+1) ﬁeld operations (N¨ usken and Ziegler ). A major advance of Kedlaya and Umans , at the heart of their subsequent results, was a modular approach to this problem. They consider it as a problem over the integers, say for a prime ﬁeld Fq, and solve it in a standard modular fashion, modulo various small primes. This leads to a method using r(dr + qr + n)(log q)o(1) bit operations. It also yields a univariate modular composition method with (n log q)1+o(1) bit operations, roughly corresponding to only O(n) ﬁeld operations. The more eﬃcient methods use a “polynomial representation of the Frobe-nius automorphism,” suggested by Erich Kaltofen, fast algorithms for it, and an interval blocking strategy. This appeared ﬁrst in von zur Gathen and Shoup . 11.4.4 Remark Many other works have contributed to the factoring problem. We only mention Serret , Arwin , Petr , Kempfert , Niederreiter [2249, 2250], Kaltofen and Lobo , Gao and von zur Gathen , Kaltofen and Shoup , von zur Gathen and Gerhard . 11.4.5 Remark Berlekamp presented a deterministic algorithm to factor in Fp[x] with O(nω+ pn2+o(1)) ﬁeld operations; see for this estimate. It was improved by Shoup to O(p1/2 log2 p + log p · n2+o(1)) operations. Berlekamp introduced the fundamental idea of probabilistic algorithms, although it became widely accepted in computer science only after the polynomial-time primality test of Solovay and Strassen . The factorization algorithms mentioned are all probabilistic, in the Las Vegas sense where the output can be veriﬁed and thus guaranteed to be correct, but the running time is a random variable (with exponentially decaying tails). Removing randomness while conserving polynomial time is still an open question, of great theoretical interest but presumably no practical import. Already Berlekamp observed that it boils down to the following. 11.4.6 Open Question Given the coeﬃcients of a polynomial which is a product of linear factors over a prime ﬁeld Fp, can one ﬁnd a nontrivial factor deterministically in polynomial time? 11.4.7 Remark Several papers provide steps in this direction, often assuming the Extended Rie-mann Hypothesis: Moenck , Adleman, Manders, and Miller , Huang [1552, 1553], von zur Gathen , R´ onyai [2476, 2477, 2478, 2479], Mignotte and Schnorr , Ev-dokimov [1019, 1018], Bach, von zur Gathen, H. Lenstra , R´ onyai and Sz´ ant´ o , Gao , Ivanyos, Karpinski, and Saxena , Ivanyos, Karpinski, R´ onyai, and Saxena , Arora, Ivanyos, Karpinski, and Saxena . 11.4.8 Remark Computations just using ﬁeld operations can be modeled by arithmetic circuits (Strassen ). All algorithms discussed here, except those of Kedlaya and Umans and F¨ urer’s multiplication, are of this type. Their running time is Ω(n log q) ﬁeld operations. 11.4.9 Open Question Do all arithmetic circuits over Fq that factor univariate polynomials of degree n require Ω(n log q) arithmetic operations? 11.4.10 Remark A positive answer is known only for n = 2 (von zur Gathen and Seroussi ). 11.4.11 Remark The survey articles of Kaltofen [1646, 1655], von zur Gathen and Panario and the textbooks by Shparlinski [2637, 2641] and von zur Gathen and Gerhard present the details of most of these algorithms, more references, and historical information. 382 Handbook of Finite Fields It is intriguing that the basics of most modern algorithms go back to Legendre, Gauß, and Galois. 11.4.12 Remark The average cost of factoring algorithms, for polynomials picked uniformly at random from those of a ﬁxed degree, is analyzed in Flajolet, Gourdon, and Panario and von zur Gathen, Panario, and Richmond . See Also §11.3 For univariate irreducibility testing. §11.5 For multivariate factorization methods. References Cited: [15, 132, 137, 158, 230, 232, 401, 498, 499, 723, 1018, 1019, 1080, 1148, 1176, 1178, 1220, 1225, 1226, 1227, 1234, 1235, 1237, 1239, 1552, 1553, 1554, 1578, 1579, 1646, 1655, 1663, 1667, 1684, 1722, 1726, 2094, 2112, 2249, 2250, 2301, 2391, 2476, 2477, 2478, 2479, 2480, 2558, 2559, 2600, 2624, 2637, 2641, 2664, 2694, 2732, 2734, 2984, 3051] 11.5 Factorization of multivariate polynomials Erich Kaltofen, North Carolina State University Gr´ egoire Lecerf, CNRS & ´ Ecole polytechnique We extend the univariate factorization techniques of the previous section to several variables. Two major ingredients are the reduction from the bivariate case to the univariate one, and the reduction from any number to two variables. We present most of the known techniques according to the representation of the input polynomial. 11.5.1 Factoring dense multivariate polynomials 11.5.1 Remark In this subsection we are concerned with diﬀerent kinds of factorizations of a multivariate polynomial f ∈Fq[x1, . . . , xn] stored in dense representation. 11.5.2 Deﬁnition Let R be any ring. A dense representation of a polynomial f ∈R[x1, . . . , xn] is the data of the vector (d1, . . . , dn) of the partial degrees of f, and the vector of the coeﬃcients of the monomials xe1 1 · · · xen n for all 0 ≤e1 ≤d1,. . . , 0 ≤en ≤dn, sorted in reverse lexicographical ordering on the exponents (e1, . . . , en), which means that (e1, . . . , en) < (e′ 1, . . . , e′ n) if, and only if, there exists j such that (en = e′ n, . . . , ej+1 = e′ j+1, and ej < e′ j). 11.5.3 Remark The representation of multivariate polynomials is an important issue, which has been discussed from the early ages of computer algebra [761, 778, 1519, 1616, 2129, 2130, 2131, 2730, 3021]. 11.5.1.1 Separable factorization 11.5.4 Remark Separable factorization can be seen as a preprocess to the other factorizations (squarefree, irreducible, and absolutely irreducible, as deﬁned below), which allows to reduce to considering separable polynomials. Algorithms 383 11.5.5 Deﬁnition Let R be an integral domain. A polynomial f ∈R[x] is primitive if the common divisors in R of all the coeﬃcients of f are invertible in R. 11.5.6 Deﬁnition Let R be a unique factorization domain of characteristic p, and let Ep represent {1} if p = 0 and {1, p, p2, p3, . . .} otherwise. If f is a primitive polynomial in R[y] of degree d ≥1, then the separable decomposition of f, written Sep(f), is deﬁned to be the set Sep(f) := {(f1, q1, m1), . . . , (fs, qs, ms)} ⊆(R[y] \ R) × Ep × N, satisfying the following properties: 1. f(y) = Qs i=1 fi(yqi)mi, 2. for all i ̸= j in {1, . . . , s}, fi(yqi) and fj(yqj) are coprime, 3. for all i ∈{1, . . . , s}, mi (mod p) ̸= 0, 4. for all i ∈{1, . . . , s}, fi is separable and primitive, 5. for all i ̸= j in {1, . . . , s}, (qi, mi) ̸= (qj, mj). The process of computing the separable decomposition is the separable factorization. 11.5.7 Example With R := F3 and f := y2(y + 1)3(y + 2)4 = y9 + 2y8 + 2y3 + y2, we have that Sep(f) = {(y, 1, 2), (y + 1, 3, 1), (y + 2, 1, 4)}. 11.5.8 Example With R := F3[x] and f := (y + 2x)7(y3 + 2x)3(y6 + x), we have that Sep(f) = {(y + 2x, 1, 7), (y + 2x3, 9, 1), (y2 + x, 3, 1)}. 11.5.9 Theorem [2108, Chap. VI, Theorem 6.3] Any primitive polynomial f ∈R[y] admits a unique (up to permutations and units in R) separable decomposition, which only depends on the coeﬃcients of f. 11.5.10 Remark Roughly speaking, the separable decomposition corresponds to sorting the roots of the given polynomial according to their multiplicity. A constructive proof of Theorem 11.5.9 can be found in [2108, Chap. VI, Theorem 6.3], and another proof using the irreducible factorization in [1882, Proposition 4]. 11.5.11 Remark Since the separable decomposition only depends on the coeﬃcients of f it can be computed in any extension of R. 11.5.12 Theorem [1882, Proposition 5] If F is a ﬁeld then the separable decomposition of a poly-nomial f ∈F[y] of degree d can be computed with O(M(d) log d) arithmetic operations in F. Let us recall that M(d) represents a bound for the complexity of multiplying two poly-nomials of degree at most d with coeﬃcients in a commutative ring with unity, in terms of the number of arithmetic operations in the latter ring. 11.5.13 Theorem [1882, Propositions 8 and 9] Let R = F[x], where F is a ﬁeld, and let f ∈F[x][y] be a primitive polynomial of degree dx in x and dy in y. 1. If F has cardinality at least dx(2dy +1)+1 then Sep(f) can be computed (deter-ministically) with O(dy(dyM(dx) log dx + dxM(dy) log dy)) or ˜ O(dxd2 y) operations in F. 2. If F has cardinality at least 4dxdy then Sep(f) can be computed with an expected number of O(dyM(dx) log dx + dxM(dy) log dy)) or ˜ O(dxdy) operations in F. Let us recall that f(d) ∈˜ O(g(d)) means that f(d) ∈g(d)(log2(3 + g(d)))O(1). With the second randomized algorithm, the ouput is always correct, and the cost estimate is the average of the number of operations in F taken over all the possible executions. 384 Handbook of Finite Fields 11.5.1.2 Squarefree factorization 11.5.14 Deﬁnition If R is a unique factorization domain then the squarefree decomposition of a ∈R, written Sqr(a), is the set of pairs (am, m), where am represents the product of all the irreducible factors of a of multiplicity m. The process of computing the squarefree decomposition is the squarefree factorization. 11.5.15 Deﬁnition For convenience, we say that a polynomial f ∈R[x1, . . . , xn] is primitive (resp. separable) in xi if it is so when seen in R[x1, . . . , xi−1, xi+1, . . . , xn][xi]. 11.5.16 Algorithm (Sketch of the algorithm squarefree factorization) Input: a polynomial f ∈Fq[x1, . . . , xn], primitive in x1, . . . , xn. Output: the squarefree decomposition Sqr(f) of f. 1. First compute the separable decomposition of f seen in Fq[x1, . . . , xn−1][xn]. Then for each separable factor g of Sep(f) compute the separable de-composition of g seen in Fq[x1, . . . , xn−2, xn][xn−1]. Then for each separa-ble factor h of Sep(g) compute the separable decomposition of h seen in Fq[x1, . . . , xn−3, xn−1, xn][xn−2], etc. At the end rewrite f as the product of poly-nomials of the form fi(xqi,1 1 , . . . , xqi,n n )mi, where the fi are separable in x1, . . . , xn, and where the qi,j are powers of p. 2. The squarefree factorization of each fi(xqi,1 1 , . . . , xqi,n n )mi is simply obtained by extracting the minj∈{1,...,n} qi,j-th root of fi(xqi,1 1 , . . . , xqi,n n ). 11.5.17 Theorem [1882, Proposition 12] Let f ∈Fq[x, y] be a polynomial of degree dx in x and dy in y. If q ≥4(3dy + 1)dx then Sqr(f) can be computed with an expected number of O(dyM(dx) log dx + dxM(dy) log dy)) or ˜ O(dxdy) operations in Fq. 11.5.18 Remark Practical multivariate squarefree factorization algorithms have been designed in to be speciﬁcally eﬃcient in small and medium sizes, when M does not behave as softly linear. Algorithms for deducing the squarefree decomposition from the separable one were proposed in and then improved in in particular cases. 11.5.1.3 Bivariate irreducible factorization 11.5.19 Deﬁnition If R is a unique factorization domain then the irreducible decomposition of a ∈R, written Irr(a), is the set of pairs (ai, mi), where ai is an irreducible factor of a of multiplicity mi. The process of computing the irreducible decomposition is the irreducible factorization. 11.5.20 Deﬁnition If F is a ﬁeld then the absolutely irreducible decomposition of f ∈F[x1, . . . , xn] is the irreducible decomposition of f in ¯ F[x1, . . . , xn], where ¯ F represents the algebraic closure of F. The process of computing the absolutely irreducible decomposition is the absolutely irreducible factorization, or absolute factorization. 11.5.21 Remark We do not discuss speciﬁc algorithms for computing the absolute factoriza-tion. In fact, whenever F is a ﬁnite ﬁeld, the absolutely irreducible decomposition of f ∈F[x1, . . . , xn] can be obtained from the irreducible decomposition over the algebraic extension of F of degree deg f. For more details and advanced algorithms we refer the reader to . Algorithms 385 11.5.22 Theorem [1883, Theorem 2] Let q = pk, and let f ∈Fq[x, y] be a polynomial of degree dx in x and dy in y. If q ≥10dxdy then Irr(f) can be computed with factoring several polynomials in Fq[y] whose degree sum does not exceed dx + dy, plus an expected number of ˜ O(k(dxdy)1.5) operations in Fp. 11.5.23 Remark If q is not suﬃciently large to apply Theorem 11.5.22 then one can compute the irreducible factorization of f over a slightly larger ﬁnite ﬁeld, and then recover the factorization over Fq by computing the norm of the factors. The algorithm underlying Theorem 11.5.22 is summarized in the following. 11.5.24 Algorithm (Sketch of the lifting and recombination technique) Input: a primitive and separable polynomial f ∈Fq[x][y], of partial degrees dx in x and dy in y. Output: the irreducible decomposition Irr(f) of f. 1. Normalization. If the cardinality of Fq is suﬃciently large then a suitable shift of the variable x reduces the problem to the normalized case deﬁned as follows: deg f(0, y) = dy and Res f(0, y), ∂f ∂y (0, y) ̸= 0. 2. Univariate factorization. Compute Irr(f(0, y)) in Fq[y]. 3. Lifting. Use the classical Hensel lifting from the previously computed irreducible factors f1(0, y), . . . , fs(0, y) of f(0, y) in order to deduce the irreducible analytic decomposition f1, . . . , fs of f in Fq[y] to a certain ﬁnite precision σ in x. 4. Recombination. Discover how the latter analytic factors f1, . . . , fs recombine into the irreducible factors. 11.5.25 Remark Since any proper factor g of f is the product of a subset of the analytic factors, the precision σ = dx is suﬃcient in Algorithm 11.5.24 to discover Irr(f) by means of exhaustive search. To be precise, it suﬃces to run over all the subsets S of {1, . . . , s} of cardinality at most s/2 and test whether the truncated polynomial of Q i∈S fi to precision dx in Fq[x][y] divides f or not. This approach was originally popularized in computer algebra by Zassenhaus in in the context of factoring in Q[y] via the p-adic completion of Q. The adaptation to two and several variables was ﬁrst pioneered in [2209, 2938, 2939]. In particular, introduced coeﬃcient ﬁeld abstractions that marked the beginning of generic programming. Von zur Gathen adopted Musser’s approach to valuation rings . The cost of this approach is, of course, exponential in s. However, as proved in the cost of the recombination process behaves in softly linear time in average over ﬁnite ﬁelds, which explains the practical eﬃciency of this approach. 11.5.26 Remark For details concerning Hensel lifting, we refer the reader to [1227, Chap. 15], that contains a variant of the multifactor Hensel lifting ﬁrst designed by Shoup for his C++ library NTL ( An improvement obtained thanks to the transposition principle is proposed in . Parallelization has been studied in . 11.5.27 Remark The ﬁrst attempt to reduce the recombination stage to linear algebra seems to be due to Sasaki et al. [2528, 2529, 2530], with a method called the trace recombination. But the ﬁrst successes in the design and proofs of complete algorithms are due to van Hoeij for the factorization in Z[x], and then to Belabas et al. for F(x)[y], with the logarithmic derivative recombination method, where the precision σ = deg f(deg f −1) + 1 is shown to be suﬃcient in general. Then a precision linear in deg f in characteristic 0 or large enough characteristic has been shown to suﬃce in [365, 1880]. 386 Handbook of Finite Fields 11.5.28 Remark In , Gao designed the ﬁrst softly quadratic time probabilistic reduction of the factorization problem from two to one variable whenever the characteristic of the coeﬃcient ﬁeld is zero or suﬃciently large. His algorithm makes use of the ﬁrst algebraic de Rham cohomology group of F[x, y, 1/f(x, y)], as previously used by Ruppert [2504, 2505] for testing the absolute irreducibility. In fact, if f factors into f1 · · · fr over the algebraic closure of F then ˆ fi ∂fi ∂x f dx + ˆ fi ∂fi ∂y f dy ! i∈{1,...,r} is a basis of the latter group, where ˆ fi := f/fi [2504, Satz 2]. In consequence, this group can be obtained by searching for closed diﬀerential 1-forms with denominators f and numerators of degrees at most deg f −1, which can be easily done by solving a linear system. A nice presentation of Ruppert’s results is made in Schinzel’s book [2542, Chapter 3]. The algorithm underlying Theorem 11.5.22 makes use of these ideas in order to show that a precision σ = dx + 1 of the series in the Hensel lifting suﬃces. 11.5.1.4 Reduction from any number to two variables 11.5.29 Remark Let f ∈F[x1, . . . , xn] continue to denote a polynomial in n variables over a ﬁeld F of total degree d. For any points (α1, . . . , αn), (β1, . . . , βn) and (γ1, . . . , γn) in F n, we deﬁne the bivariate polynomial fα,β,γ in the variables x and y by fα,β,γ := f(α1x + β1y + γ1, . . . , αnx + βny + γn). 11.5.30 Theorem (Bertini’s theorem) [2602, Chapter II, Section 6.1] If f is irreducible, then there exists a proper Zariski open subset of (F n)3 such that fα,β,γ is irreducible for any triple (α1, . . . , αn), (β1, . . . , βn), (γ1, . . . , γn) in this subset. 11.5.31 Deﬁnition For any irreducible factor g of f, a triple (α1, . . . , αn), (β1, . . . , βn), (γ1, . . . , γn) in (F n)3 is a Bertinian good point for g if g(α1x + β1y + γ1, . . . , αnx + βny + γn) is irreducible with the same total degree as g. In other words, the irreducible factors of f are in one-to-one correspondence with those of fα,β,γ. The complementary set of Bertinian good points is written B(f) and is the set of Bertinian bad points. 11.5.32 Remark For algorithmic purposes, the entries of (α1, . . . , αn), (β1, . . . , βn) and (γ1, . . . , γn) must be taken in a ﬁnite subset S of F, so that we are naturally interested in upper bounding the number of Bertinian bad points in (Sn)3. We refer to such a bound as a quantitative Bertini theorem. The density of Bertinian bad points with entries in a non-empty ﬁnite subset S of F is B(f, S) := \|B(f) ∩(Sn)3\| \|S\|3n , where \|S\| represents the cardinality of S. 11.5.33 Theorem (Quantitative Bertini theorem) [1658, Corollary 2] and [1881, Corollary 8] If F is a perfect ﬁeld of characteristic p, and according to the above notation, we have that: 1. B(f, S) ≤(3d(d −1) + 1)/\|S\| if p ≥d(d −1) + 1; 2. B(f, S) ≤2d4/\|S\| otherwise. 11.5.34 Remark What we call “Bertini’s theorem” here is a particular but central case of more gen-eral theorems such as in [2602, Chapter II, Section 6.1]. As pointed out by Kaltofen , the special application of Bertini’s theorem to reduce the factorization problem from several to two variables was already known by Hilbert [1499, p. 117]. This is why Kaltofen and some Algorithms 387 authors say “(eﬀective) Hilbert Irreducibility Theorem” instead of “Bertini’s theorem.” For more historical details about Bertini’s work, we refer the reader to [1623, 1750]. 11.5.35 Remark Bertini’s theorem was introduced in complexity theory by Heintz and Sievek-ing , and Kaltofen . It quickly became a cornerstone of many randomized fac-torization or reduction techniques including [1218, 1229, 1647, 1648, 1649]. Over the ﬁeld of complex numbers, Bajaj et al. obtained the bound B(f, S) ≤(d4−2d3+d2 +d+1)/\|S\| by following Mumford’s proof [2202, Theorem 4.17] of Bertini’s theorem. Gao proved the bound B(f, S) ≤2d3/\|S\| whenever F has characteristic 0 or larger than 2d2. Then Cheze pointed out [616, Chapter 1] that the latter bound can be reﬁned to B(f, S) ≤d(d2 −1)/\|S\| by using directly [2504, Satz C]. The paper contains a version for non-perfect ﬁelds with a bound that is exponential in d. If the cardinality \|F\| is too small, one can switch to an extension (see Remark 11.5.67 below). 11.5.36 Corollary Let S(n, d) represent a cost function for the product of two power series over a ﬁeld F in n variables truncated to precision d. Let f ∈Fq[x1, . . . , xn] be a polynomial of total degree d. If q ≥4d4 then Irr(F) can be computed with an expected number of O(1) factorizations of polynomials in Fq[x, y] of total degree d, plus an expected number of ˜ O(dS(n −1, d)) operations in Fq. 11.5.37 Remark Softly optimal series products exist in particular cases , for which the fac-torization thus reduces to the univariate case in expected softly linear time as soon as n ≥3. 11.5.38 Remark The ﬁrst deterministic polynomial time multivariate factorization algorithms are due to Kaltofen [1645, 1646]. Kaltofen constructed polynomial-time reductions to bi- (in 1981) and univariate (in 1982) factorization over an abstract ﬁeld, which were discov-ered independently of the 1982 univariate factorization algorithm over the rationals by A. K. Lenstra, H. W. Lenstra, and Lov´ asz . Kaltofen’s reduction to univariate factor-ization, however, was inspired by Zassenhaus’s algorithm . For more references to work by others (Chistov, von zur Gathen, Grigoriev, A. K. Lenstra) that immediately followed, we refer the reader to Kaltofen’s surveys [1654, 1655, 1659], and to . 11.5.39 Remark Polynomial factorization over ﬁnite ﬁelds has been implemented in Maple by Bernardin and Monagan . Other practical techniques have been reported in . At the present time, the most general algorithm is due to Steel : it handles all coeﬃcient ﬁelds being explicitly ﬁnitely generated over their prime ﬁeld, and it has been implemented within the Magma computer algebra system . Steel’s algorithm actually completes and improves a previous approach investigated by Davenport and Trager . 11.5.40 Remark It is possible, via the rank of the Petr matrix or the distinct degree factorization algorithm, to count the number of irreducible factors of a univariate polynomial over a ﬁeld Fq of characteristic p in deterministic polynomial time in log p. The same remains true for multivariate polynomials [1182, 1651], but the algorithms are not straightforward. In a multivariate deterministic distinct degree factorization is presented. There “distinct degree” is with respect to any degree order. 11.5.2 Factoring sparse multivariate polynomials 11.5.41 Remark Let F be a ﬁeld. A polynomial f in F[x1, . . . , xn] is made of a sum of terms, with each term composed of a coeﬃcient and an exponent seen as a vector in Nn. For any e = (e1, . . . , en) ∈Nn, we let fe denote the coeﬃcient of the monomial xe1 1 · · · xen n in f. If a polynomial has only a few of nonzero terms in its dense representation, one prefers to use 388 Handbook of Finite Fields the following representation. 11.5.42 Deﬁnition A sparse representation of a multivariate polynomial stores the sequence of the nonzero terms as pairs of monomials and coeﬃcients, sorted for instance in reverse lexicographical order. 11.5.43 Deﬁnition The support of f is Supp(f) := {e ∈Nn \| fe ̸= 0}. 11.5.2.1 Ostrowski’s theorem 11.5.44 Deﬁnition The Minkowski sum of two subsets Q and R of Rn, written Q + R, is Q + R := {e + f \| (e, f) ∈Q × R}. 11.5.45 Deﬁnition A polytope in Rn is integral if all of its vertices are in Zn. An integral polytope P is integrally decomposable if there exists two integral polytopes Q and R such that P = Q + R, where both Q and R have at least two points. Otherwise, P is integrally indecomposable. 11.5.46 Deﬁnition The Newton polytope of f, written N(f), is the convex hull in Rn of Supp(f). The integral convex hull of f is the subset of points in Zn lying in N(f). 11.5.47 Theorem (Ostrowski’s theorem) , translated in If f factors into gh then we have N(f) = N(g) + N(h). 11.5.2.2 Irreducibility tests based on indecomposability of polytopes 11.5.48 Remark The previous theorem leads to the following irreducibility test. 11.5.49 Corollary (Irreducibility criterion) [1175, p. 507] If f ∈F[x1, . . . , xn] is a nonzero polyno-mial not divisible by any xi, and if N(f) is integrally indecomposable, then f is irreducible over any algebraic extension of F. 11.5.50 Theorem [1175, Theorem 4.2] Let P be an integral polytope in Rn contained in a hyperplane H and let e ∈Zn be a point lying outside of H. If e1, . . . , ek are all the vertices of P, then the convex hull of P and e is integrally indecomposable if, and only if, all the entries of e −e1, e −e2, . . . , e −ek are coprime. 11.5.51 Theorem [1175, Theorem 4.11] Let P be an indecomposable integral polytope in Rn with at least two points, that is contained in a hyperplane H, and let e ∈Rn be a point outside of H. Let S be any subset of points in Zn contained in the convex hull of e and P. Then the convex hull of S and Q is integrally indecomposable. 11.5.2.3 Sparse bivariate Hensel lifting driven by polytopes 11.5.52 Remark Let f ∈Fp[x, y] be a polynomial with t nonzero terms and of total degree d such that t < d. Let r be a vector in R2, and let Γ be a subset of edges of N(f) satisfying the following properties: 1. N(f) ⊆Γ + rR≥0, 2. each of the two inﬁnite edges of Γ + rR≥0 contains exactly one point of N(f), Algorithms 389 3. no proper subset of Γ satisﬁes the previous two conditions. Assume furthermore that: 1. f factorizes into f = gh for two proper factors g and h in Fp[x, y] with tg and th terms respectively, such that max(tg, th) ≤tλ for some constant λ satisfying 1/2 ≤λ < 1. 2. For each edge γ ∈Γ we are given polynomials gγ and hγ supported by γg and γh respectively, where γg and γh are the unique vertices or edges of N(g) and N(h) respectively such that γ = γg + γh. 3. For each edge γ ∈Γ the given polynomials gγ and hγ are coprime up to monomial factors. 11.5.53 Theorem [11, Theorem 28] Under the above assumptions, there exists an integral de-composition N(f) = N(g) + N(h) such that N(g) is not a single point or a line seg-ment parallel to rR≥0. There exists at most one full factorization of f which extends the boundary factorization deﬁned by the given (gγ)γ∈Γ and (hγ)γ∈Γ. Assuming that d and p ﬁt a machine word, this factorization can be computed, or shown not to exist, using O(tλd2 + t2λd log d log log d + t4λd) bit-operations, and O(tλd) bits of memory. 11.5.54 Remark Theorem 11.5.53 extends previous results from . Although it does not provide a complete factoring algorithm, it proves to be very eﬃcient in practice for large particular problems. 11.5.2.4 Convex-dense bivariate factorization 11.5.55 Remark In the worst case, the size of the irreducible factorization is exponential in the sparse size of the polynomial f to be factored. However Theorem 11.5.47 ensures that the size of the output is upper bounded by the number π, called the convex size, of points in Zn lying inside of N(f). The next theorem to be presented reduces the bivariate sparse factorization to the usual dense case. 11.5.56 Deﬁnition The aﬃne group over Z2, written Aﬀ(Z2), is the set of the maps U U : (i, j) 7→ α β α′ β′ i j + γ γ′ , (11.5.1) with α, β, γ, α′, β′, and γ′ in Z, such that αβ′ −α′β = ±1. 11.5.57 Deﬁnition Let S be a ﬁnite subset of Z2. The set S is normalized if it belongs to N2 and if it contains at least one point in {0} × N, and also at least one point in N × {0}. 11.5.58 Theorem [256, Theorem 1.2] For any normalized ﬁnite subset S of Z2, of cardinality σ, convex size π, and included in [0, dx] × [0, dy], one can compute an aﬃne map U ∈Aﬀ(Z2) as in (11.5.1), together with U(S), with O(σ log2((dx +1)(dy +1))) bit-operations, such that U(S) is normalized and contained in a block [0, d′ x]×[0, d′ y] satisfying (d′ x +1)(d′ y +1) ≤9π. 11.5.59 Lemma For any ﬁeld F, for any f ∈F[x, y] not divisible by x and y, for any U as in Equation (11.5.1), the polynomial U(f) := X (ex,ey)∈Supp(f) f(ex,ey)xαex+βey+γyα′ex+β′ey+γ′ is irreducible in F[x, y, x−1, y−1] if and only if f is irreducible. 390 Handbook of Finite Fields 11.5.60 Remark In order to compute the irreducible factorization of F, we can compute a reduction map U as in Theorem 11.5.58 for Supp(f), then compute the irreducible factorization of U(f), and ﬁnally apply U −1 to each factor. In this way we beneﬁt from complexity bounds that only depend on the convex size π of f instead of its dense size (dx + 1)(dy + 1). 11.5.3 Factoring straight-line programs and black boxes 11.5.61 Remark The sparse representation (see Deﬁnition 11.5.42) of a polynomial allows for space eﬃcient storage of polynomials of very high degree, since the degree of the term x2500 can be represented by a 501 bit integer. Polynomials f whose sparse representation occupies (log(deg f))O(1) bit space are supersparse (lacunary) [1661, 1897]. While computing small degree factors of such polynomials over the rational numbers can be accomplished in bit time that is polynomial in the input size, over ﬁnite ﬁelds such tasks are NP- or co-NP-hard . Here we have a situation where factoring over the rational numbers is provably easier than factoring over a suﬃciently large ﬁnite ﬁeld. 11.5.62 Remark In it is shown, by transferring the construction in , that several other operations on univariate and bivariate supersparse polynomials over a suﬃciently large ﬁnite ﬁeld are NP- or co-NP-hard. For instance, in the following is proven. 11.5.63 Theorem Suppose we have a Monte Carlo polynomial-time irreducibility test for super-sparse polynomials in F2m[X, Y ] for suﬃciently large m. Then large integers can be factored in Las Vegas polynomial-time. 11.5.64 Remark A polynomial in n variables of (total) degree d can have n+d n terms, i.e., ex-ponentially many terms in the number of variables. Sparse polynomials are those that have (n + d)O(1) non-zero terms. Note that, as in all asymptotic analysis, one considers not a single polynomial but an inﬁnite set of sparse input polynomials that a given al-gorithm processes, now in polynomial time in the sparse size. By using the factorization xd −1 = (x −1)(xd−1 + · · · + 1) one can easily generate examples where the sparse size of one irreducible factor is super-polynomially larger than the input size [1230, Example 5.1]. Motivated by algebraic computation models, straight-line programs were adopted as an alternate polynomial representation, ﬁrst only for inputs , but ultimately and im-portantly as a representation of the irreducible factors themselves [1650, 1653]. Here is an example of a division-free straight-line program (single assignment program), where F is the ﬁeld generated by those operands c1, c2, . . . which are constants, while x1, x2, . . . are input variables: υ1 ←c1 × x1; υ2 ←x2 −c2; υ3 ←υ2 × υ2; υ4 ←υ3 + υ1; υ5 ←υ4 × x3; . . . υ101 ←υ100 + υ51; The variable υ101 represents a polynomial in F[x1, x2, . . .], which can be evaluated by use of the straight-line program. For instance, the determinant of an n × n matrix whose entries are n2 variables can be represented, via Gaussian elimination, by a straight-line program with divisions of length O(n3). Because those divisions can cause divisions by zero on evaluation at certain points, it is desirable to remove them from such programs : the shortest division-free straight-line program for the determinant that is known today has length O(n2.7) and uses no constants other than 1 and −1 in F . In any case, Algorithms 391 divisions can be removed by increasing the length by a factor O((deg f)1+ϵ) for any ϵ > 0. The 1986 algorithm in [1650, 1653] produces from a straight-line program of length l for a polynomial of degree d in Monte Carlo random polynomial-time straight-line programs for the irreducible factors (and their multiplicities). The factor programs themselves have length O(d2l + d3+ϵ). Over ﬁnite ﬁelds of characteristic p, for an irreducible factor g of multiplicity pmm′, where gcd(p, m′) = 1, a straight-line program for gpm is returned; see Remark 11.5.68 below. The algorithm is implemented in the Dagwood system and can factor matrix determinants. A shortcoming of the straight-line representations, which later were adopted by the TERA project, was exposed by the Dagwood program: the lengths, while polynomial in the input lengths, become quite large (over a million assignments). The construction, however, plays a key role in complexity theory . 11.5.65 Remark Since polynomials represented by straight-line programs can be converted to sparse polynomials in polynomial-time in their sparse size by the algorithm in , the straight-line factorization algorithm brought to a successful conclusion the search for polynomial-time sparse factorizers. Previous attempts based on sparse Hensel lifting [1216, 1230, 1649, 3080, 3081], retained an exponential substep for many factors, namely the computation of the so-called right-side Hensel correction coeﬃcients. The problem of computing the coeﬃcient of a given term in a sparse product is in general #P-hard. Nonetheless, if a polynomial has only a few sparse factors, such sparse lifting can be quite eﬃcient, in practice. 11.5.66 Remark Instead of straight-line programs, one can use a full-ﬂedged programmed procedure that evaluates the input polynomial. The irreducible factors are then evaluated at values for the variables by another procedure that makes (“oracle”) calls to the input evaluation procedure. Thus is the genesis of algorithms for black box polynomials [1668, 1669]. The idea is the following: Suppose one can call a black evaluation box for the polynomial f(x1, . . . , xn) ∈F[x1, . . . , xn]. First, uniformly randomly select from a suﬃciently large ﬁnite set ﬁeld elements ai, ci (2 ≤i ≤n) and bj (1 ≤j ≤n) and interpolate and factor the bivariate image ˆ f(X, Y ) = f(X + b1, c2Y + a2X + b2, . . . , cnY + anX + bn) = r Y k=1 ˆ gk(X, Y )ek. By the eﬀective Hilbert Irreducibility Theorem 11.5.33 above, the irreducible polynomials ˆ gk are with high probability bivariate images of the irreducible factors hk(x1, . . . , xn) of f. For small coeﬃcient ﬁelds we shall assume that the black box can evaluate f at elements in a ﬁnite algebraic extension E of F. Already the bivariate interpolation algorithm may require such an extension in order to have suﬃciently many distinct points. 11.5.67 Remark If one selects an extension E of degree [E : F] > deg(f) that is a prime number, all hk remain irreducible over that extension. Indeed, the Frobenius norm NormE/F(˜ h) ∈ F[x1, . . . , xn] of a possible non-trivial irreducible factor ˜ h ∈E[x1, . . . , xn] of an hk must be a power of an irreducible polynomial over F, hence a power of hk itself. For otherwise gcd(hk, NormE/F(˜ h)) would constitute a non-trivial factor of hk over F. But then deg(˜ h )·[E : F] = deg(hk) · m, where m is the exponent of that power, and because [E : F] is a prime > deg(f) ≥deg(hk), we obtain the contradiction deg(˜ h ) = deg(hk)·(m/[E : F]) ≥deg(hk). Remark 11.5.66 continued. Now the black box for evaluating all hk(ξ1, . . . , ξn) at ﬁeld elements ξi ∈F stores (“hard-wires”) the ai, bj and the factors gk(X) = ˆ gk(X, 0) in its constant pool. We note that the gk are not necessarily irreducible, but with high probability they are pairwise relatively prime [1669, Section 2, Step 3], and their leading terms only depend on the variable X. The black box ﬁrst interpolates ¯ f(X, Y ) = f(X + b1, Y (ξ2 −a2(ξ1 −b1) −b2) + a2X + b2, . . ., Y (ξn −an(ξ1 −b1) −bn) + anX + bn) (11.5.2) 392 Handbook of Finite Fields and then factors ¯ f such that ¯ f(X, Y ) = r Y k=1 ¯ hk(X, Y )ek with ¯ hk(X, 0) = gk(X). (11.5.3) We note that again the ¯ hk are not necessarily irreducible. One may Hensel-lift the factor-ization f(X + b1, a2X + b2, . . . , anX + bn) = r Y k=1 gk(X)ek (11.5.4) provided none of the multiplicities ek is divisible by p. Otherwise, one can fully factor ¯ f(X, Y ) and lump (multiply) those irreducible factors ¯ hκ(X, Y ) together where ¯ hκ(X, 0) divide one and the same gk(X). Alternatively, if pm divides ek one could lift the pm-th power of gk and take a pm-th root of the lifted factor. We have ¯ f(ξ1 −b1, 1) = f(ξ1, . . . , ξn), and for all k we obtain ¯ hk(ξ1 −b1, 1) = hk(ξ1, . . . , ξn). We observe that the scalar multiple of hk is ﬁxed in all evaluations by the choice of gk. 11.5.68 Remark Over ﬁnite coeﬃcient ﬁelds, there is no restriction on the multiplicities ek. One does not obtain a pure straight-line program for the polynomial hk because a bivariate factorization of ¯ f or a pm-th root of the lifted factor, which depend on the evaluation points ξi, are performed on each evaluation. One can obtain straight-line polynomials that equal the irreducible factors modulo (xq 1 −x1, . . . , xq n −xn) by powering by q/pm, where q < ∞is the cardinality of the coeﬃcient ﬁeld. Those straight-line programs produce correct evaluations of the irreducible factors. 11.5.69 Remark The blackbox factorization algorithm is implemented in the FoxBox system . The size blowup experienced in the straight-line factorization algorithm does not occur. In fact, the factor evaluation black box makes O(deg(f)2) calls to the black box for f and factors a bivariate polynomial, either by lifting (11.5.4) or, if multiplicities are divisible by the characteristic, by factoring ¯ f. The program is ﬁxed except for the constants ai, bj and the polynomials gk. 11.5.70 Remark We conclude that the sparse representations of the factors can be recovered by sparse interpolation over a ﬁnite ﬁeld; see and the literature cited there. Dense factors can be identiﬁed to have more than a given number of terms and skipped. See Also §3.6 For irreducible multivariate polynomials. §8.1 Where absolute factorization intervenes for testing if a univariate rational func-tion generates a permutation of a ﬁnite ﬁeld as in the algorithms of [1716, 1983]. References Cited: [10, 11, 162, 219, 237, 238, 239, 256, 360, 365, 616, 617, 761, 778, 779, 830, 1100, 1175, 1177, 1182, 1183, 1216, 1217, 1218, 1227, 1229, 1230, 1271, 1462, 1499, 1518, 1519, 1616, 1623, 1640, 1645, 1646, 1647, 1648, 1649, 1650, 1651, 1653, 1654, 1655, 1658, 1659, 1660, 1661, 1662, 1668, 1669, 1670, 1716, 1739, 1750, 1880, 1881, 1882, 1883, 1893, 1897, 1983, 2108, 2129, 2130, 2131, 2202, 2209, 2299, 2338, 2339, 2400, 2504, 2505, 2528, 2529, 2530, 2542, 2602, 2703, 2730, 2733, 2938, 2939, 3021, 3052, 3053, 3080, 3081] Algorithms 393 11.6 Discrete logarithms over ﬁnite ﬁelds Andrew Odlyzko, University of Minnesota Surveys and detailed expositions with proofs can be found in [661, 1939, 1938, 2080, 2308, 2306, 2720]. 11.6.1 Basic deﬁnitions 11.6.1 Remark Discrete exponentiation in a ﬁnite ﬁeld is a direct analog of ordinary exponentia-tion. The exponent can only be an integer, say n, but for w in a ﬁeld F, wn is deﬁned except when w = 0 and n ≤0, and satisﬁes the usual properties, in particular wm+n = wmwn and (for u and v in F) (uv)m = umvm. The discrete logarithm is the inverse function, in analogy with the ordinary logarithm for real numbers. If F is a ﬁnite ﬁeld, then it has at least one primitive element g; i.e., all nonzero elements of F are expressible as powers of g, see Chapter 2. 11.6.2 Deﬁnition Given a ﬁnite ﬁeld F, a primitive element g of F, and a nonzero element w of F, the discrete logarithm of w to base g, written as logg(w), is the least non-negative integer n such that w = gn. 11.6.3 Remark The value logg(w) is unique modulo q −1, and 0 ≤logg(w) ≤q −2. It is often convenient to allow it to be represented by any integer n such that w = gn. 11.6.4 Remark The discrete logarithm of w to base g is often called the index of w with respect to the base g. More generally, we can deﬁne discrete logarithms in groups. They are commonly called generic discrete logarithms. 11.6.5 Deﬁnition If G is a group (with multiplication as group operation), and g is an element of G of ﬁnite order m, then for any element h of ⟨g⟩, the cyclic subgroup of G generated by g, the discrete logarithm of h to base g, written as logg(h), is the least non-negative integer n such that h = gn (and therefore 0 ≤logg(h) ≤m −1). 11.6.6 Remark The deﬁnition of a group discrete logarithm allows for consideration of discrete logarithms in ﬁnite ﬁelds when the base g is not primitive, provided the argument is in the group ⟨g⟩. This situation arises in some important applications, in particular in the U.S. government standard for the Digital Signature Algorithm (DSA). DSA operations are performed in a ﬁeld Fp with p a prime (nowadays recommended to be at least 2048 bits). This prime p is selected so that p −1 is divisible by a much smaller prime r (speciﬁed in the standard to be of 160, 224, or 256 bits), and an element h of Fp is chosen to have multiplicative order r (say by ﬁnding a primitive element g of Fp and setting h = g(p−1)/r). The main element of the signature is of the form hs for an integer s, and ability to compute s would break DSA. DSA can be attacked either by using generic ﬁnite group discrete logarithm algorithms in the group ⟨h⟩or ﬁnite ﬁeld algorithms in the ﬁeld Fp (which can then easily yield a solution in ⟨h⟩). 11.6.7 Remark The basic properties of discrete logarithms given below, such as the change of base formula, apply universally. On the other hand, many of the discrete logarithm algorithms described later are valid only in ﬁnite ﬁelds. Generally speaking, discrete logarithms are comparatively easy to compute in ﬁnite ﬁelds, since they have a rich algebraic structure that can be exploited for cryptanalytic purposes. Much of the research on discrete logarithms 394 Handbook of Finite Fields in other settings has been devoted to embedding the relevant groups inside ﬁnite ﬁelds in order to apply ﬁnite ﬁeld discrete logarithm algorithms. 11.6.8 Remark This section is devoted to ﬁnite ﬁeld discrete logarithms, and only gives a few ref-erences to other ones. For elliptic curve discrete logarithms, the most prominent collection, see Section 16.4. However, other groups have also been used, for example class groups of number ﬁelds . 11.6.2 Modern computer implementations 11.6.9 Remark Most popular symbolic algebra systems contain some implementations of discrete logarithm algorithms. For example, Maple has the mlog function, while Mathematica has FieldInd. More specialized systems for number theoretic and algebraic computations, such as Magma, PARI, and Sage, also have implementations, and typically can handle larger problems. Thus for all but the largest problems that are at the edge of computability with modern methods, widely available and easy to use programs are suﬃcient. Tables of ﬁnite ﬁelds, such as those in , are now seldomly printed in books. 11.6.3 Historical remarks 11.6.10 Remark Until the mid-1970s, the main applications for discrete logarithms were similar to those of ordinary logarithms, namely in routine computations, but this time in ﬁnite ﬁelds. They allowed replacement of relatively hard multiplications by easier additions. What was frequently used was Zech’s logarithm (also called Jacobi’s logarithm, cf. ), which is a modiﬁcation of the ordinary discrete logarithm. In a ﬁnite ﬁeld F with primitive element g, Zech’s logarithm of an integer n is deﬁned as the integer Z(n) mod (q −1) which satisﬁes gZ(n) = 1 + gn. This provides a quick way to add elements given in terms of their discrete logarithms: aside from boundary cases, gm + gn = gm(1 + gn−m) = gm+Z(n−m). 11.6.11 Remark As with ordinary logarithms, where slide rules and logarithm tables have been re-placed by calculators, such routine applications of discrete logarithms in small or moderately large ﬁelds now rely on computer algebra systems. 11.6.12 Remark Interest in discrete logarithms jumped dramatically in the mid-1970s with the invention of public key cryptography, see Chapter 16. While discrete exponentiation is easy, the discrete logarithm, its inverse, appeared hard, and this motivated the invention of the Diﬃe–Hellman key exchange protocol, the ﬁrst practical public key cryptosystem. Eﬃcient algorithms for discrete logarithms in the ﬁeld over which this protocol is implemented would make it insecure. 11.6.13 Remark The Diﬃe–Hellman problem is to compute gxy, the key that the two parties to the Diﬃe–Hellman protocol obtain, from the gx and gy that are visible to the eavesdropper. Although this problem has attracted extensive attention, it has not been solved, and for the most important cases of ﬁnite ﬁeld and elliptic curve discrete logarithms, it is still unknown whether the Diﬃe–Hellman problem is as hard as the discrete logarithm one; see for recent results and references. 11.6.14 Remark It is known that single bits of discrete logarithms are about as hard to compute as the entire discrete logarithms . 11.6.15 Remark There are some rigorous lower bounds on discrete log problems, but only for groups given in ways that restrict what can be done in them [2219, 2630]. Algorithms 395 11.6.16 Remark Various cryptosystems other than the Diﬃe-Hellman one have been proposed whose security similarly depends on the intractability of the discrete logarithm problem; see Section 16.1. Many of them can be used in settings other than ﬁnite ﬁelds. 11.6.17 Remark There are close analogies between integer factorization and discrete logarithms in ﬁnite ﬁelds, and most (but not all) of the algorithms in one area have similar ones in the other. This will be seen from some of the references later. In general, considerably less attention has been devoted to discrete logarithms than to integer factorization. Hence the smaller sizes of discrete logarithm problems that have been solved result both from the greater technical diﬃculty of this problem as compared to integer factorization and from less eﬀort being devoted to it. 11.6.18 Remark Shor’s 1994 result shows that if quantum computers become practical, discrete logarithms will become easy to compute. Therefore cryptosystems based on discrete logarithms may all become suddenly insecure. 11.6.4 Basic properties of discrete logarithms 11.6.19 Remark Suppose that G is a group, and g an element of ﬁnite order m in G. If u and v are two elements of ⟨g⟩, then logg(uv) ≡logg(u) + logg(v) (mod m), logg(u−1) ≡−logg(u) (mod m). 11.6.20 Remark (Change of base formula) Suppose that G is a group, and that g and h are two elements of G that generate the same cyclic subgroup ⟨g⟩= ⟨h⟩of order m. If u is an element of ⟨g⟩, then logg(u) ≡logh(u) ∗logg(h) (mod m), and therefore logg(h) ≡1/ logh(g) (mod m). These formulas mean that one can choose the most convenient primitive element to work with in many applications. For example, in ﬁnite ﬁelds F2k, elements are usually represented as polynomials with binary coeﬃcients, and one can ﬁnd (as veriﬁed by experiment and inspired by heuristics, but not proved rigorously) primitive elements that are represented as polynomials of very low degree. This can oﬀer substantial eﬃciencies in implementations. However, it does not aﬀect the security of the system. If discrete logarithms are easy to compute in one base, they are easy to compute in other bases. Similarly, the change of the irreducible polynomial that deﬁnes the ﬁeld has little eﬀect on the diﬃculty of the discrete logarithm problem. 11.6.5 Chinese Remainder Theorem reduction: The Silver–Pohlig–Hellman algorithm 11.6.21 Remark If the order of the element g can be factored even partially, the discrete logarithm problem reduces to easier ones. This is the Silver–Pohlig–Hellman technique . Suppose that g is an element of ﬁnite order m in a group G, and m is written as m = m1m2 with gcd(m1, m2) = 1. Then the cyclic group ⟨g⟩is the direct product of the cyclic groups ⟨gm2⟩and ⟨gm1⟩of orders m1 and m2, respectively. If we determine a = loggm2 (wm2) and 396 Handbook of Finite Fields b = loggm1 (wm1), the Chinese Remainder Theorem tells us that logg(w) is determined completely, and in fact we obtain logg(w) ≡b ∗x ∗m1 + a ∗y ∗m2 (mod m), where x and y come from the Euclidean algorithm computation of gcd(m1, m2), namely 1 = xm1 + ym2. This procedure extends easily to more than two relatively prime factors. 11.6.22 Remark When m, the order of g, is a prime power, say m = pk, the computation of logg(w) reduces to k discrete logarithm computations in a cyclic group of p elements. For example, if r = pk−1 and h = gr, u = wr, then h has order p, and computing logh(u) yields the reduction of logg(w) (mod p). This process can then be iterated to obtain the reduction modulo p2, and so on. 11.6.23 Remark The above remarks, combined with results of the next section, show that when the complete factorization of the order of g can be obtained, discrete logarithms can be computed in not much more than r1/2 operations in the group, where r is the largest prime in the factorization. 11.6.24 Remark In a ﬁnite ﬁeld, any function can be represented by a polynomial. For the discrete logarithm, such polynomials do turn out to have some aesthetically pleasing properties, see [2189, 2244, 2922, 2968]. However, so far they have turned out to be of no practical use whatsoever. 11.6.6 Baby steps–giant steps algorithm 11.6.25 Remark We next consider some algorithms for discrete logarithms that work in very general groups. The basic one is the baby steps–giant steps method that combines time and space, due to Shanks . 11.6.26 Algorithm Baby steps–giant steps algorithm: Suppose that G is a group and g is an element of G of ﬁnite order m. If h ∈⟨g⟩, h = gk, and w = ⌈m1/2⌉, then k can be written as k = aw+b for some (often non-unique) a, b with 0 ≤a, b < w. To ﬁnd such a representation, compute the set A = {gjw : 0 ≤j < w} and sort it. This takes m1/2 + O(log(m)) group operations and O(m1/2 log(m)) sorting steps, which are usually very easy, since they can be performed on bit strings, or even initial segments of bit strings. Next, for 0 ≤i < w, compute hg−i and check whether it is present in A. When it is, we obtain the desired representation k = jw+i. 11.6.27 Remark The baby steps–giant steps technique has the advantage of being fully determin-istic. Its principal disadvantage is that it requires storage of approximately m1/2 group elements. A space-time tradeoﬀis available, in that one can store a smaller list (the set A in the notation above, with fewer but larger “giant steps”) but then have to do more computing (more “baby steps”). 11.6.28 Remark The baby steps–giant steps algorithm extends easily to many cases where the discrete logarithm is restricted in some way. For example, if it is known that logg(w) lies in an interval of length n, the basic approach sketched above can be modiﬁed to ﬁnd it in O(n1/2) group operations (plus the usual sorting steps). Similarly, if the discrete logarithm k is allowed to have only small digits when represented in some base (say binary digits in base 10), then the running time will be about the square root of the number of possibilities for k. For some other recent results, see . Algorithms 397 11.6.7 Pollard rho and kangaroo methods for discrete logarithms 11.6.29 Remark In 1978, Pollard invented two randomized methods for computing discrete log-arithms in any group, the rho method, and the kangaroo (or lambda) technique . Just like Pollard’s earlier rho method for integer factorization, they depend on the birthday paradox, which says that if one takes a (pseudo) random walk on a completely connected graph of n vertices, one is very likely to revisit the same vertex in about n1/2 steps. These discrete logarithm algorithms also depend, just as the original rho method does, on the Floyd algorithm (Section 3.1 of ) for detecting cycles with little memory at some cost in running time, in that they compare x2i to xi, where xi is the position of the random walk at time i. 11.6.30 Remark Since the rho and kangaroo methods for discrete logarithms are probabilistic, they cannot guarantee a solution, but heuristics suggest, and experiments conﬁrm, that both run in expected time O(m1/2), where m is the order of the group. This is the same computational eﬀort as for the baby steps–giant steps algorithm. However, the rho and kangaroo methods have two advantages. One is that they use very little memory. Another one is that, as was ﬁrst shown by van Oorschot and Wiener , they can be parallelized, with essentially linear speedup, so that k processors ﬁnd a solution about k times faster than a single one. We sketch just the standard version of the rho method, and only brieﬂy. 11.6.31 Algorithm Rho algorithm for discrete logarithms: Partition the group ⟨g⟩of order m into three roughly equal sets S1, S2, and S3, using some property that is easy to test, such as the ﬁrst few bits of a canonical representation of the elements of G. To compute logg(h), deﬁne a sequence w0, w1, . . . by w0 = g and for i > 0, wi+1 = wi2, wig, or wih, depending on whether wi ∈S1, S2, or S3. Then each wi is of the form wi = gaihbi for some integers ai, bi. If the procedure of moving from wi to wi+1 behaves like a random walk (as is expected), then in O(m1/2) steps we will ﬁnd i such that wi = w2i, and this will give a congruence ai + bi logg(h) ≡a2i + b2i logg(h) (mod m). Depending on the greatest common divisor of m and bi −b2i this congruence will typically either yield logg(h) completely, or give some stringent congruence conditions, which with the help of additional runs of the algorithm will provide a complete solution. 11.6.32 Remark The low memory requirements and parallelizability of the rho and kangaroo al-gorithms have made them the methods of choice for solving general discrete logarithm problems. There is a substantial literature on various modiﬁcations, although they do not improve too much on the original parallelization observations of . Some references are [613, 1734, 2414, 2790]. 11.6.33 Remark The rho method, as outlined above, requires knowledge of the exact order m of the group. The kangaroo method only requires an approximation to m. The kangaroo algorithm can also be applied eﬀectively when the discrete logarithm is known to lie in a restricted range. 11.6.8 Index calculus algorithms for discrete logarithms in ﬁnite ﬁelds 11.6.34 Remark The rest of this section is devoted to a brief overview of index calculus algo-rithms for discrete logarithms. Unlike the Shanks and Pollard methods of the previous two 398 Handbook of Finite Fields subsections, which take exponential time, about m1/2 for a group of order m, the index calculus techniques are subexponential, with running times closer to exp((log(m))1/2) and even exp((log(m))1/3). However, they apply directly only to ﬁnite ﬁelds. That is why much of the research on discrete logarithms in other groups of cryptographic interest, such as on elliptic curves, is devoted to ﬁnding ways to reduce those problems to ones in ﬁnite ﬁelds. 11.6.35 Remark In the case of DSA mentioned at the beginning of this section, the recommended size of the modulus p has increased very substantially, from 512 to 1024 bits when DSA was ﬁrst adopted, to the range of 2048 to 3072 bits more recently. The FIPS 186-3 standard speciﬁes bit lengths for the two primes p and r of (1024, 160), (2048, 224), (2048, 256), and (3072, 256). The relative sizes of p and r were selected to oﬀer approximately equal levels of security against index calculus algorithms (p) and generic discrete logarithm attacks (r). The reason for the much faster growth in the size of p is that with the subexponential running time estimates, the eﬀect of growing computing power is far more pronounced on the p side than on the r side. In addition, while there has been no substantial theoretical advance in index calculus algorithms in the last two decades, there have been numerous small incremental improvements, several cited later in more detailed discussions. On the other hand, there has been practically no progress in generic discrete logarithm algorithms, except for parallelization. 11.6.36 Remark The basic idea of index calculus algorithms dates back to Kraitchik, and is also key to all fast integer factorization algorithms. In a ﬁnite ﬁeld Fq with primitive element g, if we ﬁnd some elements xi, yj ∈Fq such that r Y i=1 xi = s Y j=1 yj, then r X i=1 logg xi ≡ s X j=1 logg yj (mod q −1). If enough equations are collected, this linear system can be solved for the logg xi and logg yj. Singular systems are not a problem in practice, since typically computations generate con-siderably more equations than unknowns, and one can arrange for g itself to appear in the multiplicative relations. 11.6.37 Remark To compute logg w for some particular w ∈F with index calculus algorithms, it is often necessary to run a second stage that produces a relation involving w and the previously computed discrete logarithms. In some algorithms the second stage is far easier than the initial computation, in others it is of comparable diﬃculty. 11.6.38 Remark For a long time (see for references), the best index calculus algorithms for both integer factorization and discrete logarithms had running times of the form exp((c + o(1))(log q)1/2(log log q)1/2) as p →∞ for various constants c > 0, where q denotes the integer being factored or the size of the ﬁnite ﬁeld. The ﬁrst practical method that broke through this running time barrier was Coppersmith’s algorithm for discrete logarithms in ﬁelds of size q = 2k (and more generally, of size q = pk where p is a small prime and k is large). It had running time of approximately exp(C(log q)1/3(log log q)2/3), where the C varied slightly, depending on the distance from k to the nearest power of p, and in the limit as k →∞it oscillated between two bounds . The function ﬁeld sieve of Algorithms 399 Adleman , which also applies to ﬁelds with q = pk where p is relatively small, improves on the Coppersmith method, but has similar asymptotic running time estimate. For the latest results on its developments, see [1626, 1628, 2543]. 11.6.39 Remark The running time of Coppersmith’s algorithm turned out to also apply to the number ﬁeld sieve. This method, which uses algebraic integers, was developed for integer factorization by Pollard and H. Lenstra, with subsequent contributions by many others. It was adopted for discrete log computations in prime ﬁelds by Gordon , with substantial improvements by other reseachers. For the latest estimates and references, see [711, 1627, 2544, 2545]. 11.6.9 Smooth integers and smooth polynomials 11.6.40 Remark The index calculus algorithms depend on a multiplicative splitting of some ele-ments, such as integers or polynomials, into such elements drawn from a smaller collection. This smaller collection usually is made up of elements that by some measure (norm) are small. The essence of index calculus algorithms is to select general elements from the large set at random, but as intelligently as possible in order to maximize the chances they will have the desired type of splitting. Usually elements that do have such splittings are called “smooth.” 11.6.41 Remark There are rigorous analyses that provide estimates of how often elements in various domains are “smooth.” For ordinary integers, there are the estimates of . For algebraic integers, we can use [441, 2574]. For polynomials over ﬁnite ﬁelds, recent results are . 11.6.10 Sparse linear systems of equations 11.6.42 Remark Index calculus algorithms for discrete logarithms require the solution of linear equations modulo q −1, where q is the size of the ﬁeld. As in the Silver–Pohlig–Hellman method, the Chinese Remainder Theorem (and an easy reduction of the case of a power of a prime to that of the prime itself) reduces the problem to that of solving the system modulo primes r that divide q −1. (For more extensive discussion of linear algebra over ﬁnite ﬁelds, see Section 13.4.) 11.6.43 Remark The linear algebra problems that arise in index calculus algorithms for integer factorization are very similar, but simpler, in that they are all just modulo 2. For discrete logarithm problems to be hard, they have to be resistant to the Silver–Pohlig–Hellman at-tack. Hence q −1 has to have at least one large prime factor r, and so the linear system has to be solved modulo a large prime. That increases the complexity of the linear solution computation, and thus provides slightly higher security for discrete logarithm cryptosys-tems. 11.6.44 Remark A key factor that enables the solution of the very large linear systems that arise in index calculus algorithms is that these systems are very sparse. (Those “smooth” elements do not involve too many of the “small” elements in the multiplicative relations.) Usually the structured Gaussian elimination method (proposed in and called there intelligent gaussian elimination, afterwards renamed in the ﬁrst practical demonstration of it , now sometimes called ﬁltering) is applied ﬁrst. It combines the relations in ways that re-duce the system to be solved and do not destroy the sparsity too far. Then the conjugate gradient, the Lanczos, or the Wiedemann methods (developed in [721, 2976], the ﬁrst two demonstrated in practice in ) that exploit sparsity are used to obtain the ﬁnal solution. 400 Handbook of Finite Fields 11.6.45 Remark For the extremely large linear systems that are involved in record-setting com-putations, distributed computation is required. The methods of choice, once structured Gaussian elimination is applied, are the block Lanczos and block Wiedemann methods [718, 719, 2134]. 11.6.46 Remark Some symbolic algebra systems incorporate implementations of the sparse linear system solvers mentioned above. 11.6.47 Remark As a demonstration of the eﬀectiveness of the sparse methods, the record factor-ization of RSA768 , mentioned below, produced 64 billion linear relations. These were reduced, using structured gaussian elimination, to a system of almost 200 million equations in about that many unknowns. This system was still sparse, with the average equation involving about 150 unknowns. The block Wiedemann method was then used to solve the resulting system. 11.6.11 Current discrete logarithm records 11.6.48 Remark Extreme caution should be exercised when drawing any inferences about relative performance of various integer factorization and discrete logarithm algorithms from the record results listed here. The computing resources, as well as eﬀort involved in program-ming, varied widely among the various projects. 11.6.49 Remark As of the time of writing (early 2012), the largest cryptographically hard integer (i.e., one that was chosen speciﬁcally to resist all known factoring attacks, and is a product of two roughly equal primes) that has been factored is RSA768, a 768-bit (232 decimal digit) integer from the RSA challenge list . This was the result of a large collaboration across the globe stretching over more than two years, and used the general number ﬁeld sieve. 11.6.50 Remark The largest discrete logarithm case for a prime ﬁeld Fp (with p chosen to resist simple attacks) that has been solved is for a 530-bit (160 decimal digit) prime p. This was accomplished by Kleinjung in 2007 . The number ﬁeld sieve was used. 11.6.51 Remark In ﬁelds of characteristic two, the largest case that has been solved is that of Fq with q = 2613, using the function ﬁeld sieve. (An earlier record was for q = 2607 using the Coppersmith algorithm.) This computation took several weeks on a handful of processors, and was carried out by Joux and Lercier in 2005 . 11.6.52 Remark The largest generic discrete logarithm problem that has been solved in a hard case is that of discrete logarithms over an elliptic curve modulo a 112-bit prime, thus a group of size about 2112. This is due to Bos and Kaihara , and was done in 2009. Right now, a large multi-year collaborative eﬀort is under way to break the Certicom ECC2K-130 challenge, which involves computing discrete logarithms on an elliptic curve over a ﬁeld with 2131 elements . All these eﬀorts rely on parallelized versions of the Pollard rho method. See Also Chapter 2 For basic properties of ﬁnite ﬁelds. §11.1 For basic computational techniques in ﬁnite ﬁelds. §13.4 For linear algebra over ﬁnite ﬁelds. Chapter 16 For public key cryptographic systems. Algorithms 401 References Cited: [16, 160, 353, 410, 441, 613, 661, 711, 717, 718, 719, 721, 1325, 1444, 1501, 1625, 1626, 1627, 1628, 1734, 1752, 1753, 1765, 1839, 1938, 1939, 2044, 2080, 2134, 2189, 2219, 2244, 2306, 2308, 2348, 2406, 2413, 2414, 2543, 2544, 2545, 2574, 2607, 2623, 2630, 2708, 2720, 2790, 2852, 2922, 2968, 2976] 11.7 Standard models for ﬁnite ﬁelds Bart de Smit, Universiteit Leiden Hendrik Lenstra, Universiteit Leiden 11.7.1 Deﬁnition Let p be a prime number and let n be a positive integer. An explicit model for a ﬁnite ﬁeld of size pn is a ﬁeld whose underlying additive group is Fn p = Fp×Fp×· · ·×Fp. 11.7.2 Remark Let e0, . . . , en−1 be the standard Fp-basis of Fn p. Then a ﬁeld structure on Fn p is uniquely determined by the n3 elements aijk ∈Fp such that ei · ej = n−1 X k=0 aijkek. Thus, one can specify an explicit model using O(n3 log p) bits. When we say that for an algorithm an explicit model is input or output we assume that the explicit data consisting of p and (aijk) are given as input or output. There is a deterministic polynomial time algorithm that given such data ﬁrst decides whether p is prime , and then decides whether it deﬁnes an explicit model for a ﬁnite ﬁeld [1895, Section 2]. 11.7.3 Remark Let A be a ﬁeld of characteristic p > 0 and size pn and let b0, . . . , bn−1 be a basis of A as a vector space over Fp. We then obtain an explicit model as follows. Write ψ for the unique Fp-vector space isomorphism Fn p →A sending ei to bi for 0 ≤i < n. Deﬁne a multiplication map on Fn p by v · w = ψ−1(ψ(v) · ψ(w)), for v, w ∈Fn p. Together with vector addition, this multiplication makes Fn p into a ﬁeld. 11.7.4 Remark An alternative space-eﬃcient way to give explicit models is to give an irreducible polynomial f ∈Fp[x] of degree n over Fp, which we can encode with O(n log p) bits. Using the Fp-basis 1, x, . . . , xn−1 of the ﬁeld Fp[x]/(f) we then obtain an explicit model of a ﬁeld of size pn as in Remark 11.7.3. One can convert between this representation and the representation with n3 elements by deterministic polynomial time algorithms [1895, Theorem 1.1]. Since our concern in this section is only about whether an algorithm runs in polynomial time or not, and not about the degree of the polynomial if it does, we use the more ﬂexible setup of the explicit data consisting of the n3 elements aijk in Fp. 11.7.5 Theorem There is a deterministic polynomial time algorithm such that 1. on input two explicit models for ﬁnite ﬁelds A, B of the same cardinality, it produces a ﬁeld isomorphism φA,B : A →B; 2. for any three explicit models A, B, C for ﬁnite ﬁelds of the same size we have φB,C ◦φA,B = φA,C. 11.7.6 Remark The isomorphism that the algorithm produces is given as explicit output by listing the entries of the square matrix associated to the underlying linear map over the prime 402 Handbook of Finite Fields ﬁeld; see [1895, Section 2] for a proof of this theorem without Property 2. By Property 2 this algorithm can be used for “coercion” in computer algebra . 11.7.7 Deﬁnition An algorithmic model for ﬁnite ﬁelds is a sequence (Aq)q, where q runs over all prime powers and Aq is an explicit model for a ﬁnite ﬁeld of size q, such that there is an algorithm that on input q produces Aq. 11.7.8 Example For each prime p and n ≥1 one can take the explicit model Apn to be given as in Remark 11.7.4 by the ﬁrst irreducible polynomial of degree n over Fp with respect to a lexicographic ordering of polynomials of degree n over Fp. 11.7.9 Example Conway polynomials [1452, 1966] provide an algorithmic model for ﬁnite ﬁelds that has some additional properties. However computing Conway polynomials is laborious and there is only a rather limited table of known Conway polynomials. 11.7.10 Theorem There is an algorithmic model (Sq)q such that there is a deterministic polynomial time algorithm that on input an explicit model A for a ﬁeld of size q 1. computes the model Sq; 2. computes an isomorphism of ﬁelds A →Sq. 11.7.11 Remark The theorem in fact determines (Sq)q uniquely up to “polynomial time base change.” More precisely, suppose that for every prime power q = pn we have an invert-ible n × n-matrix Mq over Fp and suppose that there is a deterministic polynomial time algorithm that on input an explicit model for a ﬁeld of size q produces Mq. Given (Sq)q as in the theorem we let (S′ q)q be the algorithmic model that one gets in the manner of Remark 11.7.3 by considering the columns of Mq as an Fp-basis of Sq for every q. Then Theorem 11.7.10 also holds for (S′ q)q. Moreover, every algorithmic model (S′ q)q with the Properties 1 and 2 of the theorem arises in this way from (Sq)q. 11.7.12 Remark Theorem 11.7.10 implies Theorem 11.7.5. To prove Theorem 11.7.10 one can show that it holds for the algorithmic model of ﬁnite ﬁelds (Sq)q, where Sq is the standard model for a ﬁnite ﬁeld of size q deﬁned below. 11.7.13 Deﬁnition (Construction of the standard model) Step 1: cyclotomic rings. Let r be a prime number and write r = r · gcd(r, 2). We write Zr for the ring of r-adic integers, Z∗ r for its group of units, and ∆r for the torsion subgroup of Z∗ r; the group ∆r is cyclic of order ϕ(r), where ϕ denotes the Euler ϕ-function. The ring Ar is the polynomial ring Z[x0, x1, x2, . . .] modulo the ideal generated by {Pr−1 j=0 xjr/r 0 , xr k+1 −xk : k ≥0}. For k ∈Z≥0, we write ζrrk for the residue class of xk in Ar, which is a unit of multiplicative order rrk. For each u ∈Z∗ r there is a unique ring automorphism of Ar that maps each ζrrk to ζ ¯ u rrk, where ¯ u = (u mod rrk); we denote this ring automorphism by σu. The ring Br is deﬁned by Br = {a ∈Ar : σu(a) = a for all u ∈∆r}. For k ∈Z>0, i ∈ {0, 1, . . . , r −1} the element ηr,k,i ∈Br is deﬁned by ηr,k,i = P u∈∆r σu(ζ1+irrk−1 rrk ). Z Z[ζr] • Z[ζrr] • • Br Ar ∆r ∆r Algorithms 403 Step 2: prime ideals. Let p, r be prime numbers with p ̸= r, and let l be the number of factors r in the integer (pϕ(r) −1)/(r2/r). Denote by Sp,r the set of prime ideals p of Br that satisfy p ∈p. This set is ﬁnite of cardinality rl, and for each p ∈Sp,r there exists a unique system (ap,j)0≤j<lr of integers ap,j ∈{0, 1, . . . , p −1} such that p is generated by p together with {ηr,k+1,i −ap,i+kr : 0 ≤k < l, 0 ≤i < r}. We deﬁne a total ordering on Sp,r by putting p < q if there exists h ∈{0, 1, . . . , lr −1} such that ap,j = aq,j for all j < h and ap,h < aq,h. The smallest element of Sp,r in this ordering is denoted by pp,r. We deﬁne Fp,r to be the ring Br/pp,r, and for k ∈Z>0 we deﬁne αp,r,k ∈Fp,r to be the residue class of ηr,k+l,0 modulo pp,r. Step 3: equal characteristic. Let p be a prime number and put Fp = Z/pZ. Let the element f = f(x, y) of the polynomial ring Fp[x, y] be deﬁned by f = xp−1−y·Pp−1 i=1 xi. We deﬁne Fp,p to be the polynomial ring Fp[x1, x2, x3, . . .] modulo the ideal generated by {f(x1, 1), f(xk+1, xk) : k > 0}. For k ∈Z>0 we denote the image of xk in Fp,p by αp,p,k. Step 4: an algebraic closure. Let p be a prime number. Then for any prime number r it is true that the ring Fp,r is a ﬁeld containing Fp; that for each k ∈Z>0, the element αp,r,k of Fp,r is algebraic of degree rk over Fp; and that one has Fp,r = Fp(αp,r,1, αp,r,2, . . .). We write ¯ Fp for the tensor product, over Fp, of the rings Fp,r, with r ranging over the set of all prime numbers. For any prime number r and k ∈Z>0, the image of αp,r,k under the natural ring homomorphism Fp,r →¯ Fp is again denoted by αp,r,k. The ring ¯ Fp is a ﬁeld containing Fp, and it is an algebraic closure of Fp. We have ¯ Fp = Fp(αp,r,k : r prime, k ∈Z>0), each αp,r,k being algebraic of degree rk over Fp. Step 5: a vector space basis. Let p be a prime number. For each s ∈Q/Z, the element ϵs ∈¯ Fp is deﬁned as follows. There exists a unique system of integers (cr,k)r,k, with r ranging over the set of prime numbers and k over Z>0, such that each cr,k belongs to {0, 1, . . . , r −1} and s equals the residue class of P r,k cr,k/rk modulo Z, the sum being ﬁnite in the sense that cr,k = 0 for all but ﬁnitely many pairs r, k. With that notation, ϵs is deﬁned to be the ﬁnite product Q r,k αcr,k p,r,k. The system (ϵs)s∈Q/Z is a vector space basis of ¯ Fp over Fp. In addition, for each s ∈Q/Z the degree of ϵs over Fp equals the order of s in the additive group Q/Z. For n ≥1 the standard model Spn for a ﬁeld of size pn is the explicit model that one obtains in the manner of Remark 11.7.3 by considering the Fp-basis ϵ0, ϵ1/n, ϵ2/n, . . . , ϵ(n−1)/n of the unique subﬁeld Fpn of ¯ Fp of size pn. 11.7.14 Example Suppose that p is an odd prime number. The standard models for ﬁelds of size pn where n is a power of 2 can be computed as follows. Put l = ord2((p2 −1)/8) and for each k ≥−l consider the image αp,2,k of η2,k+l,0 = ζ2k+l+2 + ζ−1 2k+l+2 in the ﬁeld Fp,2 = B2/pp,2. We have αp,2,−l = 0, and for each k ≥−l the element αp,2,k+1 is a root of the quadratic polynomial fk = x2 −2 −αp,2,k ∈Fp(αp,2,k)[x]. If k < 0 then fk has two roots in Fp and by the choice of the prime pp,2 in Deﬁnition 11.7.13, the element αp,2,k+1 is the smallest of the two roots if we order Fp as 0, 1, . . . , p−1. Starting from αp,2,−l = 0, this enables us to ﬁnd αp,2,−l+1, . . . , αp,2,0 ∈Fp. One can show that fk is irreducible in Fp(αp,2,k)[x] for all k ≥0. In particular, the polynomial f0 = x2 −2 −αp,2,0 ∈Fp[x] gives the standard model for a ﬁeld of size p2. Moreover, for every k ≥1 the irreducible polynomial over Fp of the generator αp,2,k of Fp2k over Fp is (x2 −2)◦k −αp,2,0, where for a polynomial f ∈Fp[x] we let f ◦k denote the k-fold composition f(f(· · · f(x) · · · )). 11.7.15 Remark Note that the deﬁnition above also provides a standard embedding Sq →Sqd for every prime power q and every integer d ≥1, which is induced by the inclusion Fq ⊂Fqd. A composition of standard embeddings is again a standard embedding. 404 Handbook of Finite Fields 11.7.16 Theorem There is a probabilistic algorithm that on input a prime p and n ≥1 computes Spn in expected time bounded by a polynomial in log p and n. 11.7.17 Remark This theorem is an easy consequence of the fact that explicit models can be made in probabilistic polynomial time , and Theorem 11.7.10 above. Under the assump-tion of the generalized Riemann hypothesis, this can also be achieved with a deterministic polynomial time algorithm . 11.7.18 Remark An algorithm as in Theorem 11.7.16 is semi-deterministic: it is a probabilistic algorithm that gives the same output when it runs twice on the same input. Thus, the output does not depend on the random numbers drawn by the algorithm. 11.7.19 Example One way to ﬁnd a non-square in Fp for a given odd prime p in semi-deterministic polynomial time is as follows. Start with the value −1. If it is a square, ﬁnd the two square roots with the probabilistic method of Tonelli-Shanks [660, 1.5.1] and select the root that has the smallest representative in the set {1, . . . , p −1}. If this is a square in Fp, repeat. Within O(log p) iterations we ﬁnd a non-square up in Fp. By considering the Fp-basis 1, √up of the ﬁeld Fp(√up) we ﬁnd a semi-deterministic algorithm that given p produces an explicit model for a ﬁeld of size p2 in polynomial time. This proves a special case of Theorem 11.7.16. The models produced in this way are not the standard models, which as we saw in Example 11.7.14 are obtained by taking inverse images of 0 under iterates of the map x 7→x2 −2 rather than inverse images of −1 under iterates of the map x 7→x2. In both cases the algorithm produces quadratic polynomials until it encounters an irreducible one, but only the method of Example 11.7.14 has the advantage that for all p all subsequent quadratic polynomials are also irreducible. 11.7.20 Remark In a similar way, one can prove that there is a semi-deterministic algorithm that given a prime power q, and a prime l and r ≥1 with lr \| q −1, computes a root of unity of order lr in Sq in expected time polynomial in l and log q. See Also Chapter 2 For basic properties of ﬁnite ﬁelds. §11.1 For basic computational techniques in ﬁnite ﬁelds. §11.3 For constructing irreducible polynomials. §13.4 For linear algebra over ﬁnite ﬁelds. References Cited: [14, 43, 361, 660, 789, 790, 1452, 1892, 1895, 1966] 12 Curves over ﬁnite ﬁelds 12.1 Introduction to function ﬁelds and curves ....... 406 Valuations and places • Divisors and Riemann–Roch theorem • Extensions of function ﬁelds • Diﬀerentials • Function ﬁelds and curves 12.2 Elliptic curves........................................ 422 Weierstrass equations • The group law • Isogenies and endomorphisms • The number of points in E(Fq) • Twists • The torsion subgroup and the Tate module • The Weil pairing and the Tate pairing • The endomorphism ring and automorphism group • Ordinary and supersingular elliptic curves • The zeta function of an elliptic curve • The elliptic curve discrete logarithm problem 12.3 Addition formulas for elliptic curves.............. 440 Curve shapes • Addition • Coordinate systems • Explicit formulas • Short Weierstrass curves, large characteristic: y2 = x3 −3x + b • Short Weierstrass curves, characteristic 2, ordinary case: y2 + xy = x3 + a2x2 + a6 • Montgomery curves: by2 = x3 + ax2 + x • Twisted Edwards curves: ax2 + y2 = 1 + dx2y2 12.4 Hyperelliptic curves ................................. 447 Hyperelliptic equations • The degree zero divisor class group • Divisor class arithmetic over ﬁnite ﬁelds • Endomorphisms and supersingularity • Class number computation • The Tate-Lichtenbaum pairing • The hyperelliptic curve discrete logarithm problem 12.5 Rational points on curves .......................... 456 Rational places • The Zeta function of a function ﬁeld • Bounds for the number of rational places • Maximal function ﬁelds • Asymptotic bounds 12.6 Towers ................................................ 464 Introduction to towers • Examples of towers 12.7 Zeta functions and L-functions .................... 469 Zeta functions • L-functions • The case of curves 12.8 p-adic estimates of zeta functions and L-functions ........................................... 479 Introduction • Lower bounds for the ﬁrst slope • Uniform lower bounds for Newton polygons • Variation of Newton polygons in a family • The case of curves and abelian varieties 12.9 Computing the number of rational points and zeta functions .............................................. 488 Point counting: sparse input • Point counting: dense input • Computing zeta functions: general case • Computing zeta functions: curve case 405 406 Handbook of Finite Fields 12.1 Introduction to function ﬁelds and curves Arnaldo Garcia, IMPA Henning Stichtenoth, Sabanci University The theory of algebraic curves is essentially equivalent to the theory of algebraic function ﬁelds. The latter requires less background and is closer to the theory of ﬁnite ﬁelds; therefore we present here the theory of function ﬁelds. At the end of the section, we give a brief introduction to the language of algebraic curves. Our exposition follows mainly the book ∗, other references are [1147, 1296, 1511, 2280, 2281, 2872]. Throughout this section, K denotes a ﬁnite ﬁeld. However, almost all results of this section hold for arbitrary perfect ﬁelds. 12.1.1 Valuations and places 12.1.1 Deﬁnition An algebraic function ﬁeld over K is an extension ﬁeld F/K with the following properties: 1. There is an element x ∈F such that x is transcendental over K and the exten-sion F/K(x) has ﬁnite degree. 2. No element z ∈F \ K is algebraic over K. The ﬁeld K is the constant ﬁeld of F. 12.1.2 Remark 1. We often use the term function ﬁeld rather than algebraic function ﬁeld. 2. Property 2 in Deﬁnition 12.1.1 is often referred to as: K is algebraically closed in F, or K is the full constant ﬁeld of F. 3. If F/K is a function ﬁeld, then the degree [F : K(z)] is ﬁnite for every z ∈F \K. 4. Every function ﬁeld F/K can be generated by two elements, F = K(x, y), where the extension F/K(x) is ﬁnite and separable. Throughout this section, F/K always means a function ﬁeld over K. 12.1.3 Example (Rational function ﬁelds) The simplest example of a function ﬁeld over K is the rational function ﬁeld F = K(x), with x being transcendental over K. The elements of K(x) are the rational functions z = f(x)/g(x) where f, g are polynomials over K and g is not the zero polynomial. 12.1.4 Example (Elliptic and hyperelliptic function ﬁelds) Let F be an extension of the rational function ﬁeld K(x) of degree [F : K(x)] = 2. For simplicity we assume that charK ̸= 2. Then there exists an element y ∈F such that F = K(x, y), and y satisﬁes an equation over K(x) of the form y2 = f(x), with f ∈K[x] square-free (i.e., f is not divisible by the square of a polynomial h ∈K[x] of degree ≥1). One shows that F is rational if deg(f) = 1 or 2. F is an elliptic function ﬁeld if deg(f) = 3 or 4, and ∗The authors thank Springer Science+Business Media for their permission to use in Section 12.1 parts from H. Stichtenoth’s book Algebraic Function Fields and Codes, GTM 254, 2009. Curves over ﬁnite ﬁelds 407 it is a hyperelliptic function ﬁeld if deg(f) ≥5. See also Deﬁnition 12.1.108 and Example 12.1.109. A detailed exposition of elliptic and hyperelliptic function ﬁelds is given in Sections 12.2 and 12.4. 12.1.5 Remark In case of charK = 2, the deﬁnition of elliptic and hyperelliptic function ﬁelds requires some modiﬁcation, see [2714, Chapters 6.1, 6.2]. 12.1.6 Deﬁnition A valuation of F/K is a map ν : F →Z∪{∞} with the following properties: 1. ν(x) = ∞if and only if x = 0. 2. ν(xy) = ν(x) + ν(y) for all x, y ∈F. 3. ν(x + y) ≥min{ν(x), ν(y)} for all x, y ∈F. 4. There exists an element z ∈F such that ν(z) = 1. 5. ν(a) = 0 for all a ∈K \ {0}. 12.1.7 Remark The symbol ∞denotes an element not in Z such that ∞+∞= ∞+n = n+∞= ∞ and ∞> m for all m, n ∈Z. It follows that ν(x−1) = −ν(x) for every nonzero element x ∈F. Property 3 above is the Triangle Inequality. The following proposition is often useful. 12.1.8 Proposition (Strict triangle inequality) Let ν be a valuation of the function ﬁeld F/K and let x, y ∈F such that ν(x) ̸= ν(y). Then ν(x + y) = min{ν(x), ν(y)}. 12.1.9 Remark For a valuation ν of F/K, consider the following subsets O, O∗, P of F: O := {z ∈F \| ν(z) ≥0}, O∗:= {z ∈F \| ν(z) = 0}, P := {z ∈F \| ν(z) > 0}. Then O is a ring, O∗is the group of invertible elements (units) of O, and P is a maximal ideal of O. In fact, P is the unique maximal ideal of O, which means that O is a local ring. The ideal P is a principal ideal, which is generated by every element t ∈F with ν(t) = 1. For distinct valuations ν1, ν2, the corresponding ideals P1 = {z ∈F \| ν1(z) > 0} and P2 = {z ∈F \| ν2(z) > 0} are distinct. 12.1.10 Deﬁnition 1. A subset P ⊆F is a place of F/K if there exists a valuation ν of F/K such that P = {z ∈F \| ν(z) > 0}. The valuation ν is uniquely determined by the place P. Therefore we write ν =: νP and say that νP is the valuation corresponding to the place P. 2. If P is a place of F/K and νP is the corresponding valuation, then the ring OP := {z ∈F \| νP (z) ≥0} is the valuation ring of F corresponding to P. 3. An element t ∈F with νP (t) = 1 is a prime element at the place P. 4. Let PF := {P \| P is a place of F}. 12.1.11 Remark Since P is a maximal ideal of its valuation ring OP , the residue class ring OP /P is a ﬁeld. The constant ﬁeld K is contained in OP , and P ∩K = {0}. Hence one has a canonical embedding K , →OP /P. We always consider K as a subﬁeld of OP /P via this embedding. 12.1.12 Deﬁnition Let P be a place of F/K. 1. The ﬁeld FP := OP /P is the residue class ﬁeld of P. 408 Handbook of Finite Fields 2. The degree of the ﬁeld extension FP /K is ﬁnite and is the degree of the place P. We write deg P := [FP : K]. 3. A place P ∈PF is rational if deg P = 1. This means that FP = K. 4. For z ∈OP , denote by z(P) ∈FP the residue class of z in FP . For z ∈F \ OP , set z(P) := ∞. The map from F to FP ∪{∞} given by z 7→z(P) is the residue class map at P. 12.1.13 Remark For a rational place P ∈PF and an element z ∈OP , the residue class z(P) is the (unique) element a ∈K such that νP (z −a) > 0. In this case, one calls the map z 7→z(P) from OP to K the evaluation map at the place P. We note that the evaluation map is K-linear. This map plays an important role in the theory of algebraic–geometry codes, see Section 15.2. 12.1.14 Example We want to describe all places of the rational function ﬁeld K(x)/K. 1. Let h ∈K[x] be an irreducible monic polynomial. Every nonzero element z ∈ K(x) can be written as z = h(x)r · f(x) g(x) with polynomials f, g ∈K[x] which are relatively prime to h, and r ∈Z. Then the map νP : K(x) →Z ∪{∞} with νP (z) := r (and νP (0) := ∞) deﬁnes a valuation of K(x)/K. The corresponding place P is P = u(x) v(x) u, v ∈K[x], h divides u but not v . The residue class ﬁeld of this place is isomorphic to K[x]/(h) and therefore we have deg P = deg(h). 2. Another valuation of K(x)/K is deﬁned by ν(z) = deg(g) −deg(f) for z = f(x)/g(x) ̸= 0. The corresponding place is called the place at inﬁnity and is denoted by P∞or (x = ∞). It follows from the deﬁnition that P∞= f(x) g(x) deg(f) < deg(g) . The place P∞has degree one, that is, it is a rational place. 3. There are no places of K(x)/K other than those described in Parts 1 and 2. 4. For a ∈K, the polynomial x −a is irreducible of degree 1 and deﬁnes a place P of degree one. We sometimes denote this place as P = (x = a). The set K ∪{∞} is therefore in 1–1 correspondence with the set of rational places of K(x)/K via a ← →(x = a). 5. The residue class map corresponding to a place P = (x = a) with a ∈K is given as follows: If z = f(x)/g(x) ∈OP then g(a) ̸= 0 and z(P) =: z(a) = f(a)/g(a) ∈K. In order to determine z(∞) := z(P) at the inﬁnite place P = P∞, we write f(x) = anxn + · · · + a0 and g(x) = bmxm + · · · + b0 with anbm ̸= 0. Then z(∞) = 0 if n < m, z(∞) = ∞if n > m, and z(∞) = an/bn if n = m. Curves over ﬁnite ﬁelds 409 12.1.2 Divisors and Riemann–Roch theorem 12.1.15 Remark [2714, Corollary 1.3.2] Every function ﬁeld F/K has inﬁnitely many places. 12.1.16 Remark The following theorem states that distinct valuations of F/K are independent of each other. 12.1.17 Theorem (Approximation theorem) [2714, Theorem 1.3.1] Let P1, . . . , Pn ∈PF be pairwise distinct places of F. Let x1, . . . , xn ∈F and r1, . . . , rn ∈Z. Then there exists an element z ∈F such that νPi(z −xi) = ri for i = 1, . . . , n. 12.1.18 Deﬁnition Let F/K be a function ﬁeld, x ∈F and P ∈PF . 1. P is a zero of x if νP (x) > 0, and the integer νP (x) is the zero order of x at P. 2. P is a pole of x if νP (x) < 0. The integer −νP (x) is the pole order of x at P. 12.1.19 Remark 1. A nonzero element a ∈K has neither zeros nor poles. 2. For all x ̸= 0 and P ∈PF , P is a pole of x if and only if P is a zero of x−1. 12.1.20 Theorem [2714, Theorem 1.4.11] For x ∈F \ K the following hold: 1. x has at least one zero and one pole. 2. The number of zeros and poles of x is ﬁnite. 3. Let P1, . . . , Pr and Q1, . . . , Qs be all zeros and poles of x, respectively. Then r X i=1 νPi(x) deg Pi = s X j=1 −νQj(x) deg Qj = [F : K(x)]. 12.1.21 Deﬁnition 1. The divisor group of F/K is the free abelian group generated by the set of places of F/K. It is denoted by Div(F). The elements of Div(F) are divisors of F. That means, a divisor of F is a formal sum D = X P ∈PF nP P with nP ∈Z and nP ̸= 0 for at most ﬁnitely many P. The set of places with nP ̸= 0 is the support of D and denoted as supp D. If supp D ⊆{P1, . . . , Pk} then D is also written as D = n1P1 + · · · + nkPk where ni = nPi. Two divisors D = P P nP P and E = P P mP P are added coeﬃcientwise, that is D +E = P P (nP +mP )P. The zero divisor is the divisor 0 = P P rP P where all rP = 0. 2. A divisor of the form D = P with P ∈PF is a prime divisor. 3. The degree of the divisor D = P P nP P is deg D := X P ∈PF nP · deg P. We note that this is a ﬁnite sum since nP ̸= 0 only for ﬁnitely many P. 410 Handbook of Finite Fields 4. A partial order on Div(F) is deﬁned as follows: if D = P P nP P and E = P P mP P, then D ≤E if and only if nP ≤mP for all P ∈PF . A divisor D ≥0 is positive (or eﬀective). 12.1.22 Remark Since every nonzero element x ∈F has only ﬁnitely many zeros and poles, the following deﬁnitions are meaningful. 12.1.23 Deﬁnition For a nonzero element x ∈F, let Z and N denote the set of zeros and poles of x, respectively. 1. The divisor (x)0 := P P ∈Z νP (x)P is the zero divisor of x. 2. The divisor (x)∞:= −P P ∈N νP (x)P is the divisor of poles of x. 3. The divisor div(x) := P P ∈PF νP (x)P = (x)0 −(x)∞is the principal divisor of x. 12.1.24 Remark 1. We note that both divisors (x)0 and (x)∞are positive divisors. By Theorem 12.1.20, deg(x)0 = deg(x)∞and hence deg(div(x)) = 0. 2. For x ∈F \ K we have deg(x)0 = deg(x)∞= [F : K(x)]. The principal divisor of a nonzero element a ∈K is the zero divisor. We observe that for the element 0 ∈K, no principal divisor is deﬁned. 3. The sum of two principal divisors and the negative of a principal divisor are principal, since div(xy) = div(x)+div(y) and div(x−1) = −div(x). Therefore the principal divisors form a subgroup of the divisor group of F. 12.1.25 Example We consider again the rational function ﬁeld F = K(x). Let f ∈K[x] be a nonzero polynomial and write f as a product of irreducible polynomials, f(x) = a · p1(x)r1 · · · pn(x)rn, where 0 ̸= a ∈K and p1, . . . , pn are pairwise distinct, monic, irreducible polynomials. Let Pi be the place of K(x) corresponding to the polynomial pi (see Example 12.1.14), and P∞ be the place at inﬁnity. Then the principal divisor of f in Div(K(x)) is div(f) = r1P1 + · · · + rnPn −dP∞where d = deg(f). As every element of K(x) is a quotient of two polynomials, we thus obtain the principal divisor for any nonzero element z ∈K(x) in this way. 12.1.26 Deﬁnition 1. Two divisors D, E ∈Div(F) are equivalent if E = D + div(x) for some x ∈F. This is an equivalence relation on the divisor group of F/K. We write D ∼E if D and E are equivalent. 2. Princ(F) := {A ∈Div(F) \|A is principal} is the group of principal divisors of F. Curves over ﬁnite ﬁelds 411 3. The factor group Cl(F) := Div(F)/Princ(F) is the divisor class group of F. 4. For a divisor D ∈Div(F) we denote by [D] ∈Cl(F) its class in the divisor class group. 12.1.27 Remark The equivalence relation ∼as deﬁned in Deﬁnition 12.1.26 is often denoted as linear equivalence of divisors. 12.1.28 Remark 1. It follows from the deﬁnitions that D ∼E if and only if [D] = [E]. 2. D ∼E implies deg D = deg E. 3. In a rational function ﬁeld K(x), the converse of Part 2 also holds. If F/K is non-rational, then there exist, in general, divisors of the same degree which are not equivalent. 12.1.29 Deﬁnition Let F/K be a function ﬁeld and let A ∈Div(F) be a divisor of F. Then the set L(A) := {x ∈F \| div(x) ≥−A } ∪{0} is the Riemann–Roch space associated to the divisor A. 12.1.30 Proposition L(A) is a ﬁnite-dimensional vector space over K. 12.1.31 Deﬁnition For a divisor A, the integer ℓ(A) := dim L(A) is the dimension of A. We point out that dim L(A) denotes here the dimension as a vector space over K. 12.1.32 Remark 1. If A ∼B then the spaces L(A) and L(B) are isomorphic (as K-vector spaces). Hence A ∼B implies ℓ(A) = ℓ(B). 2. A ≤B implies L(A) ⊆L(B) and hence ℓ(A) ≤ℓ(B). 3. deg A < 0 implies ℓ(A) = 0. 4. L(0) = K and hence ℓ(0) = 1. 12.1.33 Remark The following theorem is one of the main results of the theory of function ﬁelds. 12.1.34 Theorem (Riemann–Roch theorem) [2714, Theorem 1.5.15] Let F/K be a funcion ﬁeld. Then there exist an integer g ≥0 and a divisor W ∈Div(F) with the following property: for all divisors A ∈Div(F), ℓ(A) = deg A + 1 −g + ℓ(W −A). 12.1.35 Deﬁnition The integer g =: g(F) is the genus of F, and the divisor W is a canonical divisor of F. 12.1.36 Remark [2714, Proposition 1.6.1] 1. If W ′ ∼W, then the equation above also holds when W is replaced by W ′. 412 Handbook of Finite Fields 2. Suppose that g1, g2 ∈Z and W1, W2 ∈Div(F) satisfy the equations ℓ(A) = deg A + 1 −g1 + ℓ(W1 −A) = deg A + 1 −g2 + ℓ(W2 −A) for all divisors A. Then g1 = g2 and W1 ∼W2. 3. As a consequence of 1 and 2, the canonical divisors of F/K form a uniquely determined divisor class [W] ∈Cl(F), the canonical class of F. 12.1.37 Corollary [2714, Corollary 1.5.16] Let W be a canonical divisor and g = g(F) the genus of F. Then deg W = 2g −2 and ℓ(W) = g. Conversely, every divisor C with deg C = 2g −2 and ℓ(C) = g is canonical. 12.1.38 Remark A slightly weaker version of the Riemann–Roch theorem is often suﬃcient. 12.1.39 Theorem (Riemann’s theorem) [2714, Theorem 1.4.17] Let F/K be a function ﬁeld of genus g. Then for all divisors A ∈Div(F), ℓ(A) ≥deg A + 1 −g. Equality holds for all divisors A with deg A > 2g −2. 12.1.40 Example Consider the rational function ﬁeld F = K(x). The following hold: 1. The genus of K(x) is 0. 2. Let P∞be the inﬁnite place of K(x), see Example 12.1.14. For every k ≥0 we obtain L(kP∞) = {f ∈K[x] \| deg(f) ≤k}. This shows that Riemann–Roch spaces are natural generalizations of spaces of polynomials. 3. The divisor W = −2P∞is canonical. 12.1.41 Remark Conversely, if F/K is a function ﬁeld of genus g(F) = 0, then there exists an element x ∈F such that F = K(x). (This does not hold in general if K is not a ﬁnite ﬁeld.) 12.1.42 Remark For divisors of degree deg A > 2g −2, Riemann’s Theorem gives a precise formula for ℓ(A). On the other hand, ℓ(A) = 0 if deg A < 0. For the interval 0 ≤deg A ≤2g −2, there is no exact formula for ℓ(A) in terms of deg A. 12.1.43 Theorem (Cliﬀord’s theorem) [2714, Theorem 1.6.13] For all divisors A ∈Div(F) with 0 ≤deg A ≤2g −2, ℓ(A) ≤1 + 1 2 · deg A. 12.1.44 Remark The genus g(F) of a function ﬁeld F is its most important numerical invariant. In general it is a diﬃcult task to determine g(F). Some methods are discussed in Subsection 12.1.3. Here we give upper bounds for g(F) in some special cases. 12.1.45 Remark Assume that F = K(x, y) is a function ﬁeld over K, where x, y satisfy an equation ϕ(x, y) = 0 with an irreducible polynomial ϕ(X, Y ) ∈K[X, Y ] of degree d. Then g(F) ≤(d −1)(d −2) 2 . Equality holds if and only if the plane projective curve which is deﬁned by the aﬃne equation ϕ(X, Y ) = 0, is nonsingular. (These terms are explained in Subsection 12.1.5.) 12.1.46 Remark (Riemann’s inequality) [2714, Corollary 3.11.4] Suppose that F = K(x, y). Then g(F) ≤([F : K(x)] −1)([F : K(y)] −1). Curves over ﬁnite ﬁelds 413 12.1.3 Extensions of function ﬁelds 12.1.47 Remark In this subsection we consider the following situation: F/K and F ′/K′ are function ﬁelds with F ⊆F ′ and K ⊆K′. We always assume that K (respectively K′) is algebraically closed in F (respectively in F ′) and that the degree [F : F ′] is ﬁnite. As before, K is a ﬁnite ﬁeld. 12.1.48 Remark The extension degree [K′ : K] divides [F ′ : F]. 12.1.49 Deﬁnition Let P ∈PF and P ′ ∈PF ′. The place P ′ is an extension of P (equivalently, P ′ lies over P, or P lies under P ′) if one of the following equivalent conditions holds: 1. P ⊆P ′, 2. OP ⊆OP ′, 3. P ′ ∩F = P, 4. OP ′ ∩F = OP . We write P ′\|P to indicate that P ′ is an extension of P. 12.1.50 Remark If P ′ lies over P then the inclusion OP , →OP ′ induces a natural embedding of the residue class ﬁelds FP , →F ′ P ′. We therefore consider FP as a subﬁeld of F ′ P ′ via this embedding. 12.1.51 Deﬁnition Let P ′ be a place of F ′ lying above P. 1. There exists an integer e ≥1 such that νP ′(z) = e · νP (z) for all z ∈F. This integer e =: e(P ′\|P) is the ramiﬁcation index of P ′\|P. 2. The degree f(P ′\|P) := [F ′ P ′ : FP ] is ﬁnite and is the relative degree of P ′\|P. 12.1.52 Remark Suppose that F ′′/K′′ is another ﬁnite extension of F ′/K′. Let P, P ′, P ′′ be places of F, F ′, F ′′ such that P ′\|P and P ′′\|P ′. Then e(P ′′\|P) = e(P ′′\|P ′) · e(P ′\|P) and f(P ′′\|P) = f(P ′′\|P ′) · f(P ′\|P). 12.1.53 Theorem (Fundamental equality) [2714, Theorem 3.1.11] Let P be a place of F/K. Then there exists at least one but only ﬁnitely many places of F ′ lying above P. If P1, . . . , Pm are all extensions of P in F ′ then m X i=1 e(Pi\|P)f(Pi\|P) = [F ′ : F]. 12.1.54 Corollary Let F ′/F be an extension of degree [F ′ : F] = n, and let P ∈PF . Then 1. For every place P ′ ∈PF ′ lying over P, e(P ′\|P) ≤n and f(P ′\|P) ≤n. 2. There are at most n distinct places of F ′ lying over P. 12.1.55 Deﬁnition Let F ′/F be an extension of degree [F ′ : F] = n, and let P ∈PF . 1. A place P ′ ∈PF ′ over P is ramiﬁed if e(P ′\|P) > 1, and it is unramiﬁed if e(P ′\|P) = 1. 2. P is ramiﬁed in F ′/F if there exists an extension of P in F ′ that is ramiﬁed. Otherwise, P is unramiﬁed in F ′. 3. P is totally ramiﬁed in F ′/F if there is a place P ′ of F ′ lying over P with e(P ′\|P) = n. It is clear that P ′ is then the only extension of P in F ′. 4. P splits completely in F ′/F if P has n distinct extensions P1, . . . , Pn in F ′. It is clear that P is then unramiﬁed in F ′. 414 Handbook of Finite Fields 12.1.56 Theorem [2714, Corollary 3.5.5] If F ′/F is a ﬁnite separable extension of function ﬁelds, then at most ﬁnitely many places of F are ramiﬁed in F ′/F. 12.1.57 Remark More precise information about the ramiﬁed places in F ′/F is given in Theorem 12.1.72. 12.1.58 Deﬁnition For P ∈PF one deﬁnes the conorm of P in F ′/F as ConF ′/F (P) := X P ′\|P e(P ′\|P) · P ′. For an arbitrary divisor of F we deﬁne its conorm as ConF ′/F X P nP P ! := X P nP · ConF ′/F (P). 12.1.59 Remark ConF ′/F is a homomorphism from the divisor group of F to the divisor group of F ′, which sends principal divisors of F to principal divisors of F ′. 12.1.60 Remark For every divisor A ∈Div(F), one has deg ConF ′/F (A) = [F ′ : F] [K′ : K] · deg A. In particular, if K′ = K then deg ConF ′/F (A) = [F ′ : F] · deg A. 12.1.61 Deﬁnition Let F ′/K′ be a ﬁnite extension of F/K, let P ∈PF and OP its valuation ring. 1. An element z ∈F ′ is integral over OP if there exist elements u0, . . . , um−1 ∈OP such that zm + um−1zm−1 + · · · + u1z + u0 = 0. Such an equation is an integral equation for z over OP . 2. The set O′ P := {z ∈F ′ \| z is integral over OP } is a subring of F ′. It is the integral closure of OP in F ′. 12.1.62 Proposition [2714, Chapter 3.2, 3.3] With notation as in Deﬁnition 12.1.61, the following hold: 1. z ∈F ′ is integral over OP if and only if the coeﬃcients of the minimal polynomial of z over F are in OP . 2. O′ P = T P ′\|P OP ′. 3. There exists a basis (z1, . . . , zn) of F ′/F such that O′ P = Pn i=1 ziOP , that is, every element z ∈F ′ which is integral over OP , has a unique representation z = P xizi with xi ∈OP . Such a basis (z1, . . . , zn) is an integral basis at the place P. 4. Every basis (y1, . . . , yn) of F ′/F is an integral basis for almost all places P ∈PF (that is, for all P with only ﬁnitely many exceptions). In particular, if F ′ = F(y) then (1, y, . . . , yn−1) is an integral basis for almost all P. Curves over ﬁnite ﬁelds 415 12.1.63 Remark Using integral bases one can often determine all extensions of a place P ∈PF in F ′. In the following theorem, denote by ¯ u := u(P) ∈FP the residue class of an element u ∈OP in the residue class ﬁeld FP = OP /P. For a polynomial ψ(T) = P uiT i ∈OP [T] we set ¯ ψ(T) := P ¯ uiT i ∈FP [T]. 12.1.64 Theorem (Kummer’s theorem) [2714, Theorem 3.3.7] Suppose that F ′ = F(y) with y integral over OP . Let ϕ ∈OP [T] be the minimal polynomial of y over F and decompose ¯ ϕ into irreducible factors over FP ¯ ϕ(T) = γ1(T)ϵ1 · · · γr(T)ϵr with distinct irreducible monic polynomials γi ∈FP [T] and ϵi ≥1. Choose monic polyno-mials ϕi ∈OP [T] such that ¯ ϕi = γi. Then the following hold: 1. For each i ∈{1, . . . , r} there exists a place Pi\|P such that ϕi(y) ∈Pi. The relative degree of Pi\|P satisﬁes f(Pi\|P) ≥deg(γi). 2. If (1, y, . . . , yn−1) is an integral basis at P, then there exists for each i ∈{1, . . . , r} a unique place Pi\|P with ϕi(y) ∈Pi, and we have e(Pi\|P) = ϵi and f(Pi\|P) = deg(γi). 3. If ¯ ϕ(T) = Qn i=1(T −ai) with distinct elements a1, . . . , an ∈K, then P splits completely in F ′/F. 12.1.65 Example Consider a ﬁeld K with charK ̸= 2 and a function ﬁeld F = K(x, y), where y satisﬁes an equation y2 = f(x) with a polynomial f(x) ∈K[x] of odd degree. Then [F : K(x)] = 2, and ϕ(T) = T 2 −f(x) is the minimal polynomial of y over K(x). Let a ∈K. 1. If f(a) is a nonzero square in K (that is, f(a) = c2 with 0 ̸= c ∈K), then the place (x = a) of K(x) (see Example 12.1.14) splits into two rational places of F. 2. If f(a) is a non-square in K, then the place (x = a) has exactly one extension Q in F, and deg Q = 2. 3. If a ∈K is a simple root of the equation f(x) = 0, then the place (x = a) of K(x) is totally ramiﬁed in F/K(x), and its unique extension P ∈PF is rational. For more examples see Section 12.5. 12.1.66 Remark In what follows, we assume that F ′/F is a separable extension of function ﬁelds of degree [F ′ : F] = n. As before, P denotes a place of F and O′ P is the integral closure of OP in F ′. By TrF ′/F : F ′ →F we denote the trace mapping. For information about separable extensions and the trace map, see any standard textbook on algebra, e.g., . 12.1.67 Deﬁnition 1. For P ∈PF , the set CP := {z ∈F ′ \| TrF ′/F (zO′ P ) ⊆OP } is the complementary module of P in F ′. 2. There is an element tP ∈F ′ such that CP = tP O′ P , and we deﬁne for P ′ ∈PF ′ with P ′\|P the diﬀerent exponent of P ′ over P as d(P ′\|P) := −νP ′(tP ). We observe that the element tP is not unique, but the diﬀerent exponent is well-deﬁned (independent of the choice of tP ). 416 Handbook of Finite Fields 12.1.68 Lemma [2714, Deﬁnition 3.4.3] 1. For all P ′\|P, d(P ′\|P) ≥0. 2. For almost all P ∈PF , d(P ′\|P) = 0 holds for all extensions P ′\|P in F ′. 12.1.69 Deﬁnition The diﬀerent of a ﬁnite separable extension of function ﬁelds F ′/F is the divisor of the function ﬁeld F ′ deﬁned as Diﬀ(F ′/F) := X P ∈PF X P ′\|P d(P ′\|P)P ′. 12.1.70 Theorem [2714, Theorems 3.4.6, 3.4.13] Let F ′/K′ be a ﬁnite separable extension of F/K. 1. If W is a canonical divisor of F/K, then the divisor W ′ := ConF ′/F (W) + Diﬀ(F ′/F) is a canonical divisor of F ′/K′. 2. (Hurwitz genus formula) The genera of F ′ and F satisfy the equation 2g(F ′) −2 = [F ′ : F] K′ : K −2) + deg Diﬀ(F ′/F). 12.1.71 Remark We note that Part 2 is an immediate consequence of Part 1 since the degree of a canonical divisor of F is 2g(F) −2. Next we give some results that help to compute the diﬀerent exponents d(P ′\|P). 12.1.72 Theorem (Dedekind’s diﬀerent theorem) [2714, Theorem 3.5.1] Let F ′/F be a ﬁnite sepa-rable extension of function ﬁelds, let P ∈PF and P ′ ∈PF ′ with P ′\|P. Then 1. d(P ′\|P) ≥e(P ′\|P) −1 ≥0. 2. d(P ′\|P) = e(P ′\|P) −1 if and only if the characteristic of F does not divide e(P ′\|P). 12.1.73 Remark In other words, the diﬀerent of F ′/F contains exactly the places of F ′ which are ramiﬁed in F ′/F. In particular it follows that only ﬁnitely many places are ramiﬁed. The following deﬁnition is motivated by Dedekind’s Diﬀerent Theorem. 12.1.74 Deﬁnition Assume that P ′\|P is ramiﬁed. 1. P ′\|P is tame if the characteristic of F does not divide e(P ′\|P). 2. P ′\|P is wild if the characteristic of F divides e(P ′\|P). 12.1.75 Lemma In a tower of separable extensions F ′′ ⊇F ′ ⊇F, the diﬀerent is transitive, that is: d(P ′′\|P) = d(P ′′\|P ′) + e(P ′′\|P ′) · d(P ′\|P) for P ′′ ⊇P ′ ⊇P, and hence Diﬀ(F ′′/F) = Diﬀ(F ′′/F ′) + ConF ′′/F ′(Diﬀ(F ′/F)). 12.1.76 Proposition [2714, Theorem 3.5.10] Let F ′ = F(y) be a separable extension of degree [F ′ : F] = n. Let P ∈PF and assume that the minimal polynomial ϕ of y has all of its coeﬃcients in OP . Let P1, . . . , Pr be all extensions of P in F ′. Then one has: Curves over ﬁnite ﬁelds 417 1. 0 ≤d(Pi\|P) ≤νPi(ϕ′(y)) for i = 1, . . . , r. 2. {1, y, . . . , yn−1} is an integral basis at P if and only if d(Pi\|P) = νPi(ϕ′(y)) for i = 1, . . . , r. Here ϕ′ denotes the derivative of ϕ in the polynomial ring F[T]. 12.1.77 Remark Recall that a ﬁnite ﬁeld extension F ′/F is Galois if the automorphism group G := {σ : F ′ →F ′ \| σ is an automorphism of F ′ which is the identity on F} has order ord G = [F ′ : F]. In this case, Gal(F ′/F) := G is the Galois group of F ′/F. 12.1.78 Remark If F ′/F is Galois and P is a place of F, the Galois group Gal(F ′/F) acts on the set of extensions of P via σ(P ′) = {σ(z) \| z ∈P ′}. 12.1.79 Proposition [2714, Theorem 3.7.1] Suppose that F ′/F is a Galois extension, and let P ∈PF . 1. The Galois group acts transitively on the set of extensions of P in F ′. That is, for any two extensions P1, P2 of P in F ′, there is an automorphism σ ∈Gal(F ′/F) such that P2 = σ(P1). 2. If P1, . . . , Pr are all extensions of P in F ′, then e(Pi\|P) = e(Pj\|P), f(Pi\|P) = f(Pj\|P), and d(Pi\|P) = d(Pj\|P) holds for all i, j = 1, . . . , r. 3. Setting e(P) := e(Pi\|P) and f(P) := f(Pi\|P), we have the equality e(P) · f(P) · r = [F ′ : F]. 12.1.80 Proposition (Kummer extensions) [2714, Proposition 3.7.3] Let F ′ = F(y) be an extension of function ﬁelds of degree [F ′ : F] = n, where the constant ﬁeld of F is the ﬁnite ﬁeld Fq. Assume that yn = u ∈F and n divides (q −1). Then F ′/F is Galois, and the Galois group Gal(F ′/F) is cyclic of order n. 1. For P ∈PF deﬁne rP := gcd(n, νP (u)), the greatest common divisor of n and νP (u). Then e(P ′\|P) = n rP and d(P ′\|P) = n rP −1 for all P ′\|P. 2. Denote by K (K′, respectively) the constant ﬁeld of F (F ′, respectively). Then g(F ′) = 1 + n [K′ : K] g(F) −1 + 1 2 X P ∈PF 1 −rP n deg P ! . 3. If K = K′ and F = K(x) is a rational function ﬁeld, then g(F ′) = −n + 1 + 1 2 X P ∈PF (n −gcd(n, νP (u))) deg P. 12.1.81 Remark Let F ′/F be a Galois extension of function ﬁelds of degree [F ′ : F] = n whose Galois group is cyclic. Suppose that n divides q −1 (where the constant ﬁeld of F is Fq). Then F ′ = F(y) with some element y satisfying yn ∈F. So Proposition 12.1.80 applies. 418 Handbook of Finite Fields 12.1.82 Example Assume that the characteristic of K is odd. Let F = K(x, y) with y2 = f(x), where f ∈K[x] is a square-free polynomial of degree deg(f) = 2m + 1. This means that f = f1 · · · fs with pairwise distinct irreducible polynomials fi ∈K[x]. Let Pi ∈PK(x) be the place corresponding to fi, i = 1, . . . , s, and P∞be the pole of x in K(x). For P ∈{P1, . . . , Ps, P∞} we have gcd(2, νP (f)) = 1, and for all other places Q ∈PK(x) we have νQ(f) = 0. Then Part 3 of the Proposition above yields g(F) = (deg(f) −1)/2 = m. Hence for every integer m ≥0 there exist function ﬁelds F/K of genus g(F) = m. 12.1.83 Proposition (Artin–Schreier extensions) [2714, Proposition 3.7.8] Let F/K be a function ﬁeld, where K is a ﬁnite ﬁeld of characteristic p. Let F ′ = F(y) with yp −y = u ∈F. We assume that for all poles P of u in F, p does not divide νP (u), and that u ̸∈K. Then the following hold: 1. F ′/F is Galois of degree [F ′ : F] = p, and F, F ′ have the same constant ﬁeld. 2. Exactly the poles of u are ramiﬁed in F ′/F (in fact, they are totally ramiﬁed); all other places of F are unramiﬁed. 3. Let P be a pole of u in F and let P ′ be the unique place of F ′ lying over P. Then the diﬀerent exponent of P ′\|P is d(P ′\|P) = (p −1)(−νP (u) + 1). 4. The genus of F ′ is given by the formula g(F ′) = p · g(F) + p −1 2  −2 + X P : νP (u)<0 (−νP (u) + 1) · deg P  . 12.1.84 Remark Every Galois extension F ′/F of degree [F ′ : F] = p = charK can be written as F ′ = F(y), where y satisﬁes an equation of the form yp −y = u ∈F. If moreover F = K(x) is a rational function ﬁeld, one can choose u in such a way that for all poles P of u in K(x), p does not divide νP (u). 12.1.85 Example Suppose that charK = p and F = K(x, y), where yp −y = f(x) ∈K[x], deg(f) = m and m is not divisible by p. Then F/K(x) is Galois of degree p and K is alge-braically closed in F. The pole of x is the only place of K(x) that is ramiﬁed in F/K(x), and the genus of F is g(F) = (p −1)(m −1)/2. 12.1.86 Deﬁnition The function ﬁeld F ′/K′ is a constant ﬁeld extension of F/K, if F ′ = FK′ (that is, if K′ = K(α) then F ′ = F(α)). 12.1.87 Remark If E/K′ is a ﬁnite extension of F/K (meaning that E/F is a ﬁnite extension and K′ is the constant ﬁeld of E), we consider the intermediate ﬁeld F ⊆F ′ := FK′ ⊆E. Then F ′/K′ is a constant ﬁeld extension of F/K, and E/F ′ is an extension of function ﬁelds having the same constant ﬁeld K′. 12.1.88 Theorem [2714, Chapter 3.6] Let F ′ = FK′ be a constant ﬁeld extension of F. Then the following hold: 1. [F ′ : F] = [K′ : K], and K′ is algebraically closed in F ′. 2. F ′/F is unramiﬁed, that is, all P ∈PF are unramiﬁed in F ′/F. 3. g(F ′) = g(F). 4. For every divisor A ∈Div(F), deg ConF ′/F (A) = deg A and ℓ(ConF ′/F (A)) = ℓ(A). Curves over ﬁnite ﬁelds 419 12.1.4 Diﬀerentials 12.1.89 Remark In this subsection we consider a function ﬁeld F/K where K = Fq is a ﬁnite ﬁeld of characteristic p. The aim is to give an interpretation of the canonical divisors of F. 12.1.90 Remark The set F p := {zp \| z ∈F} is a subﬁeld of F which contains K. The extension F/F p has degree [F : F p] = p and is purely inseparable. An element z ∈F \ F p is called a separating element for F/K. For every separating element z, the extension F/K(z) is ﬁnite and separable. 12.1.91 Remark Recall that a module over a ﬁeld L is just a vector space over L. 12.1.92 Deﬁnition Let M be a module over F. A derivation of F into M is a map δ : F →M, which is K-linear and satisﬁes the product rule δ(u · v) = u · δ(v) + v · δ(u) for all u, v ∈F. 12.1.93 Remark Let δ : F →M be a derivation of F, z ∈F and n ≥0. Then δ(zn) = nzn−1 · δ(z). In particular, δ(zp) = 0 for all z ∈F. 12.1.94 Proposition [2714, Proposition 4.1.4] Let x be a separating element for F/K. Then there exists a unique derivation δx : F →F with the property δx(x) = 1. We call δx the derivation of F with respect to x. 12.1.95 Proposition [2714, Chapter 4.1] There is a one–dimensional F-module ΩF and a derivation d : F →ΩF (written as z 7→dz) with the following properties: 1. dz ̸= 0 for every separating element z ∈F. 2. dz = δx(z) · dx for every z ∈F and x ∈F \ F p. The pair (ΩF , d) is the diﬀerential module of F/K, the elements of ΩF are diﬀerentials of F/K. 12.1.96 Remark 1. If z ∈F is not separating then dz = 0. 2. Given a separating element x ∈F, every diﬀerential ω ∈ΩF has a unique representation ω = udx with u ∈F, since ΩF is a one-dimensional F-module. 3. Suppose that ω, η ∈ΩF and ω ̸= 0. Then there is a unique element u ∈F such that η = uω. We write then u = η/ω. 4. Item 2 in Proposition 12.1.95 indicates that, for a separating element x ∈F, δx(z) = dz dx for all z ∈F. 12.1.97 Remark One can attach a divisor to every nonzero diﬀerential ω ∈ΩF as follows. 12.1.98 Deﬁnition [2714, Theorem 4.3.2(e)] Let ω ∈ΩF , ω ̸= 0. 1. Let P ∈PF and let t be a P-prime element (that is, νP (t) = 1). Then t is a separating element of F/K, and we can write ω = u · dt with u ∈F. We deﬁne νP (ω) := νP (u). This deﬁnition is independent of the choice of the prime element t, and one can show that νP (ω) = 0 for almost all P ∈PF . 420 Handbook of Finite Fields 2. The divisor of ω is div(ω) := X P ∈PF νP (ω)P. 12.1.99 Remark Divisors have the property div(uω) = div(u) + div(ω) for u ∈F \ {0} and ω ∈ΩF \ {0}. Therefore div(ω) ∼div(η) for any two nonzero diﬀerentials ω, η ∈ΩF . 12.1.100 Remark Recall that the divisor of poles of an element 0 ̸= x ∈F is denoted by (x)∞. 12.1.101 Proposition [2714, Chapter 4.3] Let x ∈F be a separating element for F/K. Then div(dx) = −2(x)∞+ Diﬀ(F/K(x)). 12.1.102 Theorem [2714, Chapter 4.3] Let ω ∈ΩF be a nonzero diﬀerential of F/K. Then the divisor W := div(ω) is a canonical divisor of F. In particular, 2g(F) −2 = deg(div(ω)). 12.1.103 Deﬁnition For every divisor A ∈Div(F), we deﬁne the set ΩF (A) := {ω ∈ΩF \| div(ω) ≥A}. 12.1.104 Remark ΩF (A) is a ﬁnite-dimensional K-vector space. 12.1.105 Theorem (Riemann–Roch theorem, 2nd version) For every divisor A ∈Div(F), ℓ(A) = deg A + 1 −g(F) + dim ΩF (A), where dim ΩF (A) means the dimension as a K-vector space. 12.1.106 Corollary We have dim ΩF (0) = g(F). 12.1.107 Remark We ﬁnish this subsection with examples of function ﬁelds that will be discussed in detail in Sections 12.2 and 12.4. 12.1.108 Deﬁnition 1. A function ﬁeld F/K of genus g(F) = 1 is an elliptic function ﬁeld. 2. A function ﬁeld F/K is hyperelliptic if g(F) ≥2, and there exists an element x ∈F such that [F : K(x)] = 2. 12.1.109 Example [2714, Chapters 6.1, 6.2] Let K be a ﬁnite ﬁeld of characteristic ̸= 2, and let F/K be an elliptic or hyperelliptic function ﬁeld of genus g. Assume that F has at least one rational place P. Then there exist x, y ∈F such that F = K(x, y) and y2 = f(x) with a square-free polynomial f ∈K[x] of degree 2g + 1. The diﬀerentials ωi := xi y dx , i = 0, . . . , g −1 form a basis of ΩF (0). Curves over ﬁnite ﬁelds 421 12.1.5 Function ﬁelds and curves 12.1.110 Remark There is an alternative geometric approach to function ﬁelds via algebraic curves. We give here only a very brief (and incomplete) introduction. For more information we refer to [1147, 1427, 2281]. 12.1.111 Remark Let K be a ﬁnite ﬁeld, and denote by ¯ K the algebraic closure of K. Let K[X1, . . . , Xn] be the ring of polynomials in n variables over K. 12.1.112 Deﬁnition 1. The n-dimensional aﬃne space An = An( ¯ K) over ¯ K is the set of all n-tuples of elements of ¯ K. An element P = (a1, . . . , an) ∈An is a point, and a1, . . . , an are its coordinates. 2. Let f1, . . . , fm ∈K[X1, . . . , Xn] be polynomials. Then the set V := {P ∈ An \| f1(P) = · · · = fm(P) = 0} is the aﬃne algebraic set deﬁned by f1 = · · · = fm = 0. We say that V is deﬁned over K since the polynomials f1, . . . , fm have coeﬃcients in K. 3. Let V be as in 2. The set I(V ) := {f ∈¯ K[X1, . . . , Xn] \| f(P) = 0 for all P ∈V } is an ideal of ¯ K[X1, . . . , Xn], which is the ideal of V . 4. The algebraic set V is absolutely irreducible if I(V ) is a prime ideal of ¯ K[X1, . . . , Xn]. Then the residue class ring Γ(V ) := ¯ K[X1, . . . , Xn]/I(V ) is an integral domain, and its quotient ﬁeld ¯ K(V ) := Quot(Γ(V )) is the ﬁeld of ratio-nal functions on V . The residue class of Xi in ¯ K(V ) is the i-th coordinate func-tion on V and is denoted by xi. The subﬁeld K(V ) := K(x1, . . . , xn) ⊆¯ K(V ) is the ﬁeld of K-rational functions on V . 5. An absolutely irreducible aﬃne algebraic set V is an absolutely irreducible aﬃne algebraic curve over K (brieﬂy, an aﬃne curve over K), if the ﬁeld K(V ) as deﬁned in Part 4 has transcendence degree one over K. This means that K(V ) is an algebraic function ﬁeld over K, as in Deﬁnition 12.1.1. The curve V is a plane aﬃne curve if V ⊆A2. 6. Let V be an aﬃne curve over K. A point P ∈V is K-rational if all its coordi-nates are in K. We set V (K) := {P ∈V \| P is K-rational}. 7. Two aﬃne curves V1 and V2 are birationally equivalent if their function ﬁelds K(V1) and K(V2) are isomorphic. 12.1.113 Example Let F/K be an algebraic function ﬁeld. Then there exist elements x, y ∈F such that F = K(x, y), and there is an irreducible polynomial f ∈K[X, Y ] such that f(x, y) = 0. Let V ⊆A2 be the plane aﬃne curve deﬁned by f = 0. Then K(V ) = F. 12.1.114 Deﬁnition 1. Let V be an aﬃne curve as in Deﬁnition 12.1.112, and let P ∈V . A rational function ϕ ∈¯ K(V ) is deﬁned at P if ϕ = g(x1, . . . , xn)/h(x1, . . . , xn) with g, h ∈¯ K[x1, . . . , xn] and h(P) ̸= 0. The set OP (V ) of all rational functions on V which are deﬁned at P, is a ring and it is the local ring of V at P. 2. The point P is non-singular if its local ring is integrally closed. This means, by deﬁnition, that every z ∈¯ K(V ) which satisﬁes an integral equation over OP (V ), is in OP (V ); see Deﬁnition 12.1.61. 3. The curve V is non-singular if all of its points are non-singular. 422 Handbook of Finite Fields 12.1.115 Remark Let f ∈K[X, Y ] be an absolutely irreducible polynomial (that is, f is irreducible in ¯ K[X, Y ]). Then the equation f = 0 deﬁnes a plane aﬃne curve C ⊆A2( ¯ K). A point P ∈C is non-singular if and only fX(P) ̸= 0 or fY (P) ̸= 0, where fX(X, Y ) and fY (X, Y ) denote the partial derivatives with respect to X and Y , respectively. 12.1.116 Example Let n > 0 be relatively prime to the characteristic of K. Then the Fermat curve C which is deﬁned by the equation f(X, Y ) = Xn + Y n −1 = 0, is non-singular. 12.1.117 Remark In a sense, aﬃne curves are not “complete,” one has to add a ﬁnite number of points “at inﬁnity.” To be precise, one introduces the projective space Pn over ¯ K and the “projective closure” of an aﬃne curve in Pn. This leads to the concept of projective curves. We do not give details here and refer to textbooks on algebraic geometry, for example [1147, 1427, 2281]. 12.1.118 Remark 1. Two projective curves are birationally equivalent if their function ﬁelds are iso-morphic. 2. For every projective curve C there exists a non-singular projective curve X which is birationally equivalent to C. The curve X is uniquely determined up to isomor-phism and it is the non-singular model of C. 12.1.119 Remark There is a 1–1 correspondence between {algebraic function ﬁelds F/K, up to iso-morhism} and {absolutely irreducible, non-singular, projective curves X deﬁned over K, up to isomorphism}. Under this correspondence, extensions F ′/F of function ﬁelds correspond to coverings X ′ →X of curves, composites of function ﬁelds E = F1F2 correspond to ﬁbre products of curves, etc. What corresponds to a place P of a function ﬁeld F/K? If P is rational, then it corresponds to a K-rational point of the associated projective curve. Now let K = Fq and let P be a place of F with deg P = n. Then P corresponds to exactly n points on the associated projective curve, with coordinates in the ﬁeld Fqn. These points form an orbit under the Frobenius map, which is the map that raises the coordinates of points to the q-th power. For details, see . See Also §12.2 For more material on function ﬁelds of genus 1 (elliptic curves). §12.4 For more material on hyperelliptic function ﬁelds. §12.5 For rational places of function ﬁelds (rational points on curves). §12.6 For towers of function ﬁelds. §15.2 For applications of function ﬁelds to coding theory. References Cited: [1147, 1296, 1427, 1511, 1846, 2280, 2281, 2714, 2872] 12.2 Elliptic curves Joseph Silverman, Brown University 12.2.1 Remark Most of the background material for this section may be found in [2670, Chap-ters III, V, XI]. Other standard references for the theory of elliptic curves include the books [557, 1563, 1756, 1773, 1843, 1845, 2054, 2107, 2667, 2672, 2950] and survey arti-cles [556, 2784]. Curves over ﬁnite ﬁelds 423 12.2.1 Weierstrass equations 12.2.2 Deﬁnition A Weierstrass equation over a ﬁeld K is an equation of the form y2 + a1xy + a3y = x3 + a2x2 + a4x + a6 with a1, a2, a3, a4, a6 ∈K. (12.2.1) Associated to a Weierstrass equation are the following quantities: b2 = a2 1 + 4a4, b4 = 2a4 + a1a3, b6 = a2 3 + 4a6, b8 = a2 1a6 + 4a2a6 −a1a3a4 + a2a2 3 −a2 4, c4 = b2 2 −24b4, c6 = −b3 2 + 36b2b4 −216b6, ∆= −b2 2b8 −8b3 4 −27b2 6 + 9b2b4b6, j = c3 4/∆. The quantity ∆is the discriminant and the quantity j is the j-invariant. 12.2.3 Remark The quantities in Deﬁnition 12.2.2 satisfy the relations 4b8 = b2b6 −b2 4 and 1728∆= c3 4 −c2 6. 12.2.4 Remark The discriminant vanishes if and only if the curve deﬁned by the Weierstrass equation has a singular point, i.e, a point (x0, y0) on the curve where both partial derivatives vanish: 2y0 + a1x0 + a3 = 0 and 3x2 0 + 2a2x0 + a4 −a1y0 = 0. 12.2.5 Remark If char(K) ̸= 2, then the substitution y 7− → 1 2(y −a1x −a3) transforms the Weierstrass equation into the simpler form y2 = 4x3 + b2x2 + 2b4x + b6. If also char(K) ̸= 3, then the substitution (x, y) 7− → x−3b2 36 , y 108 yields the further simpli-ﬁcation y2 = x3 −27c4x −54c6. 12.2.6 Deﬁnition An elliptic curve E deﬁned over a ﬁeld K is a Weierstrass equation over K that is nonsingular, i.e., ∆̸= 0, together with an extra point “at inﬁnity” which is denoted O. For any extension ﬁeld L of K, the set of points of E deﬁned over L is the set E(L) = solutions (x0, y0) ∈L2 to the Weier-strass equation (12.2.1) deﬁning E ∪{O}. 12.2.7 Remark A fancier deﬁnition of an elliptic curve over K is a nonsingular projective curve of genus one deﬁned over K with a marked point whose coordinates are in K. Using the Riemann–Roch theorem, one can show that every such curve is given by a Weierstrass equation, with the marked point being the point O, which is the unique point at inﬁnity; see [2670, III.3.1]. 12.2.8 Example The real points on an elliptic curve deﬁned over R may have one or two compo-nents, as illustrated in Figure 12.2.8∗. ∗The author thanks Springer Science+Business Media for their permission to include Figures 3.1 and 3.3 from his book The Arithmetic of Elliptic Curves, GTM 106, 2009. 424 Handbook of Finite Fields y2 = x3 −3x + 3 y2 = x3 + x y2 = x3 −x Figure 12.2.1 Three elliptic curves. 12.2.9 Proposition The substitution x − →u2x + r, y − →u3y + u2sx + t, (12.2.2) transforms the Weierstrass equation (12.2.1) into a Weierstrass equation y2 + a′ 1xy + a′ 3y = x3 + a′ 2x2 + a′ 4x + a′ 6 whose coeﬃcients and associated quantities satisfy ua′ 1 = a1 + 2s u2a′ 2 = a2 −sa1 + 3r −s2 u3a′ 3 = a3 + ra1 + 2t u4a′ 4 = a4 −sa3 + 2ra2 −(t + rs)a1 + 3r2 −2st u6a′ 6 = a6 + ra4 + r2a2 + r3 −ta3 −t2 −rta1 u2b′ 2 = b2 + 12r u4b′ 4 = b4 + rb2 + 6r2 u6b′ 6 = b6 + 2rb4 + r2b2 + 4r3 u8b′ 8 = b8 + 3rb6 + 3r2b4 + r3b2 + 3r4 u4c′ 4 = c4 u6c′ 6 = c6 u12∆′ = ∆ j′ = j 12.2.10 Deﬁnition Two elliptic curves E/K and E′/K are isomorphic over K if there is a sub-stitution (12.2.2) with u ∈K∗and r, s, t ∈K that transforms the Weierstrass equation of E into the Weierstrass equation of E′. 12.2.11 Theorem [2670, III.1.4] Let K be an algebraically closed ﬁeld. Then E/K and E′/K are iso-morphic over K if and only if they have the same j-invariant, i.e., if and only if j(E) = j(E′). 12.2.12 Example According to Theorem 12.2.11, there are two F2-isomorphism classes of elliptic curves deﬁned over F2, namely those with j-invariant 0 and those with j-invariant 1. How-ever, there are ﬁve F2-isomorphism classes of elliptic curves deﬁned over F2. An example from each isomorphism class is listed in the following table. curve j y2 + y = x3 0 y2 + y = x3 + x 0 y2 + y = x3 + x + 1 0 y2 + xy = x3 + 1 1 y2 + xy + y = x3 + 1 1 Curves over ﬁnite ﬁelds 425 12.2.13 Theorem [2670, X.5.4.1] Let p ≥5, let q be a power of p, and let E1/Fq and E2/Fq be elliptic curves. Then E1 and E2 are isomorphic over Fq if and only if j(E1) = j(E2) and there exists a u ∈F∗ q such that      c4(E)c6(E) = u2c4(E′)c6(E′) if j(E1) ̸= 0 and j(E1) ̸= 1728, c4(E) = u4c4(E′) if j(E1) = 1728, c6(E) = u6c6(E′) if j(E1) = 0. 12.2.14 Deﬁnition Let Eq be the set of Fq-isomorphism classes of elliptic curves deﬁned over Fq. 12.2.15 Remark The set E2 is described in Example 12.2.12; it has ﬁve elements. For p ≥5 and q a power of p, we have #Eq = 2(q −2) + #F∗ q/F∗ q 4 + #F∗ q/F∗ q 6. 12.2.16 Remark There are curves over ﬁnite ﬁelds Fq that have no points with coordinates in Fq, but a theorem of Lang says that this does not happen for curves of genus one. 12.2.17 Theorem [2670, Exercise 10.6] Let C/Fq be a smooth projective curve of genus one. Then C(Fq) is not empty. More generally, if V/Fq is a variety that is isomorphic over Fq to an abelian variety, then V (Fq) is not empty. 12.2.18 Remark In particular, if F(X, Y, Z) ∈Fq[X, Y, Z] is homogeneous of degree 3 and if the associated curve F = 0 is nonsingular, then there are values x, y, z ∈Fq, not all zero, such that F(x, y, z) = 0. 12.2.2 The group law 12.2.19 Deﬁnition Let E/K be an elliptic curve deﬁned over a ﬁeld K, and let L/K be an extension ﬁeld. The set of points E(L) forms an abelian group using the following rules: 1. The point O is the identity element. In what follows, we use the convention that O is on every vertical line. 2. The negative of the point P = (x, y) is the point −P = (x, −y−a1x−a3). If E is given by a simpler Weierstrass equation as in Remark 12.2.5, then a1 = a3 = 0, and −P = (x, −y) is the reﬂection of P about the x-axis. 3. The sum of distinct points P and Q is obtained by intersecting the line through P and Q with E. This yields three points P, Q, and R (counted with appropriate multiplicities). Then P + Q equals −R. 4. The sum of P with itself is obtained similarly, using the tangent line to E at P. The geometric deﬁnition of the group law is illustrated in Figure 12.2.2. 12.2.20 Remark All of the group axioms are easy to verify except for associativity; see [2670, III.3.4] for a proof of associativity. 12.2.21 Algorithm Let E/K be an elliptic curve deﬁned over a ﬁeld K, and let L/K be an extension ﬁeld. The following addition algorithm gives the group structure on the set of points E(L). 1. The point O is the identity element, so P + O = O + P = P for all P ∈E(L). 426 Handbook of Finite Fields s P s Q s R s P + Q Addition of distinct points P + Q + R = O s P s R s P + P s T T + T = O Adding a point to itself Figure 12.2.2 Examples of the addition law on an elliptic curve. 2. Let P0 = (x0, y0). Then −P0 = (x0, −y0 −a1x0 −a3). 3. Let P1 = (x1, y1) and P2 = (x2, y2). If x1 = x2 and y1 + y2 + a1x2 + a3 = 0, then P1 + P2 = O. 4. Otherwise, deﬁne λ and ν by the following formulas: λ ν x1 ̸= x2 y2 −y1 x2 −x1 y1x2 −y2x1 x2 −x1 x1 = x2 3x2 1 + 2a2x1 + a4 −a1y1 2y1 + a1x1 + a3 −x3 1 + a4x1 + 2a6 −a3y1 2y1 + a1x1 + a3 Then the sum P3 = P1 + P2 is given by P3 = (x3, y3) with x3 = λ2+a1λ−a2−x1−x2 and y3 = −(λ+a1)x3−ν−a3. 12.2.22 Remark A special case of the addition algorithm is the duplication formula. Let P = (x, y) ∈ E(L). Then the x-coordinate of 2P is x(2P) = x4 −b4x2 −2b6x −b8 4x3 + b2x2 + 2b4x + b6 . 12.2.23 Remark Algorithm 12.2.21 explains how to add and double points on an elliptic curve. For cryptographic applications, it is important to do these operations as eﬃciently as possible. There are tradeoﬀs between using aﬃne versus projective coordinates. It may also be possi-ble to use the Frobenius endomorphism in place of the doubling map for “double-and-add” algorithms, and alternative equations for elliptic curves may also allow for more eﬃcient operations. For details, see Sections 12.3 and 16.4. 12.2.24 Deﬁnition For a positive integer m, the multiplication-by-m map on E(L) is the map [m] : E(L) − →E(L), m = P + P + · · · + P \| {z } m terms . Curves over ﬁnite ﬁelds 427 For m < 0 deﬁne m = −−m, and set 0 = O. The kernel of multiplication-by-m is the subgroup E(L)[m] = {P ∈E(L) : m = O}. 12.2.25 Example The elliptic curve E : y2 = x3 + x + 1 over the ﬁeld F11 has discriminant ∆= 10 and j-invariant j = 9. The points P = (2, 0), Q = (6, 5), and R = (8, 9) are in E(F11). The addition algorithm (Algorithm 12.2.21) allows us to compute quantities such as P + Q = (8, 9), Q + R = (4, 4), Q = (0, 1), P + R = (4, 5). The group E(F11) has 14 elements. The orders of the elements P, Q, and R are given by P = Q = R = O. So for example, the kernel of multiplication by 7 in E(F11) consists of the seven points E(F11) = n : 0 ≤n ≤6 , where Q = (6, 5). However, going to an extension ﬁeld, one ﬁnds that E(F11) contains 49 points; see The-orem 12.2.60. 12.2.3 Isogenies and endomorphisms 12.2.26 Deﬁnition Let E1/K and E2/K be elliptic curves. An isogeny from E1 to to E2 is a map φ : E1 →E2 that is deﬁned using rational functions and that sends the zero point on E1 to the zero point on E2. If the coeﬃcients of the rational functions are in K, the isogeny is deﬁned over K. If there is an isogeny from E1 to E2, then E1 and E2 are isogenous. 12.2.27 Deﬁnition Let E/K be an elliptic curve. An endomorphism is an isogeny from E to itself. An automorphism of E is an endomorphism of E that has an inverse, i.e., an isomorphism of E with itself. 12.2.28 Deﬁnition The set of isogenies from E1 to E2 is denoted Hom(E1, E2), the set of en-domorphisms of E is denoted End(E), and the set of automorphisms of E is de-noted Aut(E). A subscript K indicates that we take only maps that are deﬁned over K, thus HomK(E1, E2), EndK(E), and AutK(E). 12.2.29 Example The multiplication-by-m map [m] : E →E is an endomorphism of E. 12.2.30 Example Let K be a ﬁeld with char(K) ̸= 2, and let a, b ∈K satisfy b(a2 −4b) ̸= 0. Then the elliptic curves E1 : y2 = x3 + ax2 + bx, E2 : Y 2 = X3 −2aX2 + (a2 −4b)X, are isogenous via the map φ : E1 − →E2, (x, y) 7− → y2 x2 , y(b −x2) x2 . 428 Handbook of Finite Fields There is also an isogeny going the other direction, ˆ φ : E2 − →E1, (X, Y ) 7− → Y 2 4X2 , Y (a2 −4b −X2) 8X2 . A direct computation shows that ˆ φ◦φ = on E1 and φ ◦ˆ φ = on E2. The maps φ and ˆ φ are examples of dual isogenies as described in Theorem 12.2.41. 12.2.31 Example Let K be a ﬁeld of characteristic p > 0, let q be a power of p, let E/K be an elliptic curve given by a Weierstrass equation (12.2.1), and deﬁne an elliptic curve E(q)/K using the Weierstrass equation E(q) : y2 + aq 1xy + aq 3y = x3 + aq 2x2 + aq 4x + aq 6. Then the Frobenius map φq : E(K) − →E(q)(K), φq(x, y) = (xq, yq), is an isogeny from E to E(q). Note that if K = Fq, then E = E(q), so in this case φq is an endomorphism of E. Further, P ∈E(Fq) : φq(P) = P = E(Fq), and more generally, since φqn = φn q , P ∈E(Fq) : φn q (P) = P = E(Fqn). 12.2.32 Proposition [2670, II.2.3] Let E1/K and E2/K be elliptic curves and let φ : E1 →E2 be an isogeny. Then either φ(P) = O for all P ∈E1(K), or else φ E1(K) = E2(K). (The constant map φ(P) = O is the zero isogeny.) 12.2.33 Deﬁnition In general, the degree of a ﬁnite map φ : C1 →C2 between algebraic curves is the degree of the extension of function ﬁelds K(C1)/φ∗K(C2). The map is separable if the ﬁeld extension K(C1)/φ∗K(C2) is separable, and otherwise it is inseparable (which can only happen in ﬁnite characteristic). The inseparability degree of φ, denoted degi(φ), is the inseparability degree of the extension K(E1)/φ∗K(E2). Thus φ is separable if and only if degi(φ) = 1, and φ is purely inseparable if degi(φ) = deg(φ). 12.2.34 Example The Frobenius map φq deﬁned in Example 12.2.31 is purely inseparable. In gen-eral, for integers m and n, the map [m + nφq] : E(K) − →E(q)(K), m + nφq = m + n), is separable if and only if gcd(m, q) = 1. 12.2.35 Proposition [2670, III.4.10] Let φ : E1 →E2 be a nonconstant isogeny of elliptic curves deﬁned over K. If φ is separable, then φ is unramiﬁed, and for every point Q ∈E2(K) we have # P ∈E1(K) : φ(P) = Q = deg(φ). More generally, for every nonconstant isogeny φ, # P ∈E1(K) : φ(P) = Q = deg(φ) degi(φ). Curves over ﬁnite ﬁelds 429 12.2.36 Proposition [2670, II.2.12] Let φ : E1 →E2 be a nonconstant isogeny of elliptic curves deﬁned over a ﬁeld of characterstic p, and let q = degi(φ) be the inseparability degree of φ. Then φ factors as E1 φq − →E(q) 1 ψ − →E2, where φq is the Frobenius map (Example 12.2.31) and ψ is separable. 12.2.37 Remark An isogeny is only required to send zero to zero, but it turns out that this suﬃces to force it to be a homomorphism, as described in the next result. 12.2.38 Theorem [2670, III.4.8] Let E1/K and E2/K be elliptic curves and let φ : E1 →E2 be an isogeny. Then the map φ : E1(K) − →E2(K) is a homomorphism, i.e., φ(P + Q) = φ(P) + φ(Q) for all P, Q ∈E1(K). 12.2.39 Deﬁnition The sum of two isogenies φ, ψ ∈Hom(E1, E2) is the map (φ + ψ) : E1(K) − →E2(K), (φ + ψ)(P) = φ(P) + ψ(P). In this way, Hom(E1, E2) becomes a group. The product of two endomorphisms φ, ψ ∈ End(E) is the composition (φψ) : E(K) − →E(K), (φψ)(P) = φ ψ(P) . Addition and multiplication of endomorphisms give End(E) the structure of a (not necessarily commutative) ring. Similarly, multiplication (composition) gives Aut(E) the structure of a group; it is the group of units in the ring End(E). 12.2.40 Remark For every isogeny E1 →E2 there is an isogeny going in the opposite direction, as described in the next theorem. 12.2.41 Theorem [2670, III.6.1, III.6.2] Let E1/K and E2/K be a elliptic curves, and let φ : E1 → E2 be an isogeny of degree n deﬁned over K. 1 Then there is a unique isogeny ˆ φ : E2 →E1 satisfying ˆ φ ◦φ = [n] on E1. It is the dual isogeny to φ. It is deﬁned over K and has the same degree as φ. 2. The dual isogeny satisﬁes φ ◦ˆ φ = [n] on E2. 3. Let λ : E2 →E3 be another isogeny. Then [ λ ◦φ = ˆ φ ◦ˆ λ. 4. Let ψ : E1 →E2 be another isogeny. Then \ φ + ψ = ˆ φ + ˆ ψ. 5. Let m ∈Z. Then c [m] = [m]. 6. ˆ ˆ φ = φ. 12.2.42 Remark Let E/K be an elliptic curve. Recall that Div(E) is the group of formal sums P P ∈E(K) nP (P), where the nP are integers and only ﬁnitely many of them are nonzero. Also Pic(E), the Picard group of E, is the quotient of Div(E) by the subgroup of principal divisors, i.e., divisors of functions. So there is an exact sequence 1 − →K ∗− →K(E)∗− →Div(E) − →Pic(E) − →0. (For background on divisors, see Section 12.1.) On an elliptic curve, the group law can be used to describe the principal divisors, as in the next result. 430 Handbook of Finite Fields 12.2.43 Proposition [2670, III.3.5] Let E/K be an elliptic curve. A divisor D = P nP (P) ∈Div(E) is principal if and only if X P ∈E(K) nP = 0 and X P ∈E(K) nP = O. (We note that the ﬁrst sum is a sum of integers, while the second sum is a sum of points on the elliptic curve E.) 12.2.44 Proposition [2670, III.3.4] Let E/K be an elliptic curve, let Div0(E) be the group of divisors of degree 0, and let Pic0(E) be the corresponding group of divisor classes. Then there is an isomorphism E(K) − →Pic0(E), P 7− →divisor class of (P) −(O). The inverse is the map that sends the divisor class of P nP (P) to the sum of points PnP . 12.2.4 The number of points in E(Fq) 12.2.45 Remark If E/Fq is an elliptic curve deﬁned over a ﬁnite ﬁeld, then E(Fq) is a ﬁnite (abelian) group. 12.2.46 Theorem (Hasse–Weil estimate) [2670, V.3.1] Let E/Fq be an elliptic curve deﬁned over a ﬁnite ﬁeld. Then q + 1 −#E(Fq) ≤2√q. 12.2.47 Deﬁnition Let E/Fq be an elliptic curve. The quantity aq(E) = q + 1 −#E(Fq) is the trace of the Frobenius map; see Theorem 12.2.66. 12.2.48 Remark The following theorem of Birch describes how aq(E) is distributed as E ranges over all isomorphism classes of elliptic curves deﬁned over Fq. 12.2.49 Theorem , [2101, Appendix B]. For each E ∈Eq (see Deﬁnition 12.2.14), write aq(E) = 2√q cos θq(E) with 0 ≤θq(E) ≤π. Then for all 0 ≤α ≤β ≤π, lim q→∞ #{E ∈Eq : α ≤θq(E) ≤β} #Eq = 2 π Z β α sin2(t) dt. 12.2.50 Remark The number of points in E(Fq) is constrained by Theorem 12.2.46. The exact orders that occur are given in the following theorem. 12.2.51 Theorem Let q = pn be a prime power, and let b be an integer with \|b\| ≤2√q. Then there exists an elliptic curve E/Fq with #E(Fq) = q + 1 −b if and only if b satisﬁes one of the following conditions: 1. gcd(b, p) = 1; 2. n is even and b = ±2√q; 3. n is even and p ̸≡1 (mod 3) and b = ±√q; 4. n is odd and p equals 2 or 3 and b = ±p(n+1)/2; 5. n is odd and b = 0; Curves over ﬁnite ﬁelds 431 6. n is even and p ̸≡1 (mod 4) and b = 0. 12.2.52 Remark Waterhouse proved a generalization of Theorem 12.2.51 for abelian varieties of arbitrary dimension. Deuring had earlier proven the theorem in the case that q is prime, in which case every possible value of #E(Fp) allowed by the Hasse–Weil constraint (Theorem 12.2.46) occurs. See also R¨ uck’s description of all possible structures for the group E(Fq) subject to the constraints provided by Theorem 12.2.51. 12.2.53 Remark For elliptic curves over ﬁnite ﬁelds, the number of points on the curve determines its isogeny class, as in the following result. 12.2.54 Theorem Let E1/Fq and E2/Fq be elliptic curves. Then the following are equiva-lent: 1. #E1(Fq) = #E2(Fq); 2. E1 and E2 are Fq-isogenous; 3. End(E1) ⊗Q = End(E2) ⊗Q. 12.2.5 Twists 12.2.55 Deﬁnition Let E/K be an elliptic curve. A twist of E is an elliptic curve E′/K such that E′ is isomorphic to E over K, but not necessarily isomorphic over K. Two twists E′/K and E′′/K are equivalent if they are isomorphic over K. 12.2.56 Remark If E′ and E′′ are equivalent, then clearly E′(K) ∼ = E′′(K), but if E′ and E′′ are inequivalent twists of E, then E′(K) and E′′(K) may diﬀer. 12.2.57 Proposition [2670, X.5.4] Assume that char(K) ̸= 2, 3. Let E/K be an elliptic curve. The set of equivalence classes of twists of E/K is isomorphic to K∗/(K∗)d, where d = 2 if j(E) / ∈{0, 1728}, d = 4 if j(E) = 1728, and d = 6 if j(E) = 0. (Equivalently, d = # Aut(E); see Theorem 12.2.83.) Precisely, for D ∈K∗/(K∗)d, the corresponding twist E′ and iso-morphism E′ →E deﬁned over K( d √ D ) are as follows: [d = 2] E : y2 = x3 + Ax + B, E′ : y2 = x3 + D2Ax + D3B, (x, y) 7→(D−1x, D−3/2y). [d = 4] E : y2 = x3 + Ax, E′ : y2 = x3 + DAx, (x, y) 7→(D−1/2x, D−3/4y). [d = 6] E : y2 = x3 + B, E′ : y2 = x3 + DB, (x, y) 7→(D−1/3x, D−1/2y). 12.2.58 Remark Let E/K and E′/K be quadratic twists (d = 2) as in Proposition 12.2.57, E : y2 = x3 + Ax + B, E′ : y2 = x3 + D2Ax + D3B, (x, y) 7→(D−1x, D−3/2y). Let Gal K( √ D)/K = {1, σ}. Then the group E′(K) may be identiﬁed with a subgroup of E K( √ D) via E′(K) ∼ = P ∈E K( √ D) : σ(P) = −P , just as E(K) is the subgroup of E K( √ D) deﬁned by the relation σ(P) = P. With this identiﬁcation, the kernel and cokernel of the natural map E(K) ⊕E′(K) − →E K( √ D) , (P, Q) 7− →P + Q, are ﬁnite 2-groups. 12.2.59 Proposition Let p ≥5, let q be a power of p, let E/Fq be an elliptic curve, and let d > 1 divide # Aut(E). Then Proposition 12.2.57 says that for each D ∈F∗ q/(F∗ q)d, the curve E/Fq has a twist ED/Fq. The orders of the groups of points on the twists satisfy the following relations: 432 Handbook of Finite Fields 1. P D∈F∗ q/(F∗ q)d #ED(Fq) = d(q −1); 2. Q D∈F∗ q/(F∗ q)d #ED(Fq) = #E(Fqd). 12.2.6 The torsion subgroup and the Tate module 12.2.60 Theorem [2670, III.6.4] Let E/K be an elliptic curve, and let K be an algebraic closure of K. 1. Let p = char(K). Then E(K)[m] ∼ = Z/mZ × Z/mZ for all m ≥1 with p ∤m. (If char(K) = 0, then this holds for all m ≥1.) 2. If char(K) = p > 0, then one of the following is true E(K)[pr] ∼ = Z/prZ for all r ≥1, E(K)[pr] = 0 for all r ≥1. 12.2.61 Example Let E be an elliptic curve deﬁned by a Weierstrass equation (12.2.1). One can determine conditions on the coordinates of a point P = (x, y) ∈E for it to be a torsion point. For example 2 = O if and only if 2y + a1x + a3 = 0 if and only if 4x4 −b4x2 −2b6x −b8 = 0, and 3 = O if and only if 3x4 + b2x3 + 3b4x2 + 3b6x + b8 = 0. In general, there is a division polynomial ψn(x, y) ∈K[x, y] with the property that n = O if and only if ψn(x, y) = 0; see [2670, Exercise 3.7]. These polynomials can be computed recursively using the formula ψn+mψn−mψ2 r = ψn+rψn−rψ2 m −ψm+rψm−rψ2 n, valid for all n > m > r. 12.2.62 Remark Letting m = ℓi run over larger and larger powers of a prime ℓ, we obtain a module over the ring of ℓ-adic integers Zℓ. This is convenient because Zℓis a ring of characteristic zero. 12.2.63 Deﬁnition Let E/K be an elliptic curve. The ℓ-adic Tate module of E is the inverse limit Tℓ(E) = lim ←E(K)[ℓn]. If the characteristic of K is diﬀerent from ℓ, then Tℓ(E) ∼ = Zℓ× Zℓ, although the isomorphism depends on choosing a basis. 12.2.64 Remark Let E/Fq be an elliptic curve, where q is a power of p. The Frobenius map Fq : E(Fq) − →E(Fq) (Example 12.2.31) maps E(Fq)[m] to E(Fq)[m]. If p ∤m, then E(Fq)[m] is a free Z/mZ-module of rank two, so choosing a Z/mZ-basis T1, T2 for E(Fq)[m], the Frobenius map satisﬁes φq(T1) = aT1 + bT2 and φq(T2) = cT1 + dT2 for some a, b, c, d ∈Z/mZ. Thus the action of φq on E(Fq)[m] is represented by the matrix ˜ φq,m = a b c d . The matrix depends on the choice of the basis for E(Fq)[m], but its trace and determinant do not. Curves over ﬁnite ﬁelds 433 The maps ˜ φq,ℓi : E(Fq)[ℓi] →E(Fq)[ℓi] ﬁt together to give a map φq,ℓ: Tℓ(E) →Tℓ(E). Choosing a basis for Tℓ(E) ∼ = Z2 ℓallows us to evaluate the trace and the determinant of φq,ℓ as ℓ-adic numbers. 12.2.65 Remark The following theorem explains why the quantity aq(E) in Deﬁnition 12.2.47 is the trace of the Frobenius map. 12.2.66 Theorem [2670, V.2.6] Let E/Fq be an elliptic curve with q a power of the prime p. Then for every integer m with p ∤m, Tr (˜ φq,m) ≡q + 1 −#E(Fq) (mod m) and det(˜ φq,m) ≡q (mod m), and for every prime ℓ̸= p, Tr (φq,ℓ) = q + 1 −#E(Fq) and det(φq,ℓ) = q. 12.2.67 Remark More generally, any isogeny φ : E1 →E2 induces a homomorphism of the associ-ated Tate modules Tℓ(E1) →Tℓ(E2), and if φ is deﬁned over K, then the induced map is Gal(K/K)-invariant. 12.2.68 Theorem Let E1/Fq and E2/Fq be elliptic curves deﬁned over a ﬁnite ﬁeld, and let ℓ be a prime diﬀerent from the characteristic of Fq. Then the natural map HomK(E1, E2) ⊗Zℓ− →HomZℓ Tℓ(E1), Tℓ(E2) Gal(K/K) is an isomorphism. 12.2.69 Remark Theorem 12.2.68 is also true for elliptic curves deﬁned over number ﬁelds. This was conjectured by Tate and proven by Faltings . 12.2.7 The Weil pairing and the Tate pairing 12.2.70 Theorem (Weil pairing) [2670, III.8.1, III.8.2] Let E/K be an elliptic curve, let m ≥1 be an integer that is prime to the characteristic of K, and let µm ⊂K ∗denote the group of m-th roots of unity. There is a pairing em : E(K)[m] × E(K)[m] − →µm with the following properties: 1. It is bilinear, i.e., em(P + Q, R) = em(P, R)em(Q, R) and em(P, Q + R) = em(P, Q)em(P, R). 2. It is alternating, i.e., em(P, P) = 1, so in particular em(Q, P) = em(P, Q)−1. 3. It is non-degenerate, i.e., if em(P, Q) = 1 for all Q ∈E(K)[m], then P = 0. 4. It is Galois invariant, i.e., em σ(P), σ(Q) = σ em(P, Q) for all σ ∈Gal(K/K). 5. It is compatible in towers, i.e., emn(P, Q) = em(n, Q) for P ∈E(K)[mn] and Q ∈E(K)[n]. 6. It respects duality, i.e., let φ : E1 →E2 be an isogeny, then em(P, ˆ φ(Q)) = em(φ(P), Q) for P ∈E1(K)[m] and Q ∈E2(K)[m]. 12.2.71 Remark The non-degeneracy and Galois invariance of the Weil pairing imply that there exist points P, Q ∈E(K)[m] such that em(P, Q) is a primitive m-th root of unity. In particular, if E(K)[m] ⊂E(K), then µm ⊂K. 434 Handbook of Finite Fields 12.2.72 Remark The Weil pairings on E(K)[ℓi] ﬁt together to give a bilinear, alternating, non-degenerate, Galois invariant pairing eℓ∞: Tℓ(E) × Tℓ(E) − →Tℓ(µ), where Tℓ(µ) is the inverse limit of µℓi. Further, eℓ∞(P, ˆ φ(Q)) = eℓ∞(φ(P), Q). 12.2.73 Theorem Each of the following recipes computes the Weil pairing em(P, Q). 1. Choose a function fP ∈K(E) whose divisor satisﬁes div(fP ) = m(P) −m(O). Choose a function gP ∈K(E) satisfying fP ◦[m] = gm P . Then the quan-tity gP (S + Q)/gP (S) does not depend on the point S ∈E(K), and its value is em(P, Q). 2. Choose arbitrary points S, T ∈E(K) and choose functions FP and FQ in K(E) whose divisors satisfy div(FP ) = m(S + P) −m(S) and div(FQ) = m(T + Q) − m(T). Then the quantity FQ(T + P) FQ(T) FP (S + Q) FP (S) does not depend on S or T and is equal to em(P, Q). 3. Let fP ∈K(E) have divisor div(fP ) = m(P) −m(O) as in Part 1. This de-termines fP up to multiplication by a constant. The function fP has a pole of order m at O, so the function (x/y)mfP is deﬁned and nonzero at O. We ad-just the constant so that (x/y)mfP (O) = 1. Similarly, let fQ ∈K(E) have divisor div(fQ) = m(Q) −m(O), and normalize fQ in the same way. Then em(P, Q) = (−1)mfP (Q)/fQ(P). 12.2.74 Remark The fact that functions with the stated properties used in Theorem 12.2.73 exist follows from Proposition 12.2.43, which explains when a divisor is the divisor of a function. However, it is far from obvious that the formulas in Theorem 12.2.73 yield the same value; see [2670, page 462] for the equality of formulas of Parts 1 and 2, but note that the last line of [2670, page 462] should be replaced by = e(Q, P) = e(P, Q)−1. 12.2.75 Theorem (Tate Pairing) [2670, XI §9] Let E/K be an elliptic curve and let m ≥1 be an integer that is prime. There is a bilinear pairing T : E(K)[m] × E(K)/mE(K) − →K∗/(K∗)m deﬁned as follows: Let T ∈E(K)[m] and P ∈E(K). Choose a point Q ∈E(K) with m = P. Then there exists an α ∈K∗such that em σ(Q) −Q, T = ( m √α)σ/ m √α for all σ ∈Gal(K/K), and we set T(T, P) = α mod (K∗)m. The Tate pairing may be computed by choosing a function fT ∈K(E) with divisor div(fT ) = m(T) −m(O), and then T(T, P) = fT (P + Q)/fT (Q) for any Q ∈E(K) such that the functions are deﬁned and nonzero. 12.2.76 Remark If m is large, it is not clear in practice how to compute the functions used to evaluate the Weil pairing (Theorem 12.2.73) and the Tate pairing (Theorem 12.2.75). A double-and-add algorithm due to Miller allows these pairings to be computed quite eﬃ-ciently; see Theorem 16.4.38. 12.2.77 Remark There are functorial deﬁnitions of the Weil and Tate pairings from which our rather ad hoc deﬁnitions may be derived. Brieﬂy, the Weil pairing is a pairing between the m torsion on an abelian variety A and the m-torsion on its dual ˆ A ∼ = Ext(A, Gm), combined Curves over ﬁnite ﬁelds 435 with an identiﬁcation of ˆ A with the Picard group Pic0(A). The Tate pairing is a cup product pairing on Galois cohomology that uses the Weil pairing to map A[m] ⊗ˆ A[m] to µm. For details, see for example . 12.2.8 The endomorphism ring and automorphism group 12.2.78 Deﬁnition Let A be a ﬁnite Q-algebra, i.e., A is a ring that contains Q as a subring and that is a ﬁnite dimensional Q-vector space. An order of A is a subring of A that is ﬁnitely generated as a Z-module and that contains a Q-basis of A. A maximal order is an order that is contained in no other orders. 12.2.79 Example Let D ∈Z be a positive integer that is not a perfect square and let Z[δ] be the ring of integers of Q( √ −D). For example, we can take δ = −D+√−D 2 . Then every order in Q( √ −D) has the form Z + fZ[δ] for some integer f ≥1. The integer f is the conductor of the order Z + fZ[δ]. 12.2.80 Deﬁnition A (deﬁnite) quaternion algebra over Q is a non-commutative ring of the form Q+Q√ai+Q √ bj+Q √ abk, where a, b ∈Q are positive numbers and i, j, k are quantities satisfying the multiplication rules i2 = j2 = k2 = ijk = −1. 12.2.81 Remark A number ﬁeld has a unique maximal order, its ring of integers. Quaternion alge-bras may have many maximal orders. 12.2.82 Theorem [2670, III.9.4] Let E/K be an elliptic curve. The endomorphism ring End(E) of E has one of the following forms: 1. End(E) = Z; 2. End(E) is an order in a quadratic imaginary ﬁeld Q( √ −D); 3. End(E) is an order in a quaternion algebra. (This form is only possible if K is a ﬁnite ﬁeld.) 12.2.83 Theorem [2670, III.10.1] Let E/K be an elliptic curve. Then its automorphism group Aut(E) is a ﬁnite group whose order is given in the following table, where j(E) is the j-invariant of E: # Aut(E) j(E) char(K) 2 j(E) ̸= 0, 1728 — 4 j(E) = 1728 char(K) ̸= 2, 3 6 j(E) = 0 char(K) ̸= 2, 3 12 j(E) = 0 = 1728 char(K) = 3 24 j(E) = 0 = 1728 char(K) = 2 12.2.84 Example Let K be a ﬁeld whose characteristic is neither 2 nor 3, and let A, B ∈K∗with 4A3 + 27B2 ̸= 0. Then EA,B : y2 = x3 + Ax + B is an elliptic curve whose automorphism group is as follows: 1. If AB ̸= 0, then Aut(EA,B) = µ2. 2. If B = 0, then j(E) = 1728 and Aut(EA,0) = µ4, where ζ = (ζ2x, ζy) for ζ ∈µ4. 3. If A = 0, then j(E) = 0 and Aut(E0,B) = µ6, where ζ = (ζ2x, ζ3y) for ζ ∈µ6. 436 Handbook of Finite Fields 12.2.85 Remark Theorem 12.2.51 lists the possible values of #E(Fq). These determine the associ-ated endomorphism rings as described in the following theorem. 12.2.86 Theorem Let E/Fq be an elliptic curve, let φq ∈End(E) be the q-power Frobenius map, and let A = End(E) ⊗Q. Referring to the classiﬁcation in Theorem 12.2.51: In Case 1, A = Q(φq) is a quadratic imaginary ﬁeld and End(E) is an arbitrary order in A. In Case 2, A is a quaternion algebra, φq ∈Z, and End(E) is a maximal order in A. In Cases 3–6, A = Q(φq) is a quadratic imaginary ﬁeld and End(E) is an order in A whose conductor is not divisible by p. 12.2.9 Ordinary and supersingular elliptic curves 12.2.87 Theorem [2670, V.3.1] Let E/K be an elliptic curve deﬁned over a ﬁeld of characteristic p > 0. The following are equivalent: 1. E[pn] = 0 for some n ≥1, equivalently for all n ≥1. 2. The dual of the Frobenius map, ˆ φpn : E(pn) →E, is purely inseparable for some n ≥1, equivalently for all n ≥1. 3. The multiplication-by-pn map [pn] : E →E is purely inseparable for some n ≥1, equivalently for all n ≥1. 4. The endomorphism ring End(E) is an order in a quaternion algebra. 5. The formal group associated to E has height 2. (See [2670, IV §1] for the con-struction of the formal group associated to an elliptic curve and [2670, IV §7] for the deﬁnition the height of a formal group.) 12.2.88 Deﬁnition Let K be a ﬁeld of positive characteristic p. An elliptic curve E/K is supersin-gular if one (equivalently all) of the conditions in Theorem 12.2.87 are true. Otherwise E is ordinary. 12.2.89 Theorem [2670, V.3.1] If E is supersingular, then j(E) ∈Fp2. 12.2.90 Remark We comment that despite their name, supersingular elliptic curves are nonsingular curves, since they are elliptic curves, which are nonsingular by deﬁnition. 12.2.91 Remark In characteristic 2 one can check that E/F2 is supersingular if and only if j(E) = 0. There is thus only one F2-isomorphism class of such curves, a representative being y2 + y = x3. However, there are three F2-isomorphism classes of supersingular elliptic curves; see Example 12.2.12. 12.2.92 Remark The following theorem describes all supersingular curves in characteristic p ≥3. 12.2.93 Theorem [2670, V.4.1] Let q = pn with p ≥3. 1. Let E/Fq be an elliptic curve given by a Weierstrass equation of the form y2 = f(x) with f ∈Fq[x] a monic cubic polynomial. Then E is supersingular if and only if the coeﬃcient of xp−1 in f(x)(p−1)/2 is zero. 2. Let λ ∈Fp. Then the elliptic curve y2 = x(x −1)(x −λ) is supersingular if and only if λ is a root of the polynomial Hp(T) = p−1 2 X i=0 p−1 2 i 2 T i. Curves over ﬁnite ﬁelds 437 3. The polynomial Hp(T) has distinct roots in Fp. There is one supersingular elliptic curve in characteristic 3. For p ≥5, the number of Fp-isomorphism classes of supersingular elliptic curves is p −1 12 + 1 2Ap + 1 2Bp, where Ap = ( 0 if p ≡1 (mod 3), 1 if p ≡2 (mod 3), and Bp = ( 0 if p ≡1 (mod 4), 1 if p ≡3 (mod 4). Equivalently, this number is equal to j p 12 k +      0 if p ≡1 (mod 12), 1 if p ≡5 or 7 (mod 12), 2 if p ≡11 (mod 12). 12.2.94 Example The factorization of the ﬁrst few polynomials Hp(T) modulo p are listed in the following table. p Hp(T) (mod p) 3 T + 1 5 T 2 + 4T + 1 7 (T + 1)(T + 3)(T + 5) 11 (T + 1)(T + 5)(T + 9)(T 2 + 10T + 1) 13 (T 2 + 4T + 9)(T 2 + 7T + 1)(T 2 + 12T + 3) 17 (T 2 + T + 16)(T 2 + 14T + 1)(T 2 + 16T + 1)(T 2 + 16T + 16) 19 (T + 1)(T + 9)(T + 17)(T 2 + 4T + 1)(T 2 + 13T + 6)(T 2 + 18T + 16) 12.2.95 Example The elliptic curve y2 = x3 + 1 is supersingular over Fp if p ≡2 (mod 3) and ordinary if p ≡1 (mod 3). Similarly, the curve y2 = x3 + x is supersingular over Fp if p ≡3 (mod 4) and ordinary if p ≡1 (mod 4). 12.2.96 Remark [2670, V.4.2] Associated to the elliptic curve y2 = x(x −1)(x −T) is the Picard– Fuchs diﬀerential operator D = 4T(1 −T)d2 dT 2 + 4(1 −2T)d dT −1. The polynomial Hp(T) in Theorem 12.2.93, Part 3 satisﬁes DHp(T) = p p−1 2 X i=0 (p −2 −4i) p−1 2 i 2 T i. In particular, DHp(T) = 0 in characteristic p, which shows that Hp(T) has simple roots in Fp, since Hp(0) = 1 and Hp(1) = (−1)(p−1)/2 in Fp. 12.2.97 Theorem [2670, Exercise 5.9] The following mass formula for supersingular elliptic curves is due to Eichler and Deuring: X E/Fp supersingular 1 # Aut(E) = p −1 24 . 438 Handbook of Finite Fields 12.2.98 Remark Let E/Q be an elliptic curve given by a Weierstrass equation with integer coeﬃ-cients. Reducing the coeﬃcients modulo p gives an elliptic curve ˜ Ep/Fp for all primes p ∤∆. If End(E) is an order in a quadratic imaginary ﬁeld k, then ˜ Ep is supersingular if p is inert in k and ordinary if p is split in k; see Deﬁnition 12.4.3. 12.2.99 Remark The situation if End(E) = Z is more complicated. Serre and Elkies [969, 2590] have proven that SS(X) = #{p < X : ˜ Ep is supersingular} is smaller that X3/4+ϵ as X →∞. Lang and Trotter have conjectured that SS(X) is asymptotic to C √ X/ log(X) for a certain positive constant C. In the opposite direction, Elkies has proven that ˜ Ep is supersingular for inﬁnitely many p, i.e., SS(X) →∞as X →∞. 12.2.10 The zeta function of an elliptic curve 12.2.100 Remark This section describes the zeta function of an elliptic curve. See Sections 11.6 and 12.7 for zeta functions of arbitrary curves and higher dimensional algebraic varieties. 12.2.101 Deﬁnition Let E/Fq be an elliptic curve. The zeta function of E/Fq is the formal power series Z(E/Fq, T) = exp ∞ X n=1 #E(Fqn) n T n 12.2.102 Remark It might seem more natural to use the series P∞ n=1 #E(Fqn)T n, but the series deﬁning Z(E/Fq, T) has better transformation properties. 12.2.103 Theorem [2670, V.2.4] Let E/Fq be an elliptic curve, and let a = aq(E) be the trace of Frobenius for E (Deﬁnition 12.2.47). Then Z(E/Fq, T) = 1 −aT + qT 2 (1 −T)(1 −qT). (12.2.3) 12.2.104 Remark The zeta function satisﬁes Z E/Fq, 1 qT = Z(E/Fq, T). This is an instance of Poincar´ e duality. For the general statement of Poincar´ e duality and the associated functional equation for zeta functions of varieties over ﬁnite ﬁelds, see The-orems 12.7.18 and 12.7.20. 12.2.105 Remark The formula (12.2.3) for Z(E/Fq, T) is equivalent to the statement that if we factor 1 −aT + qT 2 = (1 −αT)(1 −βT) over the complex numbers, then #E(Fqn) = qn + 1 −αn −βn for all n ≥1. Further, the Hasse–Weil estimate (Theorem 12.2.46) is equivalent to the statement that α and β are complex conjugates satisfying \|α\| = \|β\| = √q. So in particular, #E(Fqn) −qn −1 = \|αn + βn\| ≤\|α\|n + \|β\|n = 2qn/2. 12.2.106 Example Consider the curve E/F2 deﬁned by the Weierstrass equation E : y2 + xy = x3 + 1. Curves over ﬁnite ﬁelds 439 Then #E(F2) = 4 and a2(E) = −1. The roots of 1 + T + 2T 2 are −1±√−7 2 , so for all n ≥1, #E(F2n) = 2n + 1 − −1 + √−7 2 n − −1 −√−7 2 n . Using this formula, it is easy to compute #E(F2n) for large values of n, for example, #E(F2173) = 2173 + 1 −67870783603944754053042229. 12.2.11 The elliptic curve discrete logarithm problem 12.2.107 Remark The elliptic curve discrete logarithm problem (ECDLP) underlies the use of elliptic curves in cryptography. In this section we discuss ECDLP and some related problems. For applications to cryptography, see Section 16.4. 12.2.108 Deﬁnition Let E/Fq be an elliptic curve and let P, Q ∈E(Fq). A (discrete) logarithm of Q to the base P is an integer N such that Q = N. The discrete logarithm, which is denoted by logP (Q), is well-deﬁned modulo the order mP of the element P in the group E(Fq), so one may view logP as a group homomorphism logP : {Q ∈E(Fq) : Q is a multiple of P} − →Z/mP Z. The elliptic curve discrete logarithm problem (ECDLP) is the problem of comput-ing logP (Q) for given points P and Q. (Note the analogy with the classical discrete logarithm problem for the multiplicative group F∗ q; see Section 11.6.) 12.2.109 Remark If the order mP of P is prime, then the fastest known general algorithm for solving the ECDLP (as of 2012) has running time on the order of √mP . This may be compared to the DLP in F∗ q, for which there are algorithms with running times that are subexponential in log q. 12.2.110 Deﬁnition Let E/Fq be an elliptic curve and let P ∈E(Fq). The (computational) el-liptic curve Diﬃe–Hellman problem (ECDHP-comp) is the following: Given the values of P, M, and N, compute the value of MN. 12.2.111 Deﬁnition The (decisional) elliptic curve Diﬃe–Hellman problem (ECDHP-dec) is the following: Given the values of P, M, and N, with better than equal probability, distinguish between the points MN and a randomly chosen point Q. 12.2.112 Deﬁnition The embedding degree of the integer m in the ﬁeld Fq is the smallest integer k such that µm ⊂F∗ qk, where µm is the group of m-th roots of unity. Equivalently, it is the smallest integer k such that qk ≡1 (mod m). 12.2.113 Remark The importance of the embedding degree is that it describes the degree of an extension ﬁeld over which the Weil and Tate pairings are deﬁned. 12.2.114 Remark , [2670, XI.6.1] Let E/Fq, m ≥1, and T ∈E(Fq)[m] be as in Theo-rem 12.2.75, and let T be the Tate pairing. Then Menezes, Okamoto, and Vanstone have noted that the ECDLP in E(Fq) can be reduced to the DLP in Fq, since if N = logP (Q), then T(P, Q) = T(P, P)N. Similarly, the decisional ECDHP is as easy to solve as comput-ing the Tate pairing, since (with rare exceptions) T([M]P, [N]P) = T(P, Q) if and only if Q = [MN]P. 440 Handbook of Finite Fields 12.2.115 Deﬁnition An elliptic curve E/Fp is anomalous if ap(E) = 1, that is, if #E(Fp) = p. 12.2.116 Remark Let p ≥3 and let E/Fp be an anomalous elliptic curve. Then Araki, Satoh, Semaev, and Smart [2535, 2581, 2682] observed that the ECDLP in E(Fp) can be solved in essentially linear time [2670, XI.6.5]. See Also §10.5 For elliptic curves and Latt es dynamical systems. §11.6 For the discrete logarithm problem for the multiplicative group F∗ q. §12.3 For eﬃcient addition formulas on elliptic curves. §12.7, §12.9 For zeta functions of higher genus curves and higher dimensional algebraic varieties. §16.4, §16.5, For cryptographic applications of elliptic curves, hyperelliptic curves, §16.6 and abelian varieties. References Cited: [284, 556, 557, 825, 968, 969, 1024, 1563, 1756, 1773, 1843, 1845, 1848, 2054, 2079, 2101, 2107, 2497, 2535, 2581, 2590, 2667, 2670, 2671, 2672, 2682, 2783, 2784, 2950, 2954] 12.3 Addition formulas for elliptic curves Daniel J. Bernstein, University of Illinois at Chicago Tanja Lange, Technische Universiteit Eindhoven 12.3.1 Curve shapes 12.3.1 Remark Section 12.2 deﬁned elliptic curves using Weierstrass equations (12.2.1). The fol-lowing deﬁnitions present other curve shapes which have algorithmic advantages. 12.3.2 Deﬁnition A short Weierstrass curve over a ﬁeld K of characteristic not equal to 2 is a curve of the form y2 = x3 + ax + b with a, b ∈K and 4a3 + 27b2 ̸= 0. 12.3.3 Deﬁnition A short Weierstrass curve over a ﬁeld K of characteristic 2 is a curve of the form y2 + xy = x3 + ax2 + b with a, b ∈K and b ̸= 0. 12.3.4 Remark The curve shape deﬁned in Deﬁnition 12.3.3 covers only ordinary curves; supersin-gular curves over a ﬁeld K of characteristic 2 have short Weierstrass equations of the form y2 + cy = x3 + ax + b with a, b, c ∈K and c ̸= 0. 12.3.5 Deﬁnition A Montgomery curve over a ﬁeld K of characteristic not equal to 2 is a curve of the form by2 = x3 + ax2 + x with a ∈K \ {−2, 2} and b ∈K \ {0}. Curves over ﬁnite ﬁelds 441 12.3.6 Deﬁnition An Edwards curve over a ﬁeld K of characteristic not equal to 2 is a curve of the form x2 + y2 = 1 + dx2y2 with d ∈K \ {0, 1}. A twisted Edwards curve over a ﬁeld K of characteristic not equal to 2 is a curve of the form ax2 + y2 = 1 + dx2y2 with a, d ∈K \ {0} and a ̸= d. 12.3.7 Deﬁnition A Hessian curve over a ﬁeld K is a curve of the form x3 + y3 + 1 = dxy with d ∈K and d3 ̸= 27. A twisted Hessian curve over a ﬁeld K is a curve of the form ax3 + y3 + 1 = dxy with a, d ∈K, a ̸= 0, and d3 ̸= 27a. 12.3.8 Remark Other curve shapes studied in the literature include binary Edwards curves, Ja-cobi quartics, Jacobi intersections, and Doche-Icart-Kohel curves. Chudnovsky and Chud-novsky’s work studied addition formulas on Hessian curves, Jacobi quartics, Jacobi intersections, and Weierstrass curves. Montgomery curves were proposed by Montgomery in in the context of the elliptic-curve method to factor integers. Edwards curves were introduced by Edwards in in the form x2 + y2 = a2(1 + x2y2), and generalized to x2 + y2 = 1 + dx2y2 by Bernstein and Lange in . Twisted Edwards curves were introduced by Bernstein, Birkner, Joye, Lange, and Peters in . 12.3.2 Addition 12.3.9 Remark Given a nonsingular projective genus one curve with a speciﬁed neutral element O, one can abstractly deﬁne addition of points P on the curve to correspond to addition of divisors P −O modulo principal divisors. However, computations use more concrete addition laws speciﬁed as rational functions of the input coordinates. 12.3.10 Remark Addition on Weierstrass curves is described concretely in Deﬁnition 12.2.19 and Algorithm 12.2.21. It is necessary to distinguish generic additions from doublings and from computations involving O as input or output; the generic addition formulas fail for those cases. There are other addition formulas that do not require so many distinctions. An addition law is strongly uniﬁed if it can be used to double generic points. It is K-complete (abbre-viated complete when K is clear from context) if it can be used to add each pair of points deﬁned over K. 12.3.11 Example Consider the twisted Edwards curve ax2 + y2 = 1 + dx2y2 over K with neutral element (0, 1). The negative of (x, y) on this curve is (−x, y). The Edwards addition law, valid for almost all points P = (x1, y1) and Q = (x2, y2) on the curve, states that the sum P + Q is x1y2 + y1x2 1 + dx1x2y1y2 , y1y2 −ax1x2 1 −dx1x2y1y2 . The Edwards addition law is strongly uniﬁed: it does not make an exception for P = Q. If a is a square in K and d is not a square in K then the Edwards addition law is K-complete. 12.3.12 Example Consider the twisted Hessian curve ax3 + y3 + 1 = dxy over K with neutral element (0, −1). The negative of (x, y) on this curve is (x/y, 1/y). The rotated Hessian addition law, valid for almost all points P = (x1, y1) and Q = (x2, y2) on the curve, states that the sum P + Q is x1 −y2 1x2y2 ax1y1x2 2 −y2 , y1y2 2 −ax2 1x2 ax1y1x2 2 −y2 . 442 Handbook of Finite Fields This addition law is strongly uniﬁed. It is K-complete if a is not a cube in K. 12.3.13 Remark The completeness of the Edwards addition law was proven by Bernstein and Lange in . Earlier addition laws, including the Weierstrass addition law and the usual (non-rotated) Hessian addition law, were not complete. 12.3.3 Coordinate systems 12.3.14 Remark Additions in aﬃne coordinates require inversions in the underlying ﬁeld K. Inver-sions are computationally expensive compared to multiplications and additions. Implemen-tations thus work with fractions, delaying the inversions for as long as possible. Mathemat-ically this means working in projective coordinates. 12.3.15 Deﬁnition The projective representations over a ﬁeld K of a vector (x, y) ∈K2 are the vectors (X, Y, Z) ∈K3 such that Z ̸= 0, X/Z = x, and Y/Z = y. 12.3.16 Remark The conditions Z ̸= 0, X/Z = x, and Y/Z = y are equivalent to (X : Y : Z) ∈P2(K) mapping to (x, y) ∈A2(K) under the natural rational map from P2(K) to A2(K). For computational purposes, however, projective representations of points are ele-ments (X, Y, Z) ∈K3, not equivalence classes (X : Y : Z) ∈P2(K). 12.3.17 Remark Weighted projective coordinates sometimes lead to more eﬃcient formulas. In (a, b, 1)-weighted coordinates, the point (X, Y, Z) ∈K3 with Z ̸= 0 represents (X/Za, Y/Zb) ∈K2; any scaled point (Xλa, Y λb, Zλ) with λ ∈K∗represents the same point in K2; one deﬁnes (X : Y : Z) as the set of such scaled points. For example, Jacobian coordinates for Weierstrass curves are (2, 3, 1)-weighted coordinates. Standard projective coordinates are (1, 1, 1)-weighted coordinates. 12.3.18 Remark Extra coordinates sometimes lead to more eﬃcient formulas. For example, ex-tended coordinates for Edwards curves represent an aﬃne point (x, y) as any element of the equivalence class (x : y : 1 : xy) in (1, 1, 1, 1)-weighted coordinates; i.e., (X, Y, Z, T) with Z ̸= 0 and XY = ZT represents (X/Z, Y/Z). 12.3.19 Example The twisted Edwards curve ax2 + y2 = 1 + dx2y2 is naturally embedded into P1 × P1 as aX2T 2 + Y 2Z2 = Z2T 2 + dX2Y 2 with x = X/Z and y = Y/T for Z, T ̸= 0. In this representation addition can be written as follows: ((X1 : Z1), (Y1 : T1)) + ((X2 : Z2), (Y2 : T2)) =          (X1Y2Z2T1 + X2Y1Z1T2 : Z1Z2T1T2 + dX1X2Y1Y2), (Y1Y2Z1Z2 −aX1X2T1T2 : Z1Z2T1T2 −dX1X2Y1Y2) if deﬁned, (X1Y1Z2T2 + X2Y2Z1T1 : aX1X2T1T2 + Y1Y2Z1Z2), (X1Y1Z2T2 −X2Y2Z1T1 : X1Y2Z2T1 −X2Y1Z1T2) if deﬁned. Bernstein and Lange showed in that these two addition laws cover all possible pairs of curve points; the outputs coincide if they are both deﬁned; each deﬁned output is on the curve; and this addition turns the set of curve points into a group. An analogous 56-monomial pair of addition laws for Weierstrass curves was presented by Bosma and Lenstra in before Edwards curves were introduced. Curves over ﬁnite ﬁelds 443 12.3.4 Explicit formulas 12.3.20 Remark One can compute the scalar multiple nP for any positive integer n by writing n as P i ni2i with ni ∈{0, 1} and computing nP as P i ni2iP. This uses about log2 n doublings and, typically, about (1/2) log2 n additions. The number of group operations to compute nP is reduced by more advanced scalar-multiplication methods such as windows, sliding windows, and fractional windows. 12.3.21 Remark Algorithms to compute elliptic-curve group operations using ﬁeld operations are referred to as explicit formulas. These algorithms try to minimize the cost of group opera-tions and of higher-level operations such as scalar multiplication. These algorithms translate addition formulas such as in Algorithm 12.2.21, and Examples 12.3.11, and 12.3.12 into se-quences of ﬁeld operations; they reduce cost by reusing intermediate results and applying various optimization techniques. Explicit formulas depend on the curve shape. A complete addition algorithm is typically built as a combination of an incomplete addi-tion formula with various special-case formulas, glued together by appropriate comparisons between P and Q, as in Algorithm 12.2.21. 12.3.22 Deﬁnition Explicit formulas to compute P, Q 7→P + Q for generic P, Q are addition formulas. Explicit formulas to compute P 7→2P for generic P are doubling formulas. Explicit formulas to compute P, Q, P −Q 7→P + Q for generic P, Q are diﬀerential addition formulas. 12.3.23 Remark One way to compute a scalar multiplication P 7→nP using diﬀerential additions, where n = Pl i=0 ni2i, is to compute the pairs (Lj, Rj) = (Pl i=j ni2i−jP, P +Pl i=j ni2i−jP) for l ≥j ≥0 recursively as follows: if nj−1 = 0 then (Lj−1, Rj−1) = (2Lj, Lj+Rj); otherwise (Lj−1, Rj−1) = (Lj + Rj, 2Rj). Note that the diﬀerences in the addition are equal to P and that the doublings are additions of points with diﬀerence O. This scalar-multiplication method was introduced for Montgomery curves in and is called the Montgomery ladder. 12.3.24 Deﬁnition Addition formulas with inputs and output in projective coordinates are pro-jective addition formulas. Addition formulas with one input and output in projective coordinates but the other input in aﬃne coordinates are mixed addition formulas. 12.3.25 Remark Aﬃne coordinates are equivalent to projective coordinates with Z = 1; mixed addition formulas eliminate multiplications by 1 and sometimes save time in other ways. The literature also contains various speedups for other types of restricted representations, such as additions of projective points having X = 1 or of two points having the same Z-coordinate. 12.3.26 Remark Often an addition formula involves some computations that depend only on one input point. A readdition of the same input point reuses those computations. 12.3.27 Remark The following subsections state the most eﬃcient explicit formulas known for the most popular curve shapes used in computations. Cost is reported as squarings (S), mul-tiplications by curve constants (D), and general multiplications (M); multiplications by curve constants are counted separately because often these constants can be chosen small. Additions and subtractions are ignored. Coordinate systems here are chosen primarily to minimize doubling cost and secondarily to minimize addition cost, where cost counts multi-plications and a fraction of squarings; the fraction is 0.8 in characteristic ̸= 2 and 0.2 in char-acteristic 2. See the Explicit Formulas Database ( 444 Handbook of Finite Fields for a much more comprehensive collection of curve shapes, coordinate systems, computer-veriﬁed addition formulas, and references. 12.3.5 Short Weierstrass curves, large characteristic: y2 = x3 −3x + b 12.3.28 Remark The following algorithms choose weights (2, 3, 1) and a = −3. These choices min-imize the cost of known doubling algorithms on short Weierstrass curves, except for a few curves (at most 6 isomorphism classes) having a = 0. All of the standard NIST curves over prime ﬁelds have a = −3, and almost all curves over prime ﬁelds have low-degree isogenies to curves with a = −3. 12.3.29 Algorithm: Doubling. Input: P1 = (X1 : Y1 : Z1). Output: P3 = (X3 : Y3 : Z3) = 2P1. δ = Z2 1; γ = Y 2 1 ; β = X1γ; α = 3(X1 −δ)(X1 + δ); X3 = α2 −8β; Z3 = (Y1 + Z1)2 −γ −δ; Y3 = α(4β −X3) −8γ2. 12.3.30 Algorithm: Addition. Input: P1 = (X1 : Y1 : Z1), P2 = (X2 : Y2 : Z2). Output: P3 = (X3 : Y3 : Z3) = P1 + P2. A = Z2 1; B = Z2 2; U1 = X1 · B; U2 = X2 · A; S1 = Y1 · Z2 · B; S2 = Y2 · Z1 · A; H = U2 −U1; I = (2H)2; J = H ·I; r = 2(S2−S1); V = U1·I; X3 = r2−J −2V ; Y3 = r·(V −X3)−2S1·J; Z3 = ((Z1 + Z2)2 −A −B) · H. 12.3.31 Remark Doubling takes 3M+5S. Addition takes 11M+5S. Readdition of P2 saves 1M+1S by caching B and Z2 ·B. Mixed addition with Z2 = 1 takes 7M+4S: it skips all operations involving Z2 and computes K = H2; I = 4K; Z3 = (Z1 + H)2 −A −K. 12.3.6 Short Weierstrass curves, characteristic 2, ordinary case: y2 + xy = x3 + a2x2 + a6 12.3.32 Remark The following algorithms for short binary Weierstrass curves choose weights (1, 2, 1), called Lopez-Dahab coordinates. The formulas use √a6 as a curve constant, as-suming implicitly that a6 is a square (which is automatic for characteristic-2 ﬁnite ﬁelds and other characteristic-2 perfect ﬁelds). For ﬁelds F2n with n odd each isomorphism class contains a curve with a2 = 1 or one with a2 = 0. Optimizations are diﬀerent, as the following algorithms show. Lopez-Dahab coordinates (X : Y : Z) are extended to include T = Z2, and further extended to include W = XZ for a2 = 0. Diﬀerential-addition formulas represent a point by its x-coordinate, and represent x in turn as a fraction X/Z. These formulas do not diﬀer between a2 = 0 and a2 = 1. 12.3.33 Algorithm: Doubling, a2 = 1. Input: P1 = (X1 : Y1 : Z1 : T1). Output: P3 = (X3 : Y3 : Z3 : T3) = 2P1. A = X2 1; B = Y 2 1 ; Z3 = T1 · A; T3 = Z2 3; X3 = (A + √a6T1)2; Y3 = B · (B + X3 + Z3) + a6T3 + T3. 12.3.34 Algorithm: Addition, a2 = 1. Input: P1 = (X1 : Y1 : Z1 : T1), P2 = (X2 : Y2 : Z2 : T2). Output: P3 = (X3 : Y3 : Z3 : T3) = P1 + P2. A = X1 · Z2; B = X2 · Z1; C = A2; D = B2; E = A + B; F = C + D; G = Y1 · T2; Curves over ﬁnite ﬁelds 445 H = Y2 ·T1; I = G+H; J = I ·E; Z3 = F ·Z1 ·Z2; T3 = Z2 3; X3 = A·(H +D)+B ·(C +G); Y3 = (A · J + F · G) · F + (J + Z3) · X3. 12.3.35 Algorithm: Mixed addition, a2 = 1. Input: P1 = (X1 : Y1 : Z1 : T1), P2 = (X2 : Y2 : 1 : 1). Output: P3 = (X3 : Y3 : Z3 : T3) = P1 + P2. A = X1 + X2 · Z1; B = Y1 + Y2 · T1; C = A · Z1; D = C · (B + C); Z3 = C2; T3 = Z2 3; X3 = B2 + C · A2 + D; Y3 = (X3 + X2 · Z3) · D + (X2 + Y2) · T3. 12.3.36 Remark For a2 = 1, doubling takes 2M + 4S + 2D; addition takes 13M + 3S; readdition does not save anything; mixed addition with Z2 = 1 takes 8M + 4S. 12.3.37 Algorithm: Doubling, a2 = 0. Input: P1 = (X1 : Y1 : Z1 : T1 : W1). Output: P3 = (X3 : Y3 : Z3 : T3 : W3) = 2P1. A = X2 1; B = Y 2 1 ; Z3 = W 2 1 ; X3 = (A + √a6T1)2; T3 = Z2 3; W3 = X3 · Z3; Y3 = B · (B + X3 + Z3) + a6T3 + W3. 12.3.38 Algorithm: Addition, a2 = 0. Input: P1 = (X1 : Y1 : Z1 : T1 : W1), P2 = (X2 : Y2 : Z2 : T2 : W2). Output: P3 = (X3 : Y3 : Z3 : T3 : W3) = P1 + P2. A = X1 · Z2; B = X2 · Z1; C = A2; D = B2; E = A + B; F = C + D; G = Y1 · T2; H = Y2 · T1; I = G + H; J = I · E; Z3 = F · Z1 · Z2; X3 = A · (H + D) + B · (C + G); Y3 = (A · J + F · G) · F + (J + Z3) · X3; T3 = Z2 3; W3 = X3 · Z3. 12.3.39 Algorithm: Mixed addition, a2 = 0. Input: P1 = (X1 : Y1 : Z1 : T1 : W1), P2 = (X2 : Y2 : 1 : 1 : 1). Output: P3 = (X3 : Y3 : Z3 : T3 : W3) = P1 + P2. A = Y1 + Y2 · T1; B = X1 + X2 · Z1; C = B · Z1; Z3 = C2; T3 = Z2 3; D = X2 · Z3; X3 = A2 + C · (A + B2 + a2C); Y3 = (D + X3) · (A · C + Z3) + (Y2 + X2) · T3; W3 = X3 · Z3. 12.3.40 Remark For a2 = 0, doubling takes 2M + 5S + 2D; addition takes 14M + 3S; readdition does not save anything; mixed addition with Z2 = 1 takes 8M + 4S + 1D. 12.3.41 Algorithm: Diﬀerential addition and doubling. Input: P2 = (X2 : Z2), P3 = (X3 : Z3) with x(P3 −P2) = x1. Output: P4 = (X4 : Z4) = 2P2, P5 = (X5 : Z5) = P2 + P3. A = X2 · Z3; B = X3 · Z2; C = X2 2; D = Z2 2; Z5 = (A + B)2; X5 = x1 · Z5 + A · B; X4 = (C + √a6D)2; Z4 = C · D. 12.3.42 Remark The combined operation of diﬀerential addition and doubling takes 5M+4S+1D. The Montgomery ladder thus takes 5M + 4S + 1D per bit of the scalar. 12.3.7 Montgomery curves: by2 = x3 + ax2 + x 12.3.43 Remark Montgomery curves are of interest primarily for their eﬃcient diﬀerential addition. Points are represented via their x-coordinate x(P) = X/Z. The formulas depend on the curve constant c = (a + 2)/4. 12.3.44 Algorithm: Diﬀerential addition and doubling. Input: P2 = (X2 : Z2), P3 = (X3 : Z3) with x(P3 −P2) = x1. Output: P4 = (X4 : Z4) = 2P2, P5 = (X5 : Z5) = P2 + P3. A = X2 + Z2; ¯ A = A2; B = X2 −Z2; ¯ B = B2; E = ¯ A −¯ B; C = X3 + Z3; D = X3 −Z3; F = D · A; G = C · B; X5 = (F + G)2; Z5 = x1 · (F −G)2; X4 = ¯ A · ¯ B; Z4 = E · ( ¯ B + cE). 446 Handbook of Finite Fields 12.3.45 Remark The combined operation of diﬀerential addition and doubling takes 5M+4S+1D. A Montgomery curve is birationally equivalent to a twisted Edwards curve. Diﬀerential addition formulas on twisted Edwards curves trade 1M for 1D; if the curve constant can be chosen small those formulas are more eﬃcient. 12.3.8 Twisted Edwards curves: ax2 + y2 = 1 + dx2y2 12.3.46 Remark The following formulas use extended coordinates: x = X/Z, y = Y/Z, xy = T/Z. Three diﬀerent formulas are stated below: a complete addition formula for the case that a is a square and d is not a square; an incomplete but faster addition formula for the case a = −1; and a faster doubling formula. 12.3.47 Algorithm: Complete addition. Input: P1 = (X1 : Y1 : Z1 : T1), P2 = (X2 : Y2 : Z2 : T2). Output: P3 = (X3 : Y3 : Z3 : T3) = P1 + P2. A = X1 · X2; B = Y1 · Y2; C = dT1 · T2; D = Z1 · Z2; E = (X1 + Y1) · (X2 + Y2) −A −B; F = D −C; G = D + C; H = B −aA; X3 = E · F; Y3 = G · H; T3 = E · H; Z3 = F · G. 12.3.48 Algorithm: Addition, a = −1. Input: P1 = (X1 : Y1 : Z1 : T1), P2 = (X2 : Y2 : Z2 : T2). Output: P3 = (X3 : Y3 : Z3 : T3) = P1 + P2. A = (Y1 −X1) · (Y2 + X2); B = (Y1 + X1) · (Y2 −X2); C = Z1 · 2 · T2; D = T1 · 2 · Z2; E = D + C; F = B −A; G = B + A; H = D −C; X3 = E · F; Y3 = G · H; T3 = E · H; Z3 = F · G. 12.3.49 Algorithm: Doubling. Input: P1 = (X1 : Y1 : Z1 : T1). Output: P3 = (X3 : Y3 : Z3 : T3) = 2P1. A = X2 1; B = Y 2 1 ; C = 2Z2 1; D = aA; E = (X1 + Y1)2 −A −B; G = D + B; F = G −C; H = D −B; X3 = E · F; Y3 = G · H; T3 = E · H; Z3 = F · G. 12.3.50 Remark Doubling takes 4M + 4S + 1D. Doubling does not use the input T; suppressing the computation of T from the previous addition or doubling reduces the eﬀective cost of doubling to 3M + 4S + 1D in typical addition chains. A complete addition takes 9M + 2D. Readdition saves 1D. Mixed addition with Z2 = 1 saves 1M. The incomplete addition formulas for a = −1 take 8M. Mixed addition saves 1M. There are also complete addition formulas for a = −1 taking 8M + 1D, where the D is a multiplication by 2d. See Also §12.2 For more material on function ﬁelds of genus 1 (elliptic curves). §12.4 For more material on hyperelliptic function ﬁelds. §12.5 For rational places of function ﬁelds (rational points on curves). §12.6 For towers of function ﬁelds. §15.2 For applications of function ﬁelds to coding theory. §16.4 For applications in cryptography. References Cited: [244, 246, 247, 249, 362, 636, 956, 2133] Curves over ﬁnite ﬁelds 447 12.4 Hyperelliptic curves Michael John Jacobson, Jr., University of Calgary Renate Scheidler, University of Calgary Recall that by Remark 12.1.119 every K-rational point on a projective curve over a ﬁeld K, and hence also every K-rational point on an aﬃne curve over K, corresponds to a degree one place of the associated function ﬁeld. This section predominantly uses the language of curves and points rather than function ﬁelds and places. 12.4.1 Hyperelliptic equations 12.4.1 Deﬁnition A hyperelliptic equation over a ﬁeld K is an equation of the form y2 + h(x)y = f(x) with h, f ∈K[x]. (12.4.1) 12.4.2 Remark The curve deﬁned by a hyperelliptic equation (12.4.1) is non-singular if and only if for no point (x0, y0) on the curve, both partial derivatives vanish, i.e., 2y0 + h(x0) = 0 and h′(x0)y0 = f ′(x0). 12.4.3 Deﬁnition A hyperelliptic curve C of genus g deﬁned over a ﬁeld K is given by a hyper-elliptic equation over K that is irreducible in K(x, y), non-singular, and satisﬁes one of the following three conditions: 1. deg(f) = 2g + 1 and deg(h) ≤g; 2. deg(f) ≤2g + 1 and h is monic of degree g + 1, or deg(f) = 2g + 2 and a. either K has characteristic diﬀerent from 2, deg(h) ≤g and the leading coeﬃcient of f is a square in K, b. or K has characteristic 2, or h is monic of degree g + 1 and the leading coeﬃcient of f is of the form s2 + s for some s ∈K∗; 3. deg(f) = 2g + 2 and a. either K has characteristic diﬀerent from 2, deg(h) ≤g, and the leading coeﬃcient of f is not a square in K, b. or K has characteristic 2, or h is monic of degree g + 1 and the leading coeﬃcient of f is not of the form s2 + s for any s ∈K∗. A curve C is imaginary or ramiﬁed in case (1), real or split in case (2), and unusual or inert in case (3). 12.4.4 Remark An elliptic curve can be thought of as an imaginary hyperelliptic curve of genus g = 1. 12.4.5 Remark Every unusual hyperelliptic curve is a real curve over a quadratic extension of K. 12.4.6 Remark It is customary, although not necessary, to take h to be identically zero if K does not have characteristic 2. This is always possible by completing the square, i.e., adding h2/4 to both sides of (12.4.1) and replacing y by y −h/2. 448 Handbook of Finite Fields 12.4.7 Deﬁnition The inﬁnite places of a hyperelliptic curve C/K as given in (12.4.1) are exactly the poles of x (see Deﬁnition 12.1.18). 12.4.8 Deﬁnition Let C/K be a hyperelliptic curve. For any extension ﬁeld L of K, the set of ﬁnite points of C deﬁned over L is the set of solutions (x0, y0) ∈L×L to the hyperelliptic equation (12.4.1) deﬁning C. The set of points at inﬁnity of C deﬁned over L is the set S =      {∞} if C/L is imaginary, {∞, ∞} if C/L is real, ∅ if C/L is unusual. The ﬁnite points and points at inﬁnity deﬁned over L together form the set of points of C deﬁned over L or the set of L-rational points of C, denoted by C(L). 12.4.9 Deﬁnition The hyperelliptic involution of a hyperelliptic curve C/K is the map ι : C(K) →C(K) that sends a ﬁnite point P = (x0, y0) of C to the point P = ι(P) = (x0, −y0 −h(x0)) of C. If C is imaginary, then ι(∞) = ∞. If C is real, then ι(∞) = ∞and ι(∞) = ∞. 12.4.10 Remark A point on a hyperelliptic curve is ramiﬁed (when viewed as a place) if it is ﬁxed by ι, and unramiﬁed otherwise. 12.4.11 Remark If C is imaginary, then the inﬁnite place of K(x) is a ramiﬁed K-rational point on C; if C is real, then it is an unramiﬁed K-rational point on C, and if C is unusual, then it is not a point on C, but rather a place of degree 2. 12.4.12 Theorem (Generalization of [2373, Section 5] and [1587, Proposition 2.1]) Let C be a hyperelliptic curve of genus g over a perfect ﬁeld K deﬁned by (12.4.1), and let x0, y0 ∈K. Substituting x = t−1 + x0, y = bz tg+1 + a into (12.4.1), with a = ( y0 if char(K) = 2, −h(x0)/2 otherwise, and b = ( 1 if h(x0) + 2y0 = 0, h(x0) + 2y0 otherwise, yields a hyperelliptic curve C′ : z2 + H(t)z = F(t) of genus g over K where H(t) = b−1tg+1(h(t−1 + x0) + 2a), F(t) = b−2t2g+2(f(t−1 + x0) −ah(t−1 + x0) −a2), and the following conditions hold: 1. If P = (x0, y0) is a ﬁnite ramiﬁed point on C, then C′ is an imaginary hyperelliptic curve. 2. If P = (x0, y0) is a ﬁnite unramiﬁed point on C, then C′ is a real hyperelliptic curve. 3. If no ﬁnite point on C has x-coordinate x0, then C′ is an unusual hyperelliptic curve. Curves over ﬁnite ﬁelds 449 12.4.13 Example A hyperelliptic curve y2 = f(x) of genus g deﬁned over R may have as few as one and as many as g + 1 connected components, as illustrated in Figure 12.4.13, depending on the number of real roots of f. (a) y2 = x5+x4−5x3−2x2+3x+1 (imaginary) (b) y2 = x6 + x4 −5x3 −2x2 + 3x + 1 (real) (c) y2 = −x6+x4−5x3−2x2+3x+1 (unusual) Figure 12.4.1 Examples of imaginary, real, and unusual hyperelliptic curves of genus 2 deﬁned over R. 12.4.2 The degree zero divisor class group 12.4.14 Remark Unlike the case of elliptic curves, the set of points of C deﬁned over any exten-sion ﬁeld of K does not form an abelian group. Instead, one needs to use divisors, as in Deﬁnition 12.1.21. 12.4.15 Deﬁnition Let C/K be a hyperelliptic curve and L an extension ﬁeld of K. A divisor of C is deﬁned over L if Dσ = P P nP P σ = D for all L-automorphisms σ on K, where P σ = (σ(x), σ(y)) if P = (x, y), ∞σ = ∞and (in case C is real) ∞σ = ∞. 12.4.16 Deﬁnition [661, Deﬁnition 14.3] Let C/K be a hyperelliptic curve. The set of degree zero divisors of C deﬁned over K, denoted by Div0 K(C), is a subgroup of the divisor group of C/K (see Deﬁnition 12.1.21). The set of principal divisors deﬁned over K, denoted by PrincK(C), is a subgroup of Div0 K(C) (see Deﬁnition 12.1.26). 450 Handbook of Finite Fields 12.4.17 Deﬁnition Let C/K be a hyperelliptic curve. The degree zero divisor class group deﬁned over K is given by Pic0 K(C) = Div0 K(C)/ PrincK(C). We denote Pic0(C) = Pic0 K(C). 12.4.18 Remark In the case of genus 1, i.e., elliptic curves, Pic0 K(C) is isomorphic to the group of points deﬁned over K on the elliptic curve; see Proposition 12.2.44. This is not true for hyperelliptic curves of genus g > 1. 12.4.19 Remark The group Pic0 K(C) is isomorphic to the group of K-rational points on the Jaco-bian variety JC(K) of C; see [661, Section 4.4.6.a]. As a result, the two terminologies and notations are sometimes used interchangeably in the literature. 12.4.20 Deﬁnition Let K = Fq. Then Pic0 Fq(C) is a ﬁnite abelian group. Its order, denoted by h, is the degree zero divisor class number, or class number, of C/Fq. 12.4.21 Deﬁnition For any divisor D of C, [D] denotes the divisor class of Pic0 K(C) represented by D. 12.4.22 Deﬁnition Let C/Fq be a real hyperelliptic curve. The order R of the subgroup of Pic0 K(C) generated by [∞−∞] is the regulator of C/K. 12.4.23 Remark The divisor R(∞−∞) is principal, and thus the divisor of a function ϵ. This function is a fundamental unit of the maximal order of the corresponding function ﬁeld; see Remark 12.1.119. 12.4.24 Remark Let C/Fq be a hyperelliptic curve and F/Fq the corresponding algebraic function ﬁeld; see Remark 12.1.119. Let C denote the ideal class group of the maximal order of F/Fq(x). Then the following relationships hold between Pic0 K(C) and C. 1. If C/K is imaginary, then h = \|C\|, and the groups Pic0 K(C) and C are isomorphic. 2. If C/K is real, then h = R\|C\|. 3. If C/K is unusual, then h = \|C\|/2. 12.4.25 Remark The ideal class number \|C\| of a real hyperelliptic curve is expected to be small (frequently 1), so we expect that R ≈h ; see also [1129, 1130, 1131] for the genus 1 case. 12.4.3 Divisor class arithmetic over ﬁnite ﬁelds 12.4.26 Remark In the case of imaginary and real hyperelliptic curves, and to a lesser extent for unusual curves, there exist algorithms for eﬃcient arithmetic in Pic0 K(C). We restrict our attention to K = Fq. 12.4.27 Deﬁnition A divisor D = P P nP P of a hyperelliptic curve C/Fq is ﬁnitely eﬀective if nP ≥0 for all ﬁnite points P of C. 12.4.28 Remark Every ﬁnitely eﬀective degree zero divisor D of a hyperelliptic curve C/Fq can be represented uniquely as D =      DS −deg(DS)∞ if C/Fq is imaginary, DS −deg(DS)∞+ n∞(∞−∞) if C/Fq is real, DS −(deg(DS)/2)∞ if C/Fq is unusual, Curves over ﬁnite ﬁelds 451 where DS = P P nP P is only supported at ﬁnite points on C and n∞∈Z; see [1588, Section 3]. 12.4.29 Deﬁnition A degree zero divisor D of a hyperelliptic curve C/Fq with a representation as given in Remark 12.4.28 is semi-reduced if for all P ∈supp(DS) we have P ̸∈supp(DS), unless P = P, in which case nP = 1. 12.4.30 Deﬁnition A semi-reduced divisor D is reduced if deg(DS) ≤g. 12.4.31 Remark If C/Fq is imaginary, then every divisor class of Pic0 Fq(C) contains a unique reduced divisor. 12.4.32 Remark If C/Fq is real, then divisor classes of Pic0 Fq(C) do not generically contain unique reduced divisors. However, each class does contain a unique reduced divisor D such that n∞lies in a speciﬁed range of length g + 1 −deg(DS). Paulus and R¨ uck proposed the interval [0, g −deg(DS)]. Galbraith, Harrison, and Mireles Morales proposed a balanced divisor representation, using the interval centered around ⌈deg(DS)/2⌉. 12.4.33 Remark If C/Fq is unusual, then every divisor class of Pic0 Fq(C) contains at most one reduced divisor. If the class contains a divisor of the form given in Remark 12.4.28, then it contains either a unique reduced divisor or q + 1 divisors of the form as in Remark 12.4.28 with deg(DS) = g+1. Arithmetic in Pic0 Fq(C) when C/Fq is unusual is not as well developed as for imaginary and real hyperelliptic curves C/Fq. For details, see . 12.4.34 Theorem [1774, Theorem 5.1] Let C/Fq be a hyperelliptic curve. If D is a semi-reduced divisor deﬁned over Fq as given in Remark 12.4.28, then DS can be represented uniquely by a pair of polynomials u, v ∈Fq[x] where u(x) = Y P ∈supp(DS) (x −xP )nP is monic and v is the unique polynomial such that deg(v) < deg(u), v(xP ) = −yP for all P = (xP , yP ) ∈supp(DS), and u \| v2 + hv −f. 12.4.35 Deﬁnition The pair [u, v] is the Mumford representation of D. 12.4.36 Remark In some sources such as , the condition v(xP ) = −yP is replaced by v(xP ) = yP . This also describes a unique representation of semi-reduced divisors in which DS from Theorem 12.4.34 is replaced by DS = P P nP P. 12.4.37 Remark Alternative representations to [u, v] can be obtained by taking [u, v′] where v′ is any polynomial with v′ ≡v (mod u) that satisﬁes the same interpolation condition. In the real case, it is sometimes computationally advantageous to use an alternative with deg(v′) = g + 1 . 12.4.38 Remark If C/Fq is imaginary, then semi-reduced divisors are uniquely determined by their Mumford representation. Thus, the Mumford representation gives an explicit, ef-ﬁcient representation of divisor classes of Pic0 Fq(C) via reduced representatives D with deg(v) < deg(u) ≤g. The identity divisor class, PrincFq(C), is represented by [1, 0], and the inverse of [u, v] is given by [u, −h −v]. Cantor’s algorithm (see also [661, Al-gorithm 14.7]) describes how to compute the reduced sum of two divisors in Mumford representation in polynomial time using only arithmetic with polynomials in Fq[x]. 452 Handbook of Finite Fields 12.4.39 Remark If C/Fq is real, then semi-reduced divisors are uniquely determined by their Mum-ford representation and n∞-value. Thus, the Mumford representation, together with the integer n∞, can be used to represent divisor classes in Pic0 Fq(C). The identity class is rep-resented by u = 1, v = 0, and n∞= 0. The inverse of the divisor class [u, v, n∞] is given by [u, −h −v, −n∞−deg(u)], after which additional adjustment steps are performed to obtain a value n∞in the required range, as described in [1155, 2373]. A modiﬁcation of Cantor’s algorithm (see, for example, ) can be used to compute the reduced sum of two reduced divisors, after which additional operations are performed to obtain a value n∞ in the required range. 12.4.40 Remark A generalization of Shanks’ NUCOMP algorithm is more eﬃcient than Cantor’s algorithm for moderate and large genus , and can be adapted for use in both the imaginary and real models. For g ≤4, optimized explicit formulas exist in the imaginary case that describe the addition and reduction algorithm in terms of operations in Fq; see [661, Chapter 14] for a survey. Explicit formulas for genus 2 also exist in the real case [987, 988]. 12.4.41 Remark The multiplication-by-m map on elliptic curves (see Deﬁnition 12.2.24) generalizes naturally to Pic0 K(C). The double-and-add method, as well as more advanced methods for scalar multiplication, can also be applied to compute this map eﬃciently; see [661, Chapter 9] for a survey. 12.4.42 Deﬁnition Let C/Fq be a real hyperelliptic curve. The infrastructure of C/Fq, denoted by R, is the ﬁnite set of all reduced principal divisors D with 0 ≥n∞> −R. 12.4.43 Remark The infrastructure is often described in terms of reduced principal ideals of the maximal order of the function ﬁeld associated to C/Fq; see, for example . 12.4.44 Deﬁnition Let C/Fq be a real hyperelliptic curve, and D ∈R. The distance of D is δ(D) = −n∞. 12.4.45 Remark Divisors in the infrastructure can be represented using a combination of the Mum-ford representation and distance. 12.4.46 Remark Distance imposes a natural ordering on the set R. The baby step operation moves cyclically from one divisor to the next in this ordering . The distance obtained by traversing one entire cycle is exactly the regulator R. 12.4.47 Remark A modiﬁcation of Cantor’s algorithm applied to two divisors in the infrastructure, where the reduction process terminates as soon as a reduced divisor is obtained, produces another infrastructure divisor . NUCOMP and explicit formulas for genus 2 [987, 988] can also be used for this purpose. 12.4.48 Deﬁnition The operation of computing the reduced sum of two divisors in R, as described in the previous remark, is a giant step. 12.4.49 Remark The divisor [1, 0] with distance 0 acts as the identity with respect to the giant step operation. The inverse of D = [u, v] ̸= [1, 0] with respect to the giant step operation is [u, −h −v] and has distance R + deg(u) −δ(D). 12.4.50 Remark Giant steps move through R in larger steps than baby steps, because the distance of a giant step applied to inputs D and D′ is δ(D) + δ(D′) −d, where 0 ≤d ≤2g; see, for example . Distances are not exactly additive due to the adjustments required to Curves over ﬁnite ﬁelds 453 achieve reduction. Thus, although R is structurally similar to a cyclic group of order R under the giant step operation, it is not a group as associativity does not hold . 12.4.51 Remark The analogue of the multiplication-by-m map in the infrastructure is computing a divisor with distance as close to m as possible but not exceeding m. This divisor can be computed in O(log m) giant steps using methods similar to the double-and-add method, and in some cases, can be sped up using the fact that computing baby steps is faster than giant steps; see for details. 12.4.4 Endomorphisms and supersingularity 12.4.52 Deﬁnition Let C/K be a hyperelliptic curve. An endomorphism of Pic0(C) is a group homomorphism of Pic0(C). An endomorphism of Pic0(C) is deﬁned over K if it is a group homomorphism of Pic0 K(C). The set of endomorphisms of Pic0(C) is de-noted by End(Pic0(C)), and the set of endomorphisms deﬁned over K is denoted by EndK(Pic0(C)). 12.4.53 Remark If C has genus 1, then Deﬁnition 12.4.52 agrees with the deﬁnition of an endomor-phism of an elliptic curve as given in Deﬁnition 12.2.27. 12.4.54 Remark As in the elliptic curve case (see Deﬁnition 12.2.39), End(Pic0(C)) and EndK(Pic0(C)) are rings. 12.4.55 Example The multiplication-by-m map [m] : Pic0(C) − →Pic0(C) is an endomorphism of Pic0(C) that is deﬁned over K. Thus, End(Pic0(C)) and EndK(Pic0(C)) always contain Z. 12.4.56 Deﬁnition [661, Deﬁnition 14.13] Let C/K be a hyperelliptic curve. If End(Pic0(C)) contains an order of a number ﬁeld of degree 2g over Q, then End(Pic0(C)) has complex multiplication. 12.4.57 Example Let C/Fq be hyperelliptic curve. As in the elliptic curve case (see Exam-ple 12.2.31), the Frobenius automorphism of Fq that sends an element a to aq extends to an endomorphism of Pic0(C) that is deﬁned over K and is diﬀerent from [m] for all m ∈Z. 12.4.58 Deﬁnition Let C/Fq be a hyperelliptic curve. The group Pic0(C) (or, more properly, the Jacobian JC) is supersingular if it is isogenous to the product of supersingular elliptic curves. 12.4.59 Remark If C/Fq is a hyperelliptic curve that is not supersingular, then C may have com-plex multiplication. The Frobenius endomorphism satisﬁes a monic polynomial equation of degree 2g with integer coeﬃcients (its characteristic polynomial, see Remark 12.4.60). If that polynomial is irreducible, then the Frobenius corresponds to an algebraic integer of degree 2g and C/Fq has complex multiplication. 12.4.5 Class number computation 12.4.60 Remark The zeta function of a hyperelliptic curve C/Fq of genus g (Deﬁnition 12.5.12) is of the form Z(C/Fq, t) = L(t) (1 −t)(1 −qt), 454 Handbook of Finite Fields where L(t) is the L-polynomial of C/Fq (Theorem 12.5.13). The reciprocal polynomial P(t) = t2gL(1/t) is the characteristic polynomial of the Frobenius endomorphism (Re-mark 12.5.15), a polynomial of degree 2g with integer coeﬃcients whose roots have absolute value √q (Theorem 12.5.17). 12.4.61 Theorem (Special case of Theorem 12.5.13) Let C/Fq be a hyperelliptic curve of genus g and class number h. Then h = L(1). 12.4.62 Remark Theorem 12.4.61 and Remark 12.4.60 immediately imply the bounds (√q−1)2g ≤ h ≤(√q + 1)2g. The interval [(√q −1)2g, (√q + 1)2g] is called the Hasse–Weil interval for hyperelliptic curves of genus g over Fq. 12.4.63 Remark There is an extensive body of literature on algorithms for computing class numbers of hyperelliptic curves over ﬁnite ﬁelds; see [815, 817, 1249, 1555, 1556, 1718, 1719, 1865, 1866, 1905, 2089, 2867] For genus 2 curves, see also . An overview of the main methods can be found in [661, Section 17.3]. 12.4.6 The Tate-Lichtenbaum pairing 12.4.64 Deﬁnition The kernel of the multiplication-by-m map on Pic0(C) is the subgroup Pic0(C)[m] = {[D] ∈Pic0(C) : m[D] = PrincK(C)}. 12.4.65 Deﬁnition Let C/Fq be a hyperelliptic curve. Let m ≥1 a prime integer with embedding degree k in Fq (see Deﬁnition 12.2.112). The Tate-Lichtenbaum pairing is deﬁned as T : Pic0 Fq(C)[m] × Pic0 Fq(C)/m Pic0 Fq(C) →F∗ qk/(F∗ qk)m ([D1], [D2]) 7→fm,D1(D2) = Y P fm,D1(P)nP , where fm,D1 is a function with divisor m[D1] and D2 = P P nP (P). 12.4.66 Remark The Tate-Lichtenbaum pairing is bilinear, i.e., T([D1] + [D2], [D3]) = T([D1], [D3])T([D2], [D3]) and T([D1], [D2] + [D3]) = T([D1], [D2])T([D1], [D3]), non-degenerate (if T([D1], [D2]) = 1 for all [D2] ∈Pic0 K(C)[m] then [D1] = PrincK(C)), and the result is independent of the divisor class representatives used. 12.4.67 Remark The Tate-Lichtenbaum pairing can be computed using an analogue of Miller’s algorithm for elliptic curves [182, Section 5]. 12.4.68 Remark Hyperelliptic curves with small embedding degree exist, i.e., for which computing the Tate-Lichtenbaum pairing is eﬃcient. For example, Galbraith proved that su-persingular hyperelliptic curves of genus g have embedding degree bounded by an integer k(g). For g ≤6, Rubin and Silverberg show that k(g) ≤7.5g. Various constructive methods for non-supersingular hyperelliptic curves also exist; see for a recent survey. Curves over ﬁnite ﬁelds 455 12.4.69 Deﬁnition Let C/Fq be a hyperelliptic curve. Let m ≥1 a prime integer with embedding degree k in Fq. Suppose that Pic0 Fq(C) contains no elements of order m2. The modiﬁed Tate-Lichtenbaum pairing is deﬁned as T ′ : Pic0 Fq(C)[m] × Pic0 Fq(C)[m] →µm ([D1], [D2]) 7→T([D1], [D2])(qk−1)/m, where µm is the group of m-th roots of unity. 12.4.70 Remark The main advantage of the modiﬁed Tate-Lichtenbaum pairing over the Tate-Lichtenbaum pairing is that it takes speciﬁc values in µm as opposed to equivalence classes. 12.4.71 Remark There are other types of pairings and algorithms to compute them, many designed to have computational advantages over the Tate-Lichtenbaum pairing. For a recent survey, see . 12.4.7 The hyperelliptic curve discrete logarithm problem 12.4.72 Remark Similar to the elliptic curve discrete logarithm problem (see Section 12.2.11), the hyperelliptic curve discrete logarithm problem (HCDLP) is the basis of many hyperelliptic curve cryptosystems. In this section we discuss the HCDLP and some related problems. For applications to cryptography, see Section 16.5. 12.4.73 Deﬁnition Let C/Fq be a hyperelliptic curve and [D1], [D2] ∈Pic0 Fq(C). The hyperelliptic curve discrete logarithm problem (HCDLP) is the problem of computing n ∈Z such that [D1] = n[D2], if it exists. 12.4.74 Remark If the order l of [D2] is prime, then the fastest known general algorithm for solving the HCDLP (as of 2011) has running time on the order of √ l. As with the ECDLP, there are a number of cases where the problem can be solved more easily; see [661, Part V] and for recent surveys. 12.4.75 Remark If the genus g is suﬃciently large compared to the ﬁnite ﬁeld order q, then the HCDLP can be solved in expected time subexponential in qg using the index-calculus method [17, 977, 980]. The current state-of-the-art is Enge and Gaudry’s result that if g/ logg q > ϑ, the expected bit complexity is Lqg[1/2, √ 2((1 + 1/2ϑ)1/2 + (1/2ϑ)1/2)] where Ln[β, c] = e((c+o(1))(log n)β(log log n)1−β). 12.4.76 Remark Index-calculus can also be used to solve the HCDLP faster than the generic methods for smaller genera [1245, 1256, 2802]. The current state-of-the-art is Gaudry, Thom´ e, Th´ eriault, and Diem’s result that the HCDLP can be solved in expected time O(g5q2−2 g +ϵ) if q > g!. This is asymptotically faster than the generic algorithms for g ≥3. 12.4.77 Remark Frey, M¨ uller, and R¨ uck [1106, 1413] showed how the modiﬁed Tate-Lichtenbaum pairing can be used to reduce the HCDLP to the DLP in the group of m-th roots of unity µm ⊂Fqk, where k is the embedding degree of m in the ﬁeld Fq (see Deﬁnition 12.2.112). If k is suﬃciently small, for example if C/Fq is supersingular, this is more eﬃcient than the generic algorithms. 12.4.78 Remark If m = pn where p is the characteristic of Fq, then an algorithm of R¨ uck can be used to solve the HCDLP in time O(n2 log p). 456 Handbook of Finite Fields 12.4.79 Remark If Fq = Fpn where n is composite, the Weil descent methodology [661, Section 22.3] can in certain cases be used to reduce the HCDLP to another instance of the HCDLP on a curve of higher genus over a smaller ﬁnite ﬁeld, where faster non-generic algorithms may apply. 12.4.80 Remark The Diﬃe-Hellman and decisional Diﬃe-Hellman problems also generalize to Pic0 Fq(C), and are also sometimes used as the underlying security assumption of certain cryptographic protocols (see Section 16.5). Both reduce to the HCDLP, but equivalence is not known. 12.4.81 Deﬁnition Let C/Fq be a real hyperelliptic curve and let D ∈R. The infrastructure discrete logarithm problem (IDLP) is the problem of computing δ(D). 12.4.82 Remark This problem has also been used as the underlying security assumption of crypto-graphic protocols . It is computationally easy to compute a divisor in the infrastructure close to a given distance , but solving the IDLP is believed to be diﬃcult. In fact, the IDLP can be reduced to the DLP in the subgroup of Pic0 Fq(C) generated by [∞−∞] [1089, 2142]. See Also §12.1 For analagous material for general function ﬁelds and curves. §12.2 For analagous material for the genus 1 case (i.e., elliptic curves). §16.5 For applications of hyperelliptic curves to cryptography. A general reference to hyperelliptic curves and their cryptographic applications. The appendix of this reference consists of an excellent introduction to the arithmetic of hyperelliptic curves. References Cited: [17, 133, 182, 497, 661, 815, 817, 977, 980, 987, 988, 1089, 1105, 1106, 1125, 1129, 1130, 1131, 1154, 1155, 1245, 1249, 1256, 1413, 1555, 1556, 1586, 1587, 1588, 1718, 1719, 1774, 1865, 1866, 1905, 2089, 2142, 2373, 2399, 2493, 2498, 2706, 2707, 2802, 2867] 12.5 Rational points on curves Arnaldo Garcia, IMPA Henning Stichtenoth, Sabanci University 12.5.1 Remark In this section we use the language of function ﬁelds rather than algebraic curves, see Section 12.1. A simple way for switching from function ﬁelds to algebraic curves is as follows. A function ﬁeld F/Fq of genus g corresponds to a curve X of genus g over Fq, that is an absolutely irreducible, non-singular, projective curve which is deﬁned over Fq. If F = Fq(x, y) and x, y satisfy the equation ϕ(x, y) = 0 for an irreducible polynomial ϕ(X, Y ) ∈ Fq[X, Y ], then X is a non-singular, projective model of the plane curve which is deﬁned by Curves over ﬁnite ﬁelds 457 ϕ(X, Y ) = 0. By abuse of notation, we say brieﬂy that the curve X is given by ϕ(x, y) = 0. Rational places of the function ﬁeld correspond to Fq-rational points of X. 12.5.1 Rational places 12.5.2 Remark Let F be a function ﬁeld over Fq. Then F has only ﬁnitely many rational places. 12.5.3 Deﬁnition Deﬁne N(F) := \|{P \| P is a rational place of F}\|. 12.5.4 Example For the rational function ﬁeld F = Fq(x) we have N(F) = q + 1. The rational places are the zeros of x −a with a ∈Fq, and the pole P∞of x. 12.5.5 Lemma [2714, Lemma 5.1] Let F ′/F be a ﬁnite extension of function ﬁelds having the same constant ﬁeld Fq. Then the following hold. 1. Let P be a place of F and P ′ a place of F ′ lying above P. If P ′ is rational, then P is rational. 2. N(F ′) ≤[F ′ : F] · N(F). 12.5.6 Remark The following special case of Kummer’s Theorem [2714, Theorem 3.3.7] is often useful to determine rational places of a function ﬁeld. 12.5.7 Lemma Let P be a rational place of F and let OP be its valuation ring. Consider a ﬁnite extension E = F(y) of F such that Fq is also the full constant ﬁeld of E. Assume that the minimal polynomial ϕ(T) of y over F has all its coeﬃcients in OP (that is, y is integral over OP ). Suppose that the reduction ¯ ϕ(T) of ϕ(T) modulo P (which is a polynomial over the residue class ﬁeld OP /P = Fq) splits over Fq as follows: ¯ ϕ(T) = (T −a1) · · · (T −as) · p1(T) · · · pr(T) with distinct elements a1, . . . , as ∈Fq and distinct irreducible polynomials p1, . . . , pr ∈ Fq[T] of degree > 1. Then there are exactly s rational places P1, . . . , Ps of E lying over P. 12.5.8 Example Assume that q = 2m with m ≥2, and consider the function ﬁeld F = Fq(x, y) with y2 + y = xq−1. The pole P∞of x is totally ramiﬁed in the extension F/Fq(x); this gives one rational place of F. Next we consider the place P = (x = a) of Fq(x) which is the zero of x −a with a ∈Fq. The reduction of the minimal polynomial ϕ(T) = T 2 + T + xq−1 modulo P is then ¯ ϕ(T) = ( T 2 + T + 1 if a ̸= 0, T 2 + T if a = 0. The polynomial T 2 + T = T(T + 1) splits over Fq into linear factors. If m is odd, then T 2 + T + 1 is irreducible over Fq, and for m even, T 2 + T + 1 splits into two distinct linear polynomials over Fq. Therefore N(F) = ( 3 if m is odd, 2q + 1 if m is even. 458 Handbook of Finite Fields 12.5.2 The Zeta function of a function ﬁeld 12.5.9 Deﬁnition Throughout this subsection we use the following notations: 1. F is an algebraic function ﬁeld over Fq of genus g(F) = g, and Fq is algebraically closed in F; 2. PF is the set of places of F/Fq; 3. N(F) is the number of rational places of F; 4. Div(F) is the divisor group of F; 5. Div0(F) := {A ∈Div(F) \| deg A = 0} is the group of divisors of degree zero, and Princ(F) ⊆Div0(F) is the group of principal divisors of F; 6. Cl0(F) := Div0(F)/Princ(F) is the class group of F. In terms of algebraic curves X, the class group corresponds to the rational points of the Jacobian of X and is then denoted as Jac(X)(Fq). 12.5.10 Lemma [2714, Proposition 5.1.3] 1. For every n ≥0, there are only ﬁnitely many divisors A ≥0 with deg A = n. 2. The class group Cl0(F) is a ﬁnite group. 12.5.11 Deﬁnition The number h := hF := ord(Cl0(F)) is the class number of F. 12.5.12 Deﬁnition The Zeta function of F is deﬁned by the power series in C below (here C is the complex number ﬁeld): Z(t) := ∞ X n=0 Antn, where An denotes the number of positive divisors D ∈Div(F) of degree n. 12.5.13 Theorem [2714, Theorem 5.1.15] 1. The power series Z(t) converges for all t ∈C with \|t\| < q−1. 2. Z(t) can be written as Z(t) = L(t) (1 −t)(1 −qt) with a polynomial L(t) = a0 + a1t + · · · + a2gt2g ∈Z[t] of degree 2g. This polynomial is the L-polynomial of F. 3. (Functional equation of the L-polynomial) The coeﬃcients of the L-polynomial of F satisfy a. a0 = 1 and a2g = qg, b. a2g−i = qg−iai for 0 ≤i ≤g. 4. N(F) = a1 + q + 1. 5. L(1) = hF is the class number of F. 12.5.14 Lemma [2714, Theorem 5.1.15] Curves over ﬁnite ﬁelds 459 1. The L-polynomial factors into linear factors over C as follows: L(t) = 2g Y j=1 (1 −ωjt) with algebraic integers ωj ∈C. As L(ω−1 j ) = 0, the complex numbers ωj are the reciprocals of the roots of L(t). 2. One can arrange ω1, . . . , ω2g in such a way that ωj · ωg+j = q for 1 ≤j ≤g. 12.5.15 Remark The reciprocal polynomial P(t) := t2g · L(1/t) has an interpretation as the char-acteristic polynomial of the Frobenius endomorphism acting on the Tate module Tℓ; see [1959, 2783]. The roots of P(t) are just the reciprocals of the roots of L(t). Therefore, the complex numbers ωj in Lemma 12.5.14 are also called the eigenvalues of the Frobenius endomorphism. 12.5.16 Remark The following theorem is fundamental for the theory of function ﬁelds over ﬁnite ﬁelds. It was ﬁrst proved by Hasse for g = 1; the generalization to all g ≥1 is due to Weil. 12.5.17 Theorem (Hasse–Weil theorem) [2714, Theorem 5.2.1] The reciprocals of the roots of the L-polynomial satisfy \|ωj\| = q1/2 for 1 ≤j ≤2g . 12.5.18 Remark The Hasse–Weil theorem is often referred to as the Riemann Hypothesis for func-tion ﬁelds over ﬁnite ﬁelds. 12.5.3 Bounds for the number of rational places 12.5.19 Remark The next result is an easy consequence of the Hasse–Weil theorem 12.5.17. 12.5.20 Theorem (Hasse–Weil bound) [2714, Theorem 5.2.3] The number N = N(F) of rational places of a function ﬁeld F/Fq of genus g satisﬁes the inequality \|N −(q + 1)\| ≤2gq1/2. 12.5.21 Remark If q is not a square, this bound can be improved as follows. 12.5.22 Theorem (Serre bound) , [2714, Theorem 5.3.1] \|N −(q + 1)\| ≤g · j 2q1/2k , where ⌊α⌋means the integer part of the real number α. 12.5.23 Deﬁnition For every g ≥0, we deﬁne Nq(g) := max{N ∈N \| there is a function ﬁeld F/Fq of genus g with N(F) = N}. 12.5.24 Remark Clearly Nq(g) ≤q + 1 + g · 2q1/2 . Further improvements of this bound can be obtained. 12.5.25 Proposition (Serre’s explicit formulas) , [2714, Proposition 5.3.4] Suppose that u1, . . . , um are non-negative real numbers, not all of them equal to zero, satisfying 460 Handbook of Finite Fields 1 + Pm n=1 un cos nθ ≥0 for all θ ∈R. Then Nq(g) ≤1 + 2g + Pm n=1 unqn/2 Pm n=1 unq−n/2 . 12.5.26 Remark The results of the examples and tables below are proved in the following way. First one derives upper bounds for Nq(g) using Serre’s explicit formulas. In some cases, these upper bounds can be improved slightly by rather subtle arguments . Lower bounds for Nq(g) are usually obtained by providing explicit examples of function ﬁelds having that number of rational places. Many methods of construction have been proposed, see [1550, 2280, 2846] for some of them. 12.5.27 Example (The case g = 1) Let q = pe with a prime number p. 1. If e is odd, e ≥3 and p divides 2q1/2 , then Nq(1) = q + 2q1/2 . 2. Nq(1) = q + 1 + 2q1/2 , otherwise. 12.5.28 Example (The case g = 2) For all prime powers q, q −2 + 2 · j 2q1/2k ≤Nq(2) ≤q + 1 + 2 · j 2q1/2k . In fact, the exact value of Nq(2) is known in all cases . 12.5.29 Example (The case g = 3) The value of Nq(3) is known for many but not for all q. For instance, one knows Nq(3) for all q ≤169 and for all q = 2k with k ≤20. For details we refer to . 12.5.30 Remark The following tables show Nq(g) for some small values of q and g. Updated tables can be found on the website see . 12.5.31 Example (Values of Nq(g) for q = 2, 4, 8 and small g) In the tables below, an entry like 21−24 means that the exact value of N4(8) is not known; one knows only that 21 ≤N4(8) ≤ 24 (at the time of printing). g 0 1 2 3 4 5 6 7 8 9 10 20 N2(g) 3 5 6 7 8 9 10 10 11 12 13 19-21 N4(g) 5 9 10 14 15 17 20 21 21-24 26 27 40-45 N8(g) 9 14 18 24 25 29 33-34 34-38 35-42 45 42-49 76-83 12.5.32 Example (Values of Nq(g) for 1 ≤g ≤4 and prime numbers q ≤43) (at the time of printing) q 2 3 5 7 11 13 17 19 23 29 31 37 41 43 Nq(1) 5 7 10 13 18 21 26 28 33 40 43 50 54 57 Nq(2) 6 8 12 16 24 26 32 36 42 50 52 60 66 68 Nq(3) 7 10 16 20 28 32 40 44 48 60 62 72 78 80 Nq(4) 8 12 18 24 33 38 46 48-50 57 67-70 72 82 88-90 92 12.5.33 Remark If the genus g(F) is large with respect to q, the Hasse–Weil bound can be improved considerably. 12.5.34 Proposition (Ihara’s bound) , [2714, Proposition 5.3.3] Suppose that Nq(g) = q +1+ 2gq1/2. Then g ≤q1/2(q1/2 −1)/2. Curves over ﬁnite ﬁelds 461 12.5.35 Example Let q be a square. Then there exists a function ﬁeld of genus g = q1/2(q1/2 −1)/2 having q + 1 + 2gq1/2 rational places. For more about function ﬁelds which attain the Hasse–Weil upper bound, see Subsection 12.5.4. 12.5.36 Example 1. For q = 22m+1 with m ≥1 , and g = 23m+1 −2m, one knows that Nq(g) = q2 +1. 2. Similarly, for q = 32m+1 with m ≥1 and g = 3m+1(34m+2 + 33m+1 −3m −1)/2 one has Nq(g) = q3 + 1. The function ﬁelds which attain the values Nq(g) in this example, correspond to the Deligne– Lusztig curves associated to the Suzuki group and to the Ree group, respectively [474, 1414, 2593]. 12.5.4 Maximal function ﬁelds 12.5.37 Deﬁnition A function ﬁeld F/Fq is maximal if g(F) > 0 and N(F) attains the Hasse-Weil upper bound N(F) = q + 1 + 2gq1/2. 12.5.38 Remark It is clear that q must be the square of a prime power, if there exists a maximal function ﬁeld F/Fq. Therefore we assume in this subsection that q = ℓ2 is a square. By Ihara’s bound 12.5.34, the genus of a maximal function ﬁeld F over Fℓ2 satisﬁes 1 ≤g(F) ≤ ℓ(ℓ−1)/2. 12.5.39 Example [2714, Lemma 6.4.4] Let H := Fℓ2(x, y) where x, y satisfy the equation yℓ+ y = xℓ+1. Then H is a maximal function ﬁeld over Fℓ2 with g(H) = ℓ(ℓ−1)/2 and N(H) = ℓ3 + 1 = ℓ2 + 1 + 2g(H)ℓ. The ﬁeld H is called the Hermitian function ﬁeld over Fℓ2. 12.5.40 Remark The rational places of the Hermitian function ﬁeld H are the following: there is a unique common pole of x and y, and for any α, β ∈Fℓ2 with αℓ+ α = βℓ+1 there is a unique common zero of y −α and x −β. In this way one obtains all 1 + ℓ3 rational places of H. 12.5.41 Remark There are generators u, v of the Hermitian function ﬁeld H which satisfy the equation uℓ+1 + vℓ+1 = 1. Hence the Hermitian function ﬁeld is a special case of a Fermat function ﬁeld, which is deﬁned by an equation un + vn = 1 with gcd(n, q) = 1. 12.5.42 Proposition 1. Suppose that F/Fℓ2 is a maximal function ﬁeld of genus g(F) = ℓ(ℓ−1)/2. Then F is isomorphic to the Hermitian function ﬁeld H . 2. There is no maximal function ﬁeld E/Fℓ2 whose genus satisﬁes 1 4(ℓ−1)2 < g(E) < 1 2ℓ(ℓ−1) for ℓodd (and 1 4ℓ(ℓ−2) < g(E) < 1 2ℓ(ℓ−1) for ℓeven) . 3. Up to isomorphism there is a unique maximal function ﬁeld E/Fℓ2 of genus g(E) = 1 4(ℓ−1)2 for ℓodd (and g(E) = 1 4ℓ(ℓ−2) for ℓeven) [3, 1144]. 12.5.43 Proposition . Let F be a maximal function ﬁeld over Fq. Then every function ﬁeld E of positive genus with Fq ⊂E ⊆F is also maximal over Fq. 12.5.44 Remark The Hermitian function ﬁeld H/Fℓ2 has a large automorphism group G. Every subgroup U ⊆G whose ﬁxed ﬁeld is not rational, provides then an example of a maximal function ﬁeld HU over Fℓ2. Most known examples of maximal function ﬁelds over Fℓ2 have been constructed in this way [474, 1205, 1280], and [1511, Chapter 10]. 462 Handbook of Finite Fields 12.5.45 Example Over the ﬁeld Fq with q = r6, consider the function ﬁeld F = Fq(x, y, z) which is deﬁned by the equations xr + x = yr+1 and y · xr2 −x xr + x = z r3+1 r+1 . Here F is the Giulietti–Korchm´ aros function ﬁeld; it is maximal over Fq of genus g(F) = (r −1)(r4 + r3 −r2)/2. It is (at the time of printing) the only known example of a maximal function ﬁeld over Fq which is not a subﬁeld of the Hermitian function ﬁeld H/Fq. 12.5.46 Remark An important ingredient in many proofs of results on maximal function ﬁelds (for example, Parts 2 and 3 of Proposition 12.5.42) is the St¨ ohr–Voloch theory which sometimes gives an improvement of the Hasse–Weil upper bound. The method of St¨ ohr– Voloch involves the construction of an auxiliary function which has zeros of high order at the Fq-rational points of the corresponding non-singular curve. We illustrate this method in the case of plane curves. Let f(X, Y ) ∈Fq[X, Y ] be an absolutely irreducible polynomial that deﬁnes a non-singular projective plane curve. Recall that an aﬃne point (a, b) with f(a, b) = 0 is non-singular if at least one of the partial derivatives fX(X, Y ) or fY (X, Y ) does not vanish at the point (a, b). The auxiliary function h(X, Y ) in this case is obtained from the equation of the tangent line as h(X, Y ) = (X −Xq)fX(X, Y )+(Y −Y q)fY (X, Y ). Suppose now that f(X, Y ) does not divide h(X, Y ). Then N(F) ≤d(d + q −1)/2 , where F = Fq(x, y) with f(x, y) = 0 is the corresponding function ﬁeld, and d denotes the degree of the polynomial f(X, Y ). As an example consider the case d = 4. The genus of F is g(F) = (d−1)(d−2)/2 = 3. The bound above gives N(F) ≤2q +6 which is better than the Hasse–Weil upper bound for all q ≤23. We note that Nq(3) = 2q + 6 for q = 5, 7, 11, 13, 17 and 19; see Example 12.5.32. 12.5.5 Asymptotic bounds 12.5.47 Remark In this subsection we give some results about the asymptotic growth of the numbers Nq(g), see 12.5.23. As was mentioned in Proposition 12.5.34, the Hasse-Weil upper bound Nq(g) ≤q + 1 + 2gq1/2 cannot be attained if the genus is large with respect to q. 12.5.48 Deﬁnition The real number A(q) := lim supg→∞Nq(g)/g is Ihara’s quantity. 12.5.49 Remark As follows from the Hasse–Weil bound, A(q) ≤2q1/2. The following bound is a signiﬁcant improvement of this estimate. 12.5.50 Theorem (Drinfeld–Vlˇ adut ¸ bound) [2714, Theorem 7.1.3], A(q) ≤q1/2 −1. 12.5.51 Remark The proof of the Drinfeld–Vlˇ adut ¸ bound is a clever application of Serre’s explicit formulas 12.5.25. If q is a square, the Drinfeld–Vlˇ adut ¸ bound is sharp. 12.5.52 Theorem [1569, 2821] A(q) = q1/2 −1 if q is a square. 12.5.53 Remark If q is a non-square, the exact value of A(q) is not known. The lower bounds for A(q), given below, are proved by providing speciﬁc sequences of function ﬁelds Fn/Fq such Curves over ﬁnite ﬁelds 463 that limn→∞N(Fn)/g(Fn) > 0. Every such sequence gives then a lower bound for A(q). For details, see Section 12.6. 12.5.54 Theorem 1. [2280, Theorem 5.2.9], There is an absolute constant c > 0 such that A(q) > c · log q for all prime powers q. 2. [265, 3079] A(q3) ≥2(q2 −1)/(q + 2). 12.5.55 Remark Recall that ⌊α⌋and ⌈α⌉denote the ﬂoor and the ceiling of a real number α. The harmonic mean of two positive real numbers α, β is given by the formula H(α, β) = 2αβ/(α +β). The following result contains Theorem 12.5.52 and Part 2 of Theorem 12.5.54 as special cases. 12.5.56 Theorem (see also Example 12.6.29) For every prime number p and every n ≥2, A(pn) ≥H(p⌊n/2⌋−1, p⌈n/2⌉−1). For example, one has for p = 2 and all suﬃciently large odd integers n, 0.9428 × (2n/2 −1) ≤A(2n) ≤2n/2 −1. 12.5.57 Example [105, 938] The best known lower bounds for A(q) for q = 2, 3, 5 were obtained from class ﬁeld towers: A(2) ≥ 0.316999... , A(3) ≥ 0.492876... , A(5) ≥ 0.727272... . 12.5.58 Remark A counterpart to Ihara’s quantity A(q) is the following quantity. 12.5.59 Deﬁnition We set A−(q) := lim infg→∞Nq(g)/g. 12.5.60 Proposition A−(q) > 0 for all q. More precisely, 1. A−(q) ≥(q1/2 −1)/4, if q is a square. 2. There is an absolute constant d > 0 such that A−(q) ≥d · log q for all q. See Also §12.2 For more material on function ﬁelds of genus 1 (elliptic curves). §12.4 For more material on hyperelliptic function ﬁelds. §12.6 For towers of function ﬁelds. §12.7 For zeta functions for curves. §15.2 For applications of function ﬁelds to coding theory. References Cited: [3, 105, 265, 474, 938, 973, 1144, 1145, 1203, 1205, 1279, 1280, 1414, 1511, 1549, 1550, 1569, 1826, 1959, 2280, 2459, 2499, 2591, 2592, 2593, 2714, 2722, 2783, 2821, 2846, 2882, 3079] 464 Handbook of Finite Fields 12.6 Towers Arnaldo Garcia, IMPA Henning Stichtenoth, Sabanci University We use terminology as in Sections 12.1 and 12.5, see also . Some methods are discussed how to obtain lower bounds for Ihara’s quantity A(q), see Deﬁnition 12.5.48. Such bounds have a great impact in applications, for instance in coding theory, see Section 15.2. 12.6.1 Introduction to towers 12.6.1 Remark Lower bounds for A(q) are usually obtained in the following way: one con-structs a sequence of function ﬁelds (Fi/Fq)i≥0 with g(Fi) →∞such that the limit limi→∞N(Fi)/g(Fi) exists. If this limit is positive, then it provides a non-trivial lower bound for A(q). 12.6.2 Remark Essentially three methods are known for constructing such sequences of function ﬁelds: modular towers, class ﬁeld towers, and explicit towers. In the following two remarks we give a very brief description of the ﬁrst two methods. 12.6.3 Remark (Modular towers) [217, 971, 972, 1569, 2821] Modular towers were introduced by Ihara, and independently by Tsfasman, Vl˘ adut ¸, and Zink. Let N be a positive integer and p a prime number not dividing N. There exists an aﬃne algebraic curve Y0(N) deﬁned over Fp such that, for any ﬁeld K of characteristic p, Y0(N) parametrizes the set of isomorphy classes of pairs (E, C), where E is an elliptic curve (see Section 12.2) and C is a cyclic subgroup of E of order N, deﬁned over K, in a functorial way. The construction of Y0(N) is independent of p and can be done in characteristic zero also. The complete curve obtained from Y0(N) is denoted X0(N). If ℓ̸= p is another prime, then the curves X0(ℓn), n = 1, 2, . . . form a tower with the maps sending (E, C) to (E, C′) where C′ is the unique subgroup of C of index ℓ. Over Fp2, the supersingular elliptic curves (see Subsection 12.2.9) together with all their cyclic subgroups of order ℓn give rational points on X0(ℓn)(Fp2), because Frobenius is multiplication by −p on those curves. This gives a tower of curves over Fp2 which attains the Drinfel’d–Vl˘ adut ¸ bound. For Fq2, with q arbitrary, a similar construction can be made using Shimura curves which parametrize abelian varieties of higher dimension with additional structure. 12.6.4 Remark (Class ﬁeld towers) [938, 2280, 2561, 2592] Starting with any function ﬁeld F0 of genus g0 ≥2 and a set S0 of rational places of F0, one deﬁnes inductively the ﬁeld Fn+1 to be the maximal abelian unramiﬁed extension of Fn in which all places of Sn split completely, and Sn+1 to be the set of all places of Fn+1 which lie over Sn. If Fn ⊊Fn+1 for all n (which is not always the case), the tower thus obtained is called a class ﬁeld tower, and its limit (see Deﬁnition 12.6.8) is at least \|S0\|/(g0 −1). The hard part is to choose F0, S0 so that the tower is inﬁnite. This is analogous to the corresponding problem in the number ﬁeld case of inﬁnite class ﬁeld towers which was solved by Golod and Shafarevich. A choice of F0, S0 then can be used to show that A(p) ≥c·log p, for p prime, with an absolute constant c > 0. This approach which is due to Serre , is so far the only way to prove that A(p) > 0 holds for prime numbers p. 12.6.5 Remark (Explicit towers of function ﬁelds) These towers were introduced by Garcia and Stichtenoth [1200, 2714]. The method, which is more elementary than modular towers and class ﬁeld towers, is presented below in some detail. Curves over ﬁnite ﬁelds 465 12.6.6 Deﬁnition A tower F over Fq is an inﬁnite sequence F = (F0, F1, F2, . . .) of function ﬁelds Fi/Fq (with Fq algebraically closed in all Fi) such that 1. F0 ⫋F1 ⫋F2 ⫋· · · ⫋Fn ⫋· · · ; 2. each extension Fn+1/Fn is ﬁnite and separable; 3. for some n ≥0, the genus g(Fn) is ≥2. 12.6.7 Remark Items 2 and 3 imply that g(Fi) →∞as i →∞. The following limit exists for every tower over Fq [2714, Lemma 7.2.3]. 12.6.8 Deﬁnition Let F = (F0, F1, . . .) be a tower of function ﬁelds over Fq. The limit λ(F) := limi→∞N(Fi)/g(Fi) is the limit of the tower F. 12.6.9 Remark We note that the inequalities 0 ≤λ(F) ≤A(q) hold for every tower over Fq. 12.6.10 Deﬁnition A tower F/Fq is asymptotically good if λ(F) > 0. It is asymptotically bad if λ(F) = 0. 12.6.11 Remark The notion of asymptotically good (bad) towers is related to the notion of asymp-totically good (bad) sequences of codes; see Section 15.2. The remark below follows imme-diately from the deﬁnitions. 12.6.12 Remark As A(q) ≥λ(F), every asymptotically good tower F over Fq provides a non-trivial lower bound for Ihara’s quantity. 12.6.13 Remark Most towers turn out to be asymptotically bad and some eﬀort is needed to ﬁnd asymptotically good ones. We discuss now some criteria which ensure that a tower is good. 12.6.14 Deﬁnition Let F = (F0, F1, . . .) be a tower over Fq. 1. A place P of F0 is ramiﬁed in F/F0, if there is some n ≥1 and some place Q of Fn lying over P with ramiﬁcation index e(Q\|P) > 1. Otherwise, P is unramiﬁed in F. 2. A rational place P of F0 splits completely in F/F0, if P splits completely in the extensions Fn/F0, for all n ≥1. 3. The set Ram(F/F0) := {P \| P is a place of F0 which is ramiﬁed in F/F0} is the ramiﬁcation locus of F over F0. 4 The set Split(F/F0) := {P \| P is a rational place of F0 splitting completely in F/F0} is the splitting locus of F over F0. 12.6.15 Remark The splitting locus is always ﬁnite (it may be empty). The ramiﬁcation locus can be ﬁnite or inﬁnite. 12.6.16 Theorem [2714, Theorem 7.2.10] Assume that the tower F = (F0, F1, . . .) over Fq has the following properties. 1. The splitting locus Split(F/F0) is non-empty. 2. The ramiﬁcation locus Ram(F/F0) is ﬁnite. 466 Handbook of Finite Fields 3. For every P ∈Ram(F/F0) there is a constant cP ∈R such that for all n ≥0 and all places Q of Fn lying over P, the diﬀerent exponent d(Q\|P) is bounded by d(Q\|P) ≤cP · (e(Q\|P) −1). Then the tower F is asymptotically good, and its limit satisﬁes the inequality λ(F) ≥ s g(F0) −1 + r, where s := Split(F/F0) and r := 1 2 X P ∈Ram(F/F0) cP · deg P. 12.6.17 Remark Of course, one should choose the constant cP as small as possible (if it exists). In general it is a diﬃcult task to prove its existence in towers having wild ramiﬁcation. 12.6.18 Remark A tower F/F0 is tame if all places P ∈Ram(F/F0) are tame in all extensions Fn/F0; that is, the ramiﬁcation index e(Q\|P) is relatively prime to q for all places Q of Fn lying over P. 12.6.19 Remark For a tame tower, the constants cP in Theorem 12.6.16 can be chosen as cP = 1. Hence a tame tower with ﬁnite ramiﬁcation locus and non-empty splitting locus is asymp-totically good, and the inequality for λ(F) given in Theorem 12.6.16 holds with r := 1 2 X P ∈Ram(F/F0) deg P. 12.6.20 Remark All known asymptotically good towers of function ﬁelds have properties 1, 2, and 3 of Theorem 12.6.16. 12.6.2 Examples of towers 12.6.21 Deﬁnition Let f(Y ) ∈Fq(Y ) and h(X) ∈Fq(X) be non-constant rational functions, and let F = (F0, F1, . . .) be a tower of function ﬁelds over Fq. The tower F is recursively deﬁned by the equation f(Y ) = h(X), if there exist elements xi ∈Fi (i = 0, 1, . . .) such that 1. F0 = Fq(x0) is a rational function ﬁeld; 2. Fi = Fi−1(xi) for all i ≥1; 3. for all i ≥1, the elements xi−1, xi satisfy the equation f(xi) = h(xi−1). 12.6.22 Example [2714, Proposition 7.3.2] Let q = ℓ2 be a square, ℓ> 2. Then the equation Y ℓ−1 = 1 −(X + 1)ℓ−1 deﬁnes an asymtotically good tame tower F over Fq. The ramiﬁcation locus of this tower is the set of all places (x0 = α) with α ∈Fℓ, and the place (x0 = ∞) splits completely. By Theorem 12.6.16 the limit satisﬁes the inequality λ(F) ≥2/(ℓ−2). For q = 9 this limit attains the Drinfeld–Vl˘ adut ¸ bound λ(F) = 2 = √ 9 −1. Curves over ﬁnite ﬁelds 467 12.6.23 Example [2714, Proposition 7.3.3] Let q = ℓe with e ≥2 and set m := (q −1)/(ℓ−1). Then the equation Y m = 1 −(X + 1)m deﬁnes an asymptotically good tame tower F over Fq with limit λ(F) ≥2/(q −2). This gives a simple proof that A(q) > 0 for all non-prime values of q. For q = 4 the tower attains the Drinfeld–Vl˘ adut ¸ bound λ(F) = 1 = √ 4 −1. 12.6.24 Example Let q = p2 where p is an odd prime. Then the equation Y 2 = X2 + 1 2X deﬁnes a tame tower F over Fq. Its ramiﬁcation locus is Ram(F/F0) = {(x0 = α) \| α4 = 1 or α = 0 or α = ∞}. There are 2(p −1) rational places of F0 which split completely in the tower. The inequality in Theorem 12.6.16 gives λ(F) ≥p −1 which coincides with the Drinfeld–Vl˘ adut ¸ bound. So, λ(F) = p −1. The fact that the splitting locus of this tower has cardinality 2(p −1) is not easy to prove. For p = 3, 5 one can check directly that the places (x0 = α) with α4 + 1 = 0 (for p = 3) and α8 −α4 + 1 = 0 (for p = 5) split completely in F. 12.6.25 Remark Now we give some examples of wild towers, that is, there are some places of F0 whose ramiﬁcation index in some extension Fn/F0 is divisible by the characteristic of Fq. In wild towers, it is usually diﬃcult to ﬁnd a bound, if it exists, for the diﬀerent exponents in terms of ramiﬁcation indices (see Theorem 12.6.16). 12.6.26 Example Let q = ℓ2 be a square and deﬁne the tower F = (F0, F1, . . .) over Fq as follows: F0 := Fq(x0) is the rational function ﬁeld, and for all n ≥0, set Fn+1 := Fn(xn+1) with (xn+1xn)ℓ+ xn+1xn = xℓ+1 n . The ramiﬁcation locus of F is Ram(F/F0) = { (x0 = 0), (x0 = ∞) }, and all other rational places of F0 split completely in the tower. We note however that Theorem 12.6.16 is not directly applicable to determine the limit λ(F). One can show that λ(F) = ℓ−1, so this tower attains the Drinfeld–Vl˘ adut ¸ bound. 12.6.27 Example The equation Y ℓ+ Y = Xℓ Xℓ−1 + 1 deﬁnes a tower over Fq with q = ℓ2, whose limit attains the Drinfeld–Vl˘ adut ¸ bound λ(F) = ℓ−1. The determination of the splitting locus and the ramiﬁcation locus for this tower is easy. The hard part is to show that cP = 2 for all ramiﬁed places (for the deﬁnition of cP see Theorem 12.6.16). 12.6.28 Example [265, 2845] Over the ﬁeld Fq with q = ℓ3, the equation Y ℓ−Y ℓ−1 = 1 −X + X−(ℓ−1) 468 Handbook of Finite Fields deﬁnes an asymptotically good tower F with limit λ(F) ≥2(ℓ2 −1) ℓ+ 2 . It follows that A(ℓ3) ≥2(ℓ2 −1) ℓ+ 2 , for all prime powers ℓ(see Theorem 12.5.54). 12.6.29 Example Let q = ℓn with n ≥2. For every partition of n into relatively prime parts, n = j + k with j ≥1, k ≥1 and gcd(j, k) = 1, a tower F over Fq is recursively deﬁned by the equation Trj Y Xℓk + Trk Y ℓj X = 1, where Tra(T) = T + T ℓ+ T ℓ2 + · · · + T ℓa−1 for any a ∈N. The limit of this tower satisﬁes the inequality λ(F) ≥2 1 ℓj −1 + 1 ℓk −1 −1 , which is the harmonic mean of ℓj −1 and ℓk −1. This tower gives the best known lower bound for Ihara’s quantity A(q), for all non-prime ﬁelds Fq (at the time of printing). 12.6.30 Remark None of the towers in Examples 12.6.22 - 12.6.24 or 12.6.26 - 12.6.29 is Galois over F0, that is, not all of the extensions Fn/F0, n ≥0 are Galois extensions. In some special cases however, one can prove that the tower ˆ F := ( ˆ F0, ˆ F1, . . .), where ˆ Fn is the Galois closure of Fn/F0, is also asymptotically good [1202, 2713]. 12.6.31 Remark There are examples of function ﬁelds with many rational points which are abelian extensions of a rational function ﬁeld (for instance, the Hermitian function ﬁeld H; see Example 12.5.39). Other abelian extensions over Fq(x) having many rational places can be obtained via the method of cyclotomic function ﬁelds . However, abelian ex-tensions F/Fq(x) of large genus have only few rational places. More precisely, if (Fi)i≥0 is a sequence of abelian extensions of a rational function ﬁeld with g(Fi) →∞, then limi→∞N(Fi)/g(Fi) = 0 . 12.6.32 Remark We conclude this section with a warning: not every irreducible equation f(Y ) = h(X) deﬁnes a recursive tower. For instance, if one replaces X +1 by X in Examples 12.6.22 and 12.6.23, one just gets a ﬁnite extension F/F0 but not a tower. Also, one has to show that Fq is algebraically closed in each ﬁeld Fi of the tower. In most of the examples above this follows from the fact that there is some place which is totally ramiﬁed in all extensions Fi/F0. Curves over ﬁnite ﬁelds 469 See Also §13.5 For discussion of Drinfeld modules. §15.2 For applications of towers to coding theory. References Cited: [217, 265, 938, 971, 972, 1107, 1200, 1201, 1202, 1203, 1204, 1569, 2280, 2561, 2592, 2713, 2714, 2821, 2845] 12.7 Zeta functions and L-functions Lei Fu, Nankai University We use the terminology in . For the deﬁnitions of schemes, morphisms between schemes, and the aﬃne Spec A for a commutative ring A, see [1427, II 2]. For the deﬁnitions of schemes or morphisms of ﬁnite type, see [1427, II 3]. For the deﬁnitions of separated, proper or projective schemes or morphisms, see [1427, II 4]. For the deﬁnition of smooth morphisms, see [1427, III 10]. Throughout this section, we assume our schemes are separated. Let X be a scheme of ﬁnite type over Z. Denote the set of Zariski closed points in X by \|X\| (observe that in the rest of the handbook this notation indicates the cardinality of the set X). For any x ∈\|X\|, the residue ﬁeld k(x) of X at x is a ﬁnite ﬁeld. Let N(x) be the number of elements of k(x). 12.7.1 Zeta functions 12.7.1 Deﬁnition The Zeta function ζX(s) of X is ζX(s) = Y x∈\|X\| 1 1 −N(x)−s . 12.7.2 Remark When X is the aﬃne scheme Spec Z, ζX(s) is just the Riemann zeta function ζ(s) = Y p 1 1 −p−s = ∞ X n=1 1 ns . 12.7.3 Remark We are concerned with the case where X is a scheme of ﬁnite type over a ﬁnite ﬁeld Fq with q elements of characteristic p. For any x ∈\|X\|, k(x) is a ﬁnite extension of Fq. Set deg(x) = [k(x) : Fq]. Then we have N(x) = qdeg(x). 12.7.4 Deﬁnition Suppose X is a scheme of ﬁnite type over Fq. We deﬁne Z(X, t) = Y x∈\|X\| 1 1 −tdeg(x) . 470 Handbook of Finite Fields 12.7.5 Remark The relation between ζX(s) and Z(X, t) is ζX(s) = Z(X, q−s). We have Z(X, t) = X α tdeg(α), where α goes over the set of eﬀective 0-cycles in X. 12.7.6 Theorem [795, Equation (1.5.2)] For any positive integer m, let X(Fqm) be the set of Fqm-rational points in X. We have an equation of formal power series t d dt ln Z(X, t) = ∞ X m=1 #X(Fqm)tm. 12.7.7 Remark For the n-dimensional aﬃne space An Fq, we have #An Fq(Fqm) = qmn, Z(An Fq, t) = 1 1 −qnt. For the n-dimensional projective space Pn Fq, we have #Pn Fq(Fqm) = 1 + qm + · · · + qmn, Z(Pn Fq, t) = 1 (1 −t)(1 −qt) · · · (1 −qnt). 12.7.8 Theorem Let X be a smooth projective scheme over Fq of dimension n. 1. Rationality: Z(X, t) is a rational function of t, that is, a quotient of polynomials with rational coeﬃcients. 2. Functional equation: Z(X, t) satisﬁes a functional equation of the form Z X, 1 qnt = ϵq nχ(X) 2 tχ(X)Z(X, t) for some constants χ(X) and ϵ = ±1. 3. Riemann hypothesis: Z(X, t) can be written in the form Z(X, t) = 2n Y i=0 Pi(X, t)(−1)i+1 such that Pi(X, t) ∈Z[t] and all reciprocal zeros of Pi(X, t) lie on the circle \|t\| = q i 2 . 12.7.9 Remark The above theorem is usually called the Weil conjecture. It was proved to be true by Dwork, Grothendieck, and Deligne. Weil points out that to prove his conjecture, one needs to construct a cohomology theory for schemes over an abstract ﬁeld. One such theory is the ℓ-adic cohomology theory constructed by Grothendieck, where ℓis a prime number distinct from the characteristic of the ground ﬁeld. Results on ℓ-adic cohomology theory used in this section can be found in [136, 797, 1570]. Curves over ﬁnite ﬁelds 471 12.7.10 Remark In the Riemann hypothesis, we consider the Archimedean absolute values of the zeros and poles of Z(X, t). One can show that for any prime number ℓ̸= p, all the zeros and poles of Z(X, t) are ℓ-adic units. (This follows from ℓ-adic cohomology theory.) It is interesting to study the p-adic absolute values of the zeros and poles of Z(X, t); see Section 12.8. 12.7.11 Deﬁnition [1570, Exp. XV] Let FrX : X →X be the morphism of schemes (FrX, Fr# X) : (X, OX) →(X, OX) such that on the underlying topological space, FrX : X →X is the identity, and Fr# X : OX →OX maps a section s of OX to sq. Fix an algebraic closure F of Fq, and let X = X ⊗Fq F. The geometric Frobenius correspondence on X is the base change FX : X →X of FrX from Fq to F. 12.7.12 Remark There is a canonical one-to-one correspondence between the set X(Fqm) of Fqm-rational points in X, and the set of ﬁxed points of F m X on X. 12.7.13 Remark Let Hi(X, Qℓ) and Hi c(X, Qℓ) be the ℓ-adic cohomology groups and the ℓ-adic cohomology groups with compact support of X, respectively. They are ﬁnite dimensional vector spaces, and they vanish if i ̸∈[0, 2dim X]. If X is proper, we have Hi(X, Qℓ) ∼ = Hi c(X, Qℓ). 12.7.14 Theorem [797, Rapport 3.2] (Lefschetz ﬁxed point theorem) We have #X(Fqm) = 2dim X X i=0 (−1)iTr F m X , Hi c(X, Qℓ) . 12.7.15 Remark The following theorem follows from Theorems 12.7.6 and 12.7.14. It proves the rationality of the function Z(X, t). 12.7.16 Theorem [797, Rapport 3.1] (Grothendieck’s formula) Let X be a scheme of ﬁnite type over Fq. We have Z(X, t) = 2dimX Y i=0 det 1 −FXt, Hi c(X, Qℓ) (−1)i+1 . 12.7.17 Remark Suppose a ﬁnite group G acts on X. Then each Hi c(X, Qℓ) is a representation G. Let Hi c(X, Qℓ) = M j∈Ii Vij be the isotypic decomposition of this representation. Then each Vij is invariant under the action of FX, and we have a further factorization for the formula of Z(X, t) in Theorem 12.7.16: Z(X, t) = 2dimX Y i=0 Y j∈Ii det(1 −FXt, Vij)(−1)i+1. In this way, we can get further information about Z(X, t). As an example, let Xλ be the Dwork hypersurface in Pn−1 Fq deﬁned by the equation xn 1 + · · · + xn n −nλx1 · · · xn = 0, 472 Handbook of Finite Fields where λ is a parameter. The group A = {(ζ1, . . . , ζn)\|ζi ∈Fq, ζn i = Y i ζi = 1}/{(ζ, . . . , ζ)\|ζ ∈Fq, ζn = 1} acts on Xλ by coordinatewise multiplication, and the symmetric group Sn acts on Xλ by permuting coordinates. Thus the ﬁnite group G = A ⋊Sn acts on Xλ. Goutet studies the factorization of Z(Xλ, t) with respect to the group action of G on Hi(Xλ, Qℓ). The function Z(Xλ, t) has also been studied by Candelas, de la Ossa, and Rodriguez-Villegas for the case n = 5, by Katz , and by Br¨ unjes for the case λ = 0. 12.7.18 Theorem [136, Exp. XVIII, Paragraph 3.2.6] (Poincar´ e duality) Suppose X is proper smooth over Fq and pure of dimension n. Then we have a perfect pairing ( , ) : Hi(X, Qℓ) × H2n−i(X, Qℓ) →Qℓ such that for any s ∈Hi(X, Qℓ) and t ∈H2n−i(X, Qℓ), we have (FX(s), FX(t)) = qn(s, t). 12.7.19 Remark Poincar´ e duality implies the functional equation for Z(X, t). 12.7.20 Theorem [795, Paragraph 2.6] Suppose X is proper smooth over Fq and pure of dimension n. We have Z X, 1 qnt = ϵq nχ(X) 2 tχ(X)Z(X, t), where χ(X) = P2n i=0(−1)idim Hi(X, Qℓ) is the Euler characteristic of X, and if N is the multiplicity of the eigenvalue q n 2 of FX acting on Hn(X, Qℓ), then we have ϵ = 1 if n is odd, (−1)N if n is even. 12.7.21 Remark Together with Theorem 12.7.16, the following theorem of Deligne proves the Rie-mann hypothesis for Z(X, t). 12.7.22 Theorem , [798, Corollaires 3.3.4-3.3.5] Suppose X is a scheme of ﬁnite type over Fq. For any eigenvalue α of FX on Hi c(X, Qℓ), α is an algebraic integer, and all the Galois conjugates of α have Archimedean absolute value q w 2 for some integer w ≤i. The equality w = i holds if X is proper smooth over Fq. 12.7.23 Remark For each 0 ≤i ≤2dim X, let bi = dim Hi c(X, Qℓ), and let αij (j = 1, . . . , bi) be all the eigenvalues of FX on Hi c(X, Qℓ). By Theorem 12.7.14, we have #X(Fq) = 2dim X X i=0 bi X j=1 (−1)iαij. Theorem 12.7.22 provides bounds for \|αij\|. To get a bound for #X(Fq), it suﬃces to ﬁnd a bound for P i bi. 12.7.24 Remark Suppose X is proper smooth over Fq, pure of dimension n, and geometrically connected (i.e., X is connected). Then H0(X, Qℓ) and H2n(X, Qℓ) are one dimensional, and FX acts on them by scalar multiplications 1 and qn, respectively. Curves over ﬁnite ﬁelds 473 12.7.25 Corollary Suppose X is proper smooth over Fq, pure of dimension n, and geometrically connected. Then we have \|#X(Fq) −(1 + qn)\| ≤ 2n−1 X i=1 biq i 2 . 12.7.26 Proposition [795, Th´ eoreme 8.1] Let X ⊂Pn+r Fq be a nonsingular complete intersection over Fq of dimension n and multi-degree (d1, . . . , dr). Let b′ = dim Hn(X, Qℓ). Set b = b′ if n is odd, and set b = b′ −1 if n is even. Then we have \|#X(Fq) −#Pn(Fq)\| ≤bq n 2 . 12.7.27 Remark By Theorems 12.7.16 and 12.7.22, we can write Z(X, t) = Q(t) P(t), where P(t) = (1 −λ1) · · · (1 −λs) and Q(t) = (1 −µ1) · · · (1 −µt) are relatively prime polynomials, and λi, µi are algebraic integers satisfying \|λi\|, \|µj\| ≤qdim X. By Theorem 12.7.6, we have #X(Fq) = λ1 + · · · + λs −µ1 −· · · −µt. To get a bound for #X(Fq), it suﬃces to ﬁnd a bound for s + t. We call s + t the total degree of Z(X, t), and denote it by totdeg Z(X, t). Note that t −s is the degree of Z(X, t), and we have s −t = 2dim X X i=0 (−1)idim Hi c(X, Qℓ). 12.7.28 Remark Using Dwork’s theory and Theorem 12.7.22, Bombieri obtains the following bound for totdeg Z(X, t); see [341, Theorems 1, 2, and Proposition in IV] and [339, Theorem 1]. 12.7.29 Theorem Let X be a closed aﬃne subvariety in AN Fq deﬁned by the vanishing of r polyno-mials f1, . . . , fr ∈Fq[t1, . . . , tN] of degrees ≤d. Then we have totdeg Z(X, t) ≤(4(d + 1) + 5)N+r. 12.7.30 Remark Adolphson and Sperber generalize Bombieri’s result as follows. 12.7.31 Proposition [22, Theorem 5.27, Corollary 6.13] Let X be a closed aﬃne subvariety in AN Fq deﬁned by the vanishing of r polynomials f1, . . . , fr ∈Fq[t1, . . . , tN] of degrees d1, . . . , dr, respectively. Set DN(x0, x1, . . . , xr) = X i0+i1+···+ir=N xi0 0 xi1 1 · · · xir r . We have 2dim X X i=0 (−1)idim Hi c(X, Qℓ) ≤2rDN(1, d1 + 1, . . . , dr + 1), totdeg Z(X, t) ≤(2e3)N(2e3 + 1)N(5 max{d1, . . . , dr} + 1)N. 12.7.32 Remark Starting from a universal bound \| P2dim X i=0 (−1)idim Hi c(X, Qℓ)\|, Katz de-duces a bound for P2dim X i=0 dim Hi c(X, Qℓ), and hence a bound for totdeg Z(X, t). In par-ticular, Katz gets the following estimate from those of Bombieri and Adolphson-Sperber. 474 Handbook of Finite Fields 12.7.33 Proposition [1706, Corollary of Theorem 1] Let X be a closed aﬃne subvariety in AN Fq deﬁned by the vanishing or r polynomials f1, . . . , fr ∈Fq[t1, . . . , tN] of degrees ≤d. Then we have 2dim X X i=0 dim Hi c(X, Qℓ) ≤ 2r−2 · 5 · (4rd + 13)N+2, 2dim X X i=0 dim Hi c(X, Qℓ) ≤ 2r+1 · 3 · (rd + 3)N+1. 12.7.2 L-functions 12.7.34 Remark For any scheme X of ﬁnite type over Fq and any ℓ-adic sheaf F, we can associate an L-function L(X, F, t). For the deﬁnition of an ℓ-adic sheaf on a scheme, we refer to [1570, Exp. VI]. We simply mention that in the case where X = Spec F for a ﬁeld F, giving an ℓ-adic sheaf on X is equivalent to giving a continuous ℓ-adic Galois representation Gal(F/F) →GL(n, Qℓ). Suppose X is a scheme of ﬁnite type over Fq. An Fqm-rational point x ∈X(Fqm) is an Fq-morphism Spec Fqm →X. Let F be an ℓ-adic sheaf on X. Then the inverse image of F on Spec Fqm deﬁnes a Galois representation which we denote by Gal(F/Fqm) →GL(F¯ x). Here F¯ x is the stalk of F at the geometric point Spec F →X over x. The Galois group Gal(F/Fqm) has a special element, the Frobenius substitution φx : Fqm →Fqm, φx(α) = αqm. Denote by Fx the inverse of φx and call it the geometric Frobenius at x. Let x ∈\|X\| be a Zariski closed point in X. Then we have a closed immersion Spec k(x) →X, and hence x deﬁnes a k(x)-rational point in X. We denote the corresponding geometric Frobenius also by Fx. 12.7.35 Deﬁnition The L-function L(X, F, t) is the formal power series with variable t and with coeﬃcients in Qℓdeﬁned by L(X, F, s) = Y x∈\|X\| 1 det(1 −Fxtdeg(x), F¯ x). 12.7.36 Remark When F is the constant ℓ-adic sheaf Qℓ, L(X, F, t) coincides with Z(X, t). 12.7.37 Theorem [797, Rapport 3] For any positive integer m, let Sm(X, F) = X x∈X(Fqm) Tr(Fx, F¯ x). We have an equation of formal power series t d dt ln L(X, F, t) = ∞ X n=1 Sm(X, F)tm. Curves over ﬁnite ﬁelds 475 12.7.38 Remark [797, Sommes trig.] Let ψ : Fq →Q ∗ ℓbe a nontrivial additive character. One can construct an ℓ-adic sheaf Lψ on A1 Fq such that for any Fqm-rational point x ∈A1 Fq(Fqm) = Fqm, we have Tr(Fx, (Lψ)¯ x) = ψ(TrFqm/Fq(x)). Let f ∈Fq[t1, . . . , tN] be a polynomial. It deﬁnes a morphism f : AN Fq →A1 Fq. For any Fqm-rational point x ∈AN Fq(Fqm) = FN qm with coordinates x = (x1, . . . , xN), the ℓ-adic sheaf f ∗Lψ has the property Tr(Fx, (f ∗Lψ)¯ x) = ψ(TrFqm/Fq(f(x1, . . . , xN))). We note that Sm(AN Fq, f ∗Lψ) = X x1,...,xN∈Fqm ψ(TrFqm/Fq(f(x1, . . . , xN))) is the classical exponential sum associated to the polynomial f. 12.7.39 Remark Let X = X ⊗Fq F and let F be the inverse image of F on X. We have the coho-mology groups Hi(X, F) and the cohomology groups with compact support Hi c(X, F). They are ﬁnite dimensional vector spaces, and they vanish if i ̸∈[0, 2dim X]. If X is proper, we have Hi(X, F) ∼ = Hi c(X, F). Moreover, we have a morphism of sheaves F ∗: F ∗ XF →F. The pair (FX, F ∗) is the geometric Frobenius correspondence for F. Denote the homomorphisms induced by this pair on cohomology groups with compact support by F : Hi c(X, F) →Hi c(X, F). 12.7.40 Theorem [797, Rapport 3.2] (Grothendieck trace formula) We have X x∈X(Fqm) Tr(Fx, F¯ x) = 2dim X X i=0 (−1)iTr F m, Hi c(X, F) . 12.7.41 Remark The following theorem follows from Theorems 12.7.37 and 12.7.40. It proves the rationality of the function L(X, F, t). 12.7.42 Theorem [797, Rapport 3.1] (Grothendieck’s formula) Let X be a scheme of ﬁnite type over Fq. We have L(X, F, t) = 2dimX Y i=0 det 1 −Ft, Hi c(X, F) (−1)i+1 . 12.7.43 Deﬁnition A number α in Qℓis pure of weight w (relative to q) if it is an algebraic number and all its Galois conjugates have Archimedean absolute value q w 2 . An ℓ-adic sheaf F on X is punctually pure of weight w if for any x ∈\|X\|, the eigenvalues of Fx on the stalk F¯ x are pure of weight w (relative to qdeg(x)). The sheaf F is mixed of weights ≤w if there exists a ﬁnite ﬁltration of F such that the successive quotients are punctually pure of weights ≤w. 12.7.44 Remark Together with 12.7.42, the following theorem of Deligne proves the Riemann hy-pothesis for L(X, F, t). 12.7.45 Theorem [798, Corollaire 3.3.4] Suppose X is a scheme of ﬁnite type over Fq and F is a mixed ℓ-adic sheaf of weights ≤w on X. Then any eigenvalue of F on Hi c(X, F) is pure of weight ≤i + w relative to q. 476 Handbook of Finite Fields 12.7.46 Remark For each 0 ≤i ≤2dim X, let bi = dim Hi c(X, F), and let αij (j = 1, . . . , bi) be all the eigenvalues of F on Hi c(X, F). By 12.7.40, we have Sm(X, F) = X x∈X(Fqm) Tr(Fx, F¯ x) = 2dim X X i=0 bi X j=1 (−1)iαm ij . Theorem 12.7.45 provides bounds for \|αij\|. To get a bound for Sm(X, F), it suﬃces to ﬁnd bounds for P i bi. 12.7.47 Remark Applying Theorem 12.7.45 to the ℓ-adic sheaf f ∗Lψ in Remark 12.7.38, we can get a bound for the exponential sum \| P x1,...,xN∈Fqm ψ(TrFqm/Fq(f(x1, . . . , xN)))\|. We have the following results. 12.7.48 Theorem [795, Th´ eor eme 8.4], [798, Paragraphs 3.7.2-3.7.4] Let f ∈Fq[t1, . . . , tN] be a polynomial of degree d, and let fd be the homogeneous part of f of degree d. Suppose fd deﬁnes a smooth hypersurface in PN−1 Fq and d is relatively prime to p. 1. Hi c(AN F , f ∗Lψ) = 0 for i ̸= N. 2. dim HN c (AN F , f ∗Lψ) = (d −1)N. 3. All eigenvalues of F on HN c (AN F , f ∗Lψ) are pure of weight N. 4. L(AN Fq, f ∗Lψ) = P(t)(−1)N+1 for a polynomial P(t) of degree (d −1)N so that all reciprocal roots of P(t) have Archimedean absolute value qN. 5. \| P x1,...,xN∈Fqm ψ(TrFqm/Fq(f(x1, . . . , xN)))\| ≤(d −1)Nq Nm 2 . 12.7.49 Theorem [23, Theorem 4.2], [812, Theorem 1.3] Let f ∈Fq[t1, . . . , tN, 1/t1, . . . , 1/tN] be a Laurent polynomial. Write f = X i1,...,iN ci1...iN ti1 1 · · · tiN N . Let ∆∞(f) be the convex hull in QN of the set {(i1, . . . , iN)\|ci1...iN ̸= 0}∪{0}. For any face τ of ∆∞(f), let fτ = P (i1,...,iN)∈τ ci1...iN ti1 1 · · · tiN N . Suppose f is nondegenerate with respect to ∆∞(f) in the sense that for any face τ of ∆∞(f) that does not contain the origin, the subscheme of GN m,Fq = (A1 Fq −{0})N deﬁned by ∂fτ ∂t1 = · · · = ∂fτ ∂tN = 0 is empty. Suppose furthermore that dim ∆∞(f) = N. 1. Hi c(GN m,F, f ∗Lψ) = 0 for i ̸= N. 2. dim HN c (GN m,F, f ∗Lψ) = N!Vol(∆∞(f)). 3. If, in addition, the origin is an interior point of ∆∞(f), then all eigenvalues of F on HN c (GN m,F, f ∗Lψ) are pure of weight N. 4. L(GN m,Fq, f ∗Lψ) = P(t)(−1)N+1 for a polynomial P(t) of degree N!Vol(∆∞(f)). If, in addition, the origin is an interior point of ∆∞(f), then all reciprocal roots of P(t) have Archimedean absolute value qN. 5. \| P x1,...,xN∈F∗ qm ψ(TrFqm/Fq(f(x1, . . . , xN)))\| ≤N!Vol(∆∞(f))q Nm 2 . 12.7.50 Remark The estimates in Theorem 12.7.29, and Propositions 12.7.31 and 12.7.33 can also be extended to the L-functions associated to exponential sums [22, 339, 341, 1706]. Curves over ﬁnite ﬁelds 477 12.7.3 The case of curves 12.7.51 Remark Suppose X is a geometrically connected smooth projective curve over Fq of genus g. Then H1(X, Qℓ) can be identiﬁed with Tℓ(JX) ⊗ZℓQℓ, where Tℓ(JX) is the Tate module of the Jacobian JX of X, and dim H1(X, Qℓ) = 2g. By Theorems 12.7.16 and 12.7.22 and Corollary 12.7.25, we have the following. 12.7.52 Theorem Suppose X is a geometrically connected smooth projective curve over Fq of genus g. We have Z(X, t) = P(t) (1 −t)(1 −qt), where P(t) = det(1 −Ft, H1(X, Qℓ)) is a polynomial with integer coeﬃcients, and all its reciprocal roots have Archimedean absolute value √q. Moreover, we have \|#X(Fq) −(1 + q)\| ≤2g√q. 12.7.53 Remark The above theorem was proved by Hasse for elliptic curves and by Weil for curves of higher genus. An elementary proof was given by Stepanov, Schmidt, and Bombieri . 12.7.54 Deﬁnition Let K(X) be the function ﬁeld of X, and let ρ : Gal(K(X)/K(X)) →GL(V ) be a continuous Galois representation, where V is a ﬁnite dimensional vector space over Qℓ. Suppose there exists a ﬁnite subset S of \|X\| such that ρ is unramiﬁed everywhere on X −S. We deﬁne the L-function L(X, ρ, t) to be L(X, ρ, t) = Y x∈\|X\| 1 det(1 −Fxtdeg(x), V Ix), where Ix is the inertia subgroup. 12.7.55 Remark The Galois representation ρ deﬁnes an ℓ-adic sheaf FV on X −S such that for any x ∈\|X −S\|, the Galois representation Gal(k(x)/k(x)) →GL(F¯ x) coincides with the Galois representation ρ\|Gal(k(x)/k(x)). Let j : X −S , →X be the open immersion. Then we have det(1 −Fxt, V Ix) = det(1 −Fxt, (j∗FV )¯ x) for any x ∈\|X\|. It follows that L(X, ρ, t) = L(X, j∗FV , t). We have H0(X, j∗FV ) ∼ = V Gal(K(X)/K(X)), H2(X, j∗FV ) ∼ = VGal(K(X)/K(X)), where K(X) is the function ﬁeld of X, and j∗FV is the inverse image of j∗FV on X. It follows from 12.7.42 that we have L(X, ρ, t) = det(1 −Ft, H1(X, j∗FV )) det 1 −Ft, V Gal(K(X)/K(X)) det 1 −qFt, VGal(K(X)/K(X)) . 12.7.56 Theorem [136, Exp. XVIII. Paragraph 3.2.6], [797, Dualit´ e 1.3] (Poincar´ e duality) We have a perfect pairing ( , ) : Hi(X, j∗FV ) × H2−i(X, j∗FV ∗) →Qℓ, 478 Handbook of Finite Fields where V ∗is the dual representation ρ∗: Gal(K(X)/K(X)) →GL(V ∗) of ρ. For any s ∈ Hi(X, j∗FV ) and t ∈H2−i(X, j∗FV ∗), we have (FX(s), FX(t)) = q(s, t). 12.7.57 Remark By Poincar´ e duality and Theorem 12.7.42, we have the following functional equa-tion for L-functions. 12.7.58 Theorem [1868, Equation (3.1.1.8)] We have L X, ρ, 1 qt = ϵ(X, ρ)(qt)χ(X,ρ)L(X, ρ∗, t), where χ(X, ρ) = 2 X i=0 (−1)idim Hi(X, j∗FV ), ϵ(X, ρ) = 2 Y i=0 det −F, Hi(X, j∗FV ) (−1)i+1 . 12.7.59 Remark Using Theorem 12.7.45 and Poincar´ e duality, one can prove the following, which gives the Riemann hypothesis for L(X, ρ, t) by Remark 12.7.55. 12.7.60 Theorem Suppose for any x ∈\|X −S\|, all eigenvalues of Fx on V are pure of weight w relative to qdeg(x). Then any eigenvalue of F on Hi(X, j∗F) is pure of weight i + w relative to q. 12.7.61 Remark For any point x ∈\|X\|, let Kx be the completion of K(X) with respect to the valuation corresponding to x, and let ρx : Gal(Kx/Kx) →GL(V ) be the restriction of the representation ρ. In Theorem 12.7.58, the Euler characteristic χ(X, ρ) and the constant ϵ(X, ρ) in the functional equation can be expressed in terms of the invariants of the Galois representations ρx (x ∈\|X\|) of the local ﬁelds Kx. 12.7.62 Theorem [1570, Exp. X, Th´ eoreme 7.1] (Grothendieck-Ogg-Shafarevich formula) We have χ(X, ρ) = (2 −2g)dim V − X x∈\|X\| deg(x)ax(ρx), where ax(ρx) is the Artin conductor of ρx. 12.7.63 Theorem [1868, Th´ eor eme 3.2.1.1] (Laumon’s product formula) Let ω be any nonzero meromorphic diﬀerential 1-form on X. Then we have ϵ(X, ρ) = q(1−g)dim(V ) Y x∈\|X\| ϵ(Kx, ρx, ω\|Spec Kx), where ϵ(Kx, ρx, ω\|Spec Kx) are the epsilon-factors deﬁned by Deligne . Curves over ﬁnite ﬁelds 479 See Also §6.2 For information on estimating exponential and character sums. §12.8 For p-adic estimates of zeta-functions and L-functions. References Cited: [22, 23, 136, 339, 340, 341, 435, 486, 794, 795, 797, 798, 812, 1345, 1427, 1570, 1706, 1710, 1868] 12.8 p-adic estimates of zeta functions and L-functions R´ egis Blache, IUFM de Guadeloupe 12.8.1 Introduction 12.8.1 Remark We know from the preceding section that the reciprocal roots and poles of zeta and L-functions deﬁned over the ﬁnite ﬁeld Fq are algebraic integers, which are units at all primes except the Archimedean ones and those lying over p. Many Archimedean estimates (the Riemann hypothesis over ﬁnite ﬁelds) have been given in Section 12.7; in this section we are interested in p-adic estimates, i.e., the p-adic Riemann hypothesis. We often refer to notations and results from Section 12.9. 12.8.2 Remark The ﬁrst result in this direction seems to be Stickelberger’s congruence which gives the valuation of Gauss sums and Jacobi sums (see Section 6.1). The modern results come from the work of Dwork , who was the ﬁrst to prove the rationality of zeta and L-functions, by p-adic means. From his pioneering work, many p-adic cohomology theories originated, such as Monsky-Washnitzer, crystalline, or rigid cohomology ; most of the results described below follow from the explicit description of these cohomologies. Note also they proved very useful for explicit calculations. 12.8.3 Remark One can describe the variation of the p-adic cohomology spaces from diﬀerential equations, such as the Picard-Fuchs one; they sometimes allow one to give an analytic expression for roots or poles of some zeta and L-functions. The best known example is the family of ordinary elliptic curves in Legendre form, which is linked to a hypergeometric function 2F1 ; the Gross-Koblitz formula (Theorem 6.1.113) links Gauss sums with the p-adic gamma function. As a consequence one gets a p-adic expression for Jacobi sums and the zeta function of a diagonal hypersurface. Other examples are cubic sums linked to the solutions of the Airy diﬀerential equation , Kloostermann sums to the Bessel diﬀerential equation , or the zeta function of a monomial deformation of a diagonal hypersurface which can be expressed from hypergeometric functions nFn−1 [1754, 3041]. 12.8.4 Remark Here we shall be concerned with p-adic valuations; we only mention brieﬂy these subjects; we neither speak about unit root functions, the reader interested in this subject should refer to [2895, 2896, 2897] and the references therein. 12.8.5 Remark Let L(T) be an L-function as in Deﬁnition 12.7.35, coming from a scheme over Fq of dimension n and a punctually pure sheaf of weight 0. From Deligne’s integrality 480 Handbook of Finite Fields theorem and Poincar´ e duality (see Section 12.7), the q-adic valuations of its reciprocal roots and poles are rational numbers lying in the interval [0, n]. 12.8.6 Deﬁnition Let F be a p-adic ﬁeld, OF its ring of integers, π a uniformizing parameter, and vq the valuation on F normalized by vq(q) = 1. If P = Pd i=0 aiT i ∈F[T] is a one variable polynomial, its q-adic Newton polygon, denoted NPq(P), is the lower convex hull of the set of points {(i, vq(ai)), 0 ≤i ≤d}. 12.8.7 Theorem Assume that P(0) = 1. Let s1, . . . , sr be the slopes of NPq(P), of respective multiplicity li (i.e., each si is the slope of a segment of horizontal length li); then the polynomial P has exactly li reciprocal roots of q-adic valuation si for any 1 ≤i ≤r. 12.8.8 Remark One can give a more general statement about the valuations of the roots of P, removing the hypothesis P(0) = 1. However, the theorem above is suﬃcient in the following results. 12.8.2 Lower bounds for the ﬁrst slope 12.8.9 Remark We give lower bounds for the ﬁrst slope of the Newton polygon of L-functions attached to families of (additive) exponential sums over aﬃne space, ﬁrst uniform, then depending on the characteristic. We end the subsection with an (incomplete) historical account on these questions. 12.8.10 Proposition Let X be a scheme of ﬁnite type over Fq and L(X, F, T) be an L-function as in Deﬁnition 12.7.35; for µ ∈R+, the following statements are equivalent: 1. The q-adic valuations of the reciprocal roots and poles of L(X, F, T) are greater than or equal to µ. 2. For any m, we have vqm(Sm(X, F)) ≥µ. 3. All slopes of the q-adic Newton polygons of the factors of L(X, F, T) are greater than or equal to µ. 12.8.11 Theorem [21, Theorem 1.2] Let f ∈Fq[x1, . . . , xN] be a polynomial, and ∆:= ∆∞(f) its Newton polytope at inﬁnity (see Theorem 12.7.49); denote by ω(∆) the smallest positive rational number such that ω(∆)∆, the dilation of ∆by the factor ω(∆), contains a lattice point with all coordinates positive (a point in ZN >0). Then every reciprocal root or pole of L(AN, f ∗Lψ, T) has q-adic valuation greater than or equal to ω(∆). 12.8.12 Remark Note that we make no assumption about the polynomial being non-degenerate with respect to its Newton polytope here. 12.8.13 Remark (see Section 7.1) One can deduce divisibility results on the numbers of points of algebraic varieties via the orthogonality relation on additive characters. Actually the Chevalley-Warning, Ax and Katz theorems are all consequences of the theorem above. 12.8.14 Deﬁnition Let D ⊂(N{0})N be a ﬁnite subset, which is not contained in some Nk, k < N; for any m ≥1, deﬁne the subset ED,p(m) of {0, . . . , pm −1}#D consisting of all (ud)d∈D such that P d∈D dud ≡0 (mod pm −1) and P d∈D dud has all coordinates positive. We set σD,p(m) := min (X d∈D σp(ud), (ud) ∈ED,p(m) ) . 12.8.15 Proposition The set n σD,p(m) m o m≥1 has a minimum. Curves over ﬁnite ﬁelds 481 12.8.16 Deﬁnition The p-density of the set D is the rational number πp(D) := 1 p−1 minm≥1 n σD,p(m) m o . 12.8.17 Theorem Let Fq[x1, . . . , xN]D be the vector space of polynomials whose monomials have their exponents in D. 1. For any f ∈Fq[x1, . . . , xN]D, the reciprocal roots and poles of L(An, f ∗Lψ, T) have q-adic valuation greater than or equal to πp(D). 2. Moreover, this bound is optimal in the sense that there exists a polynomial f in F[x1, . . . , xN]D such that a reciprocal root or pole of L(AN, f ∗Lψ, T) has q-adic valuation equal to πp(D). 12.8.18 Remark For f ∈Fq[x1, . . . , xN] a degree d polynomial, there are many lower bounds in the literature for the q-adic valuation of the exponential sum S(f) := X (x1,...,xN)∈FN q ψ(f(x1, . . . , xN)), giving in turn lower bounds for the valuations or the reciprocal roots and poles of the associated L-function. We give a brief account of these results here. 1. Sperber proves the uniform bound vq(S(f)) ≥N d ; then with Adolphson they give the bound in Theorem 12.8.11. This last bound is the best possible uniform one, since it is attained for some large enough p. 2. Later on, Moreno and Moreno take into account the characteristic via Weil descent; this leads to generally better bounds, less uniform however. Recently, Moreno, Shum, Castro, and Kumar give a bound depending on the exponents eﬀectively appearing in the polynomial, and on the cardinality of the ﬁeld (namely σD,p(m) m(p−1) , with q = pm). Note this last bound depends on too many parameters to say anything about the valuations of the reciprocal roots and poles of the L-function. 12.8.3 Uniform lower bounds for Newton polygons 12.8.19 Remark In this subsection, we describe lower bounds for the Newton polygons associated to zeta functions of smooth projective varieties, to L-functions associated to an additive character and a Laurent polynomial (toric exponential sums) or a rational function of one variable. In the case of zeta functions, these bounds come from the Hodge numbers of related varieties in characteristic 0; for this reason we shall call these bounds Hodge polygons. 12.8.20 Deﬁnition Let X be a smooth projective variety of dimension n, deﬁned over Fq. For any 0 ≤m ≤2n, deﬁne NPm(X) as the q-adic Newton polygon of the characteristic polynomial of the action of Frobenius on the m-th etale cohomology space det(1 − FXt, Hm(X, Qℓ)). Deﬁne the Hodge polygon in degree m of X as the polygon HPm(X) having slope i with multiplicity hi,m−i := dim Hm−i(X, Ωi) for 0 ≤i ≤m. 12.8.21 Theorem [29, 255, 2041] For any 0 ≤m ≤2n, the polygon NPm(X) lies on or above the polygon HPm(X), and they have the same endpoints. 12.8.22 Remark There is an analogous result in the case of a smooth complete intersection in Gm × An . 482 Handbook of Finite Fields 12.8.23 Deﬁnition Let X be a smooth projective variety of dimension n, deﬁned over Fq. The variety X is ordinary when we have NPm(X) = HPm(X) for any 0 ≤m ≤2n. 12.8.24 Deﬁnition Notations and assumptions are as in Theorem 12.7.49. We set ∆:= ∆∞(f), and denote by NPq(f) the q-adic Newton polygon of the polynomial L(GN m, f ∗Lψ, T)(−1)N+1. Denote by C(∆) := R+∆the cone of ∆in RN, M∆:= C(∆) ∩ZN the monoid associated to this cone, and A∆the algebra k[xM∆]. One can deﬁne a map from C(∆) to R+, the weight associated to ∆, by w∆(u) = min{ρ ∈R+, u ∈ρ∆}. The vertices of ∆lie in ZN, thus the image of M∆by w∆lies in Q+; more precisely there is a positive integer D(∆) such that Imw∆⊆ 1 D(∆)N. The least integer D := D(∆) having this property is the denominator of ∆. The weight w∆turns the algebra A∆into a graded algebra A∆= ⊕i≥0A∆, i D , A∆, i D = Vect xu, w∆(u) = i D to which we associate the Poincar´ e series PA∆(t) := P i≥0 dim A∆, i D ti. 12.8.25 Proposition [1802, Lemme 2.9] The series PA∆(t) is a rational function. Precisely, the series P∆(t) := (1 −tD)NPA∆(t) is a polynomial with degree less than or equal to ND, such that P∆(1) = N!Vol(∆). 12.8.26 Example Let f be a polynomial of degree d in the N variables x1, . . . , xN, containing the monomials xd 1, . . . , xd N with non-zero coeﬃcients; its Newton polytope at inﬁnity is the simplex with vertices (0, . . . , 0), (d, . . . , 0), . . . , (0, . . . , d). The associated cone is RN +, the weight is w∆(u1, . . . , uN) = P ui d , and the denominator is d. In this case the Poincar´ e series can be written PA∆(t) = 1 (1−t)N . 12.8.27 Deﬁnition Set P∆(t) := P ℓitsi. The Hodge polygon of ∆, HP(∆), is the polygon starting at the origin, and formed by the slopes si D with multiplicity ℓi. 12.8.28 Theorem [23, Theorem 3.10] For any polynomial f in Fq[t1, . . . , tN, 1 t1···tN ], non-degenerate with respect to its Newton polytope ∆, the polygon NPq(f) lies on or above HP(∆), and they have the same endpoints. 12.8.29 Remark One can show that the function L(GN m, f ∗Lψ, T)(−1)N+1 has exactly one unit root. Adolphson and Sperber give an analytic expression for this root, as an eigenvalue for the action of an operator on a p-adic Banach space. 12.8.30 Remark Adolphson and Sperber [25, 26] also consider twisted L-functions of the form L(GN m, f ∗Lψ ⊗Lχ, T), associated to the product of an additive character evaluated at a Laurent polynomial, and of a multiplicative character χ of GN m. They construct a Hodge polygon which is a lower bound for the Newton polygon as above; this Hodge polygon can be written in terms of the Hodge polygon HP(∆) and the valuations of the Gauss sums associated to the multiplicative characters appearing, as given by Stickelberger’s theorem. Curves over ﬁnite ﬁelds 483 12.8.31 Deﬁnition Let d1, . . . , ds denote positive integers. We deﬁne the Hodge polygon HP(d1, . . . , ds) as the polygon with slopes 0 and 1 with multiplicity s −1, 1 d1 , . . . , d1−1 d1 , . . . , 1 ds , . . . , ds−1 ds , each with multiplicity 1. 12.8.32 Theorem [3069, Theorem 1.1] Let f ∈Fq(x) be a rational function having s poles of prime to p orders d1, . . . , ds, and X denote the projective line with the poles removed. Then the q-adic Newton polygon of the function L(X, f ∗Lψ, T) lies on or above the Hodge polygon HP(d1, . . . , ds), and they have the same endpoints. 12.8.33 Remark Lower bounds for Newton polygons of L-functions associated to a character of order pl, evaluated at a Witt vector of functions, pure or twisted by a multiplicative character can be found in [1948, 1950]. See also , in which T-adic exponential sums are introduced, giving a framework in which to unify the study of the p-adic properties of these L-functions when l varies. 12.8.34 Remark There are also some results and conjectures on the p-adic theory of L-functions associated to multiplicative characters; see and the references therein. 12.8.4 Variation of Newton polygons in a family 12.8.35 Remark It is in general very hard to determine the exact Newton polygon of a given L-function. Instead we consider L-functions associated to data varying in a family; Grothendieck’s specialization Theorem 12.8.37 asserts that in such a family, most L-functions share the same Newton polygon, the generic Newton polygon; for instance our ﬁrst result states that smooth complete intersections are generically ordinary. We describe this last polygon (or parts of it) in the known cases (toric sums, aﬃne sums in one variable). Then we deal with its asymptotic variation with the characteristic. 12.8.36 Theorem [1571, 2910] Let S denote the scheme parametrizing smooth complete intersec-tions of dimension n and multi-degree (d1, . . . , dr) in Pn+r Fq , and X →S the universal family. There exists a Zariski dense open subset U such that for any s ∈U, Xs is ordinary. 12.8.37 Theorem (Grothendieck’s specialization theorem) Assume ft : X →A1 belongs to a family parametrized by t varying in an aﬃne scheme S/Fp. We assume the L-function L(X, f ∗ t Lψ, T) (or its inverse) to be a polynomial of constant degree when t varies, and we denote by NPq(ft) its q-adic Newton polygon (the q-adic Newton polygon of its inverse). There is a Zariski dense open subset U in S (the open stratum) and a polygon GNP(S, p) such that 1. For any t ∈U(Fq), NPq(ft) =GNP(S, p). 2. For any t ∈S(Fq), NPq(ft) lies above GNP(S, p). 12.8.38 Deﬁnition The polygon GNP(S, p) deﬁned above is the generic Newton polygon of the family ft, t ∈S. When it exists, the Hasse polynomial for the generic Newton polygon of this family is the polynomial generating the ideal deﬁning the Zariski closed subset S\U. 12.8.39 Remark One can deﬁne more general Hasse polynomials, for instance for the ﬁrst vertex, or for the ﬁrst m vertices. 12.8.40 Example Set fλ(x, y, z) := z(y2 −x(x −1)(x −λ)) for λ ∈A1 Fq(Fp){0, 1}. The generic Newton polygon has vertices (0, 0), (1, 1), (2, 3), and the Hasse polynomial for the generic Newton polygon (actually for the vertex (1, 1)) is the polynomial F(λ) = P p−1 2 i=0 p−1 2 i 2 λi. In other words, the supersingular elliptic curves in Legendre form are those for which F(λ) = 0. 484 Handbook of Finite Fields 12.8.41 Theorem [2899, 2910] Let S∆parametrize the space of Laurent polynomials over Fp with Newton polytope ∆⊂RN, non-degenerate with respect to it. 1. If N ≤3, then GNP(S∆, p) = HP(∆) for any p ≡1 (mod D(∆)). 2. If N ≥4, there exists an integer D′(∆) (in general strictly greater than D(∆)) depending only on ∆such that GNP(S∆, p) = HP(∆) for any large enough prime p such that p ≡1 (mod D′(∆)). 12.8.42 Remark The L-function associated to f has its coeﬃcients in Qp(ζp), which is a totally ramiﬁed extension of Qp of degree p −1. Since the ordinates for the vertices of the Hodge polygon are in 1 D(∆)N, a necessary condition in order to have GNP(S∆, p) = HP(∆) is p ≡1 (mod D(∆)). The theorem above shows that it is not a suﬃcient condition for N ≥4. 12.8.43 Remark When N = 1, something stronger is true: if p ≡1 (mod lcm(d, d′)), then for any f(x) = Pd i=−d′ aixi ∈Fq[x, x−1] with ada−d′ ̸= 0, we have NP(f) = HP([−d′, d]). 12.8.44 Remark More generally, this result remains true for rational functions of one variable with poles of orders d1, . . . , ds when we have p ≡1 (mod lcm(d1, . . . , ds)). 12.8.45 Theorem Let D = {1, . . . , d}; if p > 2d, the ﬁrst vertex of the generic Newton polygon GNP(SD, p) is (1, 1 p−1⌈p−1 d ⌉), with Hasse polynomial n f ⌈p−1 d ⌉o p−1, the polynomial associating to the coeﬃcients a1, . . . , ad of f the degree p −1 coeﬃcient of the ⌈p−1 d ⌉-th power of f. 12.8.46 Theorem Let D ⊂N be ﬁnite, d = max D, and consider SD, the aﬃne variety parametrizing degree d polynomials whose monomials have their exponents in D. The ﬁrst slope of the generic Newton polygon GNP(SD, p) is equal to the p-density πp(D) of the set D. 12.8.47 Theorem Let p = 2, and D = {1 ≤i ≤d, 2 ∤i}; set SD =SpecFp[{ai}i∈D, a−1 d ] (we parametrize the polynomials by their coeﬃcients). The ﬁrst vertex of the generic Newton polygon GNP(SD, p) is 1. (n, 1) if 2n −1 ≤d < 2n+1 −3, with Hasse polynomial a2n−1; 2. (2n, 2) if d = 2n+1 −3, with Hasse polynomial a3·2n−1−1; 3. in the second case, if we consider D′ = D{3·2n−1 −1}, the ﬁrst vertex of the generic Newton polygon GNP(SD′, p) is (n, 1), with Hasse polynomial a2n−1. 12.8.48 Remark The ﬁrst result in the direction of Theorem 12.8.47 can be found in where the authors determine the ﬁrst slope and necessary conditions to get it. There are also results for the ﬁrst vertex in any characteristic p , but only dealing with the cases pn −1 ≤d ≤2pn −2. 12.8.49 Theorem Let d ≥2 be an integer prime to p, and Sd = Spec Fp[a1, . . . , ad, a−1 d ] parametrize the degree d polynomials. If p ≥3d, the generic Newton polygon GNP(Sd, p) has vertices i, Yi p −1 0≤i≤d−1 , Yi = i X j=1 pj −i d , and the Hasse polynomial for the i-th vertex is a polynomial in Fp[a1, . . . , ad], homogeneous of degree Yi. 12.8.50 Remark The Newton polygons corresponding to degree 4 , degree 6 polyno-mials, and to the family xd + λx when p is large enough and p ≡−1 (mod d) are completely determined. Curves over ﬁnite ﬁelds 485 12.8.51 Remark The generic Newton polygons associated to twisted one variable sums (coming from the product of an additive character evaluated at a Laurent polynomial and of a multiplicative character evaluated at the variable) are given in for p large enough. Using the Poisson formula, they give the generic Newton polygons attached to families of polynomials P(xs), deg P = d. 12.8.52 Theorem [3067, 3068] 1. We have the limit limp→∞GNP(Sd, p) = HP(d). 2. There is a Zariski dense open subset U in SpecQ[a1, . . . , ad, a−1 d ] (parametrizing the de-gree d polynomials deﬁned over Q) such that for any f ∈U we have limp→∞NP(f mod p) = HP(d). 12.8.53 Remark In the case of twisted exponential sums, by a multiplicative character of order s (or for the family of polynomials P(xs), deg P = d), the limit no longer exists as in the ﬁrst assertion above. Actually the limit exists if we restrict to the primes in a ﬁxed residue class modulo s , and there is a result similar to the second assertion in this case. 12.8.54 Remark For L-functions associated to a one variable rational function of ﬁxed pole orders d1, . . . , ds, the generic Newton polygon tends to the Hodge polygon HP(d1, . . . , ds) when p tends to inﬁnity . 12.8.55 Problem There are conjectures by Wan asserting that under certain additional hy-potheses, Theorem 12.8.52 remains true for the space S∆of polynomials over Fp with Newton polytope at inﬁnity ∆⊂RN, non-degenerate with respect to it. Actually a conse-quence of Theorem 12.8.41 is that lim infp→∞GNP(S∆, p) = HP(∆). Some special cases of these conjectures are proved in . 12.8.5 The case of curves and abelian varieties 12.8.56 Remark We consider the Newton polygons of curves. To each curve one can associate its Jacobian variety, and more generally we consider the Newton polygons of abelian varieties. They encode useful invariants, such as the p-rank. We give the stratiﬁcation of the space of principally polarized abelian varieties by their Newton polygon as described in the work of Oort and others. Then we focus on curves; even if the situation is less well-known than in the case of abelian varieties, the subject has drawn much attention and we give the principal results. We end the section with some remarks about Artin-Schreier curves. 12.8.57 Deﬁnition The Newton polygon of a curve C deﬁned over Fq is the q-adic Newton polygon of the numerator L(C, T) of its zeta function. Let A be an abelian variety of dimension g deﬁned over Fq; for any prime ℓ̸= p, the inverse limit of the ℓ-th power torsion subgroups of A is the ℓ-adic Tate module of A, a Zℓ-module of rank 2g. Let PA(T) denote the characteristic polynomial of the action of Frobenius (q-th power) on the Qℓ-vector space Tℓ(A) ⊗Qℓ. The Newton polygon of the abelian variety A deﬁned over Fq is the q-adic Newton polygon of the polynomial T 2gPA( 1 T ). 12.8.58 Remark The Newton polygon of a curve C coincides with one of its Jacobian variety JC, as deﬁned above. 12.8.59 Remark For a curve of genus g, or an abelian variety of dimension g, the Newton polygon starts at (0, 0) and ends at (2g, g). Moreover, it follows from Poincar´ e duality that it is symmetric in the sense that if it contains the slope s with multiplicity m, it also contains the slope 1 −s with the same multiplicity. 486 Handbook of Finite Fields 12.8.60 Remark Some authors consider the Newton polygon of the polynomial PA(T); this Newton polygon is symmetric to the one we consider here, with respect to the line x = g. 12.8.61 Deﬁnition A genus g curve (an abelian variety of dimension g) is ordinary when its Newton polygon has the slopes 0 and 1 each with multiplicity g; it is supersingular when it has the slope 1 2 with multiplicity 2g. A polygon satisfying the requirements of Remark 12.8.59 is admissible. 12.8.62 Theorem Curves of genus g (resp. abelian varieties of dimension g) are generically ordinary. 12.8.63 Theorem Let Ag,1 ⊗Fp denote the space parametrizing principally polarized abelian varieties of dimension g deﬁned over Fp. It has dimension g(g+1) 2 . Let ζ be an admissible polygon; the space Wζ of principally polarized abelian varieties having their Newton polygon lying on or above ζ is closed, and has dimension dim Wζ = #{(x, y) ∈N2, y < x ≤g, (x, y) above ζ}. 12.8.64 Remark There are more precise results by Li and Oort on the supersingular stratum . 12.8.65 Problem As a consequence, every admissible polygon is the Newton polygon of an abelian variety. It is not known whether this is true for curves. Van der Geer and Van der Vlugt have shown that for p = 2, there are supersingular curves of every genus , but this is not even known in odd characteristic. 12.8.66 Deﬁnition The p-rank of the curve C (resp. of the abelian variety A) is the integer in {0, . . . , g} deﬁned as either the length of the horizontal segment of its Newton polygon, or the dimension of the Fp-vector space JC[p] (resp. A[p]). 12.8.67 Remark From Theorem 12.8.63, the space of principally polarized abelian varieties having p-rank f has codimension g −f in Ag,1 ⊗Fp; this is also true in the space Mg of genus g curves (which has dimension 3g −3) and in the space Hg of genus g hyperelliptic curves (of dimension 2g −1). 12.8.68 Deﬁnition The Hasse-Witt matrix of a non-singular curve C of genus g is the matrix of the Frobenius (p-th power) mapping on the g dimensional space H1(C, OC). 12.8.69 Remark Via Serre’s duality, the Hasse-Witt matrix is the transpose of the matrix of the Cartier-Manin operator on the space of diﬀerentials of the ﬁrst kind. 12.8.70 Theorem Let H denote the Hasse Witt matrix of C, a curve deﬁned over Fq, q = pm, and H(pi) denote the matrix obtained by raising all coeﬃcients of H to the power pi. Let Ha := HH(p) · · · H(pa−1). Then 1. The rank of Hg is the p-rank of the curve C. 2. We have the congruence L(C, T) ≡det(Ig −THm) (mod p). 12.8.71 Example With respect to the basis dual to the one given in Example 12.1.85, the Hasse-Witt matrix of the hyperelliptic curve y2 = f(x) is the matrix ({f p−1 2 }pi−j)1≤i,j≤g. 12.8.72 Remark There are more general congruences for the factor of the zeta function of an hypersurface coming from the primitive middle cohomology space . Curves over ﬁnite ﬁelds 487 12.8.73 Theorem [752, Corollary 1.8] (Deuring-Shafarevich formula) Let C and C′ be two curves with function ﬁelds E and F; assume that the extension E/F is Galois with Galois group G a p-group. If f and f ′ denote the respective p-ranks of C and C′, then the relation 1 −f = #G(1 −f ′) + X x∈C(k) (ex −1), holds, where ex is the ramiﬁcation index of the place x in the extension E/F. 12.8.74 Example Let E = k(x, y) be the extension of the rational function ﬁeld k(x) deﬁned by the (Artin-Schreier) equation yp −y = f(x); assume that f has s poles with orders prime to p. Then E/k(x) is a Galois extension with Galois group Z/pZ, and the p-rank of E is f = s −1. 12.8.75 Remark One can deduce a stratiﬁcation by the p-rank of the space of Artin-Schreier curves (i.e., Artin-Schreier coverings of the projective line) . 12.8.76 Remark We have an expression for the numerator of the zeta function of the Artin-Schreier curve C : yp −y = f(x) from the L-functions L(af, T) associated to the sums P′ x∈P1 ψ(af(x)), where the sum is taken over the points which are not poles of f. Precisely we have L(C, T) = Q a∈Fp L(af, T). The L-functions on the right are conjugated under the action of Gal(Q(ζp)/Q); thus they all have the same Newton polygon, and NPq(C) is the dilation by the factor p −1 of NPq(f). As a consequence, the above determination of (parts of) Newton polygons of L-functions associated to one variable exponential sums translate to results on the Newton polygons of Artin-Schreier curves. 12.8.77 Remark The same argument gives the Newton polygon for curves with equation A(y) = f(x), A an additive polynomial, as a dilation of the Newton polygon NPq(f). 12.8.78 Remark We focus on the case p = 2: here Artin-Schreier curves are hyperelliptic curves. From the remark above, we have NPq(C) = NPq(f) for C : y2 +y = f(x). The stratiﬁcation of Hg by the 2-rank is described in : the irreducible components of the stratum of curves with 2-rank f, Hg,f are in bijection with the partitions of g + 1 in f + 1 positive integers. Inside Hg,0, one can reduce to f a polynomial, and Theorem 12.8.47 gives the ﬁrst vertex for the generic Newton polygon in this space. One can also deduce from these results a theorem originally proved by Scholten and Zhu: there is no supersingular hyperelliptic curve of genus g = 2n −1, n ≥2 in characteristic 2 . 12.8.79 Remark This result stands in striking contrast with the situation of genus g = 2n . In this case the dimension of the space of hyperelliptic supersingular curves is greater than or equal to n; it can be as large as possible. 12.8.80 Remark For g ≤8, the supersingular hyperelliptic curves are completely determined when p = 2 in , and when p = 3 in . When p = 2 one can also give the ﬁrst vertices occuring for Newton polygons of curves in Hg,0 for g ≤9 . 12.8.81 Remark There are results asserting the non-existence of supersingular Artin-Schreier curves in odd characteristic for some inﬁnite families of genera . 488 Handbook of Finite Fields See Also §7.1 For discussion of equations over ﬁnite ﬁelds. §12.7 For discussion of zeta-functions and L-functions. §12.9 For discussion of rational points and zeta-functions. References Cited: [20, 21, 23, 25, 26, 27, 29, 31, 150, 255, 291, 292, 293, 295, 296, 553, 752, 798, 799, 940, 941, 942, 1022, 1281, 1397, 1528, 1529, 1571, 1699, 1754, 1769, 1770, 1802, 1915, 1919, 1948, 1949, 1950, 1999, 2041, 2151, 2153, 2322, 2430, 2551, 2552, 2553, 2698, 2842, 2843, 2895, 2896, 2897, 2899, 2910, 3028, 3041, 3067, 3068, 3069] 12.9 Computing the number of rational points and zeta functions Daqing Wan, University of California Irvine 12.9.1 Remark As in Section 7.1, we shall restrict to the case of hypersurfaces. We focus on theoretical and deterministic results. Probabilistic algorithms and improvements are not discussed. As always, Fq denotes a ﬁnite ﬁeld of characteristic p. The time for algorithms means the number of ﬁeld operations. 12.9.1 Point counting: sparse input 12.9.2 Deﬁnition For a polynomial f ∈Fq[x1, . . . , xn], the sparse representation of f is the sum of its non-zero terms f(x1, . . . , xn) = m X j=1 ajxVj, aj ∈F∗ q, where Vj = (v1j, . . . , vnj), xVj = xv1j 1 · · · xvnj n . The point counting problem is to compute the number #Af(Fq) of Fq-rational points of the equation f = 0. 12.9.3 Remark For this problem, we may replace xq i by xi and assume that the degree of f in each variable is at most q −1. The sparse input size of f is then mn log(q). 12.9.4 Example For non-zero elements ai ∈Fq and b ∈Fq, let f(x) = a1xq−1 1 + · · · + anxq−1 n + b, be a diagonal polynomial. Deciding if #Af(Fq) > 0 is equivalent to deciding if there is a subset {ai1, . . . , aik} of the set {a1, . . . , an} such that ai1 + · · · + aik + b = 0. Curves over ﬁnite ﬁelds 489 The latter problem is the subset sum problem over Fq, which is well known to be NP-complete. The fastest known deterministic algorithm for deciding if #Af(Fq) > 0 in this case is the baby-step-giant-step method, which runs in time O(n2n/2 log(q)). 12.9.5 Theorem Computing #Af(Fq) is NP-hard, even in the case n = 2 or deg(f) = 3. 12.9.6 Remark For a positive integer r > 1, the modular counting problem is to compute the residue class of #Af(Fq) modulo r. It is clear that 0 ≤#Af(Fq) ≤qn. Thus, if one can compute #Af(Fq) modulo r for a single large r > qn or for many small r, the Chinese remainder theorem implies that one can compute #Af(Fq) as well. This suggests that even for small r, the modular counting problem is not going to be much easier than the full counting problem. 12.9.7 Theorem Let r be a positive integer. Let q = ph. If r is not a power of p, then computing #Af(Fq) modulo r is NP-hard. If r = pb is a power of p, computing #Af(Fq) modulo r is also NP-hard, if either p ≥2n or h ≥2n or b > nh, that is, r = pb > qn. 12.9.8 Remark This complexity result shows that if r is not a power of p, one cannot expect a fast algorithm to compute #Af(Fq) modulo r. Even in the case r = pb is a power of p, any general algorithm computing #Af(Fq) modulo pb is expected to be fully exponential in each of the three parameters {p, b, h}. The next two results provide non-trivial algorithms in this direction. 12.9.9 Theorem Let q = ph and r = pb. The number #Af(Fq) modulo pb can be computed in time O(nm2qb) = O(nm2phb), where m is the number of monomials of f. 12.9.10 Theorem Let q = ph and r = pb. The number #Af(Fq) modulo pb can be computed in time O(n(8m)p(h+b)), where m is the number of monomials of f. 12.9.11 Problem Improve the exponent p(h + b) to O(p + h + b) if possible. 12.9.2 Point counting: dense input 12.9.12 Deﬁnition For a polynomial f ∈Fq[x1, . . . , xn] of degree at most d > 1, the dense repre-sentation of f is the sum of all terms of degree at most d: f(x1, . . . , xn) = X i1+···+in≤d ai1,...,inxi1 1 · · · xin n , ai1,...,in ∈Fq. 12.9.13 Remark Replacing xq i by xi and we may again assume that the degree of f in each variable is at most q −1. The dense input size of f is then (d + 1)n log(q). 12.9.14 Remark There is no known complexity result for the general point counting problem with dense input. On the contrary, there are polynomial time algorithms in various special cases. This suggests that the dense input point counting problem may have polynomial time algorithms in much greater generality. We shall describe some of these positive results below. In the special case that both n and d are ﬁxed, the sparse input size agrees with the dense input size. This is the case for elliptic curves for instance. 12.9.15 Theorem There is a p-adic algorithm which computes the number #Af(Fq) in time O(p2n+4(dn logp q)3n+7). 12.9.16 Remark This is a general purpose algorithm, which runs in polynomial time if p is small and n is ﬁxed. It assumes no conditions on the aﬃne hypersurface Af. If one assumes additional conditions on f, signiﬁcant improvements can be made. In the following, we give several such examples. 490 Handbook of Finite Fields 12.9.17 Theorem Assume that both the aﬃne hypersurface Af and its inﬁnite part are smooth of degree d not divisible by p > 2. Then, the number #Af(Fq) can be computed in time Oϵ(p2+ϵ(dn logp q)O(1)). 12.9.18 Remark It may be possible to improve the factor p2+ϵ to p0.5+ϵ. This has been done in the special case of hyperelliptic curves and more generally superelliptic curves [1429, 2109]. 12.9.19 Theorem [18, 2394, 2560] Let n = 2 and assume that the projective curve deﬁned by f is smooth. Then, the number #Af(Fq) can be computed in time O((log q)cd), where cd is a constant depending only on d. 12.9.20 Remark The constant cd is in general exponential in d. For hyperelliptic curves, the constant cd can be taken to be a polynomial in d. It is not clear if the same theorem is true for singular plane curves. 12.9.21 Remark In the case of the diagonal hypersurface f(x1, . . . , xn) = a1xd1 1 + · · · + anxdn n + b, the number #Af(Fq) has a compact expression in terms of Jacobi sums, which can be computed in polynomial time using LLL lattice basis reduction. This is worked out in some cases in . 12.9.3 Computing zeta functions: general case 12.9.22 Remark For an aﬃne hypersurface Af deﬁned by a polynomial f(x1, . . . , xn) over Fq, the zeta function Z(Af, T) = exp ∞ X k=1 #Af(Fqk) k T k ! is a rational function in T; see Section 12.8 for more details. The degrees of its numerator and the denominator can be bounded by a function depending only on the degree d of the polynomial f and the number n of variables. The output size for the zeta function is comparable to the dense input size O(dn log q) of f. Thus, in computing the zeta function, we always use the dense input size. 12.9.23 Theorem There is an algorithm, that given f ∈Fq[x1, . . . , xn] of degree d, computes the reduction of the zeta function Z(Af, T) modulo p in time bounded by a polynomial in p d n log q. 12.9.24 Remark This is a polynomial time algorithm if p is small. However, it only gives the modulo p reduction of the zeta function. For any other prime ℓ̸= p, no nontrivial general algorithm is known which computes the reduction of the zeta function modulo ℓ, except when n ≤2. By the Chinese remainder theorem, this is not much easier than computing the full zeta function in general. 12.9.25 Theorem There is an algorithm, that given f ∈Fq[x1, . . . , xn] of degree d, computes the zeta function Z(Af, T) in time bounded by a polynomial in (dnp log q)n. 12.9.26 Remark This is a polynomial time algorithm if p is small and n is ﬁxed. In the case that f is suﬃciently smooth, this result can be greatly improved as follows. 12.9.27 Theorem [1862, 1863] There is an algorithm, that given f ∈Fq[x1, . . . , xn] of degree d, computes the zeta function Z(Af, T) in time bounded by a polynomial in dnp log q, provided that the aﬃne hypersurface Af and its inﬁnite part are both smooth and d is not divisible by p > 2. Curves over ﬁnite ﬁelds 491 12.9.28 Remark In various special cases, one can expect signiﬁcantly better results. For instance, when n = 1, there is always a polynomial time algorithm which computes the zeta function of the zero-dimensional hypersurface Z(Af, T) . The case n = 2 (the curve case) has been studied most extensively. We state two such results in the next subsection. For a diagonal hypersurface f(x1, . . . , xn) = a1xd1 1 + · · · + anxdn n + b, the zeta function has an explicit expression in terms of Jacobi sums, which can then be computed in polynomial time using LLL basis reduction; see for the main ideas. 12.9.4 Computing zeta functions: curve case 12.9.29 Remark In this subsection, we restrict to the smooth plane curve Cf in P2 deﬁned by a smooth homogenous polynomial f(x1, x2, x3) of degree d over Fq. The genus of the curve is then g = (d −1)(d −2)/2. The zeta function of Cf is of the form Z(Cf, T) = Pf(T) (1 −T)(1 −qT), where Pf(T) ∈1 + TZ[T] is a polynomial of degree 2g. The special value Pf(1) is the order of the Jacobian variety of Cf. Thus, any algorithm for computing the zeta function gives an algorithm for computing the order of the Jacobian. We state two general results for curves. The ﬁrst one is ℓ-adic in nature, and the second one is p-adic in nature. 12.9.30 Theorem [18, 2394, 2560] There is an algorithm which computes the zeta function Z(Cf, T) of the curve Cf in time O((log q)cg), where cg is a constant which in general depends exponentially on g. 12.9.31 Remark This is a polynomial time algorithm for ﬁxed genus g. For hyperelliptic curves, the constant cg can be taken to be a polynomial in g. 12.9.32 Theorem [564, 1718] There is an algorithm which for ﬁxed p, computes the zeta function Z(Cf, T) of the curve Cf in time ˜ O(g6(logp q)3 + g6.5(logp q)2). 12.9.33 Remark This is a polynomial time algorithm if p is ﬁxed. The results in work for more general non-degenerate toric curves. 12.9.34 Remark There are a lot more specialized and more precise results on computing the zeta function in various special cases, such as elliptic curves, hyperelliptic curves, superelliptic curve, Cab-curves, Kummer curves, Artin-Scheirer curves, Fermat curves, etc. We refer to [552, 816, 817, 1248, 1556, 1861, 1865, 1905, 2533] for more details and for further references. For general survey, see [579, 1720, 2901]. See Also §12.5 Discusses rational points on curves. §12.7 Discusses zeta-functions and L-functions. §12.8 Discusses p-adic estimates of zeta-functions. References Cited: [18, 455, 552, 564, 579, 816, 817, 1231, 1248, 1318, 1429, 1556, 1718, 1720, 1861, 1862, 1863, 1865, 1866, 1905, 2109, 2394, 2533, 2560, 2894, 2901, 2903]. This page intentionally left blank This page intentionally left blank 13 Miscellaneous theoretical topics 13.1 Relations between integers and polynomials over ﬁnite ﬁelds ........................................... 493 The density of primes and irreducibles • Primes and irreducibles in arithmetic progression • Twin primes and irreducibles • The generalized Riemann hypothesis • The Goldbach problem over ﬁnite ﬁelds • The Waring problem over ﬁnite ﬁelds 13.2 Matrices over ﬁnite ﬁelds........................... 500 Matrices of speciﬁed rank • Matrices of speciﬁed order • Matrix representations of ﬁnite ﬁelds • Circulant and orthogonal matrices • Symmetric and skew-symmetric matrices • Hankel and Toeplitz matrices • Determinants 13.3 Classical groups over ﬁnite ﬁelds .................. 510 Linear groups over ﬁnite ﬁelds • Symplectic groups over ﬁnite ﬁelds • Unitary groups over ﬁnite ﬁelds • Orthogonal groups over ﬁnite ﬁelds of characteristic not two • Orthogonal groups over ﬁnite ﬁelds of characteristic two 13.4 Computational linear algebra over ﬁnite ﬁelds .. 520 Dense matrix multiplication • Dense Gaussian elimination and echelon forms • Minimal and characteristic polynomial of a dense matrix • Blackbox iterative methods • Sparse and structured methods • Hybrid methods 13.5 Carlitz and Drinfeld modules ...................... 535 Quick review • Drinfeld modules: deﬁnition and analytic theory • Drinfeld modules over ﬁnite ﬁelds • The reduction theory of Drinfeld modules • The A-module of rational points • The invariants of a Drinfeld module • The L-series of a Drinfeld module • Special values • Measures and symmetries • Multizeta • Modular theory • Transcendency results 13.1 Relations between integers and polynomials over ﬁnite ﬁelds Gove Eﬃnger, Skidmore College The arithmetic structures of the ring of integers Z and the ring of polynomials Fq[x], where q is a prime power, are strikingly similar. In particular, the densities of irreducible elements in these rings are virtually identical, leading to very closely analogous theorems (and conjectures) in the two settings. For a general exposition on the ideas contained in this 493 494 Handbook of Finite Fields section, see, for example, . Here we state some of these analogous deﬁnitions, theorems, and conjectures, listing ﬁrst the item involving Z and second its analogue in Fq[x]. The latter of these ﬁrst two deﬁnitions will be used throughout this section. 13.1.1 Deﬁnition For m (̸= 0) ∈Z, the absolute value of m, denoted \|m\|, is \|Z/ ⟨m⟩\|. 13.1.2 Deﬁnition For f (̸= 0) ∈Fq[x], the absolute value of f, denoted \|f\|, is \|Fq[x]/ ⟨f⟩\|. 13.1.3 Remark Suppose the degree of f in the above deﬁnition is n. Since the quotient ring consists of all polynomials of degree less than n, we see that in fact \|f\| = qn. We note also that n = logq(\|f\|). These facts will be relevant frequently in what follows. 13.1.4 Remark Throughout this section we shall use the notation f(k) ∼g(k) to mean that limk→∞f(k)/g(k) = 1. In addition, the notation log denotes the natural logarithm. 13.1.1 The density of primes and irreducibles 13.1.5 Theorem (The Prime Number Theorem) Let π(m) be the number of prime numbers less than or equal to m. Then π(m) ∼ m log(m). 13.1.6 Remark For the polynomial case, we employ the notation of Deﬁnition 2.1.23 but, in analogy with integers, add the notation πq(n) to mean the number of monic irreducible polynomials over Fq of degree less than or equal to n. The next result then follows immediately from Theorem 2.1.24. 13.1.7 Theorem (The Polynomial Prime Number Theorem) If Iq(n) is as in Deﬁnition 2.1.23, we have Iq(n) ∼qn n . 13.1.8 Remark We note that if f ∈Fq[x] is of degree n, this theorem says that Iq(n) ∼ \|f\| logq(\|f\|), in exact analogy to the Prime Number Theorem. 13.1.9 Remark The two Prime Number Theorems (integer and polynomial) both say that “near” an element, the density of primes is 1 out of the log of the absolute value of the element. In the integer case the log is natural; in the polynomial case the log is base q. 13.1.10 Remark The next result of Lenskoi gives asymptotic information for the case of counting monic irreducibles of degree less than or equal to n (see also ). 13.1.11 Theorem Let πq(n) be as deﬁned in 13.1.6 above. Then we have πq(n) ∼ q q −1 qn n . Miscellaneous theoretical topics 495 13.1.2 Primes and irreducibles in arithmetic progression 13.1.12 Theorem (Primes in Arithmetic Progression) Let the Euler function φ be as in Deﬁni-tion 2.1.43 and suppose a and d are relatively prime positive integers. By πa,d(m) we mean the number of primes less than or equal to m which are congruent to a modulo d. Then πa,d(m) ∼ m log(m)φ(m). 13.1.13 Remark This then says that for any such pair {a, d} there are inﬁnitely many primes which are congruent to a (mod d), and moreover that the primes are “ultimately uniformly distributed” among the eligible congruence classes of d. Artin proved the following analogous result for polynomials. 13.1.14 Theorem Let the “polynomial Euler function” Φq be as in Deﬁnition 2.1.111 and suppose A and D are relatively prime polynomials in Fq[x]. By Iq;A,D(n) we mean the number of monic irreducible polynomials of degree n which are congruent to A modulo D. Then Iq;A,D(n) ∼ qn nΦq(D). 13.1.15 Remark Hayes generalizes this result to a much broader class of congruence relations which he calls “arithmetically distributed” relations. As an example, he shows that the relation of two monic polynomials having the same ﬁrst k and last m coeﬃcients (see Deﬁnition 3.5.1 in ) is arithmetically distributed, and so the following theorem holds. 13.1.16 Theorem Let Iq;k,m(n) be the number of monic irreducible polynomials of degree n for which the ﬁrst k and last m coeﬃcients are prescribed. Then Iq;k,m(n) ∼ q q −1 qn−k−m n . 13.1.17 Remark This says that monic irreducibles are “ultimately uniformly distributed” with respect to their ﬁrst k and last m coeﬃcients, provided of course that the constant term is not 0. For much more information on irreducible polynomials with prescribed coeﬃcients, see Section 3.5. 13.1.3 Twin primes and irreducibles 13.1.18 Deﬁnition Two odd prime numbers are twin primes if the absolute value of their diﬀerence is as small as possible, i.e., is 2. 13.1.19 Deﬁnition Two monic irreducible polynomials over Fq are twin irreducibles if the absolute value of their diﬀerence is as small as possible, i.e., is 1 if q > 2 and is 4 if q = 2. 13.1.20 Remark This last deﬁnition implies that for q > 2, two monic irreducibles are twins provided they are identical except in their constant terms (so their diﬀerence has degree 0). For q = 2, however, all irreducible have 1 as their constant coeﬃcient and all have an odd number of terms (otherwise they are divisible by x + 1), and so in this case twins will diﬀer in their linear and quadratic terms (so their diﬀerence has degree 2). 13.1.21 Conjecture Let π2(m) be the number of twin prime pairs less than or equal to m. Then π2(m) ∼2 m (log m)2 Y odd p 1 − 1 (p −1)2 . 496 Handbook of Finite Fields 13.1.22 Remark The product over odd primes in the above conjecture is the “twin primes constant” and has value approximately 0.66016. If this conjecture were proved, it then implies the “Twin Primes Conjecture,” i.e., the existence of inﬁnitely many twin prime pairs. We have an analogous conjecture in the polynomial setting. 13.1.23 Conjecture Let I2,q(n) be the number of twin irreducible polynomials of degree n. Then I2,q(n) ∼δ q −1 2 qn n2 Y P 1 − 1 (\|P\| −1)2 , where either δ = 1 and the product is over all monic irreducibles P provided q > 2, or δ = 4 and the product excludes linear irreducibles when q = 2. 13.1.24 Remark Though unproven, these two conjectures are strongly supported by numerical evidence. For example, for polynomials of degree 16 over F3, the above formula accurately predicts the 66606 twin irreducible pairs. For more information on what is currently known along the lines of Conjecture 13.1.23, especially for the “ﬁxed n, q going to ∞” case, see Pollack and . 13.1.25 Remark Though the Twin Primes Conjecture remains unproven, it was ﬁrst observed by Hall and then further explored by Pollack and Eﬃnger , that the “Polyno-mial Twin Primes Conjecture” holds provided only that q > 2. The proof technique makes use of the fact that unlike prime numbers, irreducible polynomials have “internal struc-ture.” Speciﬁcally, “irreducibility preserving substitutions” can be exploited to guarantee the following theorem. 13.1.26 Theorem (Polynomial Twin Primes Theorem) Over every Fq with q > 2, there exist inﬁnitely many twin irreducible pairs. 13.1.27 Remark In fact one can prove the existence of “t-tuplets” (twins being 2-tuplets) in the polynomial case. For example, over F7 we can guarantee the existence of inﬁnitely many 4-tuplets. This follows from the next result. 13.1.28 Theorem Let Fq satisfy that q ≥4. If q ≡0 (mod 4) or q ≡1 (mod 4) and if p is any prime dividing q −1, then there exist exactly t = (p−1)(q−1) p irreducible polynomials of the form xpk −a over Fq for every k ≥1. If q ≡3 (mod 4), the conclusion holds for all odd primes p, but no irreducibles of the form x2k −a exist provided k ≥2. 13.1.29 Problem The polynomial twin primes conjecture remains unsolved over F2. 13.1.4 The generalized Riemann hypothesis 13.1.30 Deﬁnition If χ is a Dirichlet character (see, for example, Section 1.4.3 of ), then the corresponding Dirichlet L-function is L(s, χ) = Y p 1 −χ(p) ps −1 , where s ∈C and p ranges over all primes. 13.1.31 Remark The function L(s, χ), which converges for all Re(s) > 1, can, like the Riemann zeta function, be extended analytically to a meromorphic function on the whole complex plane. Miscellaneous theoretical topics 497 13.1.32 Conjecture (The Generalized Riemann Hypothesis) For every Dirichlet character χ, if Re(s) > 0 and if L(s, χ) = 0, then Re(s) = 1/2. 13.1.33 Deﬁnition If χ is a character “of Dirichlet type” on Fq[x] (see, for example, Chapter 5 of ), then the corresponding polynomial Dirichlet L-function is Lq(s, χ) = Y P 1 −χq(P) \|P\|s −1 , where s ∈C and P ranges over all monic irreducibles over Fq. 13.1.34 Remark The following polynomial analogue of the Generalized Riemann Hypothesis follows from the deep results of Andr´ e Weil and is a key ingredient of the results of our next subsection and of many other results in the number theory of polynomials over ﬁnite ﬁelds. For example, see for an exposition of the polynomial analogue of Artin’s conjecture on primitive roots. 13.1.35 Theorem (A Polynomial Generalized Riemann Hypothesis) The function Lq(s, χ) is a complex polynomial Fχ in q−s and when factored into Fχ(q−s) = Y (1 −γiq−s), each γi satisﬁes \|γi\| = q1/2. 13.1.5 The Goldbach problem over ﬁnite ﬁelds 13.1.36 Remark In both the integer and polynomial cases, the “two-primes” Goldbach problem of writing an appropriate element of the ring Z or Fq[x] as a sum of two irreducible elements remains unsolved. However, if one moves to the “three-primes” case, a great deal more can be said. The ﬁrst giant step forward was the pioneering work of Hardy and Littlewood in the 1920s. In the following result, Hypothesis R (a Weak Generalized Riemann Hypothesis) replaces the 1/2 in Conjecture 13.1.32 with Θ, where 1/2 ≤Θ < 3/4. 13.1.37 Theorem If Hypothesis R holds, then, as m →∞, the number N3(m) of represen-tations of the odd integer m as a sum of three odd primes satisﬁes: N3(m) ∼ m2 (log m)3 Y p>2 1 + 1 (p −1)3 Y p\|m 1 − 1 p2 −3p + 3 , where p runs over prime numbers as speciﬁed. 13.1.38 Remark Vinogradov succeeded in removing Hypothesis R from the above Hardy-Littlewood result, hence proving unconditionally that every suﬃciently large odd number is a sum of three primes. It has since been established using reﬁnements of his analysis that “suﬃciently large” can be assumed to mean above 1043000 . It has also been established that if one assumes Conjecture 13.1.32, then every odd number greater than 5 is a sum of three primes (the so-called “Complete 3-Primes Theorem under GRH”; see, for example, ). 13.1.39 Remark Turning to the polynomial case, Hayes adapted the Hardy-Littlewood “Cir-cle Method” to the function ﬁeld setting, using the completion of Fq(x) at the inﬁnite place as the analogue of the unit circle in the complex plane and employing an appropriate version 498 Handbook of Finite Fields of Theorem 13.1.35, to obtain an asymptotic “3-primes” result for polynomials. However, the irreducibles in his analysis were not monic and hence the result was not a perfect ana-logue of the integer result. Later, using adelic analysis (wherein Fq(x) is completed at all its places, not just the inﬁnite one) and again an appropriate version of Theorem 13.1.35, Hayes and Eﬃnger obtained the theorem below. First we need a deﬁnition. 13.1.40 Deﬁnition A polynomial f over Fq is even if f is divisible by an irreducible P with \|P\| = 2. Otherwise f is odd. 13.1.41 Remark Obviously, even polynomials only exist over F2 and are then ones which are divis-ible by x or x + 1. 13.1.42 Theorem Let f be an odd monic polynomial of degree n over Fq. As q →∞or n →∞, the number M3(f) of representations of f = P1 + P2 + P3, where deg(P1) = n, deg(P2) < n and deg(P3) < n, satisﬁes: M3(f) ∼q2n n3 Y \|P \|>2 1 + 1 (\|P\| −1)3 Y P \|f 1 − 1 \|P\|2 −3\|P\| + 3 , where P runs over irreducible polynomials over Fq as speciﬁed. 13.1.43 Remark Because this result is asymptotic in both q and n, careful analysis of the error terms and further investigation, both theoretical and numerical, of the low degree and small ﬁeld cases [957, 958, 960] yield the following “best possible” result. 13.1.44 Theorem (A Complete Polynomial 3-Primes Theorem) Every odd monic polynomial of degree n ≥2 over every ﬁnite ﬁeld Fq (except the case of x2 + α when q is even) is a sum of three monic irreducible polynomials, one of degree n and the others of lesser degrees. 13.1.45 Remark Hence, we get “complete 3-primes theorems” in both the integer and polynomial cases provided that we have a proven Generalized Riemann Hypothesis. In the polynomial case, thanks to Weil, we do; in the integer case we still do not. 13.1.6 The Waring problem over ﬁnite ﬁelds 13.1.46 Remark The general problem of writing an arbitrary positive integer as a sum of a limited number of k-th powers, known as the Waring Problem, was ﬁrst settled by Hilbert in 1909. 13.1.47 Theorem (The Hilbert-Waring Theorem) For every positive integer k, there exists an integer s(k) such that every positive integer m can be written as a sum of at most s(k), k-th powers. 13.1.48 Remark Tremendous eﬀort over the years has been put toward the question: Given k, what is s(k)? To that end we have following deﬁnitions, the latter having ﬁrst been introduced by Hardy and Littlewood . 13.1.49 Deﬁnition Let g(k) be smallest s such that every positive integer m is a sum of at most s, k-th powers. Let G(k) be the smallest s for which every suﬃciently large positive integer is a sum of at most s, k-th powers. 13.1.50 Remark It is known that g(2) = G(2) = 4, that g(3) = 9, and that 4 ≤G(3) ≤7. Exact formulas are now known for g(k) for all k. Although exact values of G(k) are known only for k = 2 and k = 4 (G(4) = 16), considerable progress has been made on good lower and Miscellaneous theoretical topics 499 upper bounds for G(k). For excellent surveys of the Waring Problem, see Ellison and the more current Vaughan and Wooley . 13.1.51 Remark Turning to the polynomial case, one must ﬁrst observe that no cancellation occurs in the integer Waring Problem, and so the most appropriate analogue in the polynomial case will allow as little cancellation as possible. This is achieved as follows: 13.1.52 Deﬁnition The representation f = Xk 1 + · · · + Xk s of f ∈Fq[x] as a sum of k-th powers of polynomials in Fq[x] is a strict sum provided that for every 1 ≤i ≤s, deg(Xi) ≤ ⌈deg(f)/k)⌉(equivalently, provided that k deg(Xi) < k + deg(f)). 13.1.53 Remark The following result, ﬁrst proved independently by Car , Webb , and Kubota , is the best analogue to the Hilbert-Waring Theorem. 13.1.54 Theorem (Polynomial Waring Theorem) Suppose k < p = char(Fq). Then there exists an integer s(k), independent of q, such that every f ∈Fq[x] is a strict sum of s(k) k-th powers of polynomials in Fq[x]. 13.1.55 Remark Just as in the integer case, the question becomes: Given k, what is s(k)? The following deﬁnition is from Section 1.1 of . 13.1.56 Deﬁnition Given k ≥2, let gpoly(k) be the smallest value of s such that for every q with p = char(Fq) > k, every f ∈Fq[x] is a strict sum of s k-th powers in Fq[x]. Let Gpoly(k) be the smallest s such that this same condition holds except possibly for a ﬁnite number of polynomials in the collection of all Fq[x] with p > k. 13.1.57 Remark As discussed in Section 1.2 of , the case of k = 2 is settled by Serre, with Webb showing that the only exceptions are two polynomials of degree 3 and six polynomials of degree 4 over F3 which require four squares. 13.1.58 Theorem We have gpoly(2) = 4 and Gpoly(2) = 3. 13.1.59 Remark To date no exact values of gpoly(k) or Gpoly(k) are known for k > 2. The cases of k = 3 and k = 4 have been extensively studied. See the Introduction of for a good summary of what is currently known, including cases not satisfying the hypothesis of Theorem 13.1.54 (i.e., that k < p). Following our Deﬁnition 13.1.56, however, it is known that gpoly(3) ≤9, Gpoly(3) ≤7, and that Gpoly(4) ≤11; see [514, 515]. For results on upper bounds for gpoly(k), especially for large k, see . Finally, in , the following upper bounds for general k are established for both parameters. Note that in the case of gpoly(k), there is dependence on q. 13.1.60 Theorem We have 1. Gpoly(k) ≤k(log(k −1) + 3) + 3; 2. If k ≥4, then gpoly(k) ≤gcd(q −1, k)(k3 −2k2 −k + 1). 500 Handbook of Finite Fields See Also §3.1 For discussion of counting irreducible polynomials. §3.3 For discussion of reducibility criteria for polynomials. §3.5 For discussion of prescribed coeﬃcients of irreducible polynomials. §11.2 For discussion of polynomial counting algorithms. §11.4 For discussion of univariate polynomial factoring algorithms. For discussion of polynomial analogies of Artin’s primitive root conjecture. For discussion of polynomial analogies of sequences of smooth numbers. For discussion of polynomial analogies of the 3n + 1 problem. For discussion of polynomial analogies of perfect numbers. References Cited: [134, 508, 513, 514, 515, 602, 751, 823, 957, 958, 959, 960, 961, 962, 963, 974, 1166, 1167, 1402, 1418, 1419, 1420, 1447, 1448, 1497, 1500, 1606, 1810, 1891, 2016, 2409, 2410, 2411, 2412, 2862, 2877, 2955, 2962] 13.2 Matrices over ﬁnite ﬁelds Dieter Jungnickel, University of Augsburg On one hand we collect results about the numbers of matrices of various types over Fq; on the other hand, we shall also discuss matrix representations of the ﬁeld Fqm over Fq and give a few results concerning determinants. 13.2.1 Matrices of speciﬁed rank 13.2.1 Remark As noted in Remark 2.1.90, the vector space of all m × n matrices over a ﬁeld F = Fq has dimension mn, and thus the number of m × n matrices is qmn. We shall mainly concentrate on square matrices. Clearly, the m × m matrices over Fq constitute a ring R = F(m,m) q , and the invertible matrices in R form a group G = GL(m, q), the general linear group. The elements of G are exactly those matrices in R which transform the elements of a ﬁxed ordered basis B = {α1, . . . , αm} for Fqm over Fq into an ordered basis again. Hence the order of G agrees with the total number of distinct ordered bases: \|GL(m, q)\| = qm(m−1)/2(qm −1)(qm−1 −1) · · · (q −1). (13.2.1) Trivially, the determinants of the elements of G are uniformly distributed over F ∗. In particular, the matrices in G with determinant 1 form a group SL(m, q) of order \|SL(m, q)\| = qm(m−1)/2(qm −1)(qm−1 −1) · · · (q2 −1), called the special linear group. These two groups are the most elementary instances of the classical groups, which are discussed in detail in Section 13.4. 13.2.2 Remark Generalizing Equation (13.2.1), we give a formula for the number of all m × n matrices with prescribed rank k over Fq. This is closely related to the number of subspaces of prescribed dimension in a vector space over Fq. Miscellaneous theoretical topics 501 13.2.3 Deﬁnition The number of k-dimensional subspaces of an m-dimensional vector space over Fq is denoted by m k q. These numbers are the Gaussian coeﬃcients; they constitute a q-analogue of the binomial coeﬃcients. 13.2.4 Lemma The Gaussian coeﬃcient m k q is given explicitly as follows: hm k i q = (qm −1)(qm−1 −1) · · · (qm−k+1 −1) (qk −1)(qk−1 −1) · · · (q −1) . 13.2.5 Theorem Let V = F(m,n) q be the vector space of all m × n matrices over the ﬁeld F = Fq. Then the number f(m, n, k) of matrices with rank k ≤min{m, n} in V is f(m, n, k) = hn k i q qk(k−1)/2(qm −1)(qm−1 −1) · · · (qm−k+1 −1) = hm k i q qk(k−1)/2(qn −1)(qn−1 −1) · · · (qn−k+1 −1) = qk(k−1)/2 k−1 Y i=0 qm−i −1 qn−i −1 qi+1 −1 −1 . 13.2.6 Remark Clearly, the probability that a given entry of an m × n matrix of rank k over Fq is ̸= 0 does not depend on the position of the entry. In this probability is determined to be 1 −1 q 1 −1 qk 1 − 1 qm 1 − 1 qn . In , functions counting matrices of given rank over a ﬁnite ﬁeld with speciﬁed positions equal to 0 are studied. Such matrices may be viewed as q-analogues of permutations with certain restricted values. In particular, a simple closed formula for the number of invertible matrices with zero diagonal is obtained (see below), as well as recursions to enumerate matrices with zero diagonal by rank. 13.2.7 Theorem The number of invertible m × m matrices over Fq whose diagonal consists entirely of zeros is q( m−1 2 )(q −1)m q−1 m X i=0 (−1)i m i [m −i]q ! ! , where [k]q = qk−1 + · · · + q + 1. 13.2.2 Matrices of speciﬁed order 13.2.8 Remark By basic group theory, the order of an invertible m × m matrix over Fq has to divide the order of GL(m, q) given in Equation (13.2.1). This elementary result can be strengthened in two ways. 13.2.9 Theorem The least common multiple of all orders of matrices in GL(m, q) is peM, where q is a power of the prime p, e is the least integer satisfying m ≤pe, and M is the least common multiple of q −1, q2 −1, . . . , qm −1. In particular, the order o(A) of A ∈GL(m, q) divides peM. Moreover, o(A) is always bounded by qm −1. 502 Handbook of Finite Fields 13.2.10 Example Consider F = Fqm as a vector space over K = Fq, and let γ be any element of F ∗. Then γ deﬁnes a K-linear mapping Tγ : F →F via Tγ : ξ 7→γξ for ξ ∈F, and the order of Tγ equals the order o(γ) of γ in F ∗. Now represent Tγ with respect to some basis B of F over K; then the associated matrix Aγ(B) is a matrix of order o(γ) in GL(m, q). In particular, choosing γ as a primitive element ω for F gives a matrix Aω(B) of the maximum possible order qm −1. 13.2.11 Deﬁnition An invertible m × m matrix A of the maximum possible order qm −1 over Fq is a Singer cycle, and the group generated by A is a Singer subgroup of GL(m, q). Moreover, an involutory matrix is simply a matrix of order two. 13.2.12 Theorem Any two Singer subgroups of G = GL(m, q) are conjugate in G. The number S(m, q) of Singer subgroups of G equals S(m, q) = \|GL(m, q)\| m(qm −1), and the number of Singer cycles in G is given by S(m, q)φ(qm −1) = φ(qm −1) m · m−1 Y i=1 qm −qi , where φ is the Euler function given in Deﬁnition 2.1.43. 13.2.13 Theorem The number i(m, q) of involutory m × m matrices over Fq equals i(m, q) =    gm · Pm t=0 g−1 t g−1 m−t if q is odd, gm · P⌈m/2⌉ t=0 g−1 t g−1 m−2tq−t(2m−3t) if q is even, where gt denotes the number of invertible t × t matrices over Fq as given in (13.2.1) (for t ̸= 0) and g0 = 1. 13.2.14 Remark More generally, there is a (rather involved) formula for the number of m×m matrices of order k over Fq. Clearly, the probability that a randomly chosen m × m matrix over Fq is invertible (and therefore has an order) is gm/qm2, where gm denotes the number of invertible m × m matrices over Fq as given in (13.2.1). For ﬁxed q, this probability has a limit: lim m→∞ gm qm2 = Y r≥1 1 −1 qr . For q = 2, this limit is 0.28878 . . .; for q = 3, the limit is 0.56012 . . .; and, as q →∞, the probability of being invertible goes to 1. 13.2.15 Remark On the other end of the spectrum, the number of nilpotent m × m matrices over Fq – that is, matrices A satisfying Ak = 0 for some k – is given by a very simple formula: it equals qm(m−1). Miscellaneous theoretical topics 503 13.2.3 Matrix representations of ﬁnite ﬁelds 13.2.16 Deﬁnition Any subring R of the full matrix ring K(m,m) over a ﬁeld K which is itself a ﬁeld is a matrix ﬁeld (of degree m) over K. If K is a ﬁnite ﬁeld and if R ∼ = Fq, R is a matrix representation of degree m for Fq (over K). 13.2.17 Remark The following result concerning the existence of matrix representations for Fq is an obvious consequence of Theorem 13.2.9 and Example 13.2.10. For details, see Example 13.2.19. 13.2.18 Theorem Any matrix representation for Fqm over Fq has degree at least m. Moreover, there always exists a representation of degree m. 13.2.19 Example With the notation of Example 13.2.10, the qm −1 matrices Aγ(B) with γ ∈F ∗ together with the zero matrix form a matrix representation R(B) of degree m for Fqm over Fq. Clearly, one may write R(B) as the qm −1 powers of the Singer cycle Aω(B) together with the zero matrix. Now let f(x) = xm + fm−1xm−1 + · · · + f1x + f0 be the minimal polynomial of a primitive element ω (so that f is a primitive polynomial of degree m over K), and let B = {1, ω, ω2, . . . , ωm−1} be the associated polynomial basis. Then Aω(B) is the companion matrix of f, that is, Aω(B) =          0 0 · · · 0 −f0 1 0 · · · 0 −f1 0 1 ... . . . . . . . . . . . . ... 0 −fm−2 0 0 · · · 1 −fm−1          . As this example shows, Singer cycles in GL(m, q) give rise to matrix representations for Fqm over Fq of the smallest possible degree. Essentially, all minimal degree representations may be obtained in this way. 13.2.20 Theorem Let R ⊂F(m,m) q be a matrix representation for F = Fqm over K = Fq. Then there exists a matrix A ∈R such that R = {Ak : k = 1, . . . , qm −1} ∪{0}. Moreover, A is similar to the companion matrix of a primitive polynomial of degree m over K, and det A is a primitive element of K. 13.2.21 Remark Theorem 13.2.20 characterizes the matrix representations for Fqm of smallest de-gree. There are considerably more general results due to Willett who classiﬁed all matrix ﬁelds of degree m over Fq in terms of primitive polynomials over the underlying prime ﬁeld Fp. The general result is rather technical, so we only give the special case q = p here. Proofs for all these results can also be found in Section 1.5 of . 13.2.22 Theorem Let p be a prime, and let R be any representation of Fpm as a matrix ﬁeld of degree n over Fp. Up to similarity, R has the form R = {diag 0, . . . , 0, Ak, . . . , Ak : k = 1, . . . , pm −1} ∪{0}, where A is the companion matrix of a primitive polynomial of degree m over Fp. 13.2.23 Remark It is of interest to study representations for Fqm consisting of special types of matrices. We present some results due to Seroussi and Lempel concerning symmetric 504 Handbook of Finite Fields matrix representations, that is, matrix representations consisting of symmetric matrices only. This turns out to be closely related to the duality theory for bases, see Section 5.1. With the exception of Theorem 13.2.27, proofs for the subsequent results can be found in Section 4.6 of . 13.2.24 Lemma Let B be a basis for F = Fqm over Fq, and let R(B) be the associated matrix representation. Then R(B) is symmetric if and only if there exists an element λ ∈F ∗such that the dual basis B∗of B satisﬁes B∗= λB, where λB consists of all elements λβ with β ∈B. 13.2.25 Theorem Every ﬁnite ﬁeld F = Fqm admits a basis B over Fq such that the associated matrix representation R(B) is symmetric. 13.2.26 Theorem Let B be a basis of F = Fqm over Fq with associated matrix representation R(B), and assume that q is even or that q and m are both odd. Then the dual basis B∗ of B satisﬁes B∗= λB for some λ ∈F ∗(so that R(B) is symmetric) if and only if λ is a square, say λ = µ2, and the basis µB is self-dual. 13.2.27 Theorem Let B be a basis of F = Fqm over Fq, and assume that the associated matrix representation R(B) is closed under transposition of matrices. Then R(B) is symmetric provided that either q is even or q and m are both odd. If q is odd and m is even, then R(B) is not symmetric if and only if B∗= (λB)qm/2 for some λ ∈F ∗. 13.2.4 Circulant and orthogonal matrices 13.2.28 Remark There are many results on the number of matrices of special types over Fq. In the remaining subsections, we present a selection of such results. We begin with those cases which are related to the enumeration of various types of bases as discussed in Chapter 5. Let us summarize these connections as follows: Type of basis Associated transformation matrices Reference normal bases circulant matrices 5.2.15 self-dual bases orthogonal matrices 5.1.22 self-dual normal bases orthogonal circulant matrices 5.2.28 13.2.29 Deﬁnition An m × m matrix C = (cij)i,j=1,...,m is circulant if its rows are generated by successive cyclic shifts of its ﬁrst row, that is ci+1,j+1 = cij for all i, j = 1, . . . , m, (13.2.2) where indices are computed modulo m. Thus C is speciﬁed by the entries cj = c1,j in its ﬁrst row: C =        c1 c2 · · · cm−1 cm cm c1 · · · cm−2 cm−1 cm−1 cm · · · cm−3 cm−2 . . . . . . ... . . . . . . c2 c3 · · · cm c1        . Such a matrix C is denoted as circ(c1, . . . , cm). Miscellaneous theoretical topics 505 13.2.30 Remark The following three results are well-known. They may be found, for example, in . 13.2.31 Lemma Let F be a ﬁeld. Mapping the matrix C = circ(c1, . . . , cm) to the coset of the polynomial c(x) = c0 +c1x+· · ·+cm−1xm−1 modulo xm −1 gives an isomorphism between the ring of all circulant m×m matrices over F and the ring R = F[x]/(xm−1). In particular, C is invertible if and only if the associated polynomial c(x) is a unit in R. 13.2.32 Corollary Let C(m, q) denote the multiplicative group of all invertible circulant m × m matrices over Fq. Then the order of C(m, q) equals Φq(xm −1), where Φq is the function introduced in Deﬁnition 2.1.111. 13.2.33 Theorem Let q be a power of the prime p, let m be a positive integer, and write m = pbn, where p does not divide n. Then \|C(m, q)\| = qm Y d\|n 1 −qod(q)φ(d)/od(q) , (13.2.3) where φ is the Euler function given in Deﬁnition 2.1.43 and where od(q) denotes the multi-plicative order of d modulo q. 13.2.34 Corollary Assume that q and m are co-prime. Then \|C(m, q)\| = Qr j=1 (qmj −1) , where m1, . . . , mr are the degrees of the irreducible factors of xm −1 over Fq. 13.2.35 Deﬁnition An m × m matrix A is orthogonal if it satisﬁes the condition AAT = I. We denote the multiplicative group of all orthogonal m × m matrices over Fq by O(m, q). 13.2.36 Remark The preceding terminology is somewhat ambiguous, as the term orthogonal group usually refers to the group of isometries of an orthogonal geometry, that is, of a vector space equipped with a quadratic form. In the case of ﬁnite ﬁelds, no ambiguity arises if both q and m are odd. However, if q is odd and m is even, then there are two distinct orthogonal groups (of diﬀerent orders); and if q is even, an orthogonal geometry is by deﬁnition a symplectic geometry reﬁned by an additional quadratic form, which forces m to be even. In particular, the standard text books do not contain the order of O(m, q) for even values of q. Nevertheless, the deﬁnition given in 13.2.35 makes sense for all cases. See Section 13.4 for the “classical” orthogonal groups. 13.2.37 Theorem Assume that either q is even or both q and m are odd. Then \|O(m, q)\| = γ m−1 Y i=1 qi −ϵi , (13.2.4) where ϵi = ( 1 if i is even, 0 if i is odd, and γ = ( 2 if q is even, 1 if q and n are odd. Now let q be odd and m even, say m = 2s. Then \|O(m, q)\| = qs(s−1)/2 (qs + ϵ) s−1 Y i=1 q2i −1 , (13.2.5) where ϵ = ( 1 if s is odd and q ≡3 (mod 4), −1 otherwise. 506 Handbook of Finite Fields 13.2.38 Lemma Let F be a ﬁeld, C = circ(c1, . . . , cm) a circulant matrix, and c(x) = c0 + c1x + · · · + cm−1xm−1 the associated polynomial c(x) in the ring R = F[x]/(xm −1), as in Lemma 13.2.31. Then C is an orthogonal matrix if and only if c(x)cT (x) = 1 in R, where cT (x) = c0+cm−1x+· · ·+c2xm−2+c1xm−1 is the polynomial corresponding to the transpose CT of C. 13.2.39 Remark We require some notation to state the number of orthogonal circulant m × m matrices over Fq; clearly all these matrices form a group which is denoted by OC(m, q). First assume that q and m are co-prime. (The general case is reduced recursively to this special case.) Then let x −1, possibly x + 1 (this factor arises only if q is odd and m is even) and f1, . . . , fr be the monic irreducible factors of xm −1. Some of the fi may be self-reciprocal (that is, fi = f ∗ i , see Deﬁnition 2.1.48), say f1, . . . , fs. Then the remaining fj split into pairs of the form {f, f ∧} with f ̸= f ∧, where f ∧= f ∗/f0 and where f0 is the constant term of f; say r = s + 2t, and (fs+j)∧= fs+t+j for j = 1, . . . , t. 13.2.40 Theorem [258, 471, 1633, 1990] Assume that q and m are co-prime, and write, using the notation introduced in Remark 13.2.39, deg(fi) = 2di for i = 1, . . . , s and deg(fs+j) = deg(fs+t+j) = ej for j = 1, . . . , t. Then the order of OC(m, q) is given by \|OC(m, q)\| = γ s Y i=1 qdi + 1 t Y i=1 (qej −1) , (13.2.6) where γ =      1 if q is even, 2 if q and m are odd, 4 if q is odd and m is even. Now let q be a power of the prime p. If p ̸= 2, then \|OC(pn, q)\| = qn(p−1)/2 \|OC(n, q)\| (13.2.7) for every positive integer n. Finally, \|OC(2n, q)\| =      q(n+1)/2 \|OC(n, q)\| if n is odd, 2qn/2 \|OC(n, q)\| if n ≡2 (mod 4), qn/2 \|OC(n, q)\| if n ≡0 (mod 4), (13.2.8) for every positive integer n. 13.2.41 Remark With the exception of the special case of Theorem 13.2.37 where q is odd and m is even, proofs for all the preceding results can also be found in . 13.2.5 Symmetric and skew-symmetric matrices 13.2.42 Remark In order to prove the results on (circulant) orthogonal matrices presented in the previous subsection, one needs a connection with and enumeration results for two other interesting classes of matrices, which are the topic of the present subsection. Again, proofs for these results are in . Miscellaneous theoretical topics 507 13.2.43 Deﬁnition A matrix A over Fq is symmetric if it satisﬁes the condition A = AT , and skew-symmetric if it satisﬁes A = −AT and has diagonal entries 0 only. (This extra condition has to be added in view of the special case where q is a power of 2.) 13.2.44 Deﬁnition Let A be a symmetric invertible matrix over Fq. Then any (invertible) m × m matrix M satisfying A = MM T is a factor of A. 13.2.45 Lemma The invertible symmetric m × m matrices over Fq admitting a factor are in a 1-to-1 correspondence with the cosets of O(m, q) in GL(m, q). Hence \|O(m, q)\| = \|GL(m, q)\| sf(m, q) , where sf(m, q) denotes the number of invertible symmetric matrices over Fq admitting a factor. 13.2.46 Lemma Let A be a symmetric invertible matrix over Fq. If q is odd, A admits a factor if and only if det A is a square; and if q is even, A admits a factor if and only if A has at least one non-zero diagonal entry. 13.2.47 Theorem [544, 1989] The number N(m, r) of symmetric m × m matrices of rank r over Fq equals N(m, r) = s Y i=1 q2i q2i −1 · r−1 Y i=0 qm−i −1 , where r ≤m and s = ⌊r/2⌋. In particular, the number of invertible symmetric m × m matrices over Fq is given by N(m, m) = m Y i=1 qi −δi , where δi = ( 0 if i is even, 1 otherwise. 13.2.48 Theorem [545, 1989] The number N0(m, r) of skew-symmetric m × m matrices of rank r over Fq is given by N0(m, 2s) = s Y i=1 q2i−2 q2i −1 · 2s−1 Y i=0 qm−i −1 and N0(m, 2s + 1) = 0, where 2s ≤m. In particular, there are no invertible skew-symmetric m × m matrices over Fq if m is odd; if m is even, the number of invertible skew-symmetric m × m matrices over Fq equals N0(m, m) = m−1 Y i=1 qi −δi , where δi = ( 0 if i is even, 1 otherwise. 13.2.6 Hankel and Toeplitz matrices 13.2.49 Remark (Inﬁnite) Toeplitz and Hankel matrices with complex entries play a prominent role in classical linear algebra and have many important applications; see, for instance, for an introductory treatment or for a monograph on large ﬁnite Toeplitz matrices. In this subsection, we discuss such matrices over Fq. Both classes of matrices are closely related, as a Hankel matrix may be viewed as an “upside-down” Toeplitz matrix, so it suﬃces to concentrate on one of these two classes. 508 Handbook of Finite Fields 13.2.50 Deﬁnition An m × m matrix A = (aij)i,j=1,...,m is a Toeplitz matrix if aij = akl whenever i −j = k −l. (13.2.9) Thus A is speciﬁed by the entries in its ﬁrst row and column: A =        am am−1 · · · a2 a1 am+1 am · · · a3 a2 am+2 am+1 · · · a4 a3 . . . . . . ... . . . . . . a2m−1 a2m−2 · · · am+1 am        . 13.2.51 Remark Note that indices are not computed modulo m in the deﬁning condition (13.2.9) for a Toeplitz matrix. If we would do so, we get further restrictions and arrive at an equivalent formulation for the deﬁning condition (13.2.2) for circulant matrices. Thus circulant matrices constitute a special class of Toeplitz matrices. 13.2.52 Deﬁnition An m × m matrix A = (aij)i,j=1,...,m is a Hankel matrix if aij = akl whenever i + j = k + l. Thus A is speciﬁed by the entries in its ﬁrst row and column: A =        a1 a2 · · · am−1 am a2 a3 · · · am am+1 a3 a4 · · · am+1 am+2 . . . . . . ... . . . . . . am am+1 · · · a2m−2 a2m−1        . (13.2.10) 13.2.53 Remark “Rotating” a Toeplitz matrix by 90 degrees (counter-clockwise) transforms it into a Hankel matrix. More formally, let P be the m×m matrix with entries 1 on the antidiagonal and 0 elsewhere, that is, P = δi,m−j+1 i,j=1,...,m, where δik = 0 if i ̸= k and δik = 1 if i = k. Then the bijection deﬁned by A 7→AP transforms the set of all m×m Toeplitz matrices into the set of all m×m Hankel matrices. Note that this bijection preserves the rank of matrices, so that the following result applies for Hankel matrices as well. 13.2.54 Theorem The number t(m, r, q) of m × m Toeplitz matrices over Fq with rank r is given by t(m, r, q) =      q2m−2(q −1) if r = m, q2r−2(q2 −1) if 1 ≤r ≤m −1, 1 if r = 0. 13.2.55 Corollary For any positive integer r, the number of m × m Toeplitz (or Hankel) matrices over Fq with rank r is constant for every m ≥r + 1. 13.2.56 Remark We note that Toeplitz and Hankel matrices over ﬁnite ﬁelds have some interesting applications. For instance, Toeplitz matrices are used as pre-conditioners in the process of Miscellaneous theoretical topics 509 solving linear systems having unstructured coeﬃcient matrices . In , an explicit surjective map σ from the set of all ordered pairs of coprime monic polynomials of degree m over Fq to the set of all invertible m × m Hankel matrices over Fq is constructed; this map has the additional property that, for any such matrix A, the pre-image σ−1(A) is in a one-to-one correspondence with Fq. Therefore Theorem 13.2.54 gives a proof for the following result. 13.2.57 Theorem The number of ordered pairs of coprime monic polynomials of degree m over Fq equals q2m−1(q −1). 13.2.58 Corollary The probability that two randomly chosen monic polynomials of the same posi-tive degree with coeﬃcients in Fq are coprime is 1 −(1/q). 13.2.59 Remark Further results on the greatest common divisor of polynomials are given in Section 11.2. 13.2.7 Determinants 13.2.60 Deﬁnition Let {α1, . . . , αm} be a set of m elements of Fqm. Then the determinant α1 · · · αm αq 1 · · · αq m . . . ... . . . αqm−1 1 · · · αqm−1 m is the Moore determinant of {α1, . . . , αm}. 13.2.61 Remark By Corollary 2.1.95, the set {α1, . . . , αm} is a basis for Fqm over Fq if and only if its Moore determinant is nonzero. The Moore determinant may be viewed as a ﬁnite ﬁeld analogue of the well-known Vandermonde determinant. It is used extensively in the theory of Drinfeld modules, see Section 13.3. Moore proved the following formula for his determinant, which immediately implies the validity of Corollary 2.1.95. 13.2.62 Theorem Let {α1, . . . , αm} be a set of m elements of Fqm. Then its Moore determinant det M(α1, . . . , αm) is given by det M(α1, . . . , αm) = α1 m−1 Y i=1 Y c1,...,ci∈Fq  αi+1 − i X j=1 cjαj  . 13.2.63 Remark Equation (13.2.1) immediately gives an exact formula for the probabilty that the determinant of a random m × m matrix A over Fq equals 0. From this one easily sees that the probability in question is in the order of magnitude prob(det A = 0) = 1 q + O 1 q2 . It is remarkable that one actually has the following much more general result for both determinants and permanents: prob(det A = α) = prob(per A = α) = 1 q + O 1 q2 for every α ∈Fq. This result is obtained in as a byproduct of the authors’ study of the Polya permanent problem for matrices over ﬁnite ﬁelds. 510 Handbook of Finite Fields 13.2.64 Remark Several further results concerning the enumeration of various types of matrices over ﬁnite ﬁelds can be found in the nice survey article . We have considered the number of invertible m × m matrices over Fq with the smallest (viz. 2) and the largest (viz. qm −1) possible orders in Subsection 13.2.2. Matrices of a speciﬁed order k are, of course, closely related to the solutions of the matrix equation f(X) = 0, where f(x) = xk −1. There are many papers concerning matrix equations over ﬁnite ﬁelds; see the notes to Section 6.2 in for a collection of references. See Also §5.1 For bases of ﬁnite ﬁelds. §5.2 For a discussion of normal bases. §13.3 For a discussion of classical groups over ﬁnite ﬁelds. References Cited: [69, 226, 258, 367, 412, 471, 544, 545, 910, 1064, 1070, 1207, 1269, 1354, 1517, 1631, 1633, 1665, 1989, 1990, 2093, 2140, 2166, 2293, 2584, 2980] 13.3 Classical groups over ﬁnite ﬁelds Zhe-Xian Wan, Chinese Academy of Sciences 13.3.1 Linear groups over ﬁnite ﬁelds 13.3.1 Deﬁnition Let Fq be a ﬁnite ﬁeld with q elements and n an integer > 1. The set of n × n nonsingular matrices over Fq forms a group with respect to matrix multiplication, called the general linear group of degree n over Fq and denoted by GLn(q). The set of n × n matrices over Fq of determinant 1 forms a subgroup of GLn(q), called the special linear group of degree n over Fq and denoted by SLn(q). The group GLn(q) can also be deﬁned as the group consisting of nonsingular linear transformations of an n-dimensional vector space over Fq. 13.3.2 Remark In literatures of group theory the groups GLn(q) and SLn(q) are sometimes written as GL(n, q) instead of GLn(q) and SL(n, q) instead of SLn(q). 13.3.3 Theorem The order of GLn(q) is \|GLn(q)\| = (qn −1)(qn −q) · · · (qn −qn−1); SLn(q) is a normal subgroup of GLn(q) and is of order \|SLn(q)\| = (qn −1)(qn −q) · · · (qn −qn−2)qn−1. 13.3.4 Theorem The center Zn of GLn(q) consists of those matrices λIn, where λ ∈F∗ q and In is the n×n identity matrix. The center of SLn(q) is SLn(q)∩Zn, which consists of those matrices λIn, where λ ∈F∗ q and λn = 1. Miscellaneous theoretical topics 511 13.3.5 Deﬁnition The factor group GLn(q)/Zn is the projective general linear group of degree n over Fq and denoted by PGLn(q). The factor group SLn(q)/SLn(q)∩Zn is the projective special linear group of degree n over Fq and is denoted by PSLn(q). 13.3.6 Theorem The order of PGLn(q) is \|PGLn(q)\| = (qn −1)(qn −q) · · · (qn −qn−2)qn−1; and the order of PSLn(q) is \|PSLn(q)\| = (qn −1)(qn −q) · · · (qn −qn−2)qn−1/d, where d = gcd (n, q −1). 13.3.7 Theorem The group SLn(q) is generated by the elementary matrices Tij(b) = In + bEij, 1 ≤i, j ≤n and i ̸= j, where In is the n × n identity matrix, b ∈F∗ q and Eij is the n × n matrix with 1 in the (i, j) position and 0 elsewhere. 13.3.8 Lemma Every element A of GLn(q) can be expressed in the form A = BD(µ) where B ∈SLn(q) and D(µ) is a diagonal matrix with 1, 1, . . . , 1, µ along its main diagonal. 13.3.9 Theorem The group SLn(q) is the commutator subgroup of GLn(q) unless n = 2 and q = 2. Moreover, SLn(q) is its own commutator subgroup unless n = 2 and q = 2 or 3. When n = 2 and q = 2, SL2(2) = GL2(2) ≃S3, where S3 denotes the symmetric group on three letters. When n = 2 and q = 3, \|SL2(3)\| = 24 and the commutator subgroup of SL2(3) is of order 8. 13.3.10 Theorem (Dickson) The group PSLn(q) is simple except in the cases n = 2, q = 2 or 3. 13.3.11 Remark Except for the cases n = 2, q = 2 or 3, GLn(q) has the normal series GLn(q) ⊃SLn(q) ⊃Zn ∩SLn(q) ⊃{I}, where GLn(q)/SLn(q) and Zn ∩SLn(q) are cyclic and SLn(q)/(Zn ∩SLn(q)) ≃PSLn(q) is simple. When n = 2 and q = 2, the group GL2(2) = SL2(2) ≃S3 has composition factors 2 and 3 and is not simple. When n = 2 and q = 3, the group GL2(3) has generators A = −1 0 1 1 , B = 0 1 −1 1 , C = 0 −1 1 0 , D = 1 1 1 −1 , E = −1 0 0 −1 and deﬁning relations E2 = I2, D2 = E, DE = ED, C2 = E, CE = EC, CD = EDC, B3 = E, BE = EB, BD = EDBC, BC = ECDB, A2 = I2, AE = EA, AD = CA, AC = DA, AB = ECB2AC. 512 Handbook of Finite Fields Then GL2(3) has the composition series GL2(3) = {E, D, C, B, A} ⊃SL2(3) = {E, D, C, B} ⊃{E, D, C} ⊃{E, D} ⊃Z2 = {E} ⊃{I2} and the orders of these groups are 48, 24, 8, 4, 2, respectively. Thus PSL2(3) is not simple. 13.3.12 Theorem The only isomorphisms between the groups PSLn(q) are 1. PSL2(4) ≃PSL2(5), and 2. PSL2(7) ≃PSL3(2). 13.3.13 Theorem The only isomorphisms between the groups PSLn(q) and Am are 1. PSL2(3) ≃A4, 2. PSL2(4) ≃PSL2(5) ≃A5, 3. PSL2(9) ≃A6, and 4. PSL4(2) ≃A8. 13.3.14 Deﬁnition Let Fn q be the n-dimensional row vector space over Fq consisting of all n-dimensional row vectors (x1, x2, . . . , xn), xi ∈Fq, 1 ≤i ≤n, over Fq. Let P be an m-dimensional subspace of Fn q and let v1, v2, . . . , vm be a basis of P. Then the m × n matrix      v1 v2 . . . vm      is a matrix representation of P and is denoted by P also. Two matrix representations of P diﬀer from an m × m nonsingular matrix multiplied on the left. Every element A ∈GLn(q) acts on Fn q in the following way: (x1, x2, . . . , xn) 7→(x1, x2, . . . , xn)A, which induces a transformation on the set of subspaces of Fn q : P 7→PA. The subset of subspaces of the same dimension forms an orbit. The cardinality of the orbit of m-dimensional subspaces (0 ≤m ≤n) is denoted by N(m, n). 13.3.15 Theorem We have N(m, n) = n m q, where n m q is the Gaussian coeﬃcient h n m i q = (qn −1)(qn−1 −1) · · · (qn−m+1 −1) (qm −1)(qm−1 −1) · · · (q −1) . 13.3.16 Remark For more details on Gaussian coeﬃcients, see Section 13.2. 13.3.2 Symplectic groups over ﬁnite ﬁelds 13.3.17 Deﬁnition Let n be an integer > 0. An n × n matrix K over Fq is alternate, if tK = −K and the diagonal elements are 0. When q is even, alternate matrices are simply skew-symmetric matrices. Let n = 2ν be even and K be a 2ν × 2ν nonsingular alternate Miscellaneous theoretical topics 513 matrix over Fq. The set of 2ν × 2ν matrices T satisfying TK tT = K forms a group with respect to matrix multiplication, the symplectic group of degree 2ν with respect to K over Fq and denoted by Sp2ν(q, K). Let K = (kij) be an n × n alternate matrix over Fq. The bilinear form K(x, y) deﬁned by K(x, y) = P kijxiyj is an alternate form, which is nonsingular if det K ̸= 0. Now Let n = 2ν be even and K(x, y) be a nonsingular alternate form on a 2ν-dimensional vector space V over Fq. Then Sp2ν(q, K) can also be deﬁned as the group of linear transformations T of V satisfying K(xT, yT) = K(x, y) for all x, y ∈V . 13.3.18 Theorem All elements of Sp2ν(q, K) are nonsingular matrices with determinant 1 and, hence, Sp2ν(q, K) is a subgroup of SL2ν(q). Moreover, Sp2(q, K) = SL2(q). 13.3.19 Deﬁnition Two n × n matrices A and B over Fq are cogredient, if there is an n × n nonsingular matrix P over Fq such that A = PB tP. 13.3.20 Theorem Let K1 and K2 be two cogredient 2ν × 2ν nonsingular alternate matrices, then Sp2ν(q, K1) ≃Sp2ν(q, K2). 13.3.21 Remark By the previous theorem it is suﬃcient to consider the symplectic group Sp2ν(q, K0) where K0 = 0 Iν −Iν 0 . The group Sp2ν(q, K0) is simply the symplectic group of degree 2ν over Fq and denoted by Sp2ν(q). 13.3.22 Theorem The order of Sp2ν(q) is \|Sp2ν(q)\| = qν2 ν Y i=1 (q2i −1). 13.3.23 Deﬁnition Let T ∈Sp2ν(q). If I2ν −T is of rank 1, then T is a symplectic transvection. 13.3.24 Lemma Every symplectic transvection can be expressed in the form T Iν 0 diag(λ, 0, . . . , 0) Iν T −1 where T ∈Sp2ν(q), and diag(λ, 0, . . . , 0) is a diagonal matrix with λ, 0, . . . , 0 along the main diagonal and λ ∈F∗ q. 13.3.25 Theorem The group Sp2ν(q) is generated by symplectic transvections. 13.3.26 Theorem The group Sp2ν(q) is its own commutator subgroup in all cases except ν = 1, q = 2 or 3 and ν = 2, q = 2. 13.3.27 Remark The group Sp2(2) = SL2(2) ≃S3 and its commutator subgroup is A3. The group Sp2(3) = SL2(3) is of order 24 and is solvable. 13.3.28 Theorem The group Sp4(2) ∼ = S6. 13.3.29 Theorem For ν ≥2, the symmetric group S2ν+2 is a subgroup of Sp2ν(2). 13.3.30 Theorem The center of Sp2ν(q) consists of I2ν and −I2ν. 514 Handbook of Finite Fields 13.3.31 Deﬁnition The factor group Sp2ν(q)/{±I2ν} is the projective symplectic group of degree 2ν over Fq and denoted by PSp2ν(Fq). 13.3.32 Theorem (Dickson) The group PSp2ν(q) is a simple group except for the cases PSp2(2), PSp2(3), and PSp4(2). 13.3.33 Deﬁnition Let P be an m-dimensional subspace of F2ν q . Then PK t 0P is an m×m alternate matrix. Suppose PK t 0P is cogredient to   0 Is 0 −Is 0 0 0 0 0(2ν−2s)  , then P is a subspace of type (m, s). Clearly, 0 ≤2s ≤m ≤2ν. 13.3.34 Theorem The action of GL2ν(q) on the subspaces of F2ν q induces an action of Sp2ν(q) on the subspaces of F2ν q . Two subspaces P and Q are in the same orbit of Sp2ν(q) if and only if they are of the same type. 13.3.35 Theorem Denote the cardinality of the orbit of subspaces of type (m, s) by N(m, s; 2ν). Then N(m, s; 2ν) = q2s(ν+s−m) Qν i=ν+s−m+1(q2ν −1) Qs i=1(q2i −1) Qm−2s i=1 (qi −1) . 13.3.3 Unitary groups over ﬁnite ﬁelds 13.3.36 Deﬁnition Let Fq2 be a ﬁnite ﬁeld with q2 elements, where q is a prime power. The Frobenius automorphism (refer to Remark 2.1.77) a 7→aq of Fq2 is denoted by −, i.e., a = aq for all a ∈Fq2, and is the involution of Fq2. 13.3.37 Deﬁnition Let n be an integer > 1, and H = (hij) be an n × n matrix over Fq2. The matrix (hij) is denoted by H. If tH = H, H is a Hermitian matrix. Let H be an n × n nonsingular Hermitian matrix over Fq2. The set of n × n matrices T satisfying TH tT = H forms a group with respect to matrix multiplication, the unitary group of degree n with respect to H over Fq2 and denoted by Un(q2, H). The subgroup of Un(q2, H) consisting of those T ∈Un(q2, H) with determinant 1 is the special unitary group and denoted by SUn(q2, H). Let H = (hij) be an n × n matrix over Fq2 and H(x, y) = P hijxiyj be the corresponding Hermitian form, which is nonsingular if detH ̸= 0. Let H(x, y) be a nonsingular Hermitian form on an n-dimensional vector space V over Fq2. Then Un(q2, H) can also be deﬁned as the group of linear transformations T of V satisfying H(xT, yT) = H(x, y) for x, y ∈V . 13.3.38 Theorem All elements of Un(q2, H) are nonsingular matrices and, hence, Un(q2, H) is a subgroup of GLn(q2) and SUn(q2, H) is a subgroup of SLn(q2). Miscellaneous theoretical topics 515 13.3.39 Deﬁnition Two n × n Hermitian matrices H1 and H2 over Fq2 are cogredient if there is an n × n nonsingular matrix P over Fq2 such that H1 = PH t 2 P. 13.3.40 Theorem Let H1 and H2 be two cogredient nonsingular Hermitian matrices, then Un(q2, H1) ≃Un(q2, H2), and also SUn(q2, H1) ≃SUn(q2, H2). 13.3.41 Remark When n is even, write n as n = 2ν; and when n is odd, write n as n = 2ν + 1. By the above theorem, it is suﬃcient to consider the unitary groups Un(q2, H0), SUn(q2, H0) and Un(q2, H1), SUn(q2, H1), where H0 = 0 Iν Iν 0 and H1 =   0 Iν 0 Iν 0 0 0 0 1  . Here Un(q2, H0) and Un(q2, H1) are the unitary groups of degree n over Fq2, and are denoted by U2ν(q2) and U2ν+1(q2), respectively. Similarly, we have SU2ν(q2) and SU2ν+1(q2). We use the symbols Un(q2) and SUn(q2) to cover these two cases. In the literatures of group theory these groups are sometimes written as Un(q) instead of Un(q2) and SUn(q) instead of SUn(q2). 13.3.42 Theorem The orders of Un(q2) and SUn(q2) are, respectively, \|Un(q2)\| = q 1 2 n(n−1) n Y i=1 (qi −(−1)i), and \|SUn(q2)\| = q 1 2 n(n−1) n Y i=2 (qi −(−1)i). 13.3.43 Theorem We have SU2(q2) ≃SL2(q). 13.3.44 Lemma The center Wn of Un(q2) is Wn = Un(q2) ∩Zn = {aIn : aa = 1} and the center of SUn(q2) is SUn(q2) ∩Zn = {aIn : aa = 1 and an = 1}. (Recall that Zn is the subgroup of n×n nonsingular scalar matrices over Fq2, i.e., the center of GLn(q2).) 13.3.45 Deﬁnition The factor group Un(q2)/Wn is the projective unitary group of degree n over Fq2 and denoted by PUn(q2). Similarly, the factor group SUn(q2)/(SUn(q2)∩Zn) is the projective special unitary group of degree n over Fq2 and denoted by PSUn(q2). 13.3.46 Theorem The order of PUn(q2) and PSUn(q2) are, respectively, \|PUn(q2)\| = q 1 2 n(n−1) n Y i=2 (qi −(−1)i) and \|PSUn(q2)\| = (gcd(n, q + 1))−1q 1 2 n(n−1) n Y i=2 (qi −(−1)i). 516 Handbook of Finite Fields 13.3.47 Deﬁnition Let T ∈Un(q2). If In −T is of rank 1, then T is a unitary transvection. 13.3.48 Lemma Every unitary transvection can be expressed in the form T Iν 0 diag(λ, 0, . . . , 0) Iν T −1 or T   Iν 0 0 diag(λ, 0, . . . , 0) Iν 0 0 0 1  T −1 corresponding to n = 2ν or n = 2ν +1, respectively, where T ∈Un(q2) and λ ∈F∗ q satisfying λ + λ = 0. 13.3.49 Theorem For n ≥2, the group SUn(q2) is generated by unitary transvections except SU3(22). 13.3.50 Theorem (Dickson) For n ≥2, the group PSUn(q2) is simple except for PSU2(22), PSU2(32) and PSU3(22). 13.3.51 Remark [857, 858] For the exceptional cases in Theorem 13.3.50, the group PSU2(22) ≃PSL2(2) ≃SL2(2) ≃S3, the group PSU2(32) ≃PSL2(3) ≃A4, and the group PSU3(22) is of order 23 · 32 = 72 and is solvable. 13.3.52 Theorem We have PSU4(22) ≃PSp4(3). 13.3.53 Deﬁnition Let P be an m-dimensional subspace of Fn q2 and H = H0 or H1 according to n = 2ν or n = 2ν + 1, respectively. The matrix PH tP is an m × m Hermitian matrix. Suppose PH tP is of rank r, then P is a subspace of type (m, r). Clearly, 0 < r ≤m ≤n. 13.3.54 Theorem The action of GLn(q2) on the subspaces of Fn q2 induces an action of Un(q2) on the subspaces of Fn q2. Two subspaces P and Q are in the same orbit of Un(q2) if and only if they are of the same type. 13.3.55 Theorem Denote the cardinality of the orbit of subspaces of type (m, r) by N(m, r; n). Then N(m, r; n) = qr(n+r−2m) Qn i=n+r−2m+1(qi −(−1)i) Qr i=1(qi −(−1)i) Qm−r i=1 (q2i −1) . 13.3.4 Orthogonal groups over ﬁnite ﬁelds of characteristic not two 13.3.56 Deﬁnition Let n be an integer > 1, Fq be a ﬁnite ﬁeld with q elements where q is an odd prime power; and S be an n×n nonsingular symmetric matrix over Fq. The set of n×n matrices T satisfying TS tT = S forms a group with respect to matrix multiplication, the orthogonal group of degree n with respect to S over Fq and denoted by On(q, S). Let S = (sij) be an n × n symmetric matrix over Fq and Q(x) = P sijxixj be the corresponding quadratic form, which is nonsingular if S is nonsingular. Let Q(x) be a nonsingular quadratic form on an n-dimensional vector space V over Fq. Then On(q, S) can also be deﬁned as the group of linear transformations T of V such that Q(xT) = Q(x) for all x ∈V . Miscellaneous theoretical topics 517 13.3.57 Theorem All elements of On(q, S) are nonsingular matrices of determinant ±1 and, hence, On(q, S) is a subgroup of GLn(q). 13.3.58 Theorem Let S1 and S2 be two cogredient n × n nonsingular symmetric matrices, then On(q, S1) ≃On(q, S2). Moreover, for any n × n nonsingular symmetric matrix S over Fq and any λ ∈F∗ q, On(q, S) = On(q, λS). 13.3.59 Remark Choose a ﬁxed non-square element z of F∗ q. By the previous theorem it is suﬃcient to consider the four orthogonal groups with respect to the following four n × n nonsingular symmetric matrices S2ν = 0 Iν Iν 0 , S2ν+1,1 =   0 Iν 0 Iν 0 0 0 0 1  , S2ν+1,z =   0 Iν 0 Iν 0 0 0 0 z   S2ν+2 =     0 Iν 0 0 Iν 0 0 0 0 0 1 0 0 0 0 −z    , where n = 2ν, 2ν + 1, 2ν + 1 and 2ν + 2, respectively. In order to cover these four cases, we introduce the notation S2ν+δ,∆, where δ = 0, 1 or 2 and ∆denotes its deﬁnite part, i.e., ∆=            φ if δ = 0, 1 or z if δ = 1, 1 0 0 −z if δ = 2. The orthogonal group of degree 2ν + δ with respect to S2ν+δ,∆over Fq will be denoted by O2ν+δ(q, S2ν+δ,∆). Clearly, O2ν+1(q, S2ν+1,z) = O2ν+1(q, zS2ν+1,z). Since zS2ν+1,z and S2ν+1,1 are cogredient, the groups O2ν+1(q, zS2ν+1,z) and O2ν+1(q, S2ν+1,1) are isomorphic. It follows that O2ν+1(q, S2ν+1,z) and O2ν+1(q, S2ν+1,1) are isomor-phic. Therefore actually only three types of orthogonal groups need to be considered; they are O2ν(q, S2ν), O2ν+1(q, S2ν+1,1), and O2ν+2(q, S2ν+2,∆), which simply denoted by O2ν(q), O2ν+1(q) and O2ν+2(q), respectively. We use O2ν+δ(q) to cover these three cases. 13.3.60 Remark For odd n, there is only one type of orthogonal groups, and for even n = 2ν there are two types of orthogonal groups: O2ν(q) and O2(ν−1)+2(q). In literatures of group theory they are sometimes called the plus type and the minus type, and denoted by O+ n (q) and O− n (q), respectively. 13.3.61 Remark Throughout the remainder of this section we assume ν ≥1. 13.3.62 Theorem The order of O2ν+δ(q) is \|O2ν+δ(q)\| = qν(ν+δ−1) ν Y i=1 (qi −1) ν+δ−1 Y i=0 (qi + 1). 13.3.63 Deﬁnition Let T ∈O2ν+δ(q). If T 2 = I and T −I is of rank 1, T is a symmetry. 13.3.64 Theorem Every element of O2ν+δ(q) is a product of at most 2ν + δ symmetries. 518 Handbook of Finite Fields 13.3.65 Deﬁnition Elements of O2ν+δ(q) are orthogonal matrices and those of determinant 1 are proper orthogonal matrices. All proper orthogonal matrices form a subgroup of O2ν+δ(q), the proper orthogonal group also referred to as the special orthogonal group, and denoted by SO2ν+δ(q). The commutator subgroup of O2ν+δ(q) is denoted by Ω2ν+δ(q). 13.3.66 Lemma The center of O2ν+δ(q) is {I2ν+δ, −I2ν+δ}. 13.3.67 Deﬁnition The factor group O2ν+δ(q)/{±I2ν+δ} over Fq is the projective orthogonal group of degree 2ν + δ with respect to S2ν+δ over Fq and is denoted by PO2ν+δ(q). Similarly, the factor group SO2ν+δ(q)/(SO2ν+δ(q) ∩{±I2ν+δ}) is the projective proper orthogonal group of degree 2ν + δ with respect to S2ν+δ over Fq and is denoted by PSO2ν+δ(q). We also deﬁne PΩ2ν+δ(q) = Ω2ν+δ(q)/(Ω2ν+δ(q) ∩{±I2ν+δ}). 13.3.68 Remark We have the normal series of O2ν+δ(q) O2ν+δ(q) ⊃SO2ν+δ(q) ⊃Ω2ν+δ(q) ⊃Ω2ν+δ(q) ∩{±I2ν+δ} ⊃{I}. Clearly, O2ν+δ(q) : SO2ν+δ(q) = 2. 13.3.69 Theorem We have SO2ν+δ(q)/Ω2ν+δ(q) ≃F∗ q/F∗2 q . 13.3.70 Theorem (Dickson) The group PΩ2ν+δ(q) is a simple group except the following cases: 1. ν = 2, δ = 0, 2. ν = 1, δ = 1 and q = 3. 13.3.71 Remark For the exceptional cases in the above Theorem, we have PΩ2·2(q) ≃PSL2(q) × PSL2(q), PΩ2·1+1(3) ≃PSL2(3). 13.3.72 Theorem When ν ≥2, PΩ2ν+1(q) (= Ω2ν+1(q)) and PSp2ν(q) are non-isomorphic simple groups of the same order. 13.3.73 Remark The action of GL2ν+δ(q) on the subspaces of F2ν+δ q induces an action of O2ν+δ(q) on the subspaces. Let P be an m-dimensional subspace of F2ν+δ q . PS t 2ν+δ P is cogredient to one of the following normal forms M(m, 2s, s) =   0 Is 0 Is 0 0 0 0 0(m−2s)  , M(m, 2s + 1, s, 1) =     0 Is 0 0 Is 0 0 0 0 0 1 0 0 0 0 0(m−2s−1)    , M(m, 2s + 1, s, z) =     0 Is 0 0 Is 0 0 0 0 0 z 0 0 0 0 0(m−2s−1)    , Miscellaneous theoretical topics 519 and M(m, 2s + 2, s) =         0 Is 0 0 0 Is 0 0 0 0 0 0 1 0 0 0 0 0 −z 0 0 0 0 0 0(m−2s−2)         . Then P is a subspace of type (m, 2s, s), (m, 2s + 1, s, 1), (m, 2s + 1, s, z), and (m, 2s + 2, s, diag(1, −z)), respectively. 13.3.74 Theorem Two subspaces P and Q of F2ν+δ q belong to the same orbit of O2ν+δ(q) if and only if they are of the same type. 13.3.75 Remark The cardinality of any orbit of O2ν+δ(q) has already been determined . 13.3.5 Orthogonal groups over ﬁnite ﬁelds of characteristic two 13.3.76 Deﬁnition Let q be a power of 2 and n be an integer > 1. Let G be an n×n regular matrix over Fq. An n × n matrix T over Fq is orthogonal with respect to G, if TG tT + G is an alternate matrix. The set of n × n orthogonal matrices with respect to G over Fq forms a group with respect to matrix multiplication, the orthogonal group of degree n with respect to G over Fq and denoted by On(q, G). Let G(x) be a nonsingular quadratic form on an n-dimensional vector space V over Fq. Then On(q, G) can also be deﬁned as the group of linear transformations T of V such that G(vT) = G(v) for all v ∈V . 13.3.77 Theorem All elements of On(q, G) are nonsingular matrices of determinant 1 and, hence, On(q, G) is a subgroup of SLn(q). 13.3.78 Theorem Let G1 and G2 be two cogredient n × n regular matrices, then O(q, G1) ≃ O(q, G2). 13.3.79 Remark Choose a ﬁxed element α such that α cannot be expressed in the form x2 + x where x ∈Fq. Write n = 2ν +δ where δ = 0, 1 or 2. Assume ν > 0. By the previous theorem it is suﬃcient to consider the orthogonal groups O2ν+δ(q, G), where G is an n × n matrix of one of the following forms 0 Iν 0 0 ,   0 Iν 0 0 0 0 0 0 1  ,     0 Iν 0 0 0 0 0 0 0 0 α 1 0 0 0 α    , corresponding to the cases n = 2ν, n = 2ν + 1 and n = 2ν + 2, respectively. The orthogonal group with respect to G over Fq will be denoted by O2ν+δ(q). 13.3.80 Theorem We have O2ν+1(q) ≃Sp2ν(q). 13.3.81 Remark In the following we consider only the groups O2ν(q) and O2ν+2(q). In the literature on group theory, O2ν(q) and O2ν+2(q) are sometimes denoted by O+ 2ν(q) and O− 2ν+2(q), and called the plus type and the minus type, respectively. 13.3.82 Theorem The order of O2ν+δ(q), where δ = 0 or 2, is \|O2ν+δ(q)\| = qν(ν+δ−1) ν Y i=1 (qi −1) ν+δ−1 Y i=0 (qi + 1). 520 Handbook of Finite Fields 13.3.83 Deﬁnition Let T ∈O2ν+δ(q). If I2ν+δ −T is of rank 1, T is an orthogonal transvection. 13.3.84 Theorem Every element of O2ν+δ(q) is a product of at most 2ν+δ orthogonal transvec-tions except for the case n = 4, ν = 2 and q = 2; if an element of O2ν+δ(q) is expressed as a product of an even number of orthogonal transvections, so is every such expression. 13.3.85 Deﬁnition Except for the case n = 4, ν = 2 and q = 2, an orthogonal matrix T ∈O2ν+δ(q) which is a product of an even number of orthogonal transvections is a rotation. The set of rotations forms a subgroup of O2ν+δ(q), the group of rotations and denoted by SO2ν+δ(q). The commutator subgroup of O2ν+δ(q) is denoted by Ω2ν+δ(q). 13.3.86 Lemma For n ≥4 the center of Ω2ν+δ(q) consists of the identity element only. 13.3.87 Remark Except for the case n = 4, ν = 2, and q = 2, we have the normal series of O2ν+δ(q) O2ν+δ(q) ⊃SO2ν+δ(q) ⊃Ω2ν+δ(q) ⊃{I2ν+δ}. 13.3.88 Theorem [857, 858] Except for the case n = 4, ν = 2 and q = 2, \|O2ν+δ(q)/SO2ν+δ(q)\| = \|SO2ν+δ(q)/Ω2ν+δ(q)\| = 2. 13.3.89 Theorem (Dickson) The group Ω2ν+δ(q) is a simple group except for the case n = 4, ν = 2. 13.3.90 Theorem If q ̸= 2 and n = 4, ν = 2, then Ω2·2(q) ≃SL2(q) × SL2(q). If q = 2 and n = 4, ν = 2, then SO2·2(2) = SL2(2) × SL2(2) and Ω2·2(2) is a direct product of two cyclic groups of order 3. 13.3.91 Remark The action of GL2ν+δ(q) on the subspaces of F2ν+δ q induces an action of O2ν+δ(q) on the subspaces of F2ν+δ q . The cardinality of any orbit of O2ν+δ(q) has already been de-termined . See Also §7.2 For a discussion of quadratic forms. §13.2 For a discussion of orthogonal, symmetric, and skew-symmetric matrices. References Cited: [135, 857, 858, 1589, 2786, 2920] 13.4 Computational linear algebra over ﬁnite ﬁelds Jean-Guillaume Dumas, Universit´ e de Grenoble Cl´ ement Pernet, Universit´ e de Grenoble Miscellaneous theoretical topics 521 We present algorithms for eﬃcient computation of linear algebra problems over ﬁnite ﬁelds. Implementations∗of the proposed algorithms are available through the Magma, Maple (within the LinearAlgebra[Modular] subpackage) and Sage systems; some parts can also be found within the C/C++ libraries NTL, FLINT, IML, M4RI, and the special purpose LinBox template library for exact, high-performance linear algebra computation with dense, sparse, and structured matrices over the integers and over ﬁnite ﬁelds . 13.4.1 Dense matrix multiplication 13.4.1 Deﬁnition For A ∈Fm×k q and B ∈Fk×n q with elements Ai,j and Bi,j, the matrix C = A×B has Ci,j = Pk l=1 Ai,lBl,j. We denote by MM(m, k, n) a time complexity bound on the number of ﬁeld operations necessary to compute C. 13.4.2 Remark Classical triple loop implementation of matrix multiplication makes MM(m, k, n) ≤ 2mkn. The best published estimates to date gives MM(n, n, n) ≤O (nω) with ω ≈2.3755 , though improvements to 2.3737 and 2.3727 are now claimed [2728, 2985]. For very rectangular matrices one also have astonishing results like MM(n, n, nα) ≤O n2+ϵ for a constant α > 0.294 and any ϵ > 0 . Nowadays practical implementations mostly use Strassen-Winograd’s algorithm, see Subsection 13.4.1.4, with an intermediate complexity and ω ≈2.8074. 13.4.1.1 Tiny ﬁnite ﬁelds 13.4.3 Remark The practical eﬃciency of matrix multiplication depends highly on the repre-sentation of ﬁeld elements. We thus present three kinds of compact representations for elements of a ﬁnite ﬁeld with very small cardinality: bitpacking (for F2), bit-slicing (for say F3, F5, F7, F23, or F32), and Kronecker substitution. These representations are designed to allow eﬃcient linear algebra operations, including matrix multiplication. 13.4.4 Algorithm (Greasing) Over F2, the method of the four Russians , also called Greasing, can be used as follows: 1. A 64 bit machine word can be used to represent a row vector of dimension 64. 2. Matrix multiplication of a m × k matrix A by a k × n matrix B can be done by ﬁrst storing all 2k k-dimensional linear combinations of rows of B in a table. Then the i-th row of the product is copied from the row of the table indexed by the i-th row of A. 3. By ordering indices of the table according to a binary Gray Code, each row of the table can be deduced from the previous one, using only one row addition. This brings the bit operation count to build the table from k2kn to 2kn. 4. Choosing k = log2 n in the above method implies MM(n) = O n3/ log n over F2. ∗ shoup.net/ntl, http: //m4ri.sagemath.org, 522 Handbook of Finite Fields 13.4.5 Deﬁnition Bit slicing consists in representing an n-dimensional vector of k-bit sized coeﬃcients using k binary vectors of dimension n. In particular, one can use Boolean word instruction to perform arithmetic on 64 dimensional vectors. 1. Over F3, the binary representation 0 ≡[0, 0], 1 ≡[1, 0], −1 ≡ allows to add and subtract two elements in 6 Boolean operations: Add([x0, x1], [y0, y1]) : s ←x0 ⊕y1, t ←x1 ⊕y0 Return(s ∧t, (s ⊕x1) ∨(t ⊕y1)) Sub([x0, x1], [y0, y1]) : t ←x0 ⊕y0 Return(t ∨(x1 ⊕y1), (t ⊕y1) ∧(y0 ⊕x1)) 2. Over F5 (resp. F7), a redundant representation x = x0 +2x1 +4x2 ≡[x0, x1, x2] allows to add two elements in 20 (resp. 17) Boolean operations, negate in 3 (resp. 6) Boolean operations, and double in 0 (resp. 5) Boolean operations. F3 F5 F7 Addition 6 20 17 Negation 1 5 3 Double 5 0 Table 13.4.1 Boolean operation counts for basic arithmetic using bit slicing 13.4.6 Deﬁnition Bit packing consists in representing a vector of ﬁeld elements as an integer ﬁtting in a single machine word using a 2k-adic representation: (x0, . . . , xn−1) ∈Fn q ≡X = x0 + 2kx1 + · · · + (2k)n−1xn−1 ∈Z264. Elements of extension ﬁelds are viewed as polynomials and stored as the evaluation of this polynomial at the characteristic of the ﬁeld. The latter evaluation is known as Kronecker substitution. 13.4.7 Remark We ﬁrst need a way to simultaneously reduce coeﬃcients modulo the characteristic, see . 13.4.8 Algorithm (REDQ: Q-adic REDuction) Require: three integers p, q and ˜ r = Pd i=0 e µiqi ∈Z Ensure: ρ ∈Z, with ρ = Pd i=0 µiqi where µi = e µi mod p REDQ COMPRESSION 1. s = j ˜ r p k ; 2. for i = 0 to d do 3. ui = j ˜ r qi k −p j s qi k ; 4. end for REDQ CORRECTION {only when p ∤q, otherwise µi = ui is correct} 5. µd = ud; 6. for i = 0 to d −1 do 7. µi = ui −qui+1 mod p; 8. end for Miscellaneous theoretical topics 523 9. Return ρ = Pd i=0 µiqi; 13.4.9 Remark Once we can pack and simultaneously reduce coeﬃcients of ﬁnite ﬁeld in a single machine word, the obtained parallelism can be used for matrix multiplication. Depending on the respective sizes of the matrix in the multiplication one can pack only the left operand or only the right one or both . We give here only a generic algorithm for packed matrices, which use multiplication of a right packed matrix by a non-packed left matrix. 13.4.10 Algorithm (Right packed matrix multiplication) Require: a prime p and Ac ∈Fm×k p and Bc ∈Fk×n p , stored with several ﬁeld elements per machine word Ensure: Cc = Ac × Bc ∈Fm×n p 1. A = Uncompress(Ac); {extract the coeﬃcients} 2. Cc = A × Bc; {Using e.g., Algorithm 13.4.14} 3. Return REDQ(Cc); 13.4.11 Remark Then, over extensions, fast ﬂoating point operations can be used on the Kro-necker substitution of the elements. Indeed, it is very often desirable to use ﬂoating point arithmetic, exactly. For instance ﬂoating point routines can more easily use large hard-ware registers, they can more easily optimize the memory hierarchy usage [1336, 2974] and portable implementations are more widely available. We present next the dot product and the matrix multiplication is then straightforward [929, 930, 932]. 13.4.12 Algorithm (Compressed dot product over extension ﬁelds) Require: a ﬁeld Fpk with elements represented as exponents of a generator of the ﬁeld Require: two vectors v1 and v2 of elements of Fpk Require: a suﬃciently large integer q Ensure: R ∈Fpk, with R = vT 1 · v2 {Tabulated conversion: uses tables from exponent to ﬂoating point evaluation} 1. Set e v1 and e v2 to the ﬂoating point Kronecker substitution of the elements of v1 and v2. 2. Compute ˜ r = e v1 T · e v2; {The ﬂoating point computation} 3. r = REDQ COMPRESSION(˜ r, p, q); {Computing a radix decomposition} {Variant of REDQ CORRECTION: µi = e µi mod p for ˜ r = P2k−2 i=0 e µiqi} 4. Set L = representation(Pk−2 i=0 µiXi); 5. Set H = representation(Xk−1 × P2k−2 i=k−1 µiXi−k+1); 6. Return R = H + L ∈Fpk; {Reduction in the ﬁeld} 13.4.1.2 Word size prime ﬁelds 13.4.13 Remark Over word-size prime ﬁelds one can also use the reduction to ﬂoating point routines of algorithm 13.4.12. The main point is to be able to perform eﬃciently the matrix mul-tiplication of blocks of the initial matrices without modular reduction. Thus delaying the reduction as much as possible, depending on the algorithm and internal representations, in order to amortize its cost. We present next such a delaying with the classical matrix multiplication algorithm and a centered representation . 13.4.14 Algorithm (fgemm: Finite Field GEneric Matrix Multiplication) Require: an odd prime p of size smaller than the ﬂoating point mantissa β and Fp elements stored by values between 1−p 2 and p−1 2 Require: A ∈Fm×k p and B ∈Fk×n p Ensure: C = A × B ∈Fm×n p 524 Handbook of Finite Fields 1. if n(p −1)2 < 2β+1 then 2. Convert A and B to ﬂoating point matrices Af and Bf; 3. Use ﬂoating point routines to compute Cf = Af × Bf; 4. C = Cf mod p; 5. else 6. Cut A and B into smaller blocks; 7. Call the algorithm recursively for the block multiplications; 8. Perform the block additions modulo p; 9. end if 13.4.1.3 Large ﬁnite ﬁelds 13.4.15 Remark If the ﬁeld is too large for the strategy 13.4.14 over machine words, then two main approaches would have to be considered: 1. Use extended arithmetic, either arbitrary of ﬁxed precision, if the characteristic is large, and a polynomial representation for extension ﬁelds. The diﬃculty here is to preserve an optimized memory management and to have an almost linear time extended precision polynomial arithmetic. 2. Use a residue number system and an evaluation/interpolation scheme: one can use Algorithm 13.4.14 for each prime in the residue number system (RNS) and each evaluation point. For Fpk, the number of needed primes is roughly 2 log2β(p) and the number of evaluations points is 2k −1. 13.4.1.4 Large matrices: subcubic time complexity 13.4.16 Remark With matrices of large dimension, sub-cubic time complexity algorithms, such as Strassen-Winograd’s can be used to decrease the number of operations. Algo-rithm 13.4.17 describes how to compute one recursive level of the algorithm, using seven recursive calls and 15 block additions. 13.4.17 Algorithm (Strassen-Winograd) A = A11 A12 A21 A22 ; B = B11 B12 B21 B22 ; C = C11 C12 C21 C22 ; S1 ←A21 + A22; T1 ←B12 −B11; P1 ←A11 × B11; P2 ←A12 × B21; S2 ←S1 −A11; T2 ←B22 −T1; P3 ←S4 × B22; P4 ←A22 × T4; S3 ←A11 −A21; T3 ←B22 −B12; P5 ←S1 × T1; P6 ←S2 × T2; S4 ←A12 −S2; T4 ←T2 −B21; P7 ←S3 × T3; C11 ←P1 + P2; U2 ←P1 + P6; U3 ←U2 + P7; U4 ←U2 + P5; C12 ←U4 + P3; C21 ←U3 −P4; C22 ←U3 + P5; 13.4.18 Remark In practice, one uses a threshold in the matrix dimension to switch to a base case algorithm, that can be any of the previously described ones. Following Subsection 13.4.1.2, one can again delay the modular reductions, but the intermediate computations of Strassen-Winograd’s algorithm impose a tighter bound. 13.4.19 Theorem Let A ∈Zm×k, B ∈Zk×n C ∈Zm×n and β ∈Z with ai,j, bi,j, ci,j, β ∈ {0, . . . , p −1}. Then every intermediate value z involved in the computation of A × B + βC Miscellaneous theoretical topics 525 with l (l ≥1) recursive levels of Algorithm 13.4.17 satisfy \|z\| ≤ 1 + 3l 2 2 k 2l (p −1)2. Moreover, this bound is tight. 13.4.20 Remark For instance, on a single Xeon 2.8GHz core with gcc-4.6.3, Strassen-Winograd’s variant implemented with LinBox-1.2.1 and GotoBLAS2-1.13 can be 37% faster for the multiplication of 10 000 × 10 000 matrices over F219−1, in less than 1′49”. 13.4.2 Dense Gaussian elimination and echelon forms 13.4.21 Remark We present algorithms computing the determinant and inverse of square matrices; the rank, rank proﬁle, nullspace, and system solving for arbitrary shape and rank matrices. All these problems are solved a la Gaussian elimination, but recursively in order to eﬀectively incorporate matrix multiplication. The latter is denoted generically gemm and, depending on the underlying ﬁeld, can be implemented using any of the techniques of Subsections 13.4.1.1, 13.4.1.2, or 13.4.1.3. 13.4.22 Remark A special care is given to the asymptotic time complexities: the exponent is reduced to that of matrix multiplication using block recursive algorithms, and the constants are also carefully compared. Meanwhile, this approach is also eﬀective for implementations: grouping arithmetic operations into matrix-matrix products allow to better optimize cache accesses. 13.4.2.1 Building blocks 13.4.23 Remark Algorithms 13.4.24, 13.4.25, 13.4.26, and 13.4.27 show how to reduce the computa-tion of triangular matrix systems, triangular matrix multiplications, and triangular matrix inversions to matrix-matrix multiplication. Note that they do not require any temporary storage other than the input and output arguments. 13.4.24 Algorithm [trsm: Triangular System Solve with Matrix right hand side) Require: A ∈Fm×m q non-singular upper triangular, B ∈ Fm×n q Ensure: X ∈Fm×n q s.t. AX = B 1. if m=1 then return X = A−1 1,1 × B end if 2. X2 =trsm(A3, B2); 3. B1 = B1 −A2X2; {using gemm, e.g., via Algorithm 13.4.14} 4. X1 =trsm(A1, B1); 5. return X = X1 X2 ; Using the conformal block decomposition: A1 A2 A3 X1 X2 = B1 B1 13.4.25 Algorithm (trmm: Triangular Matrix Multiplication) Require: A ∈Fm×m q upper triangular, B ∈Fm×n q Ensure: C ∈Fm×n q s.t. AB = C 1. if m=1 then return C = A1,1 × B end if 2. C1 =trmm(A1, B1); 3. C1 = C1 + A2B2; {using gemm} 4. C2 =trmm(A3, B2); 5. return C = C1 C2 ; Using the conformal block decomposition: A1 A2 A3 B1 B2 = C1 C2 526 Handbook of Finite Fields 13.4.26 Algorithm (trtri: Triangular Matrix Inversion) Require: A ∈Fn×n q upper triangular and non-singular Ensure: C = A−1 1. if m=1 then return C = A−1 1,1 end if 2. C1 = A−1 1 ; {using trtri recursively} 3. C3 = A−1 3 ; {using trtri recursively} 4. C2 = A2C3; {using trmm } 5. C2 = −C1C2; {using trmm } 6. return C = C1 C2 C3 ; Using the conformal block decomposition: A1 A2 A3 , C1 C2 C3 13.4.27 Algorithm (trtrm: Upper-Lower Triangular Matrix Multiplication) Require: L ∈Fn×n q lower triangular Require: U ∈Fn×n q upper triangular Ensure: A = UL 1. if m=1 then return A = U1,1L1,1 end if 2. A1 = U1L1; {using trtrm recursively} 3. A1 = A1 + U2L2; {using gemm} 4. A2 = U2L3; {using trmm } 5. A3 = U3L2; {using trmm } 6. A4 = U3L3; {using trtrm recursively} 7. return A = A1 A2 A3 A4 ; Using the conformal block decomposition: L1 L2 L3 , U1 U2 U3 , A1 A2 A3 A4 13.4.2.2 PLE decomposition 13.4.28 Remark Dense Gaussian elimination over ﬁnite ﬁelds can be reduced to matrix multiplica-tion, using the usual techniques for the LU decomposition of numerical linear algebra . However, in applications over a ﬁnite ﬁeld, the input matrix often has non-generic rank proﬁle and special care needs to be taken about linear dependencies and rank deﬁciencies. The PLE decomposition is thus a generalization of the PLU decomposition for matrices with any rank proﬁle. 13.4.29 Deﬁnition A matrix is in row-echelon form if all its zero rows occupy the last row positions and the leading coeﬃcient of any non-zero row except the ﬁrst one is strictly to the right of the leading coeﬃcient of the previous row. Moreover, it is in reduced row-echelon form if all coeﬃcients above a leading coeﬃcient are zeros. 13.4.30 Deﬁnition For any matrix A ∈F m×n q of rank r, there is a PLE decomposition A = PLE where P is a permutation matrix, L is a m × r lower triangular matrix and E is a r × n matrix in row-echelon form, with unit leading coeﬃcients. 13.4.31 Remark Algorithm 13.4.32 shows how to compute such a decomposition by a block recursive algorithm, thus reducing the complexity to that of matrix multiplication. 13.4.32 Algorithm (PLE decomposition) Require: A ∈Fm×n q Ensure: (P, L, E) a PLE decomposition of A 1. if n = 1 then 2. if A = 0m×1 then return (Im, I0, A); end if Miscellaneous theoretical topics 527 3. Let j be the column index of the ﬁrst non-zero entry of A and P = T1,j the transposition between indices 1 and j; 4. return (P, PA, ); 5. else 6. (P1, L1, E1) = PLE(A1); {recursively} 7. A2 = P1A2; 8. A3 = L−1 1,1A3; {using trsm} 9. A4 = A4 −L1,2A3; {using gemm} 10. (P2, L2, E2) = PLE(A4); {recursively} Split A columnwise in halves: A = A1 A2 Split A2 = A3 A4 , L1 = L1,1 L1,2 where A3 and L1,1 have r1 rows. 11. return P1 Ir1 P2 , L1,1 P2L1,2 L2 , E1 A3 E2 ; 12. end if 13.4.2.3 Echelon forms 13.4.33 Remark The row-echelon and reduced row-echelon forms can be obtained from the PLE decomposition, using additional operations: trsm, trtri, and trtrm, as shown in Algo-rithms 13.4.34 and 13.4.35. 13.4.34 Algorithm (RowEchelon) Require: A ∈Fm×n q Ensure: (X, E) such that XA = E, X is non-singular and E is in row-echelon form 1. (P, L, E) = PLE(A); 2. X1 = L−1 1 ; {using trtri} 3. X2 = −L2X1; {using trmm} Split L = L1 L2 , L1 : r × r. 4. return X = X1 X2 Im−r P T , E ; 13.4.35 Algorithm (ReducedRowEchelon) Require: A ∈Fm×n q Ensure: (Y, R) such that Y A = R, Y is non-singular and R is in reduced row-echelon form 1. (X, E) = RowEchelon(A); 2. Let Q be the permutation matrix that brings the leading row coeﬃcients of E to the diagonal; 3. Set EQ = U1 U2 ; {where U1 is r × r upper triangular} 4. Y1 = U −1 1 ; {using trtri} 5. Y1 = Y1X1; {using trtrm} 6. R = Ir U −1 1 U2 QT ; {using trsm} 7. return Y = Y1 U2 In−r P T , R ; 13.4.36 Remark Figure 13.4.2 shows the various steps between the classical Gaussian elimination (LU decomposition), the computation of the echelon form and of the reduced echelon form, together with the various problems that each of them solve. Table 13.4.3 shows the leading constant Kω in the asymptotic time complexity of these algorithms, assuming that two n × n matrices can be multiplied in Cωnω + o(nω). 13.4.37 Remark If the rank r is very small compared to the dimensions m × n of the matrix, a system Ax = b can be solved in time bounded by O (m + n)r2 [2170, Theorem 1]. 528 Handbook of Finite Fields A L E 2/3 E L -1 L -1E-1 1/3 1/3 A -1 2/3 PLE decomp A=PLE Echelon form decomp (TA=E) Reduced Echelon form decomp TA=R PLE TRTRI TRTRI TRTRM - Det - Rank - RankProﬁle - Solve - Echelon - Nullspace basis - Reduced Echelon form - Inverse Figure 13.4.2 Reductions from PLE decomposition to reduced echelon form. Algorithm Constant Kω K3 Klog2 7 gemm Cω 2 6 trsm Cω 2ω−1−2 1 4 trtri Cω (2ω−1−2)(2ω−1−1) 1 3 ≈0.33 8 5 = 1.6 trtrm, PLE Cω 2ω−1−2 − Cω 2ω−2 2 3 ≈0.66 14 5 = 2.8 Echelon Cω 2ω−2−1 − 3Cω 2ω−2 1 22 5 ≈4.4 RedEchelon Cω(2ω−1+2) (2ω−1−2)(2ω−1−1) 2 44 5 = 8.8 Table 13.4.3 Complexity of elimination algorithms 13.4.3 Minimal and characteristic polynomial of a dense matrix 13.4.38 Deﬁnition 1. A Las-Vegas algorithm is a randomized algorithm which is always correct. Its expected running time is always ﬁnite. 2. A Monte-Carlo algorithm is a randomized algorithm which is correct with a certain probability. Its running time is deterministic. 13.4.39 Remark The computation of the minimal and characteristic polynomials is closely related to that of the Frobenius normal form. 13.4.40 Deﬁnition Any matrix A ∈Fn×n q is similar to a unique block diagonal matrix F = P −1AP = diag(Cf1, . . . , Cft) where the blocks Cfi are companion matrices of the poly-nomials fi, which satisfy fi+1\|fi. The fi are the invariant factors of A and F is the Frobenius normal form of A. 13.4.41 Remark Most algorithms computing the minimal and characteristic polynomial or the Frobenius normal form rely on Krylov basis computations. 13.4.42 Deﬁnition 1. The Krylov matrix of order d for a vector v with respect to a matrix A is the matrix KA,v,d = v Av . . . Ad−1v ∈Fn×d q . 2. The minimal polynomial P A,v min of A and v is the least degree monic polynomial P such that P(A)v = 0. I~ ~ ~ - -----Miscellaneous theoretical topics 529 13.4.43 Theorem 1. AKA,v,d = KA,v,dCP A,v min , where d = deg(P A,v min). 2. For linearly independent vectors (v1, . . . , vk), if K = KA,v1,d1 . . . KA,vk,dk is non-singular. Then AK = K       CP A,v1 min B1,2 . . . B1,k B2,1 CP A,v1 min . . . B2,k . . . . . . ... . . . Bk,1 Bk,2 CP A,vk min       , where the blocks Bi,j are zero except on the last column. 3. For linearly independent vectors (v1, . . . , vk), let (d1, . . . dk) be the lexicograph-ically largest sequence of degrees such that K = KA,v1,d1 . . . KA,vk,dk is non-singular. Then K−1AK =       CP A,v1 min B1,2 . . . B1,k CP A,v1 min . . . B2,k ... . . . CP A,vk min       = H. (13.4.1) 13.4.44 Remark 1. Some choice of vectors v1, . . . , vk lead to a matrix H block diagonal: this is the Frobenius normal form . 2. The matrix obtained from Equation (13.4.1) is a Hessenberg form. It suﬃces to compute the characteristic polynomial from its diagonal blocks. 13.4.45 Theorem The Frobenius normal form can be computed: 1. by a deterministic algorithm in 6n3 + O n2 log2 n ﬁeld operations, (only (2 + 2 3)n3 + O n2 for the characteristic polynomial ); 2. by a deterministic algorithm in O (nω log n log log n), together with a trans-formation matrix (only O (nω log n) for the characteristic polynomial ); 3. by a Las-Vegas algorithm in O (nω log n) ﬁeld operations for any ﬁeld, to-gether with a transformation matrix; 4. by a Las-Vegas algorithm in O (nω) for q > 2n2, without a transformation matrix. 13.4.46 Remark The minimal and characteristic polynomials, obtained as the ﬁrst invariant factor and the product of all invariant factors, can be computed with the same complexities. 13.4.47 Remark These algorithms are all based upon Krylov bases. The algorithm in Part 1 it-eratively computes the Krylov iterates one after the other. Their cubic time complexity with a small leading constant makes them comparable to Gaussian elimination. A fast ex-ponentiation scheme by Keller-Gehrig achieves a sub-cubic time complexity for the characteristic polynomial, oﬀby a logarithmic factor of n from the matrix multiplication. The choice for the appropriate vectors that will generate the Frobenius normal form can be done either probabilistically (Las Vegas) or deterministically with an extra log log n factor. The algorithm in Part 4 uses a diﬀerent iteration where the size of the Krylov increases according to an arithmetic progression rather than geometric (as all others) and the trans-formation matrix is not computed. This allows the algorithm to match the complexity of matrix multiplication. This reduction is practical and is implemented as in LinBox. 530 Handbook of Finite Fields 13.4.48 Remark These probabilistic algorithms depend on the ability to sample uniformly from a large set of coeﬃcients from the ﬁeld. Over small ﬁelds, it is always possible to embed the problem into an extension ﬁeld, in order to make the random sampling set suﬃciently large. In the worst case, this could add a O (log(n)) factor to the arithmetic cost and prevent most of the bit-packing techniques. Instead, the eﬀort of is to handle cleanly the small ﬁnite ﬁeld case. 13.4.4 Blackbox iterative methods 13.4.49 Remark We consider now the case where the input matrix is sparse, i.e., has many zero elements, or has a structure which enables fast matrix-vector products. Gaussian elimination would ﬁll-in the sparse matrix or modify the interesting structure. Therefore one can use iterative methods instead which only use matrix-vector iterations (blackbox methods ). There are two major diﬀerences with numerical iterative routines: over ﬁnite ﬁelds there exists isotropic vectors and there is no notion of convergence, hence the iteration must proceed until exactness of the result . Probabilistic early termination can nonetheless be applied when the degree of the minimal polynomial is smaller than the dimension of the matrix [935, 945, 1664]. More generally the probabilistic nature of the algorithms presented in this section is subtle: e.g., the computation of the minimal polynomial is Monte-Carlo, but that of system solving, using the minimal polynomial, is Las Vegas (by checking consistency of the produced solution with the system). Making some of the Monte-Carlo solutions Las Vegas is a key open problem in this area. 13.4.4.1 Minimal polynomial and the Wiedemann algorithm 13.4.50 Remark The ﬁrst iterative algorithm and its analysis are due to Wiedemann . The algorithm computes the minimal polynomial in the Monte-Carlo probabilistic fashion. 13.4.51 Deﬁnition For a linearly recurring sequence S = (Si), its minimal polynomial is denoted by ΠS. 1. The minimal polynomial of a matrix is denoted ΠA = Π(Ai). 2. For a matrix A and a vector b, we note ΠA,b = Π(Ai·b). 3. For another vector u, we note Πu,A,b = Π(uT ·Ai·b). 13.4.52 Algorithm (Wiedemann minimal polynomial) Require: A ∈Fn×n q , u, b ∈Fn q Ensure: Πu,A,b 1. Compute S = (uT Aib) for i ≤2n; 2. Use the Berlekamp-Massey algorithm to compute the minimal polynomial of the scalar sequence S; 13.4.53 Deﬁnition [See Deﬁnition 2.1.111] We extend Euler’s totient function by Φq,k(f) = Q(1 −q−kdi), where di are the degrees of the distinct monic irreducible factors of the polynomial f. 13.4.54 Theorem For u1, . . . , uj selected uniformly at random, the probability that lcm(Πuj,A,b) = ΠA,b is at least Φq,k(ΠA,b). 13.4.55 Theorem For b1, . . . , bk selected uniformly at random, the probability that lcm(ΠA,bi) = ΠA is at least Φq,k(ΠA). Miscellaneous theoretical topics 531 13.4.4.2 Rank, determinant, and characteristic polynomial 13.4.56 Remark It is possible to compute the rank, determinant, and characteristic polynomial of a matrix from its minimal polynomial. All these reductions require to precondition the matrix so that the minimal polynomial of the obtained matrix will reveal the information sought, while keeping a low cost for the matrix-vector product [607, 935, 947, 1666, 2824, 2874, 2875]. 13.4.57 Theorem Let S be a ﬁnite subset of a ﬁeld F that does not include 0. Let A ∈Fm×n having rank r. Let D1 ∈Sn×n and D2 ∈Sm×m be two random diagonal matrices then deg(minpoly(D1 × At × D2 × A × D1)) = r, with probability at least 1 −11n2−n 2\|S\| . 13.4.58 Theorem Let S be a ﬁnite subset of a ﬁeld F that does not include 0. Let U ∈Sn×n be a unit upper bi-diagonal matrix where the second diagonal elements u1, . . . , un−1 are randomly selected in S. For A ∈Fn×n, the term of degree 0 of the minimal polynomial of UA is the determinant of A with probability at least 1 −n2−n 2\|S\| . 13.4.59 Remark If A is known to be non-singular the algorithm can be repeated with diﬀerent matrices U until the obtained minimal polynomial is of degree n. Then it is the characteristic polynomial of UA and the determinant is certiﬁed. Alternatively if the matrix is singular then X divides the minimal polynomial. As Wiedemann’s algorithm always returns a factor of the true minimal polynomial, and U is invertible, the algorithm can be repeated on UA until either the obtained polynomial is of degree n or it is divisible by X. Overall the determinant has a Las-Vegas blackbox solution. 13.4.60 Theorem [2874, 2875] Let S be a ﬁnite subset of a ﬁeld F that does not include 0 and A ∈ Fn×n with s1, . . . , st as invariant factors. Let U ∈Sn×k and V ∈Sk×n be randomly chosen rank k matrices in F. Then gcd(ΠA, ΠA+UV ) = sk+1 with probability at least 1 −nk+n+1 \|S\| . 13.4.61 Remark Using the divisibility of the invariant factors and the fact that their product is of degree n, one can see that the number of degree changes between successive invariant factors is of order O (√n) . Thus by a binary search over successive applications of Theorem 13.4.60 one can recover all of the invariant factors and thus the characteristic polynomial of the matrix in a Monte-Carlo fashion. 13.4.4.3 System solving and the Lanczos algorithm 13.4.62 Remark For the solution of a linear system Ax = b, one could compute the minimal polynomial ΠA,b and then derive a solution of the system as a linear combination of the Aib. The following Lanczos approach is more eﬃcient for system solving as it avoids recomputing (or storing) the latter vectors [947, 1273]. 13.4.63 Algorithm (Lanczos system solving) Require: A ∈Fm×n, b ∈Fm Ensure: x ∈Fn such that Ax = b or failure 1. Let ˜ A = D1AT D2AD1 and ˜ b = D1AT D2b + ˜ Av with D1 and D2 random diagonal matrices and v a random vector; 2. w0 = ˜ b; v1 = ˜ Aw0; t0 = vT 1 w0; γ = ˜ btw0t−1 0 ; x0 = γw0; 3. repeat 4. α = vT i+1vi+1t−1 i ; β = vT i+1vit−1 i−1; wi+1 = vi+1 −αwi −βwi−1; 5. vi+2 = ˜ Awi+1; ti+1 = wT i+1vi+2; 6. γ = ˜ btwi+1t−1 i+1; xi+1 = xi + γwi+1; 7. until wi+1 = 0 or ti+1 = 0; 8. Return x = D1(xi+1 −v); 532 Handbook of Finite Fields 13.4.64 Remark The probability of success of Algorithm 13.4.63 follows from Theorem 13.4.57. 13.4.65 Remark Over small ﬁelds, if the rank of the matrix is known, the diagonal matrices of line 1 can be replaced by sparse preconditioners with O (n log(n)) non-zero coeﬃcients to avoid the need of ﬁeld extensions [607, Corollary 7.3]. 13.4.66 Remark If the system with A and b is known to have a solution then the algorithm can be turned Las Vegas by checking that the output x indeed satisﬁes Ax = b. In general, we do not know if this algorithm returns failure because of bad random choices or because the system is inconsistent. However, Giesbrecht, Lobo, and Saunders have shown that when the system is inconsistent, it is possible to produce a certiﬁcate vector u such that uT A = 0 together with uT b ̸= 0 within the same complexity [1273, Theorem 2.4]. Overall, system solving can be performed by blackbox algorithms in a Las-Vegas fashion. 13.4.5 Sparse and structured methods 13.4.67 Remark Another approach to sparse linear system is to use Gaussian elimination with pivoting, taking into account the zero coeﬃcients. This algorithm modiﬁes the structure of the matrix and might suﬀer from ﬁll-in. Consequently the available memory is usually the bottleneck. From a triangularization one can naturally derive the rank, determinant, system solving, and nullspace. Comparisons with the blackbox approaches above can be found, e.g., in . 13.4.5.1 Reordering 13.4.68 Algorithm (Gaussian elimination with linear pivoting) Require: a matrix A ∈Fm×n Ensure: An upper triangular matrix U such that there exists a unitary lower-triangular matrix L and permutations matrices P and Q over F, with A = P ·L·U·Q 1. for all elimination steps do 2. Choose as pivot row the sparsest remaining row; 3. In this row choose the non-zero pivot with lowest number of non-zero elements in its column; 4. Eliminate using this pivot; 5. end for 13.4.69 Remark Yannakakis showed that ﬁnding the minimal ﬁll-in (or equivalently the best piv-ots) during Gaussian elimination is an NP-complete task . In numerical algorithms, heuristics have been developed and comprise minimal degree ordering, cost functions, or nested dissection; see for example [89, 1485, 3082]. These heuristics for reducing ﬁll-in in the numerical setting, often assume symmetric and invertible matrices, and do not take into account that new zeros may be produced by elimination operations (aij = aij + δi ∗akj), as is the case with matrices over ﬁnite ﬁelds. Thus, proposes the heuristic 13.4.68 to take those new zeros into account, using a local optimization of a cost function at each elimination step. 13.4.5.2 Structured matrices and displacement rank 13.4.70 Remark Originating from the seminal paper most of the algorithms dealing with structured matrices use the displacement rank approach . Miscellaneous theoretical topics 533 13.4.71 Deﬁnition For A ∈Fm×m and B ∈Fn×n, the Sylvester (respectively Stein) linear dis-placement operator ▽A,B (respectively △A,B) satisﬁes for M ∈Fm×n: ▽A,B(M) = AM −MB, △A,B(M) = M −AMB. A pair of matrices (Y, Z) ∈Fm×α × Fn×α is an (A, B)-Sylvester-generator of length α (respectively Stein) for M if ▽A,B(M) = Y ZT (respectively △A,B(M) = Y ZT ). 13.4.72 Remark The main idea behind algorithms for structured matrices is to use such generators as a compact data structure, in cases where the displacement has low rank. 13.4.73 Remark Usual choices of matrices A and B are diagonal matrices and cyclic down shift matrices. 13.4.74 Deﬁnition We denote the diagonal matrix whose (i, i) entry is xi by Dx, x ∈Fn and by Zn,ϕ, ϕ ∈F the n × n unit circulant matrix having ϕ at position (1, n), ones in the subdiagonal (i + 1, i) and zeros elsewhere. operator matrices class of structured rank of number of ﬂops A B matrices M ▽A,B(M) for computing M · v Zn,1 Zn,0 Toeplitz and its inverse ≤2 O ((m + n) log(m + n)) Zn,1 ZT n,0 Hankel and its inverse ≤2 O ((m + n) log(m + n)) Zn,0 + ZT n,0 Zn,0 + ZT n,0 Toeplitz + Hankel ≤4 O ((m + n) log(m + n)) Dx Zn,0 Vandermonde ≤1 O (m + n) log2(m + n) Zn,0 Dx inverse of Vandermonde ≤1 O (m + n) log2(m + n) ZT n,0 Dx transposed of Vandermonde ≤1 O (m + n) log2(m + n) Dx Dy Cauchy and its inverse ≤1 O (m + n) log2(m + n) Table 13.4.4 Complexity of the matrix-vector product for some structured matrices 13.4.75 Remark As computing matrix vector products with such structured matrices have close algorithmic correlation to computations with polynomials and rational functions, these ma-trices can be quickly multiplied by vectors, in nearly linear time as shown on Table 13.4.4. Therefore the algorithms of Subsection 13.4.4 can naturally be applied to structured matri-ces, to yield almost O n2 time linear algebra. 13.4.76 Remark If the displacement rank is small there exist algorithms quasilinear in n, the dimen-sion of the matrices, which over ﬁnite ﬁelds are essentially variations or extensions of the Morf/Bitmead-Anderson divide-and-conquer [290, 2154] or Cardinal’s approaches. The method is based on dividing the original problem repeatedly into two subproblems with one leading principal submatrix and the related Schur complement. This leads to O α2n1+o(1) system solvers, which complexity bound have recently been reduced to O αω−1n1+o(1) [364, 1600]. With few exceptions, all algorithms thus need matrices in generic rank proﬁle. Over ﬁnite ﬁelds this can be achieved using Kaltofen and Saunders unit upper triangu-lar Toeplitz preconditioners and by controlling the displacement rank growth and non-singularity issues . 534 Handbook of Finite Fields 13.4.6 Hybrid methods 13.4.6.1 Hybrid sparse-dense methods 13.4.77 Remark Overall, as long as the matrix ﬁts into memory, Gaussian elimination methods over ﬁnite ﬁelds are usually faster than iterative methods . There are heuristics trying to take advantage of both strategies. Among those we brieﬂy mention the most widely used: 1. Perform the Gaussian elimination with reordering 13.4.68 until the matrix is almost ﬁlled up. If the remaining non-eliminated part would ﬁt as a dense matrix, switch to the dense methods of Subsection 13.4.2. 2. Maintain two sets of rows (or columns), sparse and dense. Favor elimination on the sparse set. This is particularly adapted to index calculus . 3. Perform a preliminary reordering in order to cut the matrix into four quadrants, the upper left one being triangular. This, together with the above strategies has proven eﬀective on matrices which are already quasi-triangular, e.g., Gr¨ obner bases computations in ﬁnite ﬁelds . 4. If the rank is very small compared to the dimension of the matrix, one can use left and right highly rectangular projections to manipulate smaller structures . 5. The arithmetic cost and thus timing predictions are easier on iterative methods than on elimination methods. On the other hand the number of non-zero elements at a given point of the elimination is usually increasing during an elimination, thus providing a lower bound on the remaining time to triangularize. Thus a heuristic is to perform one matrix-vector product with the original matrix and then eliminate using Gaussian elimination. If at one point the lower bound for elimination time surpasses the predicted iterative one or if the algorithm runs out of memory, stop the elimination and switch to the iterative methods . 13.4.6.2 Block-iterative methods 13.4.78 Remark Iterative methods based on one-dimensional projections, such as Wiedmann and Lanczos algorithm can be generalized with block projections. Via eﬃcient preconditioning these extensions to the scalar iterative methods can present enhanced properties: 1. Usage of dense sub-blocks, after multiplications of blocks of vectors with the sparse matrix or the blackboxes, allows for a better locality and optimization of memory accesses, via the application of the methods of Subsection 13.4.1. 2. Applying the matrix to several vectors simultaneously introduces more paral-lelism [718, 719, 1664]. 3. The probability of success augments with the size of the considered blocks, espe-cially over small ﬁelds [1657, 2873]. 13.4.79 Deﬁnition Let X ∈Fk×n q , Y ∈Fn×k q and Hi = XAiY for i = 0, . . . , n/k. The matrix minimal polynomial of the sequence Hi is the matrix polynomial FX,A,Y ∈Fq[X]k×k of least degree, with its leading degree matrix column-reduced, that annihilates the sequence (Hi). 13.4.80 Theorem The degree d matrix minimal polynomial of a block sequence (Hi) ∈(F k×k q )Z can be computed in O k3d2 using block versions of Hermite-Pade approximation and ex-tended Euclidean algorithm or Berlkamp-Massey algorithm [719, 1657, 2873]. Further Miscellaneous theoretical topics 535 improvement by [214, 1275, 1670, 2804] bring this complexity down to O (kωd)1+o(1) , using a matrix extended Euclidean algorithm. 13.4.81 Algorithm (Nullspace vector) Require: A ∈Fn×n q Ensure: ω ∈Fn q a vector in the nullspace of A 1. Pick X ∈Fk×n q , Y ∈Fn×k q uniformly at random; 2. Compute the sequence Hi = XAiY ; 3. Compute FX,A,Y the matrix minimal polynomial; 4. Let f = frxr + · · · + fdxd be a column of FX,A,Y ; 5. Return ω = Y fr + AY fr+1 + · · · + Ad−rY fd; 13.4.82 Remark These block-Krylov techniques are used to achieve the best known time com-plexities for several computations with black-box matrices over a ﬁnite ﬁeld or the ring of integers: computing the determinant, the characteristic polynomial , and the solution of a linear system of equations . See Also Chapter 10 For generating sequences and their links to Wiedemann’s algorithm via Berlekamp-Massey’s shift register synthesis. §11.1, §11.3 For eﬀective ﬁnite ﬁeld constructions. §13.2 For structured matrices over ﬁnite ﬁelds. References Cited: [89, 124, 214, 290, 349, 364, 457, 517, 607, 718, 719, 720, 723, 929, 930, 931, 932, 933, 934, 935, 937, 944, 945, 946, 947, 1043, 1171, 1273, 1275, 1336, 1485, 1600, 1643, 1656, 1657, 1664, 1666, 1669, 1670, 1723, 1839, 1840, 2040, 2154, 2170, 2346, 2386, 2726, 2727, 2728, 2804, 2824, 2873, 2874, 2875, 2974, 2976, 2985, 2987, 3031, 3082] 13.5 Carlitz and Drinfeld modules David Goss, Ohio State University 13.5.1 Remark Much of the theory presented here will be familiar to the reader from the theory of elliptic curves. Moreover, the reader may proﬁt on ﬁrst reading this by only focusing on the basic case A = Fq[t]. We have given a very rapid introduction to Drinfeld modules and the like and have naturally omitted many topics; the subject is quite active and changing rapidly. For much of the elided material please consult [919, 920, 1333, 1451, 2793].∗ 13.5.2 Remark One of the salient points about Drinfeld modules is that they allow one to go as far as possible in replacing the integers Z as the fundamental object in arithmetic. In other words, while all of the ﬁelds involved obviously lie over Spec(Z), the theory allows one to study invariants coming from characteristic 0 (such as class groups, etc.) as well ∗This survey is dedicated to the memory of my friend, and function ﬁeld pioneer, David Hayes. 536 Handbook of Finite Fields as analogous invariants arising only in ﬁnite characteristic (such as the “class module” of Subsection 13.5.6 below). 13.5.1 Quick review 13.5.3 Remark It is well known that the most important function in classical analysis is the expo-nential function ez; in analysis in characteristic p, it is the q-th power mapping τq(z) := zq (q = pn0, n0 a positive integer) and functions created out of it. Note that τq is clearly an Fq-linear mapping by the Binomial Theorem and separable polynomials in τq are characterized among all separable polynomials as those having their zeroes form an Fq vector space. For any Fq-ﬁeld E we let E{τq} be the algebra of polynomials in τq (which is noncommutative in general); the ring E{τq} is the ring of algebraic Fq-linear endomorphisms of the additive group over E. We also let ¯ E be a ﬁxed algebraic closure of E. 13.5.4 Remark Let E be an arbitrary ﬁeld (of arbitrary characteristic). 13.5.5 Deﬁnition A non-Archimedean absolute value on E is a mapping \| \|: E →R satisfying: 1. \|x\| ≥0, 2. \|x\| = 0 if and only if x = 0, 3. \|xy\| = \|x\|\|y\|, and, 4. \|x + y\| ≤max{\|x\|, \|y\|}. 13.5.6 Example Let E = Fq(t) where t is an indeterminate. Let g ∈E and suppose that g(t) has a zero of order j at ∞; set \|g\| := q−j. 13.5.7 Remark The ﬁeld E is complete if every Cauchy sequence converges to an element of E. If the ﬁeld is not complete one can complete it by mimicking the construction of the real numbers (the ﬁeld E of Example 13.5.6 completes to the formal Laurent series ﬁeld Fq((1/t)) in 1/t). We assume that E is complete for the rest of this subsection. 13.5.8 Remark Property 4 in Deﬁnition 13.5.5 immediately implies that a series with coeﬃcients in E converges if and only if the n-th term goes to 0. 13.5.9 Remark Let F be a ﬁnite extension of E. It is well-known that \| · \| extends uniquely to F (and thus to any algebraic closure ¯ E of E). 13.5.10 Proposition Let F be a normal extension of E. Let σ: F →F be an E-automorphism. Then \|σ(x)\| = \|x\| for all x ∈F. 13.5.11 Corollary Let F/E be ﬁnite of degree d and let N F E be the norm. Then \|x\| = \|N F E (x)\|1/d for all x ∈E. 13.5.12 Remark The absolute value \| · \| is also readily seen to extend uniquely to the completion of ¯ E which remains algebraically closed. 13.5.13 Deﬁnition A power series P∞ i=0 aixi with coeﬃcients in E is entire if it converges for all x ∈E. 13.5.14 Theorem Let f(x) be an entire power series with coeﬃcients in E. Then there exists a nonnegative integer j, an element c ∈E and a (possibly empty or ﬁnite) sequence {0 ̸= λi} ⊂¯ E, where \|λi\| →∞as i →∞, with f(x) = cxj Q i(1 −x/λi). Miscellaneous theoretical topics 537 13.5.15 Remark Conversely a product, as in Theorem 13.5.14, deﬁnes an E-entire power series if {λi} is stable under E-automorphisms of ¯ E and inseparability indices are taken into account. Note that Theorem 13.5.14 immediately implies that the power series for ez is never entire as an E-power series for non-Archimedean E of characteristic 0. 13.5.16 Remark The notion of analytic continuation in complex analysis is essential; it is what allows analytic functions to be deﬁned locally. In non-Archimedean analysis, as just de-scribed, there are far too many open sets for any analogous theory. Following ideas from algebraic geometry (and Grothendieck), Tate had the fantastic idea to use a “Grothendieck topology” to very seriously cut down on the number of such open sets. This theory is called “rigid analysis” and it does in fact allow for analytic continuation, etc. More importantly, via this theory one is able to pass between (rigid) analytic sheaves and algebraic sheaves in the manner of Serre’s famous G.A.G.A. paper . In particular, one is then able to construct algebraic functions via analysis. For details on all of this, see for example . 13.5.2 Drinfeld modules: deﬁnition and analytic theory 13.5.17 Remark Let X be a smooth, complete, geometrically irreducible curve over Fq with function ﬁeld k. Note that k is precisely a global ﬁeld of characteristic p. We let ∞∈X be a ﬁxed closed point of degree d∞over Fq, and A the ring of functions holomorphic away from ∞; set K := k∞= the completion of k under the absolute value \|x\|∞= q−v(x) where v(x) is the order of zero of x at ∞. We set C∞to be the completion of a ﬁxed algebraic closure ¯ K of K. Let K1 ⊂C∞be a ﬁnite extension of K (automatically also complete) with separable closure Ksep 1 ⊂C∞. 13.5.18 Deﬁnition 1. A K1-lattice is a discrete, ﬁnitely generated, Gal(Ksep 1 /K1)-stable A-submodule M of Ksep 1 . 2. A morphism between two lattices M1 and M2 of the same A-rank is a scalar c such cM1 ⊆M2. 3. To a lattice M we attach the exponential function eM(z) := z Q 0̸=β∈M(1 −z/β). 13.5.19 Remark By Theorem 13.5.14, eM(z) is entire with K1 coeﬃcients. As M can be exhausted by ﬁnite dimensional Fq-vector spaces, one deduces that eM(z) is a limit of Fq-linear poly-nomials and so is, itself, an Fq-linear surjection from C∞to itself. We obtain an exact sequence 0 − →M − →C∞ eM(z) − →C∞− →0 . (13.5.1) Let a ∈A; by transport of structure (and matching divisors) we now obtain an exotic A-module structure φM a (z) on C∞via the functional equation eM(az) := φM a (eM(z)) = a · eM(z) Y 0̸=α∈a−1M/M (1 −eM(z)/eM(α)) . (13.5.2) The essential observation is that a 7→φM a (z) embeds A into K1{τq} as Fq-algebras; that is we are representing an element a ∈A by the nontrivial Fq-linear polynomial φM a (z) with φM a+b(z) = φM a (z) + φM b (z) and φM ab(z) = φM a (φM b (z)). 13.5.20 Deﬁnition Let E be a ﬁeld with an Fq-linear homomorphism ı: A →E. Then a Drin-feld module over (E, ı) is a homomorphism φ: A →E{τq}, a 7→φa, such that φa = ı(a)τ 0 q +{higher terms} and such that there exists a ∈A such that φa ̸= ı(a)τ 0 q . Let φ and ψ be two Drinfeld modules over E; then a morphism from φ to ψ is a polynomial P ∈E{τq} such that Pφa = ψaP for all a. Nonzero morphisms are isogenies. 538 Handbook of Finite Fields 13.5.21 Remark Via φ, E and ¯ E become A-modules with a∗z := φa(z) for a ∈A, z ∈¯ E. Let I ⊆A be an ideal and let φ[I] ⊂¯ E be those points annihilated by all i ∈I; commutativity of A implies that φ[I] is a ﬁnite A-module. Now let a ∈A ̸∈ker ı. As A is a Dedekind domain, simple counting arguments show the existence of an integer d, independent of a, such that φ[(a)] ≃A/(a)d; d is the rank of φ. Isogenies are only possible between Drinfeld modules of the same rank and isomorphisms correspond to those P of the form cτ 0 q where c ̸= 0. We then have the following fundamental result of Drinfeld. 13.5.22 Theorem The exponential construction gives an equivalence of categories between Drinfeld modules of rank d over K1 and rank d K1-lattices. 13.5.23 Example Let A = Fq[t]. The Carlitz module C is the rank 1 Drinfeld module over Fq(t) given by Ct := tτ 0 q + τq. The associated Carlitz exponential is eC(z) := P∞ j=0 zqj/Dj, where Dj is the product of all monic polynomials of degree j, and the Carlitz lattice is Aξ where ξ := (−t)q/(q−1) Q∞ j=1(1 −t1−qj)−1 is the Carlitz period. For general A, Hayes has constructed generalizations of C which are deﬁned over a Hilbert class ﬁeld of k. 13.5.24 Remark It was a fundamental observation of Carlitz that adding the elements of C[a] to k gave an abelian extension. These extensions were then shown to provide the abelian closure of k which is tamely ramiﬁed at ∞by Hayes and generalized by him to arbitrary A in in analogy with cyclotomic ﬁelds. Drinfeld gave a modular approach to the construction of these class ﬁelds in his paper ; in he obtained the full abelian closure. This full abelian closure has also recently been explicitly constructed by Zywina . 13.5.25 Remark Carlitz originally constructed his module very concretely based on the famous combinatorial formula called the Moore Determinant (a ﬁnite characteristic analog of the Wronskian as well as the Vandermonde determinant) as given in the next deﬁnition (see also Section 13.2). 13.5.26 Deﬁnition We deﬁne ∆q(w1, . . . , wn) to be the determinant of      w1 . . . wn wq 1 . . . wq n . . . . . . wqn−1 1 . . . wqn−1 n     =    τ 0 q (w1) . . . τ 0 q (wn) . . . . . . τ n−1 q (w1) . . . τ n−1 q (wn)   . 13.5.27 Remark Moore then shows the following basic result analogous to Abel’s calculation of the Wronskian. 13.5.28 Proposition ∆q(w1, . . . , wd) equals d Y i=1 Y ki−1∈Fq · · · Y k1∈Fq (wi + ki−1wi−1 + · · · + k1w1) . 13.5.29 Remark Proposition 13.5.28 allowed Carlitz to calculate various products of polynomials over ﬁnite ﬁelds, and thus to explicitly construct his exponential, lattice, and module. 13.5.30 Remark Returning to a general Drinfeld modules φ, let p now be a prime of A with completions kp, Ap. Let φ[p∞] be the set of all p-power torsion points. Miscellaneous theoretical topics 539 13.5.31 Deﬁnition We set Tp(φ) := HomA(kp/Ap, φ[p∞]); this is the p-adic Tate module of φ. 13.5.32 Remark If p ̸= ker ı, then Tp(φ) is isomorphic to Ad p as A-module; the construction of Tate modules is functorial. 13.5.3 Drinfeld modules over ﬁnite ﬁelds 13.5.33 Remark Let E be as in Deﬁnition 13.5.20 now also be ﬁnite; let φ be a Drinfeld module over E of rank d. Let FE := τ t q which is an endomorphism of φ. Set b := ker ı (which is a maximal ideal of A) and let p ̸= b be another nontrivial prime of A; set fφ(u) := det(1−uFE \| Tp(φ)). Very clever use of central simple arithmetic allows one to establish the following results. 13.5.34 Theorem The polynomial fφ(u) depends only on the isogeny class of fφ(u) and has coeﬃcients in A which are independent of the choice of p. The reciprocal roots of fφ(u) in C∞have absolute value qt/d and their product generates the ideal b[E : A/b]. 13.5.35 Deﬁnition An element α ∈C∞is a Weil number of rank d for E if and only if 1. α is integral over A, 2. there is only place in k(α) which is a zero of α and there is only one place of k(α) above ∞, 3. \|α\|∞= qt/d, and, 4. [k(α) : k] \| d. 13.5.36 Remark Clearly the set of Weil numbers of rank d for E is acted on by the group of k-automorphism of ¯ k ⊂C∞and we let Wd(E) be the set of orbits under this action. 13.5.37 Theorem The mapping from the set of isogeny classes of Drinfeld modules over E of rank d to Wd(E) given by Theorem 13.5.34 is a bijection. 13.5.4 The reduction theory of Drinfeld modules 13.5.38 Remark Let E now be a ﬁnite extension of k equipped with a Drinfeld module φ of rank d; let O be the A-integers of E. As A is a ﬁnitely generated Fq-algebra, for almost all O-primes P we can reduce the coeﬃcients of φ modulo P to obtain a Drinfeld module of rank d over O/P. For the other primes, we now localize and assume that E is equipped with a nontrivial discrete valuation v with v(A) nonnegative; let Ov ⊂E be the associated valuation ring of v-integers and let Mv ⊂Ov be the maximal ideal. 13.5.39 Deﬁnition 1. The Drinfeld module φ has stable reduction at v if there exists ψ which is isomorphic to φ such that ψ has coeﬃcients in Ov and such that the reduction ψv of ψ is a Drinfeld module (of rank d1 ≤d). 2. The module φ has good reduction at v if it has stable reduction and d1 = d. 3. The module φ has potential stable (resp. potential good) reduction if there is an extension (E1, w) of (E, v) such that φ has stable (resp. good) reduction at w. 13.5.40 Remark Drinfeld established that every Drinfeld module has potential stable reduc-tion. Let p be a prime of A not contained in Mv and view Tp(φ) as a Gal(Esep/E)-module. 13.5.41 Theorem The Drinfeld module φ has good reduction at v if and only if Tp(φ) is unramiﬁed at v as a Gal(Esep/E)-module. 540 Handbook of Finite Fields 13.5.42 Remark Theorem 13.5.41 is an obvious analog of the theorem of Ogg-N´ eron-Shafarevich and is due to Takahashi. Taguchi has established that Tp(φ) is a semisimple Gal(Esep/E)-module; this was also independently established by Tamagawa , 13.5.5 The A-module of rational points 13.5.43 Remark Let E continue to be a ﬁnite extension of k equipped with a Drinfeld module φ of rank d; using φ, E becomes a natural A-module which is denoted “φ(E)”. 13.5.44 Theorem The A-module φ(E) is isomorphic to T ⊕N where T is a ﬁnite torsion module and N is a free A-module of rank ℵ0. 13.5.45 Remark The above result, due to Poonen, applies more generally to other modules of rational points (such as a ring OS of S-integers which contains the coeﬃcients of φ). The main technique in the proof is to establish the tameness of φ(L); that is, let M ⊆φ(L) be a submodule such that k ⊗A M is ﬁnite dimensional, then M is itself ﬁnitely generated. This result is established via the use of heights and is analogous to the theorem of Mordell-Weil for elliptic curves/abelian varieties except that the rank is always inﬁnite. 13.5.6 The invariants of a Drinfeld module 13.5.46 Remark We now present very recent work of Taelman [2760, 2761] that establishes an analog of the classical class group/Tate-Shafarevich group and the group of units/Mordell-Weil group in the theory of Drinfeld modules. 13.5.47 Remark As above, let E be a ﬁnite extension of k with A-integers O; set E∞:= E ⊗k K. Let φ be a Drinfeld A-module of rank d over E whose coeﬃcients will be assumed to lie in O (N.B., this does not imply that φ has good reduction at all primes of O); let eφ(z) be the associated exponential function. The functional equation satisﬁed by eφ(z) implies that the coeﬃcients of φ lie in E. Thus eφ(z) induces a natural Fq-linear, continuous endomorphism of E∞(also denoted eφ); it is an open map as the derivative of eφ(z) is identically 1. Let ˆ M(φ/O) := e−1 φ (O) ⊂E∞. The openness of eφ(z) immediately implies the next result. 13.5.48 Proposition [2760, 2761] The A-module ˆ M(φ/O) is discrete and cocompact in E∞. 13.5.49 Deﬁnition We set M(φ/O) ⊂φ(O) to be the image of ˆ M(φ/O) under eφ(z). 13.5.50 Corollary [2760, 2761] M(φ/O) is a ﬁnitely generated A-module under the Drinfeld action. 13.5.51 Remark M(φ/O) is an analog of both the group of units of a number ﬁeld and the Mordell-Weil group of an elliptic curve. 13.5.52 Deﬁnition We set H(φ/O) := φ(E∞)/(φ(O) + eφ(E∞)). 13.5.53 Remark As O is cocompact in E∞and eφ(E∞) is open in E∞, we deduce the next result. 13.5.54 Proposition [2760, 2761] H(φ, O) is a ﬁnite A-module. 13.5.55 Remark H(φ, O) is the class module or the Tate-Shafarevich module of φ over O. 13.5.56 Example Let A = Fq[t] and let φ = C be the Carlitz module. 1. Suppose ﬁrst of all that E = k. In this case, one ﬁnds that H(C/A) = {0} and that e−1 C (A) has rank one with generator the logarithm e−1 C (1). Miscellaneous theoretical topics 541 2. For general E, all torsion elements of C(O) are contained in M(C/O). 3. When E is obtained by adjoining torsion elements of C, Anderson has pro-duced an A-module of special points (analogous to cyclotomic units) which has maximal rank inside M(C/O). 13.5.57 Remark One can reformulate the invariants of the Carlitz module in terms of Zariski sheaves and shtuka (see Remark 13.5.116) thereby reinterpreting them a la motivic cohomology, see . 13.5.7 The L-series of a Drinfeld module 13.5.58 Remark Let F∞⊂K be the ﬁeld of constants and let π ∈K be a ﬁxed element of order 1 at ∞. Every element x ∈K∗has a decomposition x = ζx,ππv∞(x)⟨x⟩π with ζx,π ∈F∗ ∞and ⟨x⟩π ≡1 (mod π); x is positive if ζx,π = 1 (generalizing the notion of “monic polynomial”). Let I be the group of A-fractional ideals with subgroups P+ ⊆P where P (resp. P+) is the group of principal (resp. and positively) generated ideals; I/P+ is ﬁnite, etc. 13.5.59 Remark Let U1 ⊂C∞be the group of elements 1 + w where \|w\|∞< 1. The binomial theorem implies that U1 is a Zp-module in the usual exponential fashion; as we can also take p-power roots uniquely in U1, it is in fact a Qp-vector space. Let I = (i) ∈P+ where i is positive and deﬁne ⟨I⟩π := ⟨i⟩π. As U1 is divisible the next result follows directly. 13.5.60 Proposition The map I 7→⟨I⟩π extends uniquely to a homomorphism I 7→⟨I⟩π of I to U1. 13.5.61 Remark Let deg(I) denote the degree over Fq of the associated divisor on Spec(A). 13.5.62 Deﬁnition We set S∞:= C∞× Zp. For a fractional ideal I and s = (x, y) ∈S∞, we set Is := xdeg(I)⟨I⟩y π. 13.5.63 Remark Let π∗∈C∞be a ﬁxed d∞-th root of π and let j ∈Z; set sj := (π−j ∗, j) ∈S∞. For positive a ∈A, we have (a)sj = aj (with the standard meaning). 13.5.64 Remark Let E, O be as above and suppose φ is a Drinfeld module over E. Let N E k be the norm mapping on O-ideals. For each O-prime p of good reduction, we let fp(u) ∈A[u] be the characteristic polynomial of the reduction as in Subsection 13.5.3. The next deﬁnition builds on classical theory. 13.5.65 Deﬁnition We formally put L(φ, s) := Q p good fp(N E k p−s)−1 for s ∈S∞. 13.5.66 Remark Theorem 13.5.34 immediately implies that L(φ, s) converges on a “half-plane” of S∞consisting of (x, y) with \|x\|∞bounded below (see for factors at the ﬁnitely many bad primes). 13.5.67 Theorem The function L(φ, s) analytically continues to a continuous (in y ∈Zp) family of entire power series in x−1 which is also continuous on S∞. 13.5.68 Remark Note that these entire power series are obtained by expanding the Euler products and “summing by degree.” 13.5.69 Corollary Let j be a nonnegative integer. Then L(φ, (x, −j)) is a polynomial in x−1. 13.5.70 Remark These special polynomials play an essential role in the theory. By , their degree grows logarithmically in j. They interpolate continuously to entire families at all places of 542 Handbook of Finite Fields k. Moreover, this logarithmic growth allows one to analogously handle all associated partial L-series obtained by summing only over a residue class . 13.5.71 Deﬁnition We set ζA(s) := P I I−1 = Q p, prime(1 −p−s)−1 where I runs over the nonzero ideals of A, etc. In a similar fashion one deﬁnes the zeta function of the A-integers in a ﬁnite extension of k. 13.5.72 Example Let A = Fq[t] and let “positive=monic.” Then L(C, s) = ζA(s −1). 13.5.8 Special values 13.5.73 Remark Let ξ be a Hayes-period (as in Example 13.5.23). Judicious use of the associated exponential function gives the next result. 13.5.74 Theorem Let j be a positive integer divisible by qd∞−1. Then 0 ̸= ζA(j)/ξj is algebraic over k. 13.5.75 Example In the case A = Fq[t], Carlitz in the 1930’s deﬁned a suitable “factorial” element and thus analogs of Bernoulli numbers; these are called “Bernoulli-Carlitz” elements. 13.5.76 Theorem Let j be as in Theorem 13.5.74. Then ζA(−j) = 0. 13.5.77 Remark Clearly, for all j ∈Z, we have ζA(jp) = ζA(j)p. For the rest of this subsection, we let A = Fq[t], etc. In this case, one knows that the order of zero (i.e., of the polynomial ζA(x, −j) at x = 1) in Theorem 13.5.76 is exactly 1. 13.5.78 Theorem The only algebraic relations on {ζA(j)}, j > 0, arise from Theorem 13.5.74 and the p-th power mapping. 13.5.79 Remark The obvious analog of Theorem 13.5.78 without the p-th power mapping is con-jectured for the Riemann zeta function. 13.5.80 Remark Let E, O be as in Subsection 13.5.6. One lets \|H(C/O)\| be the monic generator of the associated Fitting ideal and one deﬁnes a natural regulator R(C/O) using ˆ M(C/O). Taelman then establishes the next fundamental result. 13.5.81 Theorem We have ζO(1) = \|H(C/O)\|R(C/O). 13.5.82 Remark Let φ be as in Subsection 13.5.6. Using the dual representation of the Tate module of φ, one deﬁnes the L-series L(φ∨, s). Theorem 13.5.81 then generalizes to a formula for L(φ∨, 0). 13.5.83 Remark In , Thakur extends Theorem 13.5.81 to some basic examples involving certain factorial, non-rational A thus indicating that Theorem 13.5.81 should hold in general. 13.5.84 Remark In Taelman establishes the analog for the Bernoulli-Carlitz elements and the class module of the classical Theorem of Herbrand-Ribet. 13.5.9 Measures and symmetries 13.5.85 Remark Let A = Fq[t], p a prime of A of degree d, and let πp ∈A have order 1 at p; thus the completion Ap is isomorphic to F where F ≃Fqd. Set R = Ap equipped with the canonical topology. Miscellaneous theoretical topics 543 13.5.86 Deﬁnition An R-valued measure on R is a ﬁnitely additive R-valued function on the compact opens of R. 13.5.87 Remark Let f : R →R be a continuous function and µ a measure; it is easy to see that the associated “Riemann sums” will converge to an element denoted “ R R f(t) dµ(t).” Let µ and ν be two R-valued measures; it is clear that their sum is also an R-valued measure. Their product is deﬁned, as usual, by convolution as below. 13.5.88 Deﬁnition Let µ, ν be as above. Their convolution µ ∗ν is the R-valued measure on R given by R R f(t) dµ ∗ν(t) := R R R R f(x + y) dµ(x)dν(y). 13.5.89 Remark It is easy to check that the space of measures forms a commutative R-algebra under convolution. Now let Dj be the hyperdiﬀerential operator given by Djxi := i j xi−j. Notice that DiDj = i+j i Di+j. Let R{{D}} be the algebra of formal power-series in the Di with the above multiplication rule. Building certain bases for the continuous functions out of additive polynomials (following Carlitz) one can establish the next result. 13.5.90 Theorem There is an isomorphism (which depends on the basis constructed) of the R-algebra of measures and R{{D}}. 13.5.91 Remark Let y ∈Zp be written as y = P∞ i=0 ciqi, where 0 ≤ci < q. Let ρ be a permutation of the set {0, 1, 2, . . .}. 13.5.92 Deﬁnition We deﬁne ρ∗(y), y ∈Zp, by ρ∗(y) := P∞ i=0 ciqρ(i) . We let S(p) denote the induced group of bijections of Zp. 13.5.93 Theorem The map ρ∗is a homeomorphism of Zp and ρ(y0 + y1) = ρ(y0) + ρ(y1) if there is no carry-over of q-adic digits in the sum of y0 and y1. Furthermore, ρ∗stabilizes both the positive and negative integers. Finally, n ≡ρ∗(n) (mod q −1) for all integers n. 13.5.94 Remark It is remarkable that S(q) appears to act as a group of symmetries of the L-series of Drinfeld modules. We present some evidence of this here and refer the reader to for more details. Our ﬁrst result is completely general and is an application of Lucas’ formula for the reduction modulo p of binomial coeﬃcients. 13.5.95 Proposition Let ρ∗∈S(p), y ∈Zp, and j a nonnegative integer. Then we have ρ∗(y) ρ∗(j) ≡ y j (mod p). 13.5.96 Theorem Let j be a nonnegative integer. Then, as polynomials in x−1, ζA(x, −j) and ζA(x, −ρ∗(j)) have the same degree. 13.5.97 Remark Theorem 13.5.96 also depends on another old formula of Carlitz (resurrected by Thakur) that allows one to compute the relevant degrees; these degrees then turn out to be invariant of the ρ∗action as a corollary of Theorem 13.5.93. Using Lucas’ formula again, one also establishes the next result. 13.5.98 Theorem Let ρ∗be as above. Then the map Di →Dρ∗(i) gives rise to an automor-phism of R{{D}}. 13.5.99 Remark Thus, of course, S(p) also acts as automorphisms of measure algebras. Moreover, long ago Carlitz computed the denominators of his Bernoulli-Carlitz elements (as in Exam-ple 13.5.75) in analogy with the classical result of von Staudt-Clausen. It is quite remarkable that the condition that a prime f ∈A of degree d divides this denominator is invariant under S(qd). 544 Handbook of Finite Fields 13.5.100 Remark We ﬁnish this subsection by noting that, in examples including A = Fq[t], the zeroes of ζA(s) “lie on a line” (with perhaps ﬁnitely many exceptions for nonrational A) with orders agreeing with classical predictions. For this, see [2611, 2892]. 13.5.10 Multizeta 13.5.101 Remark L-series of Drinfeld modules give rise to entire power series upon summing by degree. If one takes these sums and intermixes them, one obtains “multi-L-series” whose study was initiated by Thakur (see for example ). One obtains a remarkably rich ediﬁce with analogs of many essential classical results such as shuﬄe relations and the realization of multizeta values as periods, etc. For instance, Anderson and Thakur have shown the analog of a result of Terasoma giving an interpretation of multizeta values as periods of “mixed Carlitz-Tate t-modules.” 13.5.11 Modular theory 13.5.102 Remark Let A again be a general base ring. Let M be a rank d lattice as in Deﬁnition 13.5.18. Let {m1, . . . , md} ∈M be chosen so that M = Am1+· · ·+Amd−1+Imd where I is a nonzero ideal of A (as can always be done since A is a Dedekind domain). The discreteness of M implies that (m1, . . . , md) does not belong to any hyperplane deﬁned over K = k∞. 13.5.103 Deﬁnition We deﬁne Ωd := Pd−1(C∞) \ Yd, where Yd is the subset of all points lying in K-hyperplanes. 13.5.104 Remark The space Ωd comes equipped with the structure of a rigid analytic space. It is connected (in the appropriate rigid sense) ; in particular, functions are determined by their local expansions. 13.5.105 Example Let d = 2. Then Ω2 = P1(C∞) −P1(K). This space is analogous to the classical upper and lower half-planes. 13.5.106 Remark Let M, {m1, . . . , md}, be as above; clearly M is isomorphic (Deﬁnition 13.5.18) to the lattice ˆ M := A + A m2 m1 + · · · + A md−1 m1 + I md m1 which is, itself, associated to the element (m2/m1, . . . , md/m1) ∈Ωd. It is therefore reasonable that modular spaces associated to Drinfeld modules (i.e., spaces whose points parametrize Drinfeld modules with, perhaps, some added structure on their torsion points) can be described using Ωd in a manner exactly analogous to elliptic modular curves and the classical upper half plane. Indeed, this was accomplished by Drinfeld in his original paper . 13.5.107 Remark Drinfeld’s construction can be quite roughly sketched as follows: For simplicity, assume that I = A and let J ⊂A be a nontrivial ideal. Let ΓJ ⊂GLd(A) be the principal congruence subgroup. Then ΓJ acts on Ωd in complete analogy with the standard action of congruence subgroups of SL2(Z). This action of ΓJ is rigid analytic and the quotient ΓJ\Ωd also exists as a rigid space which is denoted M d J. 13.5.108 Theorem The space M d J is a regular aﬃne variety of dimension d. It is equipped with a natural morphism to Spec(A) which is ﬂat and smooth away from the primes dividing J. 13.5.109 Remark The space M 2 J is a relative curve highly analogous to elliptic modular curves. Indeed, it also can be compactiﬁed via “Tate objects” at a ﬁnite number of cusps. 13.5.110 Example Let I = A before and let Γ = SL2(A). Let eA(z) be the exponential function associated to A considered as a rank 1 lattice (Deﬁnition 13.5.18). Set EA(z) := eA(z)−1. Miscellaneous theoretical topics 545 Then one shows that EA(z) is a uniformizing parameter at the inﬁnite cusp for Γ acting on Ω2 in analogy with e2πiz and the upper half plane classically. (We note that our notation “EA(z)” is highly nonstandard but necessary as the commonly used symbols are already taken here.) 13.5.111 Remark Given the analogy between the spaces Ωd (and especially Ω2) and the classically upper half plane, it makes sense to discuss modular forms in this context . For sim-plicity, let Γ = GL2(A). 13.5.112 Deﬁnition A rigid analytic function f on Ω2 is a modular form of weight k and type m (for a nonnegative integer k and class m of Z/(q −1)) if and only if 1. for γ = a b c d ∈Γ one has f az+b cz+d = det γ−m(cz + d)kf(z), 2. f is holomorphic at the cusps. 13.5.113 Remark Property 1 of Deﬁnition 13.5.112 implies that f(z) has a Laurent-series expansion in terms of EA(z) and Property 2 is the requirement that this expansion has no negative terms; similar expansions are mandated at the other cusps. Modular forms of a given weight and type then comprise ﬁnite dimensional vector spaces. 13.5.114 Example Let A = Fq[t]. A Drinfeld module ρ of rank 2 is determined by ρt = tτ 0+gτ +∆τ 2. Here g is a modular form of weight q −1 and type 0 and ∆is a form of weight q2 −1 and type 0. 13.5.115 Remark One can readily equip such modular forms with an action of the Hecke operators which are related to Galois representations. For this see, for example, [125, 126, 334, 1267]. 13.5.116 Remark The modular theory has been highly important in a number of ways. In analogy with work on the Korteweg-de Vries equation, Drinfeld found a way to sheaﬁfy his modules, (see, for example, for a very nice account). These sheaves, called “shtuka,” have been essential to the work of Laﬀorgue completing the Langlands program for function ﬁelds and the general linear group [1829, 1830]. 13.5.117 Remark Building on the notion of shtuka, in a fundamental paper Anderson generalized Drinfeld modules to “t-modules” by replacing the additive group with additive n-space. Such t-modules have a tensor product as well as associated exponential functions and Anderson presents an extremely useful criterion for the surjectivity of these functions. The paper has played a key role in many developments. 13.5.118 Remark Another application of the Drinfeld modular curves has been in the algebraic-geometric construction of codes due to Goppa. For example, the reader may consult [2280, 2881]. 13.5.119 Remark Finally, we mention the very recent result of Pellarin which establishes a two-variable deformation of Theorem 13.5.74. In particular, one obtains deformations of Bernoulli-Carlitz elements. 13.5.12 Transcendency results 13.5.120 Remark There is a vast theory related to the transcendency of elements arising in the theory of Drinfeld modules and the like. For instance, the Carlitz period (Example 13.5.23) 546 Handbook of Finite Fields was shown to be transcendental by Wade . Since then a vast number of other results and techniques were introduced into the theory including ideas from automata theory. For a sampling of what can be found, we refer the reader to [78, 79, 96, 581, 582, 3039, 3040]. References Cited: [78, 79, 94, 96, 97, 125, 126, 334, 335, 354, 581, 582, 919, 920, 1208, 1267, 1332, 1333, 1334, 1335, 1449, 1450, 1451, 1829, 1830, 1846, 2201, 2280, 2375, 2419, 2585, 2611, 2760, 2761, 2762, 2763, 2764, 2765, 2767, 2773, 2792, 2793, 2847, 2881, 2888, 2892, 3039, 3040, 3084] III Applications 14 Combinatorial ...................................................... 549 Latin squares • Lacunary polynomials over ﬁnite ﬁelds • Aﬃne and projective planes • Projective spaces • Block designs • Diﬀerence sets • Other combinatorial structures • (t, m, s)-nets and (t, s)-sequences • Applications and weights of multiples of primitive and other polynomials • Ramanujan and expander graphs 15 Algebraic coding theory ......................................... 659 Basic coding properties and bounds • Algebraic-geometry codes • LDPC and Gallager codes over ﬁnite ﬁelds • Turbo codes over ﬁnite ﬁelds • Raptor codes • Polar codes 16 Cryptography ...................................................... 741 Introduction to cryptography • Stream and block ciphers • Multivariate cryptographic systems • Elliptic curve cryptographic systems • Hyperel-liptic curve cryptographic systems • Cryptosystems arising from Abelian varieties • Binary extension ﬁeld arithmetic for hardware implementa-tions 17 Miscellaneous applications ...................................... 825 Finite ﬁelds in biology • Finite ﬁelds in quantum information theory • Finite ﬁelds in engineering 547 This page intentionally left blank This page intentionally left blank 14 Combinatorial 14.1 Latin squares ........................................ 550 Prime powers • Non-prime powers • Frequency squares • Hypercubes • Connections to aﬃne and projective planes • Other ﬁnite ﬁeld constructions for MOLS 14.2 Lacunary polynomials over ﬁnite ﬁelds........... 556 Introduction • Lacunary polynomials • Directions and R´ edei polynomials • Sets of points determining few directions • Lacunary polynomials and blocking sets • Lacunary polynomials and blocking sets in planes of prime order • Lacunary polynomials and multiple blocking sets 14.3 Aﬃne and projective planes ....................... 563 Projective planes • Aﬃne planes • Translation planes and spreads • Nest planes • Flag-transitive aﬃne planes • Subplanes • Embedded unitals • Maximal arcs • Other results 14.4 Projective spaces .................................... 574 Projective and aﬃne spaces • Collineations, correlations and coordinate frames • Polarities • Partitions and cyclic projectivities • k-Arcs • k-Arcs and linear MDS codes • k-Caps 14.5 Block designs ........................................ 589 Basics • Triple systems • Diﬀerence families and balanced incomplete block designs • Nested designs • Pairwise balanced designs • Group divisible designs • t-designs • Packing and covering 14.6 Diﬀerence sets ....................................... 599 Basics • Diﬀerence sets in cyclic groups • Diﬀerence sets in the additive groups of ﬁnite ﬁelds • Diﬀerence sets and Hadamard matrices • Further families of diﬀerence sets • Diﬀerence sets and character sums • Multipliers 14.7 Other combinatorial structures.................... 607 Association schemes • Costas arrays • Conference matrices • Covering arrays • Hall triple systems • Ordered designs and perpendicular arrays • Perfect hash families • Room squares and starters • Strongly regular graphs • Whist tournaments 14.8 (t, m, s)-nets and (t, s)-sequences ................... 619 (t, m, s)-nets • Digital (t, m, s)-nets • Constructions of (t, m, s)-nets • (t, s)-sequences and (T, s)-sequences • Digital (t, s)-sequences and digital (T, s)-sequences • Constructions of (t, s)-sequences and (T, s)-sequences 549 550 Handbook of Finite Fields 14.9 Applications and weights of multiples of primitive and other polynomials .............................. 630 Applications where weights of multiples of a base polynomial are relevant • Weights of multiples of polynomials 14.10 Ramanujan and expander graphs ................. 642 Graphs, adjacency matrices and eigenvalues • Ramanujan graphs • Expander graphs • Cayley graphs • Explicit constructions of Ramanujan graphs • Combinatorial constructions of expanders • Zeta functions of graphs 14.1 Latin squares Gary L. Mullen, The Pennsylvania State University 14.1.1 Deﬁnition A latin square of order n is an n × n array based upon n distinct symbols with the property that each row and each column contains each of the n symbols exactly once. 14.1.2 Example The following are latin squares of orders 3 and 5 0 1 2 1 2 0 2 0 1 , 1 2 3 4 0 3 4 0 1 2 0 1 2 3 4 2 3 4 0 1 4 0 1 2 3 . 14.1.3 Remark Given any latin square of prime power order q (with symbols from Fq, the ﬁnite ﬁeld of order q), using the Lagrange Interpolation Formula from Theorem 2.1.131, we can construct a polynomial P(x, y) of degree at most q −1 in both x and y which represents the given latin square. The ﬁeld element P(a, b) is placed at the intersection of row a and column b. For example, the two squares given in the previous example can be represented by the polynomials x + y over F3 and 2x + y + 1 over F5. 14.1.4 Deﬁnition Assume that a latin square of order n is based upon the n distinct symbols 1, 2, . . . , n. Such a latin square of order n is reduced if the ﬁrst row and ﬁrst column are in the standard order 1, 2, . . . , n. Let ln denote the number of reduced latin squares of order n. Let Ln denote the total number of distinct latin squares of order n. 14.1.5 Theorem For each n ≥1, Ln = n!(n −1)!ln. 14.1.6 Remark Using the addition table of the ring Zn of integers modulo n, it is easy to see that ln ≥1 and hence Ln ≥n!(n −1)! for each n ≥2. The total number Ln of latin squares of order n is unknown if n > 11 . The table from [706, p. 142], gives the values of ln for n ≤11. 14.1.7 Deﬁnition Two latin squares of order n are orthogonal if upon placing one of the squares on top of the other, we obtain each of the possible n2 distinct ordered pairs exactly once. In addition, a set {L1, . . . , Lt} of latin squares all of the same order is orthogonal if each distinct pair of squares is orthogonal, i.e., if Li is orthogonal to Lj whenever i ̸= j. Such a set of squares is a set of mutually orthogonal latin squares (MOLS). Combinatorial 551 14.1.8 Remark There are numerous combinatorial objects which are equivalent to sets of MOLS. These include transversal designs, orthogonal arrays, edge-partitions of a complete bipartite graph, and (k, n)-nets. We refer to Chapter III, Theorem 3.18 of for a more detailed discussion of these topics; see also Sections 14.5 and 14.7. 14.1.9 Deﬁnition Let N(n) denote the maximum number of mutually orthogonal latin squares (MOLS) of order n. 14.1.10 Theorem For n ≥2, N(n) ≤n −1. 14.1.11 Deﬁnition A set {L1, . . . , Lt} of MOLS of order n is complete if t = n −1. 14.1.1 Prime powers 14.1.12 Theorem For any prime power q, the polynomials ax + y with a ̸= 0 ∈Fq represent a complete set of q −1 MOLS of order q by placing the ﬁeld element ax+y at the intersection of row x and column y of the a-th square. 14.1.13 Remark In Subsection 14.1.5 we discuss connections of complete sets of MOLS with other combinatorial objects; in particular with aﬃne and projective planes where it is stated that the existence of a complete set of MOLS of order n is equivalent to the existence of an aﬃne, or projective, plane of order n. 14.1.14 Example The following gives a complete set of 4 MOLS of order 5, arising from the poly-nomials x + y, 2x + y, 3x + y, 4x + y over the ﬁeld F5 0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3 , 0 1 2 3 4 2 3 4 0 1 4 0 1 2 3 1 2 3 4 0 3 4 0 1 2 , 0 1 2 3 4 3 4 0 1 2 1 2 3 4 0 4 0 1 2 3 2 3 4 0 1 , 0 1 2 3 4 4 0 1 2 3 3 4 0 1 2 2 3 4 0 1 1 2 3 4 0 . 14.1.15 Theorem For q ≥5 an odd prime power, the polynomials ax + y, a ̸= 0, 1, −1 ∈Fq give a (maximal) set of q −3 MOLS of order q, each of which is diagonal, i.e., which has distinct elements on both of the main diagonals. When q ≥4 is even the same construction with a ̸= 0, 1 ∈Fq gives a (maximal) set of q −2 diagonal MOLS of order q. 14.1.16 Remark The construction of sets of inﬁnite latin squares containing nested sets of mutually orthogonal ﬁnite latin squares is discussed in [397, 398]. The construction involves use of polynomials of the form ax + y over inﬁnite algebraic extensions of ﬁnite ﬁelds. 14.1.17 Remark If q is odd, a latin square of order q −1 which is the multiplication table of the group F∗ q, is mateless; i.e., there is no latin square which is orthogonal to the given square. In fact, a latin square arising from the multiplication table of a cyclic group of even order is mateless; Fq with q odd is such an example. 14.1.18 Conjecture A complete set of n −1 MOLS of order n exists if and only if n is a prime power. 14.1.19 Remark The above conjecture is the prime power conjecture, and is discussed in many articles. In this conjecture is referred to as the next Fermat problem. 552 Handbook of Finite Fields 14.1.2 Non-prime powers 14.1.20 Deﬁnition If A is a latin square of order m and B is a latin square of order n, denote the entry at row i and column j of A by aij. Similarly we denote the (i, j) entry of B by bij. Then the Kronecker product of A and B is the mn × mn square A ⊗B, given by A ⊗B = (a11, B) (a12, B) · · · (a1m, B) (a21, B) (a22, B) · · · (a2m, B) . . . . . . . . . (am1, B) (am2, B) · · · (amm, B) where for each entry a of A, (a, B) is the n × n matrix (a, B) = (a, b11) (a, b12) · · · (a, b1n) (a, b21) (a, b22) · · · (a, b2n) . . . . . . . . . (a, bn1) (a, bn2) · · · (a, bnn) . 14.1.21 Example As an illustration of this Kronecker product construction, for m = 2, n = 3 let A = 0 1 1 0, B = 0 1 2 1 2 0 2 0 1 . Then the Kronecker product construction using A and B yields the following 6×6 square whose elements are ordered pairs: 00 01 02 10 11 12 01 02 00 11 12 10 02 00 01 12 10 11 10 11 12 00 01 02 11 12 10 01 02 00 12 10 11 02 00 01 . 14.1.22 Lemma If H and K are latin squares of orders n1 and n2, then H ⊗K is a latin square of order n1n2. 14.1.23 Lemma If H1 and H2 are orthogonal latin squares of order n1 and K1 and K2 are orthogonal latin squares of order n2, then H1 ⊗K1 and H2 ⊗K2 are orthogonal latin squares of order n1n2. 14.1.24 Corollary If there is a pair of MOLS of order n and a pair of MOLS of order m, then there is a pair of MOLS of order mn. 14.1.25 Theorem If n = q1 · · · qr, where the qi are distinct prime powers with q1 < · · · < qr, then N(n) ≥q1 −1. 14.1.26 Remark In 1922, MacNeish conjectured that N(n) = q1 −1. This has been shown to be false for all non-prime power values of n ≤62; it is in fact conjectured in that this conjecture is false at all values of n other than 6 and prime powers. Combinatorial 553 14.1.3 Frequency squares 14.1.27 Deﬁnition Let n = λm. An F(n; λ) frequency square is an n × n square based upon m distinct symbols so that each of the m symbols occurs exactly λ times in each row and column. Thus an F(n; 1) frequency square is a latin square of order n. Two F(n; λ) frequency squares are orthogonal if when one square is placed on top of the other, each of the m2 possible distinct ordered pairs occurs exactly λ2 times . A set of such squares is orthogonal if any two distinct squares are orthogonal. Such a set of mutually orthogonal squares is a set of MOFS. 14.1.28 Theorem The maximum number of MOFS of the form F(n; λ) is bounded above by (n −1)2/(m −1). 14.1.29 Theorem If q is a prime power and i ≥1 is an integer, a complete set of (qi−1)2/(q− 1), F(qi; qi−1) MOFS can be constructed using the linear polynomials a1x1 + · · · + a2ix2i over the ﬁeld Fq where 1. The vector (a1, . . . , ai) ̸= (0, . . . , 0), 2. The vector (ai+1, . . . , a2i) ̸= (0, . . . , 0), 3. The vector (a′ 1, . . . , a′ 2i) ̸= e(a1, . . . , a2i) for any e ̸= 0 ∈Fq. 14.1.4 Hypercubes 14.1.30 Deﬁnition A d-dimensional hypercube of order n is an n × · · · × n array with nd points based on n distinct symbols with the property that if any single coordinate is ﬁxed, each of the n symbols occurs exactly nd−2 times in that subarray. Such a hypercube is of type j, 0 ≤j ≤d −1 if whenever any j of the coordinates are ﬁxed, each of the n symbols appears nd−j−1 times in that subarray. Note that the deﬁnition implies that a hypercube of type j is also of types 0, 1, . . . , j −1. 14.1.31 Deﬁnition Two hypercubes are orthogonal if, when superimposed, each of the n2 ordered pairs appears nd−2 times. Again the d = 2 case reduces to that of latin squares. A set of t ≥2 hypercubes is orthogonal if every pair of distinct hypercubes is orthogonal. 14.1.32 Theorem The maximum number of mutually orthogonal hypercubes of order n ≥2, dimension d ≥2, and ﬁxed type j with 0 ≤j ≤d −1 is bounded above by 1 n −1 nd −1 − d 1 (n −1) − d 2 (n −1)2 −· · · − d j (n −1)j . 14.1.33 Corollary The maximum number of order n, dimension d, and type 1 hypercubes is bounded above by Nd(n) ≤nd −1 n −1 −d. 14.1.34 Remark In the case that d = 2, Nd(n) reduces to the familiar bound of n −1 for sets of MOLS of order n. As was the case for d = 2, the bound for d > 2 can always be realized when n is a prime power. 14.1.35 Corollary There are at most (n −1)d−1 + d d −1 (n −1)d−2 + · · · + d j + 1 (n −1)j 554 Handbook of Finite Fields cube cube cube cube cube cube cube cube cube cube 1 2 3 4 5 6 7 8 9 10 012 012 012 012 000 000 012 012 012 012 120 201 012 012 111 111 120 201 201 120 201 120 012 012 222 222 201 120 120 201 012 012 120 201 111 222 120 201 120 201 120 201 120 201 222 000 201 120 012 012 201 120 120 201 000 111 012 012 201 120 012 012 201 120 222 111 201 120 201 120 120 201 201 120 000 222 012 012 120 201 201 120 201 120 111 000 120 201 012 012 x + y 2x + y y + z y + 2z x + z x + 2z x + y 2x + y 2x + y x + y +z +2z +z +2z Figure 14.1.1 A complete set of mutually orthogonal cubes of order 3. hypercubes of order n, type j, and dimension d. 14.1.36 Theorem The polynomials a1x1 + · · · + adxd with 1. the elements ai ∈Fq for i = 1, . . . , d with at least j + 1 of the ai ̸= 0, 2. and (a′ 1, . . . , a′ d) ̸= e(a1, . . . , ad) for any e ̸= 0 ∈Fq, represent a complete set of mutually orthogonal hypercubes of dimension d, order q, and type j. 14.1.37 Remark In another deﬁnition of orthogonality for hypercubes, called equi-orthogonality is studied. In [992, 993] sets of hypercubes using various other deﬁnitions of orthogonality are considered. Such stronger deﬁnitions of orthogonality turn out to be useful in the study of MDS codes (see Section 15.1). In one deﬁnition, not only does one keep track of the total number of times that ordered pairs occur, but their locations are also taken into account. In other deﬁnitions, one studies various notions of orthogonality involving more than the usual two hypercubes at a time. In all of these deﬁnitions, polyno-mials over ﬁnite ﬁelds are used to construct complete sets of such orthogonal hypercubes of prime power orders. 14.1.38 Remark In sets of very general mutually orthogonal frequency hyperrectangles of prime power orders are constructed using linear polynomials over ﬁnite ﬁelds. 14.1.5 Connections to aﬃne and projective planes 14.1.39 Remark Aﬃne and projective planes are discussed in Section 14.3. We ﬁrst state the following fundamental result; and then discuss a few other related results. 14.1.40 Theorem , [706, Theorem III.3.20] There exists a projective plane (or an aﬃne plane) of order n if and only if there exists a complete set of MOLS of order n. 14.1.41 Deﬁnition Two complete sets of MOLS of order n are isomorphic if after permuting the rows, permuting the columns with a possibly diﬀerent permutation, and permuting the Combinatorial 555 symbols with a third possibly diﬀerent permutation of each square of the ﬁrst set, we obtain the second set of MOLS. See Part III of for further discussion of non-isomorphic sets of MOLS, aﬃne, and projective planes. 14.1.42 Conjecture If p is a prime, any two complete sets of MOLS of order p are isomorphic. 14.1.43 Remark The above conjecture is only known to be true for p = 3, 5, 7. Truth of the conjec-ture would imply that all planes of prime order are Desarguesian. 14.1.44 Theorem [2917, 2918] For q = pn, let 0 ≤k < n, N = (q −1)/(q −1, pk −1) and set e = q −N. Let u be a primitive N-th root of unity in Fq. Assume that xpk + cix is a permutation polynomial for e elements c1, . . . , ce ∈Fq, where one can assume that c1 −1 = c2. Let a ̸= 0 and c1 be such that f(x) = axpk + c1x is an orthomorphism of Fq (so f is a permutation polynomial with f(0) = 0, and f(x) −x is also a permutation). Let di = c1 −ci. Then the polynomials aujxpk +c1x+y, j = 1, . . . , N; dix+y, i = 3, . . . , e; x+y represent a complete set of q −1 MOLS of order q. 14.1.45 Corollary For each n ≥2 and any odd prime p, the above construction gives τ(n) ≥2, non-isomorphic complete sets of MOLS of order pn, where τ(n) denotes the number of positive divisors of n. 14.1.46 Example For any odd prime p, this construction gives an example of a non-Desarguesian aﬃne translation plane of order p2, constructed without the use of a right quasiﬁeld as used in . 14.1.47 Remark For q = 9, let F9 be generated by the primitive polynomial f(x) = x2 + 2x + 2 over F3. Let α be a root of f(x). The Desarguesian plane of order 9 may be constructed by using the polynomials αix + y, i = 0, . . . , 7. Since u = α2 is a primitive 4-th root of unity, the construction from the above corollary leads to the polynomials αx3 +y, α3x3 +y, α5x3 + y, α7x3+y which represent four MOLS of order 9. To extend these four MOLS to a complete set of 8 MOLS of order 9, we consider the polynomials x + y, α2x + y, α4x + y, α6x + y. Thus four of the latin squares are the same in both the Desarguesian and non-Desarguesian constructions. 14.1.6 Other ﬁnite ﬁeld constructions for MOLS 14.1.48 Remark There are other ﬁnite ﬁeld constructions for sets of MOLS; here we brieﬂy allude to a few of them which are described in much more detail in . Quasi-diﬀerence matrices and V (m, t) vectors are discussed in Section VI.17.4; self-orthogonal latin squares are considered in Section III.5.6; MOLS with holes are considered in Section III.1.7; starters are studied in VI.55.; and atomic latin squares are studied in Section III.1.6. See Also §14.3 Discusses aﬃne and projective planes. §14.5 Discusses block designs. Part III discusses latin squares. Part III, Section 3 discusses sets of MOLS. Discusses topics in discrete mathematics with topics motivated by latin squares. 556 Handbook of Finite Fields References Cited: [355, 397, 398, 706, 818, 992, 993, 1456, 1875, 1876, 1988, 2053, 2155, 2173, 2177, 2737, 2917, 2918] 14.2 Lacunary polynomials over ﬁnite ﬁelds Simeon Ball, Universitat Polit ecnica de Catalunya Aart Blokhuis, Eindhoven University of Technology 14.2.1 Introduction 14.2.1 Remark In 1970 R´ edei published his treatise L¨ uckenhafte Polynome ¨ uber endlichen K¨ orpern , soon followed by the English translation Lacunary Polynomials over Finite Fields , the title of this chapter. One of the important applications of his theory is to give information about the following two sets. 14.2.2 Deﬁnition For f : Fq →Fq, or f ∈Fq[X] deﬁne the set of directions (slopes of secants of the graph): D(f) := f(x) −f(y) x −y \| x ̸= y ∈Fq . 14.2.3 Deﬁnition For f ∈Fq[X] let P(f) := {m ∈Fq \| f(X) + mX is a permutation polynomial}. 14.2.4 Remark The sets P(f) and D(f) partition Fq. If (f(x) −f(y))/(x −y) = m then the polynomial f(x) + mx = f(y) + my, so m is a direction determined by f precisely when f(X) + mX is not a permutation polynomial (on Fq). 14.2.2 Lacunary polynomials 14.2.5 Deﬁnition Let K be a (ﬁnite) ﬁeld. A polynomial f ∈K[x] is fully reducible if K is a splitting ﬁeld for f, that is, if f factors completely into linear factors in K[X]. 14.2.6 Deﬁnition Denote by f ◦the degree of f, and by f ◦◦the second degree, the degree of the polynomial we obtain by removing the leading term. 14.2.7 Deﬁnition If f ◦◦< f ◦−1 then f is lacunary and the diﬀerence f ◦−f ◦◦is the gap. 14.2.8 Remark We want to survey what is known about lacunary polynomials (with a large gap) that are fully reducible. In many applications however the gap is not between the degree and the second degree, so instead of being of the form f(X) = Xn + h(X), where h◦≤n −2, it Combinatorial 557 is of the more general form f(X) = g(X)Xn +h(X), where h◦≤n−2, for some polynomial g. 14.2.9 Example For d \| (q −1) the ﬁeld K = Fq contains the d-th roots of unity, so the polynomial Xd −ad is fully reducible. 14.2.10 Remark In many applications the degree f ◦= q, as is the case in the following examples. 14.2.11 Example The lacunary polynomials Xq + c, Xq −X, and if q is odd then Xq ± X(q+1)/2 and Xq ± 2X(q+1)/2 + X, are fully reducible in Fq[X]. 14.2.12 Theorem Let f(X) = Xp + g(X), with g◦= f ◦◦< p, be fully reducible in Fp[X], p prime. Then either g is constant, or g = −X or g◦is at least (p + 1)/2. 14.2.13 Remark Let s(X) be the zeros polynomial of f, that is the polynomial with the same set of zeros as f, but each with multiplicity one. So s = gcd(f, Xp −X). It follows that s \| f −(Xp −X) = X + g. We may write f = s · r, where r is the fully reducible polynomial that has the zeroes of f with multiplicity one less. Hence r divides the derivative f ′ = g′. So we conclude that f = s · r \| (X + g)g′. If the right hand side is zero, then either g = −X, corresponding to the fully reducible polynomial f(X) = Xq −X, or g′ = 0 which (since g◦< p) implies g(X) = c for some c ∈K and f(X) = Xp +c = (X +c)p. If the right hand side is nonzero, then, being divisible by f, it has degree at least p, so g◦+ g◦−1 ≥p which gives g◦≥(p + 1)/2. 14.2.14 Remark In the next section we see how this result can be applied to obtain information about the number of directions determined by a function. 14.2.3 Directions and R´ edei polynomials 14.2.15 Deﬁnition Let AG(2, q) be the Desarguesian aﬃne plane of order q, where points of AG(2, q) are denoted by pairs (a, b), a, b ∈Fq. 14.2.16 Deﬁnition Let PG(2, q) be the Desarguesian projective plane of order q with homogeneous point coordinates (a : b : c) and line coordinates [u : v : w]. The point (a : b : c) is incident with the line [u : v : w] precisely when au + bv + cw = 0. The equation of the line [u : v : w] is then uX + vY + wZ = 0. 14.2.17 Remark We consider AG(2, q) as part of the projective plane PG(2, q) where [0 : 0 : 1] is the line at inﬁnity, the line with equation Z = 0. The aﬃne point (a, b) corresponds to the projective point (a : b : 1). 14.2.18 Deﬁnition Let u = (u1, u2) and v = (v1, v2) be two aﬃne points. The pair u, v determines the direction m if the line joining them has slope m, or equivalently, if (u2 −v2)/(u1 − v1) = m. 558 Handbook of Finite Fields 14.2.19 Deﬁnition Let R be a set of q points in AG(2, q). We deﬁne DR ⊆Fq ∪{∞} to be the set of directions determined by the pairs of points in R. 14.2.20 Remark The reason we take R to have size q is two-fold. Firstly, in R´ edei’s formulation of the problem R is the graph of a function f and DR = Df. Secondly, any set with more than q points determines all directions, by the pigeon hole principle: there are exactly q lines in every parallel class, so if \|R\| > q, then there is a line with at least two points of R in each parallel class. For results concerning the case \|R\| < q, see . 14.2.21 Deﬁnition With R we associate its R´ edei polynomial F(U, W) = Y (a,b)∈R (W + aU + b). 14.2.22 Lemma If the direction m ̸∈DR then F(m, W) = W q −W. 14.2.23 Lemma If the direction m ∈DR, then F(m, W) is a fully reducible lacunary polynomial of degree q, and second degree at most \|DR\| −1. 14.2.24 Theorem [322, Theorem 1] Let R be a set of q points in AG(2, q), and let N = \|DR\|. Then either N = 1, or N ≥(q + 3)/2, or 2 + (q −1)/(pe + 1) ≤N ≤(q −1)/(pe −1) for some e, 1 ≤e ≤⌊n/2⌋. 14.2.4 Sets of points determining few directions 14.2.25 Remark The third case in Theorem 14.2.24, 2 + (q −1)/(pe + 1) ≤N ≤(q −1)/(pe −1) for some e satisfying 1 ≤e ≤⌊n/2⌋, is not sharp. The following are some examples of functions that determine few directions. 14.2.26 Example The function f(X) = X(q+1)/2, where q is odd, determines (q + 3)/2 directions. 14.2.27 Example The function f(X) = Xs, where s = pe is the order of a subﬁeld of Fq, determines (q −1)/(s −1) directions. 14.2.28 Example The function f(X) = TrFq/Fs(X), the trace from Fq to the subﬁeld Fs, determines (q/s) + 1 directions. 14.2.29 Example If f(X) ∈Fq[Xs], where s is the order of a subﬁeld of Fq and is chosen maximal with this property, in other words, f is Fs-linear (apart from the constant term) but not linear over a larger subﬁeld, then (q/s) + 1 ≤N ≤(q −1)/(s −1). 14.2.30 Remark Motivated by the form of the examples the following theorem was obtained (in a number of steps) by Ball, Blokhuis, Brouwer, Storme, and Sz˝ onyi. Initial results are in , then the classiﬁcation was all but obtained in , and completed in . 14.2.31 Theorem If, for f : Fq →Fq, with f(0) = 0, the number N = \|D(f)\| > 1 of directions determined by f is less than (q + 3)/2, then for a subﬁeld Fs of Fq q s + 1 ≤N ≤q −1 s −1, and if s > 2 then f is Fs-linear. 14.2.32 Remark This result is obtained using several lemmas about fully reducible lacunary poly-nomials which are of independent interest. Combinatorial 559 14.2.33 Lemma [2445, Satz 18] Let s = pe be a power of p with 1 ≤s < q. If Xq/s + g(X) ∈Fq[X] \ Fq[Xp] is fully reducible over Fq then either s = 1 and g(X) = −X or g◦≥((q/s) + 1)/(s + 1). 14.2.34 Lemma Let s be a power of p with 1 ≤s < q and suppose that Xq/s + g(X) ∈Fq[X] \ Fq[Xp] is fully reducible over Fq. If s > 2, g◦= q/s2 and 2(g′)◦< g◦then Xq/s + g(X) is Fs-linear. 14.2.35 Remark Theorem 14.2.31 completely characterizes the case in which the number of direc-tions is small, that is less than (q + 3)/2. In the case that q = p is prime, N < (p + 3)/2 implies N = 1, and the characterization of N = (p + 3)/2 directions was given by Lov´ asz and Schrijver . 14.2.36 Theorem If f ∈Fp[X], p prime, determines (p + 3)/2 directions, then f(X) = X(p+1)/2 up to aﬃne equivalence. 14.2.37 Remark Much more can be said in this case, the following surprising theorem by G´ acs shows that there is a huge gap in the spectrum of possible number of directions. 14.2.38 Theorem If the number of directions determined by f ∈Fp[X], p prime, is more than (p + 3)/2, then it is at least 2 3(p −1) + 1. 14.2.39 Remark This bound is almost tight, there are examples that determine 2 3(p−1)+2 directions if p ≡1 (mod 3). Progress was made using G´ acs’ approach in indicating that a further gap is possible from 2p/3 to 3p/4. If there is an example with less than 3p/4 directions then lines meet the graph of f in at most 3 points or at least p/4. Futhermore, if there are 3 lines meeting the graph of f in more than 3 points then the graph of f is contained in these 3 lines. There are examples that determine 3 4(p −1) + 2 directions if p ≡1 (mod 4) and some constructions where \|D(f)\| ≈7p/9 can be found in . 14.2.40 Remark For results concerning the case q = p2, see . For related results on functions f : Fk q →Fq, with k ≥2, that determine few directions, see , and for results on functions f, g: Fq →Fq, where the set P(f, g) = {(r, s) ∈F2 q \| X + rf(X) + sg(X) is a permutation polynomial} is large . 14.2.5 Lacunary polynomials and blocking sets 14.2.41 Remark Let R be a subset of AG(2, q) of size q. The set of points of PG(2, q) B = {(a : b : 1) \| (a, b) ∈R} ∪{(1 : m : 0) \| m ∈DR} has the property that every line of PG(2, q) intersects B. 14.2.42 Deﬁnition A blocking set of PG(2, q) is a set of points B of PG(2, q) with the property that every line of is incident with a point of B. 560 Handbook of Finite Fields 14.2.43 Lemma A blocking set of PG(2, q) has at least q + 1 points and equality can only be obtained if these points all are on a line. 14.2.44 Deﬁnition A blocking set of PG(2, q) that contains a line is trivial. 14.2.45 Remark We tacitly assume that all blocking sets under consideration are minimal, so they do not contain a proper subset that is also a blocking set. For blocking sets of non-Desarguesian planes and for further reading on blocking sets see [320, 329, 430, 432, 433, 1150, 1153] and for more recent references, see Remark 14.2.54. 14.2.46 Lemma Suppose that B is a blocking set of size q + k + 1 and that (1 : 0 : 0) ∈B and assume that the line with equation Z = 0, that is [0 : 0 : 1] is a tangent to B. Then the non-horizontal lines [1 : u : v] are blocked by the aﬃne points of B and the R´ edei polynomial of the aﬃne part of B can be written as F(V, W) = (V q −V )G(V, W) + (W q −W)H(V, W), where G and H are of total degree k in the variables V and W. 14.2.47 Lemma Let F0 denote the part of F that is homogeneous of degree q + k, and let G0 and H0 be the parts of G and H that are homogeneous of total degree k. Restricting to the terms of total degree q + k we obtain the homogeneous equation F0 = V qG0 + W qH0, with F0(V, W) = Y (a:b:1)∈B (bV + W). Writing f(W) = F0(1, W) and deﬁning g and h analogously, we obtain a one-variable fully reducible lacunary polynomial in Fq[W], f(W) = g(W) + W qh(W). 14.2.48 Lemma Let f ∈Fq[X] be fully reducible, and suppose that f(X) = Xqg(X) + h(X), where g and h have no common factor. Let k be the maximum of the degrees of g and h. Then k = 0, or k = 1 and f(X) = a(Xq −X) for some a ∈F∗ q, or q is prime and k ≥(q + 1)/2, or q is a square and k ≥√q, or q = p2e+1 for some prime p and k ≥pe+1. 14.2.49 Theorem A non-trivial blocking set B in PG(2, q), q square, has at least q + √q + 1 points. If equality holds then B consists of the points of a subplane of order √q. 14.2.50 Theorem A non-trivial blocking set B in PG(2, q), q = p2e+1, p prime, q ̸= p, has at least q + pe+1 + 1 points. This bound is sharp only in the case e = 1. 14.2.51 Theorem A non-trivial blocking set B in PG(2, p), p prime, has at least 3 2(p + 1) points. If equality holds then every point of B is on precisely 1 2(p −1) tangents. 14.2.52 Remark The bound in Theorem 14.2.51 was conjectured in . 14.2.53 Remark The proof of Lemma 14.2.48 leads to the following divisibility condition f\|(Xg + h)(h′g −g′h). It would be good (and probably not infeasible) to characterize the case of equality in the case p is prime, that is ﬁnd all f, g, and h with f of degree q + (q + 1)/2, g and h of degree Combinatorial 561 at most (q + 1)/2 and f ˆ =(Xg + h)(h′g −g′h), where aˆ =b means there exists a scalar c ∈Fq such that a = cb. This is the subject of the next section. 14.2.54 Remark Blocking sets in PG(2, pn), p prime, of size less than 3(pn+1)/2 have been classiﬁed for n = 2 and n = 3 and they come from the construction in Remark 14.2.41. However, for n ≥4, there are examples known which are not of this form. These examples, called linear blocking sets, include those obtained by the construction in Remark 14.2.41. It is conjectured that all small blocking sets are linear blocking sets. More precisely, we have the following conjecture which is called the linearity conjecture. For recent articles concerning this conjecture see [1872, 1977, 1978, 2408, 2755, 2757]. 14.2.55 Conjecture If B is a blocking set in PG(2, pn), p prime, of less than 3(pn + 1)/2 points then there exists an n-dimensional subspace U of PG(3n −1, p) with the property that every point of B, when viewed as an (n −1)-dimensional subspace of PG(3n −1, p), has non-trivial intersection with U. 14.2.6 Lacunary polynomials and blocking sets in planes of prime order 14.2.56 Remark The blocking set problem in PG(2, p), p prime, leads one to search for polynomials f(X), g(X), h(X), where f = Xpg + h factors completely into linear factors and g and h have degree at most 1 2(p + 1). More precisely, given a blocking set B of size 3 2(p + 1), for each point P ∈B, and each tangent ℓpassing through P, there is a polynomial f with the above property. A factor of f of multiplicity e corresponds to a line incident with P distinct from ℓmeeting B in e + 1 points. 14.2.57 Remark The equation f ˆ =(Xg + h)(h′g −g′h) has several inﬁnite families of solutions, and some sporadic ones, not all of them necessarily corresponding to blocking sets. 14.2.58 Theorem The following list contains all non-equivalent solutions for f = Xpg + h, where f factors completely into linear factors and g and h have degree at most 1 2(p + 1), for p < 41. 1. (For odd p, say p = 2r + 1.) Take f(X) = X Q(X −a)3 where the product is over the nonzero squares a. Then f satisﬁes f(X) = X(Xr −1)3 = Xpg + h with g(X) = Xr −3, h(X) = 3Xr+1 −X. This would correspond to line intersections of the lines incident with P (with frequencies written as exponents) 1r, 22, 4r. For p = 7 this is the function for the blocking set {(1 : 0 : 0), (0 : 1 : 0), (0 : 0 : 1)} ∪{(a : b : 1) \| a, b ∈{1, 2, 4}}. 2. (For p = 4t + 1.) Take f(X) = X Q(X −a) Q(X −b)4 where the product is over the nonzero squares a and fourth powers b. Here f(X) = X(X2t −1)(Xt −1)4 = Xpg + h with g(X) = X2t −4Xt + 5 and h(X) = −5X2t+1 + 4Xt+1 −X. This would correspond to line intersections 12t, 2t+2, 6t. 3. (For p = 4t+1.) Take f(X) = Xt+1 Q(X−a) Q(X−b)2 where the product is over the nonzero squares a and fourth powers b. Here f(X) = Xt+1(X2t−1)(Xt−1)2 = Xpg +h with g(X) = Xt −2 and h(X) = 2X2t+1 −Xt+1. This would correspond to line intersections 12t, 2t, 4t (t + 2)2. For p = 13 this is a function for the blocking set {(1 : 0 : 0), (0 : 1 : 0), (0 : 0 : 1)} ∪{(1 : a : 0), (0 : 1 : a), (a : 0 : 1) \| a3 = −1} ∪{(b : c : 1) \| b3 = c3 = 1}. 4. (For p = 13.) Take f(X) = X Q(X −a)4 Q(X −1 2a) where the product is over the values a with a3 = 1. Here f(X) = X(X3 −1)4(X3 −1 8) = Xpg + h with g(X) = X3 + 4 and h(X) = 5X7 −5X4 −5X. This corresponds to line 562 Handbook of Finite Fields intersections 16, 24, 54, and indeed occurs. 5. Take f(X) = Xp −X(p+1)/2 = X(p+1)/2 Q(X −a) where the product is over the nonzero squares a. 6. Take f(X) = Xp −2X(p+1)/2 + X = X Q(X −a)2 where the product is over the nonzero squares a. 14.2.59 Remark These lacunary polynomials are just weighted subsets of the projective line, so equivalence means that f = Xpg + h is equivalent to those polynomials obtained under the maps f(X) 7→(cX +d)3(p+1)/2f((aX +b)/(cX +d)), for some a, b, c, d ∈Fp, where ad ̸= bc. 14.2.60 Theorem Let B be a non-trivial blocking set in PG(2, p) of size 3 2(p + 1), where p is a prime less than 41. Then either there is a line incident with (p + 3)/2 points of B (and hence is the example characterized in Theorem 14.2.36) or p ∈{7, 13} and there is a unique other example in both cases. 14.2.61 Conjecture The restriction p < 41 is unnecessary in the above theorem. 14.2.7 Lacunary polynomials and multiple blocking sets 14.2.62 Deﬁnition A t-fold blocking set B of PG(2, q) is a set of points such that every line is incident with at least t points of B. 14.2.63 Theorem Let B be a t-fold blocking set in PG(2, q), q = ph, p prime, of size t(q+1)+c. Let c2 = c3 = 2−1/3 and cp = 1 for p > 3. 1. If q = p2d+1 and t < q/2 −cp q2/3/2, then c ≥cp q2/3, unless t = 1 in which case B, with \|B\| < q + 1 + cp q2/3, contains a line. 2. If 4 < q is a square, t < q1/4/2 and c < cp q2/3, then c ≥t√q and B contains the union of t disjoint Baer subplanes, except for t = 1 in which case B contains a line or a Baer subplane. 3. If q = p2, p prime, and t < q1/4/2 and c < p l 1 4 + q p+1 2 m , then c ≥t√q and B contains the union of t disjoint Baer subplanes, except for t = 1 in which case B contains a line or a Baer subplane. 14.2.64 Remark For more precise results in the case t = 2 see ; for t = 3 see ; for q = p3 see [2416, 2417, 2418]; for q = p6n+3 see ; and for q = p6n see [328, 2418]. 14.2.65 Remark The proof of Theorem 14.2.63 starts with the main theorem of on fully reducible lacunary polynomials. 14.2.66 Theorem Let f ∈Fq[X], q = pn, p prime, be fully reducible, f(X) = Xqg(X)+h(X), where (g, h) = 1. Let k = max(g◦, h◦) < q. Let e be maximal such that f is a pe-th power. Then we have one of the following: 1. e = n and k = 0; 2. e ≥2n/3 and k ≥pe; 3. 2n/3 > e > n/2 and k ≥pn−e/2 −(3/2)pn−e; 4. e = n/2 and k = pe and f(X) = aTr (bX + c) + d or f(X) = a Norm(bX + c) + d for suitable constants a, b, c, d. Here Tr and Norm respectively denote the trace and norm function from Fq to F√q; 5. e = n/2 and k ≥pe l 1 4 + p (pe + 1)/2 m ; Combinatorial 563 6. n/2 > e > n/3 and k ≥pn/2+e/2 −pn−e −pe/2, or if 3e = n + 1 and p ≤3, then k ≥pe(pe + 1)/2; 7. n/3 ≥e > 0 and k ≥pe⌈(pn−e + 1)/(pe + 1)⌉; 8. e = 0 and k ≥(q + 1)/2; 9. e = 0, k = 1 and f(X) = a(Xq −X). 14.2.67 Remark Lacunary polynomials over ﬁnite ﬁelds and in particular Redei’s theorem, The-orem 14.2.12, and Blokhuis’ theorem, Theorem 14.2.51, have also been used in algebra, algebraic number theory, group theory, and group factorization. For a survey of these ap-plications, see . Polynomials over ﬁnite ﬁelds have been used to tackle a variety of problems associated with incidence geometries. Various extensions of the ideas ﬁrst used for lacunary polynomials have been studied. This has led to some interesting techniques involving ﬁeld extensions, algebraic curves which in turn have led to classiﬁcation, non-existence, and stability results concerning subsets of points of a ﬁnite projective spaces with a certain given property. For a recent survey, see . See Also Chapter 8 For more on permutation polynomials over ﬁnite ﬁelds. §14.3 For more on aﬃne and projective planes over ﬁnite ﬁelds. §14.4 For more on higher dimensional spaces over ﬁnite ﬁelds. §14.9 For more on polynomials over ﬁnite ﬁelds with restricted weights. References Cited: [184, 185, 186, 187, 188, 190, 191, 319, 320, 321, 322, 323, 328, 329, 330, 358, 430, 432, 433, 831, 1150, 1151, 1152, 1153, 1423, 1872, 1961, 1977, 1978, 2408, 2416, 2417, 2418, 2444, 2445, 2755, 2756, 2757, 2758] 14.3 Aﬃne and projective planes Gary Ebert, University of Delaware Leo Storme, Universiteit Gent All structures in this section are ﬁnite. Reference is an excellent introduction to projective and aﬃne planes. See Section VII.2 of for a concise description of the Hall, Andr´ e, Hughes, and Figueroa planes. 14.3.1 Projective planes 14.3.1 Deﬁnition A ﬁnite projective plane is a ﬁnite incidence structure of points and lines such that 1. every two distinct points together lie on a unique line; 2. every two distinct lines meet in a unique point; 3. there exists a quadrangle (four points with no three collinear). 564 Handbook of Finite Fields 14.3.2 Remark If π is a ﬁnite projective plane, then there is a positive integer n such that any line of π has exactly n + 1 points, every point lies on exactly n + 1 lines, the total number of points is n2 + n + 1, and the total number of lines is n2 + n + 1. This number n is called the order of π. 14.3.3 Construction The classical examples of ﬁnite projective planes are constructed as follows. Let V be a 3-dimensional vector space over the ﬁnite ﬁeld Fq of order q. Take as points the 1-dimensional subspaces of V and as lines the 2-dimensional subspaces of V , and let incidence be given by containment. The resulting incidence structure is a ﬁnite projective plane of order q, denoted by PG(2, q). These projective planes are Desarguesian since they satisfy the classical conﬁgurational theorem of Desargues (for instance, see ). Note that this construction shows that there exists a ﬁnite projective plane of order q for any prime power q. Alternatively, one may use homogeneous coordinates (x : y : z) = {(fx, fy, fz): f ∈ Fq{0}} for the points of PG(2, q), and [a : b : c] = {[fa, fb, fc]: f ∈Fq{0}} for the lines of PG(2, q), where the point (x : y : z) is incident with the line [a : b : c] if and only if ax + by + cz = 0. 14.3.4 Remark Some non-classical ﬁnite projective planes are discussed in Subsections 14.3.3 to 14.3.5. Many other constructions can be found in . One of the most diﬃcult problems in ﬁnite geometry is determining the spectrum of possible orders for ﬁnite projective planes. All known examples have prime power order, but it is unknown if this must be true in general. The Bruck-Ryser-Chowla Theorem (Section 14.5) excludes an inﬁnite number of positive integers as possible orders. In addition, order 10 has been excluded via a computer search . There are precisely four diﬀerent (non-isomorphic, as deﬁned in Subsection 14.3.3) projective planes of order 9, the smallest order for which non-classical examples exist . An overview of the state of knowledge concerning small projective planes follows: order n 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 existence y y y y n y y y n y ? y n ? y number 1 1 1 1 0 1 1 4 0 ≥1 ? ≥1 0 ? ≥22 14.3.2 Aﬃne planes 14.3.5 Deﬁnition A ﬁnite aﬃne plane is a ﬁnite incidence structure of points and lines such that 1. any two distinct points together lie on a unique line; 2. for any point P and any line ℓnot containing P, there exists a unique line m through P that has no point in common with ℓ(the “parallel axiom”); 3. there exists a triangle (three points not on a common line). 14.3.6 Remark If one deﬁnes a parallelism on the lines of an aﬃne plane by saying that two lines are parallel if they are equal or have no point in common, then parallelism is an equivalence relation whose equivalence classes are called parallel classes. Each parallel class of lines is a partition of the point set, and every line belongs to exactly one parallel class. 14.3.7 Remark If one removes a line ℓtogether with all its points from a projective plane π, then one obtains an aﬃne plane π0 = πℓ. Two lines of the aﬃne plane πℓare parallel if and only if the projective lines containing them meet the line ℓin the same point. We call ℓ the line at inﬁnity of π0, and the points of ℓare called the points at inﬁnity. Conversely, Combinatorial 565 to construct a projective plane from an aﬃne plane π0, create a new point for each parallel class of π0 and adjoin this new point to each line in that parallel class. Also adjoin a new line that contains all the new points and no other points. The resulting incidence structure is a projective plane π, called the projective completion of π0. The order of π0 is the order of its projective completion. 14.3.8 Construction The classical way to construct ﬁnite aﬃne planes is as follows. Take as points the ordered pairs (a, b), with a, b ∈Fq, and as lines the sets of points (x, y) satisfying an equation of the form Y = mX + b for some m, b ∈Fq or an equation of the form X = c for some c ∈Fq. The resulting structure is an aﬃne plane of order q, denoted by AG(2, q). Such an aﬃne plane is also Desarguesian since the projective completion of AG(2, q) is (isomorphic to) PG(2, q). Alternatively, AG(2, q) may be constructed from a 2-dimensional vector space V over Fq by taking as points all vectors in V and as lines all cosets of 1-dimensional subspaces, where incidence is then given by containment. 14.3.3 Translation planes and spreads 14.3.9 Deﬁnition Let π be a projective plane. A collineation (automorphism) of π is a bijective map φ on the point set of π that preserves collinearity. All collineations of π form the automorphism group Aut(π) of π under composition of maps. A collineation group of π is any subgroup of Aut(π). Two projective planes are isomorphic if there is a bijective map from the point set of one plane to the point set of the other plane that sends collinear points to collinear points. 14.3.10 Deﬁnition If φ is a collineation of a projective plane π, and φ ﬁxes all lines through a point P and all points on a line ℓ, then φ is a (P, ℓ)-perspectivity. In particular, it is a (P, ℓ)-elation if P ∈ℓ. 14.3.11 Deﬁnition A projective plane π is (P, ℓ)-transitive if for any distinct points A, B not on ℓ and collinear with P (A ̸= P ̸= B), there is a (P, ℓ)-perspectivity φ in Aut(π) such that Aφ = B. Similarly, π is (m, ℓ)-transitive if it is (P, ℓ)-transitive for all points P on m. If π is (ℓ, ℓ)-transitive for some line ℓ, then ℓis a translation line of π and π is a translation plane with respect to ℓ. 14.3.12 Remark If π is a translation plane with respect to a line ℓ, then the aﬃne plane πℓ= π \ ℓ is also a translation plane. Most often a translation plane is considered an aﬃne plane, with its line at inﬁnity the translation line. The translation group of such an aﬃne plane is the group of all (ℓ, ℓ)-elations, which acts sharply transitively on the points of the aﬃne plane πℓ. References [279, 1613] provide extensive information on translation planes. 14.3.13 Remark Translation planes are coordinatized by algebraic structures called quasiﬁelds (see Section 2.1). Every quasiﬁeld has an algebraic substructure called its kernel, which in the ﬁnite setting is necessarily a ﬁnite ﬁeld. The quasiﬁeld is then a ﬁnite dimensional vector space over its kernel, and the dimension of the translation plane is the dimension of this vector space. 566 Handbook of Finite Fields 14.3.14 Deﬁnition Let Σ = PG(2t + 1, q) be a (2t + 1)-dimensional projective space for some non-negative integer t (see Section 14.4 for the deﬁnition of projective space). A spread of Σ is a set S of t-subspaces of Σ such that any point of Σ belongs to exactly one element of S. The set-wise stabilizer of S in Aut(Σ) is the automorphism group Aut(S) of the spread. 14.3.15 Construction View the ﬁnite ﬁeld F = Fq2t+2 as a (2t+2)-dimensional vector space V over its subﬁeld Fq, and let Σ = PG(2t + 1, q) be the associated (2t + 1)-dimensional projective space. If θ is a primitive element of F and L = Fqt+1 is the subﬁeld of order qt+1, then for each positive integer i, θiL is a (t + 1)-dimensional vector subspace of V that represents a t-subspace of Σ. Moreover, S = {L, θL, θ2L, . . . , θqt+1L} is a spread of Σ. The spreads obtained in this way are regular as deﬁned below. 14.3.16 Deﬁnition A t-regulus of PG(2t + 1, q) is a set R of q + 1 mutually disjoint t-subspaces such that any line intersecting three elements of R intersects all elements of R. These lines are the transversals of R. 14.3.17 Proposition [1515, p. 200] Any three mutually skew t-subspaces of the projective space PG(2t + 1, q) determine a unique t-regulus containing them. 14.3.18 Remark The points covered by a 1-regulus R in PG(3, q) are the points of a hyperbolic quadric. The transversals to R form another 1-regulus covering the same hyperbolic quadric. This 1-regulus is the opposite regulus Ropp to R. 14.3.19 Deﬁnition Let q > 2 be a prime power. A spread S in PG(2t + 1, q) is regular if for any three elements of S, the t-regulus determined by them is contained in S. (See for an alternative deﬁnition valid for q = 2.) 14.3.20 Construction (Bruck-Bose ) Let Σ ∼ = PG(2t + 1, q) be a hyperplane of Σ = PG(2t + 2, q), for some integer t ≥0, and let S be a spread of Σ. Deﬁne A(S) to be the geometry whose points are the points of Σ\Σ, and whose lines are the (t + 1)-subspaces of Σ that intersect Σ precisely in an element of S. 14.3.21 Theorem The structure A(S) is an aﬃne translation plane of order qt+1 which is at most (t+1)-dimensional over its kernel. Conversely, any ﬁnite aﬃne translation plane can be constructed in this way for an appropriate choice of t. In particular, every ﬁnite translation plane has prime power order. In addition, A(S) is isomorphic to AG(2, qt+1) if and only if S is regular. 14.3.22 Remark The automorphism group of an aﬃne translation plane A(S) is isomorphic to the semidirect product of the translation group with the group of all nonsingular semilinear mappings of the underlying vector space which ﬁx the spread S . The aﬃne translation plane A(S) is completed to a projective plane P(S) by adding the members of the spread S as the points at inﬁnity. Projective planes P(S1) and P(S2) are isomorphic if and only if there is a collineation of Σ mapping S1 to S2 . 14.3.23 Remark Let t = 1 above. Replacing a regulus R by its opposite regulus Ropp in a regular spread S0 produces a new spread S which is not regular, provided q > 2. The associated planes A(S) are the Hall planes. Simultaneously replacing mutually disjoint reguli by their opposite reguli in a regular spread S0 produces a subregular spread S, whose associated translation planes A(S) are also called subregular. If the set of mutually disjoint reguli in S0 Combinatorial 567 is “linear” in a well-deﬁned way , then the resulting subregular planes are the Andr´ e planes which are two-dimensional over their kernels. Thus Hall planes are Andr´ e planes, but not necessarily vice versa. 14.3.24 Remark In , a method is given for obtaining a spread of PG(3, q2) from a spread of PG(3, q) for every odd prime power q. 14.3.4 Nest planes 14.3.25 Deﬁnition Let S0 be a regular spread of PG(3, q). A nest N in S0 is a set of reguli in S0 such that every line of S0 belongs to 0 or 2 reguli of N. Thus a nest is a 2-cover of the lines of S0 which are contained in the nest. 14.3.26 Remark Counting arguments show that the number t of reguli in a nest must satisfy (q + 3)/2 ≤t ≤2(q −1). In particular, we note that q must be odd for nests to exist. If a nest contains t reguli, it is a t-nest. If U denotes the t(q + 1)/2 lines of S0 contained in the reguli of a t-nest N, there is a natural potential replacement set for U. Namely, if (q + 1)/2 lines can be found in the opposite regulus to each regulus of N such that the resulting set W of t(q + 1)/2 lines are mutually disjoint, then S = (S0 \ U) ∪W is a non-regular spread of PG(3, q) and hence A(S) is a non-Desarguesian translation plane. In this case, the nest N is replaceable, and the resulting plane A(S) is a nest plane. 14.3.27 Deﬁnition An inversive plane is a 3 −(n2 + 1, n + 1, 1) design (see Section 14.5), for some integer n ≥2. That is, an inversive plane is an incidence structure of n2 + 1 points and n(n2 + 1) blocks, each block of size n + 1, such that every three points lie in a unique block. The blocks are the circles of the inversive plane. 14.3.28 Construction Let q be any prime power. Take as points the elements of Fq2 together with the symbol ∞. Take as circles the images of Fq ∪{∞} under the nonsingular linear fractional mappings on Fq2, with the usual conventions on ∞. If incidence is given by containment, this produces an inversive plane with q2 + 1 points whose circles have size q + 1. This inversive plane is Miquelian because it satisﬁes the classical conﬁgurational result of Miquel, and is denoted by M(q) . 14.3.29 Theorem There is a one-to-one correspondence between the points and circles of M(q), and the lines and reguli of a regular spread of PG(3, q). There is an associated homomorphism from the stabilizer of the regular spread to the automorphism group of M(q), whose kernel is a cyclic group of order q + 1. 14.3.30 Remark Using the above correspondence, it is usually easier to search for nests in M(q) rather than directly in a regular spread S0 of PG(3, q). Such nests can often be constructed by taking the orbit of some carefully chosen “base” circle under a natural cyclic or elemen-tary abelian subgroup of Aut(M(q)). However, to check if the resulting nest is replaceable, one must pull back to S0 and work in PG(3, q). Some nests are replaceable and some are not. Computations involving ﬁnite ﬁeld arithmetic lead to the following results. 14.3.31 Theorem [172, 173, 949, 950, 2374] For any odd prime power q ≥5, there exist replaceable t-nests for t = q −1, q, q + 1, 2(q −1). The resulting spreads determine non-Desarguesian translation planes of order q2 which are two-dimensional over their kernels. 14.3.32 Remark The nesting technique for constructing two-dimensional translation planes is quite robust. In addition to the above examples, replaceable t-nests have been constructed for 568 Handbook of Finite Fields many values of t in the range 3(q + 1)/4 −√q/2 ≤t ≤3(q + 1)/4 + √q/2; see . Moreover, the translation planes associated with nests often can be characterized by the action of certain collineation groups [1609, 1612, 1614, 1615]. 14.3.33 Remark Circle geometries and the notion of subregularity can be extended to higher dimen-sions. Using algebraic pencils of Sherk surfaces, in several inﬁnite families of non-Andr´ e subregular translation planes are constructed which are 3-dimensional over their kernels. Proofs use intricate ﬁnite ﬁeld computations involving the trace, norm, and bitrace. 14.3.5 Flag-transitive aﬃne planes 14.3.34 Deﬁnition An aﬃne plane is ﬂag-transitive if it admits a collineation group which acts transitively on incident point-line pairs. 14.3.35 Remark A straightforward counting argument shows that transitivity on lines implies tran-sitivity on ﬂags for aﬃne planes. 14.3.36 Remark By a celebrated result of Wagner , every ﬁnite ﬂag-transitive aﬃne plane is necessarily a translation plane, and hence arises from a spread S of PG(2t + 1, q), for some positive integer t, according to Theorem 14.3.21. The aﬃne plane A(S) is ﬂag-transitive if and only if the spread S admits a transitive collineation group. 14.3.37 Construction Let F = Fq2t+2 be treated as a (2t + 2)-dimensional vector space over its subﬁeld Fq, thus serving as the underlying vector space for Σ = PG(2t + 1, q). If θ is a primitive element of F, the collineation θ induced by multiplication by θ is a Singer cycle of Σ; that is, the cyclic group ⟨θ⟩acts sharply transitively on the points and hyperplanes of Σ. If G denotes the Singer subgroup of order qt+1 + 1, let O denote the partition of the points of Σ into (qt+1 −1)/(q −1) G-orbits of size qt+1 + 1 each. As shown in , these point orbits are caps when t is odd (see Section 14.4 for the deﬁnition of a cap). For future reference we let H denote the index two subgroup of G. 14.3.38 Example Let q be an odd prime power, and let t be an odd integer. Using the above model, choose b ∈F such that bqt+1−1 = −1. Let σ : F →F via σ : x 7→xq, and let E denote the subﬁeld of F whose order is qt+1. Then A1 = {x + bxσ : x ∈E} represents a t-space Γ1 of Σ that meets half the G-orbits of O in two points each (from diﬀerent H-orbits) and is disjoint from the rest. Similarly, A2 = {bx + bσ+1xσ : x ∈E} represents a t-space Γ2 of Σ that meets the G-orbits of O which are disjoint from Γ1 in two points each (from diﬀerent H-orbits). Moreover, S = ΓH 1 ∪ΓH 2 is a spread of Σ admitting a transitive collineation group, which yields a non-Desarguesian ﬂag-transitive aﬃne plane A(S) of order qt+1 with Fq in its kernel. 14.3.39 Example [1674, 1681] Let q be an odd prime power, and let t be an even integer. Using the notation of Example 14.3.38, Γ1 now meets every G-orbit of O in one point each, and hence S1 = ΓG 1 is a spread of Σ which admits a transitive (cyclic) collineation group. The resulting aﬃne plane A(S1) is a non-Desarguesian ﬂag-transitive aﬃne plane of order qt+1 with Fq in its kernel. If q ≡1 (mod 4), then Γ2 also meets each G-orbit of O in one point each, and these points lie in H-orbits that are disjoint from Γ1. Moreover, S2 = ΓH 1 ∪ΓH 2 is a spread of Σ admitting a transitive (non-cyclic) collineation group, thereby yielding another non-Desarguesian ﬂag-transitive aﬃne plane A(S2) of order qt+1 with Fq in its kernel. This plane does not admit a cyclic collineation group acting transitively on the line at inﬁnity. For q ≡3 (mod 4), one may obtain such a spread S2 by replacing Γ2 with the t-space of Σ represented by {µxqt+1 + µbqt+1(xσ)qt+1 : x ∈E}, where µ = θ(qt+1−1)/(q−1). Combinatorial 569 14.3.40 Remark The ﬁeld automorphism σ in the above examples may be replaced by any element of Gal(Fq2t+2/Fq). The resulting planes are non-Desarguesian provided σ does not induce the identity map on the subﬁeld E. Lower bounds are given in [1674, 1681] for the number of mutually non-isomorphic planes obtained as b and σ vary. 14.3.41 Example Let q be a power of 2, and let t ≥2 be an even integer. Using the notation of Example 14.3.38, let Tr denote the trace from E to Fq, and choose some element r ∈Fq2\Fq. Let Γ′ be the t-space of Σ represented by {Tr (x)+rx : x ∈E}. Then Γ′ meets every G-orbit of O in one point each, and hence S′ = (Γ′)G is a spread of Σ which admits a transitive (cyclic) collineation group. The resulting ﬂag-transitive aﬃne plane A(S′) of order qt+1 with Fq in its kernel is non-Desarguesian provided qt+1 > 8. 14.3.42 Remark Other than the Hering plane of order 27 and the L¨ uneburg planes of order 22d for odd d ≥3, all known ﬁnite ﬂag-transitive aﬃne planes arise from spreads consisting of a single G-orbit or the union of two H-orbits, where G and H are the Singer subgroups deﬁned in Construction 14.3.37. It is shown in that if q = pe for some odd prime p and some positive integer e and if gcd((qt+1 + 1)/2, (t + 1)e) = 1, then any ﬂag-transitive aﬃne plane of order qt+1 with Fq in its kernel (other than the above Hering plane) must arise in this way. More can be said for t = 1, 2. 14.3.43 Theorem If q = pe is an odd prime power such that gcd((q2 + 1)/2, e) = 1, then any two-dimensional ﬂag-transitive aﬃne plane of order q2 is isomorphic to one of the planes constructed in Example 14.3.38 with t = 1. The number of such isomorphism classes can be determined by M¨ obius inversion. For e = 1 (hence q = p prime), the above gcd condition is necessarily satisﬁed and the number of isomorphism classes is precisely (q −1)/2. 14.3.44 Theorem [169, 176] If q = pe is an odd prime power such that gcd((q3 + 1)/2, 3e) = 1, then any three-dimensional ﬂag-transitive aﬃne plane of order q3, other than Hering’s plane of order 27, is isomorphic to one of the planes constructed in Example 14.3.39 with t = 2. For e = 1 (hence q = p prime), the number of isomorphism classes of each type arising from Example 14.3.39 is precisely (q −1)/2. 14.3.45 Problem For q even, the classiﬁcation and complete enumeration of ﬁnite ﬂag-transitive aﬃne planes of dimension two or three over their kernel remains an open problem. The only known two-dimensional examples are the L¨ uneburg planes. 14.3.46 Problem The classiﬁcation of ﬁnite ﬂag-transitive aﬃne planes is one of the few open cases in the program announced in to classify all ﬁnite ﬂag-transitive linear spaces. For arbitrary dimension over the kernel, it is not known if there exist examples of ﬁnite ﬂag-transitive aﬃne planes other than the ones listed above, and the classiﬁcation seems to be quite diﬃcult. In the projective setting, it is believed that the only ﬂag-transitive projective plane is the Desarguesian one, although this remains an open problem. 14.3.6 Subplanes 14.3.47 Deﬁnition Let π be a projective plane with point set P and line set L. A projective plane π′ with point set P′ and line set L′ is a subplane of π if P′ ⊆P and L′ ⊆L, and π′ inherits its incidence relation from π. 14.3.48 Theorem Let π be a ﬁnite projective plane of order n, and let π′ be a subplane of π with order m < n. Then n = m2 or m2 + m ≤n. 14.3.49 Remark It is unknown whether equality can hold in the above inequality; if so, this would imply that the order n = m2+m of π is not a prime power. The case n = m2 is of particular 570 Handbook of Finite Fields interest. In this case, every point of π \ π′ is incident with a unique line of π′, and dually every line of π \ π′ is incident with a unique point of π′. 14.3.50 Deﬁnition A subplane π′ of order m in a projective plane π of order n = m2 is a Baer subplane of π. 14.3.51 Remark In the classical setting, the lattice of subplanes follows directly from the lattice of subﬁelds. Namely, if q = pe for some prime p and some positive integer e, then the subplanes of PG(2, q), up to isomorphism, are precisely PG(2, pk) as k varies over all positive divisors of e. So PG(2, q) has a Baer subplane if and only if q is a square. Moreover, one can easily count the number of subplanes of a given order in this classical (Desarguesian) setting. 14.3.52 Theorem [1510, Lemma 4.20] If q is any prime power and n ≥2 is any integer, then the number of subplanes of order q in PG(2, qn), all of which are isomorphic to PG(2, q), is q3(n−1)(q3n −1)(q2n −1) (q3 −1)(q2 −1) . In particular, the number of Baer subplanes in PG(2, q2) is q3(q3 + 1)(q2 + 1). 14.3.53 Remark It is currently unknown if PG(2, q2) has the greatest number of Baer subplanes among all projective planes of order q2. No counter-examples have been found. Amazingly, there are aﬃne planes of order q2 which contain more aﬃne subplanes of order q than does AG(2, q2) . 14.3.54 Deﬁnition A Baer subplane partition, or BSP for short, of PG(2, q2) is a partition of the points of PG(2, q2) into subplanes, each isomorphic to PG(2, q). 14.3.55 Example Consider the full Singer group of order q4 +q2 +1 acting sharply transitively on the points and lines of π = PG(2, q2). Then the orbits under the Singer subgroup of order q2 + q + 1 are Baer subplanes, and the orbit of any one of these Baer subplanes under the complementary Singer subgroup of order q2−q+1 forms a BSP of π. Such a BSP is classical. 14.3.56 Remark It is shown in that any spread of PG(5, q) admitting a linear cyclic sharply transitive action corresponds to a “perfect” BSP of PG(2, q2), and this spread is regular if and only if the BSP is classical. By deﬁnition, a BSP is perfect if and only if it is an orbit of some Baer subplane under an appropriate Singer subgroup, although the Baer subplane itself need not be a point orbit under a Singer subgroup. Examples 14.3.39 and 14.3.41 for t = 2 yield the following result. 14.3.57 Theorem Let q ̸= 2 be a prime power. Then there exist non-classical BSPs of PG(2, q2). 14.3.58 Remark Relatively little is known about the subplane structure of non-Desarguesian planes. There is no known example of a square order projective plane which has been shown not to contain a Baer subplane. However, it is not known if every square order projective plane must contain a Baer subplane. At the other extreme, the Hall planes, the Hughes planes, the Figueroa planes, and many two-dimensional subregular translation planes have been proven to contain subplanes of order two (that is, Fano subplanes). It has been conjectured that every ﬁnite non-Desarguesian plane must contain a subplane of order two. More surprisingly, it is shown in that the Hughes plane of order q2 (q odd) has a subplane of order 3 when q ≡2 (mod 3). Extensive, but not exhaustive, computer searches for small q have found no subplanes of order 3 in this plane when q ≡1 (mod 3). Very recently , subplanes of order 3 have been proven to exist in all odd order Figueroa planes. Combinatorial 571 14.3.7 Embedded unitals 14.3.59 Remark Reference provides an excellent introduction to the topic of unitals. Proofs and precise statements of most results in this subsection may be found in the above reference. 14.3.60 Deﬁnition A unital is a 2−(n3 + 1, n + 1, 1) design for some integer n ≥3 (that is, a geometry having n3 +1 points, with n+1 points on each line such that any two distinct points are on exactly one line). 14.3.61 Remark Here the interest is not in unitals as designs, but in unitals embedded in a projective plane of order n2. The lines (blocks) of the unital are then the lines of the ambient projective plane which meet the unital in more than one point (and hence in n + 1 points). 14.3.62 Example Let PG(2, q2) be represented using homogeneous coordinates. Then the points (x : y : z) for which xxq + yyq + zzq = 0 form a unital. This unital is a Hermitian curve. 14.3.63 Construction (Buekenhout ) Using the Bruck-Bose representation of Construc-tion 14.3.20, with t = 1 for a 2-dimensional translation plane, let S be any spread of Σ and let U be an ovoidal cone of Σ (that is, the point cone over some 3-dimensional ovoid as deﬁned in Section 14.4) that meets Σ in a line of S. Then U corresponds to a unital U in P(S) which is tangent to the line at inﬁnity. Also, if U is a nonsingular (parabolic) quadric in Σ that meets Σ in a regulus of the spread S, then U corresponds to a unital U in P(S) which meets the line at inﬁnity in q + 1 points. Of course, the second construction is valid only for those 2-dimensional translation planes of order q2 whose associated spread contains at least one regulus. 14.3.64 Remark If the ovoidal cone above is an orthogonal cone (with an elliptic quadric as base), the resulting unital in P(S) is an orthogonal Buekenhout unital. Unitals embedded in P(S) which arise from the nonsingular quadric construction are nonsingular Buekenhout unitals. 14.3.65 Remark Orthogonal Buekenhout unitals embedded in PG(2, q2) have been completely enu-merated. In particular, if q is an odd prime, then the number of mutually inequivalent or-thogonal Buekenhout unitals in PG(2, q2) is 1 2(q + 1), one of which is the Hermitian curve. The only nonsingular Buekenhout unital embedded in PG(2, q2) is the Hermitian curve. Exhaustive computer searches in [195, 2382] show that there are precisely two inequivalent unitals embedded in each of PG(2, 9) and PG(2, 16), the Hermitian curve and one other orthogonal Buekenhout unital. The enumeration of orthogonal and nonsingular Buekenhout unitals in various non-Desarguesian translation planes may be found in [179, 180]. For in-stance, if q ≥5 is a prime, then, up to equivalence, the Hall plane of order q2 has 1 2(q + 1) nonsingular Buekenhout unitals and 1 + ⌊3q 4 ⌋orthogonal Buekenhout unitals. 14.3.66 Remark In , it is shown that the Hall planes contain unitals which are not obtainable from any Buekenhout construction. This is the only inﬁnite family of unitals embedded in translation planes which has been proven to be non-Buekenhout. There are also square-order non-translation planes which are known to contain unitals, necessarily non-Buekenhout. For instance, the Hughes planes of order q2 are known to contain unitals for all odd prime powers q [2482, 2947]. In , the Figueroa planes of order q6 are shown to contain unitals for any prime power q. In fact, there is no known example of a square-order projective plane which has been shown not to contain a unital. 572 Handbook of Finite Fields 14.3.8 Maximal arcs 14.3.67 Remark Proofs of almost all results in this subsection may be found in Chapter 12 of . 14.3.68 Deﬁnition A {k; r}-arc in PG(2, q) is a set K of k points such that r is the maximum number of points in K that are collinear. A {k; 2}-arc is a k-arc. 14.3.69 Theorem Let K be a {k; r}-arc in PG(2, q). Then k ≤(q + 1)(r −1) + 1. 14.3.70 Deﬁnition The {k; r}-arcs in PG(2, q) with k = (q +1)(r −1)+1 are maximal {k; r}-arcs. 14.3.71 Example Singleton points (r = 1), the whole plane (r = q + 1), and the complement of a line (r = q) are trivial maximal {k; r}-arcs. The {q + 2; 2}-arcs in PG(2, q) for q even, also called hyperovals (see Section VII.2.9 of ), are examples of non-trivial maximal {k; r}-arcs, and have been objects of intense interest for many years. 14.3.72 Lemma If K is a non-trivial maximal {k; r}-arc in PG(2, q), then r must be a (proper) divisor of q. 14.3.73 Theorem If PG(2, q) contains a non-trivial maximal {k; r}-arc, then q must be even. 14.3.74 Construction Let X2 + βX + 1 be an irreducible quadratic polynomial over Fq, q even. Consider the algebraic pencil in PG(2, q) consisting of the conics Cλ : X2 0 + βX0X1 + X2 1 + λX2 2 = 0 for λ ∈Fq ∪{∞}. Let A be an additive subgroup of (Fq, +) of order r, and let K be the set of points which is the union of the conics Cλ for λ ∈A. Then K is a maximal {k; r}-arc of PG(2, q). 14.3.75 Construction Let q be even, and let Tr denote the absolute trace from Fq to F2. In PG(2, q), consider a set C consisting of conics Cα,β,λ : αX2 0 +X0X1+βX2 1 +λX2 2 = 0, where α, β ∈Fq with Tr (αβ) = 1 and λ ∈Fq ∪{∞}. Deﬁne the “composition” of two distinct conics from C in the following way: Cα,β,λ ⊕Cα′,β′,λ′ = Cα⊕α′,β⊕β′,λ⊕λ′, where α ⊕α′ = αλ + α′λ′ λ + λ′ , β ⊕β′ = βλ + β′λ′ λ + λ′ , λ ⊕λ′ = λ + λ′. Let F ⊂C be a set of 2d−1 non-singular conics with common nucleus (0, 0, 1), which is closed under the composition of distinct conics. Then the points of the conics in F, together with (0, 0, 1), form a maximal {k; 2d}-arc in PG(2, q). To obtain one such set F, assume q = 24m+2 and let ϵ ∈F24m+2 be such that Tr (ϵ) = 1. Let A = {x ∈F24m+2 : x2 + x ∈F22m+1}, and let r(λ) = λ3 + ϵ for all λ ∈A. Then \|A\| = 22m+2 and F = {C1,r(λ),λ : λ ∈A \ {0}} is such a subset of C which determines, as indicated above, a maximal {k; 22m+2}-arc in PG(2, 24m+2). These arcs do not arise from Construction 14.3.74. Other possibilities for F may be found in [1062, 1406, 1407]. Combinatorial 573 14.3.9 Other results 14.3.76 Remark Semiﬁelds (see Section 2.1) are algebraic structures that may be used to coor-dinatize certain translation planes, called semiﬁeld planes. These are the only translation planes which are also dual translation planes. Many new examples have recently been found. Chapter 6 of is an excellent source for many of these new developments. 14.3.77 Problem As previously mentioned, all known ﬁnite projective (and aﬃne) planes have prime power order, although it is certainly unclear whether this must be true in general. However, it is now known that if a projective plane of order n admits an abelian collineation group of order n2 or n2 −n, then n must be a prime power [326, 1634]. An equally important open problem is whether any ﬁnite projective (or aﬃne) plane of prime order p must be Desarguesian. This appears to be a very diﬃcult problem; the smallest open case is p = 11. 14.3.78 Remark A hyperbolic ﬁbration of PG(3, q) is a collection of q −1 hyperbolic quadrics and two lines that partition the points of PG(3, q). By selecting one of the two reguli ruling each hyperbolic quadric in the ﬁbration, one obtains 2q−1 spreads of PG(3, q), which in turn give rise to 2q−1 translation planes of order q2 which are at most two-dimensional over their kernels. Although there may be some isomorphisms among these planes, this is a very ro-bust method for constructing two-dimensional translation planes (see for isomorphism counts). An easy example of a hyperbolic ﬁbration is the hyperbolic pencil, which is an al-gebraic pencil of quadrics of the appropriate types. Other examples of hyperbolic ﬁbrations may be found in [170, 178]. All known hyperbolic ﬁbrations have the property that the two lines in the ﬁbration form a conjugate (skew) pair with respect to each of the polarities associated with the q −1 hyperbolic quadrics (such hyperbolic ﬁbrations are called regular in the literature), and also have the property that all q −1 hyperbolic quadrics intersect one of the two skew lines in the same pair of conjugate points with respect to the quadratic ex-tension Fq2 of Fq (such hyperbolic ﬁbrations are said to agree on one of the two skew lines). In , it is shown that all hyperbolic ﬁbrations are necessarily regular if q is even (the problem is still open for q odd), and it is also shown for any q that a hyperbolic ﬁbration which agrees on one of its two skew lines is necessarily regular. In , it is shown that there is a bijection between regular hyperbolic ﬁbrations which agree on one of their two lines and ﬂocks of a quadratic cone, once a conic of the ﬂock is speciﬁed. This further leads to a correspondence with normalized q-clans and certain types of generalized quadrangles. 14.3.79 Remark There are other survey articles on substructures in projective planes. The sec-tion on Finite Geometry in The Handbook of Combinatorial Designs and the survey article state the main results on arcs, {k; r}-arcs, caps, unitals, and blocking sets in PG(2, q), where exact deﬁnitions, tables, and supplementary results are provided. In addition, the collected work contains a great variety of results on substructures in PG(2, q), techniques for investigating these substructures, and important open problems in this area. The linearity conjecture on small minimal blocking sets in PG(2, q) is one of the most important such open problems (see Chapter 3 of for an explicit statement of this conjecture). Proving this conjecture would imply several new results on various substruc-tures in PG(2, q) as well as in PG(n, q), for n > 2. In particular, this would include new results on maximal partial spreads, minihypers, extendability of linear codes, tight sets, and Cameron-Liebler line classes. The investigation of maximal arcs in PG(2, q) for q even, inspired by the new construction method of Mathon described in Construction 14.3.75, and the investigation of embedded unitals as discussed in Section 14.3.7 are central problems on substructures in projective planes which also merit further research. 574 Handbook of Finite Fields See Also §2.1 For a discussion of traces, norms, and linearized polynomials. §9.4 For some semiﬁeld constructions. §9.5 For some examples of semiﬁeld planes constructed from planar functions. §13.3 For a discussion of the classical projective groups over ﬁnite ﬁelds. §14.4 For a discussion of projective spaces of higher dimensions. Develops the notion of inversive planes. Develops the notion of coordinatizing non-Desarguesian projective planes. References Cited: [104, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 189, 195, 205, 279, 326, 422, 425, 426, 427, 428, 452, 454, 480, 481, 555, 706, 756, 785, 788, 807, 820, 915, 948, 949, 950, 951, 1062, 1099, 1406, 1407, 1488, 1508, 1510, 1513, 1515, 1560, 1609, 1612, 1613, 1614, 1615, 1634, 1674, 1675, 1681, 1836, 1837, 1979, 1980, 2024, 2374, 2382, 2482, 2889, 2947] 14.4 Projective spaces James W.P. Hirschfeld, University of Sussex Joseph A. Thas, Ghent University 14.4.1 Projective and aﬃne spaces 14.4.1 Deﬁnition Let V = V (n + 1, F), with n ≥1, be an (n + 1)-dimensional vector space over the ﬁeld F with zero element 0. Consider the equivalence relation on the elements of V {0} whose equivalence classes are the one-dimensional subspaces of V with the zero deleted. Thus, if X, Y ∈V {0}, then X is equivalent to Y if Y = tX for some t in F0 = F{0}. 1. The set of equivalence classes is the n-dimensional projective space over F and is denoted by PG(n, F) or, when F = Fq, by PG(n, q). 2. The elements of PG(n, F) are points; the equivalence class of the vector X is the point P(X). The vector X is a coordinate vector for P(X) or X is a vector representing P(X). In this case, tX with t in F0 also represents P(X); that is, by deﬁnition, P(tX) = P(X). 3. If X = (x0, . . . , xn) for some basis, then the xi are the coordinates of the point P(X). 4. The points P(X1), . . . , P(Xr) are linearly independent if a set of vectors X1, . . . , Xr representing them is linearly independent. Combinatorial 575 14.4.2 Deﬁnition 1. For any m = −1, 0, 1, 2, . . . , n, a subspace of dimension m, or m-space, of PG(n, F) is a set of points all of whose representing vectors form, together with the zero, a subspace of dimension m + 1 of V = V (n + 1, F); it is denoted by Πm. 2. A subspace of dimension zero is a point; a subspace of dimension −1 is the empty set. A subspace of dimension one is a line, of dimension two is a plane, of dimension three is a solid. A subspace of dimension n −1 is a hyperplane. A subspace of dimension n −r is a subspace of codimension r. 14.4.3 Deﬁnition 1. The set of m-spaces of PG(n, F) is denoted PG(m)(n, F) or, when F = Fq, by PG(m)(n, q). 2. For r, s, m, n ∈N, let (a) θ(n, q) = (qn+1 −1)/(q −1), also denoted by θ(n); (b) \|PG(m)(n, q)\| = φ(m; n, q); (c) [r, s]−= Qs i=r(qi −1), for s ≥r. 14.4.4 Theorem [1510, Chapter 3] For n ≥1, m ≥0, and q any prime power, 1. \|PG(n, q)\| = θ(n, q); 2. φ(m; n, q) = [n −m + 1, n + 1]−/[1, m + 1]−. 14.4.5 Theorem [1510, Chapter 2] 1. A hyperplane is the set of points P(X) whose vectors X = (x0, . . . , xn) satisfy a linear equation u0x0 + u1x1 + · · · + unxn = 0, with U = (u0, . . . , un) in F n+1{(0, . . . , 0)}; it is denoted π(U) = Πn−1. 2. An m-space Πm is the set of points whose representing vectors X = (x0, . . . , xn) satisfy the equations XA = 0, where A is an (n + 1) × (n −m) matrix of rank n −m with coeﬃcients in F. 14.4.6 Remark [1510, Chapter 2] The vector U in the theorem is a coordinate vector of the hyper-plane; the ui are hyperplane or tangential coordinates. 14.4.7 Deﬁnition 1. If a point P lies in a subspace Πm, then P is incident with Πm or, equally well, Πm is incident with P. 2. If Πr and Πs are subspaces of PG(n, F), then the meet or intersection of Πr and Πs, written Πr ∩Πs, is the set of points common to Πr and Πs; it is also a subspace. 3. The join of Πr and Πs, written ΠrΠs, is the smallest subspace containing Πr and Πs. 14.4.8 Theorem [1510, Chapter 2] Subspaces have the following properties. 576 Handbook of Finite Fields 1. If Πr and Π′ r are both r-spaces in PG(n, F) and Π′ r ⊂Πr, then Π′ r = Πr. 2. (Grassmann Identity) If Πr ∩Πs = Πt and ΠrΠs = Πm, then r + s = m + t. 3. A subspace Πm is the join of m + 1 linearly independent points; it is also the intersection of n −m linearly independent hyperplanes. 4. Equivalently, the set of all representing vectors of the points of Πm, together with the zero vector, is the intersection of n −m hyperplanes of the vector space V , which deﬁne n −m linearly independent vectors U = (u0, . . . , un). 14.4.9 Theorem (The principle of duality) [1510, Chapter 2] For any space S = PG(n, F), there is a dual space S∗, whose points and hyperplanes are respectively the hyperplanes and points of S. For any theorem true in S, there is an equivalent theorem true in S∗. In particular, if T is a theorem in S stated in terms of points, hyperplanes, and incidence, the same theorem is true in S∗and gives a dual theorem T∗in S by substituting “hyperplane” for “point” and “point” for “hyperplane.” Thus “join” and “meet” are dual. Hence the dual of an r-space in PG(n, F) is an (n −r −1)-space. 14.4.10 Remark For small dimensions, in PG(2, F), point and line are dual; in PG(3, F), point and plane are dual, whereas the dual of a line is a line. 14.4.11 Deﬁnition 1. If H∞is any hyperplane in PG(n, F), then AG(n, F) = PG(n, F)\H∞is an aﬃne space of n dimensions over F. When F = Fq, write AG(n, F) = AG(n, q). 2. The subspaces of AG(n, F) are the subspaces of PG(n, F), apart from H∞, with the points of H∞deleted in each case. 3. This hyperplane H∞is the hyperplane at inﬁnity of AG(n, F). 14.4.2 Collineations, correlations, and coordinate frames 14.4.12 Deﬁnition 1. If S and S′ are two spaces PG(n, F), n ≥2, then a collineation α : S − →S′ is a bijection which preserves incidence; that is, if Πr ⊂Πs, then Πα r ⊂Πα s . 2. It is suﬃcient that α is a bijection such that, if Π0 ⊂Π1, then Πα 0 ⊂Πα 1 . 3. When n = 1, consider the lines S and S′ embedded in planes over F; then a collineation α : S − →S′ is a transformation induced by a collineation of the planes; that is, if S0 and S′ 0 are planes with S ⊂S0 and S′ ⊂S′ 0, and α0 : S0 − →S′ 0 is a collineation mapping S onto S′, then let α be the restriction of α0 to S. 14.4.13 Deﬁnition A projectivity α : S − →S′ is a bijection given by a matrix T, necessarily non-singular, such that P(X′) = P(X)α if and only if tX′ = XT, where t ∈F0. Write α = M(T); then α = M(λT) for any λ in F0. 14.4.14 Remark A projectivity is a collineation. Mostly the case to be considered is when S = S′. Combinatorial 577 14.4.15 Deﬁnition With respect to a ﬁxed basis of V (n+1, F), an automorphism σ of F deﬁnes an automorphic collineation σ of S = PG(n, F); in coordinates, this is given by P(X)σ = P(Xσ), where Xσ = (xσ 0, xσ 1, . . . , xσ n). 14.4.16 Theorem (The fundamental theorem of projective geometry) [1510, Chapter 2] 1. If α′ : S − →S′ is a collineation, then α′ = σα, where σ is an automorphic collineation, given by a ﬁeld automorphism σ, and α is a projectivity. In partic-ular, if K = Fq with q = ph, p prime, and P(X′) = P(X)α′, then there exists m in {1, 2, . . . , h}, tij ∈F for i, j ∈{0, 1, . . . , n}, and t ∈F0 such that tX′ = XpmT, where Xpm = (xpm 0 , . . . , xpm n ), T = (tij); that is, tx′ i = xpm 0 t0i + · · · + xpm n tni, for i = 0, 1, . . . , n. 2. If {P0, . . . , Pn+1} and {P ′ 0, . . . , P ′ n+1} are both subsets of PG(n, F) of cardinality n+2 such that no n+1 points chosen from the same set lie in a hyperplane, then there exists a unique projectivity α such that P ′ i = P α i , for all i ∈{0, 1, . . . , n+1}. 14.4.17 Remark There are cases where Theorem 14.4.16 simpliﬁes. 1. For n = 1, there is a unique projectivity transforming any three distinct points on a line to any other three. 2. When F = F2, it suﬃces to give the images of P0, . . . , Pn to determine a projec-tivity. For n = 1, the images of two points determine the projectivity. 14.4.18 Remark Part 2 of Theorem 14.4.16 emphasizes a diﬀerence between the spaces V (n+1, F) and PG(n, F). In the former, linear transformations are determined by the images of n + 1 vectors; in the latter, projectivities are determined by the images of n + 2 points. 14.4.19 Deﬁnition Let {P0, . . . , Pn+1} be any set of n + 2 points in PG(n, F), no n + 1 in a hyperplane. If P is any other point of the space, then a coordinate vector for P is determined in the following manner. Let Pi be represented by the vector Xi for some vector Xi in V (n+1, F). For any given t in F0 there exist ai in F for all i ∈{0, 1, . . . , n} such that tXn+1 = a0X0 + · · · + anXn. So, for any t, the ratios ai/aj remain ﬁxed. Thus, if P is any point with P = P(X), then X = t0a0X0 + · · · + tnanXn. Hence, with respect to {P0, . . . , Pn+1}, the point P is given by (t0, . . . , tn) where the ti are determined up to a common factor. Then {P0, . . . , Pn} is the simplex of reference and Pn+1 the unit point. Together the n + 2 points form a (coordinate) frame. 578 Handbook of Finite Fields 14.4.20 Remark In V (n + 1, F), a basis is a set of n + 1 linearly independent vectors and, in PG(n, F), a frame is a set of n + 2 points, no n + 1 in a hyperplane; that is, every set of n + 1 points is linearly independent. 14.4.21 Theorem [1510, Chapter 2] Again from Theorem 14.4.16, if two coordinate frames are given by the vectors X = (x0, . . . , xn) and Y = (y0, . . . , yn), then a change from one frame to the other is given by Y = XA, where A is an (n + 1) × (n + 1) non-singular matrix over F. If a projectivity α in the one frame is given by X′ = XT, then, since Y ′ = X′A, in the other frame it is given by Y ′ = X′A = XTA = Y A−1TA. 14.4.22 Deﬁnition Let S be a space PG(n, F) and S′ its dual space superimposed on S; that is, the points of S′ are the hyperplanes of S and the hyperplanes of S′ are the points of S. Consider a function α : S − →S′. If α is a collineation, it is a correlation of S and induces a collineation, also named α, of S′ to S; that is, as the points of S are transformed to hyperplanes, then hyperplanes are transformed to points since α preserves incidence. If α is a projectivity, then it is a reciprocity of S. In either case, if α is involutory, that is α2 = 1, where 1 is the identity, then α is a polarity of S. 14.4.23 Remark If P and P ′ are points and π is a hyperplane such that P α = π and πα = P ′, then, in a polarity, P = P ′. 14.4.24 Deﬁnition Let AG(n, F) = PG(n, F)\H∞be an aﬃne space over F. Then, in a given coordinate frame where H∞has equation x0 = 0, a point of AG(n, F) can be written P(1, x1, . . . , xn) and hence as (x1, . . . , xn). So the points of AG(n, F) are the elements of V (n, F). The xi are the aﬃne or non-homogeneous coordinates of the given point. 14.4.25 Remark It is assumed that, for any AG(n, F), the coordinate frame is this one. 14.4.26 Theorem [1510, Chapter 2] 1. The subspaces of AG(n, F) have the form X + S, where X is any vector and S is any subspace of V (n, F). 2. Three points X, Y, Z of AG(n, F) are collinear if and only if there exists λ in F{0, 1} such that X = λY + (1 −λ)Z. 14.4.27 Theorem [1510, Chapter 2] 1. If Fqn = Fq(β), then the map X = (x1, . . . , xn) 7→x = x1 + x2β + · · · + xnβn−1 gives a bijection between AG(n, q) and Fqn. 2. Distinct points X, Y, Z in AG(n, F) are collinear if and only if, in Fqn, (x −y)q−1 = (x −z)q−1. 14.4.3 Polarities 14.4.28 Theorem [1510, Chapter 2] Suppose that α is a correlation of PG(n, F); then it is the product of an automorphic collineation σ and a projectivity of PG(n, F) to its dual space given by the matrix T. Then α is a polarity if and only if σ2 = 1 and T σT ∗−1 = tI, Combinatorial 579 where T ∗denotes the transpose of T and t ∈F0. 14.4.29 Theorem [1510, Chapter 2] If α is a polarity of PG(n, F), then, with σ and T as in Theorem 14.4.28, there are the following possibilities. 1. If σ = 1 and char F ̸= 2, then T = T ∗or T = −T ∗. 2. If σ = 1 and char F = 2, then T = T ∗. 3. If σ2 = 1, σ ̸= 1, then T σ = tT ∗with tσ+1 = 1. For a given frame, T can be chosen so that t = 1; that is, T σ = T ∗. 14.4.30 Remark [1560, Chapter 2] If α is a polarity of PG(n, F) with n even, then, for a given frame, T can be chosen so that T σ = T ∗with σ2 = 1. 14.4.31 Deﬁnition Let α be a polarity of PG(n, F). 1. If σ = 1, char F ̸= 2, and T = T ∗, then α is an orthogonal or ordinary polarity or a polarity with respect to a quadric. 2. If σ = 1, char F ̸= 2, and T = −T ∗, then α is a null polarity or symplectic polarity or a polarity with respect to a linear complex. Since T is non-singular and skew-symmetric, n is odd. 3. If σ = 1, char F = 2, T = T ∗, and all elements on the main diagonal of T are zero, then α is a null polarity or symplectic polarity or a polarity with respect to a linear complex. This only occurs for n odd. 4. If σ = 1, char F = 2, T = T ∗, and some element on the main diagonal of T is not zero, then α is a pseudo-polarity. 5. If σ ̸= 1, then α is a Hermitian or unitary polarity. 14.4.32 Deﬁnition 1. In a polarity α, if P α = π and π′α = P ′, with P, P ′ points and π, π′ hyperplanes, then π is the polar (hyperplane) of P and P ′ is the pole of π′. Since α2 = 1, the converse is also true. 2. If P ′ lies in π = P α, then P lies in π′ = P ′α. In this case, P and P ′ are conjugate points, and π and π′ are conjugate hyperplanes. The point P is self-conjugate if it lies in its own polar hyperplane; the hyperplane π is self-conjugate if it contains its own pole. 14.4.33 Remark [1510, Chapter 2] The self-conjugate points P(X) of α are given by XσTX∗= 0. 14.4.34 Remark For the deﬁnition of a linear complex see [1509, Section 15.2]. 14.4.35 Deﬁnition 1. A quadric Q (or Qn) in PG(n, F), n ≥1, is the set of points P(x0, . . . , xn) satisfying a quadratic equation n X i, j = 0 i ≤j aijxixj = 0, with aij in F and not all zero. For n = 2, a quadric is a conic; for n = 3, a quadric is a quadric surface. 580 Handbook of Finite Fields 2. A Hermitian variety H (or Hn) in PG(n, F), n ≥1, is the set of points P(x0, . . . , xn) satisfying an equation n X i,j=0 aijxixσ j = 0, with aij in F and not all zero, with σ an automorphism of F of order 2, and with aσ ij = aji. For n = 2, a Hermitian variety is a Hermitian curve; for n = 3, a Hermitian variety is a Hermitian surface. 14.4.36 Deﬁnition Let F be a quadric or Hermitian variety in PG(n, F). 1. The point P of PG(n, F) is singular for F if ℓ∩F = {P} or ℓ∩F = ℓfor every line ℓthrough P. 2. If F has a singular point, then F is singular; otherwise, it is non-singular. 14.4.37 Theorem [1510, Chapters 2,5] 1. If α is an orthogonal polarity of PG(n, F), then the set of self-conjugate points of α is a non-singular quadric Q of PG(n, F). 2. If α is a symplectic polarity of PG(n, F), then every point of PG(n, F) is self-conjugate. 3. If α is a pseudo-polarity of PG(n, F), then the set of self-conjugate points of α is a subspace of PG(n, F). 4. If α is a Hermitian polarity of PG(n, F), then the set of self-conjugate points of α is a non-singular Hermitian variety H of PG(n, F). 14.4.38 Theorem [1510, Chapter 5] If α is a symplectic polarity of PG(n, F), then, in a suitable coordinate frame, the matrix of α is T =            0 1 0 0 · · · 0 0 −1 0 0 0 · · · 0 0 0 0 0 1 · · · 0 0 0 0 −1 0 · · · 0 0 . . . . . . . . . . . . . . . . . . 0 0 0 0 · · · 0 1 0 0 0 0 · · · −1 0            . 14.4.39 Theorem [1510, Chapter 5] If α is a pseudo-polarity of PG(n, q), q even, then, in a suitable coordinate frame, the matrix of α is as follows: 1. for n even, T =              1 0 0 0 0 · · · 0 0 0 0 1 0 0 · · · 0 0 0 1 0 0 0 · · · 0 0 0 0 0 0 1 · · · 0 0 0 0 0 1 0 · · · 0 0 . . . . . . . . . . . . . . . . . . . . . 0 0 0 0 0 · · · 0 1 0 0 0 0 0 · · · 1 0              ; Combinatorial 581 2. for n odd, T =            1 1 0 0 · · · 0 0 1 0 0 0 · · · 0 0 0 0 0 1 · · · 0 0 0 0 1 0 · · · 0 0 . . . . . . . . . . . . . . . . . . 0 0 0 0 · · · 0 1 0 0 0 0 · · · 1 0            . 14.4.40 Remark [1510, Chapter 5] Theorem 14.4.39 holds for every ﬁeld F of characteristic two with the property that {x2 \| x ∈F} = F. 14.4.41 Theorem [1510, Chapter 5] Let Q be a non-singular quadric of PG(n, q). The coordinate frame can be chosen so that Q has the following equation: 1. n even, x2 0 + x1x2 + x3x4 + · · · + xn−1xn = 0; 2. n odd, a. x0x1 + x2x3 + · · · + xn−1xn = 0; b. f(x0, x1) + x2x3 + · · · + xn−1xn = 0, where f is any chosen irreducible quadratic form. Hence, up to a projectivity, there is a unique non-singular quadric in PG(n, q) with n even; for n odd, there are two types of non-singular quadric. 14.4.42 Deﬁnition 1. In case 1 of Theorem 14.4.41, the quadric is parabolic. In case 2.a, it is hyperbolic; in case 2.b, it is elliptic; 2. The character w of a non-singular quadric in PG(n, q) is 0 if it is elliptic, 1 if it is parabolic, and 2 if it is hyperbolic. 14.4.43 Theorem [1510, Chapter 5] If Qn is a non-singular quadric of PG(n, q) with character w, then \|Qn\| = (qn −1)/(q −1) + (w −1)q(n−1)/2. 14.4.44 Theorem [1510, Chapter 5] Let H be a non-singular Hermitian variety of PG(n, q2). The coordinate frame can be chosen so that H has the following equation: xq+1 0 + xq+1 1 + · · · + xq+1 n = 0. 14.4.45 Theorem [1510, Chapter 5] If Hn is a non-singular Hermitian variety of PG(n, q2), then \|Hn\| = [qn+1 + (−1)n][qn −(−1)n]/(q2 −1). 582 Handbook of Finite Fields 14.4.4 Partitions and cyclic projectivities 14.4.46 Deﬁnition A spread S of PG(n, q) by r-spaces is a set of r-spaces which partitions PG(n, q); that is, every point of PG(n, q) lies in exactly one r-space of S. Hence any two r-spaces of S are disjoint. 14.4.47 Theorem [1510, Chapter 4] The following are equivalent: 1. there exists a spread S of r-spaces of PG(n, q); 2. θ(r, q) \| θ(n, q); 3. (r + 1) \| (n + 1). 14.4.48 Remark Spreads of PG(2r+1, q) by r-spaces have been much studied, particularly for their application to non-Desarguesian planes. The latter are considered in more detail in Section 14.3. 14.4.49 Deﬁnition Since Fq is a subﬁeld of Fqk for k ∈N{0}, so PG(n, q) is naturally embedded in PG(n, qk) once the coordinate frame is ﬁxed. Any PG(n, q) embedded in PG(n, qk) is a subgeometry of PG(n, qk). 14.4.50 Theorem [1510, Chapter 4] If s(n, q, qk) is the number of subgeometries PG(n, q) embedded in PG(n, qk), then s(n, q, qk) = \|PGL(n + 1, qk)\|/\|PGL(n + 1, q)\| = qn(n+1)(k−1)/2 n+1 Y i=2 [(qki −1)/(qi −1)]. 14.4.51 Corollary [1510, Chapter 4] On the line PG(1, qk), 1. s(1, q, qk) = qk−1(q2k −1)/(q2 −1); 2. s(1, q, q2) = q(q2 + 1). 14.4.52 Corollary [1510, Chapter 4] In the plane PG(2, q2), s(2, q, q2) = q3(q2 + 1)(q3 + 1). 14.4.53 Theorem [1510, Chapter 4] The following are equivalent: 1. there exists a partition of PG(n, qk) into subgeometries PG(n, q); 2. θ(n, q) \| θ(n, qk); 3. (k, n + 1) = 1. 14.4.54 Corollary [1510, Chapter 4] The line PG(1, qk) can be partitioned into sublines PG(1, q) if and only if k is odd. 14.4.55 Corollary [1510, Chapter 4] The plane PG(2, qk) can be partitioned into subplanes PG(2, q) if and only if (k, 3) = 1. 14.4.56 Corollary [1510, Chapter 4] The plane PG(2, q2) can be partitioned into q2−q+1 subplanes PG(2, q). Combinatorial 583 14.4.57 Deﬁnition A projectivity α which permutes the θ(n) points of PG(n, q) in a single cycle is a cyclic projectivity; it is a Singer cycle and the group it generates a Singer group. 14.4.58 Theorem [1510, Chapter 4] A projectivity α of PG(n, q) is cyclic if and only if the charac-teristic polynomial of an associated matrix is subprimitive; that is, the smallest power m of a characteristic root that lies in Fq is m = θ(n). 14.4.59 Corollary [1510, Chapter 4] A cyclic projectivity permutes the hyperplanes of PG(n, q) in a single cycle. 14.4.60 Corollary [1510, Chapter 4] The number of cyclic projectivities in PG(n, q) is given by σ(n, q) = qn(n+1)/2 n Y i=1 (qi −1)ϕ(θ(n))/(n + 1), where ϕ is the Euler function. 14.4.61 Corollary [1510, Chapter 4] The number of conjugacy classes of PGL(n + 1, q) consisting of cyclic projectivities is ϕ(θ(n))/(n + 1). 14.4.62 Corollary [1510, Chapter 4] In PGL(n + 1, q) there is just one conjugacy class of Singer groups. 14.4.63 Theorem [1510, Chapter 4] If (k, n + 1) = 1 and α is a projectivity of PG(n, qk) which acts as a cyclic projectivity on some PG(n, q) embedded in PG(n, qk), then, letting u = (qk(n+1) −1)(q −1)/[(qk −1)(qn+1 −1)], 1. there exists a cyclic projectivity ρ of PG(n, qk) such that ρu = α; 2. every orbit of α is a subgeometry PG(n, q); 3. if γ is any cyclic projectivity of PG(n, qk), then the orbits of γu are subgeometries PG(n, q). 14.4.64 Theorem [1510, Chapter 4] If (k, n+1) = 1, then the number of projectivities α of PG(n, qk) that act cyclically on at least one PG(n, q) of PG(n, qk) is qkn(n+1)/2 n Y i=1 (qki −1)ϕ(θ(n, q))/(n + 1). 14.4.5 k-Arcs 14.4.65 Deﬁnition 1. A k-arc in PG(n, q), n ≥2, is a set K of k points, with k ≥n + 1, such that no n + 1 of its points lie in a hyperplane. 2. An arc K is complete if it is not properly contained in a larger arc. 3. Otherwise, if K∪{P} is an arc for some point P of PG(n, q), the point P extends K. 584 Handbook of Finite Fields 14.4.66 Deﬁnition A normal rational curve in PG(n, q), n ≥2, is any set of points of PG(n, q) which is projectively equivalent to {P(tn, tn−1, . . . , t, 1) \| t ∈Fq} ∪{P(1, 0, . . . , 0, 0)}. 14.4.67 Remark [1515, Chapter 27] Any normal rational curve contains q + 1 points. For n = 2, it is a non-singular conic; for n = 3, it is a twisted cubic. Any (n + 3)-arc in PG(n, q) is contained in a unique normal rational curve of this space. 14.4.68 Problem (The three problems of Segre) I. For given n and q, what is the maximum value of k such that a k-arc exists in PG(n, q)? II. For what values of n and q with q > n + 1 is every (q + 1)-arc of PG(n, q) a normal rational curve? III. For given n and q with q > n + 1, what are the values of k such that each k-arc of PG(n, q) is contained in a normal rational curve of this space? 14.4.69 Remark For a survey of solutions to Problems I, II, III, see [1512, 1513] and [1511, Chapter 13]. 14.4.70 Theorem [1510, Chapter 8] Let K be a k-arc of PG(2, q). Then 1. k ≤q + 2; 2. for q odd, k ≤q + 1; 3. any non-singular conic of PG(2, q) is a (q + 1)-arc; 4. each (q + 1)-arc of PG(2, q), q even, extends to a (q + 2)-arc. 14.4.71 Deﬁnition 1. The (q + 1)-arcs of PG(2, q) are ovals. 2. The (q + 2)-arcs of PG(2, q), q even, are complete ovals or hyperovals. 14.4.72 Theorem [1510, Chapter 8] In PG(2, q), q odd, every oval is a non-singular conic. 14.4.73 Remark Theorem 14.4.72 is a celebrated result due to Segre . For more details on k-arcs in PG(2, q), ovals and hyperovals see [1510, Chapters 8–10]. The fundamental Theorem 14.4.76, also due to Segre , relates k-arcs of PG(2, q) to plane algebraic curves. 14.4.74 Deﬁnition Let K be a k-arc of PG(2, q). 1. A tangent of K is a line of PG(2, q) meeting K in a unique point. 2. A secant of K is a line meeting K in two points. 14.4.75 Remark At each point, K has t = q + 2 −k tangents; the total number of tangents is tk. 14.4.76 Theorem [1510, Chapter 10] 1. Let K be a k-arc in PG(2, q), with q even. Then the tk tangents of K belong to an algebraic envelope Γt of class t, that is, the dual of a plane algebraic curve of order t, with the following properties: Combinatorial 585 a. Γt is unique if k > (q + 2)/2; b. Γt contains no secant of K and so no pencil with vertex P in K, where a pencil is the set of lines through a point; c. if ∆P is the pencil of lines with vertex P in K and ℓis a tangent at P, then the intersection multiplicity of ∆P and Γt at ℓis one. 2. Let K be a k-arc in PG(2, q), with q odd. Then the tk tangents of K belong to an algebraic envelope Γ2t of class 2t with the following properties: a. Γ2t is unique if k > (2q + 4)/3; b. Γ2t contains no secant of K and so no pencil with vertex P in K; c. if ∆P is the pencil with vertex P in K and ℓis a tangent at P, then the intersection multiplicity of ∆P and Γ2t at ℓis two; d. Γ2t may contain components of multiplicity at most two, but does not consist entirely of double components. 14.4.77 Corollary [1510, Chapter 10] 1. If q is even and k > (q + 2)/2, then K is contained in a unique complete arc of PG(2, q). 2. If q is odd and k > (2q + 4)/3, then K is contained in a unique complete arc of PG(2, q). 14.4.78 Remark 1. For a survey on k-arcs in PG(n, q), n > 2, see [1512, 1513]. 2. For q odd, the main results were obtained by induction and projection of the k-arc K from one of its points P onto a hyperplane not containing P; see Thas and [1515, Chapter 27]. 3. For any q, a theorem of Bruen, Thas, and Blokhuis relates k-arcs in PG(3, q) to dual algebraic surfaces. For q even, this result enables estimates to be made for the three problems of Segre. A generalisation by Blokhuis, Bruen, and Thas now follows. 14.4.79 Theorem [1515, Chapter 27] Let K = {P1, P2, . . . , Pk} be a k-arc of PG(n, q). For distinct i1, i2, . . . , in−1 ∈{1, 2, . . . , k}, let Z{i1,i2,...,in−1} be the set of t = q + n −k hyperplanes through the (n −2)-dimensional subspace Πn−2, generated by Pi1, Pi2, . . . , Pin−1, that con-tains no other point of K. 1. a. For q even, there exists a dual algebraic hypersurface Φt of class t in PG(n, q) which contains the hyperplanes of each set Z{i1,i2,...,in−1}. b. This dual hypersurface is unique when k > (q + 2n −2)/2. 2. a. For q odd, there exists a dual algebraic hypersurface Φ2t of class 2t in PG(n, q) which contains the hyperplanes of each set Z{i1,i2,...,in−1}. b. The intersection multiplicity of Φ2t and the pencil of hyperplanes with vertex Πn−2 at each hyperplane of Z{i1,i2,...,in−1} is two. c. This dual hypersurface is unique when k > (2q + 3n −2)/3. 14.4.80 Corollary [1515, Chapter 27] 1. If q is even and k > (q + 2n −2)/2, then a k-arc K of PG(n, q) is contained in a unique complete arc. 586 Handbook of Finite Fields 2. If q is odd and k > (2q + 3n −2)/3, then a k-arc K of PG(n, q) is contained in a unique complete arc. 14.4.81 Theorem (The duality principle for k-arcs) [1515, Chapter 27] A k-arc of PG(n, q), with n ≥2 and k ≥n + 4, exists if and only if a k-arc of PG(k −n −2, q) exists. 14.4.82 Corollary [1515, Chapter 27] In PG(q −2, q), q even and q ≥4, there exist (q + 2)-arcs. 14.4.83 Conjecture [1511, Chapter 13] 1. If K is a k-arc in PG(n, q), with q −1 ≥n ≥2, then k ≤q + 1 for q odd and k ≤q + 2 for q even. 2. In PG(n, q), with q ≥n ≥2 and q even, there exist (q + 2)-arcs if and only if n ∈{2, q −2}. 14.4.6 k-Arcs and linear MDS codes 14.4.84 Deﬁnition 1. Let C be a code of length k over an alphabet A of size q with q ≥2. In other words, C is a set of ( code)words, where each word is an ordered k-tuple over A. 2. For a given m with 2 ≤m ≤k, impose the following condition: no two words in C agree in as many as m positions. Then \|C\| ≤qm. If \|C\| = qm, then C is a maximum distance separable (MDS) code. 14.4.85 Remark MacWilliams and Sloane [1991, Chapter 11] introduce the chapter on MDS codes as “one of the most fascinating in all of coding theory.” 14.4.86 Deﬁnition 1. The (Hamming) distance between two codewords X = (x1, . . . , xk) and Y = (y1, . . . , yk) is the number of indices i for which xi ̸= yi; it is denoted d(X, Y ). 2. The minimum distance of a code C, with \|C\| > 1, is d(C) = min{d(X, Y ) \| X, Y ∈C, X ̸= Y }. 14.4.87 Theorem [2849, Chapter 5] For an MDS code, d(C) = k −m + 1; see Section 15.1. 14.4.88 Remark One of the main problems concerning MDS codes is to maximize d(C), and so k, for given m and q. Another problem is to determine the structure of C in the optimal case. 14.4.89 Theorem For an MDS code, k ≤q + m −1. 14.4.90 Remark For m = 2, the MDS code C gives a set of q2 codewords of length k, no two of which agree in more than one position. This is equivalent to the existence of a net of order q and degree k; see also Section 14.1. It follows that k ≤q + 1, the case of equality corresponding to an aﬃne plane of order q; see Section 14.3. Theorem 14.4.89 follows by an inductive argument. Combinatorial 587 14.4.91 Remark 1. The case m = 3 and k = q + 2 is equivalent to the existence of an aﬃne plane of order q, with q even, containing an appropriate system of (q + 2)-arcs. For all known examples the plane is an aﬃne plane AG(2, q) with q = 2h; see Willems and Thas . 2. For m = 4 and k = q + 3, it has been shown that either q = 2 or 36 divides q; no example other than for q = 2 is known to exist; see Bruen and Silverman . 14.4.92 Remark Henceforth, only linear MDS codes are considered; that is C is an m-dimensional subspace of the vector space V (k, q), which is Fk q with the usual addition and scalar multi-plication. 14.4.93 Theorem [1511, Chapter 13] For m ≥3, linear MDS codes and arcs are equivalent objects. 14.4.94 Remark Let C be an m-dimensional subspace of V (k, q) and let G be an m × k generator matrix for C; that is, the rows of G are a basis for C. Then C is MDS if and only if any m columns of G are linearly independent; this property is preserved under multiplication of the columns by non-zero scalars. So consider the columns of G as points P1, . . . , Pk of PG(m−1, q). It follows that C is MDS if and only if {P1, . . . , Pk} is a k-arc of PG(m−1, q). This gives the relation between linear MDS codes and arcs. 14.4.95 Theorem [1511, Chapter 13] For 2 ≤m ≤k −2, the dual of a linear MDS code is again a linear MDS code. 14.4.96 Remark For 3 ≤m ≤k −3, Theorem 14.4.95 is the translation of Theorem 14.4.81 from geometry to coding theory. 14.4.7 k-Caps 14.4.97 Deﬁnition 1. In PG(n, q), n ≥3, a set K of k points no three of which are collinear is a k-cap. 2. A k-cap is complete if it is not contained in a (k + 1)-cap. 3. A line of PG(n, q) is a secant, tangent, or external line as it meets K in 2, 1 or 0 points. 14.4.98 Theorem [1509, Chapter 16] 1. For any k-cap K in PG(3, q) with q ̸= 2, the cardinality k satisﬁes k ≤q2 + 1. 2. In PG(3, 2), a k-cap satisﬁes k ≤8; an 8-cap is the complement of a plane. 14.4.99 Deﬁnition A (q2 + 1)-cap of PG(3, q), q ̸= 2 is an ovoid; the ovoids of PG(3, 2) are its elliptic quadrics. 14.4.100 Theorem [1509, Chapter 16] At each point P of an ovoid O of PG(3, q), there is a unique tangent plane π such that π ∩O = {P}. 14.4.101 Theorem [1509, Chapter 16] 1. Apart from the tangent planes, every plane meets an ovoid O in a (q + 1)-arc. 2. When q is even, the (q2 + 1)(q + 1) tangents of O are the totally isotropic lines of a symplectic polarity α of PG(3, q), that is, the lines ℓfor which ℓα = ℓ. 588 Handbook of Finite Fields 14.4.102 Theorem [1509, Chapter 16] In PG(3, q), q odd, every ovoid is an elliptic quadric. 14.4.103 Remark Theorem 14.4.102 is a celebrated result, due independently to Barlotti and Panella . Both proofs rely on Theorem 14.4.72. 14.4.104 Theorem In PG(3, q), q even, every ovoid containing at least one conic section is an elliptic quadric. 14.4.105 Theorem [1509, Chapter 16] In PG(3, q), let W(q) be the incidence structure formed by all the points and the totally isotropic lines of a symplectic polarity α. Then W(q) admits a polarity α′ if and only if q = 22e+1; in that case, the absolute points of α′, namely the points lying in their image lines, form an ovoid of PG(3, q). Such an ovoid is an elliptic quadric if and only if q = 2. 14.4.106 Deﬁnition For q = 22e+1, with e ≥1, the ovoids in Theorem 14.4.105 are Tits ovoids. 14.4.107 Theorem [1509, Chapter 16] With q = 22e+1, the canonical form of a Tits ovoid is O = {P(1, z, y, x) \| z = xy + xσ+2 + yσ} ∪{P(0, 1, 0, 0)}, where σ is the automorphism t 7→t2e+1 of Fq. 14.4.108 Theorem [1509, Chapter 16] For q = 22e+1, e ≥1, the group of all projectivities of PG(3, q) ﬁxing the Tits ovoid O is the Suzuki group Sz(q), which acts doubly transitively on O. 14.4.109 Remark The case q = 4 is the same as q odd; that is, an ovoid of PG(3, 4) is an elliptic quadric; see Barlotti or [1509, Chapter 16]. For q = 8, Segre found an ovoid other than an elliptic quadric; Fellegara showed that this example is a Tits ovoid. She also showed, using a computer program, that every ovoid in PG(3, 8) is either an elliptic quadric or a Tits ovoid. O’Keefe and Penttila [2313, 2314] showed, also using a computer program, that in PG(3, 16) every ovoid is an elliptic quadric. O’Keefe, Penttila, and Royle , also using a computer program, showed that in PG(3, 32) every ovoid is an elliptic quadric or a Tits ovoid. 14.4.110 Deﬁnition Let O be an ovoid of PG(3, q) and let B be the set of all intersections π ∩O, with π a non-tangent plane of O. Then the incidence structure formed by the triple I(O) = (O, B, ∈) is a 3 −(q2 + 1, q + 1, 1) design. A 3 −(n2 + 1, n + 1, 1) design I = (P, B, ∈) is an inversive plane of order n and the elements of B are circles; the inversive planes arising from ovoids are egglike. 14.4.111 Theorem [807, Chapter 6] Every inversive plane of even order is egglike. 14.4.112 Deﬁnition If the ovoid O is an elliptic quadric, then the inversive plane I(O), and any inversive plane isomorphic to it, is classical or Miquelian. 14.4.113 Remark By Theorem 14.4.102, an egglike inversive plane of odd order is Miquelian. For odd order, no other inversive planes are known. 14.4.114 Deﬁnition Let I be an inversive plane of order n. For any point P of I, the points of I other than P, together with the circles containing P with P removed, form a 2−(n2, n, 1) design, that is, an aﬃne plane of order n. This plane is denoted IP and is the internal plane or derived plane of I at P. Combinatorial 589 14.4.115 Remark [807, Chapter 6] For an egglike inversive plane I(O) of order q, each internal plane is Desarguesian, that is, the aﬃne plane AG(2, q) over Fq. 14.4.116 Theorem Let I be an inversive plane of odd order n. If, for at least one point P of I, the internal plane IP is Desarguesian, then I is Miquelian. 14.4.117 Remark Up to isomorphism, there is a unique inversive plane of order n for the values n = 2, 3, 4, 5, 7; see Chen , Denniston [821, 822], Witt . As a corollary of Theorem 14.4.116 and the uniqueness of the projective plane of order n for n = 3, 5, 7, a computer-free proof of the uniqueness of the inversive plane of order n is obtained for these n. 14.4.118 Remark For more information about designs, see Section 14.5. For more information about projective spaces, see [1509, 1510, 1515] and [1511, Chapter 13]. See Also §12.5 For results on curves which impinge on k-arcs. §14.2 For a technique to resolve problems on blocking sets. §14.3 For other aspects of Desarguesian planes. References Cited: [202, 324, 421, 431, 434, 608, 807, 821, 822, 1050, 1509, 1510, 1511, 1512, 1513, 1515, 1560, 1991, 2313, 2314, 2315, 2358, 2575, 2576, 2577, 2794, 2795, 2849, 2979, 2997] 14.5 Block designs Charles J. Colbourn, Arizona State University Jeﬀrey H. Dinitz, University of Vermont 14.5.1 Basics 14.5.1 Deﬁnition A balanced incomplete block design (BIBD) is a pair (V, B) where V is a v-set and B is a collection of b k-subsets of V (blocks) such that each element of V is contained in exactly r blocks and any 2-subset of V is contained in exactly λ blocks. The numbers v, b, r, k, and λ are parameters of the BIBD. Its order is v; its replication number is r; its blocksize is k; and its index is λ. 14.5.2 Proposition Trivial necessary conditions for the existence of a BIBD(v, b, r, k, λ) are 1. vr = bk, and 2. r(k −1) = λ(v −1). Parameter sets that satisfy conditions 1 and 2 are admissible. 14.5.3 Remark The three parameters v, k, and λ determine the remaining two as r = λ(v−1) k−1 and b = vr k . Hence one often writes (v, k, λ) design to denote a BIBD(v, b, r, k, λ). 590 Handbook of Finite Fields 14.5.4 Example The unique (6, 3, 2) design and the unique (7, 3, 1) design have blocks shown below as columns: 0000011122 0001123 1123423433 1242534 2345554545 3654656 14.5.5 Deﬁnition A BIBD (V, B) with parameters v, b, r, k, λ is simple if it has no repeated blocks; complete or full if it is simple and contains v k blocks; decomposable if B can be partitioned into two nonempty collections B1 and B2 so that (V, Bi) is a (v, k, λi) design for i = 1, 2; Hadamard if v = 4n −1, k = 2n −1, and λ = n −1 for some integer n ≥2; m-multiple if v, b m, r m, k, λ m are the parameters of a BIBD; nontrivial if 3 ≤k < v; quasi-symmetric if every two distinct blocks intersect in either µ1 or µ2 elements; resolvable (an RBIBD) if there exists a partition R of its set of blocks B into parallel classes, each of which in turn partitions the set V (R is a resolution); a Steiner 2-design S(2, k, v) if λ = 1; a Steiner triple system STS(v) if k = 3 and λ = 1; symmetric if v = b, or equivalently k = r; a triple system TS(v, λ) if k = 3. 14.5.2 Triple systems 14.5.6 Remark A Steiner triple system of order v can exist only when v −1 is even because every element occurs with v −1 others, and in each block in which it occurs it appears with two other elements. Moreover, every block contains three pairs and hence v 2 must be a multiple of 3. Thus, it is necessary that v ≡1, 3 (mod 6). This condition was shown to be suﬃcient in 1847. 14.5.7 Theorem A Steiner triple system of order v exists if and only if v ≡1, 3 (mod 6). 14.5.8 Theorem A TS(v, λ) exists if and only if v ̸= 2 and λ ≡0 (mod gcd(v −2, 6)). 14.5.9 Remark Existence theorems such as Theorems 14.5.7 and 14.5.8 are typically established by a combination of direct constructions to make designs for speciﬁc values of v, and recursive constructions to make solutions for large values of v from solutions with smaller values of v. Finite ﬁelds are most often used in providing direct constructions, both to provide ingredients for recursive constructions, and to produce solutions with speciﬁc properties. Examples for triple systems are developed to demonstrate these; see . 14.5.10 Construction Let p be a prime, n ≥1, and pn ≡1 (mod 6). Then there is an STS(pn). To construct one, let Fpn be a ﬁnite ﬁeld on a set X of size pn = 6t + 1 with 0 as its zero element, and ω a primitive root of unity. Then {{ωi + j, ω2t+i + j, ω4t+i + j} : 0 ≤i < t, j ∈X} Combinatorial 591 (with computations in Fpn) is the set of blocks of an STS(pn) on X. 14.5.11 Remark In order to verify Construction 14.5.10, consider two distinct elements x, y ∈Fpn. Let d = x −y (arithmetic in Fpn). Now since d ̸= 0 and ω2t −1 ̸= 0, d can be uniquely written in the form (ω2t −1)ω2jtωi for j ∈{0, 1, 2} and 0 ≤i < 2t. Since ω3t = −1, if i ≥t, we may write −d = y −x = (ω2t −1)ω2(j+1)tωi−t. Thus we suppose without loss of generality that d = x−y and i < t. Then {x, y} appears in the triple {ωi, ω2t+i, ω4t+i}+(x−ω2(j+1)t+i). Consequently, every pair of distinct elements in Fpn appears in at least one of the triples deﬁned. Because the total number of pairs in the triples deﬁned is precisely pn 2 , every pair occurs in exactly one triple. 14.5.12 Deﬁnition Two BIBDs (V1, B1), (V2, B2) are isomorphic if there exists a bijection α : V1 →V2 such that B1α = B2. An automorphism is an isomorphism of a design with itself. The set of all automorphisms of a design forms a group, the (full) automorphism group. An automorphism group is any subgroup of the full automorphism group. 14.5.13 Remark If (V, B) is a BIBD(v, b, r, k, λ) with automorphism group G, the action of G partitions B into classes (orbits). A set of orbit representatives is a set of starter blocks or base blocks. Applying the action of G to a set of base blocks yields a design, the development. 14.5.14 Remark In Construction 14.5.10, we can treat D = {{ωi, ω2t+i, ω4t+i} : 0 ≤i < t} as the base or starter triples of the design. Their development is the result of applying the action of the elementary abelian group of order pn to the base triples. The veriﬁcation requires that for every diﬀerence d ∈Fpn \ {0}, there is exactly one way to choose x, y ∈D ∈D so that d = x −y, with arithmetic in the elementary abelian group of order pn. 14.5.15 Construction Let p be a prime, n ≥1, and pn ≡7 (mod 12). Let Fpn be a ﬁnite ﬁeld on a set X of size pn = 6t + 1 = 12s + 7 with 0 as its zero element and ω a primitive root of unity. Then {{ω2i + j, ω2t+2i + j, ω4t+2i + j} : 0 ≤i < t, j ∈X} forms the blocks of an STS(pn) on X. (These are the Netto triple systems.) 14.5.16 Remark The Netto triple systems provide examples of STS(v)s that admit 2-homogeneous automorphism groups but (for v > 7) do not admit 2-transitive groups. We give another construction of the Netto triple systems. Let Γ = {a2xσ + b : a, b ∈Fpn, σ ∈Aut(Fpn)}. Let ε be a primitive sixth root of unity in Fpn. Then the orbit of {0, 1, ε} under the action of Γ is the Netto triple system of order pn. This illustrates one of the principal reasons for using large automorphism groups, and in particular for using the additive and multiplicative structure of the ﬁnite ﬁeld – a single triple represents the entire triple system. 14.5.17 Construction Let p = 2t + 1 be an odd prime. Let ω be a primitive root of unity in Zp satisfying ω ≡1 (mod 3). Then {{ωi + j, ω2t+i + j, ω4t+i + j} : 0 ≤i < t, j ∈Z3p} ∪{{j, j + p, j + 2p} : j ∈Zp} (with computations modulo 3p) is the set of blocks of an STS(3p). 592 Handbook of Finite Fields 14.5.18 Deﬁnition A set of blocks is a partial parallel class (PPC) if no two blocks in the set share an element. A PPC is an almost parallel class if it contains v−1 3 blocks; when it contains v 3 blocks, it is a parallel class or resolution class. A partition of all blocks of a TS(v, λ) into parallel classes is a resolution and the STS is resolvable. An STS(v) together with a resolution of its blocks is a Kirkman triple system, KTS(v). 14.5.19 Remark In Construction 14.5.17, the triples {{j, j+p, j+2p} : j ∈Zp} form a parallel class. Indeed we can say much more in certain cases. The method of “pure and mixed diﬀerences” is applied, using a set of elements Fq × X, for X a ﬁnite set; a pure(x) diﬀerence is the diﬀerence d = a −b associated with the pair {(a, x), (b, x)} and a mixed(x,y) diﬀerence is the diﬀerence d = a −b associated with the pair {(a, x), (b, y)}. 14.5.20 Construction If q = pα ≡1 (mod 6) is a prime power, then there exists a KTS(3q). Let t = (q −1)/6. To construct a KTS(3q), take as elements Fq × {1, 2, 3}, writing ai for (a, i). Let ω be a primitive element in Fq, and let B consist of triples: 1. C = {01, 02, 03}; 2. Bij = {ωi j, ωi+2t j , ωi+4t j }, 0 ≤i < t, j ∈{1, 2, 3}; 3. Ai = {ωi 1, ωi+2t 2 , ωi+4t 3 }, 0 ≤i < t. Each of the (nonzero) pure and mixed diﬀerences occurs exactly once in triples of B, and thus B is the set of starter triples for an STS(3q). This STS(3q) is resolvable. Indeed, R0 = C ∪{Bij : 0 ≤i < t, j ∈{1, 2, 3}} ∪{Ajt+i : j ∈{1, 3, 5}, 0 ≤i < t} forms a parallel class; when developed modulo 6t + 1, it yields a further 6t parallel classes. Each Ai, when developed modulo 6t + 1, also yields a parallel class; taking those parallel classes only from Ajt+i with j ∈{0, 2, 4} and 0 ≤i < t} thus yields a further 3t parallel classes, for a total of 9t + 1 forming the resolution. 14.5.21 Construction If q = pα ≡1 (mod 6) is a prime power, then there exists a KTS(2q+1). Let t = (q −1)/6. To construct a KTS(2q + 1), take as elements (Fq × {1, 2}) ∪{∞}. Let ω be a primitive element of Fq, and let m satisfy 2ωm = ωt + 1. Let B consist of triples 1. C = {01, 02, ∞}; 2. Bi = {ωi 1, ωi+t 1 , ωi+m 2 }, 0 ≤i < t, 2t ≤i < 3t, or 4t ≤i < 5t; 3. Ai = {ωi+m 2 , ωi+m+3t 2 , ωi+m+5t 2 }, 0 ≤i < t. Every pure and every mixed diﬀerence occurs exactly once and hence B is a set of starter triples for an STS(2q + 1). But B itself forms a parallel class, whose development modulo q yields the required q parallel classes for the KTS(2q + 1). 14.5.3 Diﬀerence families and balanced incomplete block designs 14.5.22 Deﬁnition Let B = {b1, ..., bk} be a subset of an additive group G. The G-stabilizer of B is the subgroup GB of G consisting of all elements g ∈G such that B + g = B. B is full or short according to whether GB is or is not trivial. The G-orbit of B is the set OrbGB of all distinct right translates of B, namely, OrbGB = {B + s \| s ∈S} where S is a complete system of representatives for the right cosets of GB in G. 14.5.23 Deﬁnition The multiset ∆B = {bi −bj \| i, j = 1, . . . , k, i ̸= j} is the list of diﬀerences from B. The multiplicity in ∆B of an element g ∈G is of the form µg\|GB\| for some Combinatorial 593 integer µg. The list of partial diﬀerences from B is the multiset ∂B where each g ∈G appears exactly µg times. (∆B = ∂B if and only if B is a full block.) 14.5.24 Deﬁnition Let G be a group of order v. A collection {B1, ..., Bt} of k-subsets of G forms a (v, k, λ) diﬀerence family (or diﬀerence system) if every nonidentity element of G occurs λ times in ∂B1 ∪· · · ∪∂Bt. The sets Bi are base blocks. A diﬀerence family having at least one short block is partial. 14.5.25 Remark 1. All deﬁnitions given can be extended to a multiplicative group by replacing B +g with B · g and bi −bj with bib−1 j . 2. If t = 1, then B1 is a (v, k, λ) diﬀerence set; see Section 14.6. 3. If {B1, . . . , Bt} is a (v, k, λ) diﬀerence family over G, OrbG(B1) ∪· · · ∪OrbG(Bt) is the collection of blocks of a BIBD(v, k, λ) admitting G as a sharply point-transitive automorphism group. This BIBD is cyclic (abelian, nonabelian, dihe-dral, and so on) if the group G has the respective property. In this case the diﬀerence family is a cyclic (abelian, nonabelian, dihedral, respectively) diﬀerence family. 4. A BIBD(v, k, λ) with an automorphism group G acting sharply transitively on the points is (up to isomorphism) generated by a suitable (v, k, λ) diﬀerence family. 5. Every short block of a (v, k, 1) diﬀerence family over an abelian group G is a coset of a suitable subgroup of G. 14.5.26 Theorem The set of order p subgroups of Fpn forms a (pn, p, 1) diﬀerence family generating the point-line design associated with the aﬃne geometry AG(n, p). 14.5.27 Deﬁnition 1. C = {c1, . . . , ck} is a multiple of B = {b1, . . . , bk} if, for some w, ci = w · bi for all i. 2. w is a multiplier of order n of B = {b1, . . . , bk} if wn = 1 but wi ̸= 1 for 0 < i < n, and for some g ∈G, B = w·B+g = {w·b1+g, w·b2+g, . . . , w·bk+g}. 3. w is a multiplier of a diﬀerence family D if, for each base block B ∈D, there exists C ∈D and g ∈G for which w · B + g = C. 4. If q is a prime power and D is a (q, k, λ) diﬀerence family over Fq in which one base block, B, has a multiplier of order k or k −1 and all other base blocks are multiples of B, then D is radical. 14.5.28 Theorem Suppose q ≡7 (mod 12) is a prime power and there exists a cube root of unity ω in Fq such that x = ω −1 is a primitive root. Then the following base blocks form a (7q,4,1) diﬀerence family over Z7 × Fq: 1. {(0,0), (0, (x −1)x2t−1), (0,ω(x −1)x2t−1), (0,ω2(x −1)x2t−1)} for 1 ≤t ≤ (q −7)/12, 2. {(0,0), (1, x2t), (2, ωx2t), (4, ω2x2t)} for 1 ≤t ≤(q −3)/2, x2t ̸= ω, and 3. {(0,0), (2t, ωt), (2t, x · ωt), (2t+2, 0)} for 0 ≤t ≤2. 14.5.29 Remark The (7q,4,1) diﬀerence families are obtainable by Theorem 14.5.28 for q = 7, 19, 31, 43, 67, 79, 103, 127, 151, 163, 199, 211, 367, 379, 439, 463, 487, 571, but not for q = 594 Handbook of Finite Fields 139, 223, 271, 283, 307, 331, 523, 547. A more general construction for (7q,4,1) diﬀerence families with q a prime power ≡7 (mod 12) can be found in . 14.5.30 Theorem Suppose q is a prime power, and λ(q −1) ≡0 (mod k(k −1)). Then a (q, k, λ) diﬀerence family over Fq exists if 1. λ is a multiple of k/2 or (k −1)/2; 2. λ ≥k(k −1); or 3. q > k 2 k(k−1). 14.5.31 Theorem Suppose q is an odd prime power. Then there exists a (q, k, λ) radical diﬀerence family if either: 1. λ is a multiple of k/2 and q ≡1 (mod k −1), or 2. λ is a multiple of (k −1)/2 and q ≡1 (mod k). 14.5.32 Remark For radical diﬀerence families with λ = 1, the multiplier must have odd order (that is, order k if k is odd, or order k −1 if k is even). 14.5.33 Theorem Let q = 12t + 1 be a prime power and 2e be the largest power of 2 dividing t. Then a (q, 4, 1) radical diﬀerence family in Fq exists if and only if −3 is not a 2e+2-th power in Fq. (This condition holds for q = 13, 25, 73, 97, 109, 121, 169, 181, 193, 229, 241, 277, 289, 313, 337, 409, 421, 433, 457, 529, 541, 577, 601, 625, 673, 709, 733, 757, 769, 829, 841.) 14.5.34 Theorem Let q = 20t + 1 be a prime power, and let 2e be the largest power of 2 dividing t. Then a (q, 5, 1) radical diﬀerence family in Fq exists if and only if (11 + 5 √ 5)/2 is not a 2e+1-th power in Fq. (This condition holds for q = 41, 61, 81, 241, 281, 401, 421, 601, 641, 661, 701, 761, 821, 881.) 14.5.35 Remark In , necessary and suﬃcient conditions are given for a (q, k, 1) radical diﬀerence family with k ∈{6, 7} to exist over Fq; a suﬃcient condition is also given for k ≥8. 14.5.36 Theorem [460, 1356, 2986] Among others, (q, k, 1) radical diﬀerence families exist for the following values of q and k: k = 6 q ∈{181, 211, 241, 631, 691, 1531, 1831, 1861, 2791, 2851, 3061}; k = 7 q ∈{337, 421, 463, 883, 1723, 3067, 3319, 3823, 3907, 4621, 4957, 5167, 5419, 5881, 6133, 8233, 8527, 8821, 9619, 9787, 9829}; k = 8 q ∈{449, 1009, 3137, 3697, 6217, 6329, 8233, 9869}; k = 9 q ∈{73, 1153, 1873, 2017, 6481, 7489, 7561, 8359}. 14.5.37 Theorem If there exists a (p, k, 1) radical diﬀerence family with p a prime and k odd, there exists a cyclic RBIBD(kp, k, 1) whose resolution is invariant under the action of Zkp. 14.5.4 Nested designs 14.5.38 Theorem Let p be a prime, n ≥1, pn ≡1 (mod 6), and ω be a primitive root of Fpn, pn = 6t + 1. Let S = {ω0, ω2t, ω4t}, and Si = ωiS. 1. For 0 ≤c < t, the development of {0}∪Sc under the addition and multiplication of Fpn forms a (pn, 4, 2) design in which the omission of the ﬁrst element in each block yields an STS(pn). 2. For 0 ≤c < d < t, the development of Sc ∪Sd under the addition and multipli-cation of Fpn forms a (pn, 6, 5) design. Combinatorial 595 14.5.39 Remark The STSs in Theorem 14.5.38 Part 1 have been called nested Steiner triple systems, but the standard statistical notion of nested design is diﬀerent – see Deﬁnition 14.5.40 and . 14.5.40 Deﬁnition If the blocks of a BIBD (V, D1) with v symbols in b1 blocks of size k1 are each partitioned into sub-blocks of size k2, and the b2 = b1k1/k2 sub-blocks them-selves constitute a BIBD (V, D2), then the system of blocks, sub-blocks, and symbols is a nested balanced incomplete block design (nested BIBD or NBIBD) with parameters (v, b1, b2, r, k1, k2), r denoting the common replication. Also (V, D1) and (V, D2) are the component BIBDs of the NBIBD. 14.5.41 Remark A resolvable BIBD (RBIBD) (V, D) is a nested block design (V, D1, D2) where the blocks of D1, of size k1 = v, are the resolution classes of D, and D2 = D. 14.5.42 Remark Nested block designs may have more than two blocking systems and consequently more than one level of nesting. A doubly nested block design is a system (V, D1, D2, D3) where both (V, D1, D2) and (V, D2, D3) are nested block designs. A resolvable NBIBD is a doubly nested block design. 14.5.43 Deﬁnition A multiply nested BIBD (MNBIBD) is a nested block design (V, D1, D2, . . . , Ds) with parameters (v, b1, . . . , bs, r, k1, . . . , ks) for which the systems (V, Dj, Dj+1) are NBIBDs for j = 1, . . . , s −1. 14.5.44 Theorem Let v be a prime power of the form v = a0a1a2 · · · an + 1 (a0 ≥1, an ≥1 and ai ≥2 for 1 ≤i ≤n −1 are integers). Then there is an MNBIBD with n component designs having k1 = ua1a2 · · · an, k2 = ua2a3 · · · an, . . . , kn = uan, and with a0v blocks of size k1, for any integer u with 1 ≤u ≤a0 and u > 1 if an = 1. If integer t ≥2 is chosen so that 2 ≤tu ≤a0, then there is an MNBIBD with n + 1 component designs, with the same number of big blocks but of size k0 = tk1, and with its n other block sizes being k1, . . . , kn as given. Moreover, if a0 is even and ai is odd for i ≥1, then MNBIBDs can be constructed with the same block sizes but with a0v/2 blocks of size k1. 14.5.45 Deﬁnition A nested row-column design is a system (V, D1, D2, D3) for which (1) each of (V, D1, D2) and (V, D1, D3) is a nested block design, (2) each block of D1 may be displayed as a k2 × k3 row-column array, one member of the block at each position in the array, so that the columns are the D2 sub-blocks in that block, and the rows are the D3 sub-blocks in that block. 14.5.46 Deﬁnition A (completely balanced) balanced incomplete block design with nested rows and columns, BIBRC(v, b1, k2, k3), is a nested row-column design (V, D1, D2, D3) for which each of (V, D1, D2) and (V, D1, D3) is a NBIBD. 14.5.47 Theorem If v = mpq + 1 is a prime power and p and q are relatively prime, then initial nesting blocks for a BIBRC(v, mv, sp, tq) are Al = xl−1L⊗M for l = 1, . . . , m, where Ls×t = (xi+j−2)i,j, Mp×q = (x[(i−1)q+(j−1)p]m)i,j, s and t are integers with st ≤m, and x is a primitive element of Fv. (Here ⊗is the Kronecker product.) If m is even and pq is odd, A1, . . . , Am/2 are intial nesting blocks for BIBRC(v, mv/2, sp, tq); 14.5.48 Theorem Write xui = 1−x2mi where x is a primitive element of Fv and v = 4tm+1 is a prime power. Let A be the addition table with row margin (x0, x2m, . . . , x(4t−2)m) and 596 Handbook of Finite Fields column margin (xm, x3m, . . . , x(4t−1)m), and set Al = xl−1A. If ui −uj ̸≡m (mod 2m) for i, j = 1, . . . , t, then A1, . . . , Am are initial nesting blocks for BIBRC(v, mv, 2t, 2t). Including 0 in each margin for A, if further ui ̸≡m (mod 2m) for i = 1, . . . , t, then A1, . . . , Am are initial nesting blocks for BIBRC(v, mv, 2t + 1, 2t + 1). 14.5.5 Pairwise balanced designs 14.5.49 Deﬁnition Let K be a subset of positive integers and let λ be a positive integer. A pairwise balanced design (PBD(v, K, λ) or (K, λ)-PBD) of order v with block sizes from K is a pair (V, B), where V is a ﬁnite set (the point set) of cardinality v and B is a family of subsets (blocks) of V that satisfy (1) if B ∈B, then \|B\| ∈K and (2) every pair of distinct elements of V occurs in exactly λ blocks of B. The integer λ is the index of the PBD. The notations PBD(v, K) and K-PBD of order v are often used when λ = 1. 14.5.50 Example A PBD(10, {3, 4}) is given below where the blocks are listed columnwise. 1 1 1 2 2 2 3 3 3 4 4 4 2 5 8 5 6 7 5 6 7 5 6 7 3 6 9 8 9 10 10 8 9 9 10 8 4 7 10 14.5.51 Remark Many constructions of pairwise balanced designs employ sub-structures in balanced incomplete block designs. In a (v, k, λ)-design (V, B), useful sub-structures include those speciﬁed by a set S ⊂V so that for every B ∈B, \|B ∩S\| ∈L; then setting K = {k −ℓ: ℓ∈L}, a (\|V \ S\|, K, λ)-PBD arises by removing all points of S. When the BIBD is made by a ﬁnite ﬁeld construction, such sub-structures may arise from algebraic properties of the ﬁeld. Other useful sub-structures arise from the presence of parallel classes; when a parallel class of blocks is present, a new element can be added and adjoined to each block in the parallel class to increase the size of some blocks by one. Example 14.5.50 is produced in this way from a (9,3,1)-design. This can be applied to more than one parallel class, when present [706, §IV]. 14.5.6 Group divisible designs 14.5.52 Deﬁnition Let K and G be sets of positive integers and let λ be a positive integer. A group divisible design of index λ and order v ((K, λ)-GDD) is a triple (V, G, B), where V is a ﬁnite set of cardinality v, G is a partition of V into parts (groups) whose sizes lie in G, and B is a family of subsets (blocks) of V that satisfy (1) if B ∈B then \|B\| ∈K, (2) every pair of distinct elements of V occurs in exactly λ blocks or one group, but not both, and (3) \|G\| > 1. If v = a1g1 +a2g2 +· · ·+asgs, and if there are ai groups of size gi, i = 1, 2, . . . , s, then the (K, λ)-GDD is of type ga1 1 ga2 2 . . . gas s . This is exponential notation for the group type. Alternatively, if the GDD has groups G1, G2, . . . , Gt, then the list T = [\|Gi\| : i = 1, 2, . . . , t] is the type of the GDD when more convenient. If K = {k}, then the (K, λ)-GDD is a (k, λ)-GDD. If λ = 1, the GDD is a K-GDD. Furthermore, a ({k}, 1)-GDD is a k-GDD. 14.5.53 Deﬁnition Let H be a subgroup of order h of a group G of order v. A collection {B1, ..., Bt} of k-subsets of G forms a (v, h, k, λ) diﬀerence family over G and relative to H if ∂B1 ∪ · · · ∪∂Bt covers each element of G −H exactly λ times and covers no element in H. Combinatorial 597 14.5.54 Remark 1. A (v, 1, k, λ) diﬀerence family is a (v, k, λ) diﬀerence family. 2. If {B1, . . . , Bt} is a (v, h, k, λ) diﬀerence family over G and relative to H, then OrbG(B1) ∪· · · ∪OrbG(Bt) is the collection of blocks of a (k, λ) GDD of type hv/h where the groups are the right cosets of H in G. This GDD admits G as a sharply point-transitive automorphism group. 3. A (k, λ) GDD of type hv/h with an automorphism group G acting sharply tran-sitively on the points is, up to isomorphisms, generated by a suitable (v, h, k, λ) diﬀerence family. 4. If {B1, . . . , Bt} is a (v, k, k, λ) diﬀerence family over G and relative to H, then {B1, . . . , Bt} ∪{H, . . . , H \| {z } λ times } is a (v, k, λ) diﬀerence family. 14.5.55 Theorem [4, 1357] Suppose q ≡1 (mod k−1) is a prime power. Then a (kq, k, k, 1) relative diﬀerence family over Fk × Fq exists if one of the following holds: 1. k ∈{3, 5} (for k = 5, the initial block B can be taken as {(0, 0), (1, 1), (1, −1), (4, x), (4, −x)} where x is any nonsquare in Fq such that exactly one of x−1, x+1 is a square); 2. k = 7 and q ̸= 19; 3. k = 9 and q ̸∈{17, 25, 41, 97, 113}; 4. k = 11, q < 1202, q is prime, and q ̸∈[30,192], [240,312] or [490,492]. 14.5.56 Theorem 1. Let q = 12t + 1 be a prime power, and let 3e be the largest power of 3 dividing t. If, in Fq, 3 and 2 + √ 3 are both 3e-th powers but not 3e+1-th powers and 6 is not a 3e+1-th power, then a (13q, 13, 13, 1) relative diﬀerence family exists. 2. If p and q are odd prime powers with q > p, then a (pq, p, p, (p −1)/2) relative diﬀerence family exists. If further p ≡1 (mod 4) and q ≡1 (mod p −1), then a (pq, p, p, (p −1)/4) relative diﬀerence family exists. 14.5.7 t-designs 14.5.57 Deﬁnition A t-(v, k, λ) design, a t-design in short, is a pair (X, B) where X is a v-set of points and B is a collection of k-subsets of X (blocks) with the property that every t-subset of X is contained in exactly λ blocks. The parameter λ is the index of the design. 14.5.58 Example Let q be a prime power and n > 0 be an integer. Then G = PGL(2, qn) acts sharply 3-transitively on X = Fqn ∪{∞}. If S ⊆X is the natural inclusion of Fq ∪{∞}, then the orbit of S under G is a 3-(qn + 1, q + 1, 1) design. These designs are spherical geometries; when n = 2, they are inversive planes or M¨ obius planes. 14.5.59 Theorem Let B be a subgroup of the multiplicative group of nonzero elements of Fq. Then the orbit of S = B ∪{0, ∞} under the action of PGL(2, q) is the block set of one of the designs: 598 Handbook of Finite Fields 1. 3-(qn + 1, q + 1, 1) design where q is a prime power and n ≥2 (a spherical geometry); 2. 3-(q + 1, k + 1, k(k + 1)/2) design if (k −1)\|(q −1), k̸ \|q, and k ̸∈{3, 5}; 3. 3-(q + 1, 4, 3) design for q ≡1, 5 (mod 6); 4. 3-(q + 1, 6, 5) design for q ≡1, 9, 13, 17 (mod 20). 14.5.8 Packing and covering 14.5.60 Remark Packings and coverings relax the conditions on block designs, and have been ex-tensively studied; see for a more detailed exposition. 14.5.61 Deﬁnition Let v ≥k ≥t. A t-(v, k, λ) covering is a pair (X, B), where X is a v-set of elements (points) and B is a collection of k-subsets (blocks) of X, such that every t-subset of points occurs in at least λ blocks in B. Repeated blocks in B are permitted. 14.5.62 Theorem (Sch¨ onheim bound) Cλ(v, k, t) ≥⌈v Cλ(v −1, k −1, t −1)/k⌉. Iterating this bound yields Cλ(v, k, t) ≥Lλ(v, k, t), where Lλ(v, k, t) = l v k l v−1 k−1 . . . l λ(v−t+1) k−t+1 mmm . 14.5.63 Deﬁnition Let v ≥k ≥t. A t-(v, k, λ) packing is a pair (X, B), where X is a v-set of elements (points) and B is a collection of k-subsets of X (blocks), such that every t-subset of points occurs in at most λ blocks in B. If λ > 1, then B is allowed to contain repeated blocks. 14.5.64 Remark A t-(v, k, 1) packing with b blocks is equivalent to a binary code of length v, size b, constant weight k, and minimum Hamming distance at least 2(k −t + 1); see Section 15.1. 14.5.65 Theorem (First Johnson bound) Dλ(v, k, t) ≤ j vDλ(v−1,k−1,t−1) k k . Iterating this bound yields Dλ(v, k, t) ≤Uλ(v, k, t), where Uλ(v, k, t) = j v k j v−1 k−1 . . . j λ(v−t+1) k−t+1 kkk . Further, if λ(v−1) ≡0 (mod k−1) and δ(k−1) 2 > λ δ(δ−1) 2 , where δ = λv(v−1) k−1 −kUλ(v, k, 2), then Dλ(v, k, 2) ≤Uλ(v, k, 2) −1. 14.5.66 Theorem (Second Johnson bound) Suppose d = D1(v, k, t) = qv + r, where 0 ≤r ≤ v −1. Then q(q −1)v + 2qr ≤(t −1)d(d −1), and hence D1(v, k, t) ≤ j v(k+1−t) k2−v(t−1) k . Combinatorial 599 See Also §14.1 For latin squares, MOLS, and transversal designs. §14.3 For aﬃne and projective planes. §14.4 For projective spaces. §14.6 For diﬀerence sets. §14.7 For other combinatorial structures. A textbook on combinatorial designs. Another textbook on combinatorial designs. For triple systems (§II.2); balanced incomplete block designs (§II.1,II.3,II.5); t-designs (§II.4,II.5); symmetric designs (§II.6); Hadamard designs and matrices(§V.1); diﬀerence sets (§VI.18); resolvable designs (§II.7); pairwise balanced designs and group divisible designs (§IV); coverings (§VI.11); packings (§VI.40); nested designs (§VI.36); connections with MOLS and transversal designs (§III.1,III.3). For nested designs. For an introductory textbook on combinatorial designs. References Cited: [4, 5, 261, 262, 356, 459, 460, 461, 706, 708, 807, 1268, 1356, 1357, 1459, 1559, 1741, 2106, 2159, 2222, 2440, 2719, 2986] 14.6 Diﬀerence sets Alexander Pott, Otto-von-Guericke-Universit¨ at Magdeburg 14.6.1 Basics 14.6.1 Deﬁnition Let G be an additively written group of order v. A k-subset D of G is a (v, k, λ; n)-diﬀerence set of order n = k −λ if every nonzero element of G has exactly λ representations as a diﬀerence d−d′ with elements from D. The diﬀerence set is abelian, cyclic, etc., if the group G has the respective property. The redundant parameter n is sometimes omitted, therefore the notion of (v, k, λ)-diﬀerence sets is also used. 14.6.2 Example 1. The group G itself and G{g} for an arbitrary g ∈G are (v, v, v, 0)- and (v, v − 1, v −2; 1)-diﬀerence sets. 2. The set {1, 3, 4, 5, 9} is a cyclic (11, 5, 2; 3)-diﬀerence set in the group (Z/11Z, +). These are the squares modulo 11. 3. The set {1, 2, 4} ⊂Z/7Z is a cyclic (7, 3, 1; 2)-diﬀerence set. 4. The set {(0, 1), (0, 2), (0, 3), (1, 0), (2, 0), (3, 0)} ⊂Z/4Z × Z/4Z is a (16, 6, 2; 4) diﬀerence set. 600 Handbook of Finite Fields 5. There is a non-abelian example with the same parameters: Let G = Q × Z/2Z, where Q = {±1, ±i, ±j, ±k} is the quaternion group, whose generators sat-isfy i2 = j2 = k2 = 1, ij = jk = ki = −1. Then the set D = {(1, 0), (i, 0), (j, 0), (k, 0), (1, 1), (−1, 1)} is a non-abelian (16, 6, 2; 4)-diﬀerence set. 14.6.3 Remark A thorough investigation of diﬀerence sets is contained in , see also the tables in . A short summary, including a list of small examples, is contained in . Classical textbooks are and . A recent book including a modern treatment of necessary conditions for the existence of diﬀerence sets is . 14.6.4 Remark The complement of a (v, k, λ; n)-diﬀerence set is again a diﬀerence set but with parameters (v, v −k, v −2k + λ; n). Therefore, we may assume k ≤v/2 (the case k = v/2 is actually impossible). 14.6.5 Remark Many constructions of diﬀerence sets are closely related to the connection between the additive and the multiplicative group of a ﬁnite ﬁeld: 1. The set of nonzero squares in a ﬁeld Fq, q ≡3 (mod 4), which is a multiplicative subgroup, is a diﬀerence set in the additive group of the ﬁeld, see Theorem 14.6.38. Case 2 in Example 14.6.2 is such a diﬀerence set in (F11, +). 2. The set of elements of trace 0 in F2n, which is an additive subgroup, forms a diﬀerence set in the multiplicative group of F2n; see Theorem 14.6.22. Case 3 in Example 14.6.2 is such a diﬀerence set in (F∗ 8, ·) ∼ = Z/7Z. 14.6.6 Lemma The parameters v, k and λ of a diﬀerence set satisfy λ · (v −1) = k · (k −1). 14.6.7 Remark Lemma 14.6.6 can be proved by counting diﬀerences. It also follows from Theorem 14.6.9 which shows that diﬀerence sets are the same objects as symmetric designs with a sharply transitive (regular) automorphism group. We refer the reader to Section 14.5 for the deﬁnition of symmetric designs and to for a proof of the important Theorem 14.6.9. 14.6.8 Deﬁnition The development of a diﬀerence set D is the incidence structure dev(D) whose points are the elements of G and whose blocks are the translates g+D := {g+d : d ∈D}. 14.6.9 Theorem The existence of a (v, k, λ; n)-diﬀerence set is equivalent to the existence of a symmetric (v, k, λ)-design D admitting G as a point regular automorphism group; i.e., for any two points P and Q, there is a unique group element g which maps P to Q. The design D is isomorphic with dev(D). 14.6.10 Remark Necessary conditions on the parameters v, k and λ of a symmetric design are also necessary conditions for the parameters of a diﬀerence set. In particular, the following two theorems hold. We emphasize that these are necessary conditions for symmetric designs, even if the designs are not constructed from diﬀerence sets. 14.6.11 Theorem If D is a (v, k, λ; n) diﬀerence set with v even, then n = u2 is a square. 14.6.12 Theorem [429, 631] If D is a (v, k, λ; n) diﬀerence set with v odd, then the equation nx2 + (−1)(v−1)/2λy2 = z2 must have an integral solution (x, y, z) ̸= (0, 0, 0). 14.6.13 Example Not all symmetric designs can be constructed from diﬀerence sets. There are, for instance, no diﬀerence sets with parameters (25, 9, 3; 6) or (31, 10, 3; 7), but symmetric designs with these parameters exist, see the tables in . Combinatorial 601 14.6.14 Remark The main problem about diﬀerence sets is to give necessary and suﬃcient con-ditions for their existence. These conditions sometimes depend only on the parameters (v, k, λ; n), sometimes also on the structure of the ambient group G. Another problem is to classify all (v, k, λ; n)-diﬀerence sets up to equivalence or isomorphism. In many nonexis-tence theorems, the exponent of a group plays an important role. 14.6.15 Deﬁnition The exponent exp(G) of a (multiplicatively written) ﬁnite group G is the small-est integer v∗such that gv∗= 1. 14.6.16 Remark There are many necessary conditions that the parameters of a diﬀerence set have to satisfy. An important condition is Theorem 14.6.62. Many more restrictions are in . 14.6.17 Deﬁnition Two diﬀerence sets D1 (in G1) and D2 (in G2) are equivalent if there is a group isomorphism ϕ between G1 and G2 such that Dϕ 1 = {dϕ : d ∈D1} = g + D2 for a suitable g ∈G2. The diﬀerence sets are isomorphic if the designs dev(D1) and dev(D2) are isomorphic. 14.6.18 Remark Equivalent diﬀerence sets yield isomorphic designs, but a design may give rise to several inequivalent diﬀerence sets, as the following example shows. 14.6.19 Example The following three diﬀerence sets in Z/4Z × Z/4Z with parameters (16, 6, 2; 4) are pairwise inequivalent, but the designs are all isomorphic: D1 = {(0, 0), (1, 0), (2, 0), (0, 1), (1, 2), (2, 3)}, D2 = {(0, 0), (1, 0), (2, 0), (0, 1), (0, 3), (3, 2)}, D3 = {(0, 0), (1, 0), (0, 1), (2, 1), (1, 2), (2, 3)}. 14.6.20 Deﬁnition Some types of (v, k, λ; n)-diﬀerence sets have special names. A diﬀerence set with λ = 1 is planar. The parameters can be written in terms of the order n as (n2 + n + 1, n + 1, 1; n). The corresponding design is a projective plane. Diﬀerence sets with v = 4n are Hadamard diﬀerence sets, in which case n = u2 must be a square, and the parameters are (4u2, 2u2 −u, u2 −u; u2). Diﬀerence sets with v = 4n−1, hence with parameters (4n −1, 2n −1, n −1; n), are of Paley type. Both Hadamard and Paley type diﬀerence sets are closely related to Hadamard matrices; see Constructions 14.6.44 and 14.6.52. The parameters qn−1 q−1 , qn−1−1 q−1 , qn−2−1 q−1 ; qn−2 are the Singer parameters; see Theorem 14.6.22. 14.6.21 Remark The parameters of a symmetric design, hence also the parameters of a (v, k, λ; n)-diﬀerence set satisfy 4n −1 ≤v ≤n2 + n + 1. The extremal cases are Paley type diﬀerence sets (v = 4n −1) and planar diﬀerence sets. 14.6.2 Diﬀerence sets in cyclic groups 14.6.22 Theorem Let α be a generator of the multiplicative group of Fqn, where q is a prime power. Then the set of integers {i : 0 ≤i < (qn −1)/(q −1), Tr (αi) = 0} modulo (qn −1)/(q −1) form a (cyclic) diﬀerence set with Singer parameters qn −1 q −1 , qn−1 −1 q −1 , qn−2 −1 q −1 ; qn−2 . These diﬀerence sets are Singer diﬀerence sets. 602 Handbook of Finite Fields 14.6.23 Remark 1. If q = 2, the Singer diﬀerence set is simply the set of nonzero elements in an additive subgroup of order 2n−1, interpreted as a subset of the multiplicative group of F2n. These diﬀerence sets are of Paley type. 2. In the general case, a Singer diﬀerence set can be viewed as follows. Let U := {β ∈F∗ qn : Tr (β) = 0}. This is a subgroup of the additive group of Fqn ﬁxed by multiplication with elements from Fq, hence it is a hyperplane of the vector space Fn q . The Singer diﬀerence set is the image of this hyperplane under the canonical projection F∗ qn →F∗ qn/F∗ q. 3. The design corresponding to a Singer diﬀerence set is the classical point-hyperplane design of the projective geometry PG(n −1, q). 14.6.24 Conjecture If n = 3, the Singer parameters are (q2 +q +1, q +1, 1; q). It is conjectured that the only abelian diﬀerence sets (up to equivalence) with these parameters are the Singer diﬀerence sets. Moreover, it is conjectured that planar diﬀerence sets exist only if the order q is a prime power. This holds for all orders q ≤2, 000, 000; see . 14.6.25 Construction The construction of Singer diﬀerence sets is easy if a primitive polynomial (see Section 4.1) f(x) = xn + Pn i=1 aixn−i of degree n in Fqn is known. Consider the recurrence relation γm = −Pn i=1 aiγm−i. Take arbitrary initial values, for instance γ0 = 1, γ1 = γ2 = · · · = γn−1 = 0. Then the set of integers {0 ≤i < (qn −1)/(q −1) : γi = 0} is a Singer diﬀerence set. For instance, x4 + x3 + 2 is a primitive polynomial over F3. The recurrence relation γm = 2γm−1 + γm−4 yields the sequence 10001212201112222020211201021002212022002000 . . . which gives the cyclic (40, 13, 4; 9)-diﬀerence set {1, 2, 3, 9, 17, 19, 24, 26, 29, 30, 35, 38, 39}. 14.6.26 Remark In general, there are many diﬀerence sets with Singer parameters inequivalent to Singer diﬀerence sets (Theorems 14.6.27 and 14.6.29). There are even non-abelian diﬀerence sets with Singer parameters; see Example 14.6.70. 14.6.27 Theorem Let D be an arbitrary cyclic diﬀerence set with parameters qs−1 q−1 , qs−1−1 q−1 , qs−2−1 q−1 ; qs−2 in a group G. Let G be embedded into F∗ qn/F∗ q which is possible if s\|n. Let α be a primitive element in Fqn. Then the set of integers {0 ≤i < (qn −1)(q−1) : Tr Fqn/Fqs (αi) ∈D} is a diﬀerence set with classical Singer parameters. The diﬀerence sets are Gordon-Mills-Welch diﬀerence sets corresponding to D. Note that diﬀerent embeddings of the same diﬀerence set D may result in inequivalent diﬀerence sets . 14.6.28 Remark If D is a Singer diﬀerence set, the above construction may be reformulated as follows: if s divides n and if r is relatively prime to qs −1, then the set of integers {0 ≤i < (qn −1)/(q −1) : Tr Fqs/Fq[(Tr Fqn/Fqs (αi))r] = 0} is a Gordon-Mills-Welch diﬀerence set. 14.6.29 Theorem [862, 864] Let α be a generator of the multiplicative group of F2n, and let t < n/2 be an integer relatively prime to n, and k = 4t −2t +1. Then the set D = {(x+1)k +xk +1 : x ∈F2n, x ̸= 0, 1} ⊆F∗ 2n is a Dillon-Dobbertin diﬀerence set with parameters (2n−1, 2n−1− 1, 2n−2 −1; 2n−2). If n = 3t ± 1, then the set D = F∗ 2n \ {(x + 1)k + xk : x ∈F2n} ⊆F∗ 2n is a diﬀerence set with parameters (2n −1, 2n−1 −1, 2n−2 −1; 2n−2). 14.6.30 Remark The series of Gordon-Mills-Welch diﬀerence sets and Dillon-Dobbertin dif-ference sets [862, 864] show that the number of inequivalent diﬀerence sets grows rapidly. In Combinatorial 603 these two series, inequivalent diﬀerence sets are in general also non-isomorphic; see for the Gordon-Mills-Welch case and for the Dillon-Dobbertin case. 14.6.31 Construction Cyclic diﬀerence sets can be used to construct binary sequences with 2-level autocorrelation function. Let α be the generator of the cyclic group Z/vZ, and let D be a (v, k, λ; n) diﬀerence set in Z/vZ. Deﬁne a sequence (ai) by ai = 1 if αi ∈D, otherwise ai = 0. This sequence has period v, and Ct(a) := Pn−1 i=0 (−1)ai+ai+t = v −4(k −λ) for t = 1, . . . , v −1, which are the oﬀ-phase autocorrelation coeﬃcients; see also Section 10.3. 14.6.32 Remark Cyclic Paley type diﬀerence sets yield sequences with constant oﬀ-phase autocor-relation −1. These sequences (diﬀerence sets) have numerous applications since the autocor-relation is small (in absolute value) . It is conjectured that no sequences with constant oﬀ-phase autocorrelation 0 exist if v > 4; see Remark 14.6.45. 14.6.33 Conjecture (Ryser’s Conjecture) [1842, 1909] If gcd(v, n) ̸= 1, then there is no cyclic (v, k, λ; n) diﬀerence set in a cyclic group. A strengthening of this conjecture is due to Lander: if D is a (v, k, λ; n)-diﬀerence set in an abelian group of order v, and p is a prime dividing v and n, then the Sylow p-subgroup of G is not cyclic. 14.6.34 Theorem Lander’s conjecture is true for all abelian diﬀerence sets of order n = pk, where p > 3 is prime. 14.6.35 Remark 1. The smallest open case for Lander’s conjecture is a cyclic (465, 145, 45; 100) dif-ference set. 2. More restrictions on putative counterexamples to Lander’s conjecture are con-tained in . 14.6.36 Theorem Let R = {a ∈F∗ 3m : a = x + x6 has 4 solutions with x ∈F3m} with m > 1. Then the set ρ(R) is a diﬀerence set with Singer parameters ((3m −1)/2, (3m−1 − 1)/2, (3m−2 −1)/2; 3m−2), where ρ is the canonical epimorphism F∗ 3m →F∗ 3m/F∗ 3. 14.6.37 Theorem Let q = 3e, e ≥1, m = 3k, d = q2k −qk + 1. If R = {x ∈Fqm : Tr Fqm/Fq(x + xd) = 1}, then ρ(R) is a diﬀerence set with parameters ((qm −1)/(q − 1), qm−1, qm−1 −qm−2; qm−2), where ρ is the canonical epimorphism F∗ qm →F∗ q. 14.6.3 Diﬀerence sets in the additive groups of ﬁnite ﬁelds 14.6.38 Theorem The following subsets of Fq are diﬀerence sets in the additive group of Fq. They are cyclotomic diﬀerence sets. Some of these diﬀerence sets may have Singer parameters. 1. F(2) q := {x2 : x ∈Fq{0}}, q ≡3 (mod 4) (quadratic residues, Paley diﬀerence sets); 2. F(4) q := {x4 : x ∈Fq{0}}, q = 4t2 + 1, t odd; 3. F(4) q ∪{0}, q = 4t2 + 9, t odd; 4. F(8) q = {x8 : x ∈Fq{0}}, q = 8t2 + 1 = 64u2 + 9, t, u odd; 5. F(8) q ∪{0}, q = 8t2 + 49 = 64u2 + 441, t odd, u even; 6. H(q) = {xi : x ∈Fq{0}, i ≡0, 1 or 3 (mod 6)}, q = 4t2 + 27, q ≡1 (mod 6) (Hall diﬀerence sets). 14.6.39 Remark The proofs of the statements in Theorem 14.6.38 use cyclotomic numbers . 604 Handbook of Finite Fields 14.6.40 Theorem Let q and q +2 be prime powers. Then the set D = {(x, y) : x, y are both nonzero squares or both non-squares or y = 0} is a twin prime power diﬀerence set with parameters q2 + 2q, q2 + 2q −1 2 , q2 + 2q −3 4 ; q2 + 2q + 1 4 in the group (Fq, +) × (Fq+2, +); see . 14.6.41 Deﬁnition A diﬀerence set D in the group G is skew symmetric if D is of Paley type and {0, d, −d : d ∈D} = G, hence D ∩−D = ∅. 14.6.42 Theorem The following sets are skew symmetric diﬀerence sets in the additive group of Fq, q ≡3 (mod 4): 1. {x2 : x ∈Fq, x ̸= 0} (Paley diﬀerence sets); 2. {x10 ± x6 −x2 : x ∈Fq, x ̸= 0} where q = 3h, h odd ; 3. {x4a+6 ± x2a −x2 : x ∈Fq, x ̸= 0} where q = 3h, h odd, a = 3 h+1 2 . 14.6.43 Remark A large class of skew Hadamard diﬀerence sets in elementary abelian groups of order q3 (q prime power) has been recently constructed . 14.6.4 Diﬀerence sets and Hadamard matrices 14.6.44 Construction A Hadamard diﬀerence set D in a group G of order 4u2 (see Deﬁnition 14.6.20) gives rise to a Hadamard matrix (Section 14.5) as follows: Label the rows and columns of a matrix H = (hx,y) by the elements of G, and put hx,y = 1 if x −y ∈D, otherwise hx,y = −1. This matrix is a Hadamard matrix, see Section 14.5. 14.6.45 Remark A special case of Rysers’s Conjecture 14.6.33 is that there are no cyclic Hadamard diﬀerence sets with v > 4 (if v = 4, there is a trivial cyclic (4, 1, 0; 1)-diﬀerence set). This is also known as the circulant Hadamard matrix conjecture. The smallest open case for which one cannot prove the nonexistence of a cyclic Hadamard diﬀerence set with v = 4u2 so far is u = 11715 = 3 · 5 · 11 · 71; see and also for the connection to the Barker sequence conjecture (Section 10.3). 14.6.46 Remark Hadamard diﬀerence sets in elementary abelian groups are equivalent to bent functions (Section 9.3). The bent function is the characteristic function of the Hadamard diﬀerence set. The following theorem gives an explicit construction. 14.6.47 Theorem The set {(x1, . . . , x2m) ∈F 2m 2 : x1x2 + x3x4 + · · · + x2m−1x2m = 1} ⊂F2m 2 is a Hadamard diﬀerence set with parameters (22m, 22m−1 −2m−1, 22m−2 −2m−1; 22m−2). 14.6.48 Remark There are several other constructions of diﬀerence sets with these parameters, also in other groups. Two major construction methods are the Maiorana-McFarland method (Sections 9.1 and 9.3) and the use of partial spreads (Section 9.3). 14.6.49 Theorem [781, 1803, 2830] An abelian Hadamard diﬀerence set in a group G of order 22m+2 exists if and only if exp(G) ≤2m+2. 14.6.50 Remark In the following theorem, we combine knowledge about the existence of abelian Hadamard diﬀerence sets. Many authors contributed to this theorem. Combinatorial 605 14.6.51 Theorem Let G ∼ = H × EA(w2) be an abelian group of order 4u2 with u = 2a3bw2 where w is the product of not necessarily distinct primes p ≡3 (mod 4) and EA(w2) denotes the group of order w2 which is the direct product of groups of prime order. If H is of type (2a1)(2a2) · · · (2as)(3b1)2 · · · (3br)2 with P ai = 2a+2 (a ≥0, ai ≤a+2), P bi = 2b (b ≥0), then G contains a Hadamard diﬀerence set of order u2. 14.6.52 Construction Diﬀerence sets with parameters (4n−1, 2n−1, n−1; n) (hence of Paley type) in G can be used to construct Hadamard matrices: Label the rows and columns of a matrix H by the elements of G ∪{∞}. The matrix H = (hu,v) such that hu,v = 1 if u = ∞ or v = ∞or u −v ∈D is a Hadamard matrix of order 4n. 14.6.53 Theorem For the following orders n, Paley type diﬀerence sets exist in groups of order v = 4n −1: 1. 4n −1 is a prime power (Theorem 14.6.38); 2. 4n−1 is the product q(q +2) of two prime powers q and q +2 (Theorem 14.6.40); 3. 4n −1 = 2m −1 (Theorem 14.6.22). 14.6.54 Problem It is an open question whether Paley type diﬀerence sets exist for other values. 14.6.5 Further families of diﬀerence sets 14.6.55 Theorem Let q be a prime power and d a positive integer. Let G be a group of order v = qd+1(qd + · · · + q2 + q + 2) which contains an elementary abelian subgroup E of order qd+1 in its center. View E as the additive group of Fd+1 q . Put r = (qd+1 −1)/(q −1) and let H1, . . . , Hr be the hyperplanes of order qd of E. If g0, . . . , gr are distinct coset representatives of E in G, then D = (g1 + H1) ∪(g2 + H2) ∪· · · ∪(gr + Hr) is a McFarland diﬀerence set with parameters qd+1(1 + qd+1−1 q−1 ), qd · qd+1−1 q−1 , qd · qd−1 q−1 ; q2d . 14.6.56 Remark 1. If q = 2, the McFarland construction gives Hadamard diﬀerence sets. 2. If q = 2 and G is elementary abelian, this construction is known as the Maiorana-McFarland construction of bent functions; see Section 9.3. 14.6.57 Theorem [609, 782] Let q be a prime power, and let t be any positive integer. Diﬀerence sets with parameters 4q2t q2t −1 q2 −1 , q2t−1 2q2t + q −1 q + 1 , q2t−1(q −1)q2t−1 + 1 q + 1 ; q4t−2 exist in abelian groups G in the following cases: 1. q = 3f, the Sylow 3-subgroup of G is elementary abelian; 2. q = p2f, p odd, the Sylow p-subgroup of G is elementary abelian; 3. q = 2f, the Sylow 2-subgroup of G has rank ≥2f + 1; 4. q = 2, exp(G) ≤4. If t = 1, these diﬀerence sets are Hadamard diﬀerence sets. 606 Handbook of Finite Fields 14.6.6 Diﬀerence sets and character sums 14.6.58 Remark The existence of diﬀerence sets is closely related to character sums. Most necessary conditions on the existence of diﬀerence sets are derived from character sums and number theoretic conditions. 14.6.59 Theorem [262, 2830] Let D be a (v, k, λ; n) diﬀerence set in G, and let χ be a homomorphism from G into the multiplicative group of a ﬁeld. If χ(g) = 1 for all g ∈G (in which case the homomorphism is denoted χ0), then P g∈D χ0(d) = k. If χ ̸= χ0, then X d∈D χ(d) ! · X d∈D χ(d−1) ! = n. 14.6.60 Remark Theorem 14.6.59 is very useful if χ is complex-valued. In this case, the sum χ(D) := P d∈D χ(d) is an element in the ring Z[ζv∗], where ζv∗= e2πi/v∗is a primitive v∗-th root of unity, and v∗is the exponent of G. 14.6.61 Remark If χ is complex-valued, Theorem 14.6.59 may be also viewed as an equation about the ideal generated by χ(D): For χ ̸= χ0, we have (χ(D))(χ(D)) = (n), where (·) denotes an ideal generated in Z[ζv∗] and ( ) is complex conjugation. Using results from algebraic number theory, many necessary conditions can be obtained, for instance Theorem 14.6.62. 14.6.62 Theorem Let D be an abelian (v, k, λ; n) diﬀerence set in G, and let w be a divisor of v. If p is prime, p\|n and pj ≡−1 (mod w), then an integer i exists such that p2i\|n, but p2i+1 is not a divisor of n. If w is the exponent v∗of G, then p does not divide n. 14.6.63 Example [262, 1842] There is no (40, 13, 4; 9)-diﬀerence set in Z/2Z×Z/2Z×Z/2Z×Z/5Z (use v∗= 10 and p = 3 in Theorem 14.6.62). Using a diﬀerent (though similar) theorem, one can also rule out the existence of a (40, 13, 4; 9)-diﬀerence set in Z/2Z × Z/4Z × Z/5Z. Note that a cyclic diﬀerence set with these parameters exists (Construction 14.6.25). 14.6.7 Multipliers 14.6.64 Deﬁnition Let D be a diﬀerence set in G. Then ϕ ∈Aut(G) is a multiplier of D if Dϕ := {ϕ(D) : d ∈D} = g+D for some g ∈G. If G is abelian and ϕ is the automorphism that maps h to t · h, then t is a numerical multiplier. 14.6.65 Theorem If ϕ is a multiplier of the diﬀerence set D, then there is at least one translate g + D of D which is ﬁxed by ϕ. If D is abelian and gcd(v, k) = 1, then there is a translate ﬁxed by all multipliers . 14.6.66 Remark 1. Multipliers play an important role, in particular in the theory of abelian diﬀerence sets. 2. The content of a multiplier theorem is the assertion that certain automorphisms (integers) have to be (numerical) multipliers of an abelian diﬀerence set depend-ing only on the parameters v, k, and λ. Theorem 14.6.67 is the ﬁrst multiplier theorem . 14.6.67 Theorem Let D be an abelian (v, k, λ; n)-diﬀerence set. If p is a prime which satisﬁes gcd(p, v) = 1, p\|n and p > λ, then p is a numerical multiplier. Combinatorial 607 14.6.68 Conjecture Every prime divisor p of n which is relatively prime to v is a multiplier of a (v, k, λ; n)-diﬀerence set, i.e., the condition p > λ in Theorem 14.6.67 is not necessary. 14.6.69 Remark 1. Multipliers are quite useful in constructing diﬀerence sets and in proofs of nonex-istence. 2. Several attempts have been made to weaken the assumption “p > λ” in Theorem 14.6.67 (second multiplier theorem, McFarland’s multiplier theorem) . 14.6.70 Example Multipliers may be used to construct non-abelian diﬀerence sets: The set D = {3, 6, 7, 12, 14} is a (21, 5, 1; 4)-diﬀerence set in (Z/21Z, +) with multiplier 4. Denote the automorphism x 7→x + 3 by a, and the automorphism x 7→4x + 1 by b. Then G = ⟨a, b: a7 = b3 = 1, b−1ab = a4⟩acts regularly on the points of dev(D). A diﬀerence set D′ in G corresponding to this action is D′ = {a, a2, a4, a4b, a5b}. See Also §9.2 Relative diﬀerence sets are a generalization of diﬀerence sets. An important class of relative diﬀerence sets can be described by planar functions (PN functions). §9.3 Bent functions are equivalent to elementary abelian Hadamard diﬀerence sets. §10.3 Cyclic diﬀerence sets are binary sequences with two-level autocorrelation function. §14.5 Diﬀerence sets are an important tool to construct combinatorial designs. References Cited: [116, 211, 261, 262, 429, 609, 631, 706, 781, 782, 862, 864, 871, 874, 905, 1303, 1324, 1326, 1477, 1676, 1803, 1842, 1907, 1909, 1910, 2051, 2168, 2210, 2546, 2566, 2674, 2725, 2830] 14.7 Other combinatorial structures Jeﬀrey H. Dinitz, University of Vermont Charles J. Colbourn, Arizona State University 14.7.1 Association schemes 14.7.1 Deﬁnition Let d denote a positive integer, and let X be a nonempty ﬁnite set. A d-class symmetric association scheme on X is a sequence R0, R1, . . . , Rd of nonempty subsets of the Cartesian product X × X, satisfying 1. R0 = {(x, x) \| x ∈X}, 2. X × X = R0 ∪R1 ∪· · · ∪Rd and Ri ∩Rj = ∅for i ̸= j, 3. for all i ∈{0, 1, . . . , d}, RT i = Ri where RT i := {(y, x) \| (x, y) ∈Ri}, 4. for all integers h, i, j ∈{0, 1, . . . , d}, and for all x, y ∈X such that (x, y) ∈Rh, the number ph ij := \|{z ∈X \| (x, z) ∈Ri, (z, y) ∈Rj}\| depends only on h, i, j, and not on x or y. 608 Handbook of Finite Fields 14.7.2 Example The Hamming scheme H(n, q) has the set F n of all words of length n over an alphabet F of q symbols as its vertex set. Two words are i-th associates if and only if the Hamming distance between them is i. Generally the alphabet F is F2, but other ﬁnite ﬁelds are also used. 14.7.3 Example The cyclotomic schemes are obtained as follows. Let q be a prime power and k a divisor of q −1. Let C1 be the subgroup of the multiplicative subgroup of Fq of index k, and let Ci, i = 1, 2, . . . , k be the cosets of C1 (the cyclotomic classes). The points of the scheme are the elements of Fq, and two points x, y are i-th associates if x −y ∈Ci (zero associates if x −y = 0). In order for this to be an association scheme one must have −1 ∈C1 or equivalently 2k must divide q −1 if q is odd. 14.7.2 Costas arrays 14.7.4 Deﬁnition A Costas array of order n is an n × n array of dots and blanks that satisﬁes: 1. There are n dots and n(n −1) blanks, with exactly one dot in each row and column. 2. All the segments between pairs of dots diﬀer in length or in slope. C(n) denotes the number of distinct n × n Costas arrays. 14.7.5 Construction (Welch construction) Let p be prime and α be a primitive element in the ﬁeld Fp. Let n = p −1. A Costas array of order n is obtained by placing a dot at (i, j) if and only if i = αj, for a ≤j < n + a, a a nonnegative integer, and i = 1, . . . , n. 14.7.6 Construction Let α and β be primitive elements in the ﬁeld Fq for q a prime power. Let n = q −2. Costas arrays of order n are obtained by 1. Lempel construction: Put a dot at (i, j) if and only if αi + αj = 1, 1 ≤i, j ≤n. 2. Golomb construction: Put a dot at (i, j) if and only if αi + βj = 1, 1 ≤i, j ≤n. 14.7.7 Remark Using Constructions 14.7.5 and 14.7.6, C(p −1) > 1 and C(q −2) > 1. Also, if a corner dot is present in a Costas array of order n, it can be removed along with its row and column to obtain a Costas array of order n −1. 14.7.8 Theorem If q > 2 is a prime power, then there exist primitive elements α and β in Fq such that α + β = 1. 14.7.9 Corollary Removing the corner dot at (1, 1) in the Costas array of order q −2 from Con-struction 14.7.6 Part 2 yields C(q −3) ≥1. 14.7.10 Example If there exist primitive elements α and β satisfying the conditions stated, then a Costas array of order n can be obtained by removing one or more corner dots. Conditions n α1 = 2 q −3 α1 + β1 = 1 and α2 + β2 = 1 q −4 α2 + α1 = 1 q −4 α1 + β1 = 1 and α2 + β−1 = 1 q −4 and necessarily α−1 + β2 = 1 q −5 Combinatorial 609 14.7.11 Remark All Costas arrays of order 28 are accounted for by the Golomb and Welch con-struction methods , making 28 the ﬁrst order (larger than 5) for which no sporadic Costas array exists. 14.7.12 Remark For n ≥30, n ̸∈{31, 53}, the only orders for which Costas arrays are known are orders n = p −1 or n = q −2 or orders for which some algebraic condition exists that guarantees corner dots whose removal leaves a smaller Costas array. 14.7.13 Remark The properties of a Costas array make it an ideal discrete waveform for Doppler sonar. Having one dot in each row and column minimizes reverberation. Distinct segments between pairs of dots give it a thumbtack ambiguity function because, shifted left-right in time and up-down in frequency, copies of the pattern can only agree with the original in one dot, no dots, or all n dots at once. Thus, the spike of the thumbtack makes a sharp distinction between the actual shift and all the near misses. See for a survey on Costas arrays and for up-to-date information on Costas arrays. 14.7.3 Conference matrices 14.7.14 Deﬁnition A conference matrix of order n is an n×n (0, ±1)-matrix C with zero diagonal satisfying CCT = (n −1)I. A conference matrix is normalized if all entries in its ﬁrst row and ﬁrst column are 1 (except the (1,1) entry which is 0). A square matrix A is symmetric if A = AT and skew-symmetric if A = −AT . The core of a normalized conference matrix C consists of all the rows and columns of C except the ﬁrst row and column. 14.7.15 Theorem [2851, page 360] If there exists a conference matrix of order n, then n is even; furthermore, if n ≡2 (mod 4), then, for any prime p ≡3 (mod 4), the highest power of p dividing n −1 is even. 14.7.16 Theorem Let q be an odd prime power. 1. If q ≡1 (mod 4), then there is a symmetric conference matrix of order q + 1. 2. If q ≡3 (mod 4), then there is a skew-symmetric conference matrix of order q+1. 14.7.17 Construction In the construction for Theorem 14.7.16, let q be an odd prime power and let χ denote the quadratic character on the ﬁnite ﬁeld Fq (i.e., χ(x) = 0 if x = 0, χ(x) = 1 if x is a square and χ(x) = −1 if x is a nonsquare). Number the elements of Fq : 0 = a0, a1, . . . , aq−1 and deﬁne a q × q matrix Q by qi,j := χ(ai −aj) for 0 ≤i, j < q −1. It follows that Q is symmetric if q ≡1 (mod 4) and skew-symmetric if q ≡3 (mod 4). Deﬁne the (q + 1) × (q + 1) matrix C by      0 1 1 · · · 1 ±1 . . . Q ±1      where the terms ±1 are +1 when q ≡1 (mod 4) and −1 when q ≡3 (mod 4). It follows that C is a conference matrix of order q + 1. In the special case when q is prime, Q is circulant. 14.7.18 Lemma 1. If C is a skew-symmetric conference matrix, then I + C is a Hadamard matrix. 2. If C is a symmetric conference matrix of order n, then I + C −I + C −I + C −I −C 610 Handbook of Finite Fields is a Hadamard matrix of order 2n. 14.7.19 Remark Theorem 14.7.20 follows from Theorem 14.7.16 and Lemma 14.7.18. 14.7.20 Theorem If q is a power of an odd prime, then a Hadamard matrix of order q + 1 exists if q ≡3 (mod 4), and a Hadamard matrix of order 2(q + 1) exists if q ≡1 (mod 4). 14.7.21 Construction Let A be a (0, ±1)-matrix of order n and B a ±1-matrix of order n such that AB = BA and AAT + BBT = (2n −1)I. Then the matrix C = A B BT −AT is a conference matrix of order 2n. 14.7.22 Deﬁnition When A and B are circulant matrices, the conference matrix C in Construction 14.7.21 is constructible from two circulant matrices or for short, two circulants type. 14.7.23 Theorem [1289, 2831, 2975] If q ≡1 (mod 4) is a prime power, then there is a symmetric conference matrix C of order q + 1 of two circulants type. 14.7.24 Theorem There is a symmetric conference matrix of order q2(q + 2) + 1 whenever q is a prime power, q ≡3 (mod 4), and q + 3 is the order of a conference matrix. 14.7.4 Covering arrays 14.7.25 Deﬁnition A covering array CAλ(N; t, k, v) is an N × k array containing v diﬀerent sym-bols. In every N ×t subarray, each t-tuple occurs at least λ times. Then t is the strength of the coverage of interactions, k is the number of components (degree), λ is the index, and v is the number of symbols for each component (order). Only the case when λ = 1 is treated; the subscript is then omitted in the notation. 14.7.26 Deﬁnition The size of a covering array is the covering array number CAN(t, k, v). The covering array is optimal if it has the minimum possible number of rows. 14.7.27 Construction Let q be a prime power and q ≥s ≥2. Over the ﬁnite ﬁeld Fq, let F = {f1, . . . , fqs} be the set of all polynomials of degree less than s. Let A be a subset of Fq ∪{∞}. Deﬁne an qs ×\|A\| array in which the entry in cell (j, a) is fj(a) when a ∈Fq, and is the coeﬃcient of the term of degree s −1 when a = ∞. The result is a CA(qs; s, \|A\|, q). Because every t-tuple is covered exactly once, it is in fact an orthogonal array of index one and strength s. 14.7.28 Remark Covering arrays are typically constructed by a combination of computational, direct, and recursive constructions . Finite ﬁelds arise most frequently in the direct construction of covering arrays. One example is the use of permutation vectors to construct covering arrays . A second, outlined next, uses Weil’s theorem and character theoretic arguments to establish that certain cyclotomic matrices form covering arrays. 14.7.29 Construction Let ω be a primitive element of Fq, with q ≡1 (mod v). For each q and ω, form a cyclotomic vector xq,v,ω = (xi : i ∈Fq) ∈Fq q by setting x0 = 0 and xi ≡j (mod v) when i = ωj for i ∈F⋆ q. Choosing a diﬀerent primitive element of Fq can lead to the same vector xq,v,ω, or, for some number m that is coprime to v, to a vector in which each element is multiplied by m and reduced modulo v. For our purposes, the vectors produced are equivalent, so henceforth let xq,v denote any vector so obtained. From Combinatorial 611 xq,v = (xi : i ∈Fq), form a q × q matrix Aq,v = (aij) with rows and columns indexed by Fq, by setting aij = xj−i (computing the subscript in Fq). 14.7.30 Theorem When q > t2v2t, Aq,v from Construction 14.7.29 is a covering array of strength t. 14.7.5 Hall triple systems 14.7.31 Deﬁnition A Hall triple system (HTS) is a pair (S, L) where S is a set of elements (points) and L a set of lines satisfying: 1. every line is a 3-subset of S, 2. any two distinct points lie in exactly one line, and 3. for any two intersecting lines, the smallest subsystem containing them is iso-morphic to the aﬃne plane of nine points, AG(2,3). 14.7.32 Example Let S be some (n + 1)-dimensional vector space over F3, with n ≥3. Let {e0, e1, . . . , en} be a basis for S. For any two points x = P αiei and y = P βiei set z = x ◦y when x + y + z = (α1 −β1)(α2β3 −α3β2)e0. This deﬁnes a binary operation on S. One either has x = y = z, or the three points x, y, z are pairwise distinct. The 3-subsets of the form {x, y, z} such that z = x ◦y provide S with a structure of an HTS. This HTS is referred to as H(n). 14.7.33 Example Any aﬃne space AG(n, 3) over F3 with the usual lines may be viewed as an HTS. Such an HTS is an aﬃne HTS. 14.7.34 Theorem [1403, 3037] The cardinality of any HTS is 3m for some integer m ≥2. Nonaﬃne HTS of order 3m exist for any m ≥4 and do not exist for m ∈{2, 3}. 14.7.35 Remark When m > 3, the existence of a nonaﬃne HTS of order 3m is provided by H(m−1) in Example 14.7.32. For the orders 34 and 35, there is a unique nonaﬃne HTS, namely, H(3) and H(4), respectively. 14.7.6 Ordered designs and perpendicular arrays 14.7.36 Deﬁnition An ordered design ODλ(t, k, v) is a k × λ · v t · t! array with v entries such that 1. each column has k distinct entries, and 2. each tuple of t rows contains each column tuple of t distinct entries precisely λ times. 14.7.37 Deﬁnition A perpendicular array PAλ(t, k, v) is a k × λ · v t array with v entries such that 1. each column has k distinct entries, and 2. each set of t rows contains each set of t distinct entries as a column precisely λ times. 612 Handbook of Finite Fields 14.7.38 Deﬁnition For 0 ≤s ≤t, a PAλ(t, k, v) is an s-PAλ(t, k, v) if, for each w ≤t and u ≤min(s, w), the following holds. Let E1, E2 be disjoint sets of entries, E1 ∩E2 = ∅ with \|E1 \|= u and \|E2 \|= w −u. Then the number of columns containing E1 ∪E2 and having E2 in a given set U of w −u rows is a constant, independent of the choice of E1, E2, and U. Authentication perpendicular arrays (APA) are 1-PA. 14.7.39 Deﬁnition A set S ⊆Sn of permutations is (uniformly) t-homogeneous if it is an APAλ(t, n, n); it is t-transitive if it is an ODλ(t, n, n). 14.7.40 Theorem Permutation groups yield special cases of t-transitive or t-homogeneous sets. 1. The groups PGL2(q), q a prime power, form OD1(3, q + 1, q + 1); the groups PSL2(q), q ≡3 (mod 4) are APA3(3, q + 1, q + 1). The special cases of this last family when the prime power q ≡3, 11 (mod 12) form the only known inﬁnite family of APAλ(t, n, n) with t > 2 and minimal λ. 2. The groups AGL1(q), (q a prime power), of order q · (q −1) form an OD1(2, q, q); the groups ASL1(q), (q a prime power ≡3 (mod 4)) of order q · (q −1)/2 form APA1(2, q, q). 14.7.41 Deﬁnition Let q ≡3 (mod 4) be a prime power, k odd. An APAV(q, k) (V stands for vector) is a tuple (x1, . . . , xk) where xi ∈Fq and such that for each i the xi −xj, j ̸= i are evenly distributed on squares and nonsquares . 14.7.42 Remark An APAV(q, k) implies the existence of APA1(2, k, q). In a theorem on char-acter sums based on the Hasse–Weil inequality is used to prove existence of an APAV(q, k) when q is large enough with respect to k. 14.7.43 Theorem The following exist, for a prime power q with q ≡3 (mod 4), 1. APAV(q, 7) for q ≥7, q ̸∈{11, 19}, 2. APAV(q, 9) for q ≥19, 3. APAV(q, 11) for q ≥11, q ̸∈{19, 27}, 4. APAV(q, 13) for q ≥13, q ̸∈{19, 23, 31}, and 5. APAV(q, 15) for q ≥31. 14.7.7 Perfect hash families 14.7.44 Deﬁnition Let n, q, t, and s be positive integers and suppose (to avoid trivialities) that n > q ≥t ≥2. Let V be a set of cardinality n and let W be a set of cardinality q. A function f : V →W separates a subset X of V if f is an injection when restricted to X. An (n, q, t)-perfect hash family of size s is a collection F = {f1, f2, . . . , fs} of functions from V to W with property that for all sets X ⊆V such that \|X\| = t, at least one of the functions f1, f2, . . . , fs separates X. The notation PHF(s; n, q, t) is used for an (n, q, t)-perfect hash family of size s. A perfect hash family is optimal if s is as small as possible, given n, q, t. 14.7.45 Theorem A PHF(s; n, q, t) is equivalent to an s × n array A of elements from a q-set F, such that, for any t columns of A, there exists a row of A, say r, such that the entries in the t given columns of row r of A are distinct. Combinatorial 613 14.7.46 Theorem Suppose that there exists a q-ary code C of length K, with N codewords, having minimum distance D. Then there exists a PHF(N; K, q, t), where (N −D) t 2 < N. 14.7.47 Corollary Suppose N and v are given, with v a prime power and N ≤v + 1. Then there exists a PHF(N; v⌈N/( t 2)⌉, v, t) based on a Reed–Solomon code. 14.7.48 Theorem A PHF((i + 1)2; vi+1, v, 3) exists whenever v is a prime power, v ≥3, and i ≥1. A PHF( 5 6(2i3 + 3i2 + i) + i + 1; vi+1, v, 4) exists whenever v is a prime power, v ≥4, and i ≥1. 14.7.49 Theorem For any prime power q and for any positive integers n, m, i such that n ≥m and 2 ≤i ≤qn, there exists a PHF(qn; qm+(i−1)n, qm, t) when t 2 < qm i−1. 14.7.50 Deﬁnition A PHF(N; qs, q, t) is linear if it is an N×qs array with rows indexed by elements of Fq ∪{∞} and columns indexed by the polynomials of degree less than s over Fq; each entry of the array is the evaluation of the polynomial corresponding to the column on the row index, when that index is in Fq; otherwise it is the coeﬃcient of the term of degree s −1 in the polynomial. 14.7.51 Remark In a linear PHF, columns correspond to polynomials of degree less than s over Fq. It follows directly that two columns agree in at most s −1 entries, and hence that if the linear PHF has more than (s −1) t 2 rows, it has strength at least t. By judicious selection of the particular rows (i.e., a subset A of Fq ∪{∞}), fewer rows can often be employed. The key observation, developed in [206, 302, 707], is that when A is chosen properly, a system of equations over Fq for each set of t chosen columns never admits a solution. This is developed in an algebraic setting in , in a geometric setting in , and in a graph-theoretic setting in . The results to follow all employ this basic strategy. 14.7.52 Theorem Let s ≥2 and t ≥2. When q is a suﬃciently large prime power, there is an optimal linear PHF(s(t −1); qs, q, t). 14.7.53 Theorem [206, 207, 302] 1. An optimal linear PHF(6; q2, q, 4) exists if and only if q ≥11 is a prime power and q ̸= 13. 2. An optimal linear PHF(6; q3, q, 3) exists if and only if q ≥11 is a prime power. 14.7.54 Theorem Let p be a prime. 1. A PHF(9; p4, p, 3) exists when p ≥17. 2. A PHF(8; p4, p, 3) exists when p ≥19. 3. A PHF(12; p3, p, 4) exists when p ≥17. 4. A PHF(11; p3, p, 4) exists when p ≥29. 5. A PHF(10; p3, p, 4) exists when p ≥251 and p ̸∈{257, 263}. 6. A PHF(10; p2, p, 5) exists when p ≥19. 7. A PHF(9; p2, p, 5) exists when p ≥41. 8. A PHF(8; p2, p, 5) exists when p ≥241 and p ̸∈{251, 257}. 9. A PHF(15; p2, p, 6) exists when p ≥29. 10. A PHF(14; p2, p, 6) exists when p ≥41. 11. A PHF(13; p2, p, 6) exists when p ≥73. 614 Handbook of Finite Fields 14.7.8 Room squares and starters 14.7.55 Deﬁnition A starter in the odd order abelian group G (written additively), where \|G\| = g is a set of unordered pairs S = {{si, ti} : 1 ≤i ≤(g −1)/2} that satisﬁes: 1. {si : 1 ≤i ≤(g −1)/2} ∪{ti : 1 ≤i ≤(g −1)/2} = G{0}, and 2. {±(si −ti) : 1 ≤i ≤(g −1)/2} = G{0}. 14.7.56 Deﬁnition A strong starter is a starter S = {{si, ti}} in the abelian group G with the additional property that si + ti = sj + tj implies i = j, and for any i, si + ti ̸= 0. 14.7.57 Deﬁnition A skew starter is a starter S = {{si, ti}} in the abelian group G with the additional property that si + ti = ±(sj + tj) implies i = j, and for any i, si + ti ̸= 0. 14.7.58 Example A strong starter in Z17 is {9, 10}, {3, 5}, {13, 16}, {11, 15}, {1, 6}, {2, 8}, {7, 14}, {4, 12}. 14.7.59 Deﬁnition Let S = {{si, ti} : 1 ≤i ≤(g −1)/2} and T = {{ui, vi} : 1 ≤i ≤(g −1)/2} be two starters in G. Without loss of generality, assume that si −ti = ui −vi, for all i. Then S and T are orthogonal starters if ui −si = uj −sj implies i = j, and if ui ̸= si for all i. 14.7.60 Deﬁnition Let q be a prime power that can be written in the form q = 2kt + 1, where t > 1 is odd and let ω be a primitive element in the ﬁeld Fq. Then deﬁne 1. C0 to be the multiplicative subgroup of Fq{0} of order t, 2. Ci = ωiC0, 0 ≤i ≤2k −1 to be the cosets of C0 (cyclotomic classes), and 3. ∆= 2k−1, H = ∪∆−1 i=0 Ci and Ca i = (1/(a −1))Ci. 14.7.61 Theorem Let T = {{x, ω∆x} : x ∈H}. Then T is a skew starter (the Mullin–Nemeth starter) in the additive subgroup of Fq. 14.7.62 Theorem For each a ∈C∆, let Sa = {{x, ax} : x ∈∪∆−1 i=0 Ca i }. Then for any a ∈C∆, Sa is a strong starter in the additive group of Fq. Further, Sa and Sb are orthogonal if a, b ∈C∆with a ̸= b. Hence, the set {Sa\|a ∈C∆} is a set of t pairwise orthogonal starters of order q. 14.7.63 Theorem Let p = 22n be a Fermat prime with n ≥2. There exists a strong starter in the additive group of Fp. 14.7.64 Remark No strong starter in the additive groups of F3, F5, or F9 exists. 14.7.65 Deﬁnition Let G be an additive abelian group of order g, and let H be a subgroup of order h of G, where g −h is even. A Room frame starter in G\H is a set of unordered pairs S = {{si, ti} : 1 ≤i ≤(g −h)/2} such that 1. {si : 1 ≤i ≤(g −h)/2} ∪{ti : 1 ≤i ≤(g −h)/2} = G\H, and 2. {±(si −ti) : 1 ≤i ≤(g −h)/2} = G\H. Combinatorial 615 14.7.66 Remark A starter is the special case of a frame starter when H = {0}. The concepts of strong, skew, and orthogonal for Room frame starters are as for starters replacing {0} by H and (g −1)/2 by (g −h)/2. 14.7.67 Theorem Let q ≡1 (mod 4) be a prime power such that q = 2kt + 1, where t > 1 is odd. Then there exist t orthogonal Room frame starters in (Fq × (Z2)n)({0} × (Z2)n) for all n ≥1. 14.7.68 Theorem If p ≡1 (mod 6) is a prime and p ≥19, then there exist three orthogonal frame starters in (Fp × (Z3))({0} × (Z3)). 14.7.69 Deﬁnition Let S be a set of n+1 elements (symbols). A Room square of side n (on symbol set S), RS(n), is an n × n array, F, that satisﬁes the following properties: 1. every cell of F either is empty or contains an unordered pair of symbols from S, 2. each symbol of S occurs once in each row and column of F, 3. every unordered pair of symbols occurs in precisely one cell of F. 14.7.70 Deﬁnition A Room square of side n is standardized (with respect to the symbol ∞) if the cell (i, i) contains the pair {∞, i}. 14.7.71 Deﬁnition A standardized Room square of side n is skew if for every pair of cells (i, j) and (j, i) (with i ̸= j) exactly one is ﬁlled. 14.7.72 Deﬁnition A standardized Room square of side n is cyclic if S = Zn∪{∞} and if whenever {a, b} occurs in the cell (i, j), then {a + 1, b + 1} occurs in cell (i + 1, j + 1) where all arithmetic is performed in Zn (and ∞+ 1 = ∞). 14.7.73 Example Below are skew Room squares of sides 7 and 9; the Room square of side 7 is cyclic. ∞1 49 37 28 56 ∞0 15 46 23 89 ∞2 57 34 16 34 ∞1 26 50 58 ∞3 69 24 17 61 45 ∞2 30 36 78 ∞4 19 25 02 56 ∞3 41 79 12 ∞5 38 46 52 13 60 ∞4 45 ∞6 18 39 27 63 24 01 ∞5 26 59 13 ∞7 48 04 35 12 ∞6 67 14 29 ∞8 35 23 15 68 47 ∞9 14.7.74 Theorem The existence of d pairwise orthogonal starters in an abelian group of order n implies the existence of a Room d-cube of side n. 14.7.75 Remark Construction 14.7.77 is used to establish Theorem 14.7.74 when d = 2. It is easily extended when d > 2. 616 Handbook of Finite Fields 14.7.76 Deﬁnition An adder for the starter S = {{si, ti} : 1 ≤i ≤(g −1)/2} is an ordered set AS = {a1, a2, . . . , a(g−1)/2} of (g −1)/2 distinct nonzero elements from G such that the set T = {{si + ai, ti + ai} : 1 ≤i ≤(g −1)/2} is also a starter in the group G. 14.7.77 Construction Let S = {{si, ti} : 1 ≤i ≤(n −1)/2} and T = {{ui, vi} : 1 ≤i ≤(n −1)/2} be two orthogonal starters in G (usually Zn), (n odd) with AS = {ai : 1 ≤i ≤(n−1)/2} the associated adder. Let R be an n × n array indexed by the elements of G. Each {si, ti} ∈S is placed in the ﬁrst row in cell R(0, −ai). This is cycled so that the pair {si + x, ti + x} is in the cell R(x, −ai + x), where all arithmetic is performed in G. Finally, for each x ∈G, place the pair {∞, x} in cell R(x, x). Then R is a Room square of side n. 14.7.78 Remark Construction 14.7.77 in conjunction with Theorem 14.7.61 yields skew Room squares of prime power orders q ≡3 (mod 4). This is useful in proving Theorem 14.7.79. 14.7.79 Theorem A (skew) Room square of side n exists if and only if n is odd and n ̸∈{3, 5}. 14.7.80 Theorem A cyclic skew Room square of side n exists if n = Q piαi where each pi is a non-Fermat prime or if n = pq with p, q distinct Fermat primes. 14.7.81 Deﬁnition If {S1, . . . , Sn} is a partition of a set S, an {S1, . . . , Sn}-Room frame is an \|S\| × \|S\| array, F, indexed by S, satisfying: 1. every cell of F either is empty or contains an unordered pair of symbols of S, 2. the subarrays Si × Si are empty, for 1 ≤i ≤n (these subarrays are holes), 3. each symbol x ̸∈Si occurs once in row (or column) s for any s ∈Si, and 4. the pairs occurring in F are those {s, t}, where (s, t) ∈(S × S)\ Sn i=1(Si × Si). The type of a Room frame F is the multiset {\|Si\| : 1 ≤i ≤n}. An “exponential” notation is used to describe types; a Room frame has type t1u1t2u2 · · · tkuk if there are ui Sjs of cardinality ti, 1 ≤i ≤k. 14.7.82 Remark Theorem 14.7.83 gives the connection between frame starters and Room frames. The construction for a Room frame from a pair of orthogonal frame starters is a general-ization of Construction 14.7.77. 14.7.83 Theorem Suppose a pair of orthogonal frame starters in G\H exists, where \|G\| = g and \|H\| = h. Then there exists a Room frame of type hg/h. 14.7.84 Remark Theorems 14.7.67 and 14.7.68 in conjunction with Theorem 14.7.83 yield Corollary 14.7.85. Theorem 14.7.86 details the existence of Room frames of type tu. 14.7.85 Corollary a) Let q ≡1 (mod 4) be a prime power such that q = 2kt + 1, where t > 1 is odd. Then there exist a Room frame of type (2n)q for all n ≥1. b) If p ≡1 (mod 6) is a prime and p ≥19, then there exists a Room frame of type 3p. 14.7.86 Theorem (Existence theorems for uniform Room frames) [706, §VI.50] and 1. There does not exist a Room frame of type tu if any of the following conditions hold: (i) u = 2 or 3; (ii) u = 4 and t = 2; (iii) u = 5 and t = 1; (iv) t(u −1) is odd. 2. Suppose t and u are positive integers, u ≥4 and (t, u) ̸= (1, 5), (2, 4). Then there exists a uniform Room frame of type tu if and only if t(u −1) is even. Combinatorial 617 14.7.9 Strongly regular graphs 14.7.87 Deﬁnition A strongly regular graph with parameters (v, k, λ, µ) is a ﬁnite graph on v vertices, without loops or multiple edges, regular of degree k (with 0 < k < v −1, so that there are both edges and nonedges), and such that any two distinct vertices have λ common neighbors when they are adjacent, and µ common neighbors when they are nonadjacent. 14.7.88 Remark There are many constructions for strongly regular graphs. Example 14.7.89 gives several that use ﬁnite ﬁelds. For a table of the existence of strongly regular graphs with v ≤280 see [706, pp. 852–866]. 14.7.89 Example 1. Paley(q): For prime powers q = 4t + 1, the graph with vertex set Fq where two vertices are adjacent when they diﬀer by a square. This strongly regular graph has parameters (q, 1 2q −1, 1 4(q −5), 1 4(q −1)). 2. van Lint–Schrijver(u): a graph constructed by the cyclotomic construction in , by taking the union of u classes. 3. An−1,2(q) or n 2 q: the graph on the lines in PG(n −1, q), adjacent when they have a point in common. 4. Bilin2×d(q): the graph on the 2 × d matrices over Fq, adjacent when their diﬀer-ence has rank 1. 5. Oε 2d(q): the graph on the isotropic points on a nondegenerate quadric in PG(2d− 1, q), where two points are joined when the connecting line is totally singular. 6. Sp2d(q): the graph on the points of PG(2d −1, q) provided with a nondegenerate symplectic form, where two points are joined when the connecting line is totally isotropic. 7. Ud(q): the graph on the isotropic points of PG(d −1, q2) provided with a nonde-generate Hermitian form, where two points are joined when the connecting line is totally isotropic. 8. Aﬃne diﬀerence sets : Let V be an n-dimensional vector space over Fq and let X be a set of directions (a subset of the projective space PV ). Two vectors are adjacent when the line joining them has a direction in X. Then v = qn and k = (q −1)\|X\|. This graph is strongly regular if and only if there are two integers w1, w2 such that all hyperplanes of PV miss either w1 or w2 points of X. If this is the case, then r = k −qw1, s = k −qw2 (assuming w1 < w2), and hence µ = k + (k −qw1)(k −qw2), λ = k −1 + (k −qw1 + 1)(k −qw2 + 1). 14.7.10 Whist tournaments 14.7.90 Deﬁnition A whist tournament Wh(4n) for 4n players is a schedule of games each involving two players opposing two others, such that 1. the games are arranged into 4n −1 rounds, each of n games; 2. each player plays in exactly one game in each round; 3. each player partners every other player exactly once; 4. each player opposes every other player exactly twice. 618 Handbook of Finite Fields 14.7.91 Deﬁnition Each game is denoted by an ordered 4-tuple (a, b, c, d) in which the pairs {a, c}, {b, d} are partner pairs; {a, c} is a partner pair of the ﬁrst kind, and {b, d} is a partner pair of the second kind. The other pairs are opponent pairs; in particular {a, b}, {c, d} are opponent pairs of the ﬁrst kind, and {a, d}, {b, c} are opponent pairs of the second kind. 14.7.92 Deﬁnition A whist tournament Wh(4n + 1) for 4n + 1 players is deﬁned as for 4n, except that Conditions 1, 2 are replaced by 1′. the games are arranged into 4n + 1 rounds each of n games; 2′. each player plays in one game in each of 4n rounds, but does not play in the remaining round. 14.7.93 Deﬁnition A Wh(4n) is Z-cyclic if the players are ∞, 0, 1, . . . , 4n −2 and each round is obtained from the previous one by adding 1 modulo 4n −1 to each non-∞entry. A Wh(4n + 1) is Z-cyclic if the players are 0, 1, . . . , 4n and the rounds are similarly developed modulo 4n + 1. 14.7.94 Theorem If p = 4n + 1 is prime and w is a primitive root of p, then the games (wi, wi+n, wi+2n, wi+3n), 0 ≤i ≤n −1, form the initial round of Z-cyclic Wh(4n + 1). 14.7.95 Theorem Let P denote any product of primes p with each p ≡1 (mod 4), and let q, r denote primes with both q, r ≡3 (mod 4). A Z-cyclic Wh(4n) is known to exist when: 1. 4n ≤132 (see [7, 98]); 2. 4n = 2α(α ≥2) (see ); 3. 4n = qP + 1, q ∈{3, 7, 11, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83, 103, 107, 127} (see ); 4. 4n = 3P + 1 (see ); 5. 4n = 32m+1 + 1, m ≥0 (see ); 6. 4n = qr2P + 1, q and r distinct, q < 60, r < 100 (see ). A Z-cyclic Wh(4n + 1) is known to exist when: 1. 4n + 1 = P or r2P and r ≤100 (see [7, 102]); 2. 4n + 1 ≤149 (see ); 3. 4n + 1 = 32m or 32mP (see ); 4. 4n + 1 = 3sP, s ∈{7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47} (see ). 14.7.96 Deﬁnition A triplewhist tournament TWh(v) is a Wh(v) with Condition 4 replaced by (4′′) each player has every other player once as an opponent of the ﬁrst kind and once as an opponent of the second kind. 14.7.97 Theorem A TWh(4n + 1) exists for all n ≥5, and possibly for n = 4. A TWh(4n) exists for all n ≥1 except for n = 3; see [7, 1965]. 14.7.98 Theorem A Z-cyclic TWh(4n + 1) exists when: 1. 4n + 1 is a prime p ≡1 (mod 4), p ≥29 (see ); Combinatorial 619 2. 4n + 1 = q2 where q is prime, q ≡3 (mod 4), 7 ≤q ≤83 (see ); 3. 4n + 1 = 5n or 13n (n ≥2) (see ); 4. 4n + 1 is the product of any values in the above three items (see ). 14.7.99 Theorem A Z-cyclic TWh(4n) exists when: 1. 4n = qP + 1, where P denotes any product of primes p with each p ≡1 (mod 4) and q ∈{3, 7, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83, 103, 107, 127, 131} (see ); 2. 4n = 2n (n ≥2) (see ); 3. 4n = 3m + 1 (m odd) (see ). See Also §14.1 For latin squares, MOLS, and transversal designs. §14.3 For aﬃne and projective planes. §14.4 For projective spaces. §14.5 For block designs. §14.6 For diﬀerence sets. Textbook on combinatorial designs. Advanced textbook on combinatorial designs. Another advanced textbook on combinatorial designs. For association schemes (§VI.1), Costas arrays (§VI.9), conference matrices (§V.6), covering arrays (§VI.10), Hadamard designs and matrices(§V.1); Hall triple systems (§VI.28), ordered designs and perpendicular arrays (§VI.38), perfect hash families (§VI.43), Room squares (§VI.50), strongly regular graphs (§VII.11), whist tournaments (§VI.64). Another textbook on combinatorial designs. References Cited: [6, 7, 93, 98, 99, 100, 101, 102, 167, 206, 207, 261, 262, 275, 302, 419, 462, 479, 606, 623, 704, 705, 706, 707, 897, 898, 899, 900, 916, 918, 1262, 1289, 1301, 1304, 1403, 1404, 1457, 1537, 1932, 1965, 2023, 2198, 2345, 2612, 2719, 2721, 2831, 2850, 2851, 2853, 2975, 3037] 14.8 (t, m, s)-nets and (t, s)-sequences Harald Niederreiter, KFUPM 14.8.1 (t, m, s)-nets 14.8.1 Remark The theory of (t, m, s)-nets and (t, s)-sequences is signiﬁcant for quasi-Monte Carlo methods in scientiﬁc computing (see the books and and the recent survey article ). For both (t, m, s)-nets and (t, s)-sequences, the idea is to sample the s-dimensional unit cube [0, 1]s in a uniform and equitable manner. In a nutshell, (t, m, s)-nets are ﬁnite samples (or point sets) and (t, s)-sequences are inﬁnite sequences with special 620 Handbook of Finite Fields uniformity properties. The deﬁnition of a (t, m, s)-net (see Deﬁnition 14.8.2 below) has a priori no connection with ﬁnite ﬁelds, but it turns out that most of the interesting constructions of (t, m, s)-nets use ﬁnite ﬁelds as a tool. By a point set we mean a multiset in the sense of combinatorics, i.e., a set in which multiplicities of elements are allowed and taken into account. 14.8.2 Deﬁnition [2236, 2690] For integers b ≥2 and 0 ≤t ≤m and a given dimension s ≥1, a (t, m, s)-net in base b is a point set P consisting of bm points in [0, 1]s such that every subinterval of [0, 1]s of volume bt−m which has the form s Y i=1 [aib−di, (ai + 1)b−di) with integers di ≥0 and 0 ≤ai < bdi for 1 ≤i ≤s contains exactly bt points of P. 14.8.3 Remark It is easily seen that a (t, m, s)-net in base b is also a (u, m, s)-net in base b for all integers u with t ≤u ≤m. Any point set consisting of bm points in [0, 1)s is a (t, m, s)-net in base b with t = m. Smaller values of t mean stronger uniformity properties of a (t, m, s)-net in base b. The number t is the quality parameter of a (t, m, s)-net in base b. 14.8.4 Deﬁnition A (t, m, s)-net P in base b is a strict (t, m, s)-net in base b if t is the least integer u such that P is a (u, m, s)-net in base b. 14.8.5 Example Let s = 2 and let b ≥2 and m ≥1 be given integers. For any integer n with 0 ≤n < bm, let n = Pm−1 r=0 ar(n)br with ar(n) ∈Zb = {0, 1, . . . , b −1} be the digit expansion of n in base b and put φb(n) = Pm−1 r=0 ar(n)b−r−1. Then the point set consisting of the points (nb−m, φb(n)) ∈[0, 1]2, n = 0, 1, . . . , bm −1, is a (0, m, 2)-net in base b. This point set is the Hammersley net in base b. 14.8.6 Example Let b ≥2, s ≥1, and t ≥0 be given integers. Then the point set consisting of the points (nb−1, . . . , nb−1) ∈[0, 1]s, n = 0, 1, . . . , b −1, each taken with multiplicity bt, is a (t, t + 1, s)-net in base b. 14.8.7 Remark According to Remark 14.8.3 and Example 14.8.6, a (t, m, s)-net in base b always exists for m = t and m = t + 1. For m ≥t + 2 there are combinatorial obstructions to the general existence of (t, m, s)-nets in base b. This was ﬁrst observed in . Later, a combinatorial equivalence between (t, m, s)-nets in base b and ordered orthogonal arrays as deﬁned in Deﬁnition 14.8.8 below was established. 14.8.8 Deﬁnition Let b, s, k, T, λ be positive integers with b ≥2 and sT ≥k. An ordered orthog-onal array OOAb(s, k, T, λ) is a (λbk) × (sT) matrix with entries from Zb and column labels (i, j) for 1 ≤i ≤s and 1 ≤j ≤T such that, for any integers 0 ≤d1, . . . , ds ≤T with Ps i=1 di = k, the (λbk) × k submatrix obtained by restricting to the columns (i, j), 1 ≤j ≤di, 1 ≤i ≤s, contains among its rows every element of Zk b with the same frequency λ. 14.8.9 Theorem [1873, 2183] Let b ≥2, s ≥2, k ≥2, and t ≥0 be integers. Then there exists a (t, t+k, s)-net in base b if and only if there exists an ordered orthogonal array OOAb(s, k, k− 1, bt). 14.8.10 Corollary There exists a (0, 2, s)-net in base b if and only if there exist s−2 mutually orthogonal latin squares of order b. Combinatorial 621 14.8.11 Corollary For m ≥2, a (0, m, s)-net in base b can exist only if s ≤M(b) + 2, where M(b) is the maximum cardinality of a set of mutually orthogonal latin squares of order b. In particular, if m ≥2, then a necessary condition for the existence of a (0, m, s)-net in base b is s ≤b + 1. 14.8.12 Remark The equivalence between nets and ordered orthogonal arrays enunciated in The-orem 14.8.9, when combined with extensions of standard parameter bounds for orthogonal arrays to the case of ordered orthogonal arrays, leads to lower bounds on the quality pa-rameter for nets. Examples of such bounds are the linear programming bound , the Rao bound , and the dual Plotkin bound [271, 2009]. Extensive numerical data on these bounds are available at 14.8.2 Digital (t, m, s)-nets 14.8.13 Remark Most of the known constructions of nets are based on the digital method which goes back to . In order to describe the digital method for the construction of (t, m, s)-nets in base b, we need the following ingredients. First of all, let integers b ≥2, m ≥1, and s ≥1 be given. Then we choose: 1. a commutative ring R with identity and of cardinality b; 2. bijections η(i) j : R →Zb for 1 ≤i ≤s and 1 ≤j ≤m; 3. m × m matrices C(1), . . . , C(s) over R. Now let r ∈Rm be an m-tuple of elements of R and deﬁne π(i) j (r) = η(i) j (c(i) j · r) ∈Zb for 1 ≤i ≤s, 1 ≤j ≤m, where c(i) j is the j-th row of the matrix C(i) and · denotes the standard inner product. Next we put π(i)(r) = m X j=1 π(i) j (r)b−j ∈[0, 1] for 1 ≤i ≤s and P(r) = (π(1)(r), . . . , π(s)(r)) ∈[0, 1]s. By letting r range over all bm elements of Rm, we arrive at a point set P consisting of bm points in [0, 1]s. 14.8.14 Deﬁnition If the point set P constructed in Remark 14.8.13 forms a (t, m, s)-net in base b, then P is a digital (t, m, s)-net in base b. If we want to emphasize that the construction uses the ring R, then we speak also of a digital (t, m, s)-net over R. If P is a strict (t, m, s)-net in base b, then P is a digital strict (t, m, s)-net in base b (or over R). 14.8.15 Remark The matrices C(1), . . . , C(s) in Remark 14.8.13 are generating matrices of the digital net. The quality parameter t of the digital net depends only on the generating matrices. For a convenient algebraic condition on the generating matrices to guarantee a certain value of t, we refer to Theorem 4.26 in . In the important case where the ring R is a ﬁnite ﬁeld, an even simpler condition is given in Theorem 14.8.18 below. 14.8.16 Example Let s = 2 and let b ≥2 and m ≥1 be given integers. Choose R = Zb and let the bijections η(i) j in Remark 14.8.13 be identity maps. Let C(1) be the m × m identity matrix over Zb and let C(2) = (ci,j)1≤i,j≤m be the m×m antidiagonal matrix over Zb with ci,j = 1 622 Handbook of Finite Fields if i + j = m + 1 and ci,j = 0 otherwise. Then the Hammersley net in Example 14.8.5 is easily seen to be a digital (0, m, 2)-net over Zb with generating matrices C(1) and C(2). 14.8.17 Deﬁnition Let C(1), . . . , C(s) be m × m matrices over the ﬁnite ﬁeld Fq and for 1 ≤i ≤s and 1 ≤j ≤m let c(i) j denote the j-th row of the matrix C(i). Then ϱ(C(1), . . . , C(s)) is deﬁned to be the largest nonnegative integer d such that, for any integers 0 ≤ d1, . . . , ds ≤m with Ps i=1 di = d, the vectors c(i) j , 1 ≤j ≤di, 1 ≤i ≤s, are lin-early independent over Fq (this property is assumed to be vacuously satisﬁed for d = 0). 14.8.18 Theorem The point set P constructed in Remark 14.8.13 with R = Fq and m × m generating matrices C(1), . . . , C(s) over Fq is a digital strict (t, m, s)-net over Fq with t = m −ϱ(C(1), . . . , C(s)). 14.8.19 Example Let s = 2, let b = p be a prime, and let the m × m matrices C(1) and C(2) over Fp be as in Example 14.8.16. Then it is easily seen that ϱ(C(1), C(2)) = m. Using Theorem 14.8.18, this shows again that the Hammersley net in base b = p is a digital (0, m, 2)-net over Fp. 14.8.20 Remark The equivalence between nets and ordered orthogonal arrays stated in Theorem 14.8.9 has an analog for digital nets. The special family of linear ordered orthogonal arrays was introduced in and it was shown that these arrays correspond to digital nets. 14.8.21 Remark There is a very useful duality theory for digital nets which facilitates many con-structions of good digital nets. A crucial ingredient is the weight function Vm on Fms q introduced in Deﬁnition 14.8.22 below. The main result of this duality theory is Theorem 14.8.26 below. 14.8.22 Deﬁnition Let m ≥1 and s ≥1 be integers. Put vm(a) = 0 if a = 0 ∈Fm q , and for a = (a1, . . . , am) ∈Fm q with a ̸= 0 let vm(a) be the largest value of j such that aj ̸= 0. Write a vector A ∈Fms q as the concatenation of s vectors of length m, i.e., A = (a(1), . . . , a(s)) ∈Fms q with a(i) ∈Fm q for 1 ≤i ≤s. Then the NRT weight of A is deﬁned by Vm(A) = s X i=1 vm(a(i)). 14.8.23 Remark The NRT weight is named after the work of Niederreiter and Rosenbloom and Tsfasman . The NRT space is Fms q with the metric dm(A, B) = Vm(A −B) for A, B ∈Fms q . For m = 1 the NRT space reduces to the Hamming space in coding theory. 14.8.24 Deﬁnition The minimum distance δm(N) of a nonzero Fq-linear subspace N of Fms q is given by δm(N) = min A∈N {0} Vm(A). 14.8.25 Remark Let the m × m matrices C(1), . . . , C(s) over Fq be generating matrices of a digital net P. Set up an m × ms matrix M over Fq as follows: for 1 ≤j ≤m, the j-th row of M is obtained by concatenating the transposes of the j-th columns of C(1), . . . , C(s). Let M ⊆Fms q be the row space of M and let M⊥be its dual space, i.e., M⊥= {A ∈Fms q : A · M = 0 for all M ∈M}, Combinatorial 623 where · is the standard inner product in Fms q . 14.8.26 Theorem Let m ≥1 and s ≥2 be integers. Then the point set P in Remark 14.8.25 is a digital strict (t, m, s)-net over Fq with t = m + 1 −δm(M⊥). 14.8.27 Corollary Let m ≥1 and s ≥2 be integers. Then from any Fq-linear subspace N of Fms q with dim(N) ≥ms −m we can construct a digital strict (t, m, s)-net over Fq with t = m + 1 −δm(N). 14.8.28 Remark There are digital nets for which a property analogous to that in Deﬁnition 14.8.2 holds for a wider range of subintervals of [0, 1]s. Such generalized digital nets were introduced in and are also studied in detail in Chapter 15 of . 14.8.29 Remark There is a generalization of the digital method which can be viewed as a nonlinear analog of the construction in Remark 14.8.13. For simplicity we consider only the case where R = Fq (see for a general ring R). Compared to Remark 14.8.13, the only change is that instead of linear forms c(i) j · r we now use polynomial functions, that is, for 1 ≤i ≤s and 1 ≤j ≤m we choose polynomials f (i) j over Fq in m variables and then we replace c(i) j ·r by f (i) j (r) for r ∈Fm q . The following criterion uses the concept of permutation polynomial in several variables (see Section 8.2). 14.8.30 Theorem The point set constructed in Remark 14.8.29 is a (t, m, s)-net in base q if and only if, for any integers d1, . . . , ds ≥0 with Ps i=1 di = m −t, the polynomials f (i) j , 1 ≤j ≤di, 1 ≤i ≤s, have the property that all of their nontrivial linear combinations with coeﬃcients from Fq are permutation polynomials over Fq in m variables. 14.8.3 Constructions of (t, m, s)-nets 14.8.31 Remark A general principle for the construction of (t, m, s)-nets with s ≥2 is based on the use of Proposition 14.8.50 below in conjunction with the constructions of (t, s−1)-sequences in Subsection 14.8.6. In the present subsection, we describe constructions of (t, m, s)-nets that are not derived from this principle. One of the ﬁrst constructions of this type was that of polynomial lattices in . Choose f ∈Fq[x] with deg(f) = m ≥1 and an s-tuple g = (g1, . . . , gs) ∈Fq[x]s with deg(gi) < m for 1 ≤i ≤s. Consider the Laurent series expansions gi(x) f(x) = ∞ X k=1 u(i) k x−k ∈Fq((x−1)) for 1 ≤i ≤s. Then for 1 ≤i ≤s the generating matrix C(i) = (cj,r) is the Hankel matrix given by cj,r = u(i) j+r ∈Fq for 1 ≤j ≤m, 0 ≤r ≤m −1. The bijections η(i) j in Remark 14.8.13 are chosen arbitrarily. The resulting digital net over Fq is denoted by P(g, f). 14.8.32 Deﬁnition Let s ≥2 and let f and g be as in Remark 14.8.31. Then the ﬁgure of merit ϱ(g, f) is deﬁned by ϱ(g, f) = s −1 + min s X i=1 deg(hi), where the minimum is over all nonzero s-tuples (h1, . . . , hs) ∈Fq[x]s with deg(hi) < m for 1 ≤i ≤s and f dividing Ps i=1 higi. Here we use the convention deg(0) = −1. 14.8.33 Theorem For s ≥2, the point set P(g, f) in Remark 14.8.31 is a digital strict (t, m, s)-net over Fq with t = m −ϱ(g, f). 624 Handbook of Finite Fields 14.8.34 Remark It is clear from Theorem 14.8.33 that in order to obtain a good (t, m, s)-net by this construction, i.e., a net with a small value of t, we need to ﬁnd g and f with a large ﬁgure of merit ϱ(g, f). A systematic method for the explicit construction of good polynomial lattices is the component-by-component algorithm in ; see also Chapter 10 in . 14.8.35 Remark Several constructions of digital nets are based on Corollary 14.8.27. A powerful construction of this type uses algebraic function ﬁelds (see Section 12.1 for background on algebraic function ﬁelds). We present only a simple version of this construction; more reﬁned versions can be found in . Let F be an algebraic function ﬁeld (of one variable) with full constant ﬁeld Fq, that is, Fq is algebraically closed in F. Let N(F) denote the number of rational places of F. For a given dimension s ≥2, we assume that N(F) ≥s and let P1, . . . , Ps be s distinct rational places of F. Let G be a divisor of F. For each i = 1, . . . , s, let ti ∈F be a prime element at Pi and let ni ∈Z be the coeﬃcient of Pi in G. For f in the Riemann-Roch space L(G) and a given integer m ≥1, let θ(i)(f) ∈Fm q be the vector whose coordinates are, in descending order, the coeﬃcients of tj i, j = −ni + m −1, −ni + m − 2, . . . , −ni, in the local expansion of f at Pi. Now deﬁne the Fq-linear map θ : L(G) →Fms q by θ(f) = (θ(1)(f), . . . , θ(s)(f)) for all f ∈L(G). A digital net over Fq is then obtained by applying Corollary 14.8.27 with N being the image of the map θ. A suitable choice of the divisor G leads to the following result. 14.8.36 Theorem Let s ≥2 be an integer and let F be an algebraic function ﬁeld with full constant ﬁeld Fq, genus g ≥1, and N(F) ≥s. If k and m are integers with 0 ≤k ≤g −1 and m ≥max(1, g −k −1), then there exists a digital (g −k −1, m, s)-net over Fq provided that s + m + k −g s −1 Ak(F) < h(F), where Ak(F) is the number of positive divisors of F of degree k and h(F) is the divisor class number of F. 14.8.37 Example Let q = 9 and let F be the Hermitian function ﬁeld over F9, that is, F = F9(x, y) with y3 + y = x4. Then g = 3, N(F) = 28, and h(F) = 4096. We apply Theorem 14.8.36 with s = 28, k = 0, m = 5, and we obtain a digital (2, 5, 28)-net over F9. The value t = 2 is the currently best value of the quality parameter for a (t, 5, 28)-net in base 9, according to the website which contains an extensive database for parameters of (t, m, s)-nets. 14.8.38 Remark Another construction based on Corollary 14.8.27 was introduced in . For integers m ≥1 and s ≥2, consider the Fq-linear space P = {f ∈Fqm[x] : deg(f) < s}. Fix α ∈Fqm and deﬁne the Fq-linear subspace Pα = {f ∈P : f(α) = 0} of P. Set up a map τ : P →Fms q as follows. Write f ∈P explicitly as f(x) = Ps i=1 γixi−1 with γi ∈Fqm for 1 ≤i ≤s. For each i = 1, . . . , s, choose an ordered basis Bi of Fqm over Fq and let ci(f) ∈Fm q be the coordinate vector of γi with respect to Bi. Then deﬁne τ(f) = (c1(f), . . . , cs(f)) ∈Fms q for all f ∈P. A digital net over Fq is now obtained by applying Corollary 14.8.27 with N being the image of the subspace Pα under τ. The resulting digital net is a cyclic digital net over Fq relative to the bases B1, . . . , Bs. 14.8.39 Remark A generalization of the construction in Remark 14.8.38 was presented in . For integers m ≥1 and s ≥2, consider Q = Fs qm as a vector space over Fq. Fix α ∈Q with α ̸= 0 and put Qα = {β ∈Q : α · β = 0}. Then Qα is an Fq-linear subspace of Q. Let Combinatorial 625 σ : Q →Fms q be an isomorphism between vector spaces over Fq. A digital net over Fq is now obtained by applying Corollary 14.8.27 with N being the image of the subspace Qα under σ. The resulting digital net is a hyperplane net over Fq. Detailed information on hyperplane nets and cyclic digital nets can be found in Chapter 11 of . 14.8.40 Theorem Given a linear code over Fq with length n, dimension k, and minimum distance d ≥3, we can construct a digital (n −k −d + 1, n −k, s)-net over Fq with s = ⌊(n −1)/h⌋if d = 2h + 2 is even and s = ⌊n/h⌋if d = 2h + 1 is odd. 14.8.41 Remark Further applications of coding theory to the construction of digital nets are dis-cussed in the survey articles and . We speciﬁcally mention some principles of combining several digital nets to obtain a new digital net that are inspired by coding theory. For instance, the well-known Kronecker-product construction in coding theory has an analog for digital nets . The following result is an analog of the matrix-product construction of linear codes. 14.8.42 Theorem Let h be an integer with 2 ≤h ≤q. If for k = 1, . . . , h a digital (tk, mk, sk)-net over Fq is given and if s1 ≤s2 ≤· · · ≤sh, then we can construct a digital (t, Ph k=1 mk, Ph k=1 sk)-net over Fq with t = 1 + h X k=1 mk −min 1≤k≤h (h −k + 1)(mk −tk + 1). 14.8.43 Proposition Given a (t, m, s)-net in base b, we can construct: 1. a (t, u, s)-net in base b for t ≤u ≤m; 2. a (t, m, r)-net in base b for 1 ≤r ≤s; 3. a (t + u, m + u, s)-net in base b for any integer u ≥0. 14.8.44 Remark A result of the type appearing in Proposition 14.8.43 is called a propagation rule for nets. There are also propagation rules for digital nets, in the sense that the input net and the output net are both digital nets. Furthermore, there are propagation rules that involve a base change, typically moving from a base b to a base that is a power bk with k ≥2 or vice versa. A detailed discussion of propagation rules is presented in Chapter 9 of . 14.8.4 (t, s)-sequences and (T, s)-sequences 14.8.45 Remark There is an analog of (t, m, s)-nets for sequences of points in [0, 1]s, given in Deﬁnition 14.8.46 below. First we need some notation. For an integer b ≥2 and a real number x ∈[0, 1], let x = P∞ j=1 yjb−j with all yj ∈Zb be a b-adic expansion of x, where the case yj = b −1 for all suﬃciently large j is allowed. For any integer m ≥1, we deﬁne the truncation [x]b,m = Pm j=1 yjb−j. Note that this truncation operates on the expansion of x and not on x itself, since it may yield diﬀerent results depending on which b-adic expansion of x is used. If x = (x(1), . . . , x(s)) ∈[0, 1]s and the x(i), 1 ≤i ≤s, are given by prescribed b-adic expansions, then we deﬁne [x]b,m = ([x(1)]b,m, . . . , [x(s)]b,m). 14.8.46 Deﬁnition [2236, 2690] Let b ≥2, s ≥1, and t ≥0 be integers. A sequence x0, x1, . . . of points in [0, 1]s is a (t, s)-sequence in base b if for all integers k ≥0 and m > t the points [xn]b,m with kbm ≤n < (k + 1)bm form a (t, m, s)-net in base b. Here the coordinates of all points xn, n = 0, 1, . . ., are given by prescribed b-adic expansions. 626 Handbook of Finite Fields 14.8.47 Remark It is easily seen that a (t, s)-sequence in base b is also a (u, s)-sequence in base b for all integers u ≥t. Smaller values of t mean stronger uniformity properties of a (t, s)-sequence in base b. The number t is the quality parameter of a (t, s)-sequence in base b. 14.8.48 Deﬁnition A (t, s)-sequence S in base b is a strict (t, s)-sequence in base b if t is the least integer u such that S is a (u, s)-sequence in base b. 14.8.49 Example Let s = 1 and let b ≥2 be an integer. For n = 0, 1, . . ., let n = P∞ r=0 ar(n)br with all ar(n) ∈Zb and ar(n) = 0 for all suﬃciently large r be the digit expansion of n in base b. Put φb(n) = P∞ r=0 ar(n)b−r−1. Then the sequence φb(0), φb(1), . . . is a (0, 1)-sequence in base b. This sequence is the van der Corput sequence in base b. 14.8.50 Proposition Given a (t, s)-sequence in base b, we can construct a (t, m, s + 1)-net in base b for any integer m ≥t. 14.8.51 Remark The following result is obtained by combining Corollary 14.8.11 and Proposition 14.8.50. 14.8.52 Corollary A (0, s)-sequence in base b can exist only if s ≤M(b) + 1. In particular, a necessary condition for the existence of a (0, s)-sequence in base b is s ≤b. 14.8.53 Remark It was shown in that for any integers b ≥2 and s ≥1 there exists a (t, s)-sequence in base b for some value of t. Therefore it is meaningful to deﬁne tb(s) as the least value of t for which there exists a (t, s)-sequence in base b. 14.8.54 Theorem [2283, 2565] For any integers b ≥2 and s ≥1, we have tb(s) ≥ s b −1 −cb log(s + 1) with a constant cb > 0 depending only on b. 14.8.55 Deﬁnition Let b ≥2 and s ≥1 be integers and let N0 denote the set of nonnegative integers. Let T : N →N0 be a function with T(m) ≤m for all m ∈N. Then a sequence x0, x1, . . . of points in [0, 1]s is a (T, s)-sequence in base b if for all k ∈N0 and m ∈N, the points [xn]b,m with kbm ≤n < (k + 1)bm form a (T(m), m, s)-net in base b. Here the coordinates of all points xn, n = 0, 1, . . ., are given by prescribed b-adic expansions. A (T, s)-sequence S in base b is a strict (T, s)-sequence in base b if there is no function U : N →N0 with U(m) ≤m for all m ∈N and U(m) < T(m) for at least one m ∈N such that S is a (U, s)-sequence in base b. 14.8.56 Remark If the function T in Deﬁnition 14.8.55 is such that for some integer t ≥0 we have T(m) = m for m ≤t and T(m) = t for m > t, then the concept of a (T, s)-sequence in base b reduces to that of a (t, s)-sequence in base b. 14.8.5 Digital (t, s)-sequences and digital (T, s)-sequences 14.8.57 Remark There is an analog of the digital method in Remark 14.8.13 for the construction of sequences. Let integers b ≥2 and s ≥1 be given. Then we choose: 1. a commutative ring R with identity and of cardinality b; 2. bijections ψr : Zb →R for r = 0, 1, . . ., with ψr(0) = 0 for all suﬃciently large r; 3. bijections η(i) j : R →Zb for 1 ≤i ≤s and j ≥1; 4. ∞× ∞matrices C(1), . . . , C(s) over R. Combinatorial 627 For n = 0, 1, . . ., let n = P∞ r=0 ar(n)br with all ar(n) ∈Zb and ar(n) = 0 for all suﬃciently large r be the digit expansion of n in base b. We put n = (ψr(ar(n)))∞ r=0 ∈R∞. Next we deﬁne y(i) n,j = η(i) j (c(i) j · n) ∈Zb for n ≥0, 1 ≤i ≤s, and j ≥1, where c(i) j is the j-th row of the matrix C(i). Note that the inner product c(i) j ·n is meaningful since n has only ﬁnitely many nonzero coordinates. Then we put x(i) n = ∞ X j=1 y(i) n,jb−j for n ≥0 and 1 ≤i ≤s. Finally, we deﬁne the sequence S consisting of the points xn = (x(1) n , . . . , x(s) n ) ∈[0, 1]s for n = 0, 1, . . . . 14.8.58 Deﬁnition If the sequence S constructed in Remark 14.8.57 forms a (t, s)-sequence in base b, then S is a digital (t, s)-sequence in base b. If we want to emphasize that the construction uses the ring R, then we speak also of a digital (t, s)-sequence over R. If S is a strict (t, s)-sequence in base b, then S is a digital strict (t, s)-sequence in base b (or over R). 14.8.59 Deﬁnition If the sequence S constructed in Remark 14.8.57 forms a (strict) (T, s)-sequence in base b, then S is a digital (strict) (T, s)-sequence in base b (or over R). 14.8.60 Remark The matrices C(1), . . . , C(s) in Remark 14.8.57 are generating matrices of the digital sequence. The value of t for a digital (t, s)-sequence and the function T for a digital (T, s)-sequence depend only on the generating matrices. For the determination of t in the general case, we refer to Theorem 4.35 in . For the case R = Fq, see Theorem 14.8.62 below. 14.8.61 Example Let s = 1 and let b ≥2 be an integer. Choose R = Zb and let the bijections ψr and η(i) j in Remark 14.8.57 be identity maps. Let C(1) be the ∞× ∞identity matrix over Zb. Then the van der Corput sequence in Example 14.8.49 is easily seen to be a digital (0, 1)-sequence over Zb with generating matrix C(1). 14.8.62 Theorem Let S be the sequence constructed in Remark 14.8.57 with R = Fq and ∞× ∞ generating matrices C(1), . . . , C(s) over Fq. For 1 ≤i ≤s and m ∈N, let C(i) m denote the left upper m × m submatrix of C(i). Then S is a digital strict (T, s)-sequence over Fq with T(m) = m −ϱ(C(1) m , . . . , C(s) m ) for all m ∈N, where ϱ(C(1) m , . . . , C(s) m ) is given by Deﬁnition 14.8.17. 14.8.63 Remark It was shown in that for any prime power q and any integer s ≥1, there exists a digital (t, s)-sequence over Fq for some value of t. In analogy with tb(s) in Remark 14.8.53, we deﬁne dq(s) as the least value of t for which there exists a digital (t, s)-sequence over Fq. It is trivial that tq(s) ≤dq(s), and so Theorem 14.8.54 provides also a lower bound on dq(s). 14.8.64 Problem With the previous notation, it is an open problem whether we can ever have tq(s) < dq(s). 14.8.65 Remark An analog of the duality theory for digital nets described in Subsection 14.8.2 was developed in for the case of digital (T, s)-sequences. Let the weight function vm on 628 Handbook of Finite Fields Fm q be as in Deﬁnition 14.8.22. For A = (a(1), . . . , a(s)) ∈Fms q with a(i) ∈Fm q for 1 ≤i ≤s, we put Um(A) = max 1≤i≤s vm(a(i)). 14.8.66 Deﬁnition Let s ≥2 be an integer. For each integer m ≥1, let Nm be an Fq-linear subspace of Fms q with dim(Nm) ≥ms −m. Let Nm+1,m be the projection of the set {A ∈Nm+1 : Um+1(A) ≤m}, where A = (a(1), . . . , a(s)) with all a(i) ∈Fm+1 q , on the ﬁrst m coordinates of each a(i) for 1 ≤i ≤s. Suppose that Nm+1,m is an Fq-linear subspace of Nm with dim(Nm+1,m) ≥dim(Nm) −1 for all m ≥1. Then the sequence (Nm)m≥1 of spaces is a dual space chain over Fq. 14.8.67 Theorem Let s ≥2 be an integer. Then from any dual space chain (Nm)m≥1 over Fq we can construct a digital strict (T, s)-sequence over Fq with T(m) = m + 1 −δm(Nm) for all m ≥1. 14.8.68 Remark In analogy with the generalized digital nets mentioned in Remark 14.8.28, there are generalized digital sequences as introduced in and also studied in Chapter 15 of . 14.8.69 Remark The nonlinear digital method described in Remark 14.8.29 can be used also for the construction of (t, s)-sequences . 14.8.6 Constructions of (t, s)-sequences and (T, s)-sequences 14.8.70 Remark A general family of digital (t, s)-sequences is formed by Niederreiter sequences . We describe only the simplest case of this construction. For a given dimension s ≥1, let p1, . . . , ps ∈Fq[x] be pairwise coprime polynomials over Fq. Let ei = deg(pi) ≥1 for 1 ≤i ≤s. For 1 ≤i ≤s and integers u ≥1 and 0 ≤k < ei, consider the Laurent series expansion xk pi(x)u = ∞ X r=0 a(i)(u, k, r)x−r−1 ∈Fq((x−1)). Then deﬁne c(i) j,r = a(i)(Q+1, k, r) ∈Fq for 1 ≤i ≤s, j ≥1, and r ≥0, where j−1 = Qei+k with integers Q = Q(i, j) and k = k(i, j) satisfying 0 ≤k < ei. The generating matrices of the Niederreiter sequence are now given by C(i) = (c(i) j,r)j≥1,r≥0 for 1 ≤i ≤s. The bijections ψr and η(i) j in Remark 14.8.57 are chosen arbitrarily. 14.8.71 Theorem [837, 2238] The Niederreiter sequence based on the pairwise coprime non-constant polynomials p1, . . . , ps ∈Fq[x] is a digital strict (t, s)-sequence over Fq with t = Ps i=1(deg(pi) −1). 14.8.72 Remark If q is a prime, 1 ≤s ≤q, and pi(x) = x −i + 1 ∈Fq[x] for 1 ≤i ≤s, then we obtain the digital (0, s)-sequences over Fq called Faure sequences . An analogous construction of digital (0, s)-sequences over Fq for arbitrary prime powers q and dimensions 1 ≤s ≤q was given in . Note that in view of Corollary 14.8.52, s ≤q is also a necessary condition for the existence of a (0, s)-sequence in base q. If q = 2, s ≥1 is an arbitrary dimension, p1(x) = x ∈F2[x], and p2, . . . , ps are distinct primitive polynomials over F2, then we obtain Sobol’ sequences . 14.8.73 Remark The construction of Niederreiter sequences in Remark 14.8.70 is optimized by letting p1, . . . , ps be s distinct monic irreducible polynomials over Fq of least degrees. If with this choice we put Tq(s) = Ps i=1(deg(pi) −1), then for ﬁxed q the quantity Tq(s) is of the order of magnitude s log s as s →∞. Let U(s) denote the least value of t that is Combinatorial 629 known to be achievable by Sobol’ sequences for given s. Then T2(s) = U(s) for 1 ≤s ≤7 and T2(s) < U(s) for all s ≥8. 14.8.74 Remark Substantial improvements on the construction of Niederreiter sequences in Remark 14.8.70 can be obtained by using tools from the theory of algebraic function ﬁelds over ﬁnite ﬁelds (see Section 12.1 for this theory). This leads to the family of Niederreiter-Xing sequences [2282, 3018]. Let F be an algebraic function ﬁeld with full constant ﬁeld Fq and genus g. Assume that F contains at least one rational place P∞and let D be a divisor of F with deg(D) = 2g and P∞/ ∈supp(D). Furthermore, we choose s distinct places P1, . . . , Ps of F with Pi ̸= P∞for 1 ≤i ≤s. There exist integers 0 = n0 < n1 < · · · < ng ≤2g such that ℓ(D −nuP∞) = ℓ(D −(nu + 1)P∞) + 1 for 0 ≤u ≤g. We choose wu ∈L(D −nuP∞) \ L(D −(nu + 1)P∞) for 0 ≤u ≤g. For each i = 1, . . . , s, we consider the chain L(D) ⊂L(D + Pi) ⊂L(D + 2Pi) ⊂· · · of vector spaces over Fq. By starting from the basis {w0, w1, . . . , wg} of L(D) and successively adding basis vectors at each step of the chain, we obtain for each n ≥1 a basis {w0, w1, . . . , wg, f (i) 1 , f (i) 2 , . . . , f (i) n deg(Pi)} of L(D + nPi). For a prime element z at P∞and for r = 0, 1, . . ., we put zr = zr if r / ∈{n0, n1, . . . , ng} and zr = wu if r = nu for some u ∈{0, 1, . . . , g}. Each f (i) j with 1 ≤i ≤s and j ≥1 has then a local expansion at P∞of the form f (i) j = P∞ r=0 a(i) j,rzr with all a(i) j,r ∈Fq. Let c(i) j be the sequence obtained from the sequence a(i) j,r, r = 0, 1, . . ., by deleting the terms with r = nu for some u ∈{0, 1, . . . , g}. For 1 ≤i ≤s, the generating matrix C(i) of the Niederreiter-Xing sequence is now the matrix whose j-th row is c(i) j for j ≥1. The bijections ψr and η(i) j in Remark 14.8.57 are chosen arbitrarily. 14.8.75 Theorem Let F be an algebraic function ﬁeld with full constant ﬁeld Fq and genus g which contains at least one rational place P∞. Let D be a divisor of F with deg(D) = 2g and P∞/ ∈supp(D) and let P1, . . . , Ps be distinct places of F with Pi ̸= P∞for 1 ≤i ≤s. Then the corresponding Niederreiter-Xing sequence is a digital (t, s)-sequence over Fq with t = g + Ps i=1(deg(Pi) −1). 14.8.76 Corollary For every prime power q and every dimension s ≥1, there exists a digital (Vq(s), s)-sequence over Fq, where Vq(s) = min {g ≥0 : Nq(g) ≥s + 1} and Nq(g) is the maximum number of rational places that an algebraic function ﬁeld with full constant ﬁeld Fq and genus g can have. 14.8.77 Remark It was shown in that Vq(s) = O(s) as s →∞. Since tq(s) ≤dq(s) ≤Vq(s) by Remark 14.8.63 and Corollary 14.8.76, we obtain tq(s) = O(s) and dq(s) = O(s) as s →∞. In view of Theorem 14.8.54, these asymptotic bounds are best possible. 14.8.78 Remark The only improvements on Niederreiter-Xing sequences were obtained, in some special cases, in the more recent paper . For instance, let q be an arbitrary prime power and let s = q+1. Then tq(q+1) = dq(q+1) = 1. On the other hand, the construction in yields a digital (T, q + 1)-sequence over Fq with T(m) = 0 for even m ≥2 and T(m) = 1 for odd m ≥1. 630 Handbook of Finite Fields See Also §14.1 For orthogonal arrays and related combinatorial designs. §15.1 For related constructions of linear codes. References Cited: [271, 276, 835, 836, 837, 838, 839, 1045, 1858, 1873, 1874, 2007, 2008, 2009, 2183, 2234, 2236, 2238, 2247, 2248, 2251, 2255, 2256, 2258, 2259, 2261, 2263, 2266, 2267, 2282, 2283, 2398, 2483, 2565, 2690, 3018] 14.9 Applications and weights of multiples of primitive and other polynomials Brett Stevens, Carleton University 14.9.1 Applications where weights of multiples of a base polynomial are relevant 14.9.1 Remark The performance of several applications of polynomials, frequently primitive, de-pend on the weights of multiples of the base polynomial. Many of these applications are discussed in this Handbook. 14.9.1.1 Applications from other Handbook sections 14.9.2 Remark The multiples of a polynomial f with weight w inﬂuence the statistical bias of the linear feedback shift register sequence generated from f. Fewer multiples with a given weight, w reduces the w-th moment of the Hamming weight [1622, 1944]. For more information on bias and randomness of linear feedback shift register sequences see Section 10.2. 14.9.3 Remark In Section 15.1 the use of primitive polynomials f, to generate cyclic redundancy check codes is discussed. The undetectable error patterns of these codes are precisely those whose errors correspond to multiples of f. This has the consequences that burst errors of length up to deg(f) are always detectable and that to understand how many arbitrary errors can be detected requires having knowledge of the weights of multiples of f. 14.9.4 Remark In Section 15.4, turbo codes are discussed. Turbo codes use feedback polynomials that are often primitive. The bit error rate (BER) of the turbo code’s interleaver design depends on the weights of polynomials divisible by the feedback polynomial . 14.9.5 Remark Low weight multiples of a public polynomial compromise the private key for the T CHo cryptosystem and its security therefore rests on the diﬃculty of ﬁnding low weight multiples [146, 1491]. The weight of polynomials and their multiples is important in linear feedback shift register cryptanalysis and certain attacks depend on the sparsity of the feed-back polynomial or one of its multiples . Chapter 16 discusses the many connections between ﬁnite ﬁelds and cryptography. Combinatorial 631 14.9.1.2 Application of polynomials to the construction of orthogonal arrays 14.9.6 Remark We present a discussion of applications of polynomials and the weights of their multiples to the construction and strength of orthogonal arrays. 14.9.7 Deﬁnition An orthogonal array of size N, with k constraints (or k factors or of degree k), s levels (or of order s), and strength t, denoted OA(N, k, s, t), is a k × N array (sometimes N × k) with entries from a set of s ≥2 symbols, having the property that in every t × N submatrix, every t × 1 column vector appears the same number λ = N st of times. The parameter λ is the index of the orthogonal array. An OA(N, k, s, t) is also denoted by OAλ(t, k, s). 14.9.8 Remark From the deﬁnition, an orthogonal array of strength t is also an orthogonal array of strength t′ for all 1 ≤t′ ≤t. 14.9.9 Theorem [357, 1457] Let C be a linear code over Fq with words of length n. Then the n × \|C\| array formed with the words of C as the columns is a (linear) orthogonal array of maximal strength t if and only if C⊥, its dual code, has minimum weight t + 1. 14.9.10 Remark The half of Theorem 14.9.9 that gives the strength of the orthogonal array from the minimum weight of the dual code was known as early as 1947 [1457, 1727]. Delsarte was able to generalize Theorem 14.9.9 to the case where the code and the orthogonal array are not required to be linear . We can extend Theorem 14.9.9 and exactly determine the number of times each vector appears in any (t + 1) × n submatrix of the orthogonal array. 14.9.11 Theorem Let C be a linear code of length n over Fq and assume that the words of C form the columns of an orthogonal array of strength t. Then for any t + 1-subset T = {i1, . . . it} ⊂{1, . . . , n} and for any t + 1-tuple b ∈Ft+1 q , the number of times that b appears as a column of the (t + 1) × n submatrix determined by T, λT b (C), is λT b (C) =    \|C\|/qt if there is no u ∈C⊥with support T; \|C\|/qt−1 if there exists a u ∈C⊥with support T and uij = bj for ij ∈T; 0 otherwise. 14.9.12 Theorem Let f be a primitive polynomial of degree m over Fq and let Cf n be the set of all subintervals of the shift-register sequence with length n generated by f, together with the zero vector of length n. The dual code of Cf n is given by (Cf n)⊥= {(b1, . . . , bn) : n−1 X i=0 bi+1xi is divisible by f}. 14.9.13 Remark Munemasa only proves Theorem 14.9.12 over F2 but the proof works more generally for any ﬁnite ﬁeld. 14.9.14 Remark The primitivity condition in Theorem 14.9.12 can be substantially relaxed to polynomials with distinct roots. 14.9.15 Remark The combined eﬀect of Theorems 14.9.9 and 14.9.12 is that to know the strength of the orthogonal array derived from a polynomial f, and its shift register sequences, it is essential to know about the weights of multiples of f. 632 Handbook of Finite Fields 14.9.1.3 Application of polynomials to a card trick 14.9.16 Remark Although the weight of multiples of primitive f = x5 + x2 + 1 ∈F2[x] is not relevant to this application to a card trick, the low weight of f itself facilitates the mental arithmetic so we include this application in this subsection. 14.9.17 Remark The polynomial f(x) = x5 + x2 + 1 ∈F2[x] is primitive and generates the binary shift register sequence with the property that ak+5 = ak + ak+2 0000100101100111110001101110101. 14.9.18 Remark The set of cards from a standard deck which contains the Ace, 2, 3, 4, 5, 6, and 7 of each suit and the 8 of spades, clubs, and hearts can be encoded uniquely with the non-zero binary words of length 5. The ﬁrst digit encodes the cards color, 0 for red and 1 for black. The second digit encodes whether the suit is major or minor in bridge: 0 for clubs or diamonds; 1 for hearts or spades. The remaining three digits encode the value of the card via the last three digits in the binary representation of the card’s value: 000 for 8, 001 for Ace, 010 for 2, 011 for 3, 100 for 4, 101 for 5, 110 for 6, and 111 for 7. This encoding has the property that the ﬁrst digit in a card’s code corresponds to the color of that card. Other encodings have the required properties as well . 14.9.19 Remark Using the shift register sequence from Remark 14.9.17 and the card encoding from Remark 14.9.18 we obtain the following sequence of cards: A♦, 2♦, 4♦, A♥, 2♣, 5♦, 3♥, 6♣, 4♥, A♠, 3♣, 7♦, 7♥, 7♠, 6♠, 4♠, 8♠, A♣, 3♦, 6♦, 5♥, 3♠, 7♣, 6♥, 5♠, 2♠, 5♣, 2♥, 4♣, 8♥, 8♣ 14.9.20 Remark A deck of these 31 cards arranged in this order looks upon casual inspection to be randomly ordered. The deck can be cut arbitrarily many times (since the shift register sequence is cyclic) before removing ﬁve cards in sequence from the top of the deck. With the knowledge which cards are black, the identity of all ﬁve chosen cards can be determined . 14.9.21 Remark Due to the low weight of primitive f = x5 + x2 + 1, the encoding scheme and the generating polynomial are simple enough to be quickly calculated mentally which is important for the appearance of the trick . 14.9.22 Remark Much can be done to augment the impression this trick makes on an audience. For ideas see [832, 833, 2169]. Two sets of these 31 cards with identical backs can be placed in this order repeated to give the impression of a more normal sized deck. 14.9.23 Remark The 8♦, corresponding to the binary string 00000, can be added to the deck be-tween the 8♣and A♦. This deviation from the linear shift register can simply be memorized ad-hoc or a new, nonlinear shift register sequence memorized: ak+5 = (1 + ak+1 · ak+2 · ak+3 · ak+4)(ak + ak+2) + (ak · ak+1 · ak+2 · ak+3 · ak+4), where ai is the complement of ai. 14.9.24 Remark For other mathematical card tricks see [218, 832, 833, 2169]. 14.9.25 Remark The applications discussed in Remarks 14.9.2 through 14.9.15 strongly motivate researching the distributions and patterns of weights of multiples of polynomials f over ﬁnite ﬁelds. Subsection 14.9.2 gives a summary of the knowledge in this area to date. Combinatorial 633 14.9.2 Weights of multiples of polynomials 14.9.26 Remark Polynomials in F2[x] which have large weight or large degree will sometimes be given in hexadecimal notation. For example the polynomial f = x8 + x7 + x6 + x5 + x + 1 = 1x8 + 1x7 + 1x6 + 1x5 + 0x4 + 0x3 + 0x2 + 1x1 + 1x0 is 111100011 in binary notation and, grouping these from the right into fours, is 1E3 in hexadecimal notation. The use of the two notations for polynomials will always be clear in the context. Be aware that some authors in the literature denote binary polynomials in hexidecimal after deleting the rightmost 1, since most polynomials used in applications have a constant term 1, so it can be assumed present in many contexts. 14.9.27 Deﬁnition The set of polynomials of degree d in Fq[x] is denoted by Pq,d. For f ∈F[x], the dual code of length n, (Cf n)⊥, deﬁned in Theorem 14.9.12 can be identiﬁed with all polynomials divisible by f of degree less than n. The minimum weight of a polynomial from this set is denoted by d((Cf n)⊥). This is also the minimum weight of the code (Cf n)⊥. 14.9.28 Remark We begin with some general bounds on d((Cf n)⊥), followed by results for polyno-mials f of speciﬁc degree and end with results for polynomials f of speciﬁc weights. 14.9.2.1 General bounds on d((Cf n)⊥) 14.9.29 Proposition An application of Theorem 14.9.12 with bounds on the period of polynomials gives that if f ∈Pq,m, then d((Cf n)⊥) = 2 for all n ≥qm −1. 14.9.30 Theorem Let f ∈P2,m and 0 ≤t ≤(m −1)/2. Let n1(t) be the smallest positive integer such that t+1 X j=0 n1(t) j > 2m. Set n2(0) = ∞and for t > 0, let n2(t) = 2⌊(m−1)/t⌋−1. If n1(t) < n2(t), then for all n1(t) ≤n ≤n2(t), d((Cf n)⊥) ≤2t + 2. In other words, for such n, there will always be a multiple of f of weight less than 2t + 3 and degree less than n. 14.9.31 Theorem Let e = ⌊(m −1)/t⌋and n2(t) = 2e −1. Let α be a primitive element in F2e and M (i)(x) be the minimal polynomial of αi. Let g = lcm{M (i)(x)\|0 ≤i ≤2t}, then d((Cg 2e−1)⊥) ≥2t + 2 and the BCH code (see Section 15.1) generated by g can be truncated to a code meeting the bound in Theorem 14.9.30 for any admissible n1(t) ≤n ≤ n2(t). 14.9.32 Proposition In Theorem 14.9.30, n1(t) ≤t + 2m/(t+1)(t + 1)!1/(t+1), and whenever m > (t + 1)2 + t(t + 1) log2(t + 1), we have n1(t) < n2(t). 14.9.33 Theorem If f ∈P2,m is primitive and if g = xn + xk + 1 is the trinomial multiple of f with minimum degree then n ≤2m + 2 3 . 14.9.34 Proposition If x + 1 is a factor of f ∈F2[x] then f does not divide any polynomials of odd weight. 634 Handbook of Finite Fields 14.9.35 Lemma Let f ∈F2[x] have simple roots, period n and suppose (1 + x) is a factor of f. If d((Cf n)⊥) = d then d((C(x+1)f n+1 )⊥) = d and d((C(x+1)f j )⊥) = 4 for n + 2 ≤j ≤2n. 14.9.36 Theorem Let f ∈F2[x] be an irreducible polynomial with period ρ. Then f divides a trinomial if and only if gcd(xρ + 1, (x + 1)ρ + 1) ̸= 1. 14.9.37 Theorem If f ∈P2,m is primitive and if g = xn + xk + 1 is a trinomial divisible by f then n and k belong to the same-length cyclotomic coset modulo 2m −1. 14.9.38 Theorem All primitive f ∈P2,m divide some 4-nomial of degree no bigger than $ 1 + √ 1 + 4.2m+1 2 % . 14.9.39 Theorem For a given t ≥2 and s ≥1, if m is such that 1.548m −1 ≥(t −1) ms + 1 t then there exists at least one primitive polynomial of degree m which does not divide any t-nomial of degree less than or equal to ms. 14.9.40 Theorem Let f ∈P2,m be a primitive t-nomial. Then there exists a primitive g ∈P2,m which divides some t-nomial of degree sm (s odd) when gcd(s, 2m −1) = 1. Moreover g = gcd(f(xs), x2m−1 −1) is such a polynomial. 14.9.41 Remark The previous results give information about multiples of f that can have small degree relative to the period of f. The following gives information about multiples of f that have relatively large degree. 14.9.42 Remark Let f ∈Fq[x] be primitive of degree m. Let n = qm −1 and T2 be the set of binomials of the form xi −xj, satisfying 0 ≤i < j with i ≡j (mod n). Let Ti be the set of all linear combinations of binomials from T2 which have weight i. Finally deﬁne µ : Fq[x] →Fq[x] by µ(Pd i=0 aixi) = Pd i=0 aixi (mod n). 14.9.43 Theorem Suppose that g is a polynomial of weight w and write g = g1 + g2 where g1 ∈Ti, g2 ∈Fq[x] has weight j, no two exponents of g2 are congruent modulo n, and the terms of g1 and g2 are disjoint (i.e., w = i + j). Then g is divisible by f if and only if µ(g2) is divisible by f. 14.9.44 Remark In , Theorem 14.9.43 is only stated and proved over F2. It is true for all ﬁnite ﬁelds Fq. 14.9.45 Remark Using the fact that f divides xn′ + a, with n′ = (qm −1)/(q −1) for some unique a ∈Fq, and letting T being the set of corresponding binomials, Theorem 14.9.43 can be further generalized with an increase in the complexity of the statement. 14.9.46 Remark There have also been some interesting results on enumeration and probability of multiples with given weights. We discuss this next. 14.9.47 Theorem Given t ≥2 and s ≥1, if m is such that φ(2m −1) > (t −1) ms+1 t , then the probability that a randomly chosen primitive polynomial of degree m does not divide any t-nomial of degree less than or equal to ms is at least 1 −(t −1)( ms+1 t φ(2m −1) , Combinatorial 635 where φ denotes Euler’s function. 14.9.48 Theorem There exist primitive polynomials of degree m which divide a trinomial of degree 3m and a 4-nomial of degree less than 6m. 14.9.49 Theorem Let Nm,t denote the number of t-nomial multiples with degree no more than 2m −2 of a primitive polynomial of degree m. Then Nm,2 = Nm,1 = 0 and Nm,t = 2m−2 t−2 −Nm,t−1 −t−1 t−2(2m −t + 1)Nm,t−2 t −1 . 14.9.50 Remark See for discussion and results for solving this recurrence, and for an alternative presentation. 14.9.51 Theorem Given any primitive polynomial of degree m, the sum of the degrees of all its t-nomial multiples is t −1 t (2m −1)Nm,t. 14.9.52 Theorem Given any primitive polynomial f of degree m, the average degree of its t-nomial multiples with degree no more than 2m −2 is equal to the average of the maximum of all the distinct (t −1)-tuples from 1 to 2m −2. 14.9.53 Theorem Given a primitive polynomial f of degree m, under the assumption that t-nomial multiples of f are distributed as (t −1)-tuples, there exists a t-nomial multiple g of f such that the degree of g is less than or equal to 2 m t−1 +log2(t−1)+1. 14.9.54 Remark The assumption in Theorem 14.9.53 is motivated by Theorem 14.9.52 and empirical evidence. See for precise deﬁnition of the assumption and detailed discussion. 14.9.55 Remark Theorem 14.9.53 implies that it is highly likely to get a trinomial multiple with degree no more than 2m/2+2. This is in contrast to the bound of (2m + 2)/3 from Theo-rem 14.9.33. In general Theorem 14.9.53 suggests that to avoid sparse multiples, f must be picked with very large degree. 14.9.56 Remark In , Maitra, Gupta, and Venkateswarlu extend this enumerative and proba-bilistic analysis to include the product of primitive polynomials. 14.9.2.2 Bounds on d((Cf n)⊥) for polynomials of speciﬁc degree 14.9.57 Proposition The bounds on weights of multiples of all polynomials from degree 4 to degree 16 and degrees 24 and 32 in F2[x] are given in Table 14.9.2.2. For degrees 4 ≤m ≤16, Koopman and Chakravarty exhaustively searched all polynomials of degree m and all their multiples of degrees m + 8 ≤n ≤m + 2048 . The m = 16 results are from the theoretical work of Merkey and Posner and exhaustive searches by Castagnoli, Ganz, and Graber . The bounds on weights of multiples of degree 24 polynomials, which are less complete than those for smaller m, are the work of Merkey and Posner and searches by Castagnoli, Ganz, and Graber and Ray and Koopman . In all cases for m = 24 polynomials attaining the bounds are reported to be known although the speciﬁc polynomials have not been published [559, 2083, 2439]. The even more incomplete results for m = 32 are reported in [559, 2083]. 14.9.58 Example Table 14.9.2.2 gives bounds that apply to every polynomial with the given degree. To aid the reading of Table 14.9.2.2, we give an example from it. The information from the 636 Handbook of Finite Fields deg(f) degree range of mul-tiples of f upper bound on d(Cf n)⊥ polynomial (in hexadecimal no-tation) attaining the bound 4 12 ≤n ≤15 3 f = 13 5 13 ≤n ≤15 4 f = 2B 16 ≤n ≤31 3 f = 25 6 14 ≤n ≤31 4 f = 59 32 ≤n ≤63 3 f = 43 7 15 ≤n ≤63 4 f = B7 64 ≤n ≤127 3 f = 91 8 16 ≤n ≤17 5 f = 139 18 ≤n ≤127 4 f = 12F 128 ≤n ≤255 3 f = 14D 9 n = 17 6 f = 13C 18 ≤n ≤22 5 f = 30B 23 ≤n ≤255 4 f = 297 256 ≤n ≤511 3 f = 2CF 10 18 ≤n ≤22 6 f = 51D 23 ≤n ≤31 5 f = 573 32 ≤n ≤511 4 f = 633 512 ≤n ≤1023 3 f = 64F 11 19 ≤n ≤23 7 f = AE1 24 ≤n ≤33 6 f = A65 34 ≤n ≤36 5 f = BAF 37 ≤n ≤1023 4 f = B07 1024 ≤n ≤2047 3 f = C9B 12 20 ≤n ≤23 8 f = 149F 24 ≤n ≤39 6 f = 1683 40 ≤n ≤65 5 f = 11F 1 66 ≤n ≤2047 4 f = 180F 2048 ≤n ≤2060 3 f = 16EB 13 21 ≤n ≤24 8 f = 216F n = 25 7 f = 254B 26 ≤n ≤65 6 f = 3213 66 ≤n ≤2061 4 f = 2055 14 22 ≤n ≤25 8 f = 46E3 26 ≤n ≤27 7 f = 5153 28 ≤n ≤71 6 f = 6E57 72 ≤n ≤127 5 f = 425B 128 ≤n ≤2062 4 f = 43D1 15 23 ≤n ≤27 8 f = C617 28 ≤n ≤31 7 f = B7AB 32 ≤n ≤129 6 f = AE75 128 ≤n ≤191 5 f = D51B 192 ≤n ≤2063 4 f = 92ED 16 n = 18 12 f = 15BED 19 ≤n ≤21 10 f = 1D22F n = 22 9 f = 18F 57 23 ≤n ≤31 8 f = 11F B7 32 ≤n ≤35 7 f = 126B5 36 ≤n ≤151 6 f = 13D65 152 ≤n ≤257 5 f = 15935 258 ≤n ≤32767 4 f = 1A2EB 32768 ≤n ≤65535 3 f = 1002D 24 18 ≤n ≤47 12 48 ≤n ≤50 10 51 ≤n ≤63 9 64 ≤n ≤129 8 130 ≤n ≤255 7 466 ≤n ≤211 −1 6 5793 ≤n ≤223 −1 4 32 n = 18 12 568 ≤n ≤210 −1 8 2954 ≤n ≤215 −1 6 92682 ≤n ≤231 −1 4 Table 14.9.1 Bounds on weights of multiples of degree n polynomials for 4 ≤n ≤16. Combinatorial 637 minimum distance f standard d(Cf n)⊥: 12 11 10 9 8 7 13D65 IEC TC57 after 1990 ranges of n: [17,20] [21,22] 1F29F ranges of n: 17 [18,22] 15B93 IEC TC57 before 1990 ranges of n: [17,19] [20,25] 15935 ranges of n: [17,19] [20,24] [25,26] 16F63 IEEE WG77.1 ranges of n: 17 18 [19,29] 1A2EB ranges of n: [17,18] [19,27] 1011B ranges of n: 1A097 IBM SDLC ranges of n: [17,24] 11021 CRC-CCITT ranges of n: 18005 CRC-ANSI ranges of n: minimum distance f standard d(Cf n)⊥: 6 5 4 2 13D65 IEC TC57 after 1990 ranges of n: [23,151] [151,∞] 1F29F ranges of n: [23,130] [131,258] [259,∞] 15B93 IEC TC57 before 1990 ranges of n: [26,128] [129,254] [255,∞] 15935 ranges of n: [27,51] [52,257] [258,∞] 16F63 IEEE WG77.1 ranges of n: 30 [31,255] [256,∞] 1A2EB ranges of n: [28,109] [110,32767] [32768,∞] 1011B ranges of n: [17,115] [116,28658] [28659,∞] 1A097 IBM SDLC ranges of n: [25,83] [84,32766] [32767,∞] 11021 CRC-CCITT ranges of n: [17,32767] [32768,∞] 18005 CRC-ANSI ranges of n: [17,32767] [32768,∞] Table 14.9.2 Distance proﬁles of speciﬁc degree 16 polynomials. third line of the section for polynomials of degree 11, indicates that for every binary degree 11 polynomial f ∈P2,11, there exists multiples of f which have degrees 34, 35, and 36 and weight less than or equal to 5. The polynomial cited in the last column, f(x) = BAF = x11 +x9 +x8 +x7 +x5 +x3 +x2 +x+1, meets this bound tightly; that is, all of its multiples of degree 34, 35, or 36 have weight 5 or above. 14.9.59 Remark In Table 14.9.2.2, three of the degree 16 polynomials meeting the bounds are known to be unique. For d(Cf n)⊥= 6, f = 13D65 and for d(Cf n)⊥= 4, f = 1A2EB are the unique tight polynomials, up to reciprocal. For d(Cf n)⊥= 5, f = 15935 is unique . 14.9.60 Remark In contrast to Table 14.9.2.2, Tables 14.9.2 through 14.9.4 give the distance dis-tributions of multiples of a few, speciﬁc polynomials for degrees 16, 24, and 32. 14.9.61 Remark Table 14.9.2 gives the distance proﬁles of ten speciﬁc polynomials in P2,16 found by Castagnoli, Ganz, and Graber. They exhaustively searched all degree 16 poly-nomials for those with optimum proﬁles. The polynomial f = 1F29F is the unique poly-nomial with d(Cf 130)⊥= 6 and d(Cf 258)⊥= 4. Up to reciprocal, f = 1011B is the unique polynomial with d(Cf 28658)⊥= 4 and d(Cf 115)⊥= 6. The authors of suggest that any cyclic redundancy check polynomials of degree 16 should be chosen only from the list {13D65, 1F29F, 15935, 1A2EB, 1011B}. 14.9.62 Example The third polynomial in Table 14.9.2 gives the distance distribution for the poly-nomial f(x) = 15B93 = x16 +x14 +x12 +x11 +x9 +x8 +x7 +x4 +x+1, which was the IEC TC57 standard cyclic redundancy check polynomial until 1990. All its multiples of degrees 17–19 have weight 10 or more. All its multiples of degrees 20–25 have weight 8 or more. All its multiples of degrees 26–128 have weight 6 or more. All its multiples of degree 129–254 have weight 5 or more. All its multiples of degrees 255 and higher have weight at least 2. For each degree there exist speciﬁc multiples that attain these lower bounds; for example there is a degree 17 multiple of f with weight 10. 14.9.63 Remark Table 14.9.3 gives the distance distribution for some speciﬁc polynomials of de-gree 24. All were constructed via the generalized BCH code method: multiplying together 638 Handbook of Finite Fields minimum distance f d(Cf n)⊥: 16 15 14 12 11 10 9 ref. 1323009 ranges of n: [558, 2083] 1401607 ranges of n: [558, 2083] 1805101 ranges of n: 15D6DCB ranges of n: 25 26 [27,36] 17B01BD ranges of n: [25,26] [27,41] 131FF19 ranges of n: 25 [26,33] 15BC4F5 ranges of n: [25,26] [27,28] [29,31] [32,33] [34,35] 1328B63 ranges of n: [25,30] [31,36] minimum distance f d(Cf n)⊥: 8 7 6 5 4 2 ref. 1323009 ranges of n: [25,68] [69,2048] [2049,4094] [4095,∞] [558, 2083] 1401607 ranges of n: [25,55] [56,2048] [2049,4094] [4095,∞] [558, 2083] 1805101 ranges of n: [25,1023] 15D6DCB ranges of n: [37,83] [84,2050] [2051,4098] [4099,∞] 17B01BD ranges of n: [42,95] [96,2048] [2049,4094] [4095,∞] 131FF19 ranges of n: [34,37] [38,252] [253,4097] [4098,∞] 15BC4F5 ranges of n: [36,41] [42,47] [78,217] [218,4095] [4096,∞] 1328B63 ranges of n: [37,61] [62,846] [847,23 −1] [223,∞] Table 14.9.3 Distance proﬁles of speciﬁc degree 24 polynomials. minimal polynomials of elements from F2e and small factors, x and x + 1 [558, 2083]. For discussion of BCH codes, see Section 15.1. 14.9.64 Remark Table 14.9.4 gives the distance distribution for some speciﬁc polynomials of degree 32. All were obtained via the generalized BCH code method: multiplying together minimal polynomials of elements from F2e and small factors, x and x + 1 [558, 2083]. 14.9.65 Remark For the third polynomial in Table 14.9.4, used in many standards, Jain has determined and published many of the minimum degree polynomials that establish the ranges given in Table 14.9.4. The actual polynomials are given in Table 14.9.5. Jain has determined all the polynomials that f divides which have the pattern of at most three burst errors of length 4 each and several other speciﬁc patterns of errors. 14.9.66 Remark Koopman has performed an exhaustive search over all f ∈P2,32 for 40 ≤n ≤ 131104. His primary concern was ﬁnding cyclic redundancy check polynomials which were simultaneously good at typical Ethernet maximum transmission unit (MTU) lengths, n = 12112, and much longer lengths n ≥64, 000, so although his search has in principle solved the d(Cf n)⊥problem for all n in this range he did not speciﬁcally publish these, rather he highlights the last three polynomials given in Table 14.9.4 and compares them to the others . Discussion of the beneﬁts and costs of using these various polynomials in diﬀerent scenarios appear in [558, 1590, 1792, 2083]. 14.9.2.3 Bounds on d((Cf n)⊥) for polynomials of speciﬁc weight 14.9.67 Remark We now present divisibility results that are organized by the weight of the base polynomial. 14.9.68 Theorem Let f(x) = xm + xl + 1 be a trinomial over F2 such that gcd(m, l) = 1. If g is a trinomial multiple of f of degree at most 2m, then 1. g(x) = xdeg g−mf(x); 2. g(x) = f(x)2; 3. g(x) = x5 + x4 + 1 = (x2 + x + 1)(x3 + x + 1) or; its reciprocal, Combinatorial 639 minimum distance f d(Cf n)⊥: 20 18 17 16 15 14 13 12 11 10 9 8 ref. 1404098E2 ranges of n: [33,78] [79,1023] [558, 2083] 10884C912 ranges of n: [33,79] [80,1023] [558, 2083] 104C11DB7∗ranges of n: [33,42] [43,44] [45,53] [54,66] [67,89] [90,123] [558, 1590] 1F1922815 ranges of n: [33,44] [45,48] [49,98] [99,1024] 1F4ACFB13 ranges of n: 33 [34,35] 36 37 [38,43] [44,56] [57,306] 1A833982B ranges of n: [33,35] [36,49] [50,53] [54,59] [60,90] 1572D7285 ranges of n: [33,34] 35 [36,38] [39,52] [53,68] [69,80] [81,110] 11EDC6F41 ranges of n: 33 [34,38] [39,40] [41,52] [53,79] [80,209] 1741B8CD7 ranges of n: [40,48] [49,50] [51,184] 132583499 ranges of n: [40,48] [49,58] [59,166] 120044009 ranges of n: 100210801 ranges of n: minimum distance f d(Cf n)⊥: 7 6 5 4 3 2 ref. 1404098E2 ranges of n: [1024,∞] [558, 2083] 10884C912 ranges of n: [1024,∞] [558, 2083] 104C11DB7∗ranges of n: [124, 203] [204,300] [301,3006] [3007,91639] [91640,232 −1] [232,∞] [558, 1590] 1F1922815 ranges of n: [1025,2046] [2047,∞] 1F4ACFB13 ranges of n: [307,32768] [32769,65534] [65535,∞] 1A833982B ranges of n: [91,113] [114,1092] [1093,65537] [65538,∞] 1572D7285 ranges of n: [111,266] [267,1029] [1030,65535] [65536,∞] 11EDC6F41 ranges of n: [210,5275] [5276,231 −1] [231,∞] 1741B8CD7 ranges of n: [185,16392] [16393,114695] [114696,∞] 132583499 ranges of n: [167,32769] [32770,65538] [65539,∞] 120044009 ranges of n: [40,32770] [32771,65538] [65539,∞] 100210801 ranges of n: [40,65537] [65538,∞] Table 14.9.4 Distance proﬁles of degree 32 polynomials. ∗The third polynomial f = 104C11DB7 is used in the FDDI, IEEE 802, AUTODIN-II standards. t smallest degree t-nomial divisible by f 3 x91639 + x41678 + 1 4 x3006 + x2846 + x2215 + 1 5 x300 + x155 + x117 + x89 + 1 6 x203 + x196 + x123 + x85 + x79 + 1 7 x123 + x120 + x80 + x74 + x58 + x46 + 1 8 x89 + x88 + x41 + x36 + x16 + x3 + x2 + 1 9 x66 + x57 + x37 + x32 + x31 + x16 + x7 + x5 + 1 10 x53 + x38 + x36 + x33 + x30 + x27 + x26 + x7 + x3 + 1 11 x44 + x43 + x41 + x37 + x35 + x32 + x31 + x16 + x7 + x5 + 1 12 x42 + x30 + x26 + x24 + x21 + x18 + x13 + x8 + x7 + x5 + x3 + 1 13 x42 + x40 + x37 + x35 + x33 + x29 + x28 + x20 + x18 + x15 + x6 + x1 + 1. Table 14.9.5 Smallest t-nomial divisors of f = 104C11DB7. 640 Handbook of Finite Fields 4. g(x) = x5 + x + 1 = (x2 + x + 1)(x3 + x2 + 1). 14.9.69 Theorem Let f(x) = xm + xl + xk + xj + 1 be a pentanomial over F2 such that gcd(m, l, k, j) = 1. If g is a trinomial of degree at most 2m divisible by f, with g = fh, then 1. f is one of the twenty-ﬁve polynomials given in Table 14.9.6 with the correspond-ing h; 2. m ≡1 (mod 3) and f, g, h are as follows f(x) = 1 + x + x2 + xm−3 + xm = (1 + x + x2)(1 + xm−3 + xm−2), h(x) = (1 + x) + (x3 + x4) + · · · + (xm−7 + xm−6) + xm−4, f(x)h(x) = g(x) = 1 + x2m−6 + x2m−4; or 3. f is the reciprocal of one of the polynomials listed in the previous items. No. f(x) h(x) type 1 x5 + x4 + x3 + x2 + 1 x3 + x2 + 1 p 2 x5 + x3 + x2 + x + 1 x3 + x + 1 p 3 x5 + x3 + x2 + x + 1 x4 + x + 1 p 4 x5 + x4 + x3 + x + 1 x2 + x + 1 p 5 x6 + x5 + x4 + x3 + 1 x4 + x3 + 1 r 6 x6 + x4 + x2 + x + 1 x3 + x + 1 i 7 x6 + x4 + x3 + x + 1 x2 + x + 1 p 8 x6 + x5 + x2 + x + 1 x5 + x4 + x3 + x + 1 p 9 x6 + x5 + x3 + x + 1 x2 + x + 1 r 10 x7 + x4 + x2 + x + 1 x3 + x + 1 r 11 x7 + x4 + x3 + x2 + 1 x3 + x2 + 1 p 12 x7 + x5 + x2 + x + 1 x7 + x6 + x5 + x4 + x3 + x + 1 p 13 x7 + x5 + x3 + x2 + 1 x5 + x4 + x3 + x2 + 1 r 14 x8 + x5 + x3 + x + 1 x5 + x4 + x2 + x + 1 p 15 x8 + x5 + x3 + x2 + 1 x8 + x7 + x5 + x4 + x3 + x2 + 1 p 16 x8 + x6 + x3 + x + 1 x6 + x4 + x2 + x + 1 r 17 x8 + x7 + x5 + x2 + 1 x6 + x5 + x4 + x2 + 1 r 18 x9 + x6 + x5 + x2 + 1 x8 + x5 + x4 + x2 + 1 i 19 x9 + x7 + x4 + x3 + 1 x8 + x6 + x4 + x3 + 1 i 20 x9 + x8 + x5 + x2 + 1 x6 + x5 + x4 + x2 + 1 r 21 x10 + x4 + x3 + x2 + 1 x8 + x7 + x4 + x2 + 1 i 22 x10 + x7 + x2 + x + 1 x6 + x4 + x3 + x + 1 r 23 x11 + x7 + x6 + x2 + 1 x8 + x7 + x4 + x2 + 1 r 24 x13 + x10 + x2 + x + 1 x9 + x7 + x6 + x4 + x3 + x + 1 r 25 x13 + x10 + x9 + x2 + 1 x12 + x9 + x8 + x6 + x4 + x2 + 1 p Table 14.9.6 Table of pentanomials which divide trinomials: “p” in type indicates that the given polynomial f is primitive, “i” indicates that f is irreducible, and “r” indicates that f is reducible. 14.9.70 Remark All primitive polynomials satisfy the gcd condition of Theorems 14.9.68 and 14.9.69, and thus, in particular, Theorems 14.9.68 and 14.9.69 hold for all primitive trinomials and pentanomials over F2. Combinatorial 641 14.9.71 Corollary If f(x) = xm +xl +xk +xj +1 is primitive over F2 and not one of the exceptions given in Table 14.9.6 or their reciprocals, then, for m < n ≤2m, 1. Cf n is an orthogonal array of strength at least 3; or equivalently, 2. (Cf n)⊥, the dual code of Cf n, has minimum weight at least 4; 3. the cyclic redundancy check code derived from f, of length n, can detect all errors in three or fewer positions; 4. the bias from the third moment of the Hamming weight in the linear feedback shift register sequence generated from f is small. 14.9.72 Theorem Let F be any ﬁeld and f, g, h ∈F[x], fh = g, w(f) = n > 1 and w(g) = m. If there exists an f0 ∈F[x] such that f(x) = f0(xk) for k > 1 then there exist gi ∈F[x], w(gi) = mi for 0 ≤i < k such that g(x) = k−1 X i=0 gi(xk)xi (14.9.1) m = k−1 X i=0 mi and mi ̸= 1. (14.9.2) 14.9.73 Remark Theorem 14.9.72 can be used to simplify the analysis of multiples of f. An example used in is given in Corollary 14.9.74 and was used in the proofs of Theorems 14.9.75 and 14.9.76. 14.9.74 Corollary Let F be any ﬁeld and f, g, h ∈F[x], fh = g, w(f) = n and w(g) ≤3. If there exists f0 ∈F[x] such that f(x) = f0(xk) for k > 1 then there exists g0 ∈F[x] such that g(x) = g0(xk). 14.9.75 Theorem Let f(x) = a + bxk + xm (a, b ̸= 0) be a monic trinomial over F3. If g(x) = c + xn (c ̸= 0) is a monic binomial over F3 with degree at most 3m divisible by f with g = fh, then f and g are as given in Table 14.9.7. Case f(x) g(x) 1.1 1 + bxm/2 + xm −b + x3m/2 1.2 −1 + bxm/2 + xm 1 + x2m 1.3 1 + bxm/2 + xm −1 + x3m 1.4 a + xm/3 + xm −1 + x8m/3 1.5 b + bx2m/3 + xm −1 + x8m/3 Table 14.9.7 Polynomials over F3 such that g = fh for monic trinomial f and monic binomial g. 14.9.76 Theorem Let f(x) = a + bxk + xm (a, b ̸= 0) be a monic trinomial over F3. If g(x) = c + dxl + xn (c, d ̸= 0) is a monic trinomial over F3 with degree at most 3m divisible by f with g = fh, then 1. g = f 3; 2. f and g are as in the Table 14.9.8; 3. f and g are reciprocals of polynomials listed in Table 14.9.8. 642 Handbook of Finite Fields Case f(x) g(x) 1.1 −1 + bxm/2 + xm 1 −bxm/2 + x3m 1.2 1 + bxm/2 + xm b + xm/2 + x5m/2 1.3 −1 + bxm/2 + xm b −bxm + x5m/2 1.4 −1 + bxm/2 + xm −b −x3m/2 + x5m/2 1.5 1 + bxm/2 + xm b + bx4m/2 + x5m/2 1.6 1 + bxm/2 + xm 1 + xm + x2m 1.7 −1 + bxm/2 + xm b + xm + x3m/2 1.8 −1 + bxm/2 + xm −b −bxm + x3m/2 1.9 a −xm/3 + xm −a −xm/3 + x3m 1.10 a −xm/3 + xm 1 + x2m/3 + x8m/3 1.11 a + xm/3 + xm a + ax2m/3 + x7m/3 1.12 a −xm/3 + xm a −ax4m/3 + x7m/3 1.13 a −xm/3 + xm −a + x5m/3 + x7m/3 1.14 a + xm/3 + xm 1 + ax5m/3 + x2m 1.15 a −xm/3 + xm a + ax4m/3 + x5m/3 1.16 −1 + bxm/4 + xm −b + bx6m/4 + x11m/4 1.17 1 + bxm/4 + xm 1 + bx9m/4 + x10m/4 Table 14.9.8 Table of polynomials such that g = fh with f and g monic trinomials over F3. See Also §2.2 For tables of primitives of various kinds and weights. §3.4 For results on the weights of irreducible polynomials. §4.3 For results on the weights of primitive polynomials. §10.2 For discussion on bias and randomness of linear feedback shift register sequences. §14.1 For results of latin squares which are strongly related to orthogonal arrays. §14.5 For a discussion of block designs, which include orthogonal arrays. §15.1 Uses primitive polynomials to generate cyclic redundancy check and BCH codes. §15.4 For results on turbo codes. §17.3 For more discussion of applications of ﬁnite ﬁelds and polynomials. References Cited: [146, 218, 357, 558, 559, 801, 826, 832, 833, 1362, 1457, 1491, 1590, 1591, 1622, 1727, 1792, 1793, 1834, 1944, 1994, 2074, 2083, 2169, 2204, 2354, 2439, 2513] 14.10 Ramanujan and expander graphs M. Ram Murty, Queen’s University Sebastian M. Cioab˘ a, University of Delaware In the last two decades, the theory of Ramanujan graphs has gained prominence primarily for two reasons. First, from a practical viewpoint, they resolve an extremal problem in commu-nication network theory (see for example [269, 1535]). Second, from a more aesthetic view-Combinatorial 643 point, they fuse diverse branches of pure mathematics, namely, number theory, representa-tion theory, and algebraic geometry. The purpose of this survey is to unify some of the recent developments and expose certain open problems in the area. This survey is an expanded version of and is by no means an exhaustive one and demonstrates a highly number-theoretic bias. For other surveys, we refer the reader to [1535, 1922, 1967, 1968, 2525, 2837]. For a more up-to-date survey highlighting the connection between graph theory and auto-morphic representations, we refer the reader to Li’s recent survey article . 14.10.1 Graphs, adjacency matrices, and eigenvalues 14.10.1 Deﬁnition A graph X is a pair (V, E) consisting of a vertex set V = V (X) and an edge set E = E(X) which is a multiset of unordered pairs of (not necessarily distinct) vertices. Each edge consists of two vertices that are called its endpoints. A loop is an edge whose endpoints are equal. Multiple edges are edges having the same pair of endpoints. A simple graph is a graph having no loops nor multiple edges. If a graph has loops or multiple edges, we will call it a multigraph. When two vertices u and v are endpoints of an edge, they are adjacent and write u ∼v to indicate this fact. A directed graph Y is a pair (W, F) consisting of a set of vertices W and a multiset F of ordered pairs of vertices which are called arcs. 14.10.2 Remark All the graphs in this chapter are undirected unless stated explicitly otherwise. 14.10.3 Deﬁnition The degree of a vertex v of a graph X, denoted by deg(v), is the number of edges incident with v, where we count a loop with multiplicity 1. A graph X is k-regular if every vertex has degree k. 14.10.4 Proposition (Handshaking Lemma) For any simple graph X, P v∈V (X) deg(v) = 2\|E(X)\|. If X is a k-regular graph with n vertices, then \|E(X)\| = kn/2. 14.10.5 Deﬁnition An adjacency matrix A = A(X) of a graph X with n vertices is an n×n matrix with rows and columns indexed by the vertices of X, where the (x, y)-th entry equals the number of edges between vertex x and vertex y. As X is an undirected graph with n vertices, the matrix A(X) is symmetric and therefore, its eigenvalues λ1 ≥λ2 ≥· · · ≥λn are real. The multiset of eigenvalues of X is the spectrum of X. 14.10.6 Remark We remark that the adjacency matrix deﬁned above depends on the labeling of the vertices of X. Diﬀerent labelings of the vertices of a graph X may possibly yield diﬀerent adjacency matrices. However, all these adjacency matrices are similar to each other (by permutation matrices) and thus, their spectrum is the same. 14.10.7 Remark One can deﬁne an adjacency matrix of a directed graph Y = (W, A) similarly. Given a labeling of the vertices W of Y , the (x, y)-th entry of the adjacency matrix corresponding to this labeling equals the number of arcs from x to y. Adjacency matrices of directed graphs may be non-symmetric. 14.10.8 Example The spectrum of the complete graph Kn on n vertices is (n −1)(1), (−1)(n−1), where the exponent signiﬁes the multiplicity of the respective eigenvalue. The Petersen graph has spectrum 3(1), 1(5), −2(4). 14.10.9 Theorem Let X be a graph on n vertices with maximum degree ∆and average degree d. Then d ≤λ1 ≤∆and \|λi\| ≤∆for every 2 ≤i ≤n. 644 Handbook of Finite Fields 14.10.10 Deﬁnition For a multigraph X, a walk of length r from x to y is a sequence x = v0, v1, . . . , vr = y of vertices of X such that vivi+1 is an edge of X for any i = 0, 1, . . . , r −1. The length of this walk is r. A closed walk is a walk where the start-ing vertex x is the same as the last vertex y. 14.10.11 Deﬁnition A path is a walk with no repeated vertices. A cycle is a closed walk with no repeated vertices except the starting vertex. 14.10.12 Remark A word of caution must be inserted here. In graph theory literature, the distinction between a walk and a path is as we have deﬁned it above. However, in number theory circles, the ﬁner distinction is not made and one uses the word “path” to mean a “walk”; see for example, [2523, 2789]. 14.10.13 Deﬁnition A graph X is connected if for every two distinct vertices x and y, there is a path from x to y. 14.10.14 Proposition For every graph X with adjacency matrix A and any integer r ≥1, the (x, y)-th entry of Ar equals the number of walks of length r between x and y. 14.10.15 Proposition The number of closed walks of length r in a graph X with n vertices equals λr 1 + λr 2 + · · · + λr n. 14.10.16 Deﬁnition An independent set in a graph X is a subset of vertices that are pairwise non-adjacent. A graph X is bipartite if its vertex set can be partitioned into two independent sets A and B; X is complete bipartite and denoted by K\|A\|,\|B\| if it contains all the edges between A and B. 14.10.17 Proposition A graph is bipartite if and only if it does not contain any cycles of odd length. 14.10.18 Theorem If X is a k-regular and connected graph with n vertices, then λ1 = k and the multiplicity of k is 1 with the eigenspace of k spanned by the all 1 vector of dimension n. If X is a k-regular and connected graph, then X is bipartite if and only if λn = −k. 14.10.19 Deﬁnition If X is a k-regular and connected graph, then the eigenvalue k of X is trivial. All other eigenvalues of X are non-trivial. Let λ(X) = max \|λi\|, where the maximum is taken over all non-trivial eigenvalues of X. The parameter λ(X) is the second eigenvalue of X by some authors. The second largest eigenvalue of X is λ2(X) and λ(X) ≥λ2(X). 14.10.20 Deﬁnition The distance d(x, y) between two distinct vertices x and y of a connected graph X is the length of a shortest path between x and y. The diameter D of a connected graph X is the maximum of d(x, y), where the maximum is taken over all pairs of distinct vertices x ̸= y of X. 14.10.21 Remark When k ≥3, if X is a k-regular and connected graph with n vertices and diameter D, then n ≤1 + k + k(k −1) + · · · + k(k −1)D−1 = 1 + k · (k−1)D−1 k−2 and consequently, D ≥logk−1 (n−1)(k−2) k + 1 > log(n−1) log(k−1) −log(k/(k−2)) log(k−1) . Thus, the diameter of any connected k-regular graph is at least logarithmic in the order of the graph. The next theorem implies that when the non-trivial eigenvalues of a k-regular connected graph are small, then the above inequality is tight up to a multiplicative constant. Combinatorial 645 14.10.22 Theorem If X is a connected non-bipartite k-regular graph with n vertices and di-ameter D, then: D ≤ log(n −1) log(k/λ(X)) + 1. 14.10.23 Remark Kahale obtained an upper bound on the minimum distance between i subsets of the same size of a regular graph in terms of the i-th largest eigenvalue in absolute value. Kahale also constructed examples of k-regular graphs on n vertices having λ(X) = (1+o(1))2 √ k −1 and D = 2(1+o(1)) logk−1 n showing the previous result is asymptotically best possible. Here the o(1) term tends to 0 as n goes to inﬁnity. 14.10.24 Remark A similar result can be derived for k-regular bipartite graphs; if X is a bipartite k-regular and connected graph of diameter D, we have (see Quenell ) D ≤log(n −2)/2 log(k/λ′(X)) + 2, where λ′(X) is the maximum absolute value of the eigenvalues of X that are not k nor −k. 14.10.25 Remark Chung, Faber, and Manteuﬀel and independently, Van Dam and Haemers obtained slight improvements of the previous diameter bounds. 14.10.26 Deﬁnition The chromatic number χ(X) of a graph X is the minimum number of colors that can be assigned to the vertices of a graph such that any two adjacent vertices have diﬀerent colors. The largest order of an independent set of vertices of X is the independence number of X and is denoted by α(X). 14.10.27 Remark The chromatic number of X is the minimum number of independent sets that partition the vertex set of X and consequently, χ(X) ≥\|V (X)\| α(X) . 14.10.28 Theorem If X is a k-regular non-empty graph, then α(X) ≤n(−λn) k −λn and so χ(X) ≥1 + k −λn . 14.10.29 Remark An immediate consequence of the previous result is that α(X) ≤ nλ(X) k+λ(X) and χ(X) ≥1 + k λ(X) for any non-bipartite connected k-regular graph X. These facts show that a good upper bound for the absolute values of the non-trivial eigenvalues of a regular graph will yield non-trivial bounds for the independence and chromatic number. 14.10.30 Remark The following theorem shows that the eigenvalues of a regular graph are closely related to its edge distribution. 14.10.31 Theorem If X is a k-regular connected graph with eigenvalues k = λ1 > λ2 ≥· · · ≥ λn ≥−k, let λ := max(\|λ2\|, \|λn\|). For S, T ⊂V (X), denote by e(S, T) the number of edges with one endpoint in S and another in T. Then for all S, T ⊂V (X) e(S, T) −k\|S\|\|T\| n ≤λ s \|S\|\|T\| 1 −\|S\| n 1 −\|T\| n < λ p \|S\|\|T\|. 14.10.32 Remark The previous theorem states that k-regular graphs X with small non-trivial eigen-values (compared to k) have their edges uniformly distributed (similar to random k-regular 646 Handbook of Finite Fields graphs). Such graphs are called pseudorandom graphs and are important in many situations (see Krivelevich-Sudakov’s survey ). Bilu and Linial have obtained a converse of the previous result of Alon and Chung. 14.10.2 Ramanujan graphs 14.10.33 Deﬁnition A k-regular connected multigraph X is a Ramanujan multigraph if \|λi\| ≤ 2 √ k −1 for every eigenvalue λi ̸= k. A Ramanujan graph is a Ramanujan multigraph having no loops nor multiple edges. 14.10.34 Remark We mention that the deﬁnition of a Ramanujan graph used by other authors is slightly weaker. For example, Sarnak in calls a k-regular graph Ramanujan if λ2(X) ≤2 √ k −1. 14.10.35 Example The complete graph Kn is an (n−1)-regular Ramanujan graph as its eigenvalues are (n −1)(1), (−1)(n−1), where the exponents denote the multiplicities of the eigenvalues. The complete bipartite graph Kn,n has eigenvalues n(1), 0(2n−2), −n(1) and is an n-regular Ramanujan graph. 14.10.36 Remark In [80, p.95], Alon announced a proof with Boppana of the fact that for any k-regular graph X of order n, λ2(X) ≥2 √ k −1 −O((logk n)−1), where the constant in the O term depends only on k. Many researchers refer to this result as the Alon-Boppana Theorem. Other researchers refer to the following statement proved by Nilli (pseudonym for Alon) in as the Alon-Boppana Theorem. 14.10.37 Theorem If X is a k-regular and connected graph with diameter D ≥2b + 2, then λ2(X) ≥2 √ k −1 −2 √ k −1 −1 b + 1 . 14.10.38 Remark The Alon-Boppana Theorem and Remark 14.10.21 imply that if (Xi)i≥1 is a sequence of k-regular and connected graphs with limi→+∞\|V (Xi)\| = +∞, then lim inf i→∞λ2(Xi) ≥2 √ k −1. 14.10.39 Remark The best lower bound for the second largest eigenvalue λ2(X) of a k-regular graph of diameter D is due to Friedman who showed that λ2(X) ≥2 √ k −1 cos θk,t ≥2 √ k −1 cos π t + 1 (14.10.1) where t is the largest integer such that D ≥2t and θk,t ∈ h π t+5, π t+1 i is the smallest positive solution of the equation k 2k−2 = sin(t+1)θ cos θ sin tθ . The number 2 √ k −1 cos θk,t is the largest eigenvalue of the k-regular tree Tk,t of depth t; this tree has a root vertex x and exactly k(k −1)i−1 vertices at distance i from x for each 1 ≤i ≤t. Friedman used analytic tools involving Dirichlet and Neumann eigenvalues for graphs with boundaries to prove (14.10.1). Later, Nilli gave an elementary proof of a slightly weaker bound. 14.10.40 Remark We outline here an elementary proof of the inequality λ2(X) ≥2 √ k −1 cos π t+1 for every connected k-regular graph X of diameter D ≥2t+2. The ﬁrst ingredient of the proof is that the largest eigenvalue of any subgraph induced by a ball of radius t of X is larger than the largest eigenvalue of Tk,t. The second is that if u and v are vertices at distance at least 2t+2 in X, then the subgraph induced by the vertices at distance at most t from u or v Combinatorial 647 has exactly two components X(u) and X(v). By Cauchy eigenvalue interlacing, the second largest eigenvalue of X is greater than the minimum of the largest eigenvalue of X(u) and X(v) which by the previous argument is at least 2 √ k −1 cos θk,t ≥2 √ k −1 cos π t+1. 14.10.41 Remark At this point, it is worth stating that Friedman (see also Nilli ) proved the stronger statement that if X is a k-regular graphs containing a subset of r points each of distance at least 2t from one another, then λr(X) ≥2 √ k −1 cos θk,t ≥2 √ k −1 cos π t+1. This implies that the r-th largest eigenvalue λr(X) of any connected k-regular graph X is at least 2 √ k −1 1 −π2 2f 2 + O(f −4) , where f = logk−1(n/r) 2 . 14.10.42 Remark One might wonder if the behavior of the negative eigenvalues of a connected k-regular graph X is similar to the behavior of the negatives of the positive eigenvalues of X. If X is bipartite, then the spectrum of X is symmetric with respect to 0 and this settles the previous question. In general, it turns out that additional conditions are needed in order to obtain similar results for the negative eigenvalues. This is because there are regular graphs with increasing order whose eigenvalues are bounded from below by an absolute constant. For example, the eigenvalues of a line graph are at least −2. It turns out that the number of odd cycles plays a role in the behavior of the negative eigenvalues of regular graphs. The odd girth of a graph X is the smallest length of a cycle of odd length. 14.10.43 Theorem [1126, 2290] If X is a connected k-regular graph of order n with a subset of r points each of distance at least 2t from one another, and odd girth at least 2t, then λn−r(X) ≤−2 √ k −1 cos θk,t = −2 √ k −1 cos π t + 1. (14.10.2) 14.10.44 Corollary If (Xi)i≥0 is a sequence of k-regular graphs of increasing orders such that the odd girth of Xi tends to inﬁnity as i →∞, then lim supi→∞µl(Xi) ≤−2 √ k −1, for each l ≥1, where µl(X) denotes the l-th smallest eigenvalue of X. 14.10.45 Theorem [646, 648] For an integer r ≥3, let cr(X) denote the number of cycles of length r of a graph X. If (Xi)i≥0 is a sequence of k-regular graphs of increasing orders such that limi→∞ c2r+1(Xi) \|V (Xi)\| = 0 for each r ≥1, then lim supi→∞µl(Xi) ≤−2 √ k −1, for each l ≥1. 14.10.46 Remark The diﬃculty of constructing inﬁnite families of Ramanujan graphs is also illus-trated by the following result of Serre. 14.10.47 Theorem For any ϵ > 0, there exists a positive constant c = c(ϵ, k) such that for every k-regular graph X on n vertices, the number of eigenvalues λi of X such that λi > (2 −ϵ) √ k −1 is at least c · n. 14.10.48 Remark Diﬀerent short and elementary proofs of Serre’s theorem were found indepen-dently by Nilli and Cioab˘ a [646, 648]. Nilli’s proof is similar to Friedman’s argument from while Cioab˘ a’s proof uses the fact that the trace of Al is the number of closed walks of length l. See also [646, 648] for a similar theorem to Theorem 14.10.47 for the small-est eigenvalues of regular graphs. These proofs, as well as extensions of the Alon-Bopanna theorem (see recent work of Mohar ) rely on the notion of the universal cover of a graph; see Deﬁnitions 14.10.108 and 14.10.109, and Theorem 14.10.110 for more details. 14.10.49 Remark The idea behind all the proofs of Serre’s theorem indicated above is that the universal cover of a ﬁnite k-regular graph is the rooted inﬁnite k-regular tree Tk. This implies that the number of closed walks of even length starting at some vertex of a ﬁnite k-regular graph is at least the number of closed walks of the same length starting at the root of the inﬁnite k-regular tree. The number 2 √ k −1 is the spectral radius of the adjacency operator of the inﬁnite k-regular tree (for more details see [1126, 1535]). In some circumstances, the 648 Handbook of Finite Fields lower bound for the second eigenvalue of a k-regular graph can be improved beyond 2 √ k −1 [1927, 2118]. 14.10.50 Remark Greenberg and Lubotzky (see Chapter 4 of or [646, 647] for a short ele-mentary proof) extended the Alon-Bopanna bound to any family of general graphs with isomorphic universal cover. If (Xi)i≥1 is a family of ﬁnite connected graphs with universal cover ˜ X and ρ is the spectral radius of the adjacency operator of ˜ X, then lim inf λ2(Xi) ≥ρ as i →+∞. For extensions of Alon-Boppana theorem and Serre’s theorem for irregular graphs, see [646, 647, 1534, 2118]. 14.10.51 Remark The notion of Ramanujan graph has been extended to hypergraphs and studied in this setting. Again, these notions lead to the use of the Ramanujan conjecture formulated for higher GLn in the Langlands program. 14.10.52 Deﬁnition A hypergraph X = (V, E) is a pair consisting of a vertex set V and a set of hyperedges E consisting of subsets of V . If all the edges are of the same size r, X is an r-uniform hypergraph or r-graph. In the familiar setting of a graph, an edge is viewed as a 2-element subset of V and is thus a 2-uniform hypergraph. One class of hypergraphs that are studied are the (k, r)-regular hypergraphs in which each edge contains r elements and each vertex is contained in k edges. For an ordinary graph, r = 2 and this generalizes the notion of a k-regular graph. In this special setting, the adjacency matrix A is a \|V \|×\|V \| matrix with zero diagonal entries and the (i, j)-th entry is the number of hyperedges that contain {i, j}. 14.10.53 Remark One can show easily that k(r −1) is an eigenvalue of A and this is the trivial eigenvalue. With this deﬁnition in place, a Ramanujan hypergraph is deﬁned as a ﬁnite connected (k, r)-regular hypergraph such that every eigenvalue λ of A with \|λ\| ̸= k(r −1) satisﬁes \|λ −(r −2)\| ≤2 p (k −1)(r −1). We refer the reader to the important work of Li for further details. 14.10.3 Expander graphs 14.10.54 Deﬁnition For any subset A of vertices of a graph X, the edge boundary of A, denoted ∂A, is ∂A = {xy ∈E(X) : x ∈A, y / ∈A}. That is, the edge boundary of A consists of edges with one endpoint in A and another outside A. 14.10.55 Deﬁnition The edge-expansion constant of X, denoted by h(X), is deﬁned as h(X) = min \|∂A\| \|A\| : A ⊂X, \|A\| ≤\|V (X)\| 2 . 14.10.56 Deﬁnition A family of k-regular graphs (Xi)i≥1 with \|V (Xi)\| increasing with i, is a family of expanders if there exists a positive absolute constant c such that h(Xi) > c for every i ≥1. Combinatorial 649 14.10.57 Remark Informally, a family of k-regular expanders is a family of sparse (k ﬁxed and \|V (Xi)\| →+∞as i →+∞imply that the number of edges of Xi is linear in its number of vertices), but highly connected graphs (h(Xi) > c means that in order to disconnect Xi, one must remove many edges). 14.10.58 Example Let Xm denote the following 8-regular graph on m2 vertices. The vertex set of Xm is Z/mZ × Z/mZ. The neighbors of a vertex (x, y) are (x ± y, y), (x, y ± x), (x ± y + 1, y), (x, y ± x + 1). The family (Xm)m≥4 is the ﬁrst explicit family of expanders. Mar-gulis proved that (Xm)m≥4 are expanders using representation theory. Margulis used the fact that the group SL3(Z) has Kazhdan property T. Groups having this property or the weaker property τ can be used to construct inﬁnite families of constant-degree Cay-ley graphs expanders. We refer the reader to [1535, 1967, 1968] for nice descriptions and explanations of these properties and their relation to expanders. 14.10.59 Remark Expander graphs play an important role in computer science, mathematics, and the theory of communication networks; see [269, 1535]. These graphs arise in questions about designing networks that connect many users while using only a small number of switches. 14.10.60 Theorem [80, 2117] If X is a connected k-regular graph, then q k2 −λ2 2 ≥h(X) ≥k −λ2 2 . 14.10.61 Remark The previous theorem shows that constructing an inﬁnite family of k-regular expanders (Xi)i≥1 is equivalent to constructing an inﬁnite family of k-regular graphs (Xi)i≥1 such that k −λ2(Xi) is bounded away from zero. 14.10.4 Cayley graphs 14.10.62 Deﬁnition Let G be a group written in multiplicative notation and let S be a subset of elements of G that is closed under taking inverses and does not contain the identity. The Cayley graph of G with respect to S (denoted by X(G, S)) is the graph whose vertex set is G where x ∼y if and only if x−1y ∈S. If G is abelian, then it is common to use the additive notation in the deﬁnition of X(G, S): x ∼y if and only if y −x ∈S. 14.10.63 Remark In general, if S is an arbitrary multiset of G, denote by X(G, S) the directed graph with vertex set G and arc set {(x, y) : x−1y ∈S}. If S is inverse-closed and does not contain the identity, then this graph is undirected and has no loops. 14.10.64 Theorem Let G be a ﬁnite abelian group and S a symmetric subset of G of size k. The eigenvalues of the adjacency matrix of X(G, S) are given by λχ = X s∈S χ(s) where χ ranges over all irreducible characters of G. 14.10.65 Remark For each irreducible character of G, let vχ denote the column vector (χ(g) : g ∈G). The proof of Theorem 14.10.64 follows by showing that vχ is an eigenvector of the adjacency matrix of X(G, S) corresponding to eigenvalue λχ = P s∈S χ(s). 14.10.66 Remark Notice that for the trivial character χ = 1, we have λ1 = k. If we have for all χ ̸= 1 X s∈S χ(s) < k 650 Handbook of Finite Fields then the graph is connected by our earlier remarks. Thus, to construct Ramanujan graphs, we require X s∈S χ(s) ≤2 √ k −1 for every non-trivial irreducible character χ of G. This is the strategy employed in many of the explicit constructions of Ramanujan graphs. 14.10.67 Example A simple example can be given using Gauss sums. If p ≡1 (mod 4) is a prime, let G = Z/pZ and S = {x2 : x ∈Z/pZ} be the multiset of squares. The multigraph X(G, S) is easily seen to be Ramanujan in view of the fact (see for example [2207, p. 81]) X x∈Z/pZ e2πiax2/p = √p for any a ̸= 0. By our convention in the computation of degree of a vertex, we see that X(G, S) is a p-regular graph.; see for other related examples. 14.10.68 Example When q ≡1 (mod 4) is a prime power, the Paley graph of order q is the Cayley graph X(G, S) of the additive group of a ﬁnite ﬁeld G = Fq with respect to the set S of non-zero squares. This simple and undirected graph has q vertices, is connected and regular of degree q−1 2 and its non-trivial eigenvalues are −1−√q 2 and −1+√q 2 , each of multiplicity q−1 2 . The Paley graph is Ramanujan when q ≥9. 14.10.69 Remark The proof of Theorem 14.10.64 is reminiscent of the Dedekind determinant formula in number theory. This formula computes det A, where A is the matrix whose (i, j)-th entry is f(ij−1) for any function f deﬁned on the ﬁnite abelian group G of order n. The determinant is Y χ  X g∈G f(g)χ(g)  . 14.10.70 Deﬁnition Let G be an abelian group written in the additive notation and S ⊂G. The sum graph of G with respect to S (denoted by Y (G, S)) has G as vertex set and x ∼y if and only if x + y ∈S. 14.10.71 Theorem [1922, p. 197] Let G be an abelian group. The eigenvalues of Y (G, S) are given as follows. For each irreducible character χ of G, deﬁne eχ = X s∈S χ(s). If eχ = 0, then vχ and vχ−1 are both eigenvectors with eigenvalues 0. If eχ ̸= 0, then \|eχ\|vχ ± eχvχ−1 are two eigenvectors with eigenvalues ±\|eχ\|. 14.10.72 Example Using Theorem 14.10.71, Li constructed Ramanujan graphs in the following way. Let Fq denote the ﬁnite ﬁeld of q elements. Let G = Fq2 and take for S the elements of G of norm 1. This is a symmetric subset of G and the Cayley graph X(G, S) turns out to be Ramanujan. The latter is a consequence of a theorem of Weil estimating Kloosterman sums . Combinatorial 651 14.10.73 Theorem Let G = Fq be a ﬁnite ﬁeld of q = pm elements and f(x) a polynomial with coeﬃcients in Fq and of degree 2 or 3. Let S be the multiset {f(x) : x ∈Fq}. Suppose S is symmetric. Then X(G, S) is a Ramanujan graph if the degree of f is 2 and is almost Ramanujan if the degree of f is 3. 14.10.74 Remark The required character sum estimates in Theorem 14.10.73 come from Weil’s proof of the Riemann hypothesis for the zeta functions of curves over ﬁnite ﬁelds. In particular, we have for all a ∈Fq, a ̸= 0, X x∈Fq exp(2πitrFq/Fp(af(x))/p) ≤(deg f −1)√q provided f is not identically zero; see [1922, p. 94]. In particular, if f has degree 3, we get the estimate of 2√q for the exponential sum. For example, if u ∈Z/pZ and we take S = {x3 + ux : x ∈Z/pZ}, then S is symmetric and, according to our convention, X(G, S) is a p-regular graph. In addition, it is an almost Ramanujan graph since X x∈Z/pZ exp(2πia(x3 + ux)/p) ≤2√p by virtue of the Riemann hypothesis for curves (proved by Weil). 14.10.75 Remark We observe that even though there are many constructions of Ramanujan graphs that are abelian Cayley graphs, it is actually impossible to construct an inﬁnite family of constant-degree abelian Cayley graphs that are Ramanujan. There are several proofs of this fact in the literature, see [83, 645, 1128]. Friedman, Murty, and Tillich proved that if X is a k-regular abelian Cayley graph of order n, then λ2(X) ≥k −cn−4/k, where c is some absolute positive constant. Cioab˘ a proved that for ﬁxed k ≥3 and ϵ > 0, there is a positive constant C = C(ϵ, k) such that any k-regular abelian Cayley graph on n vertices has at least Cn eigenvalues that are larger than k −ϵ. 14.10.76 Remark Lubotzky and Weiss proved the stronger result that it is impossible to construct inﬁnite families of constant-degree expanders that are Cayley graphs of solvable groups of bounded derived length. 14.10.77 Remark The eigenvalues of Cayley graphs can be calculated even in the case of non-abelian groups. This is essentially contained in a paper by Diaconis and Shahshahani . Using their results, one can easily generalize the Dedekind determinant formula as follows (and which does not seem to be widely known). Let G be a ﬁnite group and f a class function on G. Then the determinant of the matrix A whose rows (and columns) are indexed by the elements of G and whose (i, j)-th entry f(i−1j) is given by Y χ   1 χ(1) X g∈G f(g)χ(g)   χ(1) where the product is taken over the distinct irreducible characters of G. 652 Handbook of Finite Fields 14.10.78 Theorem Let G be a ﬁnite group and S a symmetric subset which is stable under conjugation. The eigenvalues of the Cayley graph X(G, S) are given by λχ = 1 χ(1) X s∈S χ(s) as χ ranges over all irreducible characters of G. Moreover, the multiplicity of λχ is χ(1)2. 14.10.79 Remark We remark that the λχ in the above theorem need not be all distinct. For example, if there is a non-trivial character χ which is trivial on S, then the multiplicity of the eigenvalue \|S\| is at least 1 + χ(1)2. We refer the reader to Babai for a more detailed proof of the above result in a slightly more general context. 14.10.80 Remark The intriguing question of what groups can be used to construct inﬁnite families of constant-degree Cayley graphs expanders was formulated as a conjecture by Babai, Kantor, and Lubotzky in 1989. 14.10.81 Conjecture Let (Gi)i≥0 be a family of non-abelian simple groups. There exist gener-ating sets Si of constant size such that (X(Gi, Si))i≥0 form a family of expanders. 14.10.82 Remark As a supporting fact of this conjecture, we mention the result of Babai, Kantor, and Lubotzky who proved constructively that any simple non-abelian group G contains a set S of at most 7 generators such that the diameter of the Cayley graph X(G, S) is at most c log \|G\|, where c > 0 is some absolute constant. 14.10.83 Remark The previous conjecture of Babai, Kantor, and Lubotzky is true and its recent resolution has been possible due to the eﬀort of several researchers. We refer the reader to the works of Kassabov, Lubotzky, and Nikolov , Breuillard, Green, and Tao , and the recent survey by Lubotzky for a thorough account of the solution of this conjecture. 14.10.5 Explicit constructions of Ramanujan graphs 14.10.84 Deﬁnition Let X be a graph. A non-backtracking walk of length r in X is a sequence x0, x1, . . . , xr of vertices of X such that xi is adjacent to xi+1 for each 0 ≤i ≤r −1 and xi−1 ̸= xi+1 for each 1 ≤i ≤r −1. For r ∈N, deﬁne the matrix Ar as follows: Ar(x, y) equals the number of non-backtracking walks of length r that start at x and end at y. 14.10.85 Proposition If X is a k-regular graph with n vertices and adjacency matrix A, then 1. A0 = In, A1 = A; 2. A2 = A2 1 −kIn; 3. Ar+1 = A1Ar −(k −1)Ar−1 for every r ≥2. 14.10.86 Proposition [780, 1969] Let Um denote the Chebychev polynomial of the second kind deﬁned by expressing sin(m+1)θ sin θ as a polynomial of degree m in cos θ: Um(cos θ) = sin(m + 1)θ sin θ . Then ⌊m 2 ⌋ X r=0 Am−2r = (k −1) m 2 Um A 2 √ k −1 . Combinatorial 653 14.10.87 Deﬁnition A graph X is vertex-transitive if the automorphism group of X acts transitively on its vertex set which means that for any x, y ∈V (X) there exists an automorphism σ of X such that σ(x) = y. 14.10.88 Proposition [780, 1969] If X is a k-regular graph with n vertices and eigenvalues k = λ1 ≥ λ2 ≥· · · ≥λn, then X x∈V ⌊l 2 ⌋ X r=0 (Al−2r)(x, x) = (k −1) l 2 n X j=1 sin(l + 1)θj sin θj . where cos θj = λj 2√k−1 for each 1 ≤j ≤n. If X is vertex-transitive of degree k, then (Aj)(x, x) = (Aj)(y, y) for any j and x, y ∈V (X) and thus, n ⌊l 2 ⌋ X r=0 (Al−2r)(x, x) = (k −1) l 2 n X j=1 sin(l + 1)θj sin θj . for every vertex x ∈V (X). 14.10.89 Remark Note that θ1 = i log √ k −1 as λ1 = k and θn = π + i log √ k −1 if λn = −k. Also, it is important to observe that θj is real if \|λj\| = \|2 √ k −1 cos θj\| ≤2 √ k −1; otherwise, θj = iIm(θj) is purely imaginary if λj > 2 √ k −1 and θj = π + iIm(θj) if λj < 2 √ k −1. 14.10.90 Remark The general idea of using quaternions (see Lubotzky, Phillips, and Sarnak or Margulis ) to construct inﬁnite families of k-regular Ramanujan graphs can be summarized in the following two steps: 1. The ﬁrst step consists of constructing the inﬁnite k-regular tree Tk as the free group of some group G of quaternions integers with some suitable set of k gener-ators Sk. Thus, Tk will be identiﬁed with the Cayley graph X(G, Sk). 2. Finite k-regular graphs are constructed from the inﬁnite k-regular tree Tk by taking suitable ﬁnite quotients of it. More precisely, by choosing appropriate normal subgroups H of G of ﬁnite index, one can construct ﬁnite k-regular graphs which are the Cayley graphs of the quotient group G/H with the set of generators being formed by the cosets of the form αH where α ∈Sk. 14.10.91 Construction [1969, 2006] Let p and q be unequal primes p, q ≡1 (mod 4). Let u be an integer so that u2 ≡−1 (mod q). By a classical formula of Jacobi, we know that there are 8(p + 1) solutions v = (a, b, c, d) such that a2 + b2 + c2 + d2 = p. Among these, there are exactly p + 1 with a > 0 and b, c, d even, as is easily shown. To each such v we associate the matrix ˜ v = a + ub c + ud −c + ud a −ub which gives p + 1 matrices in PGL2(Z/qZ). We let S be the set of these matrices ˜ v and deﬁne Xp,q =    X(PGL2(Z/qZ), S), if p q = −1, X(PSL2(Z/qZ), S), if p q = 1, (14.10.3) where p q is the Legendre symbol that equals 1 if p is a square modulo q and −1 if p is not a square modulo q. In , it is shown that the Cayley graphs Xp,q are Ramanujan graphs. As we vary q, we get an inﬁnite family of such graphs, all (p + 1)-regular. 654 Handbook of Finite Fields 14.10.92 Remark The integer quaternion algebra is H(Z) = {a0 + a1i + a2j + a3k : a0, a1, a2, a3 ∈Z}, where i2 = j2 = k2 = −1 and ij = −ji = k, jk = −kj = i, ki = −ik = j. If α = a0 + a1i + a2j + a3k, then α = a0 −a1i −a2j −a3k and N(α) = αα = a2 0 + a2 1 + a2 2 + a2 3. The units of H(Z) are ±1, ±i, ±j, ±k. Let p be a prime with p ≡1 (mod 4). By Jacobi’s Theorem, there are 8(p + 1) integer quaternions α = a0 + a1i + a2j + a3k in H(Z) such that N(α) = p. As p ≡1 (mod 4), only one of the integers ai will be odd. Let S be the set of those p + 1 elements of H(Z) such that N(α) = p, a0 > 0 is odd and a1, a2, a3 even (this last fact is denoted by α ≡1 (mod 2) from now on). As N(α) = N(α), the set S consists of s = p+1 2 conjugate pairs S = {α1, α1, . . . , αs, αs}. A reduced word of length m with letters in S is deﬁned to be a word of length m in the elements of S which does not contain subwords of the form αjαj nor αjαj. 14.10.93 Construction (Diﬀerent construction of the graphs Xp,q) Deﬁne Λ′(2) = {α : α ∈Z, α ≡1 (mod 2), N(α) = pl, l ∈Z}. As N(αβ) = N(α)N(β) and the properties of quaternion multiplication, it follows that Λ′(2) is closed under multiplication. Deﬁne α ∼β for α, β ∈ Λ′(2) whenever ±pv1α = pv2β for some v1, v2 ∈Z. This is an equivalence relation and [α] will denote the equivalence class of α ∈Λ′(2). The set of equivalence classes Λ(2) = {[α] : α ∈Λ′(2)} forms a group with the multiplication [α][β] = [αβ] and [α][α] = . One of the key observations at this point is that, by previous results, the group Λ(2) is free on [α1], [α2], . . . , [αs]. This means that the Cayley graph of Λ(2) with respect to the set [S] = {[α1], [α1], . . . , [αs], [αs]} will be the inﬁnite (p + 1)-regular tree. For m coprime with p, let Λ(2m) = {[α] : α = a0 + a1i + a2j + a3k ∈Λ′(2), 2m\|aj, 1 ≤j ≤3}. It can be shown that Λ(2m) is a normal subgroup of Λ(2) of ﬁnite index. Let q be a prime. The graphs Xp,q and the Cayley graph of Λ(2)/Λ(2q) with respect to the set of generators α1Λ(2q), α1Λ(2q), . . . , αsΛ(2q), αsΛ(2q) are isomorphic as shown by the next result. 14.10.94 Proposition Let φ : Λ(2) →PGL (2, Z/qZ) deﬁned as follows: φ([a0 + a1i + a2j + a3k]) = a0 + ua1 a2 + ua3 −a2 + ua3 a0 −ua1 where u2 ≡−1 (mod q). Then φ is a group homomorphism whose kernel is Λ(2q) and whose image is PGL (2, Z/qZ) if p q = −1 and PSL (2, Z/qZ) if p q = 1. 14.10.95 Deﬁnition Let Q = Q(x0, x1, x2, x3) denote the quadratic form Q(x0, x1, x2, x3) = x2 0 + (2q)2x2 1 + (2q)2x2 2 + (2q)2x2 3. Denote by rQ(n) the number of integer solutions of Q(x0, x1, x2, x3) = n which is the same as the number of α = a0+a1i+a2j+a3k ∈H(Z) such that 2q\|α−a0 and N(α) = n. 14.10.96 Remark Estimating rQ(n) is an important and diﬃcult problem in number theory. Ac-cording to , there is no simple explicit formula for rQ(n) as Jacobi’s formula because of additional cusp forms that appear at the higher level. The Ramanujan conjecture for weight 2 cusp forms and its proof by Eichler and Igusa yields a good approximation for rQ(n). More precisely, if p is a prime and l ≥0, then rQ(pl) = C(pl) + Oϵ(p l 2 +ϵ) Combinatorial 655 for any ϵ > 0 as l →∞. Here C(pl) =          2 P d\|pl d = 2 pl+1−1 p−1 if p q = 1, 4 P d\|pl d = 4 pl+1−1 p−1 if p q = −1 and l even , 0 if p q = −1 and l odd. 14.10.97 Remark The number theoretic facts above and the connection between the eigenvalues and the number of closed nonbacktracking walks in a regular graph were used by Lubotzky, Phillips, and Sarnak to prove the following result. 14.10.98 Theorem [1969, 2006] The graphs Xp,q are Ramanujan. 14.10.99 Remark If q p = −1, then Xp,q is bipartite of high girth; its girth is at least 4 logp q − logp 4 ≈4 3 logp \|V (Xp,q)\|. If q p = 1, then Xp,q also has high girth; its girth is at least 2 logp q ≈2 3 logp \|V (Xp,q)\|. From the results of Hoﬀman, it also follows that these graphs have large chromatic number (at least 1 + p+1 2√p). 14.10.100 Remark Morgenstern generalized Lubotzky, Phillips, and Sarnak’s construction and constructed inﬁnite families of (q + 1)-regular Ramanujan graphs for every prime power q. 14.10.6 Combinatorial constructions of expanders 14.10.101 Construction Reingold, Vadhan, and Wigderson introduced a new graph product called the zig-zag product which they used to construct inﬁnite families of constant-degree expanders. 14.10.102 Deﬁnition Let X be a k-regular graph with vertex set [n] = {1, . . . , n}. Suppose the edges incident to each vertex of X are labeled from 1 to k in some arbitrary, but ﬁxed way. The rotation map RotX : [n] × [k] →[n] × [k] is deﬁned as follows: RotX(u, i) = (v, j) if the i-th edge incident to u is the j-th edge incident to v. 14.10.103 Deﬁnition Let G1 be a D1-regular graph with vertex set [N1] with rotation map RotG1 and G2 be a D2-regular graph with vertex set [D1] with rotation map RotG2. The zig-zag product G1zG2 is the D2 2-regular graph with vertex set [N1] × [D1] whose rotation map RotG1zG2 is: 1. let (k′, i′) = RotG2(k, i); 2. let (w, l′) = RotG1(v, k′); 3. let (l, j′) = RotG2(l′, j); 4. deﬁne RotG1zG2((v, k), (i, j)) = ((w, l), (i′, j′)). 14.10.104 Deﬁnition A graph G is an (n, d, λ)-graph if G has n vertices, is d-regular, and the absolute value of any non-trivial eigenvalue of G is at most λd. 14.10.105 Theorem If G1 is an (N1, D1, µ1)-graph and G2 is an (D1, D2, µ2)-graph, then G1zG2 is an (N1D1, D2 2, µ1 + µ2 + µ2 2)-graph. 14.10.106 Construction Using the previous theorem, Reingold, Vadhan, and Wigderson con-structed inﬁnite families of constant-degree expanders. 656 Handbook of Finite Fields 14.10.107 Construction Bilu and Linial have used graph lifts to construct inﬁnite families of d-regular graphs whose non-trivial eigenvalues have absolute value at most C p d log3 d, where C is some positive absolute constant. We outline their method below. 14.10.108 Deﬁnition Given two graphs G1 and G2, a graph homomorphism from G1 to G2 is a function f : V (G1) →V (G2) which preserves adjacency, namely if xy is an edge of G1, then f(x)f(y) is an edge of G2; the function f is a graph isomorphism if it is bijective and preserves both adjacency and non-adjacency, namely xy is an edge of G1 if and only if f(x)f(y) is an edge of G2. 14.10.109 Deﬁnition A surjective homomorphism f : V (G1) →V (G2) is a covering map (see [282, 1126]) if for each vertex x of G1, the restriction of f to x and its neighbors is bijective. Given a graph G, a cover of G is a pair (H, f), where f : V (H) →V (G) is a covering map. If in addition G is connected and ﬁnite and H is ﬁnite, then for each vertex y ∈V (G), the preimage f −1(y) has the same cardinality. If \|f −1(y)\| = t for each y ∈V (G), then (H, f) is a t-cover or H is a t-lift of G. 14.10.110 Remark For every ﬁnite graph G, there is a universal cover or a largest cover ˜ G which is an inﬁnite tree whose vertices can be identiﬁed with the set of nonbacktracking walks from a ﬁxed vertex x ∈V (G). For example, the universal cover of any ﬁnite k-regular graph is the inﬁnite k-regular tree Tk. 14.10.111 Remark An important property of a t-cover (H, f) of a ﬁnite graph G is that the graph H inherits the eigenvalues of G. This is because the vertex set of H can be thought as V (G) × {1, . . . , t} with the preimage (also called the ﬁber of y) f −1(y) = {x : x ∈V (H), f(x) = y} = {(y, i) : 1 ≤i ≤t}. The edges of H are related to the edges of G as follows: each ﬁber f −1(y) induces an independent set in H; if yz ∈E(G), then the subgraph of H induced by f −1(y) ∪f −1(z) = {(y, i), (z, i) : 1 ≤i ≤t} is a perfect matching (meaning that there exists a permutation σ ∈St such that (y, i) is adjacent to (z, σ(i)) for each 1 ≤i ≤t); if yz / ∈E(G), then there are no edges between f −1(y) and f −1(z). The partition of the vertex set of H as V (H) = ∪y∈V (G)f −1(y) is equitable (see ) and its quotient matrix is the same as the adjacency matrix of G. This implies the eigenvalues of A(G) are the eigenvalues of A(H). These eigenvalues of A(H) are old and the remaining eigenvalues of A(H) are new. 14.10.112 Remark In the case of a 2-lift, the new eigenvalues can be interpreted as eigenvalues of a signed adjacency matrix as follows. If H is a 2-lift of G, then for each edge yz of G, the subgraph induced by f −1(y) ∪f −1(z) = {(y, 0), (y, 1), (z, 0), (z, 1)} in H has either (y, 0) adjacent to (z, 0) and (y, 1) adjacent to (z, 1) (in which case set s(y, z) = s(z, y) = 1) or (y, 0) adjacent to (z, 1) and (y, 1) adjacent to (z, 0) (in which case set s(y, z) = s(z, y) −1). Let s(y, z) = 0 for all other y, z ∈V (G). The symmetric {0, −1, 1} matrix As whose (y, z)-th entry is s(y, z) is the signed adjacency matrix of the G with respect to the cover H. It is known that the eigenvalues of H are union of the eigenvalues of the adjacency matrix of G and the eigenvalues of the signed adjacency matrix of G. Bilu and Linial proved that every graph G with maximum degree d has a signed adjacency matrix (which can be found eﬃciently) whose eigenvalues have absolute value at most C p d log3 d where C is some positive absolute constant. 14.10.113 Construction Bilu and Linial’s idea to construct almost Ramanujan graphs is the following: start with a k-regular Ramanujan graph G0 (for example, the complete graph Kk+1) and then construct a 2-lift of Gi (denoted by Gi+1) such that the new eigenvalues of Gi+1 are small in absolute value for i ≥0. Bilu and Linial prove that every k-regular graph G Combinatorial 657 has a 2-lift H such that the new eigenvalues of H have absolute value at most C p k log3 k where C is some positive absolute constant. In this way the sequence of k-regular graphs Gi has non-trivial eigenvalues bounded from above by C p k log3 k. 14.10.114 Remark Bilu and Linial make the following conjecture which if true, would imply the existence of inﬁnite sequences of k-regular Ramanujan graphs for every k ≥3. 14.10.115 Conjecture Every k-regular graph has a signed adjacency matrix whose eigenvalues have absolute value at most 2 √ k −1. 14.10.116 Construction A diﬀerent combinatorial construction of almost Ramanujan graphs was proposed by de la Harpe and Musitelli , and independently by Cioab˘ a and Murty [646, 649]. The idea of these constructions is that perturbing Ramanujan graphs by adding or removing perfect matchings will yield graphs with small non-trivial eigenvalues. The linear algebraic reason for this fact follows from a theorem of Weyl which bounds the eigenvalues of a sum of two Hermitian matrices in terms of the eigenvalues of the summands. De la Harpe and Musitelli note that adding a perfect matching to any 6-regular Ramanujan graph will yield a 7-regular graph whose 2nd largest eigenvalue is at most 2 √ 5 + 1 ∼ = 5.47 which is larger than the Ramanujan bound of 2 √ 6 ∼ = 4.89, but strictly less than 7. Cioab˘ a and Murty [646, 649] use known results regarding gaps between consecutive primes to observe that by adding or removing perfect matching from Ramanujan graphs, one can construct k-regular almost Ramanujan graphs for almost all k. More precisely, their result is that given ϵ > 0, for almost all k ≥3, one can construct inﬁnite families of k-regular graphs whose 2nd largest eigenvalue is at most (2 + ϵ) √ k −1. 14.10.117 Remark The following conjecture was made in ; if true, this conjecture would imply the existence of inﬁnite families of k-regular Ramanujan graphs for any k ≥3. 14.10.118 Conjecture Let X be a k-regular Ramanujan graph with an even number of vertices. Then there exists a perfect matching P with V (P) = V (X) such that the (k + 1)-regular graph obtained from the union of the edges of X and P is Ramanujan. 14.10.119 Remark In a recent outstanding work, Friedman solved a long-standing conjecture of Alon from the 1980s and proved that almost all regular graphs are almost Ramanujan. 14.10.120 Theorem Given ϵ > 0 and k ≥3, the probability that a random k-regular graph on n vertices has all non-trivial eigenvalues at most (2 + ϵ) √ k −1 goes to 1 as n goes to inﬁnity. 14.10.7 Zeta functions of graphs 14.10.121 Deﬁnition A walk in a graph X is non-backtracking if no edge is traversed and then immediately backtracked upon. A non-backtracking walk whose endpoints are equal is a closed geodesic. If γ is a closed geodesic, we denote by γr the closed geodesic obtained by repeating the walk γ r times. A walk is tailless if it is non-backtracking under any cyclic permutation of vertices. A closed geodesic which is not the power of another one and is tailless is a prime geodesic. We deﬁne an equivalence relation on the closed geodesics as follows: (x0, ..., xn) and (y0, ..., ym) are equivalent if and only if m = n and there is a d such that yi = xi+d for all i (and the subscripts are interpreted modulo n). An equivalence class of a prime geodesic is a prime geodesic cycle. 658 Handbook of Finite Fields 14.10.122 Deﬁnition Let X be a k-regular graph and denote q = k −1. The Ihara zeta function is ZX(s) = Y p 1 −q−sℓ(p)−1 where the product is over all prime geodesic cycles p and ℓ(p) is the length of p. 14.10.123 Theorem For g = (q −1)\|X\|/2, we have ZX(s) = (1 −u2)−g det(I −Au + qu2I)−1, u = q−s. Moreover, ZX(s) satisﬁes the Riemann hypothesis’ (that is, all the singular points in the region 0 < ℜ(s) < 1 lie on Re(s) = 1/2) if and only if X is a Ramanujan graph. 14.10.124 Remark Hashimoto , as well as Stark and Terras have deﬁned a zeta function for an arbitrary graph and established its rationality. The deﬁnition of this zeta function is simple enough. Let Nr be the number of closed walks γ of length r so that neither γ nor γ2 have backtracking. Then, the zeta function of the graph X is deﬁned as ZX(t) = exp ∞ X r=1 Nrtr r ! . This deﬁnition is very similar to the zeta function of an algebraic variety. 14.10.125 Remark It would be interesting to interpret the singularities of ZX(t) in terms of properties of the graph. For instance, these zeta functions have a pole at t = 1 and Hashimoto has shown that the residue at t = 1 is related to the number of spanning trees of the graph X. Thus, this number is the graph-theoretic analogue of the class number of an algebraic number ﬁeld. These constructions raise the intriguing question of whether there is a generalization of the notion of a graph to that of a “supergraph” whose zeta function would (in some cases) coincide with those higher dimensional zeta functions of varieties. Work in this direction has started . See Also §6.1, §6.2 For details on Gauss sums and other character sums. §12.7 For discussions on zeta functions and L-functions of curves. §15.1, §15.3 For details on algebraic, LDPC, and expander codes. References Cited: [80, 81, 83, 156, 157, 269, 282, 411, 637, 638, 645, 646, 647, 648, 649, 780, 786, 834, 1126, 1127, 1128, 1287, 1441, 1534, 1535, 1568, 1641, 1693, 1807, 1921, 1922, 1923, 1924, 1925, 1926, 1927, 1967, 1968, 1969, 1970, 2005, 2006, 2117, 2118, 2160, 2207, 2208, 2289, 2290, 2433, 2448, 2523, 2525, 2589, 2594, 2701, 2789, 2837, 2839] 15 Algebraic coding theory 15.1 Basic coding properties and bounds .............. 659 Channel models and error correction • Linear codes • Cyclic codes • A spectral approach to coding • Codes and combinatorics • Decoding • Codes over Z4 • Conclusion 15.2 Algebraic-geometry codes .......................... 703 Classical algebraic-geometry codes • Generalized algebraic-geometry codes • Function-ﬁeld codes • Asymptotic bounds 15.3 LDPC and Gallager codes over ﬁnite ﬁelds...... 713 15.4 Turbo codes over ﬁnite ﬁelds ...................... 719 Introduction • Convolutional codes • Permutations and interleavers • Encoding and decoding • Design of turbo codes 15.5 Raptor codes ......................................... 727 Tornado codes • LT and fountain codes • Raptor codes 15.6 Polar codes ........................................... 735 Space decomposition • Vector transformation • Decoding • Historical notes and other results 15.1 Basic coding properties and bounds Ian Blake, University of British Columbia W. Cary Huﬀman, Loyola University Chicago 15.1.1 Channel models and error correction 15.1.1 Deﬁnition A discrete memoryless channel (DMC) is a set of possible inputs {u1, u2, . . . , uK} and outputs {v1, v2, . . . , vL} and probabilities pij such that if the in-put on a given channel use is ui, then the probability that the channel output is vj is P(vj \| ui) = pij, and the channel behavior is independent from one usage to the next. Note that L X j=1 pij = 1 for 1 ≤i ≤K. Thus for a given input vector x = (x1, x2, . . . , xn) and output vector y = (y1, y2, . . . , yn) P(y \| x) = n Y i=1 P(yi \| xi). 659 660 Handbook of Finite Fields 15.1.2 Example A channel with two inputs {0, 1} and two outputs {0, 1} is a binary symmetric channel (BSC) if p00 = p11 = 1 −p, p01 = p10 = p; p is the crossover probability. 15.1.3 Example A channel with two inputs {0, 1} and three outputs {0, E, 1} is a binary erasure channel (BEC) if p00 = p11 = 1 −ϵ, p0E = p1E = ϵ, and p01 = p10 = 0. An erasure in the received word is a position containing the symbol E. 15.1.4 Remark There are many other types of channels, including q-ary channels (q inputs), burst noise channels, and the additive white Gaussian noise (AWGN) channel, which is a contin-uous channel that adds white Gaussian noise to the transmitted signal. 15.1.5 Deﬁnition A code is a subset of Fn q , the set of n-tuples over the ﬁnite ﬁeld Fq of order q. Elements (vectors) of Fn q are denoted by boldface lowercase letters. 15.1.6 Deﬁnition For x ∈Fn q , denote by ω(x) the Hamming weight of x as the number of nonzero coordinate positions of x. For x, y ∈Fn q , denote by d(x, y) the Hamming distance between x and y as the number of coordinate positions where x and y diﬀer. 15.1.7 Remark Note that d(x, y) = ω(x −y). 15.1.8 Deﬁnition A code C is an (n, M, d)q code if it has M distinct codewords of length n over the ﬁnite ﬁeld Fq such that the minimum distance between any two distinct codewords is d where d = min c1,c2∈C c1̸=c2 d(c1, c2). The rate r of such a code is the ratio of the number of information symbols transmitted per codeword to the number of symbols per codeword, assuming the codewords are chosen equally likely, i.e., r = logq(M)/n. 15.1.9 Remark An important problem of coding theory is to design codes that have a large min-imum distance for a given code size (or large size for a given minimum distance) and to have encoding and decoding algorithms that can be implemented eﬃciently. Many types of encoders and decoders are considered in the literature. 15.1.10 Remark Unless speciﬁed otherwise, a received word is the sum of a codeword transmitted over a discrete memoryless channel and an error word. All such words are vectors over a ﬁnite ﬁeld. 15.1.11 Deﬁnition A maximum likelihood decoder (MLD) is one that, on receiving a word of channel output symbols r, chooses the codeword ˆ c ∈C that maximizes the probability of r being received, i.e., maximizes the conditional probability P(r \| c): ˆ c = arg max c∈C P(r \| c). 15.1.12 Deﬁnition A maximum a posteriori (MAP) decoder is one that, on receiving the channel output word r, chooses the codeword ˆ c ∈C that maximizes the a posteriori probability P(c \| r), i.e., ˆ c = arg max c∈C P(c \| r). Algebraic coding theory 661 15.1.13 Deﬁnition A minimum distance decoder is one that, on receiving the word r, chooses a codeword ˆ c ∈C that is at minimum distance from r, i.e., ˆ c = arg min c∈C d(c, r). A bounded distance decoder attempts to ﬁnd the codeword, if it exists, within distance ⌊(d −1)/2⌋of the received word. 15.1.14 Remark The output from these decoders may not be unique. In this case, in an equally likely scenario, the decoder chooses one of the codewords satisfying the criteria. In the case of choosing codewords equally likely, MAP is equivalent to MLD. 15.1.15 Lemma On a BSC with crossover probability p < 1/2, the minimum distance decoder is equivalent to the MLD [304, 1943, 2389]. 15.1.16 Remark If C is an (n, M, d)q code, one may think of the codewords being surrounded with “spheres” of radius e = ⌊(d −1)/2⌋, i.e. the set of q-ary n-tuples at distance e or less from the center. The spheres are then nonintersecting. 15.1.17 Lemma If C is an (n, M, d)q code and a codeword c is transmitted and word r is received, with r errors and s erasures, then c is the unique codeword in C closest to the received r provided that 2r + s < d. 15.1.18 Remark Shannon showed that with every discrete memoryless channel with a ﬁnite number of input and output symbols, one may deﬁne the notion of channel capacity in bits per channel use. His celebrated channel coding theorem and its converse then state there exists a code that allows essentially error-free transfer of data over the channel as long as the rate does not exceed channel capacity. Conversely, he showed that such transmission is not possible if the channel rate exceeds capacity. More speciﬁcally, Gallager showed that if C is the channel capacity, there is an error-rate exponent function E(R) that is positive for rates 0 ≤R < C and zero elsewhere, and there exists a code of rate R < C and length n > N for which the probability Pw of word error on the channel satisﬁes Pw < exp(−NE(R)) for 0 ≤R < C. The importance of the result is that there exists a sequence of codes of increasing lengths n > N and of rates less than the channel capacity for which the probability of word error, after maximum likelihood decoding, decreases exponentially to zero. 15.1.19 Remark Algebraic coding theory arose in response to the challenge of the Shannon the-orems to deﬁne codes, as well as eﬃcient encoding and decoding algorithms, that achieve asymptotically low error rate at code rates arbitrarily close to capacity. There are numerous excellent books available on the subject. The following were particularly useful references in preparing this section: [231, 304, 311, 1558, 1639, 1943, 1945, 1991, 2047, 2136, 2389, 2390, 2405, 2484, 2511, 2849]. 15.1.2 Linear codes 15.1.20 Deﬁnition A linear code C of length n over Fq is a subspace of Fn q . If the dimension of C is k and the minimum distance is d, it is referred to as an (n, k, d)q code and the number of codewords is M = qk. 15.1.21 Remark The anomaly in notation from the nonlinear case with a designation of (n, M, d)q code, where the number of codewords is M, will be clear from the context. 662 Handbook of Finite Fields 15.1.22 Remark To determine the minimum distance of a nonlinear (n, M, d)q code C will in general require a search over the distances between all M 2 pairs of codewords. For a linear code this can be reduced to determining the weight of a minimum weight codeword since d = min x,y∈C x̸=y d(x, y) = min x∈C x̸=0 w(x). 15.1.23 Deﬁnition The scalar (or inner) product of two vectors u = (u1, u2, . . . , un), v = (v1, v2, . . . , vn) from Fn q is (u, v) = u1v1 + u2v2 + · · · + unvn ∈Fq. If (u, v) = 0, the vectors are orthogonal. 15.1.24 Deﬁnition For a linear code C ⊆Fn q , the orthogonal code or dual code to C is deﬁned as C⊥= {v ∈Fn q \| (v, c) = 0 for all c ∈C}. 15.1.25 Theorem If C is a linear code of dimension k in Fn q , then C⊥is a linear code of dimension n −k in Fn q . 15.1.26 Deﬁnition Let C be a linear code in Fn q . If C ⊆C⊥, the code C is self-orthogonal. If C = C⊥, C is self-dual. 15.1.27 Remark In Rn, the only vector orthogonal to itself is the zero vector. So nonzero linear subspaces of Rn cannot be self-orthogonal. However, the situation is markedly diﬀerent when considering subspaces of vector spaces over ﬁelds of prime characteristic; in that case, self-orthogonal and self-dual subspaces do exist. For example, C = {(0, 0), (1, 1)} is a self-dual (2, 1, 2)2 code in F2 2. 15.1.28 Remark If C is a self-dual code, its dimension is n/2 and its length n is even. 15.1.29 Deﬁnition For C an (n, k, d)q linear code, a k × n matrix G is a generator matrix of C if its row space equals C. Similarly an (n −k) × n matrix H is a parity check matrix for C if its row space equals C⊥. 15.1.30 Remark Note that some authors use the term parity cheek matrix only for the binary case, as in this case the inner product of a codeword and row of the matrix involves an even number of ones in the sum. We retain the term parity check matrix for all sizes of ﬁelds. 15.1.31 Remark As any codeword of C is orthogonal to any word in C⊥, it follows that GHT = (the k × (n −k) all zero matrix), where superscript T denotes matrix transpose. Thus c ∈C if and only if HcT = 0T where 0 is the zero row vector of length n −k. 15.1.32 Remark By using elementary row operations (which preserve the row space of a matrix) and possibly coordinate position permutations, any generator matrix of a code C can be put in the form G = [Ik \| A] , Algebraic coding theory 663 where Ik is the k ×k identity matrix and A is a k ×(n−k) matrix over Fq. A corresponding parity check matrix is one in the form H = −AT \| In−k . 15.1.33 Deﬁnition Generator and parity check matrices of a code C in the form of Remark 15.1.32 are in systematic form. 15.1.34 Remark Some authors (e.g. ) refer to the form of matrices in Remark 15.1.32 as being in standard form and reserve “systematic” for any form where the information bits appear explicitly in the codewords. 15.1.35 Remark From the deﬁnitions it is clear that a parity check matrix of a linear code C is a generator matrix of the code C⊥, and a generator matrix of C is a parity check matrix of the code C⊥. 15.1.36 Remark If C is an (n, k, d)q code with parity check matrix H, then c ∈C if and only if HcT = 0T . Thus if c is a minimum weight codeword, then HcT = 0T implies a linear combination of d columns of H is the all zero vector. It follows that C has minimum distance d if every subset of d −1 columns of H is linearly independent, and there is a dependent set of d columns. 15.1.37 Lemma In an (n, k, d)q code C, any d−1 columns of a parity check matrix for C are linearly independent, and there is at least one set of d columns that is dependent. 15.1.38 Deﬁnition Let C be an (n, M, d)q code and let Ai (resp. Bi, if C is linear) be the number of codewords in C (resp. C⊥, which exists if C is linear) of weight i. Deﬁne the weight enumerators of these codes as WC(x, y) = n X i=0 Aixn−iyi and WC⊥(x, y) = n X i=0 Bixn−iyi. The sequence {A0, A1, . . . , An} (resp. {B0, B1, . . . , Bn}) is the weight distribution of the code C (resp. C⊥). 15.1.39 Theorem (MacWilliams identities) [304, 1558, 2849] Let WC(x, y) (resp. WC⊥(x, y)) be the weight enumerator of an (n, k, d)q linear code C (resp. of the (n, n −k, d′)q dual code C⊥). Then WC⊥(x, y) = 1 qk WC(x + (q −1)y, x −y). (15.1.1) 15.1.40 Remark An alternative version of the MacWilliams identities is obtained by expanding the terms in (15.1.1) to obtain n−j X i=0 n −i j Ai = qk−j j X i=0 n −i n −j Bi for 0 ≤j ≤n. 15.1.41 Remark In Delsarte proposed an interesting method to examine code properties and in particular their combinatorial aspects. It gives fundamental insight to the problem. Only a brief look at this approach will be given. 664 Handbook of Finite Fields 15.1.42 Deﬁnition For positive integers n and λ the Krawtchouk polynomial Pk(x) is the polynomial of degree k over the rational numbers Pk(x) = k X j=0 (−1)jλk−j x j n −x k −j for 0 ≤k ≤n where x j = x(x −1)(x −2) · · · (x −j + 1)/j!. These polynomials form a set of orthogonal polynomials and have many interesting properties such as 1. Pn i=0 n i Pk(i)Pℓ(i) = δkl n k , 2. Pn i=0 Pℓ(i)Pi(k) = δkl(λ + 1)k, 3. Pn k=0 n−k n−j Pk(x) = (λ + 1)jn−x j . Later, when required, the parameter λ will be speciﬁed. 15.1.43 Deﬁnition Let A = (A0, A1, . . . , An) be an (n + 1)-tuple of rational numbers. The MacWilliams transform of A is given by A′ = (A′ 0, A′ 1, . . . , A′ n) where A′ k = n X i=0 AiPk(i) for 0 ≤k ≤n. 15.1.44 Remark If the (i, k) entry of a matrix P is deﬁned as Pk(i), 0 ≤i, k ≤n, the MacWilliams transform can be expressed as A′ = AP. 15.1.45 Remark It can be shown that (A′)′ = (λ + 1)nA for any (n + 1)-tuple A. Likewise P 2 = (λ + 1)nIn+1. 15.1.46 Deﬁnition Let C be an (n, M, d)q code. Deﬁne the distance distribution of C as Ai(C) = 1 M \|{(x, y) ∈C2 \| (x, y) = i}\| for 0 ≤i ≤n, i.e., the average number of codewords at distance i from a ﬁxed codeword. In the case C is linear, this is just the weight distribution of C. 15.1.47 Deﬁnition Deﬁne an (n + 1)-tuple A as positive if A0 = 1, Ai ≥0, and A′ i ≥0 for 0 ≤i ≤n. 15.1.48 Remark The following lemma gives motivation for the appearance of Krawtchouk polyno-mials. 15.1.49 Lemma Let Wk be the set of vectors of weight k in Fn q , and let u be a vector of weight j in Fn q . Then with λ = q −1 X x∈Wk (u, x) = Pk(j). 15.1.50 Lemma The distance distribution of any code over Fq is a positive (n + 1)-tuple for λ = q −1. Algebraic coding theory 665 15.1.51 Deﬁnition Let A(C) be the distance distribution of an (n, M, d)q code C, and deﬁne the four fundamental parameters of C (where the parameter of the Krawtchouk polynomials is λ = q −1) as 1. d = d(A(C)) is the minimum distance, 2. s = s(A(C)) is the number of nonzero distances (of C), 3. d′ = d(A′(C)) is the dual distance, 4. s′ = s(A′(C)) is the external distance. In the case the code C is linear, A′(C) is a constant multiple of the weight distribution of C⊥. 15.1.52 Remark It is important to note these four parameters are deﬁned for any code C. The parameter d′ is the smallest i such that A′ i ̸= 0, i > 0. The parameter s′ is the number of nonzero entries in the set {A′ 1, i = 1, . . . , n}. In the case the code is linear, the four parameters are the minimum distance and number of nonzero codeword weights of C and the dual code C⊥, respectively. The following three theorems from (the ﬁrst theorem was ﬁrst proved by MacWilliams) are representative of the eﬀectiveness of this approach. 15.1.53 Theorem Let A be an (n + 1)-tuple with A0 ̸= 0 and let A′ be its MacWilliams transform. Then s′ ≥⌊(d −1)/2⌋. 15.1.54 Deﬁnition A code C over Fq is distance invariant if the weight distribution of c + C is the same for all c ∈C, i.e., the number of codewords in C at distance i from c ∈C is independent of c. 15.1.55 Remark A linear code over Fq is automatically distance invariant because c + C = C for all c ∈C. 15.1.56 Theorem Let C be a code for which s ≤d′. Then C is distance invariant. 15.1.57 Remark The following theorem justiﬁes the name external distance given to the parameter s′. 15.1.58 Theorem Let C be a code with external distance s′. Then each point of Fn q is at distance at most s′ from at least one codeword. 15.1.2.1 Standard array decoding of linear codes 15.1.59 Deﬁnition Let C be an (n, k, d)q code. An encoding is an injective mapping from the set Fk q of all messages to the set C of codewords. A minimum distance decoding is a map from the set Fn q of all possible received vectors to the set C which sends the received vector r to a codeword closest to r. The codeword is unique if the number of errors in transmission is at most e = d−1 2 . 15.1.60 Remark A particular decoding algorithm that works for any linear code over Fq is the coset or standard array decoding algorithm. It is generally too ineﬃcient for practical con-sideration. Since a linear (n, k, d)q code C is a vector subspace of Fn q , it can be viewed as an (additive) subgroup of Fn q . Thus Fn q can be decomposed into the cosets of C. Such a decomposition can be used to derive a decoding algorithm. 666 Handbook of Finite Fields 15.1.61 Deﬁnition Let C be a linear (n, k, d)q code. A coset of C with representative u ∈Fn q is the set u+C = {u+c \| c ∈C}. For u1, u2 ∈Fn q , as sets u1 +C and u2 +C are either disjoint or identical. Since C is closed under addition, it follows that u + C equals u + c + C if c ∈C. There are qn−k distinct cosets of C in Fn q . 15.1.62 Deﬁnition The standard array for an (n, k, d)q code C is a qn−k × qk array of vectors of Fn q formed in the following manner: 1. In the ﬁrst row of the array, place the codewords of C in some ordering, with the all-zero codeword ﬁrst. 2. Choose a vector u1 of minimum weight in Fn q \ C and add this to each vector in the ﬁrst row to form the second row. 3. Choose a vector u2 of minimum weight in Fn q \ {C ∪{u1 + C}} and add this to each vector in the ﬁrst row to form the third row. 4. Continue the process until the vectors of Fn q are exhausted. Vectors in the ﬁrst column of the array are coset leaders. Every vector of Fn q appears exactly once in the array. 15.1.63 Remark To determine a codeword closest to a received word r ∈Fn q proceed as follows: 1. Locate the received word r in the standard array. 2. If the coset containing r has coset leader e, decode r to the codeword r −e. This procedure is referred to as standard array decoding. The coset leaders are in fact the set of correctable error patterns. If e = d−1 2 , then every vector of weight less than or equal to e will appear as a coset leader. The coset leaders of weight greater than e are then not “complete.” 15.1.64 Lemma The standard array decoding algorithm of Remark 15.1.63 ﬁnds a codeword ˆ c in C at minimum distance from the received word r. This codeword is the transmitted one if at most e = d−1 2 errors were made in transmission. 15.1.65 Remark In what follows a large number of codes are introduced. For notation, where appropriate, a linear code of length n over Fq that has some parameter (or set of parameters) m will be denoted ABCn,q(m) for some sequence of Roman letters ABC, descriptive for that particular code. The duplication of notation with (n, M, d)q and (n, k, d)q will be clear from the context. 15.1.66 Remark A useful quantity to deﬁne is πm,q = qm −1 q −1 = qm−1 + qm−2 + · · · + 1. This is the number of nonzero projective m-tuples where scalar multiples are identiﬁed. 15.1.2.2 Hamming codes 15.1.67 Deﬁnition The m-th order Hamming code over Fq, denoted Hπm,q,q(m), has a parity check matrix formed by constructing the m×πm,q matrix whose columns are the set of distinct nonzero projective q-ary m-tuples. The code Hπm,q,q(m) is a (πm,q, πm,q −m, 3)q code. Algebraic coding theory 667 15.1.68 Remark The dimension of Hπm,q,q(m) is easy to establish. That the code has minimum distance 3 is seen by ﬁnding some linear combination of three columns of the parity check matrix that equals zero. That there is no linear combination of two columns equaling zero follows from the columns being projectively distinct. 15.1.69 Remark The binary Hamming code Hπm,2,2(m) is a (2m −1, 2m −1 −m, 3)2 code. The dual of this code has all nonzero words of weight 2m−1, and hence the weight enumerator of H⊥ πm,2,2(m) is WH⊥ πm,2,2(x, y) = xn + nx(n−1)/2y(n+1)/2, n = 2m −1. Using the MacWilliams identities it is seen that the weight enumerator of Hπm,2,2(m) is WHπm,2,2(x, y) = 2−m (x + y)n + n(x + y)(n−1)/2(x −y)(n+1)/2 . In a similar manner it can be established that the dual of the code Hπm,q,q(m) over Fq has all nonzero weights equal to qm−1. The dual codes to Hamming codes are simplex codes. 15.1.2.3 Reed-Muller codes 15.1.70 Remark For vectors u, v ∈Fn 2 deﬁne the product as u ⊗v = (u1v1, u2v2, . . . , unvn). This is also the coordinate-wise “AND” operation. Similarly the usual coordinate-wise ad-dition over F2 corresponds to “XOR.” 15.1.71 Deﬁnition Deﬁne the Reed-Muller generator matrix G(r, m), 0 ≤r ≤m, as follows. 1. The zero-th row v0 is all ones. 2. The rows vi, i = 1, 2, . . . , m, are formed by alternating 2m−i 0’s with 2m−i 1s to ﬁll the vector of length 2m. 3. The matrix G(r, m) consists of the above m+1 rows together with all products of the vectors vi, i = 1, 2, . . . , m, taken j at a time for all j ≤r. G(r, m) is a (Pr j=0 m j ) × 2m matrix. 15.1.72 Deﬁnition The matrix G(r, m) is a generator matrix for the order r and degree m binary Reed-Muller code RM2m,2(r, m). 15.1.73 Lemma The parameters of RM2m,2(r, m) for 0 ≤r ≤m are n = 2m, k = r X j=0 m j , d = 2m−r. The dual of RM2m,2(r, m) is RM2m,2(m −r −1, m) for 0 ≤r < m. 15.1.74 Remark That the matrix G(r, m) has linearly independent rows can be established by showing the 2m × 2m matrix G(m, m) is nonsingular. Equivalently this can be shown by noting there exists a Boolean function, which can be realized by Boolean operations on the ﬁrst m + 1 rows, that will generate a vector with a single 1 in a given place. The codes are in fact Euclidean geometry codes discussed later. 668 Handbook of Finite Fields 15.1.75 Remark The generator matrix G(r, m) of RM2m,2(r, m) can be written as G(r, m) = G(r, m −1) G(r, m −1) 0 G(r −1, m −1) . It follows that RM2m,2(r, m) = {(u, u + v) \| u ∈RM2m−1,2(r, m −1), v ∈RM2m−1,2(r −1, m −1)}. The minimum distance of RM2m,2(r, m) can be established using Lemma 15.1.93 below. 15.1.2.4 Subﬁeld and trace codes 15.1.76 Remark Given an (n, k, d)qm code C over Fqm, there are several natural ways to obtain codes over subﬁelds. Two methods are of particular interest: subﬁeld subcodes and trace codes, introduced below. 15.1.77 Deﬁnition The subﬁeld subcode of an (n, k, d)qm code C is the set of codewords of C with all elements in Fq, i.e., Csf qm\|q = {(c1, c2, . . . , cn) ∈C \| ci ∈Fq for 1 ≤i ≤n} = C ∩Fn q . 15.1.78 Lemma The subﬁeld subcode Csf qm\|q obtained from the (n, k, d)qm code C is a linear (n, ≥ mk −(m −1)n, ≥d)q code. 15.1.79 Remark The subﬁeld subcode of an (n, M, d)qm nonlinear code is, as in the linear case, the set of codewords with all coordinates in Fq. The minimum distance of the subﬁeld sub-code is at least d. In the linear case, the bound on the dimension of the subﬁeld subcode is obtained from considering the dimension of the subspaces involved. Thus replacing ele-ments of the parity check matrix of C by m-tuple columns over Fq, the subﬁeld subcode is the set of n-tuples over Fq orthogonal to all (n −k)m rows and has dimension at least n −(n −k)m = mk −(m −1)n. 15.1.80 Deﬁnition The trace of an element η ∈Fqm is deﬁned as Tr qm\|q(η) = η + ηq + · · · + ηqm−1 and is an Fq-linear map from Fqm to Fq; see also Section 2.1. 15.1.81 Deﬁnition Given an (n, k, d)qm code C, the trace code of C is deﬁned as Ctr qm\|q = {(Tr qm\|q(c1), Tr qm\|q(c2), . . . , Tr qm\|q(cn)) \| (c1, c2, . . . , cn) ∈C}. 15.1.82 Lemma The trace code of an (n, k, d)qm code C is an (n, ≤mk, ≤d)q code. 15.1.83 Remark The dimension bound of Lemma 15.1.82 is the trivial one; since the parent code has (qm)k codewords, the trace code over Fq can have at most qmk codewords. The bound is achieved if and only if Tr qm\|q(c1) ̸= Tr qm\|q(c2) for any two distinct codewords c1, c2 ∈C. 15.1.84 Remark The following result, due to Delsarte , shows the relation between subﬁeld subcodes and trace codes. Algebraic coding theory 669 15.1.85 Lemma Let C be a linear code over Fqm. Then the dual of the subﬁeld subcode of C is the trace code of the dual code of C over Fqm, i.e., Csf qm\|q ⊥ = C⊥tr qm\|q . 15.1.86 Remark The following commutative diagram may be of value in visualizing these relation-ships: C dual − − − − → C⊥ subﬁeld   y   ytrace Csf qm\|q dual − − − − → Csf qm\|q ⊥ = C⊥tr qm\|q . 15.1.2.5 Modifying linear codes 15.1.87 Remark There are many ways of modifying a given code. The deﬁnitions given below are typical, although the terminology is not uniform in the literature. 15.1.88 Deﬁnition An (n, k, d)q code C can be modiﬁed in the following ways: 1. Expurgation: Certain codewords are deleted (no change of length). 2. Augmentation: The number of codewords is increased (no change of length). 3. Puncturing: The length of the code is reduced by deleting a coordinate position, generally without reducing the size of the code. 4. Extending: The length of the code is increased without increasing the size of the code. 5. Shortening: The length and dimension of the code are reduced. 6. Lengthening: The length and dimension of the code are increased. 15.1.89 Example Examples of the above operations on an (n, k, d)q code are given below. 1. If a row of a generator matrix is deleted, the result is an (n, k−1, d′)q code where d′ ≥d. Similarly a row could be added to a parity check matrix, which is linearly independent of the existing rows, to give the same result. 2. A row could be added to a generator matrix that is linearly independent of the existing rows to give an (n, k + 1, d′)q code where d′ ≤d. 3. If we delete a coordinate position contained in the nonzero positions of a codeword of minimum weight (assuming it is not of weight 1), the result is an (n−1, k, d−1)q code. 4. Adding an overall parity check produces an (n + 1, k, d′)q code where d′ = d or d + 1 depending on the code structure. For example, if q = 2 and the original code contains minimum weight codewords of odd weight (in which case half the words are of odd weight and half of even weight), the extended code has minimum distance d+1. This is equivalent to adding a column of zeroes to the parity check matrix and then a row of all ones. 5. If a generator matrix of the code is in systematic form and the ﬁrst row and column are deleted, the result is an (n −1, k −1, d′)q code where d′ ≥d. 6. Adding a row to a generator matrix, linearly independent of the existing rows, and then adding a column gives an (n + 1, k + 1, d′)q code where d′ ≤d + 1. 670 Handbook of Finite Fields 15.1.90 Remark The trace code and subﬁeld subcode of the previous subsection may also be thought of as ways of modifying a given code. 15.1.91 Remark Given two codes, an (n1, k1, d1)q code C1 with generator matrix G1 and an (n2, k2, d2)q code C2 with generator matrix G2, methods to produce a third code C are considered. 15.1.92 Remark The following construction proves useful in some situations, such as the construc-tion of the Reed-Muller codes of Subsection 15.1.2.3, already noted. 15.1.93 Lemma Let Ci be an (n, ki, di)q code, i = 1, 2. Then the code C = {(u, u + v) \| u ∈C1, v ∈C2} is a (2n, k1 + k2, d)q code where d = min{2d1, d2}. 15.1.94 Deﬁnition The direct sum code is the (n1 + n2, k1 + k2, d)q code where d = min{d1, d2} with generator matrix G1 ⊕G2 (block diagonal sum of matrices). 15.1.95 Deﬁnition The product code C1 ⊗C2 is the (n1n2, k1k2, d1d2)q code with generator matrix G1 ⊗G2, the tensor product of the two component generator matrices. 15.1.96 Remark One can think of the codewords of the product code in the following manner. Consider an array of k1 × k2 information symbols from Fq. The array is ﬁrst extended to a k1 × n2 array by completing each row to codewords in C2. The columns of this array are completed to an n1 × n2 array by completing each column to a codeword in C1. Notice the bottom right (n1 −k1) × (n2 −k2) array consists of checks on checks. The resulting n1 × n2 matrices are the codewords of C1 ⊗C2. The process could have been done by ﬁrst completing the columns of the information symbols with codewords in C1 and then completing rows. It is easily veriﬁed the same ﬁnal array is obtained. 15.1.97 Deﬁnition Let C1 be a linear (N, K, D)qm code and C2 a linear (n, m, d)q code. Fix a basis of Fqm over Fq and represent elements of Fqm as m-tuples over Fq. These m-tuples represent information positions in C2 which therefore give a one-to-one correspondence between elements of Fqm and codewords of C2. The concatenated code of C1 by C2 is obtained by replacing the Fqm-elements of codewords of C1 by codewords of C2 corresponding to the information m-tuples of the Fqm-elements. 15.1.98 Lemma The concatenated code of C1 by C2 is a linear (nN, kK, ≥dD)q code. 15.1.2.6 Bounds on codes 15.1.99 Remark The central problem of coding theory is to construct codes with as many codewords as possible for a given minimum distance (or as large a minimum distance as possible for a given code size), in such a way that the code can be eﬃciently encoded and decoded. This section reviews many of the bounds available. While interest is largely in linear codes, a version of a few of the bounds holds true for nonlinear codes as well. Most of the books on coding have a treatment similar to the one here. The treatment of bounds in is particularly elegant and succinct. Algebraic coding theory 671 15.1.100 Deﬁnition Deﬁne Aq(n, d) to be the size of the largest code of length n over Fq with minimum distance d, i.e., Aq(n, d) = max {M \| an (n, M, d)q code exists}. 15.1.101 Deﬁnition The sphere of radius r and center x ∈Fn q is Sq(r, n, x) = {y ∈Fn q \| d(y, x) ≤r}. The cardinality of Sq(r, n, x) is sq(r, n) = \| Sq(r, n, x) \| = r X i=0 n i (q −1)i. 15.1.102 Remark The following bound is a lower bound on the size of a maximal code. 15.1.103 Theorem (Sphere covering bound, Varshamov-Gilbert bound) [2849, 2855] Aq(n, d) ≥ qn sq(d −1, n). 15.1.104 Remark Theorem 15.1.103 is shown by surrounding each codeword of a code of maximum size with a sphere of radius d −1 and considering the union of such spheres. If the space is not exhausted, it would be possible to add a word at distance at least d from every codeword, implying the code was not optimal. For linear codes a diﬀerent argument leads to a similar result. For a given n and k, let r = n −k, and attempt to construct an r × n parity check matrix, with the property that any d −1 columns are linearly independent, by adding columns sequentially. The process may be started with the r ×r identity matrix. Suppose j −1 such columns have been found. A j-th column can be added if d−2 X i=1 j −1 i (q −1)i < qr −1. If n is the largest value of j for which this inequality holds, then an (n, k, d)q code exists. Notice that n is also the smallest value of n for which d−2 X i=1 n i (q −1)i ≥qn−k −1, an expression which can be rewritten as qk ≥qn/sq(d −2, n). This result can be compared to the result of Theorem 15.1.103 and is also referred to as the Varshamov-Gilbert bound. 15.1.105 Remark The following bounds are upper bounds on the size of codes. In many cases it is possible to ﬁnd codes that meet the bounds with equality. 672 Handbook of Finite Fields 15.1.106 Theorem (Sphere packing bound, Hamming bound) For a positive integer d with 1 ≤d ≤n, let e = ⌊d−1 2 ⌋. Then Aq(n, d) ≤ qn Pe i=0 n i (q −1)i = qn sq(e, n). 15.1.107 Remark In an optimal code with Aq(n, d) codewords, it is possible to surround codewords with nonintersecting spheres of radius e and the bound follows. 15.1.108 Deﬁnition A code whose size meets the bound of Theorem 15.1.106 with equality is perfect. 15.1.109 Remark In a perfect code the spheres of radius e around the codewords are nonintersect-ing and exhaust the space. The Hamming codes, both binary and nonbinary, are perfect codes. Besides the linear Hamming codes there are nonlinear perfect codes with the same parameters. The (23, 12, 7)2 binary Golay code and the (11, 6, 5)3 ternary Golay code (to be discussed later) are perfect. There are also trivial perfect codes: (i) the (n, n, 1)q code (the complete space), e = 0, (ii) the (n, 1, n)2 binary repetition code for n odd, e = (n −1)/2, (iii) a code of length n over Fq consisting of a single codeword, e = n, and (iv) a binary code of odd length containing only two words c and its complement c + 1, e = (n −1)/2 [1558, 2810, 2811]. 15.1.110 Theorem (Plotkin bound) If an (n, M, d)q code exists, then d ≤nM(q −1) (M −1)q . 15.1.111 Remark In the following it will be convenient to deﬁne θ = (q −1)/q. By rearranging terms in the above theorem, another expression for the Plotkin bound is Aq(n, d) ≤ d d −θn for d > θn. 15.1.112 Theorem (Griesmer bound) If an (n, k, d)q code exists, then n ≥ k−1 X i=0 d qi . 15.1.113 Theorem (Singleton bound) For a positive integer d, 1 ≤d ≤n, Aq(n, d) ≤qn−d+1. 15.1.114 Remark Puncturing the code on d −1 nonzero coordinate positions of a minimum weight codeword implies all remaining codewords are distinct and the Singleton bound follows. 15.1.115 Remark For a linear (n, k, d)q code, the Singleton bound reduces to d ≤n −k + 1. 15.1.116 Deﬁnition A code (either linear or nonlinear) that meets the Singleton bound with equality is a maximum distance separable (MDS) code. Algebraic coding theory 673 15.1.117 Remark 1. For a linear (n, k, d)q code, the Singleton bound is simply a reﬂection that every set of d−1 columns of the parity check matrix (an (n−k) × n matrix) is linearly independent, and hence the largest value of d −1 is n −k. 2. The dual of an (n, k, d = n−k +1)q MDS code is an (n, n−k, k +1)q MDS code. 3. Important examples of MDS codes are the Reed-Solomon codes discussed in the next section. Trivial examples include the (n, 1, n)q repetition code and the (n, n, 1)q full code. 15.1.118 Theorem (Elias bound) [2390, 2849] Assume that r ≤θn and r2 −2θnr + θnd > 0. Then Aq(n, d) ≤ θnd r2 −2θnr + θnd · qn sq(r, n). 15.1.2.7 Asymptotic bounds 15.1.119 Remark It is of interest to consider the constraints on codes and how they are reﬂected in the bounds discussed as the length of the code increases to inﬁnity. Notice that for a linear (n, k, d)q code, the rate is k/n, the ratio of the number of information symbols to code symbols. Asymptotically as the length increases, this rate is denoted αq(δ) where δ is the asymptotic normalized distance d/n. The following deﬁnition is valid for both linear and nonlinear codes. 15.1.120 Deﬁnition The asymptotic normalized rate is αq(δ) = lim supn→∞ 1 n logq Aq(n, δn). 15.1.121 Remark To discuss asymptotic versions of the bounds, the following deﬁnition of the q-ary entropy function is needed. 15.1.122 Deﬁnition The q-ary entropy function is deﬁned as Hq(x) = x logq(q −1) −x logq(x) −(1 −x) logq(1 −x) for 0 < x ≤θ, 0 elsewhere. 15.1.123 Remark The work of Delsarte on distance distributions for codes and their MacWilliams transforms (Deﬁnition 15.1.43) suggests the following linear programming approach to upper bounding the size of M for an (n, M, d)q code [2050, 2849]. Let A = (A0 = 1, A1, . . . , An) be the distance distribution of a putative (n, M, d)q code, where A1 = · · · = Ad−1 = 0 and Pn i=0 Ai = M. The linear program consists of maximizing Pn i=0 Ai subject to A0 = 1, A1 = · · · = Ad−1 = 0, Ai ≥0 for i = d, . . . , n, and the MacWilliams transform constraints n X i=0 AiPk(i) ≥0, k = 0, 1, . . . , n. This bound, often referred to as the linear programming or LP bound, is not very explicit, but it leads to the following two theorems which are quite strong upper bounds for binary codes. Noting that H2(x) = −x log2(x) −(1 −x) log2(1 −x), deﬁne g(x) by g(x) = H2((1 − √ 1 −x)/2). 674 Handbook of Finite Fields 15.1.124 Theorem For binary codes, if 0 < δ < 1/2, α2(δ) ≤ min 0≤u≤1−2δ{1 + g(u2) −g(u2 + 2δu + 2δ)}. 15.1.125 Remark The above bound is referred to as the MRRW bound (after the initials of the four authors of ), and it implies the following one - the bounds are actually the same over a range of values of δ. 15.1.126 Theorem For binary codes, if 0 < δ < 1/2, α2(δ) ≤g((1 −2δ)2). 15.1.127 Remark The following relationship is crucial to considering the asymptotic bounds. 15.1.128 Theorem [1558, 2389, 2849] For 0 ≤λ ≤θ, q ≥2 lim n→∞ 1 n logq sq(⌊λn⌋, n) = Hq(λ). 15.1.129 Remark The above theorem implies that, asymptotically, sq(r, n) tends to qnHq(r/n) . Apart from the MRRW bound, the following asymptotic versions of the bounds follow directly from their ﬁnite counterparts. 15.1.130 Theorem (Asymptotic bounds) [1558, 2390, 2849] 1. Asymptotic Varshamov-Gilbert bound: αq(δ) ≥1 −Hq(δ), 0 ≤δ ≤θ. 2. Asymptotic Singleton bound: αq(δ) ≤1 −δ, 0 ≤δ ≤1. 3. Asymptotic Plotkin bound: αq(δ) ≤1 −δ/θ, 0 ≤δ < θ. 4. Asymptotic Hamming bound: αq(δ) ≤1 −Hq(δ/2), 0 ≤δ ≤θ. 5. Asymptotic Elias bound: αq(δ) ≤1 −Hq(θ − p θ(θ −δ)), 0 ≤δ < θ. 15.1.3 Cyclic codes 15.1.131 Deﬁnition Let x = (x0, x1, . . . , xn−1) ∈Fqn. A cyclic (right) shift of x (with wraparound) is (xn−1, x0, x1, . . . , xn−2). Let C be a linear code. Then C is a cyclic code if every cyclic shift of a codeword in C is also a codeword in C. 15.1.132 Remark Some beneﬁts to assuming a code is linear have been seen. Restricting attention further to cyclic codes allows the formulation of eﬃcient algorithms for the construction, encoding, and decoding of them. The simple addition of requiring cyclic shifts of codewords to be codewords introduces a strong algebraic structure into the picture that allows these beneﬁts. Algebraic coding theory 675 15.1.3.1 Algebraic prerequisites 15.1.133 Deﬁnition Let R be a commutative ring with identity. A subset I of R is an ideal of R if for all a, b ∈I and r ∈R, then a −b ∈I and ra ∈I. The ideal is a principal ideal if it has a single generator a where I = ⟨a⟩= {ra \| r ∈R}. The ring R is an integral domain if whenever a, b ∈R and ab = 0, either a = 0 or b = 0. The ring R is a principal ideal domain (PID) if it is an integral domain and all its ideals are principal. The quotient ring of R by the ideal I is denoted by R/I with addition and multiplication of cosets given by (r + I) + (s + I) = (r + s) + I and (r + I)(s + I) = rs + I. 15.1.134 Example The ring Fq[x] is the set of polynomials in the indeterminate x with coeﬃcients in Fq. The rings Z, Fq[x], and Fq[x]/⟨xn −1⟩are PIDs. In Z the ideal ⟨2⟩is {2k \| k ∈Z}, i.e., the set of even integers. In Fq[x] ⟨g(x)⟩= {a(x)g(x) \| a(x) ∈Fq[x]} is the ideal generated by the polynomial g(x). 15.1.135 Remark In the quotient ring Fq[x]/⟨xn−1⟩, each coset has a unique coset representative that is either the zero polynomial or has degree less than n. For simplicity, the coset a(x)+⟨xn−1⟩ with a(x) = a0 + a1x + · · · + an−1xn−1 will be denoted a(x). Therefore the quotient ring Fq[x]/⟨xn −1⟩is the set of polynomials {a0 + a1x + · · · + an−1xn−1 \| ai ∈Fq, i = 0, 1, . . . , n −1} with addition and multiplication modulo xn −1. There is a natural map between n-tuples over Fq and polynomials in Fq[x]/⟨xn −1⟩, namely Fn q − → Fq[x]/⟨xn −1⟩ (a0, a1, . . . , an−1) 7→ a0 + a1x + · · · + an−1xn−1. Thus a linear code C can be viewed equivalently as a subspace of Fn q and an Fq-subspace of Fq[x]/⟨xn −1⟩. Notice that if (a0, a1, . . . , an−1) 7→a0 +a1x+· · ·+an−1xn−1, then the cyclic shift (an−1, a0, a1, . . . , an−2) corresponds to the polynomial x(a0 + a1x + · · · + an−1xn−1) (mod xn −1) which gives the reason for interest in the quotient ring Fq[x]/⟨xn −1⟩. 15.1.136 Lemma A linear code C is a cyclic code if and only if it is an ideal in Fq[x]/⟨xn −1⟩. 15.1.137 Remark Since Fq[x]/⟨xn −1⟩is a PID, every nonzero ideal C has a generator polynomial g(x), i.e., C = ⟨g(x)⟩, and one such generator polynomial is the unique monic codeword polynomial of least degree. It also follows that g(x) divides xn −1 (written g(x) \| (xn −1)) as otherwise xn −1 = a(x)g(x)+r(x) for some a(x) and r(x) where r(x) has degree strictly less than that of g(x). Since by deﬁnition r(x) would be in C, r(x) nonzero contradicts the fact that g(x) was of least degree. For g(x) \| (xn−1), g(x) generates an ideal in Fq[x]/⟨xn−1⟩ that is the set of polynomials divisible by g(x). More speciﬁcally ⟨g(x)⟩= {a(x)g(x) \| a(x) ∈Fq[x], deg a(x) < n −deg g(x)}. The term generator polynomial of a cyclic code is reserved for the unique monic polynomial generating the code and dividing xn−1. Thus cyclic codes are determined by factors of xn−1. Indeed, if gcd(n, q) = 1, xn −1 factors into distinct irreducible polynomials fi(x) over Fq, i = 1, 2, . . . , t, and then there are 2t cyclic codes of length n over Fq. If gcd(n, q) ̸= 1, xn −1 has repeated irreducible factors. For example if gcd(n, q) = p where Fq has characteristic p, then n = n1p and xn −1 = (xn1 −1)p. More generally, if xn −1 = Qt 1=1 f ei i (x) is the factorization, the possible number of cyclic codes is Qt i=1(ei + 1); see . Henceforth it is assumed that gcd(n, q) = 1. 676 Handbook of Finite Fields 15.1.3.2 Properties of cyclic codes 15.1.138 Remark A cyclic (n, k, d)q code C has a generator polynomial g(x) of degree n −k. If g(x) = g0+g1x+· · ·+gn−kxn−k, the code is viewed equivalently as the ideal of Fq[x]/⟨xn−1⟩ C = {a(x)g(x) \| a(x) ∈Fq[x], deg a(x) < k} and the row space of the matrix G =        g0 g1 g2 · · · gn−k 0 0 · · · 0 0 g0 g1 · · · gn−k−1 gn−k 0 · · · 0 0 0 g0 · · · gn−k−2 gn−k−1 gn−k · · · 0 . . . . . . 0 0 0 · · · g0 g1 g2 · · · gn−k        , (15.1.2) i.e., G is a generator matrix for the code C. 15.1.139 Remark As the generator polynomial g(x) \| (xn−1), deg g(x) = n−k, there is a polynomial h(x) = h0 + h1x + · · · + hkxk such that g(x)h(x) = xn −1. Since deg h(x) = k and h(x) \| (xn −1), it generates a cyclic (n, n −k, d′)q code C′ over Fq for some minimum distance d′. However C′ is not C⊥. Deﬁne the reciprocal polynomial h∗(x) = xkh(1/x), which is the polynomial obtained by reversing the coeﬃcients of h(x). Then (1/h0)h∗(x) is the generator polynomial of C⊥, which is C′ with coordinates reversed and has the same parameters as C′. A generator matrix of C⊥(and parity check matrix of C) is then given by H =        hk hk−1 hk−2 · · · h0 0 0 · · · 0 0 hk hk−1 · · · h1 h0 0 · · · 0 0 0 hk · · · h2 h1 h0 · · · 0 . . . . . . 0 0 0 · · · hk hk−1 hk−2 · · · h0        . 15.1.140 Remark Let C be an (n, k, d)q cyclic code with generator polynomial g(x) whose roots are α1, α2, . . . , αn−k. The roots will in general be in an extension ﬁeld of Fq, say Fqm where n \| (qm −1). Another way of deﬁning the code is to note that c(x) = a(x)g(x) is a codeword polynomial if and only if c(αi) = 0, i = 1, 2, . . . , n −k. Thus c = (c0, c1, . . . , cn−1) ∈C if and only if HcT = 0T where H =      1 α1 α2 1 · · · αn−1 1 1 α2 α2 2 · · · αn−1 2 . . . 1 αn−k α2 n−k · · · αn−1 n−k     . 15.1.141 Remark Let C1 and C2 be cyclic codes with generator polynomials g1(x) and g2(x), respec-tively. Then C1 ⊆C2 if and only if g2(x) \| g1(x). 15.1.142 Remark In practice there is often a preference to use systematic codes where the informa-tion symbols appear explicitly in the codeword. To achieve this one might row-reduce the generator matrix of (15.1.2) (with possible column permutations - see Remark 15.1.32) to be of the form G′ = [Ik \| A] Algebraic coding theory 677 as can be done for any linear code. One could then encode the information word m (of length k, recalling that the message length is the same as code dimension) as mG′. However if column permutations are required to obtain G′, the resulting code with generator matrix G′ may not be cyclic. For an (n, k, d)q cyclic code with generator polynomial g(x) of degree n −k, one might also do the following encoding. Divide xn−km(x) by g(x), where m(x) is the information polynomial, to obtain xn−km(x) = a(x)g(x) + r(x), deg r(x) < n −k. Thus xn−km(x) −r(x) = a(x)g(x) is a codeword with the k information symbols in the “high end” and n −k parity check symbols in the “low end.” 15.1.143 Theorem Let C be a cyclic code of length n over Fq and let v(x) ∈Fq[x]. Then C = ⟨v(x)⟩if and only if gcd(v(x), xn −1) = g(x). Equivalently v(x) generates C if and only if the n-th roots of unity that are zeros of v(x) are precisely the zeros of g(x). 15.1.144 Theorem Let Ci, i = 1, 2, be cyclic codes with generator polynomials gi(x), i = 1, 2. Then C1 ∩C2 has generator polynomial lcm(g1(x), g2(x)) and C1 + C2 has generator polynomial gcd(g1(x), g2(x)). 15.1.3.3 Classes of cyclic codes 15.1.145 Remark The additional constraint of cyclic codes (over linear) allows considerably more information on the minimum distance of the code to be obtained. The minimal polynomial over Fq of an element β in some extension ﬁeld of Fq is the monic irreducible polynomial, denoted Mβ(x), of least degree in Fq[x] that has β as a zero. 15.1.146 Remark Recall that the q-ary Hamming code Hn,q(m) is a (πm,q, πm,q −m, 3)q code where n = πm,q = (qm −1)/(q −1). When gcd(n, q −1) = 1, Hn,q(m) can be made cyclic in the following manner. Let α be a primitive element in Fqm. Then β = αq−1 is a primitive n-th root of unity. The parity check matrix of Hn,q(m) can be taken as H = 1 β β2 · · · βn−1 . Under the stated conditions the minimal polynomial of β over Fq is of degree m. It remains to verify that no two columns of H are multiples of each other over Fq to ensure a minimum distance of 3. Since the dual of this code has all codewords of weight qm−1 the weight enumerator of the q-ary Hamming code can be obtained via the MacWilliams identities, as in the binary case. This is an example of the ﬁrst class of cyclic codes to be considered. 15.1.147 Deﬁnition Let n be a positive integer relatively prime to q. For i an integer, the q-cyclotomic coset modulo n containing i is Ci = {i, iq, . . . , iqr−1} (mod n) where r is the smallest positive integer such that iqr ≡i (mod n). 15.1.148 Remark The smallest extension ﬁeld of Fq that contains a primitive n-th root of unity is Fqm where m is the size \|C1 \| of the q-cylotomic coset modulo n containing 1. If β is a primitive n-th root of unity in Fqm, then the minimal polynomial of βa over Fq is Mβa(x) = Y j∈Ca (x −βj). There is a one-to-one correspondence between the monic irreducible factors of xn −1 and the q-cyclotomic cosets modulo n. The factorization of xn −1 into irreducible factors over Fq is given by xn −1 = Y s Mβs(x) 678 Handbook of Finite Fields where s runs through a set of representatives of the distinct q-cyclotomic cosets modulo n. 15.1.149 Deﬁnition Let β be a primitive n-th root of unity in an extension ﬁeld of Fq. Let C be a cyclic code over Fq of length n with generator polynomial g(x) ∈Fq[x]. Then there is a set T ⊆{0, 1, . . . , n −1} such that the roots of g(x) are {βt \| t ∈T}. T is a deﬁning set (with respect to β) of C. 15.1.150 Remark Changing β will change T. For a given β, knowing g(x) is equivalent to knowing T because g(x) = Q t∈T (x−βt). If g(βt) = 0, then g(βtq) = 0 also, implying that T is a union of q-cyclotomic cosets modulo n. Conversely, any set T ⊆{0, 1, . . . , n −1} that is a union of q-cyclotomic cosets modulo n has the property that Q t∈T (x −βt) ∈Fq[x] and hence is the deﬁning set of a cyclic code of length n over Fq. BCH codes: 15.1.151 Deﬁnition [359, 1329, 1516, 1558, 2035] The cyclic code BCHn,q(δ) of length n and designed distance δ has a generator polynomial of the form g(x) = lcm{Mβa(x), Mβa+1(x), . . . , Mβa+δ−2(x)} for some sequence of elements βa, βa+1, . . . , βa+δ−2, where β is a primitive n-th root of unity in some extension ﬁeld of Fq. Alternately, BCHn,q(δ) has deﬁning set Ca ∪Ca+1 ∪· · · ∪Ca+δ−2 relative to β. If n = qm −1 for some positive integer m, the code is primitive, and if a = 1, it is narrow sense. 15.1.152 Remark These codes are referred to as BCHn,q(δ) codes, where BCH stands for the code name (Bose-Chaudhuri-Hocquenghem after the code originators [359, 1516]), and the subscripts on BCH are the length and ﬁeld of deﬁnition, and δ is the designed distance of the code. The relationship between the designed distance and the true minimum distance of BCHn,q(δ) is discussed below. 15.1.153 Remark The reason for requiring the generator polynomial of the BCH code to have a consecutive sequence of roots derives from properties of Vandermonde matrices as follows. Let β1, β2, . . . , βℓbe distinct elements in some extension ﬁeld of Fq. The determinant of the matrix        β1 β2 · · · βℓ β2 1 β2 2 · · · β2 ℓ β3 1 β3 2 · · · β3 ℓ . . . βℓ 1 βℓ 2 · · · βℓ ℓ        (15.1.3) is Qℓ i=1 βi times the determinant of the matrix        1 1 · · · 1 β1 β2 · · · βℓ β2 1 β2 2 · · · β2 ℓ . . . βℓ−1 1 βℓ−1 2 · · · βℓ−1 ℓ        . (15.1.4) The determinant of this matrix is Q i>j(βi −βj). If βi = βj for some i ̸= j, the determinant of (15.1.4) is zero and the matrix is singular. If the entries of the second row are distinct, Algebraic coding theory 679 the determinant is nonzero and the matrix is nonsingular. The determinant of the matrix in (15.1.3) is ℓ Y i=1 βi ! Y i>j (βi −βj). 15.1.154 Remark To complete the discussion, it is straightforward to ﬁnd the inverse of a Vander-monde matrix of the form (15.1.4) in the following manner. Deﬁne the polynomials fi(x) = ℓ−1 X j=0 fijxj = ℓ Y j=1,j̸=i x −βj βi −βj that take on the values of 0 for x = βj, j ̸= i and 1 for x = βi. The inverse of the Vandermonde matrix (15.1.4) is then the matrix [fij]. 15.1.155 Remark To consider the dimension and minimum distance of the BCH code of Deﬁnition 15.1.151, note that the parity check matrix may be written in the form H =      1 βa β2a · · · β(n−1)a 1 βa+1 β2(a+1) · · · β(n−1)(a+1) . . . . . . . . . . . . . . . 1 βa+δ−2 β2(a+δ−2) · · · β(n−1)(a+δ−2)     . The word c ∈Fn q is in the BCH code if and only if HcT = 0T . Since by the Vandermonde argument, any δ −1 columns of H are linearly independent, the code has minimum distance at least δ. In addition, as the entries of H are in Fqm, they can be expressed as column m-tuples over Fq. Therefore the rank of H over Fq is at most m(δ −1). Thus the BCH code of Deﬁnition 15.1.151 is an (n, ≥(n −m(δ −1)), ≥δ)q cyclic code. 15.1.156 Remark There are numerous techniques to improve the bound on the actual minimum distance of a BCH code, i.e., improve on the lower bound of the designed distance. Reed-Solomon (RS) codes: 15.1.157 Deﬁnition A q-ary Reed-Solomon code RSq−1,q(k) of dimension k is a BCH code of length n = q −1 and designed distance δ = n −k + 1 = q −k over Fq, i.e., for α ∈Fq a primitive element, the code has a generator polynomial of the form g(x) = a+δ−2 Y i=a (x −αi). 15.1.158 Remark The RS code deﬁned above is a (q−1, k, d = q−k)q MDS code. That the minimum distance d is exactly q−k is shown as follows. By the Singleton bound d ≤n−k+1 = q−k. Since a BCH code has minimum distance at least its designed distance δ, d ≥δ = q −k. Hence d = q −k. If 1 is not a root of g(x), then the extended narrow sense RS code, obtained by adding an overall parity check, is a (q, k, d + 1)q code and hence is also MDS. With appropriate care one can deﬁne RS codes of length less than q −1. Notice also that the code deﬁned as C = {(f(1), f(α), . . . , f(αq−2)) \| f(x) ∈Fq[x], deg f(x) ≤k −1} is the (q −1, k, q −k)q narrow sense RS code, as now outlined. The code C is certainly linear and k-dimensional as distinct polynomials of degree at most k −1 cannot produce equal 680 Handbook of Finite Fields vectors in C as otherwise the diﬀerence of these polynomials has at least q −1 > k −1 roots. Let C1 be the (q −1, k, q −k)q narrow sense RS code, which has deﬁning set T = {1, 2, . . . , q −k −1} relative to α. Since C and C1 have equal dimensions, C = C1 if C ⊆C1. Let c(x) = Pq−2 j=0 f(αj)xj ∈C for some f(x) = Pk−1 m=0 fmxm ∈Fq[x]. For i ∈T, c(αi) = q−2 X j=0 k−1 X m=0 fmαjm ! αij = k−1 X m=0 fm q−2 X j=0 α(i+m)j = k−1 X m=0 fm α(i+m)(q−1) −1 αi+m −1 noting that αi+m ̸= 1 as 1 ≤i + m ≤q −2. Since α(i+m)(q−1) = (αq−1)i+m = 1, c(αi) = 0 implying c(x) ∈C1, and so C = C1. Furthermore, in this realization, the extended code (adding a simple parity check) is realized by adding the coordinate position containing f(0). 15.1.159 Remark More generally one could deﬁne an (n, k, d = n −k + 1)q code C by choosing n distinct elements α1, α2, . . . , αn in a ﬁeld Fq and letting C = {(f(α1), f(α2), . . . , f(αn)) \| f(x) ∈Fq[x], deg f(x) ≤k −1}. The code is not, in general, cyclic but is MDS. 15.1.160 Deﬁnition A set of k column positions of a linear (n, k, d)q code is an information set if the corresponding columns of a code generator matrix are linearly independent. 15.1.161 Lemma If C is an (n, k, n −k + 1)q MDS code, then C⊥is an (n, n −k, k + 1)q MDS code. Any k columns of a generator matrix for an (n, k, d)q MDS code C are linearly independent and hence these column positions form an information set. Similarly any n −k columns of the parity check matrix for C are linearly independent and any such set of columns forms an information set for C⊥. 15.1.162 Remark If Fq has characteristic 2 (q = 2s for some s), then by ﬁxing a basis of Fq over F2 one could expand the elements of Fq in each coordinate position of each codeword of a (q −1, k, q −k)q RS code to obtain an (s(q −1), sk, ≥(n −k + 1))2 code that is eﬀective in correcting burst errors. This is simply a concatenated code with trivial inner code. 15.1.163 Remark It is easy to see that a BCH code can be viewed as a subﬁeld subcode of an RS code. 15.1.164 Remark [304, 1945, 2484] In order to study the structure of other codes, the notion of a generalized RS (GRS) code is deﬁned. Let α1, α2, . . . , αn be n ≤q distinct elements of Fq and v = (v1, v2, . . . , vn) ∈(F∗ q)n where F∗ q = Fq \ {0}. Let k ≤n. The GRS code GRSn,q(α, v, k) is deﬁned as GRSn,q(α, v, k) = {(v1f(α1), v2f(α2), . . . , vnf(αn)) \| f(x) ∈Fq[x], deg f(x) ≤k −1}. 15.1.165 Theorem The above code GRSn,q(α, v, k) is an (n, k, n−k +1)q MDS code. Further, there is a vector w ∈(F∗ q)n such that C⊥is GRSn,q(α, w, n −k), i.e., the dual of a GRS code is a GRS code, and the code GRSn,q(α, v, k)⊥is the code GRSn,q(α, w, n −k) for some w ∈(F∗ q)n. 15.1.166 Remark The proof of the above theorem is a simple generalization of the RS/BCH ar-gument. Notice that the vector of nonzero elements v has the eﬀect of multiplying each coordinate position by a nonzero element which has little eﬀect on code parameters. How-ever when subﬁeld subcodes are considered, the eﬀect can be more signiﬁcant. Algebraic coding theory 681 15.1.167 Remark A code being MDS is a strong condition. In particular, as the following theorem shows, its weight enumerator is uniquely determined. 15.1.168 Theorem If C is an (n, k, d)q MDS code, then its weight distribution {Ai, i = 0, 1, . . . , n} is given by A0 = 1, Ai = 0 for i = 1, 2, . . . , d −1, and Ai = n i (q −1) i−d X j=0 (−1)j i −1 j qi−d−j for i = d, d + 1, . . . , n. 15.1.169 Remark The existence of MDS codes is of interest. Only linear codes are considered here. As noted, there are the trivial (n, 1, n)q codes (for any alphabet of size q) and parity check (n, n −1, 2)q codes for any length n. If one considers the parity check matrix of a (q −1, k, q −k)q RS (MDS) code, it is always possible to add columns (1, 0, . . . , 0)T and (0, 0, . . . , 0, 1)T to obtain a (q + 1, k, q + 2 −k)q code for 2 ≤k ≤q −1. This “doubly extended” construction also works for BCH codes . When q is even, it is possible to obtain a triply extended (q + 2, q −1, 4)q MDS code and its dual (q + 2, 3, q)q code. 15.1.170 Conjecture It is postulated that if there is a nontrivial (n, k, n −k + 1)q MDS code over Fq, then n ≤q + 1, except when q is even and k = 3 or k = q −1 in which case n ≤q + 2. 15.1.171 Remark Reed-Solomon codes are ubiquitous in communication and storage system stan-dards of all kinds. Only a few of these are mentioned as examples: digital video broadcast-ing (DVB) (EN 300-421 (satellite) and 429 (cable)); ADSL (asymmetric digital subscriber loops) for low rate communications over telephone lines (ANSI T1.413); Internet high speed (gigabit) optical networks (ITU-T G.795 and G.709 and OC-192); wireless broadband ac-cess networks (including metropolitan (known commercially as WiMax) and local) in IEEE 802.16 (which is a family of standards covering the many types of networks in such appli-cations); deep space missions (Voyager and Mariner, among others, although more recent missions have tended to favor LDPC and turbo codes); satellite communication systems (IESS 308); two-dimensional bar codes; data storage including CDs, CD-ROMs, DRAMs, and DVDs and RAID (random arrays of inexpensive disks) systems. In each application the alphabet size and code parameters are carefully chosen for given “channel” conditions. For example ordinary music CDs use (32, 28, 5)256 and (28, 24, 5)256 codes with 8 bit symbols, obtained by shortening a (255, 251, 5)256 Reed-Solomon code, and then cross interleaved in a clever way to enable the system to withstand bursts of lengths up to 4,000 bits, caused by a scratch of up 2.5 mm on the disk surface, without replay error. The optical network standard OC-192 uses symbols from 3 to 12 bits (and RS codes of length up to 4095 symbols with a maximum of 256 parity symbols). Duadic codes: 15.1.172 Deﬁnition A vector v = (v0, v1, . . . , vn−1) ∈Fn q is even-like if Pn−1 i=0 vi = 0; otherwise v is odd-like. A code C is even-like if all of its codewords are even-like; C is odd-like if it is not even-like. 15.1.173 Remark The terms even-like and odd-like generalize the notion of even and odd weight for binary vectors; i.e., if v ∈Fn 2, v is even-like if and only if v has even weight. If C is a cyclic code of length n over Fq with deﬁning set T, then C is even-like if and only if 0 ∈T. 15.1.174 Deﬁnition Let S1 and S2 be subsets of {1, 2, . . . , n −1} that are unions of q-cyclotomic cosets modulo n. Assume further that S1 ∩S2 = ∅, S1 ∪S2 = {1, 2, . . . , n −1}, and 682 Handbook of Finite Fields that there exists b relatively prime to n such that S2 = {sb (mod n) \| s ∈S1} and S1 = {sb (mod n) \| s ∈S2}. The pair of sets S1 and S2 form a splitting of n given by b over Fq. For i ∈{1, 2}, let Di be the cyclic code of length n over Fq with deﬁning set Si. The pair D1 and D2 is a pair of odd-like duadic codes. For i ∈{1, 2}, let Ti = {0} ∪Si, and let Ci be the cyclic code of length n over Fq with deﬁning set Ti. The pair C1 and C2 is a pair of even-like duadic codes. 15.1.175 Remark Binary duadic codes were ﬁrst deﬁned in . They were later generalized to other ﬁelds in [2402, 2403, 2506, 2687]. Duadic codes include quadratic residue codes de-scribed later; quadratic residue codes exist only for prime lengths, but duadic codes can exist for composite lengths. 15.1.176 Theorem With the notation of Deﬁnition 15.1.174, the following hold. 1. Duadic codes of length n over Fq exist if and only if n is odd and q is a square modulo n. 2. For i ∈{1, 2}, Ci has dimension (n −1)/2, and there is a permutation of coordi-nates that sends C1 to C2. 3. For i ∈{1, 2}, Ci ⊆Di, Di has dimension (n + 1)/2, and there is a permutation of coordinates that sends D1 to D2. 15.1.177 Remark Part 1 of Theorem 15.1.176 can be used to ﬁnd precisely the values of n such that duadic codes of length n exist, in a manner similar to the following. Assume n = pa1 1 pa2 2 · · · par r where p1, p2, . . . , pr are distinct odd primes. Binary duadic codes exist if and only if pi ≡±1 (mod 8) for 1 ≤i ≤r. Duadic codes over F3 exist if and only if pi ≡±1 (mod 12) for 1 ≤i ≤r. Duadic codes over F4 exist if and only if n is odd. 15.1.178 Remark Much is known about the structure of duadic codes as illustrated by the next two theorems; see Chapter 6 of . 15.1.179 Theorem Let C be a cyclic (n, (n −1)/2, d)q code over Fq. Then C is self-orthogonal if and only if C is an even-like duadic code where −1 gives the splitting of n over Fq. In that case, if C is one of the duadic pair C1 and C2, C⊥ i = Di for i ∈{1, 2}. Furthermore, if n = p is a prime with p ≡−1 (mod 8), every splitting of p over F2 is given by −1 and every binary duadic code of length p is self-orthogonal. 15.1.180 Theorem Let D1 and D2 be a pair of odd-like binary duadic codes of length n where the splitting is given by −1 over F2. Then for i ∈{1, 2}, 1. the weight of every even weight codeword of Di is divisible by 4, and the weight of every odd weight codeword is congruent to n (mod 4); furthermore, 2. the extended codes b Di are self-dual. In b Di if n ≡−1 (mod 8), all codewords have weights divisible by 4, and if n ≡1 (mod 8), all codewords have even weights but some codewords have weights not divisible by 4. 15.1.181 Remark Quadratic residue codes, considered next, are special cases of duadic codes. Quadratic residue (QR) codes: 15.1.182 Deﬁnition For an odd prime p deﬁne R to be the set of integers modulo p where R = {a2 ∈Zp \| a ∈Zp and a ̸≡0 (mod p)}. The set R is the set of quadratic residues modulo p. The set N of nonzero elements of Zp that are not in R are the quadratic nonresidues modulo p. Algebraic coding theory 683 15.1.183 Remark Clearly \|R\| = \|N \| = (p −1)/2. The sets observe a parity of sorts since a residue times a residue modulo p is a residue, a residue times a nonresidue is a nonresidue, and a nonresidue times a nonresidue is a residue. This implies that, whenever q is a quadratic residue modulo p, the pair of sets R and N form a splitting of p given by any nonresidue over Fq. For ν a primitive element of Zp (i.e., ν is a generator of the multiplicative group Z∗ p = Zp \ {0}), it is clear that R = {ν2i \| i = 1, 2, . . . , (p −1)/2}, independent of the primitive element chosen. The theory of quadratic residues is a fundamental part of number theory. 15.1.184 Deﬁnition [311, 2405] Let p be an odd prime and q a prime power with gcd(p, q) = 1. Assume further that q(p−1)/2 ≡1 (mod p); hence q is a quadratic residue modulo p. Let R and N be the quadratic residues and quadratic nonresidues modulo p. Relative to a primitive p-th root of unity β in some extension ﬁeld of Fq, denote by DR and DN the (p, (p + 1)/2, d)q odd-like duadic codes over Fq with deﬁning sets R and N, respectively. Denote by CR and CN the (p, (p −1)/2, d′)q even-like duadic codes over Fq with deﬁning sets {0} ∪R and {0} ∪N, respectively. The four codes DR, DN , CR, and CN are quadratic residue (QR) codes. 15.1.185 Remark Much is known about QR codes and their relatives, their minimum distances, and automorphism groups. As noted previously, the only two perfect codes with minimum distance greater than 3 are cosets of two linear Golay codes, which are quadratic residue codes treated in the next two examples. 15.1.186 Example The binary (23, 12, 7)2 Golay code [2405, 2849]: Let p = 23 and q = 2, and note that 2 is a quadratic residue modulo 23 as 52 ≡2 (mod 23). The smallest extension ﬁeld of F2 that contains a primitive 23-rd root of unity is F211. If α is a primitive element of F211, then β = α89 is a primitive 23-rd root of unity. With the appropriate choice of primitive element, the polynomial gR(x) = Y r∈R (x −βr) = x11 + x9 + x7 + x6 + x5 + x + 1 ∈F2[x] generates the (23, 12, d)2 quadratic residue code DR. Theorem 15.1.180 implies d ≡3 (mod 4), and the minimum distance d can be determined to be 7. Notice that since P3 i=0 23 i = 211, this code is perfect. 15.1.187 Example The ternary (11, 6, 5)3 Golay code [2405, 2849]: For p = 11 and q = 3, note that 62 ≡3 (mod 11), and hence 3 is a quadratic residue modulo 11. The smallest extension ﬁeld of F3 containing a primitive 11-th root of unity is F35, and if α is a primitive root in F35, then β = α22 is a primitive 11-th root of unity. With the appropriate choice of primitive element, gR(x) = Y r∈R (x −βr) = x5 + x4 + 2x3 + x2 + 2 ∈F3[x] is a generator polynomial of the (11, 6, d)3 quadratic residue code DR; it can be shown that d = 5. Since P2 i=0 11 i 2i = 35, this code is also perfect. In fact, as noted earlier, this code and the binary (23, 12, 7)2 Golay code are the only two possible perfect linear codes with minimum distance greater than 3. This ternary perfect code was actually discovered before Golay by Virtakallio in connection with a football pool problem. 15.1.188 Remark It is interesting to note that the Golay codes were discovered very early in the history of coding and took only half a page in the Proceedings of the Institute of Radio Engineers, (IRE, precursor to the IEEE), appearing prior to the paper of Hamming 684 Handbook of Finite Fields . The original paper of Shannon actually contained an example of a Hamming code, prior to the appearance of the Hamming paper. The parameters of the Golay codes were found by examining Pascal’s triangle for sequential summations along a row that add to an appropriate power. Once the required relationship was found, Golay was able to ﬁnd generator matrices for the two (linear) codes. He also noted that 2 X i=0 90 i = 212 raising the possibility of a (90, 78, 5)2 perfect binary code, but no such code can exist [2810, 2849]. Alternant codes: 15.1.189 Remark Alternant and GRS codes bear a similar relationship as BCH and RS codes in that a BCH code of block length qm −1 over Fq is a subﬁeld subcode of an RS code of this length over Fqm. 15.1.190 Deﬁnition An alternant code An,q(α, v) is the subﬁeld subcode of GRSn,qm(α, v, k), i.e., An,q(α, v) = GRSn,qm(α, v, k) \|sf qm\|q . 15.1.191 Lemma [1991, 2484] If An,q(α, v) = GRSn,qm(α, v, k) \|sf qm\|q is an (n, k′, d′)q code, then k′ ≤k and d′ ≥n −k + 1. 15.1.192 Remark The class of alternant codes is quite large, containing all narrow sense BCH and Goppa codes. The duals of alternant codes are also of interest for which we have the following diagram : GRSn,qm(α, v, k) dual − − − − → GRSn,qm(α, w, n −k) subﬁeld   y   ytrace An,q(α, v) dual − − − − → An,q(α, v)⊥ = GRSn,qm(α, v, k)sf qm\|q = GRSn,qm(α, w, n −k)tr qm\|q . Goppa codes: 15.1.193 Remark [1319, 1320, 2047, 2849] To discuss Goppa codes it is instructive to consider a natural transition from BCH codes in the following manner. For a variable x and α a nonzero element in a ﬁeld, note that 1 1 −α−1x = 1 + α−1x + α−2x2 + · · · + α−(2t−1)x2t−1 + α−2tx2t + α−(2t+1)x2t+1 + · · · as can be veriﬁed by multiplying each side by (1−α−1x). The eﬀect of taking this equation modulo x2t is to truncate the series on the right hand side to terms of degree 2t and higher. Consider the primitive narrow sense BCH code of length n = qm −1 and designed distance 2t + 1 with deﬁning set C1 ∪C2 ∪· · · ∪C2t relative to α, a primitive element of Fqm. For convenience denote the nonzero ﬁeld elements αi as α−1 i . Then c = (c0, c1, . . . , cn−1) is a codeword of the BCH code if and only if n−1 X i=0 ciα−j i = 0 for j = 1, 2, . . . , 2t. Algebraic coding theory 685 For a received word r = (r0, r1, . . . , rn−1), the sum of a codeword and error word (presumed of weight not more than t), deﬁne the syndromes as Sj = n−1 X i=0 riα−j i for j = 1, 2, . . . , 2t, and deﬁne the syndrome polynomial as P2t j=1 Sjxj−1. Note that in the absence of errors (the received word is a codeword), this is the zero polynomial. However this polynomial can be expressed as S(x) = 2t X j=1 Sjxj−1 = 2t X j=1 xj−1 n−1 X i=0 riα−j i ! = n−1 X i=0 ri   2t X j=1 α−j i xj−1   = n−1 X i=0 ri 1 αi 2t X j=1 (α−1 i x)j−1 = − n−1 X i=0 ri x −αi (mod x2t). It follows that c = (c0, c1, . . . , cn−1) is a codeword if and only if n−1 X i=0 ci x −αi ≡0 (mod x2t). In this guise the transition from BCH to Goppa codes is more intuitive. 15.1.194 Deﬁnition [1945, 2047, 2849] Let g(x) ∈Fqm[x] be monic and let L = {α0, α1, . . . , αn−1} be a subset of Fqm such that g(αi) ̸= 0 for 0 ≤i ≤n −1. Then (c0, c1, . . . , cn−1) ∈Fn q is in the Goppa code Gn,q(L, g) if and only if n−1 X i=0 ci x −αi ≡0 (mod g(x)). The Goppa code is more often denoted Γ(L, g) in the literature, but the notation used here is consistent with our earlier code designations. 15.1.195 Remark If the Goppa polynomial is irreducible over Fqm, the code is called irreducible. By choosing the Goppa polynomial as g(x) = xd−1 and L as the set of n-th roots of unity in Fqm, the Goppa code is a BCH code with designed distance d. 15.1.196 Theorem [2047, 2849] For the Goppa code Gn,q(L, g), where \|L\| = n and deg g(x) = s, we have: 1. The minimum distance is bounded by d ≥s + 1. 2. The code dimension is bounded by k ≥n −ms. 15.1.197 Remark [1945, 2047] For any monic polynomial g(x) of degree s such that g(α) ̸= 0, it is clear from previous arguments that 1 x −α ≡− 1 g(α) g(x) −g(α) x −α (mod g(x)) is a polynomial of degree less than s. (Note that this follows as x −α is a factor of the numerator since the numerator has α as a zero.) Thus c is a codeword in Gn,q(L, g) if and 686 Handbook of Finite Fields only if its inner product with 1 g(α0) g(x) −g(α0) (x −α0) , . . . , 1 g(αn−1) g(x) −g(αn−1) (x −αn−1) is zero. To simplify this expression, let g(x) = Ps i=0 gixi and substitute this into the above equation. By expanding the fractions and considering like powers of x, one obtains a matrix which can be row reduced to      h0 h1 · · · hn−1 h0α0 h1α1 · · · hn−1αn−1 . . . h0αs−1 0 h1αs−1 1 · · · hn−1αs−1 n−1      where hi = 1/g(αi) for i = 0, 1, . . . , n−1; the Goppa code Gn,q(L, g) is the set of words over Fq orthogonal to the rows of this matrix. Notice this form of parity check matrix implies the Goppa code is an alternant code. 15.1.198 Remark To show the class of Goppa codes contains asymptotically good codes, it suﬃces to show there exists a sequence of Goppa polynomials, of increasing degrees, for which the normalized rate and distance approach the Varshamov-Gilbert bound; see Theorem 15.1.130. While not constructive, such an argument shows the class of Goppa codes includes some asymptotically good codes. The issue is discussed in . 15.1.199 Remark There are many other classes of cyclic codes not discussed here. To round out the discussion above, the generalized versions of Reed-Muller codes are mentioned, as well as codes obtained from ﬁnite Euclidean and projective geometries that will be of interest in discussing majority logic decoding. Generalized Reed-Muller (GRM) codes: 15.1.200 Remark The binary RM codes are generalized in the following manner. Construct the (m+1)×(qm −1) matrix over Fq as follows. The ﬁrst row consists of all ones and is labeled v0. Among the m-tuples of the qm −1 columns in rows v1 through vm, all nonzero m-tuples over Fq occur. A simple way of viewing this is to choose row v1 as a pseudo-noise (pn) sequence over Fq, a sequence generated by a linear feedback shift register whose feedback coeﬃcients are the coeﬃcients of a primitive polynomial of degree m over Fq. Row vi+1 is row 1 shifted by i positions for 1 ≤i ≤m−1. Let Fq[X1, . . . , Xm]ν = Fq[X]ν denote the set of polynomials of degree at most ν over Fq in m variables. Deﬁne the matrix Gν,m over Fq whose ﬁrst row is v0 and whose remaining rows are generated by all monomials in Fq[X]ν acting on rows v1, v2, . . . , vm. Let GRMqm−1,q(ν, m) be the code with generator matrix Gν,m; let GRMqm,q(ν, m) be the extended code obtained from GRMqm−1,q(ν, m) by adding an overall parity check. The exponent of a monomial is limited to degree at most q −1 in any variable since xq i = xi in Fq; thus we only consider values of ν with ν ≤m(q −1). 15.1.201 Theorem [805, 1691] The parameters of GRMqm−1,q(ν, m) are n = qm −1, k = ν X t=0 m X j=0 (−1)j m j t −jq + m −1 t −jq , d = (q −s)qm−r−1 −1, where ν = r(q −1) + s, 0 ≤s < q −1, 0 ≤r ≤m. The extended code GRMqm,q(ν, m) has length and minimum distance one greater than the above, with the same dimension. The dual code to GRMqm,q(ν, m) is GRMqm,q(µ, m) where ν + µ + 1 = m(q −1). Algebraic coding theory 687 15.1.202 Remark The expression for the code dimension in the previous theorem is the number of monomials of total degree at most ν. This is also the number of ways of placing ν balls in m cells, no cell containing more than q −1 balls. When ν < q, this expression reduces to ν+m m , and the minimum distance is (q −ν)qm−1 −1. 15.1.203 Remark To show the GRM codes are cyclic requires the following notion. The radix-q weight of an integer j is deﬁned as wq(j) = j0 + j1 + · · · where j = j0 + j1q + j2q2 + · · · with 0 ≤ji < q. 15.1.204 Theorem Let α be a primitive element of Fqm. Then the code GRMqm−1,q(ν, m), for ν ≤m(q −1), is cyclic with deﬁning set relative to α given by {j \| 0 < j < qm −1, 0 < wq(j) ≤m(q −1) −ν −1}. The code is a subcode of the primitive narrow sense BCH code with designed distance (q −s)qm−r−1 −1 where ν = r(q −1) + s as in Theorem 15.1.201. 15.1.205 Remark To discuss projective GRM codes denote by Fq[X0, . . . , Xm]0 ν = Fq[X]0 ν the set of homogeneous polynomials in m + 1 variables of degree at most ν where ν ≤(m + 1)(q −1). (The diﬀerence in the use of X, compared to the nonprojective case, is resolved by context.) There are πm+1,q projective (m + 1)-tuples, Pm q , which coordinatize the positions of the projective code. The projective codes are deﬁned as PGRMπm+1,q,q(ν, m) = {(f(x)) \| f(x) ∈Fq[X]0 ν, x ∈Pm q }. The basic parameters of the code are given in the following theorem. 15.1.206 Theorem [1827, 2696] The code PGRMπm+1,q,q(ν, m) has the following parameters: n = πm+1,q = qm+1 −1 q −1 , k = X t=ν (mod q−1) 0<t≤ν   m+1 X j=0 (−1)j m + 1 j t −jq + 1 t −jq   and d = (q −s)qm−r−1, where ν −1 = r(q −1) + s, 0 ≤s < q −1. In the case ν < q these expressions reduce to n = πm+1,q , k = r + m m , d = (q −r + 1)qm−1. Finite geometry codes: 15.1.207 Remark The binary (extended) RM codes have an interesting interpretation in terms of ﬁnite geometries for which some terminology is needed. Denote the n-dimensional projective geometry over Fq by PG(n, q) and the Euclidean geometry by EG(n, q). (Some authors denote EG(n, q) by AG(n, q); see Chapter 14.) 15.1.208 Remark Projective geometries are discussed ﬁrst; see also Section 14.4. Let α denote a primitive element of Fqn+1 and v = (qn+1 −1)/(q −1) = πn+1,q. Then β = αv is a primitive element of Fq. No two points αj with j = 0, 1, . . . , v −1 are Fq-multiples of each other, and hence these ﬁrst v powers of α can be taken as the points of PG(n, q). Let αj1, αj2, . . . , αjm+1, m ≥1, be m + 1 linearly independent points in PG(n, q), i.e., there is no linear relationship between them over Fq. The points {aj1αj1 + aj2αj2 + · · · + ajm+1αjm+1 \| aji ∈Fq}, 688 Handbook of Finite Fields with multiples over Fq identiﬁed, give the πm+1,q = (qm+1 −1)/(q−1) points of an m-ﬂat in PG(n, q). The ﬂat is a PG(m, q). In particular, the number of points on a line in PG(n, q) is q + 1. In PG(n, q) any two lines intersect in a single point, and the number of lines through a ﬁxed point in PG(n, q) is given by (qn −1)/(q −1). To construct an EG(n, q) from a PG(n, q), subtract an (n −1)-ﬂat from PG(n, q). An (n −1)-ﬂat can be thought of as the set of points in PG(n, q) orthogonal to a given line. Alternatively the points of PG(n, q) can be divided into the two groups S1 = {(1, γ2, . . . , γn+1) \| γi ∈Fq} and S2 = {(0, η2, . . . , ηn+1) \| ηj ∈Fq}. The points of S2 form an (n −1)-ﬂat and those of S1 an EG(n, q). The points of this geometry can be represented as the elements of Fqn. The points of an m-ﬂat in EG(n, q) through a given point αi is the set of points {αi + γ1αj1 + · · · + γmαjm \| γj ∈Fq} where the αjk’s are independent. Lines can be parallel in EG(n, q). In EG(2, q) the (q +1)q lines can be divided into q + 1 parallel classes, where the lines in each class are parallel, each class containing q lines. Lines in diﬀerent classes intersect. In EG(n, q) there are qn−1(qn −1)/(q −1) lines and (qn −1)/(q −1) parallel classes, each class containing qn−1 lines and each line containing q points. 15.1.209 Remark The generator matrix G(r, m) of the r-th order binary Reed-Muller code RM2m,2(r, m) (Deﬁnitions 15.1.71 and 15.1.72) has the following interpretation. The zero-th row is the incidence vector of the Euclidean space EG(m, 2). Rows 1 through m are inter-preted as incidence vectors for (m −1)-ﬂats. The product of two such rows is the incidence vector of the intersection of two such ﬂats (an (m −2)-ﬂat) etc. One aim in pursuing this geometric view for coding is the consideration of codes derived from such ﬁnite geometries and their majority logic decoding. For the remainder of this section assume q = ps for some prime p, the characteristic of Fq, and integer s. 15.1.210 Deﬁnition [1691, 2390] The r-th order Euclidean geometry code of length qm over Fp, denoted EGqm,p(r), is the largest linear code over Fp that contains in its null space the incidence vectors of all (r + 1)-ﬂats in EG(m, q). 15.1.211 Remark Note that the incidence vectors are, by deﬁnition, binary and so the codes could be deﬁned over any ﬁnite ﬁeld. However, not choosing the ﬁeld Fp would complicate the analysis considerably. It turns out that these codes are in fact extended cyclic codes as the following theorem shows. 15.1.212 Theorem [304, 1691] Let α be a primitive element of Fqm = Fpsm. Then EGqm,p(r) is the code obtained by extending the cyclic code whose deﬁning set relative to α is {j \| 0 < j < qm −1, 0 < max 0≤i<s wq(jpi) ≤(q −1)(m −r −1)}. 15.1.213 Deﬁnition The r-th order projective geometry code, denoted PG(qm−1)/(q−1),p(r), is the largest linear code containing the incidence vectors of all r-ﬂats in its null space. 15.1.214 Theorem Let β ∈Fqm = Fpsm be a primitive (qm −1)/(q −1) root of unity. The code PG(qm−1)/(q−1),p(r) is a cyclic code whose deﬁning set relative to β is {j \| 0 < j < (qm −1)/(q −1), 0 < max 0≤i<s wq(j(q −1)pi) ≤(q −1)(m −r + 1)}. Algebraic coding theory 689 15.1.215 Remark The EG and PG codes are deﬁned as the largest codes having the appropriate ﬂats in their dual codes. In the binary case, the EG codes are extended PG codes, and the PG codes can be made cyclic. The codes were investigated extensively in [1691, 1692, 1942, 2966]. Lower bounds on their minimum distances can be obtained from the number of errors they are capable of correcting with majority logic decoding; see Section 15.1.6.6. The RM, GRM, EG, and PG codes, as well as many others, can be discussed under a very general class of polynomial codes . Justesen codes: 15.1.216 Remark There have been many attempts to explicitly construct codes for which the nor-malized rate and distance functions do not both tend to zero with increasing block length. It is known that the class of BCH codes cannot achieve this. While the class of Goppa codes can be used for such a purpose, the construction is not explicit in that it calls for the construction of a sequence of suitable polynomials which are known to exist but are not given . Justesen provided the ﬁrst explicit construction. An outline of that construction is given. Let N = 2m −1, and let α be a primitive element in F2m. Let Cm,K be the (N, K, D)2m RS code given by {(f(1), f(α), . . . , f(αN−1)) \| f(x) ∈F2m[x], deg f(x) ≤K −1}. Let C′ m,K be the (2N, K, 2D)2m code given by C′ m,K = {(a0, a1, . . . , aN−1, a0, αa1, . . . , αN−1aN−1) \| (a0, a1, . . . , aN−1) ∈Cm,K}. The Justesen code C′′ m,K is found by expanding the components of each codeword, which are in F2m, into binary m-tuples with respect to some ﬁxed basis. 15.1.217 Theorem For any given code rate R < 1/2 and given m = 1, 2, . . ., choose Km to be the smallest integer K such that K/2N ≥R where N = 2m −1. The code C′′ m,Km is linear over F2 with length n = 2mN, dimension mKm, rate Km/2N ≥R, and minimum distance dn asymptotically bounded by lim inf n→∞dn/n ≥(1 −2R)H−1(1/2) ∼0.11(1 −2R). 15.1.218 Remark The normalized rate and distance of the code for a given rate R are both clearly nonzero. The construction calls for the expansion of elements of F2m into binary m-tuples as well as a primitive element α ∈F2m. To make the procedure entirely constructive (i.e., to explicitly give such an expansion) an irreducible polynomial of each degree, as the degrees tend to inﬁnity, is needed. The issue is discussed in where it is noted that an irreducible polynomial of the form x2·3ℓ+ x3ℓ+ 1 can be used, ℓ= 1, 2, . . .. Given such an irreducible polynomial for a given degree and known order, a primitive element can be determined. 15.1.4 A spectral approach to coding 15.1.219 Remark Suppose a = (a0, a1, . . . , an−1) 7→a(x) = a0 + a1x + · · · + an−1xn−1 ∈Fq[x] and n \| (qm −1), i.e., Fqm contains a primitive n-th root of unity α. The Mattson-Solomon polynomial of a is deﬁned as A(x) = n X i=1 An−ixn−i = n X i=1 a(αi)xn−i 690 Handbook of Finite Fields and the correspondence is given by Ai = a(αn−i) ← →ai = 1 nA(αi). The utility of such an approach stems from the following theorem. 15.1.220 Theorem If the Mattson-Solomon polynomial A of a vector a has r n-th roots of unity as zeros, then the weight of a must be at least n −r. 15.1.221 Remark This theorem follows as ai = A(αi)/n for 0 ≤i < n. The theorem allows a spectral approach to coding to be taken (as for example in ) where the correspondence between weights of vectors and zeros of Mattson-Solomon polynomials can be exploited. The coeﬃcients of the Mattson-Solomon polynomial A(x) can be viewed as a discrete Fourier transform of the coeﬃcients of the original polynomial a(x) and hence enjoy many useful properties, similar to those of a discrete Fourier transform. 15.1.5 Codes and combinatorics 15.1.222 Remark The relationships between codes and certain combinatorial structures are deep and of great interest. They include connections between codes and association schemes, diﬀerence sets, ﬁnite geometries and designs, among many others. Only a basic relationship between the weight classes of certain codes as incidence vectors for designs will be noted here. The reader is referred to many chapters in the handbook by Pless et al. for a recent view of the subject and to Chapter 14 of this Handbook. 15.1.223 Deﬁnition A t-(v, k, λ) design is a pair (P, B) where P is a collection of v distinct points and B is a collection of subsets of P, called blocks, each of size k, with the property that any subset of P of size t is contained in exactly λ blocks. The number of blocks of the design is given by b = λ v t / k t . 15.1.224 Remark An interesting technique for obtaining classes of t-designs is to consider the code-words of ﬁxed weight in a binary code and to determine under what conditions they might form incidence vectors for the blocks of a t-design. Such investigations have often illumi-nated the structure of a code and produced interesting classes of designs for combinatorial use. An important tool in this investigation has been the Assmus-Mattson theorem below. While quite technical, it is straightforward to apply if suﬃcient information is known about the code and its dual. The support of a vector is the list of coordinate positions where the vector is nonzero. The codewords of some ﬁxed weight hold a design if the supports of the codewords form a t-design for some t. 15.1.225 Theorem Let C be an (n, k, d)q code with dual code C⊥having minimum distance d⊥. If q = 2, let w = n; for q > 2, let w be the largest integer such that w − w + q −2 q −1 < d. Deﬁne w⊥analogously for C⊥. Suppose {Ai} and {Bi} are the weight distributions for C and C⊥respectively. Let s be the number of i with Bi ̸= 0 for 0 < i ≤n−t for some integer t. Suppose t < d and s ≤d−t. Then the codewords of weight i in C hold a t-design provided Ai ̸= 0 and d ≤i ≤w. The words of weight i in C⊥hold a t-design provided Bi ̸= 0 and d⊥≤i ≤min{n −t, w⊥}. Algebraic coding theory 691 15.1.226 Example The extended binary (24, 12, 8)2 Golay code (the extension of the quadratic residue (23, 12, 7)2 code) has weight distribution {A0 = 1, A8 = 759, A12 = 2576, A16 = 759, A24 = 1} and is self-dual. For t = 5, the number of nonzero weights less than 24−5 = 19 is s = 3. Thus the conditions and each weight class of the code satisﬁes the theorem, i.e., each weight class of the code holds a 5-design. In particular the 759 codewords of weight 8 hold a Steiner design (λ = 1), in fact a 5-(24, 8, 1) design, one of the more interesting designs known. This design is also related to the Leech lattice which leads to a particularly dense sphere packing in 24-dimensional Euclidean space. This connection is explored in detail in . 15.1.227 Example In a similar manner, the extended ternary (12, 6, 6)3 Golay code is of interest. This code has nonzero codewords only of weights 6, 9, and 12, and is self-dual. Taking t = 5, the number of nonzero code weights less than 12 −5 = 7 is s = 1 and so the weight classes of the code each hold 5-designs. This code has 264 codewords of weight 6, 440 of weight 9 and 24 of weight 12. 15.1.228 Remark The above discussion suggests that self-dual codes with few weights and large distance might be of interest in terms of applying the Assmus-Mattson theorem and such has been the case. Particular interest is in the case of binary and ternary self-dual codes. If C is a binary self-dual code, its length is even and every codeword has even weight. If C is a ternary self-orthogonal code, every weight is divisible by 3. A binary code is even if every codeword has even weight. A binary code is doubly-even if every weight is divisible by 4. Formally self-dual codes are those for which WC(x, y) = WC⊥(x, y) and all self-dual codes are formally self-dual. Many results are known about such codes and a few are touched on here [1991, 2405]. 15.1.229 Lemma Let C be a code over Fq with every weight divisible by ∆. If q = 2 and ∆= 4 or if q = 3 and ∆= 3, then C is self-orthogonal. 15.1.230 Lemma Binary self-dual doubly-even codes of length n exist if and only if 8 \| n. Ternary self-dual codes exist if and only if 4 \| n. 15.1.231 Theorem Let C be an (n, n/2, d)q code for q = 2 or q = 3. 1. If q = 2 and C is formally self-dual and even, then d ≤2⌊n/8⌋+ 2. 2. If q = 2 and C is self-dual and doubly-even, then d ≤4⌊n/24⌋+ 4. 3. If q = 3 and C is self-dual, then d ≤3⌊n/12⌋+ 3. 15.1.232 Remark Codes which achieve the upper bounds in the above theorem are referred to as extremal and often have weight classes of codewords that hold designs. The binary and ternary Golay codes are extremal. The existence of other extremal codes has long been a matter of interest. 15.1.233 Remark As noted, the work of Delsarte [801, 806] has been inﬂuential in the relationship between codes and combinatorics, particularly t-designs, orthogonal arrays and graphs. The result below is but one such example. 15.1.234 Theorem Let C be a binary code of minimum distance d and external distance s′, with d ≥s′. Then C is distance invariant and the codewords of a given weight hold a t-design with t = d −s′. 692 Handbook of Finite Fields 15.1.6 Decoding 15.1.6.1 Decoding BCH codes 15.1.235 Remark Decoding a primitive narrow sense BCH code is considered. The modiﬁcations needed for a nonnarrow sense or nonprimitive code, as well as for RS and other cyclic codes, will be clear. Also, in most cases, such errors-only decoding algorithms can be extended to errors-and-erasures algorithms. Assume the code has length n = qm −1 and designed distance d = 2t+1 and every codeword polynomial has roots αi for i = 1, 2, . . . , 2t where α is a primitive element of Fqm. The development below is fairly standard [231, 304, 1329, 1943, 2136, 2390, 2484]. It is convenient to think of codewords and received words as polynomials. Assume a codeword polynomial c(x) is sent and the polynomial r(x) = c(x) + e(x) is received, where e(x) is the error polynomial, assumed to be of weight at most t. If more than t errors occur, the algorithm may fail. For example, in the extreme case, if a certain set of d errors occurs, it may happen that an incorrect codeword is received and the algorithm will assume no errors occurred in transmission. 15.1.6.2 The Peterson-Gorenstein-Zierler decoder 15.1.236 Remark The algorithm discussed in this section was ﬁrst described by Peterson for binary codes and by Gorenstein and Zierler for nonbinary codes. Denote the i-th position of a codeword by αi for i = 0, 1, . . . , qm −2. Assume the actual number of errors that occurred (unknown) is ν ≤t = ⌊(d −1)/2⌋. Let the errors be in coordinate positions Xi = αji with error values Yi for i = 1, 2, . . . , ν. Noting that c(αj) = 0 for j = 1, 2, . . . , 2t, the information of value in the received codeword is from its syndromes evaluated as Sj = r(αj) = c(αj) + e(αj) = e(αj) = ν X i=1 YiXj i for j = 1, 2, . . . , 2t. In matrix form this is      Y1X1 + Y2X2 + · · · + YνXν Y1X2 1 + Y2X2 2 + · · · + YνX2 ν . . . Y1X2t 1 + Y2X2t 2 + · · · + YνX2t ν     =      S1 S2 . . . S2t     . (15.1.5) Denote the error locator polynomial by σ(x) = ν Y i=1 (1 −Xix) = 1 + σ1x + · · · + σν−1xν−1 + σνxν, i.e., the errors occur at the inverses of the zeros of σ(x). It follows that Sj+ν + σ1Sj+ν−1 + · · · + σνSj = 0 for j = 1, 2, . . . , 2t −ν (15.1.6) which leads to the matrix equation      S1 S2 · · · Sν−1 Sν S2 S3 · · · Sν Sν+1 . . . Sν Sν+1 · · · S2ν−2 S2ν−1           σν σν−1 . . . σ1     =      −Sν+1 −Sν+2 . . . −S2ν     . Algebraic coding theory 693 Since ν ≤t, all syndromes are known. If the number of errors made, ν, was known, the equations could be solved for the coeﬃcients σi and error locations found from the roots of σ(x). The following lemma suggests a method to ﬁnd ν. 15.1.237 Lemma The matrix      S1 S2 · · · Sℓ S2 S3 · · · Sℓ+1 . . . Sℓ Sℓ+1 · · · S2ℓ−1      is nonsingular if ℓ≤ν and singular otherwise. 15.1.238 Remark To ﬁnd the actual number of errors that occurred in transmission, ﬁrst form the matrix in Lemma 15.1.237 for ℓ= t. If nonsingular, assume t errors occurred. If singular, set ℓto t −1 and repeat until the matrix is nonsingular. The algorithm is then: 1. Compute the syndromes Si for i = 1, 2, . . . , 2t. 2. Determine the actual number of errors that occurred in transmission from Lemma 15.1.237. 3. Find the error locations by ﬁnding the roots of σ(x) (by trying all nonzero ﬁeld elements if necessary). 4. Find the error values using (15.1.5). Several steps of this algorithm are computationally expensive, such as ﬁnding successive determinants. More eﬃcient methods are introduced next. 15.1.239 Remark The above decoding technique is easily generalized to RS, GRS, alternant, and Goppa codes. 15.1.6.3 Berlekamp-Massey decoding 15.1.240 Remark [231, 2011] Equation (15.1.6) suggests the following interpretation for determining the error locator polynomial: Given the sequence of 2t syndromes, Si for i = 1, 2, . . . , 2t, determine the linear feedback shift register of minimum length such that if S1, S2, . . . are initially loaded into it, it will generate all 2t syndromes. Clearly the feedback coeﬃcients of such a shift register will be the coeﬃcients of the error locator polynomial. An eﬃcient algorithm to determine these coeﬃcients was given in Berlekamp and this algorithm was interpreted in the above manner by Massey . The details of the development are intricate and omitted here. The following lemma from gives the ﬂavor of the argument. Denote the feedback connection polynomial at the r-th stage by σ(r)(x) and the length of the register by ℓr so that the shift register generates S1, S2, . . . , Sr. The lemma decides on the length of the minimum length register in going from the (r −1)-st stage to the r-th stage. 15.1.241 Lemma Let {ℓi, σ(i)(x)} be a sequence of minimum-length shift registers such that (ℓi, σ(i)(x)) generates the sequence up to Si, for i = 1, 2, . . . , r −1. Then if σ(r)(x) ̸= σ(r−1)(x), the length of the r-th shift register is ℓr = max{ℓr−1, r −ℓr−1}. 15.1.242 Remark Once the length of the shift register for the r-th stage is determined, an eﬃcient procedure to update the coeﬃcients of σ(r)(x) is known ([304, Theorem 7.4.1], [2390, Theo-rem 9.10], and [231, Algorithm 7.4]). The ﬁnal step of the improvements for the Berlekamp-Massey algorithm involves determining the error values, once their locations are known, due 694 Handbook of Finite Fields to Forney . Let ω(x) be the error evaluator polynomial given by ω(x) ≡S(x)σ(x) (mod x2t). (15.1.7) This is the key equation for decoding. 15.1.243 Theorem The error values may be determined by the equations ω(x) = x ν X i=1 YiXi Y j̸=i (1 −Xjx) and Yi = ω(X−1 i ) X−1 i ω′(X−1 i ) where ω′ is the formal derivative of ω. 15.1.244 Remark Once the error locations are known, ω(x) can be used to compute the error values (rather than invert a ν × ν matrix over Fq). The degree of ω(x) is less than that of σ(x). Notice that this last step can be omitted for binary codes, since the error values at error locations are 1. 15.1.6.4 Extended Euclidean algorithm decoding 15.1.245 Remark Sugiyama et al. showed how the extended Euclidean algorithm (EEA) can be used to decode Goppa (and hence RS and BCH) codes. The technique was further developed in . If F is any ﬁeld, the EEA is used to determine the greatest common divisor d(x) ∈F[x] of two polynomials a(x), b(x) ∈F[x] and to also determine two polynomials s(x), t(x) ∈F[x] such that s(x)a(x) + t(x)b(x) = d(x). The application of this algorithm to solving the key equation for decoding will be seen later. To discuss this, some properties of the algorithm are required. The treatment of is followed. Three sequences of polynomials are derived: {ri(x), si(x), ti(x)} with the initial condi-tions r−1(x) = a(x) s−1(x) = 1 t−1(x) = 0 r0(x) = b(x) s0(x) = 0 t0(x) = 1 where it is assumed deg a(x) ≥deg b(x). The sequence of polynomials {ri(x)} and {qi(x)} are derived via the equation ri−2(x) = qi(x)ri−1(x) + ri(x) for i ≥1, where deg ri(x) < deg ri−1(x); thus the degrees of the ri(x)’s are strictly decreasing. Using the polynomials qi(x), two auxiliary polynomial sequences are deﬁned: si(x) = si−2(x) −qi(x)si−1(x) and ti(x) = ti−2(x) −qi(x)ti−1(x). These polynomial sequences have many properties. In particular for i ≥−1 si(x)a(x) + ti(x)b(x) = ri(x). Also deg ti(x) + deg ri−1(x) = deg a(x) for all i > 0, and since the degrees of the ri(x)’s are strictly decreasing, the degrees of the ti(x)’s are strictly increasing and in particular Algebraic coding theory 695 deg ti(x) + deg ri(x) < deg a(x). If the last nonzero remainder polynomial ri(x) is at step n, i.e., rn(x) ̸= 0 but rn+1(x) = 0, then d(x) = rn(x) = sn(x)a(x) + tn(x)b(x). 15.1.246 Remark Applying the EEA to the polynomials of the key Equation (15.1.7) with a(x) = x2t and b(x) = S(x) leads to three sequences of polynomials {ri(x)}, {si(x)}, and {ti(x)}. The degrees of the ti(x)’s are increasing and those of the ri(x)’s are decreasing. Stopping the algorithm at the point where deg tj(x) ﬁrst exceeds deg rj(x) yields the polynomials σ(x) = tj(x) and ω(x) = rj(x); also sj(x)x2t + tj(x)S(x) = rj(x). Interpreting this equation modulo x2t then solves the key Equation (15.1.7). 15.1.6.5 Welch-Berlekamp decoding of GRS codes 15.1.247 Remark The decoding method for GRS codes considered in relies on forming the syndrome polynomial where the syndromes are a form of Fourier transform of the received word. Welch-Berlekamp decoding operates directly on the received word. There are many variants of the algorithm in the literature and only the basic ideas are given here. For simplicity we consider C = GRSn,q(x, 1, k + 1) where C = {(f(x1), f(x2), . . . , f(xn)) \| f(x) ∈Fq[x], deg f(x) ≤k}, with distinct code positions xi in Fq, and let t = ⌊(n −k)/2⌋. Suppose the codeword corre-sponding to the polynomial a(x) is sent, i.e., c = (c1, c2, . . . , cn) = (a(x1), a(x2), . . . , a(xn)), and the word r = (r1, r2, . . . , rn) is received. For decoding, it suﬃces to determine the code-word polynomial a(x) from r. Suppose an unknown number e ≤t of errors has occurred in transmission and ri = ci + ei for i = 1, 2, . . . , n. Two polynomials D(x), N(x) ∈Fq[x] are sought with the properties a. deg D(x) ≤t, b. deg N(x) ≤t + k −1, c. N(xi) = riD(xi) for i = 1, 2, . . . , n. (15.1.8) 15.1.248 Theorem There exist polynomials N(x), D(x) which satisfy the above conditions and can be eﬃciently computed. Furthermore, the ratio N(x)/D(x) is the unique polynomial that gives the closest codeword to r if fewer than t errors were made in transmission. 15.1.249 Remark The thinking behind this theorem is straightforward. Let E = {i \| ri ̸= ci}, e = \|E \|, i.e., the (unknown) error positions and their number. The polynomial D(x) = Y j∈E (x −xj) is the error locator polynomial, i.e., its zeros are the error locations. The polynomial cor-responding to the codeword is a(x) = N(x)/D(x). It can be shown that the ratio of any pair of solutions of the system in Part (c) of (15.1.8), under the conditions given in Part (a) for D(x) and Part (b) for N(x), gives this polynomial. Note that N(xi) = riD(xi) for i = 1, 2, . . . , n as follows. If xi ∈E, N(xi) = 0 = riD(xi) since N(x) = a(x)D(x) and D(xi) = 0; if xi / ∈E, N(xi) = a(xi)D(xi) = ciD(xi) = riD(xi). 15.1.250 Remark When the code is presented in one of its alternative forms, for example in terms of a generator polynomial, the Welch-Berlekamp equations above are slightly modiﬁed. Such an approach is given in [578, 2164]. 696 Handbook of Finite Fields 15.1.6.6 Majority logic decoding 15.1.251 Remark Majority logic decoding involves taking a majority vote on the evaluations of certain parity check equations. It is a very simple (both to discuss and implement) decoding technique that is eﬀective when the codes possess certain structure. The Euclidean and projective geometry codes are perhaps the prime examples, but by no means the only codes in this class. Early work on this subject is due to Massey . 15.1.252 Deﬁnition If a code has J check sums (equations) which each check coordinate position j and checks other coordinate positions at most once, the checks are orthogonal on position j. 15.1.253 Lemma If it is possible to construct J parity checks orthogonal on each code coordinate position, the code can correct ⌊J/2⌋errors and has minimum distance at least J + 1. 15.1.254 Deﬁnition A set of check sums is orthogonal on a set S if each check sum contains all coordinate positions of S and checks each coordinate position not in S at most once. 15.1.255 Remark If J checks, orthogonal on a set S, can be constructed, it will be possible to determine the correct value of the sum of checks on coordinate positions for S if fewer than J/2 errors occur, by majority vote. Similarly, if checks orthogonal on certain smaller subsets can be constructed, one may be able to determine correct check sums on those smaller subsets. Proceeding iteratively until the subsets are of size one will decode the code. L-step majority logic decoding then proceeds in L steps, in this manner. Decoding to the full error correcting capability of the code with majority logic decoding may not be possible for a particular code. 15.1.256 Theorem [1942, 2390, 2966] The code EGqm,p(r) can be (r+1)-step majority logic decoded up to ⌊(dML −1)/2⌋errors where dML = qm−r −1 q −1 and the minimum distance of the code is at least this large. The code PG(qm−1)/(q−1),p(r) can be r-step majority logic decoded up to ⌊(dML −1)/2⌋ errors where dML = qm−r+1 −1 q −1 + 1. 15.1.6.7 Generalized minimum distance decoding 15.1.257 Remark In a practical communication scheme, codeword symbols are modulated in a man-ner suitable for channel transmission in that the bandwidth of the scheme must be within the bandwidth assigned. The receiver typically observes the demodulated waveform for the period of transmission for a symbol and makes a decision as to the symbol transmitted. This decision often has associated with it a reliability or conﬁdence measure as to how likely the decision made is to being correct. For example if the output of a matched ﬁlter is quantized at the decision time to two levels, this is equivalent to making a hard decision. If more levels are used, it would correspond to a soft decision. Thus in a hard decision receiver this soft information is discarded and only the hard decisions on the symbols are used. This leads to a decrease in the error performance of the system and techniques to incorporate the soft information on the received symbols to be used in the error control Algebraic coding theory 697 process have been investigated. The generalized minimum distance (GMD) technique, due to Forney , is perhaps the simplest such technique. It is discussed here for binary codes only. Assume the {0, 1} code has Hamming distance d. The generalization to the nonbinary case is straightforward. 15.1.258 Remark For the binary case assume the code is over the alphabet {−1, +1} by changing 0s in the code over F2 to −1 and 1s to +1 and assume d is the minimum Hamming distance of the code. Assume further the received word is of the form y = (y0, y1, . . . , yn−1) and deﬁne the reliability word r = (r0, r1, . . . , rn−1) where ri represents the reliability of the i-th symbol. The larger ri is in magnitude (positive or negative), the more reliable the transmitted symbol is (in the {±1} code). For maximum likelihood decoding this is the log likelihood ratio of the two possibilities, i.e., ri = log P(yi \| 1) P(yi \| 0) . Finally, for some T > 0, the vector r′ = (r′ 0, r′ 1, . . . , r′ n−1) is formed according to the equations r′ i =    +1, ri ≥T, ri/T, −T ≤ri ≤T, −1, ri ≤−T. Thus r′ is the magnitude limited version of the reliability measure formed on each coordinate position of the reliability word. The Euclidean distance from r′ to any {±1} codeword c is then \| r′ \|2 −2(r′, c)+ \| c \|2 . 15.1.259 Theorem There is at most one codeword c in the binary {±1} linear code with Hamming distance d from the received word such that (r′, c) > n −d where r′ has components in the range [−1, +1]. 15.1.260 Remark In the above theorem, the case where all components of r′ are set to +1 if ri > 0 and −1 otherwise corresponds to hard decisions. One could also choose a threshold δ > 0 and set ri to +1 if ri > δ > 0 and −1 if ri < −δ < 0 and 0 otherwise. This corresponds to the case of errors-and-erasures decoding. The (n, k, d)2 code is capable of correcting s erasures and r errors if 2r + s < d and many of the errors-only algorithms can be adapted to errors-and-erasures algorithms. GMD utilizes the fact that a code is capable of correcting more erasures than errors. For the received word r′ formulated above, where −1 ≤r′ i ≤1, denote by rℓthe codeword r′ with the least reliable ℓpositions erased (i.e., the ℓpositions with smallest \| r′ i \|) and the remaining positions quantized to hard decision (either +1 or −1). 15.1.261 Theorem [1092, 2390] If (r′, c) > n −d, then at least one of the rℓsatisﬁes (rℓ, c) > n −d for ℓ= 0, 1, . . . , d −1. 15.1.262 Remark Successively increasing the number of erasures (but not beyond (d −1)/2) and using errors-and-erasures decoding will ﬁnd the codeword closest to the quantized received word r′ under the conditions noted. 698 Handbook of Finite Fields 15.1.6.8 List decoding - decoding beyond the minimum distance bound 15.1.263 Remark It is assumed that C is a GRS code, although the results are applicable to other code classes such as algebraic geometry codes. For simplicity we consider C = GRSn,q(x, 1, k + 1) where C = {(f(x1), f(x2), . . . , f(xn)) \| f(x) ∈Fq[x], deg f(x) ≤k}, with distinct code positions xi and code rate (k + 1)/n. For such a code if the number of errors made in transmission on the channel is at most e = ⌊(d−1)/2⌋, the bounded distance decoding algorithm produces the unique correct codeword from the received word r. A list ℓdecoder with decoding radius τ produces a list of at most ℓcodewords for any received word r ∈Fn q within a radius τ of r. The decoding is successful if the transmitted codeword is in the list and fails otherwise. The relationship between the decoding radius τ and the size of the list is of interest. 15.1.264 Remark The list decoding problem is closely related to the following polynomial interpo-lation problem ; for the Lagrange Interpolation Formula, see Theorem 2.1.131. 15.1.265 Deﬁnition The polynomial interpolation problem is deﬁned as follows. For a set of pairs of elements (xi, yi) ∈Fq × Fq, the xi distinct, and for positive integers k and τ, determine a list of all polynomials f(x) ∈Fq[x] of degree at most k such that \| {i \| f(xi) = yi} \| ≥τ. 15.1.266 Remark In other words, the desire is to ﬁnd the set of all polynomials of degree at most k for which f(xi) = yi in at least τ out of the n places. The relationship of this problem to the list decoding problem is immediate. 15.1.267 Remark The technique of Sudan is to determine a nonzero bivariate polynomial Q(x, y), of a certain degree, that is zero on the set {(xi, yi) \| i = 1, 2, . . . , n} and show that factors of this polynomial of the form (y−f(x)) with deg f(x) ≤k correspond to codewords within the required distance of the received word. In the above formulation we would like τ to be as small as possible (the number of errors as large as possible). The concept of weighted degree of Q(x, y) is of interest: the (1, k) degree of a monomial xiyj is i + jk, and the (1, k) degree of Q(x, y) is m + ℓk where this is the maximum over all monomials xmyℓ of Q(x, y). This arises as we need the x-degree of Q(x, y) when a polynomial of degree k in x is substituted for y. 15.1.268 Remark The algorithm proceeds as follows: 1. Find a nonzero bivariate polynomial Q(x, y) of weighted degree at most m + ℓk (chosen later) such that Q(xi, yi) = 0 for i = 1, 2, . . . , n. 2. Factor Q(x, y) into irreducible factors and output all factors of the form y −f(x) with deg f(x) ≤k where f(xi) = yi for at least τ values of i. The number of variables qab in the linear system Pℓ b=0 Pm+(ℓ−b)k a=0 qabxa i yb i = 0 where i = 1, 2, . . . , n is (m + 1)(ℓ+ 1) + k ℓ+ 1 2 . If this quantity is greater than n, the system of n linear equations has a nontrivial solution since the number of unknowns exceeds the number of equations. The polynomial Q(x, y) is given by Q(x, y) = Pℓ b=0 Pm+(ℓ−b)k a=0 qabxayb. Algebraic coding theory 699 15.1.269 Lemma If Q(x, y) is as in Part 1 above and f(x) is such that deg f(x) ≤k with f(xi) = yi for at least τ values where τ > m + ℓk, then y −f(x) divides Q(x, y). 15.1.270 Remark In the above development if one chooses m = ⌈k/2⌉−1 and ℓ= p 2(n + 1)/k −1, it can be shown that τ ≥d p 2(n + 1)/k − k/2 or τ > √ 2kn. Thus if fewer than n− √ 2kn ≈n(1− √ 2R) errors are made in transmission, the transmitted codeword will be in the list. Notice this implies the results are valid only for fairly low rate codes. 15.1.271 Remark There are a number of items to be noted. 1. As τ decreases (number of errors increases), the size of the list tends to increase although not always strictly monotonically. For a given decoding radius there are estimates of the list size which are not discussed here. 2. A signiﬁcant improvement in the above bound was achieved by Guruswami and Sudan who increased the relative fraction of errors to ∼1 − √ R. This was achieved by allowing the bivariate polynomial Q(x, y) to have zeros of multiplicity m > 1 and optimizing the results over this parameter. Parvaresh and Vardy described a class of codes that could be list-decoded to this radius. 3. Deﬁne the entropy function hq(p) = −p logq(p) −(1 −p) logq(1 −p) and note this is diﬀerent than the q-ary entropy function of Deﬁnition 15.1.122. Denote a (p, L) list-decodable code as one capable of correcting a fraction p of errors with a list of size L. One can show that for code rates R ≤1 −hq(p) −ϵ there exists a (p, O(1/ϵ))-list decodable code while for R > 1−hq(p)+ϵ every (p, L)-list decodable code has L exponential in q. 4. The work is extended in [1376, 1377]. 15.1.7 Codes over Z4 15.1.272 Deﬁnition A Z4-linear code C of length n is an additive subgroup of Zn 4. Associated to C are two binary linear codes of length n, the residue code Res(C) = {µ(c) \| c ∈C}, where µ : Z4 → F2 is given by µ(z) = z (mod 2), and the torsion code Tor(C) = {b ∈Fn 2 \| 2b ∈C}. Let G : Z4 →F2 2 be deﬁned by G(0) = 00, G(1) = 01, G(2) = 11, and G(3) = 10; G is the Gray map. The binary code G(C) = {(G(c1), G(c2), . . . , G(cn)) \| (c1, c2, . . . , cn) ∈C} of length 2n is the Gray image of C; G(C) may not be linear over F2. 15.1.273 Remark The study of codes over Z4 began in earnest after the publication of . In that paper, a relationship, via the Gray map, was discovered between certain families of binary nonlinear codes and Z4-linear codes. That paper heightened interest in studying codes over Zr and eventually codes over other rings; see Section 2.1.7.7 for a discussion of Galois rings. 15.1.274 Remark Since a subspace of Fn q is an Fq-submodule of Fn q , a linear code of length n over Fq can be deﬁned as an Fq-submodule of Fn q . Analogously, a Z4-linear code of length n is in fact a Z4-submodule of Zn 4. In general, for any ring R, an R-linear code of length n is an R-submodule of Rn. 15.1.275 Remark Like a linear code over a ﬁeld, a Z4-linear code C of length n has a k ×n generator matrix G, where k is chosen to be minimal so that the Z4-span of the rows of G is the code 700 Handbook of Finite Fields C. By permuting coordinates, the generator matrix for C can be put in the form G = Ik1 A B1 + 2B2 O 2Ik2 2C , (15.1.9) where A, B1, B2, and C are matrices with entries from {0, 1}, Ik1 and Ik2 are k1 × k1 and k2 × k2 identity matrices, and O is the k2 × k1 zero matrix. The number of codewords in C, called the type of C, is 4k12k2. If C is a Z4-linear code with generator matrix (15.1.9), the generator matrices for Res(C) and Tor(C) are, respectively, GRes = Ik1 A B1 and GTor = Ik1 A B1 O Ik2 C . So Res(C) ⊆Tor(C), Res(C) is an (n, k1, d1)2 code for some d1, and Tor(C) is an (n, k1 + k2, d2)2 code for some d2 ≤d1. 15.1.276 Deﬁnition Analogous to the scalar product on Fn q , there is a natural scalar product on Zn 4 deﬁned by (x, y) = x1y1 + x2y2 + · · · + xnyn (mod 4), where x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) are in Zn 4. If C is a Z4-linear code of length n, deﬁne C⊥= {v ∈Zn 4 \| (v, c) = 0 for all c ∈C}, the dual of C. If C ⊆C⊥, the code C is self-orthogonal. If C = C⊥, it is self-dual. 15.1.277 Theorem If C is a self-dual Z4-linear code, then Res(C) is doubly-even and Res(C) = Tor(C)⊥. 15.1.278 Remark If C is a Z4-linear code of type 4k12k2, then C⊥is a Z4-linear code of type 4n−k1−k22k2. Furthermore, if C has generator matrix (15.1.9), C⊥has generator matrix G⊥= −(B1 + 2B2)T −CT AT CT In−k1−k2 2AT 2Ik2 O , where O is the k2 × (n −k1 −k2) zero matrix. 15.1.279 Deﬁnition The concepts of weight and distance in Fn q have analogies in Zn 4. 1. For x ∈Zn 4 and i ∈Z4, let ni(x) be the number of components of x equal to i. The Hamming weight ω(x) of x and the Hamming distance d(x, y) between x and y are deﬁned exactly as in Deﬁnition 15.1.6 for Fn q : ω(x) = n1(x) + n2(x) + n3(x) and d(x, y) = ω(x −y). The Lee weight of x is ωL(x) = n1(x) + 2n2(x) + n3(x), and the Lee distance between x and y is dL(x, y) = ωL(x −y). 2. Analogous to Deﬁnition 15.1.38, the Hamming weight distribution of a Z4-linear code C of length n is the sequence {A0, A1, . . . , An} where Ai is the number of codewords in C of Hamming weight i. The Lee weight distribution of C is the sequence {L0, L1, . . . , L2n} where Li is the number of codewords in C of Lee weight i; note that Lee weights in Zn 4 range from 0 to 2n. The minimum Hamming, respectively Lee, weight of C is the smallest Hamming, respectively Lee, weight of a nonzero codeword of C. Algebraic coding theory 701 15.1.280 Theorem The following hold. 1. The Gray map G is a distance preserving map from Zn 4 with Lee distance to F2n 2 with Hamming distance. 2. If C is a Z4-linear code, then G(C) is a distance invariant binary code. 3. If C is a Z4-linear code, then the weight distribution of the binary code G(C) is the same as the Lee weight distribution of C. 4. If C is a Z4-linear code, then G(C) is linear if and only if whenever v and w are in C, so is 2(v ∗w), where v ∗w is the componentwise product of v and w in Zn 4. 15.1.281 Example Let o8 be the Z4-linear code, called the octacode, with generator matrix G =     1 0 0 0 3 1 2 1 0 1 0 0 1 2 3 1 0 0 1 0 3 3 3 2 0 0 0 1 2 3 1 1    . The octacode is self-dual, has type 44, and has minimum Lee weight 6. Res(o8) = Tor(o8) is equivalent to the [8, 4, 4] extended binary Hamming code. By Part 3 of Theorem 15.1.280, the Gray image G(o8) is a (16, 256, 6)2 code. By [1093, 2298, 2689], G(o8) is the Nordstrom-Robinson code, a nonlinear binary code that is the largest possible binary code of length 16 and minimum Hamming distance 6. This (16, 256, 6)2 code is the extension, using an overall parity check, of a (15, 256, 5)2 code originally deﬁned by Nordstrom and Robinson in . The latter code consisted of length 15 binary vectors with the ﬁrst 8 coordinate positions arbitrary and the last 7 positions Boolean combinations of the ﬁrst 8 coordinates. 15.1.282 Remark In 1968, Preparata deﬁned a family of nonlinear (2m+1, 22m+1−2m−2, 6)2 binary codes when m is odd. These codes have twice as many codewords as a linear (2m+1, 2m+1 −2m −3, 6)2 extended binary BCH code. Each Preparata code has the largest number of codewords of any binary code of its length 2m+1 and minimum distance 6; see Chapter 17, Section 3 of . These codes lie between RM2m+1,2(m −2, m + 1) and RM2m+1,2(m −1, m + 1). In 1972, Kerdock deﬁned a family of nonlinear (2m+1, 4m+1, 2m −2(m−1)/2)2 binary codes when m is odd. These codes lie between RM2m+1,2(1, m + 1) and RM2m+1,2(2, m + 1). Amazingly, the weight distribution of the (2m+1, 22m+1−2m−2, 6)2 Preparata code is the MacWilliams transform of the weight distribu-tion of the (2m+1, 4m+1, 2m −2(m−1)/2)2 Kerdock code. When m = 3, both codes are equiv-alent to the Nordstrom–Robinson code of Example 15.1.281. In , an extended cyclic Z4-linear code of length 2m and type 4m+1, denoted b K(m+1), was deﬁned; the Gray image G( b K(m+1)) is equivalent, by permuting coordinates, to the (2m+1, 4m+1, 2m−2(m−1)/2)2 bi-nary code deﬁned by Kerdock. The Lee weight distribution {L0, L1, . . . , L2m+1} of b K(m+1), and hence the weight distribution of G( b K(m + 1)), is Li =        1 if i = 0 or i = 2m+1, 2m+1(2m −1) if i = 2m ± 2(m−1)/2, 2m+2 −2 if i = 2m, 0 otherwise. Letting P(m + 1) = b K(m + 1)⊥, the Gray image G(P(m + 1)), although generally not the same as the original Preparata code, has the same weight distribution as the (2m+1, 22m+1−2m−2, 6)2 binary Preparata code; this weight distribution can be obtained by using the MacWilliams transform on the weight distribution of G( b K(m + 1)). Preparata’s original construction begins with a single-error correcting binary BCH code and a double-error correcting subcode of this code. A linear code is created using a variation of the 702 Handbook of Finite Fields construction in Lemma 15.1.93, and then cosets of this code are adjoined to give a (2m+1 −1, 22m+1−2m−2, 5)2 code for m odd. Adding an overall parity check yields a (2m+1, 22m+1−2m−2, 6)2 code. There is a similar connection between the nonlinear binary codes of minimum distance 8 of Goethals and the nonlinear binary codes of high minimum distance of Delsarte and Goethals . 15.1.8 Conclusion 15.1.283 Remark This section has outlined certain aspects of algebraic coding theory including much of the early work in the subject. Two books [233, 305] compile and comment on initial papers from the ﬁrst 25 years of coding theory that greatly inﬂuenced the development of the discipline. The references cited in this section found in include [359, 1090, 1092, 1292, 1319, 1320, 1329, 1408, 1516, 1638, 2035, 2298, 2447, 2811]. References cited in this section found in include [359, 1292, 1329, 1408, 1516, 1638, 1942, 2011, 2035, 2298, 2427, 2447, 2855]. See Also Chapter 3 Irreducible polynomials are essential for the construction of generator polynomials for cyclic codes. Chapter 6 Character sums are used for various codes including perfect codes. §6.3.3.3 Examines the distance properties of the duals of BCH codes. §6.3.4 Kloosterman sums are used to examine the distance properties of codes. §9.1.7 Discusses the application of Boolean functions to codes including the linear Reed-Muller codes and the nonlinear Kerdock codes. §9.2.6 Examines almost perfect nonlinear (APN) and almost bent (AB) functions from a coding perspective. §9.3.8 Relates bent functions to Kerdock codes. §10.1 The Mattson-Solomon polynomial can be viewed as a discrete Fourier transform. The Gauss sum is used in the study of quadratic and generalized quadratic residue codes. §10.2 Duals of binary Hamming codes are generated by a primitive polynomial. The nonzero codewords in such a code form LFSR sequences. §11.1, §11.2 The computational techniques discussed in these subsections are useful for code implementations. Chapter 12 Is useful in the study of algebraic geometry codes. §14.4 Projective and aﬃne geometries are strongly related to RM, GRM, and polynomial codes as well as majority logic decodable codes. §14.5 Block designs can arise as the supports of codewords of ﬁxed weight in codes, particularly those satisfying the Assmus-Mattson Theorem. §14.7.1 Association schemes can be used to connect the distance distribution of a code and the dual distribution. §17.2.3 This material is useful in discussing quantum error-correcting codes. §17.3 Includes coding theory found in engineering applications. Gr¨ obner bases are useful in the theory of algebraic geometry codes. Describes the essential properties of concatenated codes. Develops the theory of Goppa and algebraic geometry codes. For a proof of the classiﬁcation of perfect codes. Algebraic coding theory 703 References Cited: [13, 142, 231, 233, 304, 305, 311, 359, 578, 801, 802, 804, 805, 806, 936, 1090, 1092, 1093, 1164, 1288, 1292, 1319, 1320, 1329, 1376, 1377, 1408, 1409, 1516, 1522, 1525, 1530, 1558, 1638, 1639, 1691, 1692, 1728, 1827, 1901, 1942, 1943, 1945, 1991, 2010, 2011, 2035, 2047, 2050, 2136, 2164, 2298, 2361, 2389, 2390, 2402, 2403, 2405, 2427, 2447, 2484, 2501, 2506, 2511, 2608, 2687, 2689, 2696, 2739, 2741, 2805, 2810, 2811, 2849, 2855, 2965, 2966, 2999] 15.2 Algebraic-geometry codes Harald Niederreiter, KFUPM 15.2.1 Classical algebraic-geometry codes 15.2.1 Remark Algebraic-geometry codes constitute a powerful family of codes which were intro-duced in their classical form by Goppa in the years 1977 to 1982 [1321, 1322, 1323]. Goppa used the language of algebraic curves over ﬁnite ﬁelds to deﬁne algebraic-geometry codes. Modern expositions prefer a description in terms of algebraic function ﬁelds over ﬁnite ﬁelds. In order to be consistent with the recent literature, we follow this practice in the present section. 15.2.2 Remark We follow the terminology for algebraic function ﬁelds in Chapter 12. Let F be an algebraic function ﬁeld (of one variable) with full constant ﬁeld Fq, that is, Fq is algebraically closed in F. A divisor of F is a ﬁnite Z-linear combination of places of F. If P denotes the set of all places of F, then a divisor G of F can be uniquely written in the form G = X P ∈P mP P, where mP ∈Z for all P ∈P and mP ̸= 0 for only ﬁnitely many P ∈P. The support of G is the set of all P ∈P with mP ̸= 0. Consequently, the support of G is a ﬁnite set. The degree deg(G) of G is deﬁned by deg(G) = X P ∈P mP deg(P), where deg(P) denotes the degree of the place P; see Section 12.1. The divisor G is positive (written G ≥0) if mP ≥0 for all P ∈P. The divisors of F form an abelian group with respect to addition, where two divisors of F are added by adding the corresponding integer coeﬃcients in the above unique representation of divisors. 15.2.3 Remark A basic object in the construction of an algebraic-geometry code is a Riemann-Roch space. For any f ∈F ∗, the principal divisor div(f) of f is given by div(f) = X P ∈P νP (f) P, where νP denotes the normalized discrete valuation of F associated with the place P. Note that the image of νP as a map is Z ∪{∞}. If f ∈F ∗is given, then νP (f) ̸= 0 for at most 704 Handbook of Finite Fields ﬁnitely many P ∈P, and so div(f) is indeed a divisor of F. Given an arbitrary divisor G of F, we now deﬁne the Riemann-Roch space L(G) = {f ∈F ∗: div(f) + G ≥0} ∪{0}. It is an important fact that L(G) is a ﬁnite-dimensional vector space over Fq. We write ℓ(G) for the dimension of this vector space. The Riemann-Roch theorem (Chapter 12) provides important information on ℓ(G). 15.2.4 Deﬁnition Let n be a positive integer and let q be an arbitrary prime power. Let F be an algebraic function ﬁeld with full constant ﬁeld Fq, genus g, and at least n rational places. Choose n distinct rational places P1, . . . , Pn of F and a divisor G of F such that none of the Pi, 1 ≤i ≤n, is in the support of G. The algebraic-geometry code C(P1, . . . , Pn; G) is deﬁned as the image of the Fq-linear map ψ : L(G) →Fn q given by ψ(f) = (f(P1), . . . , f(Pn)) for all f ∈L(G). 15.2.5 Remark Note that νPi(f) ≥0 for 1 ≤i ≤n and all f ∈L(G), since Pi is not in the support of G. Thus, f belongs to the valuation ring OPi of Pi for 1 ≤i ≤n, and so the residue class f(Pi) of f, that is, the image of f under the residue class map of the place Pi, is well deﬁned. Since Pi is a rational place, that is, a place of degree 1, we can identify f(Pi) with an element of Fq. Thus, C(P1, . . . , Pn; G) is indeed a subset of Fn q . 15.2.6 Theorem With the notation and assumptions in Deﬁnition 15.2.4, suppose that the divisor G of F satisﬁes also g ≤deg(G) < n. Then the algebraic-geometry code C(P1, . . . , Pn; G) is a linear code over Fq with length n, dimension k = ℓ(G) ≥deg(G) + 1 −g, and minimum distance d ≥n −deg(G). Moreover, if deg(G) ≥2g −1, then k = deg(G) + 1 −g. 15.2.7 Remark The condition deg(G) < n in Theorem 15.2.6 guarantees that the map ψ in Deﬁnition 15.2.4 is injective. Therefore k = ℓ(G) and the remaining information on k is obtained from the Riemann-Roch theorem (Chapter 12). 15.2.8 Remark Theorem 15.2.6 implies that k + d ≥n + 1 −g. This should be compared with the Singleton bound k + d ≤n + 1 in Section 15.1. Thus, the genus g of F controls, in a sense, the deviation of k + d from the Singleton bound. 15.2.9 Example Let F = Fq(x) be the rational function ﬁeld over Fq. Let Fq = {b1, . . . , bq} and for 1 ≤i ≤q let Pi be the rational place x −bi of Fq(x). Put G = (k −1)P∞, where k is an integer with 1 ≤k ≤q and P∞denotes the inﬁnite place of Fq(x). Then C(P1, . . . , Pq; G) is a generalized Reed-Solomon code with length q, dimension k, and minimum distance q −k + 1. Thus, algebraic-geometry codes can be considered as vast generalizations of generalized Reed-Solomon codes. 15.2.10 Example Let F be the Hermitian function ﬁeld over Fq2, that is, F = Fq2(x, y) with yq + y = xq+1. Then F has genus g = (q2 −q)/2 and exactly q3 + 1 rational places. Let Q be the rational place of F lying over the inﬁnite place of Fq2(x) and let P1, . . . , Pn with n = q3 be the remaining rational places of F. Put G = mQ with an integer m satisfying q2 −q −1 ≤m < q3. Then C(P1, . . . , Pn; G) is a linear code over Fq2 with length n = q3, Algebraic coding theory 705 dimension k = m + 1 −(q2 −q)/2, and minimum distance d ≥q3 −m. Such a code is a Hermitian code. 15.2.11 Example Take q = 2 and m = 4 in Example 15.2.10, so that G = 4Q. Then the Hermitian code C(P1, . . . , P8; G) is a linear code over F4 with length 8, dimension 4, and minimum distance d ≥4. It can be shown that actually d = 4. The code C(P1, . . . , P8; G) is optimal in the sense that there is no linear code over F4 with length 8, dimension 4, and minimum distance at least 5. 15.2.12 Remark The condition in Deﬁnition 15.2.4 that none of the Pi, 1 ≤i ≤n, is in the support of G is a conventional one in the area. However, one can get rid of this condition by replacing G by the divisor G′ = G−div(u), where u ∈F is such that νPi(u) is equal to the coeﬃcient mPi of Pi in G for 1 ≤i ≤n. Such an element u exists by the approximation theorem for valuations. Note that by construction none of the Pi, 1 ≤i ≤n, is in the support of G′, but deg(G′) = deg(G) and ℓ(G′) = ℓ(G). Thus, the linear code C(P1, . . . , Pn; G′) satisﬁes the properties stated in Theorem 15.2.6. 15.2.13 Example Let F be the Hermitian function ﬁeld over F4; see Example 15.2.10. Then F has genus 1 and exactly 9 rational places. Let P1, . . . , P9 be the rational places of F and choose a divisor G of F with deg(G) = 4. In view of Remark 15.2.12, if we replace G by a suitable divisor G′, then we need not check whether the Pi, 1 ≤i ≤9, are in the support of G or not. Thus, C(P1, . . . , P9; G′) is a linear code over F4 with length 9, dimension 4, and minimum distance d ≥5. It can be shown that actually d = 5. The linear code C(P1, . . . , P9; G′) is optimal. 15.2.14 Remark The dual codes of algebraic-geometry codes can be described in terms of diﬀeren-tials and residues for algebraic function ﬁelds . 15.2.15 Remark Certain algebraic-geometry codes can be constructed without the use of algebraic geometry or algebraic function ﬁelds, but rather by using so-called order domains . 15.2.16 Remark There exist eﬃcient decoding algorithms for algebraic-geometry codes. A survey of such decoding algorithms is presented in , and for a more recent contribution see . 15.2.17 Remark Interesting codes can be derived not only from algebraic function ﬁelds over ﬁ-nite ﬁelds, or equivalently from algebraic curves over ﬁnite ﬁelds, but also from higher-dimensional algebraic varieties over ﬁnite ﬁelds. A rather general approach to the construc-tion of such codes is described in . 15.2.2 Generalized algebraic-geometry codes 15.2.18 Remark The condition g ≤deg(G) < n in Theorem 15.2.6 implies that the number N of rational places of the algebraic function ﬁeld F must be greater than the genus g of F, in order for Theorem 15.2.6 to be applicable. However, for small values of q such as q = 2 and q = 3, the condition N > g can be satisﬁed only for small values of g. Consequently, for small values of q the construction of classical algebraic-geometry codes in Deﬁnition 15.2.4 can yield good codes only for small lengths. A possible remedy is to devise constructions that use not only rational places, but also places of higher degree. Such constructions will be discussed in this subsection. 15.2.19 Remark The construction of NXL codes due to Niederreiter, Xing, and Lam uses places of arbitrary degree. We present this construction in the more general form given in Chapter 5 of . Let F be an algebraic function ﬁeld with full constant ﬁeld Fq and 706 Handbook of Finite Fields genus g. Let G1, . . . , Gr be positive divisors ̸= 0 of F with pairwise disjoint supports. Put n = Pr i=1 si, where si = deg(Gi) ≥1 for 1 ≤i ≤r, and assume that n > g. Let E be a positive divisor of F for which the support is disjoint from the support of Gi for 1 ≤i ≤r. Furthermore, let D be a divisor of F with ℓ(D) = deg(D) + 1 −g, e.g., this holds if deg(D) ≥2g −1. Assume also that 1 ≤deg(E −D) ≤n −g. We observe that ℓ(D + Gi) = ℓ(D) + si for 1 ≤i ≤r. For each i = 1, . . . , r, we choose an Fq-basis {fi,j + L(D) : 1 ≤j ≤si} of the factor space L(D+Gi)/L(D). The n-dimensional factor space L(D+Pr i=1 Gi)/L(D) has then the Fq-basis {fi,j + L(D) : 1 ≤j ≤si, 1 ≤i ≤r} which we order in a lexicographic manner. We note further that every f ∈L(D + r X i=1 Gi −E) ⊆L(D + r X i=1 Gi) has a unique representation f = r X i=1 si X j=1 ci,j fi,j + u with all ci,j ∈Fq and u ∈L(D). 15.2.20 Deﬁnition The NXL code C(G1, . . . , Gr; D, E) is deﬁned as the image of the Fq-linear map η : L(D + r X i=1 Gi −E) →Fn q given by η(f) = (c1,1, . . . , c1,s1, . . . , cr,1, . . . , cr,sr) for all f ∈L(D + Pr i=1 Gi −E), where the ci,j are as in Remark 15.2.19. 15.2.21 Theorem Assume that the hypotheses in Remark 15.2.19 are satisﬁed and put m = deg(E −D). Then the NXL code C(G1, . . . , Gr; D, E) is a linear code over Fq with length n = Pr i=1 si, dimension k = ℓ(D + r X i=1 Gi −E) ≥n −m −g + 1, and minimum distance d ≥d0, where d0 is the least cardinality of a subset R of {1, . . . , r} for which P i∈R si ≥m. Moreover, if n −m ≥2g −1, then k = n −m −g + 1. 15.2.22 Example A simple choice for the divisors G1, . . . , Gr in the construction of NXL codes is to take distinct places P1, . . . , Pr of F. Note that these places need not be rational, but can have arbitrary degrees. This special case was considered in and is used also in the present example. Let q = 3, let F be the rational function ﬁeld over F3, and put r = 13. For P1, . . . , P13 we choose four rational places, three places of degree 2, and six places of degree 3 of F. Let D be the zero divisor and E a place of degree 7 of F. Then C(P1, . . . , P13; D, E) is a linear code over F3 with length 28, dimension k = 22, and minimum distance d ≥3. Hence k + d ≥25. By comparison, the best lower bound on k + d for a classical algebraic-geometry code over F3 of length 28 is obtained by taking g = 15, and then k + d ≥14 by Remark 15.2.8. Algebraic coding theory 707 15.2.23 Remark The following construction of XNL codes is due to Xing, Niederreiter, and Lam . It is a powerful method of combining data from algebraic function ﬁelds over ﬁnite ﬁelds with (short) linear codes in order to produce a longer linear code as the output. We present the slightly more general version of XNL codes in Chapter 5 of . 15.2.24 Deﬁnition Let F be an algebraic function ﬁeld with full constant ﬁeld Fq. Let P1, . . . , Pr be distinct places of F which can have arbitrary degrees. Let G be a divisor of F such that none of the Pi, 1 ≤i ≤r, is in the support of G. For each i = 1, . . . , r, let Ci be a linear code over Fq with length ni, dimension ki ≥deg(Pi), and minimum distance di, and let φi be an injective Fq-linear map from the residue class ﬁeld of Pi into Ci. Put n = Pr i=1 ni. Then the XNL code C(P1, . . . , Pr; G; C1, . . . , Cr) is deﬁned as the image of the Fq-linear map β : L(G) →Fn q given by β(f) = (φ1(f(P1)), . . . , φr(f(Pr))) for all f ∈L(G), where on the right-hand side we use concatenation of vectors. 15.2.25 Theorem With the notation and assumptions in Deﬁnition 15.2.24, suppose that the divisor G of F satisﬁes also g ≤deg(G) < Pr i=1 deg(Pi), where g is the genus of F. Then the XNL code C(P1, . . . , Pr; G; C1, . . . , Cr) is a linear code over Fq with length n = Pr i=1 ni, dimension k = ℓ(G) ≥deg(G) + 1 −g, and minimum distance d ≥d′, where d′ is the minimum of P i∈M di taken over all subsets M of {1, . . . , r} for which P i∈M deg(Pi) ≤deg(G), with M denoting the complement of M in {1, . . . , r}. Moreover, if deg(G) ≥2g −1, then k = deg(G) + 1 −g. 15.2.26 Corollary If in addition deg(Pi) ≥di for 1 ≤i ≤r, then the minimum distance d of the XNL code C(P1, . . . , Pr; G; C1, . . . , Cr) satisﬁes d ≥ r X i=1 di −deg(G). 15.2.27 Remark If P1, . . . , Pr are distinct rational places of F and for each i = 1, . . . , r we choose Ci to be the trivial linear code over Fq with ni = ki = di = 1 and φi the identity map on Fq, then the construction of XNL codes reduces to that of classical algebraic-geometry codes. Theorem 15.2.6 is thus a special case of Theorem 15.2.25 and Corollary 15.2.26. 15.2.28 Example Many excellent examples of XNL codes were found in and . The following simple example is typical. Let q = 2 and let F = F2(x, y) be the elliptic function ﬁeld deﬁned by y2 + y = x + x−1. We choose r = 6 and let P1, P2, P3, P4 be the four rational places and P5 and P6 the two places of degree 2 of F. The linear codes Ci have the following parameters: for 1 ≤i ≤4 we let ni = ki = di = 1 and for i = 5, 6 we let ni = 3, ki = 2, di = 2. Then for m = deg(G) = 1, . . . , 7 the corresponding XNL code is a linear code over F2 with parameters n = 10, k = m, and d ≥8 −m. The linear codes with m = 2, 3, 4 and d = 8 −m are optimal. 15.2.29 Example Let q = 3 and let F = F3(x, y) be the elliptic function ﬁeld deﬁned by y2 = x(x2 + x −1). We choose r = 9 and let P1, P2, P3, P4, P5, P6 be the six rational places and P7, P8, P9 the three places of degree 2 of F. The linear codes Ci have the following parameters: for 1 ≤i ≤6 we let ni = ki = di = 1 and for 7 ≤i ≤9 we let ni = 3, ki = 2, di = 2. Then for m = deg(G) = 1, . . . , 11 the corresponding XNL code is a linear code over F3 with parameters n = 15, k = m, and d ≥12 −m. The linear code with m = 3 and d = 9 is optimal. 708 Handbook of Finite Fields 15.2.30 Remark By the same argument as in Remark 15.2.12, the condition in Deﬁnition 15.2.24 that none of the Pi, 1 ≤i ≤r, is in the support of G can be dropped if we replace G by a suitable divisor G′. 15.2.31 Remark Decoding algorithms for generalized algebraic-geometry codes can be found in . 15.2.32 Remark As for classical algebraic-geometry codes (Remark 15.2.14), the dual codes of XNL codes can be described in terms of diﬀerentials and residues for algebraic function ﬁelds . 15.2.3 Function-ﬁeld codes 15.2.33 Remark A very general perspective on the construction of algebraic-geometry codes is that of function-ﬁeld codes. A function-ﬁeld code is a special type of subspace of an algebraic function ﬁeld over a ﬁnite ﬁeld from which linear codes can be derived. Function-ﬁeld codes were introduced in Chapter 6 of and studied in detail by Hachenberger, Niederreiter, and Xing . 15.2.34 Remark Let F be an algebraic function ﬁeld with full constant ﬁeld Fq. For a place P of F, let OP denote the valuation ring of P and let MP be the unique maximal ideal of OP . For a ﬁnite nonempty set Q of places of F, we write OQ = \ P ∈Q OP , MQ = \ P ∈Q MP . 15.2.35 Deﬁnition Let F be an algebraic function ﬁeld with full constant ﬁeld Fq and let Q be a ﬁnite nonempty set of places of F. A function-ﬁeld code (in F with respect to Q) is a nonzero ﬁnite-dimensional Fq-linear subspace V of F which satisﬁes the two conditions V ⊆OQ and V ∩MQ = {0}. 15.2.36 Theorem Let Q be a ﬁnite nonempty set of places of F and let G be a divisor of F such that ℓ(G) ≥1, none of the places in Q is in the support of G, and deg(G) < X P ∈Q deg(P). Then the Riemann-Roch space L(G) is a function-ﬁeld code in F with respect to Q. 15.2.37 Theorem Let Q = {P1, . . . , Pr} be a ﬁnite nonempty set of places of F. Let f1, . . . , fk ∈OQ be such that the k vectors (fj(P1), . . . , fj(Pr)) ∈Fq r, 1 ≤j ≤k, are linearly independent over Fq, where Fq is an algebraic closure of Fq. Then the Fq-linear subspace of F spanned by f1, . . . , fk is a function-ﬁeld code in F with respect to Q. 15.2.38 Remark For suitable Q and k in Theorem 15.2.37, appropriate elements f1, . . . , fk can be constructed by the approximation theorem for valuations. 15.2.39 Remark Any nonzero Fq-linear subspace of a function-ﬁeld code in F with respect to Q is again a function-ﬁeld code in F with respect to Q. Algebraic coding theory 709 15.2.40 Remark The most powerful method of deriving linear codes from function-ﬁeld codes is described in Deﬁnition 15.2.41 below and generalizes the construction of XNL codes in Deﬁnition 15.2.24. 15.2.41 Deﬁnition Let V be a function-ﬁeld code in F with respect to Q = {P1, . . . , Pr}, where P1, . . . , Pr are distinct places of F. For each i = 1, . . . , r, let Ci be a linear code over Fq with length ni, dimension ki ≥deg(Pi), and minimum distance di, and let φi be an injective Fq-linear map from the residue class ﬁeld of Pi into Ci. Put n = Pr i=1 ni and let the Fq-linear map γ : V →Fn q be given by γ(f) = (φ1(f(P1)), . . . , φr(f(Pr))) for all f ∈V, where on the right-hand side we use concatenation of vectors. Then the linear code CQ(V ; C1, . . . , Cr) over Fq is deﬁned as the image of V under γ. 15.2.42 Remark The following notation is convenient. For I ⊆{1, . . . , r} we write I for the com-plement of I in {1, . . . , r} and Q(I) = {Pi : i ∈I} ⊆Q. We put d′ = min I X i∈I di, where the minimum is extended over all I ⊆{1, . . . , r} for which V ∩MQ(I) ̸= {0}. The last condition is always assumed to be satisﬁed for I = {1, . . . , r}. The condition V ∩MQ = {0} in Deﬁnition 15.2.35 implies that d′ ≥1. 15.2.43 Theorem The code CQ(V ; C1, . . . , Cr) in Deﬁnition 15.2.41 is a linear code over Fq with length n, dimension k, and minimum distance d, where n = r X i=1 ni, k = dim(V ), d ≥d′, and where d′ is as in Remark 15.2.42. 15.2.44 Deﬁnition Let V be a function-ﬁeld code in F with respect to Q = {P1, . . . , Pr}. For f ∈V , the block weight of f (with respect to Q) is deﬁned to be ϑQ(f) = \|{1 ≤i ≤r : f(Pi) ̸= 0}\|. The number ϑQ(V ) = min {ϑQ(f) : f ∈V, f ̸= 0} is the minimum block weight of V (with respect to Q). 15.2.45 Theorem If the notation is arranged in such a way that d1 ≤d2 ≤· · · ≤dr, then the minimum distance d of the code CQ(V ; C1, . . . , Cr) in Deﬁnition 15.2.41 satisﬁes d ≥ ϑQ(V ) X i=1 di, where ϑQ(V ) is the minimum block weight of V . 15.2.46 Remark The codes CQ(V ; C1, . . . , Cr) derived from function-ﬁeld codes form a universal family of linear codes, in the sense that any linear code can be obtained by this construction. 710 Handbook of Finite Fields In fact, as the following theorem from Chapter 6 in shows, a special family of these codes suﬃces to represent any linear code. 15.2.47 Theorem Let C be an arbitrary linear code over Fq with length n and dimension k. Then there exists an algebraic function ﬁeld F with full constant ﬁeld Fq, a set Q of n distinct rational places of F, and a k-dimensional function-ﬁeld code V in F with respect to Q such that C is equal to the code CQ(V ; C1, . . . , Cn), where Ci is the trivial linear code over Fq of length 1 and dimension 1 for 1 ≤i ≤n. 15.2.48 Remark A stronger result than Theorem 15.2.47, according to which an even smaller family of codes can represent any linear code, was proved in . 15.2.4 Asymptotic bounds 15.2.49 Remark The asymptotic theory of codes studies the set of ordered pairs of asymptotic relative minimum distances and asymptotic information rates as the length of codes goes to ∞. This theory considers general (including nonlinear) codes as well as the special case of linear codes. Algebraic-geometry codes play a decisive role in this theory. 15.2.50 Remark We write n(C) for the length of a code C and d(C) for the minimum distance of C. Furthermore, we use logq to denote the logarithm to the base q. 15.2.51 Deﬁnition For a given prime power q, let Uq (respectively U lin q ) be the set of all ordered pairs (δ, R) ∈[0, 1]2 for which there exists a sequence C1, C2, . . . of general (respectively linear) codes over Fq such that n(Ci) →∞as i →∞and δ = lim i→∞ d(Ci) n(Ci), R = lim i→∞ logq \|Ci\| n(Ci) . 15.2.52 Proposition There exists a function αq (respectively αlin q ) on [0, 1] such that Uq = {(δ, R) : 0 ≤R ≤αq(δ), 0 ≤δ ≤1} and U lin q = {(δ, R) : 0 ≤R ≤αlin q (δ), 0 ≤δ ≤1}. 15.2.53 Proposition The functions αq and αlin q have the following properties: 1. αq(δ) ≥αlin q (δ) for 0 ≤δ ≤1; 2. αq and αlin q are nonincreasing and continuous on [0, 1]; 3. αq(0) = αlin q (0) = 1; 4. αq(δ) = αlin q (δ) = 0 for (q −1)/q ≤δ ≤1. 15.2.54 Remark Values of αq(δ) and αlin q (δ) are not known for 0 < δ < (q −1)/q. Thus, the best one can do at present from a practical point of view is to ﬁnd lower bounds on αq(δ) and αlin q (δ) for 0 < δ < (q −1)/q. Such lower bounds guarantee an asymptotic information rate R that can be achieved for a given q and a given asymptotic relative minimum distance δ. 15.2.55 Deﬁnition For 0 < δ < 1, the q-ary entropy function Hq is deﬁned by (Deﬁnition 15.1.122) Hq(δ) = δ logq(q −1) −δ logq δ −(1 −δ) logq(1 −δ). Algebraic coding theory 711 15.2.56 Remark The benchmark bound in the asymptotic theory of codes is the asymptotic Gilbert-Varshamov bound in Theorem 15.2.57 below which dates from the 1950s. 15.2.57 Theorem For any prime power q, we have αq(δ) ≥αlin q (δ) ≥RGV(q, δ) := 1 −Hq(δ) for 0 < δ < (q −1)/q. 15.2.58 Theorem For any prime power q and any real number δ with 0 < δ < (q−1)/q, there exists a sequence C1, C2, . . . of classical algebraic-geometry codes over Fq with n(Ci) →∞ as i →∞which yields αlin q (δ) ≥RGV(q, δ). 15.2.59 Remark Since no improvement on Theorem 15.2.57 was obtained for a long time, there was speculation that maybe αlin q (δ) = RGV(q, δ) for 0 < δ < (q −1)/q. However, this conjecture was disproved by the use of algebraic-geometry codes. The crucial result in this context is the TVZ bound (named after Tsfasman, Vl˘ adut ¸, and Zink ) in Theorem 15.2.60 below. For the formulation of this bound, we need the quantity A(q) = lim sup g→∞ Nq(g) g , where q is an arbitrary prime power and Nq(g) is the maximum number of rational places that an algebraic function ﬁeld with full constant ﬁeld Fq and genus g can have. We note that A(q) > 0 for all prime powers q and that A(q) = q1/2 −1 if q is a square. The proof of the TVZ bound uses appropriate sequences of classical algebraic-geometry codes for which the length goes to ∞. 15.2.60 Theorem For any prime power q, we have αq(δ) ≥αlin q (δ) ≥RTVZ(q, δ) := 1 − 1 A(q) −δ for 0 ≤δ ≤1. 15.2.61 Remark The following two theorems show that for certain suﬃciently large values of the prime power q, we can beat the asymptotic Gilbert-Varshamov bound in Theorem 15.2.57 by using suitable sequences of classical algebraic-geometry codes. Both theorems are simple consequences of the TVZ bound in Theorem 15.2.60 and information about the quantity A(q) deﬁned in Remark 15.2.59. 15.2.62 Theorem Let q ≥49 be the square of a prime power. Then there exists an open subinterval (δ1, δ2) of (0, (q −1)/q) containing (q −1)/(2q −1) such that RTVZ(q, δ) > RGV(q, δ) for all δ ∈(δ1, δ2). 15.2.63 Theorem Let q ≥343 be the cube of a prime power. Then there exists an open subinterval (δ1, δ2) of (0, (q −1)/q) containing (q −1)/(2q −1) such that RTVZ(q, δ) > RGV(q, δ) for all δ ∈(δ1, δ2). 15.2.64 Remark More generally, it was shown in that a result like Theorem 15.2.62 holds for all suﬃciently large composite nonsquare prime powers q. It is not known whether such a result holds also for all suﬃciently large primes q. 15.2.65 Remark If one considers general (i.e., not necessarily linear) codes, then global improve-ments on the TVZ bound for αq(δ) in Theorem 15.2.60 can be obtained. By a global improvement we mean an improvement for any q and δ by a positive quantity independent 712 Handbook of Finite Fields of δ. The currently best global improvement on the TVZ bound is the Niederreiter- ¨ Ozbudak bound in Theorem 15.2.66 below. 15.2.66 Theorem For any prime power q, we have αq(δ) ≥RNO(q, δ) := 1 − 1 A(q) −δ + logq 1 + 1 q3 for 0 ≤δ ≤1. 15.2.67 Remark The proof of the Niederreiter-¨ Ozbudak bound in is quite involved. A simpler proof using a new variant of the construction of algebraic-geometry codes was presented in , but this proof works only for a slightly restricted range of the parameter δ. The new construction proceeds as follows. Let n be a positive integer (which will be the length of the code) and let F be an algebraic function ﬁeld with full constant ﬁeld Fq such that F has at least n + 1 rational places. Choose n + 1 distinct rational places P0, P1, . . . , Pn of F and put Q = {P1, . . . , Pn}. Furthermore, let m, r, and s be integers with m ≥1 and 1 ≤r ≤min(n, s). We let G(Q; r, s) be the set of all positive divisors G of F of degree s such that the support of G is a subset of Q of cardinality r. Next we deﬁne S(mP0; Q; r, s) = [ G∈G(Q;r,s) {f ∈L(mP0 + G) : νP0(f) = −m}. Note that the union above is a disjoint union. A map φ : S(mP0; Q; r, s) →Fn q is now deﬁned in the following way. Take f ∈S(mP0; Q; r, s) and let G ∈G(Q; r, s) be the unique divisor such that f ∈L(mP0 + G) and νP0(f) = −m. We deﬁne φ(f) = (c1(f), . . . , cn(f)) ∈Fn q , where for i = 1, . . . , n we put ci(f) = f(Pi) if Pi is not in the support of G and ci(f) = 0 if Pi is in the support of G. The desired code C(mP0; Q; r, s) over Fq is the image of S(mP0; Q; r, s) under φ. Note that C(mP0; Q; r, s) is in general a nonlinear code. The proof of the bound αq(δ) ≥RNO(q, δ) for a restricted range of δ in is obtained by using a suitable sequence of codes of this type with increasing length. 15.2.68 Remark Recently, some local improvements on the Niederreiter-¨ Ozbudak bound in Theo-rem 15.2.66 have been obtained, that is, improvements that are valid only for a restricted range of the parameter δ (this range may depend on the value of q). We refer to the papers [2264, 2265, 3029] for these improvements, and we note that contains also improved lower bounds on αlin q (δ). An important role in this work is played by distinguished divisors in the construction of algebraic-geometry codes, that is, divisors G such that L(G −D) = {0} for a (large) ﬁnite family of positive divisors D. The point of a distinguished divisor is that it yields an improvement on the lower bound for the minimum distance in Theorem 15.2.6. See Also §15.1 For general background on coding theory. References Cited: [92, 875, 912, 1321, 1322, 1323, 1375, 1396, 1415, 1496, 1524, 2262, 2264, 2265, 2280, 2281, 2284, 2285, 2380, 2714, 2715, 2820, 2821, 3017, 3019, 3020, 3029] Algebraic coding theory 713 15.3 LDPC and Gallager codes over ﬁnite ﬁelds Ian Blake, University of British Columbia W. Cary Huﬀman, Loyola University Chicago 15.3.1 Deﬁnition A binary linear code of length n is a Gallager code provided it has an m × n parity check matrix H where every column has ﬁxed weight c and every row has weight as close to nc/m as possible. This code will be denoted Galn,2(c, H). If r = nc/m is an integer and every row of H has weight r, the Gallager code is called regular. 15.3.2 Deﬁnition A binary linear code of length n is a low density parity check (LDPC) code provided it is a regular Gallager code Galn,2(c, H). This code will be denoted by LDPCn,2(c, r, H) where c is the weight of each column of H and r is the weight of each row of H. 15.3.3 Remark Note that the m × n matrix H used to deﬁne a Gallager or LDPC code C is not assumed to have independent rows and so is technically not a parity check matrix in the sense often used. However, H can be row reduced and the zero rows removed to form a parity check matrix for C with independent rows. Thus the dimension of C is at least n−m. Generally the column and row weights are chosen to be relatively small compared to n, and thus the density of 1s in H is low. A natural generalization for Gallager and LDPC codes is to allow the row weights and column weights of H to both vary, in some controlled manner. The resulting codes are sometimes termed irregular LDPC codes. Unless speciﬁc parameters are given, for the remainder of this section, the term “LDPC” will refer to either regular or irregular LDPC codes. 15.3.4 Remark Gallager and LDPC codes were developed by R. G. Gallager in [1162, 1163]. Gallager also presented two iterative decoding algorithms designed to decode these codes of long length, several thousand bits for example. These algorithms are presented in Remarks 15.3.8 and 15.3.12. 15.3.5 Remark The reason Gallager codes are useful is because they achieve close to optimal properties. Roughly speaking, for any c ≥3 and any λ > 1, there exist Gallager codes of long enough length and rates up to 1 −1/λ such that virtually error-free transmission occurs; the speciﬁc codes that achieve this property are dependent upon the characteristics of the communication channel being used. Additionally, for an appropriate choice of c, there exist Gallager codes Galn,2(c, H), where H is m × n, of rates arbitrarily close to channel capacity and c/m arbitrarily small. The precise formulation of these statements can be found in . Furthermore, there exist irregular LDPC codes whose performance very closely approaches the Shannon channel capacity for the binary additive white Gaussian noise (AWGN) channel [642, 2457]. These codes compare favorably and can even surpass the performance of turbo codes [1987, 2457]. 15.3.6 Remark LDPC codes are part of several standards used in digital television, optical com-munication, and mobile wireless communication, including DVB-T2, DVB-S2, WiMAX-IEEE 802.16e, and IEEE 802.11n . Both the DVB-T2 standard , used in digital television, and the DVB-S2 standard , used in a variety of satellite applications, employ LDPC codes concatenated with BCH codes. In addition, a 2007 paper from the Consulta-tive Committee for Space Data Systems (CCSDS) outlines the experimental speciﬁcations for the use of LDPC codes for near-Earth and deep space communications . 714 Handbook of Finite Fields 15.3.7 Remark Gallager and LDPC codes can be deﬁned over other alphabets. Such codes were examined in Gallager’s original work ; see also . Only binary codes are considered here. 15.3.8 Remark Gallager’s ﬁrst decoding algorithm is an iterative algorithm that involves ﬂipping bits in the received vector. This algorithm works best on a binary symmetric channel (BSC) when the code rate is well below channel capacity. Assume that the codeword c is trans-mitted and y = (y1, y2, . . . , yn) is received using Galn,2(c, H). In the computation of the syndrome HyT , each received bit yi aﬀects at most c components of that syndrome, as the i-th bit is in c parity check equations. Let Si denote the set of bits involved in these c parity check equations. If among all the bits Si involved in these c parity check equations only the i-th is in error, then the c components of HyT , arising from these c parity check equations, will equal 1 indicating the parity check equations are not satisﬁed. Even if there are some other errors among the bits Si, one expects that several of these c components of HyT will equal 1. This is the basis of the Gallager Bit-Flipping Decoding Algorithm. I. Compute HyT and determine the unsatisﬁed parity check equations, i.e., the parity check equations where the components of HyT equal 1. II. For each of the n bits, compute the number of these unsatisﬁed parity check equations involving that bit. III. Flip the bits of y, from 0 to 1 or 1 to 0, that are involved in the largest number of these unsatisﬁed parity check equations; call the resulting vector y again. IV. Iteratively repeat I, II, and III until either HyT = 0T , in which case the received vector is decoded as this latest y, or until a certain number of iterations is reached, in which case the received vector is not decoded and a decoding failure is declared. 15.3.9 Remark In order to describe the second of Gallager’s decoding algorithms, we need the concept of a Tanner graph , which can be deﬁned for any code. 15.3.10 Deﬁnition The Tanner graph is a bipartite graph constructed from an m × n parity check matrix H for a binary code. The Tanner graph has two types of vertices: variable and check nodes. There are n variable nodes, one corresponding to each coordinate or column of H. There are m check nodes, one for each parity check equation or row of H. The Tanner graph has only edges between variable nodes and check nodes; a given check node is connected to precisely those variable nodes where there is 1 in the corresponding column of H. Since a code can have many diﬀerent parity check matrices, there are many diﬀerent Tanner graphs for a code. If the code is Galn,2(c, H), the degree of each variable node is c; if the code is LDPCn,2(c, r, H), the degree of each variable node is c, and the degree of each check node is r. 15.3.11 Remark The second of Gallager’s algorithms is an example of an iterative “message pass-ing” decoding algorithm. Message passing decoding algorithms have the following general characteristics. 1. Message passing algorithms are performed in iterations, also called rounds. 2. Messages are passed from a variable node to a check node and from a check node to a variable node along the edges of the Tanner graph of a code. Outgoing mes-sages sent out from a node along an adjacent edge ϵ depend on all the incoming messages to that node except the incoming message along ϵ. 3. There is an initial passing of messages from the variable nodes to the check nodes (or perhaps from the check nodes to the variable nodes). Algebraic coding theory 715 4. One iteration, or round, consists of passing messages from all check nodes to all adjacent variable nodes followed by the passing of messages from all variable nodes to all adjacent check nodes (or perhaps from variable nodes to check nodes and then back to variable nodes). 5. At the end of each round, a computation is done that will either end the algorithm or indicate that another iteration should be performed. 6. There is a preset maximum number of rounds that the algorithm will be allowed to run. 15.3.12 Remark The description of Gallager’s message passing decoding algorithm comes from and applies to binary channels in which noise bits are independent, such as the memoryless BSC or the binary AWGN channel. The algorithm falls in a class of algorithms called “sum-product algorithms” and is implemented using message passing. Let C = Galn,2(c, H) where H is an m×n matrix. In the Tanner graph T for C, number the variable nodes 1, 2, . . . , n and the check nodes 1, 2, . . . , m. Let V (j) denote the variable nodes connected to the j-th check node, and let C(k) denote the c check nodes connected to the k-th variable node. Suppose the codeword c is transmitted and y = c+e is received where e is the unknown error vector. Given the syndrome HyT = HeT = zT , the object of the decoder is to compute the conditional probabilities P(ek = 1 \| z) for 1 ≤k ≤n. From there, the most likely error vector and hence most likely codeword can be found. The algorithm is an iterative message passing algorithm in which each message is a pair of probabilities. For 1 ≤k ≤n, the initial message passed from the variable node k to the check node j ∈C(k) (in Step I of the algorithm) is the pair (q0 j,k, q1 j,k) = (p0 k, p1 k), where p0 k = P(ek = 0) and p1 k = P(ek = 1) come directly from the channel statistics. For example, if communication is over a BSC with crossover probability p, then p0 k = 1 −p and p1 k = p. Each iteration consists of ﬁrst passing messages from check nodes to variable nodes (in Step II of the algorithm) and then passing messages from variable nodes back to check nodes (in Step III of the algorithm). The message from the check node j to the variable node k for k ∈V (j) is the pair (r0 j,k, r1 j,k), where re j,k for e ∈{0, 1} is the probability that the j-th check equation is satisﬁed given that ek = e and that the other bits ei for i ∈V (j) \ {k} have probability distribution given by {q0 j,i, q1 j,i}. The message from the variable node k to the check node j ∈C(k) is the pair (q0 j,k, q1 j,k), where qe j,k for e ∈{0, 1} is the probability that ek = e given information obtained from checks C(k) \ {j}. The Gallager Message Passing Sum-Product Decoding Algorithm for binary Gallager codes is the following. I. For 1 ≤k ≤n, pass the message (q0 j,k, q1 j,k) = (p0 k, p1 k) from variable node k to check node j ∈C(k). II. Update the values of re j,k for k ∈V (j) and e ∈{0, 1} according to the equation re j,k = X i∈V (j){k}, ei∈{0,1} P(zj \| ek = e, {ei \| i ∈V (j) \ {k}}) Y i∈V (j){k} qei j,i. For 1 ≤j ≤m, pass the message (r0 j,k, r1 j,k) from check node j to variable node k ∈V (j). III. Update the values of qe j,k for j ∈C(k) and e ∈{0, 1} according to the equation qe j,k = αj,kpe k Y i∈C(k){j} re i,k where αj,k is chosen so that q0 j,k + q1 j,k = 1. For 1 ≤k ≤n, pass the message (q0 j,k, q1 j,k) from variable node k to check node j ∈C(k). 716 Handbook of Finite Fields After each pass through Step III, compute qe k = pe k Y j∈C(k) re j,k for 1 ≤k ≤n and e ∈{0, 1}. Then for 1 ≤k ≤n, set b ek = 0 if q0 k > q1 k and b ek = 1 if q1 k > q0 k. Let b e = (b e1, b e2, . . . , b en). If Hb eT = zT , decode by setting e = b e and stop the algorithm; otherwise repeat Steps II and III unless the predetermined maximum number of iterations has been reached. If the maximum number of iterations has been reached and if Hb eT never equals zT , stop the algorithm and declare a decoding failure. 15.3.13 Remark If the Tanner graph T is without cycles and the algorithm successfully halts, then the probabilities P(ek = e \| z) equal αkqe k for 1 ≤k ≤n where αk is chosen so that αkq0 k + αkq1 k = 1. If the graph has cycles, αkqe k are approximations to P(ek = e \| z). In the end, one does not care exactly what these probabilities are; one only cares about obtaining a solution to HeT = zT . Thus the algorithm can be used eﬀectively even when there are long cycles present. The algorithm has been successful for codes of length a few thousand, say n = 10, 000, particularly with c small, say c = 3. Analysis of the algorithm can be found in . 15.3.14 Remark In Step II of the algorithm, the conditional probabilities P(zj \| ek = e, {ei \| i ∈ V (j) \ {k}}) for k ∈V (j) and e ∈{0, 1} are required. These probabilities are either 0 or 1. Notice that the j-th check equation is the modulo 2 sum of the values ek for k ∈V (j). Thus P(zj \| ek = e, {ei \| i ∈V (j) \ {k}}) =    1 if zj ≡e + X i∈V (j){k} ei (mod 2), 0 otherwise. 15.3.15 Remark There is a message passing algorithm for Gallager codes on the binary erasure channel (BEC). For the BEC, there are two inputs {0, 1} to the channel, three possible outputs {0, E, 1} from channel transmission, and an associated probability ϵ. In a BEC an input symbol x is received as an output symbol y with the following probabilities P(y \| x): P(0 \| 1) = P(1 \| 0) = 0, P(0 \| 0) = P(1 \| 1) = 1 −ϵ, and P(E \| 0) = P(E \| 1) = ϵ. So the output that is received is either an erasure, denoted E, or is the original input symbol; 0 is never received as 1 and 1 is never received as 0. The message passing algorithm for the BEC, which is described and thoroughly analyzed in , employs the notation from Remark 15.3.12. Suppose that c is sent and y = (y1, y2, . . . , yn) is received. The messages on each edge of the Tanner graph are elements of {0, E, 1}. For binary Gallager codes the Message Passing Decoding Algorithm for a BEC is the following. I. For 1 ≤j ≤m, send E from every check node j to every variable node k ∈V (j). II. For 1 ≤k ≤n, determine the message to send from the variable node k to the check node j ∈C(k) as follows. If yk and all of the incoming messages to the variable node k from the check nodes i ∈C(k) \ {j} equal E, the outgoing message from the variable node k to the check node j ∈C(k) is E. Otherwise, if one of yk or the incoming messages to the variable node k from the check nodes i ∈C(k) \ {j} equals 0, respectively 1, send 0, respectively 1. III. For 1 ≤j ≤m, determine the message to send from the check node j to the variable node k ∈V (j) as follows. If any of the incoming messages to the check node j from the variable nodes i ∈V (j) \ {k} is E, the outgoing message from the check node j to the variable node k ∈V (j) is E. Otherwise, if none of the incoming messages to the check node j from the variable nodes i ∈V (j) \ {k} is Algebraic coding theory 717 E, the outgoing message from the check node j to the variable node k ∈V (j) is the modulo 2 sum of the incoming messages to the check node k from the variable nodes i ∈V (j) \ {k}. In addition, if yk = E, update yk to be the value of the incoming message. Repeat Steps II and III until all yk’s have been determined, in which case the updated y is declared to equal the transmitted codeword c, or until the number of iterations has reached a predetermined value. In the latter case, declare a decoding failure. 15.3.16 Remark In Step II of the algorithm, there is no ambiguity in assigning the value to the message to be sent. It is not possible for one of yk or the incoming messages to have the value 0 and another of yk or the incoming messages to have the value 1. In Step III, the only time the value 0 or 1 is passed as a message from the check node j to the variable node k is when there is no erasure from any other variable node sent to check node j and hence the value of the k-th variable is determined uniquely by the j-th parity check equation. 15.3.17 Remark When applying the message passing algorithm of Remark 15.3.15, the iterations may stabilize so that there are erasures left but no further iteration will remove any of these remaining erasures. This will occur because of the presence of stopping sets. 15.3.18 Deﬁnition Let V be the set of variable nodes in the Tanner graph of Galn,2(c, H). A subset S of V is a stopping set if all the check nodes connected to a variable node in S are connected to at least two variable nodes in S. 15.3.19 Theorem Suppose that a codeword of Galn,2(c, H) is sent over a BEC and that the received vector has erasures on the set E of variable nodes. Assume that the message passing decoder of Remark 15.3.15 is allowed to run on the received vector until either all erasures are corrected or the process fails to make progress. If the algorithm fails to make progress and erasures remain, then the set of variable nodes that still contain erasures is the unique maximal stopping set inside E. 15.3.20 Remark So far, the focus of this section has been on decoding Gallager codes. The algo-rithms are eﬃcient because they take advantage of the sparsity of the parity check matrix for such codes. One might ask if this sparseness of the parity check matrix can be used to eﬃciently encode Gallager codes. The answer is yes, as described in . Begin with the m × n parity check matrix H for Galn,2(c, H), which we assume has rank m. By using row and column permutations, which do not aﬀect the column or row weights, H can be transformed into an approximate upper triangular form T A B E C D . The matrix T is an upper triangular (n−k−g)×(n−k−g) matrix with 1s on the diagonal; A is (n−k −g)×g, B is (n−k −g)×k, and E, C, and D have g rows with the appropriate number of columns. There is an algorithm with O(n + g2) operations that encodes any message. 15.3.21 Remark Constructing “good” irregular LDPC codes is an important area of research where “good” means optimal or near optimal according to some measure. One approach to the construction of good LDPC codes is to use combinatorial objects, such as ﬁnite geome-tries, permutation matrices, and Latin squares to derive incidence matrices of variable versus check nodes. The properties of the combinatorial objects typically allow observa-tions on certain structural properties of the resulting graph. The objects used often give quasi-cyclic codes which have desirable encoding and decoding properties. As with most 718 Handbook of Finite Fields LDPC code constructions, performance has to be obtained through simulation. The ref-erences [1801, 1841, 3055, 3061] are representative of this approach. There are numerous decoding algorithms, some variations of those given here, and their performance analysis is also an important topic of research. When using iterative decoding, one needs to know if the algorithm will converge to a codeword and if that codeword is the original codeword transmitted. The presence of “pseudocodewords” play a role in describing convergence of an iterative decoder; see [151, 1724, 1778] and the references in these papers. The error performance of a maximum likelihood decoder is determined by the distance distribution of the codewords in the code; in the case of an iterative decoder, error performance is deter-mined by the distribution of pseudocodewords . There are three common notions of pseudocodewords: computation tree pseudocodewords, graph cover pseudocodewords, and linear programming pseudocodewords; see for connections between the three types. For instance, computation tree pseudocodewords arise as follows . From the Tanner graph of an LDPC code, the computation tree can be constructed starting from a ﬁxed variable root node. The actual structure of the tree is determined by the scheduling used by the message passing algorithm and the number of iterations used. In general a variable node from the Tanner graph will appear multiple times in the computation tree, as will the check nodes. A codeword in the LDPC code will be a binary assignment of variable nodes in the Tanner graph so that the binary sum of the neighbors of each check node is 0. The same assignment made in the computation tree will also yield, for the neighbors of each check node, a binary sum equal to 0. However, it is possible to make an assignment of the variable nodes in the computation tree so that the binary sum of the neighbors of each check node is 0, but the assignment gave diﬀerent values to some nodes of the computation tree that actually represented the same variable node of the Tanner graph. This assignment of values is a computation tree pseudocodeword. 15.3.22 Remark The special case of linear codes that can be represented by Tanner graphs with no cycles has been investigated. Such a code can be decoded with maximum likelihood by using the min-sum algorithm with complexity O(n2). However, it can be shown that such cycle-free Tanner graphs cannot support good codes, in terms of either trade-oﬀbetween rate and minimum distance or performance. Aspects of cycle-free Tanner graphs are explored further in [996, 1401, 2518]. 15.3.23 Remark The analysis of LDPC codes, particularly for ﬁnite block lengths, poses challenges. Interest is in performance under a class of belief propagation decoding algorithms, a subclass of message passing algorithms, where the messages sent between variable and check nodes represent the probability or belief a given variable node has a particular value. This usually involves either likelihoods or log likelihood ratios, conditioned on values received in the pre-vious round. While such algorithms are suboptimal, they are more eﬃcient than maximum likelihood algorithms and generally yield good performance. In the ﬁrst round of decoding, the message sent from a variable node v to its check neighbors is the log likelihood ratio of the received data, i.e., the log of the ratio of the probability the sent bit was a 0 to that it was a 1, given the received value (which may be discrete or continuous). In subsequent rounds, messages from a check node c to a neighbor variable node v is a log likelihood of the message arriving at v, involving information sent from all neighbor variable nodes other than v, in the previous round. After the ﬁrst round, messages from a variable node v to a neighbor check node c is a log likelihood of the value of the message sent from v condi-tioned on values received in the previous round from neighbor check nodes other than c. The analysis requires evaluation of probability density functions of data received by a node that involves convolutions on the order of the degree of the node, either check or variable. It can be shown that the behavior of these densities is narrowly concentrated around their expected values and thus it suﬃces to consider only the expected values of the den-Algebraic coding theory 719 sities. The updating formulae for these expected densities, referred to as density evolution, gives a method of evaluating the performance of the algorithm. The equations assume the variables involved in the convolutions are independent which is only true to the extent that the graph, grown out from a given variable node, is a tree to a certain level. At some point this ceases to be true. However, the analysis given with this assumption yields results which are accurate for most codes. The technique of density evolution introduced in has been a cornerstone of the analysis of LDPC codes. 15.3.24 Remark Many LDPC codes exhibit probability of error behavior that, as the signal to noise ratio (SNR) is increased, the error curve ﬂattens out rather than continuing as a typical “waterfall” curve. This is termed an error ﬂoor and is an undesirable feature for applications requiring very small probabilities of error, in the range of 10−12 for many standards. It is diﬃcult to simulate curves down to such a low value - even with special hardware, simulations can typically be done only to about 10−9 . Much eﬀort has been expended on ﬁnding analytical techniques to determine codes whose error ﬂoors are below such levels. The key reference for such work is where the notion of trapping sets is introduced. An (a, b) trapping set is a subgraph of the Tanner graph of the code induced by a set of variable nodes of size a with the property that the subgraph has b check nodes of odd degree. Although the sizes of the trapping sets are not the only relevant parameters for predicting error performance of the code , it can be shown they are a major inﬂuence for error ﬂoors in performance curves. In particular such ﬂoors tend to be caused by overlapping clusters of fairly small trapping sets. The determination of such trapping sets tends to be a feasible computational task, either analytically or via simulation , for many codes of interest. The design of codes without such trapping sets is a major goal of LDPC coding theorists. Trapping sets can be thought of as an analog of stopping sets for erasure codes. References Cited: [151, 573, 642, 994, 995, 996, 1046, 1162, 1163, 1401, 1724, 1778, 1801, 1841, 1987, 2455, 2456, 2457, 2458, 2518, 2779, 3055, 3061] 15.4 Turbo codes over ﬁnite ﬁelds Oscar Takeshita, Silvus Technologies 15.4.1 Introduction 15.4.1.1 Historical background Turbo codes are a class of error control codes that was ﬁrst introduced by Berrou, Glavieux, and Thitimajshima in 1993 . Turbo coding was a paradigm shift in the design of error control codes that enabled them to achieve very good performance. 15.4.1.2 Terminology 15.4.1 Remark Basic coding notation and code properties are covered in Section 15.1 of this handbook. We shall need to add some terminology in order to properly address turbo codes. 720 Handbook of Finite Fields 15.4.2 Remark In many practical communication systems, we are often interested in encoding binary (the alphabet is restricted to two symbols) messages of a given manageable length k, i.e., messages are transmitted in packets of a predetermined size. This is because of a ﬁxed amount of memory or other hardware limitations. However, the actual amount of data to be transmitted may be much more than k bits. In this case, long data streams are simply split into multiple messages and the encoding process is repeated as necessary. Those messages may be represented by a k-bit vector u. In this section, we work only over F2(= GF(2)). However, many of these ideas can be considered over general ﬁnite ﬁelds. 15.4.3 Deﬁnition Let u = (u0, u1, u2, . . . , uk−1) be a message vector of length k, where ui ∈F2. 15.4.4 Example An example of a message of length k = 5: u = (u0, u1, u2, u3, u4) = (0, 1, 1, 0, 1). 15.4.5 Deﬁnition The set of distinct messages of length k is U = Fk 2, where Fk 2 is the Cartesian product of F2 taken k times. 15.4.6 Remark Let t and s be two vectors. One is often interested in chaining them together to form a single vector r. 15.4.7 Deﬁnition Let t = (t0, t1, t2, . . . , tk1−1) and s = (s0, s1, s2, . . . , sk2−1) be two vec-tors. Deﬁne the chained vector r = t\|s = (t0, t1, t2, . . . , tk1−1, s0, s1, s2, . . . , sk2−1) = (r0, r1, r2, . . . , rk1+k2−1). 15.4.8 Remark Order matters with the chaining operator, i.e., t\|s is not the same as s\|t in general. The chaining of more than two vectors follows in a similar way. 15.4.9 Example Let three vectors r, s, and t be chained together in this order. Their chaining is denoted r\|s\|t. 15.4.10 Deﬁnition Let C be an (n, k) linear code. An encoder Ce for C is a one-to-one and onto function Ce : Fk 2 →Fn 2, Ce : u 7→v, where n ≥k, u ∈Fk 2, v ∈Fn 2, and the set of codewords C = {v ∈Fn 2\|v = Ce(u)} forms a linear subspace of Fn 2. 15.4.11 Remark A few methods for deriving codes from existing ones are explained in Section 15.1. The methods therein modify one code in order to derive a new code. Let us denote by the term compound derivation of a code any technique that derives a new code by combining two or more codes. An important classical method of compound derivation of a code by combining two codes is code concatenation . 15.4.12 Deﬁnition Let C1 be an (n1, k1) code over Fn1 2 and C2 be an (n2, k2 = n1) code over Fn2 2 . A classical (n2, k1) concatenated code C3 over Fn2 2 is deﬁned by an encoder function Ce 3 generated by the function composition Ce 3 = Ce 2 ◦Ce 1; C1 and C2 are the constituent codes of C3. Further, C1 is the outer code and C2 the inner code. 15.4.13 Remark In the deﬁnition above, the symbols for both C1 and C2 are over the same ﬁeld F2. The interested reader may refer to and learn that when the above codes are deﬁned with appropriate lengths and symbols over extension ﬁelds of diﬀerent lengths, the minimum distance of the resulting code C3 is at least the product of the minimum distances of C1 and C2. Algebraic coding theory 721 15.4.14 Remark The above deﬁnition of a concatenated code is precisely what is known as a serially-concatenated turbo code . The novelty in turbo coding is a proper choice of the constituent codes for the encoder and a decoding method for these codes. The following type of compound derived code is used to form the classical parallel-concatenated turbo code. 15.4.15 Deﬁnition Let C1 be an (n1, k) code over Fn1 2 and C2 be an (n2, k) code over Fn2 2 . Let v1 and v2 be the code vectors corresponding to a message vector u for codes C1 and C2, respectively. An (n2 + n1, k) parallel-concatenated code C3 over Fn2+n1 2 is formed by vector chaining these code vectors, i.e., encoding a message vector u with C3 produces v3 = v1\|v2. The codes C1 and C2 are the constituent codes of C3. 15.4.16 Remark Several variations of a turbo encoder have been studied, but we focus on the classi-cal parallel-concatenated version. A parallel-concatenated turbo code, which is a compound derived code, has a recursive convolutional code as one of its constituent codes. We shall see that the second constituent code itself is also a compound code derived from a recursive convolutional code and a permutation code. A permutation code is implemented in practice with a device called an interleaver. 15.4.2 Convolutional codes 15.4.17 Remark Convolutional codes are a popular type of error control code because of the sim-plicity of their encoding and the practicality of optimal decoding for many channels. They have been used in a variety of applications from home wireless networks to deep-space communication systems. Encoding is performed by polynomial operations over the ring of polynomials with coeﬃcients over F2. The Viterbi algorithm is widely used to decode convolutional codes in practice because it is known to be optimal (maximum likelihood decoding) for any memoryless channel; see Section 15.1 for a discussion of these concepts. 15.4.2.1 Non-recursive convolutional codes 15.4.18 Remark There are two major classes of convolutional codes: non-recursive convolutional codes and recursive convolutional codes. Many of the traditional applications use non-recursive convolutional codes, but turbo codes require recursive convolutional codes for reasons brieﬂy explained in Section 15.4.5. 15.4.19 Remark In convolutional coding, a message vector u is traditionally represented in poly-nomial form on the variable D. 15.4.20 Deﬁnition The polynomial representation of a message vector u = (u0, u1, u2, . . . , uk−1) is U(D) = u0 + u1D + u2D2 + · · · + uk−1Dk−1. 15.4.21 Example The polynomial representation of the information vector in Example 15.4.4 is U(D) = D + D2 + D4. 15.4.22 Deﬁnition The set of all polynomials in the polynomial ring on the variable D with coeﬃcients in F2 and whose degree is smaller than k is denoted F2[D]<k. 722 Handbook of Finite Fields 15.4.23 Deﬁnition Typically convolutional codes are encoded in “non-recursive” form. The encoder is deﬁned by a generator matrix G(D) whose entries are polynomials in F2[D]<d, where dmax = d −1 is the maximum degree among the polynomials in the matrix. 15.4.24 Example An example generator matrix with dmax = 2 and elements in F2[D]<3 is G(D) = [g1(D) g2(D)] = 1 + D + D2 1 + D2 . 15.4.25 Deﬁnition The encoding of a message U(D) is performed by multiplying it by the generator matrix. The resulting matrix V (D) = U(D)G(D) is the codeword in polynomial form. 15.4.26 Remark It is worthwhile noting that the codeword length n of convolutional codes is a function of the message space F2[D]<k. This means that a generator matrix deﬁnes a family of codes. However, in practice k is chosen to be a ﬁxed value. 15.4.27 Example We illustrate the encoding operation with an example by encoding the message in Example 15.4.21 with the generator matrix in Example 15.4.24 V (D) = U(D)G(D) = U(D) [g1(D) g2(D)] = [v1(D) v2(D)] . 15.4.28 Remark The encoding operation may be generalized to multi-dimensional message matrices and generator polynomials. However, many practical systems have a U(D) with a single message polynomial (M = N = 1) and G(D) is a simple one-by-two matrix (P = 2). This implies V (D) is typically a one-by-two matrix. 15.4.29 Deﬁnition The Hamming weight w(P(D)) of a polynomial P(D) ∈F2[D]<k is the number of non-zero monomials in P(D). 15.4.30 Remark The Hamming weight of a vector as deﬁned in Section 15.1 of the corresponding vector p gives the same value, i.e., w(p) = w(P(D)). 15.4.31 Deﬁnition The Hamming weight w(M(D)) of a matrix M(D) whose entries are polyno-mials in F2[D]<k is the sum of the Hamming weights of each of the polynomials. 15.4.2.2 Distance properties of non-recursive convolutional codes 15.4.32 Remark We examine the non-recursive convolutional code with a ﬁxed maximum degree dmax = 2 in Example 15.4.24. It is clear that its minimum distance is no larger than the weight of the generator matrix w(G(D)) = 5 regardless of the length n of the code since, by letting the message polynomial be U(D) = 1, we obtain a code polynomial V (D) = G(D). Therefore, a convolutional code with a ﬁxed generator matrix is asymptotically bad. 15.4.33 Remark A message of the form U(D) = Dx, 0 ≤x < k, generates a code polynomial whose weight is w(G(D)). 15.4.34 Remark In order to improve the minimum distance of a convolutional code, one may increase the maximum degree dmax of the generator matrix; however, it is easy to see that the minimum distance is expected to only increase linearly with dmax. The decoding of such codes, however, unfortunately becomes exponentially complex in dmax when Viterbi decoding is used. A typical value for dmax is six to keep decoding complexity manageable. 15.4.35 Theorem Non-recursive convolutional codes C with a ﬁxed generator matrix G(D) are asymptotically bad. 15.4.36 Theorem Any low-weight message polynomial encoded by a non-recursive convolutional code C produces a low-weight codeword polynomial. Algebraic coding theory 723 15.4.2.3 Recursive convolutional codes 15.4.37 Remark From the deﬁnition of encoding, it follows that v1(D) and v2(D) have g1(D) and g2(D) as their factors, respectively. 15.4.38 Theorem There is a one-to-one and onto mapping between each U(D) and v1(D) as well as U(D) and v2(D). 15.4.39 Remark We examine v1(D) = U(D)g1(D) more carefully. Let the degree of g1(D) be dmax = d −1. We may write v1(D) as v1(D) = U(D)g1(D) = Dd−1(U ′(D)) + T(D) for some U ′(D) ∈F2[D]<k and T(D) ∈F2[D]<d−1. 15.4.40 Theorem T(D) is the negative of the remainder R(D) of the division of Dd−1(U ′(D)) by g1(D). 15.4.41 Remark In characteristic two, T(D) = R(D). 15.4.42 Remark We shall now see how a recursive convolutional code may be derived from a non-recursive convolutional code. The set of codewords turns out to be identical following such a derivation. The diﬀerence is how message vectors are encoded (mapped) into codewords. 15.4.43 Deﬁnition Let a generator matrix for a non-recursive convolutional code be G(D) = [g1(D) g2(D)]. The corresponding generator matrix for an equivalent recursive convolutional code is GR(D) = h 1 g2(D) g1(D) i . 15.4.44 Remark We illustrate how to encode a message when using recursive convolutional codes. While for a non-recursive convolutional code we simply multiplied the message polynomial by the generator matrix, for a recursive convolutional code there is an additional step. 15.4.45 Remark Let U ′(D) ∈F2[D]<k be a message to be encoded. We ﬁrst form a polynomial Z(D) = Dd−1(U ′(D)) + T(D), where T(D) is the negative of the remainder of the di-vision of Dd−1(U ′(D)) by g1(D). Naturally, Z(D) is divisible by g1(D) by construction. Next we simply multiply Z(D) by the generator matrix GR(D). The resulting matrix V (D) = Z(D)GR(D) is the codeword in polynomial form. 15.4.46 Example We illustrate the encoding operation with an example by encoding the message U ′(D) = D3 +D4. The remainder of the division of D2U ′(D) = D2(D3 +D4) = D5 +D6 by g1(D) = 1 + D + D2 is R(D) = D. The polynomial Z(D) = D + D5 + D6 is then multiplied by the generator matrix GR(D) V (D) = Z(D)GR(D) = Z(D) 1 g2(D) g1(D) = [v1(D) v2(D)] . 15.4.47 Remark We note that the codeword in vector format is v = v1\|v2 = t\|u′\|v2. 15.4.48 Remark While non-recursive convolutional code encoding involves multiplication of poly-nomials, recursive convolutional encoding involves long-division. 15.4.2.4 Distance properties of recursive convolutional codes 15.4.49 Remark Recursive convolutional codes are asymptotically bad because the set of codeword polynomials is identical to an equivalent non-recursive convolutional code. 724 Handbook of Finite Fields 15.4.50 Theorem Recursive convolutional codes C with a ﬁxed generator matrix GR(D) are asymp-totically bad. 15.4.51 Remark Despite the fact that the weight distribution of the codewords in a non-recursive convolutional code and an equivalent recursive convolutional code are the same, there are fundamental diﬀerences in the relationship between the Hamming weights of the message vectors and the Hamming weights of the corresponding codewords. 15.4.52 Remark A low-weight message polynomial encoded by a recursive convolutional code C does not necessarily produce a low-weight codeword polynomial. 15.4.53 Remark A message polynomial of the form U ′(D) = Dx, 0 ≤x < k, produces codeword polynomials whose weights are lower bounded by αx for some positive constant α. Compare this with Remark 15.4.33. This is a fundamental property of recursive convolutional codes that makes them suitable for turbo coding due to the long-division encoding process. 15.4.3 Permutations and interleavers 15.4.54 Remark Permutations have long been used along with error control codes in the area of digital communications. Their main use is to reorder the elements of a vector. A device that permutes the elements of a vector is an interleaver. 15.4.55 Deﬁnition A permutation function Π on k letters is a one-to-one and onto function Π : Zk →Zk, where Zk is the set of integers modulo k. 15.4.56 Deﬁnition Let a vector u = (u0, u1, u2, . . . , uk−1) ∈Fk 2. A permuted version uΠ of u under the permutation function Π is uΠ = (uΠ(0), uΠ(1), uΠ(2), . . . , uΠ(k−1)). 15.4.57 Deﬁnition The design requirements for an interleaver in typical applications are minimal. A simple block interleaver may in general suﬃce. A block interleaver writes the elements of the vector on a matrix row-wise and then reads them back column-wise to permute the elements of the vector. 15.4.58 Example Let a vector a = (a0, a1, a2, a3, a4, a5). A 2 by 3 block interleaver ﬁrst writes the elements of a on a 2 by 3 matrix row-wise: a0 a1 a2 a3 a4 a5 . The permuted vector becomes aΠ = (a0, a3, a1, a4, a2, a5) by reading the matrix column-wise. 15.4.59 Remark Algebraic formulations of interleavers are not only elegant, but an algebraic struc-ture typically implies eﬃcient implementations. We shall describe next one of the simplest known permutation functions. 15.4.60 Deﬁnition A linear polynomial function (LPF) over Zn is deﬁned by the function L : Zn →Zn, L : x 7→f1x + f0, where f1, f0 ∈Zn. 15.4.61 Remark Let f1 be co-prime to n over Z. The linear polynomial function function L : Zn →Zn, L : x 7→f1x + f0 is a linear permutation polynomial (LPP). Algebraic coding theory 725 15.4.62 Deﬁnition A quadratic polynomial function (QPF) over Zn is deﬁned by the function Q : Zn →Zn, Q : x 7→f2x2 + f1x + f0, where f2, f1, f0 ∈Zn. 15.4.63 Proposition Let n be a power of 2. Let f1 be odd and f2 even. The QPF function Q : Zn →Zn, Q : x 7→f2x2 + f1x + f0 is a quadratic permutation polynomial (QPP). 15.4.64 Remark Necessary and suﬃcient conditions to obtain for QPPs for arbitrary n and other properties have been investigated in [2742, 2768]. 15.4.65 Deﬁnition Let Π be a permutation function on k symbols. There is an associated (n, n = k) permutation code CΠ = {uΠ\|u ∈U}. 15.4.4 Encoding and decoding 15.4.66 Deﬁnition We deﬁne a parallel-concatenated turbo code CT . Recall Deﬁnition 15.4.15. We must simply specify the constituent codes C1 and C2. Let C1 be an (n, k) recursive convolutional code with generator matrix GR(D) = [1 g2(D) g1(D)], and let C2 be a com-pound (n −k, k) derived code. The derivation of C2 is in two steps. The ﬁrst step is the derivation of an (n2 = n, k1 = k) compound code Cs by the serial concatenation of an (n1, n1 = k1 = k) permutation outer code CΠ with an (n2 = n, k2 = n1) recursive convolutional inner code identical to C1. The second step is to puncture the k message positions of Cs. 15.4.67 Example Let u be a message vector. First we encode u with the recursive convolutional code C1 and obtain the code vector v(1) = t(1)\|u\|v(1) 2 . Next we form the code vector for C2. We ﬁrst encode u with the permutation code CΠ to obtain uΠ = Π(u). Then uΠ is encoded with a recursive convolutional code C1 to obtain the code vector v(s) = t(2)\|uΠ\|v(2) 2 for Cs. The code vector for C2 becomes v(2) = t(2)\|v(2) 2 by puncturing the message symbols uΠ of Cs. Finally, the code vector for the turbo code becomes v(T) = t(1)\|u\|v(1) 2 \|t(2)\|v(2) 2 . 15.4.68 Remark The decoding of turbo codes is performed using iterative algorithms ; for iterative algorithms see Section 15.1. 15.4.5 Design of turbo codes 15.4.69 Remark We focus on the design of parallel-concatenated turbo codes. Naturally, the design narrows down to an understanding of the overall turbo code principle and to the selection of the constituent recursive convolutional codes and an interleaver. 15.4.70 Remark Recall the deﬁnition of a linear code C of length n as a linear subspace of Fn 2 from Section 15.1. Let C be a code over Fn 2 (not necessarily linear). The Holy Grail design for C has been to maximize its minimum distance as deﬁned in Section 15.1. For a linear code, this translates to maximizing its minimum Hamming weight. This task becomes even more challenging when we require a design for a family of codes that are asymptotically good, i.e., a family of codes with a ﬁxed code-rate such that the minimum distance to code length ratio does not vanish as the code length goes to inﬁnity. 15.4.71 Remark The error correction capability of a code is mostly dictated by its minimum dis-tance, with its weight distribution a secondary factor. In turbo coding, the relationship between the weight of message vectors and the weight of codeword vectors becomes crucial. 726 Handbook of Finite Fields When an error control code is used, we ultimately need not only a set of codewords C but an encoder to map messages from U to C and a decoder to map codewords (possibly with errors) back to messages. 15.4.72 Remark We observe again the codeword of a parallel-concatenated turbo code: v(T) = t(1)\|u\|v(1) 2 \|t(2)\|v(2) 2 . (15.4.1) The design principles for a turbo code are related to the following question: How do we minimize the chances of producing simultaneously low weight codewords for C1 (t(1)\|u\|v(1) 2 ) and C2 (t(2)\|v(2) 2 )? 15.4.73 Remark The ﬁrst design principle is that, since the codeword includes the message vector u, it is more important to focus on the case when w(u) is small. 15.4.74 Remark The second design principle is, given that w(u) is small, identify a good permuta-tion function that minimizes the chances of producing simultaneously low weight codewords for C1 and C2. 15.4.5.1 Design of the recursive convolutional code 15.4.75 Remark The main design parameters of constituent recursive convolutional codes are the maximum degree dmax of the generator polynomials and the choice of the generator matri-ces. Surprisingly, it has been found that dmax = 3 or 4 is suﬃcient to achieve very good performance with turbo coding. Typically, the polynomial g1(D) is chosen to be a primitive polynomial. 15.4.76 Remark The reason for choosing g1(D) to be primitive is that it maximizes the degree of polynomials of the form U(D) = 1+Dτ such that U(D) is divisible by g1(D); this is known to improve the codeword weight of recursive convolutional codes with respect to message vectors with w(U(D)) = 2. One of the latest communication systems using turbo codes is the fourth generation wireless cellular 4G LTE standard. 15.4.77 Example The 4G LTE standard uses the generator matrix with dmax = 3: GR(D) = 1 1 + D + D3 1 + D2 + D3 . 15.4.5.2 Design of interleavers 15.4.78 Remark Interleavers or permutation functions have been extensively investigated in turbo coding applications [252, 725, 755, 767, 911, 1277, 2291, 2513, 2742, 2770, 2771]. Shannon’s random coding bound theorem naturally led researchers to experiment with diﬀer-ent pseudo-random or modiﬁed pseudo-random constructions . There were two major drawbacks to early pseudo-random constructions: ﬁrst, practical implementation calls for interleavers that require little storage and computation; and second, pseudo-random con-structions often generate turbo codes that suﬀer from a performance deﬁciency known as an “error-ﬂoor”; see Remark 15.3.24. 15.4.79 Remark Some researchers attempted to use simple interleavers with turbo coding such as block interleavers. However, a block interleaver has been shown to have too much regu-larity; it has also been shown that interleavers based on LPPs behave similarly to block interleavers . A good compromise between interleaver complexity and turbo code per-formance can be achieved with QPP interleavers; the second degree polynomial brings a Algebraic coding theory 727 “non-linear” feature that is very desirable for turbo coding [2768, 2771]. The simplicity of QPP interleavers due to their algebraic structure and their very good performance resulted in them becoming part of the 4G LTE standard. See Also §15.1 For discussion of basic coding theory properties. §15.3 For discussion of LDPC and Gallager codes. , , For discussions of cycle structure of permutation functions over ﬁnite ﬁelds and their applications to turbo codes. References Cited: [225, 251, 252, 725, 755, 767, 911, 1091, 1277, 1398, 2291, 2495, 2496, 2513, 2519, 2608, 2742, 2768, 2770, 2771, 2879] 15.5 Raptor codes Ian Blake, University of British Columbia W. Cary Huﬀman, Loyola University Chicago 15.5.1 Remark This section is concerned with coding for the binary erasure channel (BEC) (see Section 15.1), except for the last subsection which includes comments on the performance of raptor codes on the binary symmetric channel (BSC). Recall that for the BEC a nonerased symbol is correct and the goal is to design codes which can be encoded and decoded eﬃ-ciently. The capacity of the BEC with erasure probability ϵ is 1−ϵ. If C is a linear (n, k, d)q code with parity check matrix H, it is capable of correcting up to d −1 erasures. If a code-word c is transmitted and word r received (whose components are in the set {0, 1, E}), the problem of correcting erasures is the problem of solving a set of linear equations of the form H′xT = yT where H′ is the matrix of columns of the original parity check matrix H of the code corresponding to erased positions of r, x is a vector of variables corresponding to the erased positions of r, whose solutions are required, and y the sum of columns of the parity check matrix corresponding to the positions of r containing 1s. This follows from the fact that a codeword c satisﬁes the equation HcT = 0T , the vector of all zeros. The com-plexity of decoding is that of matrix reduction, assumed to be O(d3) when the maximum number of erasures d −1 has occurred (unless a more sophisticated algorithm is used). It is quite remarkable that the codes in this section achieve essentially linear complexity for both encoding and decoding while achieving capacity. This is accomplished by constructing a code by means of a bipartite graph and interpreting the decoding operation as a graph algorithm. The forerunners of raptor codes are tornado and LT codes, considered in the following sections. It should be noted that packet loss on the internet, due to traﬃc con-ditions or node buﬀer overﬂow, give a natural model of an erasure channel and the results of this section yield an excellent solution for Internet multicast data transmission with no feedback channel to request missing packets. If a block code is used for such a purpose, to be eﬀective, one would have to know the erasure probability for eﬀective code design. This is usually diﬃcult to estimate and is often time varying. The LT and raptor codes described here are rateless and such an estimate is not required. The tornado codes are block codes, 728 Handbook of Finite Fields unlike LT and raptor codes, which are rateless (fountain) codes. The tornado codes are considered for the important ideas they introduced that led, in a sense, to the fountain and LT codes. However tornado codes are block codes while LT and raptor codes are rateless, having no concept of block length or dimension. All codes considered are binary. 15.5.1 Tornado codes 15.5.2 Remark The most complete description of tornado codes is given in (although not referred to as tornado codes there). Earlier work includes the papers [1972, 1974, 1973, 2619]. The paper has been inﬂuential in the last decade of progress on the problem of coding with irregular bipartite graphs. 15.5.3 Remark Consider a bipartite graph B with k information nodes on the left and βk check nodes on the right. This is a version of the Tanner graph of a code; see Section 15.3. The information nodes are associated with information symbols. Since only XOR operations are used in the encoding, the information symbols can be taken as either bits or strings of bits of the same length (packets). The manner of choosing edges in the graph is critical to code performance and is discussed later. Given the graph, the complexity of encoding and decoding is proportional to the number of edges in the graph. For the bipartite graph B, denote the associated code by C(B). The decoding is considered next. A codeword consists of the set of information and check symbols. 15.5.4 Remark Decoding process: For the code C(B) and its associated bipartite graph B, consider a received vector y, the result of transmitting a codeword through the BEC. The positions of y contain symbols from the alphabet {0, 1, E} or strings of such symbols. Associate the received symbols with the appropriate information and check graph nodes. The decoding proceeds as follows: given the correct value of a check symbol and all but one of the informa-tion symbol neighbors, set the missing information symbol to the XOR value of the check and the known information symbols. The process continues until all information nodes are retrieved or decoding failure. The algorithm is a version of the belief propagation (BP) or message passing algorithms used for noisy channels. 15.5.5 Remark The success of the decoding procedure of C(B) depends on the existence of check symbols with the required property to the completion of decoding. Before considering this condition, the construction of the codes is given. The term tornado derives from the ob-servation that one can initiate decoding on such graphs as coded symbols arrive, and the decoder stalls until, typically, the arrival of a single further symbol allows it to proceed quickly to completion. 15.5.6 Remark The bipartite graph B is constructed randomly as follows. Refer to an edge as one of degree i on the left (right) if the left (right) node it is connected to has degree i. The following approach is taken. The graph is to have e edges. Two distributions (to be discussed later) are deﬁned: a left distribution (on edges, not nodes) (λ1, . . . , λn) and a right distribution (ρ1, . . . , ρn), where λi (resp. ρi) is the fraction of edges in the graph of left degree i (resp. right degree i), for some appropriate integer n. The number of left nodes of degree i is eλi/i and k = e P i λi/i and similarly for the right nodes. The average left degree of the information nodes is 1/(P i λi/i), denoted by aℓand similarly for right degrees ar = 1/(P i ρi/i). If the graph has e edges, k left information nodes and βk right check nodes, then aℓ= arβ. As discussed below, it is possible to choose the left and right distributions to ensure complete decoding with high probability. 15.5.7 Remark Code construction: With the terminology of the previous remark, the construction of the bipartite graph, B, is described, given the left and right distributions. Consider a Algebraic coding theory 729 sequence of four columns of nodes. The ﬁrst column has k information nodes, the second and third e nodes each, and the fourth βk check nodes. Edges appear only between adjacent columns. A fraction λi of the e edges have left degree i. For each i, connect i of the nodes of the second column to a node of the ﬁrst column, eλi/i times (truncated to form integers), to give this number of left nodes of degree i. Similarly, connect eρi/i of the nodes in the third column to a node in the fourth column to give this number of right vertices of this degree. Nodes in the second and third columns are all of degree 1. To complete the process connect the e nodes of the second and third columns by a random matching (permutation). The second and third columns of nodes are removed with edges now connecting the information and check nodes. The process may yield a small number of multiple edges between an information and check node. These are replaced with a single edge with negligible eﬀect. The graph B is the random bipartite graph. 15.5.8 Remark The two distributions are chosen from the analysis of the probability of successful decoding of the algorithm of Remark 15.5.4. The decoding algorithm is a Markov process that leads to a certain diﬀerential equation. It is convenient for the analysis to deﬁne the two distribution polynomials λ(x) = X i λixi−1, ρ(x) = X i ρixi−1. The use of xi−1 rather than xi is for convenience in the analysis. The analysis leads to the following important results. 15.5.9 Theorem [1975, Proposition 2] Let B be a bipartite graph with k information nodes that is chosen at random with edge degree distributions λ(x) and ρ(x), and suppose that δ is ﬁxed such that ρ(1 −δλ(x)) > 1 −x for all 0 < x ≤1. (15.5.1) For all η > 0 there is some k0 such that for all k > k0, if the message symbols of C(B) are erased independently with probability δ, then with probability at least 1−k2/3 exp(−k3/2/2) the decoding algorithm terminates with at most ηk information symbols erased. 15.5.10 Remark While not quite enough to show the decoding terminates successfully, with a slight modiﬁcation of the conditions it can be shown: 15.5.11 Lemma [1975, Lemma 1] Let B be a bipartite graph with k left information nodes, chosen at random according to the distributions λ(x) and ρ(x) satisfying Condition (15.5.1) such that λ1 = λ2 = 0. Then there is some η > 0 such that, with probability 1 −O(k−3/2), the decoding process restricted to the subgraph induced by any η-fraction of the left nodes terminates successfully. 15.5.12 Remark Condition (15.5.1) was relaxed in to: δλ(1 −ρ(1 −x)) < x for x ∈(0, δ]. It remains to show that distributions satisfying (15.5.1) exist. Distributions that satisfy this condition are referred to as capacity achieving (CA). Numerous works have addressed the problem of ﬁnding CA sequences of distributions including [1976, 2340, 2457, 2618, 2620]. A variety of techniques are used. One such example is given in the following. 15.5.13 Example This example is taken from . Let D be a positive integer which is chosen to trade oﬀaverage left degree with how well the decoding process works. Let H(D) be the truncated Harmonic series: H(D) = PD i=1 1/i ≈ln(D). Let λi = 1 (H(D)(i −1)), i = 2, 3, . . . , D + 1, and hence λD(x) = 1 H(D) D X i=1 xi i . 730 Handbook of Finite Fields The average left degree is then aℓ= H(D)(D + 1)/D. For the right distribution choose ρi = e−ααi−1 (i −1)! , i = 1, 2, . . . , ρD(x) = eα(x−1), i.e., the Poisson distribution, where α is chosen so that ar = αeα/(eα −1) = aℓ/β, i.e., to make the fractions of nodes on the left and the right the same. It can be shown that the two distributions satisfy (15.5.1) for δ ≤β/(1 + 1/D) and hence are CA. These distributions are referred to as heavy tail/Poisson or tornado [1975, 2619]. 15.5.14 Remark It is to be noted that CA sequences lead to a graph having a high probability of successfully completing decoding on the graph. 15.5.15 Theorem [1975, Theorem 3] For any rate R, 0 < R < 1, and any ϵ with 0 < ϵ < 1, and suﬃciently large block length n, there is a linear code and decoding algorithm that, with probability 1 −O(n−3/4), is able to correct a random (1 −R)(1 −ϵ) fraction of errors in time proportional to n ln(1/ϵ). 15.5.16 Remark The original construction of tornado codes involved a cascade of m + 1 sections of random bipartite graphs, the i-th section having βik left nodes and βi+1k check nodes, with a ﬁnal code C as a good erasure correcting code. Works subsequent to typically used only a single section (and no ﬁnal erasure correcting code) and perhaps this is the important impact of the work on tornado codes. The ideas of tornado codes (although not always referred to as tornado codes) ﬁrst appeared in . Reference simpliﬁed the analysis. The notion of the graph construction and CA distribution sequences has proved important and a rather large set of papers on this subject now exists, many giving new techniques, including linear programming, for ﬁnding CA distributions. These include [2457, 2618, 2620]. The inﬂuence of the paper is evident in subsequent work on the problem. The ability to encode and decode codes on the BEC in linear time is a remarkable achievement (recall the initial comments on the equivalence of this problem to matrix reduction and Gaussian elimination). 15.5.17 Deﬁnition Deﬁne a bipartite graph B0 with k information nodes and βk check nodes, 0 < β < 1. Recursively form the bipartite graph Bi whose left nodes are the βik right nodes of Bi−1, with βi+1k right nodes, 1 ≤i ≤m, for an integer m chosen so that βm+1k is roughly √ k. Choose a ﬁnal code C as a good erasure correcting code of rate 1−β with βm+1k information nodes. The resulting cascaded code, C(B0, B1, · · · , Bm, C) has k information symbols (the leftmost k symbols as input to the graph B0) and m+1 X i=1 βik + βm+2k/(1 −β) = kβ/(1 −β) check nodes. The code C(B0, B1, . . . , Bm, C) is a tornado code. 15.5.2 LT and fountain codes 15.5.18 Remark The notion of fountain codes ﬁrst appears in the work . The term “fountain” refers to the process whereby k information symbols (either bits or packets) are encoded into coded symbols by XOR’ing subsets of information symbols in such a way that a receiver may collect any k(1+ϵ) coded symbols, for some ϵ appropriate to the system, to recover all k information symbols. The image is of a coder producing a digital fountain of coded symbols and a receiver collects a suﬃcient number of such symbols to retrieve the information. Codes Algebraic coding theory 731 with such a property can be described as universal rateless erasure codes. The codes are rateless since there is no concept of code dimension. LT codes are the ﬁrst realization of such erasure codes. The term “LT” refers to “Luby transform.” The codes were devised for a situation where packets are lost, or dropped, on the Internet and there was no eﬃcient way a receiver could request a retransmission for such a missing packet. This is a typical situation for a multicast internet protocol where a single information source transmits a large amount of data to a large user community. For a typical block erasure correcting code, each user would have to receive a speciﬁc set of codeword symbols in order to decode the remaining erasures. In the LT scenario, any suﬃciently large collection of coded symbols will do. While often thought of as erasure correcting codes, there are no erasures in this situation - in eﬀect erased packets are those that are lost. 15.5.19 Remark As a matter of notation, the terms used in this section are information symbols and coded symbols. Many papers use the terms input and output symbols which are also natural but in some situations, where more than one level of graphs is considered, they might be ambiguous. 15.5.20 Remark In constructing bipartite graphs for coding, two forms appear in the literature. In the ﬁrst form, there are n nodes on the left, and n −k check nodes on the right. This is often called the Tanner graph of the code. The other form is the one used above with k information nodes on the left and n −k check nodes on the right. 15.5.21 Deﬁnition (The LT encoding process) To encode k information symbols a probability distribution {ρi, i = 1, 2, . . . , k} is chosen. For each coded symbol, the distribution is sampled. If the integer d is produced, this number of information symbols, chosen uni-formly at random from among the k, are XOR’ed together to produce a coded symbol. Information as to how the symbol was produced is included in a symbol header. Such information is a negligible fraction for practical systems and is ignored in the analysis. Note that for the remainder of the section {ρi} denotes a node degree distribution, as opposed to the edge distribution of the previous section. 15.5.22 Deﬁnition (The LT decoding rule) From a collection of K (slightly larger than k) received coded symbols, a bipartite graph is formed with k left information nodes (initially of unknown value) and K right received code nodes. Edge connections are formed from the header information of the received coded symbols. All coded symbols of degree one are said to cover their unique information neighbors and this set of covered information symbols is the ripple. Information symbols in the ripple are determined as they are the same as the coded symbols covering them. At the ﬁrst stage an information symbol in the ripple is XOR’ed with all of its coded neighbors and all edges to the coded neighbors removed. This might produce coded nodes of degree one and their unique information symbols are added to the ripple. The process continues iteratively until either all information symbols are recovered or there are no coded symbols of degree one. If at the end some information symbols remain uncovered, decoding failure is declared. 15.5.23 Remark The decoding process above has a complexity proportional to the number of edges in the graph. The number K of coded symbols required to have a reasonably high probability of decoding success is typically in the range of 1.01 to 1.05 times k, for suﬃciently large k. From a balls in bins analysis it can be shown that for such a coding process it is necessary that at least k ln(k/δ) balls must be thrown into the k (information) bins in order to have a probability of δ that each of the bins has at least one ball (i.e., that each information symbol is used (covered) by at least one of the coded symbols - for if one is not covered it 732 Handbook of Finite Fields could not be recovered). Thus in the encoding process the average degree of an information symbol must be at least ln(k/δ) no matter what distribution is used for producing the coded symbols. 15.5.24 Remark The random behavior of the LT decoding process is determined entirely by the distribution {ρi}, the number of information symbols k, and the number of coded symbols K obtained, which are typically slightly larger than k. The desirable properties for the distribution are 1. to ensure the fewest possible number of coded symbols to be able to complete the decoding with high probability and 2. the average degree of the coded symbol nodes is as small as possible (although, as noted, at least ln(k)). 15.5.25 Remark In analyzing the LT decoding process, a desirable feature is to have the rate at which information symbols are added to the ripple to be about the same as the rate they are processed. The ripple size should be large enough to ensure the decoding process can continue to completion, but not so large that a coded symbol has too high a probability of covering an information node already in the ripple. A distribution that has desirable properties in terms of this and other properties is the soliton distribution given by: ρi = 1/k i = 1, 1/(i(i −1)) i = 2, 3, . . . , k, (15.5.2) and ρS(x) = P i ρixi. (The name derives from physics where it arises from a similar property in a refraction problem.) The expected value of this distribution is H(k) = Pk i=1 1/i, the harmonic sum to k. In , a diﬀerent analysis of the decoding process of Deﬁnition 15.5.22 is given. It is shown there that if, at each step of the decoding process one wants an expected number of degree one coded nodes, the degree distribution must satisfy the (asymptotic) equation (1 −x)η′′(x) = 1, 0 < x < 1 where η(x) = P i ηixi. The solution of this equation is almost the soliton distribution - except that η1 = 0 and the range is inﬁnite. The fact that such a distribution yields no coded symbols of degree one is a problem since the decoding algorithm cannot start. While this distribution is ideal in the sense that the expected number of coded symbols needed to recover the information in the ripple is one, in practice it performs poorly as the probability the ripple disappears before completion is high. The robust soliton distribution {µi} is proposed to correct these defects. It is deﬁned in the following manner. Let δ be the probability of decoder failure for k information symbols and K coded symbols and R = c ln(kδ) √ k for some suitable constant c: µi = (ρi + τi)/β, where β = k X i=1 (ρi + τi) (15.5.3) and where {ρi} is the ideal soliton distribution (15.5.2), {µi} is the robust soliton distribu-tion, and µRS(x) = P i µixi, with τi =    R/(ik) i = 1, . . . , k/R −1, R ln(R/δ)/k i = k/R, 0 i = k/R + 1, . . . , k. The intuition is that the probability a random walk of length k deviates by more than ln(k/δ) √ k from its mean is at most δ . With this background it is possible to show the following theorem. Algebraic coding theory 733 15.5.26 Theorem [1971, Theorems 12, 13 and 17] With the above notation, the average degree of a coded symbol is D = O(ln(k/δ)), and the number of coded symbols required to achieve a decoding failure probability of at most δ is K = k + O( √ k ln2(k/δ)). 15.5.3 Raptor codes 15.5.27 Remark While the LT codes represent a remarkable step forward for coding on an erasure channel, such as the Internet, its complexity or running time is not linear in the number of input information symbols. The following result emphasizes this. For a code with k information symbols, we say a decoding algorithm is reliable if it fails to decode with a probability at most 1/kc for some positive constant c. The overhead of the code and decoding algorithm is ϵ if the decoder needs (1+ϵ)k coded symbols in order for the decoder to succeed with high probability. The term space complexity refers to the amount of memory required to implement decoding. The reference is followed closely here, although the terms information and coded symbols are used rather than input and output symbols. The term raptor derives from RAPid TORnado although the tornado and LT codes have very diﬀerent constructions. 15.5.28 Lemma [2621, Proposition 1] If an LT code with k information symbols possesses a reliable decoding algorithm, then there is a constant c such that the graph associated to the decoder has at least ck log(k) edges. 15.5.29 Remark Recall that if K coded symbols are gathered to decode for the k information symbols, then Gaussian elimination is maximum likelihood and has a complexity of O(Kk2), i.e., from the K coded symbols a K × k matrix equation can be set up to be solved for the k information symbols. The decoding algorithm for LT codes is able to improve on this complexity considerably by choosing an appropriate distribution {ρi} and decoding algorithm, as noted above. An LT code generated in this manner is referred to as a (k, ρ(x)) LT code. 15.5.30 Remark The idea behind the raptor codes is to ﬁrst add parity check symbols to the information symbols, by use of an eﬃcient linear block erasure correcting code, to form the set of intermediate symbols and then LT encoding this set. This relieves the LT decoder from having to correct all the erasures. A few can be left to the block code to correct and this eases the burden of the LT decoder considerably and allows a linear decoding time, for a good choice of code. Let Cn be a linear code of dimension k and block length n. It is called the precode of the raptor code. The intermediate symbols are then the k information symbols and n −k parity checks of the precode Cn. For the LT code, a modiﬁcation of the above soliton like distribution is suggested : ρD(x) = 1 µ + 1 µx + D X i=2 xi (i −1)i + xD+1 D ! where D = ⌈4(1 + ϵ)/ϵ⌉for a given real number ϵ and µ = (ϵ/2) + (ϵ/2)2. Thus this is a soliton-like distribution with a positive probability of degree 1 and truncated at D + 1. The LT code is designated a (n, ρD(x)) code. The entire code is the raptor code and designated a (k, Cn, ρD(x)) raptor code. A key result on the way to the ﬁnal one is the following lemma. 15.5.31 Lemma [2621, Lemma 4] There exists a positive number c (depending on ϵ) such that with an error probability of at most e−cn any set of (1 + ϵ/2)n + 1 coded symbols of the LT code with parameters (n, ρD(x)) are suﬃcent to recover at least (1 −δ)n information symbols, where δ = (ϵ/4)(1 + ϵ). 734 Handbook of Finite Fields 15.5.32 Remark Two extreme cases of raptor codes are noted. If ρw = k w /2k, corresponding to the distribution polynomial ρ(x) = (1 + x)k/2k, it leads to the probability that any particular binary k-tuple being chosen as 1/2k, i.e., a uniform distribution over Fk 2 since a vector of weight w is chosen with probability k w /2k and each of the k w are chosen equally likely. If such a distribution was used in the LT process, performance would be very poor. The degree distribution is too large to permit the process to succeed. At the other extreme, suppose the LT process of the raptor code (k, Cn, ρ(x)) uses a trivial distribution ρ1 = 1, i.e., after the precoding, the coder chooses an intermediate symbol at random and declares it a coded symbol. Such a code is referred to as a precode only (PCO) raptor code. One can see that such a code can achieve a low overhead only for very low rate codes Cn. 15.5.33 Remark A suitable precode has the properties: 1. The rate R of Cn is (1 + ϵ/2)(1 + ϵ). 2. The BP decoder can decode Cn on a BEC with erasure probability δ = (ϵ/4)/(1 + ϵ) = (1 −R)/2, in O(n log(1/ϵ)) operations. (Note this is half of capacity for the rate of the code.) It is suggested that several types of codes meet these criteria, such as tornado codes and right regular codes (coded symbol nodes have the same degree). 15.5.34 Theorem [2621, Theorem 5] Let ϵ be a positive real number, k the number of information symbols, D = ⌈4(1 + ϵ)/ϵ⌉, R = (1 + ϵ/2)/(1 + ϵ), n = ⌈k/(1 −R)⌉, and let Cn be a code with properties described above. Then the raptor code (k, Cn, ρD(x)) has space complexity 1/R, overhead ϵ and a cost of O(log(1/ϵ)) with respect to BP decoding of both the precode and LT code. 15.5.35 Remark A problem with raptor codes is that they are not systematic. A technique is given in to generate systematic raptor codes but is not discussed here. Many applications of raptor codes in practice prefer systematic codes. Raptor codes have been incorporated into numerous standards for the reliable delivery of data objects. The codes are described in IETF RFC 5053 and IETF RFC 6330 for such applications as the DVB-H IPDC (IP datacast to handheld devices) and 3GPP TS for multimedia broadcast/multicast service (MBMS) and future standards of IEEE P2220 (a draft standard protocol for stream management in media client devices) and 3GPP eMBMS, among others. The latter standards use RaptorQ codes deﬁned over the ﬁnite ﬁeld F28. All of these standards use systematic raptor codes. The monograph has an extensive discussion of these codes for standards. 15.5.36 Remark A good erasure correcting (linear) code should have a good minimum distance. Thus a reasonable question to ask is how such a code would perform on a noisy channel, such as a binary symmetric channel (BSC). The performance of raptor codes on such a channel is considered in . Many of the results for the erasure channel are generalized there. It builds on the landmark paper which considered more general classes of low density parity check codes and introduced such fundamental concepts as density evolution. The application of these ideas to raptor codes generalizes many of the results for the erasure channel. References Cited: [470, 991, 1971, 1972, 1973, 1974, 1975, 1976, 2340, 2457, 2458, 2618, 2619, 2620, 2621, 2622] Algebraic coding theory 735 15.6 Polar codes Simon Litsyn, Tel Aviv University 15.6.1 Space decomposition 15.6.1 Deﬁnition Let F = Fq be the ﬁeld of cardinality q. For a positive integer ℓdecomposition of Fℓis deﬁned recursively. Let T0 = Fℓ, \|T0\| = qℓ, and T0 = [ a0∈F T (a0) 1 = [ a0∈F [ a1∈F T (a0,a1) 2 = · · · = [ a0∈F [ a1∈F · · · [ aℓ−1∈F T (a0,a1,...,aℓ−1) ℓ , where T (a0,a1,...,ai−1) i = qℓ−i, i = 1, . . . , ℓ. 15.6.2 Remark Each of the sets T (a0,a1,...,aℓ−1) ℓ of the last level consists of a single vector from Fℓ. 15.6.3 Example Let q = 2 and ℓ= 2. Then T0 = {00, 01, 10, 11}, T (0) 1 = {00, 11}, T (1) 1 = {01, 10}, T (0,0) 2 = {00}, T (0,1) 2 = {11}, T (1,0) 2 = {01}, T (1,1) 2 = {10}. 15.6.4 Remark A decomposition may be deﬁned by a linear transform, with T (a0,...,ai−1) i being cosets of a linear code Ti = T (0,...,0) i spanned by the ﬁrst i rows of a full-rank ℓ× ℓmatrix over F. 15.6.5 Deﬁnition Let G be an ℓ× ℓmatrix with rows gi ∈Fℓ, i = 0, 1, . . . , ℓ−1. Deﬁne T (a0,...,ai−1) i = (a0g0 + · · · + ai−1gi−1) + [ ai∈F · · · [ aℓ∈F (aigi + · · · + aℓ−1gℓ−1). 15.6.6 Example The decomposition of Example 15.6.3 is linear, use G = 0 1 1 1 . 15.6.7 Remark Cosets of extended Reed-Solomon codes can be used for space decomposition when ℓ= q . 15.6.8 Example Let α be a primitive element of F, then a decomposition of Fq can be deﬁned using the following matrix G =        0 1 1 1 . . . 1 0 1 αq−2 α2(q−2) . . . α(q−2)(q−2) . . . . . . . . . . . . . . . . . . 0 1 α α2 . . . αq−2 1 1 1 1 . . . 1        . 736 Handbook of Finite Fields 15.6.9 Deﬁnition Let di = min d T (a0,...,ai−1) i , where the minimum is taken over all possible choices of a0, . . . , ai−1, and d(·) is the minimum Hamming distance of the code. Then the distance hierarchy of a space decom-position is the vector d = (d0 = 1, d1, . . . , dℓ−1). 15.6.10 Remark The distance hierarchy of a decomposition is a parameter used to determine the decay rate of decoding error probability. 15.6.11 Example The distance hierarchy of the decomposition from Example 15.6.3 is (1, 2). The distance hierarchy of the decomposition from Example 15.6.8 is (1, 2, 3, . . . , q). 15.6.12 Deﬁnition A space decomposition is proper if for at least one vector (a0, . . . , aℓ−1) ∈Fℓ, d T (a0,...,aℓ−1) ≥2. 15.6.2 Vector transformation 15.6.13 Remark Given a space decomposition and a vector a = (a0, . . . , aℓ−1) ∈Fℓ, one may deﬁne a transform g : Fℓ→Fℓassociated to the decomposition as g(a) = T (a0,a1,...,aℓ−1) ℓ . Recall that T (a0,a1,...,aℓ−1) ℓ here is a vector from Fℓ. An extended transform of vectors of lengths greater than ℓcan be deﬁned as follows. 15.6.14 Deﬁnition Let b = (b0, . . . , bℓs−1) ∈Fℓs be a vector, and B be the matrix of size ℓ× s, B =      b0 b1 . . . bs−1 bs bs+1 . . . b2(s−1) . . . . . . . . . . . . b(ℓ−1)s b(ℓ−1)s+1 . . . bℓs−1     = (b0, . . . , bs−1), where b0, . . . , bs−1 are the columns of B. Then ˆ g(b) = (g(b′ 0), . . . , g(b′ s−1)), where b′ i is the transposition of bi. 15.6.15 Example Using the decomposition from Example 15.6.3 with s = 4 we have for b = (01011100), B = 0 1 0 1 1 1 0 0 , and ˆ g(b) = (g(01), g(11), g(00), g(10)) = (11100001). Algebraic coding theory 737 15.6.16 Deﬁnition An ℓ-step transform G : Fℓm →Fℓm is deﬁned as follows. Let b(0) = b ∈Fℓm. At the i-th step of the transform, i = 0, . . . , m −1, the vector b(i) is partitioned to the successive sub-vectors of length ℓi+1, b(i) = b(i) 0 , b(i) 1 , . . . , b(i) ℓm−i−1−1 , where b(i) j = (b(i) jℓ, b(i) jℓ+1, . . . , b(i) (j+1)ℓi+1−1), j = 0, ..., ℓm−i−1 −1. Then b(i+1) = ˆ g(b(i) 0 ), ˆ g(b(i) 1 ), . . . , ˆ g(b(i) ℓm−i−1−1) . The resulting vector of the transform is c = G(b) = b(m). 15.6.17 Example Let q = 2, ℓ= 2, m = 3 and the space decomposition is deﬁned by Example 15.6.3. Then for b = b(0) = (01011100) we have the following sequence of results: b(1) = (g(01), g(01), g(11), g(00)) = (11111000), b(2) = (g(11), g(11), g(10), g(00)) = (10100100), b(3) = (g(10), g(01), g(10), g(00)) = (01110100). Thus, G(01011100) = (01110100). 15.6.18 Deﬁnition Let n = ℓm, and J ⊆{0, 1, . . . , n −1}, \|J\| = n −k. The polar code C ⊆Fℓm of size qk is deﬁned as C = [ G(b), where the union is taken over all qk choices of b ∈Fn such that the components of b are set to zero in the coordinates having index belonging to J. 15.6.19 Remark The way to choose the set J will be discussed later. 15.6.20 Lemma If the space decomposition is linear, the polar code is a linear code. 15.6.3 Decoding 15.6.21 Lemma Encoding of a polar code requires m steps each having complexity linear in the length of the code. The complexity of encoding a polar code is O(n log n). 15.6.22 Remark Let c = G(b) be transmitted over a memoryless channel. At the output of the channel we obtain for each of the n coordinates a set of q probabilities, one for each of the ﬁeld elements, that has been transmitted at this position. Based on this we have to conclude what is the most likely code vector that had been transmitted. 15.6.23 Remark Decoding polar codes can be done recursively. The algorithm is called Successive Cancelation. The deﬁned transform G implies a natural order of the elements of b. The elements of b are processed in this order under the assumption that the previous elements of b have been determined. It can be shown that asymptotically in the length of the code 738 Handbook of Finite Fields a subset J of the elements have negligible probability of being wrongly decoded, while the rest of elements are correctly decoded only with probability tending to 1/q. This allows to employ the following encoding: place the encoded information on the positions of J, while the values of the rest of the symbols are ﬁxed to some prescribed value, e.g., to 0. The code rate, equal to the proportion \|J\|/n, asymptotically achieves the symmetric capacity of memoryless channels. A detailed description of the algorithm is given below. 15.6.24 Remark Noticing that the last step of the transform G is just a concatenation of transfor-mations g of the transposed columns of a matrix of size ℓ× ℓm−1, we may reconstruct the rows of the matrix row by row. Moreover, we use knowledge from the previously decoded rows. 15.6.25 Remark The last step of transform G is ˆ g(bm−1 0 ), i.e., if bm−1 0 = (bm−1 0 , bm−1 1 , . . . , bm−1 n−1 ), then it is a concatenation of the transformations of columns of the matrix Bm−1 =      bm−1 0 bm−1 1 . . . bm−1 ℓm−1−1 bm−1 ℓm−1 bm−1 ℓm−1+1 . . . bm−1 2ℓm−1−1 . . . . . . . . . . . . bm−1 ℓm−ℓ bm−1 ℓm−ℓ+1 . . . bm−1 ℓm−1     =      a0,0 a0,1 . . . a0,n/ℓ−1 a1,0 a1,1 . . . a1,n/ℓ−1 . . . . . . . . . . . . aℓ−1,0 aℓ−1,1 . . . aℓ−1,n/ℓ−1     . Given that we managed to decode the ﬁrst (i−1) rows, we may compute the q probabilities for the entries of the i-th row from the channel output as follows: the probability of ai,j, i = 0, . . . , ℓ−1, j = 0, . . . , n/ℓ−1, to be β ∈F, is just the probability that in the j-th segment of length ℓin the transmitted code word was a vector belonging to T (a0,j,...,a(i−2),j,β), where a0,j, . . . , a(i−2),j are known from the previous decoded rows. 15.6.26 Remark The problem of decoding a polar code of length ℓm is thus reduced to ℓdecodings of polar codes of length ℓm−1, encoding the rows of the matrix. Decoding of each row could be split into ℓdecodings of codes of length ℓm−1, etc. Finally, we arrive at decodings of single symbols, being the entries of the initial vector b. If this entry has index belonging to the set J, then it is zero, otherwise we may choose the most probable element of F as our decision. It was shown in that when the rate of the code is less than the channel capacity, there exists a choice of set J allowing negligible probabilities of errors in the entries where we make a choice. 15.6.27 Theorem The complexity of the successive cancellation decoding is O(n log n). 15.6.28 Theorem Let the polar code be based on a proper space decomposition and g be the corresponding transform. Let d = (d0, d1, . . . , dℓ−1) be its distance hierarchy. Then for any rate less than the capacity of the channel and growing code length n, the probability of decoding error decays as O(q−nE(g)), where the decomposition exponent E(g) satisﬁes E(g) ≥1 ℓ ℓ−1 X i=0 logℓdi. 15.6.29 Remark The theorem is a consequence of [122, 2161]. 15.6.30 Example For the binary decomposition from Example 15.6.3, E(g) = 0.5. For the quater-nary (q = 4, ℓ= 4) decomposition from Example 15.6.8, E(g) = 0.573120 . . .. Algebraic coding theory 739 15.6.4 Historical notes and other results 15.6.31 Remark Polar codes were proposed by Arikan and provided a scheme for achieving the symmetric capacity of binary memoryless channels (BMC) with polynomial encoding and decoding complexity. The original construction by Arikan yields a binary code of block length n = 2m, and a ﬂexible rate. 15.6.32 Remark Diﬀerent decompositions were considered in [1796, 2161, 2163, 2428, 2429, 2531, 2776]. For the binary case the decomposition from Example 15.6.3 gives the best expo-nent for all lengths up to 13, see . In non-linear decompositions of lengths 14, 15, and 16 based on partitions of a Hamming code to cosets of the Nordstrom-Robinson code, which in turn is partitioned to cosets of the ﬁrst-order Reed-Muller codes, are de-scribed. These decompositions provide a better exponent than any linear ones. However, for linear decompositions the smallest ℓfor which the exponent is greater than 0.5 is 16, for which E(g) = 0.51828; see . In , along with extended Reed-Solomon codes, nested families of algebraic-geometric codes are used to construct non-binary decomposi-tions. In it was suggested to use nested families of BCH codes and codes achieving the Gilbert-Varshamov bound to construct eﬃcient decompositions. It was shown that when ℓ increases, the best error exponent tends to 1. A construction of polar codes using several decompositions over diﬀerent ﬁelds is proposed in . 15.6.33 Remark As mentioned in [119, Section I.D] the notion of polar coding is strongly related to previous ideas in coding theory, such as multi-level coding and Reed-Muller codes. Another strong origin of polar coding is a previous paper by Arikan where the channel combining and splitting were used to demonstrate that improvements can be obtained for the sum cutoﬀrate of some appropriate channels. 15.6.34 Remark In [118, 119, 1442, 2162, 2531] the problem of optimizing the choice of the infor-mation subset was considered for diﬀerent channels. 15.6.35 Remark The tradeoﬀbetween the block length, the gap to capacity and the asymptotic decoding error probability was considered in . Decoding implementations were con-sidered in [119, 120, 1906] A list decoding for polar codes was introduced in . Using polar codes in concatenated schemes was discussed in [181, 1814]. 15.6.36 Remark Use of polar codes in other areas of information theory was considered in [2, 103, 121, 314, 753, 1520, 1685, 1794, 1797, 1993, 2532]. References Cited: [2, 103, 117, 118, 119, 120, 121, 122, 181, 314, 753, 1442, 1520, 1685, 1794, 1795, 1796, 1797, 1814, 1906, 1993, 2161, 2162, 2163, 2428, 2429, 2531, 2532, 2772, 2776] This page intentionally left blank This page intentionally left blank 16 Cryptography 16.1 Introduction to cryptography...................... 741 Goals of cryptography • Symmetric-key cryptography • Public-key cryptography • Pairing-based cryptography • Post-quantum cryptography 16.2 Stream and block ciphers .......................... 750 Basic concepts of stream ciphers • (Alleged) RC4 algorithm • WG stream cipher • Basic structures of block ciphers • RC6 • Advanced Encryption Standard (AES) RIJNDAEL 16.3 Multivariate cryptographic systems .............. 764 The basics of multivariate PKCs • Main constructions and variations • Standard attacks • The future 16.4 Elliptic curve cryptographic systems ............. 784 Cryptosystems based on elliptic curve discrete logarithms • Pairing based cryptosystems 16.5 Hyperelliptic curve cryptographic systems....... 797 Cryptosystems based on hyperelliptic curve discrete logarithms • Curves of genus 2 • Curves of genus 3 • Curves of higher genus • Key sizes • Special curves • Random curves: point counting • Pairings in hyperelliptic curves 16.6 Cryptosystems arising from Abelian varieties ... 803 Deﬁnitions • Examples • Jacobians of curves • Restriction of scalars • Endomorphisms • The characteristic polynomial of an endomorphism • Zeta functions • Arithmetic on an Abelian variety • The group order • The discrete logarithm problem • Weil descent attack • Pairings based cryptosystems 16.7 Binary extension ﬁeld arithmetic for hardware implementations ..................................... 811 Preamble and basic terminologies • Arithmetic using polynomial operations • Arithmetic using matrix operations • Arithmetic using normal bases • Multiplication using optimal normal bases • Additional notes 16.1 Introduction to cryptography Alfred Menezes, University of Waterloo 741 742 Handbook of Finite Fields 16.1.1 Goals of cryptography The fundamental goals of cryptography are to provide the following information security services: 1. Conﬁdentiality: Keeping data secret from all but those who are authorized to see it. 2. Authentication: Corroborating the source of data, and that the data has not been altered by unauthorized means. 3. Non-repudiation: Preventing the denial of a previous commitment to being the source of some data. Cryptography has been used throughout the ages to encrypt (scramble) data, so that only the intended recipient can decrypt (descramble) thereby recovering the original data. Traditional encryption schemes can be classiﬁed as symmetric-key because the sender and intended recipient share secret keying material that can be used to encrypt as well as decrypt. In 1975, Diﬃe, Hellman, and Merkle invented the concept of public-key cryptography, wherein each party A has a pair of keys — a public key that is available to everyone and a private key that is kept secret. Any party can use A’s public key to encrypt a message for A in such a way that decryption requires knowledge of the corresponding private key, thus ensuring that only A can decrypt. Furthermore, A can use her private key to generate a message-dependent signature on a message in such a way that any user in possession of A’s public key can verify that A indeed signed the message; this facilitates provision of non-repudiation services. Note that unlike the case with symmetric-key cryptography where communicating parties must possess shared secret keys, users of a public-key cryp-tosystem only need to possess authentic copies of each other’s public keys thus simplifying the management and distribution of keying material. Subsection 16.1.2 provides some examples of symmetric-key encryption schemes; a more extensive treatment can be found in Section 16.2. Subsection 16.1.3 introduces the RSA public-key encryption and signature schemes and some fundamental discrete logarithm-based cryptosystems. Discrete logarithm cryptosystems based on elliptic curves, hyperellip-tic curves, and abelian varieties are covered in Sections 16.4, 16.5, and 16.6, respectively. Subsection 16.1.4 introduces the relatively new ﬁeld of pairing-based cryptography. Finally, Subsection 16.1.5 describes a public-key encryption scheme that may be resistant to attacks by a quantum computer. 16.1.2 Symmetric-key cryptography 16.1.1 Deﬁnition Symmetric-key encryption schemes, also called ciphers, are usually classiﬁed as being a stream cipher in which encryption is performed one character (or bit) at a time, or a block cipher in which encryption is performed on a block of characters (or bits). 16.1.2 Remark Stream ciphers are generally preferred over block ciphers in applications where buﬀering is limited and message characters must be individually processed as they are received. 16.1.2.1 Stream ciphers 16.1.3 Example A classical example of a stream cipher is the simple substitution cipher. The secret key is a randomly selected permutation π of the English alphabet. An English plaintext Cryptography 743 message m = m1, m2, m3, . . . (where each mi is a letter of the English alphabet) is encrypted one letter at a time, to produce the ciphertext c = c1, c2, c3, . . . where ci = π(mi). Decryption is performed by applying the inverse permutation to the ciphertext letters: mi = π−1(ci). The total number of possible keys is 26! ≈288.4, which is suﬃciently large that exhaustively searching through the set of all keys for the secret key is computationally infeasible. However, an adversary who has captured a suﬃciently long ciphertext (say, of length 500) can easily use statistical knowledge of the English language (such as the relative frequency of letters) to determine the secret key π. 16.1.4 Example The one-time pad is a stream cipher that perfectly hides all statistical information that may be present in the plaintext message. The secret key is a randomly selected binary string k = k1, k2, k3, . . .. The plaintext is written as a binary string m = m1, m2, m3, . . .. Encryption is performed one bit at a time — the ciphertext is c = c1, c2, c3, . . ., where ci = mi ⊕ki. Here, ⊕denotes bitwise addition modulo 2. Decryption is similarly performed: mi = ci ⊕ki. As shown by Shannon , the one-time pad achieves perfect secrecy in the sense that an adversary — even one having inﬁnite computational resources — who intercepts a ciphertext c is unable to determine anything whatsoever about the plaintext m except for its length. 16.1.5 Remark While perfectly secure, the one-time pad has a serious deﬁciency that the secret key must have the same length as the plaintext message. Note that the secret key cannot be reused to encrypt a second message. Indeed, if k is used to encrypt two plaintext messages m and m′, then the resulting ciphertexts c = m ⊕k and c′ = m′ ⊕k′ can be added together to obtain c ⊕c′ = m ⊕m′, from which information about m and m′ might be deduced. 16.1.6 Remark To overcome the aforementioned deﬁciency, stream ciphers have been designed that take an initial secret key k that is relatively small (e.g., 128 bits), and use a deterministic algorithm called a keystream generator KG to produce a much longer binary string KG(k). The bits of KG(k) are then added to the plaintext bits to produce ciphertext, in this way simulating the one-time pad. Since KG(k) is no longer a truly random string, the stream cipher does not achieve perfect secrecy. However, the hope is that the bits of KG(k) are “suﬃciently random” so that a computationally bounded adversary will be unable to deduce any information about the plaintext from the ciphertext. 16.1.7 Remark Stream ciphers are further studied in Section 16.2. 16.1.2.2 Block ciphers 16.1.8 Deﬁnition A block cipher consists of a family of encryption functions Ek : {0, 1}n → {0, 1}n parameterized by an ℓ-bit key k. Each function in the family is invertible. The inverse of Ek is the decryption function Dk. 16.1.9 Remark If two parties wish to communicate securely, they ﬁrst agree upon a secret key k ∈{0, 1}ℓ. Then, to transmit a message m ∈{0, 1}n, a party computes c = Ek(m) and sends c. The recipient computes m = Dk(c). 16.1.10 Example Feistel ciphers are a general class of block ciphers. The parameters of a Feistel cipher are n (the block length), ℓ(the key length), and h (the number of rounds). The ingredients are a key scheduling algorithm that determines subkeys k1, k2, . . . , kh from k, and component functions fi : {0, 1}n/2 →{0, 1}n/2 for 1 ≤i ≤h, where fi depends on ki. A plaintext block m ∈{0, 1}n is encrypted as follows: 1. Write m = (m0, m1), where m0, m1 ∈{0, 1}n/2. 744 Handbook of Finite Fields 2. Compute mi+1 = mi−1 ⊕fi(mi) for i = 1, 2, . . . , h. 3. The ciphertext is c = (mh, mh+1). The ciphertext block c is decrypted as follows: 1. Write c = (mh, mh+1). 2. Compute mi−1 = mi+1 ⊕fi(mi) for i = h, h −1, . . . , 1. 3. The plaintext is m = (m0, m1). 16.1.11 Example The Data Encryption Standard (DES) is the most well-known example of a Feis-tel cipher . DES was developed in the mid 1970s and has been widely deployed in commercial applications. The DES parameters are n = 64, ℓ= 56, and h = 16. The set of all possible keys has cardinality only 256, whereby exhaustive key search can be performed in a few hours on a network of workstations. Hence, DES is considered to be insecure today. 16.1.12 Example Although DES is no longer recommended in practice, Triple-DES is considered secure and widely deployed. In Triple-DES, the secret key is comprised of a triple k = (k′, k′′, k′′′) of DES keys. A plaintext block m ∈{0, 1}n is encrypted as c = DESk′′′(DESk′′(DESk′(m))), where DES denotes the DES encryption function. The ciphertext block c can be decrypted as follows: m = DES−1 k′ (DES−1 k′′ (DES−1 k′′′(c))), where DES−1 denotes the DES decryption function. The secret key has bitlength 168 ren-dering exhaustive key search infeasible. Given a few plaintext-ciphertext pairs, the secret key can be recovered by a meet-in-the-middle attack that has running time approximately 2112 steps; however, this attack is considered infeasible in practice. 16.1.13 Example Block ciphers encrypt a long message n bits at a time. The drawback of this method is that identical plaintext blocks result in identical ciphertext blocks, and hence the ciphertext may leak information about the plaintext. To circumvent this weakness, long messages can be encrypted using the cipher-block-chaining (CBC) mode of encryption. A long message m is ﬁrst broken up into blocks m1, m2, . . . , mt, each of bitlength n. Then, a random initialization vector c0 ∈{0, 1}n is chosen, and ci = Ek(mi ⊕ci−1) is computed for i = 1, 2, . . . , t. The ciphertext is c = (c0, c1, . . . , ct). Decryption is accomplished by computing mi = DES−1 k (ci) ⊕ci−1 for i = 1, 2, . . . , t. 16.1.3 Public-key cryptography 16.1.3.1 RSA 16.1.14 Remark The RSA encryption and signature schemes were introduced in a 1978 paper by Rivest, Shamir, and Adleman . 16.1.15 Algorithm [RSA key generation] Each party does the following: 1. Randomly select two (distinct) primes p and q of the same bitlength. 2. Compute n = pq and φ = (p −1)(q −1). 3. Select an arbitrary integer e, 1 < e < φ, with gcd(e, φ) = 1. 4. Compute the integer d, 1 < d < φ, with ed ≡1 (mod φ). 5. The party’s public key is (n, e); her private key is d. Cryptography 745 16.1.16 Remark The adversary’s task of computing the private key d corresponding to a public key (n, e) can be shown to be equivalent to the problem of factoring n. As of 2012, factoring 1024-bit RSA moduli n is out of reach of the fastest integer factorization algorithms. However, 2048-bit moduli and 3072-bit moduli are recommended for medium- and long-term security. 16.1.17 Remark Presented next are the basic versions of the RSA encryption and signature schemes. In the signature scheme H : {0, 1}∗→[0, n −1] is a cryptographic hash function (Re-mark 16.1.23). 16.1.18 Remark In what follows, a = b mod n is understood to mean that a is the reminder of b when divided by n, and as usual, a ≡b (mod n) means a and b are congruent modulo n. 16.1.19 Algorithm (RSA encryption scheme) To encrypt a message m ∈[0, n −1] for party A, do the following: 1. Obtain an authentic copy of A’s public key (n, e). 2. Compute c = me mod n. 3. Send c to A. To decrypt c, party A does the following: 1. Compute m = cd mod n. 16.1.20 Algorithm (RSA signature scheme) To sign a message m, party A with public key (n, e) and private key d does the following: 1. Compute h = H(m). 2. Compute s = hd mod n. 3. A’s signature on m is the integer s. To verify A’s signature s on the message m, do the following: 1. Obtain an authentic copy of A’s public key (n, e). 2. Compute h = H(m). 3. Accept the signature if and only if se ≡h (mod n). 16.1.21 Remark The RSA encryption and signature schemes work because (me)d ≡m (mod n) for all m ∈[0, n −1], a property that can easily be veriﬁed using Fermat’s little theorem. 16.1.22 Remark Security is based on the intractability of the problem of computing e-th roots modulo n. While it is clear that this problem is no harder than that of factoring n, the equivalence of the two problems has not been proven. 16.1.23 Remark A hash function H : {0, 1}∗→{0, 1}ℓis an eﬃciently-computable function that meets some cryptographic requirements such as one-wayness given a randomly chosen ele-ment h ∈{0, 1}ℓit is computationally infeasible to ﬁnd any m ∈{0, 1}∗with H(m) = h) and collision resistance (it is computationaly infeasible to ﬁnd distinct m1, m2 ∈{0, 1}∗with H(m1) = H(m2)). Examples of commonly-used hash functions are SHA-1 and SHA-256 . 746 Handbook of Finite Fields 16.1.3.2 Discrete logarithm cryptosystems 16.1.24 Deﬁnition Let G be a multiplicatively-written group of prime order n, and let g be a generator of G. The discrete logarithm problem in G to the base g is the following problem: Given y ∈G, ﬁnd the integer x ∈[0, n −1] such that y = gx. 16.1.25 Remark The fastest generic algorithm known for solving the discrete logarithm problem is Pollard’s rho algorithm which has a running time of O(√n) and its parallelization by van Oorschot and Wiener . 16.1.26 Example The ﬁrst example of a discrete logarithm cryptographic system was the Diﬃe-Hellman key agreement protocol . The purpose of this protocol is to enable two parties A and B to agree upon a shared secret by exchanging messages over a communications channel whose contents are authenticated but not secret. Given a cyclic group G = ⟨g⟩of order n, party A randomly selects an integer a ∈[1, n −1] and sends ga to B. Similarly, party B randomly selects an integer b ∈[1, n −1] and sends gb to A. Both parties can compute the shared secret k = gab. An eavesdropper is faced with the task of determining gab given g, ga and gb. This is the Diﬃe-Hellman problem, whose intractability is assumed to be equal to that of the discrete logarithm problem in G . 16.1.27 Example ElGamal designed a closely-related scheme for public-key encryption . In this scheme, party A’s private key is a randomly selected integer a ∈[1, n −1] and her public key is the group element ga. To encrypt a message m ∈G for A, a party selects a random integer k ∈[1, n −1], computes c1 = gk and c2 = m(ga)k, and sends (c1, c2) to A. Party A decrypts by computing m = c2/ca 1. The basic security requirement is that an adversary should be unable to compute m given the public key ga and ciphertext (c1, c2). It is easy to see that the adversary’s task is equivalent to solving an instance of the Diﬃe-Hellman problem. 16.1.28 Remark The main criteria for selecting a suitable group G for implementing a discrete logarithm cryptosystem are that (i) the group operation can be eﬃciently computed (so that cryptographic operations can be eﬃciently performed); and (ii) the discrete logarithm problem should be intractable. Over the years, several families of groups have been proposed for cryptographic use including subgroups of: 1. The multiplicative group of a ﬁnite ﬁeld. 2. The group E(Fq) of Fq-rational points on an elliptic curve E deﬁned over a ﬁnite ﬁeld Fq [1771, 2102]. 3. The divisor class group of a genus-g hyperelliptic curve deﬁned over a ﬁnite ﬁeld Fq . 4. The group of Fq-rational points on an abelian variety deﬁned over a ﬁnite ﬁeld Fq (Section 16.6). 5. The class group of an imaginary quadratic number ﬁeld . 16.1.29 Remark Public-key cryptosystems designed using elliptic curves, hyperelliptic curves, and abelian varieties are studied in Sections 16.4, 16.5, and 16.6, respectively. An example of a discrete logarithm cryptographic scheme that employs a multiplicative subgroup of a ﬁnite ﬁeld is presented next. 16.1.3.3 DSA 16.1.30 Remark The digital signature algorithm (DSA) was proposed by the U.S. government’s National Institute of Standards and Technology in 1991 . It was the ﬁrst digital Cryptography 747 signature scheme to be recognized by a government. 16.1.31 Algorithm (DSA key generation) System parameters are (p, q, g), where p is a prime, q is a prime divisor of p −1, and g ∈Z/pZ has multiplicative order q. To generate a key pair, each party does the following: 1. Randomly select an integer a from [1, q −1]. 2. Compute y = ga mod p. 3. The party’s public key is y; her private key is a. 16.1.32 Algorithm (DSA signature scheme) To sign a message m, party A with system parameters (p, q, g), public key y, and private key a does the following: 1. Randomly select an integer k ∈[1, q −1]. 2. Compute r = (gk mod p) mod q. 3. Compute h = H(m) and s = k−1(h + ar) mod q. 4. A’s signature on m is the pair of integers (r, s). To verify A’s signature (r, s) on the message m, do the following: 1. Obtain an authentic copy of the system parameters (p, q, g) and A’s public key y. 2. Verify that r ∈[1, q −1] and s ∈[1, q −1]. 3. Compute h = H(m) and w = s−1 mod q. 4. Compute u1 = w · h mod q and u2 = r · w mod q. 5. Compute v = (gu1yu2 mod p) mod q. 6. Accept the signature if and only if v = r. 16.1.33 Remark A correctly-generated signature (r, s) on a message m will always be accepted because gu1yu2 ≡gwh+war ≡gs−1(h+ar) ≡gk (mod p). 16.1.34 Remark Security of DSA is based on the intractability of the discrete logarithm problem in the order-q multiplicative subgroup of Z/pZ. The fastest algorithms known for solving this problem are summarized in Section 11.6. As of 2012, DSA is considered to be secure if the bitlengths of the primes p and q are 1024 and 160, respectively. However, 2048-bit p and 224-bit q are recommended for medium-term security, and 3072-bit p and 256-bit q are recommended for long-term security. 16.1.4 Pairing-based cryptography 16.1.35 Deﬁnition Let G1 = ⟨g1⟩, G2 = ⟨g2⟩and G3 be multiplicatively-written groups of prime order n. A pairing on (G1, G2, G3) is a map e : G1×G2 →G3 that satisﬁes the following three conditions: 1. Bilinearity: For all r1, r2 ∈G1 and s1, s2 ∈G2, e(r1r2, s1) = e(r1, s1) · e(r2, s1) and e(r1, s1s2) = e(r1, s1) · e(r1, s2). 2. Non-degeneracy: e(g1, g2) ̸= 1. 3. Computability: e(r, s) can be eﬃciently computed for all r ∈G1 and s ∈G2. 16.1.36 Remark Since 2000, pairings have been widely used to design cryptographic protocols that attain objectives not known to be achievable using conventional methods. The ﬁrst such 748 Handbook of Finite Fields protocol was a one-round three-party key agreement scheme due to Joux . Recall that the Diﬃe-Hellman key agreement scheme is a two-party protocol where the two exchanged messages are independent of each other, and therefore can be simultaneously exchanged. Joux showed how pairings can be used to construct an analogous one-round key agreement scheme for three parties. 16.1.37 Example Suppose that e is a symmetric pairing, i.e., G1 = G2. In Joux’s protocol, each of the three communicating parties A, B, C, randomly selects integers a, b, c ∈[1, n −1], and simultaneously broadcasts the group elements ga 1, gb 1, gc 1, respectively. The shared secret is k = e(g1, g1)abc which party A, for example, can compute as k = e(gb 1, gc 1)a. A passive adversary’s task is to compute k given g1, ga 1, gb 1 and gc 1. This problem is the bilinear Diﬃe-Hellman problem, and is assumed to be no easier than the discrete logarithm problems in G1 and G3. 16.1.38 Example A fundamental pairing-based protocol is the Boneh-Franklin identity-based en-cryption scheme . The scheme has the feature that a party B can encrypt a message for a second party A using only A’s identiﬁer (such as A’s email address) and some publically-available system parameters. Party A decrypts the message using a secret key that it must obtain from a trusted third party (TTP). Unlike symmetric-key cryptography, A and B do not have to share secret keying material. Also, unlike public-key cryptography, it is not necessary for A to have a public key before B can encrypt a message for A. 16.1.39 Remark A basic version of the Boneh-Franklin scheme is described next using symmetric pairings. The scheme uses a bilinear pairing e on (G1, G1, G3) and two cryptographic hash functions H1 : {0, 1}∗→G1 and H2 : G3 →{0, 1}ℓ, where ℓis the length of the message to be encrypted. 16.1.40 Algorithm (Boneh-Franklin identity based encryption) In the setup stage, a trusted third party (TTP) generates keying material for itself. 1. The TTP randomly selects an integer t ∈[1, n −1]. 2. The TTP’s public key is T = gt 1 and its private key is t. At any time, a party A with identiﬁer IDA can request its private key dA from the TTP: 1. The TTP computes dA = H1(IDA)t and securely delivers dA to A. To encrypt a message m ∈{0, 1}ℓfor A, do the following: 1. Randomly select an integer r ∈[0, n −1]. 2. Compute R = gr 1 and C = m ⊕H2(e(H1(IDA), T)r). 3. Send (R, C) to A. To decrypt (R, C), party A does the following: 1. Obtain the private key dA from the TTP. 2. Compute m = C ⊕H2(e(dA, R)). 16.1.41 Remark Decryption works because e(dA, R) = e(H1(IDA)t, gr 1) = e(H1(IDA), gt 1)r = e(H1(IDA), T)r. 16.1.42 Remark Security is based on the hardness of the bilinear Diﬃe-Hellman problem. 16.1.43 Example Another fundamental pairing-based protocol is the Boneh-Lynn-Shacham (BLS) short signature scheme . Unlike the DSA signature scheme (Subsection 16.1.3.3), BLS signatures have only one component and thus can be shorter if suitable parameters are chosen. Cryptography 749 16.1.44 Remark The BLS signature scheme is described next using asymmetric pairings. The scheme uses a bilinear pairing e on (G1, G2, G3) and a cryptographic hash function H : {0, 1}∗→G∗ 1. 16.1.45 Algorithm (BLS key generation) To generate a key pair, each party does the following: 1. Randomly select an integer x from [1, n −1]. 2. Compute X = gx 2. 3. The party’s public key is X; her private key is x. 16.1.46 Algorithm (BLS signature scheme) To sign a message m, party A with public key X and private key x does the following: 1. Compute M = H(m) and S = M x. 2. A’s signature on m is the group element S. To verify A’s signature S on the message m, do the following: 1. Obtain an authentic copy of A’s public key X. 2. Verify that S ∈G∗ 1. 3. Compute M = H(m). 4. Accept the signature if and only if e(S, g2) = e(M, X). 16.1.47 Remark A correctly-generated signature S on a message m will always be accepted because e(S, g2) = e(M x, g2) = e(M, gx 2) = e(M, X). 16.1.48 Remark Security is based on the hardness of the following variant of the Diﬃe-Hellman problem: given M ∈G1 and X ∈G2, compute M x where X = gx 2. 16.1.49 Remark Pairings that are suitable for implementing Joux’s key agreement scheme, the Boneh-Franklin identity-based encryption scheme, and the BLS short signature scheme can be constructed from the Weil and Tate pairings deﬁned on certain elliptic curves over ﬁnite ﬁelds. For further details, see Section 16.4. 16.1.5 Post-quantum cryptography 16.1.50 Remark Shor showed that integer factorization and discrete logarithm problems can be eﬃciently solved on a quantum computer, thus rendering RSA and all discrete logarithm cryptosystems insecure. As of 2012, the feasibility of building large-scale quantum computers is far from certain. Nonetheless, cryptographers have been designing and analyzing public-key cryptosystems that potentially resist attacks by quantum computers, and which could serve as replacements to RSA and discrete logarithm cryptosystems in the event that large-scale quantum computers become a reality. Among these post-quantum cryptosystems are quantum key distribution and conventional cryptosystems based on hash functions, error-correcting codes, lattices, and multivariate quadratic equations . Cryptosystems based on multivariate quadratic equations are examined in Section 16.3. A code-based cryptosystem is described next. 16.1.51 Remark In 1978, McEliece introduced a public-key encryption scheme based on error cor-recting codes . The security of McEliece’s scheme is based on the hardness of the general decoding problem, a problem that is known to be NP-hard . 16.1.52 Algorithm (McEliece key generation) Each party does the following: 750 Handbook of Finite Fields 1. Select a k × n generator matrix G for a t-error correcting binary (n, k)-code for which there is an eﬃcient decoding algorithm. 2. Randomly select a k × k binary invertible matrix S. 3. Randomly select a n × n permutation matrix P. 4. Compute the k × n matrix b G = SGP. 5. The party’s public key is (n, k, t, b G); her private key is (S, G, P). 16.1.53 Algorithm (McEliece encryption scheme) To encrypt a message m ∈{0, 1}k for party A, do the following: 1. Obtain an authentic copy of A’s public key (n, k, t, b G). 2. Randomly select an error vector z of length n and Hamming weight t. 3. Compute c = m b G + z. 4. Send c to A. To decrypt c, party A does the following: 1. Compute b c = cP −1. 2. Use the decoding algorithm for the code generated by G to decode b c to b m. 3. Compute m = b mS−1. 16.1.54 Remark Decryption works because b c = cP −1 = (m b G + z)P −1 = (mSGP + z)P −1 = (mS)G + zP −1. Since zP −1 is a vector of Hamming weight t, the decoding algorithm for the code generated by G decodes b c to b m = mS, whence b mS−1 = m. 16.1.55 Remark McEliece’s original paper proposed using a Goppa code with parameters n = 1024, n = 524, and t = 50. However, it has recently been shown that the McEliece encryption scheme with these parameters is insecure . Research is ongoing to determine parameter sets for which one can have a high conﬁdence that the McEliece encryption scheme will remain resistant to both classical and quantum attacks for the forseeable future. See Also , , Textbooks on cryptography. , References Cited: [227, 235, 245, 250, 344, 346, 440, 965, 1048, 1065, 1067, 1068, 1521, 1624, 1694, 1771, 1772, 2039, 2048, 2080, 2102, 2413, 2462, 2609, 2623, 2720, 2852] 16.2 Stream and block ciphers Guang Gong, University of Waterloo Kishan Chand Gupta, Indian Statistical Institute Cryptography 751 16.2.1 Remark We present some algorithms for stream ciphers and block ciphers. In stream cipher cryptography a pseudorandom sequence of bits of length equal to the message length is gen-erated by a pseudorandom sequence generator (PSG). This sequence is then bitwise XOR-ed (addition modulo 2) with the message sequence and the resulting sequence is transmitted. At the receiving end, deciphering is done by generating the same pseudorandom sequence and bitwise XOR-ing the cipher bits with this sequence. The seed for the pseudorandom bit generator is the secret key. A general algorithm for a pseudorandom sequence generator is based on a recursive relation over a ﬁnite ﬁeld, a ﬁnite state machine in general and a feedback shift register sequence in particular, with a ﬁltering function or some control units. 16.2.2 Remark In block cipher cryptography, the message bits are divided into blocks and each block is separately provided as an input to a permutation, i.e., encryption, using the same key and transmitted. A block cipher basically is a permutation of a ﬁnite ﬁeld (or a ﬁnite ring), which is a composition of multiple permutations in a subﬁeld (or subring) of the ﬁnite ﬁeld (or the ﬁnite ring). Most modern day block ciphers are iterated ciphers and use substitution boxes (S-boxes) (i.e., permutations) as the nonlinear part in the scheme. 16.2.3 Remark We consider stream ciphers like RC4 and the WG stream cipher , and block ciphers like RC6 and AES . The aim is to explain the underlying ideas rather than describing complete solutions. 16.2.1 Basic concepts of stream ciphers 16.2.4 Remark Stream ciphers are a very important class of cryptographic primitives for encryp-tion, authentication, and key derivation. The basic principle behind stream cipher encryp-tion is simple. 16.2.5 Deﬁnition (One-time pad) Let zt, for t ≥0, be a random key bit sequence which is known to both the sender and the receiver. Suppose the sender wants to send a message bit sequence mt. The cipher bit sequence is computed as ct = mt ⊕zt, and transmitted to the receiver. The receiver knowing zt, computes mt = ct ⊕zt. 16.2.6 Remark This simple scheme provides the highest level of security, called perfect secrecy. It is unbreakable under the assumption that each key is used only for one encryption. 16.2.7 Remark The main problem with the one-time pad is that the key sequence is as long as the message sequence and for each encryption we need a new random key sequence which has to be shared by sender and receiver. This creates serious key management and key distribution problems. One remedy is to use a pseudorandom sequence generator (PSG) also known as keystream generator. A PSG is a deterministic algorithm which starts with a reasonably short random bit string (called a seed) and expands it into a very long bit string which is used as the keystream. The seed is the secret key shared between sender and receiver. The security of the stream cipher depends on the security of the PSG. Informally a PSG is secure if given a segment of the generated key bits it is hard to predict the next bit. Equivalently, it must be computationally very hard to distinguish the generated pseudorandom sequence from a random sequence. 16.2.8 Remark Most modern stream ciphers use an initialization vector (IV) which is not secret. The PRG is seeded by the (key, IV) pair. The same key may be used with distinct IVs and the constraint on the protocol usage is that a (key, IV) pair should not be repeated. Current stream ciphers have a similar structure which can be described by a ﬁnite state machine 752 Handbook of Finite Fields (FSM). The Ecrypt home page contains thorough information and may be referred to by anybody who is interested in the design and analysis of stream ciphers. 16.2.9 Deﬁnition (Security assumptions) It is assumed that the adversary knows everything except the secret key. This is known as Kerckhoﬀ’s principle. A few details are: 1. Algorithms in a stream cipher are public. 2. The only secret information in the system is the pre-shared key. 3. An attacker can intercept communications (ciphertext) among communicating entities. 16.2.10 Remark From Assumption 3, attackers can always obtain ciphertext. If an attacker manages to obtain a certain amount of the corresponding plaintext, then this portion of keystream is exposed. This is referred to as a known plaintext attack. Thus the security of stream cipher is reduced to randomness of PSG. The attacker’s goal may be to recover the secret key or partial information about the secret key using a portion of the known keystream, i.e., using the known portion of the output of PSG. 16.2.11 Deﬁnition (The two phases in stream cipher) A stream cipher consists of two phases: one is the key initialization phase, for which the algorithm is key initialization algorithm (KIA), and the other is the PSG running phase and the algorithm is PSG. Usually, the algorithms used in these two phases are similar. 16.2.12 Remark More speciﬁcally, KIA is the same as PSG without outputs or it may be a slightly diﬀerent function. Figure 16.2.1 shows a general model of a stream cipher. 16.2.13 Remark In the initialization phase, a key initialization algorithm (KIA) is employed, which has two inputs, one is an initial vector (IV), which is public information, and the other is a secret key, k, which is a pre-shared encryption key. The goal of KIA is to scramble key bits with IV in order to get a complex nonlinear function of k and IV. The output of KIA is provided as an initial value to the PSG. KIA only executes once for each encryption session. After the key initialization, PSG starts to output a keystream which is used in encryption. IV k KIA PSG + zi mi ci Encryption IV k KIA PSG + zi ci mi Decryption Figure 16.2.1 Two algorithms in stream ciphers: KIA and PSG. 16.2.14 Deﬁnition (Stream cipher modeled as a Finite State Machine (FSM)) In general, any PSG can be considered as a ﬁnite state machine (FSM) or some variant of it which may be deﬁned as follows. Suppose Y and Z are ﬁnite ﬁelds (or ﬁnite rings) and the elements of Z are represented by m bits. An FSM is a 5-tuple (S0, F, G, n, m) where S0 ∈Yn is the initial state, F : Yn × Z →Yn is the state update function and G : Yn × Z →Z Cryptography 753 is the output function. At time instant t ≥0, state St is represented by an n-tuple (st, st+1, . . . , st+n−1) where si ∈Y and the output zt ∈Z. The state update function and output function are given respectively by St+1 = F(St, zt) and zt+1 = G(St, zt). 16.2.15 Remark In modeling practical stream ciphers, in many cases it is seen that, for any St, F(St, zt) and G(St, zt) do not depend on zt. In this situation we write St+1 = F(St) and zt+1 = G(St). 16.2.2 (Alleged) RC4 algorithm 16.2.16 Remark RC4 was designed by Rivest in 1987 and kept as a trade secret until it was leaked in 1994. It is widely used in Internet communications. In the open literature, RC4 is one of the very few proposed keystream generators that are not based on shift registers. A design approach of RC4 which has originated from the exchange-shuﬄe paradigm, is to use a relatively big array/table that slowly changes with time under the control of itself. For a detailed discussion on RC4 see the Master’s thesis of Mantin . 16.2.17 Deﬁnition RC4 has an N-stage register S, which holds a permutation of all N = 2n possible n-bit integers, where n is typically chosen as 8. The initial state is derived from a key (whose typical size is between 40 and 256 bits) by a Key-Scheduling Algorithm (KSA), i.e., Key initialization algorithm (KIA). The PSG is referred to as the Pseudo-Random Generation Algorithm (PRGA) in RC4. 16.2.18 Remark In what follows, a = b (mod n) is understood to mean that b is the reminder of a when divided by n. 16.2.19 Deﬁnition (KSA of RC4) FSM for KSA for a given secret key K of length l bytes is a 4-tuple (Q0, FK, r, m). The secret key is used to scramble S by shuﬄing the words in S. Suppose S = (x0, x1, . . . , xN−1); we denote a state of RC4 by (i, j, S) or equivalently by (i, j, x0, x1, . . . , xN−1). Let (i, j, x0, x1, . . . , xN−1) ∈R be a state at some time instant and let (e, d, y0, y1, . . . , yN−1) = FK(i, j, x0, x1, . . . , xN−1) ∈R be the next state. In the initialization process, i and j are initialized to 0, the identity permutation (0, 1, . . . , N − 1) is loaded in the array S. Thus we have the initial state Q0 = (0, 0, 0, 1, 2, . . . , 255) of KSA. FSM for KSA: (Q0, FK, r, m) Parameters: N = 28 = 256, m = 8, r = N + 2 = 258 and R = ZN+2 N . Initial State: Q0 = (0, 0, 0, 1, 2, . . . , 255) ∈R Input: K = (k0, . . . , kl−1), ks ∈Z8 2 for s = 0, . . . , l −1. State update function: FK : R →R, given by e = i + 1 (mod N), d = (j + xe + ke (mod l) (mod N), yd = xe, ye = xd and yv = xv, for all v ̸= e or d. Final State: F 256 K (Q0) := (i, j, S) which will provide the initial state I0 = (0, 0, S) of PRGA. 754 Handbook of Finite Fields 16.2.20 Deﬁnition (PRGA of RC4.) RC4 keystream generator (PRGA) can also be represented as a ﬁnite state machine (I0, F, G, r, m) where F is the state updating function and G is the output function. Let (i, j, x0, x1, . . . , xN−1) ∈R be a state at some time instant and let (e, d, y0, y1, . . . , yN−1) = F(i, j, x0, x1, . . . , xN−1) ∈R be the next state. Figure 16.2.2 shows the state transition of PRGA. FSM for PRGA: (I0, F, G, r, m) Parameters: N = 28 = 256, m = 8, r = N + 2 = 258 and R = ZN+2 N . Initial state: I0 ∈R from KSA State update function: F : R →R, given by e = i + 1 (mod N), d = j + xe (mod N), yd = xe, ye = xd and yv = xv, for all v ̸= e or d. Output: The output function is G : R →ZN and output is given by xt where t = xe + xd (mod N). d · · · xe · · · xt · · · xd · · · x0 e j + · · · xd d = j + xe · · · xt t = xe + xd (mod N) · · · xe e = i + 1 · · · x0 i + 1 Figure 16.2.2 State transition of PRGA. 16.2.21 Remark (Attacks on RC4) RC4 has a huge internal state of 8×258 = 2064 bits. We observe that in RC4, the state update function is invertible. If the size of the internal state is s in bits (s = (N + 2)(log N) = 2064 in RC4) and the next state update function is randomly chosen, then the average cycle length is about 2s−1 . However, it is hard theoretically to determine any randomness properties for RC4. Cryptanalysis of RC4 attracted a lot of attention in the cryptographic community after it was made public in 1994. Numerous signiﬁcant weaknesses were discovered and notable weakness include weak initialization vectors, classes of weak keys, patterns that appear twice the expected number of times (the second byte bias), and biased distribution of RC4 initial permutation. Weaknesses in the key scheduling algorithm in RC4 led to a practical attack on the security protocol WEP. Currently, it has been proposed to use AES in WEP due to these weaknesses of RC4. 16.2.3 WG stream cipher 16.2.22 Remark We now introduce the WG stream cipher which was submitted to the eSTREAM project in 2005 by Nawaz and Gong . The cipher is based on WG (Welch-Gong) transformations. WG cipher has desired randomness properties, like long periods, large linear complexity, two level autocorrelation and ideal t-tuple distribution. It is resistant to Time/Memory/Data tradeoﬀattacks, algebraic attacks, and correlation attacks. The cipher can be implemented with a small amount of hardware . Cryptography 755 16.2.23 Deﬁnition A WG cipher can be regarded as a nonlinear ﬁlter generator over an exten-sion ﬁeld, ﬁltered by a WG transformation. As shown in Figure 16.2.3, it consists of a linear feedback shift register, followed by a WG permutation transform. The LFSR is based on an l degree primitive polynomial p over the ﬁnite ﬁeld F2m given by, p(x) = Pl i=0 cixi, ci ∈F2m. The LFSR generates a maximal-length sequence (an m-sequence) over F2m. This simple design generates a keystream whose period is 2n −1, where n = lm, and it is easy to analyze various cryptographic properties of the gener-ated keystream. The feedback signal Init is used only in the key initialization phase. In PSG running phase, the feedback is only from the LFSR. The output of the cipher is one bit. We denote a WG cipher with an LFSR of l stages over F2m as WG(m, l). 16.2.24 Remark The version of the WG submitted to eSTREAM is denoted by WG(29, 11). Tr WGperm al−1 al−2 · · · a1 a0 cl−1 cl−2 c1 c0 + + + + m 1 Init m Update of LFSR ak+l =        Pl−1 i=0 ciai+k + WGperm(ak+l−1), 0 ≤k < 2l (in KIA phase) Pl−1 i=0 ciai+k, k ≥2l (in PSG) Output : sk = WG(ak+l−1), k ≥2l Figure 16.2.3 A diagram for WG ciphers. 16.2.25 Deﬁnition For x ∈F2m, WGperm(x) and WG(x) are deﬁned by WGperm(x) = t(x + 1) + 1 t(x) = x + xr1 + xr2 + xr3 + xr4 WG(x) = Tr(WGperm(x)) where r1 = 2k + 1, r2 = 22k + 2k + 1, r3 = 22k −2k + 1, and r4 = 22k + 2k −1 where 3k ≡1 (mod m). 16.2.26 Remark Note that a WG transformation exists only if m ̸≡0 (mod 3) (see ). In practice, we consider a value of m to be a reasonable choice for a WG cipher where m ̸≡0 (mod 3) and either of the following holds: m is small enough to allow an eﬃcient lookup table implementation of the permutation (m ≤11), or m ≡2 (mod 3) and m has an optimal normal basis for eﬃcient implementation in hardware; see Sections 5.3 and 16.7. The suitable values of m for 7 ≤m ≤29 are 7, 8, 10, 11, 23, and 29. 16.2.27 Remark Here we compute exponents, i.e., ri’s from so that t is a permutation poly-nomial. The exponents used in which are taken from are diﬀerent. But WG sequences are identical for both representations. 756 Handbook of Finite Fields 16.2.28 Theorem The linear span of the WG cipher can be determined by the following formula LS = m × X i∈I lw(i) where w(i) is the Hamming weight of i, I = I1 ∪I2 for m = 3k −1 where I1 = {22k−1 + 2k−1 + 2 + i : 0 ≤i ≤2k−1 −3} and I2 = {22k + 3 + 2i : 0 ≤i ≤2k−1 −2}; and I = {1} ∪I3 ∪I4 for m = 3k −2 where I3 = {2k−1 + 2 + i : 0 ≤i ≤2k−1 −3} and I4 = {22k−1 + 2k−1 + 2 + i : 0 ≤i ≤2k−1 −3}. 16.2.29 Remark The linear span of WG(7, 23), WG(8, 20), WG(11, 16), and WG(29, 11) are 230, 233, 224, and 245, respectively. Here the linear spans for m = 7 and m = 8 are computed by using WG(x−1) instead of WG(x), because they have the same autocorrelation, but possess the highest algebraic degree, resulting in larger linear spans. 16.2.30 Remark (Resilient basis) The WG transformation form F2m →F2 can be regarded as a Boolean function in m variables. The exact Boolean representation depends on the basis used for computation in F2m. The basis can be selected in such a way that the corresponding Boolean representation of WG transformation is 1-order resilient. For an algorithm for ﬁnding resilient bases, see . 16.2.31 Remark The WG transformation sequences have been widely investigated in the literature on sequence design. These sequences were discovered by Golomb et al. , and the randomness properties were proved by Dobbertin and Dillon . In 2002, Gong and Youssef presented several cryptographic properties of WG transformation sequences. 16.2.32 Theorem The randomness properties of WG transformation sequences and crypto-graphic properties of the WG transformations as Boolean functions are presented in Table 16.2.4. Randomness Properties of Keystreams Cryptographic Properties of WG Period is 2n −1 1-order resilient Balanced Algebraic degree ⌈m/3⌉+ 1 Ideal 2-level autocorrelation Nonlinearity = 2m−1 −2(m−1)/2 for odd m Linear span increases exponentially with Additive autocorrelation between f(x + a) m, which can be determined exactly and f(x) has three values: 0, ±2 m+1 2 Ideal t-tuple distribution (1 ≤t ≤l) 1-order propagation property Table 16.2.4 Randomness and cryptographic properties of WG. 16.2.33 Example (A concrete design of WG(29, 11)) 1. General description : a. synchronous stream cipher submitted in Proﬁle 2 of eSTREAM (for hard-ware applications); b. key lengths of 80, 96, 112, and 128 bits; c. IVs of 32, 64 bits and also the same lengths as the key are allowed; d. estimated strength 2128 (exhaustive key search) 2. Parameters of WG(29, 11) : The LFSR is of degree 11 and generates an m-sequence over the extension ﬁeld F229. Then the elements of the m-sequence are ﬁltered by a WG transformation: F229 →F2, to produce a keystream sequence. WG transformation operations can be implemented using the optimal normal Cryptography 757 basis in the ﬁnite ﬁeld F229. The parameters for implementation are listed in Table 2. m l WG and Polynomials 29 11 WG(x) = Tr(t(x + 1) + 1) where t(x) = x + xr1 + xr2 + xr3 + xr4 and r1 = 210 + 1, r2 = 220 + 210 + 1 r3 = 220 −210 + 1, r4 = 220 + 210 −1 g(x) = x29 + x28 + x24 + x21 + x20 + x19 + x18 + x17 + x14 + x12 + x11 + x10 + x7 + x6 + x4 + x + 1 p(x) = x11 + x10 + x9 + x6 + x3 + x + γ where γ = α464730077 and g(α) = 0 Table 16.2.5 Parameters for WG(29, 11) implementation using a optimal normal basis. 3. Key Initialization Phase for WG(29, 11): An initial state of the LFSR contains 319 bits where each register holds 29 bits. For a 128-bit key and 128-bit IV (initial vector), the rule for loading the LFSR is shown in Table 3. Here x\|\|y = (x0, . . . , xr−1, y0, . . . , ys−1) is the concatenation of two vectors x = (x0, · · · , xr−1) and y = (y0, . . . , ys−1). Once the LFSR has been loaded with the key and IV, the key stream generator is run for 22 clock cycles. This is the key-initialization phase of the cipher operation. During this phase the 29 bit vector of the output of the WG permutation is added to the feedback of the LFSR which is then used to update the LFSR. Registers 29-bit Format 0,2,4,6,8 16 bits from the key\|\|8 bits from IV\|\|padding zeros 1,3,5,7,9 8 bits from the key\|\|16 bits from IV\|\|padding zeros 10 8 bits from the key\|\|8 bits from IV\|\|padding zeros Table 16.2.6 Initializing the LFSR. 4. PSG phase: After running the KIA for 2l = 22 clock cycles, PSG starts to give 1 bit output for each clock cycle. At this phase, the feedback to LFSR from the output of the WG permutation stops. 16.2.34 Remark Security against known attacks . 1. Randomness and cryptographic properties of WG(29, 11) can be computed from Table 16.2.4. 2. Time/memory/data tradeoﬀattacks: Size of the internal state is 2319. Thus this attack is not applicable. 3. Algebraic attacks: the number of linear equations is approximately 319 11 and the attack has complexity approximately 2182. 4. Correlation attacks: WG transformation is 1-order resilient, and nonlinearity is very high 228 −214. Thus the correlation between the WG transformation and any linear or aﬃne function is very small. 16.2.35 Remark We note that the WG Stream Cipher was not selected for Phase III of the e-Stream Project since, although no attacks against WG were reported, the cipher is compromised 758 Handbook of Finite Fields if a relaxation of at most 245 bits are generated from a single key. Also the hardware implementation seems to be larger than desirable; see . However, the linear span can be increased up to at least 29 × 1128 = 2101.722 using the decimation method as it is done for WG7 in . A hardware implementation has recently been reported in , which shows diﬀerent implementation methods and optimizations. 16.2.4 Basic structures of block ciphers 16.2.36 Deﬁnition A block cipher consists of a pair of encryption and decryption operators E and D. For a ﬁxed key K, E maps an n bit plaintext m = (m0, . . . , mn−1) to an n bit ciphertext c = (c0, . . . , cn−1), namely encryption, and D maps the ciphertext back to the plaintext, i.e., decryption. In other words, E : K × Fn 2 →Fn 2 (D ◦E)(m) = m where m is an n-bit message, K is the set of possible keys and Fn 2 is the set consisting of all n-bit vectors. 16.2.37 Remark E is an invertible function, i.e., E is a permutation of Fn 2, and D is the inverse of E. The encryption operator E consists of several rounds which are applied to the plaintext one after another. The secret key K is expanded using a key scheduling algorithm into a set of round keys K1, . . . , Kr. Each round function takes as input the round key and the output of the previous round and produces an output. For a ﬁxed round key, the encryption function is a bijective map. 16.2.38 Remark For a plaintext M, let M0 = M and Mi−1 denote the input to the i-th round and let Mr = C be the ﬁnal output of EK(M). If we denote by Ri the i-th round function, then we have Mi = Ri(Ki, Mi−1). This reduces the task of designing a block cipher to the task of designing the round functions and the key scheduling algorithm. Usually the round functions are identical or very similar. Two standard methods for designing round functions are Feistel Structure and Substitution-Permutation Network (SPN). 16.2.39 Deﬁnition (Feistel Structure) The Feistel Structure is a feedback shift register with time varying feedback function. In other words, the round function is a time varying feedback function. For example, DES is of a Feistel structure, the input Mi−1 to the i-th round is divided into two equal halves Li−1 and Ri−1, i.e., Mi−1 = Li−1\|\|Ri−1. The output Mi = (Li, Ri) is deﬁned as follows Li = Ri−1, and Ri = Li−1 ⊕f(Ri−1, Ki). 16.2.40 Remark Note that for the invertibility of the round function, f(·, ·) need not be invert-ible. The security of the encryption algorithm depends on the design of f(·, ·) and the key scheduling algorithm. The block cipher RC6 and many other block ciphers have this struc-ture. Here we consider RC6 as an example of a block cipher based on the Feistel structure. 16.2.41 Deﬁnition (Substitution-Permutation Network (SPN)) In an SPN, each round func-tion consists of a few successive layers. The input to a substitution layer is divided into small blocks of bits say blocks of eight bits each. An S-box is applied to each block. Each S-box is a bijective map, so that entire substitution layer is also a bijective map. Cryptography 759 The eﬀect of a substitution layer is local in the sense that an output bit in a particular position depends only on a few of the input bits in its nearby positions. This local eﬀect is compensated by having a permutation layer which permutes its input bits. The round key is usually incorporated at the beginning or at the end of the round function. 16.2.42 Remark We consider AES Rijndael as an example of a block cipher based on the SPN structure; see Subsection 16.2.6. 16.2.5 RC6 16.2.43 Remark This section is mainly from . RC6 is a symmetric key block cipher which encrypts 128-bit plaintext blocks to 128-bit ciphertext blocks and supports key sizes of 128, 192, and 256 bits. It was designed by Rivest, Robshaw, Sidney, and Yin to meet the requirements of the Advanced Encryption Standard (AES) competition . 16.2.44 Remark In general RC6 is speciﬁed as RC6-w/r/l where the word size is w bits, encryption consists of r rounds (generally r is 20), and l denotes the length of the encryption key in bytes. 16.2.45 Deﬁnition The encryption process involves three types of operations. Let x = (x0, . . . , xw−1) ∈Fw 2 and y = (y0, . . . , yw−1) ∈Fw 2 (note that x and y can be treated as binary vectors or binary numbers depending on the context). 1. Integer operations, i.e., (x+y) (mod 2w), (x−y) (mod 2w) and (x·y) (mod 2w). 2. Bitwise exclusive-or x ⊕y. 3. The rotation to the left and rotation to the right. Let L and R be the circular left and right shift operators respectively, i.e., L(x) = (x1, . . . , xw−1, x0) and R(x) = (xw−1, x0, . . . , xw−2). We denote (x ≪y) = Ly(x) = (xt, xt+1, . . . , xw−1, x0, x1, . . . , xt−1), (x ≫y) = Ry(x) = (xw−t, . . . , xw−1, x0, x1, . . . , xw−t−1), where t = σ(y0, . . . , y(lg w)−1), lg w is the logarithm of w to the base 2, and σ(y0, . . . , y(lg w)−1) is the integer representation of binary string (y0, . . . , y(lg w)−1). 16.2.46 Remark The block cipher RC6 has the following features . It is fast and simple and the best attack on RC6 appears to be exhaustive key search. 16.2.47 Deﬁnition (Encryption and decryption) RC6 is of the Feistel structure. In details, RC6 is a feedback shift register with time varying feedback and four w-bit registers which contain the input plaintext (a0, a1, a2, a3) and the output ciphertext (ar, ar+1, ar+2, ar+3) at the end of encryption. Round keys S0, . . . , S2r+3 are obtained from the key schedule algorithm, where each array element Si is of w bits (see Deﬁnition 16.2.49 and Figure 16.2.7). The state updating is done as follows: (ai, ai+1, ai+2, ai+3) = gi(ai−1, a(i−1)+1, a(i−1)+2, a(i−1)+3), for i = 1, . . . , r, where gi is deﬁned in the for loop of the Algorithm Enc(); see Figure 16.2.7. 760 Handbook of Finite Fields 16.2.48 Remark From the encryption algorithm, it can be easily seen that the process is invert-ible. The decryption algorithm is very similar to the encryption algorithm. It is a good exercise to write the decryption algorithm from the encryption algorithm. The input is (ar, ar+1, ar+2, ar+3), the round keys are in the reversed order (S2r+3, . . . , S0), ≪is re-placed by ≫and + is replaced by −in proper places of the encryption algorithm. See for a detailed description of the decryption algorithm. Algorithm Enc() for RC6-w/r/l Inputs : Plaintext (a0, a1, a2, a3), number of rounds r, round keys S0, . . . , S2r+3. Output : Ciphertext (ar, ar+1, ar+2, ar+3). Procedure : a1 = a1 + S0 a3 = a3 + S1 For i = 1 to r do t = f(a(i−1)+1) u = f(a(i−1)+3) ai−1 = ((ai−1 ⊕t) ≪u) + S2i a(i−1)+2 = ((a(i−1)+2 ⊕u) ≪t) + S2i+1 The i-th state: (ai, ai+1, ai+2, ai+3) = (a(i−1)+1, a(i−1)+2, a(i−1)+3, ai−1) End For ar = ar + S2r+2 ar+2 = ar+2 + S2r+3 End Algorithm. (Here f(x) = x(2x + 1) ≪lg w) Key Schedule for RC6-w/r/l Inputs : Secret key loaded in the array L0, . . . , Lc−1, number of rounds r. Output : Round keys S0, . . . , S2r+3. Procedure : S0 = Pw. For i = 1 to 2r + 3 do Si = Si−1 + Qw End For A = B = i = j = 0 v = 3 × max{c, 2r + 4} For s = 1 to v do A = Si = (Si + A + B) ≪3 B = Lj = (Lj + A + B) ≪(A + B) i = (i + 1) (mod (2r + 4)) j = (j + 1) (mod c) End For End Algorithm. Figure 16.2.7 Encryption and key schedule algorithm for RC6 . 16.2.49 Deﬁnition (Key schedule) The user supplies a key of l bytes, where 0 ≤l ≤255. Suﬃcient zero bytes are appended to give a key length equal to an integral number (say c) of words, and it is stored in L0, . . . , Lc−1. From this key, 2r + 4 words are derived and stored in the array S0, . . . , S2r+3. The constants P32 = B7E15163 and Q32 = 9E3779B9 (hexadecimal) are derived from the binary expansion of e −2 (e is the base of natural logarithm) and φ −1 (φ is the Golden Ratio). 16.2.50 Remark Figure 16.2.7 gives a description of the key schedule algorithm. 16.2.6 Advanced Encryption Standard (AES) RIJNDAEL 16.2.51 Remark Some of the features of AES are as follows; see for a more detailed description of AES. There are some diﬀerences between Rijndael and AES. Rijndael provides for several choices of block and key sizes. AES adopted only a subset of these parameter choices. Here we are ignoring these diﬀerences. 1. There are three allowable block lengths: 128, 192, and 256 bits. 2. There are three allowable key lengths (independent of selected block length): 128, 192, and 256 bits. Cryptography 761 3. The number of rounds is 10, 12, or 14, depending on the key length. 4. Each round consists of three functions, which are in four “layers” as a. 8-bit inverse permutation (sub-byte transform), b. 32-bit linear transformation (mix columns operation), c. 128-bit permutation (shift rows operation) and d. round key addition. 16.2.52 Deﬁnition (High level description of AES) A plaintext m of 128 bits is the initial state which is represented as a four by four array of bytes (see Figure 16.2.8). 1. For a given plaintext m, the initial state is m. Perform an AddRoundKey op-eration which is xor of the RoundKey with the initial State. 2. For each of the r −1 rounds perform a SubByte operation on State using an S-box; perform a ShiftRows on State; perform an operation called MixColumns on State; and perform an AddRoundKey operation. 3. For the r-th round, perform SubByte; perform ShiftRows and perform Ad-dRoundKey. 4. The ﬁnal State y is the ciphertext. S S S S S S S S S S S S S S S S XOR with K0 (0-th round key) Shift Rows and Mix Columns 128-bit message M 8-bit 8-bit S S S S S S S S S S S S S S S S XOR with Ki (i-th round key) Shift Rows XOR with Kr (r-th round key) 128-bit ciphertext C 8-bit 8-bit 0-th round repeat for r −1 rounds r-th round Figure 16.2.8 Round operations for AES Rijndael. 16.2.53 Remark (Algebraic structure of AES Rijndael) Rijndael uses a ﬁnite ﬁeld F28 deﬁned by the primitive polynomial p(x) = x8 + x4 + x3 + x + 1. Let α be a root of p, i.e., p(α) = 0 in F28. We use classical polynomial representation and the elements of F28 are considered as a set consisting of all polynomials of degree less than or equal to 7 with coeﬃcients from F2. So we can identify an element of F28 by an 8-bit vector. 16.2.54 Remark We introduce the following ring of matrices: M4(F28) = {X = (xij)4×4\|xij ∈F28}. 762 Handbook of Finite Fields In other words, for each matrix in M4(F28), the entries are taken from F28, i.e., each element of the matrix has 8-bit or one byte representation, and each row or column can be considered as a 32-bit word. 16.2.55 Remark For 128-bit version of Rijndael block cipher, a message M of 128 bits is parsed as 16 bytes and then further parsed as a 4 by 4 matrix: M = (m0, m1, m2, m3, m4, m5, m6, m7, m8, m9, m10, m11, m12, m13, m14, m15), where mi ∈F28, and the initial state M0 ∈M4(F28) is given as M0 =     m0 m4 m8 m12 m1 m5 m9 m13 m2 m6 m10 m14 m3 m7 m11 m15    . 16.2.56 Remark Three basic operators for AES are SubByte, ShiftRow, and MixColumn. 16.2.57 Deﬁnition SubByte is a map S : M4(F28) →M4(F28). Let c = (1, 1, 0, 0, 0, 1, 1, 0) and A =     1 0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1    . For y = (y0, y1 . . . , y7) ∈F28, deﬁne T(y) = Ayt + ct and σ(y) = y−1. So we have T −1(y) = A−1(yt + ct) and σ−1(y) = y−1. For X = (xij) ∈M4(F28), S is deﬁned as S(X) = (S(xij))4×4 where S(xij) = (T ◦σ)(xij). The inverse of S is given by S−1(X) = (S−1(xij))4×4 = ((σ−1 ◦T −1)(xij))4×4. 16.2.58 Deﬁnition ShiftRow transform R and its inverse on a state X are given as follows R(X) =     x00 x01 x02 x03 x11 x12 x13 x10 x22 x23 x20 x21 x33 x30 x31 x32    , R−1(X) =     x00 x01 x02 x03 x13 x10 x11 x12 x22 x23 x20 x21 x31 x32 x33 x30    . 16.2.59 Deﬁnition MixColumn transform L is a linear transform on F4 28. Recall that α is a root of p(x) in F28. Given a state X, L(X) = LX where LX is the matrix multiplication of L and X over F28. The linear transform L and its inverse are given as follows L =     α 1 + α 1 1 1 α 1 + α 1 1 1 α 1 + α 1 + α 1 1 α    , L−1 =     β0 β3 β2 β1 β1 β0 β3 β2 β2 β1 β0 β3 β3 β2 β1 β0    , where β0 = α3 + α2 + α, β1 = α3 + 1, β2 = α3 + α2 + 1, and β3 = α3 + α + 1. 16.2.60 Deﬁnition Composition of operators is deﬁned as follows: G(X) = (R ◦S)(X) G−1(X) = (S−1 ◦R−1)(X) H(X) = (L ◦G)(X) H−1(X) = (G−1 ◦L−1)(X) Cryptography 763 16.2.61 Remark The total number of round key bits is equal to the block length times the number of rounds plus 1. The 128-bit version needs 10 rounds. Thus 1408 bits, or 44 words of round key bits are needed. Thus, the key schedule should extend a 128-bit key to round keys, a total of 1408 key bits. Let {ki}43 i=0 be a sequence of words, ki ∈F4 28, which consists of 4 bytes. Let (k0, k1, k2, k3) be the 128-bit session key. The sequence {ki} is used as the round keys. The expansion of the key is shown in Table 16.2.9. Input: (k0, k1, k2, k3), kj ∈F4 28, 128-bit key. Output Ki = (k4i, k4i+1, k4i+2, k4i+3), i = 0, 1, . . . , 10, the i-th round key where each kj is a 32-bit word. Procedure. For i = 1, 2, . . . , 10, compute k4i = k4i−4 + (S(k4i−1,3) + αi−1, S(k4i−1,0), S(k4i−1,1), S(k4i−1,2)); k4i+j = k4i+j−4 + k4i+j−1, in F 4 28, j = 1, 2, 3. Return Ki = (k4i, k4i+1, k4i+2, k4i+3), i = 0, 1, . . . , 10 Table 16.2.9 Key scheduling, where ki = (ki0, ki1, ki2, ki3), i = 0, 1, . . . , 43 and kij ∈F28. 16.2.62 Deﬁnition (Rijndael encryption and decryption) For 128-bit version of Rijndael, a message M of 128 bits is parsed as a 4 by 4 matrix. The number of rounds is equal to 10. The process of computation of the cipher C (again written as a 4 by 4 matrix) is shown in Table 16.2.10. When viewed as an FSM, the initial state of the Rijndael block cipher is M0, the ﬁnal state is M10 (which is the output, i.e., ciphertext) and the state update function is a map : F16 28 →F16 28 as shown in Table 16.2.10. Encryption Decryption M0 = M + K0, in M4(F28) Mi = H(Mi−1) + Ki, 1 ≤i ≤9 M10 = G(M9) + K10 C0 = C + K10, C1 = G−1(C0) + K9 Ci = H−1(Ci−1) + K10−i, 2 ≤i ≤10 The ciphertext is C = M10. The plaintext is M = C10. Table 16.2.10 Encryption and decryption processes of Rijndael. See Also §16.1 For goals of cryptography and symmetric-key cryptography. For a thorough description of the design of Rijndael: AES. , , For cryptographic properties of the WG stream cipher. Contains information about the design and analysis of stream ciphers. For hardware implementations of the WG stream cipher. For information on RC4. Develops the theory of WG stream cipher. For information on the RC6 block cipher. References Cited: [762, 864, 989, 990, 1082, 1303, 1317, 1835, 1982, 2001, 2080, 2217, 2294, 2461] 764 Handbook of Finite Fields 16.3 Multivariate cryptographic systems Jintai Ding, University of Cincinnati 16.3.1 Remark Due to limited space only a few key areas more directly related to the theory of ﬁnite ﬁelds are covered. For a more complete reference, readers should consult [894, 882]. This section grows out of but with new materials from the last two years added. 16.3.2 Remark Multivariate public key cryptosystems are motivated by the need to develop new cryptosystems that have the potential to resist future quantum computer attacks. In ad-dition, multivariate public key cryptosystems are also motivated by the need to develop eﬃcient public key cryptosystems that could be used in small computing devices with lim-ited computing and memory capacities like sensors, radio-frequency identiﬁcation (RFID) tags, and other similar small devices. 16.3.3 Remark The foundation of any public key cryptosystem is a class of “trapdoor one-way functions.” The fundamental mathematical structure of such a class of functions determines all the basic characteristics of a public key cryptosystem. In the case of multivariate (public-key) cryptosystems (MPKCs), the trapdoor one-way function is usually in the form of a multivariate quadratic polynomial map over a ﬁnite ﬁeld. 16.3.4 Deﬁnition For a MPKC, the public key is, in general, given by a set of quadratic polyno-mials: P(x1, ..., xn) = (p1(x1, . . . , xn), . . . , pm(x1, . . . , xn)), where each pi is a (usually quadratic) nonlinear polynomial in X = (x1, . . . , xn): yk = pk(X) := X i Pikxi + X i Qikx2 i + X i>j Rijkxixj + Sk. (16.3.1) with all coeﬃcients and variables in Fq. 16.3.5 Remark The evaluation of these polynomials corresponds to either the encryption procedure or the veriﬁcation procedure. 16.3.6 Remark Most of the constructions are quadratic constructions due to the consideration of the eﬃciency of encryption determined by the key size. 16.3.7 Remark Inverting a multivariate quadratic map is generally equivalent to solving a set of quadratic equations over a ﬁnite ﬁeld, or the following multivariate quadratic (MQ) problem. 16.3.8 Deﬁnition (MQ problem) Solve a system p1(X) = p2(X) = · · · = pm(X) = 0, where each pi is a quadratic polynomial in X = (x1, . . . , xn). All coeﬃcients and variables are in Fq. 16.3.9 Remark MQ is an NP-complete problem, which are believed to be hard generically. A set of random quadratic polynomials cannot be a trapdoor. Since one does not deal with “random” or “generic” systems, but systems using speciﬁc trapdoors, the security MPKCs is then not guaranteed by the NP-hardness of the MQ problem, and eﬀective attacks may exist for any chosen trapdoor. The history of MPKCs therefore evolves as we understand better about how to design eﬃcient secure multivariate trapdoors. 16.3.10 Remark The mathematical structure behind a system of polynomial equations is the ideal generated by those polynomials. Multivariate cryptography is based mainly on mathemat-ics that handles polynomial ideals, namely algebraic geometry but over a ﬁnite ﬁeld. In Cryptography 765 contrast, the security of RSA-type cryptosystems relies on the hardness of integer factor-ization and is based on number theory developed in the 17th and 18th centuries. Elliptic curve cryptosystems employ the mathematical theory developed in the 19th century. This is a remark from Whitﬁeld Diﬃe at the RSA Europe conference in Paris in 2002. Algebraic geometry, the mathematics that MPKCs depend on, was developed in the 20-th century. 16.3.11 Remark This section is organized as follows: Subsection 16.3.1 provides a sketch of how MPKCs work in general; Subsection 16.3.2 describes the known trapdoor constructions in more detail; Subsection 16.3.3 describes the most important modes of attacks; the last subsection is about future research directions in this area. 16.3.1 The basics of multivariate PKCs 16.3.12 Remark After Diﬃe-Hellman , cryptographers proposed many trapdoor functions. The earliest published proposal of MPKC schemes seemed to have arisen in Japan [2029, 2822, 2823] in the early 1980s. These papers were published in Japanese, and remained largely unknown outside Japan. 16.3.13 Remark The ﬁrst article in English describing a public key cryptosystem with more than one independent variable may be the one from Ong et al , and the ﬁrst use of more than one equation is by Fell and Diﬃe . The earliest attempt bearing some resemblance to today’s MPKCs (with 4 variables) seems to be . In 1988, the ﬁrst MPKC in the current form appeared in , and the basic construction described below (Subsection 16.3.1.1) has not really changed much since. 16.3.1.1 The standard (bipolar) construction of MPKCs 16.3.14 Deﬁnition A usual MPKC has a private map Q, which is the central map and it belongs to a certain class of quadratic maps each of which can be eﬃciently inverted. 16.3.15 Deﬁnition The basic construction of the public key is derived via composition with two aﬃne maps S, T. P = T ◦Q ◦S : Fn q →Fm q , where the maps S, T are aﬃne (sometimes linear) invertible maps on Fn q and Fm q , respec-tively. 16.3.16 Remark The purpose of T and S is to hide the trap door Q. The key of a MPKC is indeed the design of the central map. 16.3.17 Remark The public key consists of the polynomials in P. The secret key consists of the information in S, T, and Q. To verify a signature or to encrypt a block, one simply computes Y = P(X), where Y = (y1, . . . , ym) and X = (x1, . . . , xn). To sign or to decrypt a block, one computes X = P −1(Y ) = S−1 ◦Q−1 ◦T −1(Y ), 766 Handbook of Finite Fields which is computed via the composition factors in turn. Notice that by the inverse of a map here, we mean ﬁnding one of possibly many pre-images, not necessarily an inverse function in the strict mathematical sense. 16.3.18 Remark The basics of MPKCs are provided below so that the reader has a basic sense about how these schemes can work in practice: Cipher block or message digest size: m elements of Fq; Plaintext block or signature size: n elements of Fq; Public key size: mn(n + 3)/2 elements of Fq; Secret key size: Usually n2 + m2 + [size of P] elements of Fq; Secret map time complexity: (n2 + m2) Fq-multiplications, plus the time it is needed to invert Q; Public map time complexity: About mn2/2 Fq-multiplications. 16.3.19 Remark In terms of computational complexity, MPKCs usually have strong advantages as we shall see below. But a disadvantage with MPKCs is that their keys are large compared to number-theory-based systems like RSA or ECC. For example, the public key size of RSA-2048 is not much more than 2048 bits, but a current version of the Rainbow signature scheme has n = 42, m = 24, q = 256, i.e., the size of the public key is 22,680 bytes. 16.3.20 Remark There are other alternative forms in which multivariate polynomials can be used for public key cryptosystems, as we discuss next. 16.3.1.2 Implicit form MPKCs 16.3.21 Deﬁnition The public key of an implicit form MPKC is a system of l equations: P(W, Z) = P(w1, . . . , wn, z1, . . . , zm) = (p1(W, Z), . . . , pl(W, Z)) = (0, . . . , 0), (16.3.2) where each pi is a polynomial in W = (w1, . . . , wn) and Z = (z1, . . . , zm). This P is built from the secret Q: Q(X, Y ) = q(x1, . . . , xn, y1, . . . , ym) = (q1(X, Y ), . . . , ql(X, Y )) = (0, . . . , 0), where qi(X, Y ) is polynomial in X = (x1, . . . , xn), Y = (y1, . . . , ym) such that 1. for any given speciﬁc element X′, we can easily solve the equation Q(X′, Y ) = (0, . . . , 0), (16.3.3) 2. for any given speciﬁc element Y ′, we can easily solve the equation Q(X, Y ′) = (0, . . . , 0), (16.3.4) 3. Equation (16.3.3) is linear and Equation (16.3.4) is nonlinear but can be solved eﬃciently. 16.3.22 Remark The public key is built as P = L ◦Q(S(W), T −1(Z)) = (0, . . . , 0), where S, T are invertible aﬃne maps and L is linear. Cryptography 767 16.3.23 Remark To verify a signature W with the digest Z, one checks that P(W, Z) = 0. If one wants to use P to encrypt the plaintext W, one would solve P(W, Z) = (0, . . . , 0), and ﬁnd the ciphertext Z. To invert (i.e., to decrypt or to sign) Z, one ﬁrst calculates Y ′ = T −1(Z), then substitutes Y ′ into the Equation (16.3.4) and solves for X. The ﬁnal plaintext or signature is given by W = S−1(X). 16.3.24 Remark In an implicit-form MPKC, the public key consists of the l polynomial components of P and the ﬁeld structure of F. The secret key mainly consists of L, S, and T. Depending on the case, the equation Q(X, Y ) = (0, . . . , 0) is either known or has parameters which are a part of the secret key. 16.3.25 Remark The maps S, T, L serve to hide the equation Q(X, Y ) = 0, which otherwise could be easily solved for any Y . Mixed schemes are relatively rare, one example being Patarin’s Dragon . 16.3.1.3 Isomorphism of polynomials 16.3.26 Remark The isomorphism of polynomials problem originated from trying to attack MPKCs by ﬁnding the secret keys. 16.3.27 Deﬁnition Let ¯ F1, ¯ F2 with ¯ Fi(x1, . . . , xn) = ( ¯ fi1, . . . , ¯ fim), (16.3.5) be two polynomial maps from Fn to Fm. The Isomorphism of Polynomials (IP) problem is to ﬁnd two invertible aﬃne linear transformations S on Fn and T over Fm (if they exist) such that ¯ F1(x1, . . . , xn) = T ◦¯ F2 ◦S(x1, . . . , xn). (16.3.6) 16.3.28 Remark The system based on the IP problem was ﬁrst proposed by Patarin , where the veriﬁcation process is performed by showing the equivalence (or isomorphism) of two diﬀerent maps. A simpliﬁed version is the isomorphism of polynomials with one secret (IP1s) problem, where we only need to ﬁnd the map S (if it exists), while the map T is known to be the identity map. This problem is used to build identiﬁcation schemes [1044, 1266, 1914, 2371, 2387]. 16.3.29 Remark Mathematically, this problem can be viewed from the perspective of the problem of classiﬁcation of quadratic maps from Fn q to Fm q under the action of the group GLn(Fq) × GLm(Fq), namely to describe precisely the orbit space of all the quadratic maps from Fn q to Fm q under the action of the group GLn(Fq) × GLm(Fq); a very hard mathematical problem we do not know much about, except the case when m = 1, which is the classiﬁcation of quadratic (or bilinear) forms, a problem we know very well. 16.3.30 Remark Since the vast majority of MPKCs are in standard form, we deal with mainly such systems in the rest of this section. 16.3.2 Main constructions and variations 16.3.2.1 Historical constructions 16.3.31 Remark The ﬁrst attempt to construct a multivariate signature [2320, 2321] utilizes a quadratic equation with two variables. y ≡x2 1 + αx2 2 (mod n), (16.3.7) 768 Handbook of Finite Fields where n = pq is an RSA modulus, a product of two large primes. The public key is essen-tially the integer n and Equation (16.3.7). Since the security is supposed to be based on the factorization of n, this system can really be viewed as a derivative of RSA, though it indeed initiated the idea of multivariate cryptosystems. This system was broken by Pollard and Schnorr in , where they gave a probabilistic algorithm to solve Equation (16.3.7) for any y without even knowing the factors of n. Assuming the generalized Riemann hy-pothesis, a solution can be found with a computational complexity of O((log n)2 log log \|k\|) in O(log n)-bit integer operations. 16.3.32 Remark Diﬃe and Fell tried to build a cryptosystem using the composition of in-vertible linear maps and simple tame maps of the form T(x1, x2) = (x1 + g(x2), x2), where g is a polynomial. Tame maps, well known in algebraic geometry, are easily invert-ible but hard to hide when composed with each other, used only two variables and equations; not surprisingly, the authors concluded that it appeared very diﬃcult to build such a cryptosystem with any real practical value that is both secure and has a public key of practical size, therefore practically useful. 16.3.33 Remark An attempt to build a true multivariate (with four variables) public key cryp-tosystem was also made by Matsumoto, Imai, Harashima, and Miyagawa , where the public keys are given by quadratic polynomials. However it was soon broken . People soon realized that more than 4 variables are needed and new mathematical ideas are needed to make MPKCs work. 16.3.2.2 Triangular constructions 16.3.34 Remark The tame maps used in are a special case of the “triangular” or de Jonquieres maps from algebraic geometry. 16.3.35 Deﬁnition A de Jonqui eres map is a polynomial map in the following form: J(x1, . . . , xn) = (x1 + g1(x2, . . . , xn), . . . , xn−1 + gn−1(xn), xn), (16.3.8) where the gi are arbitrary polynomial functions. 16.3.36 Remark A de Jonquieres map J can be eﬃciently inverted as long as gi is not too compli-cated. The invertible aﬃne linear maps over Fn q together with the de Jonqui eres maps belong to the family of tame transformations from algebraic geometry, including all transformations that are in the form of a composition of elements of these two types of transformations. Tame transformations are elements of the group of automorphisms of the polynomial ring Fq[x1, . . . , xn]. Elements in this automorphism group that are not tame are wild. Given a polynomial map, it is in general very diﬃcult to decide whether or not the map is tame, or even if there is indeed any wild map , a question closely related to the famous Jacobian conjecture. This problem was solved in 2003 when proves that the Nagata map is indeed wild. 16.3.37 Remark The ﬁrst attempt in the English literature with a clear triangular form is the birational permutations construction by Shamir . However, triangular constructions were earlier pursued in Japan under the name “sequential solution type systems” [1440, 2822, 2823]. Their construction is actually even more general in the sense that they use rational functions instead of just polynomials. These works in Japanese are not so well-known. Cryptography 769 16.3.38 Remark Triangular maps are extremely fast to evaluate and to invert if gi are very simple functions. However, they do have certain strong deﬁnitive characteristics. On the small end of a triangular system, so to speak, the variable xn is mapped to the simple function of itself. On the bigger end, the variable xi appears only once in a single equation. The other equations involve successively more variables. 16.3.39 Proposition If we write the quadratic portion of the central polynomials yi = qi(X) as bilinear forms, or take the symmetric matrix denoting the symmetric diﬀerential of the central polynomials as in qi(X + B) −qi(X) −qi(B) + qi(0) := BT MiX, (16.3.9) then the rank of the matrix Mi in general increases monotonically as i increases. If q = 2k, the equation dealing with x1 always has rank zero. Let ker Mi be the kernel of the linear map associated with Mi. Then ker M1 ⊂ker M2 ⊂· · · , which is a chain of kernels . 16.3.40 Remark This matrix rank and the chain structure of kernels is invariant under compo-sition with an invertible map, S. That is, considering yi as a function of X, the corre-sponding diﬀerential is BT M T S MiMS X. For the most part, MS is full-rank, and hence rank M T S MiMS = rank Mi. This leads to what is known as rank attacks based on linear algebra [722, 1339]. Therefore triangular/tame constructions cannot be used alone. Some ways to design around this problem are lock polynomials (Subsection 16.3.2.9), solvable segments (Subsection 16.3.2.4), and plus-minus (Subsection 16.3.2.6). 16.3.2.3 Big-ﬁeld families: Matsumoto-Imai (C∗) and HFE 16.3.41 Remark Triangular (and Oil-and-Vinegar, and variants thereof) systems are usually called “single-ﬁeld” or “small-ﬁeld” approaches to MPKC design, in contrast to the approach taken by Matsumoto and Imai in 1988 . In the “big-ﬁeld” constructions, a totally new type of mathematical construction, the central map is really a map in a larger ﬁeld L, a degree n extension of a ﬁnite ﬁeld K. One builds an invertible map Q : L →L, and picks a K-linear bijection φ : L →Kn. Then we have the following multivariate polynomial map, which should presumably be quadratic in general: Q = φ ◦Q ◦φ−1, (16.3.10) and, we “hide” this map Q by composing from both sides by two invertible aﬃne linear maps S and T in Kn. 16.3.42 Deﬁnition Matsumoto and Imai built a scheme C∗by choosing a ﬁeld K of characteristic 2 and the map Q Q : X 7− →Y = X1+qα, (16.3.11) where q is the number of elements in K, X is an element in L, and gcd(1+qα, qn−1) = 1. 16.3.43 Theorem We have Q −1(X) = Xh, (16.3.12) where h(1 + qα) ≡1(mod(qn −1)). 16.3.44 Remark In the rest of this section, for the purpose of simplicity, we simply identify the vector space Kn with the large ﬁeld L, and Q with Q, omitting the isomorphism φ from formulas. When necessary to distinguish the inner product in a vector space over K and the larger ﬁeld L, the former is denoted by a dot (·) and the latter by an asterisk (∗). 770 Handbook of Finite Fields 16.3.45 Remark The map Q is always quadratic due to the linearity of the Frobenius map Fα(X) = Xqα. 16.3.46 Remark A signiﬁcant algebraic implication of C∗and Equation (16.3.11) is Y qα−1 = Xq2α−1 or XY qα = Xq2αY. (16.3.13) Patarin used this bilinear relation to cryptanalyze the original C∗(see Subsec-tion 16.3.3.1). Though the original idea of C∗failed, it has inspired many new designs. 16.3.47 Deﬁnition An HFE (Hidden Field Equations) system, as the most signiﬁcant of the C∗ derivatives, is constructed by replacing Q, the monomial used by C∗, by the extended Dembowski-Ostrom polynomial map: Q : X ∈L = Fqn 7− →Y = X 0≤i≤j<r aijXqi+qj + X 0≤iere and Joux has an intrinsic rank of 4. 16.3.52 Remark An HFE with a high d is unbroken, although it can be really slow to decrypt/invert. Quartz probably sets a record for the slowest cryptographic algorithm when submitted to NESSIE — on a Pentium III 500MHz, it took half a minute to do a signature, but has been improved substantially since. 16.3.53 Remark Recent research progress in [878, 884, 890] shows that the HFE still has great potential if we explore the cases using ﬁnite ﬁelds with odd characteristics. 16.3.54 Remark Schemes C∗and HFE each can be modiﬁed by techniques mentioned later (Plus-Minus, vinegar variables, and internal perturbation). Also related are the ℓIC system and probabilistic big-ﬁeld based MPKCs . 16.3.2.4 Oil and vinegar (unbalanced and balanced) and variations 16.3.55 Remark The Oil and Vinegar (OV) and later unbalanced Oil and Vinegar (UOV) schemes [1737, 2367] are designed only for signatures. This construction is inspired by the idea of Cryptography 771 linearization equations (Subsection 16.3.2.3). In some sense, this construction uses a method where ones transforms an attacking method into a designing method. 16.3.56 Deﬁnition Let v < n and m = o = n −v. The variables x1, . . . , xv are vinegar variables and xv+1, . . . , xn oil variables. An oil-vinegar map Q : Kn →Km is a map in the form Y = Q(X) = (q1(X), . . . , qo(X)), where ql(X) = v X i=1 n X j=i α(l) ij xixj, l = 1, · · · , o and all coeﬃcients are randomly chosen from the base ﬁeld K. 16.3.57 Remark In each qi, there are no quadratic terms of oil variables, which means the oil variables and vinegar variables are not fully mixed (like oil and vinegar in a salad dressing), which is the origin of the name of this scheme. 16.3.58 Deﬁnition The public key map P for an Oil-Vinegar scheme is constructed as P = Q ◦S, where S is an invertible linear map. 16.3.59 Remark The change of basis by the transformation S is a process to “mix” fully oil and vinegar, so one cannot tell what are oil variables and what are the vinegar variables. With OV and UOV constructions, there is no need to compose another aﬃne map T. 16.3.60 Remark The original Oil and Vinegar signature scheme has m = o = v = n/2. When o < v, it becomes the unbalanced Oil and Vinegar signature scheme. 16.3.61 Remark The public key for an OV or an UOV is P = (p1, . . . , po), the polynomial compo-nents of P. The secret key consists of the linear map S and the map Q. 16.3.62 Remark Given a message Y = (y1, . . . , yo), to sign it, one needs to ﬁnd a vector W = (w1, . . . , wn) such that P(W) = Y . With the secret key, this can be done eﬃciently. First, one guesses values for each vinegar variable x1, . . . , xv, and obtains a set of o linear equations with the o oil variables xv+1, . . . , xn. With high probability, it has a solution. If the linear system does not have a solution, one may repeatedly assign random values to the vinegar variables until one ﬁnds a pre-image of a given element in Ko. Then one applies S−1. 16.3.63 Remark To check if W is indeed a legitimate signature for Y , one only needs to get the public map P and check if indeed P(W) = Y . 16.3.64 Remark The algebraic property that is most signiﬁcant in an unbalanced Oil-and-Vinegar system is the absence of pure oil cross-terms. Equivalently, if we have an UOV polynomial, then the quadratic part of each component qi in the central map from X to Y , when viewed as a bilinear form using a matrix, see Equation (16.3.9), looks like Mi :=             α(i) 11 · · · α(i) 1v α(i) 1,v+1, · · · α(i) 1n . . . ... . . . . . . ... . . . α(i) v1 · · · α(i) vv α(i) v,v+1, · · · α(i) vn α(i) v+1,1, · · · α(i) v+1,v, 0 · · · 0 . . . ... . . . . . . ... . . . α(i) n1 · · · α(i) nv 0 · · · 0             , (16.3.15) 772 Handbook of Finite Fields or in the block form: ∗ ∗ ∗ 0 . 16.3.2.5 UOV as a booster stage 16.3.65 Remark There have been diﬀerent attempts to make UOV more eﬃcient such as [1687, 1686], which were promptly broken . 16.3.66 Remark A new way to make the UOV type construction more eﬃcient is the Rainbow construction by stacking several layers of Unbalanced Oil-Vinegar systems together for an easily invertible central map . 16.3.67 Deﬁnition For a u-stage Rainbow 0 < v1 < v2 < · · · < vu+1 = n, the construction of central map over any ﬁnite ﬁeld is given by yk = qk(X) = vl X i=1 n X j=i α(k) ij xixj + X i<vl+1 β(k) i xi, if vl < k ≤vl+1. (16.3.16) 16.3.68 Remark In the signing process, we ﬁrst choose randomly the values for the vinegar variables x1, . . . , xv1 in the ﬁrst layer, and solve for the oil variables xv1+1, . . . , xv2. Then we use the known values of xi in the second layer and ﬁnd the values for the oil variables in the second layer. We continue like this layer by layer until we have all the xi’s. 16.3.69 Remark The components of Y in a Rainbow-type construction are typically written to have indices v1 + 1, . . . , n. In the pure Rainbow scheme, S and T and the coeﬃcients α and β are totally randomly chosen. The essential structure of the Rainbow instance is determined by 0 < v1 < v2 < · · · < vu+1 = n or the “Rainbow structure sequence” (v1, o1, o2, . . . , ou), where oi := vi+1 −vi. 16.3.70 Remark The Rainbow construction is a special case of the UOV; however, the structure of the system, consist of ou equations with the associated bilinear maps of the form ∗ ∗ ∗ 0 following m −ou equations with the associated bilinear maps of the form ∗ 0 0 0 leads to a diﬀerent attack; see Subsection 16.3.3.10). 16.3.71 Remark Aside from attacks peculiar to the UOV and Rainbow systems, the Rainbow-type constructions also share certain characteristics of triangular schemes, therefore there is a need to account for rank-based attacks (Subsection 16.3.3.8), such as the two improved attacks in [280, 895]. None of these attacks are considered essentially eﬀective. 16.3.72 Remark If one wants to make the computation of the central map and its inverse fast, another direct way is to make the Oil-vinegar polynomials sparse, such as in the case of the TTS (Tame Transformation Signatures) schemes, [604, 3023, 3024]. The TRMS of Wang et al. are also a TTS instance. Due to the sparsity, there also exist certain extra pos-sibilities of linear algebra and related vulnerabilities, principally UOV-type vulnerabilities such as described in . 16.3.2.6 Plus-Minus variations 16.3.73 Remark Minus and Plus are simple but useful ideas, earliest mentioned by Matsumoto, Patarin, and Shamir (probably found independently [2370, 2606]). Cryptography 773 16.3.74 Remark For the Minus method , several (r) polynomials are removed from the public keys. When inverting the public map, the legitimate users take random values for the missing variables. Minus is very suitable for signature schemes without any performance loss, since a document need not have a unique signature. 16.3.75 Remark For encryption, Minus causes signiﬁcant slowdown, since the missing coordinates must be searched. In theory the public map of an encryption method should be injective. If we have to search through r variables in Fq, we eﬀectively have qn+r results, only qn of which should represent valid ciphertexts, hence the expected number of guesses taken per decryption is qr. Thus, decryption is slowed by that same factor of qr. 16.3.76 Remark Minus or removing some public equations makes a C∗-based system much harder to solve. SFLASH [67, 739, 2368], a C∗−instance with (q, n, r) = (27, 37, 11), was accepted as an European security standard for low-cost smart cards by the New European Schemes for Signatures, Integrity, and Encryption . 16.3.77 Remark In 2007, the SFLASH family of cryptosystems [924, 925] was defeated. The attack uses the symmetry and the invariants of the diﬀerential of the public map P (Subsec-tion 16.3.3.4) inspired by the attack on internal perturbation. Making a secure C∗-based signature scheme may require a new modiﬁer called Projection . 16.3.78 Remark Plus is the opposite of Minus: add random central equations to the original central map, and this can be used to mask the high-end of the triangle system. For encryption methods, this again does not aﬀect performance much; for digital signatures there is a slowdown as the extra variables again need to be guessed. Regardless, Plus-Minus variations defend against attacks that are predicated on the rank of equations. 16.3.79 Remark Plus-Minus alone does not make triangular constructions secure. Paper dis-cusses this in detail and concludes exactly the opposite: Triangle-Plus-Minus constructions can be broken by very straightforward attacks using simple linear algebra. Therefore one must use more elaborate variations [280, 895, 3023]. 16.3.2.7 Internal perturbation 16.3.80 Remark Internal perturbation is a general method of improving the security of MPKCs by adding some perturbation or controlled noise. Internal perturbation was ﬁrst applied to Matsumoto-Imai systems, which produce the variation . 16.3.81 Remark Take V = (v1, . . . , vr) to be an r-tuple of random aﬃne forms in the variables X. Let f = (f1, . . . , fn) be a random r-tuple of quadratic functions in V . Let our new Q be deﬁned by X7→Y = (X)qα+1 + f(V (X)) where the power operation assumes the vector space to represent a ﬁeld. The number of Patarin relations decreases quickly down to 0 as r increases. For every Y , we may ﬁnd Q−1(Y ) by guessing at V (X) = B, ﬁnding a candidate X = (Y + B)h and checking the initial assumption that V (X) = X. Since we repeat the procedure qr times, we are almost forced to let q = 2 and make r as small as possible. 16.3.82 Remark In this system, there are possible extraneous solutions just as in the system HFE. Therefore, we must manufacture some redundancy in the form of a hash segment or check-sum. The original perturbation system was broken via a surprising diﬀerential crypt-analysis (cf. Sec. 16.3.3.4). Internal perturbation is usually coupled with the plus variation, and this is one of the best multivariate encryption schemes. 774 Handbook of Finite Fields 16.3.2.8 Vinegar as an external perturbation and projection 16.3.83 Remark The idea of vinegar variables had been introduced earlier with UOV, and was used as a defense in Quartz. The idea is to use an auxillary variable that occupies only a small subspace of the input space. It was pointed out that internal perturbation is almost exactly equal to both vinegar variables and projection, or ﬁxing the input to an aﬃne subspace. We basically set one, two, or more variables of the public key to be zero to create the new public key. In the case of signature schemes, each projected dimension will slow down the signing process by a factor of q. 16.3.84 Remark Projection is a very useful simple method. Since (Sec. 16.3.3.4) structural attacks usually start by looking for an invariant or a symmetry, it is a good idea that we should try to remove both. Restricting to a subspace of the original space breaks the symmetry. Something like the Minus modiﬁer destroys an invariant. Hence the use of projection by itself prevents some attacks, such as [924, 925, 1095]. The diﬀerential attack against C∗ (and ℓIC) derivatives uses the structure of the big ﬁeld L. Hence projection is expected to prevent such an attack . 16.3.2.9 TTM and related schemes: “lock” or repeated triangular 16.3.85 Remark MPKCs based only on triangular constructions were not pursued again until a much more complex defense against rank attacks was proposed, with the tame transforma-tion method (TTM) of Moh . 16.3.86 Remark de Jonqui eres maps can be viewed either as upper triangular or lower triangular. Moh suggested a construction where the central map Q is given by Q = Ju ◦Jl ◦I(x1, . . . , xn), (16.3.17) where Ju is a Km upper triangular de Jonquieres map and Jl is a Km lower triangular de Jonqui eres map and the linear map I is the embedding of Kn into Km: I(x1, . . . , xn) = (x1, . . . , xn, 0, 0, . . . , 0). The main achievement of such a construction is how to make Q quadratic. Moh’s trick is actually in using the map I. One can see that Jl ◦I(x1, . . . , xn) = (x1, x2 + g1(x1), . . . , xn + gn−1(x1, . . . , xn−1), gn(x1, . . . , xn), . . . , gm−1(x1, . . . , xn)), which presents the freedom to choose any gi, i = n, . . . , m −1. When decrypting, one evaluates the de Jonquieres maps from x1 up. 16.3.87 Remark The extremely low rank of central polynomials presented in published TTM in-stances [603, 2113] is the main source of weakness and therefore of eﬀective attacks. 16.3.88 Remark Courtois and Goubin used the MinRank method (Subsection 16.3.3.8) to attack this system. 16.3.89 Deﬁnition A MinRank problem is to look for non-zero matrices with minimum rank in a space of matrices. 16.3.90 Remark The Minrank problem is NP-hard in general but can be easy for special cases, in particular, when the minimum rank is very low. 16.3.91 Remark The idea of sequentially solvable equations (or stages) can also be used in conjunc-tion with other ideas. Some of the more notable attempts are from Wang, who had written Cryptography 775 about a series of schemes called “Tractable Rational Map Cryptosystems” (TRMC). TRMC v1 is essentially no diﬀerent from early TTM . The central map of TRMC v2 has a small random overdetermined block on one end and the rest of the variables are de-termined in the triangular (tame) style. Versions 3 and 4 [2930, 2932] use a similar trick as 3IC . 16.3.92 Remark The TTM construction is a truly original and very intriguing idea. So far existing constructions of the TTM cryptosystem and related schemes do not work for public-key encryption [883, 886, 887, 2226]. Most of the schemes proposed are not presented in any systematic way, and no explanation is yet given why and how they work. More sophistication is needed and we suspect that to create a successful TTM-like scheme will probably require deep insight from algebraic geometry. 16.3.93 Remark A TTS (tamed transformation signature) scheme can be viewed as a similar but simpler construction. This system is essentially the result of an application of the Minus method in for a tame transformation. A few of them were suggested mainly by Chen and Yang [604, 605]. These systems can also be defeated by the method used by Stern and Vaudenay [722, 896]. 16.3.2.10 Intermediate ﬁelds: MFE and ℓIC 16.3.94 Remark In C∗and HFE, we use one big ﬁeld L = Kn. In Rainbow/TTS or similar schemes, each component is as small as the base ﬁeld. We can use something in between, as seen in MFE (Medium Field Encryption) and ℓIC (ℓ-Invertible Cycles). These two constructions use a standard Cremona transformation from algebraic geometry. 16.3.95 Deﬁnition A standard Cremona transformation is deﬁned as: L∗:= L\{0} for some ﬁeld L: (X1, X2, X3) ∈(L∗)3 7− →(Y1, Y2, Y3) := (X1X2, X1X3, X2X3) ∈(L∗)3. (16.3.18) 16.3.96 Proposition This transformation is a bijection for any ﬁeld L, and inverts via X1 := p Y1Y2/Y3. 16.3.97 Remark MFE’s central map uses structure related to matrix multiplications to defend against linearization relations, but it does not avoid all the problems, as can be seen in Subsection 16.3.3.3. 16.3.98 Remark The ℓ-invertible cycle also uses an intermediate ﬁeld L = Kk and extends C∗ by using the following central map from (L∗)ℓto itself: Q : (X1, . . . , Xℓ) 7→ (Y1, . . . , Yℓ) (16.3.19) := (X1X2, X2X3, . . . , Xℓ−1Xℓ, XℓXqα 1 ). 16.3.99 Remark This is much faster computationally than computing the inverse of C∗. But these have so much in common with C∗that we need the same variations. In other words, we need to do 3IC−p (with minus and projection) and 2IC+i (with internal perturbation and plus), paralleling C∗−p and C∗+i (also known as PMI+) . 776 Handbook of Finite Fields 16.3.2.11 Odd characteristic 16.3.100 Remark Initially all the MPKCs were constructed over ﬁelds of characteristic 2. But in the work of , they notice that a very good idea would be the usage of ﬁelds of odd characteristic. The rationale is that in the case of relatively large odd characteristic, the ﬁeld equations in the form of xq i −xi = 0, cannot be eﬀectively used, therefore it forces one to ﬁnd all the solutions in the algebraic closure instead of the small ﬁeld itself. This would make the MPKCs much more secure in terms of direct attacks by polynomial solvers. Recent results on the degree of regularity [878, 884] further conﬁrms this notion. 16.3.2.12 Other constructions 16.3.101 Remark There are also other new constructions using diﬀerent new ideas. One interesting one is a construction using Diophantine equations over certain function rings . Another is the MQQ construction using quasi-groups, although it has been defeated [1285, 2115]. 16.3.3 Standard attacks 16.3.102 Remark To attack an MPKC directly as an MQ problem instance is usually not very eﬀective, and cryptanalysts generally try to attack MPKCs utilizing the structure of the central map by either ﬁnding the private key directly, or ﬁnding extra polynomial relations to enhance the polynomial solver. 16.3.3.1 Linearization equations 16.3.103 Deﬁnition For a given MPKC, a linearization equation is a relation between the compo-nents of the ciphertext Y and plaintext X in the following form: X aijxiyj + X bixi + X cjyj + d = 0. (16.3.20) 16.3.104 Remark The key property of these equations is that when substituted with the actual values of Y , we get an aﬃne (linear) relation between the xi’s. In general, each equation should eﬀectively eliminate one variable from the system. 16.3.105 Remark The key and ﬁrst example is the attack against C∗by Patarin . For any C∗ public key, we can compute Y from X, and substitute enough (X, Y ) pairs and solve for aij, bi, cj, and d. A basis for the solution space gives us all the linearization relations. 16.3.106 Remark If we are given any ciphertext, i.e., the values of yi, these n bilinear relations will produce linear equations satisﬁed by components of the plaintext X, and in the case of C∗, it gives us enough linearly independent linear equations to help us to ﬁnd the plaintext with the help of the public equations. In similar systems like 3IC, for example, linearization equations are also present in large numbers, which need to be eliminated to make the system secure. 16.3.3.2 Critical bilinear relations 16.3.107 Remark If the number of linearization equations is high enough, we can defeat the system eﬃciently. However, it is shown in [883, 886, 887] that even when the number of linearization equations (or special form of bilinear relations) is not so large, their existence can be lethal. Cryptography 777 16.3.108 Remark Ding and Schmidt found that the low-rank central polynomials — often rank 2 — in currently existing implementation schemes for the TTM cryptosystems make it possible to extend the linearization method by Patarin to attack all current TTM implementation schemes (Subsection 16.3.3.1). 16.3.109 Remark For the Ding-Schmidt attack , the number of linearization equations is not that high, but they manage to eliminate the “lock polynomial” that defends a TTM instance against a simple rank attack. 16.3.3.3 HOLEs (higher-order linearization equations) 16.3.110 Deﬁnition A Higher-Order Linearization Equation (HOLE) is a linearization relation that is higher degree in the components of Y only. In particular, a SOLE (second order linearization equation) would look like X i<j aijk yiyjxk + X i≤j bij yiyj + X cij yixj + X di yi + X ej xj + f = 0. 16.3.111 Remark This is the key to break the MFE systems. There are at least 8k linear dependencies derived from the SOLE out of a total of 12k variables in an MFE system, which makes the cryptanalyst’s task much easier. Paper used another trick – the fact that squaring is linear in a characteristic two ﬁeld – to get it down to 2k remaining variables at most and concluded that solving for the remaining variables is easy. 16.3.112 Remark The existence of linearization relations of higher degrees shows multivariate en-cryption schemes designed in the triangular style are full of traps and to design a secure system is very diﬃcult without an intrinsically sophisticated algebraic structure. 16.3.3.4 Diﬀerential attacks 16.3.113 Remark Structural attacks on MPKC systems to recover private keys (or equivalently useful keys) are of two related types: Invariants: invariants (mostly subspaces) that can be tracked. Symmetries: transformations that leave certain structures invariant and hence can be computed by a system of equations. These two methods are closely related since invariants are deﬁned according to symmetry. Earlier designers were not yet fully aware of the importance of symmetry. We will present the symmetry or invariants used in the new diﬀerential attacks on the C∗family of cryp-tosystems as exempliﬁed by the diﬀerential attacks developed by Stern and collaborators in . 16.3.3.5 Attacking internal perturbations 16.3.114 Remark The cryptanalysis of the PMI (perturbed Matsumoto-Imai) was a true novelty for a technique usually associated with symmetric key cryptography. 16.3.115 Remark Using the notation in a PMI system, we know that for a randomly chosen vector b, the probability is q−r that it lies in the kernel K of the linear part of V . When that happens, V (X + b) = V (X) for any X. Since q−r is not too small, if we can use this to distinguish between a vector b ∈T −1K (back-mapped into the original space) and b ̸∈T −1K, we can 778 Handbook of Finite Fields bypass the protection of the perturbation, ﬁnd our bilinear relations and accomplish the cryptanalysis. 16.3.116 Remark In , Fouque, Granboulan, and Stern built a distinguisher using a test on the kernel of the symmetric diﬀerence DP(W, b) = P(b + W) −P(b) −P(W). We say that t(b) = 1 if dim kerW DP(b, W) = 2gcd(n,α) −1, and t(b) = 0 otherwise. If b ∈K, then t(b) = 1 with probability one, otherwise it is less than one. If gcd(n, α) > 1, it is a nearly perfect distinguisher. If not, we can employ two other tricks. For one of them, we observe K is a vector space, so Pr(t(b + b′) = 0\|t(b′) = 0) will be relatively high if b ∈K and relatively low otherwise. 16.3.117 Remark There is a surprisingly simple defense dating back to (which introduced SFLASH). By using the “plus” variant, i.e., appending a random quadratic polynomial to P, enough false positives are generated to overwhelm the distinguishing test of . The extra equations also serve as a distinguisher when there are extraneous solutions. Basically, the more “plus” equations, the less discriminating power of the above mentioned test. Based on empirical results of Ding and Gower , when r = 6, a = 12 should be suﬃcient, and a = 14 would be a rather conservative estimate for the amount of “plus” needed to mask the PMI structure. 16.3.3.6 The skew symmetric transformation 16.3.118 Remark The symmetry found in can be explained by considering the case of the C∗ cryptosystem. The symmetric diﬀerential of any function G, deﬁned formally just like in Equation (16.3.21): DG(a, X) := G(X + a) −G(X) −G(a) + G(0), is bilinear and symmetric in its variables a and X. In the ﬁrst version of this attack , we look at the diﬀerential of the public map P, and look for skew-symmetric maps with respect to this bilinear function, namely, the linear maps M such that DP(c, M(W)) + DP(M(c), W) = 0. 16.3.119 Remark The reason that this works is that the central map Q and the public key, which encapsulates the vital information in the central map, unfortunately have very strong sym-metry in the sense that all the diﬀerentials from these maps share some common nontrivial skew-symmetric map M. Since Q(X) = X1+qα, its diﬀerential is DQ(a, X) = aqαX + aXqα. 16.3.120 Theorem The maps M skew-symmetric with respect to this DQ(a, X) are precisely those induced from multiplication by some element ζ satisfying the condition ζqα + ζ = 0. 16.3.121 Remark This skew-symmetry survives under change of basis. It can be seen that the skew-symmetry continues to hold even when we remove some components of P. In terms of the public key, this means that if we write DP(c, W) := (cT H1W, cT H2W, . . . , cT HmW) Cryptography 779 and try to solve M T Hi + HiM = 0 for all i = 1, . . . , m simultaneously, we should ﬁnd just k-multiples of the identity if n and α are co-prime, and a d-dimensional subspace in the space of linear maps if d = gcd(n, α) > 1. 16.3.122 Remark For a randomly chosen map G, it is expected that only trivial solutions M = u1n, where u ∈K, will satisfy this condition. This means that there is a very strong condition on C∗−cryptosystems. This symmetry can be utilized to break C∗−systems for which d = gcd(n, α) > 1. 16.3.3.7 Multiplicative symmetry 16.3.123 Remark The second symmetry is multiplicative symmetry, which comes also from the diﬀerential DP(c, W) . 16.3.124 Proposition Let ζ be an element in the big ﬁeld L. Then we have DQ(ζ · a, x) + DQ(a, ζ · x) = (ζqα + ζ)DQ(a, x). 16.3.125 Theorem Let Mζ = M−1 S ◦(X 7→ζX) ◦MS be the linear map in Kn corresponding to multiplication by ζ, then span{MT ζ Hi + HiMζ : i = 1, . . . , n} = span{Hi : i = 1, . . . , n}, i.e., the space spanned by the quadratic polynomials from the central map is invariant under the skew-symmetric action. 16.3.126 Remark The public key of C∗−inherits some of that symmetry. Note that not ev-ery skew-symmetric action by a matrix Mζ corresponding to an L-multiplication re-sults in MT ζ Hi + HiMζ being in the span of the public-key diﬀerential matrices, because S := span{Hi : i = 1, . . . , n −r} as compared to span{Hi : i = 1, . . . , n} is missing r of the basis matrices. However, as the authors of argued heuristically and backed up with empirical evidence, if we just pick the ﬁrst three MT ζ Hi + HiMζ matrices, or any three random linear combinations of the form Pn−r i=1 bi(MT ζ Hi +HiMζ) and demand that they fall in S, then there is a good chance to ﬁnd a nontrivial Mζ corresponding to a multiplication by ζ, which can be used to break the C∗−scheme. 16.3.127 Remark For a set of public keys from C∗, tests show that the above strategy almost surely eventually recovers the missing r equations and breaks the scheme. The only known attempted defense is . 16.3.3.8 Rank attacks 16.3.128 Remark Given a quadratic polynomial, we can always associate it with a symmetric matrix. By a rank attack, we mean an attack using the rank of those matrices. There are two types of rank attacks, attacks that speciﬁcally target high rank or low rank. Let Hi be the symmetric matrix corresponding to the quadratic part of public polynomials. 16.3.129 Remark Since rank attack usually means attacking via ﬁnding low rank matrices, some also call the high rank attack the dual rank attack. The high rank attack ﬁrst appeared with where Coppersmith et al. defeated a triangular construction. 16.3.130 Remark The high rank attacks of Goubin-Courtois and Yang-Chen [1339, 3023] work for “plus”-modiﬁed triangular systems; it is also easier to understand than the formulation in . Against UOV, we might possibly do even better on this attack with diﬀerentials . 780 Handbook of Finite Fields 16.3.131 Remark The ﬁrst Minrank attack is the Goubin-Courtois version against TTM. Let r be the smallest rank in linear combinations of central equations, which without loss of generality we take to be the ﬁrst central equation itself in TTM. Goubin and Courtois outline how to ﬁnd the smallest ranked combination (and hence break the Triangle-Plus-Minus system) in expected time O(q⌈m n ⌉rm3). Yang and Chen have extended the eﬀectiveness of this attack related to the number of distinct kernels of the same rank. 16.3.3.9 MinRank attacks on big-ﬁeld schemes 16.3.132 Remark The defeat of the HFE Challenge 1 by Faug ere and Joux , a direct solution of the 80 equations in 80 variables, is not the ﬁrst serious attempt on HFE systems. That credit goes to a rank-based attack by Kipnis and Shamir . The attack proceeds by moving the problem back to the extension ﬁeld, where all the underlying structure can be seen. This is a very natural approach if we intend to exploit the design structure of HFE in the attack. They transform the problem of ﬁnding the secret key into a problem of ﬁnding the minimum rank of linear combinations of certain matrices, which is exactly r (as in Subsection 16.3.2.3). This is the MinRank problem and is in general exponential, but can be easy if r is small. 16.3.133 Remark Kipnis and Shamir suggested using determinants of all (r + 1) × (r + 1) sub-matrices to derive a huge assortment of equations to solve the problem. To solve this system, they introduce an idea which they call relinearization, which led to the well-known XL paper . It has been argued that using a Lazard-Faugere solver on this system of equations is eﬀective and equally eﬀective as the direct attack, however the situation is still not very clear due to a recent observation in . 16.3.3.10 Distilling oil from vinegar and other attacks on UOV 16.3.134 Remark To forge a signature for a UOV scheme as in Subsection 16.3.2.4, one needs to solve the equation P(W) = Y . 16.3.135 Remark When o = v as with the original Oil-and-Vinegar, this is fairly easy due to the attack by Kipnis and Shamir . The basic idea is to treat each component yi = pi(W) of the public key P as a bilinear form. Namely, take their associated symmetric matrices via the symmetric diﬀerential as follows: Dpi(W, c) := pi(W + c) −pi(W) −pi(c) + pi(0) := cT HiW. (16.3.21) A basic fact of OV: each matrix Mi, see Equation (16.3.9) is in the rough form of ∗ ∗ ∗ 0 but not the matrices Hi associated to the public key. This reduces a cryptanalysis to the algebraic problem of ﬁnding a basis change for a set of bilinear forms into a common form. 16.3.136 Remark Assume v = o = n/2 = m. The vectors X that have all vinegar coordinates x1, . . . , xv equal to zero, form the Oil Space O, i.e., the collection of X-vectors looking like 0 ∗ , and similarly an X-vector in the Vinegar Space V has all oil coordinates xv+1, . . . , xn equal to zero and looks like ∗ 0 . Clearly, if each Mi is nonsingular, we have ∗ ∗ ∗ 0 0 ∗ = ∗ 0 , or MiO = V for all i. Cryptography 781 16.3.137 Proposition If Mj, Hj are invertible, M −1 j Mi O = O; H−1 j Hi (S−1O) = (S−1O). 16.3.138 Remark This proposition states that any H−1 j Hi has the common invariant subspace of S−1O. Knowing S−1O is suﬃcient to ﬁnd an equivalent form for S. Kipnis et al. claim that the same argument works if v < o; even if v > o it can be done in time directly proportional to qv−o, and hence v −o cannot be too small. However the situation about this claim is also not very clear due to recent observations in . When there are two or three times more vinegar variables than oil variables the method appears to be secure, despite the claims of . 16.3.3.11 Reconciliation 16.3.139 Remark We could also attempt to ﬁnd a sequence of change of basis that would lead to the inversion of the public map as an improvement to a brute force attack . In the case of an attack on the UOV scheme with o oil and v = n −o vinegar variables, the attack becomes a problem of solving m equations in v variables, which could be much easier than solving m equations in n variables in a direct attack. The reconciliation attack fails with a probability of approximately 1 q−1. 16.3.140 Remark Reconciliation attack can be applied to Rainbow systems, which have multiple layer structure (Subsection 16.3.2.4). This attack works for all constructions with a UOV construction in the ﬁnal stage, including all Rainbow and TTS constructions. This aﬀects how the current proposed parameters of Rainbow are selected. 16.3.3.12 Direct attacks using polynomial solvers 16.3.141 Remark To mount a direct attack, we try to solve the m equations P(W) = Z in the n variables w1, . . . , wn. If m ≥n, we are (over-)determined. If m < n, we are underdetermined. For most cases (n cannot be too large compared to m), we cannot do much more than to ﬁx values for m −n variables randomly and continue with m = n . 16.3.142 Remark Due to the NP-hardness the the MQ problem , the diﬃculty of solving “generic” or randomly chosen systems of nonlinear equations is generally conceded. But, it is very hard to quantify exactly how hard it is to solve a non-generic system. Often many techniques in algebraic cryptanalysis require solving a system of polynomial equations at the end for more or less generic systems. So we must solve the system p1 = p2 = · · · = pm = 0, where each pi is a quadratic polynomial. Coeﬃcients and variables are in the ﬁeld K = Fq. 16.3.143 Remark The standard methods for solving equations are Buchberger’s algorithm to compute a Gr¨ obner basis, and its descendents investigated by Lazard’s group . Macaulay generalized Sylvester’s matrix to multivariate polynomials . The idea is to construct a matrix whose rows are from the coeﬃcients of the multiples of the polynomials of the original system, the columns representing all the monomials up to a given degree. Lazard observed that for a large enough degree, ordering the columns according to a monomial ordering and performing usual row reduction on the matrix is equivalent to Buchberger’s algorithm. 16.3.144 Remark Faug ere proposed an improved Gr¨ obner bases algorithm called F4 . A later version, F5 made headlines when it was used for solving HFE Challenge 1 in 782 Handbook of Finite Fields 2002, but we do not really know what the real F5 is, since no one else, as far as we know, has repeated any of the many results claimed by F5. A version of F4 is implemented in the computer algebra system MAGMA and is publicly available. 16.3.145 Remark Lazard’s idea was rediscovered in 1999 by Courtois, Klimov, Patarin, and Shamir as XL. Courtois et al. proposed several adjuncts [736, 742, 743] to XL. 16.3.146 Remark Recently based on the concept of mutant, much work has been done to improve the XL algorithms, which produced a family of mutant XL algorithms [438, 877, 879, 2114, 2116]. 16.3.147 Remark For a system of equations p1(x1, . . . , xn) = · · · = pm(x1, . . . , xn) = 0, if we look at the ideal generated by pi, then each element f of the ideal can be expressed in the form: f = n X 1 fipi, which is not unique. 16.3.148 Deﬁnition For each such expression, we deﬁne the level of this expression to be the max(deg(gi) + deg(pi), 1 = 1, . . . , m). If deg(f) is lower than any of the levels of f, f is a mutant. 16.3.149 Remark The key idea of mutants is that we try to mathematically describe the degeneration of the systems, which is critical for the solving process to work. Currently, the best mutant algorithms beat all other algebraic solvers, including F4. 16.3.150 Remark A key question for understanding the security of the MPKCs is the complexity of those algebraic solvers. Using generating functions, there are heuristic estimates on the complexity of solving generic systems [201, 3022]. These works are used to estimate the complexity of algebraic attack on the HFE systems based on the concept of degree of regularity. Recently new breakthroughs have led us to mathematically prove new estimates for the degree of regularity of HFE [878, 884, 926], which provide theoretical support for the apparent security beneﬁts of using ﬁelds of odd characteristics. 16.3.151 Theorem Let P be a HFE polynomial of degree D. If Rank(P) > 1, the degree of regularity of the associated HFE system is bounded by (q −1)Rank(P) 2 + 2. In particular, this is less than or equal to (q −1)(⌊logq(D −1)⌋+ 1) 2 + 2. If Rank(P) = 1, then the degree of regularity is less than or equal to q. Here Rank is the rank of the quadratic form associated to P. 16.3.152 Remark The concept of degree of regularity and mutant are also closely related. Mutants can appear only at a degree equal to or higher than the degree of regularity . 16.3.4 The future 16.3.153 Remark What really drives the development of the designs in MPKCs are indeed new mathematical ideas that bring new mathematical structures and insights in the construction Cryptography 783 of MPKCs. The mathematical ideas we have used are just some of the very basic ideas developed in mathematics and there is great potential in advancing this idea further with some of the more sophisticated mathematical constructions in algebraic geometry. One particularly interesting problem would be to make the TTM cryptosystems work with the establishment of some new systematic approach. This deﬁnitely demands some deep insights and the usage of some intrinsically combinatorial structures from algebraic geometry. 16.3.154 Remark Though a lot has been done in studying the eﬃciency of diﬀerent attacks, we still do not fully understand the full potential or the limitations of some of the attack algorithms. We still need to understand both the theory and practice of how eﬃciently general attack algorithms work and how to implement them eﬃciently. From the theoretical point of view, to answer these problems, the foundation again lies in modern algebraic geometry. One critical step would be to prove the maximum rank conjecture postulated in , the theoretical basis used to estimate the complexity of the polynomial solving algorithms, and the F4 algorithm. Another interesting problem is to mathematically prove some of the commonly used complexity estimate formulas in . 16.3.155 Remark MPKCs interact more and more with other topics like algebraic attacks. Algebraic attacks are a very popular research topic in breaking symmetric block ciphers like AES and stream ciphers and analyzing hash functions . The origin of such an idea is from MPKCs, and in particular Patarin’s linearization equation attack method. New ideas in MPKCs will have much more broad applications in the area of algebraic attacks. The idea of multivariate constructions was also applied to the symmetric constructions [281, 893]. Similar ideas may have further applications in designing stream ciphers and block ciphers. The theory of functions on a space over a ﬁnite ﬁeld (multivariate functions) will play an increasingly important role in the uniﬁcation of the research in all these related areas. 16.3.156 Remark Research in MPKCs has already developed new challenges that need new methods and ideas. A mutually beneﬁcial interaction between MPKCs and algebraic geometry will grow rapidly. MPKCs will provide excellent motivation and critical problems for the development of the theory of functions over ﬁnite ﬁelds, and new mathematical tools and insights are critical for MPKCs’ future development. See Also §11.4 For algorithms to compute the roots of HFE polynomials. §13.4 For the fast linear algebra used in multivariate polynomial solving algorithms. §16.1 For public key cryptography and post-quantum cryptography. References Cited: [67, 128, 201, 230, 280, 281, 390, 437, 438, 467, 472, 499, 500, 603, 604, 605, 722, 736, 737, 738, 739, 740, 742, 743, 748, 853, 860, 876, 877, 878, 879, 880, 881, 882, 883, 884, 885, 886, 887, 888, 889, 890, 891, 892, 894, 893, 895, 896, 924, 925, 926, 1040, 1041, 1042, 1044, 1049, 1094, 1095, 1214, 1266, 1285, 1339, 1340, 1349, 1440, 1461, 1610, 1686, 1687, 1737, 1738, 1739, 1877, 1914, 1985, 2028, 2029, 2113, 2114, 2115, 2116, 2213, 2220, 2226, 2312, 2320, 2321, 2367, 2364, 2365, 2366, 2368, 2369, 2370, 2371, 2387, 2415, 2606, 2613, 2740, 2798, 2822, 2823, 2929, 2930, 2931, 2932, 2998, 3022, 3023, 3024] 784 Handbook of Finite Fields 16.4 Elliptic curve cryptographic systems Andreas Enge, INRIA Bordeaux–Sud-Ouest 16.4.1 Cryptosystems based on elliptic curve discrete logarithms 16.4.1 Remark The Fq-rational points on an elliptic curve E deﬁned over a ﬁnite ﬁeld Fq form a ﬁnite abelian group according to Subsection 12.2.2; its group order is close to q by Theo-rem 12.2.46. This group can be used to implement the discrete logarithm based cryptosys-tems introduced in Subsection 16.1.3.2, as ﬁrst observed in [1771, 2102]. 16.4.2 Remark For reasons of eﬃciency, elliptic curve cryptosystems are usually implemented over prime ﬁelds Fp or ﬁelds F2m of characteristic two. Supersingular curves over ﬁelds F3m of characteristic three have attracted some attention in the context of pairing based cryptography, see Section 16.4.2. 16.4.1.1 Key sizes 16.4.3 Remark To resist generic attacks on the discrete logarithm problem, elliptic curve cryp-tosystems are implemented in the prime order cyclic subgroup of maximal cardinality n inside E(Fq). For representing group elements with the minimum number of bits, it is desir-able that the curve order itself be prime. Except for special cases (see Section 16.4.1.3 and [2535, 2581, 2682]), only generic attacks are known on the elliptic curve discrete logarithm problem (ECDLP), with a running time on the order of √n. A security level of m bits, corresponding to a symmetric-key cryptosystem (see Section 16.1.2) with 2m keys, thus requires an order n of 2m bits. Extrapolating the theoretical subexponential complexity of Remark 11.6.38 for factoring or the DLP in ﬁnite ﬁelds allows to derive heuristic security estimates for the corresponding public key cryptosystems of Sections 16.1.3.1 and 16.1.3.2. Several studies have been carried out in the literature, taking added heuristics on techno-logical progress into account, see . They are summarized in the following table; the ﬁgures for the factorization based RSA system essentially carry over to systems based on discrete logarithms in ﬁnite ﬁelds; see Remark 11.6.38. The 80 bit security level is a historic ﬁgure. security (bits) symmetric ECC RSA RSA RSA [1276, 1894] [2426, §7.2.2.3] [2684, Table 7.2] 80 — 160 1513 1536 1248 112 Triple DES 224 4509 4096 2432 128 AES-128 256 6669 6000 3248 192 AES-192 384 22089 — 7936 256 AES-256 512 49562 — 15424 16.4.1.2 Cryptographic primitives 16.4.4 Remark Some cryptographic primitives (encryption, signatures, etc.) have been adapted and standardized speciﬁcally for elliptic curves. As other discrete logarithm based systems (see Section 16.1.3.2), they require a setup of public domain parameters, a cyclic subgroup G of prime order n of some curve E(Fq), with a ﬁxed base point P such that G = ⟨P⟩. Moreover, the bit patterns representing elements of Fq and E(Fq) need to be agreed upon. Cryptography 785 16.4.5 Example (Elliptic Curve Integrated Encryption Scheme, ECIES) This cryptosystem is es-sentially the same as ElGamal’s, see Example 16.1.27; but the encryption of elements of G is replaced by symmetric key encryption of arbitrary bit strings with a derived secret key. So the scheme is hybrid, using symmetric key and public key elements. An additional message authentication code (MAC) prevents alterations of the encrypted message during transmission and authenticates its sender. (A MAC is essentially a hash function, see Re-mark 16.1.23, depending additionally on a symmetric key, and can indeed be constructed from hash functions; for more details, see [2080, Subsection 9.5.2].) Besides the domain parameters for the elliptic curve group, the setup comprises a sym-metric key scheme with an encryption function Ek1 and inverse decryption function Dk1, using keys k1 of length ℓ1 bits; and a message authentication code Mk2 using keys k2 of length ℓ2. Party A has the private key a ∈[0, n −1] and the related public key Q = aP. To encrypt a message m ∈{0, 1}∗, party B selects a random integer r ∈[0, n −1], computes R = kP, S = kQ and (k1, k2) = f(S), where f : G →{0, 1}ℓ1 × {0, 1}ℓ2 is a key derivation function (for instance, a cryptographic hash function; see Remark 16.1.23). He computes c1 = Ek1(m) and c2 = Mk2(c1); the ciphertext is (R, c1, c2). To decrypt such a ciphertext, party A recovers S = aR and (k1, k2) = f(S). If Mk1(c1) ̸= c2, she rejects the ciphertext as invalid; otherwise, she obtains the clear text as m = Dk1(c1). 16.4.6 Remark The scheme has been ﬁrst described in a generic discrete logarithm setting (and in a slightly diﬀerent form) in , and standardized under the name Elliptic Curve Aug-mented Encryption Scheme in . For arguments supporting its security under suitable assumptions on the underlying primitives, see [221, 2683] and [313, Chapter III]. 16.4.7 Example (Elliptic Curve Digital Signature Algorithm, ECDSA) The algorithm is a simple transposition of the DSA of Section 16.1.3.3 to the elliptic curve setting. Besides the domain parameters for the elliptic curve group, the setup comprises a hash function H : {0, 1}∗→[0, n −1] and the reduction function f : G →[0, n −1], (x, y) 7→x (mod n). Party A has the private key a ∈[0, n −1] and the related public key Q = aP. To sign a message m, party A randomly selects an integer k ∈[1, n −1], computes R = kP, r = f(R), h = H(m), and s ≡k−1(h + ar) (mod n). The signature is the pair (r, s). To verify such a signature, party B computes h = H(m), w ≡s−1 (mod n), u1 ≡wh (mod n), u2 ≡wr (mod n), and R = u1P + u2Q. He accepts the signature as valid if and only if r = f(R). 16.4.8 Remark The scheme has been standardized in , see also , [1567, Subsections 7.2.7–7.2.8], and [2292, Section 6]. For arguments supporting its security under suitable assumptions on the underlying primitives, see [313, Chapter II] and . The fact that the function f depends only on the x-coordinate of its argument has raised doubts about the security of the scheme ; in particular, it implies weak malleability: From a signature (r, s) on a given message, another signature (r, −s) on the same message may be obtained. 16.4.1.3 Special curves 16.4.9 Remark A necessary condition for the security of an elliptic curve cryptosystem is that the order of E(Fq) be prime, or a prime multiplied by a small cofactor. Some special curves for which this condition is easily tested have been suggested in the literature. These are more and more deprecated in favor of random curves (see Section 16.4.1.4) in conventional discrete logarithm settings, . Supersingular and especially CM curves are still needed, 786 Handbook of Finite Fields however, in pairing based cryptography; see Section 16.4.2. 16.4.10 Example (Supersingular curves) The orders of supersingular elliptic curves are known by Theorem 12.2.51: Over Fp, the only occurring order is p + 1. Over Fpm with p ∈{2, 3}, the orders pm + 1 −t with t ∈{0, ±pm/2, ±p(m+1)/2, ±2pm/2} may occur depending on the parity of m. The ECDLP on supersingular curves over Fpm may be reduced to the DLP in the multiplicative group of Fp2 for curves over Fp; of Fp, Fp2, Fp3 or Fp4 for curves over F2m; and of Fp, Fp2, Fp3 or Fp6 for curves over F3m. Thus, supersingular curves are deprecated except for low security pairing based cryptosystems. 16.4.11 Example (Curves over extension ﬁelds) If E is deﬁned over a ﬁnite ﬁeld Fq with q small, then \|E(Fq)\| can be obtained by exhaustively enumerating all points; and \|E(Fqm)\| is easily computed by Remark 12.2.105. In particular, the case q = 2 has been suggested in the literature. However, since E(Fqm) contains the subgroup E(Fq) (and further subgroups if m is not prime), the group order cannot be prime anymore. 16.4.12 Remark The existence of the additional Frobenius automorphism of order m, together with the negation automorphism of order 2, may be used to speed up the generic algorithms of Sections 11.6.6 and 11.6.7 by a factor of √ 2m [1165, 2978], which reduces the eﬀective security level. 16.4.13 Remark (Weil descent) If E is deﬁned over an extension ﬁeld Fqm, then E(Fqm) can be embedded into A(Fq), where A is an abelian variety of dimension m, called the Weil restric-tion or restriction of scalars of E. There is reason to believe that the discrete logarithm problem in A(Fq) may be easier to solve than by a generic algorithm, relying on an ap-proach of representing the group A(Fq) by a set of generators (called the factor base) and relations which are solved by linear algebra, cf. Section 11.6.8, leading to a potential attack described ﬁrst in [1104, Section 3.2]. Cases where A contains the Jacobian of a hyperelliptic curve of genus close to m have been worked out for curves over ﬁelds of characteristic 2 in [1160, 1250], and ﬁelds of odd characteristic in . So far, the attack has been made eﬀective for certain curves with prime m ≤7. Another algorithm for discrete logarithms, working directly with curves over Fqm and specially adapted factor bases, is described in ; heuristically, it is faster than the generic algorithms for m ≥3 ﬁxed and q →∞. Since it involves expensive Gr¨ obner basis computations, it has been made eﬀective only for m ≤3. Combinations of these approaches are also possible and have led to an attack on curves of close to cryptographic size over Fp6 . Moreover, isogenies may be used to transport the discrete logarithm problem from a seemingly secure curve to one that may be attacked by Weil descent . It thus appears cautious to prefer for cryptographic applications curves over prime ﬁelds Fp or, if even characteristic leads to signiﬁcant performance improvements, ﬁelds F2m of prime extension degree m. 16.4.14 Example (Complex multiplication curves) All ordinary elliptic curves over a ﬁnite ﬁeld Fq = Fpm have complex multiplication by some order OD = h 1, D+ √ D 2 i Z of discriminant D < 0 in the imaginary-quadratic ﬁeld Q( √ D); see Remark 12.2.98. For small \|D\|, this can be exploited to explicitly construct curves with a known number of points as follows. 1. Let D < 0, D ≡0 or 1 (mod 4), p prime and m minimal such that 4pm = t2−v2D has a solution in integers t, v. 2. Compute the class polynomial HD ∈Z[X], the minimal polynomial of j D+ √ D 2 , where j is the absolute elliptic modular invariant function. Cryptography 787 3. HD splits completely over Fpm (and no subﬁeld), and its roots are the j-invariants of the elliptic curves deﬁned over Fpm with complex multiplication by OD. For each such j-invariant, one easily writes down a curve with pm + 1 −t points by solving the expression of j in Deﬁnition 12.2.2 for the curve coeﬃcients (up to isomorphisms and twists, see Section 12.2.5, the solution is unique). 16.4.15 Remark It is easy to see that a prime number of points is only possible for D ≡5 (mod 8). 16.4.16 Remark The degree of the class polynomial is the class number of OD, and its total bit size is of the order of O(\|D\|1+ϵ) under GRH. Several quasi-linear algorithms of complexity O(\|D\|1+ϵ) for computing class polynomials have been described in the literature, by ﬂoating point approximations of its roots , lifting to a local ﬁeld or Chinese remaindering . Nevertheless, the algorithms are restricted to small values of \|D\|, while random curves correspond to \|D\| of the order of q, so that only a negligible fraction of curves may be reached by the CM approach. 16.4.17 Remark While no attack on this particular fraction of curves has been devised so far, random curves are generally preferred where possible; note, however, that pairing-based cryptosystems require the use of either supersingular curves or ordinary curves obtained with the CM approach; see Section 16.4.2.4. 16.4.18 Example (NIST curves) The USA standard suggests a prime ﬁeld Fp and a pseu-dorandom curve (assuming that the hash function SHA-1 is secure) of prime order over Fp for p of 192, 224, 256, 384, and 521 bits. (The largest example is for the Mersenne prime p = 2521 −1.) For the binary ﬁelds F2163, F2233, F2283, F2409 and F2571, a pseudo-random curve (of order twice a prime) and a curve deﬁned over F2 (of order twice or four times a prime) are given. As recommended in Remark 16.4.13, the extension degrees are prime for curves deﬁned over F2. 16.4.19 Remark We note that the generic discrete logarithm algorithms of Subsections 11.6.6 and 11.6.7 allow for a trade-oﬀbetween precomputations and the breaking of a given discrete logarithm: In a group of size about 2m, a precomputation of 2k group elements yields additional logarithms in time 2m−k. As a precaution, one may thus wish to avoid predetermined curves, especially at lower security levels. 16.4.1.4 Random curves: point counting 16.4.20 Remark Algorithms for counting points on random elliptic curves currently come in two ﬂavours. The ﬁrst algorithm, SEA, is of polynomial complexity; for curves over extension ﬁelds Fpm, there are a variety of algorithms using p-adic numbers, with a much better polynomial exponent in m, but which are exponential in log p. 16.4.21 Algorithm (Schoof) In , Schoof describes the ﬁrst algorithm of complexity polynomial in log q for counting the number of points on an arbitrary elliptic curve E(Fq). The algorithm is deterministic and computes the trace of Frobenius aq of Deﬁnition 12.2.47 and thus the zeta function of Section 12.2.10. Given a prime ℓnot dividing q, by Theorem 12.2.66 the value of aq modulo ℓcan be determined by checking for all possible values whether the numerator of the zeta function annihilates the ℓ-torsion points. Chinese remaindering for suﬃciently many primes yields the exact value of aq, which is bounded by Theorem 12.2.46. The algorithm has a complexity of O (log q)5+ϵ) , due in part to the fact that the ℓ-torsion points generate an Fq-algebra of dimension O(ℓ2). 16.4.22 Algorithm (Schoof–Elkies–Atkin, SEA) Improvements are due to Atkin and Elkies . When there is an Fq-rational separable isogeny (see Deﬁnition 12.2.26) of degree ℓfrom 788 Handbook of Finite Fields E(Fq) to another curve, then the ℓ-torsion points may be replaced by the kernel of the isogeny, generating an algebra of dimension O(ℓ) over Fq. By the complex multiplication theory of Example 16.4.14, this happens when ℓis coprime to the conductor of the ring of endomorphisms OD of E and ℓis not inert in the quadratic number ﬁeld Q( √ D), which holds for about half of the primes. The complexity of the algorithm becomes O (log q)4+ϵ) [312, Chapter VII], [661, Section 17.2]. 16.4.23 Remark The practical bottleneck of the algorithm used to be the computation of bivariate modular polynomials, of size O(ℓ3+ϵ), needed to derive isogenies of degree ℓ. A quasi-linear algorithm is described in ; eventually limited by space, it has been used for ℓup to around 10000. A more recent algorithm computes the polynomial, reduced modulo the characteristic p of Fq and instantiated in one variable by an element of Fq, also in time O(ℓ3+ϵ), but in space O(ℓ(ℓ+ log q)); it has been used for ℓup to about 100000. Further building blocks of the SEA algorithm have also been optimized [366, 1251, 2098]. The current record is for a prime ﬁeld Fp with p having about 5000 decimal digits . 16.4.24 Remark The SEA algorithm is implemented in several major computer algebra systems, and random elliptic curves of cryptographic size with a prime number of points are easily found, be it as domain parameters, be it in a setting where each user has his own elliptic curve as part of his public key. 16.4.25 Algorithm (p-adic point counting) For an elliptic curve E over an extension ﬁeld Fpm, Satoh introduced an algorithm computing its canonical lift to a curve ˆ E over Qpm, the unramiﬁed extension of degree m of the p-adic numbers Qp. The curve ˆ E has the same endomorphism ring OD (Example 16.4.14) as E and reduces modulo the maximal ideal of Qpm to E. More precisely, an approximation to ˆ E may be computed by Newton iterations on a function derived from the modular polynomial of level p, Algorithm 16.4.21, at arbitrary p-adic precision. In a second step, the trace of the Frobenius map is computed in this characteristic 0 setting by the action of its dual isogeny (the reduction of which is separable) on a holomorphic diﬀerential; for this, the isogenies are computed explicitly. After a precomputation of O(p3+ϵ) for the p-th modular polynomial (see Algorithm 16.4.21), the complexity of the algorithm is O(p2m3+ϵ). 16.4.26 Remark Satoh’s algorithm is not immediately applicable in characteristic two. Mestre sug-gests in to use arithmetic-geometric mean (AGM) iterations, a sequence of isogenies of degree 2, to obtain the canonical lift and the trace of the Frobenius map, also in time O(m3+ϵ). 16.4.27 Remark Later work concentrates on lowering the complexity in m: to quasi-quadratic for ﬁnite ﬁelds Fqm with a Gaussian normal basis or in the general case ; or on lowering the complexity in p: to quasi-linear or even quasi-square root . The record in for a curve over F2100002 goes beyond all practical cryptographic needs. 16.4.28 Remark For a more thorough account, see [313, Chapter VI] or [661, Section 17.3]. 16.4.2 Pairing based cryptosystems 16.4.29 Remark While conventional elliptic curve cryptography relies on the map x 7→xP, which is a group homomorphism or, equivalently, a linear map of Z/nZ-modules, pairing based cryptography requires a bilinear map e : G1 × G2 →G3, see Deﬁnition 16.1.35. This introduces an additional degree of freedom and a wealth of new cryptographic primitives following the ﬁrst applications described in Section 16.1.4. Since 2007, a series of conferences, Pairing-Based Cryptography — Pairing, has been devoted to the topic [1159, 1630, 2601, 2766]. Cryptography 789 16.4.2.1 Cryptographic pairings 16.4.30 Deﬁnition Let E be an elliptic curve deﬁned over a ﬁnite ﬁeld Fq, and let n be the largest prime divisor of the cardinality of E(Fq). Assume that n does not divide q. (This is required in a cryptographic setting due to anomalous curves.) Then the embedding degree of E is the smallest integer k such that E(Fqk) contains E(Fq)[n], the n2 points of n-torsion of E(Fq); see Theorem 12.2.60, i.e., k is minimal such that E(Fqk)[n] = E(Fq)[n]. 16.4.31 Theorem [183, Theorem 1] If n does not divide q −1, then the embedding degree is the smallest integer k such that n divides qk −1. 16.4.32 Deﬁnition A cryptographic elliptic pairing is a map e : G1 × G2 →G3 satisfying the conditions of Deﬁnition 16.1.35, where E is an elliptic curve deﬁned over Fq, n is the largest prime factor of \|E(Fq)\|, n divides neither q −1 nor q, k is the embedding degree of E, and G1 ⊆E(Fq) and G2 ⊆E(Fqk) (denoted additively) and G3 ⊆F∗ qk (denoted multiplicatively) are subgroups of order n. 16.4.33 Remark In this setting, G1 = E(Fq)[n] and G3 are in fact ﬁxed, while there are n + 1 possible choices for G2; see Subsection 16.4.2.3. Diagonalizing the matrix of the Frobenius endomorphism on E(Fq)[n] by Theorem 12.2.66 yields a mathematically canonical choice also for G2, which is given by the following theorems. 16.4.34 Theorem G1 is the subgroup of E(Fqk)[n] generated by the points having eigenvalue 1 under the Frobenius endomorphism φq of Example 12.2.31. There is a unique subgroup G2 ⊆E(Fqk) of order n generated by the points having eigenvalue q under the Frobenius endomorphism. 16.4.35 Theorem Let Tr : E(Fq) →E(Fq), P 7→Pk−1 i=0 φi q(P), denote the trace endomorphism of level k on E. Then the endomorphisms Tr and π2 = id −φq, restricted to E(Fqk)[n], yield surjective group homomorphisms Tr : E(Fqk)[n] →G1 with kernel G2 and π2 : E(Fqk)[n] → G2 with kernel G1. 16.4.36 Deﬁnition For a point P on E deﬁned over some extension ﬁeld Fqm and an integer r, let fr,P be the function with divisor r(P) −(rP) −(r −1)(O) that is deﬁned over Fqm and has leading coeﬃcient 1 in O, see Deﬁnitions 12.1.21 and 12.1.23. For ﬁnite points R and S ̸= −R, denote by vR = x −x(R) the line with divisor (R) + (−R) −2(O) and by ℓR,S = (y −y(R)) −λR,S(x −x(R)) the line with divisor (R) + (S) + (−R −S) −3(O), where λR,S = ( y(S)−y(R) x(S)−x(R) if R ̸= S, 3x(R)2+2a2x(R)+a4−a1y(R) 2y(R)+a1x(R)+a3 if R = S, see Algorithm 12.2.21; additionally, ℓR,−R = vR and vO = 1. 16.4.37 Deﬁnition An addition-negation chain for an integer r is a sequence r1, . . . , rs such that r1 = 1, rs = r and each element ri is either 1. the negative of a previsously encountered one: there is 1 ≤j(i) < i such that ri = −rj(i); or 2. the sum of two previously encountered ones: there are 1 ≤j(i) ≤k(i) < i such that ri = rj(i) + rk(i). 790 Handbook of Finite Fields 16.4.38 Algorithm Require: A point P on E and an integer r Ensure: fr,P = L V , where L and V are given as products of lines Compute an addition-negation chain r1, . . . , rs for r. P1 ←P, L1 ←1, V1 ←1 for i = 2, . . . , s do j ←j(i), k ←k(i) if ri = −rj then Pi ←−Pj Li ←Vj Vi ←LjvPi else Pi ←Pj + Pk Li ←LjLkℓPj(i),Pk(i) Vi ←VjVkvPi end if end for return L = Ls, V = Vs 16.4.39 Example The Weil pairing en of Theorem 12.2.70 is a cryptographic pairing as long as G2 ̸= G1. If P, Q ∈E(Fqk)[n], then en(P, Q) = (−1)n fn,P (Q) fn,Q(P ). 16.4.40 Example Assume that E(Fqk) does not contain a point of order n2, or, equivalently, that n3 does not divide \|E(Fqk)\|. Then the map E(Fqk)[n] →E(Fqk)/nE(Fqk), Q 7→Q + nE(Fqk), is a group isomorphism, and the Tate pairing T of Theorem 12.2.75 yields a non-degenerate pairing e′ T : E(Fqk)[n] × E(Fqk)[n] →F∗ qk/(F∗ qk)n, (P, Q) 7→T(P, Q + nE(Fqk)). Since e′ T \|G1×G1 takes values in F∗ q/(F∗ q)n = {1}, the restriction e′ T \|G1×G2 is non-degenerate for any G2 ̸= G1. The reduced Tate pairing eT : G1 × G2 →G3, (P, Q) 7→ T(P, Q + nE(Fqk)) (qk−1)/n, is a cryptographic pairing for any G2 ̸= G1. It is computed as eT (P, Q) = (fn,P (Q))(qk−1)/n . 16.4.41 Remark We observe that during the computation of the reduced Tate pairing by Algo-rithm 16.4.38, all factors lying in a subﬁeld of Fqk may be omitted due to the ﬁnal expo-nentiation. In particular, if the x-coordinate of Q lies in a subﬁeld, then all vPi may be dropped, a technique known as denominator elimination; see Remark 16.4.51. 16.4.42 Deﬁnition A distortion map is an eﬀectively computable endomorphism ψ : E(Fq) → E(Fq) that restricts to an isomorphism ψ : G1 →G2 for some subgroup G2 ̸= G1. 16.4.43 Remark Since G1 is invariant under the Frobenius φq, but G2 is not, the endomorphisms ψ and φq cannot commute. So the existence of ψ implies that E is supersingular; see Theorems 12.2.82 and 12.2.87. Conversely, for supersingular curves, there are distortion maps ψ : G1 →G2 for any G2 ̸= G1 [2869, Theorem 5]. Cryptography 791 16.4.44 Example Let E be a supersingular curve with distortion map ψ and G2 = ψ(G1). Let e : G1 × G2 →G3 be a cryptographic pairing. Then e′ : G1 × G1 →G3, (P, Q) 7→e(P, ψ(Q)), is a cryptographic pairing in which both arguments come from the same group G1. This setting is sometimes called a symmetric pairing in the literature, although it does not in general satisfy e(P, Q) = e(Q, P); see also Section 16.4.2.3. 16.4.45 Remark Further work has produced a variety of pairings with a shorter loop in Algo-rithm 16.4.38, that is, deﬁned by some function fr,P with r < n. In general, this is obtained by choosing special curves and restricting to the subgroups G1 and G2 of Theorem 16.4.34. Since all involved groups are cyclic, such pairings are necessarily powers of the Tate pairing. 16.4.46 Example (Eta pairing) Let E be a supersingular curve with even k = 2a and distortion map ψ as in Deﬁnition 16.4.42. Let T = t −1, where t is the trace of the Frobenius map, see Deﬁnition 12.2.47. Then T ≡q (mod n) and n \| (T a + 1). Assume that n2 ∤(T a + 1). Then the map G1 × G1 →G3, (P, Q) 7→fT,P (ψ(Q))aT a−1 qk−1 n , is a cryptographic pairing. For a proof, see [203, Section 4] and [1494, Section III]. By Exam-ple 16.4.62, only curves over ﬁelds of characteristic two or three may satisfy the assumptions of the theorem. Notice that T is of order √q by Theorem 12.2.46, so that in the best case ρ ≈1 (see Deﬁnition 16.4.66) the loop length in Algorithm 16.4.38 is reduced by a factor of about 2, while the ﬁnal exponentiation becomes more expensive. 16.4.47 Example (Ate pairing) Let T = t−1, where t is the trace of the Frobenius map, and assume that n2 ∤(T k −1). Then the map G2 × G1 →G3, (Q, P) 7→fT,Q(P) qk−1 n , is a cryptographic pairing . Notice that the roles of G1 and G2 are inverted compared to the reduced Tate pairing of Example 16.4.40. Thus, as a price to pay for the loop shortening in Algorithm 16.4.38, the number of operations in G2 and thus Fqk increases. 16.4.48 Conjecture (Optimal ate pairing) A loop length of essentially log2 n ϕ(k) , where ϕ is Euler’s function, may be obtained for a pairing of the previous type, for instance via a product of functions fci,Q(P)qi qk−1 n with P log2 ci of the desired magnitude; concrete instances have been obtained using lattice reduction [1492, 2868]. 16.4.2.2 Pairings and twists 16.4.49 Theorem Assume that E is deﬁned over the ﬁeld Fq of characteristic at least 5 and that d ∈{2, 3, 4, 6} is such that d \| gcd(k, # Aut(E)). By Proposition 12.2.59, there is, besides E itself and up to equivalence, precisely one twist E′ of degree d such that n \| #E′(Fqk/d). As can be seen from Proposition 12.2.57, there is an isomorphism ϕ : E′ →E which is deﬁned over Fqd. The subgroup G′ 2 of order n of E′(Fqk/d) satisﬁes ϕ(G′ 2) = G2. 16.4.50 Remark Theorem 16.4.49 implies that in the presence of twists, elements of G2 are more compactly represented by elements of G′ 2; or otherwise said, any cryptographic pairing e : G1 × G2 →G3 yields an equivalent cryptographic pairing e′ : G1 × G′ 2 →G3, (P, Q′) 7→e(P, ϕ(Q′)). 792 Handbook of Finite Fields 16.4.51 Remark Theorem 16.4.49 and the explicit form of ϕ given in Proposition 12.2.57 show that the x-coordinates of elements in G2 lie in Fqk/2 for d even and that the y-coordinates lie in Fqk/3 when 3 \| d. This may allow for simpliﬁcations of Algorithm 16.4.38 in conjunction with the ﬁnal exponentiation; see Remark 16.4.41. 16.4.52 Example (Twisted ate pairing) Under the hypotheses of Theorem 16.4.49, let T = t −1, where t is the trace of the Frobenius map, and assume that n2 ∤(T k −1). Then the map G1 × G2 →G3, (P, Q) 7→fT k/d,P (Q) qk−1 n , is a cryptographic pairing . Here, the roles of G1 and G2 are again as in the reduced Tate pairing of Example 16.4.40. However, compared to the ate pairing of Example 16.4.47, the loop length in Algorithm 16.4.38 is increased by a factor of k d. Unless t is smaller than generically expected, the twisted ate pairing is in fact less eﬃcient to compute than the reduced Tate pairing. 16.4.2.3 Explicit isomorphisms 16.4.53 Remark For the sake of giving security arguments for pairing based systems, the crypto-logic literature has taken to distinguishing pairings according to the possibility of moving eﬃciently between the groups G1 and G2. For instance, if G1 = G2, then the decisional Diﬃe-Hellman problem is easy in G1: Given P, aP, bP and R ∈G1, one has R = abP if and only if e(P, R) = e(aP, bP). 16.4.54 Deﬁnition Let e be a cryptographic pairing in the sense of Deﬁnition 16.4.32. It is of 1. type 1 if G1 = G2; 2. type 2 if there is an eﬃciently computable isomorphism ψ : G2 →G1, but no such isomorphism G1 →G2 is known; 3. type 3 if no eﬃciently computable isomorphisms G1 →G2 or G2 →G1 are known. 16.4.55 Remark We note that since G1 and G2 are cyclic of the same order n, they are trivially isomorphic; but exhibiting an eﬀective isomorphism may require to compute discrete loga-rithms. In general, an eﬃciently computable isomorphism will be given by an endomorphism of the elliptic curve. 16.4.56 Example The pairing of Example 16.4.44 on supersingular curves with distortion map is of type 1. Any pairing with G2 ̸= G1, G2 is of type 2: The isomorphism is given by the trace map Tr of Theorem 16.4.35. To the best of our knowledge, pairings with G2 = G2 are of type 3; at least the trace is trivial on G2. 16.4.57 Remark The terminology type 4 has been used for pairings in which the second argument comes from the full n-torsion group; in this case, G2 can be seen as the group generated by this argument, which may vary with each use of the cryptographic primitive. As it is then unlikely that G2 = G1 or G2, a type 4 pairing essentially behaves as a type 2 pairing. 16.4.58 Remark Type 1 pairings, being restricted to supersingular curves, oﬀer a very limited choice of embedding degrees, see Example 16.4.62. Type 2 pairings are sometimes preferred in the cryptographic literature since they appear to facilitate certain security arguments. On the other hand, the existence of ψ implies that the decisional Diﬃe-Hellman problem is easy in G2, and it is apparently not possible to hash into any subgroup G2 diﬀerent from G1 and G2; see Subsection 16.4.2.5. Recent work introduces a heuristic construction to transform a Cryptography 793 cryptographic primitive in the type 2 setting, together with its security argument, into an equivalent type 3 primitive . 16.4.59 Remark Some cryptographic primitives have been formulated with a pairing on subgroups of composite order n. More precisely, n is the product of two primes that are unknown to the general public, but form part of the private key as in the RSA system of Section 16.1.3.1. Such pairings can be realized either with supersingular curves [345, Section 2.1] or using Algorithm 16.4.64; the former leads to a ρ-value (Deﬁnition 16.4.66) at least 2, the latter to a ρ-value close to 2. There is a heuristic approach to transform such cryptosystems, together with their security proofs, into the setting of prime order subgroups . 16.4.2.4 Curve constructions 16.4.60 Remark The existence of a cryptographic pairing e : G1 × G2 →G3 reduces the discrete logarithm problem in G1 or G2 to that of G3: For instance, given a point P ∈G1 and a multiple xP, choose a point Q ∈G2 such that ζ = e(P, Q) ̸= 1. Then ξ = e(P, xQ) = e(P, Q)x, and x is the discrete logarithm of ξ ∈G3 to the base ζ [1108, 2079]. 16.4.61 Remark Thus to balance the diﬃculty of the discrete logarithm problems in the elliptic curve groups G1 and G2 over Fq and G3 ⊆F∗ qk, the embedding degree k should be chosen according to the security equivalences in Section 16.4.1.1. For instance, if one follows the recommendations of , for a system of 256 bit security one would choose n ≈2512 and thus q ≈2512, and k ≈ 15425 512 ≈30. Since by Theorem 16.4.31 the embedding degree k equals the order of n in Fq, it will be close to q for random curves. Hence one needs special constructions to obtain pairing-friendly curves, curves with a prescribed, small value of k. For a comprehensive survey, see . 16.4.62 Example (Supersingular curves) As ﬁrst noticed in , the embedding degree is always exceptionally small for supersingular curves. The following table gives the possible cardi-nalities according to Theorem 12.2.51, the maximal size n of a cyclic subgroup by and the embedding degree k with respect to n. \|E(Fq)\| n k q + 1 q + 1 2 q + 1 ± √q q + 1 ± √q 3 q + 1 ± √2q q + 1 ± √2q 4 q + 1 ± √3q q + 1 ± √3q 6 q + 1 ± 2√q √q ± 1 1 16.4.63 Remark All algorithms for ﬁnding ordinary pairing-friendly curves rely on complex multi-plication constructions, cf. Example 16.4.14, and construct curves over prime ﬁelds only. 16.4.64 Algorithm A very general method is due to Cocks and Pinch [1101, Section 4.1]. It allows to ﬁx the desired group order n beforehand; choosing a low Hamming weight in the binary decomposition of n or more generally a value of n with a short addition-subtraction chain speeds up Algorithm 16.4.38. Require: An integer k ≥2, a quadratic discriminant D < 0 and a prime n such that k \| (n −1) and the Legendre symbol D n = 1 Ensure: A prime p and an elliptic curve E(Fp) (with complex multiplication by OD) having a subgroup of order n and embedding degree k repeat ζ ←an integer such that ζ modulo n is a primitive k-th root of unity in F∗ n t ←ζ + 1 794 Handbook of Finite Fields v ←an integer such that v ≡t−2 √ D (mod n) p ←t2−v2D 4 until p is an integer and prime Then p ≡t −1 (mod n) Construct the curve E over Fp with p + 1 −t points as in Example 16.4.14 16.4.65 Remark Generically, in this construction t and v will be close to n, so that p will be close to n2. This motivates the following deﬁnition. 16.4.66 Deﬁnition The ρ-value of a pairing-friendly curve is given by ρ = log p log n. 16.4.67 Remark By Theorem 12.2.46, the superior limit of ρ is at least 1 for p →∞. Values of ρ larger than 1 result in a loss of bandwidth when transmitting elements of G1, which is a log2 n-bit subgroup embedded into a ρ log2 n-bit group, and a less eﬃcient arithmetic in the elliptic curve. The security equivalences of Section 16.4.1.1 do in fact not ﬁx the value of k, but that of ρk; so diﬀerent values of k may lead to comparable security levels. 16.4.68 Remark Further research has concentrated on ﬁnding families of pairing-friendly curves, the parameters of which are given by values of polynomials. 16.4.69 Algorithm The following is a direct transcription of Algorithm 16.4.64 to polynomials. Require: An integer k ≥2 and a quadratic discriminant D < 0 Ensure: Polynomials p and n ∈Q(x) such that if the values p(x0) and n(x0) are simultaneously prime integers, then there is an elliptic curve E(Fp(x0)) (with complex multiplication by OD) having a subgroup of order n(x0) and embedding degree k n ←an irreducible polynomial in Q[x] such that the number ﬁeld K = Q[x]/(n) contains √ D and a primitive k-th root of unity z ←a polynomial in Q[x] that reduces to a primitive k-th root of unity ζ in K t ←z + 1 v ←a polynomial in Q[x] that reduces to the element ζ−1 √ D in K s ←a polynomial in Q[x] that reduces to √ D in K v ←(z−1)s D mod n p ←t2−Dv2 4 16.4.70 Remark The polynomials p and n need not represent primes or even integers; choosing small values of \|D\|, and n such that z, s ∈Z[X] may help. Let d = deg(n) be the degree of K. While it is always possible to choose n such that either z or v is of low degree (as low as 1 if n is the minimal polynomial of the corresponding algebraic number), it is a priori not clear whether both can be chosen of low degree. Generically, p is of degree 2(d −1), and the asymptotic ρ-value of the family is 2 −2 d, a small improvement over Algorithm 16.4.64. In many cases, however, actual ρ-values are much closer to 1, as demonstrated by the following example. 16.4.71 Example [413, p. 137] Let k be odd, D = −4 and K = Q(ζ, √−1) = Q[x]/(Φ4k(x)) where Φ4k(x) = Φk(−x2) is the 4k-th cyclotomic polynomial. Choose ζ(x) = −x2, t(x) = −x2 +1, s(x) = 2xk, v(x) = 1 2(xk+2 + xk), p(x) = 1 4(x2k+4 + 2x2k+2 + x2k + x4 −2x2 + 1). The polynomial p takes integral values in odd arguments and, conjecturally, represents primes if it is irreducible (since p(1) = 1, there is no local obstruction to representing primes). Cryptography 795 Asymptotically for p →∞, ρ → k+2 ϕ(k), and ρ →1 if furthermore k →∞with a ﬁxed number of prime factors. 16.4.72 Remark Similar results hold for even k, and for D = −3 since √−3 ∈Q[x]/(Φ3(x)). 16.4.73 Remark The following table, taken from , gives the current best values of ρ = deg p deg n for polynomial families of pairing-friendly curves for k ≥4. (Smaller values of k may be obtained for prime ﬁelds using supersingular curves; see Example 16.4.62.) For the constructions behind each family, see . k deg p deg n ρ kρ 4 2 2 1.00 4.0 5 14 8 1.75 8.8 6 2 2 1.00 6.0 7 16 12 1.33 9.3 8 10 8 1.25 10.0 9 8 6 1.33 12.0 10 4 4 1.00 10.0 11 24 20 1.20 13.2 12 4 4 1.00 12.0 13 28 24 1.17 15.2 14 16 12 1.33 18.7 15 12 8 1.50 22.5 16 10 8 1.25 20.0 17 36 32 1.12 13.8 18 8 6 1.33 24.0 19 40 36 1.11 21.1 20 22 16 1.38 27.5 21 16 12 1.33 28.0 22 26 20 1.30 28.6 23 48 44 1.09 25.1 24 10 8 1.25 30.0 25 52 40 1.30 32.5 26 28 24 1.17 30.3 27 20 18 1.11 30.0 k deg p deg n ρ kρ 28 16 12 1.33 37.3 29 60 56 1.07 31.1 30 12 8 1.50 45.0 31 64 60 1.07 33.1 32 34 32 1.06 34.0 33 24 20 1.20 39.6 34 36 32 1.12 38.2 35 72 48 1.50 52.5 36 14 12 1.17 42.0 37 76 72 1.06 39.1 38 40 36 1.11 42.2 39 28 24 1.17 45.5 40 22 16 1.38 55.0 41 84 80 1.05 43.0 42 16 12 1.33 56.0 43 88 84 1.05 45.0 44 46 40 1.15 50.6 45 32 24 1.33 60.0 46 50 44 1.14 52.3 47 96 92 1.04 49.0 48 18 16 1.12 54.0 49 100 84 1.19 58.3 50 52 40 1.30 65.0 16.4.2.5 Hashing into elliptic curves 16.4.74 Remark Hashing into elliptic curve groups is often required for pairing-based cryptosystems, see for instance Algorithms 16.1.40 and 16.1.46. Standard cryptographic hash functions (see Remark 16.1.23) {0, 1}∗→{0, 1}ℓof suﬃcient length ℓare easily modiﬁed to yield values in Z/nZ (by reduction modulo n) and to arbitrary ﬁnite ﬁelds (by hashing to coeﬃcients with respect to a ﬁxed basis). One would like to extend such constructions to elliptic curves. 16.4.75 Remark In the setting of Deﬁnition 16.4.32, if H : {0, 1}∗→Z/nZ is a collision-resistant hash function and G1 is generated by a point P of order n, then the function {0, 1}∗→G1, m 7→H(m)P, is trivially collision-resistant. However, this simple construction reveals the discrete logarithm of the hash value, which in general renders the cryptosystem totally insecure. 16.4.76 Remark In the following, let E be an elliptic curve deﬁned over Fq as in Deﬁnition 16.4.32, and assume that n3 ∤\|E(Fqk)\|. If one can hash into E(Fq), then one can also hash into G1 = E(Fq)[n]: It suﬃces for that to multiply the result by the cofactor \|E(Fq)\| n . The same argument holds for E(Fqk)[n]. However, it is then in general not possible to project into an arbitrary group G2. For G2 as in Theorem 16.4.35, that is, type 3 pairings as in 796 Handbook of Finite Fields Deﬁnition 16.4.54, the trace Tr : E(Fqk)[n] →G2 can be used to obtain a hash function with values in G2. Alternatively, in the presence of twists as described in Theorem 16.4.49, one may more eﬃciently hash into the subgroup G′ 2 on the twisted curve, for which the cofactor is smaller. To hash into E(k) where k = Fq or k = Fqk, one may use a hash function H : {0, 1} →k to obtain the x-coordinate of a point. As not all elements of k occur as x-coordinates, one may need several trials. A possibility is to concatenate the message m with a counter i, denoted by m\|\|i, and to increase the counter until H(m\|\|i) is the x-coordinate of a point on E. An additional hash bit may be used to determine one of the generically two points with the given x-coordinate. The algorithm is deterministic and, if H is modeled as a random function, it needs an expected number of two trials averaged over all input values. However, for \|k\| →∞, there is a doubly exponentially small fraction of the input values that will take exponential time. Several recent results exhibit special cases in which polynomial time hashing is possible uniformly for all input values. 16.4.77 Example [312, Section 4.1] If q ≡2 (mod 3), then E : y2 = x3 + 1 is a supersingular curve over Fq with q + 1 points and k = 2. Precisely, since third powering is a bijection on Fq with inverse z 7→z1/3 = z(2q−1)/3, the map Fq →E(Fq){O}, y 7→ (y2 −1)(2q−1)/3, y , is a bijection. 16.4.78 Example Let E : y2 = f(x) = x3 + a2x2 + a4x + a6 over Fq of characteristic at least 3. There are explicit rational functions u1(t), u2(t), u3(t), and v(t) such that v(t)2 = f(u1(t2))f(u2(t2))f(u3(t2)) . So for any t there is at least one i(t) such that ui(t)(t2) is a square in Fq, which yields a map Fq →E(Fq), t 7→ ui(t)(t2), f(ui(t)(t2))1/2 . In a cryptographic context, we may assume that a non-square in F∗ q is part of the input, and then Tonelli-Shanks’s algorithm computes square roots in deterministic polynomial time; see [660, Section 1.5.1] and . The argument is reﬁned in to give a deterministic procedure for computing points on the curve without knowing a non-square and to show that at least q−4 8 diﬀerent points may be reached. The case of characteristic two is also handled. 16.4.79 Example Let Fq with q ≡2 (mod 3) be of characteristic at least 5, and let E : y2 = x3+ax+b be an elliptic curve over Fq. Let v(t) = 3a−t4 6t and x(t) = v(t)2 −b −t6 27 1/3 + t2 3 . Then 0 7→O, 0 ̸= t 7→(x(t), tx(t) + v(t)) is a map Fq →E(Fq) with image size at least q 4, and conjecturally close to 5q 8 . A similar result holds for curves over F2m with odd m. 16.4.80 Remark Alternative encodings for elliptic curves in Hessian form over Fq with q ≡2 (mod 3) and odd are given in [1039, 1672]; see also . They have an image of proven size about q/2. See Also , , Give comprehensive accounts of elliptic curve cryptography. References Cited: [108, 109, 183, 203, 220, 221, 312, 313, 345, 366, 413, 420, 458, 597, 660, 661, 744, 745, 852, 970, 978, 979, 1039, 1098, 1101, 1104, 1108, 1133, 1156, 1159, 1160, 1165, 1247, 1250, 1251, 1276, 1422, 1429, 1492, 1494, 1566, 1567, 1629, 1630, 1672, 1771, 1894, 1904, 2079, 2080, 2088, 2098, 2102, 2292, 2426, 2497, 2533, 2560, 2601, 2604, 2675, 2683, 2684, 2712, 2751, 2752, 2766, 2813, 2868, 2869, 2978] Cryptography 797 16.5 Hyperelliptic curve cryptographic systems Nicolas Th´ eriault, Universidad del Bio-Bio 16.5.1 Cryptosystems based on hyperelliptic curve discrete logarithms 16.5.1 Remark As stated in Deﬁnition 12.4.17, the set of Fq-rational reduced divisors of degree zero of a hyperelliptic curve C deﬁned over a ﬁnite ﬁeld Fq form a ﬁnite abelian group, the Picard group Pic0 Fq(C). 16.5.2 Remark The Picard group can be used as a substitute for the group of points on an el-liptic curve to implement cryptosystems based on the diﬃculty of the discrete logarithm problem [1772, 661]. For eﬃciency reasons, imaginary curves are usually preferred over real curves since their group operation is slightly faster in practice. We note that for real hyperel-liptic curves, the cryptosystems rely on the infrastructure discrete logarithm problem which can be reduced to a discrete logarithm problem in the Picard group (see Deﬁnition 12.4.81 and Remark 12.4.82). 16.5.3 Remark For an imaginary hyperelliptic curve deﬁned over the ﬁnite ﬁeld Fq, the Picard group is isomorphic to the ideal class group of the curve (the quotient group of the ideals in the polynomial ring of the curve modulo the principal ideals). For a curve of genus g, this group has order close to qg (to be precise the group order is between (√q −1)2g and (√q + 1)2g, see Remark 12.4.62). In practice, Mumford’s representation (Theorem 12.4.34) is used to construct divisors (or rather the corresponding ideals). 16.5.2 Curves of genus 2 16.5.4 Remark As is the case for groups obtained from elliptic curves, the fastest known attacks on the discrete logarithm problem for hyperelliptic curves of genus two are generic attacks which require O(√n) group operations to compute the discrete logarithm. Because the group order is n ≈q2, solving the discrete logarithm problem requires O(q) group operations. 16.5.5 Remark Following the ideas of Harley , eﬃcient implementations of hyperelliptic curve group operations are usually done via explicit fomulae which replace the polynomial operations of Cantor’s algorithm with a sequence of ﬁeld operations on the coeﬃcients of these polynomials [147, 1246, 1854]. 16.5.6 Remark For eﬃciency reasons, for curves in odd characteristic the curve is usually assumed to be reduced (via isomorphisms) to the form y2 = x5 + f3x3 + f2x2 + f1x + f0. For curves in characteristic two, the curve is usually assumed to be reduced to either y2 + h1xy = x5 + f3x3 + f2x2 + f0 (with h1 = 1 in extensions of odd degree) or y2 + (x2 + h1x + h0)y = x5 + f4x4 + f1x + f0. 16.5.7 Remark Genus two curves of the form y2 + y = f(x) over a ﬁeld of characteristic two are supersingular. The discrete logarithm in these curves is much easier to compute (via the Weil or Tate pairings) than for other curves of genus two. For this reason, these curves are avoided for cryptographic applications. 16.5.8 Remark For curves in characteristic two, the binary ﬁeld structure can be used to easily solve quadratic equations. It becomes possible to compute “halving” operations, giving performance advantages over doubling operations in some cases [286, 287, 1038]. 798 Handbook of Finite Fields 16.5.9 Remark As for elliptic curves, there are various coordinate systems to represent divisors, and more precisely to represent reduced divisors of weight two (divisors of the most com-mon case). In particular, projective coordinates are sometimes used to obtain inversion-free explicit formulae. Two basic types of projective coordinates are used in practice: 1. “standard” projective coordinates [2110, 661], [U1, U0, V1, V0, Z], which corre-spond to the Mumford representation (in aﬃne coordinates) [x2 + (U1/Z)x + (U0/Z), (V1/Z)x + (V0/Z)]; 2. new projective coordinates , [U1, U0, V1, V0, Z1, Z2], which correspond to the Mumford representation (in aﬃne coordinates) [x2 + (U1/Z2 1)x + (U0/Z2 1), (V1/Z3 1Z2)x + (V0/Z3 1Z2)]. 16.5.10 Remark In some cases, extended (or redundant) coordinates are used since they can save some ﬁeld operations in further group operations. For example, the recent coordinates of Lange use the coordinates [U1, U0, V1, V0, Z, Z2] for curves over ﬁelds of characteristic two. Similarly, new coordinates are usually used in the form [U1, U0, V1, V0, Z1, Z2, Z2 1, Z2 2] (in odd characteristic) or [U1, U0, V1, V0, Z1, Z2, Z1Z2, Z2 1, Z2 2, Z2 1Z2] (in characteristic two). 16.5.3 Curves of genus 3 16.5.11 Remark Just as curves of genus two, hyperelliptic curves of genus three can be used to construct cryptosystems based on the discrete logarithm problems, however, evaluating the security of these cryptosystems is more complicated. There are two algorithms that must be considered when evaluating the diﬃculty of the discrete logarithm problem on a speciﬁc hyperelliptic curve: the index calculus algorithm and Smith’s trigonal mapping to non-hyperelliptic curves. 16.5.12 Remark Index calculus algorithms are used to map the discrete logarithm problem to computing non-trivial solutions of a system of linear equations. The algorithm proceeds in two steps: a relation search (to build the system) and a (sparse) linear algebra solver. Under the right conditions, the overall complexity of the relation search and linear algebra solver is lower than generic attacks. For curves of genus one and two, index calculus attacks appear to be slower than generic attacks, but for genus three and higher, index calculus does indeed reduce the cost of computing discrete logarithms. 16.5.13 Remark A number of variants of the index calculus relation search have been pro-posed [1245, 1256, 2212, 2802]. Currently, the most eﬃcient algorithms are those of Gaudry et al. and Nagao . Both of these algorithm require the equivalent of O(q4/3+ϵ) group operations to compute the discrete logarithm. It should be noted that index calculus algorithms have the same (estimated) running time for all genus three curves over the ﬁeld Fq. 16.5.14 Remark For curves of genus three over ﬁelds of odd characteristic, there exists a specialized attack against the discrete logarithm problem. This attack is due to Smith and uses a trigonal map to send the Picard group of a hyperelliptic curve of genus three to the Picard group of a non-hyperelliptic curve (also of genus three). As was shown by Diem [854, 856], it is easier to solve the discrete logarithm in the Picard group of a non-hyperelliptic curve than a hyperelliptic one, the running time decreasing to an equivalent of O(q1+ϵ) group operations for genus three curves. If the trigonal map in Smith’s attack is successful, the security of the curve is then no better than that of a genus two curve (but with a higher cost for the group operation). Smith’s attack depends heavily on the structure of the 2-torsion group of the curve (over the algebraic closure of the ﬁeld Fq). More speciﬁcally, it depends on the extension degree Cryptography 799 of the ﬁeld Fqk in which each of the Weierstrass points of the curve is deﬁned. As a result, not all curves over Fq are vulnerable to Smith’s attack. 16.5.15 Example For curves of the form y2 = f(x) for which f(x) splits into linear factors, Smith’s attack has a 1 −2−105 probability of success (i.e., roughly only 1 in 2105 curves will remain safe from the attack). However, the probability of success is much lower for other types of factorization of f(x), and can even reach zero. For imaginary curves, the following factorization types of f(x) (degrees of the irreducible factors of f(x)) are completely secure against Smith’s attack: , [5, 2], [5, 1, 1], [4, 3], [3, 2, 2], [3, 2, 1, 1], [3, 1, 1, 1, 1]. For real curves, there is one more factorization type of f(x) which is completely secure against the attack: [5, 3]. 16.5.16 Remark There are a number of open problems coming from Smith’s algorithm. First of all, it is not known how to adapt the trigonal map to curves over ﬁelds of characteristic two. Secondly, the trigonal map exists only for curves of genus three, but similar ideas could be important to the security of hyperelliptic curves of higher genus. 16.5.17 Remark As with genus two curves, the group operations are performed via explicit formulae rather than Cantor’s algorithm [149, 1383, 2211]. Once again, curve isomorphisms are used to reduce the curve equation and improve eﬃciency. In odd characteristic the preferred form of the equation is y2 = x7 +f5x5 +f4x4 +f3x3 +f2x2 +f1x+f0. For curves in characteristic two, similar reductions can be performed for each form of h(x). Unlike elliptic curves and curves of genus two, genus three curves of the form y2 + y = f(x) over F2n are never supersingular and can be used in cryptographic applications. These curves allow for much faster doubling operations, giving eﬃciency ad-vantages. 16.5.18 Remark As for genus two curves, for genus three curves in characteristic two, the binary ﬁeld structure can be used to easily solve quadratic (and some quartic) equations. It becomes possible to compute “halving” operations, giving performance advantages over doubling operations, especially when the degree of h(x) increases . 16.5.4 Curves of higher genus 16.5.19 Remark The highest genus that is sometimes considered for cryptographic applications is four [149, 2381]. The running time for solving the discrete logarithm problem in a hyper-elliptic curve of genus four (using the algorithm of Gaudry et al. ) is equivalent to O(q3/2+ϵ) group operations. 16.5.20 Remark For curves of small degree, the cost of computing discrete logarithm grows as O(q2−2/g+ϵ) (for a ﬁxed genus and an increasing ﬁeld size) [1256, 2212]. However, the complexity of the group operation is at least linear with respect to the genus of the curve (in practice this growth is closer to quadratic), in practice limiting the range of “small” genera which are of interest for cryptosystems. At higher genera, the situation is even more diﬃcult for cryptosystems based on the discrete logarithm. This situation is discussed in Section 16.6.10. 16.5.5 Key sizes 16.5.21 Remark The security arguments for hyperelliptic curves of genus two are very similar to those of elliptic curves, with the only distinction that the group order is close to q2 (rather than q for elliptic curves). The required bit-sizes for the ﬁnite ﬁelds are therefore half of 800 Handbook of Finite Fields what they are for elliptic curves with the same security level. Once again, the group order should be a prime or a prime with a small co-factor. Similarly, the secret key should be of size similar to the group order, which means that secret key sizes for hyperelliptic curves of genus two are in fact identical to those of elliptic curves. 16.5.22 Remark The following table summarizes these arguments, giving comparisons with sym-metric key cryptosystems and as in Remark 16.4.3, the 80 bit security level is now considered a historic ﬁgure. security level ECC ﬁeld size genus two ﬁeld size secret key size 80 160 80 160 112 224 112 224 128 256 128 256 192 384 192 384 256 512 256 512 16.5.23 Remark To choose the ﬁeld size for genus three hyperelliptic curve, the ﬁrst concern is the possibility of an index calculus attack. To have m-bits of security level, we ask that q4/3 ≈2m (using the complexity in Remark 16.5.12), hence log q ≈3m/4. It may be noted that index calculus attacks do not obtain any signiﬁcant speedups from restrictions on the key size or factorizations of the group order. In some situations it may be acceptable to allow the group order to factor into a prime with a (relatively) large cofactor, as long as the largest prime factor is of such size that the subgroup attack of Pohlig and Hellman cannot be eﬀective. The cofactor could then take up to one quarter of the bit size of the group order. Having large subgroups can still pose a problem for the security of a cryptosystem unless there is a mechanism in place to ensure the group element is indeed in the desired subgroup, otherwise an attacker could provide a group element which is outside of the (large-)prime-order subgroup, and use a subgroup attack to obtain partial information about the scalar, thus reducing the eﬀective security level. Unfortunately, verifying in which subgroup a given group element is located is quite expensive (of cost similar to the scalar multiplication itself), and has a signiﬁcant impact on the eﬃciency of the cryptosystem. For this reason, it is usually preferred to use groups whose order is (close to) a prime, even though larger cofactors may not directly decrease the security level. 16.5.24 Remark It is possible to choose keys which are signiﬁcantly smaller than the group size without necessarily weakening the cryptosystem. This can be of great importance for the eﬃciency of the cryptosystem since the cost of the scalar multiplication (the cryptographic primitive) is directly proportional to the number of bits of the secret key. However, if the secret key is too small, it may become possible to compute the secret key using a generic attack. The bit-size of the secret must therefore be equivalent to those used for elliptic curves at the same security level. 16.5.25 Remark For curves in odd characteristic, it is recommended to ask that f(x) factors into one of the factorization types given in Example 16.5.15 (to insure no trigonal mapping can be used to mount a successful attack). This ﬁnal condition (factorization of the deﬁning polynomial) can be seen as equivalent to the requirement that the group order be close to a prime for protection against generic attacks (factorization of the group order). 16.5.26 Remark The following table gives bit sizes for the ﬁeld of deﬁnition and the secret key for genus three hyperelliptic curve cryptosystems, comparing with symmetric key sizes and ECC sizes. If large cofactors are allowed in the group order, its largest prime factor should be at least of the secret key size. Cryptography 801 security level ECC ﬁeld size genus three ﬁeld size secret key size 80 160 60 160 112 224 84 224 128 256 96 256 192 384 144 384 256 512 192 512 16.5.6 Special curves 16.5.27 Deﬁnition A hyperelliptic Koblitz curve is a hyperelliptic curve deﬁned over F2 that is used over F2n for n prime (to avoid having too many subgroups). 16.5.28 Remark The following table lists all (isomorphism classes of) non-supersingular hyperel-liptic Koblitz curves of genus 2 and their characteristic polynomials. Curve C (reduced) Characteristic polynomial y2 + xy = x5 + 1 t4 + t3 + 2t + 4 y2 + xy = x5 + x2 + 1 t4 −t3 −2t + 4 y2 + (x2 + x)y = x5 + 1 t4 −t2 + 4 y2 + (x2 + x + 1)y = x5 + 1 t4 + t2 + 4 y2 + (x2 + x + 1)y = x5 + x t4 + 2t3 + 3t2 + 4t + 4 y2 + (x2 + x + 1)y = x5 + x + 1 t4 −2t3 + 3t2 −4t + 4 16.5.29 Remark Besides an easier computation of the group order (although computing the group order of a random hyperelliptic curve over a binary ﬁeld is quite eﬃcient), the main advan-tage of Koblitz curves comes from the Frobenius map over F2. 16.5.30 Remark A similar table for genus three can be found in . 16.5.31 Deﬁnition The map deﬁned by τ(x, y) = (x2, y2) for points on the curve gives a degree n endomorphism on the Picard group when applied to (the support of) divisors deﬁned over F2n. The application of τ on the Mumford representation of a divisor consists of squaring all the coeﬃcients of the polynomials. A τ-adic expansion of the scalar then gives a reduction in the cost of the scalar multiplication (as for Koblitz elliptic curves). 16.5.32 Remark Since the τ map has order n for divisors over F2n, the security level of the curve must be adjusted accordingly (generic attacks can be accelerated by a factor of O(√n). 16.5.33 Example The curve y2 + (x2 + x + 1)y = x5 + x is a genus two Koblitz curve with characteristic polynomial over F2 given by t4 + 2t3 + 3t2 + 4t + 4 = 0. Over the ﬁeld F2113, its Picard group has order 2 · 7 · 1583 · 476183 · 10218712550205474310417731984747447186313991554764219834409 (i.e., 10553167646 times a prime). Taking into account the subgroups and the degree 113 en-domorphism, this curve gives equivalent security to a random elliptic curve over a ﬁeld of 186 bits. 16.5.34 Deﬁnition A subﬁeld curve is a hyperelliptic curve of genus g deﬁned over a ﬁeld Fq, but used as a curve over Fqn . These curves are generalizations of Koblitz curves. 802 Handbook of Finite Fields 16.5.35 Remark There are two possible advantages of using subﬁeld curves. First, computing the group order is easier than for a general curve over the same ﬁeld: it can be obtained from knowledge of the characteristic polynomial of the curve over Fq . This is of particular interest in odd characteristic since computing the group order is much more expensive for these ﬁelds than for binary ﬁelds. Second, the ﬁeld structure (of Fqn as an extension of Fq) allows for more eﬃcient arithmetic than for a prime ﬁeld Fp with p ≈qn. 16.5.36 Remark One of the main drawbacks of using curves which are deﬁned over a subﬁeld (in this case of Fqn) is that the Picard group contains at least one large subgroup, namely the Picard group of the curve over the subﬁeld Fq. This opens the way for attacks on the discrete logarithm problem based on the algorithm of Pohlig and Hellman . To avoid these attacks, it is necessary to implement some techniques to ensure the group elements used are always in the larger (prime order) subgroup. Furthermore, these curves are at risk of Weil descent based attacks, in particular the variant of Gaudry . 16.5.37 Deﬁnition Given a curve C deﬁned over Fq, the trace-zero subvariety of the curve is the quotient group T = Pic0 Fqn (C)/Pic0 Fq(C). This was proposed by Lange for genus two and n = 3 as a method to construct a cryptographically viable group on a ﬁeld extension without having to deal with subgroups. If the order of the group T is prime and close to q4 (qg(n−1) in general) this group can be used for cryptosystems based on the discrete logarithm problem. 16.5.38 Remark As with subﬁeld curves, the group order of trace-zero subvarieties is computed from the group order of the Picard group of the curve over Fq. Similarly, these curves take advantage of the subﬁeld structure to make the ﬁeld arithmetic more eﬃcient. Finally, the group operations can also be tailored to work on T rather than on the general Picard group. 16.5.39 Remark Just as with any curve on a ﬁeld extension, trace-zero subvarieties can be subjected to Weil descent based attacks. For curves of genus greater than two and for n > 3 in genus two, a Weil descent attack on the whole Picard group with the simplest form of index calculus attack is suﬃcient to solve the discrete logarithm in time O(q2), signiﬁcantly faster than through generic attacks . 16.5.40 Remark For n = 3, Gaudry’s variant would allow the attack to handle the Picard group as if it were the Picard group of a genus six curve over Fq, and the discrete logarithm problem in the trace-zero subvariety can be lifted to a discrete logarithm problem in the whole Picard group. This attack has running time O(q5/3), instead of the desired O(q2) for generic attacks on the group T. Furthermore, it may be possible to adapt Gaudry’s variant to work directly on the trace-zero subvariety (by selecting a diﬀerent factor base, more appropriate to this context), treating it as a genus four curve over Fq. If such an attack is possible, it would have running time O(q3/2), making it particularly eﬀective. The construction of an appropriate factor base is an open problem. 16.5.41 Remark Because of these attacks, ﬁeld sizes for secure cryptosystems would have to be increased to preserve the security level, which in turns decreases the eﬃciency of the ﬁeld arithmetic. As a result, groups coming from trace-zero subvarieties are now mostly of the-oretical interest. 16.5.7 Random curves: point counting 16.5.42 Remark Computing the Picard group order of hyperelliptic curves in even characteristic can be done quite eﬃciently via p-adic (in this case 2-adic) algorithms. Currently, the fastest Cryptography 803 algorithms are due to Satoh and Streng . These algorithms can handle in reasonable time both curves of high genus and (more interesting for practical applications) curves of small genus over a large ﬁeld. 16.5.43 Remark Compared with the characteristic two case, computing the group order of curves in odd characteristic is mostly an open problem. In curves over prime ﬁelds, most algorithms are based on Pila’s extension of Schoof’s algorithm. For genus two curves (where such works are concentrated), the fastest algorithms are due to Gaudry and Schost [1252, 1253], and Gaudry, Kohel, and Smith for curves with complex multiplication . 16.5.44 Remark At this time, ﬁnding a random hyperelliptic curve in characteristic two that corre-sponds to required security requirements – having a given genus and ﬁeld size, with a Picard group whose order is a large prime with a small co-factor – is considered quite practical. For curves in odd characteristic, this is considerably more diﬃcult. For genus two curves, such a search is feasible although expensive ( reports such a search at the 128-bit security level taking over one million CPU-hours). For genus three (or higher), there are currently no examples of practical searches for such curves. 16.5.8 Pairings in hyperelliptic curves 16.5.45 Remark Since the Tate-Lichtenbaum pairing comes from the divisor class group structure of elliptic curves, it naturally generalizes to hyperelliptic curves, although some care has to be taken to properly adapt the deﬁnition as well as Miller’s algorithm; see Subsection 12.4.6. A survey of the diﬀerent techniques available for pairings on hyperelliptic curves can be found in . Actual implementations of cryptosystems using hyperelliptic curve pairings are much less common than elliptic curve ones, but demonstrates the potential interest of hyperelliptic curves for pairing-based cryptography. 16.5.46 Remark For curves of genus two over prime ﬁelds, it is possible to construct curves with a given group order using complex multiplication methods [2087, 2970, 2971]. Such curves may be of interest for pairing based cryptosystems on hyperelliptic curves . For cryptosys-tems based on the discrete logarithm problem, more random curves are usually preferred for fear the structure coming from complex multiplication could eventually open the way to new attacks that signiﬁcantly decreases the diﬃculty of the discrete logarithm problem in these curves (although no such attack is currently known). See Also , Give comprehensive accounts of hyperelliptic curve cryptography. References Cited: [1, 147, 149, 286, 287, 288, 313, 497, 661, 854, 855, 856, 1038, 1157, 1245, 1246, 1247, 1252, 1253, 1254, 1256, 1383, 1421, 1772, 1850, 1851, 1852, 1853, 1854, 2087, 2110, 2195, 2211, 2212, 2309, 2381, 2394, 2406, 2534, 2552, 2688, 2735, 2802, 2970, 2971] 16.6 Cryptosystems arising from Abelian varieties Kumar Murty, University of Toronto 804 Handbook of Finite Fields 16.6.1 Deﬁnitions 16.6.1 Deﬁnition An Abelian variety A over a ﬁnite ﬁeld F is a smooth projective algebraic variety deﬁned over F on which there is an algebraic group operation, also deﬁned over F. In particular, the identity element O of the group is an F-rational point. 16.6.2 Deﬁnition An Abelian subvariety of an Abelian variety A is an Abelian variety that is contained in A. Trivial Abelian subvarieties are {0} and A itself. A non-trivial Abelian subvariety is proper. 16.6.3 Deﬁnition An Abelian variety is simple over F if it contains no proper Abelian subvarieties deﬁned over F. It is absolutely simple if it is simple over the algebraic closure F of F. 16.6.4 Deﬁnition Two Abelian varieties A1 and A2 are isogenous if there is a surjective morphism A1 − →A2 with ﬁnite kernel. 16.6.2 Examples 16.6.5 Example Abelian varieties of dimension one are elliptic curves. 16.6.6 Remark Elliptic curves are the only curves that are also Abelian varieties. This follows from the result below which is a consequence of the Riemann-Roch theorem. 16.6.7 Proposition If C is an algebraic curve of genus g which is also an Abelian variety, then necessarily g = 1 and C is an elliptic curve. 16.6.8 Example Given two Abelian varieties A1 and A2, there is a natural structure of Abelian variety on the product A1 × A2. 16.6.9 Example In particular, one can take a power of an elliptic curve En = E × · · · × E and products of powers of diﬀerent elliptic curves En1 1 × · · · × Enk k . 16.6.10 Theorem An Abelian variety over F is isogenous to a product of simple Abelian varieties deﬁned over F. 16.6.3 Jacobians of curves 16.6.11 Theorem Given a curve C deﬁned over F, there is an Abelian variety J(C), the Jacobian of C, which is also deﬁned over F. It has the following properties: 1. The dimension of J(C) is the genus of C. 2. If g is the genus of C, there is a surjective map Cg = C × · · · × C − →J(C). 16.6.12 Theorem The F points on J(C) are given by D0/P where D0 is the Abelian group of divisors of degree zero and P is the subgroup of divisors of degree zero which are divisors of functions. 16.6.4 Restriction of scalars 16.6.13 Remark Let A be an Abelian variety deﬁned over an extension ﬁeld F′ over F. Suppose that [F′ : F] = r and that A is of dimension d. Then there is an Abelian variety RF′/FA Cryptography 805 (called the restriction of scalars of A) deﬁned over F of dimension rd with the property that RF′/FA(F) = A(F′). 16.6.5 Endomorphisms 16.6.14 Remark The fundamental theorem is due to Tate. 16.6.15 Theorem Let A be an Abelian variety of dimension g deﬁned over a ﬁnite ﬁeld F. Let π be the Frobenius endomorphism of A relative to F and P its characteristic polynomial. 1. The algebra F = Q[π] is the center of the semisimple algebra E = EndF(A) ⊗Q. 2. E contains a semisimple Q-subalgebra M of rank 2g which is maximal and com-mutative. 3. The following are equivalent: a. [E : Q] = 2g; b. P has no multiple roots; c. E = F; d. E is commutative. 4. The following are equivalent: a. [E : Q] = (2g)2; b. P is a power of a linear polynomial; c. F = Q; d. E is isomorphic to the algebra of g × g matrices over the unique quaternion division algebra Dp over Q which splits at all primes ℓ̸= p, ∞; e. A is F-isogenous to the g-th power of a supersingular elliptic curve, all of whose endomorphisms are deﬁned over F. 5. A is F-isogenous to a power of an F-simple Abelian variety if and only if P is a power of a Q-irreducible polynomial. When this is the case, E is a central simple algebra over F which splits at all ﬁnite primes v of F not dividing p, but does not split at any real primes of F. 16.6.6 The characteristic polynomial of an endomorphism 16.6.16 Deﬁnition Let φ : A − →A be an endomorphism. Then, there is a polynomial Pφ(T) with the property that for every integer n ≥1, we have Pφ(n) = deg(n−φ). This polynomial is the characteristic polynomial of φ. 16.6.17 Remark An important case of this is the Frobenius endomorphism x 7→xq on F. It induces an endomorphism of A. 16.6.7 Zeta functions 16.6.18 Deﬁnition Denote by Fn the unique extension of F of degree n contained in F. The zeta function Z(A, T) of an Abelian variety A deﬁned over a ﬁnite ﬁeld F is the power series Z(A, T) = exp    X n≥1 \|A(Fn)\|T n/n   . 806 Handbook of Finite Fields 16.6.19 Theorem There are polynomials P0(T), . . . , P2d(T) (where d = dim(A)) such that Z(A, T) = 2d Y i=0 Pi(T)(−1)i+1. 16.6.20 Example The zeta function of an elliptic curve E is of the form Z(E, T) = P1(T)/(1 −T)(1 −qT). Here, P1(T) is a polynomial of degree 2 and is of the form 1 −aqT + qT 2 with \|E(F)\| = 1 −aq + q. Moreover, aq is an integer with aq = π + π and π is a complex number of absolute value q1/2. We have \|E(Fn)\| = 1 −(πn + πn) + qn. 16.6.21 Remark The single integer aq determines the number of points on E over Fn for every n. 16.6.22 Theorem The polynomials Pi(T) of the previous result have the following properties: 1. We have Pi(T) ∈Z[T]. 2. The degree of Pi is 2d i . 3. The polynomial P1(T) is the characteristic polynomial of the Frobenius endo-morphism. 4. The factorization of P1 is of the form Q2d j=1(1 −πjT) where for 1 ≤j ≤d, πj+d = πj and πjπj+d = q. 5. The roots of Pi are i-fold products of the πj. 6. We have P0(T) = 1 −T and P2d(T) = 1 −qdT. 16.6.23 Theorem The polynomial P1(T) determines the isogeny class of A. 16.6.24 Remark The above theorem is a fundamental result of Tate. 16.6.25 Theorem Given a set of 2d complex numbers {πj : 1 ≤j ≤2d}, with πj+d = πj and πjπj+d = q for 1 ≤j ≤d, there is an Abelian variety A deﬁned over F for which P1(T) = Q2d j=1(1 −πjT). 16.6.26 Remark This is a fundamental result of Honda and Tate. 16.6.27 Deﬁnition The Newton polygon of an Abelian variety A is the Newton polygon of P1(T). In particular, if we write P1(T) = 1 + a1T + · · · + a2dT 2d, then the Newton polygon is the convex hull of the points (j, ordpaj) for 1 ≤j ≤2d. A slope of the Newton polygon is (d−b)/(c−a) where (a, b) and (c, d) are two vertices. The length of the slope is c−a. 16.6.28 Proposition [2438, Theorem 9.1] The slopes of the Newton polygon are (counting multi-plicities) the p-adic ordinals of the reciprocal roots of P1(T). More precisely, if λ is a slope of length m, then precisely m of the numbers ordpπi are equal to λ. 16.6.29 Remark Given the condition that for every j, we have πjπj+d = q, at most d of the πj can be prime to q (that is, have ordpπj = 0). 16.6.30 Deﬁnition An Abelian variety A is ordinary if exactly d of the πj satisfy ordpπj = 0. An Abelian variety is supersingular if none of the πj satisfy ordpπj = 0. 16.6.31 Theorem A supersingular Abelian variety A over F is isogenous to a power of a supersingular elliptic curve over F. Cryptography 807 16.6.8 Arithmetic on an Abelian variety 16.6.32 Remark There are eﬃcient algorithms to compute on elliptic curves as well as on Jacobians of hyperelliptic curves. These have been described in earlier sections. There is an interesting class of curves (called Ca,b curves) which generalizes these. We describe them below. 16.6.33 Deﬁnition Let a and b be positive integers. A Ca,b curve over F is one that is deﬁned by an equation of the form F(x, y) = αb,0xb + α0,aya + X ia+jb<ab αi,jxiyj where the coeﬃcients αi,j are elements of F and the leading coeﬃcients αb,o and α0,a are nonzero. 16.6.34 Remark Such a curve has the property that it is has exactly one F-rational point (Q say) at inﬁnity and the polar divisors of the functions x and y are aQ and bQ, respectively. 16.6.35 Example An elliptic curve is a C2,3 curve and more generally, a hyperelliptic curve given by y2 = f(x) with f a polynomial of degree 2g + 1 is a C2,2g+1 curve. 16.6.36 Example A superelliptic curve, in other words one of the form ya = f(x) where f has degree g (say) is a Ca,g curve. 16.6.37 Proposition Let C be a Ca,b curve. Any element of the Jacobian J(C) can be expressed (not necessarily uniquely) by a divisor of the form E −nQ where E is a positive divisor whose support is disjoint from Q and 0 ≤n ≤g. Here g is the genus of C. 16.6.38 Remark A divisor as in the proposition is called semi normal. It is possible that two semi normal divisors are equivalent (that is, represent the same point in the Jacobian). To get a unique representation, start with a semi normal divisor D = E −nQ with E as above. Find a nonzero function f whose polar divisor is supported at Q, whose zero divisor satisﬁes (f)0 ≥E, and which has the smallest possible order pole at Q. Then consider the divisor −D + (f). Applying this process once more yields a divisor equivalent to D which is unique. Every divisor class contains a divisor of this form, henceforth called the normal representative of the class. 16.6.39 Proposition Let C be a Ca,b curve. Every point in the Jacobian J(C) (and hence every divisor class) can be represented uniquely by a normal divisor. 16.6.40 Remark With a canonical representative of each divisor class, arithmetic is now explicit. 16.6.41 Remark By expressing the above normal form using Gr¨ obner bases, the complexity of addition is estimated to be O(g3(log q)2). 16.6.42 Remark A special case of Ca,b curves is the family of diagonal curves. They are repre-sented by an equation of the form ya = cxb + d with c, d ∈F× q . This curve has genus 1 2((a −1)(b −1) −gcd(a, b) + 1). By explicitly computing the zeta function of J(C) in terms of Jacobi sums, Blache, Cherdieu, and Sarlabous show that for F = Fp, the Jacobian J(C) is simple if p ≡1, 2, 4, 7, 8, 11, 13 (mod 15). Moreover, it is supersingular if p ≡14 (mod 15). Moreover, they develop an analogue of Cantor’s addition algorithm in J(C). 808 Handbook of Finite Fields 16.6.43 Remark Addition on the Jacobian of a general curve has been described by Volcheck but this method does not seem to be practical over ﬁelds of cryptographic size. 16.6.44 Remark Arita, Miura, and Sekiguchi have shown how to use a singular plane model of a general curve to obtain an algorithm for addition on the Jacobian. However, the complexity of this algorithm has not been analyzed. 16.6.9 The group order 16.6.45 Remark For cryptographic applications, we want the group A(F) to be “nearly” of prime order. In particular, we want A to be simple (and perhaps even absolutely simple). This rules out supersingular Abelian varieties as (by a result stated above) they are isogenous to a power of a supersingular elliptic curve. 16.6.10 The discrete logarithm problem 16.6.46 Deﬁnition Let A be an Abelian variety over the ﬁnite ﬁeld F. Let P ∈A(F) and Q an element of the subgroup of A(F) generated by P. The discrete logarithm problem for this subgroup is to determine an integer n so that Q = nP. 16.6.47 Remark In the case of hyperelliptic curves, we have the following result of Enge and Gaudry based on an index calculus approach. 16.6.48 Theorem The discrete logarithm problem on J(C) (where C is a hyperelliptic curve of genus g deﬁned over a ﬁnite ﬁeld of q elements) is of complexity O(exp{( √ 2 + o(1)) p (log qg)(log log qg)}). 16.6.49 Remark This result assumes that the group order is known and that the group itself can be computed in polynomial time. Several authors have studied analogues of this result for the Jacobian of a non-hyperelliptic curve. In particular, one has the following results. 16.6.50 Theorem Fix positive integers d (the degree) and g ≤(d −1)(d −2)/2 (the genus). Denote by S(q) the set of all instances of the discrete logarithm problem in curves of genus g over Fq represented by plane models of degree d. Then there is a subset S1(q) with \|S1(q)\|/\|S(q)\| →1 (as q →∞) such that the instances in S1(q) can be solved in an “expected” time of O(q(2− 2 d−2 )(log q)A) for some integer A ≥1. 16.6.51 Remark The analysis in involves several heuristic assumptions (including assumptions on the group order and structure) and the power of the logarithm is not speciﬁed. The result, in particular, applies to non-hyperelliptic curves of genus 3 (which by the canonical embedding, may be represented by a plane quartic (d = 4)). This particular case is further analyzed by Diem and Thom´ e . 16.6.52 Remark In the case studied by Diem and Thom´ e, the heuristic assumptions can be made explicit in graph-theoretic terms. Choose a subset F ⊂C(F) (called the factor base) of cardinality O(√q) and let L denote the complement of F in C(F). A graph G is constructed with vertices L∪{∗} where ∗is a special root vertex. The edges in this graph are determined as follows. For each pair F1, F2 ∈F, compute the line L through these points. Consider the divisor D = F1 + F2 + D1,2 representing the intersection of C with L. If P, Q are points in the support of D1,2, at least one of which is in L, construct an edge joining P and Q (or P and ∗or Q and ∗). Cryptography 809 16.6.53 Theorem Suppose that C is a non-hyperelliptic curve of genus 3 with the property that J(C)(F) is cyclic. Suppose also that the graph G constructed above has a tree rooted at ∗of depth O((log q)2) and having at least O(q5/6) elements. Then the discrete logarithm problem in J(C) can be solved in O(q(log q)A) steps. 16.6.54 Remark We say that the discrete logarithm problem for the pair (A, P) consisting of the Abelian variety A and the subgroup generated by a point P on A is diﬃcult if it is compu-tationally infeasible to solve the problem. 16.6.55 Remark Let A and P be as above and suppose that the discrete logarithm problem for (A, P) is diﬃcult. Then we can perform a key exchange using the Diﬃe Hellman protocol with the group G generated by P. 16.6.56 Remark In order for such a key exchange scheme to be useful and secure against known attacks, we require: 1. arithmetic on A should be eﬃcient; 2. the group order A(F) should be nearly prime. 16.6.57 Remark There is a vast literature on the discrete logarithm problem. The reader is referred to the surveys [1495, 1586] for some results that are not mentioned here. 16.6.58 Remark The motivation for considering higher dimensional Abelian varieties as the basis of a cryptographic scheme is based on the fact that if A has dimension d, the number of points A(F) is of order qd. Thus, we should expect the diﬃculty of the discrete logarithm problem to be of the order qd/2. In particular, a 2-dimensional Abelian variety has a discrete logarithm problem of diﬃculty O(q) whereas an elliptic curve has a discrete logarithm problem of diﬃculty O(q1/2). This means that there is the possibility of achieving the same level of security that an elliptic curve over a ﬁeld of 2163 oﬀers by using a two dimensional Abelian variety over a ﬁeld of 282. The fact that we can work over a ﬁeld of smaller size may mean that there is a reduction in the overhead of time and memory required to do computations. Whether this is actually the case is a matter of current research. 16.6.11 Weil descent attack 16.6.59 Remark Frey suggested that it might be possible to use Weil descent to attack the discrete logarithm problem on elliptic curves deﬁned over Fpn where n is composite. This idea was explicitly developed and analyzed by Gaudry, Hess, and Smart in the case p = 2. 16.6.60 Deﬁnition Let ℓand n be positive integers. Set q = 2ℓ, k = Fq, and K = Fqn. Let E be a non-supersingular elliptic curve deﬁned over K by the equation y2 + xy = x3 + ax + b with a, b ∈K and b ̸= 0. Suppose that \|E(K)\| = dr where d is small and r is prime. Let b0 = b, b1, . . . , bn−1 be the conjugates of b (under the Frobenius automorphism σ(x) = xq). Let m(b) be the dimension of the vector space over F2 spanned by the set {(1, b1/2 0 ), . . . , (1, b1/2 n−1)}. 810 Handbook of Finite Fields 16.6.61 Theorem There is a hyperelliptic curve C of genus g = 2m(b)−1 or 2m(b)−1 −1 deﬁned over k so that the discrete logarithm problem on E(K) can be solved in J(C)(k). In particular, the complexity of the discrete logarithm problem is O(qg/2). 16.6.62 Theorem Let n be an odd prime and t the multiplicative order of 2 modulo n. Set s = (n −1)/t. 1. The polynomial xn + 1 factors over F2 as (x + 1)f1f2 · · · fs where the fi are distinct irreducible polynomials of degree t. 2. The set B = {b ∈Fqn\Fq : (σ + 1)fi(σ)(b) = 0 for some 1 ≤i ≤s} has cardinality qs(qt −1). 3. For all b ∈B, the curves y2 + xy = x3 + ax2 + b has m(b) = t + 1. 16.6.63 Example 1. If n = 3 we have s = 1 and t = 2. Then B has cardinality q(q2−1) so B = Fq3\Fq and m(b) = 3. 2. If n = 5, we have s = 1 and t = 4. Then again, B = Fq5\Fq and m(b) = 5. 3. If n = 31 we have t = 5 and s = 6. Then B has cardinality 6q(q5 −1) and m(b) = 6. 16.6.64 Example The following special case was analyzed in detail by Jacobson, Menezes, and Stein . For the elliptic curve y2 + xy = x3 + x + b deﬁned over the ﬁeld F2[z]/(z155 + z62 + 1) of 2155 elements, and with b = z16 + z2 + z, instances of the discrete logarithm problem can be reduced to the discrete logarithm problem on the Jacobian J(C) of a hyperelliptic curve C of genus 31 over the ﬁeld F25. 16.6.65 Remark The case of odd characteristic was developed by Diem . In particular, he showed that the Weil descent attack is not eﬀective for elliptic curves deﬁned over Fpn when n ≥11 is prime. On the other hand, if n = 5, 7 and p is replaced by q = pm, in general, the discrete logarithm problem on an elliptic curve E over Fqn can be reduced to a discrete logarithm problem in the Jacobian of a curve of genus 5 or 7 over Fq. 16.6.66 Remark Weil descent attacks can also be mounted on the discrete logarithm problem for Jacobians of hyperelliptic curves. See the discussion in . 16.6.12 Pairings based cryptosystems 16.6.67 Remark Besides the Weil pairing described above, Frey and R¨ uck introduced the Lichtenbaum-Tate pairing. Suppose that A is the Jacobian of the curve C deﬁned over F. If ℓis a prime not dividing q (the cardinality of F) and k is the order of q modulo ℓ, there is a pairing Aℓ × A(F)/ℓA(F) − →F× qk/(F× qk)ℓ. 16.6.68 Theorem This pairing is non-degenerate. Using it, the discrete logarithm problem in Aℓ can be reduced to the corresponding problem in F× qk in probabilistic polynomial time in log q. Cryptography 811 16.6.69 Deﬁnition The number k above is the embedding degree. 16.6.70 Remark If the embedding degree is small, then solving the discrete logarithm problem in F× qk is practical. Several authors have found examples of Abelian varieties for which the embedding degree is small. 16.6.71 Theorem Given a CM ﬁeld K of degree 2g ≥4, a primitive CM-type Φ of K, a positive integer k and a prime r ≡1 (mod k) that splits completely in K, there exists a prime p and a simple, ordinary Abelian variety A deﬁned over Fp with embedding degree k with respect to r, and an element π ∈K so that \|A(Fp)\| = NK/Q(π −1). 16.6.72 Remark The construction of A is not explicit. Rather the polynomial P1(T) is constructed explicitly and an appeal is made to the theorem of Honda and Tate. Moreover, heuristic analysis suggests that P1(T) can be found in O(log r) steps. 16.6.73 Remark Galbraith, McKee, and Valenca produce examples of ordinary Abelian sur-faces with embedding degrees 5, 10, and 12. Again, the construction appeals to the Honda-Tate theorem. 16.6.74 Remark The embedding degree and security parameters in the case of supersingular Abelian varieties is analyzed carefully by Rubin and Silverberg . See Also §16.4 For elliptic curve cryptosystems. §16.5 For hyperelliptic curve cryptosystems. References Cited: [123, 148, 294, 852, 854, 856, 980, 1097, 1104, 1108, 1158, 1250, 1495, 1582, 1585, 1586, 2078, 2438, 2494, 2783] 16.7 Binary extension ﬁeld arithmetic for hardware imple-mentations M. Anwarul Hasan, University of Waterloo Haining Fan, Tsinghua University We consider algorithms and architectures for hardware realization of arithmetic opera-tions over binary extension ﬁelds F2n. In particular, we focus on arithmetic algorithms and architectures suitable for implementation in hardware for today’s cryptographic applica-tions. For representation of elements of F2n, we consider polynomial and normal bases over F2. 16.7.1 Preamble and basic terminologies 16.7.1 Remark Common choices for hardware are conﬁgurable semiconductor devices, such as ﬁeld programmable gate arrays (FPGAs) and application speciﬁc integrated circuits (ASICs). FPGAs have a very low non-recurring engineering cost and can generally be re-conﬁgured 812 Handbook of Finite Fields many times. On the other hand, ASICs require a high non-recurring engineering cost, but the unit cost is relatively lower for large quantity production, and are considered to be more suitable for applications requiring high-speed, low area, or low power designs. 16.7.2 Remark An arithmetic algorithm for hardware can be evaluated based on one or more of the following: (i) number of arithmetic operations over the underlying ﬁeld (hence it is essentially arithmetic complexity of the algorithm), (ii) amount of storage, e.g., ﬂip-ﬂops for temporary storage and memory for pre-computed values, and (iii) number of accesses to memories used. 16.7.3 Remark An arithmetic algorithm can be mapped onto diﬀerent types of hardware archi-tectures, e.g., serial, parallel, pipelined, systolic, etc. Two important ﬁgures of merit to characterize the performance of an architecture are space and time complexities. Generally, trade-oﬀs between space and time are possible. Main components of a hardware architecture include: logic gates, temporary storage (ﬂip-ﬂops, registers, etc.), and memories. 16.7.4 Deﬁnition The space complexity of an architecture is its number of logic gates (e.g., AND and XOR) and the amount of storage (i.e., ﬂip-ﬂops and memories). For simplicity, only two-input logic gates are assumed and interconnections are ignored. 16.7.5 Deﬁnition The critical path of an architecture is the path that represents the longest time delay. We approximate the delay as the delay caused by the logic gates in the critical path. In actual hardware realization, other factors such as interconnections contribute to the delay. 16.7.6 Deﬁnition The time complexity of an architecture is the amount of time needed by the architecture to complete the required arithmetic operation upon receiving any portion of the input. For a fully bit-parallel architecture, the time complexity is simply the delay in the critical path. For architecture that requires multiple clock cycles for the arithmetic operation, the time complexity is approximated as the product of the number of clock cycles and the gate delays in the critical path. 16.7.7 Remark At the architectural level, various design components can be used, for example, a two-input XOR gate for an addition over F2 and a two-input AND gate for a multiplication over F2. The time delay of a component is denoted as T along with a suitable subscript. For example, we denote the time delay for a two-input XOR gate as TX, and that of a two-input AND gate as TA. 16.7.2 Arithmetic using polynomial operations 16.7.8 Deﬁnition The elements of F2n can be represented as polynomials over F2 of degree n −1 or less. Thus, for a ∈F2n we can write a = Pn−1 i=0 aixi, where ai ∈F2 and 0 ≤i ≤n−1. The set {1, x, . . . , xn−1} is a polynomial basis of F2n over F2; see Section 2.1. 16.7.9 Deﬁnition Let a = Pn−1 i=0 aixi and b = Pn−1 i=0 bixi be two elements of F2n. Then the addition of a and b is a + b := Pn−1 i=0 (ai + bi)xi, where ai + bi is over F2. Cryptography 813 16.7.10 Proposition Using n XOR gates in parallel, a + b can be computed with a delay of one TX only. On the other hand, a + b can be be computed in bit serial fashion using only one XOR gate and a delay of nTX. 16.7.11 Deﬁnition Let a and b be two elements of F2n represented using polynomials as stated above. Then the product of a and b is ab := a(x)b(x) (mod f(x)), where f is an irre-ducible polynomial of degree n over F2 and deﬁnes the representation of the ﬁeld. 16.7.12 Remark Various algorithms exist for computing ab. One class of algorithms involves the following two steps: ﬁrst the multiplication of the two n-term polynomials, a and b, is performed, and then the resulting (2n −1)-term polynomial is reduced modulo f. In the straightforward or schoolbook method, the multiplication ab is performed by repeated shift-and-add operations and requires O(n2) additions and multiplications over F2. A more eﬃ-cient method is known as the Karatsuba algorithm and it is based on the following. 16.7.13 Theorem Assume that n is even and the n-term polynomial a is split as aHxn/2+aL, where aH and aL are each (n/2)-term polynomials in x over F2. Similarly, b is split as bHxn/2 + bL. Then ab = aHbHxn + [(aH + aL)(bH + bL) −aHbH −aLbL]xn/2 + aLbL. (16.7.1) 16.7.14 Remark We note that addition and subtraction are the same in ﬁelds of characteristic two. In case n is not even, a zero coeﬃcient can be padded at the higher degree end of a and b allowing an even splitting. This technique can be applied recursively and the following can be proven. 16.7.15 Lemma Assuming that n is a power of 2, then a recursive application of Theorem 16.7.13 for the multiplication of a and b requires no more than nlog2 3 multiplications and 6nlog2 3 −8n + 2 additions over F2. 16.7.16 Corollary [1026, 2342] A fully bit parallel implementation of the Karatsuba algorithm for the multiplication of two n-term F2 polynomials requires 6nlog2 3 −8n + 2 XOR and nlog2 3 AND gates, and its time complexity is (3 log2 n −1)TX + TA. 16.7.17 Remark Since log2 3 ≈1.58 < 2, the Karatsuba algorithm, which ﬁrst appeared in for integer multiplication, has a subquadratic arithmetic complexity. In the context of poly-nomial multiplication, a generalization of the Karatsuba algorithm can be found in . 16.7.18 Remark The main rationale behind designing a fully parallel multiplier is to achieve a higher speed. For multiplication of polynomials a and b, one way to achieve a time com-plexity that is lower than that of the Karatsuba algorithm is to split the polynomials according to the parity of the x’s exponent , i.e., a(x) = ae(y)+xao(y), where y = x2, ae(y) = Pn/2−1 i=0 a2iyi and ao(y) = Pn/2−1 i=0 a2i+1yi. Similarly, b(x) = be(y) + xbo(y). Then the product a(x)b(x) is computed using the following Karatsuba-like formula a(x)b(x) = {[ae(y)be(y) + yao(y)bo(y)]} + x{[(ae(y) + ao(y))(be(y) + bo(y))] −[ae(y)be(y) + ao(y)bo(y)]}. If n is a power of 2, then a recursive application of the above splitting can lead to a fully bit parallel polynomial multiplication hardware that has the same space complexity as the Karatsuba algorithm, but is about 30% faster - the main gate delays are 2 log2 nTX vs. (3 log2 n −1)TX. 814 Handbook of Finite Fields 16.7.19 Remark An improvement to the original F2[x] Karatsuba algorithm in terms of space complexity is given in . The main idea of this method can also be described using the following reﬁned Karatsuba identity [243, p. 325]: ab = aHbHx2n + {[(aH + aL)(bH + bL)] −[aHbH + aLbL]}xn + aLbL = (aHbHxn −aLbL)(xn −1) + (aH + aL)(bH + bL)xn. This identity is in fact closer to Karatsuba’s original squaring identity ([1684, p. 595]) than identity 16.7.1. Table 16.7.1 summarizes complexities of the Karatsuba algorithm and its two variants. Algorithm #AND #XOR Gate delay Karatsuba [32, 2342] nlog2 3 6nlog2 3 −8n + 2 (3 log2 n −1)TX + TA Overlap-free Karatsuba nlog2 3 6nlog2 3 −8n + 2 (2 log2 n)TX + TA Reﬁned Karatsuba [243, 3064] nlog2 3 5.5nlog2 3 −7n + 1.5 (3 log2 n −1)TX + TA Table 16.7.1 Asymptotic complexities of three Karatsuba algorithms. 16.7.20 Remark The Winograd short convolution algorithm may be viewed as a generalization of the original Karatsuba-based algorithm [241, 2747, 2988]: while the original Karatsuba algorithm performs evaluation and interpolation using linear factors x −∞, x and x −1, the Winograd method also uses nonlinear factors, i.e., irreducible polynomials of degrees greater than 1. For n = 3i (i > 0), the algorithm yields an asymptotic time complexity of (4 log3 n−1)TX +TA ≈(2.52 log2 n−1)TX +TA, which is better than that of the Karatsuba algorithm. However, the Winograd method has a higher asymptotic space complexity than the Karatsuba algorithm. 16.7.21 Remark Other subquadratic arithmetic complexity algorithms exist, e.g., [164, 243]. How-ever, when mapped onto hardware architectures, they yield a higher time complexity, e.g., is linear to n for a fully bit parallel implementation . 16.7.22 Theorem Let d be a binary polynomial of degree at most 2n −2 (i.e., d could be the product of two binary polynomials - each of degree at most n −1). Let Wf be the number of nonzero coeﬃcients of f of degree n. Then d (mod f) can be computed with at most (Wf −1)(n −1) bit operations. 16.7.23 Remark In addition to the class of algorithms mentioned in Remark 16.7.12 which performs polynomial multiplication ab and reduction modulo f in two separate steps, another class exists that combines these two steps. The latter class of algorithms is often used with sequential architectures. To this end, we ﬁrst look at the shift-and-reduce operation for hardware. 16.7.24 Lemma Let a and f be as deﬁned above. Then the i-th coordinate of xa modulo f is given as follows: (xa)i = an−1 i = 0, an−1 + fiai−1 1 ≤i ≤n −1. (16.7.2) 16.7.25 Remark Equation (16.7.2) can be realized using an n-stage linear feedback shift register (LFSR) that has feedback connections corresponding to the coeﬃcients of f. If the LFSR is initialized with the coordinates of a then after one shift the LFSR will contain the coordinates of xa modulo f; see Section 10.2. Cryptography 815 16.7.26 Lemma Let a polynomial f be as deﬁned above. Let a, b ∈F2n be represented as polynomials of degree at most n −1 over F2. Then ab = Pn−1 i=0 bia(i), where a(i) ∈F2n and a(i) = xia (mod f). 16.7.27 Remark Note that a(0) = a and, for 1 ≤i ≤n −1, a(i) = xa(i−1) (mod f). Thus, an LFSR can be used to obtain a(i) from a(i−1) as stated in Lemma 16.7.24. This along with Lemma 16.7.26 leads to the following algorithm for multiplication of two elements a and b of F2n. The algorithm scans the coordinates of b from the low to the high end. 16.7.28 Algorithm (Bit-level F2n multiplication - low to high) Input: a, b ∈F2n and reduction polynomial f Output: ab (mod f) 1. g ←a, c ←0 2. For i from 0 to n −1 do 3. c ←c + big 4. g ←xg (mod f) 5. Return c 16.7.29 Corollary Algorithm 16.7.28 requires O(n2) arithmetic operations over F2. For a bit-serial architecture of the algorithm, where the computations of one iteration are performed in one clock cycle, the space complexity is 2n ﬂip-ﬂops, n two-input AND gates, and n + Wf −2 two-input XOR gates. The gate delay in the critical path of the architecture is TX and the computation time is nTX. 16.7.30 Remark A multiplication algorithm that scans the coordinates of b from the high to the low can be easily obtained by writing out ab using Horner’s rule, i.e., ab = (· · · (abn−1x + abn−2)x + · · · )x + ab0 (mod f). 16.7.31 Deﬁnition Algorithm 16.7.28 can be modiﬁed so that in each iteration a digit is processed, reducing the number of iterations to ⌈n/d⌉, where d is the number of bits in each digit. We refer to the modiﬁed algorithms as digit-level multiplication algorithms. 16.7.32 Remark Compared to bit-level algorithms (e.g., Algorithm 16.7.28), the number of arith-metic operations in each iteration of a digit-level algorithm is higher and so is the space complexity of the corresponding digit-serial architecture, typically by a factor of d. This is because the x multiplication operation of Algorithm 16.7.28 is replaced by a multiplication with xd, and each AND operation by a multiplication of two polynomials of degree d −1 over F2. A number of digit-serial multiplier architectures have been proposed, see for ex-ample, and . For resource constrained applications, digit-serial multipliers may oﬀer suitable trade-oﬀs between space and time requirements. 16.7.33 Remark For applications that demand high throughput (in terms of number of operations per second), multipliers can be designed based on pipeline or systolic array architecture. Examples of such multipliers include [2055, 3035]. Pipeline and systolic array multipliers usually require extra ﬂip-ﬂips for storing intermediate results. 16.7.34 Remark Using a polynomial basis, squaring of a ∈F2n is a2 = Pn−1 i=0 aix2i (mod f). Thus, unlike a normal basis (see Subsection 16.7.4), the use of a polynomial basis to implement an F2n squaring requires bit operations or logic gates. However, the number of gates required can be quite low. For example, if the reduction polynomial f is a trinomial in some special form, then a bit parallel squaring unit can be implemented with about n/2 XOR gates . Like squaring, a square root in polynomial basis is also very eﬃcient. 816 Handbook of Finite Fields 16.7.35 Remark Let a be a nonzero element of F2n and b be the multiplicative inverse of a. Then in polynomial notation we have ab ≡1 (mod f), where f is as deﬁned earlier. Since gcd(f, a) = 1, the extended Euclidean algorithm can be used to obtain b. By simply chang-ing the initialization, the extended Euclidean algorithm can also be used for computing the division b = c/a in F2n, which in polynomial notation can be written as ab ≡c (mod f). 16.7.36 Algorithm (F∗ 2n inversion using the extended Euclidean algorithm) Input: a ∈F∗ 2n and reduction polynomial f Output: b ∈F∗ 2n such that ab ≡1 (mod f) 1. r ←f, s ←a, u ←0, v ←1 2. while s ̸= 0 do 3. q ←⌊r/s⌋ 4. (r, s) ←(s, r −q · s) 5. (u, v) ←(v, u −q · v) 6. b ←u 7. Return b 16.7.37 Remark Other algorithms for computing inverses that use polynomial operations and are suitable for hardware realization include the binary GCD and the Berlekamp-Massey algorithm [1431, 2316]. Additional inversion algorithms that rely on matrix operations over subﬁelds or multiplications and squaring operations over F2n (e.g., the Itoh-Tsujii algorithm) are mentioned in the following two subsections. 16.7.38 Remark For arithmetic over F2n, the algorithms stated earlier use operations of polynomials over F2. In the following subsection, we consider arithmetic algorithms that use matrix operations over F2. 16.7.3 Arithmetic using matrix operations 16.7.39 Remark Let a polynomial f be as deﬁned above and an element a ∈F2n in polynomial form be represented as a = Pn−1 i=0 aixi. Using the coordinates of a, we write the corresponding column vector as A = (a0, a1, . . . , an−1)T . 16.7.40 Theorem Let a, b, and c ∈F2n and c = ab. For 0 ≤i ≤n −1, let Zi be the column vector corresponding to axi. Denote the column vectors corresponding to b and c by B and C, respectively. Then, the following holds: C = ZB, (16.7.3) where Z is an n × n matrix over F2 and is given by Z = (Z0, Z1, . . . , Zn−1). 16.7.41 Remark Theorem 16.7.40 is an algorithm for multiplication of two elements of F2n using matrix operations over F2 and has two main steps: ﬁrst form the matrix Z, which is a function of input a and f, and then compute a matrix-vector product (MVP) over F2. 16.7.42 Remark The matrix Z is the Mastrovito matrix . Given Zi−1, one can obtain Zi using an LFSR operation and hence it requires Wf −2 additions over F2. Thus, the formation of matrix Z requires no more than (n −1)(Wf −2) additions over F2 . For arbitrary a and b, a straightforward approach to compute the matrix vector product ZB requires O(n2) arithmetic operations over F2. Cryptography 817 16.7.43 Deﬁnition Consider an n × n matrix T = [ti,j]n−1 i,j=0. If ti,j = ti+1,j+1 for 0 ≤i < n −1, then T is a Toeplitz matrix. Thus a Toeplitz matrix can be uniquely deﬁned by its 2n−1 entries that are in the top row and the left most column. 16.7.44 Lemma The sum of two n × n Toeplitz matrices is also an n × n Toeplitz matrix. If the matrix entries belong to F2, then the sum requires no more than 2n −1 additions over F2. 16.7.45 Lemma [1026, 2988] Let n = 2i (i > 0), T be an n × n Toeplitz matrix and V be a column vector over F2. Then the Toeplitz matrix-vector product (TMVP) TV can be computed with nlog2 3 multiplications and 5.5nlog2 3 −6n + 0.5 additions over F2. 16.7.46 Proposition The matrix T and vector V can be split as follows: T = T1 T0 T2 T1 and V = V0 V1 , where T0, T1 and T2 are (n/2) × (n/2) matrices and are individually in Toeplitz form, and V0 and V1 are (n/2) × 1 column vectors. Now the following noncommutative formula can be used to compute the TMVP TV recursively: TV = T1 T0 T2 T1 V0 V1 = P0 + P2 P1 + P2 , (16.7.4) where P0 = (T0 + T1)V1, P1 = (T1 + T2)V0 and P2 = T1(V0 + V1). 16.7.47 Theorem Let matrix Z be as deﬁned in Theorem 16.7.40. Then there exists an n×n nonsingular matrix U over F2 such that UZ is an n × n Toeplitz matrix. Furthermore, let D = UC and T = UZ. Then we have D = TB. (16.7.5) 16.7.48 Remark A class of transformation matrices U is given in . Special cases exist for which T = UZ and C = U −1D can be computed eﬃciently. In particular, when the reduction polynomial is a trinomial, these transformations can be done by simple permutations. For such special cases, the cost of multiplication of two elements of F2n is essentially the cost of the multiplication of a Toeplitz matrix and a vector over F2 . 16.7.49 Remark Let a, b, c ∈F2n and a ̸= 0. Then the division b = c/a over F2n can be performed by solving the matrix equation (16.7.3) over F2 for B, which is the column vector corresponding to the coordinates of b . Hardware architectures are available for solving such matrix equations . 16.7.50 Remark The division b = c/a over F2n can also be performed by solving equation (16.7.5) over F2 for B. Since the matrix in (16.7.5) is Toeplitz, algorithms for solving (16.7.5) are more eﬃcient that those for (16.7.3). The use of (16.7.5) however requires pre- and post-processing. As stated in Remark 16.7.48, such processing can be as simple as a permutation when f is a trinomial. 16.7.4 Arithmetic using normal bases 16.7.51 Remark Below we consider arithmetic in binary extension ﬁnite ﬁelds F2n using normal bases. Deﬁnition and various properties of normal bases can be found in Section 5.2. 818 Handbook of Finite Fields 16.7.52 Remark Let γ ∈F2n and N = {γ, γ2, γ22, . . . , γ2n−1} be a normal basis of F2n over F2. An element a ∈F2n can be represented as a = a0γ + a1γ2 + · · · + an−1γ2n−1, where ai ∈F2 and 0 ≤i ≤n −1. It is easy to see that a2 = an−1γ + a0γ2 + · · · + an−2γ2n−1. Thus the coordinates of a2 are obtained by a simple cyclic shift of the coordinates of a. In other words, ai = (a2)(i+1) = (a2j)(i+j), where subscripts are evaluated modulo n. 16.7.53 Remark Let a be a nonzero element of F2n. Then we can write a−1 = a2n−2 = a2(2n−1−1). Noting that since 2n−1 −1 = 1+2+22 +· · ·+2n−2, the inverse can be computed with O(n) multiplication and squaring operations. Using a normal basis, squaring does not require any gates. For practical applications, this method however requires a high speed multiplier. 16.7.54 Remark Consider an integer m > 1. Denote ˜ m = ⌊m/2⌋and m0 = m (mod 2). Then 2m −1 = 2m0(2 ˜ m −1)(2 ˜ m +1)+m0 = 2m0{(2 ˜ m −1)2 ˜ m +(2 ˜ m −1)}+m0. Thus, given a2 ˜ m−1, one can compute a2m−1 using 1 + m0 multiplications (disregarding other operations). By recursively applying this technique, the inverse of a nonzero a ∈F2n can be computed with ⌊log2(n−1)⌋+W(n−1)−1 multiplications over F2n, where W(n−1) is the number of nonzero bits in the binary representation of n −1. Inversion algorithms incurring only these many multiplications have been independently proposed by Itoh-Tsujii and Feng , and have been subsequently implemented in hardware for cryptographic applications, see for example [1361, 2441, 2466, 2489]. 16.7.55 Lemma Let a, b, c ∈F2n be represented with respect to a normal basis N as deﬁned above and c = ab. Let A = (a0, a1, . . . , an−1)T denote the column vector representing the coordinates of a = a0γ + a1γ2 + · · · + an−1γ2n−1. Similarly, let B be the column vector corresponding to b. Then c = n−1 X i−0 aiγ2i ! n−1 X i−0 biγ2i ! = AT GB, where G = [gi,j]n−1 i,j=0 is an n × n matrix over F2n and gi,j = γ2i+2j. 16.7.56 Remark Note that G depends only on N and is ﬁxed or constant for a given basis. As each entry of G can be represented with respect to N, one can write G = Pn−1 k=0 Gkγ2k, where Gk is an n × n matrix over F2. Using Lemma 16.7.55, for 0 ≤k ≤n −1 we can write ck = AT GkB. Consider Gn−1 and denote AT Gn−1B as g(a, b). Since c2k = (ab)2k, for 0 ≤k ≤n −1 we have cn−1−k = g(a2k, b2k). This leads to the following algorithm for multiplication using normal basis N. 16.7.57 Algorithm (Multiplication over F2n using a normal basis) Input: a, b ∈F2n represented with respect to a normal basis N Output: c = ab 1. s ←a, t ←b 2. For i from 0 to n −1 do 3. cn−1−i ←g(s, t) 4. s ←s2, t ←t2 5. Return c 16.7.58 Remark The multiplication scheme described above is due to Massey and Omura . For an arbitrary normal basis, half of the entries of Gn−1 are expected to be nonzero. Thus, up to O(n2) arithmetic operations over F2 are needed for each iteration of Algorithm 16.7.57 or O(n3) for the entire algorithm. Cryptography 819 16.7.59 Remark The number of nonzero entries in Gn−1, which is denoted by CN, determines the complexity of the normal basis multiplier. It has been shown by Mullin et al. that the number of non-zeros in Gn−1 is ≥2n −1 . The normal bases that satisfy the equality are known as optimal normal bases (ONB). There are two types of ONB (Types I and II). Section 5.3 presents more details on ONB. 16.7.60 Remark If Algorithm 16.7.57 is mapped onto a sequential architecture that requires n clock cycles, then its space complexity is O(n2) logic gates and the critical path has a gate delay of O(log2 n). The algorithm can be easily unrolled and mapped onto a fully parallel architecture, especially since the squaring operation in N does not require any logic gates. The space and the time complexities for such a fully parallel realization are O(n3) gates and O(log2 n) gate delays, respectively. 16.7.61 Remark The Massey-Omura multiplication scheme has some redundancy in the sense that there are common terms in the expressions of product coordinates ck, 0 ≤k ≤n −1 . By removing the redundancy, a bit parallel multiplier (referred to as Reyhani-Hasan-1) that oﬀers reduced space complexity and is applicable to any arbitrary normal basis has been reported in . By exchanging one AND gate for one XOR gate, the multiplier (referred to as Reyhani-Hasan-2) of reduces the number of AND gates to n(n + 1)/2. Because the complexity of subﬁeld multiplication is higher than that of subﬁeld addition, this technique is shown to be quite eﬀective for composite ﬁeld multiplications . Scheme #AND #XOR Gate delay Wang et al. 1985 nCN n(CN −1) ⌈log2 CN⌉TX + TA Reyhani-Hasan-1 n2 n(CN + n −2)/2 ⌈log2 CN⌉TX + TA Reyhani-Hasan-2 n(n + 1)/2 n(CN + 2n −3)/2 ⌈log2 CN⌉TX + TA Table 16.7.2 Complexities of quadratic F2n general normal basis parallel multipliers. 16.7.5 Multiplication using optimal normal bases 16.7.62 Proposition Let ˆ X = {x20, x21, . . . , x2n−1} be a Type I optimal normal basis of F2n over F2. Because 2 is a primitive root of prime n + 1, the following two sets are equal {20, 21, . . . , 2n−1} = {1, 2, . . . , n}. (16.7.6) Therefore, X = {x1, x2, . . . , xn} is also a basis of F2n over F2. 16.7.63 Remark Given a ﬁeld element a represented in the above two bases, i.e., a = Pn−1 i=0 ˆ aix2i and a = Pn i=1 aixi, it is easy to obtain the following coordinate transformation formula: a2i = ˆ ai, (16.7.7) where 0 ≤i ≤n −1 and the subscript 2i is reduced modulo n + 1. Equations (16.7.6) and (16.7.7) reveal that the basis conversion operation between X and ˆ X is simply a permutation and can be performed in hardware without using any logic gates. 16.7.64 Lemma Let a, b, c ∈F2n be represented with respect to basis X, c = ab and B = (b1, b2, . . . , bn)T be the column vector corresponding to coordinates of b with respect to X. Similarly, C = (c1, c2, . . . , cn)T . Then C = (Z1, . . . , Zn) B = ZB, where Zi (1 ≤i ≤n) is the column vector corresponding to the coordinates of ﬁeld element xia with respect to basis X, and Z is an n × n matrix over F2. 820 Handbook of Finite Fields 16.7.65 Proposition Using the identity xn+1 = 1 = Pn j=1 xj, we have the following decom-position of the matrix Z = Z1 + Z2: Z =          0 an an−1 · · · a3 a2 a1 0 an · · · a4 a3 a2 a1 0 · · · a5 a4 . . . . . . . . . ... . . . . . . an−2 an−3 an−4 · · · 0 an an−1 an−2 an−3 · · · a1 0          +          an an−1 an−2 · · · a2 a1 an an−1 an−2 · · · a2 a1 an an−1 an−2 · · · a2 a1 . . . . . . . . . ... . . . . . . an an−1 an−2 · · · a2 a1 an an−1 an−2 · · · a2 a1          . (16.7.8) Therefore, MVP ZB may be computed via ZB = Z1B + Z2B. 16.7.66 Remark The straightforward computation of the TMVP Z1B requires n(n−1) AND gates and n(n −2) XOR gates. Clearly, computing Z2B requires only n AND gates and n −1 XOR gates since all the rows in Z2 are the same. However, this fact was not noticed in the original Massey-Omura normal basis multiplier, where n AND gates and n × (n −1) XOR gates were used to compute MVP Z2B . After removing the above redundancy in Z2B, Hasan et al. presented a multiplier with the following complexities: n(n−1)+n = n2 AND gates, n(n−2)+(n−1)+n = n2−1 XOR gates, and a gate delay of ⌈1+log2 n⌉TX +TA . The structure of Sunar-Ko¸ c’s Type I optimal normal basis multiplier is based on their polynomial basis multiplier and its space complexity is the same as that in , but its gate delay is ⌈2 + log2 n⌉TX + TA. Another design that has the same complexities as the multiplier in is Reyhani-Hasan-1 multiplier . These multipliers all belong to the class of quadratic parallel multipliers, and their complexities are summarized in Table 16.7.3. Scheme #AND #XOR Gate delay Wang et al. n2 2n2 −2n ⌈1 + log2 n⌉TX + TA Hasan et al. n2 n2 −1 ⌈1 + log2 n⌉TX + TA Sunar-Ko¸ c n2 n2 −1 ⌈2 + log2 n⌉TX + TA Reyhani-Hasan-1 n2 n2 −1 ⌈1 + log2 n⌉TX + TA Reyhani-Hasan-2 n(n + 1)/2 1.5n2 −0.5n −1 ⌈1 + log2 n⌉TX + TA Table 16.7.3 Complexities of quadratic F2n Type I optimal normal basis parallel multipliers. 16.7.67 Remark By exchanging one AND gate for one XOR gate, Reyhani-Hasan-2 Type I optimal normal basis multiplier reduces the number of AND gates to n(n+1)/2 , while keeping the total number of AND and XOR gates unchanged. This technique is also used in the Elia-Leone-2 Type II optimal normal basis parallel multiplier listed in Table 16.7.4 . 16.7.68 Proposition Let x = y+y−1 generate a Type II optimal normal basis of F2n over F2, where y is a primitive (2n + 1)-st root of unity in F22n. Deﬁne xi = yi + y−i for 0 ≤i ≤n, we have {x20, x21, . . . , x2n−1} = {x1, x2, . . . , xn}. (16.7.9) Therefore, X = {x1, x2, . . . , xn} is also a basis of F2n over F2. 16.7.69 Proposition Given a ﬁeld element a represented with respect to the above two bases, i.e., a = Pn−1 i=0 ˆ aix2i and a = Pn i=1 aixi, the coordinate transformation formula between these two bases is given as follows: as(2i) = ˆ ai, (16.7.10) where 0 ≤i ≤n −1 and s(j) is deﬁned as the unique integer such that 0 ≤s(j) ≤n and j ≡s(j) (mod 2n + 1) or j ≡−s(j) (mod 2n + 1). Cryptography 821 16.7.70 Remark From (16.7.9) and (16.7.10), it follows that the basis conversion operation between X and ˆ X is simply a permutation and hence may be performed in hardware without using any logic gates. 16.7.71 Proposition [1884, 2750] The product ab can be computed via an MVP ZB using basis X, and the matrix Z can be decomposed as the summation of two matrices i.e., Z = Z1 + Z2: Z =        a2 a3 · · · an an a3 a4 · · · an an−1 . . . . . . ... . . . . . . an an · · · a3 a2 an an−1 · · · a2 a1        +        0 a1 · · · an−2 an−1 a1 0 · · · an−3 an−2 . . . . . . ... . . . . . . an−2 an−3 · · · 0 a1 an−1 an−2 · · · a1 0        . (16.7.11) Here, Z1 is a Hankel matrix, i.e., entries at (i, j) and (i −1, j + 1) are equal, and Z2 is a circulant matrix. 16.7.72 Remark The straightforward computation of the above matrix-vector product ZB results in the following quadratic parallel multipliers: Sunar-Ko¸ c multiplier , Elia-Leone-1 multiplier and Reyhani-Hasan-1 multiplier . Their gate counts and delay com-plexities are equal, and are summarized in Table 16.7.4. Scheme #AND #XOR Gate delay Sunar-Ko¸ c n2 3n(n −1)/2 ⌈1 + log2 n⌉TX + TA Elia-Leone-1 n2 3n(n −1)/2 ⌈1 + log2 n⌉TX + TA Elia-Leone-2 n(n + 1)/2 2n(n −1) ⌈1 + log2 n⌉TX + TA Reyhani-Hasan-1 n2 3n(n −1)/2 ⌈1 + log2 n⌉TX + TA Table 16.7.4 Complexities of quadratic F2n Type II optimal normal basis parallel multipliers. 16.7.73 Remark After being converted from a Type I optimal normal basis into basis X using (16.7.6), the computation of ab becomes a modular multiplication operation, which can be divided into two steps. The ﬁrst step, i.e., the polynomial multiplication operation step, can be performed using the Karatsuba algorithm. This results in Leone’s subquadratic multiplier . Fan-Hasan’s Type I optimal normal basis subquadratic multiplier is based on (16.7.8). For Type II optimal normal basis multiplication, Fan and Hasan use the TMVP formula (16.7.4) and the decomposition (16.7.11). By exploiting vector and matrix symmetry properties that exist in the matrix vector expressions of Types I and II optimal normal basis multiplications, Hasan et al. use the block recombination technique to design subquadratic parallel multipliers in . Table 16.7.5 gives gate counts and gate delays for the above-mentioned optimal normal basis parallel multipliers of subquadratic space complexity. ONB Scheme #AND #XOR Gate delay Leone nlog2 3 6nlog2 3 −8n + 2 ⌈3 log2 n −1⌉TX + TA Type I Fan-Hasan nlog2 3 + n 5.5nlog2 3 −4n −0.5 ⌈1 + 2 log2 n⌉TX + TA Hasan et al. 4 3 nlog2 3 + 1.5n 14 3 nlog2 3 −4n + 1 ⌈2 + 2 log2 n⌉TX + TA Type II Fan-Hasan 2nlog2 3 11nlog2 3 −12n + 1 ⌈1 + 2 log2 n⌉TX + TA Hasan et al. 2nlog2 3 + 0.5n 41 6 nlog2 3 −7.5n + 1.5 ⌈1 + 2 log2 n⌉TX + TA Table 16.7.5 Complexities of subquadratic F2n Types I and II optimal normal basis parallel multipliers. 16.7.74 Remark For Type II optimal normal bases, multiplication over F2n can be expressed in terms of one or more multiplications of n-term polynomials over F2 [1180, 1238]. Then one 822 Handbook of Finite Fields can use any suitable method for polynomial multiplication such as the Karatsuba algorithm. The scheme presented in uses two polynomial multiplications and that in uses only one. Both schemes however incur computational overhead to express the Type II optimal normal basis multiplication into polynomial multiplication(s). The overheads are O (n) for and O (n log2 n) for . In , the computational overhead of has been reduced by a factor of about two. 16.7.6 Additional notes 16.7.75 Remark For eﬃcient hardware realization of ﬁeld multiplication using polynomial bases, examples of architectures that use low weight ﬁeld deﬁning polynomials, i.e., trinomials and pentanomials, include [2453, 2465, 2749]. Other types of polynomials that have been used in multiplier architectures include all-one, nearly all-one, and equally-spaced polynomials. 16.7.76 Remark The ﬁeld F2n can be embedded into a cyclotomic ring. This embedding technique leads to a redundant representation, i.e., each ﬁeld element is represented using more that n bits. Several multiplier architectures have been reported using such redundant represen-tation [923, 2216, 3009]. The best scenario for redundant representation is when only one extra bit is needed and it occurs where a Type I optimal normal basis exists, the conditions for which are the same as those for an irreducible all-one polynomial. 16.7.77 Remark Optimal normal bases do not exist for every ﬁeld, so that alternatives that allow eﬃcient squaring operations include the use of Gaussian normal bases and Dickson polyno-mials for the representation of ﬁeld elements. Examples of multiplier architectures that use such representation include [1438, 1885]. 16.7.78 Remark Besides polynomial and normal bases, hardware multiplier architectures have been reported using bases like shifted polynomial, dual, and triangular; see for example [1025, 1060, 1435, 3008]. 16.7.79 Remark The Montgomery multiplication algorithm and Residue Number System (RNS) have been extensively studied for hardware implementations of arithmetic over prime ﬁelds; see for example [614, 2221, 2325, 2541]. These schemes have also been applied to arithmetic over binary extension ﬁelds [164, 1775, 2787]. For multiplication over F2n, the main cost of a straightforward realization of the Montgomery algorithm is a multiplication of two binary polynomials of degree n −1 , and the RNS based multiplication scheme has been shown to have O(n1.6) bit operations for a special form of reduction polynomials . 16.7.80 Remark Some cryptographic systems use exponentiation ae, where a ∈F2n and e is a nonzero positive integer of up to n bits long. A straightforward method for exponentiation is to use the well-known square-and-multiply algorithm . This method requires ⌊log2 e⌋ squaring operations and W(e) −1 multiplications over F2n, where W(e) is the number of nonzero bits in the binary representation of e. Many improvements have been proposed that require some re-coding of the exponent e and/or creation of look-up tables based on a . See Also §5.2 For deﬁnition and properties of normal bases. §5.3 For deﬁnition and properties of Types I and II optimal normal bases. Cryptography 823 References Cited: [32, 107, 164, 241, 243, 248, 574, 614, 923, 966, 1025, 1026, 1027, 1028, 1052, 1060, 1180, 1188, 1238, 1361, 1431, 1432, 1433, 1434, 1435, 1437, 1438, 1439, 1576, 1684, 1775, 1776, 1884, 1885, 1902, 2012, 2014, 2055, 2080, 2132, 2196, 2216, 2221, 2316, 2325, 2342, 2441, 2451, 2452, 2453, 2465, 2466, 2489, 2541, 2691, 2695, 2747, 2749, 2750, 2787, 2926, 2953, 2963, 2988, 3006, 3007, 3008, 3009, 3035, 3064] This page intentionally left blank This page intentionally left blank 17 Miscellaneous applications 17.1 Finite ﬁelds in biology .............................. 825 Polynomial dynamical systems as framework for discrete models in systems biology • Polynomial dynamical systems • Discrete model types and their translation into PDS • Reverse engineering and parameter estimation • Software for biologists and computer algebra software • Speciﬁc polynomial dynamical systems 17.2 Finite ﬁelds in quantum information theory..... 834 Mutually unbiased bases • Positive operator-valued measures • Quantum error-correcting codes • Period ﬁnding • Quantum function reconstruction • Further connections 17.3 Finite ﬁelds in engineering ......................... 841 Binary sequences with small aperiodic autocorrelation • Sequence sets with small aperiodic auto- and crosscorrelation • Binary Golay sequence pairs • Optical orthogonal codes • Sequences with small Hamming correlation • Rank distance codes • Space-time coding • Coding over networks 17.1 Finite ﬁelds in biology Franziska Hinkelmann, Virginia Tech Reinhard Laubenbacher, Virginia Tech 17.1.1 Polynomial dynamical systems as framework for discrete models in systems biology The goal of molecular systems biology is to understand the functionality of biological sys-tems as a whole by studying interactions among the components, e.g., genes, proteins, and metabolites. Modeling is a vital tool in achieving this goal. It allows the simulation of diﬀer-ent types of interactions within the system, leading to testable hypotheses, which can then be experimentally validated. This process leads to a better understanding of the underlying biological system. Discrete models are increasingly used in systems biology [74, 2514, 2515, 3010]. They can reveal valuable insight about the qualitative behavior of a system when it is infeasible to estimate enough parameters accurately to build a quantitative model. Many discrete models can be formulated as polynomial dynamical systems, that is, state and time discrete 825 826 Handbook of Finite Fields dynamical systems described by polynomial functions over a ﬁnite ﬁeld. This provides ac-cess to the algorithmic theory of computational algebra and the theoretical foundation of algebraic geometry, which help with all aspects of the modeling process. Polynomial dynam-ical systems provide a unifying framework for many discrete modeling types. The algebraic framework allows for eﬃcient construction and analysis of discrete models. 17.1.2 Polynomial dynamical systems 17.1.1 Remark Deﬁnitions and theorems can be found in . 17.1.2 Deﬁnition A Polynomial Dynamical System (PDS) is a time-discrete dynamical system f = (f1, . . . , fn) : Fn →Fn where each coordinate function fi is a function of the n variables x1, . . . , xn, each of which takes on values in a ﬁnite ﬁeld F, and each fi can be assumed polynomial. 17.1.3 Remark Two directed graphs are usually assigned to each such system. 17.1.4 Deﬁnition The dependency graph (or wiring diagram) D(f) of f has n vertices 1, . . . , n, corresponding to the variables x1, . . . , xn of f. There is a directed edge i →j if there exists x = (x1, . . . , xi, . . . , xn) ∈Fn such that fj(x) ̸= fj(x1, . . . , xi + 1, . . . , xn). That is, D(f) encodes the variable dependencies in f. 17.1.5 Deﬁnition The dynamics of f is encoded by its phase space (or state space), denoted by S(f). It is the directed graph with vertex set Fn and a directed edge from u to v if f(u) = v. 17.1.6 Deﬁnition For each u ∈Fn, the orbit of u is the sequence {u, f(u), f 2(u), . . .}, where f k means k-th composition of f. The sequence {u, f(u), f 2(u), . . . , f t−1(u)} is a limit cycle of length t and u is a periodic point of period t if u = f t(u) and t is the smallest such number. Since Fn is ﬁnite, every orbit must include a limit cycle. 17.1.7 Deﬁnition The point u is a ﬁxed point (or steady state) if f(u) = u. 17.1.8 Lemma (Limit cycle analysis) For a polynomial dynamical system f = (f1, . . . , fn) : Fn →Fn, states in a limit cycle of length t are elements of the algebraic variety V (f t 1 −x1, . . . , f t n −xn), deﬁned by the polynomials f t 1 −x1, . . . , f t n −xn, but not of the variety V (f s 1 −x1, . . . , f s n −xn) for any s < t. In particular, ﬁxed points are the points in the variety V (f1 −x1, . . . , fn −xn). 17.1.9 Deﬁnition A component of the phase space S(f) consists of a limit cycle and all orbits of f that contain it. Hence, the phase space is a disjoint union of components. 17.1.10 Deﬁnition A polynomial dynamical system can be iterated using diﬀerent update sched-ules: • synchronous: all variables are updated at the same time; Miscellaneous applications 827 • sequential: variables are updated sequentially according to the order spec-iﬁed by an update schedule. A sequential system with a given update schedule can be translated into a synchronous system with the same dy-namics; • asynchronous: at every time step, one variable is chosen according to a probability distribution (usually uniform) to be updated. Asynchronous systems are non-deterministic; • delays. 17.1.11 Remark Synchronous and asynchronous systems have the same ﬁxed points. 17.1.12 Remark For visualization and analysis of PDS, the Web-based tool ADAM is available, see Section 17.1.5 . PDS can be used to represent dynamic biological systems. 17.1.13 Example The lac(tose) operon is a functional unit of three genes, LacZ, LacY, and LacA, transcribed together, responsible for the metabolism of lactose in the absence of glucose in bacteria. In the presence of lactose this genetic machinery is disabled by a repressor pro-tein. The genes in the lac operon encode several proteins involved in this process. Lactose permease transports extracellular lactose into the cell, where the protein β-galactosidase breaks the lactose down into glucose, galactose, and allolactose. The allolactose binds to the repressor protein and deactivates it, which results in the transcription of the three lac genes, resulting in the production of permease and β-galactosidase. This positive feedback loop allows for a rapid increase of lactose when needed . The result is a bistable dynamical system. This process can be modeled by a polynomial dynamical system. Each variable can take on two states: 0 denotes the absence (or low concentration) of a substrate or the inactive state of a variable, and 1 denotes presence or activity. As there are only two states, the system is modeled over the ﬁnite ﬁeld F2. Permease (xP ) transports ex-ternal lactose (xeL) inside the cell, and the update function for intracellular lactose (xL) is xL(t + 1) = xeL(t) AND xP (t), or as a polynomial over F2, fL = xeLxP . Using xB for β-galactosidase, xM for mRNA, xA for allolactose, the functionality of the lac operon can be modeled by the polynomial dynamical system f = (fL, fA, fM, fP , fB) : F5 2 →F 5 2 : fL = xeLxP , fA = xLxB, fM = xA, fP = xM, fB = xM. Figure 17.1.1 denotes the dependency graph and phase space of the above model . 17.1.3 Discrete model types and their translation into PDS 17.1.14 Remark Continuous models of biological systems, such as ordinary or partial diﬀerential equation models, rely on exact rate parameters, for which it is oftentimes impossible to obtain exact measurements. When not enough information is available to build a quantita-tive model, discrete models can give valuable insight about the qualitative behavior of the system. Such models are state- and time-discrete. For example, in the most simple case, one distinguishes only between two states, ON and OFF, or active and inactive, present and absent, etc. 828 Handbook of Finite Fields Figure 17.1.1 PDS for lac operon: state space in the absence (top) and presence (bottom left) of glucose, and dependency graph (bottom right). Each 5-tuple represents the states of lactose, allolactose, mRNA, permease, and β-galactosidase, (L, A, M, P, B). Miscellaneous applications 829 17.1.15 Remark Model types include (probabilistic) Boolean networks, logical networks, Petri nets, cellular automata, and agent-based (individual-based) models, to name the most commonly found ones [343, 1859, 2215, 2512, 2617, 2704]. All these model types can be translated into PDS [1505, 2864]. 17.1.3.1 Boolean network models 17.1.16 Remark In a Boolean network model every variable is either ON or OFF, and the state of each variable at time t + 1 is determined by a Boolean expression that involves some or all of the variables at time t. Boolean models were ﬁrst introduced in 1969 by Kauﬀman for gene regulatory networks, in which each gene (variable) is either expressed (ON) or not expressed (OFF) at every time step . 17.1.17 Algorithm For Boolean network models, where there are two states (TRUE and FALSE), F = F2, 0 denotes FALSE and 1 TRUE. Table 17.1.17 lists the Boolean expressions and the corresponding polynomials. All Boolean expressions can be translated to polynomials using this correspondence. Boolean expression polynomial X x NOT X x + 1 X AND Y xy X OR Y xy + x + y Table 17.1.2 Correspondence of Boolean expressions and polynomials over F2. 17.1.3.2 Logical models 17.1.18 Remark Logical models are a generalization of Boolean models, in which variables can take on more than two states, e.g., to represent the three states low, medium, and high concentration of a substrate. The rules governing the temporal evolution are switch-like logical rules for the diﬀerent states of each variable. Updates in logical models are speciﬁed via parameters rather than by a (Boolean) expression. 17.1.19 Deﬁnition A logical model is a triple (V, E, K), where: 1. V = {v1, . . . , vn} is the set of vertices or nodes. Each vi has a maximum expres-sion level, mi. The set S = {0, . . . , m1} × · · · × {0, . . . , mn} (Cartesian product) is the state space of the logical model and its elements are states. 2. E is the set of arcs. The elements of E have the form (vi, vj, θ), where 1 ≤θ ≤mi is a threshold for (vi, vj). Let I(j) = {v : (v, vj, θ) ∈E}, the input of vj, be the set of vertices that have an edge ending in vj. Notice that an arc (vi, vj) is allowed to have multiplicity corresponding to diﬀerent thresholds; the number of thresholds is denoted by mi,j. These thresholds are indexed in increasing order, 1 ≤θi,j,1 < · · · < θi,j,mi,j ≤mi; θi,j,k is the k-th threshold for the arc (vi, vj). By convention, we deﬁne θi,j,mi,j+1 = mi +1 (actually, θi,j,mi,j+1 can be deﬁned as any number greater than mi) and θi,j,0 = 0. Let x = (x1, . . . , xn) be a state; we say that the k-th interaction for input vi of vj is active if θi,j,k ≤xi < θi,j,k+1; we denote this by Θi,j(xi) = k. 3. K = {Ki : Q vj∈I(i){0, . . . , mi,j} →{0, . . . , mi}, i = 1, . . . , n} is the set of parameters. 830 Handbook of Finite Fields 17.1.20 Example (Logical model of lambda phage ) Lambda phage is a virus that injects its DNA into a bacterial host. Once injected, it enters either a lytic cycle or a lysogenic cycle. In the lytic cycle, its DNA is replicated, the host cell lyses, and new viruses are released. In the lysogenic cycle, the virus DNA is copied into the hosts DNA where it remains without any apparent harm to the host. A number of bacterial and viral genes takes part in the decision process between lysis and lysogenisation. The core of the regulatory network that controls the life cycle consists of two regulatory genes, cI and cro. Lysogeny is maintained if cI proteins dominate, the lytic cycle if cro proteins dominate. CI inhibits cro, and vice versa. At high concentrations, cro downregulates its own production, see Figure 17.1.3. This regulatory network can be encoded in the following logical model: V = {c1, cro}, Figure 17.1.3 Logical model of lambda phage, blunt ended arrows indicate an inhibitory eﬀect. E = {(c1, cro, 1), (cro, c1, 1), (cro, cro, 2)}, and K = {Kc1, Kcro}, where Kc1 : {0, 1, 2} → {0, 1} is deﬁned as Kc1(0) = 1, Kc1(1) = Kc1(2) = 0 and Kcro : {0, 1} × {0, 1, 2} → {0, 1, 2} as Kcro(0, 2) = Kcro(1, 2) = 1, Kcro(1, 0) = Kcro(1, 1) = 0, Kcro(0, 0) = 1, and Kcro(0, 1) = 2. 17.1.21 Algorithm Logical models are translated into a PDS by the following algorithm. Let (V, E, K) be a logical model as in Deﬁnition 17.1.19. Choose a prime number p such that p ≥mi + 1 for all 1 ≤i ≤n ({0, . . . , mi} has mi + 1 elements), and let F = Fp = {0, 1, . . . , p −1} be the ﬁeld with p elements. Note that we may consider the set S to be a subset of Fn. Consider a vertex vi and let gi be its coordinate function. Our goal is to represent gi as a polynomial in terms of its inputs, say xi1, . . . , xir. That is, we need a poly-nomial function deﬁned on Fr with values in F. Denote a ∧b = min{a, b}, using the natural order on the set F, viewed as integers. To extend the domain of g from Q vj∈I(i){0, . . . , mj,i} to Fr, we deﬁne g(xi1, . . . , xir) = gi(xi1 ∧mi1, . . . , xir ∧mir) for (xi1, . . . , xir) ∈Fr. The polynomial form of gi : Fr →F is then gi(x) = X (ci1,...,cir )∈Fr gi(ci1, . . . , cir) Y vj∈I(i) (1 −(xj −cj)p−1), where the right-hand side is computed modulo p; see also the Lagrange Interpolation For-mula (Theorem 2.1.131). 17.1.22 Example The logical model of the lambda phage presented in Example 17.1.22 corresponds to the PDS over F3: c1 = x1, cro = x2, and the polynomials are f1 = −x2 2 + 1, f2 = −x2 1x2 2 + x2 1x2 + x2 1 + x2 2 −x2 −1. 17.1.23 Remark The logical model can be converted manually or with the software package ADAM . Instead of analyzing the logical model, the corresponding PDS can be analyzed for its dynamic features. Miscellaneous applications 831 17.1.3.3 Petri nets and agent-based models 17.1.24 Remark Petri nets are bipartite graphs, consisting of places and transitions. Places can be marked with tokens, usually representing concentration levels or number of molecules present. Transitions ﬁre in a non-deterministic way and move tokens between places. Petri nets have been used extensively to model chemical reaction networks, where places represent species and transitions interactions. Analysis of the Petri net can reveal which species are persistent, i.e., which species can become extinct if all species are present at the initial time. For the translation algorithm of Petri nets into polynomial dynamical systems, see . 17.1.25 Remark Agent-based models (ABM) (or individual-based models) are computational models consisting of individual agents, each agent having a set of rules that deﬁnes how it interacts with other agents and the environment. Simulation is used to assess the evolution of the system as a whole. Sophisticated agent-based models have been published that simulate biological systems including tumor growth and the immune system [90, 976, 2000]. For conversion of agent-based models into polynomial dynamical systems, see . 17.1.4 Reverse engineering and parameter estimation 17.1.26 Remark A central problem in systems biology is the construction of models based on system-level experimental data and biological input. When the wiring diagram is known but not the combinatorial eﬀect of the diﬀerent regulatory inputs, the process of identify-ing models that ﬁt the data is analogous to parameter estimation for continuous models: estimate a function that ﬁts the experimental data and satisﬁes some optimality criterion. 17.1.27 Remark Typically, the set of possible models is very large. Tools from computer algebra allow the identiﬁcation of all models that ﬁt a given experimental data set, and furthermore allow one to identify those models that satisfy a given optimality criterion [867, 1860]. Sub-section 17.1.4.1 introduces an algorithm that identiﬁes possible wiring diagrams, with the optimality criterion that every variable is aﬀected by a minimal number of other variables, and Subsections 17.1.4.2 and 17.1.6.2 introduce parameter estimation algorithms. 17.1.4.1 The minimal-sets algorithm 17.1.28 Remark The minimal-sets algorithm constructs the set of all “minimal” wiring diagrams such that a model that ﬁts the experimental data exists for each wiring dia-gram. Given m observations stating that the inputs ti result in the state si for a variable, i.e., (s1, t1), . . . , (sm, tm), where si ∈Fn and ti ∈F, the minimal-sets algorithm identi-ﬁes all inclusion minimal sets S ∈{1, . . . , n} such that there exists a polynomial function f ∈F[{xi\|i ∈S}] with f(si) = ti. The algorithm is based on the fact that one can deﬁne a simplicial complex ∆associated to the experimental data, such that the face ideal for the Alexander dual of ∆is a square-free monomial ideal M, and the generating sets for the minimal primes in the primary decomposition of the ideal M are exactly the desired minimal sets. 17.1.4.2 Parameter estimation using the Gr¨ obner fan of an ideal 17.1.29 Remark Typically, there are many models that ﬁt experimental data, even when restricting the model space to minimal models. The minimal sampling algorithm is used to sample the subspace of minimal models; it returns a set of weighted functions per node, assigning a higher weight to functions that are candidates for several monomial orders . The algorithm is based on the Gr¨ obner fan of an ideal. 832 Handbook of Finite Fields 17.1.5 Software for biologists and computer algebra software 17.1.30 Remark The methods described in this section heavily rely on computer algebra systems, e.g., Macaulay2 . A major advantage of translating other discrete modeling types into polynomial systems is that eﬃcient implementations of algorithms such as Gr¨ obner basis calculations or primary decomposition are already implemented, and can be used independently of the underlying model type. On the other hand, an algorithm implemented to analyze a Petri net cannot be re-used to analyze a logical model. As many biologists are not familiar with computer algebra systems and the mathematical theory, software packages have been developed that allow the construction and analysis of discrete models using methods described in this section, without requiring understanding of the underlying mathematics [866, 1502]. 17.1.6 Speciﬁc polynomial dynamical systems 17.1.31 Remark Sections 17.1.6.1 to 17.1.6.4 describe speciﬁc classes of polynomial dynamical sys-tems, and theorems that relate the structure of the PDS to its dynamics. 17.1.6.1 Nested canalyzing functions 17.1.32 Remark Certain polynomial functions are very unlikely to represent an interaction in a bi-ological system. For example, in a Boolean system, x + y, i.e., the exclusive OR, is unlikely to represent an actual biological process. In addition, there are classes of functions that are biologically more relevant than other functions. One such class consists of nested canalyz-ing functions, named after the genetic concept of canalization, identiﬁed by the geneticist Waddington in the 1940s. Networks consisting of nested canalyzing functions are robust and stable [1714, 2887]. 17.1.33 Deﬁnition A Boolean function f(x1, . . . , xn) is canalyzing if there exists an index i and a Boolean value a for xi such that f(x1, . . . , xi−1, a, xi+1, . . . , xn) = b is constant. That is, the variable xi, when given the canalyzing value a, determines the value of the function f, regardless of the other inputs. The output value b is the canalyzed value. 17.1.34 Deﬁnition Let f be a Boolean function in n variables. • Let σ be a permutation on {1, . . . , n}. The function f is a nested canalyzing function (NCF) in the variable order xσ(1), . . . , xσ(n) with canalyzing input values a1, . . . , an and canalyzed output values b1, . . . , bn, respectively, if it can be represented in the form f(x1, x2, . . . , xn) =                      b1 if xσ(1) = a1, b2 if xσ(1) ̸= a1 and xσ(2) = a2, b3 if xσ(1) ̸= a1 and xσ(2) ̸= a2 and xσ(3) = a3, . . . . . . bn if xσ(1) ̸= a1 and · · · and xσ(n−1) ̸= an−1 and xσ(n) = an, bn + 1 if xσ(1) ̸= a1 and · · · and xσ(n) ̸= an. • The function f is nested canalyzing if f is nested canalyzing in the variable order xσ(1), . . . , xσ(n) for some permutation σ. Miscellaneous applications 833 17.1.35 Remark Any Boolean function in n variables is a map f : {0, 1}n →{0, 1}. Denote the set of all such maps by Bn. For any Boolean function f ∈Bn, there is a unique polynomial g ∈F2[x1, . . . , xn] such that g(a1, . . . , an) = f(a1, . . . , an) for all (a1, . . . , an) ∈Fn 2 and such that the degree of each variable appearing in g is equal to 1. Namely, g(x1, . . . , xn) = X (a1,...,an)∈Fn 2 f(a1, . . . , an) n Y i=1 (1 −(xi −ai)). 17.1.36 Deﬁnition Let S be a a non-empty set whose highest element is rS. The completion of S, which is denoted by [rS], is the set [rS] := {1, 2, . . . , rS}. For S = ∅, let [r∅] := ∅. 17.1.37 Theorem Let f be a Boolean polynomial in n variables, given by f(x1, x2, . . . , xn) = X S⊆[n] cS Y i∈S xi. (17.1.1) The polynomial f is a nested canalyzing function in the order x1, x2, . . . , xn if and only if c[n] = 1, and for any subset S ⊆[n], cS = c[rS] Y i∈[rS]\S c[n]{i}. 17.1.38 Corollary The set of points in F2n 2 corresponding to coeﬃcient vectors of nested canalyzing functions in the variable order x1, . . . , xn, denoted by V ncf id , is given by V ncf id =   (c∅, . . . , c[n]) ∈F2n 2 : c[n] = 1, cS = c[rS] Y i∈[rS]\S c[n]{i}, for S ⊆[n]   . 17.1.39 Deﬁnition Let σ be a permutation on the elements of the set [n]. We deﬁne a new order relation <σ on the elements of [n] as follows: i <σ j if and only if σ−1(i) < σ−1(j). Let S be a nonempty subset of [n], say S = {i1, . . . , it}. Let rσ S := max{σ−1(i1), . . . , σ−1(it)}. The completion of S with respect to the permutation σ, de-noted by [rσ S]σ is the set [rσ S]σ := {σ(1), . . . , σ(rσ S)}. 17.1.40 Corollary Let f ∈Bn and let σ be a permutation of the set [n]. The polynomial f is a nested canalyzing function in the order xσ(1), . . . , xσ(n), with input values aσ(i) and corresponding output values bσ(i), 1 ≤i ≤n, if and only if c[n] = 1 and, for any subset S ∈[n], cS = c[rσ S]σ Y w∈[rσ S]σ\S c[n]{w}. 17.1.41 Corollary Let σ be a permutation on [n]. The set of points in F2n 2 corresponding to nested canalyzing functions in the variable order xσ(1), . . . , xσ(n), denoted by V σ id, is deﬁned by V σ id =   (c∅, . . . , c[n]) ∈F2n 2 : c[n] = 1, cS = c[rσ S]σ Y w∈[rσ S]σ\S c[n]{w}, for S ⊆[n]   . 17.1.42 Corollary Let f ∈R and let σ be a permutation of the elements of the set [n]. If f is a nested canalyzing function in the order xσ(1), . . . , xσ(n), with input values ai and corresponding 834 Handbook of Finite Fields output values bi, 1 ≤i ≤n, then ai = c[n]{σ(i)}, for 1 ≤i ≤n −1, b1 = c∅+ cσ(1)c[n]{σ(1)}, bi+1 −bi = c[i+1]σc[n]{σ(i+1)} + c[i]σ, for 1 ≤i < n −1 and bn −an = bn−1 + c[n−1]σ. 17.1.43 Remark Thus, the family of nested canalyzing polynomials in a given number of variables can be described as an algebraic variety deﬁned by a collection of binomials. This description has several useful implications. 17.1.6.2 Parameter estimation resulting in nested canalyzing functions 17.1.44 Remark For given time course data and a wiring diagram, one can identify all nested canalyzing functions (17.1.6.1) that ﬁt these information. Nested canalyzing functions can be parameterized by an ideal, and intersecting the variety of this ideal with the variety of the ideal of all functions that ﬁt the data results in the desired set of models . 17.1.6.3 Linear polynomial dynamical systems 17.1.45 Remark For linear systems, i.e., all polynomials are linear functions, the dynamics of a system can be determined completely from the structure of the polynomials [975, 2812]. 17.1.6.4 Conjunctive/disjunctive networks 17.1.46 Remark Conjunctive (respectively disjunctive) Boolean network models consist of functions constructed using only the AND (respectively OR) operator. For conjunctive or disjunctive networks with strongly connected dependency graph, the cycle structure is completely de-termined by a formula that depends on the loop number, the greatest common divisor of the lengths of the dependency graph’s simple (no repeated vertices) directed cycles. For general Boolean conjunctive or disjunctive networks, an upper and lower bound for the number of cycles of a particular length can be calculated . See Also §2.1 For the Lagrange Interpolation Formula. §10.5 For dynamical systems over ﬁnite ﬁelds. References Cited: [74, 90, 343, 748, 866, 867, 868, 975, 976, 1355, 1502, 1503, 1504, 1505, 1583, 1597, 1598, 1599, 1714, 1715, 1859, 1860, 2000, 2215, 2431, 2512, 2514, 2515, 2617, 2704, 2812, 2864, 2887, 3010] 17.2 Finite ﬁelds in quantum information theory Martin Roetteler, NEC Laboratories America Arne Winterhof, Austrian Academy of Sciences Miscellaneous applications 835 In this chapter we mention some topics of the theory of ﬁnite ﬁelds related to quantum information theory. However, we will not give any background on quantum information theory and just refer to the monographs [1742, 2286, 2360]. For a more detailed treatment of quantum algorithms for algebraic problems we refer to the survey . 17.2.1 Mutually unbiased bases 17.2.1 Deﬁnition A maximal set of mutually unbiased bases, for short MUBs, is given by a set of n2 + n vectors in Cn which are the elements of n + 1 orthonormal bases of Cn: Bh = {wh,1, . . . , wh,n}, h = 0, . . . , n. Hence, ⟨wh,i, wh,j⟩= δi,j, where δi,j = 1, i = j, 0, i ̸= j, and the deﬁning property is the mutual unbiasedness, given by \|⟨wf,i, wg,j⟩\| = 1 √n (17.2.1) for 0 ≤f, g ≤n, f ̸= g, and 1 ≤i, j ≤n, where ⟨a, b⟩= Pn u=1 aubu denotes the standard inner product of two vectors a = (a1, . . . , an), b = (b1, . . . , bn) ∈Cn. 17.2.2 Remark Mutually unbiased bases were introduced by Schwinger . They have applica-tions in quantum state determination [1577, 3005], quantum cryptography , quantum error-correcting codes [475, 476, 1337], and the mean king’s problem . 17.2.3 Theorem [1743, 3005] Let n = pr be the power of a prime p > 2 and ψ be the additive canonical character of Fn = {ξ1, . . . , ξn}. Then wh,k = 1 √n ψ(ξhξ2 u + ξkξu) u=1,...,n , h, k = 1, . . . , n, and w0,j = (δj,u)n u=1 is a maximal set of MUBs. 17.2.4 Remark Maximal sets of n + 1 MUBs in dimension n are only known to exist in any dimension n = pr which is a power of a prime p. For n = 2r a construction based on Galois rings is given in . If we relax (17.2.1) to \|⟨wf,i, wg,j⟩\| = O n−1/2(log n)1/2 we can construct sets of n+1 orthonormal bases for any dimension n . Elliptic curves over ﬁnite ﬁelds can be used to construct sets of n + 1 orthonormal bases with \|⟨wf,i, wg,j⟩\| = O n−1/2 which applies to almost all dimensions n and under some widely believed conjectures about the gaps between primes to all n. 17.2.5 Theorem Let E be an elliptic curve over a ﬁnite ﬁeld Fp of prime order p > 3 with n points. For 2 ≤d ≤n −1 denote by Ad the set of bivariate polynomials over E of degree 836 Handbook of Finite Fields at most d with f(0, 0) = 0. Let X denote the character group of E. For f ∈Fp[E] we deﬁne the set Bf = {vf,χ : χ ∈X}, where for a character χ ∈X, the vector vf,χ is given by vf,χ = 1 √n(ψ(f(P))χ(P))P ∈E where ψ denotes the additive canonical character of Fp. For 2 ≤d ≤n −1 the standard basis and the pd−1 sets Bf = {vf,χ : χ ∈X}, with f ∈Ad, are orthonormal and satisfy \|⟨vf,χ, vg,ψ⟩\| ≤2d + (2d + 1)n−1/2 n1/2 , where f, g ∈Ad, f ̸= g, and χ, ψ ∈X. 17.2.6 Remark Maximal sets of MUBs and geometries over ﬁnite ﬁelds were used to deﬁne a discrete analog of quantum mechanical phase space and the corresponding notion of Wigner transform [1272, 1360]. The latter is a real valued function that uniquely characterizes a quantum state and allows one to compute measurement statistics by performing summation along lines of the underlying ﬁnite geometry. 17.2.2 Positive operator-valued measures 17.2.7 Deﬁnition For a positive integer n let A = {vi = (vi1, . . . , vin) : i = 1, . . . , n2} be a set of n2 vectors in Cn. The set of n2 × n2 matrices E = {Ei = (vijvik/n)n2 i,k=1 : i = 1, . . . , n2} is an approximately symmetric informationally complete positive operator-valued mea-sure (ASIC-POVM) if it satisﬁes the following conditions: 1. n2 P i=1 Ei = In (completeness/POVM condition), 2. the Ei are linearly independent as elements of Cn×n (informational complete-ness), 3. \|⟨vi, vj⟩\|2 ≤(1 + o(1))n−1, 1 ≤i < j ≤n2 (approximate symmetry). 17.2.8 Remark If E satisﬁes instead of Condition 3 the stronger condition \|⟨vi, vj⟩\|2 = 1 n + 1, 1 ≤i < j ≤n2, it is a SIC-POVM. SIC-POVMs have several very desirable properties, see for a discussion in the con-text of the quantum de Finetti theorem and more generally in their Bayesian approach to quantum mechanics and its interpretation . Furthermore, see [1142, 1143] for their role in establishing the quantumness of a Hilbert space and the related question about optimal intercept-resend eavesdropping attacks on quantum cryptographic schemes. Explicit ana-lytical constructions of SIC-POVMs have been given for dimensions n = 2, 3, 4, 5, 6, 7, 8, 19, see and references therein. While it has been conjectured that SIC-POVMs exist in all Miscellaneous applications 837 dimensions and numerical evidence exists for dimensions up to 45, see , it is a diﬃcult task to explicitly construct SIC-POVMs. There are no known inﬁnite families and in fact it is not even clear if there are SIC-POVMs for inﬁnitely many n. However, ASIC-POVMS can be constructed for any power of an odd prime. 17.2.9 Theorem Let q be a power of a prime p ≥3 and ψ denote the additive canonical character of Fq. Let va,b = q−1/2 ψ(ax2 + bx)) x∈Fq ∈Cq for all (a, b) ∈Fq × F∗ q and va,0 = (δa,x)x∈Fq for all a ∈Fq. We deﬁne Ea,b = (va,b,xva,b,y)x,y∈Fq and G = X a∈Fq X b∈Fq Ea,b. Then the set {Fa,b : a, b ∈Fq}, with Fa,b = G−1/2Ea,bG−1/2, is an ASIC-POVM. 17.2.3 Quantum error-correcting codes 17.2.10 Remark Recall that for a matrix A ∈Cd×d the Hermitian conjugate is deﬁned as A† = (At)∗, where At denotes transposition and ∗denotes entry-wise complex conjugation. For two matrices A, B we denote by A ⊗B their (Kronecker) tensor product. 17.2.11 Deﬁnition (QECC characterization ) Let C ≤Cd be a subspace and let E ⊆Cd×d be a set of error operators. Then C is a quantum error-correcting code for E if the projector PC onto the code space C satisﬁes the identities PCE†FPC = αE,F PC for all E, F ∈E and for some αE,F ∈C. In this case C can detect all errors in E. 17.2.12 Remark We next give a brief description of the stabilizer formalism which allows connecting the problem of ﬁnding quantum error-correcting codes to the problem of ﬁnding isotropic subspaces over ﬁnite ﬁelds with respect to a certain symplectic inner product. 17.2.13 Deﬁnition Let Fq be the ﬁnite ﬁeld of order q = pr where p is prime. Denote by ψ an additive character of Fq. For α ∈Fq denote by eα the corresponding standard basis vector in Cq, i.e., the vector that has 1 in position α and is 0 elsewhere. We deﬁne the following unitary error operators: Xα := P x∈Fq ex+αex t for α ∈Fq and Zβ := P z∈Fq ψ(βz)ezez t for β ∈Fq. 17.2.14 Deﬁnition For each γ ∈Fq, α = (α1, α2, . . . , αn) ∈Fn q , and β = (β1, β2, . . . , βn) ∈Fn q we deﬁne the corresponding n qudit error operator Eγ,α,β = ψ(γ)(Xα1Zβ1) ⊗(Xα2Zβ2) ⊗. . . ⊗(XαnZβn). The weight of Eγ,α,β is the number of indices i for which not both αi and βi are zero. The group Gn of all Eγ,α,β has size pq2n and is the Pauli group. 17.2.15 Lemma For (α, β), (α′, β′) ∈Fn q × Fn q we have that (XαZβ)(Xα′Zβ′) = ψ((α, β) ∗(α′, β′))(Xα′Zβ′)(XαZβ), where the symplectic inner product ∗is deﬁned by (α, β) ∗(α′, β′) := n X i=1 (α′ iβi −αiβ′ i). (17.2.2) 838 Handbook of Finite Fields 17.2.16 Deﬁnition [140, 477, 1337] A stabilizer code is a quantum code C with parameters that is the joint eigenspace of a set of n −k commuting Pauli operators S1, . . . , Sn−k, where 0 ≤k ≤n. The distance d is the weight of the smallest error that cannot be detected by the code. 17.2.17 Remark A stabilizer code corresponds to an additively closed subset C of Fn q × Fn q that is contained in its dual C∗:= {v: v ∈Fn q \| for all c ∈C : TrFq/Fp(v ∗c) = 0}. 17.2.18 Remark In order to characterize the error-correcting properties of the code, it is useful to introduce the normalizer N of S in the Pauli group Gn of n qudits. The elements of the normalizer can be seen as encoded logical operations on the code. On the other hand, they also correspond to undetectable errors. 17.2.19 Deﬁnition Let S be an Abelian subgroup of Gn which has trivial intersection with the center of Gn. Furthermore, let {g1, g2, . . . , gℓ} where gi = ψ(γi)XαiZβi with γi ∈{0, . . . , p −1} and (αi, βi) ∈Fn q × Fn q be a minimal set of generators for S. Then a stabilizer matrix of the corresponding stabilizer code C is a generator matrix of the (clas-sical) additive code C ⊆F n q × F n q generated by (αi, βi). The corresponding stabilizer matrix is deﬁned as    α1 β1 . . . . . . αℓ βℓ   ∈Fℓ×2n q . 17.2.20 Remark A special class of stabilizer codes are CSS codes [478, 2702]. These codes are obtained from a pair of classical linear codes C1 = [n, k1, d1]q and C2 = [n, k2, d2]q over Fq. The codes have to satisfy the condition that C⊥ 2 ⊆C1, where C⊥ 2 denotes the dual code of C2 with respect to the standard inner product on Fn q . The basis states of the code space C are deﬁned as cw := 1 √ \|C⊥ 2 \| P c∈C⊥ 2 ec+w, where w ∈C1. Two states cw and cw′ are identical if and only if w −w′ ∈C⊥ 2 and otherwise they are orthogonal. The dimension of the code space C is qk, where k = k1 + k2 −n and the minimum distance is d ≥min(d1, d2). 17.2.21 Remark It can be shown [1338, 1353] that any stabilizer code can be encoded eﬃciently by using quantum operations from the Cliﬀord group. A generating set for this group is P y∈Fq eγyey t for γ ∈Fq{0}, 1 √q P x,z∈Fq ψ(xz)ezex t, and P x,y∈Fq exex t⊗ex+yey t, where these gates can be applied to any (pair) of the n qudits. 17.2.22 Remark Several constructions of families of quantum error-correcting codes based on ﬁnite ﬁelds are known. Starting from classical Reed-Solomon codes (see Section 15.1) over F2n, in a construction of codes with parameters was given, where K = 2n −δ and δ > (2n −1)/2 + 1 is the designed distance of the classical Reed-Solomon code [2n −1, K, δ] which is underlying the construction. For a construction of nonbinary stabilizer codes using arbitrary ﬁnite ﬁelds see . 17.2.23 Remark Various bounds on the parameters of quantum error-correcting codes are known, see . Similar to the classical Singleton bound, a quantum Singleton bound of k + 2d ≤n + 2 for any code with parameters can be shown. Codes meeting this bound with equality are quantum MDS codes and several constructions based on ﬁnite ﬁelds Fq for large q are known . 17.2.24 Remark Classical Goppa codes have been used to construct quantum error-correcting codes. A construction in is based on towers of Garcia-Stichtenoth function ﬁelds (see Section 12.6) Fi = Fq2(x0, z1, . . . , xi, zi+1), deﬁned by equations zq i +zi−xq+1 i−1 = 0 and xi = zi/xi−1, Miscellaneous applications 839 i = 0, 1, . . . These quantum codes have been shown to be asymptotically good, i.e., their parameters satisfy lim i→∞ni →∞, lim inf i→∞ki/ni > 0, and lim inf i→∞di/ni > 0, see also . 17.2.25 Remark The webpage provides tables of the best known quan-tum error-correcting codes for small parameters. The codes are speciﬁed by their stabilizer matrices and where applicable it is noted that a code is optimal, i.e., achieving the highest possible d for ﬁxed n and k or the highest possible k for ﬁxed n and d. The table also contains known bounds on the parameters and contains information about the construction of the codes. Some of these tables and constructions have also been made available in the computer algebra system Magma. 17.2.26 Remark For a more detailed survey on quantum error-correcting codes see . 17.2.4 Period ﬁnding 17.2.27 Theorem Let (en) be a periodic sequence over Fp of (unknown) period T. If the mapping n 7→en, 0 ≤n < T, is injective, T can be recovered on a quantum computer in polynomial time. 17.2.28 Remark No classical polynomial time algorithm is known for period ﬁnding. If g ∈F∗ p is an element of (unknown) order T, Shor’s algorithm determines T in quantum polynomial time. Let (fn) be another sequence over Fp of period T such that n 7→fn is injective and the (unknown) pair of positive integers (t1, t2) satisﬁes en+t1fn+t2 = enfn for all n ≥0. Then Shor’s algorithm also determines (t1, t2). Let a, b ∈F∗ p be such that the order of b is t and a = bx with 0 ≤x < t. Then x is the discrete logarithm of a to the base b. If we choose en = bn and fn = a−n, we get en+xfn+1 = bn+xa−n−1 = enfn for all n and we have (t1, t2) = (x, 1). Hence, Shor’s algorithm ﬁnds the discrete logarithm in quantum polynomial time. However, Shor’s algorithm does not provide the period of (en) if n 7→en is not injective. 17.2.29 Theorem Given two periodic sequences (en) and (fn) with periods T and t, respec-tively, we denote by D(en, fn) the number of integers n ∈[0, Tt −1] with en ̸= fn. For any constant c > 0, there is a quantum algorithm which computes in polynomial time, with probability at least 3/4, the period of any sequence (en) of period T satisfying D(en, fn) ≥ Tt (log T)c , (17.2.3) for any sequence (fn) of period t < T. 17.2.30 Remark In , for binary sequences (en) with moderately small autocorrelation (which includes all cryptographically interesting sequences) it was proved that (17.2.3) is fulﬁlled and the algorithm of Hales and Hallgren determines their period in quantum poly-nomial time. In a more general problem was studied of determining the period of a sequence over an unknown ﬁnite ﬁeld Fp, given a black-box which returns only a few most signiﬁcant bits of the sequence elements. A moderately small autocorrelation is again a suﬃcient condition such that the condition (17.2.3) is satisﬁed. 840 Handbook of Finite Fields 17.2.5 Quantum function reconstruction 17.2.31 Theorem Any monic, square-free polynomial f ∈Fp[X] of degree d can be recon-structed from an oracle Of : Fp →{−1, 0, 1} given by the Legendre symbol Of(a) = f(a) p of f(a) after O(d) quantum-queries with probability 1 + O(p−1). 17.2.32 Remark The case f(X) = X + s with an unknown s ∈Fp is a special case of the hidden shift problem. In this case in an eﬃcient quantum algorithm was presented that ﬁnds s with one query to the oracle Of. This algorithm uses the fact that the Legendre symbol is, essentially, equal to its discrete Fourier transform. 17.2.33 Remark The hidden shift problem has also been studied for functions other than the Leg-endre symbol. In the hidden shift problem for Boolean functions was studied and it was shown that for any bent function (see Section 9.3) B : Fn 2 →F2 the hidden shift s can be found by making O(n) queries to OB(x) = (−1)B(x). If furthermore an eﬃcient circuit is known to implement the dual bent function B∗, then this can be reduced to a single query. These results were recently extended to the case of random Boolean functions where the shift s can be determined from solving a system of linear equations over F2 that is obtained from a sampling procedure and to the case of functions that are close to quadratic bent functions, where closeness is measured using the Gowers U3 norm . 17.2.6 Further connections 17.2.34 Remark Some additive character sums over a ﬁnite ﬁeld (twisted Kloosterman sums) are closely connected to quantum algorithms for ﬁnding hidden nonlinear structures over ﬁnite ﬁelds . 17.2.35 Remark Schumacher and Westmoreland investigate a discrete version of quantum mechanics called modal quantum computing based on a ﬁnite ﬁeld instead of the complex numbers. Some characteristics of actual quantum mechanics are retained including the no-tions of superposition, interference, and entanglement. 17.2.36 Remark Exponential sums over ﬁnite ﬁelds are used to construct quantum ﬁnite automata . 17.2.37 Remark Classical and quantum algorithms for solvability testing and ﬁnding integer so-lutions x, y of equations of the form agx + bhy = c over a ﬁnite ﬁeld Fq are studied in . 17.2.38 Remark Given a group G, a subgroup H ≤G, and a set X, a function f : G 7→X hides the subgroup H if for all g1, g2 ∈G, f(g1) = f(g2) if and only if g1H = g2H for the cosets of H. The function f is given via an oracle. The hidden subgroup problem is the problem of ﬁnding H using information gained from evaluations of f via its oracle. In the case of an Abelian group G an eﬃcient quantum algorithm is known for solving the hidden subgroup problem. In particular, Shor’s algorithm relies on this fact. In the authors showed that the hidden subgroup problem can be also eﬃciently solved by a quantum computer in the case of the semi-direct product of the additive groups of Fp1 and Fp2 where p1 and p2 are primes with p2\|(p1 −1) and (p1 −1)/p2 is polynomial in log p2. Another related problem involving ﬁnite ﬁelds is the hidden polynomial function graph polynomial studied in . Miscellaneous applications 841 See Also §6.1, §6.3 For basics on character sums and applications. §9.1 For Boolean functions. §10.3, §17.3 For autocorrelation of sequences. §11.6, §16.1.3.2 For the discrete logarithm problem and public-key cryptography. §12.2, §12.3, §16.4 For elliptic curves. §15.1, §15.2 For classical and algebraic codes. References Cited: [88, 140, 475, 476, 477, 478, 567, 568, 601, 618, 619, 791, 981, 1054, 1142, 1143, 1260, 1272, 1337, 1338, 1351, 1352, 1353, 1360, 1400, 1577, 1729, 1742, 1743, 1744, 1759, 2027, 2138, 2286, 2360, 2397, 2401, 2450, 2487, 2488, 2509, 2562, 2572, 2623, 2656, 2657, 2702, 2840, 2841, 3005] 17.3 Finite ﬁelds in engineering Jonathan Jedwab, Simon Fraser University Kai-Uwe Schmidt, Otto-von-Guericke University 17.3.1 Binary sequences with small aperiodic autocorrelation 17.3.1 Deﬁnition A sequence of length n over an alphabet A is an n-tuple (aj) = (a0, a1, . . . , an−1), where each aj is an element of the set A. 17.3.2 Deﬁnition Let A = (aj) be a sequence over C of length n. For integer u satisfying 0 ≤ u < n, the aperiodic autocorrelation of A at shift u is Cu(A) = Pn−u−1 j=0 ajaj+u. 17.3.3 Remark The most important case, from a practical and historical viewpoint, occurs when the sequence is binary, which means that its alphabet is {1, −1}. A binary sequence for which all aperiodic autocorrelations at nonzero shifts are small in magnitude relative to the sequence length is intrinsically suited for the separation of signals from noise, and therefore has natural applications in digital communications, including radar, synchronization, and steganography. 17.3.4 Remark The overall goal is to ﬁnd binary sequences A having the property that, for each u ̸= 0 independently, \|Cu(A)\| takes its smallest possible value. An ideal binary sequence A from this viewpoint therefore satisﬁes \|Cu(A)\| = 0 or 1 for each u ̸= 0, which is known as a Barker sequence. The longest Barker sequence currently known has length 13 and there is overwhelming evidence that no longer Barker sequence exists (see for a survey). 17.3.5 Deﬁnition Write F = F2 and E = F2m. An m-sequence (yj) of length n = 2m −1 satisﬁes yj = (−1)Tr E/F (cαj) for 0 ≤j < n and for some primitive element α in E and some nonzero c in E. 842 Handbook of Finite Fields 17.3.6 Theorem Every m-sequence Y of length n = 2m −1 satisﬁes \|Cu(Y )\| < 1 + (2/π)√n + 1 log 4n/π for each u satisfying 0 < u < n. 17.3.7 Remark The above theorem shows the existence of an inﬁnite family of binary sequences for which the magnitude of the aperiodic autocorrelation (at nonzero shifts) grows no faster than order √n log n. The only known improvement of this result uses probabilistic methods and guarantees a growth rate of at most order √n log n . 17.3.8 Remark For the following deﬁnition we need quadratic characters, see Section 6.1. 17.3.9 Deﬁnition For an odd prime p, deﬁne λ : Fp →{1, −1} to be the quadratic character on F∗ p and λ(0) = 1. For real r, the r-shifted Legendre sequence of length p is the sequence (xj) of length p satisfying xj = λ(j + ⌊rp⌋) for 0 ≤j < p. 17.3.10 Theorem For all real r, the r-shifted Legendre sequence X of length p satisﬁes \|Cu(X)\| < 1 + 18√p log p for each u satisfying 0 < u < n. 17.3.11 Deﬁnition The merit factor of a binary sequence A of length n > 1 is F(A) = n2/(2 Pn−1 u=1[Cu(A)]2). 17.3.12 Theorem Let Yn be an m-sequence of length n = 2m−1. Then F(Yn) →3 as n →∞. 17.3.13 Theorem For real r satisfying \|r\| ≤1/2, let Xp be the r-shifted Legendre sequence of length p. Then 1/F(Xp) →1/6 + 8 \|r\| −1/4 2 as p →∞. 17.3.14 Remark The largest asymptotic merit factor occurring in the above theorem is 6. However, by modifying the construction as shown below, an asymptotic merit factor of approxi-mately 6.34 can be achieved, which is the largest known asymptotic merit factor for binary sequences. 17.3.15 Theorem [1604, 1605] Let t = 1.0578 . . . be the middle root of 4x3 −30x + 27 and write r = 3/4 −t/2. Let (xj) be the r-shifted Legendre sequence of length p. Deﬁne Wp to be the sequence (wj) of length ⌊tp⌋given by wj = xj for 0 ≤j < p and wj = xj−p for p ≤j < ⌊tp⌋. Then F(Wp) →6.3420 . . . , which is the largest root of 29x3 −249x2 + 417x −27. 17.3.2 Sequence sets with small aperiodic auto- and crosscorrelation 17.3.16 Deﬁnition Let A = (aj) and B = (bj) be sequences over C of length n. For integer u satisfying 0 ≤u < n, the aperiodic crosscorrelation of A and B at shift u is Cu(A, B) = Pn−u−1 j=0 ajbj+u and the periodic crosscorrelation of A and B at shift u is Ru(A, B) = Pn−1 j=0 ajbj+u, where indices are reduced modulo n. 17.3.17 Deﬁnition A set S of sequences has maximum aperiodic correlation θ if \|Cu(A, B)\| ≤θ for all A, B ∈S when either A ̸= B or u ̸= 0. 17.3.18 Remark Code-division multiple access (CDMA) is a technique that allows multiple users to communicate over the same medium. For example, direct-sequence CDMA employs a set S of sequences over C of the same length. Each user is assigned a sequence of S, and encodes information by sending a modulated version of the assigned sequence in which each sequence element is multiplied by a complex number drawn from some alphabet. In order to Miscellaneous applications 843 allow synchronization at the receiver and to minimize interference between diﬀerent users, it is necessary to minimize the maximum aperiodic correlation of S. 17.3.19 Remark It is a notoriously diﬃcult problem to design sequence sets with small maximum aperiodic correlation directly. The usual approach is therefore to design sequence sets that have good periodic crosscorrelation properties (see Section 10.3 for constructions using ﬁnite ﬁelds), and then analyze their aperiodic crosscorrelation properties using either separate methods or numerical computation. 17.3.3 Binary Golay sequence pairs 17.3.20 Deﬁnition A binary Golay sequence pair is a pair of binary sequences A, B of equal length n whose aperiodic autocorrelations satisfy Cu(A) + Cu(B) = 0 for 0 < u < n. A binary Golay sequence is a member of a binary Golay sequence pair. 17.3.21 Remark Binary Golay sequence pairs were introduced to solve a problem in infrared mul-tislit spectrometry , and have since been used in many other digital information processing applications such as optical time domain reﬂectometry and medical ul-trasound . The deﬁning aperiodic autocorrelation property can be exploited to allow very eﬃcient energy use when transmitting information, or to remove unwanted components from received signals. 17.3.22 Theorem If there exists a binary Golay sequence pair of length n and p is an odd prime factor of n, then −1 is a square in Fp and so p ≡1 (mod 4). 17.3.23 Theorem If there exist binary Golay sequence pairs of length n1 and n2, then there exists a binary Golay sequence pair of length n1n2. 17.3.24 Remark Starting from “seed” binary Golay sequence pairs of length 2, 10, and 26, the above theorem produces a binary Golay sequence pair of length 2k10ℓ26m for all non-negative integers k, ℓ, m. Once it is known that a binary Golay sequence of a particular length exists, it is then important in some applications to ﬁnd as many such sequences of this length as possible. 17.3.25 Deﬁnition Let A = ((−1)aj) be a binary sequence of length 2m. The algebraic normal form of A is the unique function fA(x1, . . . , xm) : Fm 2 →F2 satisfying fA(x1, . . . , xm) = axm+2xm−1+···+2m−1x1 for all (x1, . . . , xm) ∈Fm 2 , where indices are calculated in Z. 17.3.26 Theorem Let π be a permutation of {1, . . . , m} and let e′ 0, e0, e1, . . . , em ∈F2. The binary sequences A and B of length 2m having algebraic normal form fA(x1, . . . , xm) = Pm−1 k=1 xπ(k)xπ(k+1) + Pm k=1 ekxπ(k) + e0 and fB(x1, . . . , xm) = Pm−1 k=1 xπ(k)xπ(k+1) + Pm k=1 ekxπ(k) + e′ 0 + xπ(1) form a binary Golay sequence pair. 17.3.27 Example Take m = 3, (π(1), π(2), π(3)) = (2, 1, 3), and (e′ 0, e0, e1, e2, e3) = (0, 1, 1, 0, 1). Then fA(x1, x2, x3) = x2x1 + x1x3 + x2 + x3 + 1 is the algebraic normal form of A = (−+ + −−−−−) (writing + for 1, and −for −1), and fB(x1, x2, x3) = x2x1 + x1x3 + x3 is the algebraic normal form of B = (+ −+ −+ + −−). The sequences A and B form a binary Golay sequence pair of length 8. 17.3.28 Corollary For m > 1, there are at least 2mm! binary Golay sequences of length 2m. 844 Handbook of Finite Fields 17.3.29 Remark The 2mm! binary Golay sequences arising from the above theorem form m!/2 cosets of the ﬁrst-order Reed-Muller code RM(1, m) in the second-order Reed-Muller code RM(2, m); see Section 15.1. When these sequences are used in multicarrier transmission, the Golay property tightly controls variations in the transmitted power while the code structure allows powerful error correction . 17.3.4 Optical orthogonal codes 17.3.30 Deﬁnition An (n, w, λ) optical orthogonal code is a set C of sequences over {0, 1} of length n and Hamming weight w such that the periodic crosscorrelation satisﬁes Ru(A, B) ≤λ for all A, B ∈C when either A ̸= B or u ̸= 0. 17.3.31 Remark Optical orthogonal codes are used in optical CDMA. Each user is assigned a sequence of the code, and a 1 in the sequence corresponds to a time instant when the user is allowed to transmit a light pulse. The correlation constraint enables self-synchronization of the system and controls the interference between diﬀerent users. 17.3.32 Deﬁnition An (n, w, λ) optical orthogonal code of size M is optimal if there is no (n, w, λ) optical orthogonal code of size greater than M. 17.3.33 Remark Optical orthogonal codes are closely related to other combinatorial objects such as constant-weight codes and cyclic diﬀerence families . There are numerous construc-tions of optimal optical orthogonal codes. Two important general constructions using ﬁnite geometries are given here. 17.3.34 Construction Write F = Fq and E = Fqm, and let α be primitive in E. The points in the aﬃne geometry AG(m, q) are the elements of E; see Section 14.3. A d-ﬂat in AG(m, q) is a translate of a d-dimensional subspace of E over F. Then the lines in AG(m, q) are precisely the 1-ﬂats. Two d-ﬂats A and B are equivalent if there exists a ∈E∗such that A = {ax : x ∈B}. This partitions the d-ﬂats into orbits. Let C be a set containing exactly one representative from each orbit of a d-ﬂat that does not contain 0. With a d-ﬂat S ∈C we associate a sequence over {0, 1} of length qm −1 by placing a 1 at position i precisely when αi ∈S. 17.3.35 Remark The q-binomial coeﬃcient m k q = Qk i=1 qm−i+1−1 qi−1 equals the number of k-dimensional subspaces of Fqm over Fq; see Section 13.2. 17.3.36 Theorem The above code is a (qm −1, qd, qd−1) optical orthogonal code of size m−1 d q. If d = 1 or d = m −1, the code is optimal. 17.3.37 Example Take q = 2, m = 3, d = 1, and let α be primitive in F8. There are 21 lines in AG(3, 2) that do not contain 0. These are of the form {x, y}, where x, y ∈ F∗ 8 and x ̸= y. A list of representatives of the orbits is {1, α}, {1, α2}, {1, α3} and {(1100000), (1010000), (1001000)} is an optimal (7, 2, 1) optical orthogonal code. 17.3.38 Construction Write F = Fq, E = Fqm+1, and n = qm+1−1 q−1 . The points in the projective geometry PG(m, q) are the elements of E∗/F ∗(which is isomorphic to Z/nZ); see Section 14.4. Let α be primitive in E and let [a] denote the coset of F ∗in E∗containing a. A d-space in PG(m, q) is the image under the mapping x 7→[x] of the nonzero elements of a (d + 1)-dimensional subspace of E over F. Then the lines in PG(m, q) are precisely the 1-spaces. Two d-spaces A and B are equivalent if there exists a ∈E∗/F ∗such that Miscellaneous applications 845 A = {ax : x ∈B}. This partitions the d-spaces into orbits, whose sizes divide n. Let C be a set containing exactly one representative from each orbit of size exactly n. With a d-space S ∈C we associate a sequence over {0, 1} of length n by placing a 1 at position i precisely when [αi] ∈S. 17.3.39 Theorem The above code is a qm+1−1 q−1 , qd+1−1 q−1 , qd−1 q−1 optical orthogonal code. For d = 1, the code is optimal and has size qm−1 q2−1 for even m and qm−q q2−1 for odd m. 17.3.40 Example Take q = 2, m = 3, d = 1, and let α be the primitive element in F16 that satisﬁes α4 = α + 1. Then the 2-dimensional subspaces {0, 1, α, α4}, {0, 1, α2, α8}, {0, 1, α5, α10} of F16 map to the lines {, [α], [α4]}, {, [α2], [α8]}, {, [α5], [α10]} in PG(3, 2). Their orbits have size 15, 15, and 5, respectively, and exhaust all 35 lines in PG(3, 2). Hence, {(110010000000000), (101000001000000)} is an optimal (15, 3, 1) optical orthogonal code. 17.3.5 Sequences with small Hamming correlation 17.3.41 Deﬁnition Let A = (aj) and B = (bj) be sequences of length n. For integer u satisfying 0 ≤u < n, the Hamming correlation between A and B at shift u is Hu(A, B) = Pn−1 j=0 h(aj, bj+u), where h(x, y) = 1 if x = y and h(x, y) = 0 otherwise and indices are reduced modulo n. 17.3.42 Remark In frequency-hopping spread spectrum communications, several frequencies are shared by diﬀerent users. Let q be the number of such frequencies and let M be the number of users. The system employs a family of M “frequency-hopping” sequences of the same length, each over the same alphabet of size q. Each user is assigned a sequence of the family and switches between the q frequencies according to the elements of this sequence. In order to minimize interference between diﬀerent users, as well as to support self-synchronization of the system, the Hamming correlation between every two sequences in the family should be as small as possible. 17.3.43 Deﬁnition Let S be the set of all sequences of length n over some ﬁxed alphabet, and write H(A) = max 0<u<n Hu(A, A), H(A, B) = max 0≤u<n Hu(A, B), M(A, B) = max{H(A), H(B), H(A, B)}. 1. X ∈S is optimal if H(X) ≤H(X′) for all X′ ∈S. 2. X, Y ∈S is an optimal pair if M(X, Y ) ≤M(X′, Y ′) for all X′, Y ′ ∈S. 3. A subset F ⊂S is an optimal family if every pair of distinct sequences in F is an optimal pair. 17.3.44 Remark There are numerous constructions of optimal families. Two constructions based on ﬁnite ﬁelds are given here. 17.3.45 Construction Let k and m be integers satisfying 1 ≤k ≤m. Let α be primitive in Fpm and let β be primitive in Fpk. Write F = Fp and E = Fpm. For v ∈Fpk, deﬁne the sequence Xv = (xj) of length pm −1 over Fpk by xj = v + Pk−1 ℓ=0 Tr E/F (αj+ℓ) βℓfor 0 ≤j < pm −1. Call {Xv : v ∈Fpk} the Lempel-Greenberger family. 846 Handbook of Finite Fields 17.3.46 Remark The sequence X0 is obtained by mapping k consecutive elements of an m-sequence over Fp (as deﬁned in Section 10.3) to an element in Fpk. The other sequences are then translates of X0. 17.3.47 Theorem Each sequence in the Lempel-Greenberger family F is optimal and F is an optimal family. Speciﬁcally, H(X) = pn−k −1 and H(X, Y ) = pn−k for all distinct X, Y ∈F. 17.3.48 Construction Let p = em + 1 be an odd prime and let α be primitive in Fp. For integer i, deﬁne the i-th cyclotomic class in Fp to be Ci = {αte+i : 0 ≤t < m}. For integer v, deﬁne the sequence Xv = (xj) of length p over {∞, 0, 1, . . . , e −1} by x0 = ∞and xj = i when j ∈Cv+i for 0 ≤j < p. Call {Xv : 0 ≤v < e} the Chu-Colbourn family. 17.3.49 Theorem The Chu-Colbourn family F satisﬁes H(X) = m −1 and H(X, Y ) = m for all distinct X, Y ∈F. When e ≥3m and m ≥2, then each sequence in F is optimal and F is an optimal family. 17.3.50 Example Take e = 3, m = 2, so that p = 7. For α = 3, we have C0 = {1, 6}, C1 = {3, 4}, C2 = {2, 5}, giving X0 = (∞021120), X1 = (∞210012), X2 = (∞102201). 17.3.6 Rank distance codes 17.3.51 Deﬁnition An (m, n, d) rank distance code over Fq is a subset C of Fm×n q (the set of m×n matrices over Fq) such that the rank of X −Y is at least d for all distinct X, Y ∈C. Without loss of generality, it is assumed that m ≤n. 17.3.52 Remark Rank distance codes were introduced independently in [803, 1149, 2485]. Such codes ﬁnd applications in correcting crisscross errors in arrays, for example in memory chip arrays or magnetic tape recording . Further applications are given in the following two subsections. Motivated by these applications, the goal is to ﬁnd (m, n, d) rank distance codes that have as many elements as possible. 17.3.53 Theorem The size of an (m, n, d) rank distance code over Fq is at most qn(m−d+1). 17.3.54 Deﬁnition An (m, n, d) rank distance code over Fq of size qn(m−d+1) is a maximum rank distance code. 17.3.55 Deﬁnition The rank distribution of a rank distance code C is (ai), where ai is the number of elements in C of rank i, and its distance distribution is (bi), where bi = \|{(X, Y ) ∈ C × C : rank(X −Y ) = i}\|/\|C\|. 17.3.56 Theorem The rank distribution (ai) and the distance distribution (bi) of an (m, n, d) maximum rank distance code over Fq satisfy a0 = b0 = 1 and ai = bi = m i q i−d X j=0 (−1)jq( j 2) i j q (qn(i−j−d+1) −1) for i ∈{1, 2, . . . , m}, where m k q is the q-binomial coeﬃcient given in Remark 17.3.35. 17.3.57 Theorem Let m, n, d be positive integers satisfying d ≤m ≤n. Write F = Fq and E = Fqn, and let α be primitive in E. For λ = (λ0, λ1, . . . , λm−d) ∈Em−d+1, let Xλ = (xij) be the m × n matrix given by xij = Tr E/F (αj Pm−d k=0 λkαiqk) for i ∈{1, 2, . . . , m} and Miscellaneous applications 847 j ∈{1, 2, . . . , n}. Then {Xλ : λ ∈Em−d+1} is an (m, n, d) maximum rank distance code over Fq. 17.3.7 Space-time coding 17.3.58 Deﬁnition An (m, n, d) space-time code S over a (ﬁnite) alphabet A ⊂C is a set of m × n matrices with entries in A such that, for all distinct A, B ∈S, the rank of A −B is at least d, which is the diversity of S. Without loss of generality, it is assumed that m ≤n. 17.3.59 Remark Space-time codes are used in wireless digital communications to transmit a block of n data symbols over m transmit antennas. The motivation is that, with careful code design, the transmitter introduces spatial diversity and so can prevent signal loss caused by shadowing . The usual goal is to design space-time codes whose diversity equals the number of transmit antennas m. 17.3.60 Theorem The size of an (m, n, d) space-time code over an alphabet of size q is at most qn(m−d+1). 17.3.61 Deﬁnition A space-time code that achieves equality above is optimal. 17.3.62 Remark There are diﬀerent approaches for constructing space-time codes. Algebraic con-structions based on rank distance codes over ﬁnite ﬁelds are presented in the following. The principal diﬃculty in this approach is to ﬁnd some “rank-preserving” mapping from a ﬁnite ﬁeld to a ﬁnite subset of C. 17.3.63 Theorem Let C be an (m, n, d) maximum rank distance code over F2, where we interpret the elements of C to be in {0, 1}, as a subset of Z. Let h and ℓbe positive integers. Write θ = e2π√−1/2h, let η be a nonzero element in 2Z[θ], and let κ be a nonzero element in C. Deﬁne S = ( κ ℓ−1 X u=0 ηuθ Ph−1 v=0 2vXu,v : Xu,v ∈C ) , where the exponentiation of matrices is understood to be element-wise. Then S is an optimal (m, n, d) space-time code over an alphabet of size 2hℓ. 17.3.64 Remark The above construction admits space-time codes over alphabets commonly used in digital communications. When h = 2, κ = 1 + √−1, and η = 2, the resulting alphabet is known as 4ℓ-QAM. When ℓ= 1 and κ = 1, and η > 0, the resulting alphabet is known as 2h-PSK. 17.3.65 Remark Let K be a number ﬁeld (a ﬁnite extension of the ﬁeld of rationals Q), let R be the ring of algebraic integers in K, and let p be a prime number. The ideal pR factors uniquely into prime ideals over R. If P is such a factor, then R/P is a ﬁnite ﬁeld of size pk, where k is the inertial degree of P over pZ (see for background on number ﬁelds and prime ideal factorization). This is usually applied with K = Q, in which case R = Z and Z/pZ is a ﬁnite ﬁeld of size p. 17.3.66 Deﬁnition Let K be a number ﬁeld and R the ring of algebraic integers in K. Suppose that R contains a prime ideal P such that R/P is isomorphic to Fq. Let A ⊂C be a set containing exactly one representative from each of the q elements of R/P. Deﬁne φ : Fq →A to be the mapping induced by the isomorphism from Fq to R/P, and extend φ to act element-wise on matrices over Fq. 848 Handbook of Finite Fields 17.3.67 Theorem For each matrix X over Fq, we have rank(φ(X)) ≥rank(X). 17.3.68 Corollary Let C be an (m, n, d) maximum rank distance code over Fq. Then φ(C) is an optimal (m, n, d) space-time code over A. 17.3.69 Remark The set A is not uniquely deﬁned for a given K, but from a practical viewpoint it is advantageous to choose A such that the energy P a∈A \|a\|2 is minimized. 17.3.70 Example Write i = √−1 and take K = Q[i], so that R = Z[i]. Denote the ideal aR by (a) and let p be a prime satisfying p ≡1 (mod 4). Then (p) = (π)(π), where π = a + bi for some a, b ∈Z satisfying a2 + b2 = p, and R/(π) is isomorphic to Fp. Identify Fp with Z/pZ and deﬁne φ : Fp →C by φ(x + pZ) = x −[ x π]π, where [ · ] rounds to the nearest element (in the Euclidean metric) in Z[i]. Take A to be the image of φ. Then A minimizes the energy. Speciﬁcally for p = 5, we have (5) = (2 + i)(2 −i) and A = {0, 1, −1, i, −i}. 17.3.8 Coding over networks 17.3.71 Deﬁnition Let P(Fm q ) be the set of subspaces of Fm q over Fq. The subspace distance dS on P(Fm q ) is deﬁned by dS(U, V ) = dim(U) + dim(V ) −2 dim(U ∩V ). 17.3.72 Deﬁnition Let Pk(Fm q ) be the set of k-dimensional subspaces of Fm q over Fq. An (m, k, 2d) constant-dimension code Ωover Fq is a subset of Pk(Fm q ) such that dS(U, V ) ≥ 2d for all distinct U, V ∈Ω. 17.3.73 Remark Random linear network coding is a technique for communicating eﬃciently over networks. The transmitting node injects a basis for a k-dimensional subspace of Fm q into the network. Each intermediate node relays a randomly chosen linear combination of its incoming vectors. The receiving node waits until enough linearly independent vectors are received and then tries to reconstruct the transmitted subspace. If the transmitted subspaces are taken from an (m, k, 2d) constant-dimension code with d > 1, then the receiver can reconstruct the transmitted space even if there are (a limited number of) lost or erroneously inserted vectors. Speciﬁcally, a successful reconstruction is always possible if the transmitted subspace U and the received subspace V satisfy dS(U, V ) < d . 17.3.74 Deﬁnition The mapping Λ : Fk×(m−k) q →Pk(Fm q ) takes X ∈Fk×(m−k) q to the rowspace of [I X], where I is the identity matrix of order k. 17.3.75 Theorem We have dS(Λ(X), Λ(Y )) = 2 rank(X −Y ) for all X, Y ∈Fk×(m−k) q . 17.3.76 Corollary Let C be a (k, m −k, d) rank distance code over Fq. Then Λ(C) is an (m, k, 2d) constant-dimension code over Fq. 17.3.77 Remark If C is a (k, m−k, d) maximum rank distance code, the “lifted” code Λ(C) has size qk(m−k)−(d−1) max{k,m−k}. This code is almost optimal in the sense that every (m, k, 2d) constant-dimension code over Fq has fewer than four times as many codewords as Λ(C) . In some special cases Λ(C) can be augmented to give an optimal code, as shown below. 17.3.78 Theorem Let k and m be positive integers satisfying m = rk for integer r. Let I and Z be the identity and all-zero matrix over Fq of size k × k, respectively. Let Ω0 ⊂Pk(Fm q ) contain the rowspace of [Z · · · Z I], and for ℓ∈{1, 2, . . . , r−1}, let Ωℓ⊂Pk(Fm q ) be the set of rowspaces corresponding to {[Z · · · Z I A] : A ∈Cℓ}, where Cℓis a (k, ℓk, k) maximum Miscellaneous applications 849 rank distance code over Fq. Then S ℓΩℓis an (rk, k, 2k) constant-dimension code of size 1 + qk + q2k + · · · + q(r−1)k. 17.3.79 Remark The above code is a k-spread of Fqm, namely a set of k-dimensional subspaces of Fqm having the property that each nonzero element of Fqm belongs to exactly one such subspace. The code therefore has largest possible size. 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Zywina, Explicit class ﬁeld theory for global function ﬁelds, preprint, 2011. <538, 546> K13417 DISCRETE MATHEMATICS AND ITS APPLICATIONS Series Editor KENNETH H. ROSEN Poised to become the leading reference in the field, the Handbook of Finite Fields is exclusively devoted to the theory and applications of finite fields. More than 80 international contributors compile state-of-the-art research in this definitive handbook. Edited by two renowned researchers, the book uses a uniform style and format throughout and each chapter is self contained and peer reviewed. The first part of the book traces the history of finite fields through the eighteenth and nineteenth centuries. The second part presents theoretical properties of finite fields, covering polynomials, special functions, sequences, algorithms, curves, and related computational aspects. The final part describes various mathematical and practical applications of finite fields in combinatorics, algebraic coding theory, cryptographic systems, biology, quantum information theory, engineering, and other areas. The book provides a comprehensive index and easy access to over 3,000 references, enabling you to quickly locate up-to-date facts and results regarding finite fields. Features • Gives a complete account of state-of-the-art theoretical and applied topics in finite fields • Describes numerous applications from the fields of computer science and engineering • Presents the history of finite fields and a brief summary of basic results • Discusses theoretical properties of finite fields • Covers applications in cryptography, coding theory, and combinatorics • Includes many remarks to further explain the various results • Contains more than 3,000 references, including citations to proofs of important results • Offers extensive tables of polynomials useful for computational issues, with even larger tables available on the book’s CRC Press web page About the Editors Gary L. Mullen is a professor of mathematics at The Pennsylvania State University. Daniel Panario is a professor of mathematics at Carleton University. Mathematics
2909	https://mathspanda.com/ASFM/Lessons/Modulus-Argument_Form_of_a_Complex_Number_LESSON.pdf	www.mathspanda.com Modulus-Argument Form of a Complex Number Starter 1. (Review of last lesson) Convert to an angle in radians, expressing your answer in terms of . 2. (Review of last lesson) Find the modulus of . N.B. The argument of a complex number, , is the angle that the line between the origin and makes with the positive axis, measured anti-clockwise. It is denoted arg and is given in radians. 3. Calculate the argument of the complex numbers: (a) (b) (c) Hint: use an Argand diagram to help you. 4. You are given the modulus and argument of a complex number. Express the complex number in the form . (a) Modulus = 6, argument = (b) Modulus = 2, argument = Hint: draw an Argand diagram to help. Notes The modulus-argument form of a complex number, , consists of the modulus, , which is the distance to the origin, and the argument , which is the angle the line makes with the positive axis, measured anti-clockwise. N.B. The angle can take any real value but the principal argument, denoted by Arg , is defined as or There are two forms of a complex number: Cartesian form Modulus-argument form — we will see that this notation is rarely used Converting between Cartesian and modulus-argument forms Cartesian to modulus-argument form For the complex number The modulus is given by . To find the argument: 1. Calculate N.B. Notice that we ignore the signs of the components of the complex number when finding the initial angle. 2. Sketch a quick Argand diagram to decide which quadrant lies in and then decide what you need to do to the angle found in step 1. 240o π 4 + 2i z z x− z 1 + i 4i − 3 + i x + yi π 3 3π 2 z r θ Oz x− r ≥0 θ z 0 ≤θ < 2π −π < θ ≤π x + yi [r, θ] z = x + yi r = x2 + y2 tan−1 y x z Page of 1 3 z arg z Re Im Re Im 1st Quadrant 2nd Quadrant 4th Quadrant 3rd Quadrant www.mathspanda.com For example, if is in the 4th quadrant, subtract the angle found from (since ). E.g. 1 Convert to form. Working: Ignore the signs: is in the 3rd quadrant so we need to add ( ) to the acute angle Modulus-argument to Cartesian form E.g. 2 For the complex number , express the and coordinates in terms of and . Use the diagram to help you. This gives us the more common way to express a complex number in modulus-argument form: This is shortened to . E.g. 3 Express the complex number in Cartesian form. Video: Modulus-argument form of complex number Solutions to Starter and E.g.s Exercise p127 4E Qu 1i, 2i, 3i, 4i, 5i, 6i, 7-11 Summary Cartesian form and Modulus-argument form where and or Modulus Argument 1st quadrant: 2nd quadrant: z 2π 2π ≡360o −1 − 3i [r, θ] r = −1 − 3i = (−1)2 + (−3)2 = 2 tan−1 3 1 = π 3 −1 − 3i π 180o Arg (−1 − 3i) = π + π 3 = 4π 3 −1 − 3i ≡[2, 4π 3 ] z = [r, θ] x− y− r θ z = r cos θ + i r sin θ = r(cos θ + i sin θ) z = r cis θ 4 cis π 6 x + yi x = r cos θ y = r sin θ r cis θ r ≥0 0 ≤θ < 2π −π < θ ≤π r = x2 + y2 θ = tan−1 y x θ = π −tan−1 y x Page of 2 3 z θ Re Im r x y www.mathspanda.com 3rd quadrant: 4th quadrant: θ = π + tan−1 y x θ = 2π −tan−1 y x Page of 3 3
2910	https://www.quantstart.com/articles/Pricing-a-Call-Option-with-Two-Time-Step-Binomial-Trees/	Pricing a Call Option with Two Time-Step Binomial Trees \| QuantStart QuantStart Quantcademy BooksSuccessful Algorithmic Trading Advanced Algorithmic Trading C++ For Quantitative Finance QSTrader Articles Login QuantStart Quantcademy Books Successful Algorithmic Trading Advanced Algorithmic Trading C++ For Quantitative Finance QSTrader Articles Login Pricing a Call Option with Two Time-Step Binomial Trees Pricing a Call Option with Two Time-Step Binomial Trees In our previous articles on call option pricing we have only considered one-step models. Further, we have almost exclusively considered binomial trees as we found that trinomial trees lead to incomplete markets. In this section we are going to consider stocks that are allowed more than two future states, but over multiple time steps. The basic idea is that we are refining the movements of the stock. Consider our previous stock, valued at 100, which in state U achieved 110 and in state D achieved 90. We can modify this model to include two time-steps, the first of which allows a new up state U where the stock achieves 105 and a new down state where the stock achieves 95. In the second time step the stock can take three separate values: 110, 100 and 90, across three states U U (for up-up), M (for Middle) and D D (for down-down). A figure best illustrates the situation: Pricing via hedging The methodology for pricing in a two-step world is similar to a one-step world. Our task is to determine the price of the option at all nodes of the tree. Since we know the final outcomes of the stock on the last step of the tree, we can proceed backwards along the nodes and utilise the same hedging argument as for the one-step binomial model to price the option. Let's start with the case where the stock is either at U U or M in the final step. Hence, it is either worth 110 or 100. The hedging argument says that if we are holding Δ 1 of the stock then the value of the portfolios at U U and M will respectively be 110 Δ 1−10 and 100 Δ 1. In the first instance (U U) we are subtracting 10 because of the price of the option, whereas in state M, the value of the option is 0 because it is at the money and does not give the purchaser anything by exercising it. We can see that by setting Δ 1=1 it will cause the portfolios to be worth 100 at both U U and M. However, if it is worth 100 in U U and M, it must be worth 100 at U, otherwise there would be an arbitrage opportunity. The portfolio value at U is given by 105−C=100, which means that C=5. The situation is far simpler for the states M and D D as the option is worth 0 and hence, by a no-arbitrage argument, the value of the option must be zero at D too. We can now use the same hedging argument to obtain the option value at the first time step. To summarise: At U the portfolio is worth 105 and the option is worth 5, whereas at D the portfolio is worth 95 and the option is worth nothing. Hence, in order to be hedged perfectly, we need to hold Δ 2 of the stock such that 105 Δ 2−5=95 Δ 2. This is only the case when Δ 2=0.5, at which point the portfolios at U and D will be worth 47.5. A final application of the no-arbitrage principle states that the portfolio today must also be worth 47.5. This implies that: Δ 2⋅100−C=0.5⋅100−C=47.5 Hence C=2.5. The following figure illustrates the option prices in the tree: Pricing via risk neutrality We have already discussed risk neutral pricing in one-step binomial models. We will now apply the same method to two-step models. Recall that the intuition behind risk neutrality is harder to grasp. It involves considering investors who are fully risk neutral - that is they do not need compensation for taking on the risk of an uncertain portfolio, assuming that it has equal expected value to a riskless portfolio. Remember also that the prices obtained via both a hedging argument and risk neutrality, for a one-step binomial tree will always be equal. At the step U we know that a risk-neutral stock will go to state U U or M with p 1=0.5, since this is the only p 1 that gives the expected portfolio value of 105. The value of the option, is equal to 10⋅p 1+0⋅(1−p 1)=5 at U. For state D the option is out of the money for M and D D, hence it is worth zero and thus it is worth zero at D also. Using the same strategy, we can show that p 2, the probability of state U occurring, must also be p 2=0.5. Hence the initial value of the option is equal to 5⋅p 2+0⋅(1−p 2)=2.5. This is in agreement with the price we determined via the hedging argument. In the next article we will extend the tree to include multiple time steps (beyond the two we have considered here) » Join the QuantStart Newsletter Subscribe to get our latest content by email. Subscribe We won't send you spam. Unsubscribe at any time. Privacy Policy. Built with Kit The Quantcademy Join the Quantcademy membership portal that caters to the rapidly-growing retail quant trader community and learn how to increase your strategy profitability. Find Out More Successful Algorithmic Trading How to find new trading strategy ideas and objectively assess them for your portfolio using a Python-based backtesting engine. Find Out More Advanced Algorithmic Trading How to implement advanced trading strategies using time series analysis, machine learning and Bayesian statistics with R and Python. 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2911	http://www.multiwingspan.co.uk/as1.php?page=bin	Home Programming Computer Science Twisting Puzzles Arduino BBC micro:bit Raspberry Pi Pico Computer Science Computer Science Home Data Representation Binary Numbers Hexadecimal Numbers Character Coding Schemes Error Detection & Correction Bitmapped & Vector Graphics Image Compression Techniques Sound Real Numbers Logic Gates - Truth Tables Logic Gates - Diagrams Boolean Algebra Program Design Principles Of Computation Problem Solving Hierarchy Charts Stepwise Refinement Finite State Machines State Transition Tables Decision Tables Algorithm Design Structured English Pseudocode Program Flowcharts Dry-Running Programming Structure Charts Structured Programming Testing Finite State Machines Turing Machines Types Of Problem The Systems Life Cycle Hardware & Software Computer Architecture The Stored Program Concept Structure Of The Processor Machine Code Operations The Fetch-Execute Cycle Operating Systems Input Devices Output Devices Storage Devices Classification Of Software Networks The Internet Domain Names & IP TCP/IP Protocol Web Design Communication Methods Networks Networks 2 Server-Side Scripting Internet & Computer Security Databases Entity Relationship Modelling Database Design Structured Query Language Data Structures Stacks Queues Linear Lists Linked Lists Graphs Trees Algorithms Recursion Binary Search Insertion Sort Hashing Simulations Comparing Algorithms Other Computers & The Law Abstraction Regular Expressions Backus-Naur Form Reverse Polish Notation Programming Paradigms Object Oriented Programming Computer ScienceBinary Numbers Denary Numbers Our number system is called the denary or base 10 system. 10 digits (0 - 9) are used for counting. You should remember the place value table from your Maths lessons in Year 7. Eg 47395 \| 10000000 \| 1000000 \| 100000 \| 10000 \| 1000 \| 100 \| 10 \| Units \| --- --- --- --- \| \| 0 \| 0 \| 0 \| 4 \| 7 \| 3 \| 9 \| 5 \| As each place value moves one to the left, the power of ten is increased by 1. The units column is 100, the tens, 101, the hundreds, 102 and so on. Binary Numbers The place value columns in the binary table below are powers of 2. The only valid digits to use are 1 and 0. We call these bits. Just like in the denary table, we add up the results of multiplying each number in the table by its place value. Click on the buttons below the table to see how the numbers from 0 to 255 can be represented in pure binary. The denary value is shown below the table. \| 128 \| 64 \| 32 \| 16 \| 8 \| 4 \| 2 \| Units \| --- --- --- --- \| \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 Binary Addition Addition using binary numbers is relatively straightforward. The following rules need to be observed, 0 + 0 = 0 0 + 1 = 1 + 0 = 1 1 + 1 = 10 (0, carry 1) 1 + 1 + 1 = 11 (1, carry 1) In the following example, the binary numbers, 1011 (11 in denary) and 10010 (18 in denary) are added together. When performing binary addition, you must lay your sums out this way. Do not work things out in your head or use any mental methods that you learned in little school. The potential for error is too great not to take care with something that is quite basic. Representing Negative Numbers The number shown in the interactive binary place value table is an unsigned binary integer. That means that it is a positive whole number as far as we are concerned. In denary, when we want to represent a negative number, we simply place a minus sign before it. Binary doesn't work that way. The Two's Complement system is used to represent negative numbers in binary. The system works a bit like a milometer. If the milometer is set at 00000 and is turned back one mile, it would read 99999. A negative binary number always has a 1 as the first bit. This is often referred to as the sign bit Convert From Denary To Two's Complement To convert a negative denary number to binary, first find the binary equivalent of the positive integer. Eg -27, 27 = 00011011 Change all of the 0s to 1s and all the 1s to 0s. (Flip the bits). Eg 11100100 Add 1 to the result. Eg 11100101 The following place value table works like the previous one. This one allows you to show negative numbers. Pick a random negative number. Follow the steps described above for converting to two's complement. You should end up with the positive number. \| -128 \| 64 \| 32 \| 16 \| 8 \| 4 \| 2 \| Units \| --- --- --- --- \| \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 \| 0 Convert From Two's Complement To Denary The Two's Complement method of representing negative numbers makes sense if you think of the sign bit as representing a negative number. In the case of the binary counter above, the first column represents -128. Convert to denary as normal, adding the column heading whenever there is a 1 below it. Binary Subtraction To perform subtraction in binary, simply convert the number that you are subtracting into Two's Complement form and add it to the other number. You do not carry the one on the leftmost digit. Key Points To Remember A positive number always has 0 as the Most Significant Bit. A negative number always has 1 as the Most Significant Bit. An even number always has 0 as the Least Significant Bit. An odd number always has 1 as the Least Significant Bit. The number of digits used to represent a number is called the word length. -1 is always represented with a 1 in every bit. Fixed Point Binary Numbers So far, you have only looked at ways of representing binary integers. To understand how binary fractions work, first consider how decimal fractions work. \| 1000 \| 100 \| 10 \| Units \| \| 1/10 \| 1/100 \| 1/100 \| --- --- --- --- \| \| 0 \| 5 \| 4 \| 9 \| . \| 3 \| 6 \| 7 \| In binary, the column headings are powers of 2 rather than 10. \| 8 \| 4 \| 2 \| Units \| \| 1/2 \| 1/4 \| 1/8 \| 1/16 \| --- --- --- --- \| 1 \| 0 \| 1 \| 1 \| . \| 1 \| 0 \| 1 \| 1 \| This would store the number 8 + 2 + 1 + 0.5 +0.125 + 0.0625 = 11.6875 When storing a fraction in binary digits the binary point is not stored. We need to assume a certain number of bits before the binary point and a number after the binary point. This representation of binary numbers is called fixed point binary. Assuming 8 bits either side of the binary point and we get the following headings, In the above example the number stored would be : 32 + 8 + 2 + 1 + 0.25 + 0.03125 + 0.0078125 + 0.00390625 = 43.2869375 Unless a number is an exact power of 2 it is impossible to store it exactly using this method. Errors in conversion of decimal numbers arise unless a large number of bits are allocated to store them. You can convert any decimal fraction to a binary fraction. Multiply the fractional part of the number by 2. Take the integer part of the result (1 or 0) as the first bit. Repeat this process with the result until you run out of patience. For example, to convert 0.3568 into fixed point binary with 8 bits to the right of the binary point, 0.3568 x 2 = 0.7136 :0 0.7136 x 2 = 1.4272 :1 0.4272 x 2 = 0.8544 :0 0.8544 x 2 = 1.7088 :1 0.7088 x 2 = 1.4176 :1 0.4176 x 2 = 0.8352 :0 0.8352 x 2 = 1.6704 :1 0.6704 x 2 = 1.3408 :1 0.3568 is .01011011 When we convert the binary result back to denary, we get 0.35546875. This isn't too far away - with 16 bits we could get, 0.356796264648437. The precision increases the more bits we use. Pages designed and coded by MHA since 2003 \| Valid HTML 4.01(Strict) \| CSS
2912	https://www.scribd.com/document/342732891/Mathematical-Circles-Russian-Experience-Mathematical-World-7-D-Fomin-S-Genkin-I-Itenberg-American-Mathematical-Society-1996	Mathematical Circles (Russian Experience) - (Mathematical World 7) D. Fomin, S. Genkin, I. Itenberg-American Mathematical Society (1996) \| PDF \| Numbers \| Vertex (Graph Theory) Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for days 86%(7)86% found this document useful (7 votes) 5K views 290 pages Mathematical Circles (Russian Experience) - (Mathematical World 7) D. Fomin, S. Genkin, I. Itenberg-American Mathematical Society (1996) Mathematical Circles (Russian Experience) -(Mathematical World 7) D. Fomin, S. Genkin, I. Itenberg-American Mathematical Society (1996) Full description Uploaded by debraj123 AI-enhanced title Download Save Save Mathematical Circles (Russian Experience) -(Mathem... For Later Share 86%86% found this document useful, undefined 14%, undefined Print Embed Report 86%(7)86% found this document useful (7 votes) 5K views 290 pages Mathematical Circles (Russian Experience) - (Mathematical World 7) D. Fomin, S. Genkin, I. Itenberg-American Mathematical Society (1996) Mathematical Circles (Russian Experience) -(Mathematical World 7) D. Fomin, S. Genkin, I. Itenberg-American Mathematical Society (1996) Read more Download 86%(7)86% found this document useful (7 votes) 5K views 290 pages Mathematical Circles (Russian Experience) - (Mathematical World 7) D. Fomin, S. Genkin, I. Itenberg-American Mathematical Society (1996) Mathematical Circles (Russian Experience) -(Mathematical World 7) D. Fomin, S. Genkin, I. Itenberg-American Mathematical Society (1996) Full description Uploaded by debraj123 AI-enhanced title Go to previous items Go to next items Download Save Save Mathematical Circles (Russian Experience) -(Mathem... For Later Share 86%86% found this document useful, undefined 14%, undefined Print Embed Report Download Save Mathematical Circles (Russian Experience) -(Mathem... For Later You are on page 1/ 290 Search Fullscreen Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial Documents Teaching Methods & Materials Mathematics Footer menu Back to top About About Scribd, Inc. Everand: Ebooks & Audiobooks Slideshare Join our team! Contact us Support Help / FAQ Accessibility Purchase help AdChoices Legal Terms Privacy Copyright Do not sell or share my personal information Social Instagram Instagram Facebook Facebook Pinterest Pinterest Get our free apps About About Scribd, Inc. Everand: Ebooks & Audiobooks Slideshare Join our team! Contact us Legal Terms Privacy Copyright Do not sell or share my personal information Support Help / FAQ Accessibility Purchase help AdChoices Social Instagram Instagram Facebook Facebook Pinterest Pinterest Get our free apps Documents Language: English Copyright © 2025 Scribd Inc. We take content rights seriously. Learn more in our FAQs or report infringement here. We take content rights seriously. Learn more in our FAQs or report infringement here. Language: English Copyright © 2025 Scribd Inc. 576648e32a3d8b82ca71961b7a986505
2913	https://www.sciencedirect.com/topics/mathematics/weighted-sum	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Multi-objective Optimum Design Concepts and Methods 17.4 Weighted Sum Method The most common approach to multi-objective optimization is the weighted sum method: (17.9) Here w is a vector of weights typically set by the decision maker such that wi=1 and w > 0. If objectives are not normalized, wi need not add to 1. As with most methods that involve objective function weights, setting one or more of the weights to 0 can result in weakly Pareto optimal points. The relative value of the weights generally reflects the relative importance of the objectives. This is another common characteristic of weighted methods. If all of the weights are omitted or are set to 1, then all objectives are treated equally. The weights can be used in two ways. The user may either set w to reflect preferences before the problem is solved or systematically alter w to yield different Pareto optimal points (to generate the Pareto optimal set). In fact, most methods that involve weights can be used in both of these capacities—to generate a single solution or multiple solutions. This method is easy to use, and if all of the weights are positive, the minimum of Eq. (17.9) is always Pareto optimal. However, there are a few recognized difficulties with the weighted sum method (Koski, 1985; Das and Dennis, 1997). First, even with some of the methods discussed in the literature for determining weights, a satisfactory a priori weight selection does not necessarily guarantee that the final solution will be acceptable; one may have to re-solve the problem with new weights. In fact, this is true of most weighted methods. The second problem is that it is impossible to obtain points on nonconvex portions of the Pareto optimal set in the criterion space (Marler and Arora, 2010). Although nonconvex Pareto optimal sets are relatively uncommon, some examples are noted in the literature (Koski, 1985; Stadler and Dauer, 1992; Stadler, 1995). The final difficulty with the weighted sum method is that varying the weights consistently and continuously may not necessarily result in an even distribution of Pareto optimal points and an accurate, complete representation of the Pareto optimal set. View chapterExplore book Read full chapter URL: Book2012, Introduction to Optimum Design (Third Edition)Jasbir S. Arora Review article Objectives and methods in multi-objective routing problems: a survey and classification scheme 2021, European Journal of Operational ResearchSandra Zajac, Sandra Huber 6.1.1 WS: Weighted sum approach A well-known scalar method is to aggregate the objectives linearly in a single weighted sum function, assuming that the DM’s global utility function consists of the sum of the objectives’ partial utilities. The solution of the weighted problem is always weakly Pareto optimal, and Pareto optimal for non-negative weights or if the solution is unique (Miettinen, 1999). The main drawback of this method is that it can only yield efficient extreme solutions on the convex hull of the Pareto front. is an extreme point of if with and 0 ≤ α ≤ 1 implies that (Ehrgott, 2005). Runs with different weight combinations can result in the same efficient extreme solution so that an adequate selection of weight sets is crucial to obtain the complete Pareto front for convex problems (Mavrotas, 2009). According to Yang (2014), measuring the objectives on a common scale can be necessary so that the ranges/values of each objective are comparable and biased sampling on the Pareto front, due to badly distributed weight coefficients, is avoided. The same scale is achieved either by definition (Costa et al., 2018; Nolz et al., 2014; Tezcaner & Köksalan, 2011) or by converting the objectives to a common scale, e.g. by assuming cost factors (Dolicanin et al., 2018; Norouzi et al., 2017). Note that the objectives time and distance are often interpreted as costs in literature. Each objective is normalized to the interval [0,1] by computing percentage deviations from ideal and nadir points or other lower/upper bounds (Bronfman et al., 2015; Demir et al., 2014a; Han et al., 2014; Hu et al., 2019; Wenger & Geiger, 2008). Alternatively, the objectives can be set in relation to the best achievable objective value for the respective objective, yielding a minimum weighted sum value of 1 (Asgari et al., 2017; Muñoz-Villamizar et al., 2017). In Perugia et al. (2011), bus routes to serve the commuting needs of workers at given facilities are optimized. The existing plan provides a reference point in terms of the studied objectives that is used for scaling. Dadkar et al. (2008) consider stochastic travel times and risk exposure as relevant attributes that define path performance in a hazmat transportation problem. For normalization, the expected travel time and risk value are set in relation to the means of the respective distributions. A parameter is introduced that expresses the relative importance of risk exposure compared to travel time. In both Coelho et al. (2017), Dadkar et al. (2008), the weights do not (necessarily) total 1. However, in some of the reviewed papers the objectives are not measured on the same scale and they also have not been normalized in some way (Assadipour et al., 2015; Chen et al., 2008; Coelho et al., 2017; Demir et al., 2014a; Evers et al., 2014; Verma & Verter, 2010; Verma et al., 2012; Wang & Wang, 2010; Yi & Bauer, 2016). Typically, one (Asgari et al., 2017; Chen et al., 2008; Dolicanin et al., 2018; Hu et al., 2019; Norouzi et al., 2017; Perugia et al., 2011) or multiple weight sets are specified by the DM a priori (Assadipour et al., 2015; Bronfman et al., 2015; Coelho et al., 2017; Dadkar et al., 2008; Demir et al., 2014a; Han et al., 2014; Molenbruch et al., 2017; Muñoz-Villamizar et al., 2017; Parragh et al., 2009; Perugia et al., 2011; Verma & Verter, 2010; Verma et al., 2012; Wang & Wang, 2010; Yi & Bauer, 2016). In other approaches, the weights are determined dynamically during the search. Nolz et al. (2014) propose an ALNS for an IRP that evaluates performance with a weighted sum function. To recompute the weights, an ideal point, the two extreme points of the current Pareto frontier as well as the point defined by the current solution are required. Originating from this point, the successive search direction is randomly selected in the search space region spanned between these points. The weights are chosen in such a way that the algorithm is driven towards this search direction. In Costa et al. (2018), weights are defined by the normal line across the images in the objective space of the two extreme solutions. The authors use a local search heuristic to solve a set of weighted sum problems first which then serve as a starting point for Pareto local search. One of the solution approaches in Hu et al. (2019) dynamically adapts the weights in a weighted sum fitness function by updating the range between best and worst solution found so far. In Wenger and Geiger (2008), the interaction process of a DM with a VRP planning system is simulated. A sequence of weight vectors imitates the way the DM would change the weights in the studied VRP with six objectives. Starting from a base value for each of the weights, the weight of a randomly selected objective is increased and the others are simultaneously decreased until an extreme weight vector is reached. Results for a cost-oriented, a service-oriented and a cost-service-oriented DM are presented. Tezcaner and Köksalan (2011) focus on the difficulty for the DMs to specify weights that actually represent their true preferences. The authors narrow down the true weights of the DM interactively. The DM is iteratively asked to compare an efficient solution with its adjacent efficient solutions. The responses are then utilized to impose weight bounds that on and on restrict the set of possible weights of the underlying utility function. View article Read full article URL: Journal2021, European Journal of Operational ResearchSandra Zajac, Sandra Huber Chapter Deep learning in natural language processing 2021, Machine Reading ComprehensionChenguang Zhu 3.1.3 Parametrized weighted sum Parametrized weighted sum is a commonly used method to turn multiple word vectors into a text vector. Assume the input contains n word vectors, . A simple average pooling is equivalent to the weighted sum where the weights are all . However, if we want to dynamically tune these weights to reflect the relationship between the words, the parametrized weighted sum can be used. The first step is to define a parameter vector with the same dimension as the word vector . Then, each word vector is assigned a weight score via the inner product . As the sum of weights in average pooling is 1, we can use the softmax operation in Section 3.2 to normalize these scores to sum up to 1: Finally, the weighted sum of all word vectors is used as the text vector: The following code implements the parametrized weighted sum. import torch import torch.nn as nn import torch.nn.functional as F class WeightedSum(nn.Module): word_dim: dimension of input word vector def __init__(self, word_dim): super(WeightedSum, self).__init__() self.b=nn.Linear(word_dim, 1) # the parameter vector Input: x is the input tensor. Size: batch×seq_len×word_dim Output: res is the text tensor. Size: batch×word_dim def forward(self, x): score by inner product. Size: batch×seq_len×1 scores=self.b(x) softmax operation. Size: batch×seq_len×1 weights=F.softmax(scores, dim=1) The weighted sum is computed by matrix multiplication. Size: batch×word_dim×1 res=torch.bmm(x.transpose(1, 2), weights) Delete the last dimension. Size: batch×word_dim res=res.squeeze(2) return res The code above uses nn.Linear for inner product operation on vectors. The function torch.bmm is used to compute the multiplication between two batch matrices with dimension batch×a×b and batch×b×c. It is worth noting that the parametrized weighted sum is actually a self-attention mechanism, which will be covered in more detail in Section 3.4. View chapterExplore book Read full chapter URL: Book2021, Machine Reading ComprehensionChenguang Zhu Chapter Optimal Test Construction 2005, Encyclopedia of Social MeasurementBernard P. Veldkamp Weighted Deviations Model In the general model, all test specifications are considered to be constraints that have to be met. For some test construction problems, the test specifications are considered to be desirable properties rather than constraints. As a result, they are allowed to be violated in the test construction process. When properties are considered to be desirable properties, a weighted deviation model can be formulated. In this model, targets are defined for all test attributes. The objective function is a weighted sum of all violations or deviations. This model can be formulated as: (15a) subject to: (15b) (15c) (15d) (15e) where the variables dj denote the deviations of constraints j, j = 1,…,J, with J being the total number of constraints, and wj denotes the weight of deviation j. In this model, the difference between the target information function and the test information functions is formulated as a quantitative constraint. When some specifications are considered to be of paramount interest, their weights get high values. When other specifications are considered to be less important, the weights get low values. Because the specifications do not have to be met, the model is less restrictive. A less favorable feature of this model is that two different tests constructed by the same model might have different attributes. View chapterExplore book Read full chapter URL: Reference work2005, Encyclopedia of Social MeasurementBernard P. Veldkamp Chapter REGRESSION 2009, Introduction to Probability and Statistics for Engineers and Scientists (Fourth Edition)Sheldon M. Ross REMARKS (a) : Assuming normally distributed data, the weighted least squares estimators are precisely the maximum likelihood estimators. This follows because the joint density of the data Y 1,…, Yn is Consequently, the maximum likelihood estimators of α and β are precisely the values of α and β that minimize the weighted sum of squares . (b) : The weighted sum of squares can also be seen as the relevant quantity to be minimized by multiplying the regression equation by . This results in the equation Now, in this latter equation the error term has mean 0 and constant variance. Hence, the natural least squares estimators of αand β would be the values of A and B that minimize (c) : The weighted least squares approach puts the greatest emphasis on those data pairs having the greatest weights (and thus the smallest variance in their error term). At this point it might appear that the weighted least squares approach is not particularly useful since it requires a knowledge, up to a constant, of the variance of a response at an arbitrary input level. However, by analyzing the model that generates the data, it is often possible to determine these values. This will be indicated by the following two examples. View chapterExplore book Read full chapter URL: Book2009, Introduction to Probability and Statistics for Engineers and Scientists (Fourth Edition)Sheldon M. Ross Review article A review of multiobjective programming and its application in quantitative psychology 2011, Journal of Mathematical PsychologyHans-Friedrich Köhn 2.4.2 Weighted-sum method The original multiobjective program is re-expressed as a convex combination of the objective functions, solvable as a single objective program: () where and . If the multiobjective optimization problem is convex, any at least weakly efficient solution can be detected by systematically altering the weighting coefficients. The weighted-sum method offers the computational convenience that the scalarized problem is not harder to solve than the original multiobjective program. If the problem is not convex, the weighted-sum method is incapable of detecting all Pareto-optimal solutions—a major disadvantage, as we discuss in greater detail within the context of multiobjective combinatorial programming. View article Read full article URL: Journal2011, Journal of Mathematical PsychologyHans-Friedrich Köhn Chapter Tunneling Times in Semiconductor Heterostructures: A Critical Review 1992, Hot Carriers in Semiconductor NanostructuresA.P. Jauho 3.2 A Sum Rule Transmission and reflection are mutually exclusive events in the sense of Feynman and Hibbs : it is possible to determine whether a particle has tunneled or reflected without interfering with the scattering process. Therefore, the tunneling time τT and the reflection time τR are conditional averages (if they exist), and their weighted sum should equal the dwell time τd: (36) Expressed in words, the physical content of Eq. (36) is the following: Consider a flux of particles entering a region of space. A fraction T/(T + R) = T of these particles spends the time τT in this region, and the rest, the fraction R/(T + R) = R spends the time τR. Then the average time that a particle spends in this region of space, without specifying whether it is transmitted or reflected, is given by the weighted average, that is, by τd, Eq. (36). In the following we check the consistency of the various tunneling times with the rule (36), and review other theoretical restrictions that have been identified in the literature. View chapterExplore book Read full chapter URL: Book1992, Hot Carriers in Semiconductor NanostructuresA.P. Jauho Chapter Covariance Structure Models for Maximal Reliability of Unit-Weighted Composites 2007, Handbook of Latent Variable and Related ModelsPeter M. Bentler 7 Reliability of weighted composites The maximal reliability composite discussed above is based on simple summation of parts, since unweighted sums are typically used to obtain a total score. However, sometimes known or unknown weights wi may be used to give some variables more influence than others, resulting in the weighted composite Y = w1X1 + w2X2 + … + wpXp. Since a unit-weighted composite is a special case with wi = 1, it may be desirable to place the earlier discussion into the context of the more general case of weighted composites. First we discuss composites with known weights. It is easy to show that all of the previous results apply directly to the case of weighted composites. Suppose that w is the p-length column vector of weights, and that Dw is the diagonal matrix with w in its diagonal. Then rescaling the x variables in (1) with these weights yields the new variables y = Dwx. Their unit-weighted sum Y = 1′y = 1′ Dwx is the weighted composite. The rescaled variables possess a comparable factor analytic decomposition (12) and, with obvious notation, the covariance structure of these rescaled variables is given by (13) As a result, the latent structure of the rescaled variables parallels that of the original variables in the previous sections. Hence, the preceding results, including the theorem, apply directly to the situation of weighted composites. In the first part of this paper, we used the decomposition Σ = ΣT + ΨE to define reliability as . Using the same decomposition, we can also define reliability of a weighted composite Y as (14) The unit-weighted composite is a special case. When the error variances are not known, it is hard to estimate (14) directly. Bentler (1968) noted that factor analytic concepts can be fruitfully employed. Let the covariance matrix of the original unweighted variables Xi have the composition Σ = Σc + Ψ, where Σc is a non-negative definite covariance matrix of the common variables, and Ψ is the covariance matrix of the unique variables. Then Bentler's (1968, Eq. (12)) general formula for internal consistency reliability of a weighted composite is (15) See also Heise and Bohrnstedt (1970, Eq. (32)). Eq. (15) is a lower bound to reliability (14) under the usual factor analytic assumption Ψ = ΨE + ΨS, where unique covariance matrix Ψ is a sum of the non-negative definite covariance matrices of random errors and specific but reliable variables. Then w′Ψw = w′ΨEw + w′ΨSw, and some algebra shows that (16) Thus ρw'x ≤ ρyy, i.e., (15) is a lower bound to (14). The coefficients will be equal when there is no specificity. Actually, several of the key reliability coefficients defined in earlier parts of this chapter are just special cases of (15). Obviously, with equal weights and Σc = λλ′, a single common factor model, and Ψ = Ψu, (15) simplifies to ρ11 as discussed previously, namely reliability under a 1-factor hypothesis as popularized by McDonald (1999, p. 89) with coefficient ω. With equal weights and a general k-factor decomposition Σc = ΛΛ′, (15) becomes ρkk as given in (5) and elsewhere. With unit weights and optimization criteria to define Ψ, this is the dimension-free and greatest lower bound given in (10) and (11). View chapterExplore book Read full chapter URL: Book2007, Handbook of Latent Variable and Related ModelsPeter M. Bentler Review article Assessment on power systems non-deterministic state estimation algorithms 2023, Electric Power Systems ResearchIzar Lopez-Ramirez, ... Inmaculada Zamora 2.1 Weighted Least Squares based algorithms Since Schweppe presented the WLS algorithm for SE in power systems [14–16], it has been widely explored in deterministic SE and is one of the most common SE methods in power systems [1,2]. System state, voltage module and phase of all network buses is obtained from measurements, using the measurement Eq. (1). (1) where is the measurement vector, is the system state vector, is a nonlinear function relating measurements and state vector, is the vector of measurement errors modeled as a Gaussian distribution of zero mean and standard deviation. The residual vector is defined as the difference between the measurement and the estimated values and it is weighted by the inverse of the error standard deviation . Therefore, the WLS estimator Eq. (2) is an optimization problem, which minimizes the weighted sum of the squared residual of measurement , applying the Gauss–Newton iteration algorithm. It gives the maximum likelihood estimates as a deterministic solution . (2) Likewise, many non-deterministic SE methods apply some modifications to WLS or similar approaches. For example, the Least Square (LS) algorithm, to obtain non-deterministic solutions. View article Read full article URL: Journal2023, Electric Power Systems ResearchIzar Lopez-Ramirez, ... Inmaculada Zamora Chapter Multiobjective Optimum Design Concepts and Methods 2004, Introduction to Optimum Design (Second Edition)Jasbir S. Arora 17.5 Weighted Min-Max Method The weighted min-max method (also called the weighted Tchebycheff method) is formulated as follows: (17.6) A common approach for treatment of Eq. (17.6) is to introduce an additional unknown parameter λ as follows: minimize λ subject to additional constraints (17.7) Whereas the weighted sum method of Section 17.4 always yields Pareto optimal points but may miss certain points when the weights are varied, this method can provide all the Pareto optimal points (the complete Pareto optimal set). However, it may provide non-Pareto optimal points as well. Nonetheless, the solution to the min-max approach is always weakly Pareto optimal, and if the solution is unique, then it is Pareto optimal. Advantages of the method are: (1) it provides a clear interpretation of minimizing the largest difference between fi(x) and fi, (2) it can provide all the Pareto optimal points, (3) it always provides a weakly Pareto optimal solution, and (4) it is relatively well suited for generating the complete Pareto optimal set (with variation in the weights). Disadvantages are: (1) it requires the minimization of each objective when using the utopia point, which can be computationally expensive, (2) it requires that additional constraints be included, and (3) it is not clear exactly how to set the weights when only one solution point is desired. View chapterExplore book Read full chapter URL: Book2004, Introduction to Optimum Design (Second Edition)Jasbir S. Arora Related terms: Statistics Probability Theory Regression Model Gaussian Distribution Polynomial Sum of Squares Linear Models Linear Combination Neural Network Pareto Optimal View all Topics
2914	https://www.ncbi.nlm.nih.gov/books/NBK430909/	An official website of the United States government The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log in Account Logged in as:username Dashboard Publications Account settings Log out Access keys NCBI Homepage MyNCBI Homepage Main Content Main Navigation Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Iritis Bharat Gurnani; Jessica Kim; Koushik Tripathy; Navid Mahabadi; Mary Ann Edens. Author Information and Affiliations Authors Bharat Gurnani1; Jessica Kim2; Koushik Tripathy3; Navid Mahabadi4; Mary Ann Edens5. Affiliations 1 Gomabai Netralaya and Research Centre 2 Louisiana State University 3 ASG Eye Hospital, BT Road, Kolkata, India 4 A.T. Still University 5 Louisiana State University Last Update: August 25, 2023. Continuing Education Activity Anterior uveitis (iritis) is the inflammation of the anterior chamber and the iris. Uveitis is often idiopathic, but it may be triggered by genetic, immune, traumatic, or infectious mechanisms. The symptoms include redness, sensitivity to light, and pain. Slit-lamp examination reveals cells in the anterior chamber. Proper evaluation is necessary for recurrent anterior uveitis. The management options include topical steroids and cycloplegics. With appropriate treatment and follow-up, it has a good prognosis. This activity highlights the role of the interprofessional team in caring for patients with this condition. Describe the pathophysiology of iritis. Review the steps for evaluating iritis. Summarize the treatment options of iritis. Outline the importance of improving care coordination among interprofessional team members to improve outcomes for patients affected by iritis. Access free multiple choice questions on this topic. Introduction The International Uveitis Study Group (IUSG) recommended the anatomical classification of uveitis into anterior uveitis, intermediate uveitis, posterior uveitis, and panuveitis. The Standardization of Uveitis Nomenclature (SUN) working group endorsed the same anatomical classification. This anatomical location is based on the actual site of inflammation and is not affected by the presence or absence of structural complications, including cystoid macular edema. The primary location of inflammation in anterior uveitis is the anterior chamber (AC). The anterior uveitis includes three entities, namely iritis, iridocyclitis, and anterior cyclitis. The inflammation of the iris is termed as iritis, and that involving the anterior part of the ciliary body is known as anterior cyclitis. When both of these are involved, it is called iridocyclitis. Anterior uveitis is the most common form of uveitis. The most common clinical presentation is acute anterior uveitis (AAU). The most common etiologies include HLA-B27 and idiopathic. It is difficult to pinpoint the etiology in some cases, and it may be secondary to the cross-reactivity of microbial antigens in genetically predisposed patients. The etiology of AAU can be varied, ranging from viral infections, bacterial infections, trauma, lens-related inflammation, keratitis, and scleritis. It can exist alone as AAU or in conjuction with intermediate uveitis, posterior uveitis, or panuveitis. Chronic anterior uveitis (CAU) is usually bilateral secondary to an underlying systemic pathology and is less common than AAU. The prognosis is typically good in most idiopathic and HLAB27- related AAU cases, provided management is adequate. Outcomes are more variable in CAU and cases with an underlying ocular or systemic disorder. Etiology Most cases of iritis are idiopathic, while 20% are due to blunt trauma. Nontraumatic iritis often is associated with HLA- B27 systemic diseases, including: Juvenile Rheumatoid Arthritis (JRA) Ulcerative Colitis Reiter syndrome Sarcoidosis Behcet's disease Tubulointerstitial Nephritis and Uveitis (TINU) Systemic Lupus Erythematosus (SLE) Multiple Sclerosis Ankylosing Spondylitis Infectious Causes Tuberculosis Chlamydia Lyme's disease Herpes Simplex Toxoplasmosis Varicella-Zoster virus (herpes zoster ophthalmicus or shingles) Syphilis.. Other Causes Drug-induced Masquerade syndromes like lymphoma, leukemia, malignant melanoma Juvenile Xanthogranuloma After Idiopathic variety, HLA-B27 associated uveitic ranks as the second commonest cause of anterior uveitis. It is responsible for 40-70% of anterior uveitis cases in varied geographical regions. It is more common in males than females. The Various HLA-B27 Associated Conditions with Ocular Involvement are Listed Below Table S. No HLA-B27 associated condition Rodriguez et al., from their large-scale analysis from a tertiary eye care center, listed the prevalence of various etiologies in all the anatomic forms of uveitis. The most common conditions were: Idiopathic Seronegative spondyloarthropathies Sarcoidosis JRA SLE Behcet's HIV Ankylosing spondylitis, Inflammatory bowel disease, psoriatic arthritis, and Reiter's syndrome are grouped under seronegative spondyloarthropathies. As per Rodriguez et al., the most common form was anterior uveitis, with the most common etiology being idiopathic followed by seronegative arthropathy, JRA, herpes, sarcoidosis, SLE, and rheumatoid arthritis. The next was posterior uveitis, with common etiologies being toxoplasmosis, idiopathic, CMV, SLE, and sarcoidosis. Epidemiology Anterior uveitis is the most common form of uveitis (occurring every 12 per 100 000 cases). It predominantly occurs in young and middle-aged people. In western countries, 50% to 70% of all uveitis cases are classified as anterior uveitis. Uveitis is responsible for 10% of legal blindness in the USA, and it accounts for approximately 30,000 new patients with blindness every year. The study from northern California on the epidemiology of uveitis revealed an incidence of 52.4 per 100000 persons per year and a prevalence of 115.3 per 100000 persons. The maximum incidence and prevalence were seen in elderly patients above 65 years and lowest in the pediatric age group. Women had a higher prevalence than men. The study from Affairs Medical Centre, Northwest Pacific, reported an incidence of 25.6 cases per 100000 persons and a prevalence of 69 per 100000 persons. The study in the elderly population from the care survey medicare cohort revealed an incidence of uveitis ranging from 302-424 per 100000 persons per year and an average of 340.9 per 100000. The mean incidence of uveitis was 243.6 per 100000. The reported incidence of posterior uveitis was 76.6 per 100000, and panuveitis incidence was 41.7 per 100000. Pathophysiology Eye pain is thought to be due to irritation of the ciliary nerves and ciliary muscle spasms. Photophobia is caused by irritation of the trigeminal nerve from the ciliary spasm. Increased permeability of blood vessels in the anterior chamber allows proteinaceous transudate ("flare") and WBCs ("cells"), the characteristic 'flare and cells' seen with the slit lamp. In traumatic uveitis, there can be microbial contamination and retention of necrotic debris at the site of trauma, resulting from a florid inflammatory response in the anterior segment of the eye. In infectious uveitis, the pathophysiological mechanism is immune-mediated destruction of foreign antigens that may cause damage to the uveal tissue, vessels, and cells. There is immune complex deposition inside the uveal tract in autoimmune uveitic conditions, a form of hypersensitivity reaction. Uveal inflammation behaves similarly to inflammation in other body tissues, i.e., having a vascular and cellular response. But due to excessive vascularity and looseness of uveal tissue, there is heightened vascular response. Pathophysiologically, the uveal inflammation is subdivided into purulent (suppurative) and nonpurulent (non-suppurative) types. Wood’s Classification of Non-Suppurative Inflammation Non-granulomatous Granulomatous Suppurative Uveitis It results from exogenous infections from pyogenic microorganisms. It is a part of the spectrum of endophthalmitis or panophthalmitis and is caused by Staphylococci, Streptococci, Pseudomonas, Pneumococcus. In suppurative inflammation, there occur outpouring of purulent exudates and an increase in polymorphonuclear cells. The uveal tissue increases in size, becomes thickened, and demonstrates necrosis and pus formation in the cavities. Non-granulomatous Uveitis In non-granulomatous (NG) uveitis, diffuse inflammation is characterized by dilatation and increase in permeability of vasculature, blood-aqueous barrier breakdown, increase in exudates, increased infiltration by lymphocytes, plasma cells and, macrophages. The iris becomes edematous, muddy in color, and waterlogging occurs. As a result, there occur loss of crypts and furrows. Further, there is iridoplegia; the pupil becomes small and sluggish due to the engorgement of iris vessels. The infiltration of exudates and lymphocytes manifests as cells and flare in the anterior chamber and KPs at the back of endothelium. The exudates can also be seen in the posterior segment and manifest as posterior synechiae, where the posterior iris adheres to the anterior lens capsule. Sometimes exudates may occupy the ciliary body and result in a cyclitic membrane. In the healing phase, atrophy of the iris is evident, and there may be areas of necrosis. Further destructive changes manifest as necrosis, gliotic and fibrotic areas, resulting in adhesions, scar formation, and eventual destruction of uveal tissue. Granulomatous Uveitis In granulomatous uveitis, there is increased infiltration by lymphocytes, plasma cells, epithelioid cells, and giant cells resulting in nodule formation. The nodules may be seen at the pupillary border, such as Koeppe's nodules) and nodules may be seen at the back of the cornea as mutton fat KP's. This may be followed by necrosis, gliotic and fibrotic areas which result in adhesions, scar formation, and eventual destruction of uveal tissue. Histopathology The histopathological findings in uveitis vary with etiology. Lens induced uveitis - Macrophages, acute and chronic inflammatory cells Parasitic uveitis- Eosinophils, polymorphonuclear neutrophils, and parasites occasionally Masquerade syndrome- Cellular infiltrates in retinoblastoma, leukemia, etc. and malignant cells Phacoanaphylactic uveitis - Epithelioid cells, polymorphonuclear neutrophils, and giant cells Uveitis- Predominant lymphocytic infiltrate Late-onset Endophthalmitis -Propionibacterium on gram stain, culture plating, or PCR Endophthalmitis -Acute and chronic inflammatory cell infiltrate along with offending microorganism Lymphomas (large cell) -Cells with pleomorphism, oval or circular nuclei, and scanty cytoplasm. There is an occasional predominance of micronuclei History and Physical Classification The anatomical classification of uveitis is proposed by the Standardization of Uveitis Nomenclature (SUN) Working Group. Anatomical Anterior- Involving the iris and ciliary body Intermediate- Involving vitreous and pars plana Posterior- Involving retina and choroid Panuveitis- Anterior, intermediate and posterior The Etiological Classification of Uveitis is Proposed by the International Uveitis Study Group (IUSG) Infectious- Viral, bacterial, parasitic, Lyme disease, etc. Non-infectious- Idiopathic or associated with systemic diseases Masquerade- Neoplastic and non-neoplastic SUN Working Group Proposed the Classification Based on the Timing of Inflammation Timing of onset- Insidious or sudden Duration of inflammation- Limited (less than or equal to 3 months ) or persistent Clinical Activity Acute- Insidious or sudden Chronic- Persistent with relapse within three months after stopping the treatment Recurrent- Repeated episodes with an in-between inactive period Remission- No active episode for three months or more Anterior Uveitis Acute Pain, redness, photophobia, tearing, and decreased vision with pain developing over a few hours or days except in trauma cases. Chronic Blurred vision, mild redness, and little pain or photophobia except during an acute episode. Slit Lamp Evaluation in Uveitis Patients Circumciliary congestion or circumcorneal congestion (injection) gives the appearance of a purple hue due to involvement in conjunctival plexus and deeper vessels. It's a sign of AAU and may not be seen in CAU. Pupillary miosis is due to iridoplegia and sphincter spasms, which lead to the formation of posterior synechiae. Anterior chamber cells are estimated by SUN grading with the help of a 1x1 mm slit beam under adequate lighting and magnification. The cells and flare indicate inflammatory activity inside the anterior chamber. The cells and flare should be assessed before pupillary dilatation as dilatation invites pigment shedding into the aqueous. The cells can also be observed in the vitreous phase. SUN Grading of Cells in Anterior Chamber (1 mm x 1 mm Slit Beam) Table S. No Grade An anterior chamber or aqueous flare is the turbid appearance of the aqueous in the anterior chamber due to the proteins secondary to the breakdown of the blood-aqueous barrier. In children with JIA-associated CAU, flare is taken as the level of inflammatory activity more than cells and indicates complications in the long term. Flare is graded on a slit lamp by assessing the iris and lens details. Laser flare photometry is another modality for objective grading. SUN Classification of Flare in the Anterior Chamber Table S. No Grade Hypopyon is the presence of whitish purulent exudative material in the anterior chamber. It is composed of inflammatory material, and form appears as a horizontal level that is gravity-dependent. It is common in HLA B27 associated uveitis. The hypopyon is immobile due to high fibrin content, making it difficult to absorb. Behcet's disease patients have a characteristically mobile hypopyon due to low fibrin content. Hypopyon is inflammatory cells deposited in the anterior chamber and is composed of leucocytes. It is measured with the help of a slit lamp micrometer. The shape, color, and consistency of hypopyon are indicators of etiology. It can be mobile or fixed. To test the mobility of hypopyon, the patient is asked to lie down supine for 10 minutes. Then after 10 minutes, the level of hypopyon is tested. In the case of fixed hypopyon, there is no movement, but in the case of mobile hypopyon, the level or height is changed. Various Types of Hypopyon in Different Condition Table S. No Hypopyon characteristic Keratic precipitates (KPs) are inflammatory cellular deposits at the back of the cornea (endothelium). The cellular aggregates include lymphocytes, plasma cells, and macrophages. Epithelial cells and giant cells are also common in KPs. The KPs are seen inferior one-third of the cornea in a triangular pattern with the apex pointing upwards (Arlt's triangle). The KPs are distributed in Arlt's triangle due to aqueous convection currents and the effect of gravity. In contrast, they can be seen diffusely distributed in Fuch's heterochromic iridocyclitis (FHIC) and viral uveitis (stellate KPs). The morphological appearance of KPs indicates the type of uveitis, like small KPs in non-granulomatous (NG) uveitis and large in granulomatous uveitis. Large glistening greasy KPs are seen as granulomatous uveitis and are called mutton fat KPs. Once the acute episode resolves, KP usually disappears. Pigmented KPs can be observed (NG variety). The granulomatous KPs may also become pigmented and are classically described as the ground-glass appearance of KPs. Sometime before the appearance of KP's diffuse pigment deposition is seen at the back of the cornea called endothelial dusting. Table S. No Characteristic The fibrinous membrane is a common entity in AAU and is seen in the anterior chamber. Iris atrophy is another clinical sign of uveitic sequelae. Diffuse iris atrophy is observed in FHIC, and a patchy or sectoral pattern appears in herpetic uveitis. Sectoral iris atrophy can be seen more in herpes zoster uveitis, although both sectoral and diffuse patterns can be observed in both herpes simplex and zoster uveitis. Iris nodules are observable in both NG and the granulomatous variety. Koeppe's nodules - These are present at the pupillary margin or the location of posterior synechiae. These are ectodermal nodules usually white and may be pigmented. They can be seen in granulomatous as well as non granulomatous uveitis. Busacca's nodules - These involve the iris stroma and are predominantly seen in granulomatous variety. Roseolae - These are yellow nodules are seen in Syphilis, resulting from dilatation of iris vessels. Berlin nodules - These are seen in angles in cases with sarcoidosis. Iris pearls - These are observed in lepromatous uveitis. Russell bodies - These, also called iris crystals, are seen in FHIC and rare cases of chronic uveitis. They are thought to result from immunoglobin deposits. Posterior synechiae (PS) are another essential clinical finding in uveitic cases. They result from adhesion of the pupil margin to the anterior capsule secondary to uncontrolled inflammation. PS can be seen at the site of the Koeppe nodule. All cases should be instilled with a cycloplegic agent like atropine or homatropine to prevent PS formation. Types of Posterior Synechiae Segmental- They are adhesions of the iris at some points over the lens Annular (ring) - They are 360-degree adhesion of the iris to the anterior lens capsule Total- These are plastered posterior surface of iris to the lens Broad- They are seen in tubercular uveitis Filiform-They are seen in non-tubercular uveitis Seculusio pupillae- When synechiae cover the 360-degree pupil and prevent migration of aqueous from posterior to the anterior chamber Occlusio pupillae- Formation of membrane over the lens surface Festooned pupil- When synechiae are localized, and on dilatation, the pupil takes an irregular shape Heterochromic iris is the difference in the color of the iris between the two eyes. Heterochromia is seen in FHIC. Neovascularization of the iris (rubeosis iris) can result from chronic inflammation. The vascularization seems to be less as compared to posterior segment pathologies. FHIC shows abnormal iris vessels but does not result in synechial angle closure. Iris's new vessels are also seen in posterior uveitis. Sometimes it becomes difficult to differentiate between new and normal iris vessels (pseudo-rubeosis of the iris). The differentiating point is new vessels show irregular branching in contrast to normal vessels, which are radial. Fluorescein angiography is used to assess the leakage from new vessels. Fundus evaluation is mandatory in each case to rule out masquerades like malignant melanoma, retinal detachment, cystoid macular edema, and other posterior segment complications. Difference between Granulomatous and Non-Granulomatous Uveitis Table S. No Characteristic Difference between Acute and Chronic Uveitis Table S. No Characteristic Evaluation Since iritis involves anterior uveitis per se, the evaluation and treatment are focussed on anterior uveitis only. Further laboratory tests or imaging may be required if systemic involvement or infectious disease is suspected to be the underlying cause. In most uveitic cases, the investigations are normal, and no underlying etiology can be discerned. Tailored made investigations are needed in each patient to pinpoint the etiology based on the clinical signs. In some cases, the etiology is evident on the clinical examination, like uveitis in the case of endophthalmitis post-surgery. Most patients also need systemic evaluation and careful review by an internist. Visual Acuity Visual acuity depends on the extent of inflammatory activity and associated complications. It is generally minimally reduced in AAU. Snellen's uncorrected and best-corrected visual acuity should be documented in each patient. The improvement in visual acuity is also an indicator of resolution of uveitis and reduction in inflammation. Intraocular Pressure Intraocular pressure (IOP) can be raised in uveitic cases. The reduction in IOP is due to ciliary shutdown or reduction in aqueous secretion from the ciliary epithelium. The raised IOP can be due to angle closure, trabeculitis, steroid-induced, and other causes. The IOP can be measured by noncontact tonometry, ICARE tonometer, or Goldman applanation tonometer and should be documented on each visit. Cases of Uveitis where Investigations are Not Mandatory The first episode of unilateral mild NG AAU with no previous ocular or systemic history Classical clinical signs of a specific etiology for which investigation will not be helpful like FHIC When the systemic signs and symptoms are indicative of a particular etiology associated with uveitis like sarcoidosis, tuberculosis, multiple sclerosis, etc. Cases of Uveitis where Investigations Will be Helpful Bilateral cases of AAU Severe cases AAU Recurrent cases of AAU Chronic A.U. resistance to treatment Granulomatous AU AAU associated with I.U. or posterior uveitis or panuveitis cases Ocular and systemic signs indication underlying pathology Further laboratory tests or imaging may be required if systemic involvement or infectious disease is suspected to be the underlying cause. Imaging Chest radiography may be considered if sarcoidosis or tuberculosis is the underlying cause of uveitis. In granulomatous uveitis cases, a high index of suspicion should be kept to rule out these treatable systemic conditions as well as syphilis. X-ray of the sacroiliac joint reveals bamboo spine and sacroiliitis in cases with ankylosing spondylitis and other seronegative spondyloarthropathies. Ocular Imaging B Scan Ultrasound A B scan should be performed in the fundus that is not visualized due to small pupil, anterior chamber reaction, fibrinous membrane, or vitreous haze. Optical Coherence Tomography It's a handy tool to rule out cystoid macular edema and epiretinal membrane and look for cells and flare. Fundus Fluorescein Angiography This is an essential investigation in chronic uveitis cases to rule out posterior segment pathology like vasculitis macular ischemia delineate capillary non-perfusion areas and white dot syndrome spectrum. Fundus Autofluorescence Fundus autofluorescence (FAF) is essential in posterior segment pathologies to rule out white dot syndromes and severe inflammatory conditions. Indocyanine Green Angiography Indocyanine green angiography (ICGA) is needed in cases with uveitis-associated choroidal pathology. Ultrasound Biomicroscopy Ultrasound biomicroscopy (UBM) is helpful to rule out hypotony (ciliary shutdown), choroidal effusion, cyclodialysis cleft, and cyclitic membrane. Laboratory Workup Laboratory workup is usually not necessary. In mild unilateral non-granulomatous uveitis with trauma or no evidence of systemic disease, laboratory studies are unlikely to be helpful. If there is the presence of bilateral granulomatous or recurrent uveitis, a workup is indicated. Tests to consider include: Human Leucocyte Antigen tissue typing (HLA-B27) HLA- B27 testing should be done in any patient with recurrent A.U. or chronic N.G. A.U. The various HLA types and their associations are detailed below. HLA B27- Recurrent AAU HLA-B51 and HLA B5- Behcet's disease HLA-A29- Birdshot retinochoroiditis HLA-DR4- Sympathetic Ophthalmia and Vogt-Koyanagi-Harada disease HLA-B7 and HLA-DR- Presumed Ocular Histoplasmosis Syndrome Syphilis Serology The various test implicated in syphilis is treponemal antibody tests such as ELISA- enzyme-linked immunosorbent assay, which has high specificity and sensitivity but takes approximately three months to show positivity. Anti-cardiolipin antibody tests (non-specific) like venereal research laboratory test (VDRL) and rapid plasma regain (RPR) demonstrate positive results in the acute phase of the disease and help to monitor the disease and treatment. The patients having features of ocular syphilis should be referred to an internist to rule out sexually transmitted syphilis. Angiotensin-Converting Enzyme This is a non-specific test to rule out the cause of N.G. A.U. such as sarcoidosis, tuberculosis, and Hansen disease. Serum levels of ACE are elevated in approximately 80% of patients with sarcoidosis but come down to normal levels during remission. Serum Lysozyme Assay Lysozyme, an enzyme, is found in neutrophils and tear secretion. It has antibacterial properties and can cause bacterial cell wall breakdown. This test has comparatively less sensitivity and specificity than serum ACE in diagnosing sarcoidosis, but the combined test may prove more beneficial. Complete Blood Count A raised leucocyte count is indicative of infective may be a hematological malignancy. Eosinophils are raised in parasitic causes of uveitis. Inflammatory Markers C-reactive protein (CRP), erythrocyte sedimentation rate (ESR), and acute phase reactants may be elevated in systemic inflammatory disorders. Lyme Disease Serology Disease-specific serology is essential in endemic areas, including serology for other infections like brucellosis and leptospirosis. Antinuclear Antibody Antinuclear antibody (ANA) is the most important marker for juvenile idiopathic arthritis in children. ANA positivity is an indicator of CAU. Antineutrophil Cytoplasmic Antibody Antineutrophil cytoplasmic antibody (ANCA) is valuable for anterior uveitis associated with scleritis. It is a valuable marker for Wegner granulomatosis (c-ANCA) and Polyarteritis Nodosa (PAN) {p-ANCA}. Interferon-Gamma Release Assay Interferon-gamma Release Assay such as QuantiFERON-TB Gold for diagnosis of tuberculosis. HIV Serology HIV tridot test for suspected patients with immunosuppression. Anterior Chamber Tap An anterior chamber aqueous tap can be performed, and a sample can be sent for a polymerase chain reaction in cases of suspected viral uveitis like herpes, varicella, and rubella. PCR will also distinguish Propionibacterium acne infection in chronic uveitis in pseudophakic eyes. Iris Biopsy Iris biopsy is another critical investigation but is rarely done nowadays. Vitreous Biopsy In the case of associated posterior segment involvement, especially in the case of suspected endophthalmitis. Conjunctival Biopsy In case of suspected granulomas or infiltrative lesions, a conjunctival biopsy is indicated. Physician Referral To rule out systemic pathologies, timely referral to an internist is essential to pinpoint the etiology—for example, HRCT in pulmonary tuberculosis and MRI in cases of multiple sclerosis. Treatment / Management Treatment is primarily aimed at reducing inflammation and pain and preventing complications. First-line treatment involves topical cycloplegics (dilate the pupil, prevent the ciliary body and pupillary spasm) and topical steroids (decrease inflammation). The patient should be referred to an ophthalmologist within 24 to 48 hours. The review frequency is decided based on the clinical presentation and degree of inflammatory activity. Those with severe inflammation need review within a day or two, and those with mild AAU may be seen after a week or so. Sustained-release corticosteroid vitreous implants (fluocinolone acetonide, dexamethasone) are available to treat inflammation-induced cases of panuveitis, intermediate uveitis, and posterior uveitis. Corticosteroids should be initiated only in conjunction with the approval of an ophthalmologist because uveitis is a diagnosis of exclusion. Steroids can have adverse effects, such as causing intraocular pressure, cataract formation, steroid-induced glaucoma, and the development of herpes keratitis. Potassium-sparing drugs should be used when chronic steroid use is required to control inflammation. Approximately half of the patients with uveitis need treatment beyond corticosteroid treatment to prevent vision loss. Topical Steroids Corticosteroids decrease inflammation. Treatment should only be initiated after consultation with an ophthalmologist. Prednisolone 1% or dexamethasone 0.1% are potent steroids and are the first treatment choice for uveitis. They work by decreasing inflammation by reversing increased capillary permeability and suppressing the migration of polymorphonuclear leukocytes. The other drugs used for iritis management are difluprednate 0.05% (lower frequency), prednisolone 0.5%, loteprednol etabonate 0.5% and 0.2%, betamethasone, fluorometholone, and rimexolone (moderate to low potency). Steroid regimen can be altered based on the response and intraocular pressure elevation. Betamethasone ointment can be used at bedtime. The steroid regimen varies according to the degree of inflammation, starting from an hourly regimen for three days to every two hours for three days, then 6/5/4/3/2/1 times one week each. The treatment is usually stopped after 6 to 8 weeks. Timely and regular follow-up is mandated in each case. In CAU cases, the target is the full resolution of inflammation. Low-grade persistent inflammation can result in a higher incidence of complications which can be reduced with regular and timely steroids instillation. Steroids can act as a bilateral sword, resulting in elevated IOP (steroid response), cataract, glaucoma, keratitis, and corneal melt. Cycloplegics Cycloplegics block nerve impulses to the ciliary muscles and pupillary sphincter to decrease photophobia and pain. They are implicated in AAU and CAU exacerbations. These help to break down the already formed posterior synechiae prevent the formation of new synechiae. Homatropine Induces cycloplegia in 30 to 90 minutes. Induces mydriasis in 10 to 30 minutes. Effects last 10 to 48 hours for cycloplegia and 6 to 96 hours for mydriasis, but the duration may be less if severe anterior chamber reaction. Homatropine is an agent of choice for uveitis. Cyclopentolate 0.5% to 2% Induces cycloplegia in 25 to 75 minutes. Induces mydriasis in 30 to 60 minutes. Effects usually last one day. Less attractive for treating uveitis than homatropine. The other cycloplegics available are atropine which has a long duration of action of 7-10 days. In AAU, 2.5% or 10% phenylephrine can also break posterior synechiae. Bedtime administration of the drug should be promoted to avoid difficulties with accommodation during the daytime. Atropine has been known to cause toxic side effects like seizures amblyopia in children, and extra precautions should be taken while administering drugs to children. Mydricaine Mydricaine is a preparation of 0.3 ml containing 0.12 mg adrenaline, 1mg atropine, and 6 mg procaine. Adrenaline and atropine break the posterior synechiae and procaine to improve comfort. The drug combination is given as a subconjunctival injection. Another installation method is cotton pledget dipped in Mydricaine and kept for 5 minutes in the superior and inferior fornices for 5 minutes. Mydricaine number 1 is used pediatric version, and mydricaine number 2 is used in adults. The side effect reported with Mydricaine is transient sinus tachycardia. Tissue Plasminogen Activator In severe fibrinous anterior uveitis cases, 12.5 to 25 ug of tissue plasminogen activator is injected in the anterior chamber in 0.1 ml solution with a 30 G needle under topical anesthesia. This will dissolve the fibrin membrane and break down the posterior synechiae. Subconjunctival Steroids Betamethasone sodium phosphate and acetate combination can be given for severe A.U. and patients with poor compliance subconjunctivally. Subtenons Injection of Steroids Posterior sub tenons injection of methylprednisolone or triamcinolone acetonide can be given in cases with aggressive posterior segment inflammation and in patients with anterior uveitis having cystoid macular edema a complication. They can be instilled as an OPD procedure or given during the surgery to supplement the systemic steroids. The onset of action is four weeks, and the maximum duration is around three months. Varied complications have been reported like secondary glaucoma, globe perforation, subconjunctival hemorrhage, cataract, retrobulbar hemorrhage, and ptosis. Intravitreal Steroids Intravitreal triamcinolone acetonide (4 mg in 0.1 ml) is also implicated in cystoid macular edema after anterior uveitis, which is not responding to other therapy. The complications are similar to sub tenons administration except pseudohypopyon, exclusive to intravitreal steroids. Systemic Steroids When the inflammation is severe and response to topical treatment is inadequate, oral prednisolone is indicated. They are given in tapering doses, and each case should monitor side effects. Non- Steroidal Anti- Inflammatory Drugs (NSAIDs) Naproxen and tolmetin are implicated in CAU cases and can be used for the long term under physician observation. Antimetabolites Methotrexate is implicated in CAU secondary to juvenile idiopathic arthritic in children when steroids fail to control the inflammation. Tumor Necrosis Factor Blockers Infliximab or adalimumab may be used as second-line treatment for patients with vision-threatening chronic uveitis caused by seronegative spondyloarthropathy. Differential Diagnosis The differential diagnoses include Acute angle-closure glaucoma Conjunctivitis Subconjunctival hemorrhage Trauma Episcleritis Scleritis Dry eyes Pingueculitis Inflammed pterygium Corneal abrasion Corneal ulcer HSV keratitis Intraocular foreign body Scleritis Ulcerative keratitis Ultraviolet keratitis Prognosis The prognosis is good with appropriate treatment. To have the best prognosis, follow-up care with an ophthalmologist within 24 hours is imperative. Monitoring should include repeat slit-lamp and intraocular pressure checks every few days. When the condition is stable, monitoring can be every 1 to 6 months. If not diagnosed and treated on time and neglected cases usually develop complications like cataract, glaucoma, retinal detachment, and macular edema, and the prognosis is guarded in these cases. Complications Corneal scar (keratouveitis) Band shaped keratopathy Hyphema Non-resolving hypopyon Iris atrophy Posterior synechiae Cataract (Inflammation related, steroid-induced) Secondary glaucoma (pupillary block, inflammation, or topical corticosteroid) Occlusio pupillae Festooned pupil Seclusio pupillae Cystoid macular edema Optic neuropathy Optic disc neovascularization Macular scar Hypotony Vitreous hemorrhage (associated intermediate, posterior, or panuveitis) Retinal detachment Endophthalmitis Panophthalmitis Permanent blindness Strabismus Postoperative and Rehabilitation Care In cases undergoing cataract surgery, postoperative care is required for complicated cataract of filtration surgery for non-resolving glaucoma or progressive glaucomatous changes. For patients undergoing cataract surgery, the usual regimen is topical steroids (1% prednisolone or 0.1% dexamethasone) or steroids and antibiotic combination (0.1% dexamethasone + 0.5% gatifloxacin) 6/5/4/3/2/1 times for one week each along with a cycloplegic like topical 5% homatropine or 1% atropine two times per day for 2 to 3 weeks. The patients should be regularly and closely followed to prevent any complications. In patients undergoing filtration surgery, the usual regimen is topical steroids (prednisolone 1% or dexamethasone 0.1%) or steroids and antibiotic combination (dexamethasone 0.1% + gatifloxacin 0.5%) 8/7/6/5/4/3/2/1 times for one week each along with a cycloplegic like topical homatropine 5% or atropine 1% two times per day for 2 to 3 weeks. These patients should be closely followed up to rule out complications like macular snuff out, hypotony, uncontrolled glaucoma and check for the normal functioning bleb. The patients requiring retinal detachment repair are also treated post-operatively with the same regimen as glaucoma patients. Two to three months later, these patients will require silicon oil removal. All these patients should also be closely followed up to rule out re-detachment and other complications of retinal surgery. Consultations Any patient presenting to the clinic with pain, redness, and photophobia should be meticulously evaluated by an Ophthalmologist with suspicion of iritis. In the case of keratouveitis, the patient should be referred to a cornea specialist for higher opinion and management to rule out infective and autoimmune keratitis. Non-resolving iritis, endophthalmitis, or cases with diagnostic dilemmas must be evaluated by a retina and uvea specialist for expert opinion and further management. The patient with underlying systemic pathology leading to iritis and uveitis should be referred timely to an internist to pinpoint the etiology and targeted systemic treatment. Iritis with trabeculitis, angle-closure, or open-angle glaucoma should be managed with a glaucoma specialist. Patients with complicated cataracts requiring surgical intervention should be ideally operated on by a cataract and IOL surgeon for an excellent visual outcome. Deterrence and Patient Education Patients with iritis should be explained about the ocular pathology the importance of regular and timely treatment and follow-up. The patient should be educated about the underlying systemic pathology associated with iritis and the critical role of the internist in treating the condition. The patients should also be explained about the complications related to iritis and the side effects of irregular and long-term use of steroids. Pearls and Other Issues Anterior uveitis can present as an isolated entity or associated intermediate and posterior uveitis. There can be a spillover of anterior segment inflammation into the vitreous or posterior segment. All patients should undergo extensive evaluation to pinpoint the etiology. The internists have a key role to play in the diagnosis and management of uveitic cases. If untreated, complications can include decreased visual acuity and/or blindness, glaucoma, cataracts, and irregular pupil. Enhancing Healthcare Team Outcomes The management of a patient with iritis is interprofessional. Whenever a patient presents with eye pain, tearing, photophobia, vision loss, and red-eye in the absence of trauma, the patient must be referred to an ophthalmologist as soon as possible. The treatment of iritis primarily aims to reduce inflammation and pain and prevent complications. First-line treatment involves topical cycloplegics (dilate the pupil, prevent the ciliary body and pupillary spasm) and topical steroids (decrease inflammation). Depending on the cause, most patients respond well to treatment and retain full vision. However, at least 10 to 30% of patients may need treatment beyond steroids to prevent vision loss. [Level 5] Once discharged, the patient may follow up with the ophthalmic nurse, primary care provider, or ophthalmologist. Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. Figure Slit-lamp image of the patient of recurrent anterior uveitis depicting circumciliary congestion, corneal scarring, central band shaped keratopathy, hypopyon in anterior chamber and pseudophakia Contributed by Dr. Bharat Gurnani, MBBS, DNB, FCRS, FICO, (more...) Figure Slit- lamp image of the patient with recurrent uveitis depicting mild circumciliary congestion, nebular corneal scarring, few areas of iris atrophy along with a complicated cataract Contributed by Dr. Bharat Gurnani, MBBS, DNB, FCRS, FICO, MRCS Ed, MNAMS (more...) References 1. : Bloch-Michel E, Nussenblatt RB. International Uveitis Study Group recommendations for the evaluation of intraocular inflammatory disease. Am J Ophthalmol. 1987 Feb 15;103(2):234-5. [PubMed: 3812627] 2. : Jabs DA, Nussenblatt RB, Rosenbaum JT., Standardization of Uveitis Nomenclature (SUN) Working Group. Standardization of uveitis nomenclature for reporting clinical data. Results of the First International Workshop. Am J Ophthalmol. 2005 Sep;140(3):509-16. [PMC free article: PMC8935739] [PubMed: 16196117] 3. : Agrawal RV, Murthy S, Sangwan V, Biswas J. Current approach in diagnosis and management of anterior uveitis. Indian J Ophthalmol. 2010 Jan-Feb;58(1):11-9. 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[PMC free article: PMC4599638] [PubMed: 26491249] : Disclosure: Bharat Gurnani declares no relevant financial relationships with ineligible companies. : Disclosure: Jessica Kim declares no relevant financial relationships with ineligible companies. : Disclosure: Koushik Tripathy declares no relevant financial relationships with ineligible companies. : Disclosure: Navid Mahabadi declares no relevant financial relationships with ineligible companies. : Disclosure: Mary Ann Edens declares no relevant financial relationships with ineligible companies. Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK430909PMID: 28613659 PubReader Print View Cite this Page Gurnani B, Kim J, Tripathy K, et al. Iritis. [Updated 2023 Aug 25]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. In this Page Continuing Education Activity Introduction Etiology Epidemiology Pathophysiology Histopathology History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Complications Postoperative and Rehabilitation Care Consultations Deterrence and Patient Education Pearls and Other Issues Enhancing Healthcare Team Outcomes Review Questions References Related information PMC PubMed Central citations PubMed Links to PubMed Similar articles in PubMed Chorioretinitis.[StatPearls. 2025] Chorioretinitis. Geetha R, Tripathy K. StatPearls. 2025 Jan HLA-B27 typing in the categorisation of uveitis in a HLA-B27 rich population.[Br J Ophthalmol. 2000] HLA-B27 typing in the categorisation of uveitis in a HLA-B27 rich population. Huhtinen M, Karma A. Br J Ophthalmol. 2000 Apr; 84(4):413-6. Review Clinical Spectrum of HLA-B27-associated Ocular Inflammation.[Ocul Immunol Inflamm. 2017] Review Clinical Spectrum of HLA-B27-associated Ocular Inflammation. Pathanapitoon K, Dodds EM, Cunningham ET Jr, Rothova A. Ocul Immunol Inflamm. 2017 Aug; 25(4):569-576. Epub 2016 Jul 18. Expression profile of IL-1 family cytokines in aqueous humor and sera of patients with HLA-B27 associated anterior uveitis and idiopathic anterior uveitis.[Exp Eye Res. 2015] Expression profile of IL-1 family cytokines in aqueous humor and sera of patients with HLA-B27 associated anterior uveitis and idiopathic anterior uveitis. Zhao B, Chen W, Jiang R, Zhang R, Wang Y, Wang L, Gordon L, Chen L. Exp Eye Res. 2015 Sep; 138:80-6. Epub 2015 Jun 24. Review Clinical Features and Complications of the HLA-B27-associated Acute Anterior Uveitis: A Metanalysis.[Semin Ophthalmol. 2017] Review Clinical Features and Complications of the HLA-B27-associated Acute Anterior Uveitis: A Metanalysis. D'Ambrosio EM, La Cava M, Tortorella P, Gharbiya M, Campanella M, Iannetti L. Semin Ophthalmol. 2017; 32(6):689-701. Epub 2016 Jul 12. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Iritis - StatPearls Iritis - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... Follow NCBI Connect with NLM National Library of Medicine8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers
2915	https://www.quora.com/What-is-a-power-set	Something went wrong. Wait a moment and try again. Power Sets Axiomatic Set Theory Sets (mathematics) General Set Theory Elementary Set Theory Mathematical Sciences Set Theory (Computer Scie... Set Theory 5 What is a power set? Alan Bustany Trinity Wrangler, 1977 IMO · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) and Justin Rising , PhD in statistics · Author has 9.8K answers and 58.5M answer views · 8y Given a set, S, its power set, P(S), is the set of all subsets of S including the empty set and S itself. For example, given S={a,b},P(S)={{},{a},{b},{a,b}}. If the cardinality [ ] of S is \|S\|, then the cardinality of P(S) is \|P(S)\|=2\|S\| which is strictly greater than \|S\| by Cantor’s Theorem [ http Given a set, S, its power set, P(S), is the set of all subsets of S including the empty set and S itself. For example, given S={a,b},P(S)={{},{a},{b},{a,b}}. If the cardinality [ ] of S is \|S\|, then the cardinality of P(S) is \|P(S)\|=2\|S\| which is strictly greater than \|S\| by Cantor’s Theorem [ ]. In our example, \|S\|=2 so \|P(S)\|=22=4 as illustrated above. A set with three elements would have a power set with 23=8 members. When S is an infinite set, its power set is infinitely larger than S. For exampl... Related questions What is a power set example? What is the symbol for the power set? What is the power set of the empty set ∅ ? What is the point of power sets? What is the power set of { ∅ , { ∅ } } ? Michael Carroll Former Teacher (Mathematics - Secondary Level) · Author has 556 answers and 1.2M answer views · 7y Let’s look at this in little steps. First of all, let us start with some set which we will call set A. I will use the following notation for talking about the power set of A: P(A). P(A) will stand for another set. That is, the power set of A is also a set. What is in P(A)? We will fill this new set with all the subsets of A itself. Here is an example: Let A = {2, h, sun}. Now, let’s write out the power set of A. P(A) = { empty set, {2}, {h}, {sun}, {2,h}, {2,sun}, {h,sun}, {2,h,sun} } Why is this called the power set? Let’s find out. How many elements are in set A? There are three members in set A. How ma Let’s look at this in little steps. First of all, let us start with some set which we will call set A. I will use the following notation for talking about the power set of A: P(A). P(A) will stand for another set. That is, the power set of A is also a set. What is in P(A)? We will fill this new set with all the subsets of A itself. Here is an example: Let A = {2, h, sun}. Now, let’s write out the power set of A. P(A) = { empty set, {2}, {h}, {sun}, {2,h}, {2,sun}, {h,sun}, {2,h,sun} } Why is this called the power set? Let’s find out. How many elements are in set A? There are three members in set A. How many elements are in set P(A)? There are 8 members in set P(A). Now notice that 2^3 = 8. Is this a coincidence? No, it is not. If A has n elements, then P(A) will have 2^n elements. For this reason, another notation for P(A) is 2^A. Now, I hope this has made you curious enough to wonder if 3^A has any special meaning for sets. It does! But it is a bit too advanced to explain at this point in time. In fact, for two sets A and B, the set A^B has a special meaning, one that matches up with the sizes of A and B. Last, I would like to point out that the ideas being talked about here is one doorway into real mathematics. Mathematics is not really about arithmetic, adding, subtraction, fractions, decimals and all that stuff from primary and most of secondary school. In fact, only geometry really comes close to giving a hint as to what mathematics is really about. Mathematics is about understanding how things connect and the rules about how we describe and formulate those rules. Mathematics is to arithmetic much the same way grammar is to English. It is beautiful in and of itself. In many ways it is like poetry. Sanket Alekar I am fairly good at Math · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 633 answers and 1.5M answer views · 9y A power set of a given set S is the set of all subsets of S Example: Let S = {1,2,3,4} Power set of S = { {}, {1}, {1,2}, {1,2,3}, {1,2,4}, {1,2,3,4}, {1,3}, {1,3,4}, {1,4}, {2}, {2,3}, {2,3,4}, {2,4}, {3}, {3,4}, {4} } The number of subsets of a set is 2(number of elements of S) Why is this? Think about how you construct a subset of S. For each element in S, you have 2 choices: a) Pick it as part of your subset b) Don't pick it as part of your subset So for every element you have 2 choices. As the choices you make for each element are independent of each othe A power set of a given set S is the set of all subsets of S Example: Let S = {1,2,3,4} Power set of S = { {}, {1}, {1,2}, {1,2,3}, {1,2,4}, {1,2,3,4}, {1,3}, {1,3,4}, {1,4}, {2}, {2,3}, {2,3,4}, {2,4}, {3}, {3,4}, {4} } The number of subsets of a set is 2(number of elements of S) Why is this? Think about how you construct a subset of S. For each element in S, you have 2 choices: a) Pick it as part of your subset b) Don't pick it as part of your subset So for every element you have 2 choices. As the choices you make for each element are independent of each other, you have: 2 x 2 x 2 .... \|S\| times, where \|S\| = # of elements in S = 2\|S\| Pronoun Jow 4y Say you have the set {3,2,1}, which has 3 elements. The order of the elements in the set does not matter, but I ordered them this way on purpose and I'll explain why later. A subset of this set would have either all the elements of this set, some elements, or none of the elements of this set. I'll demonstrate by listing the subsets of the set {3,2,1} and following each subset with some notation to show which elements of the set {3,2,1} are in that subset, writing 0 if that element is not in that subset, and 1 if it is: (I'm ordering them a little weirdly here and I'll explain why later) (Each sub Say you have the set {3,2,1}, which has 3 elements. The order of the elements in the set does not matter, but I ordered them this way on purpose and I'll explain why later. A subset of this set would have either all the elements of this set, some elements, or none of the elements of this set. I'll demonstrate by listing the subsets of the set {3,2,1} and following each subset with some notation to show which elements of the set {3,2,1} are in that subset, writing 0 if that element is not in that subset, and 1 if it is: (I'm ordering them a little weirdly here and I'll explain why later) (Each subset is on the left side of each arrow) (Do not confuse the square brackets on the right side of each arrow for sets. Those are not sets!) ("//" denotes a comment about that specific subset) {} -> [0,0,0] // Empty set, with no elements {1} -> [0,0,1] {2} -> [0,1,0] {2,1} -> [0,1,1] {3} -> [1,0,0] {3,1} -> [1,0,1] {3,2} -> [1,1,0] {3,2,1} -> [1,1,1] // The set itself is a subset of itself POWER SET: (By the way, this set of subsets is the power set of the set {3,2,1}) As you can see, the set {3,2,1} has 8 subsets. Why 8? 8 is equal to 2^3, with 2 being the number of answers to the question "Is that element in the subset?", the answers being no, or 0, and yes, being 1. 3 is the number of elements in the set {3,2,1}. Okay, but why is 2^3 the formula I'm using? Why did I order the elements in the original set as {3,2,1} and not {1,2,3}? Why did I order the list of subsets like that? Look at my yes/no square bracket notation for each subset. I actually wrote the numbers 0 to 7 in binary, in ascending numerical order (this is the answer for both orderings that I chose to use), and separated the digits in each binary number with commas. Each digit represents a no or a yes for each element in the set {3,2,1}, depending on which subset is being described. There are only 2 digits in binary, AKA base 2: 0 and 1. We humans normally count in decimal, or base 10, using the 10 digits from 0 to 9. After 9, we count 10 since we don't have a unique digit for that number without using a base higher than 10 instead, then we count 11 and so on. 0 in decimal is 0 in binary. 1 in decimal is 1 in binary. 2 in decimal is... 10 in binary. We ran out of digits in binary and the number that is 2 in decimal does not have a unique digit in binary. We have to move up to the next place to continue. 3 in decimal is 11 in binary, AKA 10+1 in binary. 4 in decimal is... 100 in binary! Once again we have to move up to the next place to continue. 5 in decimal is 101 in binary, AKA 100+1 in binary, which would be 4+1 in decimal. 6 in decimal is 110 in binary, AKA 100+10 in binary. 7 in decimal is 111 in binary. Okay, but why was I starting at 0 and going to 7 earlier in this post? Why not 1 to 8? Because 8 in decimal is 1000 in binary. We have to move up to the next place to continue, which means we now have 4 digits. However, the set {3,2,1} has only 3 elements. The 4th digit in 1000, that '1', has nothing to describe, so we stop at 7, or 111 in binary. We only need 3 digits to describe 3 elements, using 1 digit per element. As for 0, that number tells us that there are no elements in that subset, making it the empty set. Yes, I said "the". The empty set is universal. How about the 2^3 formula? Look at the number 789 in decimal, AKA base 10. There's a 9 in the ones, or 10^0 (which equals 1), place, an 8 in the tens, or 10^1, place, and a 7 in the hundreds, or 10^2 (which equals 100 in decimal), place. The exponent 'x' in 10^x represents the placement of the digit, or index, and the base 10 represents... base 10, AKA decimal. That's why decimal is also called base 10. Anyway, 789 in decimal gives us seven 10^2s, eight 10^1s, and nine 10^0s. Now look at the number 1000 in binary, AKA base 2. There's a 0 in the ones, or 2^0 (which equals 1) place, a 0 in the twos, or 2^1, place, a 0 in the fours, or 2^2 (which equals 4 in decimal), place, and a 1 in the eights, or 2^3 (which equals 8 in decimal), place. Instead of 10^x, we have 2^x for binary because binary is base 2. Anyway, 1000 in binary gives us only one 2^3, or one 8 in decimal. 1000 in binary is 8 in decimal. That's where the 2^3 formula, or the 2^x formula, for x elements in a set and 2^x subsets of that set, comes from. That's also why a power set is called a power set. 2 to the power of 'x'... Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. 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It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Related questions What is power set defined with examples? What is the power set of { { a , b } , c } ? How do you find the power set of A = (2, (5))? Is the power set of the power set of a set equal to the power set of that set? What is the power set of a set? Ahaan Rungta MIT '19 · 11y The power set of a set S is the set of all subsets of S . Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Janice Retabale 5y Originally Answered: What is a power set in math? · In mathematics, the power set (or powerset) of any set S is the set of all subsets of S, including the empty set and S itself, variously denoted as P(S), 𝒫(S), ℘(S) (using the "Weierstrass p"), P(S), ℙ(S), or, identifying the powerset of S with the set of all functions from S to a given set of two elements, 2 S . Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Manish Singh 10y Originally Answered: What is power set? · In maths, power set is collection of all the subsets of a set. Let s={a,b,c} be a set.Then collection of subsets of set s is power set. It has 2^(No. Of elements of s). In this case 2^3 =8.So power set has 8 elements as shown. In maths, power set is collection of all the subsets of a set. Let s={a,b,c} be a set.Then collection of subsets of set s is power set. It has 2^(No. Of elements of s). In this case 2^3 =8.So power set has 8 elements as shown. Matias Relyea (贺彦龙） Studied Mathematics & Music (violin) at North Carolina School of Science and Mathematics (NCSSM) (Graduated 2025) · Author has 205 answers and 349.7K answer views · 4y A power set is the set of all of the subsets of a primary set A, and is denoted P(A), where A is the primary set. A subset is a set in which its elements are present in a larger set A but the converse is not true. For example, B={1,2} is a subset of A={1,2,3} because 1,2 is present in A, but A is not a subset of B because B does not contain 3. We denote this subset relationship as B⊂A. To define a power set, consider a set A={a,b,c}; to find all of its subsets, find all the combinations of elements that form sets: {{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}} - A power set is the set of all of the subsets of a primary set A, and is denoted P(A), where A is the primary set. A subset is a set in which its elements are present in a larger set A but the converse is not true. For example, B={1,2} is a subset of A={1,2,3} because 1,2 is present in A, but A is not a subset of B because B does not contain 3. We denote this subset relationship as B⊂A. To define a power set, consider a set A={a,b,c}; to find all of its subsets, find all the combinations of elements that form sets: {{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}} - insert these sets into another set, and you have the power set of A or P(A)={{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}}. Notice how as the subsets of A are inserted into P(A), they transform into elements. They still act as sets within a larger set, but by definition items within a larger set are elements, so we know that each of those subsets must be an element of A. Symbolically, we can say that {a}∈P(A) - but wait, can we say that {a}⊂P(A)? Certainly not, because we defined {a}∈P(A). Then what may a subset of P(A) look like? Theoretically, it is a set containing the element a which was previously a subset of A - then naturally, wouldn’t a subset of P(A) look like {{a}}? If we extend this and look at combinations of other elements as subsets, we realize that {{{a},{b}},{{b},{c}},{{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}}}⊂P(A) as well as a many of other combinations of elements in the form of sets. By combining their entirety, we find a power set of a power set, denoted P(P(A)). Of course, the empty set ∅ is another element, but in our situation, there is no need to include it because it would be difficult if not pointless to manually determine every element of the power set of the power set. To determine the number of elements in one such power set of a power set, denoted ∣P(P(A))∣, also known as a set’s cardinality, you must use combinatorics and counting, which involve Cartesian Products (The set C={(a,b)∋a∈A∧b∈B} where set C=A×B) among other things. As this is purely an answer explaining the power set P(A), its cardinality among other things aren’t extremely relevant. Sponsored by CDW Corporation How do updated videoconference tools support business goals? Upgrades with CDW ensure compatibility with platforms, unlock AI features, and enhance collaboration. Lawrence Stewart Research Computer Scientist · Author has 10.9K answers and 21.9M answer views · 4y Originally Answered: What is the meaning of a power set? · In words, the power set is the set of all subsets of some set. It is hard to figure out what that means, so I think of it differently. I think of a set as a collection of bits, with each bit standing for an element of the set. A set with N elements will have an associated N bit representation. If you consider an arbitrary N bit number, it represents a subset of the main set, with the “1”s standing for elements present in the subset and “0”s standing for elements not present in the subset. All 0s is the empty set, and All 1s is the full set. Now we can see that this N bit number can represent In words, the power set is the set of all subsets of some set. It is hard to figure out what that means, so I think of it differently. I think of a set as a collection of bits, with each bit standing for an element of the set. A set with N elements will have an associated N bit representation. If you consider an arbitrary N bit number, it represents a subset of the main set, with the “1”s standing for elements present in the subset and “0”s standing for elements not present in the subset. All 0s is the empty set, and All 1s is the full set. Now we can see that this N bit number can represent subset. All the possible subsets are there, and no others. This is why the power set has 2^N elements when the main set has N elements. Charles S. former mathematician, current patent lawyer · Author has 7.6K answers and 60.4M answer views · 9y Originally Answered: What are power sets used for? · “Used for” is kind of a funny phrase in this context, since set theory is a relatively abstract topic. I mean, I can tell you what a screwdriver is used for, but power sets… I dunno. That linguistic quibble aside, perhaps the main thing a power set is good for is to provide a “universe” for other things to take place in. For example, a lot of areas of math begin by singling out certain subsets of a set X as special. Topology deals with open sets, measure theory deals with measurable sets, even nonstandard analysis needs a “non-principle ultrafilter” to get going. I won’t bother to explain what “Used for” is kind of a funny phrase in this context, since set theory is a relatively abstract topic. I mean, I can tell you what a screwdriver is used for, but power sets… I dunno. That linguistic quibble aside, perhaps the main thing a power set is good for is to provide a “universe” for other things to take place in. For example, a lot of areas of math begin by singling out certain subsets of a set X as special. Topology deals with open sets, measure theory deals with measurable sets, even nonstandard analysis needs a “non-principle ultrafilter” to get going. I won’t bother to explain what a non-principle ultrafilter is, except to say that it’s a subset of a power set, as are the collection of open sets in a topological space or the collection of measurable sets in a measure space. Maybe another useful fact about power sets is that the cardinality of X is always strictly less than the cardinality of P(X). That can be useful in certain constructions. Luke Joyce 7y Power set is a one type of set.Set is a collection of well defined distinct objects where Power set is a collection of all sets in S and also include empty sets.Power set represented by P(S). For example A= {1,2,3} B= {a,b} C= {8,9,10} P(S)= { {1,2,3} {8,9,10} {a,b} {} } The power of set is calculated as 2^n where n is no.of elements Jack Ceroni Fake Qubit Entangler · Author has 158 answers and 347.2K answer views · Updated 8y A power set, denoted by P(S), where S is some set. Let’s say that: S={a0,a1,a2} A power set is essentially all of the possible subsets of some set. Let’s apply this set, S. We can say that we have: P(S) S={a0,a1,a2} P(S)={{a0},{a0,a1},{a0,a1,a2},{a1},{a1,a2},{a2},{}} I said that the last element in the defined in in the power set was, {}. This is correct, but I just wanted to say that you may also see this denoted by: ∅, which means the same thing. Both denote empty sets. Maurice Dupre Ph. D. in Mathematics, University of Pennsylvania (Graduated 1972) · Author has 1.6K answers and 452K answer views · 5y If X(1),X(2),X(3),…,X(n) are sets, then the CARTESIAN PRODUCT of these sets is the set X consisting of all sequences (x(1),x(2),x(3),…,x(n)) where for each positive integer k<n+1 it is the case that x(k) belongs to X(k). Let D be the set of positive integers less than n+1. In this setting we call D the index set. Notice that the sequence defines a function x:D→U, where U is the union of the sets X(k) for k<n+1, where the value of x at input k is x(k). Thus the cartesian product X can be viewed as the set of functions x:D→U with x(k) in X(k) for each k in D. Now, suppose that each set X(k) is t If X(1),X(2),X(3),…,X(n) are sets, then the CARTESIAN PRODUCT of these sets is the set X consisting of all sequences (x(1),x(2),x(3),…,x(n)) where for each positive integer k<n+1 it is the case that x(k) belongs to X(k). Let D be the set of positive integers less than n+1. In this setting we call D the index set. Notice that the sequence defines a function x:D→U, where U is the union of the sets X(k) for k<n+1, where the value of x at input k is x(k). Thus the cartesian product X can be viewed as the set of functions x:D→U with x(k) in X(k) for each k in D. Now, suppose that each set X(k) is the same set S. Then, it would be natural to call the cartesian product the set S raised to the power n. But, as it consists of all functions from D to S, in set theory it is natural to write instead This makes sense for any set D, even if D is infinite. Thus, for any sets A and B, B^A={f such that f:A→B}. Now suppose that we take n=2 for D and consider Here, in set theory, we construct the natural numbers by using 0 for the empty set, 1={0}, 2={0,1},3=the union of 2 and {{2}}, and in general, n+1=the union of n with {{n}}. Thus, for instance, and so on. Thus in the natural number n defined as a set, it is a set whose cardinality is exactly n. Thus 2^S={0,1}^S={f such that f:S→{0,1}}. Notice that if A is ANY subset of S, it defines a unique function A:S→{0,1}, where for any s in S we define A(s)=1 if s is in the subset A and A(s)=0 if not. If you have any function B:S→{0,1}, then it corresponds to the subset B={s in S such that B(s)=1}. Consequently, it is natural to think of 2^S as the set of all subsets of S, and it is also common to refer to the set of all subsets of S as the power set of S. Notice that if S is a finite set, then and in fact this is true in general in cardinal arithmetic, even with infinite sets. Thus, for any sets A and B, One lesson from this is, remembering mathematics is basically the art of handling information, that set theory methods are very powerful for using symbols to manipulate information, store information, and transform concepts into symbolic representation. Consequently, it has become extremely important for all forms of mathematics. This means the foundations of set theory are important for all mathematics. A good introductory textbook on set theory is Naive Set Theory by Paul R. Halmos, who was one of the great operator theorists of the twentieth century and was an extremely good writer. His textbook Finite Dimensional Vector Spaces is also still one of the best on that subject, and as well his book on Hilbert spaces. His book on measure theory as well is extremely good. Related questions What is a power set example? What is the symbol for the power set? What is the power set of the empty set ∅ ? What is the point of power sets? What is the power set of { ∅ , { ∅ } } ? What is power set defined with examples? What is the power set of { { a , b } , c } ? How do you find the power set of A = (2, (5))? Is the power set of the power set of a set equal to the power set of that set? What is the power set of a set? What is Power set of { ∅ } ? How can you break a powerful curse set from birth? 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2916	https://www.khanacademy.org/science/class-12-chemistry-india/x6a5fb67b43bb54b9:coordination-compounds/x6a5fb67b43bb54b9:isomerism-in-coordination-compounds/v/geometrical-isomerism-in-square-planar-complexes	Geometrical isomerism in square planar complexes (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content NCERT Chemistry Class 12 Course: NCERT Chemistry Class 12>Unit 5 Lesson 3: Isomerism in coordination compounds Geometrical isomerism Geometrical isomerism in square planar complexes Geometrical isomerism in octahedral complexes Isomerism in coordination compounds Optical Isomerism in Coordination Compounds Science> NCERT Chemistry Class 12> Coordination compounds> Isomerism in coordination compounds © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Geometrical isomerism in square planar complexes Google Classroom Microsoft Teams About About this video In this video, we will explore the geometrical isomers of different square planar complexes. But not that just that! Have you wondered why is it that despite having the same coordination no. of 4, only square planar complexes exhibit geometrical isomerism and not tetrahedral complexes? Let's find out in this video! Created by Revathi Ramachandran. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Ridhima 3 months ago Posted 3 months ago. Direct link to Ridhima's post “In [MABCD] type complexes...” more In [MABCD] type complexes, how do you distinguish among the isomers and identify which is cis and which is trans ? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Video transcript Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. 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2917	https://askfilo.com/user-question-answers-smart-solutions/express-the-negation-of-each-of-these-statements-in-terms-of-3239383535333733	Question asked by Filo student Express the negation of each of these statements in terms of quantifiers without using thenegation symbol.a) ∀x(x>1). b) ∀x(x ≤ 2). c) ∃x(x ≥ 4). d)∃x(x<0). e) ∀x((x<−1)∨(x>2)). f) ∃x((x < 4)∨(x > 7)) Views: 5,644 students Updated on: Sep 22, 2025 Text SolutionText solutionverified iconVerified Concepts: Quantifiers, Negation of statements Explanation: To express the negation of statements in terms of quantifiers, we use the following rules: 1. The negation of a universal quantifier (∀) becomes an existential quantifier (∃) and vice versa. 2. The negation of a statement is expressed by stating that there exists at least one case where the statement does not hold true. Here are the negations for each statement: Step by Step Solution: Step 1 The negation of ∀x(x > 1) is ∃x(x ≤ 1). Step 2 The negation of ∀x(x ≤ 2) is ∃x(x > 2). Step 3 The negation of ∃x(x ≥ 4) is ∀x(x < 4). Step 4 The negation of ∃x(x < 0) is ∀x(x ≥ 0). Step 5 The negation of ∀x((x < -1) ∨ (x > 2)) is ∃x(−1 ≤ x ≤ 2). Step 6 The negation of ∃x((x < 4) ∨ (x > 7)) is ∀x((x ≥ 4) ∧ (x ≤ 7)). Final Answer: c : ∀x(x < 4) d : ∀x(x ≥ 0) e : ∃x(−1 ≤ x ≤ 2) f : ∀x((x ≥ 4) ∧ (x ≤ 7)) a : ∃x(x ≤ 1) b : ∃x(x > 2) Students who ask this question also asked Views: 5,216 Topic: Smart Solutions View solution Views: 5,102 Topic: Smart Solutions View solution Views: 5,538 Topic: Smart Solutions View solution Views: 5,977 Topic: Smart Solutions View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| Express the negation of each of these statements in terms of quantifiers without using thenegation symbol.a) ∀x(x>1). b) ∀x(x ≤ 2). c) ∃x(x ≥ 4). d)∃x(x<0). e) ∀x((x<−1)∨(x>2)). f) ∃x((x < 4)∨(x > 7)) \| \| Updated On \| Sep 22, 2025 \| \| Topic \| All topics \| \| Subject \| Smart Solutions \| \| Class \| Class 10 \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
2918	https://my.clevelandclinic.org/health/diseases/15774-fifth-disease	Abu Dhabi\|Canada\|Florida\|London\|Nevada\|Ohio\| Home/ Health Library/ Diseases & Conditions/ Fifth Disease AdvertisementAdvertisement Fifth Disease Fifth disease is a temporary bright red skin rash that appears after a parvovirus B19 infection. It mainly affects children but can affect adults, as well. Fifth disease usually goes away on its own after several days. Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services.Policy Care at Cleveland Clinic Find a Pediatric Primary Care Provider Find a Doctor and Specialists Schedule a Pediatric Primary Care Appointment ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionLiving With Overview What is fifth disease? Fifth disease (erythema infectiosum) is a childhood condition that appears as a bright red rash on your child’s cheeks. It’s nicknamed “slapped cheek disease” because of this rash. A virus called parvovirus B19 causes fifth disease. This virus is common and very contagious. Infected people can spread it through coughing or sneezing. Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services.Policy In most cases, fifth disease isn’t a serious medical condition. It often goes away with minimal or no treatment. Why is it called fifth disease? Fifth disease got its name because it was the fifth viral skin rash known to affect children in a list of six conditions. The list is as follows: Measles. Scarlet fever. Rubella (German measles). Dukes’ disease. Erythema infectiosum (fifth disease). Roseola. How common is fifth disease? Fifth disease is one of the six most common viral rashes in children. It typically affects children between 5 and 15 years old, especially in the spring and summer months. Can adults get fifth disease? Fifth disease can affect adults, but this is rare. Once you’re exposed to the virus, your body’s immune system builds up defenses to fight it off. This means that if you had fifth disease as a child, you’ll likely be immune to it as an adult. Most adults have had fifth disease as a child. Symptoms and Causes What are the symptoms of fifth disease? A parvovirus B19 infection often starts with flu-like symptoms, which are usually mild. During this time, the virus is most contagious. These symptoms include: Fatigue. Headaches. Achiness. Low-grade fever (99° to 101° F or 37° to 38.5° C). Runny nose. Sore throat. About 20% of children who have a parvovirus B19 infection don’t have these symptoms. Still, they can pass the virus to others. Advertisement It can take several days after the onset of flu-like symptoms for the raised, bright red rash (fifth disease) to show up on your child’s face. The rash may be itchy. Children typically no longer have flu-like symptoms once the rash appears. In some cases, you may see a second rash that develops after the cheek rash. It usually looks “lacey” and may appear on your child’s: Arms. Legs. Trunk (chest and back). Buttocks. About 10% of children with fifth disease also experience joint pain and swelling. Fifth disease symptoms in adults Adults who are infected with parvovirus B19 often develop flu-like symptoms without the rash. Along with those symptoms, about 80% of adults also develop joint pain in their wrists, hands and knees. What causes fifth disease? Human parvovirus, also called parvovirus B19, causes fifth disease. This is different from the parvovirus that affects dogs and cats. Fifth disease (a red rash) typically appears four to 14 days after your child is infected with parvovirus B19. Is fifth disease contagious? Parvovirus B19 is very contagious. It mainly spreads through respiratory droplets in your mouth and nose. If an infected person talks, coughs or sneezes near your child, your child could become infected with the virus. Parvovirus B19 can also spread through blood exposure from a pregnant woman to a fetus, but this is rare. However, fifth disease — the red rash caused by parvovirus B19 — isn’t contagious. In fact, once a person infected with parvovirus B19 gets the red rash, they’re no longer contagious (they can’t spread parvovirus B19 to other people). What are the complications of fifth disease? In healthy children and adults, fifth disease very rarely causes complications. But the condition can cause problems for people who have a blood disorder or weakened immune system. This is because the virus can affect the way your body makes red blood cells. It can cause your child’s red blood cell count to drop so low that they need a blood transfusion. Children (and adults) with the following conditions are at increased risk of complications: Cancer, such as leukemia. HIV. Certain types of anemia, such as sickle cell anemia and thalassemia. A transplanted organ. If your child has any of these conditions, contact their healthcare provider as soon as the fifth disease rash appears or if they’re having flu-like symptoms, which usually come before the rash. Fifth disease complications in adults About 10% of adults who get a parvovirus B19 infection develop chronic (long-term) parvovirus-associated arthritis in several joints, or polyarthritis. Women are more at risk for this complication than men. Advertisement Fifth disease and pregnancy If you’re pregnant and develop fifth disease (parvovirus B19 infection), it can spread to the fetus and cause complications, including: Miscarriage. Stillbirth (intrauterine fetal demise). Hydrops fetalis (when large amounts of fluid build up in a fetus’s tissues and organs). These complications are rare, however. Most adults and pregnant women have already been infected with parvovirus B19, so they’re protected. The risk of fetal loss when you get a parvovirus B19 infection while pregnant is approximately 2%. The greatest risk of complications from parvovirus B19 happens during your Y second trimester. But complications can happen at all points of pregnancy. If you’re pregnant and have been exposed to someone with fifth disease, contact your healthcare provider. Diagnosis and Tests How is fifth disease diagnosed? Healthcare providers typically diagnose fifth disease based on your child’s symptoms. The “slapped cheek” rash is a strong sign of this condition. When it’s accompanied by flu-like symptoms, your child’s provider can usually diagnose fifth disease in the office without any other tests. In very rare cases, your child’s provider may order blood tests to confirm fifth disease. Management and Treatment What is the treatment for fifth disease? Fifth disease symptoms typically go away in a few weeks with minimal or no treatment. Your child’s healthcare provider may recommend over-the-counter (OTC) pain relievers that can treat fever, headaches and joint pain. These medicines include: Advertisement Acetaminophen. Nonsteroidal anti-inflammatory drugs (NSAIDs), such as ibuprofen or naproxen. Be sure to follow the dosing instructions carefully. If your child has a weakened immune system and develops fifth disease, they may need treatment in a hospital. Care at Cleveland Clinic Find a Pediatric Primary Care Provider Find a Doctor and Specialists Schedule a Pediatric Primary Care Appointment Outlook / Prognosis How long does fifth disease last? The fifth disease rash should fade within five to 10 days after it develops. If your child develops a second rash, it should go away in seven to 10 days. But in some cases, the rash can come and go for several weeks. When can my child go back to school? While your child has flu-like symptoms, they’re contagious and should stay home. Once the fifth disease rash appears, your child isn’t contagious anymore, so they can return to school or daycare if they feel OK. Prevention Can I prevent fifth disease? There isn’t a vaccine to prevent fifth disease. Because the virus spreads easily through nasal and mouth droplets, good hygiene is the best way to prevent the disease. You can reduce your family’s risk of infection by: Washing your hands frequently and thoroughly. Sneezing or coughing into the crook of your elbow. Avoiding close contact with an infected person. Living With When should I see my healthcare provider? Reach out to your healthcare provider if you think you or your child has fifth disease or has been exposed to the virus, especially if you or your child has: Advertisement Severe joint pain. An itchy rash. A pregnancy. A weakened immune system. A blood disorder. What questions should I ask my doctor? If your child has fifth disease, you may want to ask their healthcare provider the following questions: How long will we be contagious? How long should my child stay home from school? How long should I stay home from work? What steps can I take to ensure other family members don’t get infected? What can I do to make myself or my child more comfortable? What can I do to alleviate symptoms like an itchy rash or joint pain? Should I notify my child’s school (or my work) about the infection? How long will the rash last? Can it come back? What signs of complications should I look out for? A note from Cleveland Clinic Even though fifth disease can look intimidating with its distinctive red rash, it’s usually a temporary condition that goes away with little treatment. But it’s important to understand that fifth disease can spread easily. If a member of your family has any of the symptoms of the condition, call your healthcare provider. You may need to keep your family member away from others for a little while to keep the virus from spreading. Care at Cleveland Clinic As your child grows, you need healthcare providers by your side to guide you through each step. Cleveland Clinic Children’s is there with care you can trust. Find a Pediatric Primary Care Provider Find a Doctor and Specialists Schedule a Pediatric Primary Care Appointment Medically Reviewed Last reviewed on 03/21/2023. Learn more about the Health Library and our editorial process. 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2919	https://www.sciencedirect.com/science/article/pii/S1521690X15000299	Androgen receptor roles in spermatogenesis and infertility - ScienceDirect Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Keywords Introduction Rodent models of androgen signalling disruption Control of spermatogenesis by androgens Clinically relevant mutations of androgen receptor Summary Conflict of interest statement Acknowledgements References Show full outline Cited by (211) Figures (1) Tables (1) Table 1 Best Practice & Research Clinical Endocrinology & Metabolism Volume 29, Issue 4, August 2015, Pages 595-605 7 Androgen receptor roles in spermatogenesis and infertility Author links open overlay panel Laura O'Hara Ph.D.(Post-Doctoral Research Fellow)1, Lee B.Smith Ph.D.(Chair of Genetic Endocrinology) Show more Outline Add to Mendeley Share Cite rights and content Under a Creative Commons license Open access Androgens such as testosterone are steroid hormones essential for normal male reproductive development and function. Mutations of androgen receptors (AR) are often found in patients with disorders of male reproductive development, and milder mutations may be responsible for some cases of male infertility. Androgens exert their action through AR and its signalling in the testis is essential for spermatogenesis. AR is not expressed in the developing germ cell lineage so is thought to exert its effects through testicular Sertoli and peri-tubular myoid (PTM) cells. AR signalling in spermatogenesis has been investigated in rodent models where testosterone levels are chemically supressed or models with transgenic disruption of AR. These models have pinpointed the steps of spermatogenesis that require AR signalling, specifically maintenance of spermatogonial numbers, blood-testis barrier integrity, completion of meiosis, adhesion of spermatids and spermiation, together these studies detail the essential nature of androgens in the promotion of male fertility. Previous article in issue Next article in issue Keywords androgen receptor testosterone testes spermatogenesis male infertility transgenic mice Introduction The mammalian testis has two functions: spermatogenesis (the production of haploid germ cells) and steroidogenesis (the production of the steroid hormones that support male reproductive development and function). Disruption of testosterone production by hypophysectomy, Leydig cell ablation or knockout of luteinising hormone receptor demonstrate that spermatogenesis cannot proceed to completion without testosterone. The receptor for testosterone is the steroid nuclear hormone receptorandrogen receptor (AR, OMIM entry: 313700). Ar is a single copy gene on the X chromosome. The classical genomic mechanism of testosterone signalling occurs when testosterone diffuses into the cell and binds to AR, then the ligand–receptor complex translocates to the nucleus and where it binds to androgen response elements (AREs) in the regulatory regions of genes to modify their translation. Non-classical signalling occurs when the ligand–receptor complex or testosterone itself activates non-genomic cytoplasmic signalling pathways, or when in certain circumstances the AR binds to AREs in the absence of testosterone [reviewed in ∗, ∗]. Despite requiring androgens for their survival and maturation, germ cells do not express androgen receptors and germ cell-specific androgen receptor expression is not required for their normal maturation , , . However, the somatic Sertoli, PTM, Leydig, vascular endothelial and vascular smooth muscle cells of the mature testis express androgen receptor, and it is widely accepted that the requirement of testosterone for spermatogenesis is mediated by these cell types. Rodent models of androgen signalling disruption Global androgen receptor knockout models The first animal model of androgen insensitivity was described in 1970 by Lyon and Hawkes, who reported an X-linked gene for ‘testicular feminisation’ (Tfm) in the mouse . It was subsequently discovered that male Tfm mice carry a single nucleotide deletion in exon 1 of their Ar gene, the resulting frameshift introduces a premature termination codon . With the advent of Cre-loxP technology [reviewed in ], mice with a total androgen receptor knock-out (ARKO) genotype have been produced by four groups. By mating a mouse line with floxed region of the Ar gene to one expressing Cre recombinase under the control of a constitutively active gene, genetic ablation of part of the AR resulting in loss of functional protein was achieved in all cells of the resulting offspring , , , . Both ARKO and Tfm mice have small, inguinal testes, lack of Wolffian duct structures and feminisation of external genitalia. When examined histologically, some tubules in the testis lack germ cells completely and the others contain a few spermatogonia, but no post-meiotic germ cells. Normal spermatogenesis does not occur. This is analogous to testicular histology seen in CAIS patients , . Cell specific androgen receptor ablation models Models of testosterone or androgen receptor disruption can demonstrate what processes in the testis rely on AR signalling, but they cannot determine which cell type(s) mediate these processes. Recent studies of cell-specific ablation or overexpression of AR in the testis using the Cre -loxP system have elucidated both the cell and AR-specific actions of androgens in the testis . Mouse lines with cell-specific knock-outs of AR in the male reproductive system are detailed in Table 1. Through these studies it has become apparent that AR signalling in the PTM and Sertoli cells exerts the most significant effects on spermatogenesis. Table 1. Summary of Cre lines used for cell specific ablation of AR action in the male reproductive system. \| Cell type \| Promoter \| Effect on mouse fertility \| Reference \| --- --- \| \| Sertoli \| Amh-Cre \| Infertile. No post-meiotic germ cells. \| , , ∗ \| \| Peri-tubular myoid \| Myh6-Cre \| Infertile. Reduction in germ cells at all stages. \| \| \| Leydig \| Fabp4-Cre \| Fertile, but degeneration of seminiferous epithelium occurs with ageing. Leydig cells do not mature correctly. \| \| \| Spermatocytes \| Sycp1-Cre \| Fertile \| \| \| Vascular smooth muscle \| Tagln-Cre \| Fertile but testicular vasomotion impaired \| \| \| Vascular endothelial \| Tie2-Cre \| Fertile \| \| \| Prostate epithelium \| Arrpb2-Cre \| Fertile \| \| \| Prostate, vas deferens and epididymal epithelium, SV smooth muscle \| Arrpb2-Cre \| Subfertile, defects in epididymal transit of sperm \| \| \| Prostate smooth muscle \| Myh6-Cre \| Phenotype complicated by loss of AR in PTM. \| \| \| SV smooth muscle \| Myh6-Cre \| Phenotype complicated by loss of AR in PTM. \| \| \| Neuronal \| Nes-Cre \| Subfertile, variable phenotype with fewer litters and pups per litter. Altered mating behaviour. \| , \| \| Epididymal epithelium \| Foxg1-Cre ; Rnase10-Cre \| Infertile due to obstructive azoospermia. \| , \| Sertoli cells are mesoepithelial somatic cells that coordinate and structurally support the maturing germ cells. Each Sertoli cells spans the seminiferous tubule from the basement membrane to the tubule lumen, and is tightly linked to a set of germ cells at specific stages by different types of junction protein complexes. Sertoli cells have a large surface area with a well-developed cytoskeleton that assists in maintaining their shape as well as providing a scaffold for the movement of germ cells , . Since germ cells do not express AR it is likely that the action of androgens on spermatogenesis is potentiated through their binding to Sertoli cell AR. Androgen receptors are not present in the Sertoli cells of the mouse until postnatal day 4 when faint staining of occasional Sertoli cells is seen. Staining of all nuclei and a progressive increase in staining intensity is noted from day 5 . After maturation, the expression of AR in Sertoli cells fluctuates in a seminiferous tubule stage-specific manner , appearing strongest at stages VI–VII. In humans, Sertoli cell AR has been reported as either weakly detected or not present up to 5 years, but is detected from 8 years onwards and is strongly expressed in adulthood , , . Expression is also cyclical in men, appearing strongest at stage III . A model of androgen receptor overexpression in Sertoli cells (tgSCAR) from day 2 onwards results in premature maturation of Sertoli cells, reduced final Sertoli cell number and therefore a reduced final germ cell number , suggesting it has a key role in Sertoli cell maturation. Development of tgSCAR germ cells was accelerated corresponding to premature maturation of the Sertoli cells, illustrating the intimate dependence of germ cells on Sertoli cell androgen receptor signalling for their coordinated development. Sertoli-cell specific AR knockout (SCARKO) mouse lines have been created by two groups using an inserted Cre recombinase coding sequence driven by an anti-Müllerian hormone promoter , . The testes of SCARKO mice develop normally with an external phenotype similar to their control male littermates, with fully descended testes and intact Wolffian duct-derived structures, unlike the ARKO model where they are absent . However at post-natal day 50 their testes are approximately a third of the weight of their wild type littermates' and contain few post-meiotic germ cells more advanced than pachytene primary spermatocytes, indicating a meiotic block. SCARKO seminiferous tubule diameters are decreased and the tubules do not develop a normal lumen. It is important to note that mice with a Sertoli cell-specific ablation of the second zinc finger domain (vital for DNA binding) display a similar post-meiotic block phenotype, indicating that progression of spermatocytes through meiosis is an event dependent on classical genomic signalling of androgens . The two groups differed in analysis of the levels of testosterone and LH in their SCARKO models; where De Gendt and colleagues found these two hormones to be normal in adult mice, the Chang group found LH to be high and testosterone to be low. Spermatogonia and Sertoli cell numbers do not differ between SCARKOs and controls . There is controversy as to whether Leydig cell numbers in the SCARKO are maintained or are decreased which would imply that Sertoli cell AR activates a paracrine signalling pathway that affects Leydig cell division. Another study created a ‘hypomorphic’ SCARKO with Sertoli cell-specific reduction of AR action, but also a general reduction in levels of AR protein throughout the whole body due to a phenotypic effect of the placing of the loxP sites in the transgenic AR . Seminiferous tubules are surrounded by a layer of peri-tubular myoid (PTM) cells. PTM cells are contractile and produce peristaltic waves in the seminiferous tubules that propel mature spermatozoa towards the rete testis. PTM cells synthesise the basement membrane components laminin, collagen and fibronectin in co-operation with Sertoli cells , . Like other tissues in the reproductive system a reciprocal relationship exists between the mesenchymal (PTM) and epithelial (Sertoli) cells of the testis, which is vital for the maintenance and function of both cell types. The mesenchymal PTM cells are the first cell in the testis to express AR: at or before e16.5 in the mouse and at e18.5 in the rat , . Cell-specific ablation of AR in the PTM cells has been achieved using Cre/loxPtransgenic mice (PTM-ARKO) . PTM-ARKO testes are 70% smaller than controls because of a reduction in numbers of all germ cell stages. Even though these two particular cell types appear to be the most important for transmission of androgen signalling to developing germ cells, AR is present in many different cell types in the male reproductive system that may not have an immediate impact on spermatogenesis but still contribute to fertility and normal testicular function, as well as is in other non-reproductive organs throughout the body. Conditional AR ablation lines have been created and investigated to elucidate its role in these cell types . Table 1 summarises conditional ablation of AR in several different cell types of the male reproductive system and whether this ablation has any impact on spermatogenesis, testicular function or wider fertility. The results of these cell ablation experiments draws attention to the absolute requirement for AR signalling to maintain overall synergy and function of the male reproductive system, which contributes to overall male fertility. Models with decreased testosterone production Before the Cre-loxP system was widely used to interrogate cell specific AR action, changes in spermatogenesis were identified in rodent models with reduced testosterone production. Mice with a luteinising hormone receptor knockout (LuRKO) produce levels of testosterone 10 to 20 times lower than wild-type . Another model is the ‘TE’ rat that carries implants containing low-dose testosterone and estradiol to suppress LH production at the pituitary and therefore reduce testicular T levels to 3% of control . Although reduction in AR signalling in these models occurs in all cell types that express the receptor and therefore the resulting phenotype cannot be assigned to lack of AR action in a specific cell type, they have provided a number of novel observations that add to the complete picture of AR signalling in spermatogenesis. Control of spermatogenesis by androgens Experimental perturbation of testosterone and/or androgen receptor has highlighted specific steps in spermatogenesis that require AR signalling, summarised in Fig.1 and described here. 1. Download: Download high-res image (1MB) 2. Download: Download full-size image Fig.1. The anatomy of the testis and control of spermatogenesis by androgens. The testis consists of seminiferous tubules surrounded by an interstitial stroma. The stroma contains steroidogenic Leydig cells and blood vessels that are lined by vascular endothelial (VE) and vascular smooth muscle (VSM) cells. Peritubular myoid (PTM) cells line the outside of the seminiferous tubule. Sertoli cells extend from the PTM cells to the lumen and support and contact the developing germ cells. Junctions between Sertoli cells form the blood testis barrier (BTB). Maturation of germ cells in spermatogenesis progresses toward the lumen starting with the undifferentiated diploid spermatogonia on the basement membrane to spermatozoa that are released from the apical surface of the Sertoli cell. During meiosis, spermatocytes transit the BTB. Testosterone produced by the Leydig cells has autocrine effects on the Leydig cells themselves, which express AR, as well as paracrine effects on VE, VSM, PTM and Sertoli cells of the testis. It also diffuses into the interstital blood vessels to be transported to the circulatory system. There are five critical processes in spermatogenesis that are regulated by testosterone labelled 1–5 on the figure. (1) Maintenance of spermatogonial numbers, (2) maintenance of the BTB, (3) completion of meiosis by spermatocytes, (4) adherence of elongated spermatids to Sertoli cells, (5) the release of mature spermatozoa and (6) the formation of the seminiferous tubule lumen. Maintenance of spermatogonia numbers PTM-ARKO mice have a reduction of germ cell numbers at all stages, (including spermatogonia) that is not seen in any other conditional ARKO model, suggesting that maintenance of spermatogonia numbers is controlled by PTM cell AR. Since spermatogonia are juxtaposed to the basement membrane they are in intimate contact with the PTM cells. It is possible that this affect is due to direct signalling between PTM cells and spermatogonia. Since PTM cells in the PTM-ARKO also demonstrate a progressive loss of desmin and smooth muscle actin (SMA) it is likely that loss of AR is affecting the smooth muscle phenotype in PTM cells, and a disruption of laminin indicates problems with the basement membrane of the seminiferous tubules. These disruptions may impair attachment and signalling between PTM cells and spermatogonia and thus disrupt their niche . It is also possible that the decrease in spermatogonia in the PTM-ARKO is mediated through changes in Sertoli cells. The decrease in germ cells is not due to a decrease in Sertoli cell number, as this is not reduced in PTM-ARKO mice. However, levels of Sertoli cell-specific gene products including Rhox5, Epp and Tubb3 are reduced in PTM-ARKO testes, implying that AR signalling in PTM cells has an effect on the transcriptome and function of their neighbouring Sertoli cells. In vitro evidence has demonstrated that androgens can stimulate PTM cells to secrete a factor or factors dubbed “P-mod-S” that modulates Sertoli cell function, but it remains as yet unidentified, and its role is disputed by some investigators . Progression through meiosis Very few post-meiotic cells are noted in the SCARKO mouse. Morphological analysis of SCARKO testes established that germ cell entry into meiosis appeared normal, but a progressive loss of pachytene primary spermatocytes between stages VI and XII was noted. Round spermatid number was 3% of controls and no elongating spermatids were seen . In rodent models that have reduced testosterone production such as the LuRKO mouse and TE implanted rats, meiosis completes and spermatogenesis proceeds to the round spermatid stage implying that meiosis itself is not sensitive to low testosterone levels, even though subsequent stages may be. Ablation of any residual androgen action in the LuRKO mouse with flutamide (an AR antagonist) results in the reversion of the phenotype to a meiotic block which adds to the evidence that at least a low level of AR signalling is required for completion of meiosis . BTB integrity Tight junctions between adjacent Sertoli cells above the level of leptotene spermatocytes form the blood-testis barrier (BTB), a structure that partitions the adluminal compartment of the seminiferous tubule from the basal compartment so haploid germ cells can mature in an immunologically privileged site with a specialised microenvironment . During maturation, sperm move along the Sertoli cell, traversing the BTB during meiosis. It is accepted that androgen signalling contributes to the maintenance of the BTB, but some studies suggest that it is essential and that the BTB is not present , , whereas others suggest that tight junctions in the SCARKO can form, but BTB formation is delayed and incomplete . A number of genes encoding BTB proteins are down-regulated or incorrectly expressed in SCARKO mice , , . Because of the intimate association between barrier formation and meiotic progression, it has been postulated that the two processes are co-dependent, but more studies on barrier formation in other mouse models with a meiotic block is required to confirm this hypothesis . Seminiferous tubule lumen formation Sertoli cells secrete the seminiferous tubule fluid that forms the tubule lumen and acts as a vehicle to transport mature spermatozoa from the testis. It contains appropriate nutritional and hormonal factors necessary to support spermatogenesis that are unavailable from the circulation due to the presence of the BTB. The formation of the seminiferous tubule lumen depends on Sertoli cell androgen receptor as SCARKO testes do not have functional tubal lumens . When efferent ducts are ligated, control testes increase in weight but SCARKO testes do not, implying that Sertoli cell fluid secretion is affected when AR is ablated . It has been suggested that meiosis can only be completed after Sertoli cell fluid secretion has been established [reviewed in ], and also that a lumen cannot be formed when the BTB is disrupted . There appears to be a functional link between the BTB, progression of meiosis and secretion of seminiferous tubule fluid by Sertoli cells that requires further elucidation. Spermatid adhesion Because of the presence of the post-meiotic block in SCARKO mice it is impossible to use this model to pinpoint the contribution of Sertoli cell AR signalling to subsequent stages of spermatogenesis, so systemic models of androgen or androgen receptor disruption have been used instead. ‘TE implanted’ rats and a hypomorphic AR mouse model have both been used to determine that both Sertoli-spermatid adhesion and spermiation are dependent on androgens. A reduction in testosterone results in premature detachment of spermatids from Sertoli cells at the round to elongating transition stage. Large numbers of round spermatids accumulate in cauda epididymis in both TE implanted and hypomorphic AR mice . The round to elongating spermatid transition stage is also when desmosome-based connections that attach spermatids to their supporting Sertoli cell are replaced by ectoplasmic specialisations (ES), a type of adherens junction unique to the testis, promoting the theory that ES may be disrupted in models with low testosterone signalling. However, normal ES were noted in TE rats when testes were examined by electron microscopy. Spermiation Both the TE rat and hypomorphic mouse models also note spermiation failure characterised by elongating spermatid retention with subsequent phagocytosis by Sertoli cells ∗, . Release of the sperm from the seminiferous epithelium during spermiation involves the replacement of the apical ES by tubulobulbar complexes (TBC), structures thought to internalise disassembled apical ES junctions in preparation for the release of spermatids . TBC formation and removal of ES are not affected by hormone withdrawal, but final disengagement of the spermatid from the Sertoli cell is affected . Further investigation revealed that androgen action is necessary for the disengagement of a Sertoli cell complex containing α6β1-integrin and phosphorylated focal adhesion kinase (FAK) from laminin α3β3γ3 on spermatids . Control of both of spermatid adhesion and spermiation has been shown to be through the non-classical AR signalling pathway via action of Src kinase. Sertoli cells in culture expressing a mutated form of AR that can't activate Src bind fewer germ cells, and explants treated with a Src inhibitor release fewer sperm . Three days after in vivo injection of a Src inhibitor, spermatocytes and round spermatids were absent but elongating spermatids remained . It is interesting to note that AR action appears to be required for both the attachment and disengagement of maturing germ cells at different stages of spermatogenesis. Since the nature of the junctions that hold germ cells to Sertoli cells is in flux throughout spermatogenesis, it may be that Src has different roles in each of these different junction complexes that are yet to be fully characterised. Although defects in spermatid adhesion and spermiation cannot be definitively assigned to AR signalling in the Sertoli cell (because AR signalling is reduced in all testicular cell types in these models), it is likely that the Sertoli cell is responsible rather than another testicular cell type due to the isolated environment in which post-meiotic germ cells develop behind the BTB. Studying BTB remodelling, spermatid adhesion and spermiation as a coordinated process rather than isolated events may help further elucidate the androgen dependent coordination of spermatogenesis. There is already evidence that fragments of laminin chains produced during spermiation induce remodelling of the BTB, thus intimately coordinating these two androgen-dependent steps in the processes of spermatogenesis . Clinically relevant mutations of androgen receptor Mutations that result in AIS As of October 2014, more than 1100 mutations in human AR have been identified . Although some AR mutations are associated with non-testicular phenotypes such as prostate cancer, premature ovarian failure and Kennedy's disease, approximately 90% result in androgen insensitivity syndrome (AIS). AIS presents as a spectrum of masculinisation disorders in XY individuals from a fully female external phenotype present in complete androgen insensitivity syndrome (CAIS) through undervirilised male phenotype in partial androgen insensitivity syndrome (PAIS) to normal male genitals but infertility seen in mild androgen insensitivity syndrome (MAIS). PAIS and CAIS are discussed further in the chapter on androgen insensitivity syndrome by Mongan et al., in this issue of Best Practice and Research Clinical Endocrinology and Metabolism. In contrast to more severe forms of AIS, patients with MAIS have normal genitalia and may present with infertility as the first or only symptom. Investigation into Ar mutations should be prompted when a high androgen sensitivity index (ASI, the LH to testosterone ratio) is seen as there is a positive correlation between the two . The location of the mutation in the AR gene and the type of the mutation correlates with the resulting severity of the AIS phenotype. Most premature termination mutations result in a truncated AR protein and complete AIS. Single base substitutions can results in all grades of AIS, depending on what effect the substitution has on the amino acid sequence and three dimensional structure of the protein and therefore how much of the function is retained. Exon 1 of AR encodes the N-terminal domain which codes for more than half the AR protein, but only about 25% of total loss of function mutations occur in this domain. While mutations in exon 1 such as premature termination mutations or deletions generally result in CAIS due to a truncation of the protein, 22 of 39 of the single-base substitution mutations in exon 1 that result in a form of AIS results in MAIS, implying that missense mutations in this domain have a mild effect on AR function. Since the main role of the N-terminal domain is as a binding site for other transactivation proteins, mutations may impair the binding or activity of these factors and result in a protein that may have reduced efficiency in specific cell types only. For example, in one patient, the mutation prevented interaction of the AR with a TIF2, Sertoli-cell specific co-activator, thus impeding the genomic actions of AR only in Sertoli cells . The frequency of AR mutations in patients with oligospermia or azoospermia under investigation for infertility has been shown to be 2–3%, a relatively small proportion of infertile men . However mutations of genes up or downstream of AR in its signalling pathway may also be contributing to cases of infertility. Association of AR trinucleotide repeats with infertility Longer variants of a CAG polyglutamine (Q tract) and a GGN polyglycine (G tract) trinucleotide repeat in exon 1 of AR have been suggested to have an association with infertility. AR transactivation and levels of AR protein and mRNA have been shown to be reduced in vitro by increasing CAG repeat number and longer polyglutamine tracts are associated with impaired sperm production in men . Despite this, subsequent studies have been conflicting about the association [reviewed in ]. Further downstream of the polyglutamine tract is a polyglycine GGN tract (G tract) which varies in length from 10 to 30 repeats. Deletion of the polyglycine tract resulted in a 30% reduction in transactivation potential in vitro. However, there is also conflicting evidence about whether the number of GGN repeats has an association with male infertility. Summary Testosterone signalling through AR is vital for complete spermatogenesis and this is mediated by the somatic cell types of the testis that express specifically the Sertoli and PTM cells that have intimate connections with the maturing germ cells (summarised in Fig.1). Although the points at which androgen signalling affects spermatogenesis via PTM cells and Sertoli cells are well characterised, the effectors of this paracrine signalling network are not yet fully elucidated. Array studies on the SCARKO model have produced data on gene expression changes [reviewed in ], but have identified surprisingly few for a model with such a complex phenotype. More recently, immuno-precipitation of actively translated mRNAs has identified novel androgen-regulated Sertoli cell transcripts and a population of microRNAs have also been shown to change testosterone deprivation model and these data will be useful as starting points for future studies. Unravelling the combination of classical AR signalling, non-classical AR signalling and microRNA regulation in both Sertoli and PTM cells is a significant task for future research that must now focus on downstream targets of testosterone signalling if we are to fully understand and clinically support spermatogenesis in patients, an increasing number of whom are presenting with low sperm count , . Research agenda •The advent of next generation sequencing will contribute to the need for personalised genomic medicine. •Further research into the fundamental mechanisms of AR signalling and the genes and signalling pathways that it controls must occur in tandem with this so the molecular events are fully characterised. •These two research points will combine to produce tailored drug therapy for individual cases of infertility and may also contribute to the development of novel male contraceptives. Conflict of interest statement The authors have no conflicts of interest to disclose. Acknowledgements This work was funded by a Medical Research Council Program Grant Award (G1100354/1) (to LBS). Special issue articles Recommended articles References L.D. Russell, Y. Clermont Degeneration of germ cells in normal, hypophysectomized and hormone treated hypophysectomized rats Anat Rec, 187 (3) (1977), pp. 347-366 CrossrefGoogle Scholar J.M. Bartlett, J.B. Kerr, R.M. 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Targeted inactivation of the androgen receptor gene in murine proximal epididymis causes epithelial hypotrophy and obstructive azoospermia Endocrinology, 152 (2) (2011), pp. 689-696 CrossrefView in ScopusGoogle Scholar Cited by (211) Sertoli cells as key drivers of testis function 2022, Seminars in Cell and Developmental Biology Show abstract Sertoli cells are the orchestrators of spermatogenesis; they support fetal germ cell commitment to the male pathway and are essential for germ cell development, from maintenance of the spermatogonial stem cell niche and spermatogonial populations, through meiosis and spermiogeneis and to the final release of mature spermatids during spermiation. However, Sertoli cells are also emerging as key regulators of other testis somatic cells, including supporting peritubular myoid cell development in the pre-pubertal testis and supporting the function of the testicular vasculature and in contributing to testicular immune privilege. Sertoli cells also have a major role in regulating androgen production within the testis, by specifying interstitial cells to a steroidogenic fate, contributing to androgen production in the fetal testis, and supporting fetal and adult Leydig cell development and function. Here, we provide an overview of the specific roles for Sertoli cells in the testis and highlight how these cells are key drivers of testicular sperm output, and of adult testis size and optimal function of other testicular somatic cells, including the steroidogenic Leydig cells. ### Novel concepts in the aetiology of male reproductive impairment 2017, Lancet Diabetes and Endocrinology Show abstract Infertility is a widespread problem and a male contribution is involved in 20–70% of affected couples. As a man's fertility relies on the quantity and quality of his sperm, semen analysis is generally used as the proxy to estimate fertility or gain insight into the underlying reasons for infertility. Male reproductive impairment might result from factors that affect sperm production, quality, function, or transport. Although in most men the origin of infertility remains unexplained, genetic causes are increasingly being discovered. In this first of two papers in The Lancet Diabetes and Endocrinology Series on male reproductive impairment, we propose a novel, clinically based aetiological construct with a genetic focus, and consider how this might serve as a helpful way to conceptualise a diagnostic algorithm. ### Regulation of the blood-testis barrier 2016, Seminars in Cell and Developmental Biology Citation Excerpt : FSH quite clearly stimulates the organisation of other junction types in the rat inter-Sertoli cell junctional complex (ectoplasmic specialisation and adherens junctions) in vitro and in vivo but because fertility is maintained at a reduced level in knockout animals lacking FSH or its receptor [61,62], it can be concluded that FSH is probably permissive for BTB function, but not mandatory. Confirmation that androgen action was required for BTB function in vivo initially came via mouse models in which Sertoli cell androgen receptor (AR) signalling was knocked out (SCARKO) [27,63] (for reviews see [9,64,65]). Despite differences in the extent that AR action was ablated due to the targeting of different exons of the receptor [63,66], BTB formation was found to be delayed with incomplete barrier function as shown by hypertonic perfusion and lanthanum tracer permeation or by permeability assessment with a biotin tracer . Show abstract The purpose of this review is to describe the endocrine and local testicular factors that contribute to the regulation of the blood-testis barrier (BTB), using information gained from in vivo and in vitro models of BTB formation during/after puberty, and from the maintenance of BTB function during adulthood. In vivo the BTB, in part comprised of tight junctions between adjacent somatic Sertoli cells, compartmentalizes meiotic spermatocytes and post-meiotic spermatids away from the vasculature, and therefore prevents autoantibody production by the immune system against these immunogenic germ cells. This adluminal compartment also features a unique biochemical milieu required for the completion of germ cell development. During the normal process of spermatogenesis, earlier germ cells continually cross into the adluminal compartment, but the regulatory mechanisms and changes in junctional proteins that allow this translocation step without causing a ‘leak’ remain poorly understood. Recent data describing the roles of FSH and androgen on the regulation of Sertoli cell tight junctions and tight junction proteins will be discussed, followed by an examination of the role of paracrine factors, including members of the TGFβ superfamily (TGFβ3, activin A) and retinoid signalling, as potential mediators of junction assembly and disassembly during the translocation process. ### Some of the factors involved in male infertility: A prospective review 2020, International Journal of General Medicine ### The roles and mechanisms of Leydig cells and myoid cells in regulating spermatogenesis 2019, Cellular and Molecular Life Sciences ### The roles of luteinizing hormone, follicle-stimulating hormone and testosterone in spermatogenesis and folliculogenesis revisited 2021, International Journal of Molecular Sciences View all citing articles on Scopus 1 Tel.: +44 (0)131 242 9469. Copyright © 2015 The Authors. Published by Elsevier Ltd. Part of special issue Nuclear hormone receptors for the clinician Edited by Paolo Beck-Peccoz Download full issue Other articles from this issue Classical nuclear hormone receptor activity as a mediator of complex biological responses: A look at health and disease August 2015 Paul Michael Yen ### Noncoding RNAs and the control of signalling via nuclear receptor regulation in health and disease August 2015 Paul Cathcart, …, Leandro Castellano ### Specificity and sensitivity of glucocorticoid signaling in health and disease August 2015 Derek W.Cain, John A.Cidlowski View more articles Recommended articles No articles found. Article Metrics Citations Citation Indexes 210 Captures Mendeley Readers 249 Mentions References 1 View details About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. 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2920	https://leachlegacy.ece.gatech.edu/ece3050/notes/feedback/FBExamples.pdf	c ° Copyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. Collection of Solved Feedback Ampliﬁer Problems This document contains a collection of solved feedback ampliﬁer problems involving one or more active devices. The solutions make use of a graphical tool for solving simultaneous equations that is called the Mason Flow Graph (also called the Signal Flow Graph). When set up properly, the graph can be used to obtain by inspection the gain of a feedback ampliﬁer, its input resistance, and its output resistance without solving simultaneous equations. Some background on how the equations are written and how the ﬂow graph is used to solve them can be found at The gain of a feedback ampliﬁer is usually written in the form A ÷ (1 + bA), where A is the gain with feedback removed and b is the feedback factor. In order for this equation to apply to the four types of feedback ampliﬁers, the input and output variables must be chosen correctly. For ampliﬁers that employ series summing at the input (alson called voltage summing), the input variable must be a voltage. In this case, the source is modeled as a Thévenin equivalent circuit. For ampliﬁers that employ shunt summing at the input (also called current summing), the input variable must be a current. In this case, the source is modeled as a Norton equivalent circuit. When the output sampling is in shunt with the load (also called voltage sampling), the output variable must be a voltage. When the output sampling is in series with the load (also called current sampling), the output variable must be a current. These conventions are followed in the following examples. The quantity Ab is called the loop gain. For the feedback to be negative, the algebraic sign of Ab must be positive. If Ab is negative the feedback is positive and the ampliﬁer is unstable. Thus if A is positive, b must also be positive. If A is negative, b must be negative. The quantity (1 + Ab) is called the amount of feedback. It is often expressed in dB with the relation 20 log (1 + Ab). For series summing at the input, the expression for the input resistance is of the form RIN × (1 + bA), where RIN is the input resistance without feedback. For shunt summing at the input, the expression for the input resistance is of the form RIN ÷ (1 + bA). For shunt sampling at the output, the expression for the output resistance is of the form RO ÷ (1 + bA), where RO is the output resistance without feedback. To calculate this in the examples, a test current source is added in shunt with the load. For series sampling at the output, the expression for the output resistance is of the form RO × (1 + bA). To calculate this in the examples, a test voltage source is added in series with the load. Most texts neglect the feedforward gain through the feedback network in calculating the forward gain A. When the ﬂow graph is used for the analysis, this feedforward gain can easily be included in the analysis without complicating the solution. This is done in all of the examples here. The dc bias sources in the examples are not shown. It is assumed that the solutions for the dc voltages and currents in the circuits are known. In addition, it is assumed that any dc coupling capacitors in the circuits are ac short circuits for the small-signal analysis. Series-Shunt Example 1 Figure 1(a) shows the ac signal circuit of a series-shunt feedback ampliﬁer. The input variable is v1 and the output variable is v2. The input signal is applied to the gate of M1 and the feedback signal is applied to the source of M1. Fig. 1(b) shows the circuit with feedback removed. A test current source it is added in shunt with the output to calculate the output resistance RB. The feedback at the source of M1 is modeled by a Thévenin equivalent circuit. The feedback factor or feedback ratio b is the coeﬃcient of v2 in this source, i.e. b = R1/ (R1 + R3). The circuit values are gm = 0.001 S, rs = g−1 m = 1 kΩ, r0 = ∞, R1 = 1 kΩ, R2 = 10 kΩ, R3 = 9 kΩ, R4 = 1 kΩ, and R5 = 100 kΩ. The following equations can be written for the circuit with feedback removed: id1 = Gm1va Gm1 = 1 rs1 + R1kR3 va = v1 −vts1 vts1 = R1 R1 + R3 v2 id2 = −gmvtg2 1 Figure 1: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. vtg2 = −id1R2 v2 = id2RC + itRC + id1RD RC = R4k (R1 + R3) RD = R1R4 R1 + R3 + R4 The voltage va is the error voltage. The negative feedback tends to reduce va, making \|va\| →0 as the amount of feedback becomes inﬁnite. When this is the case, setting va = 0 yields the voltage gain v2/v1 = b−1 = 1 + R3/R1. Although the equations can be solved algebraically, the signal-ﬂow graph simpliﬁes the solution. Figure 2 shows the signal-ﬂow graph for the equations. The determinant of the graph is given by ∆= 1 −Gm1 × [−R2 × (−gm2) × RC + RD] × R1 R1 + R3 × (−1) Figure 2: Signal-ﬂow graph for the equations. The voltage gain v2/v1 is calculated with it = 0. It is given by v2 v1 = Gm1 × [−R2 × (−gm2) × RC + RD] ∆ = 1 rs1 + R1kR3 × (R2 × gm2 × RC + RD) 1 + 1 rs1 + R1kR3 × (R2 × gm2 × RC + RD) × R1 R1 + R3 2 This is of the form v2 v1 = A 1 + Ab where A = 1 rs1 + R1kR3 × (R2 × gm2 × RC + RD) = 4.83 b = R1 R1 + R3 = 0.1 Note that Ab is dimensionless. Numerical evaluation yields v2 v1 = 4.83 1 + 0.483 = 3.26 The output resistance RB is calculated with v1 = 0. It is given by RB = v2 it = RC ∆= RC 1 + Ab = 613 Ω Note that the feedback tends to decrease RB. Because the gate current of M1 is zero, the input resistance is RA = R5 = 100 kΩ. Series-Shunt Example 2 A series-shunt feedback BJT ampliﬁer is shown in Fig. 3(a). A test current source is added to the output to solve for the output resistance. Solve for the voltage gain v2/v1, the input resistance RA, and the output resistance RB. Assume β = 100, rπ = 10 kΩ, α = β/ (1 + β), gm = β/rπ, re = α/gm, r0 = ∞, rx = 0, R1 = 1 kΩ, R2 = 1 kΩ, R3 = 2 kΩ, R4 = 4 kΩ, and R5 = 10 kΩ. The circuit with feedback removed is shown in Fig. 3(b). The circuit seen looking out of the emitter of Q1 is replaced with a Thévenin equivalent circuit made with respect to v2. A test current source it is added to the output to solve for the output resistance. Figure 3: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. For the circuit with feedback removed, we can write ie1 = G1va va = v1 −v2 R3 R3 + R4 G1 = 1 r0 e1 + R3kR4 r0 e1 = R1 1 + β + re ic1 = αie1 ib1 = ic1 β vtb2 = −ic1R2 ie2 = −G2vtb2 G2 = 1 r0 e2 r0 e2 = R2 1 + β + re 3 ic2 = αie2 v2 = ic2Ra + itRa + ie1Rb Ra = R5k (R3 + R4) Rb = R3R5 R3 + R4 + R5 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 4. The determinant is ∆ = 1 −G1 × [α × −R2 × −G2 × α × Ra + Rb] × −R3 R3 + R4 = 1 + G1 × [α × R2 × G2 × α × Ra + Rb] × R3 R3 + R4 = 9.09 Figure 4: Signal-ﬂow graph for the equations. The voltage gain is v2 v1 = G1 × [α × −R2 × −G2 × α × Ra + Rb] ∆ = G1 × [α × R2 × G2 × α × Ra + Rb] 1 + G1 × [α × R2 × G2 × α × Ra + Rb] × R3 R3 + R4 This is of the form v2 i1 = A 1 + Ab where A = G1 × [α × R2 × G2 × α × Ra + Rb] = 1 r0 e1 + R3kR4 · α × R2 × 1 r0 e2 × α × R5k (R3 + R4) + R3R5 R3 + R4 + R5 ¸ = 24.27 b = R3 R3 + R4 = 0.333 Notice that the product Ab is positive. This must be true for the feedback to be negative. Numerical evaluation of the voltage gain yields v2 v1 = A ∆= 2.67 The resistances RA and RB are given by RA = µib1 v1 ¶−1 = µG1α/β ∆ ¶−1 = ∆× βr0 e1 α = ∆× β α (r0 e1 + R3kR4) = 1.32 MΩ RB = v2 it = Ra ∆= R5k (R3 + R4) ∆ = 412.5 Ω 4 Series-Shunt Example 3 A series-shunt feedback BJT ampliﬁer is shown in Fig. 5(a). Solve for the voltage gain v2/v1, the input resistance RA, and the output resistance RB. For J1, assume gm1 = 0.003 S, and r01 = ∞. For Q2, assume β2 = 100, rπ2 = 2.5 kΩ, α2 = β2/ (1 + β2), gm2 = β2/rπ2, re2 = α2/gm2, r02 = ∞, rx2 = 0. The circuit elements are R1 = 1 MΩ. R2 = 10 kΩ, R3 = 1 kΩ, R4 = 20 kΩ, and R5 = 10 kΩ. The circuit with feedback removed is shown in Fig. 5(b). The circuit seen looking out of the source of J1 is replaced with a Thévenin equivalent circuit made with respect to v2. A test current source it is added in shunt with the output to solve for the output resistance. Figure 5: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. For the circuit with feedback removed, we can write va = v1 −v2 R3 R3 + R4 id1 = G1va G1 = 1 rs1 + R3kR4 vtb2 = −id1R2 ie2 = −G2vtb2 G2 = 1 r0 e2 r0 e2 = R2 1 + β2 + re2 ic2 = α2ie2 v2 = ic2Ra + itRa + id1Rb Ra = R5k (R3 + R4) Rb = R3R5 R3 + R4 + R5 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 6. The determinant is ∆ = 1 −G1 × [−R2 × −G2 × α × Ra + Rb] × −R3 R3 + R4 = 1 + G1 × [R3 × G2 × α × Ra + Rb] × R3 R3 + R4 = 21.08 The voltage gain is v2 v1 = G1 × [−R2 × −G2 × α2 × Ra + Rb] ∆ = G1 × [R2 × G2 × α2 × Ra + Rb] 1 + G1 × [R3 × G2 × α × Ra + Rb] × R3 R3 + R4 5 Figure 6: Signal-ﬂow graph for the equations. This is of the form v2 i1 = A ∆= A 1 + Ab where A = G1 × [R2 × G2 × α2 × Ra + Rb] = 1 rs1 + R3kR4 · R2 × 1 r0 e1 × α1 × R5k (R3 + R4) + R3R4 R3 + R4 + R5 ¸ = 421.8 b = R3 R3 + R4 = 0.0467 Notice that the product Ab is positive. This must be true for the feedback to be negative. Numerical evaluation of the voltage gain yields v2 v1 = 421.8 1 + 421.8 × 0.0476 = 20 The resistances RA and RB are given by RA = R1 = 1 MΩ RB = v2 it = Ra ∆= R5k (R3 + R4) ∆ = 321.3 Ω Series-Shunt Example 4 A series-shunt feedback BJT ampliﬁer is shown in Fig. 7(a). A test current source is added to the output to solve for the output resistance. Solve for the voltage gain v2/v1, the input resistance RA, and the output resistance RB. Assume β = 50, rπ = 2.5 kΩ, α = β/ (1 + β), gm = β/rπ, re = α/gm, r0 = ∞, rx = 0, R1 = 1 kΩ, R2 = 100 Ω, R3 = 9.9 kΩ, R4 = 10 kΩ, and R5 = 10 kΩ. The circuit with feedback removed is shown in Fig. 8. The circuit seen looking out of the base of Q2 is a Thévenin equivalent circuit made with respect to the voltage v2. A test current source it is added in shunt with the output to solve for the output resistance. The emitter eQuivalent circuit for calculating ie1 and ie2 is shown in Fig. 7(b). For this circuit and the circuit with feedback removed, we can write ie1 = G1ve ve = v1 −v2 R2 R2 + R3 G1 = 1 re1 + r0 e2 r0 e2 = R2kR3 1 + β + re ic1 = αie1 ib1 = ic1 β vtb3 = −ic1R1 ie3 = −G2vtb3 G2 = 1 r0 e3 r0 e3 = R1 1 + β + re ic3 = αie3 v2 = ic3Ra + itRa −ib2Rb Ra = R4k (R2 + R3) Rb = R2R4 R2 + R3 + R4 6 Figure 7: (a) Ampliﬁer circuit. (b) Emitter equivalent circuit for calculating ie1 and ie2. Figure 8: Circuit with feedback removed. 7 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 9. The determinant is ∆ = 1 −G1 × µ α × −R1 × −G2 × α × Ra − Rb 1 + β ¶ × −R2 R2 + R3 = 1 + G1 × µ α × R1 × G2 × α × Ra − Rb 1 + β ¶ × R2 R2 + R3 = 8.004 Figure 9: Flow graph for the equations. The voltage gain is v2 v1 = G1 × µ α × −R1 × −G2 × α × Ra − Rb 1 + β ¶ ∆ = G1 × µ α × R1 × G2 × α × Ra − Rb 1 + β ¶ 1 + G1 × µ α × R1 × G2 × α × Ra − Rb 1 + β ¶ × R2 R2 + R3 This is of the form v2 i1 = A 1 + Ab where A = G1 × µ α × R1 × G2 × α × Ra − Rb 1 + β ¶ 1 re1 + r0 e2 · α × R1 × 1 r0 e3 × α × R4k (R2 + R3) − 1 1 + β R2R4 R2 + R3 + R4 ¸ = 700.4 b = R2 R2 + R3 = 0.01 Notice that the product Ab is positive. This must be true for the feedback to be negative. Numerical evaluation of the voltage gain yields v2 v1 = A ∆= 87.51 The resistances RA and RB are given by RA = R5k µib1 v1 ¶−1 = R5k µG1α/β ∆ ¶−1 = R5k [∆× (1 + β) × (re1 + r0 e2)] = 8.032 kΩ RB = v2 it = Ra ∆= R4k (R2 + R3) ∆ = 624.7 Ω 8 Shunt-Shunt Example 1 Figure 10(a) shows the ac signal circuit of a shunt-series feedback ampliﬁer. The input variable is v1 and the output variable is v2. The input signal and the feedback signal are applied to the base of Q1. A test current source it is added in shunt with the output to calculate the output resistance RB. For the analysis to follow convention, the input source consisting of v1 in series with R1 must be converted into a Norton equivalent. This circuit is the current i1 = v1/R1 in parallel with the resistor R1. Fig. 10(b) shows the circuit with feedback removed and the source replaced with the Norton equivalent. The feedback at the base of Q1 is modeled by a Norton equivalent circuit v2/R4 in parallel with the resistor R4. The feedback factor or feedback ratio b is the negative of the coeﬃcient of v2 in this source, i.e. b = −R−1 4 . The circuit values are β1 = 100, gm1 = 0.05 S, rx1 = 0, rib1 = β1/gm1 = 2 kΩ, r01 = ∞, gm2 = 0.001 S, rs2 = g−1 m2 = 1 kΩ, r02 = ∞, R1 = 1 kΩ, R2 = 1 kΩ, R3 = 10 kΩ, and R4 = 10 kΩ. Figure 10: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. The following equations can be written for the circuit with feedback removed: vb1 = iaRb ia = i1 + v2 R4 Rb = R1kR4krib1 ic1 = gm1vb1 id1 = −ic1 R2 rs1 + R2 v2 = id1Rc + itRc + vb1 R3 R3 + R4 Rc = R3kR4 The current ia is the error current. The negative feedback tends to reduce ia, making \|ia\| →0 as the amount of feedback becomes inﬁnite. When this is the case, setting ia = 0 yields the current gain v2/i1 = −R4. Although the equations can be solved algebraically, the signal-ﬂow graph simpliﬁes the solution. Figure 11 shows the ﬂow graph for the equations. The determinant of the graph is given by ∆= 1 −Rb × · gm1 × −R2 rs1 + R2 × Rc + R3 R3 + R4 ¸ × 1 R4 The transresistance gain is calculated with it = 0. It is given by v2 i1 = Rb × µ gm1 × −R2 rs2 + R2 × Rc + R3 R3 + R4 ¶ ∆ = − (R1kR4krib1) × · gm1 × −R2 rs2 + R2 × (R3kR4) + R3 R3 + R4 ¸ 1 + · (R1kR4krib1) × gm1 × −R2 rs2 + R2 × (R3kR4) + R3 R3 + R4 ¸ × −1 R4 9 Figure 11: Signal-ﬂow graph for the equations. This is of the form v2 i1 = A 1 + Ab where A = (R1kR4krib1) × · gm1 × −R2 rs2 + R2 × (R3kR4) + R3 R3 + R4 ¸ = −77.81 kΩ b = −1 R4 = −10−4 S Note that Ab is dimensionless and positive. Numerical evaluation yields v2 i1 = −77.81 × 103 1 + (−77.81 × 103) × (−10−4) = −8.861 kΩ The voltage gain is given by v2 v1 = v2 i1 × i1 v1 = v2 i1 × 1 R1 = −8.861 The resistance Ra is calculated with it = 0. It is given by Ra = vb1 i1 = Rb ∆= R1kR4krib1 1 + Ab = 71.17 Ω Note that the feedback tends to decrease Ra. The resistance RA is calculated as follows: RA = R1 + ¡ R−1 a −R−1 1 ¢−1 = 1.077 kΩ The resistance RB is calculated with i1 = 0. It is given by RB = v2 it = Rc ∆= R3kR4 1 + Ab = 569.4 Ω Shunt-Shunt Example 2 A shunt-shunt feedback JFET ampliﬁer is shown in Fig. 12(a). Solve for the voltage gain v2/v1, the input resistance RA, and the output resistance RB. Assume gm = 0.005 S, rs = g−1 m = 200 Ω, r0 = ∞, R1 = 3 kΩ, R2 = 7 kΩ, R3 = 1 kΩ, R4 = 10 kΩ. The circuit with feedback removed is shown in Fig. 12(b) In this circuit, the source is replaced by a Norton equivalent circuit consisting of a current i1 = v1/R1 in parallel with the resistor R1. This is necessary for the feedback analysis to conform to convention for shunt-shunt feedback.. The circuit seen looking up into R2 is replaced with a Norton equivalent circuit made with respect to v2. A test current source it is added in shunt with the output to solve for the output resistance. 10 Figure 12: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. For the circuit with feedback removed, we can write vg = i1Rb + v2 R2 Rb Rb = R1kR2 id = Gmvg Gm = 1 rs + R3 v2 = −idRc + itRc + vg R4 R2 + R4 Rc = R2kR4 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 13. The determinant is ∆ = 1 −Rb × · Gm × −Rc + R4 R2 + R4 ¸ × 1 R2 = 1 + Rb × · Gm × Rc − R4 R2 + R4 ¸ × 1 R2 = 1.853 Ω Figure 13: Signal-ﬂow graph for the equations. 11 The transresistance gain is v2 i1 = Rb × · Gm × −Rc + R4 R2 + R4 ¸ ∆ = −Rb × · Gm × Rc − R4 R2 + R4 ¸ 1 + Rb × · Gm × Rc + R4 R2 + R4 ¸ × 1 R2 This is of the form v2 i1 = A 1 + Ab where A = −Rb × · Gm × Rc − R4 R2 + R4 ¸ = −(R1kR2) × · Gm × R2kR4 − R4 R2 + R4 ¸ = −5.971 kΩ b = −1 R2 = −0.1429 mS Note that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation yields v2 i1 = A ∆= −3.22 kΩ The voltage gain is given by v2 v1 = v2 i1 × i1 v1 = A ∆× 1 R1 = −1.074 The resistances Ra, RA, and RB are given by Ra = vg i1 = Rb ∆= 1.13 kΩ RA = R1 + µ 1 Ra −1 R1 ¶−1 = 4.82 kΩ RB = v2 it = Rc ∆= 2.22 kΩ Shunt-Shunt Example 3 A shunt-shunt feedback BJT ampliﬁer is shown in Fig. 14. The input variable is the v1 and the output variable is the voltage v2. The feedback resistor is R2. The summing at the input is shunt because the input through R1 and the feedback through R2 connect in shunt to the same node, i.e. the vb1 node. The output sampling is shunt because R2 connects to the output node. Solve for the voltage gain v2/v1, the input resistance RA, and the output resistance RB. Assume β = 100, rπ = 2.5 kΩ, gm = β/rπ, α = β/ (1 + β), re = α/gm, r0 = ∞, rx = 0, VT = 25 mV. The resistor values are R1 = 1 kΩ, R2 = 20 kΩ, R3 = 500 Ω, R4 = 1 kΩ, and R5 = 5 kΩ. The circuit with feedback removed is shown in Fig. 15. A test current source it is added in shunt with the output to solve for the output resistance RB. In the circuit, the source is replaced by a Norton equivalent circuit consisting of a current i1 = v1 R1 12 Figure 14: Ampliﬁer circuit. Figure 15: Circuit with feedback removed. in parallel with the resistor R1. This is necessary for the feedback analysis to conform to convention for shunt summing. The circuit seen looking into R2 from the collector of Q3 is replaced with a Thevenin equivalent circuit made with respect with vb1. For the circuit with feedback removed, we can write ie = i1 + v2 R2 vb1 = ieRb Rb = R1kR2krπ ic1 = gmvb1 vb2 = −ic1Rc Rc = R3krπ vb3 = −ic2Rd Rd = R4krπ ic3 = gmvb3 v2 = (−ic3 + it) Re + vb1 R5 R2 + R5 Re = R2kR5 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 16. The determinant is ∆ = 1 −Rb × µ gm × −Rc × gm × −Rd × gm × −Re + R5 R2 + R5 ¶ × 1 R2 = 1 + Rb × µ gm × Rc × gm × Rd × gm × Re − R5 R2 + R5 ¶ × 1 R2 = 354.7 13 Figure 16: Signal-ﬂow graph for the equations. The transresistance gain is v2 i1 = Rb × µ gm × −Rc × gm × −Rd × gm × −Re + R2 R2 + R5 ¶ ∆ = −Rb × µ gm × Rc × gm × Rd × gm × Re − R2 R2 + R5 ¶ 1 + Rb × µ gm × Rc × gm × Rd × gm × Re − R2 R2 + R5 ¶ × 1 R2 = −Rb × µ gm × Rc × gm × Rd × gm × Re − R2 R2 + R5 ¶ 1 + · −Rb × µ gm × Rc × gm × Rd × gm × Re − R2 R2 + R5 ¶¸ × −1 R2 This is of the form ie2 v1 = A 1 + Ab where A and b are given by A = −Rb × µ gm × Rc × gm × Rd × gm × Re − R5 R2 + R5 ¶ = −R1kR2krπ × µ gm × R3krπ × gm × R4krπ × gm × R2kR5 − R5 R2 + R5 ¶ = −7.073 MΩ b = −1 R2 = −50 µS Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transresistance gain yields v2 i1 = A ∆= −19.94 kΩ The resistances Ra and RA are Ra = vb1 i1 = Rc ∆= 1.945 Ω RA = R1 + µ 1 Ra −1 R1 ¶−1 = 1.002 kΩ 14 The resistance RB is RB = v2 it = Re ∆= 11.28 Ω The voltage gain is v2 vb1 = v2 i1 × µvb1 i1 ¶−1 = µ A ∆ ¶ × 1 RA = −19.91 Shunt-Shunt Example 4 A shunt-shunt feedback BJT ampliﬁer is shown in Fig. 17. The input variable is the v1 and the output variable is the voltage v2. The feedback resistor is R2. The summing at the input is shunt because the input through R1 and the feedback through R2 connect in shunt to the same node, i.e. the ve1 node. The output sampling is shunt because R2 connects to the output node. Solve for the voltage gain v2/v1, the input resistance RA, and the output resistance RB. For Q1 and Q2, assume β = 100, rπ = 2.5 kΩ, gm = β/rπ, α = β/ (1 + β), re = α/gm, r0 = ∞, rx = 0, VT = 25 mV. For J3, assume gm3 = 0.001 S and r03 = ∞. The resistor values are R1 = 1 kΩ, R2 = 100 kΩ, R3 = 10 Ω, R4 = 30 kΩ, and R5 = 10 kΩ. Figure 17: Ampliﬁer circuit. The circuit with feedback removed is shown in Fig. 18. A test current source it is added in shunt with the output to solve for the output resistance RB. In the circuit, the source is replaced by a Norton equivalent circuit consisting of a current i1 = v1 R1 in parallel with the resistor R1. This is necessary for the feedback analysis to conform to convention for shunt summing. The circuit seen looking into R2 from the collector of Q3 is replaced with a Thévenin equivalent circuit made with respect with ve1. Figure 18: Circuit with feedback removed. 15 For the circuit with feedback removed, we can write ie = i1 + v2 R2 ve1 = ieRb Rb = R1kR2kre1 ic1 = gm1ve1 vtg3 = ic1R3 id3 = gm3vtg3 vtb2 = −id3R4 ie2 = −G1vtb2 G1 = 1 r0 e2 + R2kR5 r0 e2 = 1 R4 + re2 v2 = (−ie2 + it) Rc + ve1 R5 R2 + R5 Rc = R2kR5 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 19. The determinant is ∆ = 1 −Rb × µ gm1 × R3 × gm2 × −R4 × −G1 × −Rc + R5 R2 + R5 ¶ × 1 R2 = 1 + Rb × µ gm1 × R3 × gm2 × R4 × G1 × Rc − R5 R2 + R5 ¶ × 1 R2 = 113.0 Figure 19: Signal-ﬂow graph for the equations. The transresistance gain is v2 i1 = Rb × µ gm1 × R3 × gm2 × −R4 × −G1 × −Rc + R2 R2 + R5 ¶ ∆ = −Rb × µ gm1 × R3 × gm2 × R4 × G1 × Rc − R2 R2 + R5 ¶ 1 + Rb × µ gm1 × R3 × gm2 × R4 × G1 × Rc − R2 R2 + R5 ¶ × 1 R2 = −Rb × µ gm1 × R3 × gm2 × R4 × G1 × Rc − R2 R2 + R5 ¶ 1 + · −Rb × µ gm1 × R3 × gm2 × R4 × G1 × Rc − R2 R2 + R5 − R2 R2 + R5 ¶¸ × −1 R2 This is of the form ie2 v1 = A 1 + Ab where A and b are given by A = −Rb × µ gm1 × R3 × gm2 × R4 × G1 × Rc − R2 R2 + R5 ¶ = −R1kR2kre1 × µ gm1 × R3 × gm2 × R4 × 1 r0 e2 + R2kR5 × R2kR5 − R5 R2 + R5 ¶ = −11.2 MΩ 16 b = −1 R2 = −10 µS Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transresistance gain yields v2 i1 = A ∆= −99.11 kΩ The resistances Ra and RA are Ra = ve1 i1 = Rb ∆= 0.214 Ω RA = R1 + µ 1 Ra −1 R1 ¶−1 = 1 kΩ The resistance RB is RB = v2 it = Rc ∆= 80.49 Ω The voltage gain is v2 ve1 = v2 i1 × µve1 i1 ¶−1 = µ A ∆ ¶ × 1 RA = −99.09 Series-Series Example 1 Figure 20(a) shows the ac signal circuit of a series-series feedback ampliﬁer. The input variable is v1 and the output variable is id2. The input signal is applied to the gate of M1 and the feedback signal is applied to the source of M1. Fig. 20(b) shows the circuit with feedback removed. A test voltage source vt is added in series with the output to calculate the output resistance Rb. The feedback at the source of M1 is modeled by a Thévenin equivalent circuit. The feedback factor or feedback ratio b is the coeﬃcient of id2 in this source, i.e. b = R5. The circuit values are gm = 0.001 S, rs = g−1 m = 1 kΩ, r0 = ∞, R1 = 50 kΩ, R2 = 10 kΩ, R3 = 1 kΩ, R4 = 9 kΩ, and R5 = 1 kΩ. Figure 20: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. 17 The following equations can be written for the circuit with feedback removed: id1 = Gm1va Gm1 = 1 rs1 + R5 va = v1 −vts1 vts1 = id2R5 id2 = Gm2vb Gm2 = 1 rs2 + R3 vb = vt −vtg2 vtg2 = −id1R2 The voltage va is the error voltage. The negative feedback tends to reduce va, making \|va\| →0 as the amount of feedback becomes inﬁnite. When this is the case, setting va = 0 yields the transconductance gain id2/v1 = b−1 = R−1 5 . Although the equations can be solved algebraically, the signal-ﬂow graph simpliﬁes the solution. Figure 21 shows the signal-ﬂow graph for the equations. The determinant of the graph is given by ∆= 1 −Gm1 × (−R2) × (−1) × Gm2 × R5 × (−1) Figure 21: Flow graph for the equations. The transconductance gain id2/v1 is calculated with vt = 0. It is given by id2 v1 = Gm1 × (−R2) × (−1) × Gm2 ∆ = 1 rs1 + R5 × R2 × 1 rs1 + R5 1 + 1 rs1 + R5 × R2 × 1 rs1 + R5 × R5 This is of the form id2 v1 = A 1 + Ab where A = Gm1 × (−R2) × (−1) × Gm2 = 1 rs1 + R5 × R2 × 1 rs2 + R3 = 2.5 × 10−3 S b = R5 = 1000 Ω Note that bA is dimensionless. Numerical evaluation yields id2 v1 = 2.5 × 10−3 1 + 1000 × 2.5 × 10−3 = 7.124 × 10−4 S The resistance Rb is calculated with v1 = 0. It is given by Rb = µid2 vt ¶−1 = µGm2 ∆ ¶−1 = (1 + bA) (rs2 + R3) = 7 kΩ Note that the feedback tends to increase Rb. The resistance RB is calculated as follows: RB = (Rb −R3) kR3 = 857.1 Ω Because the gate current of M1 is zero, the input resistance is RA = R1 = 50 kΩ. 18 Series-Series Example 2 A series-series feedback BJT ampliﬁer is shown in Fig. 22. The input variable is the voltage v1 and the output variable is the voltage v2. The feedback is from ie2 to the emitter of Q1. Because the feedback does not connect to the input node, the input summing is series. The output sampling is series because the feedback is proportional to the current that ﬂows in series with the output rather than the output voltage. Solve for the transconductance gain ic3/v1, the voltage gain v2/v1, the input resistance RA, and the output resistance RB. Assume β = 100, IC1 = 0.6 mA, IC2 = 1 mA, IC3 = 4 mA, α = β/ (1 + β), gm = IC/VT , re = αVT /IC, r0 = ∞, rx = 0, VT = 25 mV, R1 = 100 Ω, R2 = 9 kΩ, R3 = 5 kΩ, R4 = 600 Ω, R5 = 640 Ω, and R6 = 100 Ω. The circuit with feedback removed is shown in Fig. 23. Figure 22: Ampliﬁer circuit. The circuit looking out of the emitter of Q1 is a Thévenin equivalent made with respect to the current ie3. The output current is proportional to this current, i.e. ic3 = αie3. Because r0 = ∞for Q3, the feedback does not aﬀect the output resistance seen looking down through R4 because it is inﬁnite. For a ﬁnite r0, a test voltage source can be added in series with R4 to solve for this resistance. It would be found that a ﬁnite r0 for Q3 considerably complicates the circuit equations and the ﬂow graph. Figure 23: Circuit with feedback removed. 19 For the circuit with feedback removed, we can write ve = v1 −ie3 R6R1 R6 + R5 + R1 ie1 = G1ve G1 = 1 re1 + R1k (R5 + R6) ic1 = αie1 ib1 = ic1 β vtb2 = −ic1R2 ie2 = G2vtb2 G2 = 1 r0 e2 r0 e2 = R2 1 + β + re ic2 = αie2 vtb3 = −ic2R3 ie3 = G3vtb3 −k1ie1 G3 = 1 r0 e3 + R6k (R1 + R5) k1 = R1 R1 + R5 + r0 e3kR6 R6 R6 + r0 e3 ic3 = αie3 v2 = −ic2R4 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 24. The determinant is ∆ = 1 − ½ G1 × [(α × −R2 × G2 × α × −R3 × G3) −k1] × −R6R1 R6 + R5 + R1 ¾ = 1 + G1 × (α × R2 × G2 × α × Ra −k1) × R6R1 R6 + R5 + R1 = 251.5 Figure 24: Signal-ﬂow graph for the circuit. The transconductance gain is ic3 v1 = G1 × (α × −R2 × G2 × α × −R3 × G3 −k1) × α ∆ = G1 × (α × R2 × G2 × α × R3 × G3 −k1) × α 1 + [G1 × (α × R2 × G2 × α × Ra −k1)] × R6R1 R6 + R5 + R1 = G1 × [(α × R2 × G2 × α × R3 × G3) −k1] × α 1 + [G1 × (α × R2 × G2 × α × Ra −k1) × α] × R6R1 R6 + R5 + R1 × 1 α This is of the form ic3 v1 = A 1 + Ab where A and b are given by A = G1 × [(α × R2 × G2 × α × R3 × G3) −k1] × α = 1 re1 + R1k (R5 + R6) × · µ α × R2 × 1 r0 e2 × α × R3 × 1 r0 e3 + R6k (R1 + R5) × α ¶ − R1 R1 + R5 + r0 e3kR6 R6 R6 + r0 e3 ¸ × α = 20.83 S 20 b = R6R1 R6 + R5 + R1 × 1 α = 12.02 Ω Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transconductance gain yields ic3 v1 = A ∆= 0.083 The voltage gain is given by v2 v1 = ic3 v1 × v2 ic3 = A ∆× −R4 = −49.7 The resistances RA and RB are given by RA = µib1 v1 ¶−1 = µG1α/β ∆ ¶−1 = ∆× (1 + β) G1 = ∆× (1 + β) × [re1 + R1k (R5 + R6)] = 3.285 MΩ RB = R4 = 600 Ω Series-Series Example 3 A series-series feedback BJT ampliﬁer is shown in Fig. 25(a). The input variable is the voltage v1 and the output variable is the voltage v2. The feedback is from ie2 to ic2 to the emitter of Q1. Because the feedback does not connect to the input node, the input summing is series. Because the feedback does not sample the output voltage, the sampling is series. That is, the feedback network samples the current in series with the outpu. Solve for the transconductance gain ie2/v1, the voltage gain v2/v1, the input resistance RA, and the output resistance RB. Assume β = 100, rπ = 2.5 kΩ, α = β/ (1 + β), re = α/gm, r0 = ∞, rx = 0, VT = 25 mV, R1 = 100 Ω, R2 = 1 kΩ, R3 = 20 kΩ, and R4 = 10 kΩ. Figure 25: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. The circuit with feedback removed is shown in Fig. 25(b). The circuit seen looking out of the emitter of Q1 is replaced with a Thévenin equivalent circuit made with respect with ic2. The output current ie2 is proportional to this current, i.e. ie2 = αic2. A test voltage source vt is added in series with the output to solve for the output resistance. The resistance seen by the test source is labeled Rb. For the circuit with feedback removed, we can write ve = v1 −ic2R2 ie1 = G1ve G1 = 1 re + R1 + R2 ic1 = αie1 ib1 = ic1 β 21 vtb2 = −ic1R3 ie2 = G2 (vt −vtb2) G2 = 1 r0 e2 + R4 r0 e2 = R3 1 + β + re ic2 = αie2 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 26. The determinant is ∆ = 1 −(G1 × α × −R3 × −G2 × α × −R2) = 1 + G1 × α × R2 × G2 × α × R2 = 1.181 Figure 26: Signal-ﬂow graph for the equations. The transconductance gain is ie2 v1 = G1 × α × −R3 × −G2 ∆ = (G1 × α × R3 × G2) 1 + (G1 × α × R3 × G2) × α × R2 This is of the form ie2 v1 = A 1 + Ab where A and b are given by A = G1 × α × R3 × G2 = 1 re1 + R1 + R2 × α × R3 × 1 r0 e2 + R4 = 0.9117 mS b = αR2 = 1.98 kΩ Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transconductance gain yields ie2 v1 = A ∆= 0.325 mS The voltage gain is given by v2 v1 = ie2 v1 × v2 ie2 = A ∆× −R4 = −3.25 The resistances RA and RB are given by RA = µib1 v1 ¶−1 = µG1α/β ∆ ¶−1 = ∆× (1 + β) (re + R1 + R2) = 602 kΩ Rb = µie2 vt ¶−1 = µGm2 ∆ ¶−1 = ∆× (R4 + r0 e2) = 28.68 kΩ RB = (Rb −R4) kR4 = 6.513 kΩ 22 Series-Series Example 4 A series-series feedback BJT ampliﬁer is shown in Fig. 27(a). The input variable is the voltage v1 and the output variable is the current ic2. The feedback is from ic2 to ie2 to the gate of J1. The input summing is series because the feedback does not connect to the same node that the source connects. The output sampling is series because the feedback is proportional to the output current ic2. Solve for the transconductance gain ic2/v1, the voltage gain v2/v1, the input resistance RA, and the output resistance RB. For J1, assume that gm1 = 0.001 S, rs1 = g−1 m1 = 1000 Ω, and r01 = ∞. For Q2, assume β2 = 100, rπ2 = 2.5 kΩ, α2 = β2/ (1 + β2), re2 = α2/gm2, r02 = ∞, rx2 = 0, VT = 25 mV. The resistor values are R1 = 1 kΩ, R2 = 10 kΩ, R3 = 1 kΩ, R4 = 10 kΩ, R5 = 1 kΩ, and R6 = 10 kΩ. Figure 27: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. The circuit with feedback removed is shown in Fig. 27(b). The circuit seen looking out of the emitter of Q1 is replaced with a Thévenin equivalent circuit made with respect with ie2. The output current is proportional to this current, i.e. ic2 = α2ie2. Because r02 = ∞, the feedback does not aﬀect the output resistance seen looking down through R6 because it is inﬁnite. For a ﬁnite r02, a test voltage source can be added in series with R6 to solve for this resistance. It would be found that a ﬁnite r02 considerably complicates the circuit equations and the ﬂow graph. For the circuit with feedback removed, we can write id1 = gm1ve ve = ie2 R5R3 R3 + R4 + R5 −v1 vtb2 = −id1R2 ie2 = G1vtb2 G1 = 1 r0 e2 + R5k (R3 + R4) r0 e2 = R2 1 + β2 + re2 ic2 = α2ie2 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 28. The determinant is ∆ = 1 − µ gm1 × −R2 × G1 × R5R3 R3 + R4 + R5 ¶ = 1 + gm1 × R2 × G1 × R5R3 R3 + R4 + R5 = 1.801 23 Figure 28: Signal-ﬂow graph for the equations. The transconductance gain is ie2 v1 = −1 × gm1 × −R2 × G1 × α2 ∆ = gm1 × R2 × G1 × α2 1 + gm1 × R2 × G1 × R5R3 R3 + R4 + R5 = (gm1 × R2 × G1 × α2) 1 + (gm1 × R2 × G1 × α2) × R5R3 R3 + R4 + R5 × 1 α2 This is of the form ie2 v1 = A 1 + Ab where A and b are given by A = gm1 × R2 × G1 × α2 = gm1 × R2 × 1 r0 e2 + R5k (R3 + R4) × α2 = 9.516 mS b = R5R3 R3 + R4 + R5 × 1 α2 = 84.17 Ω Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transconductance gain yields ic2 v1 = A ∆= 5.284 mS The voltage gain is given by v2 v1 = ic2 v1 × v2 ic2 = A ∆× −R6 = −52.84 The resistances RA and RB are given by RA = R1k µ−id1 v1 ¶−1 = R1k ³gm1 ∆ ´−1 = R1k µ ∆ gm1 ¶ = 643 Ω RB = R6 = 10 kΩ Series-Series Example 5 A series-series feedback BJT ampliﬁer is shown in Fig. 29. The input variable is the current i1 and the output variable is the current ie2. The feedback path is the path from ie2 to ic2 to ie3 to ic3 to the emitter of Q1. The input summing is series because the feedback does not connect to the input node. The output 24 sampling is series because the feedback is proportional to the output current ie2 and not the output voltage v2. Solve for the current gain gain ie2/i1, the transresistance gain v2/i1, the input resistance RA, and the output resistance RB. Assume β = 100, rπ = 2.5 kΩ, gm = β/rπ, α = β/ (1 + β), re = α/gm, r0 = ∞, rx = 0, VT = 25 mV. The resistor values are R1 = 1 kΩ, R2 = 100 Ω, R3 = 10 kΩ, R4 = 100 Ω, R5 = 1 kΩ, and R6 = 10 kΩ. Figure 29: Ampliﬁer circuit. The circuit with feedback removed is shown in Fig. 30. The source is replaced with a Thevenin equivalent circuit consisting of a voltage v1 = i1R1 in series with the resistor R1. This is necessary for the feedback analysis to conform to convention for series summing at the input. The circuit seen looking out of the emitter of Q1 is replaced with a Thévenin equivalent circuit made with respect with ic3. The latter is proportional to the output current ie2. The relation is ic3 ie2 = ic3 ie3 × ie3 ic2 × ic2 ie2 = α × R3 R3 + R4 + re3 × α Note that re3 in this equation is the small-signal resistance seen looking into the emitter of Q3. Figure 30: Circuit with feedback removed. 25 For the circuit with feedback removed, we can write ve = v1 −ic3R2 ie1 = G1ve G1 = 1 r0 e1 + R2 r0 e1 = R1 1 + β + re ic1 = αie1 ib1 = ic1 β vtb2 = −ic1R6 ie2 = G2 (vt −vtb2) G2 = 1 r0 e2 + R5 r0 e2 = R6 1 + β + re ic2 = αie2 ie3 = ic2 R3 R3 + R4 + re ic3 = αie3 Figure 31: Signal-ﬂow graph for the equations. The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 31. The determinant is ∆ = 1 −G1 × α × −R6 × −G2 × α × R3 R3 + R4 + re × α × R2 = 1 + G1 × α × R6 × G2 × α × R3 R3 + R4 + re × α × R2 = 7.335 The transconductance gain is ie2 v1 = 1 × G1 × α × −R6 × −G2 ∆ = G1 × α × R6 × G2 1 + G1 × α × R6 × G2 × α × R3 R3 + R4 + re × α × R2 This is of the form ie2 v1 = A 1 + Ab where A and b are given by A = G1 × α × R6 × G2 = 1 r0 e1 + R2 × α × R6 × 1 r0 e2 + R5 = 65.43 mS b = α × R3 R3 + R4 + re × α × R2 = 96.82 Ω Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transconductance gain yields ie2 v1 = A ∆= 8.92 mS 26 The current gain is given by ie2 i1 = ie2 v1 × v1 i1 = A ∆× R1 = 8.92 The transresistance gain is Rm = v2 i1 = ie2 i1 × v2 ie2 = A ∆× R1 × −R5 = −8.92 kΩ The resistances Ra and RA Ra = µib1 v1 ¶−1 = µG1α/β ∆ ¶−1 = ∆× [R1 + rπ + (1 + β) R2] = 113.3 kΩ RA = vb1 i1 = (Ra −R1) kR1 = 991.2 Ω The voltage gain is v2 vb1 = v2 i1 × µvb1 i1 ¶−1 = µ A ∆× R1 ¶ × 1 RA = −990 The resistances Rb and RB are Rb = µie2 vt ¶−1 = µG2 ∆ ¶−1 = ∆× (r0 e2 + R5) = 10.98 kΩ RB = (Rb −R5) kR5 = 889.5 Ω Shunt-Series Example 1 Figure 32(a) shows the ac signal circuit of a shunt-series feedback ampliﬁer. The input variable is v1 and the output variable is id2. The input signal and the feedback signal are applied to the source of M1. A test voltage source vt is added in series with the output to calculate the output resistance Rb. For the analysis to follow convention, the input source consisting of v1 in series with R1 must be converted into a Norton equivalent. This circuit is the current i1 = v1 R1 in parallel with the resistor R1. Fig. 32(b) shows the circuit with feedback removed and the source replaced with the Norton equivalent. A test source vt is added in series with the output to calculate the resistance Rb. The feedback at the source of M1 is modeled by a Norton equivalent circuit id2 in parallel with the resistor R4. The feedback is from the output current id2 to the source of M1. The circuit values are gm = 0.001 S, rs = g−1 m = 1 kΩ, r0 = ∞, R1 = 10 kΩ, R2 = 20 kΩ, R3 = 1 kΩ, R4 = 1 kΩ, and R5 = 1 kΩ. The following equations can be written for the circuit with feedback removed: vs1 = iaRc ia = i1 + id2 Rc = R1kR4krs1 id1 = −gm1vs1 id2 = Gm2vb Gm2 = 1 rs2 + R3 vb = vt −vtg2 vtg2 = −id1R2 The current ia is the error current. The negative feedback tends to reduce ia, making \|ia\| →0 as the amount of feedback becomes inﬁnite. When this is the case, setting ia = 0 yields the current gain id2/i1 = −1. Although the equations can be solved algebraically, the signal-ﬂow graph simpliﬁes the solution. Fig. 33 shows the ﬂow graph for the equations. The determinant of the graph is given by ∆= 1 −Rc × (−gm1) × (−R2) × (−1) × Gm2 × 1 The current gain is calculated with vt = 0. It is given by id2 i1 = Rc × (−gm1) × (−R2) × (−1) × Gm2 ∆ = − R1kR4krs1 × gm1 × R2 × 1 rs2 + R3 1 + Rc × gm1 × R2 × 1 rs2 + R3 × 1 27 Figure 32: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. Figure 33: Flow graph for the equations. 28 This is of the form id2 i1 = A 1 + Ab where A = Rc × (−gm1) × (−R2) × (−1) × Gm2 = −(R1kR4krs1) × gm1 × R2 × 1 rs2 + R3 = −0.3333 b = −1 Note that Ab is dimensionless. Numerical evaluation yields id2 i1 = 2.5 × 10−3 1 + 1000 × 2.5 × 10−3 = −0.7692 The voltage gain is given by v2 v1 = id2 i1 × i1 v1 × v2 id2 = id2 i1 × 1 R1 × (−R3) = 0.7692 The resistance Ra is calculated with vt = 0. It is given by Ra = vs1 i1 = Rc ∆= R1kR4krs1 1 + 1 × Rc × gm1 × R2 × 1 rs2 + R3 = 76.92 Ω Note that the feedback tends to decrease Ra. The resistance RA is calculated as follows: RA = R1 + ¡ R−1 a −R−1 1 ¢−1 = 1.083 kΩ The resistance Rb is calculated with i1 = 0. It is given by Rb = µid2 vt ¶−1 = µGm2 ∆ ¶−1 = (1 + Ab) (rs2 + R3) = 8.667 kΩ Note that the feedback tends to increase Rb. The resistance RB is calculated as follows: RB = (Rb −R3) kR3 = 884.6 Ω Shunt-Series Example 2 A shunt-series feedback BJT ampliﬁer is shown in Fig. 34(a). The input variable is the voltage v1 and the output variable is the current ie2. The feedback is from ie2 to ic2 to the source of M1. The input summing is shunt because the feedback connects to the same node that the source connects. The output sampling is series because the feedback is proportional to the output current ie2. Solve for the voltage gain v2/v1, the input resistance RA, and the output resistance RB. For M1, assume that gm1 = 0.001 S, rs1 = g−1 m1 = 1000 Ω, and r01 = ∞. For Q2, assume β2 = 100, rπ2 = 2.5 kΩ, α2 = β2/ (1 + β2), re2 = α2/gm2, r02 = ∞, rx2 = 0, VT = 25 mV. The resistor values are R1 = 10 kΩ, R2 = 100 kΩ, R3 = 100 kΩ, R4 = 10 kΩ, and R5 = 1 kΩ. The circuit with feedback removed is shown in Fig. 34(b). The source is replaced with a Norton equivalent circuit. The current i1 is given by i1 = v1 R1 The circuit seen looking into R2 from the vs1 node is replaced with a Norton equivalent circuit made with respect with ic2. The output current is proportional to this current, i.e. ic2 = α2ie2. A test voltage source vt is added in series with ie2 to calculate the resistance Rb. For the circuit with feedback removed, we can write vs1 = iaRc ia = i1 − R4 R2 + R4 ic2 Rc = R1k (R2 + R4) krs1 id1 = gm1vs1 vtb2 = id1R3 29 Figure 34: (a) Shunt-series ampliﬁer. (b) Ampliﬁer with feedback removed. ie2 = G1vtb2 −vt Rd G1 = 1 r0 e2 + R5 r0 e2 = R3 1 + β2 + re2 Rd = R5 + r0 e2 ic2 = α2ie2 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 35. The determinant is ∆ = 1 − µ 1 × Rc × gm1 × R3 × G1 × α2 × −R4 R2 + R4 ¶ = 1 + Rc × gm1 × R3 × G1 × α2 × R4 R2 + R4 = 5.025 Figure 35: Signal-ﬂow graph for the equations. The current gain is ie2 i1 = Rc × gm1 × R3 × G1 ∆ = Rc × gm1 × R3 × G1 1 + Rc × gm1 × R3 × G1 × α2 × R4 R2 + R4 = µ R1k (R2 + R4) krs1 × gm1 × R3 × 1 r0 e2 + R4 ¶ 1 + µ R1k (R2 + R4) krs1 × gm1 × R3 × 1 r0 e2 + R4 ¶ × α2 × R4 R2 + R4 30 This is of the form ie2 i1 = A 1 + Ab where A and b are given by A = Rc × gm1 × R3 × G1 = R1k (R2 + R4) krs1 × gm1 × R3 × 1 r0 e2 + R4 = 44.75 b = α2 × R4 R2 + R4 = 0.09 Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the current gain yields ie2 i1 = A ∆= 8.90 The resistances Ra and RA are Ra = vs1 i1 = Rc ∆= 179.3 Ω RA = R1 + ¡ R−1 a −R−1 1 ¢−1 = 10.13 kΩ The resistance Rb and RB are Rb = µ−ie2 vt ¶−1 = µ 1 ∆ 1 Rd ¶−1 = ∆Rd = 10.12 kΩ RB = (Rb −R5) kR5 = 901.3 Ω The voltage gain is given by v2 v1 = ie2 i1 × v2 ie2 × i1 v1 = A ∆× R2 × 1 R1 = 89.0 Shunt-Series Example 3 A shunt-series feedback BJT ampliﬁer is shown in Fig. 36(a). The input variable is the voltage v1 and the output variable is the current ic2. The feedback is from ic2 to ie2 to ic3 to the emitter of Q1. The input summing is shunt because the feedback connects to the same node that the source connects. The output sampling is series because the feedback is proportional to the output current ic2. Solve for the voltage gain v2/v1, the input resistance RA, and the output resistance RB. Assume β = 100, rπ = 2.5 kΩ, α = β/ (1 + β), re = α/gm, r0 = ∞, rx = 0, VT = 25 mV. The resistor values are R1 = R3 = 1 kΩand R2 = R4 = R5 = 10 kΩ. The circuit with feedback removed is shown in Fig. 36(b). The source is replaced with a Norton equivalent circuit consisting of the current i1 = v1 R1 in parallel with the resistor R1. The feedback is modeled by a Norton equivalent circuit consisting of the current ic3. Because r03 = ∞, the output resistance of this source is an open circuit. The output current is proportional to this current. Because r02 = ∞, the feedback does not aﬀect the output resistance seen looking down through R5 because it is inﬁnite. For a ﬁnite r02, a test voltage source can be added in series with R5 to solve for this resistance. It would be found that a ﬁnite r02 considerably complicates the circuit equations and the ﬂow graph. 31 Figure 36: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. For the circuit with feedback removed, we can write ia = i1 −ic3 ve1 = iaRb Rb = R1kR2kre1 ic1 = −gm1ve1 ic2 = −β2ic1 ie2 = ie2 α2 vtb3 = ie2R4 ie3 = G1vtb3 G1 = 1 r0 e3 + R3 r0 e3 = R4 1 + β3 + re3 ic3 = α3ie3 The equations can be solved algebraically or by a ﬂow graph. The ﬂow graph for the equations is shown in Fig. 37. The determinant is ∆ = 1 − µ Rb × −gm1 × −β1 × 1 α2 × R4 × G1 × α3 × −1 ¶ = 1 + Rb × gm1 × β1 × 1 α2 × R4 × G1 × α3 = 858.7 Figure 37: Signal-ﬂow graph for the equations. 32 The transconductance gain is ic2 i1 = 1 × Rb × −gm1 × −β2 ∆ = Rb × gm1 × β2 1 + Rb × gm1 × β2 × 1 α2 × R4 × G1 × α3 = (Rb × gm1 × β2) 1 + (Rb × gm1 × β2) × µ 1 α2 × R4 × G1 × α3 ¶ This is of the form ic2 i1 = A 1 + Ab where A and b are given by A = Rb × gm1 × β2 = R1kR2kre1 × gm1 × β2 = 96.39 b = 1 α2 × R4 × G1 × α3 = 1 α2 × R4 × 1 r0 e3 + R3 × α3 = 8.899 Notice that the product Ab is dimensionless and positive. The latter must be true for the feedback to be negative. Numerical evaluation of the transconductance gain yields ic2 i1 = A ∆= 0.112 The voltage gain is given by v2 v1 = i1 v1 × ic2 i1 × v2 ic2 = 1 R1 × A ∆× −R5 = −1.122 The resistances Ra, RA, and RB are given by Ra = Rb ∆= 0.028 Ω RA = R1 + ¡ R−1 a −R−1 1 ¢−1 = 1 kΩ RB = R5 = 10 kΩ Shunt-Series Example 4 Figure 38(a) shows the ac signal circuit of a shunt-series feedback ampliﬁer. The input variable is v1 and the output variable is id2. The input signal and the feedback signal are applied to the gate of M1. For the analysis to follow convention, the input source consisting of v1 in series with R1 must be converted into a Norton equivalent. The feedback is from the output current id2 to the source of M2 and to the gate of M1. The circuit values are gm = 0.001 S, rs = g−1 m = 1 kΩ, r0 = ∞, R1 = 1 kΩ, R2 = 100 kΩ, R3 = 10 kΩ, R4 = 1 kΩ, R5 = 1 kΩ, and R6 = 100 Ω. The circuit with feedback removed is shown in Fig. 38(b). The source is replaced with a Norton equivalent circuit consisting of the current i1 = v1 R1 in parallel with the resistor R1. The feedback is modeled by a Norton equivalent circuit consisting of the current k1id2. The output current is proportional to this current. Because r02 = ∞, the feedback does not aﬀect the output resistance seen looking up from signal ground into the lower terminal of R3 because it is inﬁnite. For a ﬁnite r02, a test voltage source can be added in series with R3 to solve for this resistance. It would be found that a ﬁnite r02 considerably complicates the circuit equations and the ﬂow graph. 33 Figure 38: (a) Ampliﬁer circuit. (b) Circuit with feedback removed. The following equations can be written for the circuit with feedback removed: ia = i1 + k1id2 k1 = R5 R5 + R6 vg1 = iaRc Rc = R1kRb Rb = R5 + R6 id1 = gm1vg1 id2 = G1 (vtg2 −vts2) vtg2 = −id1R2 vts2 = k2vg1 k2 = R5 R5 + R6 G1 = 1 rs2 + Rts2 Rts2 = R4 + R5kR6 The current ia is the error current. The negative feedback tends to reduce ia, making \|ia\| →0 as the amount of feedback becomes inﬁnite. When this is the case, setting ia = 0 yields the current gain id2/i1 = −1/k1. Although the equations can be solved algebraically, the signal-ﬂow graph simpliﬁes the solution. Fig. 39 shows the ﬂow graph for the equations. The determinant of the graph is given by ∆ = 1 −(Rc × gm1 × −R2 × G1 × k1) = 1 + Rc × gm1 × R2 × G1 × k1 Figure 39: Signal-ﬂow graph for the equations. The current gain is given by id2 i1 = Rc × gm1 × −R2 × G1 ∆ = − µ Rc × gm1 × R2 × 1 rs2 + R4 + R5kR6 ¶ 1 + · − µ Rc × gm1 × R2 × 1 rs2 + R4 + R5kR6 ¶¸ × (−k1) This is of the form id2 i1 = A 1 + Ab 34 where A = − µ Rc × gm1 × R2 × 1 rs2 + R4 + R5kR6 ¶ = −25.02 b = −k1 = −0.909 Note that Ab is dimensionless. Numerical evaluation yields id2 i1 = −25.02 1 + (−25.02) × (−0.909) = −1.054 The voltage gain is given by v2 v1 = id2 i1 × i1 v1 × v2 id2 = id2 i1 × 1 R1 × −R3 = −10.54 The resistance Ra is Ra = vs1 i1 = Rc ∆= R1k (R5 + R6) ∆ = 22.03 Ω Note that the feedback tends to decrease Ra. The resistance RA is RA = R1 + ¡ R−1 a −R−1 1 ¢−1 = 1.023 kΩ The resistance RB is RB = R3 = 10 kΩ This is not a function of the feedback because r02 has been assumed to be inﬁnite. Shunt-Series Example 5 Figure 40(a) shows the ac signal circuit of a shunt-series feedback ampliﬁer. The input variable is v1 and the output variable is id2. The input signal and the feedback signal are applied to the base Q1. For the analysis to follow convention, the input source consisting of v1 in series with R1 must be converted into a Norton equivalent. The feedback is from the output current ic2 to the current ie2 to the current ie3 to the current ic3. The resistor values are R1 = 1 kΩ, R2 = 10 kΩ, R3 = 10 kΩ, and R4 = 10 kΩ. Assume β = 100, rπ = 2.5 kΩ, α = β/ (1 + β), re = α/gm, r0 = ∞, rx = 0, VT = 25 mV. Figure 40: (a) Ampliﬁer circuit. (b) Circuit with the the source replaced with a Norton equivalent. The circuit with feedback removed is shown in Fig. 40(b). The source is replaced with a Norton equivalent circuit consisting of the current i1 = v1 R1 in parallel with the resistor R1. The feedback is modeled by a Norton equivalent circuit consisting of the current ic3. The output current is proportional to this current. Because r02 = ∞, the feedback does not aﬀect the output resistance seen looking up from signal ground into the lower terminal of R4 because it is 35 inﬁnite. For a ﬁnite r02, a test voltage source can be added in series with R4 to solve for this resistance. It would be found that a ﬁnite r02 considerably complicates the circuit equations and the ﬂow graph. The following equations can be written for the circuit with feedback removed: ia = i1 + ic3 vb1 = iaRb Rb = R1krπ1 ic1 = gm1vb1 vtb2 = −ic1R3 ie2 = G1vtb2 G1 = 1 r0 e2 + R2 + re3 r0 e2 = R3 1 + β2 + re2 ic2 = α2ie2 ie3 = ie2 ic3 = α3ie3 The current ia is the error current. The negative feedback tends to reduce ia, making \|ia\| →0 as the amount of feedback becomes inﬁnite. When this is the case, setting ia = 0 yields the current gain id2/i1 = −1/k1. Although the equations can be solved algebraically, the signal-ﬂow graph simpliﬁes the solution. Fig. 41 shows the ﬂow graph for the equations. The determinant of the graph is given by ∆ = 1 −(Rb × gm1 × −R3 × G1 × 1 × α3 × 1) = 1 + Rb × gm1 × R3 × G1 × α3 = 28.88 Figure 41: Signal-ﬂow graph for the equations. The current gain is given by id2 i1 = 1 × Rb × gm1 × −R3 × G1 × α2 ∆ = −(Rb × gm1 × R3 × G1 × α2) 1 + [−(Rb × gm1 × R3 × G1 × α2)] × µ −α3 α2 ¶ This is of the form id2 i1 = A 1 + Ab where A = −(Rb × gm1 × R3 × G1 × α2) = −27.88 b = −α3 α2 = −1 Note that Ab is dimensionless and the product is positive. The latter is a result of the feedback being negative. Numerical evaluation yields id2 i1 = −27.88 1 + (−27.88) × (−1) = −0.965 The voltage gain is given by v2 v1 = id2 i1 × i1 v1 × v2 id2 = id2 i1 × 1 R1 × −R4 = 9.654 The resistance Ra is Ra = vs1 i1 = Rb ∆= R1krπ1 ∆ = 24.74 Ω 36 Note that the feedback tends to decrease Ra. The resistance RA is RA = R1 + ¡ R−1 a −R−1 1 ¢−1 = 1.025 kΩ The resistance RB is RB = R4 = 10 kΩ This is not a function of the feedback because r02 has been assumed to be inﬁnite. 37
2921	https://wayground.com/admin/quiz/5aeafdd2b9c1e8001b818bc3/mean-median-mode-range	Mean Median Mode Range 7th Grade Quiz \| Wayground (formerly Quizizz) Enter code Login/Signup Find similar resources box and whisker plots [x] mean absolute deviation [x] dot plots and histograms [x] mean median mode and range vocabulary [x] 5 number summary [x] Mean Median Mode Range 7th Grade • 10 Qs Try it as a student Similar activities Mean, Median, Mode Quiz 6th - 8th Grade • 10 Qs Measures of Center and Changes in a Data Set 5th - 8th Grade • 10 Qs Mean Median Mode Range Outlier II 6th - 7th Grade • 10 Qs Mean, Median, Mode, and Range 5th - 7th Grade • 10 Qs Mode, Median and Range of a Data Set 6th Grade - University • 13 Qs Mean Median Mode Range 6th - 8th Grade • 15 Qs Statistics 7th Grade • 12 Qs Mean Median Mode Range 6th - 7th Grade • 13 Qs View more activities Mean Median Mode Range Quiz • Mathematics • 7th Grade • Medium • CCSS 6.SP.B.5C, 6.SP.B.4, HSS.ID.A.1 Edit Worksheet Share Save Preview Use this activity Standards-aligned Used 106+ times FREE Resource 10 questions Show all answers [x] 1. MULTIPLE CHOICE QUESTION 2 mins • 1 pt Preview Edit Find the mean of this data set: (5, 8, 1, 7, 4, 3, 2, 2) 7 4 2 12 Tags CCSS.6.SP.B.5C 2. MULTIPLE CHOICE QUESTION 2 mins • 1 pt Preview Edit What is the MEDIAN of this data set: (4, 7, 2, 1, 9, 3, 6)? 9 2 3 4 Tags CCSS.6.SP.B.5C 3. MULTIPLE CHOICE QUESTION 2 mins • 1 pt Preview Edit What is the MODE in this data set: (3, 7, 2, 3, 9, 4, 12, 3)? 3 9 12 7 Tags CCSS.6.SP.B.5C 4. MULTIPLE CHOICE QUESTION 2 mins • 1 pt Preview Edit Find the RANGE of this data set: (10, 3, 11, 17, 4, 3, 10) 20 7 14 1 Tags CCSS.6.SP.B.4 CCSS.HSS.ID.A.1 5. MULTIPLE CHOICE QUESTION 2 mins • 1 pt Preview Edit Find the MEAN and MODE of the following data set: (8, 12, 16, 8, 6, 7, 9) Mean: 9 Mode: 8 Mean: 8 Mode: 9 Mean: 9.4 Mode:8 Mean: 7.6 Mode: 8 Tags CCSS.6.SP.B.5C 6. MULTIPLE CHOICE QUESTION 2 mins • 1 pt Preview Edit Find the MEDIAN and the RANGE of this data set: ( 12, 17, 15, 9, 14, 19, 21) Median: 14 Range: 12 Median: 9 Range: 21 Median: 15 Range: 9 Median: 15 Range: 12 Tags CCSS.6.SP.B.5C 7. MULTIPLE CHOICE QUESTION 2 mins • 1 pt Preview Edit TRUE/FALSE: To find the range of a data set, you subtract the smallest number from the biggest number? TRUE FALSE Tags CCSS.6.SP.B.4 CCSS.HSS.ID.A.1 8. MULTIPLE CHOICE QUESTION 2 mins • 1 pt Preview Edit TRUE/FALSE: You can only have ONE mode in a data set. TRUE FALSE Tags CCSS.6.SP.B.5C 9. MULTIPLE CHOICE QUESTION 5 mins • 1 pt Preview Edit The temperatures in Disney last week were: (98, 89, 76, 93, 89). Find the MEAN, MEDIAN, MODE, RANGE for the above data set. A B Tags CCSS.6.SP.B.5C 10. MULTIPLE CHOICE QUESTION 5 mins • 1 pt Preview Edit Think of 5 test scores you have received in any of your classes this year. (Real/fake). Use those to find the MEAN, MEDIAN, MODE, RANGE for that data set. 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2922	https://web.evanchen.cc/upload/1802/r06.pdf	Notes for 18.02 Recitation 6 18.02 Recitation MW9 Evan Chen 23 September 2024 Run, you clever boy, and remember. — Clara Oswald, in Doctor Who This handout (and any other DLC’s I write) are posted at • Optional midterm review, 4-270, Thu 4:30pm-6:30pm. Led by me, Vinjay, and Sebastian. • Please fill out the survey at when you can. • I made a Discord server. If you didn’t get the link emailed to you, ask me to join. • Remember that September 30 is the last day to switch sections freely on Canvas. §1 It’s a miracle that multiplication in ℂ has geometric meaning Let ℂ denote the set of complex numbers (just as ℝ denotes the real numbers). It’s important that realize that, until we add in complex multiplication, ℂ is just an elaborate ℝ2 cosplay. Concept For ℝ2 For ℂ Notation 𝐯 𝑧 Components (𝑥 𝑦) 𝑥+ 𝑦𝑖 Length Length \|𝐯\| Abs val \|𝑧\| Direction (slope, maybe?) argument 𝜃 Length 1 unit vector 𝑒𝑖𝜃= cos 𝜃+ 𝑖sin 𝜃 Multiply NONE ✨ 𝑧1𝑧2 ✨ • All the way back in R01, when I warned you about type safety, I repeatedly stressed you that you cannot multiply two vectors in ℝ𝑛 to get another vector. You had a “dot product”, but it spits out a number. (Honestly, you shouldn’t think of dot product as a “product”; the name sucks.) • Of course, the classic newbie mistake (which you better not make on your midterm) is to define a product on vectors component-wise: why can’t ( 𝑥1 ⋮ 𝑥𝑛 ) and ( 𝑦1 ⋮ 𝑦𝑛 ) have “product” ( 𝑥1𝑦1 ⋮ 𝑥𝑛𝑦𝑛 )? Well, in 18.02, every vector definition needed a corresponding geometric picture for us to consider it worthy of attention (see table at start of r03.pdf). This definition has no geometric meaning. • However, there is a big miracle for ℂ. For complex numbers, you can define multiplication by (𝑎+ 𝑏𝑖)(𝑐+ 𝑑𝑖) = (𝑎𝑐−𝑏𝑑) + (𝑎𝑑+ 𝑏𝑐)𝑖 and there is an amazing geometric interpretation. 1 Notes for 18.02 Recitation 6 — Evan Chen Unfortunately, AFAIK there is no English word for “complex number whose absolute value is one” (err, CNWAVIO?), the same way there is for “unit vector”. For 18.02, we instead use 𝑒𝑖𝜃≔cos 𝜃+ 𝑖sin 𝜃 as the “word”; whenever you see 𝑒𝑖𝜃, draw it as unit vector cos 𝜃+ 𝑖sin 𝜃. It’s worth pointing out the notation 𝑒𝑖𝜃 should strike you as nonsense. What meaning does it have to raise a number to an imaginary power? Does 𝑖𝑖 have a meaning? Does cos(𝑖) have a meaning? (If you want to know, check Section 4.1 in the post-recitation notes.) But for 18.02, when starting out, I would actually think of the notation 𝑒𝑖𝜃 as a mnemonic, i.e. a silly way to remember the following result: Theorem 1.1. If you multiply two CNWAVIO’s, you get the CNWAVIO with the angles added: 𝑒𝑖𝜃1𝑒𝑖𝜃2 = 𝑒𝑖(𝜃1+𝜃2) ⟺cos(𝜃1 + 𝜃2) + 𝑖sin(𝜃1 + 𝜃2) = (cos 𝜃1 + 𝑖sin 𝜃1)(cos 𝜃2 + 𝑖sin 𝜃2). More generally, multiplying two complex numbers multiplies the norms and adds the angles. This is IMO the biggest miracle in all of precalculus. Corollary: 𝑒𝑖𝑛𝜃= (𝑒𝑖𝜃) 𝑛⟺(cos 𝜃+ 𝑖sin 𝜃)𝑛= cos(𝑛𝜃) + 𝑖sin(𝑛𝜃), allows taking 𝑛th roots; Maulik showed 𝑧2 = 2𝑖 in class. §2 Rectangular vs polar Every complex number can be written in either rectangular form (𝑎+ 𝑏𝑖 for 𝑎, 𝑏∈ℝ) or polar form (𝑟𝑒𝑖𝜃). Depending on what you are trying to do, some forms are easier to work with than others. Operation In rectangular In polar 𝑧1 ± 𝑧2 ✅ Component-wise like in ℝ2 ❌ Unless 𝑧1 is a real multiple of 𝑧2 𝑧1𝑧2 ✅ Expanding ✅ by Theorem 1.1 𝑧1/𝑧2 ✅ Use 1 𝑐+𝑑𝑖= 𝑐−𝑑𝑖 𝑐2+𝑑2 trick then multiply ✅ by Theorem 1.1 𝑛th root of 𝑧1 ❌ Not recommended for 𝑛> 1 ✅ by Theorem 1.1 corollary §3 Recitation problems from official course 1. For each of the following points, convert it from Cartesian to polar or vice versa: • (𝑥, 𝑦) = (− √ 3, 1) • (𝑟, 𝜃) = (3, 𝜋/6) • (𝑥, 𝑦) = (− √ 6, − √ 2) 2. Show that sin(𝜃) = 1 2𝑖(𝑒𝑖𝜃−𝑒−𝑖𝜃) and cos(𝜃) = 1 2(𝑒𝑖𝜃+ 𝑒−𝑖𝜃). Use this to write (sin(𝜃))3 in terms sin(3𝜃) and sin(𝜃). 3. Consider the complex number 𝑓(𝑡) = 𝑡+2𝑖 1−3𝑖 where 𝑡 is real. • Find the real and imaginary part of 𝑓(𝑡). • Find 𝑓(𝑡) and \|𝑓(𝑡)\|2. 4. Use polar form to find the fourth powers of 2 + 2𝑖 and −3 + 𝑖 √ 3. Graph these numbers and their fourth powers on the complex plane. 5. (If you have time) Consider the matrix 𝐴= (0 1 −1 0 ). In class, working with real numbers, this had no eigenvectors. But now we can treat it as a matrix with complex number entries. Find complex number eigenvalues for 𝐴 and for each one, find an eigenvector in ℂ2, i.e. a two-component vector (𝑧 𝑤) where 𝑧, 𝑤 are complex numbers. 2 Notes for 18.02 Recitation 6 — Evan Chen §4 Post-recitation notes §4.1 The importance of definitions; also cos(𝑖) and 𝑖𝑖 (not for exam) When learning mathematics, I believe definitions are actually more important than theorems. A lot of confusion comes from not having been given careful definitions of the objects. (See evanchen.cc/handouts/NaturalProof/NaturalProof.pdf for more on that.) So in general any time you are confused about whether an operation is “legal” — and this is true in all of math, not just 18.02 — the first thing to really check whether you have been given a precise definition. The endless Internet debates on whether 0 is even or whether 0.999… = 1 or whether 1 𝑥 is a continuous function (hint: yes) are all examples of people who don’t know the definitions of objects they’re dsicussing. §4.1.1 Real exponents, real base With that in mind, let’s fix 𝑎> 0 a positive real number and think about what 𝑎𝑟 should mean. Definition 4.1 (18.100 definition). • When 𝑛> 0 is an integer, then 𝑎𝑛≔𝑎× … × 𝑎, where 𝑎 is repeated 𝑛 times. • Then we let 𝑎−𝑛≔ 1 𝑎𝑛 for each integer 𝑛> 0. • When 𝑚 𝑛 is a rational number, 𝑎 𝑚 𝑛 means the unique 𝑏> 0 such that 𝑎𝑚= 𝑏𝑛. (In 18.100, one proves this 𝑏 is unique and does exist.) • It’s less clear what 𝑎𝑥 means when 𝑥∈ℝ, like 𝑥= √ 2 or 𝑥= 𝜋. I think usually one takes a limit of rational numbers 𝑞 close to 𝑥 and lets 𝑎𝑥≔lim𝑞→𝑥𝑎𝑞. (In 18.100, one proves this limit does in fact exist.) §4.1.2 Complex exponents, real base But when 𝑧∈ℂ, what does 𝑎𝑧 mean? There’s no good way to do this. You likely don’t find an answer until 18.112, but I’ll tell you now. In 18.100 you will also prove that the Taylor series 𝑒𝑥= ∑ 𝑘≥0 𝑟𝑘 𝑘! is correct, where 𝑒≔∑𝑘≥0 1 𝑘! is Euler’s constant. So then when you start 18.112, we will flip the definition on its head: Definition 4.2 (18.112 definition). If 𝑧∈ℂ, we define 𝑒𝑧≔∑ 𝑘≥0 𝑧𝑘 𝑘! . Then for 𝑎> 0, we let 𝑎𝑧= 𝑒𝑧log 𝑎. To summarize: in 18.100, we defined exponents in the way you learned in grade school and then proved there was a Taylor series. But in 18.112, you start with the Taylor series and then prove that the rules in grade school you learned still applied. And checking this consistency requires work. Because we threw away Definition 4.1, identities like 𝑒𝑧1+𝑧2 = 𝑒𝑧1𝑒𝑧2 and (𝑒𝑧1)𝑧2 = 𝑒𝑧1𝑧2 are no longer “free”: they have to be proved rigorously too. (To 3 Notes for 18.02 Recitation 6 — Evan Chen be fair, they need to be proved in 18.100 too, but there it’s comparatively easier.) I think you shouldn’t be surprised they’re true; we know it’s true for ℝ, so it’s one heck of a good guess. But you shouldn’t take these on faith. At least get your professor to acknowledge they require a (non-obvious) proof, even if you aren’t experienced enough to follow the proof yourself yet. Anyway, if we accept this definition, then Euler’s formula makes more sense: Theorem 4.3 (Euler). We have 𝑒𝑖𝜃= cos 𝜃+ 𝑖sin 𝜃. The point is that cosine and sine also have a Taylor series that is compatible with definition: cos(𝑥) = 1 −𝑥2 2! + 𝑥4 4! −𝑥6 6! + … sin(𝑥) = 𝑥−𝑥3 3! + 𝑥5 5! −𝑥7 7! + …. (1) And if you put these together, you can verify Theorem 4.3, up to some technical issues with infinite sums. I think Maulik even showed this in class: cos(𝜃) + 𝑖sin(𝜃) = (1 −𝜃2 2! + 𝜃4 4! −…) + (𝜃−𝜃3 3! + 𝜃5 5! −…)𝑖 = 1 + (𝜃𝑖) + (𝜃𝑖)2 2! + (𝜃𝑖)3 3! + 𝜃𝑖4 4! + (𝜃𝑖)5 5! = 𝑒𝑖𝜃. §4.1.3 Complex exponents, complex base But what about 𝑖𝑖? Our Definition 4.2 above only worked for positive real numbers 𝑎> 0. Here, it turns out you’re out of luck. There isn’t any way to define 𝑖𝑖 in a way that makes internal sense. The problem is that there’s no way to take a single log of a complex number, so the analogy with log 𝑎 breaks down. Put another way: there’s no good way to assign a value to log(𝑖), because 𝑒𝑖𝜋/2 = 𝑒5𝑖𝜋/2 = … are all equal to 𝑖. You might hear this phrased “complex-valued logarithms are multivalued”. You can have some fun with this paradox: 𝑖= 𝑒𝑖𝜋/2 ⟹𝑖𝑖= 𝑒−𝜋/2 𝑖= 𝑒5𝑖𝜋/2 ⟹𝑖𝑖= 𝑒−5𝜋/2. Yeah, trouble. §4.1.4 Trig functions with complex arguments On the other hand, cos(𝑖) can be defined: use the Taylor series Equation 1, like we did for 𝑒𝑧. To spell it out: Definition 4.4 (18.112 trig definitions). If 𝑧 is a complex number, we define 4 Notes for 18.02 Recitation 6 — Evan Chen cos(𝑧) ≔1 −𝑧2 2! + 𝑧4 4! −𝑧6 6! + … sin(𝑧) ≔𝑧−𝑧3 3! + 𝑧5 5! −𝑧7 7! + …. If you do this, then Definition 4.2 implies the following identities are kosher: Proposition 4.5. Under Definition 4.4, we have the identities cos(𝑧) ≔𝑒𝑖𝑧+ 𝑒−𝑖𝑧 2 sin(𝑧) ≔𝑒𝑖𝑧−𝑒−𝑖𝑧 2𝑖 . Proof. If you write out 𝑒𝑖𝑧= ∑(𝑖𝑧)𝑘 𝑘! and 𝑒−𝑖𝑧= ∑(−𝑖𝑧)𝑘 𝑘! and add them, the odd 𝑘’s cancel and the even 𝑘’s don’t, which gives you 𝑒𝑖𝑧+ 𝑒−𝑖𝑧= 2 −2 ⋅𝑧2 2! + 2 ⋅𝑧4 4! −2 ⋅𝑧6 6! + …. So dividing by 2, we see cos(𝑧) on the right-hand side, as needed. The argument with sin is similar, but this time the even 𝑘’s cancel and you divide by 2𝑖 instead. □ So for example, from Proposition 4.5, we conclude for example that cos(𝑖) = 𝑒+ 1 𝑒 2 . Strange but true. §4.2 The future: what are 18.100 and 18.112 anyway? (not for exam) First I need to tell you what analysis is. When students in USA ask me what analysis is, I sometimes say “calculus but you actually prove things”. But that’s actually a bit backwards; it turns out that in much parts of the world, there is no topic called “calculus”.¹ It would be more accurate to say calculus is analysis with proofs, theorems, and coherent theorem statements deleted, and it only exists in some parts of the world (which is why mathematicians will tend to look down on it). With that out of the way, • 18.100 is real analysis, i.e. analysis of functions over ℝ • 18.112 is complex analysis, i.e. analysis of functions over ℂ. If you ever take either class, I think the thing to know about them is: Complex analysis is the good twin and real analysis is the evil one: beautiful formulas and elegant theorems seem to blossom spontaneously in the complex domain, while toil and pathol/ ogy rule the reals — Charles Pugh, in Real Mathematical Analysis Personally, I think it’s insane that MIT uses 18.100 as their “intro to proofs” topic. (This is why 18.100 is a prerequisite for 18.701, abstract algebra, which makes no sense either.) ¹See 5 Notes for 18.02 Recitation 6 — Evan Chen §5 Exponentiation (for exam) This section is dedicated to 𝑧𝑛 and is on-syllabus for exam. Specifically, you ought to be able to solve equations like 𝑧5 = 243𝑖. This section shows you how. In this whole section, you always prefer to work in polar form. So if you get input in rectangular form, you should first convert to rectangular form. Conversely, if the answer is asked for in rectangular form, you should work with polar form anyway, and only convert to rectangular output at the end. §5.1 Raising to the 𝑛th power Before being able to extract 𝑛th roots, I need to make sure you know how to do 𝑛th powers. This is easy: (𝑟(cos 𝜃+ 𝑖sin 𝜃))𝑛= 𝑟𝑛(cos 𝑛𝜃+ 𝑖sin 𝑛𝜃). For example, (3(cos 18° + 𝑖sin 18°))5 = 243(cos 90° + 𝑖sin 90°) = 243𝑖. §5.2 Extracting 𝑛th roots If you can run the process in forwards, then you should be able to run the process backwards too. First, I will tell you what the answer looks like: Theorem 5.1. Consider solving the equation 𝑧𝑛= 𝑤, where 𝑤 is a given nonzero complex number, for 𝑧. Then you should always output exactly 𝑛 answers. Those 𝑛 answers all have magnitude \|𝑤\| 1 𝑛 and arguments spaced apart by 360° 𝑛. I think it’s most illustrative if I show you the five answers to 𝑧5 = 243𝑖 to start. Again, first we want to convert everything to polar coordinates: 𝑧5 = 243𝑖= 243(cos 90° + 𝑖sin 90°). At this point, we know that if \|𝑧5\| = 243, then \|𝑧\| = 3; all the answers should have absolute 3. So the idea is to find the angles. Here are the five answers: 𝑧1 = 3(cos 18° + 𝑖sin 18°) ⟹(𝑧1)5 = 243(cos 90° + 𝑖sin 90°) 𝑧2 = 3(cos 90° + 𝑖sin 90°) ⟹(𝑧2)5 = 243(cos 450° + 𝑖sin 450°) 𝑧3 = 3(cos 162° + 𝑖sin 162°) ⟹(𝑧3)5 = 243(cos 810° + 𝑖sin 810°) 𝑧4 = 3(cos 234° + 𝑖sin 234°) ⟹(𝑧4)5 = 243(cos 1170° + 𝑖sin 1170°) 𝑧5 = 3(cos 306° + 𝑖sin 306°) ⟹(𝑧5)5 = 243(cos 1530° + 𝑖sin 1530°). Here’s a picture of the five numbers: 6 Notes for 18.02 Recitation 6 — Evan Chen Figure 1: The five answers to 𝑧5 = 243𝑖, each of length 3. On the right column, all the numbers are equal. Notice something interesting happening on the right-hand side. The numbers cos 90° + 𝑖sin 90° and cos 450° + 𝑖sin 450°, etc. are all the same number; if you draw them in the plane, they’ll point to the same thing. However, they give five different answers on the left. But if you continue the pattern one more, you start getting a cycle 𝑧6 = 3(cos 378° + 𝑖sin 378°) ⟹(𝑧6)5 = 243(cos 1890° + 𝑖sin 1890°). This doesn’t give you a new answer, because 𝑧6 = 𝑧1. In general, if 𝑤 has argument 𝜃, then the arguments of 𝑧 satisfying 𝑧𝑛= 𝑤 start at 𝜃 𝑛 and then go up in increments of 360° 𝑛. (For example, they started at 90° 5 = 18° for answers to 𝑧5 = 243𝑖.) So you can describe the general recipe as: Recipe for 𝑛th roots 1. Convert 𝑤 to polar form; say it has angle 𝜃. 2. One of the 𝑛 answers will be \|𝑤\| 1 𝑛(cos 𝜃 𝑛+ 𝑖sin 𝜃 𝑛). 3. The other 𝑛−1 answers are obtained by increasing the angle in increments of 360° 𝑛. • Example 1: Solve 𝑧5 = 243𝑖. Answer 1: we first convert to polar form as 243𝑖= 243(cos 90° + 𝑖sin 90°) 7 Notes for 18.02 Recitation 6 — Evan Chen and see that 243 1 5 = 3, and 𝜃= 90°. The first angle is 𝜃 5 = 18°. So the five answers are 𝑧1 = 3(cos 18° + 𝑖sin 18°) 𝑧2 = 3(cos 90° + 𝑖sin 90°) 𝑧3 = 3(cos 162° + 𝑖sin 162°) 𝑧4 = 3(cos 234° + 𝑖sin 234°) 𝑧5 = 3(cos 306° + 𝑖sin 306°). (As it happens, 𝑧2 = 3𝑖, which is easy to check by hand works.) • Example 2: Solve 𝑧4 = 8 + 8 √ 3𝑖. Answer 2: We first convert to polar form as 8 + 8 √ 3𝑖= 16(cos 60° + 𝑖sin 60°) and see that 16 1 4 = 2, and 𝜃= 60°. The first angle is 𝜃 4 = 15°. So the four answers are 𝑧1 = 2(cos 15° + 𝑖sin 15°) 𝑧2 = 2(cos 105° + 𝑖sin 105°) 𝑧3 = 2(cos 195° + 𝑖sin 195°) 𝑧4 = 2(cos 285° + 𝑖sin 285°). • Example 3: Solve 𝑧3 = −1000. Answer 3: We first convert to polar form as −1000 = 1000(cos 180° + 𝑖sin 180°) and see that 1000 1 3 = 10, and 𝜃= 180°. The first angle is 𝜃 3 = 60°. So the three answers are 𝑧1 = 10(cos 60° + 𝑖sin 60°) 𝑧2 = 10(cos 180° + 𝑖sin 180°) 𝑧3 = 10(cos 300° + 𝑖sin 300°). (As it happens, 𝑧2 = −10, as expected, since (−10)3 = −1000.) 8
2923	https://www.st-andrews.ac.uk/~wjh/neurotut/RestingPotential.html	What causes this voltage? A microelectrode is a specialized electrode for recording voltages from the inside of cells. It typically consists of a glass tube drawn out to a very fine point and filled with an electrolyte solution such as potassium chloride. The sharp tip of the electrode penetrates through the cell membrane. The blunt end of the electrode is connected by a wire to a voltage measuring instrument. The Membrane Potential Instruments for recording voltages from cells have to have a very high input impedance. This means that they can measure the voltage without drawing much current from the cell, and thus without changing the very voltage that they are trying to measure. All living cells maintain a potential difference across their cell membranes, with the inside usually negative relative to the outside. In nerve cells the value of the resting potential varies between about -40 and -90 mV. The membrane potential can be measured by penetrating the cell with a microelectrodeA microelectrode is a specialized electrode for recording voltages from the inside of cells. It typically consists of a glass tube drawn out to a very fine point and filled with an electrolyte solution such as potassium chloride. The sharp tip of the electrode penetrates through the cell membrane. The blunt end of the electrode is connected by a wire to a voltage measuring instrument.. This is connected to a specialized voltmeterInstruments for recording voltages from cells have to have a very high input impedance. This means that they can measure the voltage without drawing much current from the cell, and thus without changing the very voltage that they are trying to measure. which measures the potential difference between the inside of the cell and the surrounding environment. This voltage is the membrane potential. This tutorial describes the cellular mechanisms that generate the resting membrane potential. Early in the 20th century, Bernstein proposed that the resting membrane potential was due to 3 factors: the cell membrane is selectively permeable to potassium ions,the intracellular potassium concentration is high,the extracellular potassium concentration is low.This is known as the potassium electrode hypothesis. In the next section we will consider how these factors would cause a membrane potential with the inside negative. This will lead to an understanding of the Nernst equation and the ionic equilibrium (reversal) potentials. Sneak preview - Bernstein was almost, but not completely, right. Bernstein’s Potassium Hypothesis The Membrane Potential Simple Goldman equation Na-K Exchange Pump Key points Bernstein’s Potassium Hypothesis Action potential Unusual Mechanisms The Goldman Equation Explaining Bernstein Chloride ions Steady state Key Points Concentration changes The Nernst Equation Gradient run-down Deriving the Nernst equation Electrogenic pumps Donnan Equilibrium Using the Nernst equation Donnan equation Leakage channels Expaining Donnan Reversal potential Summary Frog and squid Concentration measurements show cells do indeed have a K⁺ gradient across the membrane in the appropriate direction (high inside, low outside). 2. Flux measurements using radioactive potassium show that the membrane is permeable to K⁺. 3. These facts are consistent with Bernstein’s hypothesis for why there is a membrane potential with the inside negative. 4. Note that it is the side with the high concentration of positive ions that becomes the negative side of the membrane (because the positive ions flow away from the high concentration side). This often causes confusion. 5. Not many ions have to flow to set up the potential, so the concentration on the two sides is virtually unchanged. This will be explained in more detail later; for now just trust me … 6. The potential is set up almost instantly, although the cartoon illustration shows it taking quite some time. Now let’s quantify this ... Potassium electrode hypothesis: Key points electrical gradient When the electrical gradient exactly balances the chemical gradient, equilibrium is achieved. There is no further net movement of ions. K can flow! The electrical gradient opposes the movement of positively-charged potassium ions. K+ gradient Potassium ions diffuse down their concentration gradient. They carry positive charge with them (K⁺). Explaining Bernstein 4 HOWEVER, as far as K+ ions are concerned, the two compartments are now connected through pores in the membrane (shown as just a single gap in the diagram). What happens now? Reset Positive charge accumulates in the outside compartment, building up an electrical gradient (voltage) across the membrane. Ca I suggest you look at the pre-built examples first: K [Xout] VmV = K⁺ Ca⁺⁺ Cl⁻ Examples inside Using the Nernst Equation Then try it out yourself: mM outside Cl log10 -58 Note: if the membrane is only permeable to one ion, then the membrane potential will be the same as the Nernst potential. This is approximately true for glial cells, which are almost exclusively permeable to potassium. However, most cell membranes are permeable to more than one ion type, which makes things more complicated. This is explored further in the Donnan and Goldman sections of the tutorial. Select the ion to which the membrane is permeable, and the intracellular and extracellular concentrations. z [Xin] When the driving force is zero, potassium ions are in equilibrium across the membrane and just move randomly back and forth. As the electrical gradient builds up, it counteracts the chemical gradient, and the driving force decreases. The difference between the chemical gradient and the electrical gradient is called the driving force. In our scenario there is initially a large driving force (shown as an arrow through the pore). This is because the chemical gradient is not balanced by any electrical gradient. Driving force Explaining Bernstein 5 The Experimental Situation Explaining Bernstein 1 We start off with a simple container divided into two compartments.The left-hand compartment represents the inside of a neuron, while the right-hand compartment represents the outside, i.e. extracellular space. The barrier represents the cell membrane. Assume that both compartments contain only water, and that the barrier is impermeable to allIf the barrier were truly completely impermeable to everything (i.e. a perfect insulator) then the voltmeter inputs would be “floating” and it would probably show random values due to picking up weak electric fields in the environment. We can avoid this by assuming that the barrier is very slightly permeable to water. ions. A voltmeter measures the potential difference between them (in-out). Since the two compartments have identical (empty) contents, this is zero (0 mV). If the barrier were truly completely impermeable to everything (i.e. a perfect insulator) then the voltmeter inputs would be “floating” and it would probably show random values due to picking up weak electric fields in the environment. We can avoid this by assuming that the barrier is very slightly permeable to water. Add Some Ions We now add some KCl to both sides. We put a relatively high concentration (100 mM) into the inside, and a low concentration (10 mM) into the outside. The KCl dissociates into K⁺ and Cl⁻. There is now a chemical concentration gradient between the two compartments, but since the membrane is impermeable to K⁺ and Cl⁻, this has no effect on the membrane potential, which remains at 0. K+ Cl- gradient Explaining Bernstein 2 Cl- Explaining Bernstein 3 Change the Barrier We now replace the impermeable barrier with a semi-permeable membrane, which is permeable to potassium (K⁺) ions only. So far as the chloride (Cl⁻) ions are concerned, the two compartments are totally separated by the barrier, and so the concentration gradient in Cl⁻ ions has no effect. The Nernst equation is fundamental for understanding not only the mechanism generating the resting potential, but also that of action potentials (spikes) and synaptic potentials. We have seen empirically that equilibrium will occur if a gradient in chemical concentration, that would normally cause an ion to move from the high concentration to the low concentration, is balanced by an electrical gradient that opposes the movement of charge. Nernst equation The Nernst equation is a formula that relates the numerical values of the concentration to the electrical gradient that balances it. electrical gradient The Nernst Equation chemical gradient Electrical energy Deriving the Nernst equation 1 V RT Equilibrium is achieved when the energy released by a particle sliding down a concentration gradient is equal to the energy required to push a charged particle up an electrical gradient. The equations describing these energies are derived from thermodynamic principles (which we won’t go into …) Rearranging the equations gives us the Nernst equation. = [X2] δWe where δWe is the energy change associated with moving n moles of a charged particle with valency z in an electric field of strength V (volts), and F is Faraday's number (C mol⁻¹). ln where δWc is the energy change associated with moving n moles from concentration X1 to concentration X2 (mol l⁻¹), R is the gas constant (8.31 J K⁻¹ mol⁻¹), T is the absolute temperature (°K), ln is the natural logarithm (base e). Chemical energy z F n The Conditions for Equilibrium δWc [X1] (Note it is minus 58) Simplify Deriving the Nernst equation 3 However, R and F are constants, and if we assume that T is room temperature (21℃), convert from natural logarithm to log base 10, state V in millivolts rather than volts, make compartment 1 the inside of a neuron and compartment 2 the outside, the equation simplifies to: V = The “official” Nernst equation is: The Nernst equation defines the equilibrium potential for an ion - i.e. the electrical potential that exactly balances a concentration gradient for that ion. This is also called the reversal potential, for reasons that will become clear later ... Each ion will have its own individual equilibrium potential, depending on its concentration gradient across the membrane. Using the Nernst Equation The potassium concentration ratio is 10:1. The log of 10 is 1, potassium is monovalent and carries a single +ve charge, so the solution to the Nernst equation is simply -58 mV. 10 For a monovalent ion a 10-to-1 inside-to-outside concentration ratio is exactly balanced by a -58 mV membrane potential. K specific Example 1 of 5 VmV = -58 log(10) = -58 x 1 = -58 back Next Return 100 Na Na specific 46 If we reverse the direction of the concentration gradient, so that the ratio is now 1:10 (i.e. 0.1), then the absolute value of the membrane potential stays the same, but its polarity reverses (the log of 0.1 is -1). Note that the ion has changed to sodium but the Nernst equation doesn’t care - it is still a monovalent ion. (We assume the membrane is now permeable to sodium.) 460 Example 3 of 5 VmV = -58 log(0.1) = -58 x -1 = 58 The polarity of the Nernst equation depends on the direction of the concentration gradient. 58 Example 2 of 5 VmV = -58 log(460/46) = -58log(10) = -58 x 1 = -58 The ratio of 460:46 is still 10:1, and the Nernst potential is still -58 mV. It doesn’t matter what the absolute values of the concentrations are, it is the ratio that counts. Ca This is a bit counter-intuitive, but it takes more energy to move two charges in an electric field than just one. This means that a half-strength field will balance a given concentration gradient involving a divalent ion compared to the field strength required to balance the same concentration gradient involving a monovalent ion. Example 4 of 5 log(0.1) = -29 x -1 = 29 For a divalent ion (z = 2 in the Nernst equation) the membrane potential is halfThis is a bit counter-intuitive, but it takes more energy to move two charges in an electric field than just one. This means that a half-strength field will balance a given concentration gradient involving a divalent ion compared to the field strength required to balance the same concentration gradient involving a monovalent ion. what it is for a monovalent ion with the same concentration ratio. The ratio is still 1:10, but for Ca²⁺ the Nernst potential is 29 mV. Ca specific 29 2 Example 5 of 5 log(0.1) = 58 x -1 = -58 We are back to a monovalent ion, but the chloride ion carries a negative charge (z = -1), so the membrane potential is reversed compared to an ion with the same concentration gradient but a positive charge. Cl The ration is still 1:10, but for Cl⁻ the Nernst potential is -58 mV. Cl specific -1 The ions flow in the opposite direction if they are negatively charged, like chloride. -75 Positive If the membrane potential is at the equilibrium potential of an ion (in this case -75 mV), the ions move randomly back and forth across the membrane through any open channels selective for the ion. If the membrane potential is below the equilibrium potential, then positivelyThe ions flow in the opposite direction if they are negatively charged, like chloride. charged ions like potassiumn will flow into the cell, but if the membrane potential is above the equilibrium potential, the ions will flow out of the cell. Thus the direction of current carried by an ion reverses when the membrane potential is on either side of its equilibrium potential, which is why this is also called the reversal potential. Negative Equilibrium = Reversal Potential Reversal Potential -50 A battery is inserted to make the inside of the cell more positive, taking the membrane potential above the potassium equilibrium potential. The electrical gradient is now weaker than the chemical gradient, there is an outward-directed driving force, and positively-charged potassium ions flow down the concentration gradient out of the cell. This flux will continue unless and until the concentration gradient decreases to achieve a new Nernst equilibrium potential. Above equilibrium potential Reversal Potential A battery is inserted to make the inside of the cell more negative, taking the membrane potential below the potassium equilibrium potential. The electrical gradient is now stronger than the chemical gradient, there is an inward-directed driving force, and positively-charged potassium ions flow up their concentration gradient into of the cell. This flux will continue unless and until the concentration gradient increases to achieve a new Nernst equilibrium potential. ? If the membrane potential of a cell is caused by potassium ions moving down their concentration gradient, does this alter the internal concentration of potassium? In other words, how many ions have to leave the cell in order to set up the equilibrium potential? How Many Ions Move? Concentration Changes 1 K⁺ leaves The Membrane is a Capacitor Concentration Changes 2 This question can be approached by treating the cell membrane as a capacitor, in which the intracellular cytoplasm and extracellular fluid are the conducting electrolytes, and the phospholipid component of the cell membrane is the insulating dialectric. The typical value of capacitance for a biological membrane is 1 µF cm⁻². For a spherical cell with a diameter of 50 µm the surface area (4πr2) is 7.85 x 10-5 cm², so the capacitance is 78.5 pF. If the membrane potential is -70 mV, the charge on the capacitor is: Q = VC = 70 x 10-3 × 7.85 x 10-11 Thus Q = 5.50 x 10-12 coulombs. One K⁺ ion has a charge of 1.6 x 10-19 coulombs, so the number of ions that move down the concentration gradient is about 34,340,000 i.e. about 34 million. Concentration Changes 3 38 million K⁺ions leave About 34 million Ions Leave Concentration Changes 4 0.006% of K⁺ions leave Does this Alter the Concentration Gradient? A cell with diameter 50 µm has a volume (4πr³/3) of 6.54 x 10-11 litres and its intracellular K⁺ concentration (for a mammal) is about 140 mM, so it contains 9.16 x 10-12 moles of K⁺ ions. One mole consists of 6.02 x 1023 ions (Avogadro's number), so the cell contains about 5.52 x 1012 ions. The loss of 3.43 x107 ions represents a concentration change of about 0.0006% So the answer is HARDLY AT ALL. Note: the external volume is relatively huge, so the change in external concentration is infinitesimal. Nernst equation: Key points The Nernst equation defines the electrical potential across a membrane that will balance a particular chemical concentration gradient of an ion. So if the membrane potential is at the Nernst value, there will be no net movement of that ion across the membrane, even if the membrane is permeable to the ion. 2. This potential is called the equilibrium or reversal potential for that ion. It depends on the concentration gradient (ratio) and the valency of the ion (and the temperature), but does not depend on the degree of permeability of the membrane to the ion, nor on the absolute concentrations. 3. Each type of ion will have its own equilibrium potential, and this is likely to be different to that of other ion types. 4. If a membrane is permeable to only ONE type of ion, then the membrane potential will automatically move to the equilibrium potential for that ion. 5. Relatively few ions have to move to set up the potential, so the concentration gradient is not significantly perturbed in achieving equilibrium. However, what if the cell membrane is permeable to more than one type of ion …? The intracellular composition is thus: K⁺Cl⁻ + K⁺A⁻ This can result in a Donnan equilibrium, which both maintains opposite K⁺ and Cl⁻ gradients across the membrane, and sets up a resting potential. The Donnan equilibrium applies to ions whose distribution is passive and unregulated, i.e. there are no metabolic pumps maintaining the intracellular concentrations at fixed values. Donnan Equilibrium Many cells have membranes with quite a high permeability to both potassium and chloride ions. In the absence of anything else, this would prevent the establishment of the membrane potential, because Cl⁻ would simply move down the concentration gradient along with the K⁺, and counteract the positive charge. The concentrations would just equilibrate across the membrane. However, most cells also have a relatively high intracellular concentration of large molecular weight anions A⁻ (e.g. negatively charged proteins). There is no guarantee that equilibrium can be reached. In many cells metabolic pumps maintain intracellular ion concentrations within fairly narrow ranges. In these cases the resting potential is a steady state situation, rather than a chemical equilibrium, and the Donnan rule does not apply. This is explored further in the Goldman section of the tutorial. If the membrane is permeable to both K⁺ and Cl⁻ , both will try to move across the membrane until their concentration gradients are balanced by the membrane potential as given by the Nernst equation. There can be only one membrane potential across a patch of membrane at any one time, so ifThere is no guarantee that equilibrium can be reached. In many cells metabolic pumps maintain intracellular ion concentrations within fairly narrow ranges. In these cases the resting potential is a steady state situation, rather than a chemical equilibrium, and the Donnan rule does not apply. This is explored further in the Goldman section of the tutorial. equilibrium is reached: [K⁺in] This is the Donnan rule of equilibrium. [Cl⁻out] [K⁺out] [Cl⁻in] (Note the K-Cl gradient inversion due to the opposite charge of K⁺ and Cl⁻.) 1 F Thus at equilibrium the product of the concentrations of diffusible ions on one side of the membrane equals the product of the concentrations of the diffusible ions on the other: Losing the common factors gives: The Donnan Equation = V = [K⁺in] × [Cl⁻in] = [K⁺out] × [Cl⁻out] in If the membrane is permeable to both K⁺ and Cl⁻, and no impermeant molecules are involved, then equal numbers of K⁺ and Cl⁻ ions cross the membrane, until the concentrations are equal on either side. Because exactly equal numbers of positive and negative charges move across the membrane, no membrane potential is set up. The final state obeys the Donnan rule, but it is not very interesting because there is no concentration gradient and no membrane potential. out Explaining Donnan 1 Run-Down! Now consider the situation where some intracellular Cl⁻ ions are replaced by the impermeant anion A⁻. The K⁺ and Cl⁻ ions can still flow down their concentration gradients and leave the cell, but A⁻ cannot pass through the membrane. In the diagram each symbol represents 10 mM of substance - e.g. there are 12 K⁺s giving a total of 120 mM. We start off with everything inside the cell, nothing outside. What happens now? K+ gradient A⁻ Explaining Donnan 2 Cl- gradient Impermeant Intracellular Anions Details So more K⁺ leaves the cell. Its positive charge attracts Cl⁻ ions, and so more Cl⁻ moves out of the cell in order to maintain space-charge neutrality (it has to be Cl⁻ because A⁻ cannot cross the membrane). This means that the Cl⁻ gradient now starts to reverse from its original condition. K⁺ is still "winning" the osmotic battle at this stage, because its gradient (85:35) is greater than the reverse Cl⁻ gradient (25:35). So K⁺ will continue to leave the cell, taking more Cl⁻ with it ... Eventually, an equilibrium is reached with: Inside Outside + - + - K80 40 Cl 20 40 A 60 0 total+80-80 +40-40 Positive and negative charges again balance on both sides of the membrane, so we still have space-charge neutrality. The concentration gradient (in:out) for K⁺ is 80:40, and for Cl⁻ is 20:40, which gives a 2:1 ratio for both ions, and thus a Nernst potential of -17 mV for both ions. Everything balances! Counting ions at the start, we find that positive and negative charges balance on each side of the membrane. Inside Outside + - + - K120 0 Cl60 0 A60 0 total+120-120 00 and thus we have space-charge neutralityThe principle of space-charge neutrality says that in a given volume the total positive charge is equal to the total negative charge. The only exception is very close to the cell membrane, where separation of charge gives rise to the membrane potential. However, the number of uncompensated ions needed to set up this potential is a very small fraction (<<1%) of the total available, so produces only a tiny deviation from the principle.. However, both K⁺ and Cl⁻ can cross the membrane ... Explaining Donnan 3 and so both K⁺ and Cl⁻ flow down their concentration gradients out of the cell. When 30 mM K⁺ and 30 mM Cl⁻ have moved to the outside of the cell, Cl⁻ is in chemical equilibrium with no gradient (30:30), but K⁺ still has a concentration gradient (90:30) driving it out of the cell ... The principle of space-charge neutrality says that in a given volume the total positive charge is equal to the total negative charge. The only exception is very close to the cell membrane, where separation of charge gives rise to the membrane potential. However, the number of uncompensated ions needed to set up this potential is a very small fraction (<<1%) of the total available, so produces only a tiny deviation from the principle. K 120 Cl 60 K 0 Cl 0 A 60 How Many Ions Move? How do we know that 40 mM K⁺ and Cl⁻ had to move in the previous example? Consider the initial and final concentrations: Initial [K⁺in] = 120, [Cl⁻in] = 60 [K⁺out] = 0, [Cl⁻out] = 0 To maintain approximate space-charge neutrality, an almost equal amount [x] of K⁺ and Cl⁻ move across the membrane, therefore: Final [K⁺in] = 120 - x, [Cl⁻in] = 60 - x [K⁺out] = x, [Cl⁻out] = x The Donnan rule states that: [K⁺in] x [Cl⁻in] = [K⁺out] x [Cl⁻out] so at equilibrium: (120 - x) x (60 - x) = x2 and thus 7200 - 180x +x2 = x2 So x = 40. Conclusion 40 mM of KCl had to move across the membrane to set up the Donnan equilibrium. Note: the final K⁺ and Cl⁻ ion distribution is independent of the starting distribution, so long as the ion totals are the same. Squid giant axon in out ENernst [K⁺] 400 20 -75 [Cl⁻] 108 560 -41 [K⁺] x [Cl⁻] 4320011200 resting potential about -60 mV. The squid axon does not obey the Donnan rule. The internal and external [K⁺] x [Cl⁻] products are very different, and the Nernst equilibrium potentials are different for K⁺ and Cl⁻, and different from the resting potential. Is the Donnan Rule Obeyed? Data source: Aidley, D.J. (1989) The physiology of excitable cells. 2nd edition, Cambridge University Press. Frog and Squid [Kin] x [Clin] = [Kout] x [Clout] Frog muscle in out ENernst [K⁺] 124 2.25 -101 [Cl⁻] 1.5 77.5 -99 [K⁺] x [Cl⁻] 186174 resting potential -90 to -100 mV. Frog muscle does approximately obey the Donnan rule. The internal and external [K⁺] x [Cl⁻] products are about equal, and the Nernst equilibrium potentials for both K⁺ and Cl⁻ are approximately equal to the resting potential. Donnan equilibrium: Key points The Donnan rule states that the product of the concentration of diffusible ions on one side of the membrane equals the product of the concentration of diffusible ions on the other. 2. The Donnan rule only applies to cells where the ions are passively distributed, i.e. there are no metabolic pumps using energy to regulate the ion concentration inside the cell. 3. The Donnan rule is important because, where it applies, it explains the origin of concentration gradients, and hence the resting potential. 4. The Donnan rule is approximately obeyed by frog muscle fibres (which were important preparations in early physiological experiments). 5. The Donnan rule is not obeyed by many nerve cells, where the resting potential is significantly different from the Nernst equilibrium potential of the major ions involved in setting it up. What is happening when the Donnan rule is NOT obeyed? The Goldman Equation Regulation is usually achieved by metabolic pumps such as the Na/K-ATPase (also known as the sodium-potassium exchange pump, or simply the sodium pump). So far we have considered the situation where the membrane is permeable only to potassium and chloride ions, and their distribution is entirely passive. However, most real nerve membranes are also somewhat permeable to sodium ions, which usually have a high extracellular concentration and a low intracellular concentration. Also, most nerves (and many other cells) regulateRegulation is usually achieved by metabolic pumps such as the Na/K-ATPase (also known as the sodium-potassium exchange pump, or simply the sodium pump). the intracellular concentration of ions, so that they remain at a relatively fixed value. We next consider what happens in this situation. This will lead to an understanding of the Goldman-Hodgkin-Katz constant field equation, usually know as the Goldman equation for short. The Goldman equation describes a steady-state condition, unlike the Nernst equation, which describes an equilibrium condition. The difference will be explained shortly ... Explaining Steady-State 1 Na+ gradient Na We start off with our familiar 2-compartment model with a 10:1 ratio of [K⁺] between the inside and outside compartments, but this time there is also a 1:10 ratio of [Na⁺] in the opposite direction. The membrane will eventually be permeable to both K⁺ and Na⁺ ions, but to begin with, imagine the membrane is impermeable (all channels are shut). The membrane potential is therefore 0. K Explaining Steady-State 2 Now we open the K channels. Just as before, the membrane potential goes to the Nernst equilibrium potential for the K⁺ gradient, which is -58 mV. We mark this as EK on the meter. K Permeable EK Explaining Steady-State 3 Na Permeable ENa Next we close the K channels and open the Na channels. The membrane potential goes to the Nernst equilibrium potential for the Na⁺ gradient, which is +58 mV. We mark this as ENa on the meter. \| Explaining Steady-State 4 Now we open both sets of channels. The membrane is permeable to K⁺, which tends to set the membrane potential to EK. However, the membrane is also permeable to Na⁺, which tends to set the membrane potential to ENa. There can only be one membrane potential, and it will be a “compromise” between the two equilibrium potentials. With symmetrical gradients and equal permeability to both ions, the compromise potential is half way between the two equilibrium potentials, i.e. 0 mV. What effect does this have on ion flow? Na and K Permeable Steady-State Fluxes Explaining Steady-State 5 Since the membrane potential is not at either equilibrium potential, neither K⁺ nor Na⁺ are in equilibrium. Each has a steady flux down its concentration gradient. (Note: In the absence of any restorative pumps, eventually the concentration gradients will run down, but this takes some time.) Each ion experiences a driving force, which is the difference between its equilibrium potential (which itself would balance the concentration gradient) and the actual membrane potential. driving force = Em - Eeq The driving forces are equal and opposite, so the fluxesFlux in this case is the equivalent of current (charge per second). are equal and opposite. Neither side has a net change in charge, so there is no change in membrane potential. Flux in this case is the equivalent of current (charge per second). Unequal Na and K Permeability Explaining Steady-State 6 Let us now suppose there are twice as many K channels as Na channels. The membrane is thus highly permeable to K⁺, which tends to set the membrane potential to EK. However, the membrane is also somewhat permeable to Na⁺, which tends to drag the membrane potential a bit towards ENa. The “compromise” potential is now between the two equilibrium potentials, but weighted towards the K end of the scale. What effect does this have on the fluxes? Each ion flux is driven by its driving force, which depends on how far the membrane potential is from the equilibrium potential. There is thus a strong driving force on the Na⁺ ions, but only a weak driving force on the K⁺ ions. On the other hand, the membrane permeability to K⁺ is high, but the membrane permeability to Na⁺ is low (note there are 2 K channels to 1 Na channel. The result is that the Na⁺ flux (i.e. the number of Na⁺ ions crossing the membrane per second) is the same as the K⁺ flux. So again there is no change in the transmembrane charge distribution, and the membrane potential remains constant. This is therefore a steady-stateIt is steady-state, but only in the short term. Eventually the gradients will run down, unless maintained by metabolic pumps. That will be explained in more detail in a later section in the tutorial. condition. It is steady-state, but only in the short term. Eventually the gradients will run down, unless maintained by metabolic pumps. That will be explained in more detail in a later section in the tutorial. Explaining Steady-State 7 Imbalance in Flux Self-Corrects Explaining Steady-State 8 Why are the Na and K fluxes equal? If more Na⁺ entered the inside compartment than K⁺ left it, then the inside compartment would become more postive. This would reduce the driving force on Na⁺ (because the membrane potential had moved closer to the Na equilibrium potential), and increase the driving force on K⁺. The changes in driving force would reduce the inflow of Na⁺ ions and increase the outflow of K⁺ ions, thus counteracting the original imbalance. VmV = -58 log where α is the Na:K permeability ratio. α [Kin] + α [Nain] I suggest you look at the pre-built examples first: [Kout] + α [Naout] Then try it out for yourself. We have seen at a qualitative level that when the membrane is permeable to more than one ion species, the membrane potential is a weighted average of the Nernst equilibrium potentials for each permeant ion type. This is quantified by the Goldman equation. If we consider only Na and K, a simplified form of the Goldman equation is: Simple Form of the Equation With a completely symmetrical situation, the membrane potential is, not surprisingly, 0. With 10:1 concentration ratios, the equilibrium potentials are +/- 58 mV. 200 Example 1 of 4 20 The Goldman Equation 1 Example 2 of 4 With Na⁺ permeability 5 times greater than K⁺ permeability (α = 5), the membrane potential moves towards the Na⁺ equilibrium potential, even though the concentration gradients are unchanged. 5 Increasing α obviously increases the influence of the Na⁺ gradient in the equation. Example 3 of 4 From the equation it is clear that increasing the absolute Na⁺ concentrations by a factor of 5 has the same effect on the membrane potential as increasing α by a factor of 5. The Na⁺ permeability is back equal to the K⁺ permeability, but now the absolute Na⁺ concentrations have increased 5 times. The membrane potential moves towards the Na⁺ equilibrium potential, even though the concentration gradients are unchanged. 1000 Example 4 of 4 Increasing the Na⁺ concentration gradient increases the Na⁺ equilibrium potential, and this has the effect of increasing the membrane potential. Thus in summary, when the membrane is permeable to more than one ion, the membrane potential depends on: the relative permeabilities, the absolute concentrations,the concentration gradients. Relative Na⁺ : K⁺ permeability returns to its resting level, and so does the membrane potential. +50 — 440 An increase in K⁺ permeability and a decrease Na⁺ permeability (α = 0.01) causes the membrane potential to go even more negative than the resting potential, towards the K⁺ equilibrium potential. -50 — 400 Also known as a nerve impulse, or simply a spike. 0.04 In the resting state, the Na⁺ permeability is much lower than the K⁺ permeability (α = 0.04), and so the membrane potential is close to the K⁺ equilibrium potential. mV 0 — Action Potential This is the preparation that Hodgkin and Huxley used in their classic study of the mechanism of the action potential, which won them a Nobel prize in 1963. 50 A massive increase in Na⁺ permeability (α = 20) causes the membrane potential to shift towards the Na⁺ equilibrium potential. As a real example, let’s consider the squid giant axonThis is the preparation that Hodgkin and Huxley used in their classic study of the mechanism of the action potential, which won them a Nobel prize in 1963., which has approximately the concentration gradients shown here. By varying the value of α (the permeability of Na relative to K), the membrane can generate the waveform of an action potentialAlso known as a nerve impulse, or simply a spike.. 0.01 PK[Kin] + PNa[Nain] + PCl[Clout] Cl⁻ 0.45 where Pion is the membrane permeability to that ion. The consequences of this depend on whether the neurons regulates the intracellular chloride concentration [Cl⁻in]. Some neurons do, some do not. K⁺ 1 So far we have ignored chloride ions in the Goldman equation. However, many neurons have a relatively high chloride permeability. To take account of this, we need a more elaborate form of the Goldman equation: Na⁺ 0.04 Ion Permeability of the Resting Squid Giant Axon Relative to K PK[Kout] + PNa[Naout] + PCl[Clin] Membranes are Often Chloride Permeable Regulated and Unregulated Chloride Chloride Is Regulated If intracellular Cl⁻ concentration is regulated, then it is unlikely that it will be exactly at the value necessary for Cl⁻ to be in Nernstian equilibrium. In this case the full version of the Goldman equation must be used to calculate the steady-state membrane potential. Thus at steady-state, Cl⁻ is not in equilibrium, there is a flux of Cl⁻ across the membrane, and Cl⁻ ions do contribute to the resting membrane potential. However, the regulated intracellular Cl⁻ concentration is usually such that the Cl⁻ equilibrium potential is quite close to the resting membrane potential. Chloride Is Not Regulated If intracellular Cl⁻ concentration is not regulated, then Cl⁻ ions will flow into or out of the neuron until the intracellular Cl⁻ concentration arrives at a value that has a Nernst equilibrium potential equal to the membrane potential set by the simple Goldman equation, which does not include chloride. Thus at steady-state, Cl⁻ is in equilibrium across the membrane and does not contribute to the resting membrane potential. The Goldman equation: Key points Each ion has its own equilibrium (Nernst) potential, which is probably different from that of all the other ion types. 2. The actual membrane potential is a “compromise” between the various equilibrium potentials, each weighted by the membrane permeability and absolute concentration of the ion in question. This is described by the Goldman equation. 3. In the resting state potassium usually dominates the Goldman equation. Sodium permeability is low, so it only makes a small contribution. Chloride permeability is intermediate, but its equilibrium potential is usually quite close to the potassium equilibrium potential. 4. There will be a flux across the membrane of any ion whose equilibrium potential is not equal to the actual membrane potential. 5. If the membrane potential is stable, then the total net flow of ionic charge across the membrane is zero. Leakage Channels Not Much Known! The hydrophobic phospholipid component of the cell membrane is virtually impermeable to charged particles such as ions. However, all membranes contain proteinaceous ion channels that allow certain ions to pass through. The most basic of these are called the leakage channels. These form a sort of “background” permeability which is usually unchanging. Their absolute permeability is low compared to voltage- or ligand-gated channels (generating action and synaptic potentials respectively), but it is the leakage channels that are largely responsible for generating the resting potential. ? Leakage channels are predominantly permeable to potassium, but have some sodium permeability as well. They mainly belong to a heterogenous collection of what are called tandem pore domainAlso sometimes called "2 pore domain" channels, these channels do not have two pores that cross the membrane. They have two pore loops in each segment of their transmembrane sub-unit structure. channels, and fall into classes with names such as TREK, TASK, TWIK or TALK These channels do not have two pores that cross the membrane. They have two pore loops in each segment of their transmembrane sub-unit structure. Gradient Run-Down Run-Down Through Leakage? The internal concentration now equals the external concentration. This should not happen! In the situation described by the Goldman equation, where the membrane is permeable to both K⁺ and Na⁺, the membrane potential is usually not at the equilibrium potential for either ion. Therefore there is a continuous flux of both ions through the leakage channels, and, unless something else happened, the concentration gradients would run down. Because the internal volume is small compared to the external volume, it would be the internal concentration that changed to become equal to the external concentration. Also known as the Na-K ATPase, or simply the sodium pump. What Stops Run-Down? Concentration gradients are maintained by active ion pumps. There are several different types of pump, but one of the most important is the Na-K exchangeAlso known as the Na/K-ATPase or simply the sodium pump. pump. This uses metabolic energyIn neurons, the pump accounts for up to 75% of energy expenditure. obtained by hydrolysing ATP to pump Na⁺ and K⁺ ions against their concentration gradients. The pump has a transport ratio of 3:2, with 3 Na⁺ ions being pumped out of the cell for every 2 K⁺ ions pumped into it. Na-K Exchange Pump 1 The Pump Balances the Passive Flux Na Na Na K K Na-K Exchange Pump 2 In the steady-state condition when the resting potential is stable, the 3 Na⁺ : 2 K⁺ ratio of the pump is exactly balanced by a 3 Na⁺ : 2 K⁺ flux ratio through the leakage channels. If this were not so, and more positive ions continuously left the cell than entered it, the membrane potential would continuously increase. (Note: it is the passive flux that adjusts to meet the pump ratio, not vice versa. The pump ratio is fixed by the molecular structure of the pump. The pump pushes out more positive charge than it brings in, so it has a negative influence on the membrane potential. The quantitative effect depends on the pump transport ratio r, which is 1.5 (3 Na⁺ : 2 K⁺). This can be expressed in a modified Goldman equation by increasing the influence of K⁺: r [Kin] + α [Nain] r [Kout] + α [Naout] For the squid giant axon concentration and permeability values: V = -64 mV (with pump: r = 1.5) compared to V = -60 mV (without pump: r = 1) The Pump Contributes to the Membrane Potential The Pump is Electrogenic The electrogenic Na-K exchange pump thus makes a contribution (usually just a few millivolts) to the resting membrane potential. The Na-K exchange pump maintains the Na and K concentration gradients across the cell membrane. Without the pump, the gradients would inevitably run down due to flux through the leakage channels. (Note that the ion fluxes during action potentials and synaptic potentials would greatly increase the rate of run-down, and the pump rate has to increase in neurons with a high activity level to compensate.) 2. The pump hydrolyses ATP to provide the energy needed to move ions across the membrane against their concentration gradient. 3. The pump is electrogenic, and makes a direct contribution to the membrane potential, but it is usually not the dominant immediate cause of the membrane potential. 4. When the membrane potential is stable, the pump current is exactly balanced by the ion flux through the leakage channels. The Na-K Exchange Pump: Key points Neurons have high [K⁺in] and low [K⁺out], and a high membrane permeability to K⁺. 2. K⁺ tends to leave the neuron down its concentration gradient, producing an electrical gradient across the membrane with the inside negative relative to the outside. 3. The Nernst equation defines the equilibrium potential at which the electrical gradient balances the concentration gradient. Not many ions have to cross the membrane to produce equilibrium. 4. In a purely passive system, [K⁺] and [Cl⁻] gradients and a membrane potential can be set up by the Donnan equilibrium, if there are impermeant anions within the cell. 5. Most neurons have low [Na⁺in] and high [Na⁺out], and a low but significant resting permeability to Na⁺. 6. The actual resting membrane potential is a compromise between the negative equilibrium potential for K⁺ and and the positive equilibrium potential for Na⁺, weighted towards K because of its greater permeability. Cl⁻ plays a role if it is actively regulated, but not if it is passively distributed. The Goldman equation describes the resulting membrane potential. 7. Neither K⁺ nor Na⁺ are in equilibrium across the membrane. Cl⁻ may or may not be in equilibrium. 8. The passive flux of K⁺ and Na⁺ is balanced by active pumping by the Na-K exchange pump, so the pump is essential to maintain the concentration gradients. The pump is electrogenic and makes a contribution to the resting membrane potential, but this is usually quite small. 9. The main immediate cause of the resting membrane potential in most neurons is the differential distribution of sodium and potassium ions across the membrane, and the different permeability of the membrane to those ions. Overall Summary The Origin of the Resting Membrane Potential home page Dr W. J. Heitler School of Psychology and Neuroscience University of St Andrews Scotland All living cells have a potential difference (voltage) across their cell membranes, with the inside usually negative relative to the outside. In nerve cells the value of the resting potential varies between about -40 and -90 mV. The voltage can be measured by penetrating the cell with a microelectrodeA microelectrode is a specialized electrode for recording voltages from the inside of cells. It typically consists of a glass tube drawn out to a very fine point and filled with an electrolyte solution such as potassium chloride. The sharp tip of the electrode penetrates through the cell membrane. The blunt end of the electrode is connected by a wire to a voltage measuring instrument.. This is connected to a specialized voltmeterInstruments for recording voltages from cells have to have a very high input impedance. This means that they can measure the voltage without drawing much current from the cell, and thus without changing the very voltage that they are trying to measure. which measures the potential difference between the inside of the cell and the surrounding environment. This is the membrane potential. This tutorial describes the cellular mechanisms that generate the resting membrane potential. Early in the 20th century, Bernstein proposed that the resting membrane potential was due to 3 factors: the cell membrane is selectively permeable to potassium ions,the intracellular potassium concentration is high,the extracellular potassium concentration is low.This is known as the potassium electrode hypothesis. In the next section we will show how these factors would cause a membrane potential with the inside negative. This will lead to an understanding of the Nernst equation and the ionic equilibrium (reversal) potentials. Sneak preview - Bernstein was almost, but not completely, right. In our scenario there is initially a large driving force (shown as an arrow through the pore). This is because the chemical gradient is not balanced by any electrical gradient. The difference between the chemical gradient and the electrical gradient is called the driving force. and rearranging ... Gives the Nernst equation. Deriving the Nernst equation 2 At equilibrium the energies are equal, so the two equations are equal. Losing common factors n and δW Deriving the Nernst equation 3 K⁺ Ca⁺⁺ Cl⁻ If we reverse the direction of the concentration gradient, so that the ratio is now 1:10 (i.e. 0.1), then the absolute value of the membrane potential stays the same, but its polarity reversesNote that the ion has changed to sodium but the Nernst equation doesn’t care - it is still a monovalent ion. (We assume the membrane is now permeable to sodium.) (the log of 0.1 is -1). If the membrane potential is -70 mV, the charge on the capacitor is: Q = VC = 70 x 10-3 × 7.8 x 10-11 Thus Q = 5.50 x 10-12 coulombs. One K⁺ ion has a charge of 1.6 x 10-19 coulombs, so the number of ions that move down the concentration gradient is about 34,340,000 i.e. about 34 million. A cell with diameter 50 µm has a volume (4πr³/3) of 6.54 x 10-11 litres and its intracellular K⁺ concentration (for a mammal) is about 140 mM, so it contains 9.16 x 10-12 moles of K⁺ ions. One mole consists of 6.02 x 1023 ions (Avogadro's number), so the cell contains about 5.52 x 1012 ions. The loss of 3.43 x107 ions represents a concentration change of about 0.006% So the answer is HARDLY AT ALL. Note: the external volume is relatively huge, so the change in external concentration is infinitesimal. The Nernst equation defines the electrical potential across a membrane that will balance a particular chemical concentration gradient of an ion. So if the membrane potential is at the Nernst value, there will be no net movement of that ion across the membrane, even if the membrane is permeable to the ion. 2. This potential is called the equilibrium or reversal potential for that ion. It depends on the concentration gradient (ratio) and the valency of the ion (and the temperature), but does not depend on the degree of permeability of the membrane to the ion, nor on the absolute concentrations. 3. Each type of ion will have its own equilibrium potential, and this is likely to be different to that of other ion types. 4. If a membrane is permeable to only ONE type of ion, then the membrane potential will automatically move to the equilibrium potential for that ion. 5. Relatively few ions have to move to set up the potential, so the concentration gradient is not significantly perturbed in achieving equilibrium. However, what if the cell membrane is permeable to more than one type of ion …? Many cells have membranes with quite a high permeability to both potassium and chloride ions. In the absence of anything else, this would prevent the establishment of the membrane potential, because Cl⁻ would simply move down the concentration gradient along with the K⁺, and counteract the positive charge. The concentrations would just equilibrate across the membrane. However, most cells also have a relatively high intracellular concentration of large molecular weight anions A⁻ (e.g. negatively charged proteins). The intracellular composition is thus: K⁺Cl⁻ + K⁺A⁻ This can result in a Donnan equilibrium, which both maintains opposite K⁺ and Cl⁻ gradients across the membrane, and sets up a resting potential. The Donnan equilibrium applies to ions whose distribution is passive and unregulated, i.e. there are no metabolic pumps maintaining the intracellular concentrations at fixed values. Now consider the situation where some intracellular Cl⁻ ions are replaced by the impermeant anion A⁻. The K⁺ and Cl⁻ ions can still flow down their concentration gradients and leave the cell, but A⁻ cannot pass through the membrane. In the diagram each symbol represents 10 mM of substance - e.g. there are 12 K⁺s giving a total of 120 mM. We start off with everything inside the cell, nothing outside. What happens now? Eventually, an equilibrium is reached with: Inside Outside + - + - K80 40 Cl 20 40 A 60 0 total+80-80 +40-40 We still have space-charge neutrality, and the concentration gradient (in:out) for K⁺ is 80:40, and for Cl⁻ is 20:40, which gives a 2:1 ratio for both ions, and thus a Nernst potential of -17 mV for both ions. Everything balances! so at equilibrium: (120 - x) x (60 - x) = x2 and thus 7200 - 180x +x2 = x2 So x = 40. Conclusion 40 mM of KCl had to move across the membrane to set up the Donnan equilibrium. Note: the final K⁺ and Cl⁻ ion distribution is independent of the starting distribution, so long as the ion totals are the same. The Donnan rule states that the product of the concentration of diffusible ions on one side of the membrane equals the product of the concentration of diffusible ions on the other. 2. The Donnan rule only applies to cells where the ions are passively distributed, i.e. there are no metabolic pumps using energy to regulate the ion concentration inside the cell. 3. The Donnan rule is important because, where it applies, it explains the origin of concentration gradients, and hence the resting potential. 4. The Donnan rule is approximately obeyed by frog muscle fibres (which were important preparations in early physiological experiments). 5. The Donnan rule is not obeyed by many nerve cells, where the resting potential is significantly different from the Nernst equilibrium potential of the major ions involved in setting it up. What is happening when the Donnan rule is NOT obeyed? Donnan equilibrium: Key points So far we have considered the situation where the membrane is permeable only to potassium and chloride ions, and their distribution is entirely passive. However, most real nerve membranes are also somewhat permeable to sodium ions, which usually have a high extracellular concentration and a low intracellular concentration. Also, most nerves (and many other cells) regulateRegulation is usually achieved by metabolic pumps such as the Na/K-ATPase (also known as the sodium-potassium exchange pump, or simply the sodium pump). the intracellular concentration of ions, so that they remain at a relatively fixed value. We next consider what happens in this situation. This will lead to an understanding of the Goldman-Hodgkin-Katz constant field equation, usually know as the Goldman equation for short. The Goldman equation describes a steady-state condition, unlike the Nernst equation, which describes an equilibrium condition. The difference will be explained shortly ... Since the membrane potential is not at either equilibrium potential, neither K⁺ nor Na⁺ are in equilibrium. Each has a steady flux down its concentration gradient. (Note: In the absence of any restorative pumps, eventually the concentration gradients will run down, but this takes some time.) Each ion experiences a driving force, which is the difference between its equilibrium potential (which itself would balance the concentration gradient) and the actual membrane potential. driving force = Em - Eeq The driving forces are equal and opposite, so the fluxesFlux in this case is the equivalent of current (charge per second). are equal and opposite. Neither side has a net change in charge, so there is no change in membrane potential. previous Each ion has its own equilibrium (Nernst) potential, which is probably different from that of all the other ion types. 2. The actual membrane potential is a “compromise” between the various equilibrium potentials, each weighted by the membrane permeability and absolute concentration of the ion in question. This is described by the Goldman equation. 3. In the resting state potassium usually dominates the Goldman equation. Sodium permeability is low, so it only makes a small contribution. Chloride permeability is intermediate, but its equilibrium potential is usually quite close to the potassium equilibrium potential. 4. There will be a flux across the membrane of any ion whose equilibrium potential is not equal to the actual membrane potential. 5. If the membrane potential is stable, then the total net flow of ionic charge across the membrane is zero. For the squid giant axon concentration and permeability values: V = -64 mV (with pump: r = 1.5) V = -60 mV (without pump: r = 1) Overall Summary In most neurons the electrogenic Na/K pump makes only a minor contribution to the resting potential. However, the actual amount is variable, and in some specialised neurons it may be considerable. One unusal example occurs in olfactory receptor neurons in frogs (and proably mammals too). These have an extremely high membrane resistance, which makes them very sensitive because a small stimulus current produces a large voltage response. The high membrane resistance is because they have virtually no resting potassium conductance. So the Bernstein hypothesis is completely wrong for these neurons! Instead, in these neurons the resting potential is almost entirely due to the electrogenic Na/K pump. The pump on its own would produce a membrane potential of about -140 mV. However, the pump-induced hyperpolarization activates a mixed cation channel which produces a depolarizing current (Ih, sometimes called the “funny current” because it is unusual in being activated by hyperpolarization) that works against the pump current. The end result is a resting potential of about -80 mV, which is within the normal range. But it is produced by a very abnormal mechanism! Trotier & Døving (1996) Functional role of receptor neurons in encoding olfactory information. J. Neurobiol. 30; 58-66. Unusual Mechanisms In most neurons the electrogenic Na/K pump makes only a minor direct contribution to the resting potential. However, the actual amount is variable, and in some specialised neurons it may be considerable. One such example occurs in olfactory receptor neurons in frogs (and probably mammals too). These have an extremely high membrane resistance, which makes them very sensitive because a small stimulus current produces a large voltage response. The high membrane resistance is because they have virtually no resting potassium conductance. So the Bernstein hypothesis is completely wrong for these neurons! Instead, in these neurons the resting potential is almost entirely due to the electrogenic Na/K pump. The pump on its own would produce a membrane potential of about -140 mV. However, the pump-induced hyperpolarization activates a mixed cation channel which produces a depolarizing current (Ih, sometimes called the “funny current” because it is unusual in being activated by hyperpolarization) that works against the pump current. The end result is a resting potential of about -80 mV, which is within the normal range. But it is produced by a very abnormal mechanism!
2924	https://socialsci.libretexts.org/Bookshelves/Economics/Applied_Economics/Managerial_Economics_Principles_(LibreTexts)/07%3A_Firm_Competition_and_Market_Structure/7.05%3A_Seller_Concentration	7.5: Seller Concentration - Social Sci LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 7: Firm Competition and Market Structure Managerial Economics Principles (LibreTexts) { } { "7.01:_Why_Perfect_Competition_Usually_Does_Not_Happen" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.02:_Monopoly" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.03:_Oligopoly_and_Cartels" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.04:_Production_Decisions_in_Noncartel_Oligopolies" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.05:_Seller_Concentration" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.06:_Competing_in_Tight_Oligopolies-_Pricing_Strategies" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.07:_Competing_in_Tight_Oligopolies-_Nonpricing_Strategies" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.08:_Buyer_Power" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Introduction_to_Managerial_Economics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Key_Measures_and_Relationships" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Demand_and_Pricing" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Cost_and_Production" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Economics_of_Organization" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Market_Equilibrium_and_the_Perfect_Competition_Model" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Firm_Competition_and_Market_Structure" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Market_Regulation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sun, 23 Jul 2023 23:58:00 GMT 7.5: Seller Concentration 44802 44802 admin { } Anonymous Anonymous User 2 false false [ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:anonymous", "program:hidden", "licenseversion:30", "source@ "concentration ratios", "market shares", "HHI", "Herfindahl-Hirschmann Index" ] [ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:anonymous", "program:hidden", "licenseversion:30", "source@ "concentration ratios", "market shares", "HHI", "Herfindahl-Hirschmann Index" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Economics 4. Applied Economics 5. Managerial Economics Principles (LibreTexts) 6. 7: Firm Competition and Market Structure 7. 7.5: Seller Concentration Expand/collapse global location Managerial Economics Principles (LibreTexts) Front Matter 1: Introduction to Managerial Economics 2: Key Measures and Relationships 3: Demand and Pricing 4: Cost and Production 5: Economics of Organization 6: Market Equilibrium and the Perfect Competition Model 7: Firm Competition and Market Structure 8: Market Regulation Back Matter 7.5: Seller Concentration Last updated Jul 23, 2023 Save as PDF 7.4: Production Decisions in Noncartel Oligopolies 7.6: Competing in Tight Oligopolies- Pricing Strategies picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report Submit a Peer Review View on CommonsDonate Page ID 44802 Anonymous LibreTexts ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents No headers Sellers in oligopolies can limit competition by driving out competitors, blocking entry by new competitors, or cooperating with other sellers with market power to keep prices higher than would be the case in a market with strong price competition. In order for sellers to exercise market power, either the market will have fairly few selling firms or there will be some selling firms that account for a large portion of all the market sales. When this happens, the market is said to have high seller concentration. Although high seller concentration in itself is not sufficient for exercise of seller power, it is generally a necessary condition and constitutes a potential for the exercise of seller power in the future. In this section, we will consider two numerical measures of market concentration: concentration ratios and the Herfindahl-Hirschmann Index ( HHI ). Both measures of seller concentration are based on seller market shares. A firm ’s market share is the percentage of all market sales that are purchased from that firm . The highest possible market share is 100%, which is the market share of a monopolist . Market shares may be based either on the number of units sold or in terms of monetary value of sales. The latter use of monetary value is convenient when there are variations in the good or service sold and different prices are charged. Concentration ratios are the result of sorting all sellers on the basis of market share, selecting a specified number of the firms with the highest market shares , and adding the market shares for those firms. For example, the concentration ratio CR 4 is the sum of the market shares for the four largest firms in terms of volume in a market and CR 8 is the sum of the eight largest firms in terms of volume. The U.S. Census Bureau periodically publishes concentration ratios for different industries in the United States.See U.S. Census Bureau (2010). Suppose a market has 10 sellers with market shares (ranked from high to low) of 18%, 17%, 15%, 13%, 12%, 8%, 7%, 5%, 3%, and 2%. The CR 4 ratio for this market would be 63 (18 + 17 + 15 + 13), and the CR 8 ratio would be 95 (18 + 17 + 15 + 13 + 12 + 8 + 7 + 5). Although concentration ratios are easy to calculate and easily understood, there are two shortcomings. First, the number of firms in the ratio is arbitrary. There is no reason that a four- firm concentration ratio indicates concentration potential any better than a three- firm or five- firm concentration ratio. Second, the ratio does not indicate whether there are one or two very large firms that clearly dominate all other firms in market share or the market shares for the firms included in the concentration ratio are about the same. An alternative concentration measure that avoids these problems is the HHI. This index is computed by taking the market shares of all firms in the market, squaring the individual market shares , and finally summing them. The squaring has the effect of amplifying the larger market shares . The highest possible value of the HHI is 10,000, which occurs in the case of a monopoly (10,000 = 100 2). If, on the other hand, you had a market that had 100 firms that each had a market share of 1%, the HHI would be 100 (1 = 1 2, summed 100 times). For the previous 10- firm example, the HHI would be 1302. Although there is no inherent reason for squaring market shares , the HHI includes all firms in the computation (avoiding the issue of how many firms to include) and reflects the variation in magnitude of market shares . As far as interpreting these concentration measures, the following statements provide some guidance on the potential for market power by sellers: If CR 4 is less than 40 or the HHI is less than 1000, the market has fairly low concentration and should be reasonably competitive. If CR 4 is between 40 and 60 or the HHI is between 1000 and 2000, there is a loose oligopoly that probably will not result in significant exercise of market power by sellers. If CR 4 is above 60 or the HHI is above 2000, then there is a tight oligopoly that has significant potential for exercise of seller power. If CR 1 is above 90 or the HHI is above 8000, one firm will be a clear leader and may function effectively as a monopoly. Again, a high concentration measure indicates a potential for exploitation of seller power but not proof it will actually happen. Another important caution about these measures is that the scope of the market needs to be considered. In the case of banking services, even with the mergers that have resulted in higher seller concentration, if you look at measures of bank concentration at the national level, there seems be a loose oligopoly . However, if you limit the scope to banking in a single city or region, it is very likely that only few banks serve those areas. There can be modest concentrations when examining national markets but high concentration at the local level. This page titled 7.5: Seller Concentration is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform. Back to top 7.4: Production Decisions in Noncartel Oligopolies 7.6: Competing in Tight Oligopolies- Pricing Strategies Was this article helpful? Yes No Recommended articles 8.3: Measuring Market Structure 7.1: Why Perfect Competition Usually Does Not HappenThe perfect competition model (and its variants like monopolistic competition and contestable markets) represents an ideal operation of a market. As w... 7.2: MonopolyOften, the main deterrent to a highly competitive market is market power possessed by sellers. In this section, we will consider the strongest form of... 7.3: Oligopoly and CartelsUnless a monopoly is allowed to exist due to a government license or protection from a strong patent, markets have at least a few sellers. When a mark... Article typeSection or PageAuthoranonymousLicenseCC BY-NC-SALicense Version3.0OER program or PublisherThe Publisher Who Must Not Be NamedShow TOCno Tags concentration ratios Herfindahl-Hirschmann Index HHI market shares source@ © Copyright 2025 Social Sci LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more calculating the average of nested array in particular manner by python? Ask Question Asked 5 years, 11 months ago Modified5 years, 11 months ago Viewed 1k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. i need a way to calculate the average of the nested array ( even or odd ) in this following manner: let's say we have this array ( list ) ( even 44 ) : python mylist = [ [1,6,5,6], [2,5,6,8], [7,2,8,1], [4,4,7,3] ] the output must be like this python mylist = [[3,6], [4,4] ] based on this calculation python 1 + 6 + 2 + 5 / 4 = 3 5 + 6 + 6 + 8 / 4 = 6 7 + 2 + 4 + 4 / 4 = 4 8 + 1 + 7 + 3 / 4 = 4 the same thing if we have an odd nested array like this python mylist = [[7,9,1], [4,2,1], [3,2,3]] the output would be : python mylist = [[5,1], [2,3] ] based in same calculation above .. python 7 + 9 + 4 + 2 / 4 = 5 1 + 1 / 2 = 1 3 + 2 / 2 = 2 3 / 1 = 3 so how we can implement this process by python, notice that i know how to do the normal average for each array like just increasing the numbers for each array line by line and divided by its count .. python mylist = [[70,80,90], [30,40,50], [0,10,20], [10,40,40]] avglist = [] for x in mylist: temp = 0 counter = 0 for y in x: temp = temp + y counter = counter + 1 avglist.append(temp/counter) print() print() print(avglist) but in that problem .. i face a problem to how to jump to next array then came back to first array and so forth .... notice: it has to be a square array ( row length = column length ) python arrays square Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Oct 28, 2019 at 5:06 Hussam GoldHussam Gold asked Oct 28, 2019 at 4:17 Hussam GoldHussam Gold 13 5 5 bronze badges 7 Can you use numpy or is it a constraint to use a list?pissall –pissall 2019-10-28 04:17:44 +00:00 Commented Oct 28, 2019 at 4:17 it has to be a list ....Hussam Gold –Hussam Gold 2019-10-28 04:18:21 +00:00 Commented Oct 28, 2019 at 4:18 @pissall the problem is on the algorithm in itself not the mean .. thank u for replying Hussam Gold –Hussam Gold 2019-10-28 04:24:57 +00:00 Commented Oct 28, 2019 at 4:24 @HussamGold in your first even 4x4 example, is the calculation 1 + 6 + 5 + 6 / 4 = 3 suppose to be (1 + 6 + 2 + 5) / 4 = 3? Also, just to clarify, you're not actually calculating the means, but rather the floor of the mean right?Jethro Cao –Jethro Cao 2019-10-28 04:45:46 +00:00 Commented Oct 28, 2019 at 4:45 @JethroCao u r right sorry for mistake .. yeah in way to become 3 instead of decimal number Hussam Gold –Hussam Gold 2019-10-28 04:47:33 +00:00 Commented Oct 28, 2019 at 4:47 \|Show 2 more comments 3 Answers 3 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Okay so here's my try. It's a bit verbose, but I think it's very easy to follow. ```python helper func to split nested list (NxN matrix) into 4 quadrants def split_mat(mat): n = len(mat) s = math.ceil(n/2) up_left = [mat[i][j] for i in range(0, s) for j in range(0, s)] up_right = [mat[i][j] for i in range(0, s) for j in range(s, n)] bt_left = [mat[i][j] for i in range(s, n) for j in range(0, s)] bt_right = [mat[i][j] for i in range(s, n) for j in range(s, n)] return [up_left, up_right, bt_left, bt_right] then the averages you want to calculate becomes trivial def avg_mat(mat): quadrants = split_mat(mat) avgs = [sum(q)//len(q) for q in quadrants] return , [avgs, avgs]] ``` ```python even_list = [ [1,6,5,6], [2,5,6,8], [7,2,8,1], [4,4,7,3]] print(avg_mat(even_list)) ``` ```python odd_list = [ [7,9,1], [4,2,1], [3,2,3]] print(avg_mat(odd_list)) ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Oct 28, 2019 at 5:26 Jethro CaoJethro Cao 1,060 2 2 gold badges 10 10 silver badges 23 23 bronze badges Comments Add a comment This answer is useful 1 Save this answer. Show activity on this post. A bit golfy, sorry, typing on phone means short names :). I take advantage of integer division to simply the logic, and use totals plus division over intermediate lists. ```python from itertools import product def quad_avgs(M): sums = L, S, s = len(M), len(M) / 2, len(M) - len(M) / 2 for r, c in product(range(L), repeat=2): sums[r / S][c / S] += M[r][c] return for r in [0, 1]] ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Oct 28, 2019 at 6:43 CireoCireo 4,457 2 2 gold badges 22 22 silver badges 27 27 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. I present a frame challenge: instead of counting each quadrant one at a time, count all quadrants at once. Assuming we are always computing four numbers (the sums of each quadrant, with a bias towards the top-left quadrant if the sizes are unequal), we can simply prepare them in advance, and then scan over each cell of the matrix, determine which quadrant it's in, and increment that sum: python def calculation(matrix): # prepare empty lists for each quadrant # we will add each element from the matrix to the list that corresponds to its quadrant # then at the end we will take the average of each. sums = { (True, True): [], # top-left (True, False): [], # top-right (False, True): [], # bottom-left (False, False): [], # bottom-right } # scan over each cell by row and column index for row in range(len(matrix)): for col in range(len(matrix[row])): # use boolean checks to index into sums depending on quadrant sums[(row < len(matrix) / 2, # is it in the top half? col < len(matrix[row]) / 2 # is it in the left half? )].append(matrix[row][col]) # calculate and return averages (using integer division instead of true division) return [[sum(sums[True, True]) // len(sums[True, True]), # top-left sum(sums[True, False]) // len(sums[True, False])], # top-right [sum(sums[False, True]) // len(sums[False, True]), # bottom-left sum(sums[False, False]) // len(sums[False, False])]] # bottom-right Test cases: ```python mylist = [ ... [1,6,5,6], ... [2,5,6,8], ... [7,2,8,1], ... [4,4,7,3] ... ] calculation(mylist) mylist = [[7,9,1], ... [4,2,1], ... [3,2,3]] calculation(mylist) ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Oct 28, 2019 at 5:14 Green Cloak GuyGreen Cloak Guy 24.8k 4 4 gold badges 39 39 silver badges 58 58 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. 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2926	https://www.youtube.com/watch?v=9Sck-rejNjU	Annuity - future value (monthly savings) Brian Veitch 7200 subscribers 20 likes Description 10548 views Posted: 2 Aug 2012 In this video we discuss a simple annuity, a monthly savings. 10 comments Transcript: Intro hello welcome back to our series on finance math in this video we're going to talk about annuities specifically a monthly savings um so let's kind of Define annuities here first an annuity has to have the following conditions it's a sequence of equal siiz payments the payments are going to be over a regular time interval like you know monthly or you can make yearly payments uh quarterly payments and so on but it has to be the same every single time and then there's a couple other different things like are you making the payment uh at the end are you making it in um the beginning um so but we're not going to you know focus on that right now it's just it's a sequence of equal siiz payments at regular time intervals now an example of one type of annuity would be like a monthly Savings Program so say you get a paycheck and the bank takes out $100 every single month well that would be an annuity and we can figure out how much you would have after however many years you are um saving so let's go ahead and do an Formula example but first I do want to write down the formula we will be using again because different textbooks will use different formulas and different people use different formulas um they're all going to look a little different but again they all are actually the same thing just maybe different notations or um you know different letters or maybe they simplify the formula a little bit but the formula that I'm going to use is this it's going to be a again a for like the accumulated amount or in the previous video I we now started calling it future value and that's going to be M uh and then two parentheses 1 + r/ m to the Mt minus one time M over R so now we have a couple diff or um couple letters we're not familiar with well I guess really just the M the Big M here I have another M okay well this would be a capital M this be a little M usually would actually use start using n's here um but I'll stick with little M for now all right so let's label all this stuff okay so we have a is future value Big M is our month or um well will be a monthly payment because that's the example we're going to do but it's going to be the payment uh R is interest rate m is your compounding period T is time and years now the reason the formulas look a little different um from textbook to textbook is in some what they'll do is they'll simplify instead of using Mt they might use a different letter but Mt repres reprs the number of payments okay um but let's go ahead and try an Example example so in this example I want to find the amount of an annuity consisting of 12 monthly payments of $250 each that earn that earns earn uh an interest rate of 5% compounded monthly okay so things that we know we know M all right m is 20 50 we know R which is 005 uh let's see little m is 12 because I said compounded monthly uh what else do we need and T now T sometimes they'll tell you and sometimes you have to kind of work it out but if you're making 12 monthly payments how how many years have passed well one so T is one and I think that's it so we can go right into the formula now so we have a equals 250 uh parentheses we got 1 + 05 over 12 raised to the well 12 1 - 1 12 over .05 and I do want to stress this one more time that formula and different textbooks will look different um but they will give you the same answer okay I promise they will give you the same answer uh what I want to do is I'm going to use this I'm going to use the graphing calculator so you can see how it's done so let's pull the calculator out okay now I kind of set this I set this up so I pretty much um type in an exactly the way I see it so it'll be 250 2 parentheses 1 + 05 / 12 and then exponent and then in parentheses 12 1 minus 1 parentheses and then times uh 12 divided by the interest rate 05 and then that's it so 3,69 and71rus using fancy words like annuity you know 12 monthly payments and so on U but what this is is this is a monthly savings program and this is how much you would have after you know saving for one total year you made 12 monthly payments so this is how much you would have so let's try another one okay so our next example uh you begin to save for a vacation by depositing $400 into an account earning 3.75% per year compounded monthly so how much will you have after five years oh so depositing $400 sorry per month I got to make that clear so per month uh component monthly how much do you have after 5 years okay so our formula is this again sorry the uh I have little r or Little M and Big M so I'll try to keep them separate okay so let's see we have a monthly payment which is $400 we have an interest rate of 3.75 so that's 0.375 divided by 12 because it's compounding monthly raised to the 12 time now m and then T is the number of years which is 5 years minus one and then it's going to be 12 over the interest rate which is 0.375 all right so let me pull out that calculator again and we will compute this so it's going to be 400 two parentheses 1 + 0 sorry .375 / 12 so I'm going to skip ahead all right and we press enter and we'll get $ 26,35 to39 so let me write that down we got 26,000 $352 39
2927	https://math.stackexchange.com/questions/3404451/how-to-get-a-serie-of-five-minute-for-each-minute-in-an-hour	ceiling and floor functions - How to get a serie of five minute for each minute in an hour? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to get a serie of five minute for each minute in an hour? Ask Question Asked 5 years, 11 months ago Modified5 years, 11 months ago Viewed 28 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. This is an application on programming but it nature is a simple arithmetic progression I need to update a database at each 5 min interval, if is 13:03 I want it to fill 13:00 in database, if is 13:06 -> 13:05, always rounding down to five multiple. The way that I got to do this is date = date.replace(minutes=(date.minutes/5)5 which seems silly to me, like if I'm missing something I thought that this would make sense m 5 5⇒5 m 5⇒m m 5 5⇒5 m 5⇒m But m m is any number [0..59][0..59] while m 5 5 m 5 5 is always a multiple of 5 between [0..59][0..59] so I cannot simplify m 5 5 m 5 5 and expect to get a multiple of 5 What I'm missing? ceiling-and-floor-functions arithmetic-progressions Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Oct 22, 2019 at 16:26 geckosgeckos 225 1 1 silver badge 6 6 bronze badges 4 Floor functions. ⌊m 5⌋⋅5⌊m 5⌋⋅5. In javascript for instance, this would be Math.floor(date.minutes/5)5JMoravitz –JMoravitz 2019-10-22 16:28:28 +00:00 Commented Oct 22, 2019 at 16:28 1 If you can guarantee that minutes is always a non-negative integer, you could do something similar using modular arithmetic. For example, date.minutes - (date.minutes % 5). (How negative numbers behave with the % operator is inconsistent across certain languages and so you should verify the exact behavior if you wish to use it in those scenarios)JMoravitz –JMoravitz 2019-10-22 16:31:53 +00:00 Commented Oct 22, 2019 at 16:31 Thanks @JMoravitz, yeah they are always positive integers between [0..59]geckos –geckos 2019-10-22 17:43:01 +00:00 Commented Oct 22, 2019 at 17:43 1 a nitpick, 0 0 is not a positive number.JMoravitz –JMoravitz 2019-10-22 19:00:44 +00:00 Commented Oct 22, 2019 at 19:00 Add a comment\| 0 Sorted by: Reset to default You must log in to answer this question. 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2928	https://www.vocabulary.com/dictionary/duck	SKIP TO CONTENT /dÉk/ IPA guide Other forms: ducks; ducked; ducking A duck is a bird that spends much of its life on or near water. You can sometimes see whole families of ducks paddling across lakes and ponds. Ducks can be wild or domesticated if you keep ducks in your yard, you'll need to provide them with at least a small body of water. When the word duck is a verb, it means "to bow or dip suddenly," like when your kite plunges suddenly and you duck to avoid getting hit in the head. You can duck figuratively, too: "Don't you dare duck out of helping me clean up after the party!" Definitions of duck small wild or domesticated web-footed broad-billed swimming bird usually having a depressed body and short legs see moresee less types: show 37 types... hide 37 types... drake adult male of a wild or domestic duck quack-quack child's word for a duck duckling young duck diving duck any of various ducks of especially bays and estuaries that dive for their food dabbler, dabbling duck any of numerous shallow-water ducks that feed by upending and dabbling Anas platyrhynchos, mallard wild dabbling duck from which domestic ducks are descended; widely distributed Anas rubripes, black duck a dusky duck of northeastern United States and Canada teal any of various small short-necked dabbling river ducks of Europe and America Anas penelope, widgeon, wigeon freshwater duck of Eurasia and northern Africa related to mallards and teals Anas clypeata, broadbill, shoveler, shoveller freshwater duck of the northern hemisphere having a broad flat bill Anas acuta, pin-tailed duck, pintail long-necked river duck of the Old and New Worlds having elongated central tail feathers sheldrake Old World gooselike duck slightly larger than a mallard with variegated mostly black-and-white plumage and a red bill Oxyura jamaicensis, ruddy duck reddish-brown stiff-tailed duck of North America and northern South America Bucephela albeola, bufflehead, butterball, dipper small North American diving duck; males have bushy head plumage Bucephela clangula, goldeneye, whistler large-headed swift-flying diving duck of Arctic regions Aythya valisineria, canvasback, canvasback duck North American wild duck valued for sport and food Aythya ferina, pochard heavy-bodied Old World diving duck having a grey-and-black body and reddish head Aythya americana, redhead North American diving duck with a grey-and-black body and reddish-brown head bluebill, broadbill, scaup, scaup duck diving ducks of North America having a bluish-grey bill wild duck an undomesticated duck (especially a mallard) Aix sponsa, summer duck, wood duck, wood widgeon showy North American duck that nests in hollow trees Aix galericulata, mandarin duck showy crested Asiatic duck; often domesticated Cairina moschata, muscovy duck, musk duck large crested wild duck of Central America and South America; widely domesticated sea duck any of various large diving ducks found along the seacoast: eider; scoter; merganser Anas crecca, green-winged teal, greenwing common teal of Eurasia and North America Anas discors, blue-winged teal, bluewing American teal Anas querquedula, garganey small Eurasian teal American widgeon, Anas americana, baldpate a widgeon the male of which has a white crown shelduck female sheldrake Barrow's goldeneye, Bucephala islandica North American goldeneye diving duck Aythya marila, greater scaup large scaup of North America having a greenish iridescence on the head of the male Aythya affinis, lake duck, lesser scaup, lesser scaup duck common scaup of North America; males have purplish heads wood drake male wood duck eider, eider duck duck of the northern hemisphere much valued for the fine soft down of the females scooter, scoter large black diving duck of northern parts of the northern hemisphere Clangula hyemalis, old squaw, oldwife a common long-tailed sea duck of the northern parts of the United States fish duck, merganser, sawbill, sheldrake large crested fish-eating diving duck having a slender hooked bill with serrated edges type of: anseriform bird chiefly web-footed swimming birds 2. noun flesh of a duck (domestic or wild) see moresee less types: duckling flesh of a young domestic duck type of: poultry flesh of chickens or turkeys or ducks or geese raised for food 3. verb submerge or plunge suddenly see moresee less type of: dive, plunge, plunk drop steeply 4. verb dip into a liquid synonyms: dip, douse see moresee less type of: dip, douse, dunk, plunge, souse immerse briefly into a liquid so as to wet, coat, or saturate 5. verb to move (the head or body) quickly downwards or away “Before he could duck, another stone struck him” see moresee less type of: move move so as to change position, perform a nontranslational motion 6. verb avoid or try to avoid fulfilling, answering, or performing (duties, questions, or issues) synonyms: circumvent, dodge, elude, evade, fudge, hedge, parry, put off, sidestep, skirt see moresee less types: beg, beg the question dodge, avoid answering, or take for granted quibble evade the truth of a point or question by raising irrelevant objections type of: avoid stay clear from; keep away from; keep out of the way of someone or something 7. noun a heavy cotton fabric of plain weave; used for clothing and tents see moresee less type of: cloth, fabric, material, textile artifact made by weaving or felting or knitting or crocheting natural or synthetic fibers 8. noun (cricket) a score of nothing by a batsman synonyms: duck's egg see moresee less type of: score a number that expresses the accomplishment of a team or an individual in a game or contest Pronunciation US /dÉk/ UK /dÉk/ Cite this entry Style: MLA MLA APA Chicago Copy citation DISCLAIMER: These example sentences appear in various news sources and books to reflect the usage of the word duck'. Views expressed in the examples do not represent the opinion of Vocabulary.com or its editors. Send us feedback Word Family Vocabulary lists containing duck Birds, Birds, Birds, List 2 We're honoring our feathered friends. So be bald and dive into this list of trilling vocabulary all about birds. Here are links to our lists in the collection: List 1, List 2 MORE VOCABULARY LISTS 2 million people are mastering new words. Master a word Sign up now (its free!) Whether youre a teacher or a learner, Vocabulary.com can put you or your class on the path to systematic vocabulary improvement. Get started
2929	https://amslaurea.unibo.it/id/eprint/7387/1/Riccardo_Nanni_Tesi.pdf	ALMA Mater Studiorum Università' degli Studi di Bologna SCUOLA DI SCIENZE Corso di Laurea Triennale in Astronomia Dipartimento di Fisica e Astronomia L'atomo di idrogeno: righe, serie e sua importanza in astrofisica Elaborato Finale Candidato: Relatore: Riccardo Nanni Prof. Daniele Dallacasa Sessione II Anno Accademico 2013/2014 Indice Introduzione 2 1 Righe e serie dell’idrogeno e l’atomo di Bohr 3 1.1 Le serie dell’idrogeno . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 L’atomo di Bohr . . . . . . . . . . . . . . . . . . . . . . . . . 4 2 La teoria delle transizioni di riga 6 2.1 I livelli energetici e i numeri quantici . . . . . . . . . . . . . . 7 2.2 Transizioni permesse e coeﬃcienti di Einstein . . . . . . . . . 8 2.3 Le righe proibite e la riga 21 cm . . . . . . . . . . . . . . . . . 10 3 Importanza in astroﬁsica 13 3.1 Le regioni HII . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.2 L’idrogeno e le atmosfere stellari . . . . . . . . . . . . . . . . . 14 3.3 L’idrogeno atomico HI . . . . . . . . . . . . . . . . . . . . . . 16 Bibliograﬁa 17 1 Introduzione L’idrogeno è l’elemento chimico più semplice, leggero e abbondante nell’uni-verso. L’atomo è composto da un nucleo, nella maggior parte dei casi formato da un unico protone o al più da un protone e un neutrone (che formano l’iso-topo meno stabile detto deuterio), e da un elettrone che orbita attorno al nucleo. Per tale motivo viene classiﬁcato come il primo elemento della tavola periodica, con simbolo H e con numero atomico pari ad 1 (Z = 1) e stesso nu-mero di massa (o numero di massa pari a 2 per il deuterio A = 2). Dal punto di vista isotopico l’idrogeno è composto per il 99.985% da prozio (idrogeno con A=1) e per il 0,015% da deuterio (A=2). Tutti gli altri isotopi sono instabili e meno abbondanti in natura. Data la sua semplicità l’idrogeno è il primo elemento formatosi dopo il Big Bang e da ciò ne deriva la sua abbondanza nell’universo e dunque la sua importanza in astroﬁsica. Attualmente si stima che l’universo sia composto circa per il 70% di idrogeno (ionizzato, neutro o molecolare) e quindi lo studio di tale elemen-to, del suo comportamento e del suo spettro fornisce informazioni peculiari riguardanti la natura dell’universo stesso. Lo spettro dell’idorgeno (come tutti gli spettri atomici) è discontinuo e solcato da righe di assorbimento o emissione che forniscono informazioni ﬁsiche riguardanti la regione da cui proviene una tale radiazione (densità, temperature, velocità...). Particolar-mente interessanti sono gli studi riguardanti la radiazione proveniente da regioni HII (regioni ad idrogeno caldo ionizzato), generalmente legate alla formazione stellare, o da regioni HI (idrogeno neutro), particolarmente utili per studiare la dinamica delle galassie a spirale. Inoltre le reazioni tra nuclei di idrogeno sono il motore fondamentale che regola la vita delle stelle mentre la molecola di idrogeno H2, come l’atomo omonimo, è la molecola più diﬀusa nell’universo. Nel primo capitolo viene spiegata la prima trattazione teorica del modello atomico dell’idrogeno, ovvero l’atomo di Bohr, e le principali righe e serie di assorbimento ed emissione. Il secondo capitolo fa riferimento alla teoria delle transizioni permesse e proibite per questo atomo. Il terzo riguarda la sua presenza e importanza nell’ambiente astroﬁsico. 2 Capitolo 1 Righe e serie dell’idrogeno e l’atomo di Bohr Il modello dell’atomo di Bohr è uno dei primi modelli semi-classici, riguardan-ti la quantizzazione dell’energia, che descrive la struttura atomica. Sebbene sia un modello ormai superato e inesatto quantisticamente, fornisce comunque dei risultati giusti per quanto riguarda l’atomo di idrogeno ed ha aperto la strada a tutta la formulazione della meccanica quantistica. 1.1 Le serie dell’idrogeno L’idea di Bohr nasce a seguito dello studio dello spettro di radiazione di un gas di idrogeno. Lo studio degli spettri atomici (la spettroscopia) aveva messo in evidenza il presentarsi di una radiazione non continua, ma solcata da righe luminose e righe scure corrispondenti a determinate frequenze (o lunghezze d’onda) a seconda del materiale emittente osservato. In particolare nel 1885 il ﬁsico svizzero Balmer, osservando la radiazione emessa da un gas di idrogeno opportunamente scaldato, scoprì che le righe osservate avevano frequenze date dalla serie: νn = cR 1 4 −1 n2 con n = 3, 4, 5... (1.1) che venne poi in seguito battezzata serie di Balmer. Nella formula (1.1) n è un numero intero positivo, R è la costante di Rydberg che vale R = 109677.576 cm−1, νn è la frequenza dell’ennesima riga e c è la velocità della luce, dal valore di c ≈3 ∗1010 cm s . 3 4CAPITOLO 1. RIGHE E SERIE DELL’IDROGENO E L’ATOMO DI BOHR Altra peculiarità delle righe di questa serie è che appartengono alla ra-diazione visibile (380 nm ≤λ ≤760 nm) e vennero catalogate secondo questa classiﬁcazione: • H-α per n = 3, λ = 656.3 nm • H-β per n = 4, λ = 486.1 nm • H-γ per n = 5, λ = 434.0 nm, ecc. Poco tempo dopo vennero trovate altre serie di righe dell’idrogeno, come la serie di Lymann che cade nell’ultravioletto: νn = cR 1 −1 n2 con n = 2, 3, 4... (1.2) Quella di Paschen nell’infrarosso: νn = cR 1 9 −1 n2 con n = 4, 5, 6... (1.3) Quella di Brackett: νn = cR 1 16 −1 n2 con n = 5, 6, 7... (1.4) E quella di Pfound: νn = cR 1 25 −1 n2 con n = 6, 7, 8... (1.5) Fino ad ottenere una regola generale valida per tutte le serie: νm,n = cR 1 m2 −1 n2 con m = 1, 2, 3... n = m + 1, m + 2... (1.6) la quale però mancava di una valida interpretazione ﬁsica. 1.2 L’atomo di Bohr Una possibile interpetazione ﬁsica del fenomeno venne fornita dal ﬁsico danese Niels Bohr nel 1913. Secondo l’idea di Bohr l’atomo di idrogneo sarebbe composto da un nucleo centrale carico positivamente (il protone) e da un 1.2. L’ATOMO DI BOHR 5 elettrone che gli orbita attorno a seguito del campo elettrico centrale prodot-to dal protone (un tipico problema a due corpi). Particolarmente innovativi erano i postulati introdotti da Bohr riguardo una tale struttura: • L’atomo ha un insieme discreto di livelli di energia con energia E1, E2, E3... • Un elettrone in un livello energetico descrive un’orbita circolare senza perdere energia, contraddicendo la meccanica classica. • Le orbite stazionarie vengono determinate dalla condizione che il modu-lo del momento angolare orbitale (l) dell’elettrone sia quantizzato, cioè sia un multiplo intero di ¯ h (costante di Planck divisa per 2π). • Nel passaggio da un’orbita stazionaria (m) ad un’altra (n) l’elettrone assorbe o emette un fotone di energia hν = \|Em −En\| Dunque ogni orbita ha una sua energia e un suo momento angolare dato da l = n¯ h. L’elettrone compie queste orbite senza perdere energia e po-tendo scambiare solo quantità discrete di quest’ultima per cambiare distan-za dal centro, contrariamente a quanto aﬀerma la meccanica classica che non prevede una quantizzazione dell’energia. Le righe osservate negli spettri atomici corrispondono alle transizioni elettroniche da un livello all’altro. Dalla conservazione dell’energia meccanica, della quantizzazione del mo-mento angolare e dalla condizione di orbita stabile: En = Tn + Vn = 1 2mev2 e −e2 re vn = ¯ h mern mev2 n = e2 rn (1.7) si perviene all’equazione (1.6) che descrive le righe dell’idrogeno: νm,n = Em −En h = cR 1 m2 −1 n2 (1.8) Dunque le frequenze osservate nello spettro di emissione dell’idrogeno sono tutte e sole quelle corrispondenti ai salti da un’orbita n-esima ad un’orbita m-esima. Le energie degli stati legati, dove l’elettrone è in orbita attorno al nucleo, sono tutte negative mentre il livello zero di energia è posto pari all’energia di fuga dell’elettrone dall’atomo. Energie positive rappresentano invece elettroni liberi dal nucleo. Nel caso dell’idrogeno l’energia di ioniz-zazione è pari a Eion = 13.6 eV . Con questo modello Bohr era riuscito a spiegare molto bene la natura degli spettri a righe degli atomi, sebbene il modello stesso fosse incomple-to, parzialmente incorretto e ancora lungi dall’essere pienamente quantistico. Capitolo 2 La teoria delle transizioni di riga Secondo il modello atomico di Bohr l’elettrone dell’atomo di idrogeno orbita attorno al nucleo su orbite circolari senza emettere radiazione. L’emissione o assorbimento di radiazione avviene solo nella transizione elettronica da un’orbita ad un’altra. Ogni orbita corrisponde ad un determinato valore di energia e il complesso forma uno spettro discreto di livelli energetici che vengono identifcati mediante opportuni numeri quantici. In questo capitolo si parlerà dei livelli energetici e della teoria di transizione con particolari riferimenti al caso dell’idrogeno. Figura 2.1: Transizioni tra livelli energetici 6 2.1. I LIVELLI ENERGETICI E I NUMERI QUANTICI 7 2.1 I livelli energetici e i numeri quantici Il sistema dell’atomo di idrogeno rappresenta un tipico problema a due corpi (risolvibile con soluzione esatta analiticamente), che può essere trattato come un sistema a massa ridotta in cui il nucleo centrale è il punto rispetto al quale viene studiato il moto dell’elettrone. Il sistema a massa ridotta dell’atomo di idrogeno può essere descritto in termini di funzioni d’onda sferiche Ψ che soddisfano l’equazione di Schrödinger indipendente dal tempo: HΨ = EΨ (2.1) dove H è l’operatore Hamiltoniano ed E è l’energia e le funzioni d’onda sferiche possono essere separate in funzioni angolari e radiali: Ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ) Gli autovalori o i livelli di energia associati alla funzione d’on-da che soddisfa l’equazione di Schrödinger sono quantizzati da tre numeri quantici (n, l, m) che danno luogo in una serie di livelli di energia discreti determinati solo dalla variabile n, che per l’idrogeno sono: E = −13.6Z2 n2 (2.2) Questi livelli di energia sono detti degeneri poiché la diﬀerenza di energia tra livelli con stesso n è generalmente trascurabile. Solo in atomi complessi con molti elettroni l’interazione spin-orbita di quest’ultimi porta ad una se-parazione tra livelli con stesso n e diverso l (struttura ﬁne). La stessa cosa avviene nel caso in cui gli atomi sono immersi in un forte campo magnetico che porterebbe alla separazione dei livelli con stesso n ma diverso m (ef-fetto Zeeman). Tuttavia l’idrogeno presenta generalmente livelli fortemente degeneri. I numeri atomici sono quindi utilizzati per descrivere i livelli energetici e sono generalmente associati a quantità ﬁsiche: • n è il numero quantico principale, assume valori interi da 1 ad ∞e deﬁnisce l’energia e le dimensioni delle orbite: an = a0 n2 Z con a0 = 0.53 ˚ A ovvero il raggio atomico di Bohr. • l è il numero quantico azimutale, quantizza il modulo quadro del mo-mento angolare orbitale degli elettroni attorno al nucleo e assume valori interi positivi da 0 ﬁno a n-1. • m è il numero quantico magnetico che descrive la componente lungo un asse del momento angolare. Assume valori interi da -l a +l e siccome in simmetria sferica non esistono direzioni privilegiate, livelli energetici con stesso n e diverso m hanno praticamente stessa energia. 8 CAPITOLO 2. LA TEORIA DELLE TRANSIZIONI DI RIGA • s è il numero quantico di spin che quantizza il momento angolare in-trinseco di spin degli elettroni, grandezza quantistica che non ha un analogo classico. Per gli elettroni vale 1 2. • ms è il numero quantico magnetico di spin che è associato ad una componente dello spin. Per gli elettroni assume valori −1 2 e + 1 2. Inoltre i vari momenti angolari (orbitali e di spin) possono essere combinati tra loro per dare momenti angolari totali (J) per i vari atomi. 2.2 Transizioni permesse e coeﬃcienti di Ein-stein Per capire i tipi di transizione permessi tra livelli atomici basta considerare il caso semplice di due livelli atomici. L’elettrone che passa da un livello più esterno al nucleo (livello eccitato ad energia più alta) ad un livello più interno emette una radiazione con energia pari alla diﬀerenza di energia tra i due livelli. Viceversa assorbe radiazione per passare in un’orbita più esterna (ad energia più alta). L’intensità di emissione di radiazione da parte di una carica elettrica accelerata, vincolata in una regione di dimensioni dell’ordine di a ≈10−8 cm, è data da una formula che può essere sviluppata in termini di dipolo, quadrupolo, ecc... elettrici: I = 2e2 3c3 d2x dt2 !2 + O (ka)4 (2.3) valida se ka ≪1 che permette di trattare la transizione stessa come se fosse un oscillatore legato. Essendo a dell’ordine del ˚ A e la lunghezza d’onda ge-neralmente nel visibile (o UV) i termini superiori al dipolo sono spesso trascu-rati e le righe permesse (più probabili e intense) sono dovute a transizioni di dipolo. Le transizioni tra livelli sono descritte in termini di probabilità di tran-sizione mediante i coeﬃcienti di Einstein; la forza di transizione è misurata dalla probabilità che un atomo eccitato emetta un fotone nell’unità di tempo. Per esempio se ci sono N(0) atomi nello stato eccitato al tempo t = 0 e la probabilità di transizione nell’unità di tempo è A (s−1) allora ad un istante t generico gli atomi eccitati sono: N(t) = N(0)e−At. La probabilità (s−1) di emissione spontanea di un fotone di un atomo eccitato in una transizione tra un livello m ad un livello n (con m > n) è: Amn = 64π4 3hc3 ! ν3 mnSmn gm [s−1] (2.4) 2.2. TRANSIZIONI PERMESSE E COEFFICIENTI DI EINSTEIN 9 Amn prende il nome di coeﬃciente di Einstein di emissione spontanea, che dipende fortemente dalla frequenza di transizione ν3 mn; gm è il peso statistico di un livello energetico (il numero di stati quantici per ogni livello) e Smn è la forza della riga espressa in termini dell’elemento di matrice di dipolo elettrico mediante la relazione: µ2 mn = Smn gm = e2\|⟨Ψn\|r\|Ψm⟩\|2 (2.5) Più A è grande più la transizione è probabile, quindi se l’elettrone ha più livelli in cui decadere predilige quelli con salto maggiore ovvero con grande diﬀerenza di frequenza, data la dipendenza di A dal cubo di ν. Gli elementi della matrice di dipolo elettrico esprimono la probabilità di transizione da un livello ad un altro e possono essere interpretati come forza dell’oscillatore per questa transizione. Se tali elementi sono nulli la transizione si dice proibita. Oltre all’emissione sponteanea un elettrone può assorbire un fotone di opportuna frequenza per passare ad un’orbita più esterna. La probabilità [s-1] di assorbimento di un opportuno fotone per la transizione è BnmU(νnm) dove U(νnm) è la densità del campo di radiazione e Bnm è il coeﬃciente di Einstein per l’assorbimento, legato ad Amn attraverso la relazione: Amn = 8πh c3 ! ν3 nmBnm [s−1] (2.6) Per simmetria segue che: gnBnm = gmBmn (2.7) dove BmnU(νnm) è la probabilità che un fotone di opportuna frequenza sti-moli l’emissione di un altro fotone, coerente al primo, da parte di un atomo eccitato. Per questo Bmn è detto coeﬃciente di Einstein di emissione sti-molata. Quando un fotone ha un’energia pari alla diﬀerenza tra due livelli atomici, il più alto dei quali è occupato, allora la probabilità di emissione stimolata risulta massima e a ﬁne processo si ha una radiazione ampliﬁcata e diseccitazione atomica. All’equilibrio termico e con una radiazione incidente opportuna, il rapporto tra le popolazioni di due livelli atomici dice quale fenomeno, tra l’assorbimento e l’emissione stimolata, prevale. Per concludere questa sezione è bene riportare quelle che sono le regole di selezione delle transizioni per le righe permesse. Tali regole derivano dal-la meccanica quantistica e tengono conto sia dei principi di conservazione (energia, momento angolare, ecc..) sia della probabilità di transizione per l’approssimazione di dipolo. Queste sono: 10 CAPITOLO 2. LA TEORIA DELLE TRANSIZIONI DI RIGA Regole Condizioni ∆n = arbitrario ∆n ̸= 0 ∆l = ±1 cambio di parità ∆L = 0, ±1 L = 0 ⇒L = 0 proibita ∆J = 0, ±1 J = 0 ⇒J = 0 proibita ∆S = 0 Figura 2.2: Transizioni tra due livelli nei tre casi possibili 2.3 Le righe proibite e la riga 21 cm Come detto nella sezione precedente, il fatto che una transizione sia detta proibita non implica che non avvenga o che non la si possa ossservare. Infat-ti, per quanto più deboli e improbabili, si possono avere anche transizioni di quadrupolo, di dipolo magnetico, ecc... che portano a particolari osservazioni spesso rilevanti in astroﬁsica. In questi due casi le regole di selezione sono diverse dal caso del dipolo. Per il quadrupolo elettrico si ha: Regole Condizioni ∆n = arbitrario ∆n ̸= 0 ∆l = 0, ±2 cambio di parità ∆L = 0, ±1, ±2 L = 0 ⇒L = 0 proibita ∆J = 0, ±1, ±2 J = 0 ⇒J = 0 proibita ∆S = 0 E per il dipolo magnetico: 2.3. LE RIGHE PROIBITE E LA RIGA 21 CM 11 Regole Condizioni ∆n, ∆l = 0 ∆J = 0, ±1 J = 0 ⇒J = 0 proibita ∆S, ∆L = 0 Inoltre il rapporto tra i coeﬃcienti di Einstein nei due casi citati (AQ, AM) rispetto al dipolo (AD) fornisce la predominanza in termini probabilistici del termine che dà la transizione: AQ AD ≈3 ∗10−8 AM AD ≈5 ∗10−5 (2.8) Dal punto di vista dell’idrogeno la transizione di dipolo magnetico è la più interessante. Infatti l’idrogeno neutro (temperature che vanno da 101 a 104) si trova generalmente nel suo stato fondamentale e quindi non è in grado di emettere, e assorbe solo la radiazione dell’ordine del UV (serie di Lymann) o superiore. Tuttavia il suo stato fondamentale consiste di due livelli iperﬁni che corrispondono alle conﬁgurazioni con spin dell’elettrone e del protone paralleli (livello superiore n1) e antiparalleli (livello inferiore n0) e con dif-ferenza di energia, tra i due, corrispondente ad una lunghezza d’onda pari a λ = 21 cm. Il rapporto tra i pesi statistici delle due popolazioni dei livelli è: n2 n1 = 3. La probabilità di transizione tra i due livelli è fortemente proibita con coeﬃciente di emissione spontanea pari a: AH = 2.87 ∗10−15 [s−1], che corrisponde ad una vita media per il livello superiore di τ = 1 AH = 107 anni. Tale riga risulta dunque inosservabile in laboratorio, ma non in ambiente astroﬁsico in cui giocano un ruolo fondamentale l’abbondanza di idrogeno e la densità. Le collisioni tra atomi accorciano il tempo di transizione a 300 anni, quindi il rapporto x tra il tempo collisionale e quello radiativo è dato da: x = a ∗10−7T nH ≈4 ∗10−3 (2.9) per valori di T = 100 K e nH = 10 cm-3 ovvero c’è un prevalenza delle transizioni collisionali. Inoltre è possibile deﬁnire una densità di colonna per l’idrogneo NHI come il prodotto tra lo spessore della nube di idrogeno (l) e la densità di atomi lungo la linea di vista (nH) e una temperatura di brillanza (TBH) data da: TBH(ν) = Ts(1 −e−τH(ν)) ≈TsτH(ν) se τH ≪1 (2.10) dove τH(ν) è lo spessore ottico ad una data frequenza ν e Ts è la temperatura di spin che coincide con la temperatura cinetica (Tc) quando le collisioni sono 12 CAPITOLO 2. LA TEORIA DELLE TRANSIZIONI DI RIGA il fattore dominante nel determinare lo stato degli spin. Si dimostra che: τH(ν) = 2.58 ∗10−15NHφ(ν) Ts (2.11) con φ(ν) proﬁlo della riga. Integrando sul proﬁlo di riga si ottiene: TBH = 2.58 ∗10−15NH (2.12) ovvero, in queste condizioni, la temperatura di brillanza non dipende dalla temperatura cinetica Tc ma solo dalla densità di colonna. Quindi una misura della temperatura di brillanza fornisce una stima del numero totale di atomi di idrogeno che emettono la riga 21 cm in una certa direzione. Tale riga appartiene alla banda radio e, poiché il mezzo interstellare è completamente trasparente alla radiazione a 21 cm, è possibile esaminare in questo modo la struttura della Galassia anche in regioni che sono troppo lon-tane per potervi eﬀettuare degli studi ottici o in regioni dove l’assorbimento nell’ottico è molto elevato. Con essa è possibile persino tracciare la struttura delle galassie a spirale, la curva di rotazione e ricavare la relativa scoperta di materia oscura proprio perché la distribuzione di HI si estende ben oltre i conﬁni dell’osservazione ottica e risulta quindi comoda per gli studi dinamici delle galassie. La larghezza della riga è dovuta principalmente ai moti termici degli atomi che hanno una distribuzione di velocità di tipo Maxwelliano. Dunque, a causa del moto di agitazione termica, il proﬁlo della riga dovrebbe essere gaussiano. Questo è valido per le righe in assorbimento ma non per quelle in emissione; per questo si suppone che lo spessore ottico per l’idrogeno caldo diﬀuso sia maggiore di 1. La riga si può vedere anche in assorbimento quando una nube di idrogeno si trova interposta sulla linea di vista fra l’osservatore ed una radiosorgente con temperatura di brillanza TBS che emette con uno spettro continuo. Se la radiosorgente di sincrotrone è localizzata dietro una nube di idrogeno, alla frequenza νH la brillanza è: TB1 = TBSe−τH(ν) + Tc(1 −e−τH(ν)) (2.13) mentre in una direzione che esclude la radiosorgente ma attraversa ancora la nube è data da: TB2 = Tc(1 −e−τH(ν)) (2.14) Ad una frequenza ν leggermente diversa da νH la nube è completamente trasparente alla radiazione della sorgente di fondo e la temperatura di bril-lanza in direzione della sorgente è T ′ B1 ≈TBS. Dalla diﬀerenza fra le temperature si ha: TB1 −T ′ B1 −TB2 = −TBS(1−e−τH(ν)) da cui si può ricavare τH e quindi NH. Capitolo 3 Importanza in astroﬁsica Come accennato nell’introduzione, l’idrogeno è l’elemento più abbondante nell’universo e si trova in fase neutra, ionizzata o molecolare. Di seguito vengono presentati alcuni dei più interessanti ambienti astroﬁsici ricchi di idrogeno. 3.1 Le regioni HII L’idrogeno ionizzato costituisce l’1% della Galassia e per essere tale necessita di meccanismi che portino alla ionizzazione. Come detto in precedenza, l’e-nergia di ionizzazione è circa 13.6 eV il che signiﬁca che sono necessari fotoni con energia pari o superiore a tale valore per avere l’idrogeno ionizzato, ossia fotoni ultravioletti o più energetici (λ ≤916 ˚ A). Stelle in grado di produrre radiazione con energie simili sono generalmente stelle calde con temperatura superﬁciale ≥5 ∗104 K di tipo spettrale O e B. Quindi l’idrogeno ioniz-zato è un buon tracciante per la distribuzione delle stelle molto calde di formazione recente, le quali fungono da sorgente di radiazione per la ioniz-zazione del gas circostante. Il fotone UV ionizza l’idrogeno liberando una coppia elettrone-protone che può perdere energia: per collisione con altre particelle, per irraggiamento di Bremsstrahlung degli elettroni o per ricom-binazione della coppia. La ricombinazione porta generalmente ad un atomo eccitato che decade a livelli di energia più bassi emettendo fotoni. L’eﬀetto complessivo dà uno spettro continuo (dovuto alla Bremsstrahlung) solcato da righe (dovute alla ricombinazine e decadimento). Tuttavia le regioni ad idrogeno ionizzato spesso contengono grandi quantità di polveri che ne ren-dono diﬃcile o impossibile l’osservazione in banda visuale e modiﬁcano lo spettro. 13 14 CAPITOLO 3. IMPORTANZA IN ASTROFISICA Nelle galassie a spirale (nostra compresa) l’idrogeno ionizzato si trova sotto forma di nubi discrete (a temperature di ≈104 K), dette regioni HII, lungo i bracci di spirale nelle vicinanze di stelle di recente formazione, le quali, a volte, sono in grado di costituire complessi molto grandi e densi come la Nebulosa di Orione o la Nebulosa dell’Aquila. Le regioni HII sono classiﬁcate Warm Ionized Medium (WIM) come fase del mezzo interstellare. Dalla ricombinazione della coppia protone-elettrone è più probabile ot-tenere un atomo eccitato che non nello stato fondamentale. Dunque per ogni fotone UV che aveva originato la ionizzazione vengono prodotti almeno due fotoni secondari con energia inferiore: uno, di frequenza qualsiasi, dovuto alla cattura elettronica da parte del protone (che contribuisce all’emissione continua) e uno dovuto alla diseccitazione (che contribuisce con una riga di frequenza ben determinata dalla transizione). In questo modo i fotoni UV eccitanti vengono degradati mentre quelli secondari non sono generalmente in grado di eccitare altri atomi. Le righe di ricombinazione generalmente cadono e vengono studiate nell’ottico, come la serie di Balmer, o nell’UV, come la serie di Lymann, che però non è osservabile da Terra a causa dell’opacità at-mosferica. Tuttavia la ricombinazione può produrre anche righe nella banda radio se l’atomo si diseccita mediante transizioni fra livelli energetici elevati e quindi aventi piccola diﬀerenza di energia. Inﬁne lo studio del rapporto fra l’intensità delle righe di ricombinazione e l’emissione termica del continuo fornisce una misura del rapporto fra elettroni legati ed elettroni liberi nel gas e permette quindi di analizzare le variazioni della temperatura delle regioni HII nella Galassia. Gli spettri ottici delle regioni HII, conosciuta la temperatura, permettono di calcolare la metal-licità che risulta decrescere dal centro della Galassia verso la periferia. Ciò suggerisce che nelle regioni periferiche si ha minor formazione stellare. Il rap-porto He+ H+ , invece, è uniforme su tutta la Galassia, sintomo che la quantità di elio non è stata sensibilmente alterata dai processi di formazione stellare, e quindi è presumibilmente rappresentativa di quella primordiale. 3.2 L’idrogeno e le atmosfere stellari L’idrogeno è anche l’elemento principale che costituisce le stelle; è il car-burante utilizzato per le reazioni termonucleari nella prima fase evolutiva del nucleo stellare ed è abbondante nell’atmosfera stessa. Per tale motivo le prime classiﬁcazioni spettrali delle stelle vennero costruite valutando la presenza e intensità delle righe di assorbimento dell’idrogeno nello spettro continuo di corpo nero delle stelle. La classiﬁcazione attuale si basa sulla temperatura superﬁciale della stella, la quale inﬂuisce sulle righe di assorbi-3.2. L’IDROGENO E LE ATMOSFERE STELLARI 15 mento degli elementi nell’atmosfera, agendo sull’allargamento delle righe e determinando il grado di eccitazione e ionizzazione degli elementi (descritto dall’equazione di Boltzmann e di Saha). Inoltre la temperatura determina il massimo di emissione nel proﬁlo di corpo nero. In particolare è possibile valutare la temperatura stellare osservando le righe dello spettro e sapendo che esse dipendono dallo stato di eccitazione atomico. Per esempio, se la temperatura dell’atmosfera di una stella è tale da portare gli atomi di idrogeno al primo stato eccitato, lo spettro continuo subisce un calo di intensità, detto Balmer Jump, nelle lunghezze del visibile a seguito dell’assorbimento dei fotoni visibili da parte dell’idrogeno stesso. Dall’intensità del Balmer Jump è possibile ricavare la frazione di idrogeno nel primo livello eccitato e dunque la temperatura (mediante la formula di Boltzamnn). Figura 3.1: Un esempio di spettro con Balmer Jump (Corso di Astroﬁsica Stellare, Prof. Ferraro) Stelle di tipo A ed F presentano forti righe dell’idrogeno mentre stelle più calde di 104 K lo ionizzano impedendone l’assorbimento di riga e stelle troppo fredde non sono in grado di eccitarlo in quantità rilevante. 16 CAPITOLO 3. IMPORTANZA IN ASTROFISICA 3.3 L’idrogeno atomico HI Caratteristico di due fasi del mezzo interstellare (il Warm Neutral Medium [WNM] e il Cold Neutral Medium [CNM]), l’idrogeno atomico, citato nel paragrafo 2.3, è diﬀuso più o meno ovunque sul piano galattico delle galassie a spirale, in regioni in cui la temperatura oscilla tra 101 K e 102 K (CNM) e tra 103 K e 104 (WNM). Presenta bassa densità nel WNM pari a 0.1 cm-3 e superirore nel CNM dove oscilla tra i 1 e 10 cm-3 ed è quest’ultima a fare da discriminante per la temperatura del mezzo: regioni con densità di colonna superiore ad un certo valore critico (NHI = 2 ∗1020 cm−2) sono in grado di raﬀreddarsi e diventare CNM, se invece hanno NHI inferiore non ci riescono. Una possi-bile spiegazione di questo fatto puramente sperimentale potrebbe derivare dalla autoschermatura da parte degli atomi stessi. I procesi di riscaldamen-to del HI sono principalmente dovuti alla radiazione UV e X diﬀusa, per quanto riguarda il WNM. Dunque, a basse densità di colonna, i fotoni UV possono penetrare nelle nubi e riscaldarle e, volendo, ionizzare l’idrogeno. Solo le nubi che riescono ad autoschermarsi contro la radiazione UV eﬃcen-temente possono rimanere a basse temperature e fare parte del CNM. Per poter fare ciò necessitano di una densità di colonna che è stata misurata sperimentalmente (mediante il Westerbork Synthesis Radio Telescope) pari a NHI = 2 ∗1020 cm−2. Al di sotto di tale valore suﬃcienti fotoni UV pe-netrano negli interni delle nubi e inibiscono la formazione del CNM. Tuttavia questa scoperta e relativa ipotesi di spiegazione sono piuttosto recenti e da sviluppare. Sicuramente il ruolo schermante da parte delle nubi è di vitale importanza soprattutto per la formazione delle molecole, che verrebbero distrutte dalla radiazione UV se non fossero immerse nel CNM. Tra queste troviamo la molecola più diﬀusa, l’idrogeno molecolare (H2). Su tale molecola ci sarebbe molto da dire ma è ancora in fase di studio intenso, proprio perché non è sempre possibile studiarla direttamente; l’idrogeno molecolare infatti irradia debolmente nell’UV (non osservabile da terra) per transizioni elettroniche e nell’infrarosso per transizioni roto-vibrazionali. Risulta più facile studiare la molecola del CO che fa da tracciante dell’idrogeno molecolare in quanto la sua principale sorgente di eccitazione è costituita proprio dalle collisioni con l’H2. Tuttavia, con l’avvento dello studio della radiazione ad IR mediante satelliti in orbita, l’idrogeno molecolare potrebbe aprire un nuovo grande futuro sulle osservazioni astroﬁsiche. Bibliograﬁa M. A. Dopita, R. S. Sutherland, Astrophysics of the Diﬀuse Universe, Springer, 2003. G. B. Rybicki, A. P. Lightman, Radiative Processes in Astrophysics, WILEY-VCH, 2004. H. Bradt. Astrophysics Processes, Cambridge University Press, 2008. C. Fanti, R. Fanti, Lezioni di Radioastronomia, Appunti, 2012. D. Dallacasa, Processi di radiazione e MHD, Dispense. F. Ravanini, Istituzioni di ﬁsica teorica, Appunti. F. R. Ferraro, Astroﬁsica stellare, Dispense. N. Kanekar, R. Braun, N. Roy, An HI Column Density Threshold for Cold Gas Formation in the Galaxy, Articolo di Astrophysical Journal, 2011. 17
2930	https://courses.lumenlearning.com/suny-osalgebratrig/chapter/domain-and-range/	Domain and Range Learning Objectives In this section, you will: Find the domain of a function defined by an equation. Graph piecewise-defined functions. If you’re in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all time—I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. (Figure) shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these. Figure 1. Based on data compiled by www.the-numbers.com. Finding the Domain of a Function Defined by an Equation In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0. We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as another “holding area” for the machine’s products. See (Figure). Figure 2. We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [ when the set includes the endpoint and a parenthesis ( to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write(0, 100].We will discuss interval notation in greater detail later. Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that would make the radicand negative. Before we begin, let us review the conventions of interval notation: The smallest number from the interval is written first. The largest number in the interval is written second, following a comma. Parentheses, ( or ), are used to signify that an endpoint value is not included, called exclusive. Brackets, [ or ], are used to indicate that an endpoint value is included, called inclusive. See (Figure) for a summary of interval notation. Figure 3. Finding the Domain of a Function as a Set of Ordered Pairs Find the domain of the following function:{(2, 10),(3, 10),(4, 20),(5, 30),(6, 40)}. Show Solution First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs. {2,3,4,5,6} Try It Find the domain of the function: {(−5,4),(0,0),(5,−4),(10,−8),(15,−12)} Show Solution {−5,0,5,10,15} How To Given a function written in equation form, find the domain. Identify the input values. Identify any restrictions on the input and exclude those values from the domain. Write the domain in interval form, if possible. Finding the Domain of a Function Find the domain of the functionf(x)=x2−1. Show Solution The input value, shown by the variablexin the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers. In interval form, the domain offis(−∞,∞). Try It Find the domain of the function:f(x)=5−x+x3. Show Solution (−∞,∞) How To Given a function written in an equation form that includes a fraction, find the domain. Identify the input values. Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve forx. If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve. Write the domain in interval form, making sure to exclude any restricted values from the domain. Finding the Domain of a Function Involving a Denominator Find the domain of the functionf(x)=x+12−x. Show Solution When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve forx. 2−x=0−x=−2x=2 Now, we will exclude 2 from the domain. The answers are all real numbers wherex<2orx>2as shown in (Figure). We can use a symbol known as the union,∪,to combine the two sets. In interval notation, we write the solution:(−∞,2)∪(2,∞). Figure 4. Try It Find the domain of the function:f(x)=1+4x2x−1. Show Solution (−∞,12)∪(12,∞) How To Given a function written in equation form including an even root, find the domain. Identify the input values. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve forx. The solution(s) are the domain of the function. If possible, write the answer in interval form. Finding the Domain of a Function with an Even Root Find the domain of the functionf(x)=√7−x. Show Solution When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve forx. 7−x≥0−x≥−7x≤7 Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to7,or(−∞,7]. Try It Find the domain of the functionf(x)=√5+2x. Show Solution [−52,∞) Can there be functions in which the domain and range do not intersect at all? Yes. For example, the functionf(x)=−1√xhas the set of all positive real numbers as its domain but the set of all negative real numbers as its range. As a more extreme example, a function’s inputs and outputs can be completely different categories (for example, names of weekdays as inputs and numbers as outputs, as on an attendance chart), in such cases the domain and range have no elements in common. Using Notations to Specify Domain and Range In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities, or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation. For example,{x\|10≤x<30}describes the behavior ofxin set-builder notation. The braces{}are read as “the set of,” and the vertical bar \| is read as “such that,” so we would read{x\|10≤x<30}as “the set of x-values such that 10 is less than or equal tox,andxis less than 30.” (Figure) compares inequality notation, set-builder notation, and interval notation. Figure 5. To combine two intervals using inequality notation or set-builder notation, we use the word “or.” As we saw in earlier examples, we use the union symbol,∪,to combine two unconnected intervals. For example, the union of the sets{2,3,5} and{4,6} is the set{2,3,4,5,6}.It is the set of all elements that belong to one or the other (or both) of the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending order of numerical value. If the original two sets have some elements in common, those elements should be listed only once in the union set. For sets of real numbers on intervals, another example of a union is {x\| \|x\|≥3}=(−∞,−3]∪[3,∞) Set-Builder Notation and Interval Notation Set-builder notation is a method of specifying a set of elements that satisfy a certain condition. It takes the form{x\|statement about x}which is read as, “the set of allxsuch that the statement aboutxis true.” For example, [latex]\left{x\|4 Interval notation is a way of describing sets that include all real numbers between a lower limit that may or may not be included and an upper limit that may or may not be included. The endpoint values are listed between brackets or parentheses. A square bracket indicates inclusion in the set, and a parenthesis indicates exclusion from the set. For example, (4,12] How To Given a line graph, describe the set of values using interval notation. Identify the intervals to be included in the set by determining where the heavy line overlays the real line. At the left end of each interval, use [ with each end value to be included in the set (solid dot) or ( for each excluded end value (open dot). At the right end of each interval, use ] with each end value to be included in the set (filled dot) or ) for each excluded end value (open dot). Use the union symbol∪to combine all intervals into one set. Describing Sets on the Real-Number Line Describe the intervals of values shown in (Figure) using inequality notation, set-builder notation, and interval notation. Figure 6. Show Solution To describe the values,x,included in the intervals shown, we would say, “xis a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.” \| \| \| --- \| \| Inequality \| 1≤x≤3orx>5 \| \| Set-builder notation \| {x\|1≤x≤3orx>5} \| \| Interval notation \| [1,3]∪(5,∞) \| Remember that, when writing or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set. Try It Given (Figure), specify the graphed set in words set-builder notation interval notation Figure 7. values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3; {x\|x≤−2or−1≤x<3} ; (−∞,−2]∪[−1,3) Finding Domain and Range from Graphs Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See (Figure). Figure 8. We can observe that the graph extends horizontally from−5to the right without bound, so the domain is[−5,∞).The vertical extent of the graph is all range values5and below, so the range is(−∞,5].Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range. Finding Domain and Range from a Graph Find the domain and range of the functionf whose graph is shown in (Figure). Figure 9. Show Solution We can observe that the horizontal extent of the graph is –3 to 1, so the domain off is(−3,1]. The vertical extent of the graph is 0 to –4, so the range is[−4,0).See (Figure). Figure 10. Finding Domain and Range from a Graph of Oil Production Find the domain and range of the functionfwhose graph is shown in (Figure). Figure 11. (credit: modification of work by the U.S. Energy Information Administration) Show Solution The input quantity along the horizontal axis is “years,” which we represent with the variabletfor time. The output quantity is “thousands of barrels of oil per day,” which we represent with the variablebfor barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as1973≤t≤2008and the range as approximately180≤b≤2010. In interval notation, the domain is [1973, 2008], and the range is about [180, 2010]. For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines. Try It Given (Figure), identify the domain and range using interval notation. Figure 12. Show Solution domain =[1950,2002] range = [47,000,000,89,000,000] Can a function’s domain and range be the same? Yes. For example, the domain and range of the cube root function are both the set of all real numbers. Finding Domains and Ranges of the Toolkit Functions We will now return to our set of toolkit functions to determine the domain and range of each. Figure 13. For the constant functionf(x)=c,the domain consists of all real numbers; there are no restrictions on the input. The only output value is the constantc,so the range is the set{c}that contains this single element. In interval notation, this is written as[c,c],the interval that both begins and ends withc. Figure 14. For the identity functionf(x)=x, there is no restriction on x. Both the domain and range are the set of all real numbers. Figure 15. For the absolute value functionf(x)=\|x\|,there is no restriction onx.However, because absolute value is defined as a distance from 0, the output can only be greater than or equal to 0. Figure 16. For the quadratic functionf(x)=x2,the domain is all real numbers since the horizontal extent of the graph is the whole real number line. Because the graph does not include any negative values for the range, the range is only nonnegative real numbers. Figure 17. For the cubic functionf(x)=x3,the domain is all real numbers because the horizontal extent of the graph is the whole real number line. The same applies to the vertical extent of the graph, so the domain and range include all real numbers. Figure 18. For the reciprocal functionf(x)=1x,we cannot divide by 0, so we must exclude 0 from the domain. Further, 1 divided by any value can never be 0, so the range also will not include 0. In set-builder notation, we could also write{x\| x≠0},the set of all real numbers that are not zero. Figure 19. For the reciprocal squared functionf(x)=1x2,we cannot divide by 0, so we must exclude 0 from the domain. There is also no x that can give an output of 0, so 0 is excluded from the range as well. Note that the output of this function is always positive due to the square in the denominator, so the range includes only positive numbers. Figure 20. For the square root functionf(x)=√x,we cannot take the square root of a negative real number, so the domain must be 0 or greater. The range also excludes negative numbers because the square root of a positive numberxis defined to be positive, even though the square of the negative number−√xalso gives usx. Figure 21. For the cube root functionf(x)=3√x,the domain and range include all real numbers. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function). How To Given the formula for a function, determine the domain and range. Exclude from the domain any input values that result in division by zero. Exclude from the domain any input values that have nonreal (or undefined) number outputs. Use the valid input values to determine the range of the output values. Look at the function graph and table values to confirm the actual function behavior. Finding the Domain and Range Using Toolkit Functions Find the domain and range off(x)=2x3−x. Show Solution There are no restrictions on the domain, as any real number may be cubed and then subtracted from the result. The domain is(−∞,∞)and the range is also(−∞,∞). Finding the Domain and Range Find the domain and range off(x)=2x+1. Show Solution We cannot evaluate the function at−1because division by zero is undefined. The domain is(−∞,−1)∪(−1,∞).Because the function is never zero, we exclude 0 from the range. The range is(−∞,0)∪(0,∞). Finding the Domain and Range Find the domain and range off(x)=2√x+4. Show Solution We cannot take the square root of a negative number, so the value inside the radical must be nonnegative. x+4≥0 when x≥−4 The domain off(x)is[−4,∞). We then find the range. We know thatf(−4)=0,and the function value increases asxincreases without any upper limit. We conclude that the range offis[0,∞). Analysis (Figure) represents the functionf. Figure 22. Try It Find the domain and range off(x)=−√2−x. Show Solution domain:(−∞,2];range:(−∞,0] Graphing Piecewise-Defined Functions Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value functionf(x)=\|x\|.With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0. If we input 0, or a positive value, the output is the same as the input. f(x)=xifx≥0 If we input a negative value, the output is the opposite of the input. f(x)=−xifx<0 Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function. A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain. We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain “boundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to $10,000 are taxed at 10%, and any additional income is taxed at 20%. The tax on a total incomeSwould be0.1SifS≤$10,000and$1000+0.2(S−$10,000)ifS>$10,000. Piecewise Function A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this: f(x)={formula 1 if x is in domain 1formula 2 if x is in domain 2formula 3 if x is in domain 3 In piecewise notation, the absolute value function is \|x\|={x if x≥0−x if x<0 How To Given a piecewise function, write the formula and identify the domain for each interval. Identify the intervals for which different rules apply. Determine formulas that describe how to calculate an output from an input in each interval. Use braces and if-statements to write the function. Writing a Piecewise Function A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed $50 fee for a group of 10 or more people. Write a function relating the number of people,n,to the cost,C. Show Solution Two different formulas will be needed. For n-values under 10,C=5n.For values ofnthat are 10 or greater,C=50. [latex]C\left(n\right)=\left{\begin{array}{ccc}5n& \text{if}& 0 Analysis The function is represented in (Figure). The graph is a diagonal line fromn=0ton=10and a constant after that. In this example, the two formulas agree at the meeting point wheren=10,but not all piecewise functions have this property. Figure 23. Working with a Piecewise Function A cell phone company uses the function below to determine the cost,C,in dollars forggigabytes of data transfer. [latex]C\left(g\right)=\left{\begin{array}{ccc}25& \text{if}& 0 Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data. Show Solution To find the cost of using 1.5 gigabytes of data,C(1.5),we first look to see which part of the domain our input falls in. Because 1.5 is less than 2, we use the first formula. C(1.5)=$25 To find the cost of using 4 gigabytes of data,C(4),we see that our input of 4 is greater than 2, so we use the second formula. C(4)=25+10(4−2)=$45 Analysis The function is represented in (Figure). We can see where the function changes from a constant to a shifted and stretched identity atg=2.We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain. Figure 24. How To Given a piecewise function, sketch a graph. Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do not graph two functions over one interval because it would violate the criteria of a function. Graphing a Piecewise Function Sketch a graph of the function. [latex]f\left(x\right)=\left{\begin{array}{ccc}{x}^{2}& \text{if}& x\le 1\ 3& \text{if}& 12\end{array}[/latex] Show Solution Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality. (Figure) shows the three components of the piecewise function graphed on separate coordinate systems. Figure 25. (a)f(x)=x2 if x≤1;(b)f(x)=3 if 1< x≤2;(c)f(x)=x if x>2 Now that we have sketched each piece individually, we combine them in the same coordinate plane. See (Figure). Figure 26. Analysis Note that the graph does pass the vertical line test even atx=1andx=2because the points (1,3) and (2,2) are not part of the graph of the function, though (1,1) and (2,3) are. Try It Graph the following piecewise function. [latex]f\left(x\right)=\left{\begin{array}{ccc}{x}^{3}& \text{if}& x<-1\ -2& \text{if}& -14\end{array}[/latex] Show Solution Can more than one formula from a piecewise function be applied to a value in the domain? No. Each value corresponds to one equation in a piecewise formula. Access these online resources for additional instruction and practice with domain and range. Domain and Range of Square Root Functions Determining Domain and Range Find Domain and Range Given the Graph Find Domain and Range Given a Table Find Domain and Range Given Points on a Coordinate Plane Key Concepts The domain of a function includes all real input values that would not cause us to attempt an undefined mathematical operation, such as dividing by zero or taking the square root of a negative number. The domain of a function can be determined by listing the input values of a set of ordered pairs. See (Figure). The domain of a function can also be determined by identifying the input values of a function written as an equation. See (Figure), (Figure), and (Figure). Interval values represented on a number line can be described using inequality notation, set-builder notation, and interval notation. See (Figure). For many functions, the domain and range can be determined from a graph. See (Figure) and (Figure). An understanding of toolkit functions can be used to find the domain and range of related functions. See (Figure), (Figure), and (Figure). A piecewise function is described by more than one formula. See (Figure) and (Figure). A piecewise function can be graphed using each algebraic formula on its assigned subdomain. See (Figure). Section Exercises Verbal Why does the domain differ for different functions? Show Solution The domain of a function depends upon what values of the independent variable make the function undefined or imaginary. How do we determine the domain of a function defined by an equation? Explain why the domain off(x)=3√xis different from the domain off(x)=√x. Show Solution There is no restriction onxforf(x)=3√xbecause you can take the cube root of any real number. So the domain is all real numbers,(−∞,∞).When dealing with the set of real numbers, you cannot take the square root of negative numbers. Sox-values are restricted forf(x)=√xto nonnegative numbers and the domain is[0,∞). When describing sets of numbers using interval notation, when do you use a parenthesis and when do you use a bracket? How do you graph a piecewise function? Show Solution Graph each formula of the piecewise function over its corresponding domain. Use the same scale for thex-axis andy-axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate−∞or ∞.Combine the graphs to find the graph of the piecewise function. Algebraic For the following exercises, find the domain of each function using interval notation. f(x)=−2x(x−1)(x−2) f(x)=5−2x2 Show Solution (−∞,∞) f(x)=3√x−2 f(x)=3−√6−2x Show Solution (−∞,3] f(x)=√4−3x f(x)=√x2+4 Show Solution (−∞,∞) f(x)=3√1−2x f(x)=3√x−1 Show Solution (−∞,∞) f(x)=9x−6 f(x)=3x+14x+2 Show Solution (−∞,−12)∪(−12,∞) f(x)=√x+4x−4 f(x)=x−3x2+9x−22 Show Solution (−∞,−11)∪(−11,2)∪(2,∞) f(x)=1x2−x−6 f(x)=2x3−250x2−2x−15 Show Solution (−∞,−3)∪(−3,5)∪(5,∞) 5√x−3 2x+1√5−x Show Solution (−∞,5) f(x)=√x−4√x−6 f(x)=√x−6√x−4 Show Solution [6,∞) f(x)=xx f(x)=x2−9xx2−81 Show Solution (−∞,−9)∪(−9,9)∪(9,∞) Find the domain of the functionf(x)=√2x3−50xby: using algebra. graphing the function in the radicand and determining intervals on the x-axis for which the radicand is nonnegative. Graphical For the following exercises, write the domain and range of each function using interval notation. Show Solution domain:(2,8],range[6,8) Show Solution domain:\,\left[-4,\text{ 4],}\,range:\,\left[0,\text{ 2]} Show Solution domain:[−5, 3),range:[0,2] Show Solution domain:(−∞,1],range:[0,∞) Show Solution domain:[−6,−16]∪[16,6];range:[−6,−16]∪[16,6] Show Solution domain:[−3, ∞);range:[0,∞) For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation. f\left(x\right)=\left{\begin{array}{lll}x+1\hfill & \text{if}\hfill & x<-2\hfill \ -2x-3\hfill & \text{if}\hfill & x\ge -2\hfill \end{array} f\left(x\right)=\left{\begin{array}{lll}2x-1\hfill & \text{if}\hfill & x<1\hfill \ 1+x\hfill & \text{if}\hfill & x\ge 1\hfill \end{array} Show Solution domain:(−∞,∞) f\left(x\right)=\left{\begin{array}{c}x+1\,\,\text{if}\,\,x<0\ x-1\,\,\text{if}\,\,\,x>0\end{array} f\left(x\right)=\left{\begin{array}{ccc}3& \text{if}& x<0\ \sqrt{x}& \text{if}& x\ge 0\end{array} Show Solution domain:(−∞,∞) f\left(x\right)=\left{\begin{array}{c}{x}^{2}\text{ if }x<0\ 1-x\text{ if }x>0\end{array} [Math Processing Error] Show Solution domain:(−∞,∞) f\left(x\right)=\left{\begin{array}{ccc}x+1& \text{if}& x<1\ {x}^{3}& \text{if}& x\ge 1\end{array} [Math Processing Error] Show Solution domain:(−∞,∞) Numeric For the following exercises, given each function f,evaluate f(−3),f(−2),f(−1), and f(0). f\left(x\right)=\left{\begin{array}{lll}x+1\hfill & \text{if}\hfill & x<-2\hfill \ -2x-3\hfill & \text{if}\hfill & x\ge -2\hfill \end{array} f\left(x\right)=\left{\begin{array}{cc}1& \text{if }x\le -3\ 0& \text{if }x>-3\end{array} Show Solution f(−3)=1;f(−2)=0;f(−1)=0;f(0)=0 f\left(x\right)=\left{\begin{array}{cc}-2{x}^{2}+3& \text{if }x\le -1\ 5x-7& \text{if }x>-1\end{array} For the following exercises, given each functionf,evaluatef(−1),f(0),f(2),andf(4). f\left(x\right)=\left{\begin{array}{lll}7x+3\hfill & \text{if}\hfill & x<0\hfill \ 7x+6\hfill & \text{if}\hfill & x\ge 0\hfill \end{array} Show Solution f(−1)=−4;f(0)=6;f(2)=20;f(4)=34 f\left(x\right)=\left{\begin{array}{ccc}{x}^{2}-2& \text{if}& x<2\ 4+\|x-5\|& \text{if}& x\ge 2\end{array} f\left(x\right)=\left{\begin{array}{ccc}5x& \text{if}& x<0\ 3& \text{if}& 0\le x\le 3\ {x}^{2}& \text{if}& x>3\end{array} Show Solution f(−1)=−5;f(0)=3;f(2)=3;f(4)=16 For the following exercises, write the domain for the piecewise function in interval notation. f\left(x\right)=\left{\begin{array}{c}x+1\,\,\,\,\,\text{ if}\,\,x<-2\ -2x-3\,\,\text{if}\,\,x\ge -2\end{array} f\left(x\right)=\left{\begin{array}{c}{x}^{2}-2\,\,\,\,\,\text{ if}\,\,x<1\ -{x}^{2}+2\,\,\text{if}\,\,x>1\end{array} Show Solution domain:(−∞,1)∪(1,∞) f\left(x\right)=\left{\begin{array}{c}2x-3\ -3{x}^{2}\end{array}\,\,\begin{array}{c}\text{if}\,\,\,x<0\ \text{if}\,\,\,x\ge 2\end{array} Technology Graphy=1x2on the viewing window[−0.5,−0.1]and[0.1,0.5].Determine the corresponding range for the viewing window. Show the graphs. Show Solution window: [−0.5,−0.1]; [−0.5,−0.1]; [−0.5,−0.1]; range: [4, 100] [4, 100] [4, 100] window: [0.1, 0.5]; [0.1, 0.5]; [0.1, 0.5]; range: [4, 100] [4, 100] [4, 100] Graphy=1xon the viewing window[−0.5,−0.1]and[0.1, 0.5].Determine the corresponding range for the viewing window. Show the graphs. Extension Suppose the range of a functionfis[−5, 8].What is the range of\|f(x)\|? Show Solution [0, 8] Create a function in which the range is all nonnegative real numbers. Create a function in which the domain isx>2. Show Solution Many answers. One function isf(x)=1√x−2. Real-World Applications The heighthof a projectile is a function of the timetit is in the air. The height in feet fortseconds is given by the functionh(t)=−16t2+96t. What is the domain of the function? What does the domain mean in the context of the problem? Show Solution The domain is[0, 6];it takes 6 seconds for the projectile to leave the ground and return to the ground The cost in dollars of makingxitems is given by the functionC(x)=10x+500. The fixed cost is determined when zero items are produced. Find the fixed cost for this item. What is the cost of making 25 items? Suppose the maximum cost allowed is $1500. What are the domain and range of the cost function,C(x)? Glossary interval notation : a method of describing a set that includes all numbers between a lower limit and an upper limit; the lower and upper values are listed between brackets or parentheses, a square bracket indicating inclusion in the set, and a parenthesis indicating exclusion piecewise function : a function in which more than one formula is used to define the output set-builder notation : a method of describing a set by a rule that all of its members obey; it takes the form{x\|statement about x} Candela Citations CC licensed content, Shared previously Algebra and Trigonometry. Authored by: Jay Abramson, et. al. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at The Numbers: Where Data and the Movie Business Meet. “Box Office History for Horror Movies.” Accessed 3/24/2014 ↵ ↵ Licenses and Attributions CC licensed content, Shared previously Algebra and Trigonometry. Authored by: Jay Abramson, et. al. Provided by: OpenStax CNX. Located at: License: CC BY: Attribution. License Terms: Download for free at
2931	https://www.youtube.com/watch?v=udRQwdTBGaI	Relative motion: Example Eyere Solutions 2520 subscribers 39 likes Description 2913 views Posted: 21 May 2022 This video describes the solution to a problem regarding the motion of two vehicles moving relative to each other. Three formulations were solved, and solution was sought for acceleration, velocity and time the two vehicle collides with each other. varied conditions describing the motion geometry of the vehicles were given for the different samples. The problem is an example from the topic: systems of particles and sub-topic: independent motion of a system of particles or relative motion engineeringmechanics mechanics scienceandtechnology motion 9 comments Transcript: [Music] [Applause] [Music] hello welcome to the area solutions channel in this video we'll be looking at an example in the topic motion of severe particles and we shall be walking through a particular problem that bothers on relative motion and to do that we're going to walk through this example that involves two vehicles a and b which are driving towards each other on the adjacent sides of a load and at time t equal to zero the distance between them was 0.5 kilometers and their speeds where one of them ua is equal to 144 kilometer per hour while the other one was moving in the opposite direction at 72 kilometer per hour and they were at points pa and pb respectively given that a passes point pb 10 seconds after b was there and b passes point p 20 seconds after a was there were to determine the uniform accelerations of both vehicles then we're to find the time when the vehicle passes each other and the speed both scars we're moving at that time when the passes each other so to work through this problem let's um start by bringing that our coordinate system and say from origin along the x direction and initially the vehicle one of them was at the origin point zero and the second was at another position and then the vehicle that was at the origin was moving at a speed of 144 kilometer per house in the direction of the x-axis so we take that as positive and if we convert that to si unit we have 40 meters per second and in the same vein the second was moving in the opposite direction there's a direction away from the direction of the x-axis at 72 kilometer per hour and if we take that to sr units we get minus 20 meter per seconds mind you it is minus because it's moving in the opposite direction and the first vehicle was at zero position so we see the initial position of vehicle a is there and the distance between both vehicles is 0.5 kilometers so we can as well say the second vehicle that's vehicle b was at same point 0.5 kilometers from the origin at time zero which is also equal to 500 meters then also we said we're told that after 10 seconds vehicle a was at the point pb where vehicle b was initially so we had that that the final position of vehicle a would be 0.5 kilometers which is 500 meters while at equal to 20 seconds there it could be got to the point where vehicle a was initially at that point we know that same zero position so we have that s final that's the final position of vehicle b will be zero and that will take place when the time is 20 seconds and of course we're told to look for the acceleration component of each of the vehicle so we know there are different relations that the different equations that relates um the different variables that describe a particular motion undirected linear motion and we have that one of them is that s is equal to u plus half a t square and let's not forget we are supposed to add the initial position of the object or the particle when we are doing this analysis so we have that our f x final is equal to x initial plus u t plus half a t square or for vector a so if we input all the variables that we have into the equation we can easily evaluate our acceleration from that equation to be equal to 2 meter per second and if we do same for vector b if we bring in our equation where we have s final is equal to s initial plus u t plus half a t square and if we input all the variables that we have into the equation we can easily evaluate the acceleration for the second variable b to be minus 0.5 meter per second so by sodium we've got in our acceleration for the vehicle a to be 2 meters per second square why the acceleration of the vehicle b is minus 0.5 meter per second square that is to say the acceleration is in the direction that is away from the x axis and then also we are told to look for the time the variable passes each shoulder at that time we need to call our relations as our equation for relative motion relative position of two vehicles let's say it's x a b s a b will be equal to s b minus a and we're looking for the time that the two vehicle passes passes each other we're going to have that this particular relative position will be zero because the position of s a will be equal to the position of s b so we have s a b to be equal to zero so if we such that s s b minus s a will be equal to zero if we bring in our previous equation back into this equation for relative motion we have one singular equation that relates all the distance that relates the position of the two vehicles together so if we call in our variables that already had the velocity the position initial position and the acceleration as well as the same for vehicle b if we input them into the singular equation for relative motion and we evaluate it we'll be able to get one simple quadratic equation which we can solve using any suitable method whether the formula method or factorization whatever method is suitable to get a value for t and we have two answers one is minus 55.24 and the other is 7.24 and of course we know that time cannot be negative so we have our time t to be equal to 7.24 seconds and that's the time that the two vehicles we pass each other given the condition given the initial conditions and all the variables of the rest of the motion of their respective vehicles then finally we're told to look for the speed of that vehicles when they collided to find the speed of the vehicles we've been given we've estimated the value for time we bring in our equation that relates velocities acceleration and time and we have all the variables for the first vehicle that's a vehicle a we have our initial velocity we have our acceleration that is constant 2 meter per second square so if we use our equation we can easily work out our value of velocity putting in the value of time t equal to 7.24 seconds and we get our velocity at that time to be 54.48 meter per second if we do same for vehicle b we impute the variables that we have into the equation for velocity we can evaluate it simply to get our velocity of vector b to be minus 23.62 meter per second mind you the value of minus 23.62 indicates that the vehicle is moving in the direction that is away from the x-axis so we've gotten our velocities for vehicle a and velocity for vehicle b when the two vehicles collided against themselves against each other now i'm considering another kind of problem that looks like the previous one but with similar procedure we're going to solve this for this case the same two vectors a and b we're driving in the same direction on the same side of the road however this time around they were moving in the same direction as against the previous example that we solved during they were moving against each other we're told that at time t equal to zero the distance between them was 0.5 kilometers and their speed were 144 kilometer per hour and then 72 kilometers per hour respectively and there we have points pa and pb as well then given that a passes points pb 10 seconds after b was there and we covered the same distance after 20 seconds that is to say there could be would travel the same distance which is 0.5 kilometers after 20 seconds we're told to determine the uniform accelerations of the vehicles a and b when the vehicles passes the shoulder and the speed of both cars at that time now for this case the difference between the previous one we solved and this one is that for of course they are both moving in the same direction so for the first vehicle we have our velocity to be 40 meter per second we'll convert to this unit we have our initial position to be zero and time is equal to 10 seconds then we have our final position of the first vehicle to be 500 meters whether to look for the acceleration we just bring in our equation for calculating acceleration just like the previous case we can evaluate easily to get our value of acceleration to be 2 meter per second square however we're told that the second vehicle was moving of course because it is in the same direction this time around let's note that our velocity is going to be 20 meter per second as against the previous case that we have minus 20 meter per second because the vehicle was moving in the opposite direction of the x-axis for this case we had our initial velocity for vector b to be 20 meter per second and let's note it is positive because the vector is moving in the same direction as the s as this so it's going to be positive unlike the previous example where we had negative values because the vehicle was moving in the opposite direction of the first vehicle and we have our initial position of course it's 500 meters because both vehicles they are 500 meters away and we have the time to be 20 seconds we're told at 10 at 20 seconds this vehicle covered an equivalent distance that vehicle a covered after 10 seconds which is 500 meters so we have our final position to be 500 meters which it was initially plus 500 meters that is going to travel making the current position of vehicle b to be 1000 meters and once that is done if you call in your equation that relates initial positions to velocity time and acceleration and we input all the variables that we have we can easily evaluate our value of acceleration to be 0.5 meter per second square so for this case we have our acceleration of the equator to be 2 meter per second square by our acceleration of b to be minus to be positive 0.5 meter per second square positive means it is it is the same it is in the direction of the positive x as this so we're looking for the time the two vehicle passes the shoulder and to evaluate that we know that our relative position s b minus s a is going to be zero and of course if it is going to be zero we have the values we have the equations for s b and s a and we have the values of all variables for vehicle a and all variables for vehicle b so if we input all this into the singular equation into the equation for relative motion and evaluate we get one singular quadratic equation which you can which we can solve using any suitable method whether the formula method or by factorizing to get our time to be minus forty two point three nine and fifteen point seven three seconds clearly because time cannot be negative we have our value for time to be fifteen points seven three seconds and finally to estimate the speed of the cars when they collided we have our equation for velocity to be to be v equal to u plus 80. we have all our variables u and a and the values of time so if we input that into like into our equation we get our value for velocity to be 71.45 meter per second squared and likewise we have our variables for vehicle b if we put all that into the equation for velocity which is equal to u plus 80 and if we evaluate it we get our velocity to be 27.86 meter per second so we've gotten our velocity for vehicle a and our velocity of vehicle b also we could be given the problem in another form where the two vehicles are moving in the same direction at time t equal to zero the distance between them is 0.5 kilometers as well the equation starts moving from rest while there could be was already moving at a constant speed and they were both at point p and pb respectively given that a passes point b when pb 10 seconds after b was there were two to determine the uniform accelerations of the vehicle a when the vehicles passes the shoulder and the speed of boot vehicle the difference between this problem and the previous ones is that vehicle a starts from rest where vehicle b is moving with a constant speed so this would eventually affect the equations if eventually affect our equations and some of the variables has been changed so to work through this problem we're going to bring in all that we have for this case our initial velocity zero for vehicle a and the initial position is zero as well and we have at 10 seconds it has traveled through the 0.5 kilometers which is 500 meters and if we're to look for acceleration we bring in our equation for calculating our position at every point in time and if we input all our variables that we've already identified into the equation and simplify we get our value for acceleration to be one meter per second squared if we do same for the for vehicle b we're told that it's moving at a constant speed for this case is going to be 20 meter per second because it's in the direction of the x axis and we said because the speed is constant so our acceleration b or vector b is going to be zero so we've got in our values for accelerations for both vehicles to be one meter per second square and zero meter per second square respectively for a and b then the time the vehicle passes his shoulder is just to call in our equation for relative motion s b minus s is equal to zero because the difference in their position at that point in time when they are passing each other is going to be zero and if we input our equations for s b and s into the relative motion equation we get one singular equation so if we bring in our values for both vectors both vector a and vector b i will put them into the relative equation motion and we'll get one singular equation which we can simplify to get a quadratic equation and once we solve this quadratic equation using any suitable method maybe formula method or whatever method that is most suitable you get our values for time to be test 4.49 and minus 14.49 seconds and because time cannot be negative our value for time would be 34.49 seconds when they collided we have our time t equal to 4.49 seconds however our velocity should be v equal to u plus 80. if we bring in our values of initial velocity and acceleration for vehicle a and we'll put them into our equation for velocity and simplify we'll get our velocity to be 34.49 meter per second if we do same for for vehicle b we remember because vehicle b was moving at a constant speed so a is equal to zero meter per second and if you put that into the equation v equal to u plus 80 and simplify you get v is equal to 20 meter per second at every point in time the velocity or speed of car b is 20 meter per second because it's moving at a constant speed so so far we've seen how the problem may change based on what is given to us and then right now i want to present this practice problem for you to walk through it this this is similar to the first example we solved however for this case some of the variables were altered for this particular problem we're given the velocities we're given the accelerations of boot vehicles and we asked to look for the distance between the two vehicle pa and pb initially when the vehicle and the time when the vehicle passes each other as well as the speed of both vehicles at that time and finally the time it took for vehicle b to get to point a so these are some of the things what to look for for this practice problem so i want you to try your hands on this and just as i said it's similar to the previous example that we solved i want to thank you for watching please do well to subscribe to my channel and see
2932	https://www.studocu.com/ko/document/sungkyunkwan-university/%EC%A0%84%EC%9E%90%EA%B8%B0%ED%95%99/problems-electromagnetics-text-book-by-yeon-ho-lee-solution-chap2/53659811	Electromagnetics Problem Solutions - Chap. 2 (Course Code: EM-202) - Studocu Skip to document Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes New Home My Library AI Notes Ask AI AI Quiz Chats Recent You don't have any recent items yet. My Library Courses You don't have any courses yet. Add Courses Books You don't have any books yet. Studylists You don't have any Studylists yet. Create a Studylist Home My Library Discovery Discovery Universities High Schools Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI Premium Electromagnetics Problem Solutions - Chap. 2 (Course Code: EM-202) Electromagnetics Text Book by Yeon Ho Lee (Solution chap.2) Original title: Problems - Electromagnetics Text Book by Yeon Ho Lee (Solution chap.2) Course 전자기학 (dontknow) 82 documents University Sungkyunkwan University Academic year:2014/2015 Uploaded by: Anonymous Student Sungkyunkwan University Recommended for you 6 Midterm Exam 18 4월 2017, 문제 전자기학 Practice materials 100% (2) 73 Cheng - Field and Wave Electromagnetics 2ed solution manual 전자기학 Other 97% (91) 93 David K. Cheng - Fundamentals of Engineering Electromagnetics [Solutions Manual] (1992, Prentice Hall) - libgen 전자기학 Other 100% (10) 15 Hayt8e SM Ch3 전자기학 Other 100% (6) 29 Hayt8e SM Ch6 전자기학 Other 100% (4) Comments Please sign in or register to post comments. Report Document Students also viewed Electromagnetics Problems (Solutions) - Chapter 4 (Course Code: EM101) Electromagnetics - Chapter 6 Problem Set Solutions - AET 301 Electromagnetics Solutions for Chapter 7 - Problems & Analysis Electromagnetics Problems: Solutions for Chapter 9 (Course Code: EM101) Solutions to Electromagnetics Problems - Yeon Ho Lee (Chap. 8) Electromagnetics - Problems & Solutions for Chapter 1 (Course Code: EM101) Related documents 2020(2)II Quiz3 sol2020(2)II Quiz3 sol2020(2)II Quiz3 sol2020(2)II Quiz3 sol2020(2)II Quiz3 sol2020(2)II Quiz3 sol2020(2)II Quiz3 sol2020(2)II Quiz3 sol 2020(2)II Quiz4 sol2020(2)II Quiz4 sol2020(2)II Quiz4 sol2020(2)II Quiz4 sol2020(2)II Quiz4 sol2020(2)II Quiz4 sol2020(2)II Quiz4 sol2020(2)II Quiz4 sol2020(2)II Quiz4 sol2020(2)II Quiz4 sol Chapter 1ddfdsdffsjjsisksk ejieiejfkfkfkd ProblemsProblems solution chapter1_10 ProblemsProblems solution chapter1_10 Electromagnetics - Problem Set Solutions for Chapter 10 Preview text # Problems for Chapter 2 # 2-1 For an ellipse ( x 2 /16)  ( y 2 / 4)  1 , determine a unit tangent vector at point p 1 : (2, 3, 0)on the ellipse # by expressing the ellipse in parametric representation with a parameter # (a) t, as in cos t and sin t # (b) x, the space coordinate # Fig. 2-20 An ellipse(Problem 2-1) # ANS # (a) Parametric representation of the ellipse r  t  x ax  y ay  4 cos t ax 2 sintay (1) # Unit tangent vector is     2 2 4 sin 2 cos 4 sin 2 cos x y T t t t t        r a a a r # Point p 1 corresponds to t   / 3. The unit tangent vector at p 1 is 2 3 13 x y T    a a # a (2) # (b) Using x as parameter, the parametric representation is   2 2 1 x y x 16 y x r x  x a  y a  xa   a  y  0  (3) # Derivative of Eq. (3) with respective to x   2 / 2 16 x y x x x     # r a a (4) # Inserting the x -coordinate of point p 1 , or x  2 , in Eq. (4)   1 12 r x  a x  ay # Unit tangent vector at p 1 is 2 3 13 x y T      r a a a r # (5) # As t increases in (1), the point moves counterclockwise on the ellipse. On the other hand, as x increases in # (3), the point moves clockwise. For this reason, the signs in (2) and (5) are opposite to each other. # 2-2 A closed loop C forms a circle of radius 0, centered at a point ( x  2 , y  1 ) in the xy-plane as shown in # Fig. 2-21. Find # (a) parametric representation of C # (b) expression for dl on C # Fig. 2-21 A circle centered at point ( x  2, y 1)(Problem 2-2) # ANS # (a) r(  ) (2  0 sin ) ax  (1  0 cos )ay # (b) (0 cos ) x (0 sin ) y d d d d d d          r l a a # 2-3 Given a vector field E  ( x  y) ( 2 a x ay ), find the line integral of E from point A : ( 1,0,0) to point # B : (1,2,0)along a path (a) C 1 and (b) C 2 as shown in Fig. 2-22. # Fig. 2-22 Two paths of integration(Problem 2-3) # ANS # (a) Equation of path C 1 y  x 1 # Line integral of E along C 1               1,2,0 2 1,0, 1 2 1 0 B d A x y x y dx x dy y dz z x y dx x y dy                  C E  l a a  a a a We have used identities sin 2 t   1  cos 2t / 2and  t cos t dt  cos t t sint. 2-5 Given a vector field A  R sin  cos  aR  cos  cos  a  sina in spherical coordinates, determine the close line integral of A around path C shown in Fig. 2-23. Fig. 2-23 A closed path(Problem 2-5) ANS Closed line integral of A  C A  d l   A  d l   A  d l   A  d l A dl Rewriting it using d l  dR aR  Rd  a  R sin da in spherical coordinates                 3 60 2 30 2 30 3 60 sin cos sin sin sin cos sin sin o o o o R R R R R R R R d R dR R d R dR R d                                  A l a a a a a a a a C       Inserting proper values of R,  and  in the above equation     3 60 2 30 2 30 3 60 sin 90 cos 30 sin 3 sin 90 sin 90 cos 60 sin 2 sin 90 o o o o R o o o R R o o o R d R dR d R dR d                         C A  l   3 5 1 3 1 5 3 1 3 2 2 2 2 2 2 2 2 2 3 3 4                           2-6 Determine the line integral of a vector field A ax along an arc that is defined by R  4 ,   30 oand 0     / 2in spherical coordinates. ANS Differential length vector along the arc is d l  R sin d a Line integral of A is   /2 / 0 4 sin 30 2 0 sin 2 o d x d d    C A  l   a   a       # 2-7 Find the area of a circular strip shown in Fig. 2-24. # Fig. 2-24 A circular strip (Problem 2-7) # ANS # Differential area vector on the strip 2 sin d s  R  d d aR # Area of the strip       1 1 cos 2/5 2 cos 3/5 0 2 3 sin 5 2 10 5 5 d R d d                    S s     # 2-8 Given that A   a  3  cos  a  2 z az in cylindrical coordinates, determine the closed surface integral of # A over the bounding surface of a half cylinder defined by   4 , 30 o    210 oand 0  z 5 as show in # Fig. 1-33. # ANS # Closed surface integral of A               4 30 210 0 5 7 /6 5 2 4 5 4 5 /6 0 4 0 0 3 cos 30 0 03 cos 210 S o o z z z z z z o z o z z z d d dz d dz d dz d d d d d dz d dz d dz                                                            A s A a A a A a A a A a            4 7 /6 4 7 / 0 /6 2 0 0 /6 2 5 80 60 3 60 3 80 0 d d d d                               # 2-9 For a vector field 2 2 3 R cos R  # D  a  a given in spherical coordinates, determine the closed surface # integral of D over the bounding surface of a volume defined by 0  R 4 and 30 o    60 oas shown in # Fig, 2-25. 1 1 sin cos sin R R V V V V R R  R                  a a a a a # (c) Coordinate transformation into Cartesian system sin cos cos cos sin cos 0 sin sin cos sin cos sin 0 cos sin 0 0 1 x y z A A A                                                  # Rewriting it, we have V  a z # The same result as in part (a) # 2-12 Two families of curves are given by f ( x y, ) x 2  y 2  c 1 and g x y( , )  ( x  6) 2  y 2  c 2 in the z  0 # plane, where c 1 and c 2 are constants. Determine the smaller angle between the two curves at point # p : (3,4,0). # ANS # The gradient of the two scalar fields are # f  2 x ax  2 yay , g  2  x  6 ax  2 yay , # At point p, the normal vectors to the two curves are  f p A  6 ax  8 ay  g p B   6 ax  8 ay # The smaller angle between A and B is 1 1 2 2 36 64 cos cos 73. 6 8 o AB          A B # 2-13 An ellipsoid is defined by x 2  y 2  z 2 / 4  5. Find, at point p 1 : (2, 3 / 2, 1)on the ellipsoid, # (a) outward unit normal vector # (b) expression for the tangent plane # ANS # (a) Consider a family of ellipsoids : 2 2 2 4 z f  x  y  # The gradient of f : 1 2 2 x y 2 z f  x a  y a  za # At point p 1 : 1 4 3 1 p x y 2 z  f  a  a  a # Unit normal vector at p 1 : 2 1 4 3 n 77 x y 2 z          a a a a # It is outward, away from the origin. (b) The tangent plane is conveniently defined as an   r  r 1  0 , where r 1  2 a x   3 / 2ay az is the position vector of p 1 , and r is the position vector in general:  1        2 4 3 12 3 / 2 1 0 n 77 x y 2 z x y z        x   y   z     a  r r a a a  a a a Conducting the dot product     3 1 4 2 3 1 0 2 2 x y z            Rewriting the equation, we have 1 4 3 10 2 x  y  z 2-14 A family of surfaces is defined as f ( x y z, , )  x 2  y 2  4 z 2  c in Cartesian coordinates. A curve crosses the family of surfaces at right angles, and passes through point p : (2, 6, 32). Find the parametric representation of the curve by using x as parameter. ANS A vector normal to the family of surfaces f  2 x ax  2 y ay  8 zaz or 4 2 x y z y z f x x x           a a a (1) Parametric representation of a curve, in general r  x ax  y  x  ay z  xaz (2) Differentiating (2) with respect to x, we obtain the tangent to the curve as ' x y z dy dz dx dx r  a  a  a (3) Two vectors in (1) and (3) are parallel to each other dy y dx x  (4) dz 4 z dx x  (5) Solving (4) and (5) separately by using the method of separating variables, we obtain y  c x 1 (6) 4 z  c x 2 (7) Inserting (6) and (7) into (2), with unknown constants c 1 and c 2 4 r  x ax  c x 1 ay c x 2 az We can determine c 1 and c 2 by making the curve pass through point p :  2, 6, 32. The parametric representation of the curve is therefore 3 2 4 r  x ax  x ay  x az 2-16 Consider a scalar field f ( )r k  rin Cartesian coordinates, where r is position vector, and a constant vector k  kx ax  ky ay kz az. (a) Find f (b) What is the geometric shape that is defined by f ( )r c 1 , a constant. (c) What is the significance of c 1 in part (b)? ANS (a) f   k x x  k yy  k zz  k (b) An infinite plane perpendicular to k (c) The perpendicular distance, from the origin to the plane, multiplied by k 2-17 Under the condition x  xo, a smooth function f ( x )can be expanded by Taylor series as f ( x )  f x( o )  f ( xo )( x  xo), where f  is the first derivative of f. Verify the Taylor series by use of the gradient of f. ANS Differential of f is x y z ( o ) x ( o) df df df df df f d x x x x dx dy dz dx                l a a a  a Thus f ( x ) f x( o ) df df( x xo) dx     The Taylor series is thus verified. 2-18 For the scalar fields U, V and W given in Problem 2-10, determine the line integral of the gradient of each field from point p 1 : 2,2,0 to point p 2 : 0, 0,2 in Cartesian coordinates along the intersection between a surface x 2  y 2  2 z 2  8 and the   450 plane: (a) 2 1 p p U  dlin Cartesian coordinates (b) 2 1 p p  V  dl in cylindrical coordinates (c) 2 1 p p W  dlin spherical coordinates ANS (a)         2 1 0 4 2 2 1 3 2 10 3 0 10 6 p p U d U p U p e e              l (b) In cylindrical coordinates, two points are given by 1 :  8, 45 ,0 p o , and   p 2 : 0,0,       2 1 0 2 1 ln 5 5 8 sin 45 ln 5 10 p p V d  V p  V p         l (c) In spherical coordinates, two points are 1 :  8, 90 , 45 p o o , and   2 : 2, 0, 45 p o     2 1 2 1 2 sin 0 cos 45 sin 90 cos 45 2 8 1 8 2 p o o o o p W d W p W p                     l 2-19 Determine the divergence of the following vector fields: (a) 2 2 2   1 x y z x y z      A a a a , ( x 2  y 2  z 2  0 ) (b) B  ln  a   cos  a z 3 az (c) 13 R Re Rsin  Rsin cos  2 R  C  a   a     a ANS (a)     2 2 2 3/ x y z x y z         A (b)   2 1 ln     1  sin   3 z   B (c) 14 2 e Rcos Rsin sin 2 R  C        2-20 Take the divergence of a vector field D xax in two different ways and compare the results following these steps: (a) Find   Din Cartesian coordinates (b) Transform D into spherical coordinates and then take the divergence ANS (a)  D 1 (b) Coordinate transformation sin cos sin sin cos cos cos cos sin sin 0 sin cos 0 0 D R x D D                                         Rewriting it D  R, ,     x sin  cos  a R x cos  cos  a   xsina Using x  Rsin  cos  , ,  sin 2 cos 2 sin 2 cos 2 sin sin 2 R 2 2 R R D R    R   a    a   a The divergence of D # Fig. 2-27 A region defined by two hemispherical surfaces (Problem 2-22) # ANS # (a) Closed surface integral of D               0 2 2 0 0 2 1 0 sin sin sin sin cos cos inner R outer R plane plane sphere sphere R R d ds ds ds ds R d d R d d R R R dRd R                                                      S D  s D  a D  a D  a D  a   2 1 0 3 3 4 8 4 3 2 2 R R dRd                  # (b) Divergence of D 2     2 1 1 sin cos sin sin sin sin R R R R R R                 D # Volume integral of D     2 2 1 0 2 2 2 1 0 sin sin sin sin sin sin sin 4 3 R R R R d R dRd d R R R dRd d                                               V  D v # Two results in parts (a) and (b) are equal. Divergence theorem is therefore verified. # 2-23 A vector field is defined as D  3 z 2 a zin Cartesian coordinates. Verify divergence theorem over the # volume bounded by a cone of half angle   30 oand the z  2 plane, by computing (a) closed surface integral  S D ds (b) volume integral  V Ddv # ANS # (a) Closed surface integral of D is separated into two parts: one over the conical surface and the other over the # top plate  S D  d s   cone D  d s topD ds # The first one is done in spherical coordinates, while the second is done in cylindrical coordinates. # Coordinate transformation of D into spherical coordinates 2 sin cos sin sin cos 0 cos cos cos sin sin 0 sin cos 0 3 D R D D z                                         # Rewrite it using z  Rcos D  3 R 2 cos 2   cos  aR  sina # Closed surface integral of D is therefore   2/cos 30 2 2 0 0 2 tan 30 2 0 0 3 cos cos sin sin 3 o o cone top R R R z z d d d R R dR d z d d                                 D s D s D s a a a a a S       # Inserting   30 oand z  2 into the equation   2 2 4/ 2 3 2/ 3 3 cos 30 sin 30 0 0 120 8 o o R d R R dR d d d                       S D  s     # (b) Divergence of D in cylindrical coordinates  D 6 z # Volume integral of D is done in cylindrical coordinates 2 2 tan 30 0 0 0 2 0 6 2 8 z z o V z z z dv z dzd d z dz                         D # The divergence theorem is verified. # 2-24 Take the curl of a vector field A  x 2 a y zaz in two different ways and compare the results following # these steps: # (a) Find   A in Cartesian coordinates # (b) Transform A into cylindrical coordinates and then take the curl. # ANS # (a) In Cartesian coordinates 2 2 0 x y z x y z x z x z            a a a A a # (b) Coordinate transformation of A into cylindrical coordinates         2 cos sin 0 0 sin cos 0 z 0 0 A A x A z                              # Rewrite the equation using x   cos 2 cos 2 sin 2 cos 3 A     a    a zaz # In cylindrical coordinates # Parametric representation of C r (   ) sin  ax  2   cosay # Differential length vector on C cos x sin y d d d d d d            r l a a # The closed line integral is   2 2     2 0 2 2 0 2 cos 2 sin lim cos sin sin 2 cos 2 lim 2 x y d d x d y d                                   a a C H l C a a C      2-27 For the vector field A   x  2 y 2  ax   2 xy y 2 ay given in Cartesian coordinates, verify Stokes’s # theorem over a triangle residing in the xy-plane as shown in Fig. 2-28. # Fig. 2-28 A triangular region in the xy-plane (Problem 2-27) # ANS # Denoting the vertex on the x-axis by A and that on the y-axis by B, the line segment AB is expressed as y   2 x 2 # Closed line integral of A along the perimeter C               1 0 0 2 1 2 0 2 2 2 0 2 1 2 0 2 2 2 x B y x x A x y y y x x y y x x y y d dx dx dy dy x y dx x y dx xy y dy xy y dy                                     C A  l A  a A  a a A  a     y  0 y   2 x 2 1 1 2 x   y x  0   1 0 2 2 2 0 1 0 2 1 2 2 2 2 1 2 1 1 8 8 4 4 2 2 3 3 x x y y x xdx x x x dx y y y y dy y y dy                                           # The curl of A 2 2 6 2 2 0 x y z x y z y z x y xy y             a a a A a # Surface integral of   A         1 2 2 0 0 1 0 6 6 12 1 4 x y x z z x y x d y dxdy y dxdy x dx                   S A  s S a  a # Stokes’s theorem is verified. 2-28 Given that B  (3 )  1 a   cos 2  a    1  sin a zin cylindrical coordinates, verify Stokes’s theorem # over the outer surface of a hollow cylinder that is closed except for a circular opening on top as shown in Fig. # 2-29. # Fig. 2-29 A hollow cylinder with an opening on top (Problem 2-28) # ANS # The curl of B in cylindrical coordinates # where R  2 / cos 30 o 4 / 3 and   30 oare used # (b) The curl of H in spherical coordinates         2 2 2 2 2 2 sin 1 sin 0 3 sin 1 6 sin cos 6 sin sin 6 cos 6 sin R R R R R R R R R R R R R                            a a a H a a a a # Surface normal to S is in  a direction, according to the right-hand rule. # The surface integral of   H over S 1         1 1 2/cos 30 2 0 0 2 4/ 3 0 sin 6 sin 6 sin 30 2 8 o R o R d R dRd R dRd RdR                           S H  s S H  a # Stokes’s theorem is verified. 2-30 Consider the vector field E  x y z, ,  A cos k  r, where A and k are constant vectors, and r is position # vector in Cartesian coordinates. Under what condition is E # (a) solenoidal? # (b) irrotational? # ANS # (a) Divergence of E   E    A  k  sin k r # In order to make   E 0 at all times, we should have k A 0 # (b) Curl of E   E    k A  sin k r # In order to make   E  0 at all times, we should have k  A 0 END of CHApPTER 2 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ # 2-30 Find the Laplacian of the following scalar fields: # (a) V  x z 2 yz 2 # (b) V   2 zcos # (c) V  R 2 sin 2  cos # ANS # (a)  2 V  2 z  2 y # (b)  2 V  3 zcos # (c)  2 V 3 cos # 2-15 A scalar field V and its partial derivatives are known at point p 1 with position vector r 1. The differential # length vector dl is from point p 1 to a nearby point p 2. Express V ( r 2 )in terms of V ( r 1 )and the partial # derivatives at p 1 in # (a) Cartesian coordinates # (b) cylindrical coordinates # (c) spherical coordinates # ANS # From calculus V  r 2   V  r 1   dV  V  r 1  V dl # (a) In Cartesian coordinates x y z  x y z V V V V d dx dy dz x y z V V V dx dy dz x y z                             l a a a  a a a  2   1  V V V V V dx dy dz x y z           r r # (b) In cylindrical coordinates  1  1 1 z z V V V V d d d dz z V V V d d dz z                                       l a a a  a a a  2   1  V V V d V d Vdz z            r r # (c) In spherical coordinates Electromagnetics Problem Solutions - Chap. 2 (Course Code: EM-202) Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save Premium Electromagnetics Problem Solutions - Chap. 2 (Course Code: EM-202) Course: 전자기학 (dontknow) 82 documents University: Sungkyunkwan University Info More info Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save View full document ProprietaryofProf.Lee,YeonHo,2014 P r o b l e m sc h a p t e r2\|1 Problems for Chapter 2 2-1 For an ellipse 2 2 (/16)(/4)1 x y , determine a unit tangent vecto r at point 1:(2,3,0)p on the ellipse by expressing the ellipse i n para met ric representation with a p aram eter (a) t, as in cos t and sin t (b)x, the space coordinate Fig. 2-20 An ellipse(Pr oblem 2-1) ANS (a) Parametric representation of the ellipse  4 cos 2 sin x y x y t x y t tr a a a a(1) Unit tangent vector is  2 2 4 sin 2 cos 4 sin 2 cos x y T t t t t     a a r a r P o i n t 1 p corresponds to /3 t. The unit tangent vector at 1 p is 2 3 13 x y T  a a a(2) (b) Using x as parameter, the parametric represen tation is  2 2 1 16 x y x y x x x y xr a a a a   0 y (3) Derivative of Eq. (3) with respectiv e to x  2 /2 16 x y x x x  r a a(4) Inserting the x -coordinate of point 1 p, or 2 x  , in Eq. (4)  1 12 x y x r a a Unit tangent vector at 1 p is 2 3 13 x y T     a a r a r(5) A s t increases in (1), the point moves counterclockwise o n the elli pse. O n the other h and, as x increases in (3), the point moves clockwise. For this rea son, the signs in (2) and (5) ar e opposite to each other. ProprietaryofProf.Lee,YeonHo,2014 P r o b l e m sc h a p t e r2\|2 2-2 A closed loop C forms a circle of radius 0.5, centered at a point (2 x  ,1 y  ) in th e xy-plane as shown in Fig. 2-21. Find (a) parametric representation of C (b) expression for d l on C Fig. 2-21 A circle centered at point(2,1)x y (Problem 2-2) ANS (a) ()(2 0.5 sin)(1 0.5 co s) x y r a a (b) (0.5 cos)(0.5 sin) x y d d d d d d   r l a a 2-3 Given a vector field 2 ()() x y x yE a a, find the line integral of E f ro m point:(1,0,0)A to point : ( 1,2,0 ) B along a path (a) 1 C and (b) 2 C as shown in Fig. 2-22. Fig. 2-22 Two paths of integration(Problem 2-3) ANS (a) Equation of path 1 C 1 y x Line integral of E along 1 C      1,2,0 2 1,0,0 1 2 2 2 1 0 B x y x y z A d x y dx dy dz x y dx x y dy        E l a a a a a C  This is a preview Do you want full access?Go Premium and unlock all 28 pages Access to all documents Get Unlimited Downloads Improve your grades Upload Share your documents to unlock Unlock for $0 Get 7 days of free Premium Already Premium?Log in Why is this page out of focus? This is a Premium document. Become Premium to read the whole document. ProprietaryofProf.Lee,YeonHo,2014 P r o b l e m sc h a p t e r2\|4 We have used identities  2 sin 1 cos 2/2 t t and cos cos sin t t dt t t t . 2-5 Given a vector field sin cos c os cos sin R R  A a a a in spherical coordinates, determine the close line integral of A around path C shown in Fig. 2-23. F ig. 2-23 A closed path(Problem 2-5) ANS Closed line integral of A d d d d d  A l A l A l A l A l C   Rewriting it using sin R d dR Rd R d   l a a a in spherical coo rdinates     3 60 2 30 2 30 3 60 sin cos sin sin s in cos sin sin o o o o R R R R R R R R d R dR R d R dR R d           A l a a a a a a a a C    Inserting proper values of R,  and  in the above equation   3 60 2 30 2 30 3 60 sin 90 c os 30 sin 3 sin 90 s in 90 cos 60 sin 2 sin 90 o o o o R o o o R R o o o R d R dR d R dR d         A l C    3 5 1 3 1 5 3 1 3 2 2 2 2 2 2 2 2 2 3 3 1 4          2-6 Determine the line integral of a vector field x A a along an arc that is defined by 4 R, 30 o  and 0/2 in spherical coordinates. ANS Differential length vector along the arc is sin d R d  l a Line integral of A is  /2/2 0 0 4 sin 3 0 2 sin 2 o x d d d      A l a a C   ProprietaryofProf.Lee,YeonHo,2014 P r o b l e m sc h a p t e r2\|5 2-7 Find the area of a circular str ip shown in Fig. 2-24. Fig. 2-24 A circular strip (Probl em 2-7) ANS Differential area vector on the strip 2 sin R d R d ds a Area of the strip    1 1 cos 2/5 2 2 2 cos 3/5 0 2 3 sin 5 2 10 5 5 d R d d          s S 2-8 Given that 3 cos 2 z z  A a a a in cylindrical coordinates, determine the closed surface inte g ral o f A over the bounding surf ace of a half cylinder defined by 4  , 30 210 o o  and 0 5 z as show in Fig. 1-33. ANS Closed surface integral of A        4 30 210 0 5 7/6 5 4 5 4 5 2 /6 0 0 0 0 0 4 3 cos 30 3 cos 210 o o S z z z z z z z o o z z z d d dz d dz d dz d d d d d dz d dz d dz            A s A a A a A a A a A a     4 7/6 4 7/6 0/6 0/6 2 0 2 5 80 60 3 60 3 80 0 d d d d      2-9 For a vector field 2 2 3 cos R R   D a a gi ven in spherical coordinates, determ ine the closed surface integral of D over the bounding surface of a volume defined by 0 4 R   and 30 60 o o  as shown in Fig, 2-25. This is a preview Do you want full access?Go Premium and unlock all 28 pages Access to all documents Get Unlimited Downloads Improve your grades Upload Share your documents to unlock Unlock for $0 Get 7 days of free Premium Already Premium?Log in Why is this page out of focus? This is a Premium document. Become Premium to read the whole document. ProprietaryofProf.Lee,YeonHo,2014 P r o b l e m sc h a p t e r2\|7 1 1 sin cos sin R R V V V V R R R        a a a a a (c) Coordinate transformation into Cartesian system sin cos cos cos sin cos 0 sin sin cos sin co s sin 0 cos sin 0 0 1 x y z A A A         Rewriting it, we have z Va The same result as in part (a) 2-12 Two families of curves are given by 2 2 1 (,) f x y x y c   and 2 2 2 (,)(6)g x y x y c   in the 0 z plane, where 1 c and 2 c are constants. Determ ine the smaller angle between the two cu rves at point :(3,4,0)p. ANS The gradient of the two scalar fields are 2 2 x y f x ya a,  2 6 2 x y g x ya a, A t p o i n t p, the nor mal vectors to the two curves are 6 8 x y p fA a a 6 8 x y p gB a a The smaller angle between A and B is 1 1 2 2 36 64 cos cos 73.7 6 8 o AB     A B 2-13 An ellipsoid i s defined by 2 2 2/4 5 x y z. Fin d, at point 1: ( 2,3/2,1 ) p on the ellipsoi d, (a) outward unit normal v ector (b) expression for the tangent plane ANS (a) Consider a family of ellipsoids : 2 2 2 4 z f x y The gra dient of f: 1 2 2 2 x y z f x y za a a A t p o i n t 1 p: 1 1 4 3 2 x y z p fa a a Unit nor mal ve ctor at 1 p : 2 1 4 3 2 77 n x y z     a a a a It is outward, away fr om the origin. This is a preview Do you want full access?Go Premium and unlock all 28 pages Access to all documents Get Unlimited Downloads Improve your grades Upload Share your documents to unlock Unlock for $0 Get 7 days of free Premium Already Premium?Log in Why is this page out of focus? This is a Premium document. Become Premium to read the whole document. 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2933	https://www.csun.edu/sites/default/files/micro6_0.pdf	Microeconomics Topic 6: “Be able to explain and calculate average and marginal cost to make production decisions.” Reference: Gregory Mankiw’s Principles of Microeconomics, 2nd edition, Chapter 13. Long-Run versus Short-Run In order to understand average cost and marginal cost, it is first necessary to understand the distinction between the “long run” and the “short run.” Short run: a period of time during which one or more of a firm’s inputs cannot be changed. Long run: a period of time during which all inputs can be changed. For example, consider the case of Bob’s Bakery. Bob’s uses two inputs to make loaves of bread: labor (bakers) and capital (ovens). (This is obviously a simplification, because the bakery uses other inputs such as flour and floor space. But we will pretend there are just two inputs to make the example easier to understand.) Bakers can be hired or fired on very short notice. But new ovens take 3 months to install. Thus, the short run for Bob’s Bakery is any period less than 3 months, while the long run is any period longer than 3 months. The concepts of long run and short run are closely related to the concepts of fixed inputs and variable inputs. Fixed input: an input whose quantity remains constant during the time period in question. Variable input: an input whose quantity can be altered during the time period in question. In the case of Bob’s Bakery, ovens are a fixed input during any period less than 3 months, whereas labor is a variable input. Fixed Cost, Variable Cost, and Total Cost In the short run, a firm will have both fixed inputs and variable inputs. These correspond to two types of cost: fixed cost and variable cost. Fixed cost (FC): the cost of all fixed inputs in a production process. Another way of saying this: production costs that do not change with the quantity of output produced. Variable cost (VC): the cost of all variable inputs in a production process. Another way of saying this: production costs that change with the quantity of output produced. In the case of Bob’s Bakery, the cost of renting ovens is a fixed cost in the short run, while the cost of hiring labor is a variable cost. Since fixed inputs cannot be changed in the short run, fixed cost cannot be changed either. That means fixed cost is constant, no matter what quantity the firm chooses to produce in the short run. Variable cost, on the other hand, does depend on the quantity the firm produces. Variable cost rises when quantity rises, and it falls when quantity falls. When you add fixed and variable costs together, you get total cost. Total cost (TC): the total cost of producing a given amount of output. TC = FC + VC Note: the total cost curve has the same shape as the variable cost curve because total costs rise as output increases. In the case of Bob’s Bakery, suppose the firm’s rental payments on ovens add up to $40 a day; then FC = 40. And suppose that if the firm produces 100 loaves in a day, its labor cost (wages for bakers) is $500; then VC = 500. The firm’s total cost is TC = 40 + 500 = 540. Suppose that when the firm produces 150 loaves a day, its labor cost rises to $700; then the new VC = 700 and the new TC = 40 + 700 = 740. This information is summarized in the table below. Bob’s Bakery’s Total, Fixed, and Variable Costs Quantity (per day) Total Cost Fixed Cost Variable Cost 100 540 40 500 150 740 40 700 Average Cost or Average Total Cost Average cost (AC), also known as average total cost (ATC), is the average cost per unit of output. To find it, divide the total cost (TC) by the quantity the firm is producing (Q). Average cost (AC) or average total cost (ATC): the per-unit cost of output. ATC = TC/Q Since we already know that TC has two components, fixed cost and variable cost, that means ATC has two components as well: average fixed cost (AFC) and average variable cost (AVC). The AFC is the fixed cost per unit of output, and AVC is the variable cost per unit of output. ATC = AFC + AVC AFC = FC/Q AVC = VC/Q In the case of Bob’s Bakery, we said earlier that the firm can produce 100 loaves with FC = 40, VC = 500, and TC = 540. Therefore, ATC = TC/Q = 540/100 = 5.4. Also, AFC = 40/100 = 0.4 and AVC = 500/100 = 5. Notice that we can use AFC and AVC to find ATC a different way: ATC = AFC + AVC = 0.4 + 5 = 5.4, which is the same answer we got before. If Bob’s Bakery produced 150 loaves instead of 100, the calculations would be the same, except we’d use Q = 150, VC = 700, and TC = 740 instead. FC would still be 40. This information is summarized in the table below. Bob’s Bakery’s Total and Average Costs Quantity (per day) TC FC VC ATC AFC AVC 100 540 40 500 5.40 0.40 5.00 150 740 40 700 4.93 0.27 4.67 It’s easy to find ATC using TC and Q, like we just did. But you should also be able to find Q using TC and ATC, or find TC using Q and ATC. Since we know that ATC = TC/Q, we also know that TC = ATC × Q and Q = TC/ATC. For example, suppose you know that Bob’s Bakery has TC = 740 and ATC = 4.93. Since ATC = TC/Q, the following equation must hold: 4.93 = 740/Q If you solve the equation, you’ll find that Q = 150 (approximately). Marginal Cost Often, we are interested in knowing what happens to a firm’s costs if output is increased by just a small amount. This is not the same as the average cost, because the next unit of output the firm produces might be more or less costly to produce than previous units. Marginal cost (MC): the additional cost that results from increasing output by one unit. Another way of saying this: the additional cost per additional unit of output. We use the symbol ∆ (the Greek letter delta) to designate the change in a variable. For instance, if total cost (TC) rose from 75 to 100, we would say ∆TC = 100 - 75 = 25. Using this symbol, we can write the mathematical formula for marginal cost: MC = ∆TC/∆Q Notice that we divide by the change in quantity (∆Q). Often, we set ∆Q = 1, because marginal cost is defined as the additional cost from one more unit of output. But sometimes we don’t know how much the added cost from just one more unit is, so we calculate marginal cost for a larger change in quantity. In the case of Bob’s Bakery, we said that TC = 540 when Q = 100, and TC = 740 when Q = 150. So ∆TC = 740 - 540 = 200, ∆Q = 150 - 100 = 50, and therefore MC = 200/50 = 4. We say that the marginal cost is 4 for units between 100 and 150. This is assuming we don’t have information about how much it would cost to increase output by just one, from 100 to 101 loaves of bread. Notice that the MC differs from ATC. At Bob’s Bakery, the original ATC was 5.4, and the new ATC (after increasing quantity to 150) was 4.93. Neither of these is equal to the MC of 4 that we just calculated. The table is the same as the last one, but with a new column for MC. Bob’s Bakery’s Total, Average, and Marginal Costs Quantity (per day) TC FC VC ATC AFC AVC MC 100 540 40 500 5.40 0.40 5.00 4 150 740 40 700 4.93 0.27 4.67 Notice that MC is listed between lines. That’s because MC shows the change that results from going from the first line to the second line. But sometimes, for simplicity, MC is just shown on the second line. In calculating MC, notice that you use four different numbers: the old TC, the new TC, the old Q, and the new Q. You should also be able to find any one of these numbers if you know the other three plus the marginal cost. For example, suppose you know that Bob’s Bakery had a total cost of $540 at its old quantity of 100, and you know that Bob’s has a total cost of $740 at its new quantity, but you don’t know the firm’s new quantity. But you do know its marginal cost per unit in making the change was 4. Since MC = ∆TC/∆Q, the following equation must hold: 4 = (740 - 540)/(new Q - 100) If you solve the equation for new Q, you’ll find that it’s 150. The Typical Shapes of Cost Curves Sometimes you will need to use curves representing the different kinds of cost. Here are the typical curve shapes. Since FC is constant, the FC curve is always horizontal like the one shown in the figure below. In our example, FC is always $40 --whether the firm produces 10 or 100 units of output. Fixed Cost Curve Quantity $ FC $40 0 On the other hand, VC increases as quantity increases. As a result, any VC curve is upward sloping like the one shown in the figure below. The TC curve looks very much like the VC curve, except that it’s higher because it includes the FC as well. Both AVC and ATC curve tend to have a U-shape, as shown in the figure below. That is, both AVC and ATC tends to fall at first and then rise as the output level increases. Of course, ATC is higher because it includes fixed costs. Although it is not shown in the figure above, we could also draw the AFC curve. Since AFC = FC/Q and FC is constant, AFC gets smaller and smaller as Q gets larger. That means the AFC curve is always downward sloping. MC also tends to have a U-shape. However, the bottom of the U occurs earlier, so the upward-sloping part of the MC is more important. The figure below shows a typical MC curve, along with ATC and AVC. Notice that MC crosses both ATC and AVC at their lowest points. Total and Variable Cost Curves Quantity $ VC 0 TC ATC and AVC Curves Quantity $ ATC 0 AVC ATC and AVC Curves Quantity $ ATC 0 AVC MC The figure above is the most important one for you to remember, because ATC, AVC, and MC are useful in understanding how a firm maximizes profit. For more on this, see the notes on Micro Topic 7.
2934	https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/resources/lec31/	Browse Course Material Course Info Instructor Prof. David Jerison Departments Mathematics As Taught In Fall 2006 Level Undergraduate Topics Mathematics Calculus Differential Equations Learning Resource Types assignment Problem Sets grading Exams with Solutions notes Lecture Notes theaters Lecture Videos Download Course search GIVE NOW about ocw help & faqs contact us 18.01 \| Fall 2006 \| Undergraduate Single Variable Calculus Lecture Notes Lecture 31: Parametric Equations, Arclength, Surface Area Description: Lecture notes on parametric equations, arclength, and surface area. Resource Type: Lecture Notes pdf 958 kB Lecture 31: Parametric Equations, Arclength, Surface Area Download File Course Info Instructor Prof. David Jerison Departments Mathematics As Taught In Fall 2006 Level Undergraduate Topics Mathematics Calculus Differential Equations Learning Resource Types assignment Problem Sets grading Exams with Solutions notes Lecture Notes theaters Lecture Videos Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Continue
2935	https://en.wikipedia.org/wiki/Singular_point_of_an_algebraic_variety	Published Time: 2004-09-13T12:01:38Z Singular point of an algebraic variety - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Definition 2 Singular points of smooth mappings 3 Nodes 4 See also 5 References Singular point of an algebraic variety [x] 7 languages Català Español Galego 한국어 日本語 Português Română Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Point without a tangent space This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources:"Singular point of an algebraic variety"–news·newspapers·books·scholar·JSTOR(September 2008) (Learn how and when to remove this message) In the mathematical field of algebraic geometry, a singular point of an algebraic varietyV is a point P that is 'special' (so, singular), in the geometric sense that at this point the tangent space at the variety may not be regularly defined. In case of varieties defined over the reals, this notion generalizes the notion of local non-flatness. A point of an algebraic variety that is not singular is said to be regular. An algebraic variety that has no singular point is said to be non-singular or smooth. The concept is generalized to smooth schemes in the modern language of scheme theory. The plane algebraic curve (a cubic curve) of equation y 2 − x 2(x + 1) = 0 crosses itself at the origin (0, 0). The origin is a double point of this curve. It is singular because a single tangent may not be correctly defined there. Definition [edit] A plane curve defined by an implicit equation F(x,y)=0,{\displaystyle F(x,y)=0,} where F is a smooth function is said to be singular at a point if the Taylor series of F has order at least 2 at this point. The reason for this is that, in differential calculus, the tangent at the point (x 0, y 0) of such a curve is defined by the equation (x−x 0)F x′(x 0,y 0)+(y−y 0)F y′(x 0,y 0)=0,{\displaystyle (x-x_{0})F'{x}(x{0},y_{0})+(y-y_{0})F'{y}(x{0},y_{0})=0,} whose left-hand side is the term of degree one of the Taylor expansion. Thus, if this term is zero, the tangent may not be defined in the standard way, either because it does not exist or a special definition must be provided. In general for a hypersurface F(x,y,z,…)=0{\displaystyle F(x,y,z,\ldots )=0} the singular points are those at which all the partial derivatives simultaneously vanish. A general algebraic varietyV being defined as the common zeros of several polynomials, the condition on a point P of V to be a singular point is that the Jacobian matrix of the first-order partial derivatives of the polynomials has a rank at P that is lower than the rank at other points of the variety. Points of V that are not singular are called non-singular or regular. It is always true that almost all points are non-singular, in the sense that the non-singular points form a set that is both open and dense in the variety (for the Zariski topology, as well as for the usual topology, in the case of varieties defined over the complex numbers). In case of a real variety (that is the set of the points with real coordinates of a variety defined by polynomials with real coefficients), the variety is a manifold near every regular point. But it is important to note that a real variety may be a manifold and have singular points. For example the equation y 3 + 2 x 2 y − x 4 = 0 defines a real analytic manifold but has a singular point at the origin. This may be explained by saying that the curve has two complex conjugatebranches that cut the real branch at the origin. Singular points of smooth mappings [edit] As the notion of singular points is a purely local property, the above definition can be extended to cover the wider class of smooth mappings (functions from M to Rn where all derivatives exist). Analysis of these singular points can be reduced to the algebraic variety case by considering the jets of the mapping. The k th jet is the Taylor series of the mapping truncated at degree k and deleting the constant term. Nodes [edit] Further information: Acnode and Crunode In classical algebraic geometry, certain special singular points were also called nodes. A node is a singular point where the Hessian matrix is non-singular; this implies that the singular point has multiplicity two and the tangent cone is not singular outside its vertex. See also [edit] Milnor map Resolution of singularities Singular point of a curve Singularity theory Smooth scheme Zariski tangent space References [edit] ^Hartshorne, Robin (1977). Algebraic Geometry. Berlin, New York: Springer-Verlag. p.33. ISBN978-0-387-90244-9. MR0463157. Zbl0367.14001. ^Milnor, John (1969). Singular Points of Complex Hypersurfaces. Annals of Mathematics Studies. Vol.61. Princeton University Press. pp.12–13. ISBN0-691-08065-8. Retrieved from " Categories: Algebraic varieties Singularity theory Hidden categories: Articles with short description Short description is different from Wikidata Articles needing additional references from September 2008 All articles needing additional references This page was last edited on 7 July 2025, at 09:23(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Singular point of an algebraic variety 7 languagesAdd topic
2936	https://archive.lib.msu.edu/crcmath/math/math/h/h348.htm	\| \| \| --- \| \| \| \| Homothetic Center The meeting point of lines that connect corresponding points from Homothetic figures. In the above figure, is the homothetic center of the Homothetic figures and . For figures which are similar but do not have Parallel sides, a Similitude Center exists (Johnson 1929, pp. 16-20). Given two nonconcentric Circles, draw Radii Parallel and in the same direction. Then the line joining the extremities of the Radii passes through a fixed point on the line of centers which divides that line externally in the ratio of Radii. This point is called the external homothetic center, or external center of similitude (Johnson 1929, pp. 19-20 and 41). If Radii are drawn Parallel but instead in opposite directions, the extremities of the Radii pass through a fixed point on the line of centers which divides that line internally in the ratio of Radii (Johnson 1929, pp. 19-20 and 41). This point is called the internal homothetic center, or internal center of similitude (Johnson 1929, pp. 19-20 and 41). The position of the homothetic centers for two circles of radii , centers , and segment angle are given by solving the simultaneous equations \| \| \| \| \| \| for , where \| \| \| \| \| \| and the plus signs give the external homothetic center, while the minus signs give the internal homothetic center. \| \| \| --- \| \| \| \| As the above diagrams show, as the angles of the parallel segments are varied, the positions of the homothetic centers remain the same. This fact provides a (slotted) Linkage for converting circular motion with one radius to circular motion with another. The six homothetic centers of three circles lie three by three on four lines (Johnson 1929, p. 120), which ``enclose'' the smallest circle. The homothetic center of triangles is the Perspective Center of Homothetic Triangles. It is also called the Similitude Center (Johnson 1929, pp. 16-17). See also Apollonius' Problem, Perspective, Similitude Center References Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929. Weisstein, E. W. ``Plane Geometry.'' Mathematica notebook PlaneGeometry.m. \| \| \| --- \| \| \| \| © 1996-9 Eric W. Weisstein 1999-05-25
2937	https://archive.cdc.gov/www_cdc_gov/ncbddd/birthdefects/surveillancemanual/chapters/chapter-5/chapter5-1.html	Skip directly to site content Skip directly to search Español \| Other Languages Birth Defects Surveillance Toolkit 5.1 Congenital Rubella Syndrome (CRS) PAGE 2 of 5 ‹View Table of Contents Related Pages Background Rubella virus (RV) is a togavirus of the genus Rubivirus, transmitted through droplets shed from the respiratory secretions of infected persons. Teratogenic effects of infection during pregnancy can cause harm to the embryo and developing fetus. CRS is heavily underreported, as many countries lack the capacity to conduct surveillance for this condition. It is estimated that there are approximately 105 000 CRS cases worldwide each year. Countries with sustained high rates of immunization have greatly reduced or eliminated rubella and CRS. Main clinical manifestations in the mother The incubation period of RV infection is 14 days (range, 12–23 days). Clinical symptoms include mild illness with low-grade fever (< 39 °C), headache, conjunctivitis and rhinitis. A characteristic feature is post-auricular, occipital and posterior cervical adenopathy (swelling of the lymph nodes), which precedes a red, maculopapular rash by 5–10 days. The rash occurs in 50–80% of rubella-infected persons, begins on the face and neck, and progresses to the lower parts of the body, lasting about three days. In 70% of women, joint pain (arthralgia) also occurs. Main clinical manifestations in the infant If primary rubella infection occurs during pregnancy, the virus can infect the placenta and fetus, causing a constellation of specific malformations labelled CRS. The classic triad of clinical manifestations associated with CRS among surviving neonates are hearing impairment, congenital heart defects – in particular, branch pulmonary artery stenosis and patent ductus arteriosus – and eye anomalies such as cataract(s), pigmentary retinopathy (salt and pepper type), chorioretinitis or congenital glaucoma. Additional clinical signs include skin purpura (blueberry muffin skin lesions), splenomegaly (enlargement of the spleen), microcephaly (small head circumference), developmental delay, meningoencephalitis, low birth weight, radiolucent bone disease and jaundice within 24 hours after birth (Fig. 5.1). The periconception period and early pregnancy (8–10 weeks) are the most vulnerable time frames and pose the greatest risk of CRS, which is as high as 90%. CRS can result in fetal death. For infants with CRS, hearing impairment, eye symptoms and developmental delay might not be detected until later. If maternal rubella infection is diagnosed beyond 18 weeks of gestation, the fetus might be infected but does not typically develop signs and symptoms of CRS. Infants with laboratory evidence of rubella and without any signs or symptoms of CRS are classified as having congenital rubella infection (CRI) only. Fig. 5.1. Clinical findings in the infant Infant with typical cloudiness of the eye lenses; that is, cataracts, in a case of CRS. Photograph source: CDC public health image library/Dr Andre J. Lebrum Congenital glaucoma (and cataract) in a 7-month-old infant with CRS. The left eye displays a congenital cataract; the right eye is normal. The infant was operated on on day 3 of life to correct the congenital cataract. Photograph source: CDC public health image library/Dr Andre J. Lebrum Infant with congenital rubella and “blueberry muffin” skin lesions. Lesions are sites of extramedullary hematopoiesis and can be associated with several different congenital viral infections and hematologic diseases. Photograph source: CDC public health image library/Dr Andre J. Lebrum. Radiolucent bone disease. X-ray of the lower limbs in a newborn with CRS. The ends of the long bones are ragged and streaky (like celery stalks) – changes due to active rubella infection. Photograph source: Government of Canada web page (Public health/ Rubella). Diagnosis Laboratory: Rubella immunoglobulin M (IgM) antibody detected (infants < 6 months old) in serum; or Sustained rubella immunoglobulin G (IgG) antibody level detected in serum; present on at least two occasions between 6 and 12 months of age (in the absence of having received rubella vaccine or being exposed to rubella); or Rubella virus detection (nucleic acid amplification tests [NAATs] or rubella virus isolation) from a clinical sample. The optimal sample is a throat swab; however, nasal swabs, blood, urine or cerebrospinal fluid are also acceptable. Nucleic acid amplification tests (NAATs) include, but are not limited to, reverse transcriptase polymerase chain reaction (RT-PCR), real-time RT-PCR, quantitative RT-PCR, and next-generation sequencing. Infants: In CRS and CRI cases, IgM might be negative at birth and a suspected infant should be tested again at 1 month of age. Although IgM might persist up to 12 months of age, in 50% of the cases, IgM is negative at 6 months of age, and should be complemented with IgG testing. The IgG testing should include serial testing to ensure sustained levels of IgG after 10 months (when maternal antibodies have waned) and before vaccination. Testing should also include virus detection (through NAATs or virus isolation) over several months for virus shedding until testing negative at two occasions at least one month apart. Infants with CRS or CRI have been demonstrated to shed virus up to 27 months after birth and might be the source of rubella outbreaks. Pregnant women: Pregnant women exposed to rubella should undergo IgM testing of serum. If IgM for rubella is detected in a pregnant woman with no history of illness or contact with a rubella illness case, further laboratory investigation to rule out a false-positive test result is warranted. In some countries, pregnant women are routinely screened for rubella IgG antibody. If IgG antibody is negative, vaccination is recommended after delivery because the vaccine contains live virus. Rubella vaccination during pregnancy should be avoided. Case definition Clinical diagnosis alone is unreliable and should be verified with laboratory testing. Routine surveillance of CRS is focused on infants < 12 months of age (Table 5.1). The algorithm for CRS case confirmation in infants < 6 months and 6–12 months is presented in Fig. 5.2. Table 5.1. CRS case definition Table 5.1. CRS case definition/caption> \| Suspected CRS case \| Any infant < 12 months with suspicion of CRS. The following clinical manifestation should lead to suspicion of CRS: (1) Congenital heart disease; and/or (2) suspicion of hearing impairment; and/or (3) one or more of the following eye signs: (a) cataract;a or (b) congenital glaucoma;b or (c) pigmentary retinopathy (salt and pepper). In addition, if the infant’s mother has a history of suspected or confirmed rubella during pregnancy, a clinician should suspect CRS, even if the infant shows no signs of CRS. \| \| Clinically confirmed CRS case \| An infant of < 12 months in whom a qualified clinician detects: At least two of the complications listed in group A or one in group A and one in group B: Group A: Cataract(s), congenital glaucoma, congenital heart disease, hearing impairment, pigmentary retinopathy. Group B: Purpura,c splenomegaly,d microcephaly, meningoencephalitis, radiolucent bone disease, jaundice that begins within 24 hours after birth. \| \| Laboratory-confirmed CRS case \| An infant who is a suspected case with one condition from group A (as above) and meets the laboratory criteria for CRS laboratory confirmation. \| \| CRI \| An infant who does not have group A clinical signs of CRS but who meets the laboratory criteria for CRS is classified as having congenital rubella CRI. \| a cataract, opaque or white lens; b congenital glaucoma, enlarged eyeball; c purpura, blueberry muffin skin lesions; d splenomegaly, enlarged spleen Fig. 5.2. CRS case confirmation CRS case confirmation in infants <6 monthsa a Testing should also include virus detection (through NAAT or virus isolation) over several months for virus shedding until testing negative at two occasions at least one month apart. CRS case confirmation in infants 6 -12 months Relevant ICD-10 codes P35.0 Congenital rubella syndrome (CRS)Q02 MicrocephalyQ12.0 Congenital cataractQ15.0 Congenital glaucomaQ25.0 Patent ductus arteriosusQ25.6 Stenosis of pulmonary artery Checklist Examine the neonate for: Eye: Glaucoma, cataracts, chorioretinitis or pigmentary retinopathy (salt and pepper) and the sclera for jaundice Skin: Jaundice that begins within 24 hours after birth and purpura Abdomen: Splenomegaly Cardiac: Murmur Neurological system: Developmental delay, meningoencephalitis. Additional clinical examinations: Skeletal radiograph: Radiolucent bone disease Hearing screening test: Hearing impairment (failed hearing screening must be followed with diagnostic testing to verify hearing loss) Echocardiography: CHD. Inquire about maternal medical health and pregnancy history to ascertain rubella infection and vaccination status. Lack of a history of known infection should not preclude suspicion of CRS. Collect neonatal samples for laboratory testing (IgM and IgG). Determine whether the case is suspected, clinically confirmed, laboratory-confirmed, CRI, or excluded. Obtain photographs of any malformations noted Record diagnostic information and report. Table of Contents 5. Congenital Infectious Syndromes ›5.1 Congenital Rubella Syndrome (CRS) 5.2 Congenital Syphilis 5.3 Congenital Cytomegalovirus (cCMV) 5.4 Congenital Zika Syndrome (CZS) Last Reviewed: December 8, 2020 Source: Division of Birth Defects and Developmental Disabilities, NCBDDD, Centers for Disease Control and Prevention Help Center Learn how to navigate this website. Our Partners International Clearinghouse for Birth Defects Surveillance and Research (ICBDSR) World Health Organization (WHO)
2938	https://math.stackexchange.com/questions/171568/finding-the-radical-or-squarefree-part-of-an-integer	Skip to main content Finding the radical (or squarefree part) of an integer Ask Question Asked Modified 1 year, 9 months ago Viewed 2k times This question shows research effort; it is useful and clear 11 Save this question. Show activity on this post. Given a number x=pe11⋯penn with different primes pi and exponents ei≥1, is there an efficient way to find p1⋯pn? I ask this because for polynomials it's easy: with K a field of characteristic 0 and f=ge11⋯genn∈K[X] irreducible we have g1⋯gn=fgcd(f,f′) where f′ is the formal derivative. This proof can't be used for integers, unless there's a trick that I don't see. elementary-number-theory algorithms factoring Share CC BY-SA 4.0 Follow this question to receive notifications edited Nov 8, 2023 at 19:33 Bill Dubuque 283k4242 gold badges338338 silver badges1k1k bronze badges asked Jul 16, 2012 at 16:06 DogDog 11344 bronze badges 3 1 I think it will work with the arithmetic derivative in place of the formal derivative, but then efficient computation is another story. – anon Commented Jul 16, 2012 at 16:11 @anon Oh this is interesting, I haven't seen that. You're right, it is much more difficult to compute, since it doesn't satisfy a nice rule for sums. – Dog Commented Jul 16, 2012 at 16:16 See my Q & A here for a fast algorithm for 64 bit integers. stackoverflow.com/questions/78656578/… – Simon Goater Commented Jun 23, 2024 at 9:18 Add a comment \| 1 Answer 1 Reset to default This answer is useful 16 Save this answer. Show activity on this post. Currently, no feasible (polynomial time) algorithm is known for recognizing squarefree integers or for computing the squarefree part of an integer. In fact it may be the case that this problem is no easier than the general problem of integer factorization. Computing the radical rad(n) is equivalent to computing the squarefree part sf(n) because rad(n)=sf(n)sf(n/rad(n)) This problem is important because one of the main tasks of computational algebraic number theory reduces to it (in deterministic polynomial time). Namely the problem of computing the ring of integers of an algebraic number field depends upon the square-free decomposition of a polynomial discriminant when computing an integral basis, e.g. S.7.3 p.429 or This is due to Chistov . See also Problems 7,8, p.9 in , which lists 36 open problems in number theoretic complexity. The primary reason that such problems are simpler in function fields versus number fields is due to the availability of derivatives. This opens up a powerful toolbox that is not available in the number field case. For example once derivatives are available so are Wronskians - which provide powerful measures of dependence in transcendence theory and diophantine approximation. A simple yet stunning example is the elementary proof of the polynomial case of Mason's ABC theorem, which yields as a very special case a high-school-level proof of FLT for polynomials, cf. my MO post and my old sci.math post . Such observations have motivated searches for "arithmetic analogues of derivations". For example, see Buium's paper by that name in Jnl. Algebra, 198, 1997, 290-99, and see his book Arithmetic differential equations. A. L. Chistov. The complexity of constructing the ring of integers of a global field. Dokl. Akad. Nauk. SSSR, 306:1063--1067, 1989. English Translation: Soviet Math. Dokl., 39:597--600, 1989. 90g:11170 Lenstra, H. W., Jr. Algorithms in algebraic number theory. Bull. Amer. Math. Soc. (N.S.) 26 (1992), no. 2, 211--244. 93g:11131 Pohst, M.; Zassenhaus, H. Algorithmic algebraic number theory. Cambridge University Press, Cambridge, 1997. Adleman, Leonard M.; McCurley, Kevin S. Open problems in number-theoretic complexity. II. Algorithmic number theory (Ithaca, NY, 1994), 291--322, Lecture Notes in Comput. Sci., 877, Springer, Berlin, 1994. 95m:11142 Dubuque, Bill. sci.math.research post, 1996/07/17 poly FLT, abc theorem, Wronskian formalism [was: Entire solutions of f^2+g^2=1] Share CC BY-SA 4.0 Follow this answer to receive notifications edited Nov 8, 2023 at 19:34 answered Jul 16, 2012 at 16:23 Bill DubuqueBill Dubuque 283k4242 gold badges338338 silver badges1k1k bronze badges 1 Here is a link to that page in Pohst-Zassenhaus. – J. M. ain't a mathematician Commented Jul 16, 2012 at 16:36 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions elementary-number-theory algorithms factoring See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Linked 0 Is there a function that returns x with all repeated prime factors removed? 37 Why is it "easier" to work with function fields than with algebraic number fields? 4 A question about the name of a property of positive numbers 2 Computing the radical of an integer's equality 1 Prime-base products: Plot 0 Method to solve x2=0modm,m composite Related 3 A question on perfect square 7 n-th prime number is less than 4n 0 Squares and Non-squares in Gaussian integers 2 How do we apply induction to this proof of the Fundamental Theorem of Arithmetic? 1 Proof verification of GCD prime factorization 1 Canonical Prime Factorisation 7 Show that there are only finitely many pairs of positive integers (n,m) such that d(m!)=n!. 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2939	https://www.youtube.com/watch?v=MCPn7Bnuh2o	Introduction to number theory lecture 37 Continued fractions Richard E Borcherds 75600 subscribers 188 likes Description 7750 views Posted: 23 Mar 2022 This lecture is part of my Berkeley math 115 course "Introduction to number theory" For the other lectures in the course see We show how to used continued fractions to find rationalapproximations to real numbers and to soleve the Pellian equation. The textbook is "An introduction to the theory of numbers" by Niven, Zuckerman, and Montgomery (5th edition). 2 comments Transcript: this lecture is part of berkeley math 115 an introductory undergraduate course on number theory and will be mostly about continued fractions and so um we're going to be using them to solve equations i'll just give a quick bit of background what we want to do is to solve equations um in several variables where f is some sort of polynomial and the complexity depends on the degree of f which can be one two three four and so on ones of degree one are called linear because they sort of give the equations of lines one's a degree two are called quadratic because they've got squares in them and squares are called quadratic things degrees three are called cubics um degree four are called quadra um quadrix and so on and unfortunately this terminology is a bit of a mess we can also talk about the number of variables and if there's one variable this is kind of very easy to solve because it's just a polynomial and you can in one variable so you can find its roots over the reals and so on so um if it has two variables this is called a binary um polynomial if it's three variables it's called ternary and so on um now degree one um polynomials are rather easy to solve using euclid's algorithm and polynomials in one variable are rather easy to solve so the next first sort of non-trivial case is um degree two polynomials in two variables um and these sort of look like a x squared plus b x y plus c y squared plus d x plus e y plus f equals zero and by making changes of variables um you can sort of get rid of the linear terms you just change x to x plus a constant to complete the square and get rid of that and you do the same for y um so what we have to do is to solve an equation form ax squared plus b x y plus c y squared equals some constant so this bit here as i said is a binary quadratic form and the next few lectures are going to be mostly about binary quadratic forms so a very famous example of this is the pellion equation which you want to solve x squared minus d y squared equals one um that it is named after this guy called pell who had absolutely nothing to do with solving the equation and the name is some sort of historical accident but it's it's kind of stuck so that's what everybody calls it um so it was actually solved many centuries before pell i think one of the first people to solve it was this indian mathematician um brahmagupta he was in about 650 ads so this was really very early it was you know almost a thousand years before europeans managed to solve this equation and he said one who can solve the equation x squared minus 92 y squared equals one within a year is a mathematician if you try and solve it by trial and error you're probably not going to manage it at least not if you do it by hand so we want to find a way to solve equations like this and if you think about this this this equation says that x squared minus 92 y squared is approximately zero um so that means x over y squared is approximately 92 so x over y is approximately the square root of 92. so what we have to do is to find a rational number um rational number x over y close to this number the square root of 92 and this is a very general problem in mathematics you want to find a good rational approximation to some real number and there's a good technique of doing this using continued fractions which i'll illustrate as follows so as you take the number pi and we want to find a good rational approximation to it with x and y integers well we can take pi well let's take the integer part of it as approximately 3 plus something where this is you know five 0.14159 and so on and we're going to write pi as a continued fraction now continued fraction means you want to write pi as um some integer plus one over some integer plus one over some integer plus one over some integer and so on and it's fairly obvious how we can do this we take this number here and we write one over this bit here is equal to and well it's going to be about seven plus some remainder term and then you take one over this bit here and it will be it's about 15 plus something and then this something one over this something is about one plus something and 1 over something turns out to be about 292 plus something and so pi is going to be approximately 3 plus 1 over 7 plus 1 over 15 plus 1 over 1 plus 1 over 292 and so on and when you've got a continued fraction like this it's quite easy to find a good rational approximation for example here's the first rational approximation what you notice is that 1 over 15 plus something is approximately zero i mean well you know because 15 is quite a large number so we find that pi is approximately 3 plus 7 which is the well-known classical approximation to it but you can do better than that you notice this number 291 over 292 is is is even closer to zero so pi is going to be approximately three plus one over seven plus one over fifteen plus one over one and if you work out what this is it turns out to be three five five over one one three which is a very famous um approximation to pi found several hundred years ago and it's it it it's quite good because it's got three digits here but it gives it gives pi correct to to more than so it's got six digits here but it actually gives pi correctly to more than six digits so this is about 3.1415929 or something and pi is actually two 3.1415926 so you see it's got one two three four five six seven digits correct even though you've only used six digits in the approximation so anyway so continued fractions are a really neat way of finding good rational approximations and numbers um now let's go back and solve the pending equation x squared minus 92 y squared equals one um and first of all we notice that 92 is actually divisible by 4 so we could try solving x squared minus 23y squared equals 1 which is going to be a bit easier because 23 is smaller and if we're lucky y will be even and then we'll have a solution to to this by dividing y by 2. um and if it isn't even well we will see what we can do about that so let's let's first solve this well we expand 23 is a continued fraction so you get out your pocket calculation you find the square root of 23 is about 4 plus some error term that i'm not going to worry about and 1 over this error term is about 1 plus something and 1 over this something is about 3 plus something and you continue like this you find the square root of 23 can be expanded as something like 4 plus 1 over 1 plus 1 over 3 plus 1 over 1 plus 1 over 8 and so on and you keep going like this and you can um take various bits of this and see how how good they are as a solution to this for instance you might say three is quite large so we tried saying root 23 is about four plus one over one well that's that's going to give us um um this is going to give us the number five so we get 5 squared minus 23 times 1 squared equals well that 2 well 2 is a little bit too big we actually want this to be equal to 1 so so this approximation isn't quite good enough well how about we try um you know there's this quite large number 8 here so so let's try setting this equal to zero so we get that the square root of 23 is about four plus one over one plus one over three plus one over one and now you can work this out well that's um 4 plus 1 over 1 plus 1 over 4 and 1 plus 1 over 4 is 5 over 4 so that's 4 plus 4 fifths which is equal to 24 over 5. and now if we try this we find 24 squared minus 23 times 5 squared is well 24 is is is 576 and if you subtract 23 times 25 that's 575 so this is equal to 1. so we've solved our original equation here and all we need is for this number 5 to be even and then we can solve x squared minus 92 squared y squared equals 1. well um unfortunately as you probably noticed the number 5 is not even so um what do we do about this well it turns out we can actually find lots of other solutions this equation if we've got one of them so here we've got 24 minus 23 times 5 squared equals 1. and i'm going to write this as 24 minus 5 square root of 23 times 24 plus 5 times the square root of 23 is equal to 1. and now you notice if i've got a minus b root 23 times a plus b root 23 that's going to give me a solution of a squared minus 23 b squared equals 1. um and what i can do is i can just square this so we get 24 minus 5 root 23 squared times 24 plus 5 root 23 squared is also equal to 1. and now what i do is i i i can just write this as something plus something times um the square root of 23. um and if you square this out it turns out to be 1 one five sorry one one five one um minus 240 times the square root of 23 times 1151 plus times the square root of 23. now this is equal to 1. and now you see we've now got another solution of x squared minus 23y squared equals 1 we've got 1 1 5 1 squared minus 23 times 240 squared is equal to 1. and now this is now become even so we can solve our original equation 1151 squared minus 92 times 120 squared is equal to one here i've just multiplied this by four and divided this by four so we found a solution of of um the the original equation um so um now what i want to do is to do a rather more complicated example just to show you um how bad things can get so this time i'm going to do x squared minus 67 y squared equals 1. and i'm going to do it a little bit more explicitly um so so let's write out the continued fraction well we have the square root of 67 and now i'm going to do everything explicitly with with integer arithmetic so the square root of 67 um is equal to um let's leave a bit of space here 8 plus the square root of 67 minus 8 divided by 1. and so this is the start of the continued fraction and then i take the inverse of this so 1 over square root of 67 minus 8 is equal to square root of 67 plus 8 over 3 which is equal to 5 plus the square root of 67 minus 7 over 3. um and um if you work out the what you get as a solution to the equation here here we here we've got the square root of 67 is about 8 so we try 8 squared minus 67 times 1 squared and this is equal to -3 and that's no good we want this to be equal to 1 and here if we look at the continued fraction 8 we would have eight plus a fifth and that if we work that out that's giving us 41 squared minus 67 times five squared uh that doesn't work either that's equal to six which isn't one so so this is this is using the approximation root seven is about eight this is using the approximation square root of 67 is about eight plus one over five which should be um 41 over five so we get 41 and five there so and that doesn't work so we don't give up we go into the next term which is 3 over the square root of 67 minus 7 which is equal to the square root of 67 plus 7 divided by 6 which is equal to 2 plus the square root of 67 minus 5 over 6. and let's see what we get here well um here we now get 90 squared minus 67 times 11 squared is equal to well minus 7. that's no good either and here we're using the the approximation of 8 plus 1 over 5 plus 1 over 2 for the square root of 67 which is 90 divided by 11. well if that doesn't work we just keep going so the next one is 6 over the square root of 67 minus 5 and i'm going to miss out some of the calculation because it's getting a bit boring this is going to be 1 plus square root of 67 minus 2 over 7. and if you work out what we get here we actually get a 9. um let me write these out that's 1 over 31 squared minus 67 times 16 squared which is equal to that and now working out the continued fractions are getting a bit tedious but there's there's a quicker way of doing it what you notice is there's actually a recursion relation for each of these numbers so 90 is equal to 2 times 41 plus 8. so the 41 and the 8 are here and the 2 is here and similarly 11 is 2 times 5 plus 1 so the 5 and the 1 are here and and this 2 is here so we're getting a sort of recursion relation for these numbers depending on this number here for example here we would use this number one and we would get 131 is equal to 1 times 90 plus 41. so there's the 90 and the 41 and similarly we get 16 is equal to 1 times 11 plus 5. so there's the 11 there's the 5 and there's the 1. so so we can work out these numbers here using using a simple recursion relation um and let's do a few more so we get 7 over square root of 67 minus 2 which is 1 plus the square root of 67 minus 7 over 9 and here we get 221 squared and here we get 27 and we get a minus 2 new notice the 221 is 1 times this plus this and 27 is 1 times that plus that the next one is 9 over the square root of 67 minus 7 and here we get 7 plus the square root of 67 minus 7 over 2 and here we get a 9 so that's no good we do a few more 2 over the square root of 67 minus 7 is equal to 1 plus the square root of 67 minus 2 over 9. and that's no good we get a minus 7 there and we get a 9 over the square root of 67 minus 2 and this is equal to 1 plus square root of 67 minus 5 over 7. and does that work no we get a 6 there and the next one is 7 over the square root of 67 minus 5 and this is equal to 2 plus the square root of 67 minus 7 over 6. and we get a minus 3 here and you wonder how long is this going to go on for well if you stare at it you see there's something funny going on look at these numbers here minus 3 6 minus seven nine minus two nine minus seven six minus three can you see a pattern well one pattern is that it's kind of symmetric so um these numbers here are all the same and and this is sort of seems to be in the middle of something so we seem to be coming to the end because you you know we started with minus three six minus seven nine and now we're going back down again and this suggests we should be getting to something something interesting fairly soon and so the next one is 6 over the square root of 67 minus 7 and this is giving us 5 plus the square root of 67 minus eight over three and now the number we get is one and that shouldn't be too surprising because actually i missed out a number at the beginning here we actually had a solution one squared minus 67 times zero squared is equal to one i mean this equation does have a rather trivial solution x equals one of course which is not the one we're interested in um so um um this one here corresponds to this one here now we've finally found a non-trivial solution and now if you work at all these numbers here which i won't be bothered doing because it's not terribly interesting you find this number here is is is actually four eight eight four 2 squared minus 67 times um 5967 squared is equal to 1. so we found a solution of this equation which by hand and as you see you you wouldn't be able to find this by hand by trial and errors it is just too big so continued fractions give a very efficient way of finding quite large solutions of of pelian equations um and if you look at this you can see there are several other patterns going on for example um these numbers you know six minus seven nine minus two nine well if you look the denominators here you're getting six seven nine two nine seven so the same numbers except with with signs every second step and as we as we said before there's another pattern that if you look at these numbers here they're giving the recursion relations for these numbers here so you can work these out by some simple recursion relations and if you write them out you you can find several other patterns going on and really the best thing to do to understand these is is not to watch somebody else do it but to do an example yourself so let me give you uh an example if you want to practice on so so here's a sort of exercise let's do x squared minus 61 squared 61 times y squared equals one and we want to find a non-trivial solution i should warn you that the solution to this is actually really rather large and if you manage to do this by hand you you are being you are extremely patient um but it does work in the end um by the way um the re that there's a problem we didn't um solve which does show this process actually terminates um i'm not going to prove that because the proof is a little bit messy to follow but becomes kind of obvious if you do this calculation by hand and sort of understand what's going on the the point is you can find all these numbers here are bounded by roughly the square root of 67. um so they're only a finite number of possibilities for what you get here so eventually this process must cycle and once it cycles it just keeps on cycling and if it's cycling you can cycle it backwards and you can cite it backwards and get to the one that you start with up here so if you cycle forwards you must also get to a one eventually so rather than give a proof that this terminates i will say if you actually try it on an exam a couple of examples you will sort of understand it so well that you can kind of see that it always terminates um if you just say a little bit more about how you can find an infinite number of solutions now the the problem with this example here is that although it has an infinite number of solutions the infinite number of solutions get rather big for instance the next one is 477 108 1927 squared minus five eight two eight eight zero four two eight squared times 67 is equal to one and i i don't want to have to deal with numbers like this so i'm going to do a much smaller example to illustrate how to find infinitely many solutions so let's just do x squared minus three y squared equals one and this is a very simple solution um x equals two y equals one so now i want to show you how to generate an infinite number of solutions from this so um we just copy what we did for 92 we we observe that 2 plus root 3 times 2 minus root 3 is equal to 1. and now we can just raise both sides to the power of n 2 plus root 3 to the n times 2 minus root 3 to the n is equal to 1. and this will be a plus b root 3 and this will be a minus b root three and this will be equal to one so we find that a squared minus three b squared equals one so this gives it gives us a new solution of this for every value of n so let's just do the first few so 2 plus root 3 squared is equal to 7 plus 4 root 3 and we find 7 squared minus 3 times 4 squared is equal to 1 that's 49 minus 3 times 16 which is 48 and 2 plus root 3 cubed is equal to 26 plus 15 times the square root of 3 so we find 26 squared minus 3 times 15 squared is equal to one so that's you know that's 576 that's three times 225 and so on so um um we can generally find an infinite number of solutions of x squared minus d y squared equals one unless we obviously can't so we certainly want to take d to be not a square for example otherwise there are no solutions at all but if d is a square then the continued fraction thing kind of breaks down because the square root of d is already an integer um so that shows um how to use continued fractions to solve this equation what we're going to do in the next lecture is look in a bit more detail at more general quadratic form so in general we want to solve the equation ax squared plus bxy plus c y squared equals some constant and to do this we will be studying these binary quadratic forms in more detail
2940	https://artofproblemsolving.com/wiki/index.php/Concurrence?srsltid=AfmBOorTkVJbpCpgd2ZDDbTOApbS9env76MRa2vCb5FXV1kiVLEeRZWt	Art of Problem Solving Concurrence - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Concurrence Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Concurrence Several (that is, three or more) lines or curves are said to be concurrent at a point if they all contain that point. The point is said to be the point of concurrence. Contents 1 Proving concurrence 2 Problems 2.1 Introductory 2.2 Intermediate 2.3 Olympiad Proving concurrence In analytical geometry, one can find the point of concurrency of any two lines by solving the system of equations of the lines. To see if it shares the point of concurrency with other lines/curves requires only to test that point. Ceva's Theorem gives a criteria for three cevians of a triangle to be concurrent. In particular, the three altitudes, angle bisectors, medians, symmedians, and perpendicular bisectors (which are not cevians) of any triangle are concurrent, at the orthocenter, incenter, centroid, Lemoine point, and circumcenter, respectively. Concurrence of lines can occasionally be proved by showing that a certain point is a center of a particular homothecy. Problems Introductory Are the lines , , and concurrent? If so, find the the point of concurrency. (Source) Intermediate In triangle , , , and are on the sides , , and , respectively. Given that , , and are concurrent at the point , and that , find . (Source) Olympiad Hallie is teaching geometry to Warren. She tells him that the three medians, the three angle bisectors, and the three altitudes of a triangle each meet at a point (the centroid, incenter, and orthocenter respectively). Warren gets a little confused and draws a certain triangle ABC along with the median from vertex A, the altitude from vertex B, and the angle bisector from vertex C. Hallie is surprised to see that the three segments meet at a point anyway! She notices that all three sides measure an integer number of inches, that the side lengths are all distinct, and that the side across from vertex C is 13 inches in length. How long are the other two sides? (Mathcamp Qualifying Quiz, #10, also see the Solution) This article is a stub. Help us out by expanding it. Retrieved from " Categories: Definition Geometry Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
2941	https://zh.hinative.com/questions/15919444	品质积分: 3307 回答数: 762 被“赞”数: 1942 空调和冷气有什么区别？如果难以说明的话，请教我一下例句。品质积分: 827 回答数: 304 被“赞”数: 183 空调有制热和制冷模式，当使用制冷模式的时候你可以说是冷气，当然，仍然可以说是空调这个答案有帮助吗？品质积分: 123 回答数: 29 被“赞”数: 20 One more way to explain：「空调」大陆用得多。→ Mandarin：「你开了空调没有」「冷气」感觉是香港的粤语中用得多。Cantonese→「你开左冷气未啊」这个答案有帮助吗？这个语言水平的标志表示您有兴趣的语言的水平。如果设置您的语言水平，其他用户会参照您的语言水平来对您的提问进行回答。使用这个语言回答可能无法被理解可以提出简单的问题并且理解比较简单的回答内容可以提出各种一般性的问题并且理解较长的回答可以理解长而且复杂的回答比起点赞和贴图更能体现您的感激之情赠送礼物会让你更容易收到回答！如果你在赠送礼物之后提问，你的问题将出现在对方的问题列表的显著位置。免费向母语使用者提问用这个应用程序轻易解决问题！
2942	https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_25?srsltid=AfmBOorfnFdgzfgHRSE66IZ8RvxoyGUfHPRU7ZCcaZUk01ayD0eT48WI	Art of Problem Solving 2006 AMC 12A Problems/Problem 25 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2006 AMC 12A Problems/Problem 25 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2006 AMC 12A Problems/Problem 25 Contents [hide] 1 Problem 2 Solution 1 3 Solution 2 4 Solution 3 5 Solution 4 6 Solution 5 (Possibly Clever bash) 7 Video Solutions 8 See also Problem How many non-emptysubsets of have the following two properties? No two consecutive integers belong to . If contains elements, then contains no number less than . Solution 1 This question can be solved fairly directly by casework and pattern-finding. We give a somewhat more general attack, based on the solution to the following problem: How many ways are there to choose elements from an ordered element set without choosing two consecutive members? You want to choose numbers out of with no consecutive numbers. For each configuration, we can subtract from the -th element in your subset. This converts your configuration into a configuration with elements where the largest possible element is , with no restriction on consecutive numbers. Since this process is easily reversible, we have a bijection. Without consideration of the second condition, we have: Now we examine the second condition. It simply states that no element in our original configuration (and hence also the modified configuration, since we don't move the smallest element) can be less than , which translates to subtracting from the "top" of each binomial coefficient. Now we have, after we cancel all the terms where , Solution 2 Another way of visualizing the solution above would be to use 's and 's. Denote as the numbers we have chosen and as other numbers. Taking an example, assuming we are picking two numbers, we imagine the shape . This notation forces a number between the two chosen numbers, which blocks the two numbers we picked from being consecutive. Now we consider the orientations with this shape. We have remaining numbers. We need to find the number of ways to place the remaining 's. We can find this by utilizing stars and bars, with the following marker being placed to represent groups: . Now, we have to place numbers within groups, which is . The same concept can be used for the remaining numbers. The rest of the solution continues as above. Solution by: Everyoneintexas Solution 3 We have the same setup as in the previous solution. Note that if , the answer will be 0. Otherwise, the elements we choose define boxes (which divide the nonconsecutive numbers) into which we can drop the remaining elements, with the caveat that each of the middle boxes must have at least one element (since the numbers are nonconsecutive). This is equivalent to dropping elements into boxes, where each box is allowed to be empty. And this is equivalent to arranging objects, of which are dividers, which we can do in ways. Now, looking at our original question, we see that the thing we want to calculate is just Solution 4 Let be the numbers of elements in a subset. First we examine the second condition. No elements less than can be put in a subset of size , therefore the "lowest" element that can go into the subset is , whereas the "highest" element that can go into the subset is . This is a total of or possible elements. Now we consider the first condition. No consecutive elements are allowed. This means that if an element is put into the subset, both and are no longer possibilities. Assume that all subsets are ordered from least to greatest (we are looking for the number of combinations, so we can order these combinations however we want). Then the first element will be (as a reminder, the lowest possible element in a subset is ), the second element will be at least , and so on. After elements are chosen, we will have skipped elements (these are the elements that were "eliminated" as they were consecutive). Therefore, we ignore exactly elements (if we ignore more or less elements, then changes) Since we must ignore elements, we can simply remove those beforehand. ( possible elements Now we look for the bounds of . We are looking for non-empty subsets, so . If is too large, there will not be enough non-consecutive terms between and . Specifically, the highest element in a subset using "optimal" selection will be or . If , that means s is too large. Therefore ; solving for yields . Now we know that . We want to know the number of ways to arrange "balls" into identical "boxes" with at most ball per box, for to . This is equivalent to , or . Solution 5 (Possibly Clever bash) We will split the problem into cases, and maybe one could then generalize this to arbitrary . (). Then this is easy. We have . (). Now we have something tricky. To get a good grasp on this case, let us consider the smallest element; , in the first spot. We then have numbers left. However, we cannot have the digit . Hence we have to choose numbers from integers left. This can be done in ways. (). As in the last case, we can use the smallest element to get a good grasp on this case. Put the digit in the first spot. Then there are integers left for the second spot. But in total, we have to avoid elements for each subset of cardinality . So we have to choose elements from elements, which can be done in ways. (). As in the previous case, we have to avoid numbers, and choose from them. Which is choosing elements from elements, which can be done in ways. (). Now you are accustomed to the strategy. We remove integers leaving ways. Adding up all of the cases yields , as in . SHEN KISLAY KAI~~2023 Note also that we only went up to cardinality or , or else for greater , we would always have consecutive numbers. EDIT: I have coincidentally found a video solution using the exact same method: (Also if you think I am copying, why would I have then posted the video, hm? Explain that)! Video Solutions (by Challenge 25) See also 2006 AMC 12A Problems 2006 AMC 12A (Problems • Answer Key • Resources) Preceded by Problem 24Followed by Last Question 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 12 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Intermediate Combinatorics Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
2943	https://promova.com/confusing-words/can-able-to	Can and Able To \| Meaning, Examples & Difference \| Promova For Business Group lessons English tutors App For Business Group lessons English tutors App Get the App: Free for Ukrainians Sign Up Free for Ukrainians Sign Up Can vs Able To Type your word here Try:Affect vs Effect What’s the difference between them? Can Meaning: Can is an auxiliary verb that is used to express possibility or permission. Examples: Can I borrow your pen? Can you help me with this task? I can't believe it's already Friday! Able To Meaning: Able To: having the power, skill, or means to do something; capable of doing something. Examples: I am able to help you with your project. She was able to find the answer to the question quickly. He was able to finish the task before the deadline. Learn similar and opposite words to spot the difference Synonyms Antonyms Can Able Capable Possess Have the capacity Be equipped with the ability Cannot Shouldn't Disallow Deny Refuse Able To Capable Competent Adept Skilled Up to the task Incapable of Unable to Powerless to Incompetent at Unqualified for Tricks for mastery Useful tips to understand the difference between confusing words "Can", "Able To". Remember that can is an auxiliary verb that is used to express possibility or permission, whereas able to expresses having the power, skill, or means to do something. Use the phrase 'can do' to help you remember that can is used to express possibility or permission. Use the phrase 'able and willing' to help you remember that able to expresses having the power, skill, or means to do something. Pay attention to the context in which the words are used to determine their meaning. Practice English with the Promova app and avoid misusing confusing words Download the app Frequently asked questions When should 'can' be used? Can is used to express ability, permission, or possibility. It is typically used as an auxiliary verb to form questions or to make requests. It can also be used to ask for or give permission, or to indicate a strong possibility. When is the appropriate context for using the phrase 'able to'? Able To is used to express having the power, skill, or means to do something; capable of doing something. It can also be used to express the notion of having the opportunity or freedom to do something. Do the two words share the same pronunciation? No, can is pronounced /ken/, while able to is pronounced /eibl tu/. What are some common mistakes people make when using these words? One common mistake people make when using these words is confusing 'can' with 'may'. Can is used to express ability, permission, or possibility while may is used to express permission or possibility in the form of a polite request. Moreover, people tend to use 'able to' instead of 'can' which is incorrect as 'able to' is used to express the notion of having the opportunity or freedom to do something. Fill in the gaps to check yourself Test Questions Correct Answers With hard work and dedication, anyone ____ achieve their goals. Shes not feeling well today, so she ____ attend the meeting. Its amazing how some animals ____ communicate with humans through gestures and sounds. He ____ play the piano beautifully since he was a child. Im sorry, but I ____ not help you with that problem right now; Im busy. Despite the difficult circumstances, they were ____ complete the project on time. With hard work and dedication, anyone can achieve their goals. Explanation: Can is used to indicate the general ability to achieve goals with effort. Shes not feeling well today, so she isnt able to attend the meeting. Explanation: Isnt able to indicates her current inability to attend the meeting due to not feeling well. Its amazing how some animals are able to communicate with humans through gestures and sounds. Explanation: Are able to emphasizes the remarkable ability of certain animals to communicate with humans. He has been able to play the piano beautifully since he was a child. Explanation: Has been able to indicates his continuous ability to play the piano beautifully from childhood to the present. Im sorry, but I cannot help you with that problem right now; Im busy. Explanation: Cannot expresses the inability to help due to being busy. Despite the difficult circumstances, they were able to complete the project on time. Explanation: Able to indicates their achievement of completing the project despite challenges. Get a gift by subscribing to our newsletter! Download the PDF with a list of commonly confused words made as flashcards for comfortable learning. [x] I accept the Privacy Policy Get my gift List of Commonly Confused Words Finding your way around the English language can be hard, especially since there are so many confusing words and rules. So, a list of the most confusing words in English is an extremely useful tool for improving language accuracy and sharing the ideas clearly. 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2944	https://www.youtube.com/watch?v=Z0tNDfhrOkI	Testing for Symmetry with Respect to the x-axis, y-axis, and Origin Cole's World of Mathematics 46100 subscribers 3532 likes Description 335999 views Posted: 1 Nov 2015 This video goes through three examples: one testing for symmetry around the x-axis, one testing for symmetry around the y-axis, and one testing for symmetry around the origin. mathematics #algebra #maths Math Tutorials on this channel are targeted at college-level mathematics courses including calculus, pre-calculus, college algebra, trigonometry, probability theory, TI-84 tutorials, introductory college algebra topics, and remedial math topics from algebra 1 and 2 for the struggling college adult (age 18 and older). Don’t forget guys, if you like this video please “Like” and “Share” it with your friends to show your support - it really helps me out! SUBSCRIBE HERE for more Math Tutorials --► Interested in purchasing the TI-84 Plus CE Graphing Calculator? 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Thank you for the support! 253 comments Transcript: Introduction okay today I'm going to be going through the rules for testing for symmetry for a variety of different equations um and I'll do one of each to demonstrate how we're going to do this all right so let's say I've got an um function and I want to test to see if it's symmetric around the x-axis what I'm going to do is I'm going to replace all my y's with a negative y I'm going to go through I'm going to simplify the equation and then I'm going to check when I get down at the bottom to see if it matches the original function if it does then it will be symmetrical to the x-axis and if it does not then it will not be symmetrical all right so my first um Testing for Symmetry equation here I want to test to see if it is symmetrical to the x axis is a y^2 = x 5 - 7 x all right so I going to replace every one of my y's with a negative y so negative y^ squared and then the rest of the equation I'm just going to keep because I'm not doing anything to it all right now I'm going to go through and simplify I only have to simplify something on the left because that's the only thing that I have changed so far all right so a negative y^2 well negative time negative is going to be a positive so this is going to simplify to a y^2 all right again nothing is changing there on the right hand side so I have an x to the 5th minus a 7x right there all right now Does it match once you have simplified all right you've plugged that in and you've simplified down to there then what you're going to ask yourself is does this what you simplified match the original equation all right and in this case y^2 = x 5 - 7 x it does match okay does it match that's the question you're asking yourself does it match all right in this instance yes it does so therefore this equation is this function is symmetrical to the x-axis so um you could write a little concluding sentence here therefore um yes it's symmetric to the x-axis all right now let's suppose um I'm running it on a different equation here just so you can see some different examples of the plugging in and simplifying all right let's suppose I've got this cubic equation y = x 3 + 4 and I want to know if it is symmetrical to the Y AIS all right if I want to know if it's symmetrical to the Y axis I'm going to take every one of my x's and replace them with a Negative X and then simplify and then ask myself doesn't match that original function that I started with all right so y equals I'm going to take that X I'm going to replace it with a negative x to the thir power and then plus 4 okay Simplify now I'm going to go through and I'm going to simplify there's nothing to simplify on the left all right now a negative raised to the third power all right anything raised to an odd power any negative number raised to OD power is going to be negative so then this is going to simplify to a x 3r all right and then the plus4 doesn't simplify all right so I have simplified that equation down as far as I can all right so now I'm going to ask myself does what I've just simplified match my original equation all right does it match all right and as we look at this then we can see the Y is match the fours would match but that x to the 3 does not okay it's positive in the positive Co ient there in the original one and then what I've simplified here has a negative all right so I can conclude that this Cubit equation is not symmetrical to the Y AIS so therefore no it's not symmetric to the Y AIS and if you happen to have your family of Curves uh memorized and you happen to know that an X the 3 basically looks something like that then you're going to realize that well no it's not symmetric to the y- axis all right um we can also test for symmetry around the origin all right and I chose a rational function for this example all right to test to see if something is symmetrical to the origin you're going to replace all the X's with a Negative X you're going to replace the Y's all the Y's with a negative y all right so my original function here y equal x over x^2 + 5 so I'm going to replace the Y here with a a y I'm going to replace both of those X's with a Negative X so on top I'll have a negative X down here on the bottom I'm going to have ax^ 2ar + 5 all right now I'm going to go through and I'm going to attempt to simplify all right first off and the first thing I want to simplify is that denominator right there I've got Negative X and it's being squared any negative number squared is going to give me a positive there on the bottom so I'll have a y = ax over an x^2 + 5 all right now I've got one more simplifying here I can do I've got negative on the left hand side I've got negative on the right hand side so I can divide both sides of the equation by negative one which in essence is just going to cross those two things out all right now I can look at that I've got a y = x over x^2 + 5 for my final simplified answer all right and again I'm going to ask myself does my simplified answer match my original equation all right does it match all right and again on this one as we look things over yes it does match so therefore this one is going to be symmetric to the origin so therefore yes it is symmetric to the origin all right so rules for testing for symmetry there prettyy straightforward three different types of functions on how you would apply it
2945	https://blngcc.files.wordpress.com/2008/11/viktor-prasolov-problems-in-plane-and-solid-geometry.pdf	PROBLEMS IN PLANE AND SOLID GEOMETRY v.1 Plane Geometry Viktor Prasolov translated and edited by Dimitry Leites Abstract. This book has no equal. The priceless treasures of elementary geometry are nowhere else exposed in so complete and at the same time transparent form. The short solutions take barely 1.5 −2 times more space than the formulations, while still remaining complete, with no gaps whatsoever, although many of the problems are quite diﬃcult. Only this enabled the author to squeeze about 2000 problems on plane geometry in the book of volume of ca 600 pages thus embracing practically all the known problems and theorems of elementary geometry. The book contains non-standard geometric problems of a level higher than that of the problems usually oﬀered at high school. The collection consists of two parts. It is based on three Russian editions of Prasolov’s books on plane geometry. The text is considerably modiﬁed for the English edition. Many new problems are added and detailed structuring in accordance with the methods of solution is adopted. The book is addressed to high school students, teachers of mathematics, mathematical clubs, and college students. Contents Editor’s preface 11 From the Author’s preface 12 Chapter 1. SIMILAR TRIANGLES 15 Background 15 Introductory problems 15 §1. Line segments intercepted by parallel lines 15 §2. The ratio of sides of similar triangles 17 §3. The ratio of the areas of similar triangles 18 §4. Auxiliary equal triangles 18 19 §5. The triangle determined by the bases of the heights 19 §6. Similar ﬁgures 20 Problems for independent study 20 Solutions 21 CHAPTER 2. INSCRIBED ANGLES 33 Background 33 Introductory problems 33 §1. Angles that subtend equal arcs 34 §2. The value of an angle between two chords 35 §3. The angle between a tangent and a chord 35 §4. Relations between the values of an angle and the lengths of the arc and chord associated with the angle 36 §5. Four points on one circle 36 §6. The inscribed angle and similar triangles 37 §7. The bisector divides an arc in halves 38 §8. An inscribed quadrilateral with perpendicular diagonals 39 §9. Three circumscribed circles intersect at one point 39 §10. Michel’s point 40 §11. Miscellaneous problems 40 Problems for independent study 41 Solutions 41 CHAPTER 3. CIRCLES 57 Background 57 Introductory problems 58 §1. The tangents to circles 58 §2. The product of the lengths of a chord’s segments 59 §3. Tangent circles 59 §4. Three circles of the same radius 60 §5. Two tangents drawn from one point 61 3 4 CONTENTS ∗∗∗ 61 §6. Application of the theorem on triangle’s heights 61 §7. Areas of curvilinear ﬁgures 62 §8. Circles inscribed in a disc segment 62 §9. Miscellaneous problems 63 §10. The radical axis 63 Problems for independent study 65 Solutions 65 CHAPTER 4. AREA 79 Background 79 Introductory problems 79 §1. A median divides the triangle into triangles of equal areas 79 §2. Calculation of areas 80 §3. The areas of the triangles into which a quadrilateral is divided 81 §4. The areas of the parts into which a quadrilateral is divided 81 §5. Miscellaneous problems 82 82 §6. Lines and curves that divide ﬁgures into parts of equal area 83 §7. Formulas for the area of a quadrilateral 83 §8. An auxiliary area 84 §9. Regrouping areas 85 Problems for independent study 86 Solutions 86 CHAPTER 5. TRIANGLES 99 Background 99 Introductory problems 99 1. The inscribed and the circumscribed circles 100 100 100 §2. Right triangles 101 §3. The equilateral triangles 101 101 §4. Triangles with angles of 60◦and 120◦ 102 §5. Integer triangles 102 §6. Miscellaneous problems 103 §7. Menelaus’s theorem 104 105 §8. Ceva’s theorem 106 §9. Simson’s line 107 §10. The pedal triangle 108 §11. Euler’s line and the circle of nine points 109 §12. Brokar’s points 110 §13. Lemoine’s point 111 CONTENTS 5 111 Problems for independent study 112 Solutions 112 Chapter 6. POLYGONS 137 Background 137 Introductory problems 137 §1. The inscribed and circumscribed quadrilaterals 137 138 138 §2. Quadrilaterals 139 §3. Ptolemy’s theorem 140 §4. Pentagons 141 §5. Hexagons 141 §6. Regular polygons 142 142 143 §7. The inscribed and circumscribed polygons 144 144 §8. Arbitrary convex polygons 144 §9. Pascal’s theorem 145 Problems for independent study 145 Solutions 146 Chapter 7. LOCI 169 Background 169 Introductory problems 169 §1. The locus is a line or a segment of a line 169 170 §2. The locus is a circle or an arc of a circle 170 170 §3. The inscribed angle 171 §4. Auxiliary equal triangles 171 §5. The homothety 171 §6. A method of loci 171 §7. The locus with a nonzero area 172 §8. Carnot’s theorem 172 §9. Fermat-Apollonius’s circle 173 Problems for independent study 173 Solutions 174 Chapter 8. CONSTRUCTIONS 183 §1. The method of loci 183 §2. The inscribed angle 183 §3. Similar triangles and a homothety 183 §4. Construction of triangles from various elements 183 §5. Construction of triangles given various points 184 §6. Triangles 184 §7. Quadrilaterals 185 §8. Circles 185 6 CONTENTS §9. Apollonius’ circle 186 §10. Miscellaneous problems 186 §11. Unusual constructions 186 §12. Construction with a ruler only 186 §13. Constructions with the help of a two-sided ruler 187 §14. Constructions using a right angle 188 Problems for independent study 188 Solutions 189 Chapter 9. GEOMETRIC INEQUALITIES 205 Background 205 Introductory problems 205 §1. A median of a triangle 205 §2. Algebraic problems on the triangle inequality 206 §3. The sum of the lengths of quadrilateral’s diagonals 206 §4. Miscellaneous problems on the triangle inequality 207 207 §5. The area of a triangle does not exceed a half product of two sides 207 §6. Inequalities of areas 208 §7. Area. One ﬁgure lies inside another 209 209 §8. Broken lines inside a square 209 §9. The quadrilateral 210 §10. Polygons 210 211 §11. Miscellaneous problems 211 211 Problems for independent study 212 Supplement. Certain inequalities 212 Solutions 213 Chapter 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE 235 §1. Medians 235 §2. Heights 235 §3. The bisectors 235 §4. The lengths of sides 236 §5. The radii of the circumscribed, inscribed and escribed circles 236 §6. Symmetric inequalities between the angles of a triangle 236 §7. Inequalities between the angles of a triangle 237 §8. Inequalities for the area of a triangle 237 238 §9. The greater angle subtends the longer side 238 §10. Any segment inside a triangle is shorter than the longest side 238 §11. Inequalities for right triangles 238 §12. Inequalities for acute triangles 239 §13. Inequalities in triangles 239 Problems for independent study 240 Solutions 240 Chapter 11. PROBLEMS ON MAXIMUM AND MINIMUM 255 CONTENTS 7 Background 255 Introductory problems 255 §1. The triangle 255 256 §2. Extremal points of a triangle 256 §3. The angle 257 §4. The quadrilateral 257 §5. Polygons 257 §6. Miscellaneous problems 258 §7. The extremal properties of regular polygons 258 Problems for independent study 258 Solutions 259 Chapter 12. CALCULATIONS AND METRIC RELATIONS 271 Introductory problems 271 §1. The law of sines 271 §2. The law of cosines 272 §3. The inscribed, the circumscribed and escribed circles; their radii 272 §4. The lengths of the sides, heights, bisectors 273 §5. The sines and cosines of a triangle’s angles 273 §6. The tangents and cotangents of a triangle’s angles 274 §7. Calculation of angles 274 274 §8. The circles 275 275 §9. Miscellaneous problems 275 §10. The method of coordinates 276 Problems for independent study 277 Solutions 277 Chapter 13. VECTORS 289 Background 289 Introductory problems 289 §1. Vectors formed by polygons’ (?) sides 290 §2. Inner product. Relations 290 §3. Inequalities 291 §4. Sums of vectors 292 §5. Auxiliary projections 292 §6. The method of averaging 293 §7. Pseudoinner product 293 Problems for independent study 294 Solutions 295 Chapter 14. THE CENTER OF MASS 307 Background 307 §1. Main properties of the center of mass 307 §2. A theorem on mass regroupping 308 §3. The moment of inertia 309 §4. Miscellaneous problems 310 §5. The barycentric coordinates 310 8 CONTENTS Solutions 311 Chapter 15. PARALLEL TRANSLATIONS 319 Background 319 Introductory problems 319 §1. Solving problems with the aid of parallel translations 319 §2. Problems on construction and loci 320 320 Problems for independent study 320 Solutions 320 Chapter 16. CENTRAL SYMMETRY 327 Background 327 Introductory problems 327 §1. Solving problems with the help of a symmetry 327 §2. Properties of the symmetry 328 §3. Solving problems with the help of a symmetry. Constructions 328 Problems for independent study 329 Solutions 329 Chapter 17. THE SYMMETRY THROUGH A LINE 335 Background 335 Introductory problems 335 §1. Solving problems with the help of a symmetry 335 §2. Constructions 336 336 §3. Inequalities and extremals 336 §4. Compositions of symmetries 336 §5. Properties of symmetries and axes of symmetries 337 §6. Chasles’s theorem 337 Problems for independent study 338 Solutions 338 Chapter 18. ROTATIONS 345 Background 345 Introductory problems 345 §1. Rotation by 90◦ 345 §2. Rotation by 60◦ 346 §3. Rotations through arbitrary angles 347 §4. Compositions of rotations 347 348 348 Problems for independent study 348 Solutions 349 Chapter 19. HOMOTHETY AND ROTATIONAL HOMOTHETY 359 Background 359 Introductory problems 359 §1. Homothetic polygons 359 §2. Homothetic circles 360 §3. Costructions and loci 360 CONTENTS 9 361 §4. Composition of homotheties 361 §5. Rotational homothety 361 362 362 §6. The center of a rotational homothety 362 §7. The similarity circle of three ﬁgures 363 Problems for independent study 364 Solutions 364 Chapter 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT 375 Background 375 §1. The least and the greatest angles 375 §2. The least and the greatest distances 376 §3. The least and the greatest areas 376 §4. The greatest triangle 376 §5. The convex hull and the base lines 376 §6. Miscellaneous problems 378 Solutions 378 Chapter 21. DIRICHLET’S PRINCIPLE 385 Background 385 §1. The case when there are ﬁnitely many points, lines, etc. 385 §2. Angles and lengths 386 §3. Area 387 Solutions 387 Chapter 22. CONVEX AND NONCONVEX POLYGONS 397 Background 397 §1. Convex polygons 397 397 §2. Helly’s theorem 398 §3. Non-convex polygons 398 Solutions 399 Chapter 23. DIVISIBILITY, INVARIANTS, COLORINGS 409 Background 409 §1. Even and odd 409 §2. Divisibility 410 §3. Invariants 410 §4. Auxiliary colorings 411 §5. More auxiliary colorings 412 412 §6. Problems on colorings 412 413 Solutions 413 Chapter 24. INTEGER LATTICES 425 §1. Polygons with vertices in the nodes of a lattice 425 §2. Miscellaneous problems 425 Solutions 426 10 CONTENTS Chapter 25. CUTTINGS 431 §1. Cuttings into parallelograms 431 §2. How lines cut the plane 431 Solutions 432 Chapter 26. SYSTEMS OF POINTS AND SEGMENTS. EXAMPLES AND COUNTEREXAMPLES 437 §1. Systems of points 437 §2. Systems of segments, lines and circles 437 §3. Examples and counterexamples 438 Solutions 438 Chapter 27. INDUCTION AND COMBINATORICS 445 §1. Induction 445 §2. Combinatorics 445 Solutions 445 Chapter 28. INVERSION 449 Background 449 §1. Properties of inversions 449 §2. Construction of circles 450 §3. Constructions with the help of a compass only 450 §4. Let us perform an inversion 451 §5. Points that lie on one circle and circles passing through one point 452 §6. Chains of circles 454 Solutions 455 Chapter 29. AFFINE TRANSFORMATIONS 465 §1. Aﬃne transformations 465 §2. How to solve problems with the help of aﬃne transformations 466 Solutions 466 Chapter 30. PROJECTIVE TRANSFORMATIONS 473 §1. Projective transformations of the line 473 §2. Projective transformations of the plane 474 §3. Let us transform the given line into the inﬁnite one 477 §4. Application of projective maps that preserve a circle 478 §5. Application of projective transformations of the line 479 §6. Application of projective transformations of the line in problems on construction 479 §7. Impossibility of construction with the help of a ruler only 480 Solutions 480 Index 493 EDITOR’S PREFACE 11 Editor’s preface The enormous number of problems and theorems of elementary geometry was considered too wide to grasp in full even in the last century. Even nowadays the stream of new problems is still wide. (The majority of these problems, however, are either well-forgotten old ones or those recently pirated from a neighbouring country.) Any attempt to collect an encyclopedia of all the problems seems to be doomed to failure for many reasons. First of all, this is an impossible task because of the huge number of the problems, an enormity too vast to grasp. Second, even if this might have been possible, the book would be terribly overloaded, and therefore of no interest to anybody. However, in the book Problems in plane geometry followed by Problems in solid geometry this task is successfully perfomed. In the process of writing the book the author used the books and magazines published in the last century as well as modern ones. The reader can judge the completeness of the book by, for instance, the fact that American Mathematical Monthly yearly1 publishes, as “new”, 1–2 problems already published in the Russian editions of this book. The book turned out to be of interest to a vast audience: about 400 000 copies of the ﬁrst edition of each of the Parts (Parts 1 and 2 — Plane and Part 3 — Solid) were sold; the second edition, published 5 years later, had an even larger circulation, the total over 1 000 000 copies. The 3rd edition of Problems in Plane Geometry was issued in 1996 and the latest one in 2001. The readers’ interest is partly occasioned by a well-thought classiﬁcation system. The collection consists of three parts. Part 1 covers classical subjects of plane geometry. It contains nearly 1000 problems with complete solutions and over 100 problems to be solved on one’s own. Still more will be added for the English version of the book. Part 2 includes more recent topics, geometric transformations and problems more suitable for contests and for use in mathematical clubs. The problems cover cuttings, colorings, the pigeonhole (or Dirichlet’s) principle, induction, and so on. Part 3 is devoted to solid geometry. A rather detailed table of contents serves as a guide in the sea of geometric problems. It helps the experts to easily ﬁnd what they need while the uninitiated can quickly learn what exactly is that they are interested in in geometry. Splitting the book into small sections (5 to 10 problems in each) made the book of interest to the readers of various levels. FOR THE ENGLISH VERSION of the book about 150 new problems are already added and several hundred more of elementary and intermideate level problems will be added to make the number of more elementary problems suﬃcient to use the book in the ordinary school: the Russian editions are best suited for coaching for a mathematical Olympiad than for a regular class work: the level of diﬃculty increases rather fast. Problems in each section are ordered diﬃculty-wise. The ﬁrst problems of the sections are simple; they are a match for many. Here are some examples: 1Here are a few samples: v. 96, n. 5, 1989, p. 429–431 (here the main idea of the solution is the right illustration — precisely the picture from the back cover of the 1st Russian edition of Problems in Solid Geometry, Fig. to Problem 13.22); v. 96, n. 6, p. 527, Probl. E3192 corresponds to Problems 5.31 and 18.20 of Problems in Plane Geometry — with their two absolutely diﬀerent solutions, the one to Problem 5.31, unknown to AMM, is even more interesting. 12 CONTENTS Plane 1.1. The bases of a trapezoid are a and b. Find the length of the segment that the diagonals of the trapezoid intersept on the trapezoid’s midline. Plane 1.52. Let AA1 and BB1 be the altitudes of △ABC. Prove that △A1B1C is similar to △ABC. What is the similarity coeﬃcient? Plane 2.1. A line segment connects vertex A of an acute △ABC with the center O of the circumscribed circle. The altitude AH is dropped from A. Prove that ∠BAH = ∠OAC. Plane 6.1. Prove that if the center of the circle inscribed in a quadrilateral coincides with the intersection point of the quadrilateral’s diagonals, then the quadrilateral is a rhombus. Solid 1. Arrange 6 match sticks to get 4 equilateral triangles with side length equal to the length of a stick. Solid 1.1. Consider the cube ABCDA1B1C1D1 with side length a. Find the angle and the distance between the lines A1B and AC1. Solid 6.1. Is it true that in every tetrahedron the heights meet at one point? The above problems are not diﬃcult. The last problems in the sections are a challenge for the specialists in geometry. It is important that the passage from simple problems to complicated ones is not too long; there are no boring and dull long sequences of simple similar problems. (In the Russian edition these sequences are, perhaps, too short, so more problems are added.) The ﬁnal problems of the sections are usually borrowed from scientiﬁc journals. Here are some examples: Plane 10.20. Prove that la + lb + mc ≤ √ 3p, where la, lb are the lengths of the bisectors of the angles ∠A and ∠B of the triangle △ABC, mc is the length of the median of the side AB, and p is the semiperimeter. Plane 19.55. Let O be the center of the circle inscribed in △ABC, K the Lemoine’s point, P and Q Brocard’s points. Prove that P and Q belong to the circle with diameter KO and that OP = OQ. Plane 22.29. The numbers α1, . . . , αn, whose sum is equal to (n−2)π, satisfy inequalities 0 < αi < 2π. Prove that there exists an n-gon A1 . . . An with the angles α1, . . . , αn at the vertices A1, . . . , An, respectively. Plane 24.12. Prove that for any n there exists a circle on which there lie precisely n points with integer coordinates. Solid 4.48. Consider several arcs of great circles on a sphere with the sum of their angle measures < π. Prove that there exists a plane that passes through the center of the sphere but does not intersect any of these arcs. Solid 14.22. Prove that if the centers of the escribed spheres of a tetrahedron belong to the circumscribed sphere, then the tetrahedron’s faces are equal. Solid 15.34. In space, consider 4 points not in one plane. How many various parallelip-ipeds with vertices in these points are there? From the Author’s preface The book underwent extensive revision. The solutions to many of the problems were rewritten and about 600 new problems were added, particularly those concerning the ge-ometry of the triangle. I was greatly inﬂuenced in the process by the second edition of the book by I. F. Sharygin Problems on Geometry. Plane geometry, Nauka, Moscow,1986 and a wonderful and undeservedly forgotten book by D. Efremov New Geometry of the Triangle, Matezis, Odessa, 1902. The present book can be used not only as a source of optional problems for students but also as a self-guide for those who wish (or have no other choice but) to study geometry FROM THE AUTHOR’S PREFACE 13 independently. Detailed headings are provided for the reader’s convenience. Problems in the two parts of Plane are spread over 29 Chapters, each Chapter comprising 6 to 14 sections. The classiﬁcation is based on the methods used to solve geometric problems. The purpose of the division is basically to help the reader ﬁnd his/her bearings in this large array of problems. Otherwise the huge number of problems might be somewhat depressingly overwhelming. Advice and comments given by Academician A. V. Pogorelov, and Professors A. M. Abramov, A. Yu. Vaintrob, N. B. Vasiliev, N. P. Dolbilin, and S. Yu. Orevkov were a great help to me in preparing the ﬁrst Soviet edition. I wish to express my sincere gratitude to all of them. To save space, sections with background only contain the material directly pertinent to the respective chapter. It is collected just to remind the reader of notations. Therefore, the basic elements of a triangle are only deﬁned in chapter 5, while in chapter 1 we assume that their deﬁnition is known. For the reader’s convenience, cross references in this translation are facilitated by a very detailed index. Chapter 1. SIMILAR TRIANGLES Background 1) Triangle ABC is said to be similar to triangle A1B1C1 (we write △ABC ∼△A1B1C1) if and only if one of the following equivalent conditions is satisﬁed: a) AB : BC : CA = A1B1 : B1C1 : C1A1; b) AB : BC = A1B1 : B1C1 and ∠ABC = ∠A1B1C1; c) ∠ABC = ∠A1B1C1 and ∠BAC = ∠B1A1C1. 2) Triangles AB1C1 and AB2C2 cut oﬀfrom an angle with vertex A by parallel lines are similar and AB1 : AB2 = AC1 : AC2 (here points B1 and B2 lie on one leg of the angle and C1 and C2 on the other leg). 3) A midline of a triangle is the line connecting the midpoints of two of the triangle’s sides. The midline is parallel to the third side and its length is equal to a half length of the third side. The midline of a trapezoid is the line connecting the midpoints of the trapezoid’s sides. This line is parallel to the bases of the trapezoid and its length is equal to the halfsum of their lengths. 4) The ratio of the areas of similar triangles is equal to the square of the similarity coeﬃcient, i.e., to the squared ratio of the lengths of respective sides. This follows, for example, from the formula SABC = 1 2AB · AC sin ∠A. 5) Polygons A1A2 . . . An and B1B2 . . . Bn are called similar if A1A2 : A2A3 : · · · : AnA1 = B1B2 : B2B3 : · · · : BnB1 and the angles at the vertices A1, . . . , An are equal to the angles at the vertices B1, . . . , Bn, respectively. The ratio of the respective diagonals of similar polygons is equal to the similarity coeﬃ-cient. For the circumscribed similar polygons, the ratio of the radii of the inscribed circles is also equal to the similarity coeﬃcient. Introductory problems 1. Consider heights AA1 and BB1 in acute triangle ABC. Prove that A1C · BC = B1C · AC. 2. Consider height CH in right triangle ABC with right angle ∠C. Prove that AC2 = AB · AH and CH2 = AH · BH. 3. Prove that the medians of a triangle meet at one point and this point divides each median in the ratio of 2 : 1 counting from the vertex. 4. On side BC of △ABC point A1 is taken so that BA1 : A1C = 2 : 1. What is the ratio in which median CC1 divides segment AA1? 5. Square PQRS is inscribed into △ABC so that vertices P and Q lie on sides AB and AC and vertices R and S lie on BC. Express the length of the square’s side through a and ha. §1. Line segments intercepted by parallel lines 1.1. Let the lengths of bases AD and BC of trapezoid ABCD be a and b (a > b). 15 16 CHAPTER 1. SIMILAR TRIANGLES a) Find the length of the segment that the diagonals intercept on the midline. b) Find the length of segment MN whose endpoints divide AB and CD in the ratio of AM : MB = DN : NC = p : q. 1.2. Prove that the midpoints of the sides of an arbitrary quadrilateral are vertices of a parallelogram. For what quadrilaterals this parallelogram is a rectangle, a rhombus, a square? 1.3. Points A1 and B1 divide sides BC and AC of △ABC in the ratios BA1 : A1C = 1 : p and AB1 : B1C = 1 : q, respectively. In what ratio is AA1 divided by BB1? 1.4. Straight lines AA1 and BB1 pass through point P of median CC1 in △ABC (A1 and B1 lie on sides BC and CA, respectively). Prove that A1B1 ∥AB. 1.5. The straight line which connects the intersection point P of the diagonals in quadri-lateral ABCD with the intersection point Q of the lines AB and CD bisects side AD. Prove that it also bisects BC. 1.6. A point P is taken on side AD of parallelogram ABCD so that AP : AD = 1 : n; let Q be the intersection point of AC and BP. Prove that AQ : AC = 1 : (n + 1). 1.7. The vertices of parallelogram A1B1C1D1 lie on the sides of parallelogram ABCD (point A1 lies on AB, B1 on BC, etc.). Prove that the centers of the two parallelograms coincide. 1.8. Point K lies on diagonal BD of parallelogram ABCD. Straight line AK intersects lines BC and CD at points L and M, respectively. Prove that AK2 = LK · KM. 1.9. One of the diagonals of a quadrilateral inscribed in a circle is a diameter of the circle. Prove that (the lengths of) the projections of the opposite sides of the quadrilateral on the other diagonal are equal. 1.10. Point E on base AD of trapezoid ABCD is such that AE = BC. Segments CA and CE intersect diagonal BD at O and P, respectively. Prove that if BO = PD, then AD2 = BC2 + AD · BC. 1.11. On a circle centered at O, points A and B single out an arc of 60◦. Point M belongs to this arc. Prove that the straight line passing through the midpoints of MA and OB is perpendicular to that passing through the midpoints of MB and OA. 1.12. a) Points A, B, and C lie on one straight line; points A1, B1, and C1 lie on another straight line. Prove that if AB1 ∥BA1 and AC1 ∥CA1, then BC1 ∥CB1. b) Points A, B, and C lie on one straight line and A1, B1, and C1 are such that AB1 ∥ BA1, AC1 ∥CA1, and BC1 ∥CB1. Prove that A1, B1 and C1 lie on one line. 1.13. In △ABC bisectors AA1 and BB1 are drawn. Prove that the distance from any point M of A1B1 to line AB is equal to the sum of distances from M to AC and BC. 1.14. Let M and N be the midpoints of sides AD and BC in rectangle ABCD. Point P lies on the extension of DC beyond D; point Q is the intersection point of PM and AC. Prove that ∠QNM = ∠MNP. 1.15. Points K and L are taken on the extensions of the bases AD and BC of trapezoid ABCD beyond A and C, respectively. Line segment KL intersects sides AB and CD at M and N, respectively; KL intersects diagonals AC and BD at O and P, respectively. Prove that if KM = NL, then KO = PL. 1.16. Points P, Q, R, and S on sides AB, BC, CD and DA, respectively, of convex quadrilateral ABCD are such that BP : AB = CR : CD = α and AS : AD = BQ : BC = β. Prove that PR and QS are divided by their intersection point in the ratios β : (1 −β) and α : (1 −α), respectively. §2. THE RATIO OF SIDES OF SIMILAR TRIANGLES 17 §2. The ratio of sides of similar triangles 1.17. a) In △ABC bisector BD of the external or internal angle ∠B is drawn. Prove that AD : DC = AB : BC. b) Prove that the center O of the circle inscribed in △ABC divides the bisector AA1 in the ratio of AO : OA1 = (b + c) : a, where a, b and c are the lengths of the triangle’s sides. 1.18. The lengths of two sides of a triangle are equal to a while the length of the third side is equal to b. Calculate the radius of the circumscribed circle. 1.19. A straight line passing through vertex A of square ABCD intersects side CD at E and line BC at F. Prove that 1 AE2 + 1 AF 2 = 1 AB2. 1.20. Given points B2 and C2 on heights BB1 and CC1 of △ABC such that AB2C = AC2B = 90◦, prove that AB2 = AC2. 1.21. A circle is inscribed in trapezoid ABCD (BC ∥AD). The circle is tangent to sides AB and CD at K and L, respectively, and to bases AD and BC at M and N, respectively. a) Let Q be the intersection point of BM and AN. Prove that KQ ∥AD. b) Prove that AK · KB = CL · LD. 1.22. Perpendiculars AM and AN are dropped to sides BC and CD of parallelogram ABCD (or to their extensions). Prove that △MAN ∼△ABC. 1.23. Straight line l intersects sides AB and AD of parallelogram ABCD at E and F, respectively. Let G be the intersection point of l with diagonal AC. Prove that AB AE + AD AF = AC AG. 1.24. Let AC be the longer of the diagonals in parallelogram ABCD. Perpendiculars CE and CF are dropped from C to the extensions of sides AB and AD, respectively. Prove that AB · AE + AD · AF = AC2. 1.25. Angles α and β of △ABC are related as 3α + 2β = 180◦. Prove that a2 + bc = c2. 1.26. The endpoints of segments AB and CD are gliding along the sides of a given angle, so that straight lines AB and CD are moving parallelly (i.e., each line moves parallelly to itself) and segments AB and CD intersect at a point, M. Prove that the value of AM·BM CM·DM is a constant. 1.27. Through an arbitrary point P on side AC of △ABC straight lines are drawn parallelly to the triangle’s medians AK and CL. The lines intersect BC and AB at E and F, respectively. Prove that AK and CL divide EF into three equal parts. 1.28. Point P lies on the bisector of an angle with vertex C. A line passing through P intercepts segments of lengths a and b on the angle’s legs. Prove that the value of 1 a + 1 b does not depend on the choice of the line. 1.29. A semicircle is constructed outwards on side BC of an equilateral triangle ABC as on the diameter. Given points K and L that divide the semicircle into three equal arcs, prove that lines AK and AL divide BC into three equal parts. 1.30. Point O is the center of the circle inscribed in △ABC. On sides AC and BC points M and K, respectively, are selected so that BK · AB = BO2 and AM · AB = AO2. Prove that M, O and K lie on one straight line. 1.31. Equally oriented similar triangles AMN, NBM and MNC are constructed on segment MN (Fig. 1). Prove that △ABC is similar to all these triangles and the center of its curcumscribed circle is equidistant from M and N. 1.32. Line segment BE divides △ABC into two similar triangles, their similarity ratio being equal to √ 3. Find the angles of △ABC. 18 CHAPTER 1. SIMILAR TRIANGLES Figure 1 (1.31) §3. The ratio of the areas of similar triangles 1.33. A point E is taken on side AC of △ABC. Through E pass straight lines DE and EF parallel to sides BC and AB, respectively; D and E are points on AB and BC, respectively. Prove that SBDEF = 2√SADE · SEFG. 1.34. Points M and N are taken on sides AB and CD, respectively, of trapezoid ABCD so that segment MN is parallel to the bases and divides the area of the trapezoid in halves. Find the length of MN if BC = a and AD = b. 1.35. Let Q be a point inside △ABC. Three straight lines are pass through Q par-allelly to the sides of the triangle. The lines divide the triangle into six parts, three of which are triangles of areas S1, S2 and S3. Prove that the area of △ABC is equal to ¡√S1 + √S2 + √S3 ¢2. 1.36. Prove that the area of a triangle whose sides are equal to the medians of a triangle of area S is equal to 3 4S. 1.37. a) Prove that the area of the quadrilateral formed by the midpoints of the sides of convex quadrilateral ABCD is half that of ABCD. b) Prove that if the diagonals of a convex quadrilateral are equal, then its area is the product of the lengths of the segments which connect the midpoints of its opposite sides. 1.38. Point O lying inside a convex quadrilateral of area S is reﬂected symmetrically through the midpoints of its sides. Find the area of the quadrilateral with its vertices in the images of O under the reﬂections. §4. Auxiliary equal triangles 1.39. In right triangle ABC with right angle ∠C, points D and E divide leg BC of into three equal parts. Prove that if BC = 3AC, then ∠AEC + ∠ADC + ∠ABC = 90◦. 1.40. Let K be the midpoint of side AB of square ABCD and let point L divide diagonal AC in the ratio of AL : LC = 3 : 1. Prove that ∠KLD is a right angle. 1.41. In square ABCD straight lines l1 and l2 pass through vertex A. The lines intersect the square’s sides. Perpendiculars BB1, BB2, DD1, and DD2 are dropped to these lines. Prove that segments B1B2 and D1D2 are equal and perpendicular to each other. 1.42. Consider an isosceles right triangle ABC with CD = CE and points D and E on sides CA and CB, respectively. Extensions of perpendiculars dropped from D and C to AE intersect the hypotenuse AB at K and L. Prove that KL = LB. 1.43. Consider an inscribed quadrilateral ABCD. The lengths of sides AB, BC, CD, and DA are a, b, c, and d, respectively. Rectangles are constructed outwards on the sides of §5. THE TRIANGLE DETERMINED BY THE BASES OF THE HEIGHTS 19 the quadrilateral; the sizes of the rectangles are a × c, b × d, c × a and d × b, respectively. Prove that the centers of the rectangles are vertices of a rectangle. 1.44. Hexagon ABCDEF is inscribed in a circle of radius R centered at O; let AB = CD = EF = R. Prove that the intersection points, other than O, of the pairs of circles circumscribed about △BOC, △DOE and △FOA are the vertices of an equilateral triangle with side R. 1.45. Equilateral triangles BCK and DCL are constructed outwards on sides BC and CD of parallelogram ABCD. Prove that AKL is an equilateral triangle. 1.46. Squares are constructed outwards on the sides of a parallelogram. Prove that their centers form a square. 1.47. Isosceles triangles with angles 2α, 2β and 2γ at vertices A′, B′ and C′ are con-structed outwards on the sides of triangle ABC; let α + β + γ = 180◦. Prove that the angles of △A′B′C′ are equal to α, β and γ. 1.48. On the sides of △ABC as on bases, isosceles similar triangles AB1C and AC1B are constructed outwards and an isosceles triangle BA1C is constructed inwards. Prove that AB1A1C1 is a parallelogram. 1.49. a) On sides AB and AC of △ABC equilateral triangles ABC1 and AB1C are constructed outwards; let ∠C1 = ∠B1 = 90◦, ∠ABC1 = ∠ACB1 = ϕ; let M be the midpoint of BC. Prove that MB1 = MC1 and ∠B1MC1 = 2ϕ. b) Equilateral triangles are constructed outwards on the sides of △ABC. Prove that the centers of the triangles constructed form an equilateral triangle whose center coincides with the intersection point of the medians of △ABC. 1.50. Isosceles triangles AC1B and AB1C with an angle ϕ at the vertex are constructed outwards on the unequal sides AB and AC of a scalene triangle △ABC. a) Let M be a point on median AA1 (or on its extension), let M be equidistant from B1 and C1. Prove that ∠B1MC1 = ϕ. b) Let O be a point of the midperpendicular to segment BC, let O be equidistant from B1 and C1. Prove that ∠B1OC = 180◦−ϕ. 1.51. Similar rhombuses are constructed outwards on the sides of a convex rectangle ABCD, so that their acute angles (equal to α) are adjacent to vertices A and C. Prove that the segments which connect the centers of opposite rhombuses are equal and the angle between them is equal to α. §5. The triangle determined by the bases of the heights 1.52. Let AA1 and BB1 be heights of △ABC. Prove that △A1B1C ∼△ABC. What is the similarity coeﬃcient? 1.53. Height CH is dropped from vertex C of acute triangle ABC and perpendiculars HM and HN are dropped to sides BC and AC, respectively. Prove that △MNC ∼△ABC. 1.54. In △ABC heights BB1 and CC1 are drawn. a) Prove that the tangent at A to the circumscribed circle is parallel to B1C1. b) Prove that B1C1 ⊥OA, where O is the center of the circumscribed circle. 1.55. Points A1, B1 and C1 are taken on the sides of an acute triangle ABC so that segments AA1, BB1 and CC1 meet at H. Prove that AH · A1H = BH · B1H = CH · C1H if and only if H is the intersection point of the heights of △ABC. 1.56. a) Prove that heights AA1, BB1 and CC1 of acute triangle ABC bisect the angles of △A1B1C1. 20 CHAPTER 1. SIMILAR TRIANGLES b) Points C1, A1 and B1 are taken on sides AB, BC and CA, respectively, of acute triangle ABC. Prove that if ∠B1A1C = ∠BA1C1, ∠A1B1C = ∠AB1C1 and ∠A1C1B = ∠AC1B1, then points A1, B1 and C1 are the bases of the heights of △ABC. 1.57. Heights AA1, BB1 and CC1 are drawn in acute triangle ABC. Prove that the point symmetric to A1 through AC lies on B1C1. 1.58. In acute triangle ABC, heights AA1, BB1 and CC1 are drawn. Prove that if A1B1 ∥AB and B1C1 ∥BC, then A1C1 ∥AC. 1.59. Let p be the semiperimeter of acute triangle ABC and q the semiperimeter of the triangle formed by the bases of the heights of △ABC. Prove that p : q = R : r, where R and r are the radii of the circumscribed and the inscribed circles, respectively, of △ABC. §6. Similar ﬁgures 1.60. A circle of radius r is inscribed in a triangle. The straight lines tangent to the circle and parallel to the sides of the triangle are drawn; the lines cut three small triangles oﬀthe triangle. Let r1, r2 and r3 be the radii of the circles inscribed in the small triangles. Prove that r1 + r2 + r3 = r. 1.61. Given △ABC, draw two straight lines x and y such that the sum of lengths of the segments MXM and MYM drawn parallel to x and y from a point M on AC to their intersections with sides AB and BC is equal to 1 for any M. 1.62. In an isosceles triangle ABC perpendicular HE is dropped from the midpoint of base BC to side AC. Let O be the midpoint of HE. Prove that lines AO and BE are perpendicular to each other. 1.63. Prove that projections of the base of a triangle’s height to the sides between which it lies and on the other two heights lie on the same straight line. 1.64. Point B lies on segment AC; semicircles S1, S2, and S3 are constructed on one side of AC, as on diameter. Let D be a point on S3 such that BD ⊥AC. A common tangent line to S1 and S2 touches these semicircles at F and E, respectively. a) Prove that EF is parallel to the tangent to S3 passing through D. b) Prove that BFDE is a rectangle. 1.65. Perpendiculars MQ and MP are dropped from an arbitrary point M of the circle circumscribed about rectangle ABCD to the rectangle’s two opposite sides; the perpendic-ulars MR and MT are dropped to the extensions of the other two sides. Prove that lines PR ⊥QT and the intersection point of PR and QT belongs to a diagonal of ABCD. 1.66. Two circles enclose non-intersecting areas. Common tangent lines to the two circles, one external and one internal, are drawn. Consider two straight lines each of which passes through the tangent points on one of the circles. Prove that the intersection point of the lines lies on the straight line that connects the centers of the circles. Problems for independent study 1.67. The (length of the) base of an isosceles triangle is a quarter of its perimeter. From an arbitrary point on the base straight lines are drawn parallel to the sides of the triangle. How many times is the perimeter of the triangle greater than that of the parallelogram? 1.68. The diagonals of a trapezoid are mutually perpendicular. The intersection point divides the diagonals into segments. Prove that the product of the lengths of the trapezoid’s bases is equal to the sum of the products of the lengths of the segments of one diagonal and those of another diagonal. 1.69. A straight line is drawn through the center of a unit square. Calculate the sum of the squared distances between the four vertices of the square and the line. SOLUTIONS 21 1.70. Points A1, B1 and C1 are symmetric to the center of the circumscribed circle of △ABC through the triangle’s sides. Prove that △ABC = △A1B1C1. 1.71. Prove that if ∠BAC = 2∠ABC, then BC2 = (AC + AB)AC. 1.72. Consider points A, B, C and D on a line l. Through A, B and through C, D parallel straight lines are drawn. Prove that the diagonals of the parallelograms thus formed (or their extensions) intersect l at two points that do not depend on parallel lines but depend on points A, B, C, D only. 1.73. In △ABC bisector AD and midline A1C1 are drawn. They intersect at K. Prove that 2A1K = \|b −c\|. 1.74. Points M and N are taken on sides AD and CD of parallelogram ABCD such that MN ∥AC. Prove that SABM = SCBN. 1.75. On diagonal AC of parallelogram ABCD points P and Q are taken so that AP = CQ. Let M be such that PM ∥AD and QM ∥AB. Prove that M lies on diagonal BD. 1.76. Consider a trapezoid with bases AD and BC. Extensions of the sides of ABCD meet at point O. Segment EF is parallel to the bases and passes through the intersection point of the diagonals. The endpoints of EF lie on AB and CD. Prove that AE : CF = AO : CO. 1.77. Three straight lines parallel to the sides of the given triangle cut three triangles oﬀ it leaving an equilateral hexagon. Find the length of the side of the hexagon if the lengths of the triangle’s sides are a, b and c. 1.78. Three straight lines parallel to the sides of a triangle meet at one point, the sides of the triangle cutting oﬀthe line segments of length x each. Find x if the lengths of the triangle’s sides are a, b and c. 1.79. Point P lies inside △ABC and ∠ABP = ∠ACP. On straight lines AB and AC, points C1 and B1 are taken so that BC1 : CB1 = CP : BP. Prove that one of the diagonals of the parallelogram whose two sides lie on lines BP and CP and two other sides (or their extensions) pass through B1 and C1 is parallel to BC. Solutions 1.1. a) Let P and Q be the midpoints of AB and CD; let K and L be the intersection points of PQ with the diagonals AC and BD, respectively. Then PL = a 2 and PK = 1 2b and so KL = PL −PK = 1 2(a −b). b) Take point F on AD such that BF ∥CD. Let E be the intersection point of MN with BF. Then MN = ME + EN = q · AF p + q + b = q(a −b) + (p + q)b p + q = qa + pb p + q . 1.2. Consider quadrilateral ABCD. Let K, L, M and N be the midpoints of sides AB, BC, CD and DA, respectively. Then KL = MN = 1 2AC and KL ∥MN, that is KLMN is a parallelogram. It becomes clear now that KLMN is a rectangle if the diagonals AC and BD are perpendicular, a rhombus if AC = BD, and a square if AC and BD are of equal length and perpendicular to each other. 1.3. Denote the intersection point of AA1 with BB1 by O. In △B1BC draw segment A1A2 so that A1A2 ∥BB1. Then B1C B1A2 = 1 + p and so AO : OA1 = AB1 : B1A2 = B1C : qB1A2 = (1 + p) : q. 22 CHAPTER 1. SIMILAR TRIANGLES 1.4. Let A2 be the midpoint of A1B. Then CA1 : A1A2 = CP : PC1 and A1A2 : A1B = 1 : 2. So CA1 : A1B = CP : 2PC1. Similarly, CB1 : B1A = CP : 2PC1 = CA1 : A1B. 1.5. Point P lies on the median QM of △AQD (or on its extension). It is easy to verify that the solution of Problem 1.4 remains correct also for the case when P lies on the extension of the median. Consequently, BC ∥AD. 1.6. We have AQ : QC = AP : BC = 1 : n because △AQP ∼△CQB. So AC = AQ + QC = (n + 1)AQ. 1.7. The center of A1B1C1D1 being the midpoint of B1D1 belongs to the line segment which connects the midpoints of AB and CD. Similarly, it belongs to the segment which connects the midpoints of BC and AD. The intersection point of the segments is the center of ABCD. 1.8. Clearly, AK : KM = BK : KD = LK : AK, that is AK2 = LK · KM. 1.9. Let AC be the diameter of the circle circumscribed about ABCD. Drop perpen-diculars AA1 and CC1 to BD (Fig. 2). Figure 2 (Sol. 1.9) We must prove that BA1 = DC1. Drop perpendicular OP from the center O of the circumscribed circle to BD. Clearly, P is the midpoint of BD. Lines AA1, OP and CC1 are parallel to each other and AO = OC. So A1P = PC1 and, since P is the midpoint of BD, it follows that BA1 = DC1. 1.10. We see that BO : OD = DP : PB = k, because BO =PD. Let BC = 1. Then AD = k and ED = 1 k. So k = AD = AE + ED = 1 + 1 k, that is k2 = 1 + k. Finally, observe that k2 = AD2 and 1 + k = BC2 + BC · AD. 1.11. Let C, D, E and F be the midpoints of sides AO, OB, BM and MA, respectively, of quadrilateral AOMB. Since AB = MO = R, where R is the radius of the given circle, CDEF is a rhombus by Problem 1.2. Hence, CE ⊥DF. 1.12. a) If the lines containing the given points are parallel, then the assertion of the problem is obviously true. We assume that the lines meet at O. Then OA : OB = OB1 : OA1 and OC : OA = OA1 : OC1. Hence, OC : OB = OB1 : OC1 and so BC1 ∥CB1 (the ratios of the segment should be assumed to be oriented). b) Let AB1 and CA1 meet at D, let CB1 and AC1 meet at E. Then CA1 : A1D = CB : BA = EC1 : C1A. Since △CB1D ∼△EB1A, points A1, B1 and C1 lie on the same line. 1.13. A point that lies on the bisector of an angle is equidistant from the angle’s legs. Let a be the distance from point A1 to lines AC and AB, let b be the distance from point B1 to lines AB and BC. Further, let A1M : B1M = p : q, where p + q = 1. Then the distances SOLUTIONS 23 from point M to lines AC and BC are equal to qa and pb, respectively. On the other hand, by Problem 1.1 b) the distance from point M to line AB is equal to qa + pb. 1.14. Let the line that passes through the center O of the given rectangle parallel to BC intersect line segment QN at point K (Fig. 3). Figure 3 (Sol. 1.14) Since MO ∥PC, it follows that QM : MP = QO : OC and, since KO ∥BC, it follows that QO : OC = QK : KN. Therefore, QM : MP = QK : KN, i.e., KM ∥NP. Hence, ∠MNP = ∠KMO = ∠QNM. 1.15. Let us draw through point M line EF so that EF ∥CD (points E and F lie on lines BC and AD). Then PL : PK = BL : KD and OK : OL = KA : CL = KA : KF = BL : EL. Since KD = EL, we have PL : PK = OK : OL and, therefore, PL = OK. 1.16. Consider parallelogram ABCD1. We may assume that points D and D1 do not coincide (otherwise the statement of the problem is obvious). On sides AD1 and CD1 take points S1 and R1, respectively, so that SS1 ∥DD1 and RR1 ∥DD1. Let segments PR1 and QS1 meet at N; let N1 and N2 be the intersection points of the line that passes through N parallel to DD1 with segments PR and QS, respectively. Then − − → N1N = β− − → RR1 = αβ− − → DD1 and − − → N2N = α− − → SS1 = αβ− − → DD1. Hence, segments PR and QS meet at N1 = N2. Clearly, PN1 : PR = PN : PR1 = β and QN2 : QS = α. Remark. If α = β, there is a simpler solution. Since BP : BA = BQ : BC = α, it follows that PQ ∥AC and PQ : AC = α. Similarly, RS ∥AC and RS : AC = 1 −α. Therefore, segments PR and QS are divided by their intersection point in the ratio of α : (1 −α). 1.17. a) From vertices A and C drop perpendiculars AK and CL to line BD. Since ∠CBL = ∠ABK and ∠CDL = ∠KDA, we see that △BLC ∼△BKA and △CLD ∼ △AKD. Therefore, AD : DC = AK : CL = AB : BC. b) Taking into account that BA1 : A1C = BA : AC and BA1 + A1C = BC we get BA1 = ac b+c. Since BO is the bisector of triangle ABA1, it follows that AO : OA1 = AB : BA1 = (b + c) : a. 1.18. Let O be the center of the circumscribed circle of isosceles triangle ABC, let B1 be the midpoint of base AC and A1 the midpoint of the lateral side BC. Since △BOA1 ∼ △BCB1, it follows that BO : BA1 = BC : BB1 and, therefore, R = BO = a2 √ 4a2−b2. 24 CHAPTER 1. SIMILAR TRIANGLES 1.19. If ∠EAD = ϕ, then AE = AD cos ϕ = AB cos ϕ and AF = AB sin ϕ. Therefore, 1 AE2 + 1 AF 2 = cos2 ϕ + sin2 ϕ AB2 = 1 AB2. 1.20. It is easy to verify that AB2 2 = AB1 · AC = AC1 · AB = AC2 2. 1.21. a) Since BQ : QM = BN : AM = BK : AK, we have: KQ ∥AM. b) Let O be the center of the inscribed circle. Since ∠CBA + ∠BAD = 180◦, it follows that ∠ABO + ∠BAO = 90◦. Therefore, △AKO ∼△OKB, i.e., AK : KO = OK : KB. Consequently, AK·KB = KO2 = R2, where R is the radius of the inscribed circle. Similarly, CL · LD = R2. 1.22. If angle ∠ABC is obtuse (resp. acute), then angle ∠MAN is also obtuse (resp. acute). Moreover, the legs of these angles are mutually perpendicular. Therefore, ∠ABC = ∠MAN. Right triangles ABM and ADN have equal angles ∠ABM = ∠ADN, therefore, AM : AN = AB : AD = AB : CB, i.e., △ABC ∼△MAN. 1.23. On diagonal AC, take points D′ and B′ such that BB′ ∥l and DD′ ∥l. Then AB : AE = AB′ : AG and AD : AF = AD′ : AG. Since the sides of triangles ABB′ and CDD′ are pairwise parallel and AB = CD, these triangles are equal and AB′ = CD′. Therefore, AB AE + AD AF = AB′ AG + AD′ AG = CD′ + AD′ AG = AC AG. 1.24. Let us drop from vertex B perpendicular BG to AC (Fig. 4). Figure 4 (Sol. 1.24) Since triangles ABG and ACE are similar, AC · AG = AE · AB. Lines AF and CB are parallel, consequently, ∠GCB = ∠CAF. We also infer that right triangles CBG and ACF are similar and, therefore, AC · CG = AF · BC. Summing the equalities obtained we get AC · (AG + CG) = AE · AB + AF · BC. Since AG + CG = AC, we get the equality desired. 1.25. Since α + β = 90◦−1 2α, it follows that γ = 180◦−α −β = 90◦+ 1 2α. Therefore, it is possible to ﬁnd point D on side AB so that ∠ACD = 90◦−1 2α, i.e., AC = AD. Then △ABC ∼△CBD and, therefore, BC : BD = AB : CB, i.e., a2 = c(c −b). 1.26. As segments AB and CD move, triangle AMC is being replaced by another triangle similar to the initial one. Therefore, the quantity AM CM remains a constant. Analogously, BM DM remains a constant. 1.27. Let medians meet at O; denote the intersection points of median AK with lines FP and FE by Q and M, respectively; denote the intersection points of median CL with lines EP and FE by R and N, respectively (Fig. 5). Clearly, FM : FE = FQ : FP = LO : LC = 1 : 3, i.e., FM = 1 3FE. Similarly, EN = 1 3FE. SOLUTIONS 25 Figure 5 (Sol. 1.27) 1.28. Let A and B be the intersection points of the given line with the angle’s legs. On segments AC and BC, take points K and L, respectively, so that PK ∥BC and PL ∥AC. Since △AKP ∼△PLB, it follows that AK : KP = PL : LB and, therefore, (a −p)(b −p) = p2, where p = PK = PL. Hence, 1 a + 1 b = 1 p. 1.29. Denote the midpoint of side BC by O and the intersection points of AK and AL with side BC by P and Q, respectively. We may assume that BP < BQ. Triangle LCO is an equilateral one and LC ∥AB. Therefore, △ABQ ∼△LCQ, i.e., BQ : QC = AB : LC = 2 : 1. Hence, BC = BQ + QC = 3QC. Similarly, BC = 3BP. 1.30. Since BK : BO = BO : AB and ∠KBO = ∠ABO, it follows that △KOB ∼ △OAB. Hence, ∠KOB = ∠OAB. Similarly, ∠AOM = ∠ABO. Therefore, ∠KOM = ∠KOB + ∠BOA + ∠AOM = ∠OAB + ∠BOA + ∠ABO = 180◦, i.e., points K, O and M lie on one line. 1.31. Since ∠AMN = ∠MNC and ∠BMN = ∠MNA, we see that ∠AMB = ∠ANC. Moreover, AM : AN = NB : NM = BM : CN. Hence, △AMB ∼△ANC and, therefore, ∠MAB = ∠NAC. Consequently, ∠BAC = ∠MAN. For the other angles the proof is similar. Let points B1 and C1 be symmetric to B and C, respectively, through the midperpen-dicular to segment MN. Since AM : NB = MN : BM = MC : NC, it follows that MA · MC1 = AM · NC = NB · MC = MB1 · MC. Therefore, point A lies on the circle circumscribed about trapezoid BB1CC1. 1.32. Since ∠AEB+∠BEC = 180◦, angles ∠AEB and ∠BEC cannot be diﬀerent angles of similar triangles ABE and BEC, i.e., the angles are equal and BE is a perpendicular. Two cases are possible: either ∠ABE = ∠CBE or ∠ABE = ∠BCE. The ﬁrst case should be discarded because in this case △ABE = △CBE. In the second case we have ∠ABC = ∠ABE + ∠CBE = ∠ABE + ∠BAE = 90◦. In right triangle ABC the ratio of the legs’ lengths is equal to 1 : √ 3; hence, the angles of triangle ABC are equal to 90◦, 60◦, 30◦. 1.33. We have SBDEF 2SADE = SBDE SADE = DB AD = EF AD = q SEF C SADE . Hence, SBDEF = 2 p SADE · SEFC. 1.34. Let MN = x; let E be the intersection point of lines AB and CD. Triangles EBC, EMN and EAD are similar, hence, SEBC : SEMN : SEAD = a2 : x2 : b2. Since SEMN −SEBC = SMBCN = SMADN = SEAD −SEMN, it follows that x2 −a2 = b2 −x2, i.e., x2 = 1 2(a2 + b2). 26 CHAPTER 1. SIMILAR TRIANGLES 1.35. Through point Q inside triangle ABC draw lines DE, FG and HI parallel to BC, CA and AB, respectively, so that points F and H would lie on side BC, points E and I on side AC, points D and G on side AB (Fig. 6). Figure 6 (Sol. 1.35) Set S = SABC, S1 = SGDQ, S2 = SIEQ, S3 = SHFQ. Then r S1 S + r S2 S + r S3 S = GQ AC + IE AC + FQ AC = AI + IE + EC AC = 1, i.e., S = (√S1 + √S2 + √S3)2. 1.36. Let M be the intersection point of the medians of triangle ABC; let point A1 be symmetric to M through the midpoint of segment BC. The ratio of the lengths of sides of triangle CMA1 to the lengths of the corresponding medians of triangle ABC is to 2 : 3. Therefore, the area to be found is equal to 9 4SCMA1. Clearly, SCMA1 = 1 3S (cf. the solution of Problem 4.1). 1.37. Let E, F, G and H be the midpoints of sides AB, BC, CD and DA, respectively. a) Clearly, SAEH + SCFG = 1 4SABD + 1 4SCBD = 1 4SABCD. Analogously, SBEF + SDGH = 1 4SABCD; hence, SEFGH = SABCD −1 4SABCD −1 4SABCD = 1 2SABCD. b) Since AC = BD, it follows that EFGH is a rhombus (Problem 1.2). By heading a) we have SABCD = 2SEFGH = EG · FH. 1.38. Let E, F, G and H be the midpoints of sides of quadrilateral ABCD; let points E1, F1, G1 and H1 be symmetric to point O through these points, respectively. Since EF is the midline of triangle E1OF1, we see that SE1OF1 = 4SEOF. Similarly, SF1OG1 = 4SFOG, SG1OH1 = 4SGOH, SH1OE1 = 4SHOE. Hence, SE1F1G1H1 = 4SEFGH. By Problem 1.37 a) SABCD = 2SEFGH. Hence, SE1F1G1H1 = 2SABCD = 2S. 1.39. First solution. Let us consider square BCMN and divide its side MN by points P and Q into three equal parts (Fig. 7). Then △ABC = △PDQ and △ACD = △PMA. Hence, triangle △PAD is an isosceles right triangle and ∠ABC + ∠ADC = ∠PDQ + ∠ADC = 45◦. Second solution. Since DE = 1, EA = √ 2, EB = 2, AD = √ 5 and BA = √ 10, it follows that DE : AE = EA : EB = AD : BA and △DEA ∼△AEB. Therefore, ∠ABC = ∠EAD. Moreover, ∠AEC = ∠CAE = 45◦. Hence, ∠ABC + ∠ADC + ∠AEC = (∠EAD + ∠CAE) + ∠ADC = ∠CAD + ∠ADC = 90◦. SOLUTIONS 27 Figure 7 (Sol. 1.39) 1.40. From point L drop perpendiculars LM and LN on AB and AD, respectively. Then KM = MB = ND and KL = LB = DL and, therefore, right triangles KML and DNL are equal. Hence, ∠DLK = ∠NLM = 90◦. 1.41. Since D1A = B1B, AD2 = BB2 and ∠D1AD2 = ∠B1BB2, it follows that △D1AD2 = △B1BB2. Sides AD1 and BB1 (and also AD2 and BB2) of these triangles are perpendicular and, therefore, B1B2 ⊥D1D2. 1.42. On the extension of segment AC beyond point C take point M so that CM = CE (Fig. 8). Figure 8 (Sol. 1.42) Then under the rotation with center C through an angle of 90◦triangle ACE turns into triangle BCM. Therefore, line MB is perpendicular to line AE; hence, it is parallel to line CL. Since MC = CE = DC and lines DK, CL and MB are parallel, KL = LB. 1.43. Let rectangles ABC1D1 and A2BCD2 be constructed on sides AB and BC; let P, Q, R and S be the centers of rectangles constructed on sides AB, BC, CD and DA, respectively. Since ∠ABC + ∠ADC = 180◦, it follows that △ADC = △A2BC1 and, there-fore, △RDS = △PBQ and RS = PQ. Similarly, QR = PS. Therefore, PQRS is a parallelogram such that one of triangles RDS and PBQ is constructed on its sides outwards and on the other side inwards; a similar statement holds for triangles QCR and SAP as well. Therefore, ∠PQR + ∠RSP = ∠BQC + ∠DSA = 180◦because ∠PQB = ∠RSD and ∠RQC = ∠PSA. It follows that PQRS is a rectangle. 1.44. Let K, L and M be the intersection points of the circumscribed circles of triangles FOA and BOC, BOC and DOE, DOE and FOA, respectively; 2α, 2β and 2γ the angles at the vertices of isosceles triangles BOC, DOE and FOA, respectively (Fig. 9). 28 CHAPTER 1. SIMILAR TRIANGLES Figure 9 (Sol. 1.44) Point K lies on arc ⌣OB of the circumscribed circle of the isosceles triangle BOC and, therefore, ∠OKB = 90◦+ α. Similarly, ∠OKA = 90◦+ γ. Since α + β + γ = 90◦, it follows that ∠AKB = 90◦+ β. Inside equilateral triangle AOB there exists a unique point K that serves as the vertex of the angles that subtend its sides and are equal to the given angles. Similar arguments for a point L inside triangle COD show that △OKB = △CLO. Now, let us prove that △KOL = △OKB. Indeed, ∠COL = ∠KBO; hence, ∠KOB + ∠COL = 180◦−∠OKB = 90◦−α and, therefore, ∠KOL = 2α + (90◦−α) = 90◦+ α = ∠OKB. It follows that KL = OB = R. Similarly, LM = MK = R. 1.45. Let ∠A = α. It is easy to verify that both angles ∠KCL and ∠ADL are equal to 240◦−α (or 120◦+ α). Since KC = BC = AD and CL = DL, it follows that △KCL = △ADL and, therefore, KL = AL. Similarly, KL = AK. 1.46. Let P, Q and R be the centers of the squares constructed on sides DA, AB and BC, respectively, in parallelogram ABCD with an acute angle of α at vertex A. It is easy to verify that ∠PAQ = 90◦+ α = ∠RBQ; hence, △PAQ = △RBQ. Sides AQ and BQ of these triangles are perpendicular, hence, PQ ⊥QR. 1.47. First, observe that the sum of the angles at vertices A, B and C of hexagon AB′CA′BC′ is equal to 360◦because by the hypothesis the sum of its angles at the other vertices is equal to 360◦. On side AC′, construct outwards triangle △AC′P equal to triangle △BC′A′ (Fig. 10). Figure 10 (Sol. 1.47) Then △AB′P = △CB′A′ because AB′ = CB′, AP = CA′ and ∠PAB′ = 360◦−∠PAC′ −∠C′AB′ = 360◦−∠A′BC′ −∠C′AB′ = ∠A′CB′. Hence, △C′B′A′ = △C′B′P and, therefore, 2∠A′B′C′ = ∠PB′A′ = ∠AB′C because ∠PB′A = ∠A′B′C. SOLUTIONS 29 1.48. Since BA : BC = BC1 : BA1 and ∠ABC = ∠C1BA1, it follows that △ABC ∼ △C1BA1. Similarly, △ABC ∼△B1A1C. Since BA1 = A1C, it follows that △C1BA1 = △B1A1C. Therefore, AC1 = C1B = B1A1 and AB1 = B1C = C1A1. It is also clear that quadrilateral AB1A1C1 is a convex one. 1.49. a) Let P and Q be the midpoints of sides AB and AC. Then MP = 1 2AC = QB1, MQ = 1 2AB = PC1 and ∠C1PM = ∠C1PB + ∠BPM = ∠B1QC + ∠CQM = ∠B1QM. Hence, △MQB1 = △C1PM and, therefore, MC1 = MB1. Moreover, ∠PMC1 + ∠QMB1 = ∠QB1M + ∠QMB1 = 180◦−∠MQB1 and ∠MQB1 = ∠A + ∠CQB1 = ∠A + (180◦−2ϕ). Therefore, ∠B1MC1 = ∠PMQ + 2ϕ −∠A = 2ϕ. (The case when ∠C1PB + ∠BPM > 180◦ is analogously treated.) b) On sides AB and AC, take points B′ and C′, respectively, such that AB′ : AB = AC′ : AC = 2 : 3. The midpoint M of segment B′C′ coincides with the intersection point of the medians of triangle ABC. On sides AB′ and AC′, construct outwards right triangles AB′C1 and AB1C′ with angle ϕ = 60◦as in heading a). Then B1 and C1 are the centers of right triangles constructed on sides AB and AC; on the other hand, by heading a), MB1 = MC1 and ∠B1MC1 = 120◦. Remark. Statements of headings a) and b) remain true for triangles constructed in-wards, as well. 1.50. a) Let B′ be the intersection point of line AC and the perpendicular to line AB1 erected from point B1; deﬁne point C′ similarly. Since AB′ : AC′ = AC1 : AB1 = AB : AC, it follows that B′C′ ∥BC. If N is the midpoint of segment B′C′, then, as follows from Problem 1.49, NC1 = NB1 (i.e., N = M) and ∠B1NC1 = 2∠AB′B1 = 180◦−2∠CAB1 = ϕ. b) On side BC construct outwards isosceles triangle BA1C with angle 360◦−2ϕ at vertex A1 (if ϕ < 90◦construct inwards a triangle with angle 2ϕ). Since the sum of the angles at the vertices of the three constructed isosceles triangles is equal to 360◦, it follows that the angles of triangle A1B1C1 are equal to 180◦−ϕ, 1 2ϕ and 1 2ϕ (cf. Problem 1.47). In particular, this triangle is an isosceles one, hence, A1 = O. 1.51. Let O1, O2, O3 and O4 be the centers of rhombuses constructed on sides AB, BC, CA and DA, respectively; let M be the midpoint of diagonal AC. Then MO1 = MO2 and ∠O1MO2 = α (cf. Problem 1.49). Similarly, MO3 = MO4 and ∠O3MO4 = α. Therefore, under the rotation through an angle of α about point M triangle △O1MO3 turns into △O2MO4. 1.52. Since A1C = AC\| cos C\| , B1C = BC\| cos C\| and angle ∠C is the common angle of triangles ABC and A1B1C, these triangles are similar; the similarity coeﬃcient is equal to \| cos C\|. 1.53. Since points M and N lie on the circle with diameter CH, it follows that ∠CMN = ∠CHN and since AC ⊥HN, we see that ∠CHN = ∠A. Similarly, ∠CNM = ∠B. 1.54. a) Let l be the tangent to the circumscribed circle at point A. Then ∠(l, AB) = ∠(AC, CB) = ∠(C1B1, AC1) and, therefore, l ∥B1C1. b) Since OA ⊥l and l ∥B1C1, it follows that OA ⊥B1C1. 1.55. If AA1, BB1 and CC1 are heights, then right triangles AA1C and BB1C have equal angles at vertex C and, therefore, are similar. It follows that △A1BH ∼△B1AH, consequently, AH · A1H = BH · B1H. Similarly, BH · B1H = CH · C1H. If AH · A1H = BH · B1H = CH · C1H, then △A1BH ∼△B1AH; hence, ∠BA1H = ∠AB1H = ϕ. Thus, ∠CA1H = ∠CB1H = 180◦−ϕ. 30 CHAPTER 1. SIMILAR TRIANGLES Similarly, ∠AC1H = ∠CA1H = 180◦−ϕ and ∠AC1H = ∠AB1H = ϕ. Hence, ϕ = 90◦, i.e., AA1, BB1 and CC1 are heights. 1.56. a) By Problem 1.52 ∠C1A1B = ∠CA1B1 = ∠A. Since AA1 ⊥BC, it follows that ∠C1A1A = ∠B1A1A. The proof of the fact that rays B1B and C1C are the bisectors of angles A1B1C1 and A1C1B1 is similar. b) Lines AB, BC and CA are the bisectors of the outer angles of triangle A1B1C1, hence, A1A is the bisector of angle ∠B1A1C1 and, therefore, AA1 ⊥BC. For lines BB1 and CC1 the proof is similar. 1.57. From the result of Problem 1.56 a) it follows that the symmetry through line AC sends line B1A1 into line B1C1. 1.58. By Problem 1.52 ∠B1A1C = ∠BAC. Since A1B1 ∥AB, it follows that ∠B1A1C = ∠ABC. Hence, ∠BAC = ∠ABC. Similarly, since B1C1 ∥BC, it follows that ∠ABC = ∠BCA. Therefore, triangle ABC is an equilateral one and A1C1 ∥AC. 1.59. Let O be the center of the circumscribed circle of triangle ABC. Since OA ⊥B1C1 (cf. Problem 1.54 b), it follows that SAOC1 + SAOB1 = 1 2(R · B1C1). Similar arguments for vertices B and C show that SABC = qR. On the other hand, SABC = pr. 1.60. The perimeter of the triangle cut oﬀby the line parallel to side BC is equal to the sum of distances from point A to the tangent points of the inscribed circle with sides AB and AC; therefore, the sum of perimeters of small triangles is equal to the perimeter of triangle ABC, i.e., P1 + P2 + P3 = P. The similarity of triangles implies that ri r = Pi P . Summing these equalities for all the i we get the statement desired. 1.61. Let M = A. Then XA = A; hence, AYA = 1. Similarly, CXC = 1. Let us prove that y = AYA and x = CXC are the desired lines. On side BC, take point D so that AB ∥MD, see Fig. 11. Let E be the intersection point of lines CXC and MD. Then, XMM + YMM = XCE + YMM. Since △ABC ∼△MDC, it follows that CE = YMM. Therefore, CE = YMM. Hence, XMM + YMM = XCE + CE = XCC = 1. Figure 11 (Sol. 1.61) 1.62. Let D be the midpoint of segment BH. Since △BHA ∼△HEA, it follows that AD : AO = AB : AH and ∠DAH = ∠OAE. Hence, ∠DAO = ∠BAH and, therefore, △DAO ∼△BAH and ∠DOA = ∠BAH = 90◦. 1.63. Let AA1, BB1 and CC1 be heights of triangle ABC. Let us drop from point B1 perpendiculars B1K and B1N to sides AB and BC, respectively, and perpendiculars B1L and B1M to heights AA1 and CC1, respectively. Since KB1 : C1C = AB1 : AC = LB1 : A1C, it follows that △KLB1 ∼△C1A1C and, therefore, KL ∥C1A1. Similarly, MN ∥C1A1. Moreover, KN ∥C1A1 (cf. Problem 1.53). It follows that points K, L, M and N lie on one line. 1.64. a) Let O be the midpoint of AC, let O1 be the midpoint of AB and O2 the midpoint of BC. Assume that AB ≤BC. Through point O1 draw line O1K parallel to EF (point K lies on segment EO2). Let us prove that right triangles DBO and O1KO2 are equal. Indeed, SOLUTIONS 31 O1O2 = DO = 1 2AC and BO = KO2 = 1 2(BC −AB). Since triangles DBO and O1KO2 are equal, we see that ∠BOD = ∠O1O2E, i.e., line DO is parallel to EO2 and the tangent drawn through point D is parallel to line EF. b) Since the angles between the diameter AC and the tangents to the circles at points F, D, E are equal, it follows that ∠FAB = ∠DAC = ∠EBC and ∠FBA + ∠DCA = ∠ECB, i.e., F lies on line segment AD and E lies on line segment DC. Moreover, ∠AFB = ∠BEC = ∠ADC = 90◦and, therefore, FDEB is a rectangle. 1.65. Let MQ and MP be perpendiculars dropped on sides AD and BC, let MR and MT be perpendiculars dropped on the extensions of sides AB and CD (Fig. 12). Denote by M1 and P1 the other intersection points of lines RT and QP with the circle. Figure 12 (Sol. 1.65) Since TM1 = RM = AQ and TM1 ∥AQ, it follows that AM1 ∥TQ. Similarly, AP1 ∥ RP. Since ∠M1AP1 = 90◦, it follows that RP ⊥TQ. Denote the intersection points of lines TQ and RP, M1A and RP, P1A and TQ by E, F, G, respectively. To prove that point E lies on line AC, it suﬃces to prove that rectangles AFEG and AM1CP1 are similar. Since ∠ARF = ∠AM1R = ∠M1TG = ∠M1CT, we may denote the values of these angles by the same letter α. We have: AF = RA sin α = M1A sin2 α and AG = M1T sin α = M1C sin2 α. Therefore, rectangles AFEG and AM1CP1 are similar. 1.66. Denote the centers of the circles by O1 and O2. The outer tangent is tangent to the ﬁrst circle at point K and to the other circle at point L; the inner tangent is tangent to the ﬁrst circle at point M and to the other circle at point N (Fig. 13). Figure 13 (Sol. 1.66) Let lines KM and LN intersect line O1O2 at points P1 and P2, respectively. We have to prove that P1 = P2. Let us consider points A, D1, D2 — the intersection points of KL with MN, KM with O1A, and LN with O2A, respectively. Since ∠O1AM +∠NAO2 = 90◦, 32 CHAPTER 1. SIMILAR TRIANGLES right triangles O1MA and ANO2 are similar; we also see that AO2 ∥KM and AO1 ∥LN. Since these lines are parallel, AD1 : D1O1 = O2P1 : P1O1 and D2O2 : AD2 = O2P2 : P2O1. The similarity of quadrilaterals AKO1M and O2NAL yields AD1 : D1O1 = D2O2 : AD2. Therefore, O2P1 : P1O1 = O2P2 : P2O1, i.e., P1 = P2. CHAPTER 2. INSCRIBED ANGLES Background 1. Angle ∠ABC whose vertex lies on a circle and legs intersect this circle is called inscribed in the circle. Let O be the center of the circle. Then ∠ABC = ½ 1 2∠AOC if points B and O lie on one side of AC 180◦−1 2∠AOC otherwise. The most important and most often used corollary of this fact is that equal chords subtend angles that either are equal or the sum of the angles is equal to 180◦. 2. The value of the angle between chord AB and the tangent to the circle that passes through point A is equal to half the angle value of arc ⌣AB. 3. The angle values of arcs conﬁned between parallel chords are equal. 4. As we have already said, if two angles subtend the same chord, either they are equal or the sum of their values is 180◦. In order not to consider various variants of the positions of points on the circle let us introduce the notion of an oriented angle between lines. The value of the oriented angle between lines AB and CD (notation: ∠(AB, CD)) is the value of the angle by which we have to rotate line AB counterclockwise in order for it to become parallel to line CD. The angles that diﬀer by n · 180◦are considered equal. Notice that, generally, the oriented angle between lines CD and AB is not equal to the oriented angle between lines AB and CD (the sum of ∠(AB, CD) and ∠(CD, AB) is equal to 180◦which, according to our convention, is the same as 0◦). It is easy to verify the following properties of the oriented angles: a) ∠(AB, BC) = −∠(BC, AB); b) ∠(AB, CD) + ∠(CD, EF) = ∠(AB, EF); c) points A, B, C, D not on one line lie on one circle if and only if ∠(AB, BC) = ∠(AD, DC). (To prove this property we have to consider two cases: points B and D lie on one side of AC; points B and D lie on diﬀerent sides of AC.) Introductory problems 1. a) From point A lying outside a circle rays AB and AC come out and intersect the circle. Prove that the value of angle ∠BAC is equal to half the diﬀerence of the angle measures of the arcs of the circle conﬁned inside this angle. b) The vertex of angle ∠BAC lies inside a circle. Prove that the value of angle ∠BAC is equal to half the sum of angle measures of the arcs of the circle conﬁned inside angle ∠BAC and inside the angle symmetric to it through vertex A. 2. From point P inside acute angle ∠BAC perpendiculars PC1 and PB1 are dropped on lines AB and AC. Prove that ∠C1AP = ∠C1B1P. 3. Prove that all the angles formed by the sides and diagonals of a regular n-gon are integer multiples of 180◦ n . 4. The center of an inscribed circle of triangle ABC is symmetric through side AB to the center of the circumscribed circle. Find the angles of triangle ABC. 33 34 CHAPTER 2. INSCRIBED ANGLES 5. The bisector of the exterior angle at vertex C of triangle ABC intersects the circum-scribed circle at point D. Prove that AD = BD. §1. Angles that subtend equal arcs 2.1. Vertex A of an acute triangle ABC is connected by a segment with the center O of the circumscribed circle. From vertex A height AH is drawn. Prove that ∠BAH = ∠OAC. 2.2. Two circles intersect at points M and K. Lines AB and CD are drawn through M and K, respectively; they intersect the ﬁrst circle at points A and C, the second circle at points B and D, respectively. Prove that AC ∥BD. 2.3. From an arbitrary point M inside a given angle with vertex A perpendiculars MP and MQ are dropped to the sides of the angle. From point A perpendicular AK is dropped on segment PQ. Prove that ∠PAK = ∠MAQ. 2.4. a) The continuation of the bisector of angle ∠B of triangle ABC intersects the circumscribed circle at point M; O is the center of the inscribed circle, Ob is the center of the escribed circle tangent to AC. Prove that points A, C, O and Ob lie on a circle centered at M. b) Point O inside triangle ABC is such that lines AO, BO and CO pass through the centers of the circumscribed circles of triangles BCO, ACO and ABO, respectively. Prove that O is the center of the inscribed circle of triangle ABC. 2.5. Vertices A and B of right triangle ABC with right angle ∠C slide along the sides of a right angle with vertex P. Prove that in doing so point C moves along a line segment. 2.6. Diagonal AC of square ABCD coincides with the hypothenuse of right triangle ACK, so that points B and K lie on one side of line AC. Prove that BK = \|AK −CK\| √ 2 and DK = AK + CK √ 2 . 2.7. In triangle ABC medians AA1 and BB1 are drawn. Prove that if ∠CAA1 = ∠CBB1, then AC = BC. 2.8. Each angle of triangle ABC is smaller than 120◦. Prove that inside △ABC there exists a point that serves as the vertex for three angles each of value 120◦and subtending the side of the triangle diﬀerent from the sides subtended by the other angles. 2.9. A circle is divided into equal arcs by n diameters. Prove that the bases of the perpendiculars dropped from an arbitrary point M inside the circle to these diameters are vertices of a regular n-gon. 2.10. Points A, B, M and N on a circle are given. From point M chords MA1 and MB1 perpendicular to lines NB and NA, respectively, are drawn. Prove that AA1 ∥BB1. 2.11. Polygon ABCDEF is an inscribed one; AB ∥DE and BC ∥EF. Prove that CD ∥AF. 2.12. Polygon A1A2 . . . A2n as an inscribed one. We know that all the pairs of its opposite sides except one are parallel. Prove that for any odd n the remaining pair of sides is also parallel and for any even n the lengths of the exceptional sides are equal. 2.13. Consider triangle ABC. Prove that there exist two families of equilateral triangles whose sides (or extensions of the sides) pass through points A, B and C. Prove also that the centers of triangles from these families lie on two concentric circles. §3. THE ANGLE BETWEEN A TANGENT AND A CHORD 35 §2. The value of an angle between two chords The following fact helps to solve problems from this section. Let A, B, C, D be points on a circle situated in the order indicated. Then ∠(AC, BD) = ⌣AB+ ⌣CD 2 and ∠(AB, CD) = \| ⌣AD−⌣CB\| 2 . To prove this, we have to draw a chord parallel to another chord through the endpoint of one of the chords. 2.14. Points A, B, C, D in the indicated order are given on a circle. Let M be the midpoint of arc ⌣AB. Denote the intersection points of chords MC and MD with chord AB by E and K. Prove that KECD is an inscribed quadrilateral. 2.15. Concider an equilateral triangle. A circle with the radius equal to the triangle’s height rolls along a side of the triangle. Prove that the angle measure of the arc cut oﬀthe circle by the sides of the triangle is always equal to 60◦. 2.16. The diagonals of an isosceles trapezoid ABCD with lateral side AB intersect at point P. Prove that the center O of the inscribed circle lies on the inscribed circle of triangle APB. 2.17. Points A, B, C, D in the indicated order are given on a circle; points A1, B1, C1 and D1 are the midpoints of arcs ⌣AB, ⌣BC, ⌣CD and ⌣DA, respectively. Prove that A1C1 ⊥B1D1. 2.18. Point P inside triangle ABC is taken so that ∠BPC = ∠A + 60◦, ∠APC = ∠B + 60◦and ∠APB = ∠C + 60◦. Lines AP, BP and CP intersect the circumscribed circle of triangle ABC at points A′, B′ and C′, respectively. Prove that triangle A′B′C′ is an equilateral one. 2.19. Points A, C1, B, A1, C, B1 in the indicated order are taken on a circle. a) Prove that if lines AA1, BB1 and CC1 are the bisectors of the angles of triangle ABC, then they are the heights of triangle A1B1C1. b) Prove that if lines AA1, BB1 and CC1 are the heights of triangle ABC, then they are the bisectors of the angles of triangle A1B1C1. 2.20. Triangles T1 and T2 are inscribed in a circle so that the vertices of triangle T2 are the midpoints of the arcs into which the circle is divided by the vertices of triangle T1. Prove that in the hexagon which is the intersection of triangles T1 and T2 the diagonals that connect the opposite vertices are parallel to the sides of triangle T1 and meet at one point. §3. The angle between a tangent and a chord 2.21. Two circles intersect in points P and Q. Through point A on the ﬁrst circle lines AP and AQ are drawn. The lines intersect the second circle in points B and C. Prove that the tangent at A to the ﬁrst circle is parallel to line BC. 2.22. Circles S1 and S2 intersect at points A and P. Tangent AB to circle S1 is drawn through point A, and line CD parallel to AB is drawn through point P (points B and C lie on S2, point D on S1). Prove that ABCD is a parallelogram. 2.23. The tangent at point A to the inscribed circle of triangle ABC intersects line BC at point E; let AD be the bisector of triangle ABC. Prove that AE = ED. 2.24. Circles S1 and S2 intersect at point A. Through point A a line that intersects S1 at point B and S2 at point C is drawn. Through points C and B tangents to the circles are drawn; the tangents intersect at point D. Prove that angle ∠BDC does not depend on the choice of the line that passes through A. 2.25. Two circles intersect at points A and B. Through point A tangents AM and AN, where M and N are points of the respective circles, are drawn. Prove that: 36 CHAPTER 2. INSCRIBED ANGLES a) ∠ABN + ∠MAN = 180◦; b) BM BN = ¡AM AN ¢2. 2.26. Inside square ABCD a point P is taken so that triangle ABP is an equilateral one. Prove that ∠PCD = 15◦. 2.27. Two circles are internally tangent at point M. Let AB be the chord of the greater circle which is tangent to the smaller circle at point T. Prove that MT is the bisector of angle AMB. 2.28. Through point M inside circle S chord AB is drawn; perpendiculars MP and MQ are dropped from point M to the tangents that pass through points A and B respectively. Prove that the value of 1 PM + 1 QM does not depend on the choice of the chord that passes through point M. 2.29. Circle S1 is tangent to sides of angle ABC at points A and C. Circle S2 is tangent to line AC at point C and passes through point B, circle S2 intersects circle S1 at point M. Prove that line AM divides segment BC in halves. 2.30. Circle S is tangent to circles S1 and S2 at points A1 and A2; let B be a point of circle S, let K1 and K2 be the other intersection points of lines A1B and A2B with circles S1 and S2, respectively. Prove that if line K1K2 is tangent to circle S1, then it is also tangent to circle S2. §4. Relations between the values of an angle and the lengths of the arc and chord associated with the angle 2.31. Isosceles trapezoids ABCD and A1B1C1D1 with parallel respective sides are in-scribed in a circle. Prove that AC = A1C1. 2.32. From point M that moves along a circle perpendiculars MP and MQ are dropped on diameters AB and CD, respectively. Prove that the length of segment PQ does not depend on the position of point M. 2.33. In triangle ABC, angle ∠B is equal to 60◦; bisectors AD and CE intersect at point O. Prove that OD = OE. 2.34. In triangle ABC the angles at vertices B and C are equal to 40◦; let BD be the bisector of angle B. Prove that BD + DA = BC. 2.35. On chord AB of circle S centered at O a point C is taken. The circumscribed circle of triangle AOC intersects circle S at point D. Prove that BC = CD. 2.36. Vertices A and B of an equilateral triangle ABC lie on circle S, vertex C lies inside this circle. Point D lies on circle S and BD = AB. Line CD intersects S at point E. Prove that the length of segment EC is equal to the radius of circle S. 2.37. Along a ﬁxed circle another circle whose radius is half that of the ﬁxed one rolls on the inside without gliding. What is the trajectory of a ﬁxed point K of the rolling circle? §5. Four points on one circle 2.38. From an arbitrary point M on leg BC of right triangle ABC perpendicular MN is dropped on hypothenuse AP. Prove that ∠MAN = ∠MCN. 2.39. The diagonals of trapezoid ABCD with bases AD and BC intersect at point O; points B′ and C′ are symmetric through the bisector of angle ∠BOC to vertices B and C, respectively. Prove that ∠C′AC = ∠B′DB. 2.40. The extensions of sides AB and CD of the inscribed quadrilateral ABCD meet at point P; the extensions of sides BC and AD meet at point Q. Prove that the intersection points of the bisectors of angles ∠AQB and ∠BPC with the sides of the quadrilateral are vertices of a rhombus. §6. THE INSCRIBED ANGLE AND SIMILAR TRIANGLES 37 2.41. The inscribed circle of triangle ABC is tangent to sides AB and AC at points M and N, respectively. Let P be the intersection point of line MN with the bisector (or its extension) of angle ∠B. Prove that: a) ∠BPC = 90◦; b) SABP : SABC = 1 : 2. 2.42. Inside quadrilateral ABCD a point M is taken so that ABMD is a parallelogram. Prove that if ∠CBM = ∠CDM, then ∠ACD = ∠BCM. 2.43. Lines AP, BP and CP intersect the circumscribed circle of triangle ABC at points A1, B1 and C1, respectively. On lines BC, CA and AB points A2, B2 and C2, respectively, are taken so that ∠(PA2, BC) = ∠(PB2, CA) = ∠(PC2, AB). Prove that △A2B2C2 ∼△A1B1C1. 2.44. About an equilateral triangle APQ a rectangular ABCD is circumscribed so that points P and Q lie on sides BC and CD, respectively; P ′ and Q′ are the midpoints of sides AP and AQ, respectively. Prove that triangles BQ′C and CP ′D are equilateral ones. 2.45. Prove that if for inscribed quadrilateral ABCD the equality CD = AD + BC holds, then the intersection point of the bisectors of angles ∠A and ∠B lies on side CD. 2.46. Diagonals AC and CE of a regular hexagon ABCDEF are divided by points M and N, respectively, so that AM : AC = CN : CE = λ. Find λ if it is known that points B, M and N lie on a line. 2.47. The corresponding sides of triangles ABC and A1B1C1 are parallel and sides AB and A1B1 lie on one line. Prove that the line that connects the intersection points of the circumscribed circles of triangles A1BC and AB1C contains point C1. 2.48. In triangle ABC heights AA1, BB1 and CC1 are drawn. Line KL is parallel to CC1; points K and L lie on lines BC and B1C1, respectively. Prove that the center of the circumscribed circle of triangle A1KL lies on line AC. 2.49. Through the intersection point O of the bisectors of triangle ABC line MN is drawn perpendicularly to CO so that M and N lie on sides AC and BC, respectively. Lines AO and BO intersect the circumscribed circle of triangle ABC at points A′ and B′, respectively. Prove that the intersection point of lines A′N and B′M lies on the circumscribed circle. §6. The inscribed angle and similar triangles 2.50. Points A, B, C and D on a circle are given. Lines AB and CD intersect at point M. Prove that AC · AD AM = BC · BD BM . 2.51. Points A, B and C on a circle are given; the distance BC is greater than the distance from point B to line l tangent to the circle at point A. Line AC intersects the line drawn through point B parallelly to l at point D. Prove that AB2 = AC · AD. 2.52. Line l is tangent to the circle of diameter AB at point C; points M and N are the projections of points A and B on line l, respectively, and D is the projection of point C on AB. Prove that CD2 = AM · BN. 2.53. In triangle ABC, height AH is drawn and from vertices B and C perpendiculars BB1 and CC1 are dropped on the line that passes through point A. Prove that △ABC ∼ △HB1C1. 2.54. On arc ⌣BC of the circle circumscribed about equilateral triangle ABC, point P is taken. Segments AP and BC intersect at point Q. Prove that 1 PQ = 1 PB + 1 PC . 38 CHAPTER 2. INSCRIBED ANGLES 2.55. On sides BC and CD of square ABCD points E and F are taken so that ∠EAF = 45◦. Segments AE and AF intersect diagonal BD at points P and Q, respectively. Prove that SAEF SAP Q = 2. 2.56. A line that passes through vertex C of equilateral triangle ABC intersects base AB at point M and the circumscribed circle at point N. Prove that CM · CN = AC2 and CM CN = AM · BM AN · BN . 2.57. Consider parallelogram ABCD with an acute angle at vertex A. On rays AB and CB points H and K, respectively, are marked so that CH = BC and AK = AB. Prove that: a) DH = DK; b) △DKH ∼△ABK. 2.58. a) The legs of an angle with vertex C are tangent to a circle at points A and B. From point P on the circle perpendiculars PA1, PB1 and PC1 are dropped on lines BC, CA and AB, respectively. Prove that PC2 1 = PA1 · PB1. b) From point O of the inscribed circle of triangle ABC perpendiculars OA′, OB′, OC′ are dropped on the sides of triangle ABC opposite to vertices A, B and C, respectively, and perpendiculars OA′′, OB′′, OC′′ are dropped to the sides of the triangle with vertices at the tangent points. Prove that OA′ · OB′ · OC′ = OA′′ · OB′′ · OC′′. 2.59. Pentagon ABCDE is inscribed in a circle. Distances from point E to lines AB, BC and CD are equal to a, b and c, respectively. Find the distance from point E to line AD. 2.60. In triangle ABC, heights AA1, BB1 and CC1 are drawn; B2 and C2 are the midpoints of heights BB1 and CC1, respectively. Prove that △A1B2C2 ∼△ABC. 2.61. On heights of triangle ABC points A1, B1 and C1 that divide them in the ratio 2 : 1 counting from the vertex are taken. Prove that △A1B1C1 ∼△ABC. 2.62. Circle S1 with diameter AB intersects circle S2 centered at A at points C and D. Through point B a line is drawn; it intersects S2 at point M that lies inside S1 and it intersects S1 at point N. Prove that MN 2 = CN · ND. 2.63. Through the midpoint C of an arbitrary chord AB on a circle chords KL and MN are drawn so that points K and M lie on one side of AB. Segments KN and ML intersect AB at points Q and P, respectively. Prove that PC = QC. 2.64. a) A circle that passes through point C intersects sides BC and AC of triangle ABC at points A1 and B1, respectively, and it intersects the circumscribed circle of triangle ABC at point M. Prove that △AB1M ∼△BA1M. b) On rays AC and BC segments AA1 and BB1 equal to the semiperimeter of triangle ABC are drawn. Let M be a point on the circumscribed circle such that CM ∥A1B1. Prove that ∠CMO = 90◦, where O is the center of the inscribed circle. §7. The bisector divides an arc in halves 2.65. In triangle ABC, sides AC and BC are not equal. Prove that the bisector of angle ∠C divides the angle between the median and the height drawn from this vertex in halves if and only if ∠C = 90◦. 2.66. It is known that in a triangle the median, the bisector and the height drawn from vertex C divide the angle ∠C into four equal parts. Find the angles of this triangle. 2.67. Prove that in triangle ABC bisector AE lies between median AM and height AH. §9. THREE CIRCUMSCRIBED CIRCLES INTERSECT AT ONE POINT 39 2.68. Given triangle ABC; on its side AB point P is chosen; lines PM and PN parallel to AC and BC, respectively, are drawn through P so that points M and N lie on sides BC and AC, respectively; let Q be the intersection point of the circumscribed circles of triangles APN and BPM. Prove that all lines PQ pass through a ﬁxed point. 2.69. The continuation of bisector AD of acute triangle ABC inersects the circumscribed circle at point E. Perpendiculars DP and DQ are dropped on sides AB and AC from point D. Prove that SABC = SAPEQ. §8. An inscribed quadrilateral with perpendicular diagonals In this section ABCD is an inscribed quadrilateral whose diagonals intersect at a right angle. We will also adopt the following notations: O is the center of the circumscribed circle of quadrilateral ABCD and P is the intersection point of its diagonals. 2.70. Prove that the broken line AOC divides ABCD into two parts whose areas are equal. 2.71. The radius of the circumscribed circle of quadrilateral ABCD is equal to R. a) Find AP 2 + BP 2 + CP 2 + DP 2. b) Find the sum of squared lengths of the sides of ABCD. 2.72. Find the sum of squared lengths of the diagonals of ABCD if the length of segment OP and the radius of the circumscribed circle R are known. 2.73. From vertices A and B perpendiculars to CD that intersect lines BD and AC at points K and L, respectively, are drawn. Prove that AKLB is a rhombus. 2.74. Prove that the area of quadrilateral ABCD is equal to 1 2(AB · CD + BC · AD). 2.75. Prove that the distance from point O to side AB is equal to half the length of side CD. 2.76. Prove that the line drawn through point P perpendicularly to BC divides side AD in halves. 2.77. Prove that the midpoints of the sides of quadrilateral ABCD and the projections of point P on the sides lie on one circle. 2.78. a) Through vertices A, B, C and D tangents to the circumscribed circle are drawn. Prove that the quadrilateral formed by them is an inscribed one. b) Quadrilateral KLMN is simultaneously inscribed and circumscribed; A and B are the tangent points of the inscribed circle with sides KL and LM, respectively. Prove that AK · BM = r2, where r is the radius of the inscribed circle. §9. Three circumscribed circles intersect at one point 2.79. On sides of triangle ABC triangles ABC′, AB′C and A′BC are constructed outwards so that the sum of the angles at vertices A′, B′ and C′ is a multiple of 180◦. Prove that the circumscribed circles of the constructed triangles intersect at one point. 2.80. a) On sides (or their extensions) BC, CA and AB of triangle ABC points A1, B1 and C1 distinct from the vertices of the triangle are taken (one point on one side). Prove that the circumscribed circles of triangles AB1C1, A1BC1 and A1B1C intersect at one point. b) Points A1, B1 and C1 move along lines BC, CA and AB, respectively, so that all triangles A1B1C1 are similar and equally oriented. Prove that the intersection point of the circumscribed circles of triangles AB1C1, A1BC1 and A1B1C remains ﬁxed in the process. 2.81. On sides BC, CA and AB of triangle ABC points A1, B1 and C1 are taken. Prove that if triangles A1B1C1 and ABC are similar and have opposite orientations, then circumscribed circles of triangles AB1C1, ABC1 and A1B1C pass through the center of the circumscribed circle of triangle ABC. 40 CHAPTER 2. INSCRIBED ANGLES 2.82. Points A′, B′ and C′ are symmetric to a point P relative sides BC, CA and AB, respectively, of triangle ABC. a) The circumscribed circles of triangles AB′C′, A′BC′, A′B′C and ABC have a common point; b) the circumscribed circles of triangles A′BC, AB′C, ABC′ and A′B′C′ have a common point Q; c) Let I, J, K and O be the centers of the circumscribed circles of triangles A′BC, AB′C, ABC′ and A′B′C′, respectively. Prove that QI : OI = QJ : OJ = QK : OK. §10. Michel’s point 2.83. Four lines form four triangles. Prove that a) The circumscribed circles of these triangles have a common point. (Michel’s point.) b) The centers of the circumscribed circles of these triangles lie on one circle that passes through Michel’s point. 2.84. A line intersects sides (or their extensions) AB, BC and CA of triangle ABC at points C1, B1 and A1, respectively; let O, Oa, Ob and Oc be the centers of the circumscribed circles of triangles ABC, AB1C1, A1BC1 and A1B1C, respectively; let H, Ha, Hb and Hc be the respective orthocenters of these triangles. Prove that a) △OaObOc ∼△ABC. b) the midperpendiculars to segments OH, OaHa, ObHb and OcHc meet at one point. 2.85. Quadrilateral ABCD is an inscribed one. Prove that Michel’s point of lines that contain its sides lies on the segment that connects the intersection points of the extensions of the sides. 2.86. Points A, B, C and D lie on a circle centered at O. Lines AB and CD intersect at point E and the circumscribed circles of triangles AEC and BED ilntersect at points E and P. Prove that a) points A, D, P and O lie on one circle; b) ∠EPO = 90◦. 2.87. Given four lines prove that the projections of Michel’s point to these lines lie on one line. See also Problem 19.45. §11. Miscellaneous problems 2.88. In triangle ABC height AH is drawn; let O be the center of the circumscribed circle. Prove that ∠OAH = \|∠B −∠C\|. 2.89. Let H be the intersection point of the heights of triangle ABC; let AA′ be a diameter of its circumscribed circle. Prove that segment A′H divides side BC in halves. 2.90. Through vertices A and B of triangle ABC two parallel lines are drawn and lines m and n are symmetric to them through the bisectors of the corresponding angles. Prove that the intersection point of lines m and n lies on the circumscribed circle of triangle ABC. 2.91. a) Lines tangent to circle S at points B and C are drawn from point A. Prove that the center of the inscribed circle of triangle ABC and the center of its escribed circle tangent to side BC lie on circle S. b) Prove that the circle that passes through vertices B and C of any triangle ABC and the center O of its inscribed circle intercepts on lines AB and AC chords of equal length. 2.92. On sides AC and BC of triangle ABC squares ACA1A2 and BCB1B2 are con-structed outwards. Prove that lines A1B, A2B2 and AB1 meet at one point. SOLUTIONS 41 2.93. Circles S1 and S2 intersect at points A and B so that the tangents to S1 at these points are radii of S2. On the inner arc of S1 a point C is taken; straight lines connect it with points A and B. Prove that the second intersection points of these lines with S2 are the endpoints of a diameter. 2.94. From the center O of a circle the perpendicular OA is dropped to line l. On l, points B and C are taken so that AB = AC. Through points B and C two sections are drawn one of which intersects the circle at points P and Q and the other one at points M and N. Lines PM and QN intersect line l at points R and S, respectively. Prove that AR = AS. Problems for independent study 2.95. In triangle ABC heights AA1 and BB1 are drawn; let M be the midpoint of side AB. Prove that MA1 = MB1. 2.96. In convex quadrilateral ABCD angles ∠A and ∠C are right ones. Prove that AC = BD sin ABC. 2.97. Diagonals AD, BE and CF of an inscribed hexagon ABCDEF meet at one point. Prove that AB · CD · EF = BC · DE · AF. 2.98. In a convex quadrilateral AB = BC = CD, let M be the intersection point of diagonals, K is the intersection point of bisectors of angles ∠A and ∠D. Prove that points A, M, K and D lie on one circle. 2.99. Circles centered at O1 and O2 intersect at points A and B. Line O1A intersects the circle centered at O2 at point N. Prove that points O1, O2, B and N lie on one circle. 2.100. Circles S1 and S2 intersect at points A and B. Line MN is tangent to circle S1 at point M and to S2 at point N. Let A be the intersection point of the circles, which is more distant from line MN. Prove that ∠O1AO2 = 2∠MAN. 2.101. Given quadrilateral ABCD inscribed in a circle and such that AB = BC, prove that SABCD = 1 2(DA+CD)·hb, where hb is the height of triangle ABD dropped from vertex B. 2.102. Quadrilateral ABCD is an inscribed one and AC is the bisector of angle ∠DAB. Prove that AC · BD = AD · DC + AB · BC. 2.103. In right triangle ABC, bisector CM and height CH are drawn from the vertex of the right angle ∠C. Let HD and HE be bisectors of triangles AHC and CHB. Prove that points C, D, H, E and M lie on one circle. 2.104. Two circles pass through the vertex of an angle and a point on its bisector. Prove that the segments cut by them on the sides of the angle are equal. 2.105. Triangle BHC, where H is the orthocenter of triangle ABC is complemented to the parallelogram BHCD. Prove that ∠BAD = ∠CAH. 2.106. Outside equilateral triangle ABC but inside angle ∠BAC, point M is taken so that ∠CMA = 30◦and ∠BMA = α. What is the value of angle ∠ABM? 2.107. Prove that if the inscribed quadrilateral with perpendicular diagonals is also a circumscribed one, then it is symmetric with respect to one of its diagonals. Solutions 2.1. Let us draw diameter AD. Then ∠CDA = ∠CBA; hence, ∠BAH = ∠DAC because ∠BHA = ∠ACD = 90◦. 2.2. Let us make use of the properties of oriented angles: ∠(AC, CK) = ∠(AM, MK) = ∠(BM, MK) = ∠(BD, DK) = ∠(BD, CK), 42 CHAPTER 2. INSCRIBED ANGLES i.e., AC ∥BD. 2.3. Points P and Q lie on the circle with diameter AM. Therefore, ∠QMA = ∠QPA as angles that intersect the same arc. Triangles PAK and MAQ are right ones, therefore, ∠PAK = ∠MAQ. 2.4. a) Since ∠AOM = ∠BAO + ∠ABO = ∠A + ∠B 2 and ∠OAM = ∠OAC + ∠CAM = ∠A 2 + ∠CBM = ∠A + ∠B 2 , we have MA = MO. Similarly, MC = MO. Since triangle OAOb is a right one and ∠AOM = ∠MAO = ϕ, it follows that ∠MAOb = ∠MObA = 90◦−ϕ and, therefore, MA = MOb. Similarly, MC = MOb. b) Let P be the center of the circumscribed circle of triangle ACO. Then ∠COP = 180◦−∠CPO 2 = 90◦−∠OAC. Hence, ∠BOC = 90◦+ ∠OAC. Similarly, ∠BOC = 90◦+ ∠OAB and, therefore, ∠OAB = ∠OAC. We similarly establish that point O lies on the bisectors of angles ∠B and ∠C. 2.5. Points P and C lie on the circle with diameter AB, and, therefore, ∠APC = ∠ABC, i.e., the value of angle ∠APC is a constant. Remark. A similar statement is true for any triangle ABC whose vertices are moving along the legs of angle ∠MPN equal to 180◦−∠C. 2.6. Points B, D and K lie on the circle with diameter AC. Let, for deﬁniteness sake, ∠KCA = ϕ ≤45◦. Then BK = AC sin(45◦−ϕ) = AC(cos ϕ −sin ϕ) √ 2 and DK = AC sin(45◦+ ϕ) = AC(cos ϕ + sin ϕ) √ 2 . Clearly, AC cos ϕ = CK and AC sin ϕ = AK. 2.7. Since ∠B1AA1 = ∠A1BB1, it follows that points A, B, A1 and B1 lie on one circle. Parallel lines AB and A1B1 intercept on it equal chords AB1 and BA1. Hence, AC = BC. 2.8. On side BC of triangle ABC construct outwards an equilateral triangle A1BC. Let P be the intersection point of line AA1 with the circumscribed circle of triangle A1BC. Then point P lies inside triangle ABC and ∠APC = 180◦−∠A1PC = 180◦−∠A1BC = 120◦. Similarly, ∠APB = 120◦. 2.9. The bases of perpendiculars dropped from point M on the diameters lie on the circle S with diameter OM (where O is the center of the initial circle). The intersection points of the given diameters with circle S distinct from O divide the circle into n arcs. Since the angles 180◦ n intersect all the circles that do not contain point O, the angle measure of each of these arcs is equal to 360◦ n . Therefore, the angle measure of the arc on which point O lies is equal to 360◦−(n −1) · 360◦ n = 360◦ n . Thus, the bases of the perpendiculars divide the circle S into n equal arcs. 2.10. Clearly, ∠(AA1, BB1) = ∠(AA1, AB1) + ∠(AB1, BB1) = ∠(MA1, MB1) + ∠(AN, BN). SOLUTIONS 43 Since MA1 ⊥BN and MB1 ⊥AN, it follows that ∠(MA1, MB1) = ∠(BN, AN) = −∠(AN, BN). Therefore, ∠(AA1, BB1) = 0◦, i.e., AA1 ∥BB1. 2.11. Since AB ∥DE, it follows that ∠ACE = ∠BFD and since BC ∥EF, it follows that ∠CAE = ∠BDF. Triangles ACE and BDF have two pairs of equal angles and, therefore, their third angles are also equal. The equality of these angles implies the equality of arcs ⌣AC and ⌣DF, i.e., chords CD and AF are parallel. 2.12. Let us carry out the proof by induction on n. For the quadrilateral the statement is obvious; for the hexagon it had been proved in the preceding problem. Assume that the statement is proved for the 2(n −1)-gon; let us prove the statement for the 2n-gon. Let A1 . . . A2n be a 2n-gon in which A1A2 ∥An+1An+2, . . . , An−1An ∥A2n−1A2n. Let us consider 2(n −1)-gon A1A2 . . . An−1An+1 . . . A2n−1. By the inductive hypothesis for n odd we have An−1An+1 = A2n−1A1, and for n even we have An−1An+1 ∥A2n−1A1. Let us consider triangles An−1AnAn+1 and A2n−1A2nA1. Let n be even. Then vectors {An−1An} and {A2n−1A2n}, as well as {An−1An+1} and {A2n−1A1} are parallel and directed towards each other; hence, ∠AnAn−1An+1 = ∠A1A2n−1A2n and AnAn+1 = A2nA1 as chords that cut equal arcs, as required. Let n be odd. Then An−1An+1 = A2n−1A1, i.e., A1An−1 ∥An+1A2n−1. In hexagon An−1AnAn+1A2n−1A2nA1 we have A1An−1 ∥An+1A2n−1 and An−1An ∥A2n−1A2n; hence, thanks to the preceding problem AnAn+1 ∥A2nA1, as required. 2.13. Let lines FG, GE and EF pass through points A, B and C, respectively, so that triangle EFG is an equilateral one, i.e., ∠(GE, EF) = ∠(EF, FG) = ∠FG, GE) = ±60◦. Then ∠(BE, EC) = ∠(CF, FA) = ∠(AG, GB) = ±60◦. Selecting one of the signs we get three circles SE, SF and SG on which points E, F and G should lie. Any point E of circle SE uniquely determines triangle EFG. Let O be the center of triangle EFG; let P, R and Q be the intersection points of lines OE, OF and OG with the corresponding circles SE, SF and SG. Let us prove that P, Q and R are the centers of equilateral triangles constructed on sides of triangle ABC (outwards for one family and inwards for the other one), and point O lies on the circumscribed circle of triangle PQR. Clearly, ∠(CB, BP) = ∠(CE, EP) = ∠(EF, EO) = ∓30◦ and ∠(BP, CP) = ∠(BE, EC) = ∠(GE, EF) = ±60◦. Hence, ∠(CB, CP) = ∠(CB, BP) + ∠(BP, CP) = ±30◦. Therefore, P is the center of an equilateral triangle with side AB. For points Q and R the proof is similar. Triangle PQR is an equilateral one and its center coincides with the intersection point of medians of triangle ABC (cf. Problem 1.49 b)). As is not diﬃcult to verify, ∠(PR, RQ) = ∓60◦= ∠(OE, OG) = ∠(OP, OQ), i.e., point O lies on the circumscribed circle of triangle PQR. 44 CHAPTER 2. INSCRIBED ANGLES 2.14. Clearly, 2(∠KEC + ∠KDC) = (⌣MB+ ⌣AC) + (⌣MB+ ⌣BC) = 360◦, since ⌣MB =⌣AM. 2.15. Denote the angle measure of the arc intercepted on the circle by the sides of triangle ABC by α. Denote the angle measure of the arc intercepted by the extensions of the sides of the triangle on the circle by α′. Then 1 2(α + α′) = ∠BAC = 60◦. But α = α′ because these arcs are symmetric through the line that passes through the center of the circle parallel to side BC. Hence, α = α′ = 60◦. 2.16. Since ∠APB = 1 2(⌣AB+ ⌣CD) = ∠AOB, point O lies on the circumscribed circle of triangle APB. 2.17. Let O be the point where lines A1C1 and B1D1 meet; let α, β, γ and δ be angle measures of arcs AB, BC, CD and DA. Then ∠A1OB1 = ⌣A1B+ ⌣BB1+ ⌣C1D+ ⌣DD1 2 = α + β + γ + δ 4 = 90◦. 2.18. By summing up the equalities we get ⌣C′A+ ⌣CA′ = 2(180◦−∠APC) = 240◦−2∠B and ⌣AB′+ ⌣BA′ = 240◦−2∠C. Then by subtracting from their sum the equality ⌣BA′+ ⌣CA′ = 2∠A we get ⌣C′B′ =⌣C′A+ ⌣AB′ = 480◦−2(∠A + ∠B + ∠C) = 120◦. Similarly, ⌣B′A′ =⌣C′A′ = 120◦. 2.19. a) Let us prove, for example, that AA1 ⊥C1B1. Let M be the intersection point of these segments. Then ∠AMB1 = ⌣AB1+ ⌣A1B+ ⌣BC1 2 = ∠ABB1 + ∠A1AB + ∠BCC1 = ∠B + ∠A + ∠C 2 = 90◦. b) Let M1 and M2 be the intersection points of segments AA1 with BC and BB1 with AC. Right triangles AM1C and BM2C have a common angle ∠C; hence, ∠B1BC = ∠A1AC. Consequently, ⌣B1C =⌣A1C and ∠B1C1C = ∠A1C1C, i.e., CC1 is the bisector of angle ∠A1C1B1. 2.20. Denote the vertices of triangle T1 by A, B and C; denote the midpoints of arcs ⌣BC, ⌣CA, ⌣AB by A1, B1, C1, respectively. Then T2 = A1B1C1. Lines AA1, BB1, CC1 are the bisectors of triangle T1; hence, they meet at one point, O. Let lines AB and C1B1 intersect at point K. It suﬃces to verify that KO ∥AC. In triangle AB1O, line B1C1 is a bisector and height, hence, this triangle is an isosceles one. Therefore, triangle AKO is also an isosceles one. Lines KO and AC are parallel, since ∠KOA = ∠KAO = ∠OAC. 2.21. Let l be tangent to the ﬁrst circle at point A. Then ∠(l, AP) = ∠(AQ, PQ) = ∠(BC, PB), hence, l ∥BC. 2.22. Since ∠(AB, AD) = ∠(AP, PD) = ∠(AB, BC), we have BC ∥AD. 2.23. Let, for deﬁniteness, point E lie on ray BC. Then ∠ABC = ∠EAC and ∠ADE = ∠ABC + ∠BAD = ∠EAC + ∠CAD = ∠DAE. SOLUTIONS 45 2.24. Let P be the other intersection point of the circles. Then ∠(AB, DB) = ∠(PA, PB) and ∠(DC, AC) = ∠(PC, PA). By summing these equalities we get ∠(DC, DB) = ∠(PC, PB) = ∠(PC, CA) + ∠(BA, PB). The latter two angles subtend constant arcs. 2.25. a) Since ∠MAB = ∠BNA, the sum of angles ∠ABN and ∠MAN is equal to the sum of the angles of triangle ABN. b) Since ∠BAM = ∠BNA and ∠BAN = ∠BMA, it follows that △AMB ∼△NAB and, therefore, AM : NA = MB : AB and AM : NA = AB : NB. By multiplying these equalities we get the desired statement. 2.26. Point P lies on the circle of radius BC with center B and line DC is tangent to this circle at point C. Hence, ∠PCD = 1 2∠PBC = 15◦. 2.27. Let A1 and B1 be intersection points of lines MA and MB, respectively, with the smaller circle. Since M is the center of homothety of the circles, A1B1 ∥AB. Hence, ∠A1MT = ∠A1TA = ∠B1A1T = ∠B1MT. 2.28. Let ϕ be the angle between chord AB and the tangent that passes through one of the chord’s endpoints. Then AB = 2R sin ϕ, where R is the radius of circle S. Moreover, PM = AM sin ϕ and QM = BM sin ϕ. Hence, 1 PM + 1 QM = µAM + BM sin ϕ ¶ AM · BM = 2R AM · BM . The value AM · BM does not depend on the choice of chord AB. 2.29. Let line AM intersect circle S2 at point D. Then ∠MDC = ∠MCA = ∠MAB; hence, CD ∥AB. Further, ∠CAM = ∠MCB = ∠MDB; hence, AC ∥BD. Therefore, ABCD is a parallelogram and its diagonal AD divides diagonal BC in halves. 2.30. Let us draw line l1 tangent to S1 at point A1. Line K1K2 is tangent to S1 if and only if ∠(K1K2, K1A1) = ∠(K1A1, l1). It is also clear that ∠(K1A1, l1) = ∠(A1B, l1) = ∠(A2B, A1A2). Similarly, line K1K2 is tangent to S2 if and only if ∠(K1K2, K2A2) = ∠(A1B, A1A2). It remains to observe that if ∠(K1K2, K1A1) = ∠(A2B, A1A2), then ∠(K1K2, K2A2) = ∠(K1K2, A2B) = ∠(K1K2, A1B) + ∠(A1B, A1A2) + ∠(A1A2, A2B) = ∠(A1B, A1A2). 2.31. Equal angles ABC and A1B1C1 intersect chords AC and A1C1, hence, AC = A1C1. 2.32. Let us denote the center of the circle by O. Points P and Q lie on the circle with diameter OM, i.e., points O, P, Q and M lie on a circle of radius 1 2R. Moreover, either ∠POQ = ∠AOD or ∠POQ = ∠BOD = 180◦−∠AOD, i.e., the length of chord PQ is a constant. 2.33. Since ∠AOC = 90◦+ 1 2∠B (cf. Problem 5.3), it follows that ∠EBD + ∠EOD = 90◦+ 3 2∠B = 180◦ and, therefore, quadrilateral BEOD is an inscribed one. Equal angles ∠EBO and ∠OBD subtend chords EO and OD, hence, EO = OD. 2.34. On the extension of segment BD beyond point D take a point Q such that ∠ACQ = 40◦. Let P be the intersection point of lines AB and QC. Then ∠BPC = 60◦ and D is the intersection point of the bisectors of angles of triangle BCP. By Problem 2.33 AD = DQ. Moreover, ∠BQC = ∠BCQ = 80◦. Therefore, BC = BD + DQ = BD + DA. 46 CHAPTER 2. INSCRIBED ANGLES 2.35. It suﬃces to verify that the exterior angle ACD of triangle BCD is twice greater than the angle at vertex B. Clearly, ∠ACD = ∠AOD = 2∠ABD. 2.36. Let O be the center of circle S. Point B is the center of the circumscribed circle of triangle ACD, hence, ∠CDA = 1 2∠ABC = 30◦and, therefore, ∠EOA = 2∠EDA = 60◦, i.e., triangle EOA is an equilateral one. Moreover, ∠AEC = ∠AED = ∠AOB = 2∠AOC; hence, point E is the center of the circumscribed circle of triangle AOC. Therefore, EC = EO. 2.37. Let us consider two positions of the moving circle: at the ﬁrst moment, when point K just gets to the ﬁxed circle (the tangent point of the circles at this moment will be denoted by K1) and at some other (second) moment. Let O be the center of the ﬁxed circle, O1 and O2 be the positions of the center of the moving circle at the ﬁrst and the second moments, respectively, K2 be the position of point K at the second moment. Let A be the tangent point of the circles at the second moment. Since the moving circle rolls without gliding, the length of arc ⌣K1A is equal to the length of arc ⌣K2A. Since the radius of the moving circle is one half of the radius of the ﬁxed circle, ∠K2O2A = 2∠K1OA. Point O lies on the moving circle, hence, ∠K2OA = 1 2∠K2O2A = ∠K1OA, i.e., points K2, K1 and O lie on one line. The trajectory of point K is the diameter of the ﬁxed circle. 2.38. Points N and C lie on the circle with diameter AM. Angles ∠MAN and ∠MCN subtend the same arc and therefore, are equal. 2.39. The symmetry through the bisector of angle ∠BOC sends lines AC and DB into each other and, therefore, we have to prove that ∠C′AB′ = ∠B′DC′. Since BO = B′O, CO = C′O and AO : DO = CO : BO, it follows that AO · B′O = DO · C′O, i.e., the quadrilateral AC′B′D is an inscribed one and ∠C′AB′ = ∠B′DC′. 2.40. Denote the intersection points and angles as indicated on Fig. 14. Figure 14 (Sol. 2.40) It suﬃces to verify that x = 90◦. The angles of quadrilateral BMRN are equal to 180◦−ϕ, α + ϕ, β + ϕ and x, hence, the equality x = 90◦is equivalent to the equality (2α+ϕ)+(2β +ϕ) = 180◦. It remains to notice that 2α+ϕ = ∠BAD and 2β +ϕ = ∠BCD. 2.41. a) It suﬃces to prove that if P1 is the point on the bisector (or its extension) of angle ∠B that serves as the vertex of an angle of 90◦that subtends segment BC, then P1 lies on line MN. Points P1 and N lie on the circle with diameter CO, where O is the intersection point of bisectors, hence, ∠(P1N, NC) = ∠(P1O, OC) = 1 2(180◦−∠A) = ∠(MN, NC). SOLUTIONS 47 b) Since ∠BPC = 90◦, it follows that BP = BC · cos ∠B 2 ; hence, SABP : SABC = µ BP · sin ∠B 2 ¶ : (BC sin B) = 1 : 2. 2.42. Take point N so that BN ∥MC and NC ∥BM. Then NA ∥CD, ∠NCB = ∠CBM = ∠CDM = ∠NAB, i.e., points A, B, N and C lie on one circle. Hence, ∠ACD = ∠NAC = ∠NBC = ∠BCM. 2.43. Points A2, B2, C and P lie on one circle, hence, ∠(A2B2, B2P) = ∠(A2C, CP) = ∠(BC, CP). Similarly, ∠(B2P, B2C2) = ∠(AP, AB). Therefore, ∠(A2B2, B2C2) = ∠(BC, CP) + ∠(AP, AB) = ∠(B1B, B1C1) + ∠(A1B1, B1B) = ∠(A1B1, B1C1). We similarly verify that all the other angles of triangles A1B1C1 and A2B2C2 are either equal or their sum is equal to 180◦; therefore, these triangles are similar (cf. Problem 5.42). 2.44. Points Q′ and C lie on the circle with diameter PQ, hence, ∠Q′CQ = ∠Q′PQ = 30◦. Therefore, ∠BCQ′ = 60◦. Similarly, ∠CBQ′ = 60◦and, therefore, triangle BQ′C is equilateral one. By similar reasons triangle CP ′D is an equilateral one. 2.45. Let ∠BAD = 2α and ∠CBA = 2β; for deﬁniteness we will assume that α ≥β. On side CD take point E so that DE = DA. Then CE = CD −AD = CB. The angle at vertex C of an isosceles triangle BCE is equal to 180◦−2α; hence, ∠CBE = α. Similarly, ∠DAE = β. The bisector of angle B intersects CD at a point F. Since ∠FBA = β = ∠AED, quadrilateral ABFE is an inscribed one and, therefore, ∠FAE = ∠FBE = α −β. It follows that ∠FAD = β + (α −β) = α, i.e., AF is the bisector of angle ∠A. 2.46. Since ED = CB, EN = CM and ∠DEC = ∠BCA = 30◦(Fig. 15), it follows that △EDN = △CBM. Let ∠MBC = ∠NDE = α, ∠BMC = ∠END = β. Figure 15 (Sol. 2.46) It is clear that ∠DNC = 180◦−β. Considering triangle BNC we get ∠BNC = 90◦−α. Since α + β = 180◦−30◦= 150◦, it follows that ∠DNB = ∠DNC + ∠CNB = (180◦−β) + (90◦−α) = 270◦−(α + β) = 120◦. Therefore, points B, O, N and D, where O is the center of the hexagon, lie on one circle. Moreover, CO = CB = CD, i.e., C is the center of this circle, hence, λ = CN : CE = CB : CA = 1 : √ 3. 2.47. Let D be the other intersection point of the circumscribed circles of triangles A1BC and AB1C. Then ∠(AC, CD) = ∠(AB1, B1D) and ∠(DC, CB) = ∠(DA1, A1B). 48 CHAPTER 2. INSCRIBED ANGLES Hence, ∠(A1C1, C1B1) = ∠(AC, CB) = ∠(AC, CD) + ∠(DC, CB) = ∠(AB1, B1D) + ∠(DA1, A1B) = ∠(A1D, DB1), i.e., points A1, B1, C1 and D lie on one circle. Therefore, ∠(A1C1, C1B) = ∠(A1B1, B1D) = ∠(AC, CD). Taking into account that A1C1 ∥AC, we get the desired statement. 2.48. Let point M be symmetric to point A1 through line AC. By Problem 1.57 point M lies on line B1C1. Therefore, ∠(LM, MA1) = ∠(C1B1) = ∠(C1C, CB) = ∠(LK, KA1), i.e., point M lies on the circumscribed circle of triangle A1KL. It follows that the center of this circle lies on line AC — the midperpendicular to segment A1M. 2.49. Let PQ be the diameter perpendicular to AB and such that Q and C lie on one side of AB; let L be the intersection point of line QO with the circumscribed circle; let M ′ and N ′ be the intersection points of lines LB′ and LA′ with sides AC and BC, respectively. It suﬃces to verify that M ′ = M and N ′ = N. Since ⌣PA+ ⌣AB′+ ⌣B′Q = 180◦, it follows that ⌣B′Q = ∠A and, there-fore, ∠B′LQ = ∠M ′AO. Hence, quadrilateral AM ′OL is an inscribed one and ∠M ′OA = ∠M ′LA = 1 2∠B. Therefore, ∠CMO = 1 2(∠A + ∠B), i.e., M ′ = M. Similarly, N ′ = N. 2.50. Since △ADM ∼△CBM and △ACM ∼△DBM, it follows that AD : CB = DM : BM and AC : DB = AM : DM. It remains to multiply these equalities. 2.51. Let D1 be the intersection point of line BD with the circle distinct from point B. Then ⌣AB =⌣AD1; hence, ∠ACB = ∠AD1B = ∠ABD1. Triangles ACB and ABD have a common angle, ∠A, and, moreover, ∠ACB = ∠ABD; hence, △ACB ∼△ABD. Therefore, AB : AC = AD : AB. 2.52. Let O be the center of the circle. Since ∠MAC = ∠ACO = ∠CAO, it follows that △AMC = △ADC. Similarly, △CDB = △CNB. Since △ACD ∼△CDB, it follows that CD2 = AD · DB = AM · NB. 2.53. Points B1 and H lie on the circle with diameter AB, hence, ∠(AB, BC) = ∠(AB, BH) = ∠(AB1, B1H) = ∠(B1C1, B1H). Similarly, ∠(AC, BC) = ∠(B1C1, C1H). 2.54. On an extension of segment BP beyond point P take point D such that PD = CP. Then triangle CDP is an equilateral one and CD ∥QP. Therefore, BP : PQ = BD : DC = (BP + CP) : CP, i.e., 1 PQ = 1 CP + 1 BP . 2.55. Segment QE subtends angles of 45◦with vertices at points A and B, hence, quadrilateral ABEQ is an inscribed one. Since ∠ABE = 90◦, it follows that ∠AQE = 90◦. Therefore, triangle AQE is an isosceles right triangle and AE AQ = √ 2. Similarly, AF AP = √ 2. 2.56. Since ∠ANC = ∠ABC = ∠CAB, it follows that △CAM ∼△CNA and, there-fore, CA : CM = CN : CA, i.e., CM · CN = AC2 and AM : NA = CM : CA. Similarly, BM : NB = CM : CB. Therefore, AM · BM AN · BN = CM 2 CA2 = CM 2 CM · CN = CM CN . 2.57. Since AK = AB = CD, AD = BC = CH and ∠KAD = ∠DCH, it follows that △ADK = △CHD and DK = DH. Let us show that points A, K, H, C and D lie on one circle. Let us circumscribe the circle about triangle ADC. Draw chord CK1 in this circle parallel to AD and chord AH1 parallel to DC. Then K1A = DC and H1C = AD. Hence, K1 = K and H1 = H, i.e., the constructed circle passes through points K and H and angles SOLUTIONS 49 ∠KAH and ∠KDH are equal because they subtend the same arc. Moreover, as we have already proved, KDH is an isosceles triangle. 2.58. a) ∠PBA1 = ∠PAC1 and ∠PBC1 = ∠PAB1 and, therefore, right triangles PBA1 and PAC1, PAB1 and PBC1 are similar, i.e., PA1 : PB = PC1 : PA and PB1 : PA = PC1 : PB. By multiplying these equalities we get PA1 · PB1 = PC2 1. b) According to heading a) OA′′ = √ OB′ · OC′, OB′′ = √ OA′ · OC′, OC′′ = √ OA′ · OB′. By multiplying these equalities we get the desired statement. 2.59. Let K, L, M and N be the bases of perpendiculars dropped from point E to lines AB, BC, CD and DA, respectively. Points K and N lie on the circle with diameter AE, hence, ∠(EK, KN) = ∠(EA, AN). Similarly, ∠(EL, LM) = ∠(EC, CM) = ∠(EA, AN) and, therefore, ∠(EK, KN) = ∠(EL, LM). Similarly, ∠(EN, NK) = ∠(EM, ML) and ∠(KE, EN) = ∠(LE, EM). It follows that △EKN ∼△ELM and, therefore, EK : EN = EL : EM, i.e., EN = EK·EM EL = ac b . 2.60. Let H be the intersection point of heights, M the midpoint of side BC. Points A1, B2 and C2 lie on the circle with diameter MH, hence, ∠(B2A1, A1C2) = ∠(B2M, MC2) = ∠(AC, AB). Moreover, ∠(A1B2, B2C2) = ∠(A1H, HC2) = ∠(BC, AB) and ∠(A1C2, C2B2) = ∠(BC, AC). 2.61. Let M be the intersection point of medians, H the intersection point of heights of triangle ABC. Points A1, B1 and C1 are the projections of point M on the heights and, therefore, these points lie on the circle with diameter MH. Hence, ∠(A1B1, B1C1) = ∠(AH, HC) = ∠(BC, AB). By writing similar equalities for the other angles we get the desired statement. 2.62. Let lines BM and DN meet S2 at points L and C1, respectively. Let us prove that lines DC1 and CN are symmetric through line AN. Since BN ⊥NA, it suﬃces to verify that ∠CNB = ∠BND. But arcs ⌣CB and ⌣BD are equal. Arcs ⌣C1M and ⌣CL are symmetric through line AN, hence, they are equal and, therefore, ∠MDC1 = ∠CML. Besides, ∠CNM = ∠MND. Thus, △MCN ∼△DMN, i.e., CN : MN = MN : DN. 2.63. Let us drop from point Q perpendiculars QK1 and QN1 to KL and NM, respec-tively, and from point P perpendiculars PM1 and PL1 to NM and KL, respectively. Clearly, QC PC = QK1 PL1 = QN1 PM1, i.e., QC2 PC2 = QK1·QN1 PL1·PM1. Since ∠KNC = ∠MLC and ∠NKC = ∠LMC, it follows that QN1 : PL1 = QN : PL and QK1 : PM1 = QK : PM. Therefore, QC2 PC2 = QK · QN PL · PM = AQ · QB PB · AP = (AC −QC) · (AC + QC) (AC −PC) · (AC + PC) = AC2 −QC2 AC2 −PC2. This implies that QC = PC. 2.64. a) Since ∠CAM = ∠CBM and ∠CB1M = ∠CA1M, it follows that ∠B1AM = ∠A1BM and ∠AB1M = ∠BA1M. b) Let M1 be a point of the circle S with diameter CO such that CM1 ∥A1B1; let M2 be an intersection point of circle S with the circumscribed circle of triangle ABC; let A2 and B2 be the tangent points of of the inscribed circle with sides BC and AC, respectively. It suﬃces to verify that M1 = M2. By Problem a) △AB2M2 ∼△BA2M2, hence, B2M2 : A2M2 = AB2 : BA2. Since CA1 = p −b −BA2 and CB1 = AB2, it follows that B2M1 A2M1 = sin B2CM1 sin A2CM1 = sin CA1B1 sin CB1A1 = CB1 CA1 = AB2 BA2 . On arc ⌣A2CB2 of circle S, there exists a unique point X for which B2X : A2X = k (Problem 7.14), hence, M1 = M2. 50 CHAPTER 2. INSCRIBED ANGLES 2.65. Let O be the center of the circumscribed circle of the triangle, M the midpoint of side AB, H the base of height CH, D the midpoint of the arc on which point C does not lie and with endpoints A and B. Since OD ∥CH, it follows that ∠DCH = ∠MDC. The bisector divides the angle between the median and the height in halves if and only if ∠MCD = ∠DCH = ∠MDC = ∠ODC = ∠OCD, i.e., M = O and AB is the diameter of the circle. 2.66. Let α = ∠A < ∠B. By the preceding problem ∠C = 90◦. Median CM divides triangle ABC into two isosceles triangles. Since ∠ACM = ∠A = α, ∠MCB = 3α, it follows that α + 3α = 90◦, i.e., α = 22.5◦. Therefore, ∠A = 22.5◦, ∠B = 67.5◦, ∠C = 90◦. 2.67. Let D be a point at which line AE intersects the circumscribed circle. Point D is the midpoint of arc ⌣BC. Therefore, MD ∥AH, moreover, points A and D lie on diﬀerent sides of line MH. It follows that point E lies on segment MH. 2.68. Clearly, ∠(AQ, QP) = ∠(AN, NP) = ∠(PM, MB) = ∠(QP, QB). Therefore, point Q lies on the circle such that segment AB subtends an angle of 2∠(AC, CB) with vertex at Q and line QP divides arc ⌣AB of this circle in halves. 2.69. Points P and Q lie on the circle with diameter AD; this circle intersects side BC at point F. (Observe that F does not coincide with D if AB ̸= AC.) Clearly, ∠(FC, CE) = ∠(BA, AE) = ∠(DA, AQ) = ∠(DF, FQ), i.e., EC ∥FQ. Similarly, BE ∥FP. To complete the proof it suﬃces to notice that the areas of triangles adjacent to the lateral sides of the trapezoid are equal. 2.70. Let ∠AOB = α and ∠COD = β. Then α 2 + β 2 = ∠ADP + ∠PAD = 90◦. Since 2SAOB = R2 sin α and 2SCOD = R2 sin β, where R is the radius of the circumscribed circle, it follows that SAOB = SCOD. Similarly, SBOC = SAOD. 2.71. Let ∠AOB = 2α and ∠COD = 2β. Then α+β = ∠ADP +∠PAD = 90◦. Hence, (AP 2 + BP 2) + (CP 2 + DP 2) = AB2 + CD2 = 4R2(sin2 α + cos2 α) = 4R2. Similarly, BC2 + AD2 = 4R2. 2.72. Let M be the midpoint of AC, N the midpoint of BD. We have AM 2 = AO2 − OM 2 and BN 2 = BO2 −ON 2; hence, AC2 + BD2 = 4(R2 −OM 2) + 4(R2 −ON 2) = 8R2 −4(OM 2 + ON 2) = 8R2 −4OP 2 since OM 2 + ON 2 = OP 2. 2.73. The correspondiong legs of acute angles ∠BLP and ∠BDC are perpendicular, hence, the angles are equal. Therefore, ∠BLP = ∠BDC = ∠BAP. Moreover, AK ∥BL and AL ⊥BK. It follows that AKLB is a rhombus. 2.74. In the circumscribed circle take a point D′ so that DD′ ∥AC. Since DD′ ⊥BD, it follows that BD′ is a diameter and, therefore, ∠D′AB = ∠D′CB = 90◦. Hence, SABCD = SABCD = 1 2(AD′ · AB + BC · CD′) = 1 2(AB · CD + BC · AD). 2.75. Let us draw diameter AE. Since ∠BEA = ∠BCP and ∠ABE = ∠BPC = 90◦, it follows that ∠EAB = ∠CBP. The angles that intersect chords EB and CD are equal, hence, EB = CD. Since ∠EBA = 90◦, the distance from point O to AB is equal to 1 2EB. 2.76. Let the perpendicular dropped from point P to BC intersect BC at point H and AD at point M (Fig. 16). SOLUTIONS 51 Figure 16 (Sol. 2.76) Therefore, ∠BDA = ∠BCA = ∠BPH = ∠MPD. Since angles MDP and MPD are equal, MP is a median of right triangle APD. Indeed, ∠APM = 90◦−∠MPD = 90◦−∠MDP = ∠PAM, i.e., AM = PM = MD. 2.77. The midpoints of the sides of quadrilateral ABCD are vertices of a rectangle (cf. Problem 1.2), hence, they lie on one circle. Let K and L be the midpoints of sides AB and CD, let M be the intersection point of lines KP and CD. By Problem 2.76 PM ⊥CD; hence, M is the projection of point P on side CD and point M lies on the circle with diameter KL. For the other projections the proof is similar. 2.78. a) It is worth to observe that since points A, B, C and D divide the circle into arcs smaller than 180◦each, then the quadrilateral constructed contains this circle. The angle ϕ between the tangents drawn through points A and B is equal to 180◦−∠AOB and the angle ψ between the tangents drawn through points C and D is equal to 180◦−∠COD. Since ∠AOB + ∠COD = 180◦, it follows that ϕ + ψ = 180◦. Remark. Conversely, the equality ϕ + ψ = 180◦implies that ∠AOB + ∠COD = 180◦, i.e., AC ⊥BD. b) Let O be the center of the inscribed circle. Since ∠AKO + ∠BMO = 90◦, it follows that ∠AKO = ∠BOM and △AKO ∼△BOM. Therefore, AK · BM = BO · AO = r2. 2.79. First, let us suppose that the circumscribed circles of triangles A′BC and AB′C are not tangent to each other and P is their common point distinct from C. Then ∠(PA, PB) = ∠(PA, PC) + ∠(PC, PB) = ∠(B′A, B′C) + ∠(A′C, A′B) = ∠(C′A, C′B), i.e., point P lies on the circumscribed circle of triangle ABC′. If the the circumscribed circles of triangles A′BC and AB′C are tangent to each other, i.e., P = C, then our arguments require an insigniﬁcant modiﬁcations: instead of line PC we have to take the common tangent. 2.80. a) By applying the statement of Problem 2.79 to triangles AB1C1, A1BC1 and A1B1C constructed on the sides of triangle A1B1C1 we get the desired statement. b) Let P be the intersection point of the indicated circles. Let us prove that the value of the angle ∠(AP, PC) is a constant. Since ∠(AP, PC) = ∠(AP, AB) + ∠(AB, BC) + ∠(BC, PC) 52 CHAPTER 2. INSCRIBED ANGLES and angle ∠(AB, BC) is a constant, it remains to verify that the sum ∠(AP, AB)+∠(BC, PC) is a constant. Clearly, ∠(AP, AB) + ∠(BC, CP) = ∠(AP, AC1) + ∠(CA1, CP) = ∠(B1P, B1C1) + ∠(B1A1, B1P) = ∠(B1A1, B1C1) and the value of the latter angle is constant by hypothesis. We similarly prove that the values of angles ∠(AP, PB) and ∠(BP, PC) are constants. Hence, point P remains ﬁxed. 2.81. As follows from Problem 2.80 b) it suﬃces to carry out the proof for one such triangle A1B1C1 only; for instance, for the triangle with vertices in the midpoints of sides of triangle ABC. Let H be the intersection point of heights of triangle A1B1C1, i.e., the center of the circumscribed circle of triangle ABC. Since A1H ⊥B1C1 and B1H ⊥A1C1, it follows that ∠(A1H, HB1) = ∠(B1C1, A1C1) = ∠(A1C, CB1), i.e., point H lies on the circumscribed circle of triangle A1B1C. A similar argument shows that point H lies on the circumscribed circles of triangles A1BC1 and AB1C1. 2.82. a) Let X be the intersection point of the circumscribed circles of triangles ABC and AB′C′. Then ∠(XB′, XC) = ∠(XB′, XA) + ∠(XA, XC) = ∠(C′B′, C′A) + ∠(BA, BC). Since AC′ = AP = AB′, triangle C′AB′ is an isosceles one and ∠C′AB′ = 2∠A; hence, ∠(C′B′, C′A) = ∠A −90◦. Therefore, ∠(XB′, XC) = ∠A −90◦+ ∠B = 90◦−∠C = ∠(A′B′, A′C), i.e., point X lies on the circumscribed circle of triangle A′B′C. For the circumscribed circle of triangle A′BC′ the proof is similar. b) Let X be the intersection point of the circumscribed circles of triangles A′B′C′ and A′BC. Let us prove that X lies on the circumscribed circle of triangle ABC′. Clearly, ∠(XB, XC′) = ∠(XB, XA′) + ∠(XA′, XC′) = ∠(CB, CA′) + ∠(B′A′, B′C′). Let A1, B1 and C1 be the midpoints of segments PA′, PB′ and PC′. Then ∠(CB, CA′) = ∠(CP, CA1) = ∠(B1P, B1A1), ∠(B′A′, B′C′) = ∠(B1A1, B1C1) and ∠(AB, AC′) = ∠(AP, AC1) = ∠(B1P, B1C1). It follows that ∠(XB, XC′) = ∠(AB, AC′). We similarly prove that point X lies on the circumscribed circle of triangle AB′C. c) Since QA′ is the common chord of circles centered at O and I, it follows that QA′ ⊥OI. Similarly, QB′ ⊥OJ and QC ⊥IJ. Therefore, sides of angles OJI and B′QC, as well as sides of angles OIJ and A′QC, are mutually perpendicular, hence, sin OJI = sin B′QC and sin OIJ = sin A′QC. Therefore, OI : OJ = sin OJI : sin OIJ = sin B′QC : sin A′QC. It is also clear that QI QJ = sin QJI sin QIJ = sin( 1 2QJC) sin( 1 2QIC) = sin QB′C sin QA′C . Taking into account that sin B′QC : sin QB′C = B′C : QC and sin A′QC : sin QA′C = A′C : QC we get OI OJ : QI QJ = B′C QC : A′C QC = 1. SOLUTIONS 53 2.83. a) The conditions of the problem imply that no three lines meet at one point. Let lines AB, AC and BC intersect the fourth line at points D, E, and F, respectively (Fig. 17). Figure 17 (Sol. 2.83) Denote by P the intersection point of circumscribed circles of triangles ABC and CEF distinct from point C. Let us prove that point P belongs to the circumscribed circle of triangle BDF. For this it suﬃces to verify that ∠(BP, PF) = ∠(BD, DF). Clearly, ∠(BP, PF) = ∠(BP, PC) + ∠(PC, PF) = ∠(BA, AC) + ∠(EC, EF) = ∠(BD, AC) + ∠(AC, DF) = ∠(BD, DF). We similarly prove that point P belongs to the circumscribed circle of triangle ADE. b) Let us make use of notations of Fig. 17. Thanks to heading a), the circumscribed circles of triangles ABC, ADE and BDF pass through point P and, therefore, we can consider them as the circumscribed circles of triangles ABP, ADP and BDP respectively. Therefore, their centers lie on a circle that passes through point P (cf. Problem 5.86). We similarly prove that the centers of any of the three of given circles lie on a circle that passes through point P. It follows that all the four centers lie on a circle that passes through point P. 2.84. a) Let P be Michel’s point for lines AB, BC, CA and A1B1. The angles between rays PA, PB, PC and the tangents to circles Sa, Sb, Sc are equal to ∠(PB1, B1A) = ∠(PC1, C1A), ∠(PC1, C1B) = ∠(PA1, A1B), ∠(PA1, A1C) = ∠(PB1, B1C), respectively. Since ∠(PC1, C1A) = ∠(PC1, C1B) = ∠(PA1, A1C) = ϕ, it follows that after a rotation through an angle of ϕ about point P lines PA, PB and PC turn into the tangents to circles Sa, Sb and Sc, respectively, and, therefore, after a rotation through an angle of 90◦−ϕ these lines turn into lines POa, POb and POc respectively. Moreover, POa PA = POb PB = POc PC = 1 2 sin ϕ. Therefore, the composition of the rotation through an angle of 90◦−ϕ and the homothety (see ???) with center P and coeﬃcient 1 2 sin ϕ sends triangle ABC to OaObOc. b) The transformation considered in the solution of heading a) sends the center O of the circumscribed circle of triangle ABC into the center O′ of the circumscribed circle of triangle OaObOc and the orthocenter H of triangle ABC to orthocenter H′ of triangle OaObOc. Let us complement triangle OO′H′ to parallelogram OO′H′M. Since OH OM = OH O′H′ = 2 sin ϕ and ∠HOM = ∠(HO, O′H′) = 90◦−ϕ, it follows that MH = MO, i.e., point M lies on the midperpendicular of segment OH. It remains to notice that for the inscribed quadrilateral OOaObOc point M is uniquely determined: taking instead of point O any of the points Oa, Ob or Oc we get the same point M (cf. Problem 13.33). 2.85. We may assume that rays AB and DC meet at point E and rays BC and AD meet at point F. Let P be the intersection point of circumscribed circles of triangles BCE 54 CHAPTER 2. INSCRIBED ANGLES and CDF. Then ∠CPE = ∠ABC and ∠CPF = ∠ADC. Hence, ∠CPE + ∠CPF = 180◦, i.e., point P lies on segment EF. 2.86. a) Since ∠(AP, PD) = ∠(AP, PE) + ∠(PE, PD) = ∠(AC, CD) + ∠(AB, BD) + ∠(AO, OD), points A, P, D and O lie on one circle. b) Clearly, ∠(EP, PO) = ∠(EP, PA) + ∠(PA, PO) = ∠(DC, CA) + ∠(DA, DO) = 90◦, because the arcs intersected by these angles constitute a half of the circle. 2.87. Let us make use of notations of Fig. 17. The projections of point P on lines CA and CB coincide with its projection to CE and CF, respectively. Therefore, Simson’s lines of point P relative triangles ABC and CEF coincide (cf. Problem 5.85 a). 2.88. Let point A′ be symmetric to point A through the midperpendicular to segment BC. Then ∠OAH = 1 2∠AOA′ = ∠ABA′ = \|∠B −∠C\|. 2.89. Since AA′ is a diameter, A′C ⊥AC; hence, BH ∥A′C. Similarly, CH ∥A′B. Therefore, BA′CH is a parallelogram. 2.90. Let l be a line parallel to the two given lines, D the intersection point of lines m and n. Then ∠(AD, DB) = ∠(m, AB) + ∠(AB, n) = ∠(AC, l) + ∠(l, CB) = ∠(AC, CB) and, therefore, point D lies on the circumscribed circle of triangle ABC. 2.91. a) Let O be the midpoint of the arc of circle S that lies inside triangle ABC. Then ∠CBO = ∠BCO and due to a property of the angle between a tangent and a chord, ∠BCO = ∠ABO. Therefore, BO is the bisector of angle ABC, i.e., O is the center of the inscribed circle of triangle ABC. We similarly prove that the midpoint of the arc of circle S that lies outside triangle ABC is the center of its escribed circle. b) We have to prove that the center of the considered circle S lies on the bisector of angle BAC. Let D be the intersection point of the bisector of the angle with the circumscribed circle of triangle ABC. Then DB = DO = DC (cf. Problem 2.4 a), i.e., D is the center of circle S. 2.92. If angle ∠C is a right one, then the solution of the problem is obvious: C is the intersection point of lines A1B, A2B2, AB1. If ∠C ̸= 90◦, then the circumscribed circles of squares ACA1A2 and BCB1B2 have in addition to C one more common point, C1. Then ∠(AC1, A2C1) = ∠(A2C1, A1C1) = ∠(A1C1, C1C) = ∠(C1C, C1B1) = ∠(C1B1, C1B2) = ∠(C1B2, C1B) = 45◦ (or −45◦; it is only important that all the angles are of the same sign). Hence, ∠(AC, C1B1) = 4 · 45◦= 180◦, i.e., line AB1 passes through point C1. Similarly, A2B2 and A1B pass through point C1. 2.93. Let P and O be the centers of circles S1 and S2, respectively; let α = ∠APC, β = ∠BPC; lines AC and BC intersect S2 at points K and L, respectively. Since ∠OAP = ∠OBP = 90◦, it follows that ∠AOB = 180◦−α −β. Furthermore, ∠LOB = 180◦−2∠LBO = 2∠CBP = 180◦−β. Similarly, ∠KOA = 180◦−α. Therefore, ∠LOK = ∠LOB + ∠KOA −∠AOB = 180◦, SOLUTIONS 55 i.e., KL is a diameter. 2.94. Let us consider points M ′, P ′, Q′ and R′ symmetric to points M, P, Q and R, respectively, through line OA. Since point C is symmetric to point B through OA, it follows that line P ′Q′ passes through point C. The following equalities are easy to verify: ∠(CS, NS) = ∠(Q′Q, NQ) = ∠(Q′P, NP ′) = ∠(CP ′, NP ′); ∠(CR′, P ′R′) = ∠(MM ′, P ′M ′) = ∠(MN, P ′N) = ∠(CN, P ′N). From these equalities we deduce that points C, N, P ′, S and R′ lie on one circle. But points S, R′ and C lie on one line, therefore, S = R′. CHAPTER 3. CIRCLES Background 1. A line that has exactly one common point with a circle is called a line tangent to the circle. Through any point A outside the circle exactly two tangents to the circle can be drawn. Let B and C be the tangent points and O the center of the circle. Then: a) AB = AC; b) ∠BAO = ∠CAO; c) OB ⊥AB. (Sometimes the word “tangent” is applied not to the whole line AB but to the segment AB. Then property a), for example, is formulated as: the tangents to one circle drawn from one point are equal.) 2. Let lines l1 and l2 that pass through point A intersect a circle at points B1, C1 and B2, C2, respectively. Then AB1 · AC1 = AB2 · AC2. Indeed, △AB1C2 ∼△AB2C1 in three angles. (We advise the reader to prove this making use of the properties of the inscribed angles and considering two cases: A lies outside the circle and A lies inside the circle.) If line l2 is tangent to the circle, i.e., B2 = C2, then AB1 · AC1 = AB2 2. The proof runs along the same lines as in the preceding case except that now we have to make use of the properties of the angle between a tangent and a chord. 3. The line that connects the centers of tangent circles passes through their tangent point. 4. The value of the angle between two intersecting circles is the value of the angle between the tangents to these circles drawn through the intersection point. It does not matter which of the two of intersection points we choose: the corresponding angles are equal. The angle between tangent circles is equal to 0◦. 5. In solutions of problems from §6 a property that has no direct relation to circles is used: the heights of a triangle meet at one point. The reader can ﬁnd the proof of this fact in solutions of Problems 5.45 and 7.41 or can take it for granted for the time being. 6. It was already in the middle of the V century A.D. that Hyppocratus from island Chios (do not confuse him with the famous doctor Hyppocratus from island Kos who lived somewhat later) and Pythagoreans began to solve the quadrature of the circle problem. It is formulated as follows: with the help of a ruler and compass construct a square of the same area as the given circle. In 1882 the German mathematician Lindemann proved that number π is transcendental, i.e., is not a root of a polynomial with integer coeﬃcients. This implies, in particular, that the problem on the quadrature of the circle is impossible to solve as stated (using other tools one can certainly solve it). It seems that it was the problem on Hyppocratus’ crescents (Problem 3.38) that induced in many a person great expectations to the possibility of squaring the circle: the area of the ﬁgure formed by arcs of circles is equal to the area of a triangle. Prove this statement and try to understand why such expectations were not grounded in this case. 57 58 CHAPTER 3. CIRCLES Introductory problems 1. Prove that from a point A outside a circle it is possible to draw exactly two tangents to the circle and the lengths of these tangents (more exactly, the lengths from A to the tangent points) are equal. 2. Two circles intersect at points A and B. Point X lies on line AB but not on segment AB. Prove that the lengths of all the tangents drawn from point X to the circles are equal. 3. Two circles whose radii are R and r are tangent from the outside (i.e., none of them lies inside the other one). Find the length of the common tangent to these circles. 4. Let a and b be the lengths of the legs of a right triangle, c the length of its hypothenuse. Prove that: a) the radius of the inscribed circle of this triangle is equal to 1 2(a + b −c); b) the radius of the circle tangent to the hypothenuse and the extensions of the legs is equal to 1 2(a + b + c). §1. The tangents to circles 3.1. Lines PA and PB are tangent to a circle centered at O; let A and B be the tangent points. A third tangent to the circle is drawn; it intersects with segments PA and PB at points X and Y , respectively. Prove that the value of angle XOY does not depend on the choice of the third tangent. 3.2. The inscribed circle of triangle ABC is tangent to side BC at point K and an escribed circle is tangent at point L. Prove that CK = BL = 1 2(a + b −c), where a, b, c are the lengths of the triangle’s sides. 3.3. On the base AB of an isosceles triangle ABC a point E is taken and circles tangent to segment CE at points M and N are inscribed into triangles ACE and ECB, respectively. Find the length of segment MN if the lengths of segments AE and BE are known. 3.4. Quadrilateral ABCD is such that there exists a circle inscribed into angle ∠BAD and tangent to the extensions of sides BC and CD. Prove that AB + BC = AD + DC. 3.5. The common inner tangent to circles whose radii are R and r intersects their common outer tangents at points A and B and is tangent to one of the circles at point C. Prove that AC · CB = Rr. 3.6. Common outer tangents AB and CD are drawn to two circles of distinct radii. Prove that quadrilateral ABCD is a circumscribed one if and only if the circles are tangent to each other. 3.7. Consider parallelogram ABCD such that the escribed circle of triangle ABD is tangent to the extensions of sides AD and AB at points M and N, respectively. Prove that the intersection points of segment MN with BC and CD lie on the inscribed circle of triangle BCD. Figure 18 (3.7) §3. TANGENT CIRCLES 59 3.8. On each side of quadrilateral ABCD two points are taken; these points are con-nected as shown on Fig. 18. Prove that if all the ﬁve dashed quadrilaterals are circumscribed ones, then the quadrilateral ABCD is also a circumscribed one. §2. The product of the lengths of a chord’s segments 3.9. Through a point P lying on the common chord AB of two intersecting circles chord KM of the ﬁrst circle and chord LN of the second circle are drawn. Prove that quadrilateral KLMN is an inscribed one. 3.10. Two circles intersect at points A and B; let MN be their common tangent. Prove that line AB divides MN in halves. 3.11. Line OA is tangent to a circle at point A and chord BC is parallel to OA. Lines OB and OC intersect the circle for the second time at points K and L, respectively. Prove that line KL divides segment OA in halves. 3.12. In parallelogram ABCD, diagonal AC is longer than diagonal BD; let M be a point on diagonal AC such that quadrilateral BCDM is an inscribed one. Prove that line BD is a common tangent to the circumscribed circles of triangles ABM and ADM. 3.13. Given circle S and points A and B outside it. For each line l that passes through point A and intersects circle S at points M and N consider the circumscribed circle of triangle BMN. Prove that all these circles have a common point distinct from point B. 3.14. Given circle S, points A and B on it and point C on chord AB. For every circle S′ tangent to chord AB at point C and intersecting circle S at points P and Q consider the intersection point M of lines AB and PQ. Prove that the position of point M does not depend on the choice of circle S′. §3. Tangent circles 3.15. Two circles are tangent at point A. A common (outer) tangent line is drawn to them; it is tangent to the circles at points C and D, respectively. Prove that ∠CAD = 90◦. 3.16. Two circles S1 and S2 centered at O1 and O2 are tangent to each other at point A. A line that intersects S1 at point A1 and S2 at point A2 is drawn through point A. Prove that O1A1 ∥O2A2. 3.17. Three circles S1, S2 and S3 are pairwise tangent to each other at three distinct points. Prove that the lines that connect the tangent point of circles S1 and S2 with the other two tangent points intersect circle S3 at points that are the endpoints of its diameter. 3.18. Two tangent circles centered at O1 and O2, respectively, are tangent from the inside to the circle of radius R centered at O. Find the perimeter of triangle OO1O2. 3.19. Circles S1 and S2 are tangent to circle S from the inside at points A and B so that one of the intersection points of circles S1 and S2 lies on segment AB. Prove that the sum of the radii of circles S1 and S2 is equal to the radius of circle S. 3.20. The radii of circles S1 and S2 tangent at point A are equal to R and r (R > r). Find the length of the tangent drawn to circle S2 from point B on circle S1 if AB = a (consider the cases of the inner and outer tangent). 3.21. A point C is taken on segment AB. A line that passes through point C intersects circles with diameters AC and BC at points K and L and the circle with diameter AB at points M and N, respectively. Prove that KM = LN. 3.22. Given four circles S1, S2, S3 and S4 such that Si and Si+1 are tangent from the outside for i = 1, 2, 3, 4 (S5 = S1). Prove that the tangent points are the vertices of an inscribed quadrilateral. 60 CHAPTER 3. CIRCLES 3.23. a) Three circles centered at A, B and C are tangent to each other and line l; they are placed as shown on Fig. 19. Let a, b and c be radii of circles centered at A, B and C, respectively. Prove that 1 √c = 1 √a + 1 √ b. Figure 19 (3.23) b) Four circles are pairwise tangent from the outside (at 6 distinct points). Let a, b, c and d be their radii; α = 1 a, β = 1 b, γ = 1 c and δ = 1 d. Prove that 2(α2 + β2 + γ2 + δ2) = (α + β + γ + δ)2. §4. Three circles of the same radius 3.24. Three circles of radius R pass through point H; let A, B and C be points of their pairwise intersection distinct from H. Prove that a) H is the intersection point of heights of triangle ABC; b) the radius of the circumscribed circle of the triangle ABC is also equal to R. Figure 20 (3.24) 3.25. Three equal circles intersect as shown on Fig. 20 a) or b). Prove that ⌣AB1+ ⌣ BC1± ⌣CA1 = 180◦, where the minus sign is taken in case b) and plus in case a). Figure 21 (3.26) 3.26. Three circles of the same radius pass through point P; let A, B and Q be points of their pairwise intersections. A fourth circle of the same radius passes through point Q §6. APPLICATION OF THE THEOREM ON TRIANGLE’S HEIGHTS 61 and intersects the other two circles at points C and D. The triangles ABQ and CDP thus obtained are acute ones and quadrilateral ABCD is a convex one (Fig. 21). Prove that ABCD is a parallelogram. §5. Two tangents drawn from one point 3.27. Tangents AB and AC are drawn from point A to a circle centered at O. Prove that if segment AO subtends a right angle with vertex at point M, then segments OB and OC subtend equal angles with vertices at M. 3.28. Tangents AB and AC are drawn from point A to a circle centered at O. Through point X on segment BC line KL perpendicular to XO is drawn so that points K and L lie on lines AB and AC, respectively. Prove that KX = XL. 3.29. On the extension of chord KL of a circle centered at O a point A is taken and tangents AP and AQ to the circle are drawn from it; let M be the midpoint of segment PQ. Prove that ∠MKO = ∠MLO. 3.30. From point A tangents AB and AC to a circle and a line that intersects the circle at points D and E are drawn; let M be the midpoint of segment BC. Prove that BM 2 = DM · ME and either ∠DME = 2∠DBE or ∠DME = 2∠DCE; moreover, ∠BEM = ∠DEC. 3.31. Quadrilateral ABCD is inscribed in a circle so that tangents to this circle at points B and D intersect at a point K that lies on line AC. a) Prove that AB · CD = BC · AD. b) A line parallel to KB intersects lines BA, BD and BC at points P, Q and R, respectively. Prove that PQ = QR. ∗∗∗ 3.32. A circle S and a line l that has no common points with S are given. From point P that moves along line l tangents PA and PB to circle S are drawn. Prove that all chords AB have a common point. Let point P lie outside circle S; let PA and PB be tangents to the circle. Then line AB is called the polar line of point P relative circle S. 3.33. Circles S1 and S2 intersect at points A and B so that the center O of circle S1 lies on S2. A line that passes through point O intersects segment AB at point P and circle S2 at point C. Prove that point P lies on the polar line of point C relative circle S1. §6. Application of the theorem on triangle’s heights 3.34. Points C and D lie on the circle with diameter AB. Lines AC and BD, AD and BC meet at points P and Q, respectively. Prove that AB ⊥PQ. 3.35. Lines PC and PD are tangent to the circle with diameter AB so that C and D are tangent points. Prove that the line that connects P with the intersection point of lines AC and BD is perpendicular to AB. 3.36. Given diameter AB of a circle and point C outside AB. With the help of the ruler alone (no compasses) drop the perpendicular from C to AB if: a) point C does not lie on the circle; b) point C lies on the circle. 3.37. Let Oa, Ob and Oc be the centers of circumscribed circles of triangles PBC, PCA and PAB. Prove that if points Oa and Ob lie on lines PA and PB, then point Oc lies on line PC. 62 CHAPTER 3. CIRCLES §7. Areas of curvilinear ﬁgures 3.38. On the hypothenuse and legs of a rectangular triangle semicircles are constructed as shown on Fig. 22. Prove that the sum of the areas of the crescents obtained (shaded) is equal to the area of the given triangle. Figure 22 (3.38) 3.39. In a disc two perpendicular diameters, i.e., four radii, are constructed. Then there are constructed four disks whose diameters are these radii. Prove that the total area of the pairwise common parts of these four disks is equal to the area of the initial (larger) disk that lies outside the considered four disks (Fig. 23). Figure 23 (3.39) 3.40. On three segments OA, OB and OC of the same length (point B lies outside angle AOC) circles are constructed as on diameters. Prove that the area of the curvilinear triangle bounded by the arcs of these circles and not containing point O is equal to a half area of the (common) triangle ABC. 3.41. On sides of an arbitrary acute triangle ABC as on diameters circles are constructed. They form three “outer” curvilinear triangles and one “inner” triangle (Fig. 24). Prove that if we subtract the area of the “inner” triangle from the sum of the areas of “outer” triangles we get the doubled area of triangle ABC. §8. Circles inscribed in a disc segment In this section a segment is always a disc segment. 3.42. Chord AB divides circle S into two arcs. Circle S1 is tangent to chord AB at point M and one of the arcs at point N. Prove that: a) line MN passes through the midpoint P of the second arc; b) the length of tangent PQ to circle S1 is equal to that of PA. 3.43. From point D of circle S the perpendicular DC is dropped to diameter AB. Circle S1 is tangent to segment CA at point E and also to segment CD and to circle S. Prove that DE is a bisector of triangle ADC. §10. THE RADICAL AXIS 63 Figure 24 (3.41) 3.44. Two circles inscribed in segment AB of the given circle intersect at points M and N. Prove that line MN passes through the midpoint C of arc AB complementary for the given segment. 3.45. A circle tangent to sides AC and BC of triangle ABC at points M and N, respectively is also tangent to its circumscribed circle (from the inside). Prove that the midpoint of segment MN coincides with the center of the inscribed circle of triangle ABC. 3.46. Triangles ABC1 and ABC2 are inscribed in circle S so that chords AC2 and BC1 intersect. Circle S1 is tangent to chord AC2 at point M2, to chord BC1 at point N1 and to circle S (???where?). Prove that the centers of the inscribed circles of triangles ABC1 and ABC2 lie on segment M2N1. §9. Miscellaneous problems 3.47. The radii of two circles are equal to R1 and R2 and the distance between the centers of the circles is equal to d. Prove that these circles are orthogonal if and only if d2 = R2 1 + R2 2. 3.48. Three circles are pairwise tangent from the outside at points A, B and C. Prove that the circumscribed circle of triangle ABC is perpendicular to all the three circles. 3.49. Two circles centered at O1 and O2 intersect at points A and B. A line is drawn through point A; the line intersects the ﬁrst circle at point M1 and the second circle at point M2. Prove that ∠BO1M1 = ∠BO2M2. §10. The radical axis 3.50. Circle S and point P are given on the plane. A line drawn through point P intersects the circle at points A and B. Prove that the product PA · PB does not depend on the choice of a line. This product taken with the plus sign if point P is outside the circle and with minus sign if P is inside of the circle is called the degree of point P with respect to circle S. 3.51. Prove that for a point P outside circle S its degree with respect to S is equal to the square of the length of the tangent drawn to the circle from point P. 3.52. Prove that the degree of point P with respect to circle S is equal to d2 −R2, where R is the radius of S and d is the distance from P to the center of S. 3.53. Two nonconcentric circles S1 and S2 are given in plane. Prove that the locus of points whose degree with respect to S1 is equal to the degree with respect to S2 is a line. This line is called the radical axis of circles S1 and S2. 64 CHAPTER 3. CIRCLES 3.54. Prove that the radical axis of two intersecting circles passes through the intersection points. 3.55. Given three circles in plane whose centers do not lie on one line. Let us draw radical axes for each pair of these circles. Prove that all the three radical axes meet at one point. This point is called the radical center of the three circles. 3.56. Consider three pairwise intersecting circles in plane. Through the intersection points of any two of them a line is drawn. Prove that either these three lines meet at one point or are parallel. 3.57. Two nonconcentric circles S1 and S2 are given. Prove that the set of centers of circles that intersect both these circles at a right angle is their radical axis (without their common chord if the given circles intersect). 3.58. a) Prove that the midpoints of the four common tangents to two nonintersecting circles lie on one line. b) Through two of the tangent points of common exterior tangents with two circles a line is drawn, see Fig. . Prove that the circles cut on this line equal chords. 3.59. On sides BC and AC of triangle ABC, points A1 and B1, respectively, are taken; let l be the line that passes through the common points of circles with diameters AA1 and BB1. Prove that: a) Line l passes through the intersection point H of heights of triangle ABC; b) line l passes through point C if and only if AB1 : AC = BA1 : BC. 3.60. The extensions of sides AB and CD of quadrilateral ABCD meet at point F and the extensions of sides BC and AD meet at point E. Prove that the circles with diameters AC, BD and EF have a common radical axis and the orthocenters of triangles ABE, CDE, ADF and BCF lie on it. 3.6l. Three circles intersect pairwise at points A1 and A2, B1 and B2, C1 and C2. Prove that A1B2 · B1C2 · C1A2 = A2B1 · B2C1 · C2A1. 3.62. On side BC of triangle ABC point A′ is taken. The midperpendicular to segment A′B intersects side AB at point M and the midperpendicular to segment A′C intersects side AC at point N. Prove that point symmetric to point A′ through line MN lies on the circumscribed circle of triangle ABC. 3.63. Solve Problem 1.66 making use of the properties of the radical axis. 3.64. Inside a convex polygon several pairwise nonintersecting disks of distinct radii are placed. Prove that it is possible to cut the polygon into smaller polygons so that all these small polygons are convex and each of them contains exactly one of the given disks. 3.65. a) In triangle ABC, heights AA1, BB1 and CC1 are drawn. Lines AB and A1B1, BC and B1C1, CA and C1A1 intersect at points C′, A′ and B′, respectively. Prove that points A′, B′ and C′ lie on the radical axis of the circle of nine points (cf. Problem 5.106) and on that of the circumscribed circle. b) The bisectors of the outer angles of triangle ABC intersect the extensions of the opposite sides at points A′, B′ and C′. Prove that points A′, B′ and C′ lie on one line and this line is perpendicular to the line that connects the centers of the inscribed and circumscribed circles of triangle ABC. 3.66. Prove that diagonals AD, BE and CF of the circumscribed hexagon ABCDEF meet at one point. (Brianchon’s theorem.) 3.67. Given four circles S1, S2, S3 and S4 such that the circles Si and Si+1 are tangent from the outside for i = 1, 2, 3, 4, where S5 = S1. Prove that the radical axis of circles S1 and S3 passes through the intersection point of common outer tangents to S2 and S4. SOLUTIONS 65 3.68. a) Circles S1 and S2 intersect at points A and B. The degree of point P of circle S1 with respect to circle S2 is equal to p, the distance from point P to line AB is equal to h and the distance between the centers of circles is equal to d. Prove that \|p\| = 2dh. b) The degrees of points A and B with respect to the circumscribed circles of triangles BCD and ACD are equal to pa and pb, respectively. Prove that \|pa\|SBCD = \|pb\|SACD. Problems for independent study 3.69. An easy chair of the form of a disc sector of radius R is swinging on a horizontal table. What is the trajectory of the vertex of the sector? 3.70. From a point A outside a circle of radius R two tangents AB and AC are drawn, B and C are tangent points. Let BC = a. Prove that 4R2 = r2 + r2 a + 1 2a2, where r and ra are the radii of the inscribed and escribed circles of triangle ABC. 3.71. Two circles have an inner tangent. The line that passes through the center of a smaller circle intersects the greater one at points A and D and the smaller one at points B and C. Find the ratio of the radii of the circles if AB : BC : CD = 2 : 3 : 4. 3.72. The centers of three circles each of radius R, where 1 < R < 2, form an equilateral triangle with side 2. What is the distance between the intersection points of these circles that lie outside the triangle? 3.73. A point C is taken on segment AB and semicircles with diameters AB, AC and BC are constructed (on one side of line AB). Find the ratio of the area of the curvilinear triangle bounded by these semicircles to the area of the triangle formed by the midpoints of the arcs of these semicircles. 3.74. A circle intersects side BC of triangle ABC at points A1 and A2, side AC at points B1 and B2, side AB at points C1 and C2. Prove that AC1 C1B · BA1 A1C · CB1 B1A = µAC2 C2B · BA2 A2C · CB2 B2A ¶−1 . 3.75. From point A tangents AB and AC to a circle are drawn (B and C are tangent points); PQ is a diameter of the circle; line l is tangent to the circle at point Q. Lines PA, PB and PC intersect line l at points A1, B1 and C1. Prove that A1B1 = A1C1. Solutions 3.1. Let line XY be tangent to the given circle at point Z. The corresponding sides of triangles XOA and XOZ are equal and, therefore, ∠XOA = ∠XOZ. Similarly, ∠ZOY = ∠BOY . Therefore, ∠XOY = ∠XOZ + ∠ZOY = ∠AOZ + ∠ZOB 2 = ∠AOB 2 . 3.2. Let M and N be the tangent points of the inscribed circle with sides AB and BC. Then BK + AN = BM + AM = AB, hence, CK + CN = a + b −c. Let P and Q be the tangent points of the escribed circle with the extensions of sides AB and BC. Then AP = AB + BP = AB + BL and AQ = AC + CQ = AC + CL. Hence, AP + AQ = a + b + c. Therefore, BL = BP = AP −AB = 1 2(a + b −c). 3.3. By Problem 3.2 CM = 1 2(AC + CE −AE) and CN = 1 2(BC + CE −BE). Taking into account that AC = BC we get MN = \|CM −CN\| = 1 2\|AE −BE\|. 3.4. Let lines AB, BC, CD and DA be tangent to the circle at points P, Q, R and S, respectively. Then CQ = CR = x, hence, BP = BC +CQ = BC +x and DS = DC +CR = DC +x. Therefore, AP = AB +BP = AB +BC +x and AS = AD +DS = AD +DC +x. Taking into account that AP = AS, we get the statement desired. 66 CHAPTER 3. CIRCLES 3.5. Let line AB be tangent to the circles centered at O1 and O2 at points C and D, respectively. Since ∠O1AO2 = 90◦, the right triangles AO1C and O2AD are similar. Therefore, O1C : AC = AD : DO2. Moreover, AD = CB (cf. Problem 3.2). Therefore, AC · CB = Rr. 3.6. Let lines AB and CD intersect at point O. Let us assume for deﬁniteness that points A and D lie on the ﬁrst circle while points B and C lie on the second one. Suppose also that OB < OA (Fig. 25). Figure 25 (Sol. 3.6) The intersection point M of bisectors of angles ∠A and ∠D of quadrilateral ABCD is the midpoint of the arc of the ﬁrst circle that lies inside triangle AOD and the intersection point N of bisectors of angles ∠B and ∠C is the midpoint of the arc of the second circle that lies outside triangle BOC, cf. Problem 2.91 a). Quadrilateral ABCD is a circumscribed one if and only if points M and N coincide. 3.7. Let R be the tangent point of the escribed circle with side BD, let P and Q be the intersection points of segment MN with BC and CD, respectively (Fig. 26). Figure 26 (Sol. 3.7) Since ∠DMQ = ∠BPN, ∠DQM = ∠BNP and ∠DMQ = ∠BNP, it follows that triangles MDQ, PBN and PCQ are isosceles ones. Therefore, CP = CQ, DQ = DM = DR and BP = BN = BR. Therefore, P, Q and R are the tangent points of the inscribed circle of triangle BCD with its sides (cf. Problem 5.1). 3.8. Denote some of the tangent points as shown on Fig. 27. The sum of the lengths of one pair of the opposite sides of the inner quadrilateral is equal to the sum of the lengths SOLUTIONS 67 of the pair of its other sides. Let us extend the sides of this quadrilateral to tangent points with inscribed circles of the other quadrilaterals (ST is one of the obtained segments). Figure 27 (Sol. 3.8) Then both sums of lengths of pairs of opposite segments increase by the same number. Each of the obtained segments is the common tangent to a pair of “corner” circles; each segment can be replaced with another common outer tangent of equal length (i.e., replace ST with QR). To prove the equality AB + CD = BC + AD, it remains to make use of equalities of the form AP = AQ. 3.9. Let P be the intersection point of diagonals of convex quadrilateral ABCD. Quadri-lateral ABCD is an inscribed one if and only if △APB ∼△DPC, i.e., PA·PC = PB ·PD. Since quadrilaterals ALBN and AMBK are inscribed ones, PL·PN = PA·PB = PM ·PK. Hence, quadrilateral KLMN is an inscribed one. 3.10. Let O be the intersection point of line AB and segment MN. Then OM 2 = OA · OB = ON 2, i.e., OM = ON. 3.11. Let, for deﬁniteness, rays OA and BC be codirected, M the intersection point of lines KL and OA. Then ∠LOM = ∠LCB = ∠OKM and, therefore, △KOM ∼△OLM. Hence, OM : KM = LM : OM, i.e., OM 2 = KM · LM. Moreover, MA2 = MK · ML. Therefore, MA = OM. 3.12. Let O be the intersection point of diagonals AC and BD. Then MO · OC = BO · OD. Since OC = OA and BO = OD, we have MO · OA = BO2 and MO · OA = DO2. These equalities mean that OB is tangent to the circumscribed circle of triangle ABM and OD is tangent to the circumscribed circle of triangle ADM. 3.13. Let C be the intersection point of line AB with the circumscribed circle of triangle BMN distinct from point B; let AP be the tangent to circle S. Then AB·AC = AM ·AN = AP 2 and, therefore, AC = AP 2 AB , i.e., point C is the same for all lines l. Remark. We have to exclude the case when the length of the tangent drawn to S from A is equal to AB. 3.14. Clearly, MC2 = MP · MQ = MA · MB and point M lies on ray AB if AC > BC and on ray BA if AC < BC. Let, for deﬁniteness sake, point M lie on ray AB. Then (MB + BC)2 = (MB + BA) · MB. Therefore, MB = BC2 AB−2BC and we deduce that the position of point M does not depend on the choice of circle S′. 3.15. Let M be the intersection point of line CD and the tangent to circles at point A. Then MC = MA = MD. Therefore, point A lies on the circle with diameter CD. 68 CHAPTER 3. CIRCLES 3.16. Points O1, A and O2 lie on one line, hence, ∠A2AO2 = ∠A1AO1. Triangles AO2A2 and AO1A1 are isosceles ones, hence, ∠A2AO2 = ∠AA2O2 and ∠A1AO1 = ∠AA1O1. Therefore, ∠AA2O2 = ∠AA1O1, i.e., O1A1 ∥O2A2. 3.17. Let O1, O2 and O3 be the centers of circles S1, S2 and S3; let A, B, C be the tangent points of circles S2 and S3, S3 and S1, S1 and S2, respectively; A1 and B1 the intersection points of lines CA and CB, respectively, with circle S3. By the previous problem B1O3 ∥CO1 and A1O3 ∥CO2. Points O1, C and O2 lie on one line and, therefore, points A1, O3 and B1 also lie on one line, i.e., A1B1 is a diameter of cicle S3. 3.18. Let A1, A2 and B be the tangent points of the circles centered at O and O1, O and O2, O1 and O2, respectively. Then O1O2 = O1B + BO2 = O1A1 + O2A2. Therefore, OO1 + OO2 + O1O2 = (OO1 + O1A1) + (OO2 + O2A2) = OA1 + OA2 = 2R. 3.19. Let O, O1 and O2 be centers of circles S, S1 and S2; let C be the common point of circles S1 and S2 that lies on segment AB. Triangles AOB, AO1C and CO2B are isosceles ones; consequently, OO1CO2 is a parallelogram and OO1 = O2C = O2B; hence, AO = AO1 + O1O = AO1 + O2B. 3.20. Let O1 and O2 be the centers of circles S1 and S2; let X be the other intersection point of line AB with circle S2. The square of the length of the tangent in question is equal to BA · BX. Since AB : BX = O1A : O1O2, it follows that AB · BX = AB2·O1O2 R = a2(R±r) R , where the minus sign is taken for the inner tangent and the plus sign for the outer tangent. 3.21. Let O, O1 and O2 be the centers of the circles with diameters AB, AC and BC, respectively. It suﬃces to verify that KO = OL. Let us prove that △O1KO = △O2OL. Indeed, O1K = 1 2AC = O2O, O1O = 1 2BC = O2L and ∠KO1O = ∠OO2L = 180◦−2α, where α is the value of the angle between lines KL and AB. 3.22. Let Oi be the center of circle Si and Ai the tangent point of circles Si and Si+1. Quadrilateral O1O2O3O4 is a convex one; let α1, α2, α3 and α4 be the values of its angles. It is easy to verify that ∠Ai−1AiAi+1 = 1 2(αi + αi+1) and, therefore, ∠A1 + ∠A3 = 1 2(α1 + α2 + α3 + α4) = ∠A2 + ∠A4. 3.23. a) Let A1, B1 and C1 be the projections of points A, B and C, respectively, to line l; let C2 be the projection of point C to line AA1. By Pythagorus theorem CC2 2 = AC2 −AC2 2, i.e., A1C2 1 = (a + c)2 −(a −c)2 = 4ac. Similarly, B1C2 1 = 4bc and A1B2 1 = 4ab. Since A1C1 + C1B1 = A1B1, it follows that √ac + √ bc = √ ab, i.e., 1 √ b + 1 √a = 1 √c. Figure 28 (Sol. 3.23 b)) SOLUTIONS 69 b) Let A, B, C be the centers of “outer” circles, D the center of the “inner” circle (Fig. 28). The semiperimeter of triangle BDC is equal to b + c + d, and, therefore, cos2 µ∠BDC 2 ¶ = d(b + c + d) (b + d)(c + d), sin2 µ∠BDC 2 ¶ = bc (b + d)(c + d) (cf. Problem 12.13). As is easy to see the law of cosines is equivalent to the statement: α′ + β′ + γ′ = 180◦= ⇒sin2 α′ −sin2 β′ −sin2 γ′ + 2 sin β′ sin γ′ cos α′ = 0. (∗) Substituting the values α′ = 1 2∠BDC, β′ = 1 2∠ADC and γ′ = 1 2∠ADB into formula (∗), we get bc (b + d)(c + d) − ac (a + d)(c + d) − ab (a + d)(b + d) + 2 a p bcd(b + c + d) (a + d)(b + d)(c + d) = 0, i.e., a + d a −b + d b −c + d c + 2 r d(b + c + d) bc = 0. Dividing this by d we get α −β −γ −δ + 2 p βγ + γδ + δβ = 0. Therefore, (α + β + γ + δ)2 = (α −β −γ −δ)2 + 4(αβ + αγ + αδ)+ 4(βγ + γδ + δβ) + 4(αβ + αγ + αδ) = 2(α + β + γ + δ)2 −2(α2 + β2 + γ2 + δ2), i.e., 2(α2 + β2 + γ2 + δ2) = (α + β + γ + δ)2. 3.24. Let A1, B1 and C1 be the centers of the given circles (Fig. 29). Then A1BC1H is a rhombus and, therefore, BA1 ∥HC1. Similarly, B1A ∥HC1; hence, B1A ∥BA1 and B1ABA1 is a parallelogram. Figure 29 (Sol. 3.24) a) Since A1B1 ⊥CH and A1B1 ∥AB, it follows that AB ⊥CH. We similarly prove that BC ⊥AH and CA ⊥BH. b) In the same way as we have proved that B1A ∥BA1, we can prove that B1C ∥BC1 and A1C ∥AC1; moreover, the lengths of all these six segments are equal to R. Let us complement the triangle BA1C to a rhombus BA1CO. Then AB1CO is also a rhombus. Therefore, AO = BO = CO = R, i.e., O is the center of the circumscribed circle of triangle ABC and its radius is equal to R. 70 CHAPTER 3. CIRCLES 3.25. It is easy to verify that ⌣AB1± ⌣B1A1 = ⌣AC1+ ⌣C1A1, ⌣BC1+ ⌣C1B1 = ⌣BA1± ⌣B1A1 ⌣C1A1± ⌣CA1 = ⌣C1B1± ⌣B1C, where the minus sign is only taken in case b). Adding up these equalities we get ⌣AB1+ ⌣BC1± ⌣CA1 =⌣AC1+ ⌣BA± ⌣CB1. On the other hand, the doubled values of the angles of triangle ABC are equal to ⌣BA1± ⌣ CA1, ⌣AB1± ⌣CB1 and ⌣BC1+ ⌣AC1, and their sum is equal to 360◦. 3.26. Since ⌣AP+ ⌣BP+ ⌣PQ = 180◦(cf. Problem 3.25), it follows that ⌣AB = 180◦−⌣PQ. Similarly, ⌣CD = 180◦−⌣PQ, i.e., ⌣AB =⌣CD and, therefore, AB = CD. Moreover, PQ ⊥AB and PQ ⊥CD (cf. Problem 3.24) and, therefore, AB ∥CD. 3.27. Points M, B and C lie on the circle with diameter AO. Moreover, chords OB and OC of the circle are equal. 3.28. Points B and X lie on the circle with diameter KO, and, therefore, ∠XKO = ∠XBO. Similarly, ∠XLO = ∠XCO. Since ∠XBO = ∠XCO, triangle KOL is an isosceles one and OX is its height. 3.29. It suﬃces to verify that AK·AL = AM·AO. Indeed, if such is the case, then points K, L, M and O lie on one circle and, therefore, ∠MKO = ∠MLO. Since △AOP ∼△APM, it follows that AM · AO = AP 2; it is also clear that AK · AL = AP 2. 3.30. Let O be the center of the circle; let points D′ and E′ be symmetric to points D and E through line AO. By Problem 28.7 the lines ED′ and E′D meet at point M. Hence, ∠BDM = ∠EBM and ∠BEM = ∠DBM and, therefore, △BDM ∼△EBM. It follows that BM : DM = EM : BM. Moreover, if line ED separates points B and M, then ∠DME =⌣DE = 2∠DCE. The equality ∠BEM = ∠DBM implies that ∠BEM = ∠DBC = ∠DEC. 3.31. a) Since △KAB ∼△KBC, we have AB : BC = KB : KC. Similarly, AD : DC = KD : KC. Taking into account that KB = KD we get the desired statement. b) The problem of this heading reduces to that of the previous one, since PQ BQ = sin ∠PBQ sin ∠BPQ = sin ∠ABD sin ∠KBA = sin ∠ABD sin ∠ADB = AD AB , QR BQ = CD CB . 3.32. Let us drop perpendicular OM to line l from center O of circle S. Let us prove that point X at which AB and OM intersect remains ﬁxed. Points A, B and M lie on the circle with diameter PO. Hence, ∠AMO = ∠ABO = ∠BAO and, therefore, △AMO ∼△XAO, because these triangles have a common angle at vertex O. It follows that AO : MO = XO : AO, i.e., OX = OA2 MO is a constant. 3.33. Since ∠OBP = ∠OAB = ∠OCB, we deduce that △OBP ∼△OCB and, therefore, OB2 = OP · OC. Let us draw tangent CD to circle S1 from point C. Then OD2 = OB2 = OP · OC. Therefore, △ODC ∼△OPD and ∠OPD = ∠ODC = 90◦. 3.34. Lines BC and AD are heights of triangle APB and, therefore, line PQ that passes through their intersection point Q is perpendicular to line AB. 3.35. Denote the intersection points of lines AC and BD, BC and AD by K and K1, respectively. Thanks to the above problem, KK1 ⊥AB and, therefore, it suﬃces to show that the intersection point of tangents at points C and D lies on line KK1. Let us prove that the tangent at point C passes through the midpoint of segment KK1. Let M be the intersection point of the tangent at point C and segment KK1. The respective sides of acute angles ∠ABC and ∠CKK1 are perpendicular and, therefore, the angles are SOLUTIONS 71 equal. Similarly, ∠CAB = ∠CK1K. It is also clear that ∠KCM = ∠ABC and, therefore, triangle CMK is an isosceles one. Similarly, triangle CMK1 is an isosceles one and KM = CM = K1M, i.e., M is the midpoint of segment KK1. We similarly prove that the tangent at point D passes through the midpoint of segment KK1. 3.36. a) Line AC intersects the circle at points A and A1, line BC does same at points B and B1. If A = A1 (or B = B1), then line AC (or BC) is the perpendicular to be constructed. If this is not the case, then AB1 and BA1 are heights of triangle ABC and the line to be constructed is the line that passes through point C and the intersection point of lines AB1 and BA1. b) Let us take point C1 that does not lie on the circle and drop from it perpendicular to AB. Let the perpendicular intersect the circle at points D and E. Let us construct the intersection point P of lines DC and AB and then the intersection point F of line PE with the circle. The symmetry through AB sends point C to point F. Therefore, CF is the perpendicular to be constructed. 3.37. Since PA ⊥ObOc, line PA passes through point Oa if and only if line POa passes through the intersection point of heights of triangle OaObOc. Similar statements are true for points B and C as well. The hypothesis of the problem implies that P is the intersection point of heights of triangle OaObOc and, therefore, POc ⊥OaOb. 3.38. Let 2a and 2b be the lengths of the legs, 2c the length of the hypothenuse. The sum of the areas of the “crescents” is equal to πa2+πb2+SABC −πc2. But π(a2+b2−c2) = 0. 3.39. It suﬃces to carry out the proof for each of the four parts into which the diameters divide the initial disc (Fig. 30). Figure 30 (Sol. 3.39) In the disc, consider the segment cut oﬀby the chord intercepted by the central angle of 90◦; let S and s be the areas of such segments for the initial disc and any of the four constructed disks, respectively. Clearly, S = 4s. It remains to observe that the area of the part shaded once is equal to S −2s = 2s and the area of the part shaded twice is equal to 2s. 3.40. Denote the intersection points of circles constructed on segments OB and OC, OA and OC, OA and OB as on diameters by A1, B1, C1, respectively (Fig. 31). Since ∠OA1B = ∠OA1C = 90◦, it follows that points B, A1 and C lie on one line and since all the circles have equal radii, BA1 = A1C. Points A1, B1, C1 are the midpoints of sides of triangle ABC, therefore, BA1 = C1B1 and BC = A1B1. Since the disks are of the same radius, the equal chords BA1 and C1B1 cut oﬀthe disks parts of equal area and equal chords C1B and B1A1 also cut oﬀthe disc’s parts of equal area. Therefore, the area of curvilinear triangle A1B1C1 is equal to the area of parallelogram A1B1C1B, i.e., is equal to half the area of triangle ABC. 72 CHAPTER 3. CIRCLES Figure 31 (Sol. 3.40) Figure 32 (Sol. 3.41) 3.41. The considered circles pass through the bases of the triangle’s heights and, there-fore, their intersection points lie on the triangle’s sides. Let x, y, z and u be the areas of the considered curvilinear triangles; let a, b, c, d, e and f be the areas of the segments cut oﬀ the circles by the sides of the triangle; let p, q and r be the areas of the parts of the triangle that lie outside the inner curvilinear triangle (see Fig. 32). Then x + (a + b) = u + p + q + (c + f), y + (c + d) = u + q + r + (e + b), z + (e + f) = u + r + p + (a + d) By adding up these equalities we get x + y + z = 2(p + q + r + u) + u. 3.42. a) Let O and O1 be the centers of circles S and S1. The triangles MO1N and PON are isosceles ones and ∠MO1N = ∠PON. Therefore, points P, M and N lie on one line. b) It is clear that PQ2 = PM · PN = PM · (PM + MN). Let K be the midpoint of chord AB. Then PM 2 = PK2 + MK2 and PM · MN = AM · MB = AK2 −MK2. Therefore, PQ2 = PK2 + AK2 = PA2. 3.43. By Problem 3.42 b) BE = BD. Hence, ∠DAE + ∠ADE = ∠DEB = ∠BDE = ∠BDC + ∠CDE. Since ∠DAB = ∠BDC, it follows that ∠ADE = ∠CDE. 3.44. Let O1 and O2 be the centers of the inscribed circles, CP and CQ the tangents to them. Then CO2 1 = CP 2+PO2 1 = CP 2+O1M 2 and since CQ = CA = CP (by Problem 3.42 SOLUTIONS 73 b), we have CO2 2 = CQ2 + QO2 2 = CP 2 + O2M 2. It follows that CO2 1 −CO2 2 = MO2 1 −MO2 2 and, therefore, line CM is perpendicular to O1O2 (see Problem 7.6). Therefore, line MN passes through point C. Remark. If the circles do not intersect but are tangent to each other the statement is still true; in this case, however, one should replace line MN with the tangent to the circles at their common point. 3.45. Let A1 and B1 be the midpoints of arcs ⌣BC and ⌣AC;let O the center of the inscribed circle. Then A1B1 ⊥CO (cf. Problem 2.19 a) and MN ⊥CO, consequently, MN ∥A1B1. Let us move points M ′ and N ′ along rays CA and CB, respectively, so that M ′N ′ ∥A1B1. Only for one position of points M ′ and N ′ does point L at which lines B1M ′ and A1N ′ intersect lie on the circumscribed circle of triangle ABC. On the other hand, if segment MN passes through point O, then point L lies on this circle (cf. Problem 2.49). 3.46. The solution of this problem generalizes the solution of the preceding problem. It suﬃces to prove that the center O1 of the inscribed circle of triangle ABC1 lies on segment M2N1. Let A1 and A2 be the midpoints of arcs ⌣BC1 and ⌣BC2; let B1 and B2 be the midpoints of arcs ⌣AC1 and ⌣AC2; let PQ be the diameter of circle S perpendicular to chord AB and let points Q and C1 lie on one side of line AB. Point O1 is the intersection point of chords AA1 and BB1 and point L of tangent of circles S and S1 is the intersection point of lines A1N1 and B2M2 (Fig. 33). Figure 33 (Sol. 3.46) Let ∠C1AB = 2α, ∠C1BA = 2β, ∠C1AC2 = 2ϕ. Then ⌣A1A2 = 2ϕ =⌣B1B2, i.e., A1B2 ∥B1A2. For the angles between chords we have: ∠(A1B2, BC1) = 1 2(⌣B2C1+ ⌣A1B) = β −ϕ + α, ∠(BC1, AC2) = 1 2(⌣C1C2+ ⌣AB) = 2ϕ + 180◦−2α −2β. Consequently, chord M2N1 constitutes equal angles with tangents BC1 and AC2, each angle equal to α + β −ϕ. Therefore, M2N1 ∥A1B2. Now, suppose that points M ′ 2 and N ′ 1 are moved along chords AC2 and BC1 so that M ′ 2N ′ 1 ∥A1B2. Let lines A1N ′ 1 and B2M ′ 2 meet at point L′. Point L′ lies on circle S for one position of points M ′ 2 and N ′ 1 only. Therefore, it suﬃces to indicate on arc ⌣AB a point L1 such that if M ′′ 2 , N ′′ 1 are the intersection points of chords AC2 and L1B2, BC1 and L1A1, 74 CHAPTER 3. CIRCLES respectively, then M ′′ 2 N ′′ 1 ∥A1B2 and point O1 lies on segment M ′′ 2 N ′′ 1 . Let Q1 be a point on circle S such that 2∠(PQ, PQ1) = ∠(PC2, PC1) and L1 the intersection point of line Q1O1 with S. Let us prove that L1 is the desired point. Since ⌣B1Q = 2α, it follows that ⌣B2Q1 = 2(α −2ϕ) =⌣C2A1. Hence, quadrilateral AM ′′ 2 O1L1 is an inscribed one and, therefore, ∠M ′′ 2 O1A = ∠M ′′ 2 L1A = ∠B2A1A, i.e., M ′′ 2 O1 ∥B2A1. Similarly, N ′′ 1 O1 ∥B2A1. 3.47. Let circles centered at O1 and O2 pass through point A. The radii O1A and O2A are perpendicular to the tangents to circles at point A and, therefore, the circles are orthogonal if and only if ∠O1AO2 = 90◦, i.e., ∠O1O2 2 = O1A2 + O2A2. 3.48. Let A1, B1 and C1 be the centers of the given circles so that points A, B and C lie on segments B1C1, C1A1 and A1B1, respectively. Since A1B = A1C, B1A = B1C and CA = C1B, it follows that A, B and C are the tangent points of the inscribed circle of triangle A1B1C1 with its sides (cf. Problem 5.1). Therefore, the radii A1B, BC and C1A of the given circles are tangent to the circumscribed circle of triangle ABC. 3.49. It is easy to verify that the angle of rotation from vector − − → OiB to vector − − − → OiMi (coun-terclockwise) is equal to 2∠(AB, AMi). It is also clear that ∠(AB, AM1) = ∠(AB, AM2). 3.50. Let us draw through point P another line that intersects the circle at points A1 and B1. Then △PAA1 ∼△PB1B and, therefore, PA : PA1 = PB1 : PB. 3.51. Let us draw through point P tangent PC. Since △PAC ∼△PCB, it follows that PA : PC = PC : PB. 3.52. Let the line that passes through point P and the center of the circle intersect the circle at points A and B. Then PA = d+R and PB = \|d−R\|. Therefore, PA·PB = \|d2−R2\|. It is also clear that the signs of the expression d2 −R2 and of the degree of point P with respect to to S are the same. 3.53. Let R1 and R2 be the radii of the circles. Let us consider the coordinate system in which the coordinates of the centers of the circles are (−a, 0) and (a, 0). By Problem 3.52 the degrees of the point with coordinates (x, y) with respect to the given circles are equal to (x + a)2 + y2 −R2 1 and (x −a)2 + y2 −R2 2, respectively. By equating these expressions we get x = R2 1−R2 2 4a . This equation determines the perpendicular to the segment that connects the centers of the circles. 3.54. The degrees of the intersection point of the circles with respect to each one of the cicles are equal to zero and, therefore, the point belongs to the radical axis. If there are two intersection points, then they uniquely determine the radical axis. 3.55. Since the centers of the circles do not lie on one line, the radical axis of the ﬁrst and the second circles intersects with the radical axis of the second and third circles. The degrees of the intersection point with respect to all three circles are equal and, therefore, this intersection point lies on the radical axis of the ﬁrst and third circles. 3.56. By Problem 3.54 the lines that contain chords are radical axes. By Problem 3.55 the radical axes meet at one point if the centers of the circles do not lie on one line. Otherwise they are perpendicular to this line. 3.57. Let O1 and O2 be the centers of given circles, r1 and r2 their radii. The circle S of radius r centered at O is orthogonal to circle Si if and only if r2 = OO2 i −r2 i , i.e., the squared radius of S is equal to the degree of point O with respect to circle Si. Therefore, the locus of the centers of the circles to be found is the set of the points of the radical axis whose degrees with respect to the given circles are positive. 3.58. a) The indicated points lie on the radical axis. SOLUTIONS 75 b) The tangent points of the outer tangents with the circles are vertices of trapezoid ABCD with base AB. The midpoints of lateral sides AD and BC belong to the radical axis and, therefore, the midpoint O of diagonal AC also belongs to the radical axis. If line AC intersects the circles at points A1 and C1, then OA1 · OA = OC1 · OC; consequently, OA1 = OC1 and AA1 = CC1. 3.59. a) Let SA and SB be circles with diameters AA1 and BB1; let S be the circle with diameter AB. The common chords of circles S and SA, S and SB are heights AHa and BHb and, therefore, these heights (or their extensions) intersect at point H. By Problem 3.56 the common chord of circles SA and SB passes through the intersection point of chords AHa and BHb. b) The common chord of circles SA and SB passes through the intersection point of lines A1Ha and B1Hb (i.e., through point C) if and only if CB1 · CHb = CA1 · CHa (here we should consider the lengths of segments as oriented). Since CHb = a2 + b2 −c2 2b and CHa = a2 + b2 −c2 2a , we deduce that CB1 b = CA1 a . 3.60. In triangle CDE, draw heights CC1 and DD1; let H be their intersection point. The circles with diameters AC and BD pass through points C1 and D1, respectively, there-fore, the degree of point H with respect to each of these circles is equal to its degree with respect to the circle with diameter CD (this circle passes through points C1 and D1). We similarly prove that the degrees of point H with respect to to circles with diameters AC, BD and EF are equal, i.e., the radical axes of these circles pass through point H. For the intersection points of heights of the other three triangles the proof is carried out in a similar way. Remark. The centers of the considered circles lie on the Gauss’ line (cf. Problem 4.55) and, therefore, their common radical axis is perpendicular to the Gauss line. 3.61. Lines A1A2, B1B2 and C1C2 meet at a point O (cf. Problem 3.56). Since △A1OB2 ∼△B1OA2, it follows that A1B2 : A2B1 = OA1 : OB1. Similarly, B1C2 : B2C1 = OB1 : OC1 and C1A2 : C2A1 = OC1 : OA1. By multiplying these equalities we get the statement desired. 3.62. Denote by B′ and C′ the intersection points of lines A′M and A′N, respectively, with the line drawn through point A parallel to BC (Fig. 34). Figure 34 (Sol. 3.62) Since triangles A′BM and A′NC are isosceles ones, △ABC = △A′B′C′. Since AM · BM = A′M · B′M, the degrees of point M with respect to circles S and S′ circumscribed about triangles ABC and A′B′C′, respectively, are equal. This is true for point N as well 76 CHAPTER 3. CIRCLES and, therefore, line MN is the radical axis of circles S and S′. Circles S and S′ have equal radii and, therefore, their radical axis is their axis of symmetry. The symmetry through line MN sends a point A′ that lies on circle S′ into a point that lies on circle S. 3.63. Let AC and BD be the tangents; E and K the intersection points of lines AC and BD, AB and CD, respectively; O1 and O2 the centers of the circles (Fig. 35). Figure 35 (Sol. 3.63) Since AB ⊥O1E, O1E ⊥O2E and O2E ⊥CD, it follows that AB ⊥CD and, therefore, K is the intersection point of circles S1 and S2 with diameters AC and BD, respectively. Point K lies on the radical axis of circles S1 and S2; it remains to verify that line O1O2 is this radical axis. The radii O1A and O1B are tangent to S1 and S2, respectively, and, therefore, point O1 lies on the radical axis. Similarly, point O2 also lies on the radical axis. 3.64. Denote the given circles by S1, . . . , Sn. For each circle Si consider the set Mi that consists of all the points X whose degree with respect to Si does not exceed their degrees with respect to the other circles S1, . . . , Si−1, Si+1, . . . , Sn. The set Mi is a convex one. Indeed, let Mij be the set of points X whose degree with respect to Si does not exceed the degree with respect to Sj. The set Mij is a half plane that consists of the points that lie on the same side of the radical axis of circles Si and Sj as Si does. The set Mi is the intersection of the convex sets Mij for all j and, therefore, is a convex set itself. Moreover, since each of the sets Mij contains circle Si, then Mi also contains Si. Since for each point of the plane at least one of the degrees with respect to S1, . . . , Sn is the least one, the sets Mi cover the whole plane. Now, by considering the parts of the sets Mi that lie inside the initial polygon we get the partition statement desired. 3.65. a) Points B1 and C1 lie on the circle with diameter BC and, therefore, the degrees of point A′ with respect to the circumscribed circles of triangles A1B1C1 and ABC are equal to the degrees of point A′ with respect to this circle. This means that point A′ lies on the radical axis of the Euler circle and the circumscribed circle of triangle ABC. For points B′ and C′ the proof is similar. b) Let us consider triangle A1B1C1 formed by the outer bisectors of triangle ABC (tri-angle A1B1C1 is an acute one). Thanks to heading a) points A′, B′ and C′ lie on the radical axis of the circumscribed circles of triangles ABC and A1B1C1. The radical axis of these circles is perpendicular to the line that connects their centers, i.e., the Euler line of triangle A1B1C1. It remains to notice that the intersection point of the heights of triangle A1B1C1 is the intersection point of the bisectors of triangle ABC, cf. Problem 1.56 a). 3.66. Let a convex hexagon ABCDEF be tangent to the circle at points R, Q, T, S, P, U (point R lies on AB, point Q lies on BC, etc.). SOLUTIONS 77 Take a number a > 0 and construct points Q′ and P ′ on lines BC and EF so that QQ′ = PP ′ = a and vectors − − → QQ′ and − − → PP ′ are codirected with vectors − − → CB and − → EF. Let us similarly construct points R′, S′, T ′, U ′ (see Fig. 36, where RR′ = SS′ = TT ′ = UU ′ = a). Let us construct circle S1 tangent to lines BC and EF at points Q′ and P ′. Let us similarly construct circles S2 and S3. Figure 36 (Sol. 3.66) Let us prove that points B and E lie on the radical axis of circles S1 and S2. We have BQ′ = QQ′ −BQ = RR′ −BR = BR′ (if QQ′ < BQ, then BQ′ = BQ −QQ′ = BR −RR′ = BR′) and EP ′ = EP + PP ′ = ES + SS′ = ES′. We similarly prove that lines FC and AD are the radical axes of circles S1 and S3, S2 and S3, respectively. Since the radical axes of three circles meet at one point, lines AD, BE and CF meet at one point. 3.67. Let Ai be the tangent point of circles Si and Si+1 and X be the intersection point of lines A1A4 and A2A3. Then X is the intersection point of the common outer tangents to circles S2 and S4 (cf. Problem 5.60). Since quadrilateral A1A2A3A4 is an inscribed one (by Problem 3.22), XA1 · XA4 = XA2 · XA3; consequently, point X lies on the radical axis of circles S1 and S3. 3.68. a) Let us consider the coordinate system whose origin O is at the center of the segment that connects the centers of the circles and the Ox-axis is directed along this segment. Let (x, y) be the coordinates of point P; let R and r be the radii of circles S1 and S2, respectively; a = 1 2d. Then (x + a)2 + y2 = R2 and p = (x −a)2 + y2 −r2 = ((x + a)2 + y2 −R2) −4ax −r2 + R2 = R2 −r2 −4ax. Let (x0, y0) be the coordinates of point A. Then (x0 + a)2 + y2 0 −R2 = (x0 −a)2 + y2 0 −r2, i.e., x0 = R2 −r2 4a . Therefore, 2dh = 4a\|x0 −x\| = \|R2 −r2 −4ax\| = \|p\|. b) Let d be the distance between the centers of the circumscribed circles of triangles ACD and BCD; let ha and hb be the distances from points A and B to line CD. By heading a) \|pa\| = 2dha and \|pb\| = 2dhb. Taking into account that SBCD = 1 2hbCD and SACD = 1 2haCD we get the statement desired. CHAPTER 4. AREA Background 1. One can calculate the area S of triangle ABC with the help of the following formulas: a) S = 1 2aha, where a = BC and ha is the length of the height dropped to BC; b) S = 1 2bc sin ∠A, where b, c are sides of the triangle, ∠A the angle between these sides; c) S = pr, where p is a semiperimeter, r the radius of the inscribed circle. Indeed, if O is the center of the inscribed circle, then S = SABO + SAOC + SOBC = 1 2(c + b + a)r = pr. 2. If a polygon is cut into several polygons, then the sum of their areas is equal to the area of the initial polygon. Introductory problems 1. Prove that the area of a convex quadrilateral is equal to 1 2d1d2 sin ϕ, where d1 and d2 are the lengths of the diagonals and ϕ is the angle between them. 2. Let E and F be the midpoints of sides BC and AD of parallelogram ABCD. Find the area of the quadrilateral formed by lines AE, ED, BF and FC if it is known that the area of ABCD is equal to S. 3. A polygon is circumscribed about a circle of radius r. Prove that the area of the polygon is equal to pr, where p is the semiperimeter of the polygon. 4. Point X is inside parallelogram ABCD. Prove that SABX + SCDX = SBCX + SADX. 5. Let A1, B1, C1 and D1 be the midpoints of sides CD, DA, AB, BC, respectively, of square ABCD whose area is equal to S. Find the area of the quadrilateral formed by lines AA1, BB1, CC1 and DD1. §1. A median divides the triangle into triangles of equal areas 4.1. Prove that the medians divide any triangle into six triangles of equal area. 4.2. Given triangle ABC, ﬁnd all points P such that the areas of triangles ABP, BCP and ACP are equal. 4.3. Inside given triangle ABC ﬁnd a point O such that the areas of triangles BOL, COM and AON are equal (points L, M and N lie on sides AB, BC and CA so that OL ∥BC, OM ∥AC and ON ∥AB; see Fig. 37). 4.4. On the extensions of the sides of triangle ABC points A1, B1 and C1 are taken so that − − → AB1 = 2− → AB, − − → BC1 = 2− − → BC and − − → CA1 = 2− → AC. Find the area of triangle A1B1C1 if it is known that the area of triangle ABC is equal to S. 4.5. On the extensions of sides DA, AB, BC, CD of convex quadrilateral ABCD points A1, B1, C1, D1 are taken so that − − → DA1 = 2− − → DA, − − → AB1 = 2− → AB, − − → BC1 = 2− − → BC and − − → CD1 = 2− − → CD. Find the area of the obtained quadrilateral A1B1C1D1 if it is known that the area of quadrilateral ABCD is equal to S. 79 80 CHAPTER 4. AREA Figure 37 (4.3) 4.6. Hexagon ABCDEF is inscribed in a circle. Diagonals AD, BE and CF are diameters of this circle. Prove that SABCDEF = 2SACE. 4.7. Inside a convex quadrilateral ABCD there exists a point O such that the areas of triangles OAB, OBC, OCD and ODA are equal. Prove that one of the diagonals of the quadrilateral divides the other diagonal in halves. §2. Calculation of areas 4.8. The height of a trapezoid whose diagonals are mutually perpendicular is equal to 4. Find the area of the trapezoid if it is known that the length of one of its diagonals is equal to 5. 4.9. Each diagonal of convex pentagon ABCDE cuts oﬀit a triangle of unit area. Calculate the area of pentagon ABCDE. 4.10. In a rectangle ABCD there are inscribed two distinct rectangles with a common vertex K lying on side AB. Prove that the sum of their areas is equal to the area of rectangle ABCD. 4.11. In triangle ABC, point E is the midpoint of side BC, point D lies on side AC; let AC = 1, ∠BAC = 60◦, ∠ABC = 100◦, ∠ACB = 20◦and ∠DEC = 80◦(Fig. 38). Find S△ABC + 2S△CDE. Figure 38 (4.11) 4.12. Triangle Ta = △A1A2A3 is inscribed in triangle Tb = △B1B2B3 and triangle Tb is inscribed in triangle Tc = △C1C2C3 so that the sides of triangles Ta and Tc are pairwise parallel. Express the area of triangle Tb in terms of the areas of triangles Ta and Tc. Figure 39 (4.12) 4.13. On sides of triangle ABC, points A1, B1 and C1 that divide its sides in ratios BA1 : A1C = p, CB1 : B1A = q and AC1 : C1B = r, respectively, are taken. The §4. THE AREAS OF THE PARTS INTO WHICH A QUADRILATERAL IS DIVIDED 81 intersection points of segments AA1, BB1 and CC1 are situated as depicted on Fig. 39. Find the ratio of areas of triangles PQR and ABC. §3. The areas of the triangles into which a quadrilateral is divided 4.14. The diagonals of quadrilateral ABCD meet at point O. Prove that SAOB = SCOD if and only if BC ∥AD. 4.15. a) The diagonals of convex quadrilateral ABCD meet at point P. The areas of triangles ABP, BCP, CDP are known. Find the area of triangle ADP. b) A convex quadrilateral is divided by its diagonals into four triangles whose areas are expressed in integers. Prove that the product of these integers is a perfect square. 4.16. The diagonals of quadrilateral ABCD meet at point P and S2 ABP + S2 CDP = S2 BCP + S2 ADP. Prove that P is the midpoint of one of the diagonals. 4.17. In a convex quadrilateral ABCD there are three inner points P1, P2, P3 not on one line and with the property that SABPi + SCDPi = SBCPi + SADPi for i = 1, 2, 3. Prove that ABCD is a parallelogram. §4. The areas of the parts into which a quadrilateral is divided 4.18. Let K, L, M and N be the midpoints of sides AB, BC, CD and DA, respectively, of convex quadrilateral ABCD; segments KM and LN intersect at point O. Prove that SAKON + SCLOM = SBKOL + SDNOM. 4.19. Points K, L, M and N lie on sides AB, BC, CD and DA, respectively, of parallel-ogram ABCD so that segments KM and LN are parallel to the sides of the parallelogram. These segments meet at point O. Prove that the areas of parallelograms KBLO and MDNO are equal if and only if point O lies on diagonal AC. 4.20. On sides AB and CD of quadrilateral ABCD, points M and N are taken so that AM : MB = CN : ND. Segments AN and DM meet at point K, and segments BN and CM meet at point L. Prove that SKMLN = SADK + SBCL. 4.21. On side AB of quadrilateral ABCD, points A1 and B1 are taken, on side CD points C1 and D1 are taken so that AA1 = BB1 = pAB and CC1 = DD1 = pCD, where p < 0.5. Prove that SA1B1C1D1 SABCD = 1 −2p. 4.22. Each of the sides of a convex quadrilateral is divided into ﬁve equal parts and the corresponding points of the opposite sides are connected as on Fig. 40. Prove that the area of the middle (shaded) quadrilateral is 25 times smaller than the area of the initial quadrilateral. Figure 40 (4.22) 82 CHAPTER 4. AREA 4.23. On each side of a parallelogram a point is taken. The area of the quadrilateral with vertices at these points is equal to half the area of the parallelogram. Prove that at least one of the diagonals of the quadrilateral is parallel to a side of the parallelogram. 4.24. Points K and M are the midpoints of sides AB and CD, respectively, of convex quadrilateral ABCD, points L and N lie on sides BC and AD so that KLMN is a rectangle. Prove that SABCD = SKLMN. 4.25. A square is divided into four parts by two perpendicular lines whose intersection point lies inside the square. Prove that if the areas of three of these parts are equal, then the area of all four parts are equal. §5. Miscellaneous problems 4.26. Given parallelogram ABCD and a point M, prove that SACM = \|SABM ± SADM\|. 4.27. On sides AB and BC of triangle ABC, parallelograms are constructed outwards; let P be the intersection point of the extensions of the sides of these parallelograms parallel to AB and BC. On side AC, a parallelogram is constructed whose other side is equal and parallel to BP. Prove that the area of this parallelogram is equal to the sum of areas of the ﬁrst two parallelograms. 4.28. Point O inside a regular hexagon is connected with the vertices. The six triangles obtained in this way are alternately painted red and blue. Prove that the sum of areas of red triangles is equal to the sum of areas of blue ones. 4.29. The extensions of sides AD and BC of convex quadrilateral ABCD meet at point O; let M and N be the midpoints of sides AB and CD; let P and Q be the midpoints of diagonals AC and BD. Prove that: a) SPMQN = 1 2\|SABD −SACD\|; b) SOPQ = 1 2SABCD. 4.30. On sides AB and CD of a convex quadrilateral ABCD points E and F are taken. Let K, L, M and N be the midpoints of segments DE, BF, CE and AF, respectively. Prove that quadrilateral KLMN is a convex one and its area does not depend on the choice of points E and F. 4.31. The midpoints of diagonals AC, BD, CE, . . . of convex hexagon ABCDEF are vertices of a convex hexagon. Prove that the area of the new hexagon is 1 4 of that of the initial one. 4.32. The diameter PQ and the chord RS perpendicular to it intersect in point A. Point C lies on the circle, point B lies inside the circle and we know that BC ∥PQ and BC = RA. From points A and B perpendiculars AK and BL are dropped to line CQ. Prove that SACK = SBCL. 4.33. Through point O inside triangle ABC segments are drawn parallel to its sides (Fig. 41). Segments AA1, BB1 and CC1 divide triangle ABC into four triangles and three quadrilaterals. Prove that the sum of areas of the triangles adjacent to vertices A, B and C is equal to the area of the fourth triangle. 4.34. On the bisector of angle ∠A of triangle ABC a point A1 is taken so that AA1 = p −a = 1 2(b + c −a) and through point A1 line la perpendicular to the bisector is drawn. If we similarly construct lines lb and lc, then triangle ABC will be divided into parts among §7. FORMULAS FOR THE AREA OF A QUADRILATERAL 83 Figure 41 (4.33) which there are four triangles. Prove that the area of one of these triangles is equal to the sum of areas of the three other triangles. See also problems 3.38–3.41, 13.52–13.56, 16.5, 24.5. §6. Lines and curves that divide ﬁgures into parts of equal area 4.35. Segment MN parallel to side CD of quadrilateral ABCD divides its area in halves (points M and N lie on sides BC and AD). The lengths of segments drawn from points A and B parallel to CD till they intersect with lines BC and AD are equal to a and b, respectively. Prove that MN 2 = 1 2(ab + c2), where c = CD. 4.36. Each of certain three lines divides the area of a ﬁgure in halves. Prove that the area of the part of the ﬁgure conﬁned inside the triangle formed by these lines does not exceed 1 4 of the area of the whole ﬁgure. 4.37. Line l divides the area of a convex polygon in halves. Prove that this line divides the projection of the given polygon onto a line perpendicular to l in the ratio that does not exceed 1 + √ 2. 4.38. Prove that any convex polygon can be cut by two mutually perpendicular lines in four ﬁgures of equal area. 4.39. a) Prove that any line that divides the area and the perimeter of the triangle in halves passes through the center of the inscribed circle. b) Prove a similar statement for any circumscribed polygon. 4.40. Points A and B of circle S1 are connected by an arc of circle S2 that divides the area of the disk bounded by S1 into equal parts. Prove that the length of the arc of S2 that connects A and B is greater than that of the diameter of S1. 4.41. Curve Γ divides a square into two parts of equal area. Prove that on Γ we can select two points A and B so that line AB passes through the center O of the square. See also problems 6.51, 6.52, 16.8, 18.29. §7. Formulas for the area of a quadrilateral 4.42. The diagonals of quadrilateral ABCD meet at point P. The distances from points A, B and P to line CD are equal to a, b and p, respectively. Prove that the area of quadrilateral ABCD is equal to ab·CD 2p . 4.43. Quadrilateral ABCD is inscribed into a circle of radius R; let ϕ be the angle between the diagonals of ABCD. Prove that the area S of ABCD is equal to 2R2 · sin ∠A · sin ∠B · sin ϕ. 84 CHAPTER 4. AREA 4.44. Prove that the area of a quadrilateral whose diagonals are not perpendicular is equal to 1 4 tan ϕ · \|a2 + c2 −b2 −d2\|, where a, b, c and d are the lengths of the consecutive sides and ϕ is the angle between the diagonals. 4.45. a) Prove that the area of a convex quadrilatral ABCD can be computed with the help of the formula S2 = (p −a)(p −b)(p −c)(p −d) −abcd cos2 µ∠B + ∠D 2 ¶ , where p is the semiperimeter, a, b, c, d are the lengths of the quadrilateral’s sides. b) Prove that if quadrilateral ABCD is an inscribed one, then S2 = (p −a)(p −b)(p −c)(p −d). c) Prove that if quadrilateral ABCD is a circumscribed one, then S2 = abcd sin2 µ∠B + ∠D 2 ¶ . See also Problem 11.34. §8. An auxiliary area 4.46. Prove that the sum of distances from an arbitrary point within an equilateral triangle to the triangle’s sides is constant (equal to the length of the triangle’s height). 4.47. Prove that the length of the bisector AD of triangle ABC is equal to 2bc b+c cos 1 2α. 4.48. Inside triangle ABC, point O is taken; lines AO, BO and CO meet the sides of the triangle at points A1, B1 and C1, respectively. Prove that: a) OA1 AA1 + OB1 BB1 + OC1 CC1 = 1; b) AC CB · BA1 A1C · CB1 B1A = 1. 4.49. A (2n −1)-gon A1 . . . A2n−1 and a point O are given. Lines AkO and An+k−1An+k meet at point Bk. Prove that the product of ratios An+k−1Bk An+kBk for k = 1, . . . , n is equal to 1. 4.50. A convex polygon A1A2 . . . An is given. On side A1A2 points B1 and D2 are taken, on side A2A3 points B2 and D3, etc. so that if we construct parallelograms A1B1C1D1, . . . , AnBnCnDn, then lines A1C1, . . . , AnCn would meet at one point O. Prove that A1B1 · A2B2 · · · · · AnBn = A1D1 · A2D2 · · · · · AnDn. 4.51. The lengths of the sides of a triangle form an arithmetic progression. Prove that the (length of the) radius of the inscribed circle is equal to one third of the length of one of the triangle’s heights. 4.52. The distances from point X on side BC of triangle ABC to lines AB and AC are equal to db and dc, respectively. Prove that db dc = BX·AC CX·AB. 4.53. A polygon circumscribed about a circle of radius r is divided into triangles (in an arbitrary way). Prove that the sum of radii of the inscribed circles of these triangles is greater than r. 4.54. Through point M inside parallelogram ABCD lines PR and QS parallel to sides BC and AB are drawn (points P, Q, R and S lie on sides AB, BC, CD and DA, respec-tively). Prove that lines BS, PD and MC meet at one point. 4.55. Prove that if no side of a quadrilateral is parallel to any other side, then the midpoint of the segment that connects the intersection points of the opposite sides lies on the line that connects the midpoints of the diagonals. (The Gauss line.) 4.56. In an acute triangle ABC heights BB1 and CC1 are drawn and points K and L are taken on sides AB and AC so that AK = BC1 and AL = CB1. Prove that line AO, §9. REGROUPING AREAS 85 where O is the center of the circumscribed circle of triangle ABC, divides segment KL in halves. 4.57. Medians AA1 and CC1 of triangle ABC meet at point M. Prove that if quadri-lateral A1BC1M is a circumscribed one, then AB = BC. 4.58. Inside triangle ABC a point O is taken. Denote the distances from O to sides BC, CA, AB of the triangle by da, db, dc, respectively, and the distances from point O to vertices A, B, C by Ra, Rb, Rc, respectively. Prove that: a) aRa ≥cdc + bdb; b) daRa + dbRb + dcRc ≥2(dadb + dbdc + dcda); c) Ra + Rb + Rc ≥2(da + db + dc); d) RaRbRc ≥R 2r(da + db)(db + dc)(dc + da). See also problems 5.5, 10.6. §9. Regrouping areas 4.59. Prove that the area of a regular octagon is equal to the product of the lengths of its greatest and smallest diagonals. 4.60. From the midpoint of each side of an acute triangle perpendiculars are dropped to two other sides. Prove that the area of the hexagon bounded by these perpendiculars is equal to a half area of the initial triangle. 4.61. Sides AB and CD of parallelogram ABCD of unit area are divided into n equal parts; sides AD and BC are divided into m equal parts. The division points are connected as indicated on a) Fig. 42 a); b) Fig. 42 b). Figure 42 (4.61) What are the areas of small parallelograms obtained in this way? 4.62. a) Four vertices of a regular 12-gon lie in the midpoints of a square (Fig. 43). Prove that the area of the shaded part is equal to 1 12 that of the 12-gon. Figure 43 (4.62) b) Prove that the area of a 12-gon inscribed in the unit circle is equal to 3. 86 CHAPTER 4. AREA Problems for independent study 4.63. The sides of an inscribed quadrilateral ABCD satisfy the relation AB · BC = AD · DC. Prove that the areas of triangles ABC and ADC are equal. 4.64. Is it possible to use two straight cuts passing through two vertices of a triangle to divide the triangle into four parts so that three triangles (of these parts) were of equal area? 4.65. Prove that all the convex quadrilaterals with common midpoints of sides are of equal area. 4.66. Prove that if two triangles obtained by extention of sides of a convex quadrilateral to their intersection are of equal area, then one of the diagonals divides the other one in halves. 4.67. The area of a triangle is equal to S, its perimeter is equal to P. Each of the lines on which the sides of the triangle lie are moved (outwards) by a distance of h. Find the area and the perimeter of the triangle formed by the three obtained lines. 4.68. On side AB of triangle ABC, points D and E are taken so that ∠ACD = ∠DCE = ∠ECB = ϕ. Find the ratio CD : CE if the lengths of sides AC and BC and angle ϕ are known. 4.69. Let AA1, BB1 and CC1 be the bisectors of triangle ABC. Prove that SA1B1C1 SABC = 2abc (a + b) · (b + c) · (c + a). 4.70. Points M and N are the midpoints of lateral sides AB and CD of trapezoid ABCD. Prove that if the doubled area of the trapezoid is equal to AN · NB + CM · MD, then AB = CD = BC + AD. 4.71. If a quadrilateral with sides od distinct lengths is inscribed into a circle of radius R, then there exist two more quadrilaterals not equal to it with the same lengths of sides inscribed in the same circle. These quadrilaterals have not more than three distinct lengths of diagonals: d1, d2 and d3. Prove that the area of the quadrilateral is equal to d1d2d3 4R . 4.72. On sides AB, BC and CA of triangle ABC points C1, A1 and B1 are taken; points C2, A2 and B2 are symmetric to these points through the midpoints of the corresponding sides. Prove that SA1B1C1 = SA2B2C2. 4.73. Inside triangle ABC, point P is taken. The lines that pass through P and vertices of the triangle intersect the sides at points A1, B1 and C1. Prove that the area of the triangle determined by the midpoints of segments AA1, BB1 and CC1 is equal to 1 4 of the area of triangle A1B1C1. Solutions 4.1. The triangles adjacent to one side have equal bases and a common height and, therefore, are of equal area. Let M be the intersection point of the medians of triangle ABC. Line BM divides each of the triangles ABC and AMC into two triangles of equal area; consequently, SABM = SBCM. Similarly, SBCM = SCAM. 4.2. The equality of areas of triangles ABP and BCP implies that the distances from points A and C to line BP are equal. Therefore, either line BP passes through the midpoint of segment AC or it is parallel to it. The points to be found are depicted on Fig. 44. 4.3. Denote the intersection point of line LO with side AC by L1. Since SLOB = SMOC and △MOC = △L1OC, it follows that SLOB = SL1CO. The heights of triangles LOB and L1OC are equal and, therefore, LO = L1O, i.e., point O lies on the median drawn from vertex A. SOLUTIONS 87 Figure 44 (Sol. 4.2) We similarly prove that point O lies on the medians drawn from vertices B and C, i.e., O is the intersection point of the medians of the triangle. These arguments also demonstrate that the intersection point of the medians of the triangle possesses the necessary property. 4.4. Since SA1BB1 = SA1AB = SABC, it follows that SAA1B1 = 2S. Similarly, SBB1C1 = SCC1A1 = 2S. Therefore, SABC = 7S. 4.5. Since AB = BB1, it follows that SBB1C = SBAC. Since BC = CC1, we have SB1C1C = SBB1C = SBAC and SBB1C1 = 2SBAC. Similarly, SDD1A1 = 2SACD and, conse-quently, SBB1C1 + SDD1A1 = 2SABC + 2SACD = 2SABCD. Similarly, SAA1B1 + SCC1D1 = 2SABCD, consequently, SA1B1C1D1 = SABCD + SAA1B1 + SBB1C1 + SCC1D1 + SDD1A1 = 5SABCD. 4.6. Let O be the center of the circumscribed circle. Since AD, BE and CF are diameters, SABO = SDEO = SAEO, SBCO = SEFO = SCEO, SCDO = SAFO = SACO. It is also clear that SABCDEF = 2(SABO + SBCO + SCDO) and SACE = SAEO + SCEO + SACO. Therefore, SABCDEF = 2SACE. 4.7. Let E and F be the midpoints of diagonals AC and BD, respectively. Since SAOB = SAOD, point O lies on line AF. Similarly, point O lies on line CF. Suppose that the intersection point of the diagonals is not the midpoint of either of them. Then the lines AF and CF have a unique common point, F; hence, O = F. We similarly prove that O = E. Contradiction. 4.8. Let the length of diagonal AC of trapezoid ABCD with base AD be equal to 5. Let us complement triangle ACB to parallelogram ACBE. The area of trapezoid ABCD is equal to the area of the right triangle DBE. Let BH be a height of triangle DBE. Then EH2 = BE2−BH2 = 52−42 = 32 and ED = BE2 EH = 25 3 . Therefore, SDBE = 1 2ED·BH = 50 3 . 4.9. Since SABE = SABC, it follows that EC ∥AB. The remaining diagonals are also parallel to the corresponding sides. Let P be the intersection point of BD and EC. If SBPC = x, then SABCDE = SABE + SEPB + SEDC + SBPC = 3 + x. (we have SEPB = SABE = 1 because ABPE is a parallelogram). Since SBPC : SDPC = BP : DP = SEPB : SEPD, it follows that x : (1 −x) = 1 : x and, therefore, x = √ 5−1 2 and SABCDE = √ 5+5 2 . 4.10. The centers of all the three rectangles coincide (see Problem 1.7) and, therefore, two smaller rectangles have a common diagonal, KL. Let M and N be the vertices of these rectangles that lie on side BC. Points M and N lie on the circle with diameter KL. Let O be 88 CHAPTER 4. AREA the center of the circle, O1 the projection of O to BC. Then BO1 = CO1 and MO1 = NO1 and, therefore, BM = NC. To prove that SKLM + SKLN = SKBCL it suﬃces to verify that (SKBM + SLCM) + (SKBN + SLCN) = SKBCL = 1 2BC(KB + CL) = 1 2BC · AB. It remains to observe that KB · BM + KB · BN = KB · BC, LC · CM + LC · CN = LC · BC, KB · BC + LC · BC = AB · BC. 4.11. Let us drop perpendicular l from point C to line AB. Let points A′, B′ and E′ be symmetric to points A, B and E, respectively, through line l. Then triangle AA′C is an equilateral one and ∠ACB = ∠BCB′ = ∠B′CA′ = 20◦. Triangles EE′C and DEC are isosceles ones with the angle of 20◦at the vertex and a common lateral side EC. Therefore, SABC + 2SEDC = SABC + 2SEE′C. Since E is the midpoint of BC, it follows that 2SEE′C = SBE′C = 1 2SBB′C. Hence, SABC + 2SEDC = SAA′C 2 = √ 3 8 . 4.12. Let the areas of triangles Ta, Tb and Tc be equal to a, b and c, respectively. Triangles Ta and Tc are homothetic and, therefore, the lines that connect their respective vertices meett at one point, O. The similarity coeﬃcient k of these triangles is equal to pa c. Clearly, SA1B3O : SC1B3O = A1O : C1O = k. Writing similar equations for $??? and adding them, we get a : b = k and, therefore, b = √ac. 4.13. Making use of the result of Problem 1.3 it is easy to verify that BQ BB1 = p + pq 1 + p + pq, B1R BB1 = qr1 + q + qr, CR CC1 = q + qr 1 + q + qr, CP CC1 = pr 1 + r + pr. (1) It is also clear that SP QR SRB1C = QR RB1 · PR RC and SRB1C SABC = B1C AC · B1R BB1. Hence, SPQR SABC = QR BB1 · PR RC · B1C AC = QR BB1 · PR CC1 · CC1 CR · B1C AC . Taking into account that QR BB1 = 1 − p+pq 1+p+pq − qr 1+q+rq = 1 1+p+pq − rq 1+q+rq = (1+q)(1−pqr) (1+p+pq)(1+q+qr) and PR CC1 = (1 + r)(1 −pqr) (1 + q + qr)(1 + r + pr) we get SPQR SABC = (1 −pqr)2 (1 + p + pq)(1 + q + qr)(1 + r + pr). 4.14. If SAOB = SCOD, then AO · BO = CO · DO. Hence, △AOD ∼△COB and AD ∥BC. These arguments are invertible. 4.15. a) Since SADP : SABP = DP : BP = SCDP : SBCP, we have SADP = SABP · SCDP SBCP . SOLUTIONS 89 b) Thanks to heading a) SADP · SCBP = SABP · SCDP. Therefore, SABP · SCBP · SCDP · SADP = (SADP · SCBP)2. 4.16. After division by 1 4 sin2 ϕ, where ϕ is the angle between the diagonals, we rewrite the given equality of the areas in the form (AP · BP)2 + (CP · DP)2 = (BP · CP)2 + (AP · DP)2, i.e., (AP 2 −CP 2)(BP 2 −DP 2) = 0. 4.17. Suppose that quadrilateral ABCD is not a parallelogram; for instance, let lines AB and CD intersect. By Problem 7.2 the set of points P that lie inside quadrilateral ABCD for which SABP + SCDP = SBCP + SADP = 1 2SABCD is a segment. Therefore, points P1, P2 and P3 lie on one line. Contradiction. 4.18. Clearly, SAKON = SAKO + SANO = 1 2(SAOB + SAOD). Similarly, SCLOM = 1 2(SBCO + SCOD). Hence, SAKON + SCLOM = 1 2SABCD. 4.19. If the areas of the parallelograms KBLO and MDNO are equal, then OK · OL = OM · ON. Taking into account that ON = KA and OM = LC, we get KO : KA = LC : LO. Therefore, △KOA ∼△LCO which means that point O lies on diagonal AC. These arguments are invertible. 4.20. Let h1, h and h2 be the distances from points A, M and B to line CD, respectively. By Problem 1.1 b) we have h = ph2 + (1 −p)h1, where p = AM AB . Therefore, SDMC = h · DC 2 = h2p · DC + h1(1 −p) · DC 2 = SBCN + SADN. Subtracting SDKN + SCLN from both sides of this equality we get the desired statement. 4.21. Thanks to Problem 4.20, SABD1 + SCDB1 = SABCD. Hence, SA1B1C1D1 = SA1B1D1 + SC1D1B1 = (1 −2p)SABD1 + (1 −2p)SCDB1 = (1 −2p)SABCD. 4.22. By Problem 4.21 the area of the middle quadrilateral of those deteremined by segments that connect points of sides AB and CD is 1 5 of the area of the initial quadrilateral. Since each of the considered segments is divided by segments that connect the corresponding points of the other pair of opposite sides into 5 equal parts (see Problem 1.16). By making use once again of the result of Problem 4.21, we get the desired statement. 4.23. On sides AB, BC, CD and AD points K, L, M and N, respectively, are taken. Suppose that diagonal KM is not parallel to side AD. Fix points K, M and N and let us move point L along side BC. In accordance with this movement the area of triangle KLM varies strictly monotonously. Moreover, if LN ∥AB, then the equality SAKN + SBKL + SCLM + SDMN = 1 2SABCD holds, i.e., SKLMN = 1 2SABCD. 4.24. Let L1 and N1 be the midpoints of sides BC and AD, respectively. Then KL1MN1 is a parallelogram and its area is equal to a half area of quadrilateral ABCD, cf. Problem 1.37 a). Therefore, it suﬃces to prove that the areas of parallelograms KLMN and KL1MN1 are equal. If these parallelograms coincide, then there is nothing more to prove and if they do not coincide, then LL1 ∥NN1 and BC ∥AD because the midpoint of segment KM is 90 CHAPTER 4. AREA their center of symmetry. In this case the midline KM of trapezoid ABCD is parallel to bases BC and AD and therefore, heights of triangles KLM and KL1M dropped to side KM are equal, i.e., the areas of parallelograms KLMN and KL1MN1 are equal. 4.25. Let the given lines l1 and l2 divide the square into four parts whose areas are equal to S1, S2, S3 and S4 so that for the ﬁrst line the areas of the parts into which it divides the square are equal to S1 + S2 and S3 + S4 and for the second line they are equal to S2 + S3 and S1 + S4. Since by assumption S1 = S2 = S3, it follows that S1 + S2 = S2 + S3. This means that the image of line l1 under the rotation about the center of the square through an angle of +90◦or −90◦is not just parallel to line l2 but coincides with it. It remains to prove that line l1 (hence, line l2) passes through the center of the square. Suppose that this is not true. Let us consider the images of lines l1 and l2 under rotations through an angle of ±90◦and denote the areas of the parts into which they divide the square as plotted on Fig. 45 (on this ﬁgure both distinct variants of the disposition of the lines are plotted). Figure 45 (Sol. 4.25) Lines l1 and l2 divide the square into four parts whose areas are equal to a, a+b, a+2b+c and a + b, where numbers a, b and c are nonzero. It is clear that three of the four numbers indicated cannot be equal. Contradiction. 4.26. All the three triangles considered have a common base AM. Let hb, hc and hd be the distances from points B, C and D, respectively, to line AM. Since − → AC = − → AB + − − → AD, it follows that hc = \|hb ± hd\|. 4.27. We may assume that P is a common point of parallelograms constructed on sides AB and BC, i.e., these parallelograms are of the form ABPQ and CBPR. It is clear that SACRQ = SABPQ + SCBPR. 4.28. Let the length of the hexagon’s side be equal to a. The extensions of red sides of the hexagon form an equilateral triangle with side 3a and the sum of areas of red triangles is equal to a half product of a by the sum of distances from point O to a side of this triangle and, therefore, it is equal to 3 √ 3 4 a2, cf. Problem 4.46. The sum of areas of blue triangles is similarly calculated. 4.29. a) The area of parallelogram PMQN is equal to 1 4BC · AD sin α, where α is the angle between lines AD and BC. The heights of triangles ABD and ACD dropped from vertices B and C are equal to OB sin α and OC sin α, respectively; hence, \|SABD −SACD\| = \|OB −OC\| · AD sin α 2 = BC · AD sin α 2 . b) Let, for deﬁniteness, rays AD and BC intersect. Since PN ∥AO and QN ∥CO, point N lies inside triangle OPQ. Therefore, SOPQ = SPQN + SPON + SQON = 1 2SPMQN + 1 4SACD + 1 4SBCD = 1 4(SABD −SACD + SACD + SBCD = 1 4SABCD. SOLUTIONS 91 4.30. Segments KM and LN are the midlines of triangles CED and AFB and, therefore, they have a common point — the midpoint of segment EF. Moreover, KM = 1 2CD, LN = 1 2AB and the angle between lines KM and LN is equal to the angle α between lines AB and CD. Therefore, the area of quadrilateral KLMN is equal to 1 8AB · CD sin α. Figure 46 (Sol. 4.31) 4.31. Denote the midpoints of the diagonals of hexagon ABCDEF as shown on Fig. 46. Let us prove that the area of quadrilateral A1B1C1D1 is 1 4 of the area of quadrilateral ABCD. To this end let us make use of the fact that the area of the quadrilateral is equal to a half product of the lengths of the diagonals by the sine of the angle between them. Since A1C1 and B1D1 are the midlines of triangles BDF and ACE, we get the desired statement. We similarly prove that the area of quadrilateral D1E1F1A1 is 1 4 of the area of quadrilat-eral DEFA. 4.32. Let α = ∠PQC. Then 2SACK = CK · AK = (AP cos α) · (AQ sin α) = AR2 sin α · cos α = BC2 sin α · cos α = BL · CL = 2SBCL. 4.33. Let Sa, Sb and Sc be the areas of the triangles adjacent to vertices A, B and C; let S be the area of the fourth of the triangles considered. Clearly, SACC1 + SBAA1 + SCBB1 = SABC −S + Sa + Sb + Sc. Moreover, SABC = SAOC + SAOB + SBOC = SACC1 + SBAA1 + SCBB1. 4.34. Let O be the center of the inscribed circle of triangle ABC, let B1 be the tangent point of the inscribed circle and side AC. Let us cut oﬀtriangle ABC triangle AOB1 and reﬂect AOB1 symmetrically through the bisector of angle OAB1. Under this reﬂection line OB1 turns into line la. Let us perform a similar operation for the remaining triangles. The common parts of the triangles obtained in this way are three triangles of the considered partition and the uncovered part of triangle ABC is the fourth triangle. It is also clear that the area of the uncovered part is equal to the sum of areas of the parts covered twice. 4.35. Let, for deﬁniteness, rays AD and BC meet at point O. Then SCDO : SMNO = c2 : x2, where x = MN and SABO : SMNO = ab : x2 because OA : ON = a : x and OB : OM = b : x. It follows that x2 −c2 = ab −x2, i.e., 2x2 = ab + c2. 4.36. Denote the areas of the parts of the ﬁgure into which it is divided by lines as shown on Fig. 47. Let us denote by S the area of the whole ﬁgure. Since S3 + (S2 + S7) = 1 2S = S1 + S6 + (S2 + S7), 92 CHAPTER 4. AREA Figure 47 (Sol. 4.36) it follows that S3 = S1 + S6. Adding this equality to the equality 1 2S = S1 + S2 + S3 + S4 we get 1 2S = 2S1 + S2 + S4 + S6 ≥2S1, i.e., S1 ≤1 4S. 4.37. Let us denote the projection of line l by B and the endpoints of the projection of the polygon by A and C. Let C1 be a point of the polygon whose projection is C. Then line l intersects the polygon at points K and L; let points K1 and L1 be points on lines C1K and C1L that have point A as their projection (Fig. 48). Figure 48 (Sol. 4.37) One of the parts into which line l divides the polygon is contained in trapezoid K1KLL1, the other part contains triangle C1KL. Therefore, SK1KLL1 ≥SC1KL, i.e., AB · (KL + K1L1) ≥BC · KL. Since K1L1 = KL · AB+BC BC , we have AB · ¡ 2 + AB BC ¢ ≥BC. Solving this inequality we get BC AB ≤1 + √ 2. Similarly, AB BC ≤1 + √ 2. (We have to perform the same arguments but interchange A and C.) 4.38. Let S denote the area of the polygon, l an arbitrary line. Let us introduce a coordinate system in which line l is Ox-axis. Let S(a) be the area of the part of the polygon below the line y = a. Clearly, S(a) varies continuously from 0 to S as a varies from −∞to +∞and, therefore (by Calculus, see, e.g., ??), S(a) = 1 2S for some a, i.e., the line y = a divides the area of the polygon in halves. Similarly, there exists a line perpendicular to l and this perpendicular also divides the area of the polygon in halves. These two lines divide the polygon into parts whose areas are equal to S1, S2, S3 and S4 (see Fig. 49). Since S1 + S2 = S3 + S4 and S1 + S4 = S2 + S3, we have S1 = S3 = A and S2 = S4 = B. The rotation of line l by 90◦interchanges points A SOLUTIONS 93 Figure 49 (Sol. 4.38) and B. Since A and B vary continuously under the rotation of l, it follows that A = B for a certain position of l, i.e., the areas of all the four ﬁgures are equal at this moment. 4.39. a) Let the line that divides the area and the perimeter of triangle ABC in halves intersect sides AC and BC at points P and Q, respectively. Denote the center of the inscribed circle of triangle ABC by O and the radius of the inscribed circle by r. Then SABQOP = 1 2r(AP + AB + BQ) and SOQCP = 1 2r(QC + CP). Since line PQ divides the perimeter in halves, AP +AB+BQ = QC+CP and, therefore, SABQOP = SOQCP. Moreover, SABQP = SQCP by the hypothesis. Therefore, SOQP = 0, i.e., line QP passes through point O. b) Proof is carried out similarly to that of heading a). 4.40. By considering the image of circle S2 under the symmetry through the center of circle S1 and taking into account the equality of areas, it is possible to prove that diameter AA1 of circle S1 intersects S2 at a point K distinct from A and so that AK > A1K. The circle of radius KA1 centered at K is tangent to S1 at point A1 and, therefore, BK > A1K, i.e., BK + KA > A1A. It is also clear that the sum of the lengths of segments BK and KA is smaller than the length of the arc of S2 that connects points A and B. 4.41. The case when point O belongs to Γ is obvious; therefore, let us assume that O does not belong to Γ. Let Γ′ be the image of the curve Γ under the symmetry through point O. If curves Γ and Γ′ do not intersect, then the parts into which Γ divides the square cannot be of equal area. Let X be the intersection point of Γ and Γ′; let X′ be symmetric to X through point O. Since under the symmetry through point O curve Γ′ turns into Γ, it follows that X′ belongs to Γ. Hence, line XX′ is the desired one. 4.42. Let the areas of triangles APB, BPC, CPD and DPA be equal to S1, S2, S3 and S4, respectively. Then a p = S3+S4 S3 and b·CD 2 = S3 + S2; consequently, ab · CD 2p = (S3 + S4)(S3 + S2) S3 . Taking into account that S2S4 = S1S3 we get the desired statement. 4.43. By applying the law of sines to triangles ABC and ABD we get AC = 2R sin ∠B and BD = 2R · sin ∠A. Therefore, S = 1 2AC · BD sin ϕ = 2R2 sin ∠A · sin ∠B · sin ϕ. 4.44. Since the area of the quadrilateral is equal to 1 2d1d2 sin ϕ, where d1 and d2 are the lengths of the diagonals, it remains to verify that 2d1d2 cos ϕ = \|a2 + c2 −b2 −d2\|. Let O be the intersection point of the diagonals of quadrilateral ABCD and ϕ = ∠AOB. Then AB2 = AO2 + BO2 −2AO · OB cos ϕ; BC2 = BO2 + CO2 + 2BO · CO cos ϕ. 94 CHAPTER 4. AREA Hence, AB2 −BC2 = AO2 −CO2 −2BO · AC cos ϕ. Similarly, CD2 −AD2 = CO2 −AO2 −2DO · AC cos ϕ. By adding these equalities we get the desired statement. Remark. Since 16S2 = 4d2 1d2 2 sin2 ϕ = 4d2 1d2 2 −(2d1d2 cos ϕ)2, it follows that 16S2 = 4d2 1d2 2 −(a2 + c2 −b2 −d2)2. 4.45. a) Let AB = a, BC = b, CD = c and AD = d. Clearly, S = SABC + SADC = 1 2ab sin ∠B + cd sin ∠D; a2 + b2 −2ab cos ∠B = AC2 = c2 + d2 −2cd cos ∠D. Therefore, 16S2 = 4a2b2 −4a2b2 cos2 ∠B + 8abcd sin ∠B sin ∠D + 4c2d2 −4c2d2 cos2 ∠D, (a2 + b2 −c2 −d2)2 + 8abcd cos ∠B cos ∠D = 4a2b2 · cos2 ∠B + 4c2d2 cos2 ∠D. By inserting the second equality into the ﬁrst one we get 16S2 = 4(ab + cd)2 −(a2 + b2 −c2 −d2)2 −8abcd(1 + cos ∠B cos ∠D −sin ∠B sin ∠D). Clearly, 4(ab + cd)2 −(a2 + b2 −c2 −d2)2 = 16(p −a)(p −b)(p −c)(p −d); 1 + cos ∠B cos ∠D −sin ∠B sin ∠D = 2 cos2 ∠B+∠D 2 . b) If ABCD is an inscribed quadrilateral, then ∠B + ∠D = 180◦and, therefore, cos2 ∠B+∠D 2 = 0. c) If ABCD is a circumscribed quadrilateral, then a + c = b + d and, therefore, p = a + c = b + d and p −a = c, p −b = d, p −c = a, p −d = b. Hence, S2 = abcd µ 1 −cos2 ∠B + ∠D 2 ¶ = abcd sin2 ∠B + ∠D 2 . If ABCD is simultaneously an inscribed and circumscribed quadrilateral, then S2 = abcd. 4.46. Let us drop perpendiculars OA1, OB1 and OC1 to sides BC, AC and AB, respec-tively, of an equilateral triangle ABC from a point O inside it. In triangle ABC, let a be the length of the side, h the length of the height. Clearly, SABC = SBCO + SACO + SABO. Therefore, ah = a · OA1 + a · OB1 + a · OC1, i.e., h = OA1 + OB1 + OC1. 4.47. Let AD = l. Then 2SABD = cl sin α 2 , 2SACD = bl sin α 2 , 2SABD = bc sin α. Hence, cl sin α 2 + bl sin α 2 = bc sin α = 2bc sin α 2 cos α 2 . 4.48. a) Let the distances from points A and O to line BC be equal to h and h1, respectively. Then SOBC : SABC = h1 : H = OA1 : AA1. Similarly, SOAC : SABC = OB1 : BB1 and SOAB : SABC = OC1 : CC1. By adding these equalities and taking into account that SOBC + SOAC + SOAB = SABC we get the desired statement. b) Let the distances from points B and C to line AA1 be equal to db and dc, respectively. Then SABO : SACO = db : dc = BA1 : A1C. Similarly, SACO : SBCO = AC1 : C1B and SBCO : SABO = CB1 : B1A. It remains to multiply these equalities. SOLUTIONS 95 4.49. It is easy to verify that the ratio of the lengths of segments An+k−1Bk and An+kBk is equal to the ratio of areas of triangles An+k−1OAk and AkOAn+k. By multiplying these equalities we get the desired statement. 4.50. Since AiBiCiDi is a parallelogram and point O lies on the extension of its diagonal AiCi, it follows that SAiBiO = SAiDiO and, therefore, AiBi : AiDi = hi : hi−1, where hi is the distance from point O to side AiAi+1. It remains to multiply these equalities for i = 1, . . . , n. 4.51. Let the lengths of sides of triangle ABC be equal to a, b and c, where a ≤b ≤c. Then 2b = a + c and 2SABC = r(a + b + c) = 3rb, where r is the radius of the inscribed circle. On the other hand, 2SABC = hbb. Therefore, r = 1 3hb. 4.52. It suﬃces to observe that db · AB = 2SAXB = BX · AX sin ϕ, where ϕ = ∠AXB and dc · AC = 2SAXC = CX · AX sin ϕ. 4.53. Let r1, . . . , rn be the radii of the inscribed circles of the obtained triangles, let P1, . . . , Pn their perimeters and S1, . . . , Sn their areas. Let us denote the area and the perimeter of the initial polygon by S and P, respectively. It is clear that Pi < P (cf. Problem 9.27, b). Hence, r1 + · · · + rn = 2S1 P1 + · · · + 2Sn Pn > 2S1 P + · · · + 2Sn P = 2 S P = r. 4.54. Let us draw lines Q1S1 and P1R1 parallel to lines QS and PR through the inter-section point N of lines BS and CM (points P1, Q1, R1 and S1 lie on sides AB, BC, CD and DA, respectively). Let F and G be the intersection points of lines PR and Q1S1, P1R1 and QS, respectively. Since point M lies on diagonal NC of parallelogram NQ1CR1, it follows that SFQ1QM = SMRR1G (by Problem 4.19) and, therefore, SNQ1QG = SNFRR1. Point N lies on diagonal BS of parallelogram ABQS and, therefore, SAP1NS1 = SNQ1QG = SNFRR1. It follows that point N lies on diagonal PD of parallelogram APRD. 4.55. Let E and F be the intersection points of the extensions of sides of the given quadrilateral. Denote the vertices of the quadrilateral so that E is the intersection point of the extensions of sides AB and CD beyond points B and C and F is the intersection point of rays BC and AD. Let us complement triangles AEF and ABD to parallelograms AERF and ABLD, respectively. The homothety with center A and coeﬃcient 2 sends the midpoint of the diagonal BD, the midpoint of the diagonal AC and the midpoint of segment EF to points L, C and R, respectively. Therefore, it suﬃces to prove that points L, C and R lie on one line. This is precisely the fact proved in the preceding problem. 4.56. It suﬃces to verify that SAKO = SALO, i.e., AO·AL sin ∠OAL = AO·AK sin ∠OAK. Clearly, AL = CB1 = BC cos ∠C, sin ∠OAL = cos ∠B, AK = BC1 = BC cos ∠B, sin ∠OAK = cos ∠C. 4.57. Since quadrilateral A1BC1M is a circumscribed one, then, ﬁrst, the sums of the lengths of its opposite sides are equal: a 2 + mc 3 = c 2 + ma 3 and, second, its inscribed circle is simultaneously the inscribed circle of triangles AA1B and CC1B. Since these triangles have equal areas, their perimeters are equal: c + ma + a 2 = a + mc + c 2. 96 CHAPTER 4. AREA By multiplying the ﬁrst equality by 3 and adding to the second one we get the desired statement. 4.58. First, let us prove a general inequality that will be used in the proof of headings a)–d): (∗) BC1 · Ra ≥B1K · Ra + C1L · Ra = 2SAOB1 + 2SAOC1 = AB1 · dc + AC1 · db. On rays AB and AC take arbitrary points B1 and C1 and drop from them perpendiculars B1K and C1L to line AO. Since B1C1 ≥B1K + C1L, inequality (∗) follows. a) Setting B1 = B and C1 = C we get the desired statement. b) By multiplying both sides of the inequality aRa ≥cdc + bdb by da a we get daRa ≥c adadc + b adadb. Taking the sum of this inequality with the similar inequalities for dbRb and dcRc and taking into account that x y + y x ≥2 we get the desired statement. c) Take points B1 and C1 such that AB1 = AC and AC1 = AB. Then aRa ≥bdc + cdb, i.e., Ra ≥b adc + c adb. Taking the sum of this inequality with similar inequalities for Rb and Rc and taking into account that x y + y x ≥2 we get the desired statement. d) Take points B1 and C1 such that AB1 = AC1 = 1; then B1C1 = 2 sin 1 2∠A and, therefore, 2 sin 1 2Ra ≥dc + db. By multiplying this inequality by similar inequalities for Rb and Rc and taking into account that sin 1 2∠A sin 1 2∠B sin 1 2∠C = r 4R (by Problem 12.36 a)) we get the desired statement. 4.59. Let us cut triangles oﬀa regular octagon and replace the triangles as shown on Fig. 50. As a result we get a rectangle whose sides are equal to the longest and shortest diagonals of the octagon. Figure 50 (Sol. 4.59) 4.60. Let A1, B1 and C1 be the midpoints of sides BC, CA and AB, respectively, of triangle ABC. The drawn segments are heights of triangles AB1C1, A1BC1 and A1B1C, respectively. Let P, Q and R be the respective intersection points of the heights of these triangles and O the intersection point of the heights of triangle A1B1C1 (Fig. 51). The considered hexagon consists of triangle A1B1C1 and triangles B1C1P, C1A1Q and A1BR. Clearly, △B1C1P = △C1B1O, △C1A1Q = △A1C1O and △A1B1R = △B1A1O. Therefore, the area of the considered hexagon is equal to the doubled area of triangle A1B1C1. It remains to observe that SABC = 4SA1B1C1. 4.61. a) Let us cut two parts oﬀthe parallelogram (Fig. 52 a)) and replace these parts as shown on Fig. 52 b). We get a ﬁgure composed of mn + 1 small parallelograms. Therefore, the area of a small parallelogram is equal to 1 mn+1. b) Let us cut oﬀthe parallelogram three parts (Fig. 53 a)) and replace these parts as indicated on Fig. 53 b). We get a ﬁgure that consists of mn −1 small parallelograms. Therefore, the area of a small parallelogram is equal to 1 mn−1. SOLUTIONS 97 Figure 51 (Sol. 4.60) Figure 52 (Sol. 4.61 a)) Figure 53 (Sol. 4.61 b)) 4.62. a) Let us cut the initial square into four squares and consider one of them (Fig. 54). Let point B′ be symmetric to point B through line PQ. Let us prove that △APB = △OB′P. Indeed, triangle APB is an isosceles one and angle at its base is equal to 15◦(Problem 2.26), hence, triangle BPQ is an isosceles one. Therefore, ∠OPB′ = ∠OPQ −∠B′PQ = 75◦−60◦= 15◦ and ∠POB′ = 1 2∠OPQ = 15◦. Moreover, AB = OP. We similarly prove that △BQC = △OB′Q. It follows that the area of the shaded part on Fig. 43 is equal to the area of triangle OPQ. Figure 54 (Sol. 4.62) b) Let the area of the regular 12-gon inscribed in a circle of radius 1 be equal to 12x. Thanks to heading a) the area of the square circumscribed around this circle is equal to 98 CHAPTER 4. AREA 12x + 4x = 16x; on the other hand, the area of the square is equal to 4; hence, x = 1 4 and 12x = 3. CHAPTER 5. TRIANGLES Background 1) The inscribed circle of a triangle is the circle tangent to all its sides. The center of an inscribed circle is the intersection point of the bisectors of the triangle’s angles. An escribed circle of triangle ABC is the circle tangent to one side of the triangle and extensions of the other two sides. For each triangle there are exactly three escribed circles. The center of an escribed circle tangent to side AB is the intersection point of the bisector of angle C and the bisectors of the outer angles A and B. The circumscribed circle of a triangle is the circle that passes through the vertices of the triangle. The center of the circumscribed circle of a triangle is the intersection point of the midperpendiculars to the triangle’s sides. 2) For elements of a triangle ABC the following notations are often used: a, b and c are the lengths of sides BC, CA and AB, respectively; α, β and γ are the values of angles at vertices A, B, C; R is the radius of the circumscribed cirlce; r is the radius of the inscribed circle; ra, rb and rc are the radii of the escribed circles tangent to sides BC, CA and AB, respectively; ha, hb and hc the lengths of the heights dropped from vertices A, B and C, respectively. 3) If AD is the bisector of angle A of triangle ABC (or the bisector of the outer angle A), then BD : CD = AB : AC (cf. Problem 1.17). 4) In a right triangle, the median drawn from the vertex of the right angle is equal to a half the hypothenuse (cf. Problem 5.16). 5) To prove that the intersection points of certain lines lie on one line Menelaus’s theorem (Problem 5.58) is often used. 6) To prove that certain lines intersect at one point Ceva’s theorem (Problem 5.70) is often used. Introductory problems 1. Prove that the triangle is an isosceles one if a) one of its medians coincides with a height; b) if one of its bisectors coincides with a height. 2. Prove that the bisectors of a triangle meet at one point. 3. On height AH of triangle ABC a point M is taken. Prove that AB2 −AC2 = MB2 −MC2. 4. On sides AB, BC, CA of an equilateral triangle ABC points P, Q and R, respectively, are taken so that AP : PB = BQ : QC = CR : RA = 2 : 1. Prove that the sides of triangle PQR are perpendicular to the respective sides of triangle ABC. 99 100 CHAPTER 5. TRIANGLES 1. The inscribed and the circumscribed circles 5.1. On sides BC, CA and AB of triangle ABC, points A1, B1 and C1, respectively, are taken so that AC1 = AB1, BA1 = BC1 and CA1 = CB1. Prove that A1, B and C1 are the points at which the inscribed circle is tangent to the sides of the triangle. 5.2. Let Oa, Ob and Oc be the centers of the escribed circles of triangle ABC. Prove that points A, B and C are the bases of heights of triangle OaObOc. 5.3. Prove that side BC of triangle ABC subtends (1) an angle with the vertex at the center O of the inscribed circle; the value of the angle is equal to 90◦+ 1 2∠A and (2) an angle with the vertex at the center Oa of the escribed circle; the value of the angle is equal to 90◦−1 2∠A. 5.4. Inside triangle ABC, point P is taken such that ∠PAB : ∠PAC = ∠PCA : ∠PCB = ∠PBC : ∠PBA = x. Prove that x = 1. 5.5. Let A1, B1 and C1 be the projections of an inner point O of triangle ABC to the heights. Prove that if the lengths of segments AA1, BB1 and CC1 are equal, then they are equal to 2r. 5.6. An angle of value α = ∠BAC is rotated about its vertex O, the midpoint of the basis AC of an isosceles triangle ABC. The legs of this angle meet the segments AB and BC at points P and Q, respectively. Prove that the perimeter of triangle PBQ remains constant under the rotation. 5.7. In a scalene triangle ABC, line MO is drawn through the midpoint M of side BC and the center O of the inscribed circle. Line MO intersects height AH at point E. Prove that AE = r. 5.8. A circle is tangent to the sides of an angle with vertex A at points P and Q. The distances from points P, Q and A to a tangent to this circle are equal to u, v and w, respectively. Prove that uv w2 = sin2 1 2∠A. 5.9. Prove that the points symmetric to the intersection point of the heights of triangle ABC through its sides lie on the circumscribed circle. 5.10. From point P of arc BC of the circumscribed circle of triangle ABC perpendiculars PX, PY and PZ are dropped to BC, CA and AB, respectively. Prove that BC PX = AC PY + AB PZ . 5.11. Let O be the center of the circumscribed circle of triangle ABC, let I be the center of the inscribed circle, Ia the center of the escribed circle tangent to side BC. Prove that a) d2 = R2 −2Rr, where d = OI; b) d2 a = R2 + 2Rra, where da = OIa. 5.12. The extensions of the bisectors of the angles of triangle ABC intersect the circum-scribed circle at points A1, B1 and C1; let M be the intersection point of bisectors. Prove that a) MA·MC MB1 = 2r; b) MA1·MC1 MB = R. 5.13. The lengths of the sides of triangle ABC form an arithmetic progression: a, b, c, where a < b < c. The bisector of angle ∠B intersects the circumscribed circle at point B1. Prove that the center O of the inscribed circle divides segment BB1 in halves. 5.14. In triangle ABC side BC is the shortest one. On rays BA and CA, segments BD and CE, respectively, each equal to BC, are marked. Prove that the radius of the 101 circumscribed circle of triangle ADE is equal to the distance between the centers of the inscribed and circumscribed circles of triangle ABC. §2. Right triangles 5.15. In triangle ABC, angle ∠C is a right one. Prove that r = a+b−c 2 and rc = a+b+c 2 . 5.16. In triangle ABC, let M be the midpoint of side AB. Prove that CM = 1 2AB if and only if ∠ACB = 90◦. 5.17. Consider trapezoid ABCD with base AD. The bisectors of the outer angles at vertices A and B meet at point P and the bisectors of the angles at vertices C and D meet at point Q. Prove that the length of segment PQ is equal to a half perimeter of the trapezoid. 5.18. In an isosceles triangle ABC with base AC bisector CD is drawn. The line that passes through point D perpendicularly to DC intersects AC at point E. Prove that EC = 2AD. 5.19. The sum of angles at the base of a trapezoid is equal to 90◦. Prove that the segment that connects the midpoints of the bases is equal to a half diﬀerence of the bases. 5.20. In a right triangle ABC, height CK from the vertex C of the right angle is drawn and in triangle ACK bisector CE is drawn. Prove that CB = BE. 5.21. In a right triangle ABC with right angle ∠C, height CD and bisector CF are drawn; let DK and DL be bisectors in triangles BDC and ADC. Prove that CLFK is a square. 5.22. On hypothenuse AB of right triangle ABC, square ABPQ is constructed outwards. Let α = ∠ACQ, β = ∠QCP and γ = ∠PCB. Prove that cos β = cos α · cos γ. See also Problems 2.65, 5.62. §3. The equilateral triangles 5.23. From a point M inside an equilateral triangle ABC perpendiculars MP, MQ and MR are dropped to sides AB, BC and CA, respectively. Prove that AP 2 + BQ2 + CR2 = PB2 + QC2 + RA2, AP + BQ + CR = PB + QC + RA. 5.24. Points D and E divide sides AC and AB of an equilateral triangle ABC in the ratio of AD : DC = BE : EA = 1 : 2. Lines BD and CE meet at point O. Prove that ∠AOC = 90◦. 5.25. A circle divides each of the sides of a triangle into three equal parts. Prove that this triangle is an equilateral one. 5.26. Prove that if the intersection point of the heights of an acute triangle divides the heights in the same ratio, then the triangle is an equilateral one. 5.27. a) Prove that if a + ha = b + hb = c + hc, then triangle ABC is a equilateral one. b) Three squares are inscribed in triangle ABC: two vertices of one of the squares lie on side AC, those of another one lie on side BC, and those of the third lie one on AB. Prove that if all the three squares are equal, then triangle ABC is an equilateral one. 5.28. The circle inscribed in triangle ABC is tangent to the sides of the triangle at points A1, B1, C1. Prove that if triangles ABC and A1B1C1 are similar, then triangle ABC is an equilateral one. 5.29. The radius of the inscribed circle of a triangle is equal to 1, the lengths of the heights of the triangle are integers. Prove that the triangle is an equilateral one. 102 CHAPTER 5. TRIANGLES See also Problems 2.18, 2.26, 2.36, 2.44, 2.54, 4.46, 5.56, 7.45, 10.3, 10.77, 11.3, 11.5, 16.7, 18.9, 18.12, 18.15, 18.17-18.20, 18.22, 18.38, 24.1. §4. Triangles with angles of 60◦and 120◦ 5.30. In triangle ABC with angle A equal to 120◦bisectors AA1, BB1 and CC1 are drawn. Prove that triangle A1B1C1 is a right one. 5.31. In triangle ABC with angle A equal to 120◦bisectors AA1, BB1 and CC1 meet at point O. Prove that ∠A1C1O = 30◦. 5.32. a) Prove that if angle ∠A of triangle ABC is equal to 120◦then the center of the circumscribed circle and the orthocenter are symmetric through the bisector of the outer angle ∠A. b) In triangle ABC, the angle ∠A is equal to 60◦; O is the center of the circumscribed circle, H is the orthocenter, I is the center of the inscribed circle and Ia is the center of the escribed circle tangent to side BC. Prove that IO = IH and IaO = IaH. 5.33. In triangle ABC angle ∠A is equal to 120◦. Prove that from segments of lengths a, b and b + c a triangle can be formed. 5.34. In an acute triangle ABC with angle ∠A equal to 60◦the heights meet at point H. a) Let M and N be the intersection points of the midperpendiculars to segments BH and CH with sides AB and AC, respectively. Prove that points M, N and H lie on one line. b) Prove that the center O of the circumscribed circle lies on the same line. 5.35. In triangle ABC, bisectors BB1 and CC1 are drawn. Prove that if ∠CC1B1 = 30◦, then either ∠A = 60◦or ∠B = 120◦. See also Problem 2.33. §5. Integer triangles 5.36. The lengths of the sides of a triangle are consecutive integers. Find these integers if it is known that one of the medians is perpendicular to one of the bisectors. 5.37. The lengths of all the sides of a right triangle are integers and the greatest common divisor of these integers is equal to 1. Prove that the legs of the triangle are equal to 2mn and m2 −n2 and the hypothenuse is equal to m2 + n2, where m and n are integers. A right triangle the lengths of whose sides are integers is called a Pythagorean triangle. 5.38. The radius of the inscribed circle of a triangle is equal to 1 and the lengths of its sides are integers. Prove that these integers are equal to 3, 4, 5. 5.39. Give an example of an inscribed quadrilateral with pairwise distinct integer lengths of sides and the lengths of whose diagonals, the area and the radius of the circumscribed circle are all integers. (Brakhmagupta.) 5.40. a) Indicate two right triangles from which one can compose a triangle so that the lengths of the sides and the area of the composed triangle would be integers. b) Prove that if the area of a triangle is an integer and the lengths of the sides are consecutive integers then this triangle can be composed of two right triangles the lengths of whose sides are integers. 5.41. a) In triangle ABC, the lengths of whose sides are rational numbers, height BB1 is drawn. Prove that the lengths of segments AB1 and CB1 are rational numbers. §6. MISCELLANEOUS PROBLEMS 103 b) The lengths of the sides and diagonals of a convex quadrilateral are rational numbers. Prove that the diagonals cut it into four triangles the lengths of whose sides are rational numbers. See also Problem 26.7. §6. Miscellaneous problems 5.42. Triangles ABC and A1B1C1 are such that either their corresponding angles are equal or their sum is equal to 180◦. Prove that the corresponding angles are equal, actually. 5.43. Inside triangle ABC an arbitrary point O is taken. Let points A1, B1 and C1 be symmetric to O through the midpoints of sides BC, CA and AB, respectively. Prove that △ABC = △A1B1C1 and, moreover, lines AA1, BB1 and CC1 meet at one point. 5.44. Through the intersection point O of the bisectors of triangle ABC lines parallel to the sides of the triangle are drawn. The line parallel to AB meets AC and BC at points M and N, respectively, and lines parallel to AC and BC meet AB at points P and Q, respectively. Prove that MN = AM + BN and the perimeter of triangle OPQ is equal to the length of segment AB. 5.45. a) Prove that the heigths of a triangle meet at one point. b) Let H be the intersection point of heights of triangle ABC and R the radius of the circumscribed circle. Prove that AH2 + BC2 = 4R2 and AH = BC\| cot α\|. 5.46. Let x = sin 18◦. Prove that 4x2 + 2x = 1. 5.47. Prove that the projections of vertex A of triangle ABC on the bisectors of the outer and inner angles at vertices B and C lie on one line. 5.48. Prove that if two bisectors in a triangle are equal, then the triangle is an isosceles one. 5.49. a) In triangles ABC and A′B′C′, sides AC and A′C′ are equal, the angles at vertices B and B′ are equal, and the bisectors of angles ∠B and ∠B′ are equal. Prove that these triangles are equal. (More precisely, either △ABC = △A′B′C′ or △ABC = △C′B′A′.) b) Through point D on the bisector BB1 of angle ABC lines AA1 and CC1 are drawn (points A1 and C1 lie on sides of triangle ABC). Prove that if AA1 = CC1, then AB = BC. 5.50. Prove that a line divides the perimeter and the area of a triangle in equal ratios if and only if it passes through the center of the inscribed circle. 5.51. Point E is the midpoint of arc ⌣AB of the circumscribed circle of triangle ABC on which point C lies; let C1 be the midpoint of side AB. Perpendicular EF is dropped from point E to AC. Prove that: a) line C1F divides the perimeter of triangle ABC in halves; b) three such lines constructed for each side of the triangle meet at one point. 5.52. On sides AB and BC of an acute triangle ABC, squares ABC1D1 and A2BCD2 are constructed outwards. Prove that the intersection point of lines AD2 and CD1 lies on height BH. 5.53. On sides of triangle ABC squares centered at A1, B1 and C1 are constructed outwards. Let a1, b1 and c1 be the lengths of the sides of triangle A1B1C1; let S and S1 be the areas of triangles ABC and A1B1C1, respectively. Prove that: a) a2 1 + b2 1 + c2 1 = a2 + b2 + c2 + 6S. b) S1 −S = 1 8(a2 + b2 + c2). 5.54. On sides AB, BC and CA of triangle ABC (or on their extensions), points C1, A1 and B1, respectively, are taken so that ∠(CC1, AB) = ∠(AA1, BC) = ∠(BB1, CA) = 104 CHAPTER 5. TRIANGLES α. Lines AA1 and BB1, BB1 and CC1, CC1 and AA1 intersect at points C′, A′ and B′, respectively. Prove that: a) the intersection point of heights of triangle ABC coincides with the center of the circumscribed circle of triangle A′B′C′; b) △A′B′C′ ∼△ABC and the similarity coeﬃcient is equal to 2 cos α. 5.55. On sides of triangle ABC points A1, B1 and C1 are taken so that AB1 : B1C = cn : an, BC1 : CA = an : bn and CA1 : A1B = bn : cn (here a, b and c are the lengths of the triangle’s sides). The circumscribed circle of triangle A1B1C1 singles out on the sides of triangle ABC segments of length ±x, ±y and ±z, where the signs are chosen in accordance with the orientation of the triangle. Prove that x an−1 + y bn−1 + z cn−1 = 0. 5.56. In triangle ABC trisectors (the rays that divide the angles into three equal parts) are drawn. The nearest to side BC trisectors of angles B and C intersect at point A1; let us deﬁne points B1 and C1 similarly, (Fig. 55). Prove that triangle A1B1C1 is an equilateral one. (Morlie’s theorem.) Figure 55 (5.56) 5.57. On the sides of an equilateral triangle ABC as on bases, isosceles triangles A1BC, AB1C and ABC1 with angles α, β and γ at the bases such that α + β + γ = 60◦are constructed inwards. Lines BC1 and B1C meet at point A2, lines AC1 and A1C meet at point B2, and lines AB1 and A1B meet at point C2. Prove that the angles of triangle A2B2C2 are equal to 3α, 3β and 3γ. §7. Menelaus’s theorem Let − → AB and − − → CD be colinear vectors. Denote by AB CD the quantity ± AB CD, where the plus sign is taken if the vectors − → AB and − − → CD are codirected and the minus sign if the vectors are directed opposite to each other. 5.58. On sides BC, CA and AB of triangle ABC (or on their extensions) points A1, B1 and C1, respectively, are taken. Prove that points A1, B1 and C1 lie on one line if and only if BA1 CA1 · CB1 AB1 · AC1 BC1 = 1. (Menelaus’s theorem) 5.59. Prove Problem 5.85 a) with the help of Menelaus’s theorem. 105 5.60. A circle S is tangent to circles S1 and S2 at points A1 and A2, respectively. Prove that line A1A2 passes through the intersection point of either common outer or common inner tangents to circles S1 and S2. 5.61. a) The midperpendicular to the bisector AD of triangle ABC intersects line BC at point E. Prove that BE : CE = c2 : b2. b) Prove that the intersection point of the midperpendiculars to the bisectors of a triangle and the extensions of the corresponding sides lie on one line. 5.62. From vertex C of the right angle of triangle ABC height CK is dropped and in triangle ACK bisector CE is drawn. Line that passes through point B parallel to CE meets CK at point F. Prove that line EF divides segment AC in halves. 5.63. On lines BC, CA and AB points A1, B1 and C1, respectively, are taken so that points A1, B1 and C1 lie on one line. The lines symmetric to lines AA1, BB1 and CC1 through the corresponding bisectors of triangle ABC meet lines BC, CA and AB at points A2, B2 and C2, respectively. Prove that points A2, B2 and C2 lie on one line. 5.64. Lines AA1, BB1 and CC1 meet at one point, O. Prove that the intersection points of lines AB and A1B1, BC and B1C1, AC and A1C1 lie on one line. (Desargues’s theorem.) 5.65. Points A1, B1 and C1 are taken on one line and points A2, B2 and C2 are taken on another line. The intersection pointa of lines A1B2 with A2B1, B1C2 with B2C1 and C1A2 with C2A1 are C, A and B, respectively. Prove that points A, B and C lie on one line. (Pappus’ theorem.) 5.66. On sides AB, BC and CD of quadrilateral ABCD (or on their extensions) points K, L and M are taken. Lines KL and AC meet at point P, lines LM and BD meet at point Q. Prove that the intersection point of lines KQ and MP lies on line AD. 5.67. The extensions of sides AB and CD of quadrilateral ABCD meet at point P and the extensions of sides BC and AD meet at point Q. Through point P a line is drawn that intersects sides BC and AD at points E and F. Prove that the intersection points of the diagonals of quadrilaterals ABCD, ABEF and CDFE lie on the line that passes through point Q. 5.68. a) Through points P and Q triples of lines are drawn. Let us denote their intersection points as shown on Fig. 56. Prove that lines KL, AC and MN either meet at one point or are parallel. Figure 56 (5.68) b) Prove further that if point O lies on line BD, then the intersection point of lines KL, AC and MN lies on line PQ. 5.69. On lines BC, CA and AB points A1, B1 and C1 are taken. Let P1 be an arbitrary point of line BC, let P2 be the intersection point of lines P1B1 and AB, let P3 be the 106 CHAPTER 5. TRIANGLES intersection point of lines P2A1 and CA, let P4 be the intersection point of P3C1 and BC, etc. Prove that points P7 and P1 coincide. See also Problem 6.98. §8. Ceva’s theorem 5.70. Triangle ABC is given and on lines AB, BC and CA points C1, A1 and B1, respectively, are taken so that k of them lie on sides of the triangle and 3 −k on the extensions of the sides. Let R = BA1 CA1 · CB1 AB1 · AC1 BC1 . Prove that a) points A1, B1 and C1 lie on one line if and only if R = 1 and k is even. (Menelaus’s theorem.) b) lines AA1, BB1 and CC1 either meet at one point or are parallel if and only if R = 1 and k is odd. (Ceva’s theorem.) 5.71. The inscribed (or an escribed) circle of triangle ABC is tangent to lines BC, CA and AB at points A1, B1 and C1, respectively. Prove that lines AA1, BB1 and CC1 meet at one point. 5.72. Prove that the heights of an acute triangle intersect at one point. 5.73. Lines AP, BP and CP meet the sides of triangle ABC (or their extensions) at points A1, B1 and C1, respectively. Prove that: a) lines that pass through the midpoints of sides BC, CA and AB parallel to lines AP, BP and CP, respectively, meet at one point; b) lines that connect the midpoints of sides BC, CA and AB with the midpoints of segments AA1, BB1, CC1, respectively, meet at one point. 5.74. On sides BC, CA, and AB of triangle ABC, points A1, B1 and C1 are taken so that segments AA1, BB1 and CC1 meet at one point. Lines A1B1 and A1C1 meet the line that passes through vertex A parallel to side BC at points C2 and B2, respectively. Prove that AB2 = AC2. 5.75. a) Let α, β and γ be arbitrary angles such that the sum of any two of them is not less than 180◦. On sides of triangle ABC, triangles A1BC, AB1C and ABC1 with angles at vertices A, B, and C equal to α, β and γ, respectively, are constructed outwards. Prove that lines AA1, BB1 and CC1 meet at one point. b) Prove a similar statement for triangles constructed on sides of triangle ABC inwards. 5.76. Sides BC, CA and AB of triangle ABC are tangent to a circle centered at O at points A1, B1 and C1. On rays OA1, OB1 and OC1 equal segments OA2, OB2 and OC2 are marked. Prove that lines AA2, BB2 and CC2 meet at one point. 5.77. Lines AB, BP and CP meet lines BC, CA and AB at points A1, B1 and C1, respectively. Points A2, B2 and C2 are selected on lines BC, CA and AB so that BA2 : A2C = A1C : BA1, CB2 : B2A = B1A : CB1, AC2 : C2B = C1B : AC1. Prove that lines AA2, BB2 and CC2 also meet at one point, Q (or are parallel). Such points P and Q are called isotomically conjugate with respect to triangle ABC. 5.78. On sides BC, CA, AB of triangle ABC points A1, B1 and C1 are taken so that lines AA1, BB1 and CC1 intersect at one point, P. Prove that lines AA2, BB2 and CC2 symmetric to these lines through the corresponding bisectors also intersect at one point, Q. §9. SIMSON’S LINE 107 Such points P and Q are called isogonally conjugate with respect to triangle ABC. 5.80. The opposite sides of a convex hexagon are pairwise parallel. Prove that the lines that connect the midpoints of opposite sides intersect at one point. 5.81. From a point P perpendiculars PA1 and PA2 are dropped to side BC of triangle ABC and to height AA3. Points B1, B2 and C1, C2 are similarly deﬁned. Prove that lines A1A2, B1B2 and C1C2 either meet at one point or are parallel. 5.82. Through points A and D lying on a circle tangents that intersect at point S are drawn. On arc ⌣AD points B and C are taken. Lines AC and BD meet at point P, lines AB and CD meet at point Q. Prove that line PQ passes through point S. 5.83. a) On sides BC, CA and AB of an isosceles triangle ABC with base AB, points A1, B1 and C1, respectively, are taken so that lines AA1, BB1 and CC1 meet at one point. Prove that AC1 C1B = sin ∠ABB1 · sin ∠CAA1 sin ∠BAA1 · sin ∠CBB1 . b) Inside an isosceles triangle ABC with base AB points M and N are taken so that ∠CAM = ∠ABN and ∠CBM = ∠BAN. Prove that points C, M and N lie on one line. 5.84. In triangle ABC bisectors AA1, BB1 and CC1 are drawn. Bisectors AA1 and CC1 intersect segments C1B1 and B1A1 at points M and N, respectively. Prove that ∠MBB1 = ∠NBB1. See also Problems 10.56, 14.7, 14.38. §9. Simson’s line 5.85. a) Prove that the bases of the perpendiculars dropped from a point P of the circumscribed circle of a triangle to the sides of the triangle or to their extensions lie on one line. This line is called Simson’s line of point P with respect to the triangle. b) The bases of perpendiculars dropped from a point P to the sides (or their extensions) of a triangle lie on one line. Prove that point P lies on the circumscribed circle of the triangle. 5.86. Points A, B and C lie on one line, point P lies outside this line. Prove that the centers of the circumscribed circles of triangles ABP, BCP, ACP and point P lie on one circle. 5.87. In triangle ABC the bisector AD is drawn and from point D perpendiculars DB′ and DC′ are dropped to lines AC and AB, respectively; point M lies on line B′C′ and DM ⊥BC. Prove that point M lies on median AA1. 5.88. a) From point P of the circumscribed circle of triangle ABC lines PA1, PB1 and PC1 are drawn at a given (oriented) angle α to lines BC, CA and AB, respectively, so that points A1, B1 and C1 lie on lines BC, CA and AB, respectively. Prove that points A1, B1 and C1 lie on one line. b) Prove that if in the deﬁnition of Simson’s line we replace the angle 90◦by an angle α, i.e., replace the perpendiculars with the lines that form angles of α, their intersection points with the sides lie on the line and the angle between this line and Simson’s line becomes equal to 90◦−α. 5.89. a) From a point P of the circumscribed circle of triangle ABC perpendiculars PA1 and PB1 are dropped to lines BC and AC, respectively. Prove that PA·PA1 = 2Rd, where R is the radius of the circumscribed circle, d the distance from point P to line A1B1. b) Let α be the angle between lines A1B1 and BC. Prove that cos α = PA 2R . 108 CHAPTER 5. TRIANGLES 5.90. Let A1 and B1 be the projections of point P of the circumscribed circle of triangle ABC to lines BC and AC, respectively. Prove that the length of segment A1B1 is equal to the length of the projection of segment AB to line A1B1. 5.91. Points P and C on a circle are ﬁxed; points A and B move along the circle so that angle ∠ACB remains ﬁxed. Prove that Simson’s lines of point P with respect to triangle ABC are tangent to a ﬁxed circle. 5.92. Point P moves along the circumscribed circle of triangle ABC. Prove that Simson’s line of point P with respect to triangle ABC rotates accordingly through the angle equal to a half the angle value of the arc circumvent by P. 5.93. Prove that Simson’s lines of two diametrically opposite points of the circumscribed circle of triangle ABC are perpendicular and their intersection point lies on the circle of 9 points, cf. Problem 5.106. 5.94. Points A, B, C, P and Q lie on a circle centered at O and the angles between vector − → OP and vectors − → OA, − − → OB, − → OC and − → OQ are equal to α, β, γ and 1 2(α + β + γ), respectively. Prove that Simson’s line of point P with respect to triangle ABC is parallel to OQ. 5.95. Chord PQ of the circumscribed circle of triangle ABC is perpendicular to side BC. Prove that Simson’s line of point P with respect to triangle ABC is parallel to line AQ. 5.96. The heights of triangle ABC intersect at point H; let P be a point of its circum-scribed circle. Prove that Simson’s line of point P with respect to triangle ABC divides segment PH in halves. 5.97. Quadrilateral ABCD is inscribed in a circle; la is Simson’s line of point A with respect to triangle BCD; let lines lb, lc and ld be similarly deﬁned. Prove that these lines intersect at one point. 5.98. a) Prove that the projection of point P of the circumscribed circle of quadrilateral ABCD onto Simson’s lines of this point with respect to triangles BCD, CDA, DAB and BAC lie on one line. (Simson’s line of the inscribed quadrilateral.) b) Prove that by induction we can similarly deﬁne Simson’s line of an inscribed n-gon as the line that contains the projections of a point P on Simson’s lines of all (n −1)-gons obtained by deleting one of the vertices of the n-gon. See also Problems 5.10, 5.59. §10. The pedal triangle Let A1, B1 and C1 be the bases of the perpendiculars dropped from point P to lines BC, CA and AB, respectively. Triangle A1B1C1 is called the pedal triangle of point P with respect to triangle ABC. 5.99. Let A1B1C1 be the pedal triangle of point P with respect to triangle ABC. Prove that B1C1 = BC·AP 2R , where R is the radius of the circumscribed circle of triangle ABC. 5.100. Lines AP, BP and CP intersect the circumscribed circle of triangle ABC at points A2, B2 and C2; let A1B1C1 be the pedal triangle of point P with respect to triangle ABC. Prove that △A1B1C1 ∼△A2B2C2. 5.101. Inside an acute triangle ABC a point P is given. If we drop from it perpendiculars PA1, PB1 and PC1 to the sides, we get △A1B1C1. Performing for △A1B1C1 the same operation we get △A2B2C2 and then we similarly get △A3B3C3. Prove that △A3B3C3 ∼ △ABC. §11. EULER’S LINE AND THE CIRCLE OF NINE POINTS 109 5.102. A triangle ABC is inscribed in the circle of radius R centered at O. Prove that the area of the pedal triangle of point P with respect to triangle ABC is equal to 1 4 ¯ ¯ ¯1 −d2 R2 ¯ ¯ ¯ SABC, where d = \|PO\|. 5.103. From point P perpendiculars PA1, PB1 and PC1 are dropped on sides of triangle ABC. Line la connects the midpoints of segments PA and B1C1. Lines lb and lc are similarly deﬁned. Prove that la, lb and lc meet at one point. 5.104. a) Points P1 and P2 are isogonally conjugate with respect to triangle ABC, cf. Problem 5.79. Prove that their pedal triangles have a common circumscribed circle whose center is the midpoint of segment P1P2. b) Prove that the above statement remains true if instead of perpendiculars we draw from points P1 and P2 lines forming a given (oriented) angle to the sides. See also Problems 5.132, 5.133, 14.19 b). §11. Euler’s line and the circle of nine points 5.105. Let H be the point of intersection of heights of triangle ABC, O the center of the circumscribed circle and M the point of intersection of medians. Prove that point M lies on segment OH and OM : MH = 1 : 2. The line that contains points O, M and H is called Euler’s line. 5.106. Prove that the midpoints of sides of a triangle, the bases of heights and the midpoints of segments that connect the intersection point of heights with the vertices lie on one circle and the center of this circle is the midpoint of segment OH. The circle deﬁned above is called the circle of nine points. 5.107. The heights of triangle ABC meet at point H. a) Prove that triangles ABC, HBC, AHC and ABH have a common circle of 9 points. b) Prove that Euler’s lines of triangles ABC, HBC, AHC and ABH intersect at one point. c) Prove that the centers of the circumscribed circles of triangles ABC, HBC, AHC and ABH constitute a quadrilateral symmetric to quadrilateral HABC. 5.108. What are the sides the Euler line intersects in an acute and an obtuse triangles? 5.109. a) Prove that the circumscribed circle of triangle ABC is the circle of 9 points for the triangle whose vertices are the centers of escribed circles of triangle ABC. b) Prove that the circumscribed circle divides the segment that connects the centers of the inscribed and an escribed circles in halves. 5.110. Prove that Euler’s line of triangle ABC is parallel to side BC if and only if tan B tan C = 3. 5.111. On side AB of acute triangle ABC the circle of 9 points singles out a segment. Prove that the segment subtends an angle of 2\|∠A −∠B\| with the vertex at the center. 5.112. Prove that if Euler’s line passes through the center of the inscribed circle of a triangle, then the triangle is an isosceles one. 5.113. The inscribed circle is tangent to the sides of triangle ABC at points A1, B1 and C1. Prove that Euler’s line of triangle A1B1C1 passes through the center of the circumscribed circle of triangle ABC. 5.114. In triangle ABC, heights AA1, BB1 and CC1 are drawn. Let A1A2, B1B2 and C1C2 be diameters of the circle of nine points of triangle ABC. Prove that lines AA2, BB2 and CC2 either meet at one point or are parallel. 110 CHAPTER 5. TRIANGLES See also Problems 3.65 a), 13.34 b). §12. Brokar’s points 5.115. a) Prove that inside triangle ABC there exists a point P such that ∠ABP = ∠CAP = ∠BCP. b) On sides of triangle ABC, triangles CA1B, CAB1 and C1AB similar to ABC are constructed outwards (the angles at the ﬁrst vertices of all the four triangles are equal, etc.). Prove that lines AA1, BB1 and CC1 meet at one point and this point coincides with the point found in heading a). This point P is called Brokar’s point of triangle ABC. The proof of the fact that there exists another Brokar’s point Q for which ∠BAQ = ∠ACQ = ∠CBQ is similar to the proof of existence of P given in what follows. We will refer to P and Q as the ﬁrst and the second Brokar’s points. 5.116. a) Through Brokar’s point P of triangle ABC lines AB, BP and CP are drawn. They intersect the circumscribed circle at points A1, B1 and C1, respectively. Prove that △ABC = △B1C1A1. b) Triangle ABC is inscribed into circle S. Prove that the triangle formed by the inter-section points of lines PA, PB and PC with circle S can be equal to triangle ABC for no more than 8 distinct points P. (We suppose that the intersection points of lines PA, PB and PC with the circle are distinct from points A, B and C.) 5.117. a) Let P be Brokar’s point of triangle ABC. Let ϕ = ∠ABP = ∠BCP = ∠CAP. Prove that cot ϕ = cot α + cot β + cot γ. The angle ϕ from Problem 5.117 is called Brokar’s angle of triangle ABC. b) Prove that Brokar’s points of triangle ABC are isogonally conjugate to each other (cf. Problem 5.79). c) The tangent to the circumscribed circle of triangle ABC at point C and the line passing through point B parallel to AC intersect at point A1. Prove that Brokar’s angle of triangle ABC is equal to angle ∠A1AC. 5.118. a) Prove that Brokar’s angle of any triangle does not exceed 30◦. b) Inside triangle ABC, point M is taken. Prove that one of the angles ∠ABM, ∠BCM and ∠CAM does not exceed 30◦. 5.119. Let Q be the second Brokar’s point of triangle ABC, let O be the center of its circumscribed circle; A1, B1 and C1 the centers of the circumscribed circles of triangles CAQ, ABQ and BCQ, respectively. Prove that △A1B1C1 ∼△ABC and O is the ﬁrst Brokar’s point of triangle A1B1C1. 5.120. Let P be Brokar’s point of triangle ABC; let R1, R2 and R3 be the radii of the circumscribed circles of triangles ABP, BCP and CAP, respectively. Prove that R1R2R3 = R3, where R is the radius of the circumscribed circle of triangle ABC. 5.121. Let P and Q be the ﬁrst and the second Brokar’s points of triangle ABC. Lines CP and BQ, AP and CQ, BP and AQ meet at points A1, B1 and C1, respectively. Prove that the circumscribed circle of triangle A1B1C1 passes through points P and Q. 5.122. On sides CA, AB and BC of an acute triangle ABC points A1, B1 and C1, respectively, are taken so that ∠AB1A1 = ∠BC1B1 = ∠CA1C1. Prove that △A1B1C1 ∼ △ABC and the center of the rotational homothety that sends one triangle into another coincides with the ﬁrst Brokar’s point of both triangles. See also Problem 19.55. 111 §13. Lemoine’s point Let AM be a median of triangle ABC and line AS be symmetric to line AM through the bisector of angle A (point S lies on segment BC). Then segment AS is called a simedian of triangle ABC; sometimes the whole ray AS is referred to as a simedian. Simedians of a triangle meet at the point isogonally conjugate to the intersection point of medians (cf. Problem 5.79). The intersection point of simedians of a triangle is called Lemoine’s point. 5.123. Let lines AM and AN be symmetric through the bisector of angle ∠A of triangle ABC (points M and N lie on line BC). Prove that BM·BN CM·CN = c2 b2. In particular, if AS is a simedian, then BS CS = c2 b2. 5.124. Express the length of simedian AS in terms of the lengths of sides of triangle ABC. Segment B1C1, where points B1 and C1 lie on rays AC and AB, respectively, is said to be antiparallel to side BC if ∠AB1C1 = ∠ABC and ∠AC1B1 = ∠ACB. 5.125. Prove that simedian AS divides any segment B1C1 antiparallel to side BC in halves. 5.126. The tangent at point B to the circumscribed circle S of triangle ABC intersects line AC at point K. From point K another tangent KD to circle S is drawn. Prove that BD is a simedian of triangle ABC. 5.127. Tangents to the circumscribed circle of triangle ABC at points B and C meet at point P. Prove that line AP contains simedian AS. 5.128. Circle S1 passes through points A and B and is tangent to line AC, circle S2 passes through points A and C and is tangent to line AB. Prove that the common chord of these circles is a simedian of triangle ABC. 5.129. Bisectors of the outer and inner angles at vertex A of triangle ABC intersect line BC at points D and E, respectively. The circle with diameter DE intersects the circumscribed circle of triangle ABC at points A and X. Prove that AX is a simedian of triangle ABC. 5.130. Prove that Lemoine’s point of right triangle ABC with right angle ∠C is the midpoint of height CH. 5.131. Through a point X inside triangle ABC three segments antiparallel to its sides are drawn, cf. Problem 5.125?. Prove that these segments are equal if and only if X is Lemoine’s point. 5.132. Let A1, B1 and C1 be the projections of Lemoine’s point K to the sides of triangle ABC. Prove that K is the intersection point of medians of triangle A1B1C1. 5.133. Let A1, B1 and C1 be the projections of Lemoine’s point K of triangle ABC on sides BC, CA and AB, respectively. Prove that median AM of triangle ABC is perpendic-ular to line B1C1. 5.134. Lines AK, BK and CK, where K is Lemoine’s point of triangle ABC, intersect the circumscribed circle at points A1, B1 and C1, respectively. Prove that K is Lemoine’s point of triangle A1B1C1. 5.135. Prove that lines that connect the midpoints of the sides of a triangle with the midpoints of the corresponding heights intersect at Lemoine’s point. See also Problems 11.22, 19.54, 19.55. 112 CHAPTER 5. TRIANGLES Problems for independent study 5.136. Prove that the projection of the diameter of a circumscribed circle perpendicular to a side of the triangle to the line that contains the second side is equal to the third side. 5.137. Prove that the area of the triangle with vertices in the centers of the escribed circles of triangle ABC is equal to 2pR. 5.138. An isosceles triangle with base a and the lateral side b, and an isosceles triangle with base b and the lateral side a are inscribed in a circle of radius R. Prove that if a ̸= b, then ab = √ 5R2. 5.139. The inscribed circle of right triangle ABC is tangent to the hypothenuse AB at point P; let CH be a height of triangle ABC. Prove that the center of the inscribed circle of triangle ACH lies on the perpendicular dropped from point P to AC. 5.140. The inscribed circle of triangle ABC is tangent to sides CA and AB at points B1 and C1, respectively, and an escribed circle is tangent to the extension of sides at points B2 and C2. Prove that the midpoint of side BC is equidistant from lines B1C1 and B2C2. 5.141. In triangle ABC, bisector AD is drawn. Let O, O1 and O2 be the centers of the circumscribed circles of triangles ABC, ABD and ACD, respectively. Prove that OO1 = OO2. 5.142. The triangle constructed from a) medians, b) heights of triangle ABC is similar to triangle ABC. What is the ratio of the lengths of the sides of triangle ABC? 5.143. Through the center O of an equilateral triangle ABC a line is drawn. It intersects lines BC, CA and AB at points A1, B1 and C1, respectively. Prove that one of the numbers 1 OA1, 1 OB1 and 1 OC1 is equal to the sum of the other two numbers. 5.144. In triangle ABC heights BB1 and CC1 are drawn. Prove that if ∠A = 45◦, then B1C1 is a diameter of the circle of nine points of triangle ABC. 5.145. The angles of triangle ABC satisfy the relation sin2 ∠A + sin2 ∠B + sin2 ∠C = 1. Prove that the circumscribed circle and the circle of nine points of triangle ABC intersect at a right angle. Solutions 5.1. Let AC1 = AB1 = x, BA1 = BC1 = y and CA1 = CB1 = z. Then a = y + z, b = z + x and c = x + y. Subtracting the third equality from the sum of the ﬁrst two ones we get z = a+b−c 2 . Hence, if triangle ABC is given, then the position of points A1 and B1 is uniquely determined. Similarly, the position of point C1 is also uniquely determined. It remains to notice that the tangency points of the inscribed circle with the sides of the triangle satisfy the relations indicated in the hypothesis of the problem. 5.2. Rays COa and COb are the bisectors of the outer angles at vertex C, hence, C lies on line OaOb and ∠OaCB = ∠ObCA. Since COc is the bisector of angle ∠BCA, it follows that ∠BCOc = ∠ACOc. Adding these equalities we get: ∠OaCOc = ∠OcCOb, i.e., OcC is a height of triangle OaObOc. We similarly prove that OaA and ObB are heights of this triangle. 5.3. Clearly, ∠BOC = 180◦−∠CBO −∠BCO = 180◦−∠B 2 −∠C 2 = 90◦+ ∠A 2 and ∠BOaC = 180◦−∠BOC, because ∠OBOa = ∠OCOa = 90◦. 5.4. Let AA1, BB1 and CC1 be the bisectors of triangle ABC and O the intersection point of these bisectors. Suppose that x > 1. Then ∠PAB > ∠PAC, i.e., point P lies SOLUTIONS 113 inside triangle AA1C. Similarly, point P lies inside triangles CC1B and BB1A. But the only common point of these three triangles is point O. Contradiction. The case x < 1 is similarly treated. 5.5. Let da, db and dc be the distances from point O to sides BC, CA and AB. Then ada + bdb + cdc = 2S and aha = bhb = chc = 2S. If ha −da = hb −db = hc −dc = x, then (a + b + c)x = a(ha −da) = b(hb −db) + c(hc −dc) = 6S −2S = 4S. Hence, x = 4S 2p = 2r. 5.6. Let us prove that point O is the center of the escribed circle of triangle PBQ tangent to side PQ. Indeed, ∠POQ = ∠A = 90◦−1 2∠B. The angle of the same value with the vertex at the center of the escribed circle subtends segment PQ (Problem 5.3). Moreover, point O lies on the bisector of angle B. Hence, the semiperimeter of triangle PBQ is equal to the length of the projection of segment OB to line CB. 5.7. Let P be the tangent point of the inscribed circle with side BC, let PQ be a diameter of the inscribed circle, R the intersection point of lines AQ and BC. Since CR = BP (cf. Problem 19.11 a)) and M is the midpoint of side BC, we have: RM = PM. Moreover, O is the midpoint of diameter PQ, hence, MO ∥QR and since AH ∥PQ, we have AE = OQ. 5.8. The given circle can be the inscribed as well as the escribed circle of triangle ABC cut oﬀby the tangent from the angle. Making use of the result of Problem 3.2 we can verify that in either case uv w2 = (p −b)(p −c) sin ∠B sin ∠C h2 a . It remains to notice that ha = b sin ∠C = c sin ∠B and (p−b)(p−c) bc = sin2 1 2∠A (Problem 12.13). 5.9. Let A1, B1 and C1 be points symmetric to point H through sides BC, CA and AB, respectively. Since AB ⊥CH and BC ⊥AH, it follows that ∠(AB, BC) = ∠(CH, HA) and since triangle AC1H is an isosceles one, ∠(CH, HA) = ∠(AC1, C1C). Hence, ∠(AB, BC) = ∠(AC1, C1C), i.e., point C1 lies on the circumscribed circle of triangle ABC. We similarly prove that points A1 and B1 lie on this same circle. 5.10. Let R be the radius of the circumscribed circle of triangle ABC. This circle is also the circumscribed circle of triangles ABP, APC and PBC. Clearly, ∠ABP = 180◦−∠ACP = α, ∠BAP = ∠BCP = β and ∠CAP = ∠CBP = γ. Hence, PX = PB sin γ = 2R sin β sin γ, PY = 2R sin α sin γ and P = 2R sin α sin β. It is also clear that BC = 2R sin ∠BAC = 2R sin(β + γ), AC = 2R sin(α −γ), AB = 2R sin(α + β). It remains to verify the equality sin(β + γ) sin β sin γ = sin(α −γ) sin α sin γ + sin(α + β) sin α sin β which is subject to a direct calculation. 5.11. a) Let M be the intersection point of line AI with the circumscribed circle. Drawing the diameter through point I we get AI · IM = (R + d)(R −d) = R2 −d2. Since IM = CM (by Problem 2.4 a)), it follows that R2 −d2 = AI · CM. It remains to observe that AI = r sin 1 2 ∠A and CM = 2R sin 1 2∠A. 114 CHAPTER 5. TRIANGLES b) Let M be the intersection point of line AIa with the circumscribed circle. Then AIa·IaM = d2 a−R2. Since IaM = CM (by Problem 2.4 a)), it follows that d2 a−R2 = AIa·CM. It remains to notice that AIa = ra sin 1 2 ∠A and CM = 2R sin 1 2∠A. 5.12. a) Since B1 is the center of the circumscribed circle of triangle AMC (cf. Problem 2.4 a)), AM = 2MB1 sin ∠ACM. It is also clear that MC = r sin ∠ACM . Hence, MA·MC MB1 = 2r. b) Since ∠MBC1 = ∠BMC1 = 180◦−∠BMC and ∠BC1M = ∠A, it follows that MC1 BC = BM BC · MC1 BM = sin ∠BCM sin ∠BMC · sin ∠MBC1 sin ∠BC1M = sin ∠BCM sin ∠A . Moreover, MB = 2MA1 sin ∠BCM. Therefore, MC1·MA1 MB = BC 2 sin ∠A = R. 5.13. Let M be the midpoint of side AC, and N the tangent point of the inscribed circle with side BC. Then BN = p −b (see Problem 3.2), hence, BN = AM because p = 3 2b by assumption. Moreover, ∠OBN = ∠B1AM and, therefore, △OBN = △B1AM, i.e., OB = B1A. But B1A = B1O (see Problem 2.4 a)). 5.14. Let O and O1 be the centers of the inscribed and circumscribed circles of triangle ABC. Let us consider the circle of radius d = OO1 centered at O. In this circle, let us draw chords O1M and O1N parallel to sides AB and AC, respectively. Let K be the tangent point of the inscribed circle with side AB and L the midpoint of side AB. Since OK ⊥AB, O1L ⊥AB and O1M ∥AB, it follows that O1M = 2KL = 2BL −2BK = c −(a + c −b) = b −a = AE. Similarly, O1N = AD and, therefore, △MO1N = △EAD. Consequently, the radius of the circumscribed circle of triangle EAD is equal to d. 5.15. Let the inscribed circle be tangent to side AC at point K and the escribed circle be tangent to the extension of side AC at point L. Then r = CK and rc = CL. It remains to make use of the result of Problem 3.2. 5.16. Since 1 2AB = AM = BM, it follows that CM = 1 2AB if and only if point C lies on the circle with diameter AB. 5.17. Let M and N be the midpoints of sides AB and CD. Triangle APB is a right one; hence, PM = 1 2AB and ∠MPA = ∠PAM and, therefore, PM ∥AD. Similar arguments show that points P, M and Q lie on one line and PQ = PM + MN + NQ = AB + (BC + AD) + CD 2 . 5.18. Let F be the intersection point of lines DE and BC; let K be the midpoint of segment EC. Segment CD is simultaneosly a bisector and a height of triangle ECF, hence, ED = DF and, therefore, DK ∥FC. Median DK of right triangle EDC is twice shorter its hypothenuse EC (Problem 5.16), hence, AD = DK = 1 2EC. 5.19. Let the sum of the angles at the base AD of trapezoid ABCD be equal to 90◦. Denote the intersection point of lines AB and CD by O. Point O lies on the line that passes through the midpoints of the bases. Let us draw through point C line CK parallel to this line and line CE parallel to line AB (points K and E lie on base AD). Then CK is a median of right triangle ECD, hence, CK = ED 2 = AD−BC 2 (cf. Problem 5.16). 5.20. It is clear that ∠CEB = ∠A + ∠ACE = ∠BCK + ∠KCE = ∠BCE. 5.21. Segments CF and DK are bisectors in similar triangles ACB and CDB and, therefore, AB : FB = CB : KB. Hence, FK ∥AC. We similarly prove that LF ∥CB. SOLUTIONS 115 Therefore, CLFK is a rectangle whose diagonal CF is the bisector of angle LCK, i.e., the rectangle is a square. 5.22. Since sin ∠ACQ AQ = sin ∠AQC AC , it follows that sin α a = sin(180◦−α −90◦−ϕ) a cos ϕ = cos(α + ϕ) a cos ϕ , where a is the (length of the) side of square ABPQ and ϕ = ∠CAB. Hence, cot α = 1+tan ϕ. Similarly, cot γ = 1 + tan(90◦−ϕ) = 1 + cot ϕ. It follows that tan α + tan γ = 1 1 + tan ϕ + 1 1 + cot ϕ = 1 and, therefore, cos α cos γ = cos α sin γ + cos γ sin α = sin(α + γ) = cos β. 5.23. By Pythagoras theorem AP 2 + BQ2 + CR2 + (AM 2 −PM 2) + (BM 2 −QM 2) + (CM 2 −RM 2) and PB2 + QC2 + RA2 = (BM 2 −PM 2) + (CM 2 −QM 2) + (AM 2 −RM 2). These equations are equal. Since AP 2 + BQ2 + CR2 = (a −PB)2 + (a −QC)2 + (a −RA)2 = 3a2 −2a(PB + QC + RA) + PB2 + QC2 + RA2, where a = AB, it follows that PB + QC + RA = 3 2a. 5.24. Let point F divide segment BC in the ratio of CF : FB = 1 : 2; let P and Q be the intersection points of segment AF with BD and CE, respectively. It is clear that triangle OPQ is an equilateral one. Making use of the result of Problem 1.3 it is easy to verify that AP : PF = 3 : 4 and AQ : QF = 6 : 1. Hence, AP : PQ : QF = 3 : 3 : 1 and, therefore, AP = PQ = OP. Hence, ∠AOP = 180◦−∠APO 2 = 30◦and ∠AOC = ∠AOP + ∠POQ = 90◦. 5.25. Let A and B, C and D, E and F be the intersection points of the circle with sides PQ, QR, RP, respectively, of triangle PQR. Let us consider median PS. It connects the midpoints of parallel chords FA and DC and, therefore, is perpendicular to them. Hence, PS is a height of triangle PQR and, therefore, PQ = PR. Similarly, PQ = QR. 5.26. Let H be the intersection point of heights AA1, BB1 and CC1 of triangle ABC. By hypothesis, A1H · BH = B1H · AH. On the other hand, since points A1 and B1 lie on the circle with diameter AB, then AH · A1H = BH · B1H. It follows that AH = BH and A1H = B1H and, therefore, AC = BC. Similarly, BC = AC. 5.27. a) Suppose that triangle ABC is not an equilateral one; for instance, a ̸= b. Since a + ha = a + b sin γ and b + hb = b + a sin γ, it follows that (a −b)(1 −sin γ) = 0; hence, sin γ = 0, i.e., γ = 90◦. But then a ̸= c and similar arguments show that β = 90◦. Contradiction. b) Let us denote the (length of the) side of the square two vertices of which lie on side BC by x. The similarity of triangles ABC and APQ, where P and Q are the vertices of the square that lie on AB and AC, respectively, yields x a = ha−x ha , i.e., x = aha a+ha = 2S a+ha. Similar arguments for the other squares show that a + ha = b + hb = c + hc. 5.28. If α, β and γ are the angles of triangle ABC, then the angles of triangle A1B1C1 are equal to β+γ 2 , γ+α 2 and α+β 2 . Let, for deﬁniteness, α ≥β ≥γ. Then α+β 2 ≥α+γ 2 ≥β+γ 2 . Hence, α = α+β 2 and γ = β+γ 2 , i.e., α = β and β = γ. 116 CHAPTER 5. TRIANGLES 5.29. In any triangle a height is longer than the diameter of the inscribed circle. There-fore, the lengths of heights are integers greater than 2, i.e., all of them are not less than 3. Let S be the area of the triangle, a the length of its longest side and h the corresponding height. Suppose that the triangle is not an equilateral one. Then its perimeter P is shorter than 3a. Therefore, 3a > P = Pr = 2S = ha, i.e., h < 3. Contradiction. 5.30. Since the outer angle at vertex A of triangle ABA1 is equal to 120◦and ∠A1AB1 = 60◦, it follows that AB1 is the bisector of this outer angle. Moreover, BB1 is the bisector of the outer angle at vertex B, hence, A1B1 is the bisector of angle ∠AA1C. Similarly, A1C1 is the bisector of angle ∠AA1B. Hence, ∠B1A1C1 = ∠AA1C + ∠AA1B 2 = 90◦. 5.31. Thanks to the solution of the preceding problem ray A1C1 is the bisector of angle ∠AA1B. Let K be the intersection point of the bisectors of triangle A1AB. Then ∠C1KO = ∠A1KB = 90◦+ ∠A 2 = 120◦. Hence, ∠C1KO + ∠C1AO = 180◦, i.e., quadrilateral AOKC1 is an inscribed one. Hence, ∠A1C1O = ∠KC1O = ∠KAO = 30◦. 5.32. a) Let S be the circumscribed circle of triangle ABC, let S1 be the circle symmetric to S through line BC. The orthocenter H of triangle ABC lies on circle S1 (Problem 5.9) and, therefore, it suﬃces to verify that the center O of circle S also belongs to S1 and the bisector of the outer angle A passes through the center of circle S1. Then POAH is a rhombus, because PO ∥HA. Let PQ be the diameter of circle S perpendicular to line BC; let points P and A lie on one side of line BC. Then AQ is the bisector of angle A and AP is the bisector of the outer angle ∠A. Since ∠BPC = 120◦= ∠BOC, point P is the center of circle S1 and point O belongs to circle S1. b) Let S be the circumscribed circle of triangle ABC and Q the intersection point of the bisector of angle ∠BAC with circle S. It is easy to verify that Q is the center of circle S1 symmetric to circle S through line BC. Moreover, points O and H lie on circle S1 and since ∠BIC = 120◦and ∠BIaC = 60◦(cf. Problem 5.3), it follows that IIa is a diameter of circle S1. It is also clear that ∠OQI = ∠QAH = ∠AQH, because OQ ∥AH and HA = QO = QH. Hence, points O and H are symmetric through line IIa. 5.33. On side AC of triangle ABC, construct outwards an equilateral triangle AB1C. Since ∠A = 120◦, point A lies on segment BB1. Therefore, BB1 = b + c and, moreover, BC = a and B1C = b, i.e., triangle BB1C is the desired one. 5.34. a) Let M1 and N1 be the midpoints of segments BH and CH, respectively; let BB1 and CC1 be heights. Right triangles ABB1 and BHC1 have a common acute angle — the one at vertex B; hence, ∠C1HB = ∠A = 60◦. Since triangle BMH is an isosceles one, ∠BHM = ∠HBM = 30◦. Therefore, ∠C1HM = 60◦−30◦= 30◦= ∠BHM, i.e., point M lies on the bisector of angle ∠C1HB. Similarly, point N lies on the bisector of angle ∠B1HC. b) Let us make use of the notations of the preceding problem and, moreover, let B′ and C′ be the midpoints of sides AC and AB. Since AC1 = AC cos ∠A = 1 2AC, it follows that C1C′ = 1 2\|AB −AC\|. Similarly, B1B′ = 1 2\|AB −AC\|, i.e., B1B′ = C1C′. It follows that the parallel lines BB1 and B′O, CC1 and C′O form not just a parallelogram but a rhombus. Hence, its diagonal HO is the bisector of the angle at vertex H. SOLUTIONS 117 5.35. Since ∠BB1C = ∠B1BA + ∠B1AB > ∠B1BA = ∠B1BC, it follows that BC > B1C. Hence, point K symmetric to B1 through bisector CC1 lies on side BC and not on its extension. Since ∠CC1B = 30◦, we have ∠B1C1K = 60◦and, therefore, triangle B1C1K is an equilateral one. In triangles BC1B1 and BKB1 side BB1 is a common one and sides C1B1 and KB1 are equal; the angles C1BB1 and KBB1 are also equal but these angles are not the ones between equal sides. Therefore, the following two cases are possible: 1) ∠BC1B1 = ∠BKB1. Then ∠BB1C1 = ∠BB1K = 60◦ 2 = 30◦. Therefore, if O is the intersection point of bisectors BB1 and CC1, then ∠BOC = ∠B1OC1 = 180◦−∠OC1B1 −∠OB1C1 = 120◦. On the other hand, ∠BOC = 90◦+ ∠A 2 (cf. Problem 5.3), i.e., ∠A = 60◦. 2) ∠BC1B1 + ∠BKB1 = 180◦. Then quadrilateral BC1B1K is an inscribed one and since triangle B1C1K is an equilateral one, ∠B = 180◦−∠C1B1K = 120◦. 5.36. Let BM be a median, AK a bisector of triangle ABC and BM ⊥AK. Line AK is a bisector and a height of triangle ABM, hence, AM = AB, i.e., AC = 2AM = 2AB. Therefore, AB = 2, BC = 3 and AC = 4. 5.37. Let a and b be legs and c the hypothenuse of the given triangle. If numbers a and b are odd, then the remainder after division of a2 + b2 by 4 is equal to 2 and a2 + b2 cannot be a perfect square. Hence, one of the numbers a and b is even and another one is odd; let, for deﬁniteness, a = 2p. The numbers b and c are odd, hence, c + b = 2q and c −b = 2r for some q and r. Therefore, 4p2 = a2 = c2 −b2 = 4qr. If d is a common divisor of q and r, then a = 2√qr, b = q −r and c = q + r are divisible by d. Therefore, q and r are relatively prime, ??? since p2 = qr, it follows that q = m2 and r = n2. As a result we get a = 2mn, b = m2 −n2 and c = m2 + n2. It is also easy to verify that if a = 2mn, b = m2 −n2 and c = m2 + n2, then a2 + b2 = c2. 5.38. Let p be the semiperimeter of the triangle and a, b, c the lengths of the triangle’s sides. By Heron’s formula S2 = p(p −a)(p −b)(p −c). On the other hand, S2 = p2r2 = p2 since r = 1. Hence, p = (p −a)(p −b)(p −c). Setting x = p −a, y = p −b, z = p −c we rewrite our equation in the form x + y + z = xyz. Notice that p is either integer or half integer (i.e., of the form 2n+1 2 , where n is an integer) and, therefore, all the numbers x, y, z are simultaneously either integers or half integers. But if they are half integers, then x + y + z is a half integer and xyz is of the form m 8 , where m is an odd number. Therefore, numbers x, y, z are integers. Let, for deﬁniteness, x ≤y ≤z. Then xyz = x + y + z ≤3z, i.e., xy ≤3. The following three cases are possible: 1) x = 1, y = 1. Then 2 + z = z which is impossible. 2) x = 1, y = 2. Then 3 + z = 2z, i.e., z = 3. 3) x = 1, y = 3. Then 4 + z = 3z, i.e., z = 2 < y which is impossible. Thus, x = 1, y = 2, z = 3. Therefore, p = x + y + z = 6 and a = p −x = 5, b = 4, c = 3. 5.39. Let a1 and b1, a2 and b2 be the legs of two distinct Pythagorean triangles, c1 and c2 their hypothenuses. Let us take two perpendicular lines and mark on them segments OA = a1a2, OB = a1b2, OC = b1b2 and OD = a2b1 (Fig. 57). Since OA · OC = OB · OD, quadrilateral ABCD is an inscribed one. By Problem 2.71 4R2 = OA2 + OB2 + OC2 + OD2 = (c1c2)2, 118 CHAPTER 5. TRIANGLES i.e., R = c1c2 2 . Magnifying, if necessary, quadrilateral ABCD twice, we get the quadrilateral to be found. Figure 57 (Sol. 5.39) 5.40. a) The lengths of hypothenuses of right triangles with legs 5 and 12, 9 and 12 are equal to 13 and 15, respectively. Identifying the equal legs of these triangles we get a triangle whose area is equal to 12(5+9) 2 = 84. b) First, suppose that the length of the shortest side of the given triangle is an even number, i.e., the lengths of the sides of the triangle are equal to 2n, 2n + 1, 2n + 2. Then by Heron’s formula 16S2 = (6n + 3)(2n + 3)(2n + 1)(2n −1) = 4(3n2 + 6n + 2)(4n2 −1) + 4n2 −1. We have obtained a contradiction since the number in the right-hand side is not divisible by 4. Consecutively, the lengths of the sides of the triangle are equal to 2n −1, 2n and 2n + 1, where S2 = 3n2(n2 −1). Hence, S = nk, where k is an integer and k2 = 3(n2 −1). It is also clear that k is the length of the height dropped to the side of length 2n. This height divides the initial triangle into two right triangles with a common leg of length k and hypothenuses of length 2n + 1 and 2n −1 the squares of the lengths of the other legs of these triangles are equal to (2n ± 1)2 −k2 = 4n2 ± 4n + 1 −3n2 + 3 = (n ± 2)2. 5.41. a) Since AB2 −AB2 1 = BB2 1 = BC2 −(AC ± AB1)2, we see that AB1 = ±AB2+AC2−BC2 2AC . b) Let diagonals AC and BD meet at point O. Let us prove, for example, that the number q = BO OD is a rational one (then the number OD = BD q+1 is also a rational one). In triangles ABC and ADC draw heights BB1 and DD1. By heading a) the numbers AB1 and CD1 — the lengths of the corresponding sides — are rational and, therefore, the number B1D1 is also rational. Let E be the intersection point of line BB1 and the line that passes through point D parallel to AC. In right triangle BDE, we have ED = B1D1 and the lengths of leg ED and hypothenuse BD are rational numbers; hence, BE2 is also a rational number. From triangles ABB1 and CDD1 we derive that numbers BB2 1 and DD2 1 are rational. Since BE2 = (BB1 + DD1)2 = BB2 1 + DD2 1 + 2BB1 · DD1, number BB1 · DD1 is rational. It follows that the number BO OD = BB1 DD1 = BB1 · DD1 DD2 1 is a rational one. SOLUTIONS 119 5.42. Triangles ABC and A1B1C1 cannot have two pairs of corresponding angles whose sum is equal to 180◦since otherwise their sum would be equal to 360◦and the third angles of these triangles should be equal to zero. Now, suppose that the angles of the ﬁrst triangle are equal to α, β and γ and the angles of the second one are equal to 180◦−α, β and γ. The sum of the angles of the two triangles is equal to 360◦, hence, 180◦+ 2β + 2γ = 360◦, i.e., β + γ = 90◦. It follows that α = 90◦= 180◦−α. 5.43. Clearly, − − → A1C = − − → BO and − − → CB1 = − → OA, hence, − − − → A1B1 = − → BA. Similarly, − − − → B1C1 = − − → CB and − − − → C1A1 = − → AC, i.e., △ABC = △A1B1C1. Moreover, ABA1B1 and ACA1C1 are parallelograms. It follows that segments BB1 and CC1 pass through the midpoint of segment AA1. 5.44. Since ∠MAO = ∠PAO = ∠AOM, it follows that AMOP is a rhombus. Similarly, BNOQ is a rhombus. It follows that MN = MO + ON = AM + BN and OP + PQ + QO = AP + PQ + QB = AB. 5.45. a) Through vertices of triangle ABC let us draw lines parallel to the triangle’s opposite sides. As a result we get triangle A1B1C1; the midpoints of the sides of the new triangle are points A, B and C. The heights of triangle ABC are the midperpendiculars to the sides of triangle A1B1C1 and, therefore, the center of the circumscribed circle of triangle A1B1C1 is the intersection point of heights of triangle ABC. b) Point H is the center of the circumscribed circle of triangle A1B1C1, hence, 4R2 = B1H2 = B1A2 + AH2 = BC2 + AH2. Therefore, AH2 = 4R2 −BC2 = µ 1 sin2 α −1 ¶ BC2 = (BC cot α)2. 5.46. Let AD be the bisector of an equilateral triangle ABC with base AB and angle 36◦at vertex C. Then triangle ACD is an isosceles one and △ABC ∼△BDA. Therefore, CD = AD = AB = 2xBC and DB = 2xAB = 4x2BC; hence, BC = CD + DB = (2x + 4x2)BC. 5.47. Let B1 and B2 be the projections of point A to bisectors of the inner and outer angles at vertex B; let M the midpoint of side AB. Since the bisectors of the inner and outer angles are perpendicular, it follows that AB1BB2 is a rectangular and its diagonal B1B2 passes through point M. Moreover, ∠B1MB = 180◦−2∠MBB1 = 180◦−∠B. Hence, B1B2 ∥BC and, therefore, line B1B2 coincides with line l that connects the midpoints of sides AB and AC. We similarly prove that the projections of point A to the bisectors of angles at vertex C lie on line l. 5.48. Suppose that the bisectors of angles A and B are equal but a > b. Then cos 1 2∠A < cos 1 2∠B and 1 c + 1 b > 1 c + 1 a, i.e., bc b+c < ac a+c. By multiplying these inequalities we get a contradiction, since la = 2bc cos ∠A 2 b+c and lb = 2ac cos ∠B 2 a+c (cf. Problem 4.47). 5.49. a) By Problem 4.47 the length of the bisector of angle ∠B of triangle ABC is equal to 2ac cos ∠B 2 a+c and, therefore, it suﬃces to verify that the system of equations ac a + c = p, a2 + c2 −2ac cos ∠B = q 120 CHAPTER 5. TRIANGLES has (up to a transposition of a with c) a unique positive solution. Let a + c = u. Then ac = pu and q = u2 −2pu(1 + cos β). The product of the roots of this quadratic equation for u is equal to −q and, therefore, it has one positive root. Clearly, the system of equations a + c = u, ac = pu has a unique solution. b) In triangles AA1B and CC1B, sides AA1 and CC1 are equal; the angles at vertex B are equal, and the bisectors of the angles at vertex B are also equal. Therefore, these triangles are equal and either AB = BC or AB = BC1. The second equality cannot take place. 5.50. Let points M and N lie on sides AB and AC. If r1 is the radius of the circle whose center lies on segment MN and which is tangent to sides AB and AC, then SAMN = qr1, where q = AM+AN 2 . Line MN passes through the center of the inscribed circle if and only if r1 = r, i.e., SAMN q = SABC p = SBCNM p−q . 5.51. a) On the extension of segment AC beyond point C take a point B′ such that CB′ = CB. Triangle BCB′ is an isosceles one; hence, ∠AEB = ∠ACB = 2∠CBB′ and, therefore, E is the center of the circumscribed circle of triangle ABB′. It follows that point F divides segment AB′ in halves; hence, line C1F divides the perimeter of triangle ABC in halves. b) It is easy to verify that the line drawn through point C parallel to BB′ is the bisector of angle ACB. Since C1F ∥BB′, line C1F is the bisector of the angle of the triangle with vertices at the midpoints of triangle ABC. The bisectors of this new triangle meet at one point. 5.52. Let X be the intersection point of lines AD2 and CD1; let M, E1 and E2 be the projections of points X, D1 and D2, respectively, to line AC. Then CE2 = CD2 sin γ = a sin γ and AE1 = c sin α. Since a sin γ = c sin α, it follows that CE2 = AE1 = q. Hence, XM AM = D2E2 AE2 = a cos γ b + q and XM CM = c cos α b + q . Therefore, AM : CM = c cos α : a cos γ. Height BH divides side AC in the same ratio. 5.53. a) By the law of cosines B1C2 1 = AC2 1 + AB2 1 −2AC1 · AB1 · cos(90◦+ α), i.e., a2 1 = c2 2 + b2 2 + bc sin α = b2 + c2 2 + 2S. Writing similar equalities for b2 1 and c2 1 and taking their sum we get the statement desired. b) For an acute triangle ABC, add to S the areas of triangles ABC1, AB1C and A1BC; add to S1 the areas of triangles AB1C1, A1BC1 and A1B1C. We get equal quantities (for a triangle with an obtuse angle ∠A the area of triangle AB1C1 should be taken with a minus sign). Hence, S1 = S + a2 + b2 + c2 4 −ab cos γ + ac cos β + bc cos α 4 . It remains to notice that ab cos γ + bc cos α + ac cos β = 2S(cot γ + cot α + cot β) = a2 + b2 + c2 2 ; cf. Problem 12.44 a). SOLUTIONS 121 5.54. First, let us prove that point B′ lies on the circumscribed circle of triangle AHC, where H is the intersection point of heights of triangle ABC. We have ∠(AB′, B′C) = ∠(AA1, CC1) = ∠(AA1, BC) + ∠(BC, AB) + ∠(AB, CC1) = ∠(BC, AB). But as follows from the solution of Problem 5.9 ∠(BC, AB) = ∠(AH, HC) and, therefore, points A, B′, H and C lie on one circle and this circle is symmetric to the circumscribed circle of triangle ABC through line AC. Hence, both these circles have the same radius, R, consequently, B′H = 2R sin B′AH = 2R cos α. Similarly, A′H = 2R cos α = C′H. This completes solution of heading a); to solve heading b) it remains to notice that △A′B′C′ ∼△ABC since after triangle A′B′C′ is rotated through an angle of α its sides become parallel to the sides of triangle ABC. 5.55. Let a1 = BA1, a2 = A1C, b1 = CB1, b2 = B1A, c1 = AC1 and c2 = C1B. The products of the lengths of segments of intersecting lines that pass through one point are equal and, therefore, a1(a1 + x) = c2(c2 −z), i.e., a1x + c2z = c2 2 −a2 1. We similarly get two more equations for x, y and z: b1y + a2x = a2 2 −b2 1 and c1z + b2y = b2 2 −c2 1. Let us multiply the ﬁrst equation by b2n; multiply the second and the third ones by c2n and a2n, respectively, and add the equations obtained. Since, for instance, c2bn −c1an = 0 by the hypothesis, we get zero in the right-hand side. The coeﬃcient of, say, x in the left-hand side is equal to a1b2n + a2c2n = acnb2n + abnc2n bn + cn = abncn. Hence, abncnx + bancny + canbnz = 0. Dividing both sides of this equation by (abc)n we get the statement desired. 5.56. Let in the initial triangle ∠A = 3α, ∠B = 3β and ∠C = 3γ. Let us take an equilateral triangle A2B2C2 and construct on its sides as on bases isosceles triangles A2B2R, B2C2P and C2A2Q with angles at the bases equal to 60◦−γ, 60◦−α, 60◦−β, respectively (Fig. 58). Let us extend the lateral sides of these triangles beyond points A2, B2 and C2; denote the intersection point of the extensions of sides RB2 and QC2 by A3, that of PC2 and RA2 by B3, that of QA2 and PB2 by C3.Through point B2 draw the line parallel to A2C2 and denote by M and N the its intersection points with lines QA3 and QC3, respectively. Clearly, B2 is the midpoint of segment MN. Let us compute the angles of triangles B2C3N and B2A3M: ∠C3B2N = ∠PB2M = ∠C2B2M = ∠C2B2P = α; ∠B2NC3 = 180◦−∠C2A2Q = 120◦+ β; hence, ∠B2C3N = 180◦−α −(120◦+ β) = γ. Similarly, ∠A3B2M = γ and ∠B2A3M = α. Hence, △B2C3N ∼△A3B2M. It follows that C3B2 : B2A3 = C3N : B2M and since B2M = B2N and ∠C3B2A3 = ∠C3NB2, it follows that C3B2 : B2A3 = C3N : NB2 and △C3B2A3 ∼△C3NB2; hence, ∠B2C3A3 = γ. 122 CHAPTER 5. TRIANGLES Figure 58 (Sol. 5.56) Similarly, ∠A2C3B3 = γ and, therefore, ∠A3C3B3 = 3γ = ∠C and C3B3, C3A2 are the trisectors of angle C3 of triangle A3B3C3. Similar arguments for vertices A3 and B3 show that △ABC ∼△A3B3C3 and the intersection points of the trisectors of triangle A3B3C3 are vertices of an equilateral triangle A2B2C2. 5.57. Point A1 lies on the bisector of angle ∠BAC, hence, point A lies on the extension of the bisector of angle ∠B2A1C2. Moreover, ∠B2AC2 = α = 180◦−∠B2A1C2 2 . Hence, A is the center of an escribed circle of triangle B2A1C2 (cf. Problem 5.3). Let D be the intersection point of lines AB and CB2. Then ∠AB2C2 = ∠AB2D = 180◦−∠B2AD −∠ADB2 = 180◦−γ −(60◦+ α) = 60◦+ β. Since ∠AB2C = 180◦−(α + β) −(β + γ) = 120◦−β, it follows that ∠CB2C2 = ∠AB2C −∠AB2C2 = 60◦−2β. Similarly, ∠AB2A2 = 60◦−2β. Hence, ∠A2B2C2 = ∠AB2C −∠AB2A2 −∠CB2C2 = 3β. Similarly, ∠B2A2C2 = 3α and ∠A2C2B2 = 3γ. 5.58. Let the projection to a line perpendicular to line A1B1 send points A, B and C to A′, B′ and C′, respectively; point C1 to Q and points A1 and B1 into one point, P. Since A1B A1C = PB′ PC′ , B1C B1A = PC′ PA′ and C1A C1B = QA′ QB′, it follows that A1B A1C · B1C B1A · C1A C1B = PB′ PC′ · PC′ PA′ · QA′ QB′ = PB′ PA′ · QA′ QB′ = b′ a′ · a′ + x b′ + x, where \|x\| = PQ. The equality b′ a′ · a′+x b′+x = 1 is equivalent to the fact that x = 0. (We have to take into account that a′ ̸= b′ since A′ ̸= B′.) But the equality x = 0 means that P = Q, i.e., point C1 lies on line A1B1. 5.59. Let point P lie on arc ⌣BC of the circumscribed circle of triangle ABC. Then BA1 CA1 = −BP cos ∠PBC CP cos ∠PCB , CB1 AB1 = −CP cos ∠PCA AP cos PAC , AC1 BC1 = −AP cos ∠PAB PB cos ∠PBA. SOLUTIONS 123 By multiplying these equalities and taking into account that ∠PAC = ∠PBC, ∠PAB = ∠PCB and ∠PAC + ∠PBA = 180◦ we get BA1 CA1 · CB1 AB1 · AC1 BC1 = 1. 5.60. Let O, O1 and O2 be the centers of circles S, S1 and S2; let X be the intersection point of lines O1O2 and A1A2. By applying Menelaus’s theorem to triangle OO1O2 and points A1, A2 and X we get O1X O2X · O2A2 OA2 · OA1 O1A1 = 1 and, therefore, O1X : O2X = R1 : R2, where R1 and R2 are the radii of circles S1 and S2, respectively. It follows that X is the intersection point of the common outer or common inner tangents to circles S1 and S2. 5.61. a) Let, for deﬁniteness, ∠B < ∠C. Then ∠DAE = ∠ADE = ∠B + ∠A 2 ; hence, ∠CAE = ∠B. Since BE AB = sin ∠BAE sin ∠AEB and AC CE = sin ∠AEC sin ∠CAE , it follows that BE CE = c sin ∠BAE b sin ∠CAE = c sin(∠A + ∠B) b sin ∠B = c sin ∠C b sin ∠B = c2 b2. b) In heading a) point E lies on the extension of side BC since ∠ADC = ∠BAD+∠B > ∠CAD. Therefore, making use of the result of heading a) and Menelaus’s theorem we get the statement desired. 5.62. Since ∠BCE = 90◦−∠B 2 , we have: ∠BCE = ∠BEC and, therefore, BE = BC. Hence, CF : KF = BE : BK = BC : BK and AE : KE = CA : CK = BC : BK. Let line EF intersect AC at point D. By Menelaus’s theorem AD CD · CF KF · KE AE = 1. Taking into account that CF : KF = AE : KE we get the statement desired. 5.63. Proof is similar to that of Problem 5.79; we only have to consider the ratio of oriented segments and angles. 5.64. Let A2, B2 and C2 be the intersection points of lines BC with B1C1, AC with A1C1, AB with A1B1, respectively. Let us apply Menelaus’s theorem to the following triangles and points on their sides: OAB and (A1, B1, C2), OBC and (B1, C1, A2), OAC and (A1, C1, B2). Then AA1 OA1 · OB1 BB1 · BC2 AC2 = 1, OC1 CC1 · BB1 OB1 · CA2 BA2 = 1, OA1 AA1 · CC1 OC1 · AB2 CB2 = 1. By multiplying these equalities we get BC2 AC2 · AB2 CB2 · CA2 BA2 = 1. Menelaus’s theorem implies that points A2, B2, C2 lie on one line. 5.65. Let us consider triangle A0B0C0 formed by lines A1B2, B1C2 and C1A2 (here A0 is the intersection point of lines A1B2 and A2C1, etc), and apply Menelaus’s theorem to this triangle and the following ﬁve triples of points: (A, B2, C1), (B, C2, A1), (C, A2, B1), (A1, B1, C1) and (A2, B2, C2). 124 CHAPTER 5. TRIANGLES As a result we get B0A C0A · A0B2 B0B2 · C0C1 A0C1 = 1, C0B A0B · B0C2 C0C2 · A0A1 B0A1 = 1, A0C B0C · C0A2 A0A2 · B0B1 C0B1 = 1, B0A1 A0A1 · C0B1 B0B1 · A0C1 C0C1 = 1, (2) A0A2 C0A2 · B0B2 A0B2 · C0C2 B0C2 = 1. (3) By multiplying these equalities we get B0A C0A · C0B A0B · A0C B0C = 1 and, therefore, points A, B and C lie on one line. 5.66. Let N be the intersection point of lines AD and KQ, P ′ the intersection point of lines KL and MN. By Desargue’s theorem applied to triangles KBL and NDM we derive that P ′, A and C lie on one line. Hence, P ′ = P. 5.67. It suﬃces to apply Desargues’s theorem to triangles AED and BFC and Pappus’ theorem to triples of points (B, E, C) and (A, F, D). 5.68. a) Let R be the intersection point of lines KL and MN. By applying Pappus’ theorem to triples of points (P, L, N) and (Q, M, K), we deduce that points A, C and R lie on one line. b) By applying Desargues’s theorem to triangles NDM and LBK we see that the inter-section points of lines ND with LB, DM with BK, and NM with LK lie on one line. 5.69. Let us make use of the result of Problem 5.68 a). For points P and Q take points P2 and P4, for points A and C take points C1 and P1 and for K, L, M and N take points P5, A1, B1 and P3, respectively. As a result we see that line P6C1 passes through point P1. 5.70. a) This problem is a reformulation of Problem 5.58 since the number BA1 : CA1 is negative if point A1 lies on segment BC and positive otherwise. b) First, suppose that lines AA1, BB1 and CC1 meet at point M. Any three (nonzero) vectors in plane are linearly dependent, i.e., there exist numbers λ, µ and ν (not all equal to zero) such that λ− − → AM + µ− − → BM + ν− − → CM = 0. Let us consider the projection to line BC parallel to line AM. This projection sends points A and M to A1 and points B and C into themselves. Therefore, µ− − → BA1 + ν− − → CA1 = 0, i.e., BA1 CA1 = −ν µ. Similarly, CB1 AB1 = −λ ν and AC1 BC1 = −µ λ. By multiplying these three equalities we get the statement desired. If lines AA1, BB1 and CC1 are parallel, in order to get the proof it suﬃces to notice that BA1 CA1 = BA C1A and CB1 AB1 = C1B AB . Now, suppose that the indicated relation holds and prove that then lines AA1, BB1 and CC1 intersect at one point. Let C∗ 1 be the intersection point of line AB with the line that passes through point C and the intersection point of lines AA1 and BB1. For point C∗ 1 the same relation as for point C1 holds. Therefore, C∗ 1A : C∗ 1B = C1A : C1B. Hence, C∗ 1 = C1, i.e., lines AA1, BB1 and CC1 meet at one point. SOLUTIONS 125 It is also possible to verify that if the indicated relation holds and two of the lines AA1, BB1 and CC1 are parallel, then the third line is also parallel to them. 5.71. Clearly, AB1 = AC1, BA1 = BC1 and CA1 = CB1, and, in the case of the inscribed circle, on sides of triangle ABC, there are three points and in the case of an escribed circle there is just one point on sides of triangle ABC. It remains to make use of Ceva’s theorem. 5.72. Let AA1, BB1 and CC1 be heights of triangle ABC. Then AC1 C1B · BA1 A1C · CB1 B1A = b cos ∠A a cos ∠B · c cos ∠B b cos ∠C · a cos ∠C c cos ∠A = 1. 5.73. Let A2, B2 and C2 be the midpoints of sides BC, CA and AB. The considered lines pass through the vertices of triangle A2B2C2 and in heading a) they divide its sides in the same ratios in which lines AP, BP and CP divide sides of triangle ABC whereas in heading b) they divide them in the inverse ratios. It remains to make use of Ceva’s theorem. 5.74. Since △AC1B2 ∼△BC1A1 and △AB1C2 ∼△CB1A1, it follows that AB2 ·C1B = AC1 · BA1 and AC2 · CB1 = A1C · B1A. Hence, AB2 AC2 = AC1 C1B · BA1 A1C · CB1 B1A = 1. 5.75. Let lines AA1, BB1 and CC1 intersect lines BC, CA and AB at points A1, B2 and C2. a) If ∠B + β < 180◦and ∠C + γ < 180◦, then BA2 A2C = SABA1 SACA1 = AB · BA1 sin(∠B + β) AC · CA1 sin(∠C + γ) = AB AC · sin γ sin β · sin(∠B + β) sin(∠C + γ). The latter expression is equal to BA2 : A2C in all the cases. Let us write similar expressions for CB2 : B2A and AC2 : C2B and multiply them. Now it remains to make use of Ceva’s theorem. b) Point A2 lies outside segment BC only if precisely one of the angles β and γ is greater than the corresponding angle ∠B or ∠C. Hence, BA2 A2C = AB AC · sin γ sin β · sin(∠B −β) sin(∠C −γ). 5.76. It is easy to verify that this problem is a particular case of Problem 5.75. Remark. A similar statement is also true for an escribed circle. 5.77. The solution of the problem obviously follows from Ceva’s theorem. 5.78. By applying the sine theorem to triangles ACC1 and BCC1 we get AC1 C1C = sin ∠ACC1 sin ∠A and CC1 C1B = sin ∠B sin ∠C1CB, i.e., AC1 C1B = sin ∠ACC1 sin ∠C1CB · sin ∠B sin ∠A. Similarly, BA1 A1C = sin ∠BAA1 sin ∠A1AC · sin ∠C sin ∠B and CB1 B1A = sin ∠CBB1 sin ∠B1BA · sin ∠A sin ∠C . To complete the proof it remains to multiply these equalities. Remark. A similar statement is true for the ratios of oriented segments and angles in the case when the points are taken on the extensions of sides. 126 CHAPTER 5. TRIANGLES 5.79. We may assume that points A2, B2 and C2 lie on the sides of triangle ABC. By Problem 5.78 AC2 C2B · BA2 A2C · CB2 B2A = sin ∠ACC2 sin ∠C2CB · sin ∠BAA2 sin ∠A2AC · sin ∠CBB2 sin ∠B2BA. Since lines AA2, BB2 and CC2 are symmetric to lines AA1, BB1 and CC1, respectively, through the bisectors, it follows that ∠ACC2 = ∠C1CB, ∠C2CB = ∠ACC1 etc., hence, sin ∠ACC2 sin ∠C2CB · sin ∠BAA2 sin ∠A2AC · sin ∠CBB2 sin ∠B2BA = sin ∠C1CB sin ∠ACC1 · sin ∠A1AC sin ∠BAA1 · sin ∠B1BA sin ∠CBB1 = C1B AC1 · A1C BA1 · B1A CB1 = 1. Therefore, AC2 C2B · BA2 A2C · CB2 B2A = 1, i.e., lines AA2, BB2 and CC2 meet at one point. Remark. The statement holds also in the case when points A1, B1 and C1 are taken on the extensions of sides if only point P does not lie on the circumscribed circle S of triangle ABC; if P does lie on S, then lines AA2, BB2 and CC2 are parallel (cf. Problem 2.90). 5.80. Let diagonals AD and BE of the given hexagon ABCDEF meet at point P; let K and L be the midpoints of sides AB and ED, respectively. Since ABDE is a trapezoid, segment KL passes through point P (by Problem 19.2). By the law of sines sin ∠APK : sin ∠AKP = AK : AP and sin ∠BPK : sin ∠BKP = BK : BP. Since sin ∠AKP = sin ∠BKP and AK = BK, we have sin ∠APK : sin ∠BPK = BP : AP = BE : AD. Similar relations can be also written for the segments that connect the midpoints of the other two pairs of the opposite sides. By multiplying these relations and applying the result of Problem 5.78 to the triangle formed by lines AD, BE and CF, we get the statement desired. 5.81. Let us consider the homothety with center P and coeﬃcient 2. Since PA1A3A2 is a rectangle, this homothety sends line A1A2 into line la that passes through point A3; lines la and A3P are symmetric through line A3A. Line A3A divides the angle B3A3C3 in halves (Problem 1.56 a)). We similarly prove that lines lb and lc are symmetric to lines B3P and C3P, respectively, through bisectors of triangle A3B3C3. Therefore, lines la, lb and lc either meet at one point or are parallel (Problem 1.79) and, therefore, lines A1A2, B1B2 and C1C2 meet at one point. 5.82. By Problems 5.78 and 5.70 b)) we have sin ∠ASP sin ∠PSD · sin ∠DAP sin ∠PAS · sin ∠SDP sin ∠PDA = 1 = sin ∠ASQ sin ∠QSD · sin ∠DAQ sin ∠QAS · sin ∠SDQ sin ∠QDA. But ∠DAP = ∠SDQ, ∠SDP = ∠DAQ, ∠PAS = ∠QDA and ∠PDA = ∠QAS. Hence, sin ∠ASP sin ∠PSD = sin ∠ASQ sin ∠QSD. SOLUTIONS 127 This implies that points S, P and Q lie on one line, since the function sin(α−x) sin x is monotonous with respect to x: indeed, d dx µsin(α −x) sin x ¶ = −sin α sin2 x. 5.83. a) By Ceva’s theorem AC1 C1B = CA1 A1B · AB1 B1C and by the law of sines CA1 = CA sin ∠CAA1 sin ∠AA1B , A1B = AB sin ∠BAA1 sin ∠AA1B , AB1 = AB sin ∠ABB1 sin ∠AB1B , B1C = BC sin ∠CBB1 sin ∠AB1B . Substituting the last four identities in the ﬁrst identity and taking into account that AC = BC, we get the statement desired. b) Let us denote the intersection points of lines CM and CN with base AB by M1 and N1, respectively. We have to prove that M1 = N1. From heading a) it follows that AM1 : M1B = AN1 : N1B, i.e., M1 = N1. 5.84. Let segments BM and BN meet side AC at points P and Q, respectively. Then sin ∠PBB1 sin PBA = sin ∠PBB1 sin ∠BPB1 · sin ∠APB sin ∠PBA = PB BB1 · AB PA. If O is the intersection point of bisectors of triangle ABC, then AP PB1 · B1O OB · BC1 C1A = 1 and, therefore, sin ∠PBB1 sin ∠PBA = AB BB1 · B1O OB · BC1 C1A. Observe that BC1 : C1A = BC : CA and perform similar calculations for sin ∠QBB1 : sin ∠QBC; we deduce that sin ∠PBB1 sin ∠PBA = sin ∠QBB1 sin ∠QBC . Since ∠ABB1 = ∠CBB1, we have: ∠PBB1 = ∠QBB1. 5.85. a) Let point P lie on arc ⌣AC of the circumscribed circle of triangle ABC; let A1, B1 and C1 be the bases of perpendiculars dropped from point P to lines BC, CA and AB. The sum of angles at vertices A1 and C1 of quadrilateral A1BC1P is equal to 180◦, hence, ∠A1PC1 = 180◦−∠B = ∠APC. Therefore, ∠APC1 = ∠A1PC, where one of points A1 and C1 (say, A1) lies on a side of the triangle and the other point lies on the extension of a side. Quadrilaterals AB1PC1 and A1B1PC are inscribed ones, hence, ∠AB1C1 = ∠APC1 = ∠A1PC = ∠A1B1C and, therefore, point B1 lies on segment A1C1. b) By the same arguments as in heading a) we get ∠(AP, PC1) = ∠(AB1, B1C) = ∠(CB1, B1A1) = ∠(CP, PA1). Add ∠(PC1, PC) to ∠(AP, PC1); we get ∠(AP, PC) = ∠(PC1, PA1) = ∠(BC1, BA1) = ∠(AB, BC), i.e., point P lies on the circumscribed circle of triangle ABC. 5.86. Let A1, B1 and C1 be the midpoints of segments PA, PB and PC, respectively; let Oa, Ob and Oc be the centers of the circumscribed circles of triangles BCP, ACP and ABP, respectively. Points A1, B1 and C1 are the bases of perpendiculars dropped from point 128 CHAPTER 5. TRIANGLES P to sides of triangle OaObOc (or their extensions). Points A1, B1 and C1 lie on one line, hence, point P lies on the circumscribed circle of triangle OaOcOc (cf. Problem 5.85, b). 5.87. Let the extension of the bisector AD intersect the circumscribed circle of triangle ABC at point P. Let us drop from point P perpendiculars PA1, PB1 and PC1 to lines BC, CA and AB, respectively; clearly, A1 is the midpoint of segment BC. The homothety centered at A that sends P to D sends points B1 and C1 to B′ and C′ and, therefore, it sends point A1 to M, because M(???) lies on line B1C1 and PA1 ∥DM. 5.88. a) The solution of Problem 5.85 can be adapted without changes to this case. b) Let A1 and B1 be the bases of perpendiculars dropped from point P to lines BC and CA, respectively, and let points A2 and B2 from lines BC and AC, respectively, be such that ∠(PA2, BC) = α = ∠(PB2, AC). Then △PA1A2 ∼△PB1B2 hence, points A1 and B1 turn under a rotational homothety centered at P into A2 and B2 and ∠A1PA2 = 90◦−α is the angle of the rotation. 5.89. a) Let the angle between lines PC and AC be equal to ϕ. Then PA = 2R sin ϕ. Since points A1 and B1 lie on the circle with diameter PC, the angle between lines PA1 and A1B1 is also equal to ϕ. Hence, PA1 = d sin ϕ and, therefore, PA · PA1 = 2Rd. b) Since PA1 ⊥BC, it follows that cos α = sin ϕ = d PA1. It remains to notice that PA1 = 2Rd PA . 5.90. Points A1 and B1 lie on the circle with diameter PC, hence, A1B1 = PC sin ∠A1CB1 = PC sin ∠C. Let the angle between lines AB and A1B1 be equal to γ and C1 be the projection of point P to line A1B1. Lines A1B1 and B1C1 coincide, hence, cos γ = PC 2R (cf. Problem 5.89). Therefore, the length of the projection of segment AB to line A1B1 is equal to AB cos γ = (2R sin ∠C)PC 2R = PC sin ∠C. 5.91. Let A1 and B1 be the bases of perpendiculars dropped from point P to lines BC and AC. Points A1 and B1 lie on the circle with diameter PC. Since sin ∠A1CB1 = sin ∠ACB, the chords A1B1 of this circle are of the same length. Therefore, lines A1B1 are tangent to a ﬁxed circle. 5.92. Let A1 and B1 be the bases of perpendiculars dropped from point P to lines BC and CA. Then ∠(A1B1, PB1) = ∠(A1C, PC) = ⌣BP 2 . It is also clear that for all points P lines PB1 have the same direction. 5.93. Let P1 and P2 be diametrically opposite points of the circumscribed circle of triangle ABC; let Ai and Bi be the bases of perpendiculars dropped from point Pi to lines BC and AC, respectively; let M and N be the midpoints of sides AC and BC, respectively; let X be the intersection point of lines A1B1 and A2B2, respectively. By Problem 5.92 A1B1 ⊥ A2B2. It remains to verify that ∠(MX, XN) = ∠(BC, AC). Since AB2 = B1C, it follows that XM is a median of right triangle B1XB2. Hence, ∠(XM, XB2) = ∠(XB2, B2M). Similarly, ∠(XA1, XN) = ∠(A1N, XA1). Therefore, ∠(MX, XN) = ∠(XM, XB2) + ∠(XB2, XA1) + ∠(XA1, XN) = ∠(XB2, B2M) + ∠(A1N, XA1) + 90◦. Since ∠(XB2, B2M) + ∠(AC, CB) + ∠(NA1, A1X) + 90◦= 0◦, we have: ∠(MN, XN) + ∠(AC, CB) = 0◦. SOLUTIONS 129 5.94. If point R on the given circle is such that ∠(− → OP, − → OR) = 1 2(β +γ), then OR ⊥BC. It remains to verify that ∠(OR, OQ) = ∠(PA1, A1B1). But ∠(OR, OQ) = 1 2α and ∠(PA1, A1B1) = ∠(PB, BC1) = ∠(− → OP, − → OA) 2 = α 2 . 5.95. Let lines AC and PQ meet at point M. In triangle MPC draw heights PB1 and CA1. Then A1B1 is Simson’s line of point P with respect to triangle ABC. More-over, by Problem 1.52 ∠(MB1, B1A1) = ∠(CP, PM). It is also clear that ∠(CP, PM) = ∠(CA, AQ) = ∠(MB1, AQ). Hence, A1B1 ∥AQ. 5.96. Let us draw chord PQ perpendicular to BC. Let points H′ and P ′ be symmetric to points H and P, respectively, through line BC; point H′ lies on the circumscribed circle of triangle ABC (Problem 5.9). First, let us prove that AQ ∥P ′H. Indeed, ∠(AH′, AQ) = ∠(PH′, PQ) = ∠(AH′, P ′H). Simson’s line of point P is parallel to AQ (Problem 5.95), i.e., it passes through the midpoint of side PP ′ of triangle PP ′H and is parallel to side P ′H; hence, it passes through the midpoint of side PH. 5.97. Let Ha, Hb, Hc and Hd be the orthocenters of triangles BCD, CDA, DAB and ABC, respectively. Lines la, lb, lc and ld pass through the midpoints of segments AHa, BHb, CHc and DHd, respectively (cf. Problem 5.96). The midpoints of these segments coincide with point H such that 2− − → OH = − → OA + − − → OB + − → OC + − − → OD, where O is the center of the circle (cf. Problem 13.33). 5.98. a) Let B1, C1 and D1 be the projections of point P to lines AB, AC and AD, respectively. Points B1, C1 and D1 lie on the circle with diameter AP. Lines B1C1, C1D1 and D1B1 are Simson’s lines of point P with respect to triangles ABC, ACD and ADB, respectively. Therefore, projections of point P to Simson’s lines of these triangles lie on one line — Simson’s line of triangle B1C1D1. We similarly prove that any triple of considered points lies on one line. b) Let P be a point of the circumscribed circle of n-gon A1 . . . An; let B2, B3, . . . , Bn be the projections of point P to lines A1A2, . . . , A1An, respectively. Points B2, . . . , Bn lie on the circle with diameter A1P. Let us prove by induction that Simson’s line of point P with respect to n-gon A1 . . . An coincides with Simson’s line of point P with respect to (n −1)-gon B2 . . . Bn (for n = 4 this had been proved in heading a)). By the inductive hypothesis Simson’s line of the (n−1)-gon A1A3 . . . An coincides with Simson’s line of (n −2)-gon B3 . . . Bn. Hence, the projections of point P to Simson’s line of (n −1)-gons whose vertices are obtained by consecutive deleting points A2, . . . , An from the collection A1, . . . , An ????? lie on Simson’s line of the (n−1)-gon B2 . . . Bn. The projection of point P to Simson’s line of the (n −1)-gon A2 . . . An lies on the same line, because our arguments show that any n −1 of the considered n points of projections lie on one line. 5.99. Points B1 and C1 lie on the circle with diameter AP. Hence, B1C1 = AP sin ∠B1AC1 = AP ¡ BC 2R ¢ . 5.100. This problem is a particular case of Problem 2.43. 5.101. Clearly, ∠C1AP = ∠C1B1P = ∠A2B1P = ∠A2C2P = ∠B3C2P = ∠B3A3P. (The ﬁrst, third and ﬁfth equalities are obtained from the fact that the corresponding quadri-laterals are inscribed ones; the remaining equalities are obvious.) Similarly, ∠B1AP = ∠C3A3P. Hence, ∠B3A3C3 = ∠B3A3P + ∠C3A3P = ∠C1AP + ∠BAP = ∠BAC. 130 CHAPTER 5. TRIANGLES Similarly, the equalities of the remaining angles of triangles ABC and A3B3C3 are similarly obtained. 5.102. Let A1, B1 and C1 be the bases of perpendiculars dropped from point P to lines BC, CA and AB, respectively; let A2, B2 and C2 be the intersection points of lines PA, PB and PC, respectively, with the circumscribed circle of triangle ABC. Further, let S, S1 and S2 be areas of triangles ABC, A1B1C1 and A2B2C2, respectively. It is easy to verify that a1 = a·AP 2R (Problem 5.99) and a2 = a·B2P CP . Triangles A1B1C1 and A2B2C2 are similar (Problem 5.100); hence, S1 S2 = k2, where k = a1 a2 = AP·CP 2R·B2P . Since B2P · BP = \|d2 −R2\|, we have: S1 S2 = (AP · BP · CP)2 4R2(d2 −R2)2 . Triangles A2B2C2 and ABC are inscribed in one circle, hence, S2 S = a2b2c2 abc (cf. Problem 12.1). It is also clear that, for instance, a2 a = B2P CP = \|d2 −R2\| BP · CP . Therefore, S2 : S = \|d2 −R2\|3 : (AP · BP · CP)2. Hence, S1 S = S1 S2 · S2 S = \|d2 −R2\| 4R2 . 5.103. Points B1 and C1 lie on the circle with diameter PA and, therefore, the midpoint of segment PA is the center of the circumscribed circle of triangle AB1C1. Consequenly, la is the midperpendicular to segment B1C1. Hence, lines la, lb and lc pass through the center of the circumscribed circle of triangle A1B1C1. 5.104. a) Let us drop from points P1 and P2 perpendiculars P1B1 and P2B2, respectively, to AC and perpendiculars P1C1 and P2C2 to AB. Let us prove that points B1, B2, C1 and C2 lie on one circle. Indeed, ∠P1B1C1 = ∠P1AC1 = ∠P2AB2 = ∠P2C2B2; and, since ∠P1B1A = ∠P2C2A, it follows that ∠C1B1A = ∠B2C2A. The center of the circle on which the indicated points lie is the intersection point of the midperpendiculars to segments B1B2 and C1C2; observe that both these perpendiculars pass through the midpoint O of segment P1P2, i.e., O is the center of this circle. In particular, points B1 and C1 are equidistant from point O. Similarly, points A1 and B1 are equidistant from point O, i.e., O is the center of the circumscribed circle of triangle A1B1C1. Moreover, OB1 = OB2. b) The preceding proof passes virtually without changes in this case as well. 5.105. Let A1, B1 and C1 be the midpoints of sides BC, CA and AB. Triangles A1B1C1 and ABC are similar and the similarity coeﬃcient is equal to 2. The heights of triangle A1B1C1 intersect at point O; hence, OA1 : HA = 1 : 2. Let M ′ be the intersection point of segments OH and AA1. Then OM ′ : M ′H = OA1 : HA = 1 : 2 and AM ′ : M ′A1 = OA1 : HA = 1 : 2, i.e., M ′ = M. 5.106. Let A1, B1 and C1 be the midpoints of sides BC, CA and AB, respectively; let A2, B2 and C2 the bases of heights; A3, B3 and C3 the midpoints of segments that connect the intersection point of heights with vertices. Since A2C1 = C1A = A1B1 and A1A2 ∥B1C1, point A2 lies on the circumscribed circle of triangle A1B1C1. Similarly, points B2 and C2 lie on the circumscribed circle of triangle A1B1C1. Now, consider circle S with diameter A1A3. Since A1B3 ∥CC2 and A3B3 ∥AB, it follows that ∠A1B3A3 = 90◦and, therefore, point B3 lies on S. We similarly prove that points C1, SOLUTIONS 131 B1 and C3 lie on S. Circle S passes through the vertices of triangle A1B1C1; hence, it is its circumscribed circle. The homothety with center H and coeﬃcient 1 2 sends the circumscribed circle of trian-gle ABC into the circumscribed circle of triangle A3B3C3, i.e., into the circle of 9 points. Therefore, this homothety sends point O into the center of the circle of nine points. 5.107. a) Let us prove that, for example, triangles ABC and HBC share the same circle of nine points. Indeed, the circles of nine points of these triangles pass through the midpoint of side BC and the midpoints of segments BH and CH. b) Euler’s line passes through the center of the circle of 9 points and these triangles share one circle of nine points. c) The center of symmetry is the center of the circle of 9 points of these triangles. 5.108. Let AB > BC > CA. It is easy to verify that for an acute and an obtuse triangles the intersection point H of heights and the center O of the circumscribed circle are positioned precisely as on Fig. 59 (i.e., for an acute triangle point O lies inside triangle BHC1 and for an acute triangle points O and B lie on one side of line CH). Figure 59 (Sol. 5.108) Therefore, in an acute triangle Euler’s line intersects the longest side AB and the shortest side AC, whereas in an acute triangle it intersects the longest side AB, and side BC of intermediate length. 5.109. a) Let Oa, Ob and Oc be the centers of the escribed circles of triangle ABC. The vertices of triangle ABC are the bases of the heights of triangle OaObOc (Problem 5.2) and, therefore, the circle of 9 points of triangle OaObOc passes through point A, B and C. b) Let O be the intersection point of heights of triangle OaObOc, i.e., the intersection point of the bisectors of triangle ABC. The circle of 9 points of triangle OaObOc divides segment OOa in halves. 5.110. Let AA1 be an height, H the intersection point of heights. By Problem 5.45 b) AH = 2R\| cos ∠A\|. The medians are divided by their intersection point in the ratio of 1:2, hence, Euler’s line is parallel to BC if and only if AH : AA1 = 2 : 3 and vectors − − → AH and − − → AA1 are codirected, i.e., 2R cos ∠A : 2R sin ∠B sin ∠C = 2 : 3. Taking into account that cos ∠A = −cos(∠B + ∠C) = sin ∠B sin ∠C −cos ∠B cos ∠C we get sin ∠B sin ∠C = 3 cos ∠B cos ∠C. 5.111. Let CD be a height, O the center of the circumscribed circle, N the midpoint of side AB and let point E divide the segment that connects C with the intersection point of the 132 CHAPTER 5. TRIANGLES heights in halves. Then CENO is a parallelogram, hence, ∠NED = ∠OCH = \|∠A −∠B\| (cf. Problem 2.88). Points N, E and D lie on the circle of 9 points, hence, segment ND is seen from its center under an angle of 2∠NED = 2\|∠A −∠B\|. 5.112. Let O and I be the centers of the circumscribed and inscribed circles, respectively, of triangle ABC, let H be the intersection point of the heights; lines AI and BI intersect the circumscribed circle at points A1 and B1. Suppose that triangle ABC is not an isosceles one. Then OI : IH = OA1 : AH and OI : IH = OB1 : BH. Since OB1 = OA1, we see that AH = BH and, therefore, AC = BC. Contradiction. 5.113. Let O and I be the centers of the circumscribed and inscribed circles, respectively, of triangle ABC, H the orthocenter of triangle A1B1C1. In triangle A1B1C1, draw heights A1A2, B1B2 and C1C2. Triangle A1B1C1 is an acute one (e.g., ∠B1A1C1 = ∠B+∠C 2 < 90◦), hence, H is the center of the inscribed circle of triangle A2B2C2 (cf. Problem 1.56, a). The corresponding sides of triangles ABC and A2B2C2 are parallel (cf. Problem 1.54 a) and, therefore, there exists a homothety that sends triangle ABC to triangle A2B2C2. This homothety sends point O to point I and point I to point H; hence, line IH passes through point O. 5.114. Let H be be the intersection point of the heights of triangle ABC, let E and M be the midpoints of segments CH and AB, see Fig. 60. Then C1MC2E is a rectangle. Figure 60 (Sol. 5.114) Let line CC2 meet line AB at point C3. Let us prove that AC3 : C3B = tan 2α : tan 2β. It is easy to verify that C3M : C2E = MC2 : EC, EC = R cos γ, MC2 = C1E = 2R sin α sin β −R cos γ and C2E = MC1 = R sin(β −α) Hence, C3M = R sin(β −α)(2 sin β sin α −cos γ) cos γ = R sin(β −α) cos(β −α) cos γ . Therefore, AC3 C3B = AM + MC3 C3M + MB = sin 2γ + sin 2(α −β) sin 2γ −sin 2(α −β) = tan 2α tan 2β . Similar arguments show that AC3 C3B · BA3 A3C · CB3 B3A = tan 2α tan 2β · tan 2β tan 2γ · tan 2γ tan 2α = 1. 5.115. Let us solve a more general heading b). First, let us prove that lines AA1, BB1 and CC1 meet at one point. Let the circumscribed circles of triangles A1BC and AB1C SOLUTIONS 133 intersect at point O. Then ∠(BO, OA) = ∠(BO, OC) + ∠(OC, OA) = ∠(BA1, A1C) + ∠(CB1, B1A) = = ∠(BA, AC1) + ∠(C1B, BA) = ∠(C1B, AC1), i.e., the circumscribed circle of triangle ABC1 also passes through point O. Hence, ∠(AO, OA1) = ∠(AO, OB) + ∠(BO, OA1) = ∠(AC1, C1B + ∠(BC, CA1) = 0◦, i.e., line AA1 passes through point O. We similarly prove that lines BB1 and CC1 pass through point O. Now, let us prove that point O coincides with point P we are looking for. Since ∠BAP = ∠A−∠CAP, the equality ∠ABP = ∠CAP is equivalent to the equality ∠BAP +∠ABP = ∠A, i.e., ∠APB = ∠B + ∠C. For point O the latter equality is obvious since it lies on the circumscribed circle of triangle ABC1. 5.116. a) Let us prove that ⌣AB =⌣B1C1, i.e., AB = B1C1. Indeed, ⌣AB =⌣ AC1+ ⌣C1B and ⌣C1B =⌣AB1; hence, ⌣AB =⌣AC1+ ⌣AB1 =⌣B1C1. b) Let us assume that triangles ABC and A1B1C1 are inscribed in one circle, where triangle ABC is ﬁxed and triangle A1B1C1 rotates. Lines AA1, BB1 and CC1 meet at one point for not more than one position of triangle A1B1C1, see Problem 7.20 b). We can obtain 12 distinct families of triangles A1B1C1: triangles ABC and A1B1C1 can be identiﬁed after a rotation or an axial symmetry; moreover, there are 6 distinct ways to associate symbols A1, B1 and C1 to the vertices of the triangle. From these 12 families of triangles 4 families can never produce the desired point P. For similarly oriented triangles the cases △ABC = △A1C1B1, △ABC = △C1B1A1, △ABC = △B1A1C1 are excluded: for example, if △ABC = △A1C1B1, then point P is the intersection point of line BC = B1C1 with the tangent to the circle at point A = A1; in this case triangles ABC and A1B1C1 coincide. For diﬀerently oriented triangles the case △ABC = △A1B1C1 is excluded: in this case AA1 ∥BB1 ∥CC1. Remark. Brokar’s points correspond to diﬀerently oriented triangles; for the ﬁrst Brokar’s point △ABC = △B1C1A1 and for the second Brokar’s point we have △ABC = △C1A1B1. 5.117. a) Since PC = AC sin ∠CAP sin ∠APC and PC = BC sin ∠CBP sin ∠BPC , it follows that sin ϕ sin β sin γ = sin(β −ϕ) sin α sin β . Taking into account that sin(β −γ) = sin β cos ϕ −cos β sin ϕ we get cot ϕ = cot β + sin β sin α sin γ. It remains to notice that sin β = sin(α + γ) = sin α cos γ + sin γ cos α. b) For the second Brokar’s angle we get precisely the same expression as in heading a). It is also clear that both Brokar’s angles are acute ones. c) Since ∠A1BC = ∠BCA and ∠BCA1 = ∠CAB, it follows that △CA1B ∼△ABC. Therefore, Brokar’s point P lies on segment AA1 (cf. Problem 5.115 b)). 5.118. a) By Problem 10.38 a) cot ϕ = cot α + cot β + cot γ ≥ √ 3 = cot 30◦; hence, ϕ ≤30◦. 134 CHAPTER 5. TRIANGLES b) Let P be the ﬁrst Brokar’s point of triangle ABC. Point M lies inside (or on the boundary of) one of the triangles ABP, BCP and CAP. If, for example, point M lies inside triangle ABP, then ∠ABM ≤∠ABP ≤30◦. 5.119. Lines A1B1, B1C1 and C1A1 are the midperpendiculars to segments AQ, BQ and CQ, respectively. Therefore, we have, for instance, ∠B1A1C1 = 180◦−∠AQC = ∠A. For the other angles the proof is similar. Moreover, lines A1O, B1O and C1O are the midperpendiculars to segments CA, AB and BC, respectively. Hence, acute angles ∠OA1C1 and ∠ACQ, for example, have pair-wise perpendicular sides and, consecutively, they are equal. Similar arguments show that ∠OA1C1 = ∠OB1A1 = ∠OC1B1 = ϕ, where ϕ is the Brokar’s angle of triangle ABC. 5.120. By the law of sines R1 = AB 2 sin ∠APB , R2 = BC 2 sin ∠BPC and R3 = CA 2 sin ∠CPA. It is also clear that sin ∠APB = sin ∠A, sin ∠BPC = sin ∠B and sin ∠CPA = sin ∠C. 5.121. Triangle ABC1 is an isosceles one and the angle at its base AB is equal to Brokar’s angle ϕ. Hence, ∠(PC1, C1Q) = ∠(BC1, C1A) = 2ϕ. Similarly ∠(PA1, A1Q) = ∠(PB1, B1Q) = ∠(PC1, C1Q) = 2ϕ. 5.122. Since ∠CA1B1 = ∠A + ∠AB1A1 and ∠AB1A1 = ∠CA1C1, we have ∠B1A1C1 = ∠A. We similarly prove that the remaining angles of triangles ABC and A1B1C1 are equal. The circumscribed circles of triangles AA1B1, BB1C1 and CC1A1 meet at one point O. (Problem 2.80 a). Clearly, ∠AOA1 = ∠AB1A1 = ϕ. Similarly, ∠BOB1 = ∠COC1 = ϕ. Hence, ∠AOB = ∠A1OB1 = 180◦−∠A. Similarly, ∠BOC = 180◦−∠B and ∠COA = 180◦− ∠C, i.e., O is the ﬁrst Brokar’s point of both triangles. Hence, the rotational homothety by angle ϕ with center O and coeﬃcient AO A1O sends triangle A1B1C1 to triangle ABC. 5.123. By the law of sines AB BM = sin ∠AMB sin ∠BAM and AB BN = sin ∠ANB sin ∠BAN . Hence, AB2 BM · BN = sin ∠AMB sin ∠ANB sin ∠BAM sin ∠BAN = sin ∠AMC sin ∠ANC sin ∠CAN sin ∠CAM = AC2 CM · CN . 5.124. Since ∠BAS = ∠CAM, we have BS CM = SBAS SCAM = AB · AS AC · AM , i.e., AS AM = 2b·BS ac . It remains to observe that, as follows from Problems 5.123 and 12.11 a), BS = ac2 b2+c2 and 2AM = √ 2b2 + 2c2 −a2. 5.125. The symmetry through the bisector of angle A sends segment B1C1 into a segment parallel to side BC, it sends line AS to line AM, where M is the midpoint of side BC. 5.126. On segments BC and BA, take points A1 and C1, respectively, so that A1C1 ∥ BK. Since ∠BAC = ∠CBK = ∠BA1C1, segment A1C1 is antiparallel to side AC. On the other hand, by Problem 3.31 b) line BD divides segment A1C1 in halves. 5.127. It suﬃces to make use of the result of Problem 3.30. 5.128. Let AP be the common chord of the considered circles, Q the intersection point of lines AP and BC. Then BQ AB = sin ∠BAQ sin ∠AQB and AC CQ = sin ∠AQC sin ∠CAQ. SOLUTIONS 135 Hence, BQ CQ = AB sin ∠BAP AC sin ∠CAP . Since AC and AB are tangents to circles S1 and S2, it follows that ∠CAP = ∠ABP and ∠BAP = ∠ACP and, therefore, ∠APB = ∠APC. Hence, AB AC = AB AP · AP AC = sin ∠APB sin ∠ABP · sin ∠ACP sin ∠APC = sin ∠ACP sin ∠ABP = sin ∠BAP sin ∠CAP . It follows that BQ CQ = AB2 AC2. 5.129. Let S be the intersection point of lines AX and BC. Then AS AB = CS CX and AS AC = BSBX and, therefore, CS BS = AC AB · XC XB . It remains to observe that XC XB = AC AB (see the solution of Problem 7.16 a)). 5.130. Let L, M and N be the midpoints of segments CA, CB and CH. Since △BAC ∼ △CAH, it follows that △BAM ∼△CAN and, therefore, ∠BAM = ∠CAN. Similarly, ∠ABL = ∠CBN. 5.131. Let B1C1, C2A2 and A3B3 be given segments. Then triangles A2XA3, B1XB3 and C1XC2 are isosceles ones; let the lengths of their lateral sides be equal to a, b and c. Line AX divides segment B1C1 in halves if and only if this line contains a simedian. Hence, if X is Lemoin’s point, then a = b, b = c and c = a. And if B1C1 = C2A2 = A3B3, then b + c = c + a = a + b and, therefore, a = b = c. 5.132. Let M be the intersection point of medians of triangle ABC; let a1, b2, c1 and a2, b2, c2 be the distances from points K and M, respectively, to the sides of the triangle. Since points K and M are isogonally conjugate, a1a2 = b1b2 = c1c2. Moreover, aa2 = bb2 = cc2 (cf. Problem 4.1). Therefore, a a1 = b b1 = c c1. Making use of this equality and taking into account that areas of triangles A1B1K, B1C1K and C1A1K are equal to a1b1c 4R , b1c1a 4R and c1a1b 4R , respectively, where R is the radius of the circumscribed circle of triangle ABC, we deduce that the areas of these triangles are equal. Moreover, point K lies inside triangle A1B1C1. Therefore, K is the intersection point of medians of triangle A1B1C1 (cf. Problem 4.2). 5.133. Medians of triangle A1B1C1 intersect at point K (Problem 5.132); hence, the sides of triangle ABC are perpendicular to the medians of triangle A1B1C1. After a rotation through an angle of 90◦the sides of triangle ABC become pairwise parallel to the medians of triangle A1B1C1 and, therefore, the medians of triangle ABC become parallel to the corresponding sides of triangle A1B1C1 (cf. Problem 13.2). Hence, the medians of triangle ABC are perpendicular to the corresponding sides of triangle A1B1C1. 5.134. Let A2, B2 and C2 be the projections of point K to lines BC, CA and AB, respectively. Then △A1B1C1 ∼△A2B2C2 (Problem 5.100) and K is the intersection point of medians of triangle A2B2C2 (Problem 5.132). Hence, the similarity transformation that sends triangle A2B2C2 to triangle A1B1C1 sends point K to the intersection point M of medians of triangle A1B1C1. Moreover, ∠KA2C2 = ∠KBC2 = ∠B1A1K, i.e., points K and M are isogonally conjugate with respect to triangle A1B1C1 and, therefore, K is Lemoin’s point of triangle A1B1C1. 5.135. Let K be Lemoin’s point of triangle ABC; let A1, B1 and C1 be the projections of point K on the sides of triangle ABC; let L be the midpoint of segment B1C1 and N the intersection point of line KL and median AM; let O be the midpoint of segment AK (Fig. 61). Points B1 and C1 lie on the circle with diameter AK, hence, by Problem 5.132 OL ⊥B1C1. Moreover, AN ⊥B1C1 (Problem 5.133) and O is the midpoint of segment AK, consequently, OL is the midline of triangle AKN and KL = LN. Therefore, K is the midpoint of segment A1N. It remains to notice that the homothety with center M that sends N to A sends segment NA1 to height AH. 136 CHAPTER 5. TRIANGLES Figure 61 (Sol. 5.135) Chapter 6. POLYGONS Background 1) A polygon is called a convex one if it lies on one side of any line that connects two of its neighbouring vertices. 2) A convex polygon is called a circumscribed one if all its sides are tangent to a circle. A convex quadrilateral is a circumscribed one if and only if AB + CD = BC + AD. A convex polygon is called an inscribed one if all its vertices lie on one circle. A convex quadrilateral is an inscribed one if and only if ∠ABC + ∠CDA = ∠DAB + ∠BCD. 3) A convex polygon is called a regular one if all its sides are equal and all its angles are also equal. A convex n-gon is a regular one if and only if under a rotation by the angle of 2π n with center at point O it turns into itself. This point O is called the center of the regular polygon. Introductory problems 1. Prove that a convex quadrilateral ABCD can be inscribed into a circle if and only if ∠ABC + ∠CDA = 180◦. 2. Prove that a circle can be inscribed in a convex quadrilateral ABCD if and only if AB + CD = BC + AD. 3. a) Prove that the axes of symmetry of a regular polygon meet at one point. b) Prove that a regular 2n-gon has a center of symmetry. 4. a) Prove that the sum of the angles at the vertices of a convex n-gon is equal to (n −2) · 180◦. b) A convex n-gon is divided by nonintersecting diagonals into triangles. Prove that the number of these triangles is equal to n −2. §1. The inscribed and circumscribed quadrilaterals 6.1. Prove that if the center of the circle inscribed in a quadrilateral coincides with the intersection point of the quadrilateral’s diagonals, then this quadrilateral is a rhombus. 6.2. Quadrilateral ABCD is circumscribed about a circle centered at O. Prove that ∠AOB + ∠COD = 180◦. 6.3. Prove that if there exists a circle tangent to all the sides of a convex quadrilateral ABCD and a circle tangent to the extensions of all its sides then the diagonals of such a quadrilateral are perpendicular. 6.4. A circle singles out equal chords on all the four sides of a quadrilateral. Prove that a circle can be inscribed into this quadrilateral. 6.5. Prove that if a circle can be inscribed into a quadrilateral, then the center of this circle lies on one line with the centers of the diagonals. 137 138 CHAPTER 6. POLYGONS 6.6. Quadrilateral ABCD is circumscribed about a circle centered at O. In triangle AOB heights AA1 and BB1 are drawn. In triangle COD heights CC1 and DD1 are drawn. Prove that points A1, B1, C1 and D1 lie on one line. 6.7. The angles at base AD of trapezoid ABCD are equal to 2α and 2β. Prove that the trapezoid is a circumscribed one if and only if BC AD = tan α tan β. 6.8. In triangle ABC, segments PQ and RS parallel to side AC and a segment BM are drawn as plotted on Fig. 62. Trapezoids RPKL and MLSC are circumscribed ones. Prove that trapezoid APQC is also a circumscribed one. Figure 62 (6.8) 6.9. Given convex quadrilateral ABCD such that rays AB and CD intersects at a point P and rays BC and AD intersect at a point Q. Prove that quadrilateral ABCD is a circumscribed one if and only if one of the following conditions hold: AB + CD = BC + AD, AP + CQ = AQ + CP BP + BQ = DP + DQ. 6.10. Through the intersection points of the extension of sides of convex quadrilateral ABCD two lines are drawn that divide it into four quadrilaterals. Prove that if the quadri-laterals adjacent to vertices B and D are circumscribed ones, then quadrilateral ABCD is also a circumscribed one. 6.11. Prove that the intersection point of the diagonals of a circumscribed quadrilateral coincides with the intersection point of the diagonals of the quadrilateral whose vertices are the tangent points of the sides of the initial quadrilateral with the inscribed circle. 6.12. Quadrilateral ABCD is an inscribed one; Hc and Hd are the orthocenters of triangles ABD and ABC respectively. Prove that CDHcHd is a parallelogram. 6.13. Quadrilateral ABCD is an inscribed one. Prove that the centers of the inscribed circles of triangles ABC, BCD, CDA and DAB are the vertices of a rectangle. 6.14. The extensions of the sides of quadrilateral ABCD inscribed in a circle centered at O intersect at points P and Q and its diagonals intersect at point S. a) The distances from points P, Q and S to point O are equal to p, q and s, respectively, and the radius of the circumscribed circle is equal to R. Find the lengths of the sides of triangle PQS. b) Prove that the heights of triangle PQS intersect at point O. 6.15. Diagonal AC divides quadrilateral ABCD into two triangles whose inscribed circles are tangent to diagonal AC at one point. Prove that the inscribed circles of triangle ABD and BCD are also tangent to diagonal BD at one point and their tangent points with the sides of the quadrilateral lie on one circle. §2. QUADRILATERALS 139 6.16. Prove that the projections of the intersection point of the diagonals of the inscribed quadrilateral to its sides are vertices of a circumscribed quadrilateral only if the projections do not lie on the extensions of the sides. 6.17. Prove that if the diagonals of a quadrilateral are perpendicular, then the pro-jections of the intersection points of the diagonals on its sides are vertices of an inscribed quadrilateral. See also Problem 13.33, 13.34, 16.4. §2. Quadrilaterals 6.18. The angle between sides AB and CD of quadrilateral ABCD is equal to ϕ. Prove that AD2 = AB2 + BC2 + CD2 −2(AB · BC cos B + BC · CD cos C + CD · AB cos ϕ). 6.19. In quadrilateral ABCD, sides AB and CD are equal and rays AB and DC intersect at point O. Prove that the line that connects the midpoints of the diagonals is perpendicular to the bisector of angle AOD. 6.20. On sides BC and AD of quadrilateral ABCD, points M and N, respectively, are taken so that BM : MC = AN : ND = AB : CD. Rays AB and DC intersect at point O. Prove that line MN is parallel to the bisector of angle AOD. 6.21. Prove that the bisectors of the angles of a convex quadrilateral form an inscribed quadrilateral. 6.22. Two distinct parallelograms ABCD and A1B1C1D1 with corresponding parallel sides are inscribed into quadrilateral PQRS (points A and A1 lie on side PQ, points B and B1 lie on side QR, etc.). Prove that the diagonals of the quadrilateral are parallel to the corresponding sides of the parallelograms. 6.23. The midpoints M and N of diagonals AC and BD of convex quadrilateral ABCD do not coincide. Line MN intersects sides AB and CD at points M1 and N1. Prove that if MM1 = NN1, then AD ∥BC. 6.24. Prove that two quadrilaterals are similar if and only if four of their corresponding angles are equal and the corresponding angles between the diagonals are also equal. 6.25. Quadrilateral ABCD is a convex one; points A1, B1, C1 and D1 are such that AB ∥C1D1 and AC ∥B1D1, etc. for all pairs of vertices. Prove that quadrilateral A1B1C1D1 is also a convex one and ∠A + ∠C1 = 180◦. 6.26. From the vertices of a convex quadrilateral perpendiculars are dropped on the diagonals. Prove that the quadrilateral with vertices at the basis of the perpendiculars is similar to the initial quadrilateral. 6.27. A convex quadrilateral is divided by the diagonals into four triangles. Prove that the line that connects the intersection points of the medians of two opposite triangles is perpendicular to the line that connects the intersection points of the heights of the other two triangles. 6.28. The diagonals of the circumscribed trapezoid ABCD with bases AD and BC intersect at point O. The radii of the inscribed circles of triangles AOD, AOB, BOC and COD are equal to r1, r2, r3 and r4, respectively. Prove that 1 r1 + 1 r3 = 1 r2 + 1 r4. 6.29. A circle of radius r1 is tangent to sides DA, AB and BC of a convex quadrilateral ABCD; a circle of radius r2 is tangent to sides AB, BC and CD; the radii r3 and r4 are similarly deﬁned. Prove that AB r1 + CD r3 = BC r2 + AD r4 . 6.30. A quadrilateral ABCD is convex and the radii of the circles inscribed in triangles ABC, BCD, CDA and DAB are equal. Prove that ABCD is a rectangle. 140 CHAPTER 6. POLYGONS 6.31. Given a convex quadrilateral ABCD and the centers A1, B1, C1 and D1 of the circumscribed circles of triangles BCD, CDA, DAB and ABC, respectively. For quadrilat-eral A1B1C1D1 points A2, B2, C2 and D2 are similarly deﬁned. Prove that quadrilaterals ABCD and A2B2C2D2 are similar and their similarity coeﬃcient is equal to 1 4 \|(cot A + cot C)(cot B + cot D)\| . 6.32. Circles whose diameters are sides AB and CD of a convex quadrilateral ABCD are tangent to sides CD and AB, respectively. Prove that BC ∥AD. 6.33. Four lines determine four triangles. Prove that the orthocenters of these triangles lie on one line. §3. Ptolemy’s theorem 6.34. Quadrilateral ABCD is an inscribed one. Prove that AB · CD + AD · BC = AC · BD (Ptolemy’s theorem). 6.35. Quadrilateral ABCD is an inscribed one. Prove that AC BD = AB · AD + CB · CD BA · BC + DA · DC . 6.36. Let α = π 7. Prove that 1 sin α = 1 sin 2α + 1 sin 3α. 6.37. The distances from the center of the circumscribed circle of an acute triangle to its sides are equal to da, db and dc. Prove that da + db + dc = R + r. 6.38. The bisector of angle ∠A of triangle ABC intersects the circumscribed circle at point D. Prove that AB + AC ≤2AD. 6.39. On arc ⌣CD of the circumscribed circle of square ABCD point P is taken. Prove that PA + PC = √ 2PB. 6.40. Parallelogram ABCD is given. A circle passing through point A intersects seg-ments AB, AC and AD at points P, Q and R, respectively. Prove that AP · AB + AR · AD = AQ · AC. 6.41. On arc ⌣A1A2n+1 of the circumscribed circle S of a regular (2n + 1)-gon A1 . . . A2n+1 a point A is taken. Prove that: a) d1 + d3 + · · · + d2n+1 = d2 + d4 + · · · + d2n, where di = AAi; b) l1 +· · ·+l2n+1 = l2 +· · ·+l2n, where li is the length of the tangent drawn from point A to the circle of radius r tangent to S at point Ai (all the tangent points are simultaneously either inner or outer ones). 6.42. Circles of radii x and y are tangent to a circle of radius R and the distance between the tangent points is equal to a. Calculate the length of the following common tangent to the ﬁrst two circles: a) the outer one if both tangents are simultaneously either outer or inner ones; b) the inner one if one tangent is an inner one and the other one is an outer one. 6.43. Circles α, β, γ and δ are tangent to a given circle at vertices A, B, C and D, respectively, of convex quadrilateral ABCD. Let tαβ be the length of the common tangent to circles α and β (the outer one if both tangent are simultaneously either inner or outer §5. HEXAGONS 141 ones and the inner one if one tangent is an inner one and the other one is an outer one); tβγ, tγδ, etc. are similarly determined. Prove that tαβtγδ + tβγtδα = tαγtβδ (The generalized Ptolemy’s theorem) See also Problem 9.67. §4. Pentagons 6.44. In an equilateral (non-regular) pentagon ABCDE we have angle ∠ABC = 2∠DBE. Find the value of angle ∠ABC. 6.45. a) Diagonals AC and BE of a regular pentagon ABCDE intersect at point K. Prove that the inscribed circle of triangle CKE is tangent to line BC. b) Let a be the length of the side of a regular pentagon, d the length of its diagonal. Prove that d2 = a2 + ad. 6.46. Prove that a square can be inscribed in a regular pentagon so that the vertices of the square would lie on four sides of the pentagon. Figure 63 (6.46) 6.47. Regular pentagon ABCDE with side a is inscribed in circle S. The lines that pass through the pentagon’s vertices perpendicularly to the sides form a regular pentagon with side b (Fig. 63). A side of a regular pentagon circumsribed about circle S is equal to c. Prove that a2 + b2 = c2. See also Problems 2.59, 4.9, 9.23, 9.44, 10.63, 10.67, 13.10, 13.56, 20.11. §5. Hexagons 6.48. The opposite sides of a convex hexagon ABCDEF are pairwise parallel. Prove that: a) the area of triangle ACE constitutes not less than a half area of the hexagon. b) the areas of triangles ACE and BDF are equal. 6.49. All the angles of a convex hexagon ABCDEF are equal. Prove that \|BC −EF\| = \|DE −AB\| = \|AF −CD\|. 6.50. The sums of the angles at vertices A, C, E and B, D, F of a convex hexagon ABCDEF with equal sides are equal. Prove that the opposite sides of this hexagon are parallel. 6.51. Prove that if in a convex hexagon each of the three diagonals that connect the opposite vertices divides the area in halves then these diagonals intersect at one point. 6.52. Prove that if in a convex hexagon each of the three segments that connect the midpoints of the opposite sides divides the area in halves then these segments intersect at one point. 142 CHAPTER 6. POLYGONS See also problems 2.11, 2.20, 2.46, 3.66, 4.6, 4.28, 4.31, 5.80, 9.45 a), 9.76–9.78, 13.3, 14.6, 18.22, 18.23. §6. Regular polygons 6.53. The number of sides of a polygon A1 . . . An is odd. Prove that: a) if this polygon is an inscribed one and all its angles are equal, then it is a regular polygon; b) if this polygon is a circumscribed one and all its sides are equal, then it is a regular polygon. 6.54. All the angles of a convex polygon A1 . . . An are equal; an inner point O of the polygon is the vertex of equal angles that subtend all the polygon’s sides. Prove that the polygon is a regular one. 6.55. A paper band of constant width is tied in a simple knot and then tightened in order to make the knot ﬂat, cf. Fig. 64. Prove that the knot is of the form of a regular pentagon. Figure 64 (6.55) 6.56. On sides AB, BC, CD and DA of square ABCD equilateral triangles ABK, BCL, CDM and DAN are constructed inwards. Prove that the midpoints of sides of these triangles (which are not the sides of a square) and the midpoints of segments KL, LM, MN and NK form a regular 12-gon. 6.57. Does there exist a regular polygon the length of one of whose diagonal is equal to the sum of lengths of some other two diagonals? 6.58. A regular (4k + 2)-gon is inscribed in a circle of radius R centered at O. Prove that the sum of the lengths of segments singled out by the legs of angle ∠AkOAk+1 on lines A1A2k, A2A2k−1, . . . , AkAk+1 is equal to R. 6.59. In regular 18-gon A1 . . . A18, diagonals AaAd, AbAe and AcAf are drawn. Let k = a −b, p = b −c, m = c −d, q = d −e, n = e −f and r = f −a. Prove that the indicated diagonals intersect at one point in any of the following cases and only in these cases: a) − − − − → k, m, n = − − − → p, q, r; b) − − − − → k, m, n = − − − → 1, 2, 7 and − − − → p, q, r = − − − → 1, 3, 4; c) − − − − → k, m, n = − − − → 1, 2, 8 and − − − → p, q, r = − − − → 2, 2, 3. Remark. The equality − − − − → k, m, n = − − − → x, y, z means that the indicated tuples of numbers coincide; the order in which they are written in not taken into account. 6.60. In a regular 30-gon three diagonals are drawn. For them deﬁne tuples − − − − → k, m, n and − − − → p, q, r as in the preceding problem. Prove that if − − − − → k, m, n = − − − − → 1, 3, 14 and − − − → p, q, r = − − − → 2, 2, 8, then the diagonals intersect at one point. 143 6.61. In a regular n-gon (n ≥3) the midpoints of all its sides and the diagonals are marked. What is the greatest number of marked points that lie on one circle? 6.62. The vertices of a regular n-gon are painted several colours so that the points of one colour are the vertices of a regular polygon. Prove that among these polygons there are two equal ones. 6.63. Prove that for n ≥6 a regular (n −1)-gon is impossible to inscribe in a regular n-gon so that on every side of the n-gon except one there lies exactly one vertex of the (n −1)-gon. 6.64. Let O be the center of a regular n-gon A1 . . . An and X an arbitrary point. Prove that − − → OA1 + · · · + − − → OAn = − → 0 and − − → XA1 + · · · + − − → XAn = n− − → XO. 6.65. Prove that it is possible to place real numbers x1, . . . , xn all distinct from zero in the vertices of a regular n-gon so that for any regular k-gon all vertices of which are vertices of the initial n-gon the sum of the numbers at the vertices of the k-gon is equal to zero. 6.66. Point A lies inside regular 10-gon X1 . . . X10 and point B outside it. Let a = − − → AX1 + . . . − − − → AX10 and b = − − → BX1 + . . . − − − → BX10. Is it possible that \|a\| > \|b\|? 6.67. A regular polygon A1 . . . An is inscribed in the circle of radius R centered at O; let X be an arbitrary point. Prove that A1X2 + · · · + AnX2 = n(R2 + d2), where d = OX. 6.68. Find the sum of squares of the lengths of all the sides and diagonals of a regular n-gon inscribed in a circle of radius R. 6.69. Prove that the sum of distances from an arbitrary X to the vertices of a regular n-gon is the least if X is the center of the n-gon. 6.70. A regular n-gon A1 . . . An is inscribed in the circle of radius R centered at O; let ei = − − → OAi and x = − − → OX be an arbitrary vector. Prove that X (ei, x)2 = nR2 · OX2 2 . 6.71. Find the sum of the squared distances from the vertices of a regular n-gon inscribed in a circle of radius R to an arbitrary line that passes through the center of the n-gon. 6.72. The distance from point X to the center of a regular n-gon is equal to d and r is the radius of the inscribed circle of the n-gon. Prove that the sum of squared distances from point X to the lines that contain the sides of the n-gon is equal to n ³ r2 + d2 2 ´ . 6.73. Prove that the sum of squared lengths of the projections of the sides of a regular n-gon to any line is equal to 1 2na2, where a is the length of the side of the n-gon. 6.74. A regular n-gon A1 . . . An is inscribed in a circle of radius R; let X be a point on this circle. Prove that XA4 1 + · · · + XA4 n = 6nR4. 6.75. a) A regular n-gon A1 . . . An is inscribed in the circle of radius 1 centered at 0, let ei = − − → OAi and u an arbitrary vector. Prove that P(u, ei)ei = 1 2nu. b) From an arbitrary point X perpendiculars XA1, . . . , XAn are dropped to the sides (or their extensions) of a regular n-gon. Prove that P − − → XAi = 1 2n− − → XO, where O is the center of the n-gon. 6.76. Prove that if the number n is not a power of a prime, then there exists a convex n-gon with sides of length 1, 2, . . . , n, all the angles of which are equal. 144 CHAPTER 6. POLYGONS See also Problems 2.9, 4.59, 4.62, 6.36, 6.41, 6.45–6.47, 9.83, 9.84, 11.46, 11.48, 17.31, 18.30, 19.47, 23.8, 24.2. §7. The inscribed and circumscribed polygons 6.77. On the sides of a triangle three squares are constructed outwards. What should be the values of the angles of the triangle in order for the six vertices of these squares distinct from the vertices of the triangle belong to one circle? 6.78. A 2n-gon A1 . . . A2n is inscribed in a circle. Let p1, . . . , p2n be the distances from an arbitrary point M on the circle to sides A1A2, A2A3, . . . , A2nA1. Prove that p1p3 . . . p2n−1 = p2p4 . . . p2n. 6.79. An inscribed polygon is divided by nonintersecting diagonals into triangles. Prove that the sum of radii of all the circles inscribed in these triangles does not depend on the partition. 6.80. Two n-gons are inscribed in one circle and the collections of the length of their sides are equal but the corresponding sides are not necessarily equal. Prove that the areas of these polygons are equal. 6.81. Positive numbers a1, . . . , an are such that 2ai < a1 + · · · + an for all i = 1, . . . , n. Prove that there exists an inscribed n-gon the lengths of whose sides are equal to a1, . . . , an. 6.82. A point inside a circumscribed n-gon is connected by segments with all the vertices and tangent points. The triangles formed in this way are alternately painted red and blue. Prove that the product of areas of red triangles is equal to the product of areas of blue triangles. 6.83. In a 2n-gon (n is odd) A1 . . . A2n circumscribed about a circle centered at O the diagonals A1An+1, A2An+2, . . . , An−1A2n−1 pass through point O. Prove that the diagonals AnA2n also passes through point O. 6.84. A circle of radius r is tangent to the sides of a polygon at points A1, . . . , An and the length of the side on which point Ai lies is equal to ai. The distance from point X to the center of the circle is equal to d. Prove that a1XA2 1 + · · · + anXA2 n = P(r2 + d2), where P is the perimeter of the polygon. 6.85. An n-gon A1 . . . An is circumscribed about a circle; l is an arbitrary tangent to the circle that does not pass through any vertex of the n-gon. Let ai be the distance from vertex Ai to line l and bi the distance from the tangent point of side AiAi+1 with the circle to line l. Prove that: a) the value b1...bn a1...an does not depend on the choice of line l; b) the value a1a3...a2m−1 a2a4...a2m does not depend on the choice of line l if n = 2m. 6.86. Certain sides of a convex polygon are red; the other ones are blue. The sum of the lengths of the red sides is smaller than the semiperimeter and there is no pair of neighbouring blue sides. Prove that it is impossible to inscribe this polygon in a circle. See also Problems 2.12, 4.39, 19.6. §8. Arbitrary convex polygons 6.87. What is the greatest number of acute angles that a convex polygon can have? PROBLEMS FOR INDEPENDENT STUDY 145 6.88. How many sides whose length is equal to the length of the longest diagonal can a convex polygon have? 6.89. For which n there exists a convex n-gon one side of which is of length 1 and the lengths of the diagonals are integers? 6.90. Can a convex non-regular pentagon have exactly four sides of equal length and exactly four diagonals of equal lengths? Can the ﬁfth side of such a pentagon have a common point with the ﬁfth diagonal? 6.91. Point O that lies inside a convex polygon forms, together with each two of its vertices, an isosceles triangle. Prove that point O is equidistant from the vertices of this polygon. See also Problems 4.49, 4.50, 9.82, 9.85, 9.86, 11.35, 13.14, 14.26, 16.8, 17.33, 17.34, 19.9, 23.13, 23.15. §9. Pascal’s theorem 6.92. Prove that the intersection points of the opposite sides (if these sides are not parallel) of an inscribed hexagon lie on one line. (Pascal’s theorem.) 6.93. Point M lies on the circumscribed cirlce of triangle ABC; let R be an arbitrary point. Lines AR, BR and CR intersect the circumscribed circle at points A1, B1 and C1, respectively. Prove that the intersection points of lines MA1 and BC, MB1 and CA, MC1 and AB lie on one line and this line passes through point R. 6.94. In triangle ABC, heights AA1 and BB1 and bisectors AA2 and BB2 are drawn; the inscribed circle is tangent to sides BC and AC at points A3 and B3, respectively. Prove that lines A1B1, A2B2 and A3B3 either intersect at one point or are parallel. 6.95. Quadrilateral ABCD is inscribed in circle S; let X be an arbitrary point, M and N be the other intersection points of lines XA and XD with circle S. Lines DC and AX, AB and DX intersect at points E and F, respectively. Prove that the intersection point of lines MN and EF lies on line BC. 6.96. Points A and A1 that lie inside a circle centered at O are symmetric through point O. Rays AP and A1P1 are codirected, rays AQ and A1Q1 are also codirected. Prove that the intersection point of lines P1Q and PQ1 lies on line AA1. (Points P, P1, Q and Q1 lie on the circle.) 6.97. On a circle, ﬁve points are given. With the help of a ruler only construct a sixth point on this circle. 6.98. Points A1, . . . , A6 lie on one circle and points K, L, M and N lie on lines A1A2, A3A4, A1A6 and A4A5, respectively, so that KL ∥A2A3, LM ∥A3A6 and MN ∥A6A5. Prove that NK ∥A5A2. Problems for independent study 6.99. Prove that if ABCD is a rectangle and P is an arbitrary point, then AP 2 +CP 2 = DP 2 + BP 2. 6.100. The diagonals of convex quadrilateral ABCD are perpendicular. On the sides of the quadrilateral, squares centered at P, Q, R and S are constructed outwards. Prove that segment PR passes through the intersection point of diagonals AC and BD so that PR = 1 2(AC + BD). 6.101. On the longest side AC of triangle ABC, points A1 and C1 are taken so that AC1 = AB and CA1 = CB and on sides AB and BC points A2 and C2 are taken so that AA1 = AA2 and CC1 = CC2. Prove that quadrilateral A1A2C2C1 is an inscribed one. 146 CHAPTER 6. POLYGONS 6.102. A convex 7-gon is inscribed in a circle. Prove that if certain three of its angles are equal to 120◦each, then some two of its sides are equal. 6.103. In plane, there are given a regular n-gon A1 . . . An and point P. Prove that from segments A1P, . . . , AnP a closed broken line can be constructed. 6.104. Quadrilateral ABCD is inscribed in circle S1 and circumscribed about circle S2; let K, L, M and N be tangent points of its sides with circle S2. Prove that KM ⊥LN. 6.105. Pentagon ABCDE the lengths of whose sides are integers and AB = CD = 1 is circumscribed about a circle. Find the length of segment BK, where K is the tangent point of side BC with the circle. 6.106. Prove that in a regular 2n-gon A1 . . . A2n the diagonals A1An+2, A2n−1A3 and A2nA5 meet at one point. 6.107. Prove that in a regular 24-gon A1 . . . A24 diagonals A1A7, A3A11 and A5A21 intersect at a point that lies on diameter A4A16. Solutions 6.1. Let O be the center of the inscribed circle and the intersection point of the diagonals of quadrilateral ABCD. Then ∠ACB = ∠ACD and ∠BAC = ∠CAD. Hence, triangles ABC and ADC are equal, since they have a common side AC. Therefore, AB = DA. Similarly, AB = BC = CD = DA. 6.2. Clearly, ∠AOB = 180◦−∠BAO −∠ABO = 180◦−∠A + ∠B 2 and ∠COD = 180◦−∠C+∠D 2 . Hence, ∠AOB + ∠COD = 360◦−∠A + ∠B + ∠C + ∠D 2 = 180◦. 6.3. Let us consider two circles tangent to the sides of the given quadrilateral and their extensions. The lines that contain the sides of the quadrilateral are the common inner and outer tangents to these circles. The line that connects the midpoints of the circles contains a diagonal of the quadrilateral and besides it is an axis of symmetry of the quadrilateral. Hence, the other diagonal is perpendicular to this line. 6.4. Let O be the center of the given circle, R its radius, a the length of chords singled out by the circle on the sides of the quadrilateral. Then the distances from point O to the sides of the quadrilateral are equal to q R2 −a2 4 , i.e., point O is equidistant from the sides of the quadrilateral and is the center of the inscribed circle. 6.5. For a parallelogram the statement of the problem is obvious therefore, we can assume that lines AB and CD intersect. Let O be the center of the inscribed circle of quadrilateral ABCD; let M and N be the midpoints of diagonals AC and BD. Then SANB + SCND = SAMB + SCMD = SAOB + SCOD = SABCD 2 . It remains to make use of the result of Problem 7.2. 6.6. Let the inscribed circle be tangent to sides DA, AB and BC at points M, H and N, respectively. Then OH is a height of triangle AOB and the symmetries through lines AO and BO sends point H into points M and N, respectively. Hence, by Problem 1.57 points A1 and B1 lie on line MN. Similarly, points C1 and D1 lie on line MN. 6.7. Let r be the distance from the intersection point of bisectors of angles A and D to the base AD, let r′ be the distance from the intersection point of bisectors of angles B and SOLUTIONS 147 C to base BC. Then AD = r(cot α + cot β) and BC = r′(tan α + tan β). Hence, r = r′ if and only if BC AD = tan α + tan β cot α cot β = tan α · tan β. 6.8. Let ∠A = 2α, ∠C = 2β and ∠BMA = 2ϕ. By Problem 6.7, PK RL = tan α tan ϕ and LS MC = cot ϕ tan β. Since PQ RS = PK RL and RS AC = LS MC, it follows that PQ AC = PK RL LS MC = tan α tan β. Hence, trapezoid APQC is a circumscribed one. 6.9. First, let us prove that if quadrilateral ABCD is a circumscribed one, then all the conditions take place. Let K, L, M and N be the tangent points of the inscribed circle with sides AB, BC, CD and DA. Then AB + CD = AK + BK + CM + DM = AN + BL + CL + DN = BC + AD, AP + CQ = AK + PK + QL −CL = AN + PM + QN −CM = AQ + CP, BP + BQ = AP −AB + BC + CQ = (AP + CQ) + (BC −AB) = AQ + CP + CD −AD = DP + DQ. Now, let us prove, for instance, that if BP +BQ = DP +DQ, then quadrilateral ABCD is a circumscribed one. For this let us consider the circle tangent to side BC and rays BA and CD. Assume that line AD is not tangent to this circle; let us shift this line in order for it to touch the circle (Fig. 65). Figure 65 (Sol. 6.9) Let S be a point on line AQ such that Q′S ∥DD′. Since BP + BQ = DP + DQ and BP + BQ′ = D′P + D′Q′, it follows that QS + SQ′ = QQ′. Contradiction. In the other two cases the proof is similar. 6.10. Let rays AB and DC intersect at point P, let rays BC and AD intersect at point Q; let the given lines passing through points P and Q intersect at point O. By Problem 6.9 we have BP + BQ = OP + OQ and OP + OQ = DP + DQ. Hence, BP + BQ = DP + DQ and, therefore, quadrilateral ABCD is a circumscribed one. 6.11. Let sides AB, BC, CD and DA of quadrilateral ABCD be tangent of the inscribed circle at points E, F, G and H, respectively. First, let us show that lines FH, EG and AC intersect at one point. Denote the points at which lines FH and EG intersect line AC by M and M ′, respectively. Since ∠AMH = ∠BFM as angles between the tangents and chord HF, it follows that sin ∠AHM = sin ∠CFM. Hence, AM · MH FM · MC = SAMH SFMC = AH · MH FC · FM , 148 CHAPTER 6. POLYGONS i.e., AM MC = AH FC . Similarly, AM ′ M ′C = AE CG = AH FC = AM MC ; hence, M = M ′, i.e., lines FH, EG and AC intersect at one point. Similar arguments show that lines FH, EG and BD intersect at one point and therefore, lines AC, BD, FH and EG intersect at one point. 6.12. Segments CHd and DHc are parallel because they are perpendicular to line BC. Moreover, since ∠BCA = ∠BDA = ϕ, the lengths of these segments are equal to AB\| cot ϕ\|, cf. Problem 5.45 b). 6.13. Let Oa, Ob, Oc and Od be the centers of the inscribed circles of triangles BCD, ACD, ABD and ABC, respectively. Since ∠ADB = ∠ACB, it follows that ∠AOcB = 90◦+ ∠ADB 2 = 90◦+ ∠ACB 2 = ∠AOdB, cf. Problem 5.3. Therefore, quadrilateral ABOdOc is an inscribed one, i.e., ∠OcOdB = 180◦−∠OcAB = 180◦−∠A 2 . Similarly, ∠OaOdB = 180◦−∠C 2 . Since ∠A + ∠C = 180◦, it follows that ∠OcOdB + ∠OaObB = 270◦and, therefore, ∠OaOdOc = 90◦. We similarly prove that the remaining angles of quadrilateral OaObOcOd are equal to 90◦. 6.14. a) Let rays AB and DC intersect at point P and rays BC and AD intersect at point Q. Let us prove that point M at which the circumscribed circles of triangles CBP and CDQ intersect lies on segment PQ. Indeed, ∠CMP + ∠CMQ = ∠ABC + ∠ADC = 180◦. Hence, PM + QM = PQ and since PM · PQ = PD · PC = p2 −R2 and QM · PQ = QD · QA = q2 −R2, it follows that PQ2 = PM ·PQ+QM ·PQ = p2 +q2 −2R2. Let N be the intersection point of the circumscribed circles of triangles ACP and ABS. Let us prove that point S lies on segment PN. Indeed, ∠ANP = ∠ACP = 180◦−∠ACD = 180◦−∠ABD = ∠ANS. Hence, PN −SN = PS and since PN · PS = PA · PB = p2 −R2 and SN · PS = SA · SC = R2 −s2, it follows that PS2 = PN · PS −SN · PS = p2 + s2 −2R2. Similarly, QS2 = q2 + s2 −2R2. b) By heading a) PQ2 −PS2 = q2 −s2 = OQ2 −OS2. Hence, OP ⊥QS, cf. Problem 7.6. We similarly prove that OQ ⊥PS and OS ⊥PQ. 6.15. Let the inscribed circles of triangles ABC and ACD be tangent to diagonal AC at points M and N, respectively. Then AM = AC + AB −BC 2 andquadAN = AC + AD −CD 2 , cf. Problem 3.2. Points M and N coincide if and only if AM = AN, i.e., AB + CD = BC + AD. Thus, if points M and N coincide, then quadrilateral ABCD is a circumscribed SOLUTIONS 149 one and similar arguments show that the tangent points of the inscribed circles of triangles ABD and BCD with the diagonal BD coincide. Let the inscribed circle of triangle ABC be tangent to sides AB, BC and CA at points P, Q and M, respectively and the inscribed circle of triangle ACD be tangent to sides AC, CD and DA at points M, R and S, respectively. Since AP = AM = AS and CQ = CM = CR, it follows that triangles APS, BPQ, CQR and DRS are isosceles ones; let α, β, γ and δ be the angles at the bases of these isosceles triangles. The sum of the angles of these triangles is equal to 2(α + β + γ + δ) + ∠A + ∠B + ∠C + ∠D; hence, α + β + γ + δ = 180◦. Therefore, ∠SPQ + ∠SRQ = 360◦−(α + β + γ + δ) = 180◦, i.e., quadrilateral PQRS is an inscribed one. 6.16. Let O be the intersection point of diagonals AC and BD; let A1, B1, C1 and D1 be the projections of O to sides AB, BC, CD and DA, respectively. Points A1 and D1 lie on the circle with diameter AO, hence, ∠OA1D1 = ∠OAD1. Similarly, ∠OA1B1 = ∠OBB1. Since ∠CAD = ∠CBD, we have: ∠OA1D1 = ∠OA1B1. We similarly prove that B1O, C1O and D1O are the bisectors of the angles of quadrilateral A1B1C1D1, i.e., O is the center of its inscribed circle. Figure 66 (Sol. 6.17) 6.17. Let us make use of the notations on Fig. 66. The condition that quadrilateral A1B1C1D1 is an inscribed one is equivalent to the fact that (α+β)+(γ+δ) = 180◦and the the fact that AC and BD are perpendicular is equivalent to the fact that (α1 +δ1)+(β1 +γ1) = 180◦. It is also clear that α = α1, β = β1, γ = γ1 and δ = δ1. 6.18. By the law of cosines AD2 = AC2 + CD2 −2AC · CD cos ACD, AC2 = AB2 + BC2 −2AB · BC cos B. Since the length of the projection of segment AC to line l perpendicular to CD is equal to the sum of the lengths of projections of segments AB and BC to line l, AC cos ACD = AB cos ϕ + BC cos C. 6.19. Let ∠AOD = 2α; then the distances from point O to the projections of the midpoints of diagonals AC and BD to the bisector of angle ∠AOD are equal to OA+OC 2 cos α and OB+OD 2 cos α, respectively. Since OA + OC = AB + OB + OC = CD + OB + OC = OB + OD, these projections coincide. 150 CHAPTER 6. POLYGONS 6.20. Let us complement triangles ABM and DCM to parallelograms ABMM1 and DCMM2. Since AM1 : DM2 = BM : MC = AN : DN, it follows that △ANM1 ∼ △DNM2. Hence, point N lies on segment M1M2 and MM1 : MM2 = AB : CD = AN : ND = M1N : M2N, i.e., MN is the bisector of angle M1MM2. 6.21. Let a, b, c and d be (the lengths of) the bisectors of the angles at vertices A, B, C and D. We have to verify that ∠(a, b) + ∠(c, d) = 0◦. Clearly, ∠(a, b) = ∠(a, AB) + ∠(AB, b) and ∠(c, d) = ∠(c, CD) + ∠(CD, d). Since quadrilateral ABCD is a convex one and ∠(a, AB) = ∠(AD, AB) 2 , ∠(AB, b) = ∠(AB, BC) 2 , ∠(c, CD) = ∠(CB, CD) 2 , ∠(CD, d) = ∠(CD, DA) 2 , it follows that ∠(a, b) + ∠(c, d) = ∠(AD, AB) + ∠(AB, BC) + ∠(CB, CD) + ∠(CD, DA) 2 = 360◦ 2 = 0◦ (see Background to Chapter 2). 6.22. Let, for deﬁniteness, AB > A1B1. The parallel translation by vector − − → CB sends triangle SD1C1 to S′D′ 1C′ 1 and segment CD to BA. Since QA1 : QA = A1B1 : AB = S′D′ 1 : S′A, we see that QS′ ∥A1D′ 1. Hence, QS ∥AD. Similarly, PR ∥AB. 6.23. Suppose that lines AD and BC are not parallel. Let M2, K, N2 be the midpoints of sides AB, BC, CD, respectively. If MN ∥BC, then BC ∥AD, because AM = MC and BN = ND. Therefore, let us assume that lines MN and BC are not parallel, i.e., M1 ̸= M2 and N1 ̸= N2. Clearly, − − − → M2M = 1 2 − − → BC = − − → NN2 and − − − → M1M = − − → NN1. Hence, M1M2 ∥N1N2. Therefore, KM ∥AB ∥CD ∥KN, i.e., M = N. Contradiction. 6.24. By a similarity transformation we can identify one pair of the corresponding sides of quadrilaterals, therefore, it suﬃces to consider quadrilaterals ABCD and ABC1D1 whose points C1 and D1 lie on rays BC and AD and such that CD ∥C1D1. Denote the intersection points of diagonals of quadrilaterals ABCD and ABC1D1 by O and O1, respectively. Suppose that points C and D lie closer to points B and A, then points C1 and D1, respectively. Let us prove then that ∠AOB > ∠AO1B. Indeed, ∠C1BA > ∠CAB and ∠D1BA > ∠DBA, hence, ∠AO1B = 180◦−∠C1AB −∠D1BA < 180◦−∠CAB −∠DBA = ∠AOB. We have obtained a contradiction and, therefore, C1 = C, D1 = D. 6.25. Any quadrilateral is determined up to similarity by the directions of its sides and diagonals. Therefore, it suﬃces to construct one example of a quadrilateral A1B1C1D1 with the required directions of sides and diagonals. Let O be the intersection point of diagonals AC and BD. On ray OA, take an arbitrary point D1 and draw D1A1 ∥BC, A1B1 ∥CD and B1C1 ∥DA (Fig. 67). SOLUTIONS 151 Figure 67 (Sol. 6.25) Since OC1 : OB1 = OD : OA, OB1 : OA1 = OC : OD and OA1 : OD1 = OB : OC, it follows that OC1 : OD1 = OB : OA, consequently, C1D1 ∥AB. The obtained plot shows that ∠A + ∠C1 = 180◦. 6.26. Let O be the intersection point of the diagonals of quadrilateral ABCD. Without loss of generality we may assume that α = ∠AOB < 90◦. Let us drop perpendiculars AA1, BB1, CC1, DD1 to the diagonals of quadrilateral ABCD. Since OA1 = OA cos α, OB1 = OB cos α, OC1 = OC cos α, OD1 = OD cos α, it follows that the symmetry through the bisector of angle AOB sends quadrilateral ABCD into a quadrilateral homothetic to quadrilateral A1B1C1D1 with coeﬃcient − − → BC1cos α. 6.27. Let the diagonals of quadrilateral ABCD intersect at point O; let Ha and Hb be the orthocentres of triangles AOB and COD; let Ka and Kb be the midpoints of sides BC and AD; let P be the midpoint of diagonal AC. The intersection point of medians of triangles AOD and BOC divide segments KaO and KbO in the ratio of 1 : 2 and, therefore, we have to prove that HaHb ⊥KaKb. Since OHa = AB\| cot ϕ\| and OHb = CD\| cot ϕ\|, where ϕ = ∠AOB, cf. Problem 5.45 b), then OHa : OHb = PKa : PKb. The correspondiong legs of angles ∠HaOHb and ∠KaPKb are perpendicular; moreover, vectors − − → OHa and − − → OHb are directed towards lines AB and CD for ϕ < 90◦and away from these lines for ϕ > 90◦. Hence, ∠HaOHb = ∠KaPKb and △HaOHb ∼△KaPKb. It follows that HaHb ⊥KaKb. 6.28. Let S = SAOD, x = AO, y = DO, a = AB, b = BC, c = CD, d = DA and k the similarity coeﬃcient of triangles BOC and AOD. Then 2 µ 1 r1 + 1 r3 ¶ = d + x + y S + kd + kx + ky k2S , 2 µ 1 r2 + 1 r4 ¶ = a + x + ky kS + c + kx + y kS because SBOC = k2S and SAOB = SCOD = kS. Since x + y S + x + y k2S = x + ky kS + kx + y kS , it remains to notice that a + c = b + d = kd + d. 6.29. It is easy to verify that AB = r1 µ cot A 2 + cot B 2 ¶ and CD = r3 µ cot C 2 + cot D 2 ¶ . Hence, AB r1 + CD r3 = cot A 2 + cot B 2 + cot C 2 + cot D 2 = BC r2 + AD r4 . 152 CHAPTER 6. POLYGONS 6.30. Let us complete triangles ABD and DBC to parallelograms ABDA1 and DBCC1. The segments that connect point D with the vertices of parallelogram ACC1A1 divide it into four triangles equal to triangles DAB, CDA, BCD and ABC and, therefore, the radii of the inscribed circles of these triangles are equal. Let us prove that point D coincides with the intersection point O of the diagonals of the parallelogram. If D ̸= 0, then we may assume that point D lies inside triangle AOC. Then rADC < rAOC = rA1OC1 < rA1DC1 = rABC, cf. Problem 10.86. We have obtained a contradiction, hence, D = O. Since p = − − → BCSr and the areas and radii of the inscribed circles of triangles into which the diagonals divide the parallelogram ACC1A1 are equal, the triangles’ perimeters are equal. Hence, ACC1A1 is a rhombus and ABCD is a rectangular. 6.31. Points C1 and D1 lie on the midperpendicular to segment AB, hence, AB ⊥C1D1. Similarly, C1D1 ⊥A2B2 and, therefore, AB ∥A2B2. We similarly prove that the remaining corresponding sides and the diagonals of quadrilaterals ABCD and A2B2C2D2 are parallel. Therefore, these quadrilaterals are similar. Let M be the midpoint of segment AC. Then B1M = \|AM cot D\| and D1M = \|AM cot B\|, where B1D1 = \| cot B + cot D\| · 1 2AC. Let us rotate quadrilateral A1B1C1D1 by 90◦. Then making use of the result of Problem 6.25 we see that this quadrilateral is a convex one and cot A = −cot C1, etc. Therefore, A2C2 = \| cot A + cot C\| · 1 2B1D1 = 1 4\|(cot A + cot C)(cot B + cot D)\| · AC. 6.32. Let M and N be the midpoints of sides AB and CD, respectively. Let us drop from point D perpendicular DP to line MN and from point M perpendicular MQ to line CD. Then Q is the tangent point of line CD and a circle with diameter AB. Right triangles PDN and OMN are similar, hence, DP = ND · MQ MN = ND · MA MN . Similarly, the distance from point A to line MN is equal to − − → BCND · MAMN. Therefore, AD ∥MN. Similarly, BC ∥MN. 6.33. It suﬃces to verify that the orthocentres of any three of the four given triangles lie on one line. Let a certain line intersect lines B1C1, C1A1 and A1B1 at points A, B and C, respectively; let A2, B2 and C2 be the orthocentres of triangles A1BC, AB1C and ABC1, respectively. Lines AB2 and A2B are perpendicular to line A1B1 and, therefore, they are parallel. Similarly, BC2 ∥B2C and CA2 ∥C2A. Points A, B and C lie on one line and, therefore, points A2, B2 and C2 also lie on one line, cf. Problem 1.12 b). 6.34. On diagonal BD, take point M so that ∠MCD = ∠BCA. Then △ABC ∼ △DMC, because angles ∠BAC and ∠BDC subtend the same arc. Hence, AB · CD = AC ·MD. Since ∠MCD = ∠BCA, then ∠BCM = ∠ACD and △BCM ∼△ACD because angles ∠CBD and ∠CAD subtend one arc. Hence, BC · AD = AC · BM. It follows that AB · CD + AD · BC = AC · MD + AC · BM = AC · BD. 6.35. Let S be the area of quadrilateral ABCD, let R be the radius of its circumscribed circle. Then S = SABC + SADC = AC(AB · BC + AD · DC) 4R , cf. Problem 12.1. Similarly, S = BD(AB · AD + BC · CD) 4R . SOLUTIONS 153 By equating these equations for S we get the desired statement. 6.36. Let regular hexagon A1 . . . A7 be inscribed in a circle. By applying Ptolemey’s theorem to qudrilateral A1A3A4A5 we get A1A3 · A5A4 + A3A4 · A1A5 = A1A4 · A3A5, i.e., sin 2α sin α + sin α sin 3α = sin 3α sin 2α. 6.37. Let A1, B1 and C1 be the midpoints of sides BC, CA and AB, respectively. By Ptolemy’s theorem AC1 · OB1 + AB1 · OC1 = AO · B1C1, where O is the center of the circumscribed circle. Hence, cdb+bdc = aR. Similarly, adc+cda = bR and adb + bda = cR. Moreover, ada + bdb + cdc = 2S = (a + b + c)r. By adding all these equalities and dividing by a + b + c we get the desired statement. 6.38. By Ptolemy’s theorem AB · CD + AC · BD = AD · BC. Taking into account that CD = BD ≥1 2BC we get the desired statement. 6.39. By applying Ptolemy’s theorem to quadrilateral ABCP and dividing by the lengths of the square’s side we get the desired statement. 6.40. By applying Ptolemy’s theorem to quadrilateral APQR we get AP · RQ + AR · QP = AQ · PR. Since ∠ACB = ∠RAQ = ∠RPQ and ∠RQP = 180◦−∠PAR = ∠ABC, it follows that △RQP ∼△ABC and, therefore, RQ : QP : PR = AB : BC : CA. It remains to notice that BC = AD. 6.41. a) Let us express Ptolemy’s theorem for all quadrilaterals with vertices at point A and three consecutive vertices of the given polygon; then let us group in the obtained equalities the factors in which di with even indices enter in the right-hand side. By adding these equalities we get (2a + b)(d1 + · · · + d2n+1) = (2a + b)(d2 + · · · + d2n), where a is the side of the given polygon and b is its shortest diagonal. b) Let R be the radius of circle S. Then li = di q R±r R , cf. Problem 3.20. It remains to make use of the result of heading a). Figure 68 (Sol. 6.42) 6.42. Let both tangent be exterior ones and x ≤y. The line that passes through the center O of the circle of radius x parallel to the segment that connects the tangent points intersects the circle of radius y −x (centered in the center of the circle of radius y) at points A and B (Fig. 68). 154 CHAPTER 6. POLYGONS Then OA = a(R+x) R and OB = OA + a(y −x) R = a(R + y) R . The square of the length to be found of the common outer tangent is equal to OA · OB = ³ a R ´2 (R + x)(R + y). Similar arguments show that if both tangent are inner ones, then the square of the lengths of the outer tangent is equal to ¡ a R ¢2 (R −x)(R −y) and if the circle of radius x is tangent from the outside and the circle of radius y from the inside, then the square of the length of the inner tangent is equal to ¡ a R ¢2 (R −y)(R + x). Remark. In the case of an inner tangency of the circles we assume that R > x and R > y. 6.43. Let R be the radius of the circumscribed circle of quadrilateral ABCD; let ra, rb, rc and rd be the radii of circles α, β, γ and δ, respectively. Further, let a = √R ± ra, where the plus sign is taken if the tangent is an outer one and the minus sign if it is an inner one; numbers b, c and d are similarly deﬁned. Then tαβ = ab·AB R , cf. Problem 6.42, etc. Therefore, by multiplying the equality AB · CD + BC · DA = AC · BD by abcd R we get the desired statement. 6.44. Since ∠EBD = ∠ABE + ∠CBD, it is possible to take a point P on side ED so that ∠EBP = ∠ABE = ∠AEB, i.e., BP ∥AE. Then ∠PBD = ∠EBD −∠EBP = ∠CBD = ∠BDC, i.e., BP ∥CD. Therefore, AE ∥CD and since AE = CD, CDEA is a parallelogram. Hence, AC = ED, i.e., triangle ABC is an equilateral one and ∠ABC = 60◦. 6.45. a) Let O be the center of the circumscribed circle of triangle CKE. It suﬃces to verify that ∠COK = 2∠KCB. It is easy to calculate both these angles: ∠COK = 180◦−2∠OKC = 180◦−∠EKC = 180◦−∠EDC = 72◦ and ∠KCB = 180◦−∠ABC 2 = 36◦. b) Since BC is a tangent to the circumscribed circle of triangle CKE, then BE · BK = BC2, i.e., d(d −a) = a2. 6.46. Let the perpendiculars erected to line AB at points A and B intersect sides DE and CD at points P and Q, respectively. Any point of segment CQ is a vertex of a rectangle inscribed in pentagon ABCDE (the respective sides of this pentagon are parallel to AB and AP); as this point moves from Q to C the ratio of the lengths of the sides of the rectangles varies from AP AB to 0. Since angle ∠AEP is an obtuse one, AP > AE = AB. Therefore, for a certain point of segment QC the ratio of the lengths of the sides of the rectangle is equal to 1. 6.47. Let points A1, . . . , E1 be symmetric to points A, . . . , E through the center of circle S; let P, Q and R be the intersection points of lines BC1 and AB1, AE1 and BA1, BA1 and CB1, see Fig. 69. Then PQ = AB = a and QR = b. Since PQ ∥AB and ∠ABA1 = 90◦, it follows that PR2 = PQ2 + QR2 = a2 + b2. Line PR passes through the center of circle S and ∠AB1C = 4 · 18◦= 72◦, hence, PR is a side of a regular pentagon circumscribed about the circle with center B1 whose radius B1O is equal to the radius of circle S. SOLUTIONS 155 Figure 69 (Sol. 6.47) Figure 70 (Sol. 6.48) 6.48. Through points A, C and E draw lines l1, l2 and l3 parallel to lines BC, DE and FA, respectively. Denote the intersection points of lines l1 and l2, l2 and l3, l3 and l1 by P, Q, R, see Fig. 70. Then SACE = SABCDEF −SPQR 2 + SPQR = SABCDEF + SPQR 2 ≥SABCDEF 2 . Similarly, SBDF = 1 2(SABCDEF + SP ′Q′R′). Clearly, PQ = \|AB −DE\|, QR = \|CD −AF\|, PR = \|EF −BC\|, hence, triangles PQR and P ′Q′R′ are equal. Therefore, SACE = SBDF. 6.49. Let us construct triangle PQR as in the preceding problem. This triangle is an equilateral one and PQ = \|AB −DE\|, QR = \|CD −AF\|, RP = \|EF −BC\|. Hence, \|AB −DE\| = \|CD −AF\| = \|EF −BC\|. 6.50. The sum of the angles at vertices A, C and E is equal to 360◦, hence, from isosceles triangles ABF, CBD and EDF we can construct a triangle by juxtaposing AB to CB, ED to CD and EF to AF. The sides of the obtained triangle are equal to the respective sides of triangle BDF. Therefore, the symmetry through lines FB, BD and DF sends points A, C and E, respectively, into the center O of the circumscribed circle of triangle BDF, and, therefore, AB ∥OF ∥DE. 6.51. Let us suppose that the diagonals of the hexagon form triangle PQR. Denote the vertices of the hexagon as follows: vertex A lies on ray QP, vertex B on RP, vertex C on RQ, etc. Since lines AD and BE divide the area of the hexagon in halves, then SAPEF + SPED = SPDCB + SABP and SAPEF + SABP = SPDCB + SPED. 156 CHAPTER 6. POLYGONS Hence, SABP = SPED, i.e., AP · BP = EP · DP = (ER + RP)(DQ + QP) > ER · DQ. Similarly, CQ · DQ > AP · FR and FR · ER > BP · CQ. By multiplying these inequalities we get AP · BP · CQ · DQ · FR · ER > ER · DQ · AP · FR · BP · CQ which is impossible. Hence, the diagonals of the hexagon intersect at one point. Figure 71 (Sol. 6.52) 6.52. Denote the midpoints of the sides of convex hexagon ABCDEF as plotted on Fig. 71. Let O be the intersection point of segments KM and LN. Let us denote the areas of triangles into which the segments that connect point O with the vertices and the midpoints of the sides divide the hexagon as indicated on the same ﬁgure. It is easy to verify that SKONF = SLOMC, i.e., a + f = c + d. Therefore, the broken line POQ divides the hexagon into two parts of equal area; hence, segment PQ passes through point O. 6.53. a) Let O be the center of the circumscribed circle. Since ∠AkOAk+2 = 360◦−2∠AkAk+1Ak+2 = ϕ is a constant, the rotation through an angle of ϕ with center O sends point Ak into Ak+2. For n odd this implies that all the sides of polygon A1 . . . An are equal. b) Let a be the length of the side of the given polygon. If one of its sides is divided by the tangent point with the inscribed circle into segments of length x and a −x, then its neighbouring sides are also divided into segments of length x and a −x (the neighbouring segments of neighbouring sides are equal), etc. For n odd this implies that all the sides of polygon A1 . . . An are divided by the tangent points with the inscribed circle in halves; therefore, all the angles of the polygon are equal. 6.54. The sides of polygon A1 . . . An are parallel to respective sides of a regular n-gon. On rays OA1, . . . , OAn mark equal segments OB1, . . . , OBn. Then polygon B1 . . . Bn is a regular one and the sides of polygon A1 . . . An form equal angles with the respective sides of polygon B1 . . . Bn. Therefore, OA1 : OA2 = OA2 : OA3 = · · · = OAn : OA1 = k, i.e., OA1 = kOA2 = k2OA3 = · · · = knOA1; thus, k = 1. 6.55. Denote the vertices of the pentagon as indicated on Fig. 72. Notice that if in a triangle two heights are equal, then the sides on which these heights are dropped are also equal. SOLUTIONS 157 Figure 72 (Sol. 6.55) From consideration of triangles EAB, ABC and BCD we deduce that EA = AB, AB = BC and BC = CD. Therefore, trapezoids EABC and ABCD are isosceles ones, i.e., ∠A = ∠B = ∠C. By considering triangles ABD and BCE we get AD = BD and BE = CE. Since triangles EAB, ABC, BCD are equal, it follows that BE = AC = BD. Hence, AD = BE and BD = CE, i.e., trapezoids ABDE and CDEB are isosceles ones. Therefore, ED = AB = BC = CD = AE and ∠E = ∠A = ∠B = ∠C = ∠D, i.e., ABCDE is a regular pentagon. 6.56. Triangles BAM and BCN are isosceles ones with angle 15◦at the base, cf. Problem 2.26, and, therefore, triangle BMN is an equilateral one. Let O be the centre of the square, P and Q the midpoints of segments MN and BK (Fig. 73). Since OQ is the midline of triangle MBK, it follows that OQ = 1 2BM = MP = OP and ∠QON = ∠MBA = 15◦. Therefore, ∠POQ = ∠PON −∠QON = 30◦. The remaining part of the proof is carried out similarly. Figure 73 (Sol. 6.56) 6.57. Let us consider a regular 12-gon A1 . . . A12 inscribed in a circle of radius R. Clearly, A1A7 = 2R, A1A3 = A1A11 = R. Hence, A1A7 = A1A3 + A1A11. 6.58. For k = 3 the solution of the problem is clear from Fig. 74. Indeed, A3A4 = OQ, KL = QP and MN = PA14 and, therefore, A3A4 + KL + MN = OQ + QP + PA14 = OA14 = R. Proof is carried out in a similar way for any k. 6.59. In the proof if suﬃces to apply the result of Problems 5.78 and 5.70 b) to triangle AaAcAe and lines AaAd, AcAf and AeAb. Solvling heading b) we have to notice additionally that sin 20◦sin 70◦= sin 20◦cos 20◦= sin 40◦ 2 = sin 30◦sin 40◦ and in the solution of heading c) that sin 10◦sin 80◦= sin 30◦sin 20◦. 158 CHAPTER 6. POLYGONS Figure 74 (Sol. 6.58) 6.60. As in the preceding problem we have to verify the equality sin 2α sin 2α sin 8α = sin α sin 3α sin 14α, where α = 180◦ 30 = 6◦. Clearly, sin 14α = cos α, hence, 2 sin α sin 3α sin 14α = sin 2α sin 3α. It remains to verify that sin 3α = 2 sin 2α sin 8α = cos 6α −cos 10α = 1 −2 sin2 3α −1 2, i.e., 4 sin2 18◦+ 2 sin 18◦= 1, cf. Problem 5.46. 6.61. First, let n = 2m. The diagonals and sides of a regular 2m-gon have m distinct lengths. Therefore, the marked points lie on m −1 concentric circles (having n points each) or in the common center of these circles. Since distinct circles have not more than two common points, the circle that does not belong to this family of concentric circles contains not more than 1 + 2(m −1) = 2m −1 = n −1 of marked points. Now, let n = 2m + 1. There are m distinct lengths among the lengths of the diagonals and sides of a regular (2m + 1)-gon. Hence, the marked points lie on m concentric circles (n points on each). A circle that does not belong to this family of concentric circles contains not more than 2m = n −1 marked points. In either case the greatest number of marked points that lie on one circle is equal to n. 6.62. Denote the center of the polygon by O and the vertices of the polygon by A1, . . . , An. Suppose that there are no equal polygons among the polygons of the same colour, i.e., they have m = m1 < m2 < m3 < · · · < mk sides, respectively. Let us consider a transformation f deﬁned on the set of vertices of the n-gon as the one that sends vertex Ak to vertex Amk : f(Ak) = Amk (we assume that Ap+qn = Ap). This transformation sends the vertices of a regular m-gon into one point, B, hence, the sum of vectors − − − − → Of(Ai), where Ai are the vertices of an m-gon, is equal to m− − → OB ̸= − → 0 . Since ∠AmiOAmj = m∠AiOAj, the vertices of any regular polygon with the number of sides greater than m pass under the considered transformation into the vertices of a regular polygon. Therefore, the sum of vectors − − − − → Of(Ai) over all vertices of an n-gon and similar sums over the vertices of m2-, m3-, . . . , mk-gons are equal to zero. We have obtained a contradiction with the fact that the sum of vectors − − − − → Of(Ai) over the vertices of an m-gon is not equal to zero. Therefore, among the polygons of one color there are two equal ones. 6.63. Let a regular (n −1)-gon B1 . . . Bn−1 be inscribed into a regular n-gon A1 . . . An. We may assume that A1 and B1 are the least distant from each other vertices of these polygons and points B2, B3, B4 and B5 lie on sides A2A3, A3A4, A4A5 and A5A6. Let αi = ∠Ai+1BiBi+1 and βi = ∠BiBi+1Ai+1, where i = 1, 2, 3, 4. By the sine theorem SOLUTIONS 159 A2B2 : B1B2 = sin α1 : sin ϕ and B2A3 : B2B3 = sin β2 : sin ϕ, where ϕ is the angle at a vertex of a regular n-gon. Therefore, sin α1 + sin β2 = an sin ϕ an−1 , where an and an−1 are the (lengths of the) sides of the given polygons. Similar arguments show that sin α1 + sin β2 = sin α2 + sin β3 = sin α3 + sin β4. Now, observe that sin αi + sin βi+1 = 2 sin αi + βi+1 2 cos αi −βi+1 2 and compute αi + βi+1 and αi −βi+1. Since αi + βi = 2π n and αi+1 + βi = 2π n−1, it follows that αi+1 = αi + 2π n(n−1) and βi+1 = βi − 2π n(n−1); therefore, αi + βi+1 = 2π n − 2π n(n −1) is a constant and αi −βi+1 = αi−1 −βi + 4π n(n −1). Hence, cos θ = cos µ θ + 2π n(n −1) ¶ = cos µ θ + 4π (n −1)n ¶ for θ = α1 −β2 2 . We have obtained a contradiction because on an interval shorter than 2π the cosine cannot attain the same value at three distinct points. Remark. A square can be inscribed in a regular pentagon, cf. Problem 6.46. 6.64. Let a = − − → OA1 + · · · + − − → OAn. A rotation about point O by 360◦ n sends point Ai to Ai+1 and, therefore, sends vector a into itself, i.e., a = 0. Since − − → XAi = − − → XO + − − → OAi and − − → OA1 + · · · + − − → OAn = − → 0 , it follows that − − → XA1 + · · · + − − → XAn = n− − → XO. 6.65. Through the center of a regular polygon A1 . . . An, draw line l that does not pass through the vertices of the polygon. Let xi be equal to the length of the projection of vector − − → OAi to a line perpendicular to l. Then all the xi are nonzero and the sum of numbers xi assigned to the vertices of a regular k-gon is equal to zero since the corresponding sum of vectors − − → OAi vanishes, cf. Problem 6.64. 6.66. By Problem 6.64 a = 10− → AO and b = 10− − → BO, where O is the center of polygon X1 . . . X10. Clearly, if point A is situated rather close to a vertex of the polygon and point B rather close to the midpoint of a side, then AO > BO. 6.67. Since AiX2 = \|− − → AiO + − − → OX\|2 = AiO2 + OX2 + 2(− − → AiO, − − → OX) = R2 + d2 + 2(− − → AiO, − − → OX), it follows that X AiX2 = n(R2 + d2) + 2( X − − → AiO, − − → OX) = n(R2 + d2), cf. Problem 6.64. 160 CHAPTER 6. POLYGONS 6.68. Denote by Sk the sum of squared distances from vertex Ak to all the other vertices. Then Sk = AkA2 1 + AkA2 2 + · · · + AkA2 n = AkO2 + 2(− − → AkO, − − → OA1) + A1O2 + . . . + AkO2 + 2(− − → AkO, − − → OAn) + AnO2 = 2nR2 because Pn i=1 − − → OAi = − → 0 . Hence, Pn i=1 Sk = 2n2R2. Since each squared side and diagonal enters this sum twice, the sum to be found is equal to n2R2. 6.69. Consider the rotation of the given n-gon about the n-gon’s center O that sends Ak to A1. Let Xk be the image of point X under the rotation. This rotation sends segment AkX to A1Xk. Therefore, A1X + · · · + AnX = A1X1 + · · · + A1Xn. Since n-gon X1 . . . Xn is a regular one, − − − → A1X1 + · · · + − − − → A1Xn = n− − → A1O, cf. Problem 6.64. Therefore, A1X1 + · · · + AnXn ≥n− − → A1O. 6.70. Let Bi be the projection of point X to line OAi. Then (ei, x) = (− − → OAi, − − → OBi + − − → BiX) = (− − → OAi, − − → OBi) = ±R · OBi. Points B1, . . . , Bn lie on the circle with diameter OX and are vertices of a regular n-gon for n odd and vertices of an n 2-gon counted twice for n even, cf. Problem 2.9. Therefore, P OB2 i = 1 2n · OX2, cf. Problem 6.67. 6.71. Let e1, . . . , en be the vectors that go from the center of the given n-gon into its vertices; x a unit vector perpendicular to line l. The sum to be found is equal to P(ei, x)2 = 1 2n · R2, cf. Problem 6.70. 6.72. Let e1, . . . , en be the unit vectors directed from the center O of a regular n-gon into the midpoints of its sides; x = − − → OX. Then the distance from point X to the i-th side is equal to \|(x, ei) −r\|. Hence, the sum to be found is equal to X ((x, ei)2 −2r(x, ei) + r2) = X (x, ei)2 + nr2. By Problem 6.70 P(x, ei)2 = 1 2nd2. 6.73. Let x be the unit vector parallel to line l and ei = − − − − → AiAi+1. Then the squared length of the projection of side AiAi+1 to line l is equal to (x, ei)2. By Problem 6.70 P(x, ei)2 = 1 2na2. 6.74. Let a = − − → OX, ei = − − → OAi. Then XA4 i = \|a + ei\|4 = (\|a\|2 + 2(a, ei) + \|ei\|2)2 = 4(R2 + (a, ei))2 = 4(R4 + 2R2(a, ei) + (a, ei)2). Clearly, P(a, ei) = (a, P ei) = 0. By Problem 6.70 P(a, ei)2 = 1 2nR4; hence, the sum to be found is equal to 4 ³ nR4 + nR4 2 ´ = 6nR4. SOLUTIONS 161 6.75. a) First, let us prove the required relation for u = e1. Let ei = (sin ϕi, cos ϕi), where cos ϕ1 = 1. Then X (e1, ei)ei = X cos ϕiei = X (sin ϕi cos ϕi, cos2 ϕi) = X µsin 2ϕi 2 , 1 + cos 2ϕi 2 ¶ = ³ 0, n 2 ´ = ne1 2 . For u = e2 the proof is similar. It remains to notice that any vector u can be represented in the form u = λe1 + µe2. b) Let B1, . . . , Bn be the midpoints of sides of the given polygon, ei = − − → OBi OBi, u = − − → XO. Then − − → XAi = − − → OBi + (u, ei)ei. Since P − − → OBi = − → 0 , it follows that X − − → XAi = X (u, ei)ei = nu 2 = n− − → XO 2 . 6.76. Let e0, . . . , en−1 be the vectors of sides of a regular n-gon. It suﬃces to prove that by reordering these vectors we can get a set of vectors − − − − − − − → a1, . . . , an such that Pn k=1 kak = 0. A number n which is not a power of a prime can be represented in the form n = pq, where p and q are relatively prime. Now, let us prove that the collection e0, ep, . . . , e(q−1)p; eq, eq+p, . . . , eq+(q−1)p; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e(p−1)q, e(p−1)q+p, . . . , e(p−1)q+(q−1)p is the one to be found. First, notice that if x1q + y1p ≡x2q + y2p ( mod pq), then x1 ≡x2 ( mod p) and y1 ≡y2 ( mod q); therefore, in the considered collection each of the vectors e0, . . . , en−1 is encountered exactly once. The endpoints of vectors eq, eq+p, . . . , eq+(q−1)p with a common beginning point des-tinguish a regular q-gon and, therefore, their sum is equal to zero. Moreover, vectors e0, ep, . . . , e(q−1)p turn into eq, eq+p, . . . , eq+(p−1)q under the rotation by an angle of ϕ = 2π p . Hence, if e0 + 2ep + · · · + qe(q−1)p = b, then (q + 1)eq + (q + 2)eq+p + · · · + 2qeq+(q−1)p = q(eq + · · · + eq+(q−1)p) + eq + 2eq+p + · · · + qeq+(q−1)p = Rϕb, where Rϕb is the vector obtained from b after the rotation by ϕ = 2π p . Similar arguments show that for the considered set of vectors we have n X k=1 kak = b + Rϕb + · · · + R(p−1)ϕb = 0. 6.77. Suppose that on the sides of triangle ABC squares ABB1A1, BCC2B2, ACC3A3 are constructed outwards and vertices A1, B1, B2, C2, C3, A3 lie on one circle S. The mid-perpendiculars to segments A1B1, B2C2, A3C3 pass through the center of circle S. It is clear 162 CHAPTER 6. POLYGONS that the midperpendiculars to segments A1B1, B2C2, A3C3 coincide with the midperpendic-ulars to sides of triangle ABC and therefore, the center of circle S coincides with the center of the circumscribed circle of the triangle. Denote the center of the circumscribed circle of triangle ABC by O. The distance from O to line B2C2 is equal to R cos ∠A + 2R sin ∠A, where R is the radius of the circumscribed circle of triangle ABC. Hence, OB2 2 = (R sin ∠A)2 + (R cos ∠A + 2R sin ∠A)2 = R2(3 + 2(sin ∠2A −cos 2∠A)) = R2(3 −2 √ 2 cos(45◦+ 2∠A)). Clearly, in order for the triangle to possess the desired property, it is necessary and suﬃcient that OB2 2 = OC2 3 = OA2 1, i.e., cos(45◦+ 2∠A) = cos(45◦+ 2∠B) = cos(45◦+ 2∠C). This equality holds for ∠A = ∠B = ∠C = 60◦. If, contrarywise, ∠A ̸= ∠B, then (45◦+ 2∠A) + (45◦+ 2∠B) = 360◦, i.e., ∠A + ∠B = 135◦. Hence, ∠C = 45◦and ∠A = ∠C = 45◦, ∠B = 90◦(or ∠B = 45◦, ∠A = 90◦). We see that the triangle should be either an equilateral or an isosceles one. 6.78. In any triangle we have hc = ab 2R (Problem 12.33); hence, pk = MAk·MAk+1 2R . There-fore, p1p3 . . . p2n−1 = MA1 · MA2 . . . MA2n (2R)n = p2p4 . . . p2n. 6.79. Let ABC be a triangle inscribed in circle S. Denote the distances from the center O of S to sides BC, CA and AB by a, b and c, respectively. Then R + r = a + b + c if point O lies inside triangle ABC and R + r = −a + b + c if points A and O lie on various sides of line BC, cf. Problem 12.38. Each of the diagonals of the partition belongs to two triangles of the partition. For one of these triangles point O and the remaining vertex lie on one side of the diagonal, for the other one the points lie on diﬀerent sides. A partition of an n-gon by nonintersecting diagonals into triangles consists of n −2 triangles. Therefore, the sum (n−2)R+r1 +· · ·+rn−2 is equal to the sum of distances from point O to the sides of an n-gon (the distances to the sides are taken with the corresponding signs). This implies that the sum r1 + · · · + rn−2 does not depend on the partition. 6.80. Let polygon A1 . . . An be inscribed in a circle. Let us consider point A′ 2 symmetric to point A2 through the midperpendicular to segment A1A3. Then polygon A1A′ 2A3 . . . An is an inscribed one and its area is equal to the area of polygon A1 . . . An. Therefore, we can transpose any two sides. Therefore, we can make any side, call it X, a neighbouring side of any given side, Y ; next, make any of the remaining sides a neighbour of X, etc. Therefore, the area of an n-gon inscribed into the given circle only depends on the set of lengths of the sides but not on their order. 6.81. Without loss of generality we may assume that an is the greatest of the numbers a1, . . . , an. Let n-gon A1 . . . An be inscribed into a circle centered at O. Then AiAi+1 : A1An = sin ∠AiOAi+1 2 : sin ∠A1OAn 2 . Therefore, let us proceed as follows. From the relation sin ϕi 2 : sin ϕ 2 = ai : an the angle ϕi is uniquely determined in terms of ϕ if ϕi < π. On a circle of radius 1, ﬁx a point An and SOLUTIONS 163 consider variable points A1, . . . , An−1, A′ n such that ⌣AnA1 = ϕ, ⌣A1A2 = ϕ1, . . . , ⌣An−2An−1 = ϕn−2 and ⌣An−1A′ n = ϕn−1. Denote these points in two distinct ways as plotted on Fig. 75. (The ﬁrst way — Fig. 75 a) — corresponds to an n-gon that contains the center of the circle, and the second way — Fig. 75 b) — corresponds to an n-gon that does not contain the center of the circle). It remains to prove that as ϕ varies from 0 to π, then in one of these cases point A′ n coincides with An (indeed, then up to a similarity we get the required n-gon). Suppose that in the ﬁrst case points A′ n and An never coincide for 0 ≤ϕ ≤π, i.e., for ϕ = π we have ϕ1 + · · · + ϕn−1 < π. Figure 75 (Sol. 6.81) Fig. 75 b) requires certain comments: sin α ≈α for small values of α; hence, the conditions of the problem imply that for small angles point An does indeed lie on arc ⌣A1A′ n because ϕ1 + · · · + ϕn−1 > ϕ. Thus, for small angles ϕ1 + · · · + ϕn−1 > ϕ and if ϕ = π, then by the hypothesis ϕ1 + · · · + ϕn−1 < π = ϕ. Hence, at certain moment ϕ = ϕ1 + · · · + ϕn−1, i.e., points An and A′ n coincide. 6.82. Let h1, . . . , hn be the distances from the given point to the corresponding sides; let a1, . . . , an be the distances from the vertices of the polygon to tangent points. Then the product of areas of red as well as blue triangles is equal to a1...anh1...hn 2n . 6.83. Let OHi be a height of triangle OAiAi+1. Then ∠Hi−1OAi = ∠HiOAi = ϕi. The conditions of the problem imply that ϕ1 + ϕ2 = ϕn+1 + ϕn+2, ϕn+2 + ϕn+3 = ϕ2 + ϕ3, ϕ3 + ϕ4 = ϕn+3 + ϕn+4, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . , ϕn−2 + ϕn−1 = ϕ2n−2 + ϕ2n−1 (expressing the last equality we have taken into account that n is odd) and ϕn−1 + 2ϕn + ϕn+1 = ϕ2n−1 + 2ϕ2n + ϕ1. Adding all these equalities we get ϕn−1 + ϕn = ϕ2n−1 + ϕ2n, as required. 6.84. Let O be the center of the given circle. Then − − → XAi = − − → XO + − − → OAi and, therefore, XA2 i = XO2 + OA2 i + 2(− − → OX, − − → OAi) = d2 + r2 + 2(− − → XO, − − → OAi). Since a1 − − → OA1 + · · · + An − − → OAn = − → 0 (cf. Problem 13.4), it follows that a1XA2 1 + · · · + anXA2 n = (a1 + · · · + an)(d2 + r2). 164 CHAPTER 6. POLYGONS 6.85. By Problem 5.8 bi−1bi a2 i = sin2 ∠Ai 2 . To solve heading a) it suﬃces to multiply all these equalities and to solve heading b) we have to divide the product of all equalities with even index i by the product of all equalities with odd index i. 6.86. Let BC be a blue side, AB and CD be the sides neighbouring with BC. By the hypothesis sides AB and CD are red ones. Suppose that the polygon is a circumscribed one; let P, Q, R be the tangent points of sides AB, BC, CD, respectively, with the inscribed circle. Clearly, BP = BQ, CR = CQ and segments BP, CR only neighbour one blue segment. Therefore, the sum of the lengths of the red sides is not smaller than the sum of the lengths of the blue sides. We have obtained a contradiction with the fact that the sum of the lengths of red sides is smaller than the semiperimeter. Therefore, a circle cannot be inscribed into the polygon. 6.87. Let the given n-gon have k acute angles. Then the sum of its angles is smaller than k · 90◦+ (n −k) · 180◦. On the other hand, the sum of the angles of the n-gon is equal to (n −2) · 180◦. Hence, (n −2) · 180◦< k · 90◦+ (n −k) · 180◦, i.e., k < 4. Since k is an integer, k ≤3. Figure 76 (Sol. 6.87) For any n ≥3 there exists a convex n-gon with three acute angles (Fig. 76). 6.88. Suppose that the lengths of nonadjacent sides AB and CD are equal to the length of the greatest diagonal. Then AB + CD ≥AC + BD. But by Problem 9.14 AB + CD < AC + BD. We have obtained a contradiction and therefore, the sides whose length is equal to the length of the longest diagonal should be adjacent ones, i.e., there are not more than two of such sides. Figure 77 (Sol. 6.88) An example of a polygon with two sides whose lengths are equal to the length of the longest diagonal is given on Fig. 77. Clearly, such an n-gon exists for any n > 3. 6.89. Let us prove that n ≤5. Let AB = 1 and C the vertex not adjacent to either A or B. Then \|AC−BC\| < AB = 1. Hence, AC = BC, i.e., point C lies on the midperpendicular SOLUTIONS 165 to side AB. Therefore, in addition to vertices A, B, C the polygon can have only two more vertices. Figure 78 (Sol. 6.89) An example of a pentagon with the required property is given on Fig. 78. Let us elucidate its construction. Clearly, ACDE is a rectangle, AC = ED = 1 and ∠CAD = 60◦. Point B is determined from the condition BE = BD = 3. An example of a quadrilateral with the desired property is rectangle ACDE on the same ﬁgure. 6.90. An example of a pentagon satisfying the conditions of the problem is plotted on Fig. 79. Let us clarify its construction. Take an equilateral right triangle EAB and draw midperpendiculars to sides EA, AB; on them construct points C and D, respectively, so that ED = BC = AB (i.e., lines BC and ED form angles of 30◦with the corresponding midperpendiculars). Clearly, DE = BC = AB = EA < EB < DC and DB = DA = CA = CE > EB. Now, let us prove that the ﬁfth side and the ﬁfth diagonal cannot have a common point. Suppose that the ﬁfth side AB has a common point A with the ﬁfth diagonal. Then the ﬁfth diagonal is either AC or AD. Let us consider these two cases. Figure 79 (Sol. 6.90) In the ﬁrst case △AED = △CDE; hence, under the symmetry through the midper-pendicular to segment ED point A turns into point C. This symmetry preserves point B because BE = BD. Therefore, segment AB turns into CB, i.e., AB = CB. Contradiction. 166 CHAPTER 6. POLYGONS In the second case △ACE = △EBD; hence, under the symmetry through the bisector of angle ∠AED segment AB turns into DC, i.e., AB = CD. Contradiction. 6.91. Let us consider two neighbouring vertices A1 and A2. If ∠A1OA2 ≥90◦, then OA1 = OA2 because neither right nor acute angle can be adjacent to the base of an isosceles triangle. Figure 80 (Sol. 6.91) Now, let ∠A1OA2 < 90◦. Let us draw through point O lines l1 and l2 perpendicular to lines OA1 and OA2, respectively. Denote the regions into which these lines divide the plane as indicated on Fig. 80. If in region 3 there is a vertex, Ak, then A1O = AkO = A2O because ∠A1OAk ≥90◦and ∠A2OAk ≥90◦. If region 3 has no vertices of the polygon, then in region 1 there is a vertex Ap and in region 2 there is a vertex Aq (if neither of the regions 1 or 2 would have contained vertices of the polygon, then point O would have been outside the polygon). Since ∠A1OAq ≥90◦, ∠A2OAp ≥90◦and ∠ApOAq ≥90◦, it follows that A1O = AqO = ApO = A2O. It remains to notice that if the distances from point O to any pair of the neighbouring vertices of the polygon are equal, then all the distances from point O to the vertices of the polygon are equal. 6.92. Let us prove that if A, B, C, D, E, F are points on the circle placed in an arbitrary order; lines AB and DE, BC and EF, CD and FA, intersect at points G, H, K, respectively. Then points G, H and K lie on one line. Let a, b, . . . , f be oriented angles between a ﬁxed line and lines OA, OB, . . . , OF, respectively, where O is the center of the circumscribed circle of the hexagon. Then ∠(AB, DE) = a + b −d −e 2 , ∠(CD, FA) = c + d −f −a 2 , ∠(EF, BC) = e + f −b −c 2 and, therefore, the sum of these angles is equal to 0. Let Z be the intersection point of circumscribed circles of triangles BDG and DFK. Let us prove that point B, F, Z and H lie on one circle. For this we have to verify that ∠(BZ, ZF) = ∠(BH, HF). Clearly, ∠(BZ, ZF) = ∠(BZ, ZD) + ∠(DZ, ZF), ∠(BZ, ZD) = ∠(BG, GD) = ∠(AB, DE), ∠(DZ, ZF) = ∠(DK, KF) = ∠(CD, FA) and, as we have just proved, ∠(AB, DE) + ∠(CD, FA) = −∠(EF, BC) = ∠(BC, EF) = ∠(BH, HF). SOLUTIONS 167 Now, let us prove that points H, Z and G lie on one line. For this it suﬃces to verify that ∠(GZ, ZB) = ∠(HZ, ZB). Clearly, ∠(GZ, ZB) = ∠(GD, DB) = ∠(ED, DB), ∠(HZ, ZB) = ∠(HF, FB) = ∠(ED, DB). We similarly prove that points K, Z and G lie on one line: ∠(DZ, ZG) = ∠(DB, BG) = ∠(DB, BA); ∠(DZ, ZK) = ∠(DF, FK) = ∠(DB, BA) We have deduced that points H and K lie on line GZ, consequently, points G, H and K lie on one line. 6.93. Let A2, B2 and C2 be the indicated intersection points of lines. By applying Pascal’s theorem to points M, A1, A, C, B, B1 we deduce that A2, B2 and R lie on one line. Similarly, points A2, C2 and R lie on one line. Hence, points A2, B2, C2 and R lie on one line. 6.94. Points A1 and B1 lie on circle S of diameter AB. Let A4 and B4 be the intersection points of lines AA2 and BB2 with line A3B3. By Problem 2.41 a) these points lie on circle S. Lines A1B and A4A intersect at point A2 and lines BB4 and AB1 at point B2. Therefore, applying Pascal’s theorem to points B1, A1, B, B4, A4, A we see that the intersection point of lines B1A1 and B4A4 (the latter line coincides with A3B3) lies on line A2B2. 6.95. Let K be the intersection point of lines BC and MN. Apply Pascal’s theorem to points A, M, N, D, C, B. We see that points E, K, F lie on one line and, therefore, K is the intersection point of lines MN and EF. 6.96. Let rays PA and QA intersect the circle at points P2 and Q2, i.e., P1P2 and Q1Q2 are diameters of the given circle. Let us apply Pascal’s theorem to hexagon PP2P1QQ2Q1. Lines PP2 and QQ2 intersect at point A and lines P1P2 and Q1Q2 intersect at point O, hence, the intersection point of lines P1Q and Q1P lies on line AO. 6.97. Let given points A, B, C, D, E lie on one line. Suppose that we have constructed point F of the same circle. Denote by K, L, M the intersection points of lines AB and DE, BC and EF, CD and FA, respectively. Then by Pascal’s theorem points K, L, M lie on one line. The above implies the following construction. Let us draw through point E an arbitrary line a and denote its intersection point with line BC by L. Then construct the intersection point K of lines AB and DE and the intersection point M of lines KL and CD. Finally, let F be the intersection point of lines AM and a. Let us prove that F lies on our circle. Let F1 be the intersection point of the circle and line a. From Pascal’s theorem it follows that F1 lies on line AM, i.e., F1 is the intersection point of a and AM. Hence, F1 = F. 6.98. Let P and Q be the intersection points of line A3A4 with A1A2 and A1A6, respec-tively, and R and S be the intersection points of line A4A5 with A1A6 and A1A2, respectively. Then A2K : A3L = A2P : A3P, A3L : A6M = A3Q : A6Q, A6M : A5N = A6R : A5R. Therefore, the desired relation A2K : A5N = A2S : A5S takes the form A2P A3P · A3Q A6Q · A6R A5R · A5S A2S = 1. Let T be the intersection point of lines A2A3 and A5A6; by Pascal’s theorem points S, Q and T lie on one line. By applying Menelau’s theorem (cf. Problem 5.58) to triangle PQS and points T, A2, A3 and also to triangle RQS and points T, A5, A6 we get A2P A2S · A3Q A3P · TS TQ = 1 and TQ TS · A5S A5R · A6R A6Q = 1. 168 CHAPTER 6. POLYGONS By multiplying these equalities we get the statement desired. (The ratio of segments should be considered oriented ones.) Chapter 7. LOCI Background 1) A locus is a ﬁgure consisting of all points having a cirtain property. 2) A solution of a problem where a locus is to be found should contain the proof of the following facts: a) the points with a required property belong to ﬁgure Φ which is the answer to the problem; b) All points of Φ have the required property. 3) A locus possessing two properties is the intersection of two ﬁgures: (1) the locus of points possessing the ﬁrst property and (2) the locus of points possessing the other property. 4) Three most important loci: a) The locus of points equidistant from points A and B is the midperpendicular to segment AB; b) The locus of points whose distance from a given point O is equal to R is the circle of radius R centered at O; c) The locus of vertices of a given angle that subtend given segment AB is the union of two arcs of circles symmetric through line AB (points A and B do not belong to the locus). Introductory problems 1. a) Find the locus of points equidistant from two parallel lines. b) Find the locus of points equidistant from two intersecting lines. 2. Find the locus of the midpoints of segments with the endpoints on two given parallel lines. 3. Given triangle ABC, ﬁnd the locus of points X satisfying inequalities AX ≤BX ≤CX. 4. Find the locus of points X such that the tangents drawn from X to the given circle have a given length. 5. A point A on a circle is ﬁxed. Find the locus of points X that divide chords with A as an endpoint in the ratio of 1 : 2 counting from point A. §1. The locus is a line or a segment of a line 7.1. Two wheels of radii r1 and r2 roll along line l. Find the set of intersection points M of their common inner tangents. 7.2. Sides AB and CD of quadrilateral ABCD of area S are not parallel. Inside the quadrilateral ﬁnd the locus of points X for which SABX + SCDX = 1 2S. 7.3. Given two lines that meet at point O. Find the locus of points X for which the sum of the lengths of projections of segments OX to these lines is a constant. 7.4. Given rectangle ABCD, ﬁnd the locus of points X for which AX+BX = CX+DX. 7.5. Find the locus of points M that lie inside rhombus ABCD and with the property that ∠AMD + ∠BMC = 180◦. 169 170 CHAPTER 7. LOCI 7.6. Given points A and B in plane, ﬁnd the locus of points M for which the diﬀerence of the squared lengths of segments AM and PM is a constant. 7.7. A circle S and a point M outside it are given. Through point M all possible circles S1 that intersect S are drawn; X is the intersection point of the tangent at M to S1 with the extension of the common chord of circles S and S1. Find the locus of points X. 7.8. Given two nonintersecting circles, ﬁnd the locus of the centers of circles that divide the given circles in halves (i.e., that intersect the given circles in diametrically opposite points). 7.9. A point A inside a circle is taken. Find the locus of the intersection points of tangents to circles drawn through the endpoints of possible chords that contain point A. 7.10. a) Parallelogram ABCD is given. Prove that the quantity AX2 + CX2 −BX2 −DX2 does not depend on the choice of point X. b) Quadrilateral ABCD is not a parallelogram. Prove that all points X that satisfy the relation AX2 + CX2 = BX2 + DX2 lie on one line perpendicular to the segment that connects the midpoints of the diagonals. See also Problems 6.14, 15.14. §2. The locus is a circle or an arc of a circle 7.11. A segment moves along the plane so that its endpoints slide along the legs of a right angle ∠ABC. What is the trajectory traversed by the midpoint of this segment? (We naturally assume that the length of the segment does not vary while it moves.) 7.12. Find the locus of the midpoints of the chords of a given circle, provided the chords pass through a given point. 7.13. Given two points, A and B and two circles that are tangent to line AB: one circle is tangent at A and the other one at B, and the circles are tangent to each other at point M. Find the locus of points M. 7.14. Two points, A and B in plane are given. Find the locus of points M for which AM : BM = k. (Apollonius’s circle.) 7.15. Let S be Apollonius’s circle for points A and B where point A lies outside circle S. From point A tangents AP and AQ to circle S are drawn. Prove that B is the midpoint of segment PQ. 7.16. Let AD and AE be the bisectors of the inner and outer angles of triangle ABC and Sa be the circle with diameter DE; circles Sb and Sc are similarly deﬁned. Prove that: a) circles Sa, Sb and Sc have two common points, M and N, such that line MN passes through the center of the circumscribed circle of triangle ABC; b) The projections of point M (and N) to the sides of triangle ABC distinguish an equilateral triangle. 7.17. Triangle ABC is an equilateral one, M is a point. Prove that if the lengths of segments AM, BM and CM form a geometric progression, then the quotient of this progression is smaller than 2. See also Problems 14.19 a), 18.14. §6. A METHOD OF LOCI 171 §3. The inscribed angle 7.18. Points A and B on a circle are ﬁxed and a point C runs along the circle. Find the set of the intersection points of a) heights; b) bisectors of triangles ABC. 7.19. Point P runs along the circumscribed circle of square ABCD. Lines AP and BD intersect at point Q and the line that passes through point Q parallel to AC intersects line BP at point X. Find the locus of points X. 7.20. a) Points A and B on a circle are ﬁxed and points A1 and B1 run along the same circle so that the value of arc ⌣A1B1 remains a constant; let M be the intersection point of lines AA1 and BB1. Find the locus of points M. b) Triangles ABC and A1B1C1 are inscribed in a circle; triangle ABC is ﬁxed and triangle A1B1C1 rotates. Prove that lines AA1, BB1 and CC1 intersect at one point for not more than one position of triangle A1B1C1. 7.21. Four points in the plane are given. Find the locus of the centers of rectangles formed by four lines that pass through the given points. 7.22. Find the locus of points X that lie inside equilateral triangle ABC and such that ∠XAB + ∠XBC + ∠XCA = 90◦. See also Problems 2.5, 2.37. §4. Auxiliary equal triangles 7.23. A semicircle centered at O is given. From every point X on the extension of the diameter of the semicircle a ray tangent to the semicircle is drawn. On the ray segment XM equal to segment XO is marked. Find the locus of points M obtained in this way. 7.24. Let A and B be ﬁxed points in plane. Find the locus of points C with the following property: height hb of triangle ABC is equal to b. 7.25. A circle and a point P inside it are given. Through every point Q on the circle the tangent is drawn. The perpendicular dropped from the center of the circle to line PQ and the tangent intersect at a point M. Find the locus of points M. §5. The homothety 7.26. Points A and B on a circle are ﬁxed. Point C runs along the circle. Find the set of the intersection points of the medians of triangles ABC. 7.27. Triangle ABC is given. Find the locus of the centers of rectangles PQRS whose vertices Q and P lie on side AC and vertices R and S lie on sides AB and BC, respectively. 7.28. Two circles intersect at points A and B. Through point A a line passes. It intersects the circles for the second time at points P and Q. What is the line plotted by the midpoint of segment PQ while the intersecting line rotates about point A. 7.29. Points A, B and C lie on one line; B is between A and C. Find the locus of points M such that the radii of the circumscribed circles of triangles AMB and CMB are equal. See also Problems 19.10, 19.21, 19.38. §6. A method of loci 7.30. Points P and Q move with the same constant speed v along two lines that intersect at point O. Prove that there exists a ﬁxed point A in plane such that the distances from A to P and Q are equal at all times. 7.31. Through the midpoint of each diagonal of a convex quadrilateral a line is drawn parallel to the other diagonal. These lines meet at point O. Prove that segments that connect 172 CHAPTER 7. LOCI O with the midpoints of the sides of the quadrilateral divide the area of the quadrilateral into equal parts. 7.32. Let D and E be the midpoints of sides AB and BC of an acute triangle ABC and point M lies on side AC. Prove that if MD < AD, then ME > EC. 7.33. Inside a convex polygon points P and Q are taken. Prove that there exists a vertex of the polygon whose distance from Q is smaller than that from P. 7.34. Points A, B and C are such that for any fourth point M either MA ≤MB or MA ≤MC. Prove that point A lies on segment BC. 7.35. Quadrilateral ABCD is given; in it AB < BC and AD < DC. Point M lies on diagonal BD. Prove that AM < MC. §7. The locus with a nonzero area 7.36. Let O be the center of rectangle ABCD. Find the locus of points M for which AM ≥OM, BM ≥OM, CM ≥OM and DM ≥OM. 7.37. Find the locus of points X from which tangents to a given arc AB of a circle can be drawn. 7.38. Let O be the center of an equilateral triangle ABC. Find the locus of points M satisfying the following condition: any line drawn through M intersects either segment AB or segment CO. 7.39. In plane, two nonintersecting disks are given. Does there necessarily exist a point M outside these disks that satisﬁes the following condition: each line that passes through M intersects at least one of these disks? Find the locus of points M with this property. See also Problem 18.11. §8. Carnot’s theorem 7.40. Prove that the perpendiculars dropped from points A1, B1 and C1 to sides BC, CA, AB of triangle ABC intersect at one point if and only if A1B2 + C1A2 + B1C2 = B1A2 + A1C2 + C1B2. (Carnot’s formula) 7.41. Prove that the heights of a triangle meet at one point. 7.42. Points A1, B1 and C1 are such that AB1 = AC1, BC = BA1 and CA1 = CB1. Prove that the perpendiculars dropped from points A1, B1 and C1 to lines BC, CA and AB meet at one point. 7.43. a) The perpendiculars dropped from the vertices of triangle ABC to the corre-sponding sides of triangle A1B1C1 meet at one point. Prove that the perpendiculars dropped from the vertices of triangle A1B1C1 to the corresponding sides of triangle ABC also meet at one point. b) Lines drawn through vertices of triangle ABC parallelly to the corresponding sides of triangle A1B1C1 intersect at one point. Prove that the lines drawn through the vertices of triangle A1B1C1 parallelly to the corresponding sides of triangle ABC also intersect at one point. 7.44. On line l points A1, B1 and C1 are taken and from the vertices of triangle ABC perpendiculars AA2, BB2 and CC2 are dropped to this line. Prove that the perpendiculars dropped from points A1, B1 and C1 to lines BC, CA and AB, respectively, intersect at one point if and only if A1B1 : B1C1 = A2B2 : B2C2. The ratios of segments are oriented ones. PROBLEMS FOR INDEPENDENT STUDY 173 7.45. Triangle ABC is an equilateral one, P an arbitrary point. Prove that the perpen-diculars dropped from the centers of the inscribed circles of triangles PAB, PBC and PCA to lines AB, BC and CA, respectively, meet at one point. 7.46. Prove that if perpendiculars raised at the bases of bisectors of a triangle meet at one point, then the triangle is an isosceles one. §9. Fermat-Apollonius’s circle 7.47. Prove that the set of points X such that k1A1X2 + · · · + knAnX2 = c is either a) a circle or the empty set if k1 + · · · + kn ̸= 0; b) a line, a plane or the empty set if k1 + · · · + kn = 0. 7.48. Line l intersects two circles at four points. Prove that the quadrilateral formed by the tangents at these points is a circumscribed one and the center of its circumscribed circle lies on the line that connects the centers of the given circles. 7.49. Points M and N are such that AM : BM : CM = AN : BN : CN. Prove that line MN passes through the center O of the circumscribed circle of triangle ABC. See also Problems 7.6, 7.14, 8.59–8.63. Problems for independent study 7.50. On sides AB and BC of triangle ABC, points D and E are taken. Find the locus of the midpoints of segments DE. 7.51. Two circles are tangent to a given line at two given points A and B; the circles are also tangent to each other. Let C and D be the tangent points of these circles with another outer tangent. Both tangent lines to the circles are outer ones. Find the locus of the midpoints of segments CD. 7.52. The bisector of one of the angles of a triangle has inside the triangle a common point with the perpendicular erected from the midpoint of the side opposite the angle. Prove that the triangle is an isosceles one. 7.53. Triangle ABC is given. Find the locus of points M of this triangle for which the condition AM ≥BM ≥CM holds. When the obtained locus is a) a pentagon; b) a triangle? 7.54. Square ABCD is given. Find the locus of the midpoints of the sides of the squares inscribed in the given square. 7.55. An equilateral triangle ABC is given. Find the locus of points M such that triangles AMB and BCM are isosceles ones. 7.56. Find the locus of the midpoints of segments of length 2 √ 3 whose endpoints lie on the sides of a unit square. 7.57. On sides AB, BC and CA of a given triangle ABC points P, Q and R, respectively, are taken, so that PQ ∥AC and PR ∥BC. Find the locus of the midpoints of segments QR. 7.58. Given a semicircle with diameter AB. For any point X on this semicircle, point Y on ray XA is taken so that XY = XB. Find the locus of points Y . 7.59. Triangle ABC is given. On its sides AB, BC and CA points C1, A1 and B1, respectively, are selected. Find the locus of the intersection points of the circumscribed circles of triangles AB1C1, A1BC1 and A1B1C. 174 CHAPTER 7. LOCI Solutions 7.1. Let O1 and O2 be the centers of the wheels of radii r1 and r2, respectively. If M is the intersection point of the inner tangents, then OM : O2M = r1 : r2. It is easy to derive from this condition that the distance from point M to line l is equal to 2r1r2 r1+r2. Hence, all the intersection points of the common inner tangents lie on the line parallel to l and whose distance from l is equal to 2r1r2 r1+r2. 7.2. Let O be the intersection point of lines AB and CD. On rays OA and OD, mark segments OK and OL equal to AB and CD, respectively. Then SABX + SCDX = SKOX + SLOX = SKOL ± SKXL. Therefore, the area of triangles KXL is a constant, i.e., point X lies on a line parallel to KL. 7.3. Let a and b be unit vectors parallel to the given lines; x = − − → OX. The sum of the lengths of the projections of vector x to the given lines is equal to \|(a, x)\| + \|(b, x)\| = \|(a ± b, x)\|, where the change of sign occurs on the perpendiculars to the given lines erected at point O. Therefore, the locus to be found is a rectangle whose sides are parallel to the bisectors of the angles between the given lines and the vertices lie on the indicated perpendiculars. 7.4. Let l be the line that passes through the midpoints of sides BC and AD. Suppose that point X does not lie on l; for instance, points A and X lie on one side of l. Then AX < DX and BX < CX and, therefore, AX + BX < CX + DX. Hence, l is the locus to be found. 7.5. Let N be a point such that − − → MN = − − → DA. Then ∠NAM = ∠DMA and ∠NBM = ∠BMC and, therefore, quadrilateral AMBN is an inscribed one. The diagonals of the inscribed quadrilateral AMBN are equal, hence, either AM ∥BN or BM ∥AN. In the ﬁrst case ∠AMD = ∠MAN = ∠AMB and in the second case ∠BMC = ∠MBN = ∠BMA. If ∠AMB = ∠AMD, then ∠AMB + ∠BMC = 180◦and point M lies on diagonal AC and if ∠BMA = ∠BMC, then point M lies on diagonal BD. It is also clear that if point M lies on one of the diagonals, then ∠AMD + ∠BMC = 180◦. 7.6. Introduce a coordinate system selecting point A as the origin and directing Ox-axis along ray AB. Let (x, y) be the coordinates of M. Then AM 2 = x2 + y2 and BM 2 = (x −a)2 + y2, where a = AB. Hence, AM 2 −BM 2 = 2ax −a2. This quantity is equal to k for points M whose coordinates are (a2+k 2a , y). All such points lie on a line perpendicular to AB. 7.7. Let A and B be the intersection points of circles S and S1. Then XM 2 = XA·XB = XO2 −R2, where O and R are the center and the radius, respectively, of circle S. Hence, XO2−XM 2 = R2 and, therefore, points X lie on the perpendicular to line OM (cf. Problem 7.6). 7.8. Let O1 and O2 be the centers of the given circles, R1 and R2 their respective radii. The circle of radius r centered at X intersects the ﬁrst circle in the diametrically opposite points if and only if r2 = XO2 1 + R2 1; hence, the locus to be found consists of points X such that XO2 1 + R2 1 = XO2 2 + R2 2. All such points X lie on a line perpendicular to O1O2, cf. Problem 7.6. 7.9. Let O be the center of the circle, R its radius, M the intersection point of the tangents drawn through the endpoints of the chord that contains point A, and P the midpoint of this chord. Then OP · OM = R2 and OP = OA cos ϕ, where ϕ = ∠AOP. Hence, AM 2 = OM 2 + OA2 −2OM · OA cos ϕ = OM 2 + OA2 −2R2, SOLUTIONS 175 and, therefore, the quantity OM 2 −AM 2 = 2R2 −OA2 is a constant. It follows that all points M lie on a line perpendicular to OA, cf. Problem 7.6. 7.10. Let P and Q be the midpoints of diagonals AC and BD. Then AX2 + CX2 = 2PX2 + AC2 2 and BX2 + DX2 = 2QX2 + BD2 2 (cf. Problem 12.11 a)) and, therefore, in heading b) the locus to be found consists of points X such that PX2 −QX2 = 1 4(BD2 −AC2) and in heading a) P = Q and, therefore, the considered quantity is equal to 1 2(BD2 −AC2). 7.11. Let M and N be the midpoints of the given segment, O its midpoint. Point B lies on the circle with diameter MN, hence, OB = 1 2MN. The trajectory of point O is the part of the circle of radius 1 2MN centered at B conﬁned inside angle ∠ABC. 7.12. Let M be the given point, O the center of the given circle. If X is the midpoint of chord AB, then XO ⊥AB. Therefore, the locus to be found is the circle with diameter MO. 7.13. Let us draw through point M a common tangent to the circles. Let O be the intersection point of this tangent with line AB. Then AO = MO = BO, i.e., O is the midpoint of segment AB. Point M lies on the circle with center O and radius 1 2AB. The locus of points M is the circle with diameter AB (points A and B excluded). 7.14. For k = 1 we get the midperpendicular to segment AB. In what follows we will assume that k ̸= 1. Let us introduce a coordinate system in plane so that the coordinates of A and B are (−a, 0) and (a, 0), respectively. If the coordinates of point M are (x, y), then AM 2 BM 2 = (x + a)2 + y2 (x −a)2 + y2. The equation AM2 BM2 = k2 takes the form x + 1 − k2 1 −k2a + y2 = 2ka 1 −k2 2 . This is an equation of the circle with center (−1 + k2 1−k2a, 0) and radius 2ka \|1−k2\|. 7.15. Let line AB intersect circle S at points E and F so that point E lies on segment AB. Then PE is the bisector of triangle APB, hence, ∠EPB = ∠EPA = ∠EFP. Since ∠EPF = 90◦, it follows that PB ⊥EF. 7.16. a) The considered circles are Apollonius’s circles for the pairs of vertices of triangle ABC and, therefore, if X is a common point of circles Sa and Sb, then XB : XC = AB : AC and XC : XA = BC : BA, i.e., XB : XA = CB : CA and, therefore, point X belongs to circle Sc. It is also clear that if AB > BC, then point D lies inside circle Sb and point A outside it. It follows that circles Sa and Sb intersect at two distinct points. To complete the proof, it remains to make use of the result of Problem 7.49. b) According to heading a) MA = λ a, MB = λ b and MC = λ c . Let B1 and C1 be the projections of point M on lines AC and AB, respectively. Points B1 and C1 lie on the circle with diameter MA, hence, B1C1 = MA sin ∠B1AC1 = λ a a 2R = λ 2R, where R is the radius of the circumscribed circle of triangle ABC. Similarly, A1C1 = A1B1 = λ 2R. 176 CHAPTER 7. LOCI 7.17. Let O1 and O2 be points such that − − → BO1 = 4 3 − → BA and − − → CO2 = 4 3 − − → CB. It is easy to verify that if BM > 2AM, then point M lies inside circle S1 of radius 2 3AB with center O1 (cf. Problem 7.14) and if CM > 2BM, then point M lies inside circle S2 of radius 2 3AB centered at O2. Since O1O2 > BO1 = 4 3AB and the sum of the radii of circles S1 and S2 is equal to 4 3AB, it follows that these circles do not intersect. Therefore, if BM = qAM and CM = qBM, then q < 2. 7.18. a) Let O be the intersection point of heights AA1 and BB1. The points A1 and B1 lie on the circle with diameter CO. Therefore, ∠AOB = 180◦−∠C. Hence, the locus to be found is the circle symmetric to the given one through line AB (points A and B should be excluded). b) If O is the intersection point of the bisectors of triangle ABC, then ∠AOB = 90◦+ 1 2∠C. On each of the two arcs ⌣AB the angles C are constant and, therefore, the desired locus of the vertices of angles of 90◦+ 1 2∠C that subtend segment AB is the union of two arcs (points A and B should be excluded). 7.19. Points P and Q lie on the circle with diameter DX, hence, ∠(QD, DX) = ∠(QP, PX) = ∠(AP, PB) = 45◦, i.e., point X lies on line CD. 7.20. a) If point A1 traverses along the circle an arc of value 2ϕ, then point B1 also traverses an arc of value 2ϕ, consequently, lines AA1 and BB1 turn through an angle of ϕ and the angle between them will not change. Hence, point M moves along a circle that contains points A and B. b) Let at some moment lines AA1, BB1 and CC1 meet at point P. Then, for instance, the intersection point of lines AA1 and BB1 moves along the circumscribed circle of triangle ABP. It is also clear that the circumscribed circles of triangles ABP, BCP and CAP have a unique common point, P. 7.21. Suppose that points A and C lie on opposite sides of a rectangle. Let M and N be the midpoints of segments AC and BD, respectively. Let us draw through point M line l1 parallel to the sides of the rectangle on which points A and C lie and through point N line l2 parallel to the sides of the rectangle on which points B and D lie. Let O be the intersection point of lines l1 and l2. Clearly, point O lies on circle S constructed on segment MN as on a diameter. On the other hand, point O is the center of the rectangle. Clearly, the rectangle can be constructed for any point O that lies on circle S. It remains to notice that on the opposite sides of the rectangle points A and B or A and D can also lie. Hence, the locus to be found is the union of three circles. 7.22. It is easy to verify that the points of heights of triangle ABC possess the required property. Suppose that a point X not belonging to any of the heights of triangle ABC possesses the required property. Then line BX intersects heights AA1 and CC1 at points X1 and X2. Since ∠XAB + ∠XBC + ∠XCA = 90◦= ∠X1AB + ∠X1BC + ∠X1CA, it follows that ∠XAB −∠X1AB = ∠X1CA −∠XCA, i.e., ∠(XA, AX1) = ∠(X1C, CX). Therefore, point X lies on the circumscribed circle of triangle AXC′, where point C′ is symmetric to C through line BX. We similarly prove that point X2 lies on the circle and, therefore, line BX intersects this circle at three distinct points. Contradiction. SOLUTIONS 177 7.23. Let K be the tangent point of line MX with the given semicircle and P the projection of point M to the diameter. In right triangles MPX and OKX, the hypothenuses are equal and ∠PXM = ∠OXK; hence, these triangles are equal. In particular, MP = KO = R, where R is the radius of the given semicircle. It follows that point M lies on line l parallel to the diameter of the semicircle and tangent to the semicircle. Let AB be the segment of line l whose projection is the diameter of the semicircle. From a point on l that does not belong to segment AB a tangent to the given semicircle cannot be drawn because the tangent drawn to the circle should be tangent to the other semicircle as well. The locus to be found is punctured segment AB: without points A, B, and the midpoint. 7.24. Let H be the base of height hb of triangle ABC and hb = b. Denote by B′ the intersection point of the perpendicular to line AB drawn through point A and the perpendicular to line AH drawn through point C. Right triangles AB′C and BAH are equal, becuase ∠AB′C = ∠BAH and ∠AC = BH. Therefore, AB′ = AB, i.e., point C lies on the circle with diameter AB′. Figure 81 (Sol. 7.24) Let S1 and S2 be the images of circle S with diameter AB under the rotations through angles of ±90◦with center at A (Fig. 81). We have proved that point C ̸= A belongs to the union of circles S1 and S2. Conversely, let a point C, C ̸= A, belong to either of the circles S1 or S2; let AB′ be a diameter of the corresponding circle. Then ∠AB′C = ∠HAB and A′B = AB; hence, AC = HB. 7.25. Let O be the center of the circle, N the intersection point of lines OM and QP. Let us drop from point M perpendicular MS to line OP. Since △ONQ ∼△OQM and △OPN ∼△OMS, we derive that ON : OQ = OQ : OM and OP : ON = OM : OS. By multiplying these equalities we get OP : OQ = OQ : OS. Hence, OS = OQ2 : OP is a constant. Since point S lies on line OP, its position does not depend on the choice of point Q. The locus to be found is the line perpendicular to line OP and passing through point S. 7.26. Let O be the midpoint of segment AB, and M the intersection point of the medians of triangle ABC. The homothety with center O and coeﬃcient 1 3 sends point C to point M. Therefore, the intersection point of the medians of triangle ABC lies on circle S which is the image of the initial circle under the homothety with center O and coeﬃcient 1 3. To get the desired locus we have to delete from S the images of points A and B. 178 CHAPTER 7. LOCI 7.27. Let O be the midpoint of height BH; let M, D and E be the midpoints of segment AC, and sides RQ and PS, respectively (Fig. 82). Figure 82 (Sol. 7.27) Points D and E lie on lines AO and CO, respectively. The midpoint of segment DE is the center of rectangle PQRS. Clearly, this midpoint lies on segment OM. The locus in question is segment OM without its endpoints. 7.28. Let O1 and O2 be the centers of the given circles (point P lies on the circle centered at O1); O the midpoint of segment O1O2; P ′, Q′ and O′ the projections of points O1, O2 and O to line PQ. As line PQ rotates, point O′ runs the circle S with diameter AO. Clearly, the homothety with center A and coeﬃcient 2 sends segment P ′Q′ to segment PQ, i.e., point O′ turns into the midpoint of segment PQ. Hence, the locus in question is the image of circle S under this homothety. 7.29. Let P and Q be the centers of the circumscribed circles of triangles AMB and CMB. Point M belongs to the locus to be found if BPMQ is a rhombus, i.e., point M is the image of the midpoint of segment PQ under the homothety with center B and coeﬃcient 2. Since the projections of points P and Q to line AC are the midpoints of segments AB and BC, respectively, the midpoints of all segments PQ lie on one line. (The locus to be found is the above-obtained line without the intersection point with line AC.) 7.30. Point P passes through point O at time t1, it passes point Q at time t2. At time 1 2(t1 + t2) the distances from O to points P and Q are equal to 1 2\|t1 + t2\|v. At this moment erect the perpendiculars to the lines at points P and Q. It is easy to verify that the intersection point of these perpendiculars is the required one. 7.31. Denote the midpoints of diagonals AC and BD of quadrilateral ABCD by M and N, respectively. Clearly, SAMB = SBMC and SAMD = SDMC, i.e., SDABM = SBCDM. Since the areas of quadrilaterals DABM and BCDM do not vary as point M moves parallelly to BD, it follows that SDABO = SCDAO. Similar arguments for point N show that SABCO = SCDAO. Hence, SADO + SABO = SBCO + SCDO and SABO + SBCO = SCDO + SADO and, therefore, SADO = SBCO = S1 and SABO = SCDO = S2, i.e., the area of each of the four parts into which the segments that connect point O with the midpoints of sides of the quadrilateral divide it is equal to 1 2(S1 + S2). 7.32. Let us drop height BB1 from point B. Then AD = B1D and CE = B1E. Clearly, if MD < AD, then point M lies on segment AB1, i.e., outside segment B1C. Therefore, ME > EC. 7.33. Suppose that the distance from any vertex of the polygon to point Q is not shorter than to point P. Then all the vertices of the polygon lie in the same half plane determined by the perpendicular to segment PQ at point P; point Q lies in the other half plane. Therefore, point Q lies outside the polygon. This contradicts the hypothesis. SOLUTIONS 179 7.34. Let us ﬁnd the locus of points M for which MA > MB and MA > MC. Let us draw midperpendiculars l1 and l2 to segments AB and AC. We have MA > MB for the points that lie inside the half-plane bounded by line l1 and the one without point A. There-fore, the locus in question is the intersection of half-planes (without boundaries) bounded by lines l1 and l2 and not containing point A. If points A, B and C do not lie on one line, then this locus is always nonempty. If A, B, C lie on one line but A does not lie on segment BC, then this locus is also nonempty. If point A lies on segment BC, then this locus is empty, i.e., for any point M either MA ≤MB or MA ≤MC. 7.35. Let O be the midpoint of diagonal AC. The projections of points B and D to line AC lie on segment AO, hence, the projection of point M also lies on segment AO. 7.36. Let us draw the midperpendicular l to segment AO. Clearly, AM ≥OM if and only if point M lies on the same side of line l as O (or lies on line l itself). Therefore, the locus in question is the rhombus formed by the midperpendiculars to segments OA, OB, OC and OD. 7.37. The locus to be found is shaded on Fig. 83 (the boundary belongs to the locus). Figure 83 (Sol. 7.37) 7.38. Let A1 and B1 be the midpoints of sides CB and AC, respectively. The locus to be found is the interior of quadrilateral OA1CB1. Figure 84 (Sol. 7.39) 7.39. Let us draw the common tangents to given disks (Fig. 84). It is easy to verify that the points that belong to the shaded domains (but not to their boundaries) satisfy the required condition and the points that do not belong to these domains do not satisfy this condition. 7.40. Let the perpendiculars dropped from points A1, B1, C1 to lines BC, CA, AB, respectively, intersect at point M. Since points B1 and M lie on one perpendicular to line AC, we have B1A2 −B1C2 = MA2 −MC2. 180 CHAPTER 7. LOCI Similarly, C1B2 −C1A2 = MB2 −MA2 and A1C2 −A1B2 = MC2 −MB2. Adding these equalities we get (∗) A1B2 + C1A2 + B1C2 = B1A2 + A1C2 + C1B2. Conversely, let (∗) hold. Denote the intersection point of the perpendiculars dropped from points A1 and B1 to lines BC and AC, respectively, by M. Let us draw through point M line l perpendicular to line AB. If C′ 1 is a point on line l, then by the above A1B2 + C′ 1A2 + B1C2 = B1A2 + A1C2 + C′ 1B2. Hence, C′ 1A2 −C′ 1B2 = C1A2 −C1B2. By Problem 7.6 the locus of points X for which XA2 −XB2 = k is a line perpendicular to segment AB. Therefore, the perpendicular dropped from point C1 to line AB passes through point M, as required. 7.41. Set A1 = A, B1 = B and C1 = C. From the obvious identity AB2 + CA2 + BC2 = BA2 + AC2 + CB2 we derive that the heights dropped from points A, B and C to sides BC, CA and AB, respectively, intersect at one point. 7.42. It suﬃces to make use of the result of Problem 7.40. 7.43. a) This problem is an obvious corollary of Problem 7.40. b) Let the rotation by 90◦about a point send triangle A1B1C1 to triangle A2B2C2. The perpendiculars to sides of triangle A2B2C2 are parallel to the corresponding sides of triangle A1B1C1 and, therefore, the perpendiculars dropped from the vertices of triangle ABC to the corresponding sides of triangle A2B2C2 intersect at one point. It follows that the perpendiculars dropped from the vertices of triangle A2B2C2 to the corresponding sides of triangle ABC intersect at one point. It remains to notice that the rotation by 90◦that sends triangle A2B2C2 to triangle A1B1C1 sends these perpendiculars to the lines that pass through the sides of triangle A1B1C1 parallelly the corresponding sides of triangle ABC. 7.44. We have to ﬁnd out when the identity AB2 1 + BC2 1 + CA2 1 = BA2 1 + CB2 1 + AC2 1 holds. By subtracting AA2 2 + BB2 2 + CC2 2 from both sides of this identity we get A2B2 1 + B2C2 1 + C2A2 1 = B2A2 1 + C2B2 1 + A2C2 1, i.e., (b1 −a2)2 + (c1 −b2)2 + (a1 −c2)2 = (a1 −b2)2 + (b1 −c2)2 + (c1 −a2)2, where ai, bi and ci are the coordinates of points Ai, Bi and Ci on line l. After simpliﬁcation we get a2b1 + b2c1 + c2a1 = a1b2 + b1c2 + c1a2 and, therefore, (b2 −a2)(c1 −b1) = (b1 −a1)(c2 −b2), i.e., A2B2 : B2C2 = A1B1 : B1C1. 7.45. We may assume that the length of a side of the given equilateral triangle is equal to 2. Let PA = 2a, PB = 2b and PC = 2c; let A1, B1 and C1 be the projections of the SOLUTIONS 181 centers of the inscribed circles of triangles PBC, PCA and PAB to lines BC, CA and AB, respectively. By Problem 3.2 we have AB2 1 + BC2 1 + CA2 1 = (1 + a −c)2 + (1 + b −a)2 + (1 + c −b)2 = 3 + (a −c)2 + (b −a)2 + (c −b)2 = BA2 1 + CB2 1 + AC2 1. 7.46. The segments into which the bisectors divide the sides of the triangle are easy to calculate. As a result we see that if the perpendiculars raised from the bases of the bisectors intersect, then (ac b + c)2 + (ab a + c)2 + (bc a + b)2 = (ab b + c)2 + (bc a + c)2 + (ac a + b)2, i.e., 0 = a2c −b b + c + b2a −c a + c + c2b −a a + b = −(b −a)(a −c) a2 + b2 + c2 (a + b)(a + c)(b + c). 7.47. Let (ai, bi) be the coordinates of point Ai and (x, y) the coordinates of point X. Then the equation satisﬁed by point X takes the form c = X ki((x −ai)2 + (x −bi)2) = ( X ki)(x2 + y2) −(2 X kiai)x −(2 X kibi)y + X ki(a2 i + b2 i ). If the coeﬃcient of x2 + y2 is nonzero, then this equation determines either a circle or the empty set and if it is zero, then the equation determines either a line, or a plane, or the empty set. Remark. If in case a) points A1, . . . , An lie on one line l, then this line can be taken for Ox-axis. Then bi = 0 and, therefore, the coeﬃcient of y is equal to zero, i.e., the center of the circle lies on l. 7.48. Let line l cut on the given circles arcs ⌣A1B1 and ⌣A2B2 whose values are 2α1 and 2α2, respectively; let O1 and O2 be the centers of the circles, R1 and R2 their respective radii. Let K be the intersection point of the tangents at points A1 and A2. By the law of sines KA1 : KA2 = sin α2 : sin α1, i.e., KA1 sin α1 = KA2 sin α2. Since KO2 1 = KA2 1 + R2 1 and KO2 2 = KA2 2 + R2 2, it follows that (sin2 α2)KO2 1 −(sin2 α2)KO2 2 = (R1 sin α1)2 −(R2 sin α2)2 = q. We similarly prove that the other intersection points of the tangents belong to the locus of points X such that (sin2 α1)XO2 1 −(sin2 α2)XO2 2 = q. This locus is a circle whose center lies on line O1O2 (cf. Remark to Problem 7.47). 7.49. Let AM : BM : CM = p : q : r. All the points X that satisfy (q2 −r2)AX2 + (r2 −p2)BX2 + (p2 −q2)CX2 = 0 lie on one line (cf. Problem 7.47) and points M, N and O satisfy this relation. Chapter 8. CONSTRUCTIONS §1. The method of loci 8.1. Construct triangle ABC given a, ha and R. 8.2. Inside triangle ABC construct point M so that SABM : SBCM : SACM = 1 : 2 : 3. 8.3. Through given point P inside a given circle draw a chord so that the diﬀerence of the lengths of the segments into which P divides the chord would be equal to the given value a. 8.4. Given a line and a circle without common points, construct a circle of a given radius r tangent to them. 8.5. Given point A and circle S draw a line through point A so that the chord cut by circle S on this line would be of given length d. 8.6. Quadrilateral ABCD is given. Inscribe in it a parallelogram with given directions of sides. §2. The inscribed angle 8.7. Given a, mc and angle ∠A, construct triangle ABC. 8.8. A circle and two points A and B inside it are given. Inscribe a right triangle in the circle so that the legs would pass through the given points. 8.9. The extensions of sides AB and CD of rectangle ABCD intersect a line at points M and N, respectively, and the extensions of sides AD and BC intersect the same line at points P and Q, respectively. Construct rectangle ABCD given points M, N, P, Q and the length a of side AB. 8.10. Construct a triangle given its bisector, median and height drawn from one vertex. 8.11. Construct triangle ABC given side a, angle ConstructatriangleA and the radius r of the inscribed circle. §3. Similar triangles and a homothety 8.12. Construct a triangle given two angles ∠A, ∠B and the perimeter P. 8.13. Construct triangle ABC given ma, mb and mc. 8.14. Construct triangle ABC given ha, hb and hc. 8.15. In a given acute triangle ABC inscribe square KLMN so that vertices K and N lie on sides AB and AC and vertices L and M lie on side BC. 8.16. Construct triangle ABC given ha, b −c and r. Cf. also Problems 19.15-19.20, 19.39, 19.40. §4. Construction of triangles from various elements In the problems of this section it is necessary to construct triangle ABC given the elements indicated below. 8.17. c, ma and mb. 8.18. a, b and ha. 183 184 CHAPTER 8. CONSTRUCTIONS 8.19. hb, hc and ma. 8.20. ∠A, hb and hc. 8.21. a, hb and mb. 8.22. ha, ma and hb. 8.23. a, b and mc. 8.24. ha, ma and ∠A. 8.25. a, b and lc. 8.26. ∠A, ha and p. See also Problems 17.6-17.8. §5. Construction of triangles given various points 8.27. Construct triangle ABC given (1) line l containing side AB and (2) bases A1 and B1 of heights dropped on sides BC and AC, respectively. 8.28. Construct an equilateral triangle given the bases of its bisectors. 8.29. a) Construct triangle ABC given three points A′, B′, C′ at which the bisectors of the angles of triangle ABC intersect the circumscribed circle (both triangles are supposed to be acute ones). b) Construct triangle ABC given three points A′, B′, C′ at which the heights of the triangle intersect the circumscribed circle (both triangles are supposed to be acute ones). 8.30. Construct triangle ABC given three points A′, B′, C′ symmetric to the center O of the circumscribed circle of this triangle through sides BC, CA, AB, respectively. 8.31. Construct triangle ABC given three points A′, B′, C′ symmetric to the intersection point of the heights of the triangle through sides BC, CA, AB, respectively (both triangles are supposed to be acute ones). 8.32. Construct triangle ABC given three points P, Q, R at which the height, the bisector and the median drawn from vertex C, respectively, intersect the circumscribed circle. 8.33. Construct triangle ABC given the position of points A1, B1, C1 that are the centeres of the escribed circles of triangle ABC. 8.34. Construct triangle ABC given the center of the circumscribed circle O, the inter-section point of medians, M, and the base H of height CH. 8.35. Construct triangle ABC given the centers of the inscribed, the circumscribed, and one of the escribed circles. §6. Triangles 8.36. Construct points X and Y on sides AB and BC, respectively, of triangle ABC so that AX = BY and XY ∥AC. 8.37. Construct a triangle from sides a and b if it is known that the angle opposite one of the sides is three times the angle opposite the other side. 8.38. In given triangle ABC inscribe rectangle PRQS (vertices R and Q lie on sides AB and BC and vertices P and S lie on side AC) so that its diagonal would be of a given length. 8.39. Through given point M draw a line so that it would cut from the given angle with vertex A a triangle ABC of a given perimeter 2p. 8.40. Construct triangle ABC given its median mc and bisector lc if ∠C = 90◦. 8.41. Given triangle ABC such that AB < BC, construct on side AC point D so that the perimeter of triangle ABD would be equal to the length of side BC. §8. CIRCLES 185 8.42. Construct triangle ABC from the radius of its circumscribed circle and the bisector of angle ∠A if it is known that ∠B −∠C = 90◦. 8.43. On side AB of triangle ABC point P is given. Draw a line (distinct from AB) through point P that cuts rays CA and CB at points M and N, respectively, such that AM = BN. 8.44. Construct triangle ABC from the radius of the inscribed circle r and (nonzero) lengths of segments AO and AH, where O is the center of the inscribed circle and H the orthocenter. See also Problems 15.12 b), 17.12-17.15, 18.10, 18.29. §7. Quadrilaterals 8.45. Construct a rhombus two sides of which lie on two given parallel lines and two other sides pass through two given points. 8.46. Construct quadrilateral ABCD given the lengths of the four sides and the angle between AB and CD. 8.47. Through vertex A of convex quadrilateral ABCD draw a line that divides ABCD into two parts of equal area. 8.48. In a convex quadrilateral three sides are equal. Given the midpoints of the equal sides construct the quadrilateral. 8.49. A quadrilateral is both inscribed and circumscribed. Given three of its vertices, construct its fourth vertex. 8.50. Given vertices A and C of an isosceles circumscribed trapezoid ABCD (AD ∥BC) and the directions of its bases, construct vertices B and D. 8.51. On the plane trapezoid ABCD is drawn (AD ∥BC) and perpendicular OK from the intersection point O is dropped on base AD; the midpoint EF is drawn. Then the trapezoid itself was erased. How to recover the plot of the trapezoid from the remaining segments OK and EF? 8.52. Construct a convex quadrilateral given the lengths of all its sides and one of the midlines. 8.53. (Brachmagupta.) Construct an inscribed quadrilateral given its four sides. See also Problems 15.10, 15.13, 16.17, 17.4, 17.5. §8. Circles 8.54. Inside an angle two points A and B are given. Construct a circle that passes through these points and intercepts equal segments on the sides of the angle. 8.55. Given circle S, point A on it and line l. Construct a circle tangent to the given circle at point A and tangent to the given line. 8.56. a) Two points, A, B and line l are given. Construct a circle that passes through point A, B and is tangent to l. b) Two points A, B and circle S are given. Construct a circle that passes through points A and B and is tangent to S. 8.57. Three points A, B and C are given. Construct three circles that are pairwise tangent at these points. 8.58. Construct a circle the tangents to which drawn from three given points A, B and C have given lengths a, b and c, respectively. See also Problems 15.8, 15.9, 15.11, 15.12 a), 16.13, 16.14, 16.18–16.20, 18.24. 186 CHAPTER 8. CONSTRUCTIONS §9. Apollonius’ circle 8.59. Construct triangle ABC given a, ha and b c. 8.60. Construct triangle ABC given the length of bisector CD and the lengths of segments AD and BD into which the bisector divides side AB. 8.61. On a line four points A, B, C, D are given in the indicated order. Construct point M — the vertex of equal angles that subtend segments AB, BC, CD. 8.62. Two segments AB and A′B′ are given in plane. Construct point O so that triangles AOB and A′OB′ would be similar (equal letters stand for the corresponding vertices of similar triangles). 8.63. Points A and B lie on a diameter of a given circle. Through A and B draw two equal chords with a common endpoint. §10. Miscellaneous problems 8.64. a) On parallel lines a and b, points A and B are given. Through a given point C draw line l that intersects lines a and b at points A1 and B1, respectively, and such that AA1 = BB1. b) Through point C draw a line equidistant from given points A and B. 8.65. Construct a regular decagon. 8.66. Construct a rectangle with the given ratio of sides knowing one point on each of its sides. 8.67. Given diameter AB of a circle and point C on the diameter. On this circle, construct points X and Y symmetric through line AB and such that lines AX and Y C are perpendicular. See also Problems 15.7, 16.15, 16.16, 16.21, 17.9–17.11, 17.27–17.29, 18.41. §11. Unusual constructions 8.68. With the help of a ruler and a compass divide the angle of 19◦into 19 equal parts. 8.69. Prove that an angle of value n◦, where n is an integer not divisible by 3, can be divided into n equal parts with the help of a compass and ruler. 8.70. On a piece of paper two lines are drawn. They form an angle whose vertex lies outside this piece of paper. With the help of a ruler and a compass draw the part of the bisector of the angle that lies on this piece of paper. 8.71. With the help of a two-sided ruler construct the center of the given circle whose diameter is greater than the width of the ruler. 8.72. Given points A and B; the distance between them is greater than 1 m. The length of a ruler is 10 cm. With the help of the ruler only construct segment AB. (Recall that with the help of a ruler one can only draw straight lines.) 8.73. On a circle of radius a a point is given. With the help of a coin of radius a construct the point diametrically opposite to the given one. §12. Construction with a ruler only In the problems of this section we have to perform certain constructions with the help of a ruler only, without a compass or anything else. With the help of one ruler it is almost im-possible to construct anything. For example, it is even impossible to construct the midpoint of a segment (Problem 30.59). But if certain additional lines are drawn on the plane, it is possible to perform certain constructions. In particular, if an additional circle is drawn on the plane and its center is §13. CONSTRUCTIONS WITH THE HELP OF A TWO-SIDED RULER 187 marked, then with the help of a ruler one can perform all the constructions that can be performed with the help of a ruler and a compass. One has, however, to convene that a circle is “constructed” whenever its center and one of its points are marked. Remark. If a circle is drawn on the plane but its center is not marked then to construct its center with the help of a ruler only is impossible (Problem 30.60). 8.74. Given two parallel lines and a segment that lies on one of the given lines. Divide the segment in halves. 8.75. Given two parallel lines and a segment that lies on one of the given lines. Double the segment. 8.76. Given two parallel lines and a segment that lies on one of the given lines. Divide the segment into n equal parts. 8.77. Given two parallel lines and point P, draw a line through P parallel to the given lines. 8.78. A circle, its diameter AB and point P are given. Through point P draw the perpendicular to line AB. 8.79. In plane circle S and its center O are given. Then with the help of a ruler only one can: a) additionally given a line, draw a line through any point parallel to the given line and drop the perpendicular to the given line from this point; b) additionally given a line a point on it and a length of a segment, on the given line, mark a segment of length equal to the given one and with one of the endpoints in the given point; c) additionally given lengths of a, b, c of segments, construct a segment of length ab c ; d) additionally given line l, point A and the length r of a segment, construct the inter-section points of line l with the circle whose center is point A and the radius is equal to r; e) additionally given two points and two segments, construct the intersection points of the two circles whose centers are the given points and the radii are the given segments. See also Problem 6.97. §13. Constructions with the help of a two-sided ruler In problems of this section we have to perform constructions with the help of a ruler with two parallel sides (without a compass or anything else). With the help of a two-sided ruler one can perform all the constructions that are possible to perform with the help of a compass and a ruler. Let a be the width of a two-sided ruler. By deﬁnition of the two-sided ruler with the help of it one can perform the following elementary constructions: 1) draw the line through two given points; 2) draw the line parallel to a given one and with the distance between the lines equal to a; 3) through two given points A and B, where AB ≥a, draw a pair of parallel lines the distance between which is equal to a (there are two pairs of such lines). 8.80. a) Construct the bisector of given angle ∠AOB. b) Given acute angle ∠AOB, construct angle ∠BOC whose bisector is ray OA. 8.81. Erect perpendicular to given line l at given point A. 8.82. a) Given a line and a point not on the line. Through the given point draw a line parallel to the given line. 188 CHAPTER 8. CONSTRUCTIONS b) Construct the midpoint of a given segment. 8.83. Given angle ∠AOB, line l and point P on it, draw through P lines that form together with l an angle equal to angle ∠AOB. 8.84. Given segment AB, a non-parallel to it line l and point M on it, construct the intersection points of line l with the circle of radius AB centered at M. 8.85. Given line l and segment OA, parallel to l, construct the intersection points of l with the circle of radius OA centered at O. 8.86. Given segments O1A1 and O2A2, construct the radical axis of circles of radii O1A1 and O2A2 centered at O1 and O2, respectively. §14. Constructions using a right angle In problems of this section we have to perform the constructions indicated using a right angle. A right angle enables one to perform the following elementary constructions: a) given a line and a point not on it, place the right angle so that one of its legs lies on the given line and the other leg runs through the given point; b) given a line and two points not on it, place the right angle so that its vertex lies on the given line and thelegs pass through two given points (if, certainly, for the given line and points such a position of the right angle exists). Placing the right angle in one of the indicated ways we can draw rays corresponding to its sides. 8.87. Given line l and point A not on it, draw a line parallel to l. 8.88. Given segment AB, construct a) the midpoint of AB; b) segment AC whose midpoint is point B. 8.89. Given angle ∠AOB, construct a) an angle of value 2∠AOB; b) an angle of value 1 2∠AOB. 8.90. Given angle ∠AOB and line l, draw line l1 so that the angle between lines l and l1 is equal to ∠AOB. 8.91. Given segment AB, line l and point O on it, construct on l point X such that OX = AB. 8.92. Given segment OA parallel to line l, construct the locus of points in which the disc segment of radius OA centered at O intersects l. Problems for independent study 8.93. Construct a line tangent to two given circles (consider all the possible cases). 8.94. Construct a triangle given (the lengths of) the segments into which a height divides the base and a median drawn to a lateral side. 8.95. Construct parallelogram ABCD given vertex A and the midpoints of sides BC and CD. 8.96. Given 3 lines, a line segment and a point. Construct a trapezoid whose lateral sides lie on the given lines, the diagonals intersect at the given point and one of the bases is of the given length. 8.97. Two circles are given. Draw a line so that it would be tangent to one of the circles and the other circle would intersept on it a chord of a given length. 8.98. Through vertex C of triangle ABC draw line l so that the areas of triangles AA1C and BB1C, where A1 and B1 are projections of points A and B on line l, are equal. SOLUTIONS 189 8.99. Construct triangle ABC given sides AB and AC if it is given that bisector AD, median BM, and height CH meet at one point. 8.100. Points A1, B1 and C1 that divide sides BC, CA and AB, respectively, of triangle ABC in the ratio of 1 : 2 are given. Recover triangle ABC from this data. Solutions 8.1. Let us construct segment BC of length a. The center O of the circumscribed circle of triangle ABC is the intersection point of two circles of radius R centered at B and C. Select one of these intersection points and construct the circumscribed circle S of triangle ABC. Point A is the intersection point of circle S and a line parallel line BC and whose distance from BC is equal to ha (there are two such lines). 8.2. Let us construct points A1 and B1 on sides BC and AC, respectively, so that BA1 : A1C = 1 : 3 and AB1 : B1C = 1 : 2. Let point X lie inside triangle ABC. Clearly, SABX : SBCX = 1 : 2 if and only if point X lies on segment BB1 and SABX : SACX = 1 : 3 if and only if point X lies on segment AA1. Therefore, the point M to be constucted is the intersection point of segments AA1 and BB1. 8.3. Let O be the center of the given circle, AB a chord that passes through point P and M the midpoint of AB. Then \|AP −BP\| = 2PM. Since ∠PMO = 90◦, point M lies on circle S with diameter OP. Let us construct chord PM of circle S so that PM = 1 2a (there are two such chords). The chord to be constructed is determined by line PM. 8.4. Let R be the radius of the given circle, O its center. The center of the circle to be constructed lies on circle S of radius R + r centered at O. On the other hand, the center to be constructed lies on line l passing parallelly to the given line at distance r (there are two such lines). Any intersection point of S with l can serve as the center of the circle to be constructed. 8.5. Let R be the radius of circle S and O its center. If circle S intersepts on the line that passes through point A chord PQ and M is the midpoint of PQ, then OM 2 + OQ2 −NQ2 = R2 −d2 4 . Therefore, the line to be constructed is tangent to the circle of radius q R2 −d2 4 centered at O. 8.6. On lines AB and CD take points E and F so that lines BF and CE would have had prescribed directions. Let us considered all possible parallelograms PQRS with prescribed directions of sides whose vertices P and R lie on rays BA and CD and vertex Q lies on side BC (Fig. 85). Figure 85 (8.6) 190 CHAPTER 8. CONSTRUCTIONS Let us prove that the locus of vertices S is segment EF. Indeed, SR EC = PQ EC = BQ BC = FR FC, i.e., point S lies on segment EF. Conversely if point S′ lies on segment EF then let us draw lines S′P ′, P ′Q′ and Q′R′ so that S′P ′ ∥BF, P ′Q′ ∥EC and Q′R′ ∥BF, where P ′, Q′ and R′ are some points on lines AB, BC, CD, respectively. Then S′P ′ BF = P ′E BE = Q′C BC = Q′R′ BF , i.e., S′P ′ = Q′R′ and P ′Q′R′S′ is a parallelogram. This implies the following construction. First, construct points E and F. Vertex S is the intersection point of segments AD and EF. The continuation of construction is obvious. 8.7. Suppose that triangle ABC is constructed. Let A1 and C1 be the midpoints of sides CD and AB, respectively. Since C1A1 ∥AC, it follows that ∠A1C1B = ∠A. This implies the following construction. First, let us construct segment CD of length a and its midpoint, A1. Point C1 is the intersection point of the circle of radius mc centered at C and the arcs of the circles whose points are vertices of the angles equal to ∠A that segment A1B subtends. Construct point C1, then mark on ray BC1 segment BA = 2BC1. Then A is the vertex of the triangle to be constructed. 8.8. Suppose that the desired triangle is constructed and C is the vertex of its right angle. Since ∠ACB = 90◦, point C lies on circle S with diameter AB. Hence, point C is the intersection point of circle S and the given circle. Constructing point C and drawing lines CA and AB, we ﬁnd the remaining vertices of the triangle to be constructed. 8.9. Suppose that rectangle ABCD is constructed. Let us drop perpendicular PR from point P to line BC. Point R can be constructed because it lies on the circle with diameter PQ and PR = AB = a. Constructing point R, let us construct lines BC and AD and drop on them perpendiculars from points M and N, respectively. 8.10. Suppose that triangle ABC is constructed, AH is its height, AD its bisector, AM its median. By Problem 2.67 point D lies between M and H. Point E, the intersection point of line AD with the perpendicular drawn from point M to side BC, lies on the circumscribed circle of triangle ABC. Hence, the center O of the circumscribed circle lies on the intersection of the midperpendicular to segment AE and the perpendicular to side BC drawn through point M. The sequence of constructions is as follows: on an arbitrary line (which in what follows turns out to be line BC) construct point H, then consecutively construct points A, D, M, E, O. The desired vertices B and C of triangle ABC are intersection points of the initial line with the circle of radius OA centered at O. 8.11. Suppose that triangle ABC is constructed and O is the center of its inscribed circle. Then ∠BOC = 90◦+ 1 2∠A (Problem 5.3). Point O is the vertex of an angle of 90◦+ 1 2∠A that subtends segment BC; the distance from O to line BC is equal to r, hence, BC(??) can be constructed. Further, let us construct the inscribed circle and draw the tangents to it from points B and C. 8.12. Let us construct any triangle with angles ∠A and ∠B and ﬁnd its perimeter P1. The triangle to be found is similar to the constructed triangle with coeﬃcient P P1. 8.13. Suppose that triangle ABC is constructed. Let AA1, BB1 and CC1 be its medians, M their intersection point, M ′ the point symmetric to M through point A1. Then MM ′ = 2 3ma, MC = 2 3mc and M ′C = 2 3mb; hence, triangle MM ′C can be constructed. Point A is symmetric to M ′ through point M and point B is symmetric to C through the midpoint of segment MM ′. 8.14. Clearly, BC : AC : AB = S ha : S hb : S hc = 1 ha : 1 hb : 1 hc . SOLUTIONS 191 Let us take an arbitrary segment B′C′ and construct triangle A′B′C′ so that B′C′ : A′C′ = hb : ha and B′C′ : A′B′ = hc : ha. Let h′ a be the height of triangle A′B′C′ dropped from vertex A′. The triangle to be found is similar to triangle A′B′C′ with coeﬃcient ha h′ a. 8.15. On side AB, take an arbitrary point K′ and drop from it perpendicular K′L′ to side BC; then construct square K′L′M ′N ′ that lies inside angle ∠ABC. Let line BN ′ intersect side AC at point N. Clearly, the square to be constructed is the image of square K′L′M ′N ′ under the homothety with center B and coeﬃcient BN : BN ′. 8.16. Suppose that the desired triangle ABC is constructed. Let Q be the tangent point of the inscribed circle with side BC; let PQ be a diameter of the circle, R the tangent point of an escribed circle with side BC. Clearly, BR = a + b + c 2 −c = a + b −c 2 and BQ = a + c −b 2 . Hence, RQ = \|BR −BQ\| = \|b −c\|. The inscribed circle of triangle ABC and the escribed circle tangent to side BC are homothetic with A being the center of homothety. Hence, point A lies on line PR (Fig. 86). Figure 86 (Sol. 8.16) This implies the following construction. Let us construct right triangle PQR from the known legs PQ = 2r and RQ = \|b −c\|. Then draw two lines parallel to line RQ and whose distances from RQ are equal to ha. Vertex A is the intersection point of one of these lines with ray RP. Since the length of diameter PQ of the inscribed circle is known, it can be constructed. The intersection points of the tangents to this circle drawn from point A with line RQ are vertices B and C of the triangle. 8.17. Suppose that triangle ABC is constructed. Let M be the intersection point of medians AA1 and BB1. Then AM = 2 3ma and BM = 2 3mb. Triangle ABM can be constructed from the lengths of sides AB = c, AM and BM. Then on rays AM and BM segments AA1 = ma and BB1 = mb should be marked. Vertex C is the intersection point of lines AB1 and A1B. 8.18. Suppose triangle ABC is constructed. Let H be the base of the height dropped from vertex A. Right triangle ACH can be constructed from its hypothenuse AC = b and leg AH = ha. Then on line CH construct point B so that CB = a. 8.19. Suppose that triangle ABC is constructed. Let us draw from the midpoint A1 of side BC perpendiculars A1B′ and A1C′ to lines AC and AB, respectively. Clearly, AA1 = ma, A1B′ = 1 2hb and A1C′ = 1 2hc. This implies the following construction. First, let us construct segment AA1 of length ma. Then construct right triangles AA1B′ and AA1C′ from the known legs and hypothenuse so that they would lie on distinct sides of line AA1. It remains to construct points B and C on sides AC′ and AB′ of angle C′AB′ so that segment BC would be divided by points A1 in halves. For this let us mark on ray AA1 segment AD = 2AA1 and then draw through point D the lines parallel to the legs of 192 CHAPTER 8. CONSTRUCTIONS Figure 87 (Sol. 8.19) angle ∠C′AB′. The intersection points of these lines with the legs of angle ∠C′AB′ are the vertices of the triangle to be constructed (Fig. 87). 8.20. Let us construct angle ∠B′AC′ equal to ∠A. Point B is constructed as the intersection of ray AB′ with a line parallel to ray AC′ and passsing at distance hb from it. Point C is similarly constructed. 8.21. Suppose that triangle ABC is constructed. Let us drop height BH from point B and draw median BB1. In right triangles CBH and B1BH, leg BH and hypothenuses CB and B1B are known; hence, these segments can be constructed. Then on ray CB1 we mark segment CA = 2CB1. The problem has two solutions because we can construct triangles CBH and B1BH either on one or on distinct sides of line BH. 8.22. Suppose that triangle ABC is constructed. Let M be the midpoint of segment BC. From point A drop height AH and from point M drop perpendicular MD to side AC. Clearly, MD = 1 2hb. Hence, triangles AMD and AMH can be constructed. Vertex C is the intersection point of lines AD and MH. On ray CM, mark segment CB = 2CM. The problem has two solutions because triangles AMD and AMH can be constructed either on one or on distinct sides of line AM. 8.23. Suppose that triangle ABC is constructed. Let A1, B1 and C1 be the midpoints of sides BC, CA and AB, respectively. In triangle CC1B1 all the sides are known: CC1 = mc, C1B1 = 1 2a and CB1 = 1 2b; hence, it can be constructed. Point A is symmetric to C through point B1 and point B is symmetric to A through C1. 8.24. Suppose that triangle ABC is constructed, AM is its median, AH its height. Let point A′ be symmetric to A through point M. Let us construct segment AA′ = 2ma. Let M be the midpoint of AA′. Let us construct right triangle AMH with hypothenuse AM and leg AH = ha. Point C lies on an arc of the circle whose points are the vertices of the angles that subtend segment AA′; the values of these angles are equal to 180◦−∠A because ∠ACA′ = 180◦−∠CAB. Hence, point C is the intersection point of this arc and line MH. Point B is symmetric to C through point M. 8.25. Suppose triangle ABC is constructed. Let CD be its bisector. Let us draw line MD parallel to side BC (point M lies on side AC). Triangle CMD is an isosceles one because ∠MCD = ∠DCB = ∠MDC. Since MC : AM = DB : AD = CB : AC = a : b and AM + MC = b, it follows that MC = ab a+b. Let us construct an isosceles triangle CMD from its base CD = lc and lateral sides MD = MC = ab a+b. Further, on ray CM, mark segment CA = b and on the ray symmetric to ray CM through line CD mark segment CB = a. 8.26. Suppose that triangle ABC is constructed. Let S1 be the escribed circle tangent to side BC. Denote the tangent points of circle S1 with the extensions of sides AB and AC by K and L, respectively, and the tangent point of S1 with side BC by M. Since AK = AL, AL = AC + CM and AK = AB + BM, SOLUTIONS 193 it follows that AK = AL = p. Let S2 be the circle of radius ha centered at A. Line BC is a common inner tangent to circles S1 and S2. This implies the following construction. Let us construct angle ∠KAL whose value is equal to that of A so that KA = LA = p. Next, construct circle S1 tangent to the sides of angle ∠KAL at points K and L and circle S2 of radius ha centered at A. Then let us draw a common inner tangent to circles S1 and S2. The intersection points of this tangent with the legs of angle ∠KAL are vertices B and C of the triangle to be constructed. 8.27. Points A1 and B1 lie on the circle S with diameter AB. The center O of this circle lies on the midperpendicular to chord A1B1. This implies the following construction. First, let us construct point O which is the intersection point of the midperpendicular to segment A1B1 with line l. Next, construct the circle of radius OA1 = OB1 centered at O. The vertices A and B are the intersection points of circle S with line l. Vertex C is the intersection point of lines AB1 and BA1. 8.28. Let AB = BC and A1, B1, C1 the bases of the bisectors of triangle ABC. Then ∠A1C1C = ∠C1CA = ∠C1CA1, i.e., triangle CA1C1 is an isosceles one and A1C = A1C1. This implies the following construction. Let us draw through point B1 line l parallel to A1C1. On l, construct point C such that CA1 = C1A1 and ∠C1A1C > 90◦. Point A is symmetric to point C through point B1 and vertex B is the intersection point of lines AC1 and A1C. 8.29. a) By Problem 2.19 a) points A, B and C are the intersection points of the extensions of heights of triangle A′B′C′ with its circumscribed circle. b) By Problem 2.19 b) points A, B and C are the intersection points of the extensions of bisectors of the angles of triangle A′B′C′ with its circumscribed circle. 8.30. Denote the midpoints of sides BC, CA, AB of the triangle by A1, B1, C1, respec-tively. Since BC ∥B1C1 ∥B′C′ and OA1 ⊥BC, it follows that OA′ ⊥B′C′. Similarly, OB′ ⊥A′C′ and OC′ ⊥A′B′, i.e., O is the intersection point of the heights of triangle A′B′C′. Constructing point O, let us draw the midperpendiculars to segments OA′, OB′, OC′. These lines form triangle ABC. 8.31. Thanks to Problem 5.9 our problem coincides with Problem 8.29 b). 8.32. Let O be the center of the circumscribed circle, M the midpoint of side AB and H the base of the height dropped from point C. Point Q is the midpoint of arc ⌣AB, therefore, OQ ⊥AB. This implies the following construction. First, the three given points determine the circumscribed circle S of triangle PQR. Point C is the intersection point of circle S and the line drawn parallelly to OQ through point P. Point M is the intersection point of line OQ and line RC. Line AB passes through point M and is perpendicular to OQ. 8.33. By Problem 5.2, points A, B and C are the bases of the heights of triangle A1B1C1. 8.34. Let H1 be the intersection point of heights of triangle ABC. By Problem 5.105, OM : MH1 = 1 : 2 and point M lies on segment OH1. Therefore, we can construct point H1. Then let us draw line H1H and erect at point H of this line perpendicular l. Dropping perpendicular from point O to line l we get point C1 (the midpoint of segment AB). On ray C1M, construct point C so that CC1 : MC1 = 3 : 1. Points A and B are the intersection points of line l with the circle of radius CO centered at O. 8.35. Let O and I be the centers of the circumscribed and inscribed circles, Ic the center of the escribed circle tangent to side AB. The circumscribed circle of triangle ABC divides segment IIc (see Problem 5.109 b)) in halves and segment IIc divides arc ⌣AB in halves. It is also clear that points A and B lie on the circle with diameter IIc. This implies the following construction. 194 CHAPTER 8. CONSTRUCTIONS Let us construct circle S with diameter IIc and circle S1 with center O and radius OD, where D is the midpoint of segment IIc. Circles S and S1 intersect at points A and B. Now, we can construct the inscribed circle of triangle ABC and draw tangents to it at points A and B. 8.36. Suppose that we have constructed points X and Y on sides AB and BC, respec-tively, of triangle ABC so that AX = BY and XY ∥AC. Let us draw Y Y1 parallel to AB and Y1C1 parallel to BC (points Y1 and C1 lie on sides AC and AB, respectively). Then Y1Y = AX = BY , i.e., BY Y1C is a rhombus and BY1 is the bisector of angle ∠B. This implies the following construction. Let us draw bisector BY1, then line Y1Y parallel to side AB (we assume that Y lies on BC). Now, it is obvious how to construct point X. 8.37. Let, for deﬁniteness, a < b. Suppose that triangle ABC is constructed. On side AC, take point D such that ∠ABD = ∠BAC. Then ∠BDC = 2∠BAC and ∠CBD = 3∠BAC −∠BAC = 2∠BAC, i.e., CD = CB = a. In triangle BCD all the sides are known: CD = CB = a and DB = AD = b −a. Constructing triangle BCD, draw ray BA that does not intersect side CD so that ∠DBA = 1 2∠DBC. Vertex A to be constructed is the intersection point of line CD and this ray. 8.38. Let point B′ lie on line l that passes through point B parallelly to AC. Sides of triangles ABC and AB′C intersept equal segments on l. Hence, rectangles P ′R′Q′S′ and PRQS inscribed in triangles ABC and AB′C, respectively, are equal if points R, Q, R′ and Q′ lie on one line. On line l, take point B′ so that ∠B′AC = 90◦. It is obvious how to inscribe rectangle P ′R′Q′S′ with given diagonal P ′Q′ in triangle AB′C (we assume that P ′ = A). Draw line R′Q′; we thus ﬁnd vertices R and Q of the rectangle to be found. 8.39. Suppose that triangle ABC is constructed. Let K and L be points at which the escribed circle tangent to side BC is tangent to the extensions of sides AB and AC, respectively. Since AK = AL = p, this escribed circle can be constructed; it remains to draw the tangent through the given point M to the constructed circle. 8.40. Let the extension of the bisector CD intersect the circumscribed circle of triangle ABC (with right angle ∠C) at point P, let PQ be a diameter of the circumscribed circle and O its center. Then PD : PO = PQ : PC, i.e., PD · PC = 2R2 = 2m2 c. Therefore, drawing a tangent of length √ 2mc to the circle with diameter CD, it is easy to construct a segment of length PC. Now, the lengths of all the sides of triangle OPC are known. 8.41. Let us construct point K on side AC so that AK = BC −AB. Let point D lie on segment AC. The equality AD + BD + AB = BC is equivalent to the equality AD + BD = AK. For point D that lies on segment AK the latter equality takes the form AD + BD = AD + DK and for point D outside segment AK it takes the form AD + BD = AD −DK. In the ﬁrst case BD = DK and the second case is impossible. Hence, point D is the intersection point of the midperpendicular to segment BK and segment AC. 8.42. Suppose that triangle ABC is constructed. Let us draw diameter CD of the circumscribed circle. Let O be the center of the circumscribed circle, L the intersection point of the extension of the bisector AK with the circumscribed circle (Fig. 88). Since ∠ABC −∠ACB = 90◦, it follows that ∠ABD = ∠ACB; hence, ⌣DA =⌣AB. It is also clear that ⌣BL =⌣LC. Therefore, ∠AOL = 90◦. This implies the following construction. Let us construct circle S with center O and a given radius. On circle S select an arbitrary point A. Let us construct a point L on circle S so that ∠AOL = 90◦. On segment AL, construct segment AK whose length is equal to SOLUTIONS 195 Figure 88 (Sol. 8.42) that of the given bisector. Let us draw through point K line l perpendicular to OL. The intersection points of l with circle S are vertices B and C of triangle ABC to be constructed. 8.43. On sides BC and AC, take points A1 and B1 such that PA1 ∥AC and PB1 ∥BC. Next, on rays A1B and B1A mark segments A1B2 = AB1 and B1A2 = BA1. Let us prove that line A2B2 is the one to be found. Indeed, let k = AP AB. Then BA2 BP = (1 −k)a ka = (1 −k)a + (1 −k)b ka + kb = CA2 CB2 , i.e., △A2B1P ∼△A2CB2 and line A2B2 passes through point P. Moreover, AA2 = \|(1 − k)a −kb\| = BB2. 8.44. Suppose that triangle ABC is constructed. Let B1 be the tangent point of the inscribed circle with side AC. In right triangle AOB1 leg OB1 = r and hypothenuse AO are known, therefore, we can construct angle ∠OAB1, hence, angle ∠BAC. Let O1 be the center of the circumscribed circle of triangle ABC, let M be the midpoint of side BC. In right triangle BO1M leg O1M = 1 2AH is known (see solution to Problem 5.105) and angle ∠BO1M is known (it is equal to either ∠A or 180◦−∠A); hence, it can be constructed. Next, we can determine the length of segment OO1 = p R(R −2r), cf. Problem 5.11 a). Thus, we can construct segments of length R and OO1 = d. After this take segment AO and construct point O1 for which AO1 = R and OO1 = d (there could be two such points). Let us draw from point A tangents to the circle of radius r centered at O. Points B and C to be found lie on these tangents and their distance from point O1 is equal to R; obviously, points B and C are distinct from point A. 8.45. Let the distance between the given parallel lines be equal to a. We have to draw parallel lines through points A and B so that the distance between the lines is equal to a. To this end, let us construct the circle with segment AB as its diameter and ﬁnd the intersection points C1 and C2 of this circle with the circle of radius a centered at B. A side of the rhombus to be constructed lies on line AC1 (another solution: it lies on AC2). Next, let us draw through point B the line parallel to AC1 (resp. AC2). 8.46. Suppose that quadrilateral ABCD is constructed. Let us denote the midpoints of sides AB, BC, CD and DA by P, Q, R and S, respectively, and the midpoints of diagonals AC and BD by K and L, respectively. In triangle KSL we know KS = 1 2CD, LS = 1 2AB and angle ∠KSL equal to the angle between the sides AB and CD. Having constructed triangle KSL, we can construct triangle KRL because the lengths of all its sides are known. After this we complement triangles KSL and KRL to parallelograms KSLQ and KRLP, respectively. Points A, B, C, D are vertices of parallelograms PLSA, QKPB, RLQC, SKRD (Fig. 89). 8.47. Let us drop perpendiculars BB1 and DD1 from vertices B and D, respectively, to diagonal AC. Let, for deﬁniteness, DD1 > BB1. Let us construct a segment of length 196 CHAPTER 8. CONSTRUCTIONS Figure 89 (Sol. 8.46) a = DD1 −BB1; draw a line parallel to line AC and such that the distance between this line and AC is equal to a and which intersects side CD at a point, E. Clearly, SAED = ED CDSACD = BB1 DD1 SACD = SABC. Therefore, the median of triangle AEC lies on the line to be constructed. 8.48. Let P, Q, R be the midpoints of equal sides AB, BC, CD of quadrilateral ABCD. Let us draw the midperpendiculars l1 and l2 to segments PQ and QR. Since AB = BC = CD, it is clear that points B and C lie on lines l1 and l2 and BQ = QC. This implies the following construction. Let us draw the midperpendiculars l1 and l2 to segments PQ and QR, respectively. Then through point Q we draw a segment with endpoints on lines l1 and l2 so that Q were its midpoint, cf. Problem 16.15. 8.49. Let vertices A, B and C of quadrilateral ABCD which is both inscribed and circumscribed be given and AB ≥BC. Then AD −CD = AB −BC ≥0. Hence, on side AD we can mark segment DC1 equal to DC. In triangle AC1C the lengths of sides AC and AC1 = AB −BC are known and ∠AC1C = 90◦+ 1 2∠D = 180◦−1 2∠B. Since angle ∠AC1C is an obtuse one, triangle AC1C is uniquely recoverable from these elements. The remaining part of the construction is obvious. 8.50. Let ABCD be a circumscribed equilateral trapezoid with bases AD and BC such that AD > BC; let C1 be the projection of point C to line AD. Let us prove that AB = AC1. Indeed, if P and Q are the tangent points of sides AB and AD with the inscribed circle, then AB = AP + PB = AQ + 1 2BC = AQ + QC1 = AC1. This implies the following construction. Let C1 be the projection of point C to base AD. Then B is the intersection point of line BC and the circle of radius AC1 centered at A. A trapezoid with AD < BC is similarly constructed. 8.51. Let us denote the midpoints of bases AD and BC by L and N and the midpoint of segment EF by M. Points L, O, N lie on one line (by Problem 19.2). Clearly, point M also lies on this line. This implies the following construction. Let us draw through point K line l perpendicular to line OK. Base AD lies on l. Point L is the intersection point of l and line OM. Point N is symmetric to point L through point M. Let us draw lines through point O parallel to lines EN and FN. The intersection points of the lines we have just drawn are vertices A and D of the trapezoid. Vertices B and C are symmetric to vertices A and D through points E and F, respectively. 8.52. Suppose that we have constructed quadrilateral ABCD with given lengths of sides and a given midline KP (here K and P are the midpoints of sides AB and CD, respectively). Let A1 and B1 be the points symmetric to points A and B, respectively, through point P. Triangle A1BC can be constructed because its sides BC, CA1 = AD and BA1 = 2KP are SOLUTIONS 197 known. Let us complement triangle A1BC to parallelogram A1EBC. Now we can construct point D because CD and ED = BA are known. Making use of the fact that − − → DA = − − → A1C we construct point A. 8.53. Making use of the formulas of Problems 6.34 and 6.35 it is easy to express the lengths of the diagonals of the inscribed triangle in terms of the lengths of its sides. The obtained formulas can be applied for the construction of the diagonals (for convenience it is advisable to introduce an arbitrary segment e as the measure of unit length and construct segments of length pq, p q and √p as pq e , pe q and √pe). 8.54. A circle intercepts equal segments on the legs of an angle if and only if the center of the circle lies on the bisector of the angle. Therefore, the center of the circle to be found is the intersection point of the midperpendicular to segment AB and the bisector of the given angle. 8.55. Let us suppose that we have constructed circle S′ tangent to the given circle S at point A and the given line l at a point, B. Let O and O′ be the centers of circles S and S′, respectively (Fig. 90). Clearly, points O, O′ and A lie on one line and O′B = O′A. Hence, we have to construct point O′ on line OA so that O′A = O′B, where B is the base of the perpendicular dropped from point O′ to line l. Figure 90 (Sol. 8.55) To this end let us drop perpendiculr OB′ on line l. Next, on line AO mark segment OA′ of length OB′. Let us draw through point A line AB parallel to A′B′ (point B lies on line l). Point O′ is the intersection point of line OA and the perpendicular to l drawn through point B. 8.56. a) Let l1 be the midperpendicular to segment AB, let C be the intercection point of lines l1 and l; let l′ be the line symmetric to l through line l1. The problem reduces to the necessity to construct a circle that passes through point A and is tangent to lines l and l′, cf. Problem 19.15. b) We may assume that the center of circle S does not lie on the midperpendicular to segment AB (otherwise the construction is obvious). Let us take an arbitrary point C on circle S and construct the circumscribed circle of triangle ABC; this circle intersects S at a point D. Let M be the intersection point of lines AB and CD. Let us draw tangents MP and MQ to circle S. Then the circumscribed circles of triangles ABP and ABQ are the ones to be found since MP 2 = MQ2 = MA · MB. 8.57. Suppose we have constructed circles S1, S2 and S3 tangent to each other pairwise at given points: S1 and S2 are tangent at point C; circles S1 and S3 are tangent at point B; circles S2 and S3 are tangent at point A. Let O1, O2 and O3 be the centers of circles S1, S2 and S3, respectively. Then points A, B and C lie on the sides of triangle O1O2O3 and O1B = O1C, O2C = O2A and O3A = O3B. Hence, points A, B and C are the tangent points of the inscribed circle of triangle O1O2O3 with its sides. 198 CHAPTER 8. CONSTRUCTIONS This implies the following construction. First, let us construct the circumscribed circle of triangle ABC and draw tangents to it at points A, B and C. The intersection points of these tangents are the centers of circles to be found. 8.58. Suppose that we have constructed circle S whose tangents AA1, BB1 and CC1, where A1, B1 and C1 are the tangent points, are of length a, b and c, respectively. Let us construct circles Sa, Sb and Sc with the centers A, B and C and radii a, b and c, respectively (Fig. 91). If O is the center of circle S, then segments OA1, OB1 and OC1 are radii of circle S and tangents to circles Sa, Sb and Sc as well. Hence, point O is the radical center (cf. §3.10) of circles Sa, Sb and Sc. Figure 91 (Sol. 8.58) This implies the following construction. First, construct circles Sa, Sb and Sc. Then let us construct their radical center O. The circle to be found is the circle with center O and the radius whose length is equal to that of the tangent drawn from point O to circle Sa. 8.59. First, let us construct segment BC of length a. Next, let us construct the locus of points X for which CX : BX = b : c, cf. Problem 7.14. For vertex A we can take any of the intersection points of this locus with a line whose distance from line BC is equal to ha. 8.60. Given the lengths of segments AD′ and BD, we can construct segment AB and point D on this segment. Point C is the intersection point of the circle of radius CD centered at D and the locus of points X for which AX : BX = AD : BD. 8.61. Let X be a point that does not lie on line AB. Clearly, ∠AXB = ∠BCX if and only if AX : CX = AB : CB. Hence, point M is the intersection point of the locus of points X for which AX : CX = AB : CB and the locus of points Y for which BY : DY = BC : DC (it is possible for these loci not to intersect). 8.62. We have to construct a point O for which AO : A′O = AB : A′B′ and BO : B′O = AB : A′B′. Point O is the intersection point of the locus of points X for which AX : A′X = AB : A′B′ and the locus of points Y for which BY : B′Y = AB : A′B′. 8.63. Let O be the center of the given circle. Chords XP and XQ that pass through points A and B are equal if and only if XO is the bisector of angle PXO, i.e., AX : BX = AO : BO. The point X to be found is the intersection point of the corresponding Apollonius’s circle with the given circle. 8.64. a) If line l does not intersect segment AB, then ABB1A1 is a parallelogram and l ∥AB. If line l intersects segment AB, then AA1BB1 is a parallelogram and l passes through the midpoint of segment AB. b) One of the lines to be found is parallel to line AB and another one passes through the midpoint of segment AB. 8.65. Let us construct a circle of radius 1 and in it draw two perpendicular diameters, AB and CD. Let O be the center of the circle, M the midpoint of segment OC, P the intersection SOLUTIONS 199 point of line AM and the circle with diameter OC (Fig. 92). Then AM 2 = 1 + 1 4 = 5 4 and, therefore, AP = AM −PM = √ 5−1 2 = 2 sin 18◦(cf. Problem 5.46), i.e., AP is the length of a side of a regular decagon inscribed in the given circle. Figure 92 (Sol. 8.65) 8.66. Suppose we have constructed rectangle PQRS so that the given points A, B, C, D lie on sides PQ, QR, RS, SP, respectively, and PQ : QR = a, where a is the given ratio of sides. Let F be the intersection point of the line drawn through point D perpendicularly to line AC and line QR. Then DF : AC = a. This implies the following construction. From point D draw a ray that intersects segment AC at a right angle and on this ray construct a point F so that DF = a · AC. Side QR lies on line BF. The continuation of the construction is obvious. 8.67. Suppose that points X and Y with the required properties are constructed. Denote the intersection point of lines AX and Y C by M, that of lines AB and XY by K. Right triangles AXK and Y XM have a common acute angle ∠X, hence, ∠XAK = ∠XY M. Angles ∠XAB and ∠XY B subtend the same arc, hence, ∠XAB = ∠XY B. Therefore, ∠XY M = ∠XY B. Since XY ⊥AB, it follows that K is the midpoint of segment CB. Conversely, if K is the midpoint of segment CB, then ∠MY X = ∠BY X = ∠XAB. Triangles AXK and Y XM have a common angle ∠X and ∠XAK = ∠XY M; hence, ∠Y MX = ∠AKX = 90◦. This implies the following construction. Through the midpoint K of segment CB draw line l perpendicular to line AB. Points X and Y are the intersection points of line l with the given circle. 8.68. If we have an angle of value α, then we can construct angles of value 2α, 3α, etc. Since 19 · 19◦= 361◦, we can construct an angle of 361◦that coincides with the angle of 1◦. 8.69. First, let us construct an angle of 36◦, cf. Problem 8.65. Then we can construct the angle of 36◦−30◦ 2 = 3◦. If n is not divisible by 3, then having at our disposal angles of n◦ and 3◦we can construct an angle of 1◦. Indeed, if n = 3k + 1, then 1◦= n◦−k · 3◦and if n = 3k + 2, then 1◦= 2n◦−(2k + 1) · 3◦. 8.70. The sequence of constructions is as follows. On the piece of paper take an arbitrary point O and perform the homothety with center O and suﬃciently small coeﬃcient k so that this homothety sends the image of the intersection point of the given lines on the piece of paper. Then we can construct the bisector of the angle between the images of the lines. Next, let us perform the homothety with the same center and coeﬃcient 1 k which yields the desired segment of the bisector. 8.71. Let us construct with the help of a two-sided ruler two parallel chords AB and CD. Let P and Q be the intersection points of lines AC with BD and AD with BC, respectively. Then line PQ passes through the center of the given circle. Constructing similarly one more such line we ﬁnd the center of the circle. 200 CHAPTER 8. CONSTRUCTIONS 8.72. Let us draw through point A two rays p and q that form a small angle inside which point B lies (the rays can be constructed by replacing the ruler). Let us draw through point B segments PQ1 and P1Q (Fig. 93). If PQ < 10 cm and P1Q1 < 10 cm, then we can construct point O at which lines PQ and P1Q1 intersect. Figure 93 (Sol. 8.72) Through point O draw line P2Q2. If PQ2 < 10 cm and P2Q < 10 cm; then we can construct point B′ at which lines PQ2 and P2Q intersect. If BB′ < 10 cm, then by Problem 5.67 we can construct line BB′; this line passes through point A. 8.73. The construction is based on the fact that if A and B are the intersection points of equal circles centered at P and Q, then − → PA = − − → BQ. Let S1 be the initial circle, A1 the given point. Let us draw circle S2 through point A1 and circle S3 through the intersection point A2 of circles S1 and S2; circle S4 through the intersection point A3 of circles S2 and S3 and, ﬁnally, circle S5 through the intersection points B1 and A4 of circles S1 and S3, respectively, with circle S4. Let us prove that the intersection point B2 of circles S5 and S1 is the one to be found. Let Oi be the center of circle Si. Then − − − → A1O1 = − − − → O2A2 = − − − → A3O3 = − − − → O4A4 = − − − → B1O5 = − − − → O1B2. Remark. There are two intersection points of circles S1 and S4; for point B1 we can take any of them. 8.74. Let AB be the given segment, P an arbitrary point not on the given lines. Let us construct the intersection points C and D of the second of the given lines with lines PA and PB, respectively, and the intersection point Q of lines AD and BC. By Problem 19.2 line PQ passes through the midpoint of segment AB. 8.75. Let AB be the given segment; let C and D be arbitrary points on the second of given lines. By the preceding problem we can construct the midpoint, M, of segment CD. Let P be the intersection point of lines AM and BD; let E be the intersection point of lines PC and AB. Let us prove that EB is the segment to be found. Since △PMC ∼△PAE and △PMD ∼△PAB, it follows that AB AE = AB AP : AE AP = MD MP : MC MP = MD MC = 1. 8.76. Let AB be the given segment; let C and D be arbitrary points on the second of the given lines. By the preceding problem we can construct points D1 = D, D2, . . . , Dn such that all the segments DiDi+1 are equal to segment CD. Let P be the intersection point of lines AC and BDn and let B1, . . . , Bn−1 be the intersection points of line AB with lines PD1, . . . , PDn−1, respectively. Clearly, points B1, . . . , Bn−1 divide segment AB in n equal parts. SOLUTIONS 201 8.77. On one of the given lines take segment AB and construct its midpoint, M (cf. Problem 8.74). Let A1 and M1 be the intersection points of lines PA and PM with the second of the given lines, Q the intersection point of lines BM1 and MA1. It is easy to verify that line PQ is parallel to the given lines. 8.78. In the case when point P does not lie on line AB, we can make use of the solution of Problem 3.36. If point P lies on line AB, then we can ﬁrst drop perpendiculars l1 and l2 from some other points and then in accordance with Problem 8.77 draw through point P the line parallel to lines l1 and l2. 8.79. a) Let A be the given point, l the given line. First, let us consider the case when point O does not lie on line l. Let us draw through point O two arbitrary lines that intersect line l at points B and C. By Problem 8.78, in triangle OBC, heights to sides OB and OC can be dropped. Let H be their intersection point. Then we can draw line OH perpendicular to l. By Problem 8.78 we can drop the perpendicular from point A to OH. This is the line to be constructed that passes through A and is parallel to l. In order to drop the perpendicular from A to l we have to erect perpendicular l′ to OH at point O and then drop the perpendicular from A to l′. If point O lies on line l, then by Problem 8.78 we can immediately drop the perpendicular l′ from point A to line l and then erect the perpendicular to line l′ from the same point A. b) Let l be the given line, A the given point on it and BC the given segment. Let us draw through point O lines OD and OE parallel to lines l and BC, respectively (D and E are the intersection points of these lines with circle S). Let us draw through point C the line parallel to OB to its intersection with line OE at point F and through point F the line parallel to ED to its intersection with OD at point G and, ﬁnally, through point G the line parallel to OA to its intersection with l at point H. Then AH = OG = OF = BC, i.e., AH is the segment to be constructed. c) Let us take two arbitrary lines that intersect at point P. Let us mark on one of them segment PA = a and on the other one segments PB = b and PC = c. Let D be the intersection point of line PA with the line that passes through B and is parallel to AC. Clearly, PD = ab c . d) Let H be the homothety (or the parallel translation) that sends the circle with center A and radius r to circle S (i.e., to the given circle with the marked center O). Since the radii of both circles are known, we can construct the image of any point X under the mapping H. For this we have to draw through point O the line parallel to line AX and mark on it a segment equal to rs·AX r , where rs is the radius of circle S. We similarly construct the image of any point under the mapping H−1. Hence, we can construct the line l′ = H(l) and ﬁnd its intersection points with circle S and then construct the images of these points under the map H−1. e) Let A and B be the centers of the given circles, C one of the points to be constructed, CH the height of triangle ABC. From Pythagoras theorem for triangles ACH and BCH we deduce that AH = b2+c2−a2 2c . The quantities a, b and c are known, hence, we can construct point H and the intersection points of line CH with one of the given circles. 8.80. a) Let us draw lines parallel to lines OA and OB, whose distance from the latter lines is equal to a and which intersect the legs of the angles. The intersection point of these lines lies on the bisector to be constructed. b) Let us draw the line parallel to OB, whose distance from OB is equal to a and which intersects ray OA at a point M. Let us draw through points O and M another pair of parallel lines the distance between which is equal to a; the line that passes through point O contains the leg of the angle to be found. 202 CHAPTER 8. CONSTRUCTIONS 8.81. Let us draw through point A an arbitrary line and then draw lines l1 and l2 parallel to it and whose distance from this line is equal to a; these lines intersect line l at points M1 and M2. Let us draw through points A and M1 one more pair of parallel lines, la and lm, the distance between which is equal to a. The intersection point of lines l2 and lm belongs to the perpendicular to be found. 8.82. Let us draw a line parallel to the given one at a distance of a. Now, we can make use of the results of Problems 8.77 and 8.74. 8.83. Let us draw through point P lines PA1 and PB1 so that PA1 ∥OA and PB1 ∥OB. Let line PM divide the angle between lines l and PA1 in halves. The symmetry through line PM sends line PA1 to line l and, therefore, line PB1 turns under this symmetry into one of the lines to be constructed. 8.84. Let us complement triangle ABM to parallelogram ABMN. Through point N draw lines parallel to the bisectors of the angles between lines l and MN. The intersection points of these lines with line l are the ones to be found. 8.85. Let us draw line l1 parallel to line OA at a distance of a. On l, take an arbitrary point B. Let B1 be the intersection point of lines OB and l1. Through point B1 draw the line parallel to AB; this line intersects line OA at point A1. Now, let us draw through points O and A1 a pair of parallel lines the distance between which is equal to a. There could be two pairs of such lines. Let X and X1 be the intersection points of the line that passes through point O with lines l and l1. Since OA1 = OX1 and △OA1X1 ∼△OAX, point X is the one to be found. 8.86. Let us erect perpendiculars to line O1O2 at points O1 and O2 and on the perpen-diculars mark segments O1B1 = O2A2 and O2B2 = O1A1. Let us construct the midpoint M of segment B1B2 and erect the perpendicular to B1B2 at point M. This perpendicu-lar intersects line O1O2 at point N. Then O1N 2 + O1B2 1 = O2N 2 + O2B2 2 and, therefore, O1N 2 −O1A2 1 = O2N 2 −O2A2 2, i.e., point N lies on the radical axis. It remains to erect the perpendicular to O1O2 at point N. 8.87. First, let us construct an arbitrary line l1 perpendicular to line l and then draw through point A the line perpendicular to l1. 8.88. a) Let us draw through points A and B lines AB and BQ perpendicular to line AB and then draw an arbitrary perpendicular to line AP. As a result we get a rectangle. It remains to drop from the intersection point of its diagonals the perpendicular to line AB. b) Let us raise from point B perpendicular l to line AB and draw through point A two perpendicular lines; they intersect line l at points M and N. Let us complement right triangle MAN to rectangle MANR. The base of the perpendicular dropped from point R to line AB is point C to be found. 8.89. a) Let us drop perpendicular AP from point A to line OB and construct segment AC whose midpoint is points P. Then angle ∠AOC is the one to be found. b) On line OB, take points B and B1 such that OB = OB1. Let us place the right angle so that its sides would pass through points B and B1 and the vertex would lie on ray OA. If A is the vertex of the right angle, then angle ∠AB1B is the one to be found. 8.90. Let us draw through point O line l′ parallel to line l. Let us drop perpendiculars BP and BQ from point B to lines l′ and OA, respectively, and then drop perpendicular OX from point O to line PQ. Then line XO is the desired one (cf. Problem 2.3); if point Y is symmetric to point X through line l′, then line Y O is also the one to be found. 8.91. Let us complement triangle OAB to parallelogram OABC and then construct segment CC1 whose midpoint is point O. Let us place the right angle so that its legs pass through points C and C1 and the vertex lies on line l. Then the vertex of the right angle coincides with point X to be found. SOLUTIONS 203 8.92. Let us construct segment AB whose midpoint is point O and place the right angle so that its legs passes through points A and B and the vertex lies on line l. Then the vertex of the right angle coincides with the point to be found. Chapter 9. GEOMETRIC INEQUALITIES Background 1) For elements of a triangle the following notations are used: a, b, c are the lengths of sides BC, CA, AB, respectively; α, β, γ the values of the angles at vertices A, B, C, respectively; ma, mb, mc are the lengths of the medians drawn from vertices A, B, C, respectively; ha, hb, hc are the lengths of the heights dropped from vertices A, B, C, respectively; la, lb, lc are the lengths of the bisectors drawn from vertices A, B, C, respectively; r and R are the radii of the inscribed and circumscribed circles, respectively. 2) If A, B, C are arbitrary points, then AB ≤AC + CB and the equality takes place only if point C lies on segment AB (the triangle inequality). 3) The median of a triangle is shorter than a half sum of the sides that conﬁne it: ma < 1 2(b+c) (Problem 9.1). 4) If one convex polygon lies inside another one, then the perimeter of the outer polygon is greater than the perimeter of the inner one (Problem 9.27 b). 5) The sum of the lengths of the diagonals of a convex quadrilateral is greater than the sum of the length of any pair of the opposite sides of the quadrilateral (Problem 9.14). 6) The longer side of a triangle subtends the greater angle (Problem 10.59). 7) The length of the segment that lies inside a convex polygon does not exceed either that of its longest side or that of its longest diagonal (Problem 10.64). Remark. While solving certain problems of this chapter we have to know various alge-braic inequalities. The data on these inequalities and their proof are given in an appendix to this chapter; one should acquaint oneself with them but it should be taken into account that these inequalities are only needed in the solution of comparatively complicated problems; in order to solve simple problems we will only need the inequality √ ab ≤1 2a + b and its corollaries. Introductory problems 1. Prove that SABC ≤1 2AB · BC. 2. Prove that SABCD ≤1 2(AB · BC + AD · DC). 3. Prove that ∠ABC > 90◦if and only if point B lies inside the circle with diameter AC. 4. The radii of two circles are equal to R and r and the distance between the centers of the circles is equal to d. Prove that these circles intersect if and only if \|R −r\| < d < R + r. 5. Prove that any diagonal of a quadrilateral is shorter than the quadrilateral’s semiperime-ter. §1. A median of a triangle 9.1. Prove that 1 2(a + b −c) < mc < 1 2(a + b). 205 206 CHAPTER 9. GEOMETRIC INEQUALITIES 9.2. Prove that in any triangle the sum of the medians is greater than 3 4 of the perimeter but less than the perimeter. 9.3. Given n points A1, . . . , An and a unit circle, prove that it is possible to ﬁnd a point M on the circle so that MA1 + · · · + MAn ≥n. 9.4. Points A1, . . . , An do not lie on one line. Let two distinct points P and Q have the following property A1P + · · · + AnP = A1Q + · · · + AnQ = s. Prove that A1K + · · · + AnK < s for a point K. 9.5. On a table lies 50 working watches (old style, with hands); all work correctly. Prove that at a certain moment the sum of the distances from the center of the table to the endpoints of the minute’s hands becomes greater than the sum of the distances from the center of the table to the centers of watches. (We assume that each watch is of the form of a disk.) §2. Algebraic problems on the triangle inequality In problems of this section a, b and c are the lengths of the sides of an arbitrary triangle. 9.6. Prove that a = y + z, b = x + z and c = x + y, where x, y and z are positive numbers. 9.7. Prove that a2 + b2 + c2 < 2(ab + bc + ca). 9.8. For any positive integer n, a triangle can be composed of segments whose lengths are an, bn and cn. Prove that among numbers a, b and c two are equal. 9.9. Prove that a(b −c)2 + b(c −a)2 + c(a −b)2 + 4abc > a3 + b3 + c3. 9.10. Let p = a b + b c + c a and q = a c + c b + b a. Prove that \|p −q\| < 1. 9.11. Five segments are such that from any three of them a triangle can be constructed. Prove that at least one of these triangles is an acute one. 9.12. Prove that (a + b −c)(a −b + c)(−a + b + c) ≤abc. 9.13. Prove that a2b(a −b) + b2c(b −c) + c2a(c −a) ≥0. §3. The sum of the lengths of quadrilateral’s diagonals 9.14. Let ABCD be a convex quadrilateral. Prove that AB + CD < AC + BD. 9.15. Let ABCD be a convex quadrilateral and AB + BD ≤AC + CD. Prove that AB < AC. 9.16. Inside a convex quadrilateral the sum of lengths of whose diagonals is equal to d, a convex quadrilateral the sum of lengths of whose diagonals is equal to d′ is placed. Prove that d′ < 2d. 9.17. Given closed broken line has the property that any other closed broken line with the same vertices (?) is longer. Prove that the given broken line is not a self-intersecting one. 9.18. How many sides can a convex polygon have if all its diagonals are of equal length? 9.19. In plane, there are n red and n blue dots no three of which lie on one line. Prove that it is possible to draw n segments with the endpoints of distinct colours without common points. 9.20. Prove that the mean arithmetic of the lengths of sides of an arbitrary convex polygon is less than the mean arithmetic of the lengths of all its diagonals. THE AREA OF A TRIANGLE 207 9.21. A convex (2n + 1)-gon A1A3A5 . . . A2n+1A2 . . . A2n is given. Prove that among all the closed broken lines with the vertices in the vertices of the given (2n + 1)-gon the broken line A1A2A3 . . . A2n+1A1 is the longest. §4. Miscellaneous problems on the triangle inequality 9.22. In a triangle, the lengths of two sides are equal to 3.14 and 0.67. Find the length of the third side if it is known that it is an integer. 9.23. Prove that the sum of lengths of diagonals of convex pentagon ABCDE is greater than its perimeter but less than the doubled perimeter. 9.24. Prove that if the lengths of a triangle’s sides satisfy the inequality a2 + b2 > 5c2, then c is the length of the shortest side. 9.25. The lengths of two heights of a triangle are equal to 12 and 20. Prove that the third height is shorter than 30. 9.26. On sides AB, BC, CA of triangle ABC, points C1, A1, B1, respectively, are taken so that BA1 = λ · BC, CB1 = λ · CA and AC1 = λ · AB, where 1 2 < λ < 1. Prove that the perimeter P of triangle ABC and the perimeter P1 of triangle A1B1C1 satisfy the following inequality: (2λ −1)P < P1 < λP. 9.27. a) Prove that under the passage from a nonconvex polygon to its convex hull the perimeter diminishes. (The convex hull of a polygon is the smallest convex polygon that contains the given one.) b) Inside a convex polygon there lies another convex polygon. Prove that the perimeter of the outer polygon is not less than the perimeter of the inner one. 9.28. Inside triangle ABC of perimeter P, a point O is taken. Prove that 1 2P < AO + BO + CO < P. 9.29. On base AD of trapezoid ABCD, a point E is taken such that the perimeters of triangles ABE, BCE and CDE are equal. Prove that BC = 1 2AD. See also Problems 13.40, 20.11. §5. The area of a triangle does not exceed a half product of two sides 9.30. Given a triangle of area 1 the lengths of whose sides satisfy a ≤b ≤c. Prove that b ≥ √ 2. 9.31. Let E, F, G and H be the midpoints of sides AB, BC, CD and DA of quadrilateral ABCD. Prove that SABCD ≤EG · HF ≤(AB + CD)(AD + BC) 4 . 9.32. The perimeter of a convex quadrilateral is equal to 4. Prove that its area does not exceed 1. 9.33. Inside triangle ABC a point M is taken. Prove that 4S ≤AM · BC + BM · AC + CM · AB, where S is the area of triangle ABC. 9.34. In a circle of radius R a polygon of area S is inscribed; the polygon contains the center of the circle and on each of its sides a point is chosen. Prove that the perimeter of the convex polygon with vertices in the chosen points is not less than 2S R . 9.35. Inside a convex quadrilateral ABCD of area S point O is taken such that AO2 + BO2 + CO2 + DO2 = 2S. Prove that ABCD is a square and O is its center. 208 CHAPTER 9. GEOMETRIC INEQUALITIES §6. Inequalities of areas 9.36. Points M and N lie on sides AB and AC, respectively, of triangle ABC, where AM = CN and AN = BM. Prove that the area of quadrilateral BMNC is at least three times that of triangle AMN. 9.37. Areas of triangles ABC, A1B1C1, A2B2C2 are equal to S, S1, S2, respectively, and AB = A1B1 + A2B2, AC = A1C1 + A2B2, BC = B1C1 + B2C2. Prove that S ≤4√S1S2. 9.38. Let ABCD be a convex quadrilateral of area S. The angle between lines AB and CD is equal to α and the angle between AD and BC is equal to β. Prove that AB · CD sin α + AD · BC sin β ≤2S ≤AB · CD + AD · BC. 9.39. Through a point inside a triangle three lines parallel to the triangle’s sides are drawn. Figure 94 (9.39) Denote the areas of the parts into which these lines divide the triangle as plotted on Fig. 94. Prove that a α + b β + c γ ≥3 2. 9.40. The areas of triangles ABC and A1B1C1 are equal to S and S1, respectively, and we know that triangle ABC is not an obtuse one. The greatest of the ratios a1 a , b1 b and c1 c is equal to k. Prove that S1 ≤k2S. 9.41. a) Points B, C and D divide the (smaller) arc ⌣AE of a circle into four equal parts. Prove that SACE < 8SBCD. b) From point A tangents AB and AC to a circle are drawn. Through the midpoint D of the (lesser) arc ⌣BC the tangent that intersects segments AB and AC at points M and N, respectively is drawn. Prove that SBCD < 2SMAN. 9.42. All sides of a convex polygon are moved outwards at distance h and extended to form a new polygon. Prove that the diﬀerence of areas of the polygons is more than Ph + πh2, where P is the perimeter. 9.43. A square is cut into rectangles. Prove that the sum of areas of the disks circum-scribed about all these rectangles is not less than the area of the disk circumscribed about the initial square. 9.44. Prove that the sum of areas of ﬁve triangles formed by the pairs of neighbouring sides and the corresponding diagonals of a convex pentagon is greater than the area of the pentagon itself. 9.45. a) Prove that in any convex hexagon of area S there exists a diagonal that cuts oﬀthe hexagon a triangle whose area does not exceed 1 6S. b) Prove that in any convex 8-gon of area S there exists a diagonal that cuts oﬀit a triangle of area not greater than 1 8S. See also Problem 17.19. §8. BROKEN LINES INSIDE A SQUARE 209 §7. Area. One ﬁgure lies inside another 9.46. A convex polygon whose area is greater than 0.5 is placed in a unit square. Prove that inside the polygon one can place a segment of length 0.5 parallel to a side of the square. 9.47. Inside a unit square n points are given. Prove that: a) the area of one of the triangles some of whose vertices are in these points and some in vertices of the square does not exceed 1 2(n+1); b) the area of one of the triangles with the vertices in these points does not exceed 1 n−2. 9.48. a) In a disk of area S a regular n-gon of area S1 is inscribed and a regular n-gon of area S2 is circumscribed about the disk. Prove that S2 > S1S2. b) In a circle of length L a regular n-gon of perimeter P1 is inscribed and another regular n-gon of perimeter P2 is circumscribed about the circle. Prove that L2 < P1P2. 9.49. A polygon of area B is inscribed in a circle of area A and circumscribed about a circle of area C. Prove that 2B ≤A + C. 9.50. In a unit disk two triangles the area of each of which is greater than 1 are placed. Prove that these triangles intersect. 9.51. a) Prove that inside a convex polygon of area S and perimeter P one can place a disk of radius S P . b) Inside a convex polygon of area S1 and perimeter P1 a convex polygon of area S2 and perimeter P2 is placed. Prove that 2S1 P1 > S2 P2. 9.52. Prove that the area of a parallelogram that lies inside a triangle does not exceed a half area of the triangle. 9.53. Prove that the area of a triangle whose vertices lie on sides of a parallelogram does not exceed a half area of the parallelogram. 9.54. Prove that any acute triangle of area 1 can be placed in a right triangle of area √ 3. 9.55. a) Prove that a convex polygon of area S can be placed in a rectangle of area not greater than 2S. b) Prove that in a convex polygon of area S a parallelogram of area not less than 1 2S can be inscribed. 9.56. Prove that in any convex polygon of area 1 a triangle whose area is not less than a) 1 4; b) 3 8 can be placed. 9.57. A convex n-gon is placed in a unit square. Prove that there are three vertices A, B and C of this n-gon, such that the area of triangle ABC does not exceed a) 8 n2; b) 16π n3 . See also Problem 15.6. §8. Broken lines inside a square 9.58. Inside a unit square a non-self-intersecting broken line of length 1000 is placed. Prove that there exists a line parallel to one of the sides of the square that intersects this broken line in at least 500 points. 9.59. In a unit square a broken line of length L is placed. It is known that each point of the square is distant from a point of this broken line less than by ε. Prove that L ≥ 1 2ε −1 2πε. 9.60. Inside a unit square n2 points are placed. Prove that there exists a broken line that passes through all these points and whose length does not exceed 2n. 9.61. Inside a square of side 100 a broken line L is placed. This broken line has the following property: the distance from any point of the square to L does not exceed 0.5. 210 CHAPTER 9. GEOMETRIC INEQUALITIES Prove that on L there are two points the distance between which does not exceed 1 and the distance between which along L is not less than 198. §9. The quadrilateral 9.62. In quadrilateral ABCD angles ∠A and ∠B are equal and ∠D > ∠C. Prove that AD < BC. 9.63. In trapezoid ABCD, the angles at base AD satisfy inequalities ∠A < ∠D < 90◦. Prove that AC > BD. 9.64. Prove that if two opposite angles of a quadrilateral are obtuse ones, then the diagonal that connects the vertices of these angles is shorter than the other diagonal. 9.65. Prove that the sum of distances from an arbitrary point to three vertices of an isosceles trapezoid is greater than the distance from this point to the fourth vertex. 9.66. Angle ∠A of quadrilateral ABCD is an obtuse one; F is the midpoint of side BC. Prove that 2FA < BD + CD. 9.67. Quadrilateral ABCD is given. Prove that AC · BD ≤AB · CD + BC · AD. (Ptolemy’s inequality.) 9.68. Let M and N be the midpoints of sides BC and CD, respectively, of a convex quadrilateral ABCD. Prove that SABCD < 4SAMN. 9.69. Point P lies inside convex quadrilateral ABCD. Prove that the sum of distances from point P to the vertices of the quadrilateral is less than the sum of pairwise distances between the vertices of the quadrilateral. 9.70. The diagonals divide a convex quadrilateral ABCD into four triangles. Let P be the perimeter of ABCD and Q the perimeter of the quadrilateral formed by the centers of the inscribed circles of the obtained triangles. Prove that PQ > 4SABCD. 9.71. Prove that the distance from one of the vertices of a convex quadrilateral to the opposite diagonal does not exceed a half length of this diagonal. 9.72. Segment KL passes through the intersection point of diagonals of quadrilateral ABCD and the endpoints of KL lie on sides AB and CD of the quadrilateral. Prove that the length of segment KL does not exceed the length of one of the diagonals of the quadrilateral. 9.73. Parallelogram P2 is inscribed in parallelogram P1 and parallelogram P3 whose sides are parallel to the corresponding sides of P1 is inscribed in parallelogram P2. Prove that the length of at least one of the sides of P1 does not exceed the doubled length of a parallel to it side of P3. See also Problems 13.19, 15.3 a). §10. Polygons 9.74. Prove that if the angles of a convex pentagon form an arithmetic progression, then each of them is greater than 36◦. 9.75. Let ABCDE is a convex pentagon inscribed in a circle of radius 1 so that AB = A, BC = b, CD = c, DE = d, AE = 2. Prove that a2 + b2 + c2 + d2 + abc + bcd < 4. 9.76. Inside a regular hexagon with side 1 point P is taken. Prove that the sum of the distances from point P to certain three vertices of the hexagon is not less than 1. 9.77. Prove that if the sides of convex hexagon ABCDEF are equal to 1, then the radius of the circumscribed circle of one of triangles ACE and BDF does not exceed 1. 9.78. Each side of convex hexagon ABCDEF is shorter than 1. Prove that one of the diagonals AD, BE, CF is shorter than 2. 211 9.79. Heptagon A1 . . . A7 is inscribed in a circle. Prove that if the center of this circle lies inside it, then the value of any angle at vertices A1, A3, A5 is less than 450◦. 9.80. a) Prove that if the lengths of the projections of a segment to two perpendicular lines are equal to a and b, then the segment’s length is not less than a+b √ 2 . b) The lengths of the projections of a polygon to coordinate axes are equal to a and b. Prove that its perimeter is not less than √ 2(a + b). 9.81. Prove that from the sides of a convex polygon of perimeter P two segments whose lengths diﬀer not more than by 1 3P can be constructed. 9.82. Inside a convex polygon A1 . . . An a point O is taken. Let αk be the value of the angle at vertex Ak, xk = OAk and dk the distance from point O to line AkAk+1. Prove that P xk sin αk 2 ≥P dk and P xk cos αk 2 ≥p, where p is the semiperimeter of the polygon. 9.83. Regular 2n-gon M1 with side a lies inside regular 2n-gon M2 with side 2a. Prove that M1 contains the center of M2. 9.84. Inside regular polygon A1 . . . An point O is taken. Prove that at least one of the angles ∠AiOAj satisﬁes the inequalities π ¡ 1 −1 n ¢ ≤ ∠AiOAj ≤π. 9.85. Prove that for n ≥7 inside a convex n-gon there is a point the sum of distances from which to the vertices is greater than the semiperimeter of the n-gon. 9.86. a) Convex polygons A1 . . . An and B1 . . . Bn are such that all their corresponding sides except for A1An and B1Bn are equal and ∠A2 ≥∠B2, . . . , ∠An−1 ≥∠Bn−1, where at least one of the inequalities is a strict one. Prove that A1An > B1Bn. b) The corresponding sides of nonequal polygons A1 . . . An and B1 . . . Bn are equal. Let us write beside each vertex of polygon A1 . . . An the sign of the diﬀerence ∠Ai −∠Bi. Prove that for n ≥4 there are at least four pairs of neighbouring vertices with distinct signs. (The vertices with the zero diﬀerence are disregarded: two vertices between which there only stand vertices with the zero diﬀerence are considered to be neighbouring ones.) See also Problems 4.37, 4.53, 13.42. §11. Miscellaneous problems 9.87. On a segment of length 1 there are given n points. Prove that the sum of distances from a point of the segment to these points is not less than 1 2n. 9.88. In a forest, trees of cylindrical form grow. A communication service person has to connect a line from point A to point B through this forest the distance between the points being equal to l. Prove that to acheave the goal a piece of wire of length 1.6l will be suﬃcient. 9.89. In a forest, the distance between any two trees does not exceed the diﬀerence of their heights. Any tree is shorter than 100 m. Prove that this forest can be fenced by a fence of length 200 m. 9.90. A (not necessarily convex) paper polygon is folded along a line and both halves are glued together. Can the perimeter of the obtained lamina be greater than the perimeter of the initial polygon? 9.91. Prove that a closed broken line of length 1 can be placed in a disk of radius 0.25. 9.92. An acute triangle is placed inside a circumscribed circle. Prove that the radius of the circle is not less than the radius of the circumscribed circle of the triangle. 212 CHAPTER 9. GEOMETRIC INEQUALITIES Is a similar statement true for an obtuse triangle? 9.93. Prove that the perimeter of an acute triangle is not less than 4R. See also problems 14.23, 20.4. Problems for independent study 9.94. Two circles divide rectangle ABCD into four rectangles. Prove that the area of one of the rectangles, the one adjacent to vertices A and C, does not exceed a quarter of the area of ABCD. 9.95. Prove that if AB + BD = AC + CD, then the midperpendicular to side BC of quadrilateral ABCD intersects segment AD. 9.96. Prove that if diagonal BD of convex quadrilateral ABCD divides diagonal AC in halves and AB > BC, then AD < DC. 9.97. The lengths of bases of a circumscribed trapezoid are equal to 2 and 11. Prove that the angle between the extensions of its lateral sides is an acute one. 9.98. The bases of a trapezoid are equal to a and b and its height is equal to h. Prove that the length of one of its diagonals is not less than q h2+(b+a)2 4 . 9.99. The vertices of an n-gon M1 are the midpoints of sides of a convex n-gon M. Prove that for n ≥3 the perimeter of M1 is not less than the semiperimeter of M and for n ≥4 the area of M1 is not less than a half area of M. 9.100. In a unit circle a polygon the lengths of whose sides are conﬁned between 1 and √ 2 is inscribed. Find how many sides does the polygon have. Supplement. Certain inequalities 1. The inequality between the mean arithmetic and the mean geometric of two numbers √ ab ≤1 2(a + b), where a and b are positive numbers, is often encountered. This inequality follows from the fact that a −2 √ ab + b = (√a − √ b)2 ≥0, where the equality takes place only if a = b. This inequality implies several useful inequalities, for example: x(a −x) ≤ ¡x+a−x 2 ¢2 = a2 4 ; a + 1 a ≥2 q a · 1 a = 2 fora > 0. 2. The inequality between the mean arithmetic and the mean geometric of n positive numbers (a1a2 . . . an) 1 n ≤a1+···+an n is sometimes used. In this inequality the equality takes place only if a1 = · · · = an. First, let us prove this inequality for the numbers of the form n = 2m by induction on m. For m = 1 the equality was proved above. Suppose that it is proved for m and let us prove it for m + 1. Clearly, akak+2m ≤ ¡ak+ak+2m 2 ¢2. Therefore, (a1a2 . . . a2m+1) 1 2m+1 ≤(b1b2 . . . b2m) 1 2m , where bk = 1 2(ak + ak+2m) and by the inductive hypothesis (b1 . . . b2m) 1 2m ≤1 2m(b1 + · · · + b2m) = 1 2m+1(a1 + · · · + a2m+1). Now, let n be an arbitrary number. Then n < 2m for some m. Suppose an+1 = · · · = a2m = a1+···+an n = A. Clearly, (a1 + · · · + an) + (an+1 + · · · + a2m) = nA + (2m −n)A = 2mA SOLUTIONS 213 and a1 . . . a2m = a1 . . . an · A2m−n. Hence, a1 . . . an · A2m−n ≤ µ2mA 2m ¶2m = A2m, i.e. a1 . . . an ≤An. The equality is attained only for a1 = · · · = an. 3. For arbitrary numbers a1, . . . , an we have (a + · · · + an)2 ≤n(a2 1 + · · · + a2 n). Indeed, (a1 + · · · + an)2 = X a2 i + 2 X i<j aiaj ≤ X a2 i + X i<j (a2 i + a2 j) = n X a2 i . 4. Since R α 0 cos t dt = sin α and R α 0 sin t dt = 1 −cos α, it follows that starting from the inequality cos t ≤1 we get: ﬁrst, sin α ≤α, then 1 −cos α ≤α2 2 (i.e. cos α ≥1 −α2 2 ), next, sin α ≥α −α3 6 , cos α ≤1 −α2 2 + α4 24, etc. (the inequalities are true for all α ≥0). 5. Let us prove that tan α ≥α for 0 ≤α < π 2. Let AB be the tangent to the unit circle centered at O; let B be the tangent point, C the intersection point of ray OA with the circle and S the area of the disk sector BOC. Then α = 2S < 2SAOB = tan α. 6. On the segment [0, π 2] the function f(x) = x sin x monotonously grows because f ′(x) = tan x−x cos x sin2 x > 0. In particular, f(α) ≤f ¡ π 2 ¢ , i.e., α sin α ≤π 2 for 0 < α < π 2. 7. If f(x) = a cos x + b sin x, then f(x) ≤ √ a2 + b2. Indeed, there exists an angle ϕ such that cos ϕ = a √ a2+b2 and sin ϕ = b √ a2+b2; hence, f(x) = √ a2 + b2 cos(ϕ −x) ≤ √ a2 + b2. The equality takes place only if ϕ = x + 2kπ, i.e., cos x = a √ a2+b2 and sin x = b √ a2+b2. Solutions 9.1. Let C1 be the midpoint of side AB. Then CC1 +C1A > CA and BC1 +C1C > BC. Therefore, 2CC1 + BA > CA + BC, i.e., mc > 1 2(a + b −c). Let point C′ be symmetric to C through point C1. Then CC1 = C1C′ and BC′ = CA. Hence, 2mc = CC′ < CB + BC′ = CB + CA, i.e., mc < 1 2(a + b). 9.2. The preceding problem implies that ma < 1 2(b+c), mb < 1 2(a+c) and mc < 1 2(a+b) and, therefore, the sum of the lengths of medians does not exceed the perimeter. Let O be the intersection point of medians of triangle ABC. Then BO + OA > BA, AO + OC > AC and CO + OB > CB. Adding these inequalities and taking into account that AO = 2 3ma, BO = 2 3mb, CO = 2 3mc we get ma + mb + mc > 3 4(a + b + c). 9.3. Let M1 and M2 be diametrically opposite points on a circle. Then M1Ak +M2Ak ≥ M1M2 = 2. Adding up these inequalities for k = 1, . . . , n we get (M1A1 + · · · + M1An) + (M2A1 + · · · + M2An) ≥2n. Therefore, either M1A1+· · ·+M1An ≥n and then we set M = M1 or M2A1+· · ·+M2An ≥n and then we set M = M2. 9.4. For K we can take the midpoint of segment PQ. Indeed, then AiK ≤1 2(AiP +AiQ) (cf. Problem 9.1), where at least one of the inequalities is a strict one because points Ai cannot all lie on line PQ. 214 CHAPTER 9. GEOMETRIC INEQUALITIES 9.5. Let Ai and Bi be the positions of the minute hands of the i-th watch at times t and t + 30 min, let Oi be the center of the i-th watch and O the center of the table. Then OOi ≤1 2(OAi + OBi) for any i, cf. Problem 9.1. Clearly, at a certain moment points Ai and Bi do not lie on line OiO, i.e., at least one of n inequalities becomes a strict one. Then either OO1 + · · · + OOn < OA1 + · · · + OAn or OO1 + · · · + OOn < OB1 + · · · + OBn. 9.6. Solving the system of equations x + y = c, x + z = b, y + z = a we get x = −a + b + c 2 , y = a −b + c 2 , z = a + b −c 2 . The positivity of numbers x, y and z follows from the triangle inequality. 9.7. Thanks to the triangle inequality we have a2 > (b −c)2 = b2 −2bc + c2, b2 > a2 −2ac + c2, c2 > a2 −2ab + b2. Adding these inequalities we get the desired statement. 9.8. We may assume that a ≥b ≥c. Let us prove that a = b. Indeed, if b < a, then b ≤λa and c ≤λa, where λ < 1. Hence, bn + cn ≤2λnan. For suﬃciently large n we have 2λn < 1 which contradicts the triangle inequality. 9.9. Since c(a −b)2 + 4abc = c(a + b)2, it follows that a(b −c)2 + b(c −a)2 + c(a −b)2 + 4abc −a3 −b3 −c3 = a((b −c)2 −a2) + +b((c −a)2 −b2) + c((a + b)2 −c2) = (a + b −c)(a −b + c)(−a + b + c). The latter equality is subject to a direct veriﬁcation. All three factors of the latter expression are positive thanks to the triangle inequality. 9.10. It is easy to verify that abc\|p −q\| = \|(b −c)(c −a)(a −b)\|. Since \|b −c\| < a, \|c −a\| < b and \|a −b\| < c, we have \|(b −c)(c −a)(a −b)\| < abc. 9.11. Let us index the lengths of the segments so that a1 ≤a2 ≤a3 ≤a4 ≤a5. If all the triangles that can be composed of these segments are not acute ones, then a2 3 ≥a2 1 + a2 2, a2 4 ≥a2 2 + a2 3 and a2 5 ≥a2 3 + a2 4. Hence, a2 5 ≥a2 3 + a2 4 ≥(a2 1 + a2 2) + (a2 2 + a2 3) ≥2a2 1 + 3a2 2. Since a2 1 + a2 2 ≥2a1a2, it follows that 2a2 1 + 3a2 2 > a2 1 + 2a1a2 + a2 2 = (a1 + a2)2. We get the inequality a2 5 > (a1 + a2)2 which contradicts the triangle inequality. 9.12. First solution. Let us introduce new variables x = −a + b + c, y = a −b + c, z = a + b −c. Then a = 1 2(y + z), b = 1 2(x + z), c = 1 2(x + y), i.e., we have to prove that either xyz ≤1 8(x + y)(y + z)(x + z) or 6xyz ≤x(y2 + z2) + y(x2 + z2) + z(x2 + y2). The latter inequality follows from the fact that 2xyz ≤x(y2 + z2), 2xyz ≤y(x2 + z2) and 2xyz ≤z(x2 + y2), because x, y, z are positive numbers. SOLUTIONS 215 Second solution. Since 2S = ab sin γ and sin γ = c 2R, it follows that abc = 2SR. By Heron’s formula (a + b −c)(a −b + c)(−a + b + c) = 8S2 p . Therefore, we have to prove that 8S2 p ≤4SR, i.e., 2S ≤pR. Since S = pr, we infer that 2r ≤R, cf. Problem 10.26. 9.13. Let us introduce new variables x = −a + b + c 2 , y = a −b + c 2 , z = a + b −c 2 . Then numbers x, y, z are positive and a = y + z, b = x + z, c = x + y. Simple but somewhat cumbersome calculations show that a2b(a −b) + b2c(b −c) + c2a(c −a) = 2(x3z + y3x + z3y −xyz(x + y + z)) = 2xyz µx2 y + y2 z + z2 x −x −y −z ¶ . Since 2 ≤x y + y x, it follows that 2x ≤x µx y + y x ¶ = x2 y + y. Similarly, 2y ≤y µy z + z y ¶ = y2 z + z; 2z ≤z ³z x + x z ´ = z2 x + x. Adding these inequalities we get x2 y + y2 z + z2 x ≥x + y + z. 9.14. Let O be the intersection point of the diagonals of quadrilateral ABCD. Then AC + BD = (AO + OC) + (BO + OD) = (AO + OB) + (OC + OD) > AB + CD. 9.15. By the above problem AB + CD < AC + BD. Adding this inequality to the inequality AB + BD ≤AC + CD we get 2AB < 2AC. 9.16. First, let us prove that if P is the perimeter of convex quadrilateral ABCD and d1 and d2 are the lengths of its diagonals, then P > d1 + d2 > 1 2P. Clearly, AC < AB + BC and AC < AD + DC; hence, AC < AB + BC + CD + AD 2 = P 2 . Similarly, BD < 1 2P. Therefore, AC + BD < P. On the other hand, adding the inequalities AB + CD < AC + BD and BC + AD < AC + BD (cf. Problem 9.14) we get P < 2(AC + BD). Let P be the perimeter of the outer quadrilateral, P ′ the perimeter of the inner one. Then d > 1 2P and since P ′ < P (by Problem 9.27 b)), we have d′ < P ′ < P < 2d. 9.17. Let the broken line of the shortest length be a self-intersecting one. Let us consider two intersecting links. The vertices of these links can be connected in one of the following three ways: Fig. 95. Let us consider a new broken line all the links of which are the same 216 CHAPTER 9. GEOMETRIC INEQUALITIES as of the initial one except that the two solid intersecting links are replaced by the dotted links (see Fig. 95). Figure 95 (Sol. 9.17) Then we get again a broken line but its length is less than that of the initial one because the sum of the lengths of the opposite sides of a convex quadrilateral is less than the sum of the length of its diagonals. We have obtained a contradiction and, therefore, the closed broken line of the least length cannot have intersecting links. 9.18. Let us prove that the number of sides of such a polygon does not exceed 5. Suppose that all the diagonals of polygon A1 . . . An are of the same length and n ≥6. Then segments A1A4, A1A5, A2A4 and A2A5 are of equal length since they are the diagonals of this polygon. But in convex quadrilateral A1A2A4A5 segments A1A5 and A2A4 are opposite sides whereas A1A4 and A2A5 are diagonals. Therefore, A1A5 + A2A4 < A1A4 + A2A5. Contradiction. It is also clear that a regular pentagon and a square satisfy the required condition. 9.19. Consider all the partitions of the given points into pairs of points of distinct colours. There are ﬁnitely many such partitions and, therefore, there exists a partition for which the sum of lengths of segments given by pairs of points of the partition is the least one. Let us show that in this case these segments will not intersect. Indeed, if two segments would have intersected, then we could have selected a partition with the lesser sum of lengths of segments by replacing the diagonals of the convex quadrilateral by its opposite sides as shown on Fig. 96. Figure 96 (Sol. 9.19) 9.20. Let ApAp+1 and AqAq+1 be nonadjacent sides of n-gon A1 . . . An (i.e., \|p −q\| ≥2). Then ApAp+1 + AqAq+1 < ApAq + Ap+1Aq+1. Let us write all such inequalities and add them. For each side there exist precisely n −3 sides nonadjacent to it and, therefore, any side enters n −3 inequality, i.e., in the left-hand side of the obtained sum there stands (n −3)p, where p is the sum of lengths of the n-gon’s sides. Diagonal AmAn enters two inequalities for p = n, q = m and for p = n−1, q = m−1; hence, in the right-hand side stands 2d, where d is the sum of lengths of diagonals. Thus, (n −3)p < 2d. Therefore, p n < d n(n−3)/2, as required. 9.21. Let us consider an arbitrary closed broken line with the vertices in vertices of the given polygon. If we have two nonintersecting links then by replacing these links by the SOLUTIONS 217 diagonals of the quadrilateral determined by them we enlarge the sum of the lengths of the links. In this process, however, one broken line can get split into two nonintersecting ones. Let us prove that if the number of links is odd then after several such operations we will still get in the end a closed broken line (since the sum of lengths of the links increases each time, there can be only a ﬁnite number of such operations). One of the obtained closed broken lines should have an odd number of links but then any of the remaining links does not intersect at least one of the links of this broken line (cf. Problem 23.1 a)); therefore, in the end we get just one broken line. Figure 97 (Sol. 9.21) Now, let us successively construct a broken line with pairwise intersecting links (Fig. 97). For instance, the 10-th vertex should lie inside the shaded triangle and therefore, the position of vertices is precisely as plotted on Fig. 97. Therefore, to convex polygon A1A3A5 . . . A2n+1A2 . . . A2n the broken line A1A2A3 . . . A2n+1A1 corresponds. 9.22. Let the length of the third side be equal to n. From the triangle inequality we get 3.14 −0.67 < n < 3.14 + 0.67. Since n is an integer, n = 3. 9.23. Clearly, AB + BC > AC, BC + CD > BD, CD + DE > CE, DE + EA > DA, EA + AB > EB. Adding these inequalities we see that the sum of the lengths of the pentagon’s diagonals is shorter than the doubled perimeter. Figure 98 (Sol. 9.23) The sum of the the diagonals’ lengths is longer than the sum of lengths of the sides of the “rays of the star” and it, in turn, is greater than the perimeter of the pentagon (Fig. 98). 9.24. Suppose that c is the length of not the shortest side, for instance, a ≤c. Then a2 ≤c2 and b2 < (a + c)2 ≤4c2. Hence, a2 + b2 < 5c2. Contradiction. 9.25. Since c > \|b −a\| and a = 2S ha , c = 2S hc , it follows that 1 hc > ¯ ¯ ¯ 1 ha −1 hb ¯ ¯ ¯. Therefore, in our case hc < 20·12 8 = 30. 218 CHAPTER 9. GEOMETRIC INEQUALITIES 9.26. On sides AB, BC, CA take points C2, A2, B2, respectively, so that A1B2 ∥AB, B1C2 ∥BC, CA2 ∥CA (Fig. 99). Then A1B1 < A1B2 + B2B1 = (1 −λ)AB + (2λ −1)CA. Similarly, BC1 < (1 −λ)BC + (2λ −1)AB and C1A1 < (1 −λ)CA + (2λ −1)BC. Adding these inequalities we get P1 < λP. Figure 99 (Sol. 9.26) Clearly, A1B1 + AC > B1C, i.e., A1B1 + (1 −λ)BC > λ · CA. Similarly, B1C1 + (1 −λ)CA > λ · AB and C1A1 + (1 −λ)AB > λ · BC. Adding these inequalities we get P1 > (2λ −1)P. 9.27. a) Passing from a nonconvex polygon to its convex hull we replace certain broken lines formed by sides with segments of straight lines (Fig. 100). It remains to take into account that any brokenline is longer than the line segment with the same endpoints. Figure 100 (Sol. 9.27 a)) b) On the sides of the inner polygon construct half bands directed outwards; let the parallel sides of half bands be perpendicular to the corresponding side of the polygon (Fig. 101). Denote by P the part of the perimeter of the outer polygon corresponding to the boundary of the polygon contained inside these half bands. Then the perimeter of the inner polygon does not exceed P whereas the perimeter of the outer polygon is greater than P. 9.28. Since AO + BO > AB, BO + OC > BC and CO + OA > AC, it follows that AO + BO + CO > AB + BC + CA 2 . Since triangle ABC contains triangle ABO, it follows that AB+BO+OA < AB+BC +CA (cf. Problem 9.27 b)), i.e., BO + OA < BC + CA. Similarly, AO + OC < AB + BC and CO + OB < CA + AB. Adding these inequalities we get AO + BO + CO < AB + BC + CA. SOLUTIONS 219 Figure 101 (Sol. 9.27 b)) 9.29. It suﬃces to prove that ABCE and BCDE are parallelograms. Let us complement triangle ABE to parallelogram ABC1E. Then perimeters of triangles BC1E and ABE are equal and, therefore, perimeters of triangles BC1E and BCE are equal. Hence, C1 = C because otherwise one of the triangles BC1E and BCE would have lied inside the other one and their perimeters could not be equal. Hence, ABCE is a parallelogram. We similarly prove that BCDE is a parallelogram. 9.30. Clearly, 2 = 2S = ab sin γ ≤ab ≤b2, i.e., b ≥ √ 2. 9.31. Since EH is the midline of triangle ABD, it follows that SAEH = 1 4SABD. Similarly, SCFG = 1 4SCBD. Therefore, SAEH + SCFG = 1 4SABCD. Similarly, SBFE + SDGH = 1 4SABCD. It follows that SABCD = 2SEFGH = EG · HF sin α, where α is the angle between lines EG and HF. Since sin α ≤1, then SABCD ≤EG · HF. Adding equalities − − → EG = − − → EB + − − → BC + − → CG and − − → EG = − → EA + − − → AD + − − → DG we obtain 2− − → EG = (− − → EB + − → EA) + (− − → BC + − − → AD) + (− − → DG + − → CG) = − − → BC + − − → AD. Therefore, EG ≤1 2(BC + AD). Similarly, HF ≤1 2(AB + CD). It follows that SABCD ≤EG · HF ≤(AB + CD)(BC + AD) 4 . 9.32. By Problem 9.31 SABCD ≤1 4(AB + CD)(BC + AD). Since ab ≤1 4(a + b)2, it follows that SABCD ≤ 1 16(AB + CD + AD + BC)2 = 1. 9.33. From points B and C drop perpendiculars BB1 and CC1 to line AM. Then 2SAMB + 2SAMC = AM · BB1 + AM · CC1 ≤AM · BC because BB1 + CC1 ≤BC. Similarly, 2SBMC + 2SBMA ≤BM · AC and 2SCMA + 2SCMB ≤CM · AB. Adding these inequalities we get the desired statement. 9.34. Let on sides A1A2, A2A3, . . . , AnA1 points B1, . . . , Bn, respectively, be selected; let O be the center of the circle. Further, let Sk = SOBkAk+1Bk+1 = OAk+1 · BkBk+1 sin ϕ 2 , 220 CHAPTER 9. GEOMETRIC INEQUALITIES where ϕ is the angle between OAk+1 and BkBk+1. Since OAk+1 = R and sin ϕ ≤1, it follows that Sk ≤1 2R · BkBk+1. Hence, S = S1 + · · · + Sn ≤R(B1B2 + · · · + BnB1) 2 , i.e., the perimeter of polygon B1B2 . . . Bn is not less than 2S R . 9.35. We have 2SAOB ≤AO · OB ≤1 2(AO2 + BO2), where the equality is only possible if ∠AOB = 90◦and AO = BO. Similarly, 2SBOC ≤BO2 + CO2 2 , 2SCOD ≤CO2 + DO2 2 and 2SDOA ≤DO2 + AO2 2 . Adding these inequalities we get 2S = 2(SAOB + SBOC + SCOD + SDOA) ≤AO2 + BO2 + CO2 + DO2, where the equality is only possible if AO = BO = CO = DO and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90◦, i.e., ABCD is a square and O is its center. 9.36. We have to prove that SABC SAMN ≥4. Since AB = AM + MB = AM + AN = AN + NC = AC, it follows that SABC SAMN = AB · AC AM · AN = (AM + AN)2 AM · AN ≥4. 9.37. Let us apply Heron’s formula S2 = p(p −a)(p −b)(p −c). Since p−a = (p1−a1)+(p2−a2) and (x+y)2 ≥4xy, it follows that (p−a)2 ≥4(p1−a1)(p2−a2). Similarly, (p −b)2 ≥4(p1 −b1)(p2 −b2), (p −c)2 ≥4(p1 −c1)(p2 −c2) and p2 ≥4p1p2. Multiplying these inequalities we get the desired statement. 9.38. For deﬁniteness, we may assume that rays BA and CD, BC and AD intersect (Fig. 102). If we complement triangle ADC to parallelogram ADCK, then point K occurs inside quadrilateral ABCD. Therefore, 2S ≥2SABK + 2SBCK = AB · AK sin α + BC · CK sin β = AB · CD sin α + BC · AD sin β. The equality is obtained if point D lies on segment AC. Figure 102 (Sol. 9.38) SOLUTIONS 221 Let point D′ be symmetric to point D through the midperpendicular to segment AC. Then 2S = 2SABCD′ = 2SABD′ + 2SBCD′ ≤ AB · AD′ + BC · CD′ = AB · CD + BC · AD. 9.39. Thanks to the inequality between the mean geometric and the mean arithmetic, we have a α + b β + c γ ≥3 3 q abc αβγ = 3 2 because α = 2 √ bc, β = 2√ca and γ = 2 √ ab, cf. Problem 1.33. 9.40. The inequalities α < α1, β < β1 and γ < γ1 cannot hold simultaneously. Therefore, for instance, α1 ≤α ≤90◦; hence, sin α1 ≤sin α. It follows that 2S1 = a1b1 sin α1 ≤ k2ab sin α = 2k2S. 9.41. a) Let chords AE and BD intersect diameter CM at points K and L, respectively. Then AC2 = CK · CM and BC2 = CL · CM. It follows that CK CL = AC2 BC2 < 4. Moreover, AE BD = AE AC < 2. Therefore, SACE SBCD = AE·CK BD·CL < 8. b) Let H be the midpoint of segment BC. Since ∠CBD = ∠BCD = ∠ABD, it follows that D is the intersection point of the bisectors of triangle ABC. Hence, AD DH = AB BH > 1. Therefore, SMAN > 1 4SABC and SBCD = BC · DH 2 < BC · AH 4 = SABC 4 . 9.42. Let us cut oﬀthe obtained polygon rectangles with side h constructed outwards on the sides of the initial polygon (Fig. 103). Then beside the initial polygon there will be left several quadrilaterals from which one can compose a polygon circumscribed about a circle of radius h. The sum of the areas of these quadrilaterals is greater than the area of the circle of radius h, i.e., greater than πh2. It is also clear that the sum of areas of the cut oﬀrectangles is equal to Ph. Figure 103 (Sol. 9.42) 9.43. Let s, s1, . . . , sn be the areas of the square and the rectangles that constitute it, respectively; S, S1, . . . , Sn the areas of the disks circumscribed about the square and the rectangles, respectively. Let us prove that sk ≤2Sk π . Indeed, if the sides of the rectangle are equal to a and b, then sk = ab and Sk = πR2, where R2 = a2 4 + b2 4 . Therefore, sk = ab ≤ a2+b2 2 = 2πR2 π = 2Sk π . It follows that 2S π = s = s1 + · · · + sn ≤2(S1 + · · · + Sn) π . 222 CHAPTER 9. GEOMETRIC INEQUALITIES 9.44. Let, for deﬁniteness, ABC be the triangle of the least area. Denote the intersection point of diagonals AD and EC by F. Then SABCDE < SAED + SEDC + SABCF. Since point F lies on segment EC and SEAB ≥SCAB, it follows that SEAB ≥SFAB. Similarly, SDCB ≥SFCB. Therefore, SABCF = SFAB+SFCB ≤SEAB+SDCB. It follows that SABCDE < SAED + SEDC + SEAB + SDCB and this is even a stronger inequality than the one required. 9.45. a) Denote the intersection points of diagonals AD and CF, CF and BE, BE and AD by P, Q, R, respectively (Fig. 104). Quadrilaterals ABCP and CDEQ have no common inner points since sides CP and QC lie on line CF and segments AB and DE lie on distinct sides of it. Similarly, quadrilaterals ABCP, CDEQ and EFAR have no pairwise common inner points. Therefore, the sum of their areas does not exceed S. Figure 104 (Sol. 9.45 a)) It follows that the sum of the areas of triangles ABP, BCP, CDQ, DEQ, EFR, FAR does not exceed S, i.e., the area of one of them, say ABP, does not exceed 1 6S. Point P lies on segment CF and, therefore, one of the points, C or F, is distant from line AB not further than point P. Therefore, either SABC ≤SABP ≤1 6S or SABF ≤SABP ≤1 6S. b) Let ABCDEFGH be a convex octagon. First, let us prove that quadrilaterals ABEF, BCFG, CDGH and DEHA have a common point. Clearly, a convex quadrilateral KLMN (Fig. 105) is the intersection of ABEF and CDGH. Segments AF and HC lie inside angles ∠DAH and ∠AHE, respectively; hence, point K lies inside quadrilateral DEHA. We similarly prove that point M lies inside quadrilateral DEHA, i.e., the whole segment KM lies inside it. Similarly, segment LN lies inside quadrilateral BCFG. The intersection point of diagonals KM and LN belongs to all our quadrilaterals; denote it by O. Figure 105 (Sol. 9.45 b)) Let us divide the 8-gon into triangles by connecting point O with the vertices. The area of one of these triangles, say ABO, does not exceed 1 8S. Segment AO intersects side KL at a point P, therefore, SABP < SABO ≤1 8S. Since point P lies on diagonal CH, it follows that either SABC ≤SABP ≤1 8S or SABH ≤SABP ≤1 8S. 9.46. Let us draw through all the vertices of the polygon lines parallel to one pair of sides of the square thus dividing the square into strips. Each such strip cuts oﬀthe polygon SOLUTIONS 223 either a trapezoid or a triangle. It suﬃces to prove that the length of one of the bases of these trapezoids is greater than 0.5. Suppose that the length of each base of all the trapezoids does not exceed 0.5. Then the area of each trapezoid does not exceed a half height of the strip that conﬁnes it. Therefore, the area of the polygon, equal to the sum of areas of trapezoids and triangles into which it is cut, does not exceed a half sum of heights of the strips, i.e., does not exceed 0.5. Contradiction. 9.47. a) Let P1, . . . , Pn be the given points. Let us connect point P1 with the vertices of the square. We will thus get four triangles. Next, for k = 2, . . . , n let us perform the following operation. If point Pk lies strictly inside one of the triangles obtained earlier, then connect it with the vertices of this triangle. If point Pk lies on the common side of two triangles, then connect it with the vertices of these triangles opposite to the common side. Each such operation increases the total number of triangles by 2. As a result we get 2(n + 1) triangles. The sum of the areas of these triangles is equal to 1, therefore, the area of any of them does not exceed 1 2(n+1). b) Let us consider the least convex polygon that contains the given points. Let is have k vertices. If k = n then this k-gon can be divided into n −2 triangles by the diagonals that go out of one of its vertices. If k < n, then inside the k-gon there are n −k points and it can be divided into triangles by the method indicated in heading a). We will thus get k + 2(n −k −1) = 2n −k −2 triangles. Since k < n, it follows that 2n −k −2 > n −2. The sum of the areas of the triangles of the partition is less than 1 and there are not less than n −2 of them; therefore, the area of at least one of them does not exceed 1 n−2. 9.48. a) We may assume that the circumscribed n-gon A1 . . . An and the inscribed n-gon B1 . . . Bn are placed so that lines AiBi intersect at the center O of the given circle. Let Ci and Di be the midpoints of sides AiAi+1 and BiBi+1, respectively. Then SOBiCi = p · OBi · OCi, SOBiDi = p · OBi · ODi and SOAiCi = p · OAi · OCi, where p = 1 2 sin ∠AiOCi. Since OAi : OCi = OBi : ODi, it follows that S2 OBiCi = SOBiDiSOAiCi. It remains to notice that the area of the part of the disk conﬁned inside angle ∠AiOCi is greater than SOBiCi and the areas of the parts of the inscribed and circum-scribed n-gons conﬁned inside this angle are equal to SOBiDi and SOAiCi, respectively. b) Let the radius of the circle be equal to R. Then P1 = 2nR sin π n, P2 = 2nR tan π n and L = 2πR. We have to prove that sin x tan x > x2 for 0 < x ≤1 3π. Since µsin x x ¶2 ≥ µ 1 −x2 6 ¶2 = 1 −x2 3 + x4 36 and 0 < cos x ≤1 −x2 2 + x4 24 (see Supplement to this chapter), it remains to verify that 1 −x2 3 + x4 36 ≥1 −x2 2 + x4 24, i.e., 12x2 > x4. For x ≤1 3π this inequality is satisﬁed. 9.49. Let O be the center of homothety that sends the inscribed circle into the circum-scribed one. Let us divide the plane by rays that exit from point O and pass through the vertices of the polygon and the tangent points of its sides with the inscribed circle (Fig. 106). It suﬃces to prove the required inequality for the parts of disks and the polygon conﬁned inside each of the angles formed by these rays. Let the legs of the angle intersect the inscribed circle at points P, Q and the circumscribed circle at points R, S so that P is the tangency point and S is a vertex of the polygon. The areas of the parts of disks are greater than the areas of triangles OPQ and ORS and, therefore, it suﬃces to prove that 2SOPS ≤SOPQ + SORS. Since 2SOPS = 2SOPQ + 2SPQS and SORS = SOPQ + SPQS + SPRS, 224 CHAPTER 9. GEOMETRIC INEQUALITIES Figure 106 (Sol. 9.49) it remains to prove that SPQS ≤SPRS. This inequality is obvious, because the heights of triangles PQS and PRS dropped to bases PQ and RS, respectively, are equal and PQ < RS. 9.50. It suﬃces to prove that both triangles contain the center O of the disk. Let us prove that if triangle ABC placed in the disk of radius 1 does not contain the center of the disk, then its area is less than 1. Indeed, for any point outside the triangle there exists a line that passes through two vertices and separating this point from the third vertex. Let, for deﬁniteness, line AB separate points C and O. Then hc < 1 and AB < 2, hence, S = 1 2hc · AB < 1. 9.51. a) On the sides of the polygon, construct inwards rectangles whose other side is equal to R = S P . The rectangles will not cover the whole polygon (these rectangles overlap and can stick out beyond the limits of the polygon whereas the sum of their areas is equal to the area of the polygon). An uncovered point is distant from every side of the polygon further than by R, consequently, the disk of radius R centered at this point entirely lies inside the polygon. b) Heading a) implies that in the inner polygon a disk of radius S2 P2 can be placed. Clearly, this disk lies inside the outer polygon. It remains to prove that if inside a polygon a disk of radius R lies, then R ≤2S P . For this let us connect (with lines) the center O of the disk with the vertices of the polygon. These lines split the polygon into triangles whose respective areas are equal to 1 2hiai, where hi is the distance from point O to the i-th side and ai is the length of the i-th side. Since hi ≥R, we deduce that 2S = P hiai ≥P Rai = RP. 9.52. First, let us consider the case when two sides of a parallelogram lie on lines AB and AC and the fourth vertex X lies on side BC. If BX : CX = x : (1 −x), then the ratio of the area of the parallelogram to the area of the triangle is equal to 2x(1 −x) ≤1 2. Figure 107 (Sol. 9.52) In the general case let us draw parallel lines that contain a pair of sides of the given parallelogram (Fig. 107). The area of the given parallelogram does not exceed the sum of areas of the shaded parallelograms which fall in the case considered above. If lines that contain a pair of sides of the given parallelogram only intersect two sides of the triangle, then we can restrict ourselves to one shaded parallelogram only. SOLUTIONS 225 9.53. First, let us consider the following case: two vertices A and B of triangle ABC lie on one side PQ of the parallelogram. Then AB ≤PQ and the height dropped to side AB is not longer than the height of the parallelogram. Therefore, the area of triangle ABC does not exceed a half area of the parallelogram. Figure 108 (Sol. 9.53) If the vertices of the triangle lie on distinct sides of the parallelogram, then two of them lie on opposite sides. Let us draw through the third vertex of the triangle a line parallel to these sides (Fig. 108). This line cuts the parallelogram into two parallelograms and it cuts the triangle into two triangles so that two vertices of each of these triangles lie on sides of the parallelogram. We get the case already considered. 9.54. Let M be the midpoint of the longest side BC of the given acute triangle ABC. The circle of radius MA centered at M intersects rays MB and MC at points B1 and C1, respectively. Since ∠BAC < 90◦, it follows that MB < MB1. Let, for deﬁniteness, ∠AMB ≤∠AMC, i.e., ∠AMB ≤90◦. Then AM 2 + MB2 ≤AB2 ≤BC2 = 4MB2, i.e., AM ≤ √ 3BM. If AH is a height of triangle ABC, then AH · BC = 2 and, therefore, SAB1C1 = B1C1 · AH 2 = AM · AH ≤ √ 3BM · AH = √ 3. 9.55. a) Let AB be the longest of the diagonals and sides of the given polygon M. Polygon M is conﬁned inside the strip formed by the perpendiculars to segment AB passing through points A and B. Let us draw two baselines to M parallel to AB. Let them intersect polygon M at points C and D. As a result we have conﬁned M into a rectangle whose area is equal to 2SABC + 2SABD ≤2S. Figure 109 (Sol. 9.55) b) Let M be the initial polygon, l an arbitrary line. Let us consider the polygon M1 one of whose sides is the projection of M to l and the lengths of the sections of polygons M and M1 by any line perpendicular to l are equal (Fig. 109). It is easy to verify that M1 is also a convex polygon and its area is equal to S. Let A be the most distant from l point of M1. The line equidistant from point A and line l intersects the sides of M1 at points B and C. 226 CHAPTER 9. GEOMETRIC INEQUALITIES Let us draw base lines through points B and C. As a result we will circumscribe a trapezoid about M1 (through point A a base line can also be drawn); the area of this trapezoid is no less than S. If the height of the trapezoid, i.e., the distance from A to l is equal to h then its area is equal to h · BC and, therefore, h · BC ≥S. Let us consider sections PQ and RS of polygon M by lines perpendicular to l and passing through B and C. The lengths of these sections are equal to 1 2h and, therefore, PQRS is a parallelogram whose area is equal to 1 2BC · h ≥1 2S. 9.56. a) Let us conﬁne the polygon in the strip formed by parallel lines. Let us shift these lines parallelly until some vertices A and B of the polygon lie on them. Then let us perform the same for the strip formed by lines parallel to AB. Let the vertices that lie on these new lines be C and D (Fig. 110). The initial polygon is conﬁned in a parallelogram and, therefore, the area of this parallelogram is not less than 1. On the other hand, the sum of areas of triangles ABC and ADB is equal to a half area of the parallelogram and, therefore, the area of one of these triangles is not less than 1 4. Figure 110 (Sol. 9.56 a)) b) As in heading a) let us conﬁne the polygon in a strip formed by parallel lines so that some vertices, A and B, lie on these lines. Let d be the width of this strip. Let us draw three lines that divide this strip into equal strips of width 1 4d. Let the ﬁrst and the third lines intersect sides of the polygon at points K, L and M, N, respectively (Fig. 111). Figure 111 (Sol. 9.56 b)) Let us extend the sides on which points K, L, M and N lie to the intersection with the sides of the initial strip and with the line that divides it in halves. In this way we form two trapezoids with the midlines KL and MN and heights of length 1 2d each. Since these trapezoids cover the whole polygon, the sum of their areas is not less than its area, i.e., 1 2(d · KL + d · MN) ≥1. The sum of areas of triangles AMN and BKL contained in the initial polygon is equal to 1 8(3d · MN + 3d · KL) ≥3 4. Therefore, the area of one of these triangles is not less than 3 8. 9.57. Let us prove that there exists even three last vertices satisfying the required condition. Let αi be the angle between the i-th and (i + 1)-th sides βi = π −αi; let ai be the length of the i-th side. SOLUTIONS 227 a) The area of the triangle formed by the i-th and (i + 1)-th sides is equal to Si = aiai+1 sin αi 2 . Let S be the least of these areas. Then 2S ≤aiai+1 sin αi; hence, (2S)n ≤(a2 1 . . . a2 n)(sin α1 . . . sin αn) ≤a2 1 . . . a2 n. By the inequality between the mean arithmetic and the mean geometric we have (a1 . . . an) 1 n ≤ a1+···+an n and, therefore, 2S ≤(a1 . . . an) 2 n ≤(a1 + · · · + an)2 n2 . Since ai ≤pi+qi, where pi and qi are projections of the i-th side to a vertical and a horizontal sides of the square, it follows that a1 + · · · + an ≤(p1 + · · · + pn) + (q1 + · · · + qn) ≤4. Hence, 2S ≤16n2, i.e., S ≤ 8 n2. b) Let us make use of the inequality 2S ≤(a1 . . . an) 2 n(sin α1 . . . sin αn) 1 n ≤16 n2(sin α1 . . . sin αn) proved above. Since sin αi = sin βi and β1 + · · · + βn = 2π, it follows that (sin α1 . . . sin αn) 1 n = (sin β1 . . . sin βn) 1 n ≤β1 + · · · + βn n = 2π n . Hence, 2S ≤32π n3 , i.e., S ≤16π n3 . 9.58. Let li be the length of the i-th link of the broken line; ai and bi the lengths of its projections to the sides of the square. Then li ≤ai + bi. It follows that 1000 = l1 + · · · + ln ≤(a1 + · · · + an) + (b1 + · · · + bn), i.e., either a1 + · · · + an ≥500 or b1 + · · · + bn ≥500. If the sum of the lengths of the links’ projections on a side of length 1 is not less than 500, then not fewer than 500 distinct lengths of the broken line are projected into one of the points of this side, i.e., the perpendicular to the side that passes through this point intersects the broken line at least at 500 points. 9.59. The locus of points distant from the given segment not further than by ε is depicted on Fig. 112. The area of this ﬁgure is equal to πε2+2εl, where l is the length of the segment. Figure 112 (Sol. 9.59) Let us construct such ﬁgures for all N links of the given broken lines. Since neighbouring ﬁgures have N −1 common disks of radius ε centered at vertices of the broken line which are not its endpoints, it follows that the area covered by these ﬁgures does not exceed Nπε2 + 2ε(l1 + · · · + ln) −(N −1)πε2 = 2εL + πε2. This ﬁgure covers the whole square since any point of the square is distant from a point of the broken line by less than ε. Hence, 1 ≤2εL + πε2, i.e., L ≥ 1 2ε −πε 2 . 9.60. Let us divide the square into n vertical strips that contain n points each. Inside each strip let us connect points downwards thus getting n broken lines. These broken lines can be connected into one broken line in two ways: Fig. 113 a) and b). 228 CHAPTER 9. GEOMETRIC INEQUALITIES Figure 113 (Sol. 9.60) Let us consider the segments that connect distinct bands. The union of all such segments obtained in both ways is a pair of broken lines such that the sum of the lengths of the horizontal projections of each of them does not exceed 1. Therefore, the sum of the lengths of horizontal projections of the connecting segments for one of these ways does not exceed 1. Let us consider such a connection. The sum of the lengths of the horizontal projections for connecting links does not exceed 1 and for all the other links it does not exceed (n − 1)(h1 + · · · + hn), where hi is the width of the i-th strip. Clearly, h1 + · · · + hn = 1. The sum of the vertical projections of all links of the broken line does not exceed n. As a result we deduce that the sum of the vertical and horizontal projections of all the links does not exceed 1 + (n −1) + n = 2n and, therefore, the length of the broken line does not exceed 2n. 9.61. Let M and N be the endpoints of the broken line. Let us traverse along the broken line from M to N. Let A1 be the ﬁrst of points of the broken line that we meet whose distance from a vertex of the square is equal to 0.5. Let us consider the vertices of the square neighboring to this vertex. Let B1 be the ﬁrst after A1 point of the broken line distant from one of these vertices by 0.5. Denote the vertices of the square nearest to points A1 and B1 by A and B, respectively (Fig. 114). Figure 114 (Sol. 9.61) Denote the part of the broken line from M to A1 by L1 and the part from A1 to N by L2. Let X and Y be the sets of points that lie on AD and distant not further than by 0.5 from L1 and L2, respectively. By hypothesis, X and Y cover the whole side AD. Clearly, A ∈X and D ̸∈X; hence, D ∈Y , i.e., both sets, X and Y , are nonempty. But each of these sets consists of several segments and, therefore, they should have a common point P. Therefore, on L1 and L2, there are points F1 and F2 for which PF1 ≤0.5 and PF2 ≤0.5. Let us prove that F1 and F2 are the points to be found. Indeed, F1F2 ≤F1P + PF2 ≤1. On the other hand, while traversing from F1 to F2 we should pass through point B; and we SOLUTIONS 229 have F1B1 ≥99 and F2B1 ≥99 because point B1 is distant from side BC no further than by 0.5 while F1 and F2 are distant from side AD not further than by 0.5. 9.62. Let ∠A = ∠B. It suﬃces to prove that if AD < BC; then ∠D > ∠C. On side BC, take point D1 such that BD1 = AD. Then ABD1D is an isosceles trapezoid. Hence, ∠D > ∠D1DA = ∠DD1B ≥∠C. 9.63. Let B1 and C1 be the projections of points B and C on base AD. Since ∠BAB1 < ∠CDC1 and BB1 = CC1, it follows that AB1 > DC1 and, therefore, B1D < AC1. It follows that BD2 = B1D2 + B1B2 < AC2 1 + CC2 1 = AC2. 9.64. Let angles ∠B and ∠D of quadrilateral ABCD be obtuse ones. Then points B and D lie inside the circle with diameter AC. Since the distance between any two points that lie inside the circle is less than its diameter, BD < AC. 9.65. In an isosceles trapezoid ABCD diagonals AC and BD are equal. Therefore, BM + (AM + CM) ≥BM + AC = BM + BD ≥DM. 9.66. Let O be the midpoint of segment BD. Point A lies inside the circle with diameter BD, hence, OA < 1 2BD. Moreover, FO = 1 2CD. Therefore, 2FA ≤2FO + 2OA < CD + BD. 9.67. On rays AB, AC and AD mark segments AB′, AC′ and AD′ of length 1 AB, 1 AC and 1 AD. Then AB : AC = AC′ : AB′, i.e., △ABC ∼△AC′B′. The similarity coeﬃcient of these triangles is equal to 1 AB·AC and therefore, B′C′ = BC AB·AC. Analogously, C′D′ = CD AC·AD and B′D′ = BD AB·AD. Substituting these expressions in the inequality B′D′ ≤B′C′ + C′D′ and multiplying both sides by AB · AC · AD, we get the desired statement. 9.68. Clearly, SABCD = SABC + SACD = 2SAMC + 2SANC = 2(SAMN + SCMN). If segment AM intersects diagonal BD at point A1, then SCMN = SA1MN < SAMN. There-fore, SABCD < 4SAMN. 9.69. Diagonals AC and BD intersect at point O. Let, for deﬁniteness, point P lie in side of AOB. Then AP + BP ≤AO + BO < AC + BD (cf. the solution of Problem 9.28) and CP + DP < CB + BA + AD. 9.70. Let ri, Si and pi be the radii of the inscribed circles, the areas and semiperimeters of the obtained triangles, respectively. Then Q ≥2 X ri = 2 X µSi pi ¶ > 4 X µSi P ¶ = 4S P . 9.71. Let AC ≤BD. Let us drop from vertices A and C perpendiculars AA1 and CC1 to diagonal BD. Then AA1 + CC1 ≤AC ≤BD and, therefore, either AA1 ≤ 1 2BD or CC1 ≤1 2BD. 9.72. Let us draw through the endpoints of segment KL lines perpendicular to it and consider projections to these lines of the vertices of the quadrilateral. Consider also the intersection points of lines AC and BD with these lines, cf. Fig. 115. Let, for deﬁniteness, point A lie inside the strip determined by these lines and point B outside it. Then we may assume that D lies inside the strip, because otherwise BD > KL and the proof is completed. Since AA′ BB′ ≤A1K B1K = C1L D1L ≤CC′ DD′, 230 CHAPTER 9. GEOMETRIC INEQUALITIES Figure 115 (Sol. 9.72) then either AA′ ≤CC′ (and, therefore, AC > KL) or BB′ ≥DD′ (and, therefore, BD > KL). 9.73. Let us introduce the notations as plotted on Fig. 116. All the parallelograms considered have a common center (thanks to Problem 1.7). The lengths of the sides of parallelogram P3 are equal to a + a1 and b + b1 and the lengths of the sides of parallelogram P1 are equal to a + a1 + 2x and b + b1 + 2y, consequently, we have to verify that either a + a1 + 2x ≤2(a + a1) or b + b1 + 2y ≤2(b + b1), i.e., either 2x ≤a + a1 or 2y ≤b + b1. Figure 116 (Sol. 9.73) Suppose that a + a1 < 2x and b + b1 < 2y. Then √aa1 ≤1 2(a + a1) < x and √bb1 < y. On the other hand, the equality of the areas of shaded parallelograms (cf. Problem 4.19) shows that ab = xy = a1b and, therefore, √aa1 √bb1 = xy. Contradiction. 9.74. Let the angles of the pentagon be equal to α, α + γ, α + 2γ, α + 3γ, α + 4γ, where α, γ ≥0. Since the sum of the angles of the pentagon is equal to 3π, it follows that 5α + 10γ = 3π. Since the pentagon is a convex one, each of its angles is less than π, i.e., either α + 4γ < π or −51 2α −10γ > −1 25π. Taking the sum of the latter inequality with 5α + 10γ = 3π we get 5α 2 > π 2, i.e., α > π 5 = 36◦. 9.75. Clearly, 4 = AE2 = \|− → AB + − − → BC + − − → CD + − − → DE\|2 = \|− → AB + − − → BC\|2 + 2(− → AB + − − → BC, − − → CD + − − → DE) + \|− − → CD + − − → DE\|2. Since ∠ACE = 90◦, we have (− → AB + − − → BC, − − → CD + − − → DE) = (− → AC, − − → CE) = 0. Hence, 4 = \|− → AB + − − → BC\|2 + \|− − → CD + − − → DE\|2 = AB2 + BC2 + CD2 + DE2 + 2(− → AB, − − → BC) + 2(− − → CD, − − → DE), i.e., it suﬃces to prove that abc < 2(− → AB, − − → BC) and bcd < 2(− − → CD, − − → DE). SOLUTIONS 231 Since 2(− → AB, − − → BC) = 2ab cos(180◦−∠ABC) = 2ab cos AEC = ab · CE and c < CE, it follows that abc < 2(− → AB, − − → BC). The second inequality is similarly proved, because in notations A1 = E, B1 = D, C1 = C, a1 = d, b1 = c, c1 = b the inequality bcd < 2(− − → CD, − − → DE) takes the form a1b1c1 < 2(− − − → A1B1, − − − → B1C1). 9.76. Let B be the midpoint of side A1A2 of the given hexagon A1 . . . A6 and O its center. We may assume that point P lies inside triangle A1OB. Then PA3 ≥1 because the distance from point A3 to line BO is equal to 1; since the distances from points A4 and A5 to line A3A6 are equal to 1, we deduce that PA4 ≥1 and PA5 ≥1. 9.77. Suppose that the radii of the circumscribed circles of triangles ACE and BDF are greater than 1. Let O be the center of the circumscribed circle of triangle ACE. Then ∠ABC > ∠AOC, ∠CDE > ∠COE and ∠EFA > ∠EOA and, therefore, ∠B +∠D+∠F > 2π. Similarly, ∠A + ∠C + ∠E > 2π, i.e., the sum of the angles of hexagon ABCDEF is greater than 4π. Contradiction. Remark. We can similarly prove that the radius of the circumscribed circle of one of triangles ACE and BDF is not less than 1. 9.78. We may assume that AE ≤AC ≤CE. By Problem 9.67 AD · CE ≤AE · CD + AC · DE < AE + AC ≤2CE, i.e., AD < 2. 9.79. Since ∠A1 = 180◦−1 2 ⌣A2A7, ∠A3 = 180◦−1 2 ⌣A4A2 and ∠A5 = 180◦−1 2 ⌣ A6A4, it follows that ∠A1 + ∠A3 + ∠A5 = 2 · 180◦+ 360◦−⌣A2A7−⌣A4A2−⌣A6A4 2 = 2 · 180◦+ ⌣A7A6 2 . Since the center of the circle lies inside the hexagon, it follows that ⌣A7A6 < 180◦and, therefore, ∠A1 + ∠A3 + ∠A5 < 360◦+ 90◦= 450◦. 9.80. a) We have to prove that if c is the hypothenuse of the right triangle and a and b are its legs, then c ≥a+b √ 2 , i.e., (a + b)2 ≤2(a2 + b2). Clearly, (a + b)2 = (a2 + b2) + 2ab ≤(a2 + b2) + (a2 + b2) = 2(a2 + b2). b) Let di be the length of the i-th side of the polygon; xi and yi the lengths of its projections to coordinate axes. Then x1 + · · · + xn ≥2a, y1 + · · · + yn ≥2b. By heading a) di ≥xi+yi √ 2 . Therefore, d1 + · · · + dn ≥x1 + · · · + xn + y1 + · · · + yn √ 2 ≥ √ 2(a + b). 9.81. Let us take a segment of length P and place the sides of the polygon on the segment as follows: on one end of the segment place the greatest side, on the other end place the second long side; place all the other sides between them. Since any side of the polygon is shorter than 1 2P, the midpoint O of the segment cannot lie on these two longest sides. The length of the side on which point O lies, does not exceed 1 3P (otherwise the ﬁrst two sides would also have been longer than 1 3P and the sum of the three sides would have been greater than P) and, therefore, one of its vertices is distant from O not further than by 1 6P. This vertex divides the segment into two segments to be found since the diﬀerence of their lengths does not exceed 2 6P = 1 3P. 232 CHAPTER 9. GEOMETRIC INEQUALITIES 9.82. Let βk = ∠OAkAk+1. Then xk sin βk = dk = xk+1 sin(αk+1 −βk+1). Hence, 2 P dk = P xk(sin(αk −βk) + sin βk) = 2 P xk sin αk 2 cos ¡ αk 2 −βk ¢ ≤2 P xk sin αk 2 . It is also clear that AkAk+1 = xk cos βk + xk+1 cos(αk+1 −βk+1). Therefore, 2p = P AkAk+1 = P xk(cos(αk −βk) + cos βk) = 2 P xk cos αk 2 cos ¡ αk 2 −βk ¢ ≤2 P xk cos αk 2 . In both cases the equality is only attained if αk = 2βk, i.e., O is the center of the inscribed circle. 9.83. Suppose that the center O of polygon M2 lies outside polygon M1. Then there exists a side AB of polygon M1 such that polygon M1 and point O lie on distinct sides of line AB. Let CD be a side of M1 parallel to AB. The distance between lines AB and CD is equal to the radius of the inscribed circle S of polygon M2 and, therefore, line CD lies outside S. On the other hand, segment CD lies inside M2. Therefore, segment CD is shorter than a half side of polygon M2, cf. Problem 10.66. Contradiction. 9.84. Let A1 be the nearest to O vertex of the polygon. Let us divide the polygon into triangles by the diagonals that pass through vertex A1. Point O lies inside one of these triangles, say, in triangle A1AkAk+1. If point O lies on side A1Ak, then ∠A1OAk = π and the problem is solved. Therefore, let us assume that point O lies strictly inside triangle A1AkAk+1. Since A1O ≤ AkO and A1O ≤Ak+1O, it follows that ∠A1AkO ≤∠AkA1O and ∠A1Ak+1O ≤∠Ak+1A1O. Hence, ∠AkOA1 + ∠Ak+1OA1 = (π −∠OA1Ak −∠OAkA1) + (π −∠OA1Ak+1 −∠OAk+1A1) ≥ 2π −2∠OA1Ak −2∠OA1Ak+1 = 2π −2∠AkA1Ak+1 = 2π −2π n , i.e., one of the angles ∠AkOA1 and ∠Ak+1OA1 is not less than π ¡ 1 −1 n ¢ . 9.85. Let d be the length of the longest diagonal (or side) AB of the given n-gon. Then the perimeter of the n-gon does not exceed πd (Problem 13.42). Let A′ i be the projection of Ai to segment AB. Then either P AA′ i ≥1 2nd or P BA′ i ≥1 2nd (Problem 9.87); let, for deﬁniteness, the ﬁrst inequality hold. Then P AAi > P AA′ i ≥1 2nd > πd ≥P because 1 2n ≥3.5 > π. Any point of the n-gon suﬃciently close to vertex A possesses the required property. 9.86. a) First, suppose that ∠Ai > ∠Bi and for all the other considered pairs of angles an equality takes place. Let us arrange polygons so that vertices A1, . . . , Ai coincide with B1, . . . , Bi. In triangles A1AiAn and A1AiBn sides AiAn and AiBn are equal and ∠A1AiAn > ∠A1AiBn; hence, A1An > A1Bn. If several angles are distinct, then polygons A1 . . . An and B1 . . . Bn can be included in a chain of polygons whose successive terms are such as in the example considered above. b) As we completely traverse the polygon we encounter the changes of minus sign by plus sign as often as the opposite change. Therefore, the number of pairs of neighbouring vertices with equal signs is an even one. It remains to verify that the number of sign changes cannot be equal to 2 (the number of sign changes is not equal to zero because the sums of the angles of each polygon are equal). Suppose the number of sign changes is equal to 2. Let P and Q, as well as P ′ and Q′ be the midpoints of sides of polygons A1 . . . An and B1 . . . Bn on which a change of sign occurs. We can apply the statement of heading a) to pairs of polygons M1 and M ′ 1, M2 and M ′ 2 SOLUTIONS 233 Figure 117 (Sol. 9.86) (Fig. 117); we get PQ > P ′Q′ in the one case, and PQ < P ′Q′ in the other one, which is impossible. 9.87. Let A and B be the midpoints of the segment; X1, . . . , Xn the given points. Since AXi + BXi = 1, it follows that P AXi + P BXi = n. Therefore, either P AXi ≥1 2n or P BXi ≥1 2n. Figure 118 (Sol. 9.88) 9.88. Let us draw a wire along segment AB circumventing the encountered trees along the shortest arc as on Fig. 118. It suﬃces to prove that the way along an arc of the circle is not more than 1.6 times longer than the way along the line. The ratio of the length of an arc with the angle value 2ϕ to the chord it subtends is equal to ϕ sin ϕ. Since 0 < ϕ ≤π 2, it follows that ϕ sin ϕ ≤π 2 < 1.6. 9.89. Let the trees of height a1 > a2 > · · · > an grow at points A1, . . . , An. Then by the hypothesis A1A2 ≤\|a1 −a2\| = a1 −a2, . . . , An−1An ≤an−1 −an. It follows that the length of the broken line A1A2 . . . An does not exceed (a1 −a2) + (a2 −a3) + · · · + (an−1 −an) = a1 −an < 100 m. This broken line can be fenced by a fence, whose length does not exceed 200 m (Fig. 119). Figure 119 (Sol. 9.89) 9.90. In the obtained pentagon, distinguish the parts that were glued (on Fig. 120 these parts are shaded). All the sides that do not belong to the shaded polygons enter the 234 CHAPTER 9. GEOMETRIC INEQUALITIES perimeters of the initial and the obtained polygons. The sides of the shaded polygons that lie on the line along which the folding was performed enter the perimeter of the obtained polygon whereas all the other sides enter the perimeter of the initial polygon. Figure 120 (Sol. 9.90) Since for any polygon the sum of its sides that lie on a line is less than the sum of the other sides, the perimeter of the initial polygon is always longer than the perimeter of the obtained one. 9.91. On the broken line, take two points A and B, that divide its perimeter in halves. Then AB ≤1 2. Let us prove that all the points of the broken line lie inside the circle of radius 1 4 centered at the midpoint O of segment AB. Let M be an arbitrary point of the broken line and point M1 be symmetric to M through point O. Then MO = M1M 2 ≤M1A + AM 2 = BM + AM 2 ≤1 4 because BM + AM does not exceed a half length of the broken line. 9.92. Let acute triangle ABC be placed inside circle S. Let us construct the circum-scribed circle S1 of triangle ABC. Since triangle ABC is an acute one, the angle value of the arc of circle S1 that lies inside S is greater than 180◦. Therefore, on this arc we can select diametrically opposite points, i.e., inside circle S a diameter of circle S1 is contained. It follows that the radius of S is not shorter than the radius of S1. A similar statement for an acute triangle is false. An acute triangle lies inside a circle constructed on the longest side a as on diameter. The radius of this circle is equal to 1 2a and the radius of the circle circumscribed about the triangle is equal to a 2 sin α. Clearly, 1 2a < a 2 sin α. 9.93. First solution. Any triangle of perimeter P can be placed in a disk of radius 1 4P and if an acute triangle is placed in a disk of radius R1, then R1 ≥R (Problem 9.92). Hence, 1 4P = R1 ≥R. Second solution. If 0 < x < π 2, then sin x > 2x π . Hence, a + b + c = 2R(sin α + sin β + sin γ) > 2R(2α + 2β + 2γ) π = 4R. Chapter 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE This chapter is in close connection with the preceding one. For background see the preceding chapter. §1. Medians 10.1. Prove that if a > b, then ma < mb. 10.2. Medians AA1 and BB1 of triangle ABC intersect at point M. Prove that if quadrilateral A1MB1C is a circumscribed one, then AC = BC. 10.3. Perimeters of triangles ABM, BCM and ACM, where M is the intersection point of medians of triangle ABC, are equal. Prove that triangle ABC is an equilateral one. 10.4. a) Prove that if a, b, c are the lengths of sides of an arbitrary triangle, then a2 + b2 ≥1 2c2. b) Prove that m2 a + m2 b ≥2 8c2. 10.5. Prove that m2 a + m2 b + m2 c ≤27 4 R2. b) Prove that ma + mb + mc ≤9 2R. 10.6. Prove that \|a2−b2\| 2c < mc ≤a2+b2 2c . 10.7. Let x = ab + bc + ca, x1 = mamb + mbmc = mcma. Prove that 9 20 < x1 x < 5 4. See also Problems 9.1, 10.74, 10.76, 17.17. §2. Heights 10.8. Prove that in any triangle the sum of the lengths of its heights is less than its semiperimeter. 10.9. Two heights of a triangle are longer than 1. Prove that its area is greater than 1 2. 10.10. In triangle ABC, height AM is not shorter than BC and height BH is not shorter than AC. Find the angles of triangle ABC. 10.11. Prove that 1 2r < 1 ha + 1 hb < 1 r. 10.12. Prove that ha + hb + hc ≥9r. 10.13. Let a < b. Prove that a + ha ≤b + hb. 10.14. Prove that ha ≤√rbrc. 10.15. Prove that ha ≤a 2 cot α 2 . 10.16. Let a ≤b ≤c. Prove that ha + hb + hc ≤3b(a2 + ac + c2) 4pR . See also Problems 10.28, 10.55, 10.74, 10.79. §3. The bisectors 10.17. Prove that la ≤ p p(p −a). 235 236 CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE 10.18. Prove that ha la ≥ q 2r R . 10.19. Prove that a) l2 a + l2 b + l2 c ≤p2; b) la + lb + lc ≤ √ 3p. 10.20. Prove that la + lb + mc ≤ √ 3p. See also Problems 6.38, 10.75, 10.94. §4. The lengths of sides 10.21. Prove that 9r 2S ≤1 a + 1 b + 1 c ≤9R 4S . 10.22. Prove that 2bc cos α b+c < b + c −a < 2bc a . 10.23. Prove that if a, b, c are the lengths of sides of a triangle of perimeter 2, then a2 + b2 + c2 < 2(1 −abc). 10.24. Prove that 20Rr −4r2 ≤ab + bc + ca ≤4(R + r)2. §5. The radii of the circumscribed, inscribed and escribed circles 10.25. Prove that rrc ≤c2 4 . 10.26. Prove that r R ≤2 sin α 2 ¡ 1 −sin α 2 ¢ . 10.27. Prove that 6r ≤a + b. 10.28. Prove that ra ha + rb hb + rc hc ≥3. 10.29. Prove that 27Rr ≤2p2 ≤1 227R2. 10.30. Let O be the centre of the inscribed circle of triangle ABC and OA ≥OB ≥OC. Prove that OA ≥2r and OB ≥r √ 2. 10.31. Prove that the sum of distances from any point inside of a triangle to its vertices is not less than 6r. 10.32. Prove that 3 ³ a ra + b rb + c rc ´ ≥4 ¡ ra a + rb b + rc c ¢ . 10.33. Prove that: a) 5R −r ≥ √ 3p; b) 4R −ra ≥(p −a) h√ 3 + a2+(b−c)2 2S i . 10.34. Prove that 16Rr −5r2 ≤p2 ≤4R2 + 4Rr + 3r2. 10.35. Prove that r2 a + r2 b + r2 c ≥1 427R2. See also Problems 10.11, 10.12, 10.14, 10.18, 10.24, 10.55, 10.79, 10.82, 19.7. §6. Symmetric inequalities between the angles of a triangle Let α, β and γ be the angles of triangle ABC. In problems of this section you have to prove the inequalities indicated. Remark. If α, β and γ are the angles of a triangle, then there exists a triangle with angles π−α 2 , π−β 2 and π−γ 2 . Indeed, these numbers are positive and their sum is equal to π. It follows that if a symmetric inequality holds for sines, cosines, tangents and cotangents of the angles of any triangle then a similar inequality in which sin x is replaced with cos x 2, cos x with sin x 2, tan x with cot x 2 and cot x with tan x 2 is also true. The converse passage from inequalities for halved angles to inequalities with whole angles is only possible for acute triangles. Indeed, if α′ = 1 2(π −α), then α = π −2α′. Therefore, for an acute triangle with angles α′, β′, γ′ there exists a triangle with angles π −2α′, π −2β′ and π −2γ′. Under such a passage sin x 2 turns into cos x, etc., but the inequality obtained can only be true for acute triangles. §8. INEQUALITIES FOR THE AREA OF A TRIANGLE 237 10.36. a) 1 < cos α + cos β + cos γ ≤3 2. b) 1 < sin α 2 + sin β 2 + sin γ 2 ≤3 2. 10.37. a) sin α + sin β + sin γ ≤3 2 √ 3. b) cos α 2 + cos β 2 + cos γ 2 ≤3 2 √ 3. 10.38. a) cot α + cot β + cot γ ≥ √ 3. b) tan α 2 + tan β 2 + tan γ 2 ≥ √ 3. 10.39. cot α 2 + cot β 2 + cot γ 2 ≥3 √ 3. b) For an acute triangle tan α + tan β + tan γ ≥3 √ 3. 10.40. a) sin α 2 sin β 2 sin γ 2 ≤1 8. b) cos α cos β cos γ ≤1 8. 10.41. a) sin α sin β sin γ ≤3 √ 3 8 ; b) cos α 2 cos β 2 cos γ 2 ≤3 8 √ 3. 10.42. a) cos2 α + cos2 β + cos2 γ ≥3 4. b) For an obtuse triangle cos2 α + cos2 β + cos2 γ > 1. 10.43. cos α cos β + cos β cos γ + cos γ cos α ≤3 4. 10.44. For an acute triangle sin 2α + sin 2β + sin 2γ ≤sin(α + β) + sin(β + γ) + sin(γ + α). §7. Inequalities between the angles of a triangle 10.45. Prove that 1 −sin α 2 ≤2 sin β 2 sin γ 2. 10.46. Prove that sin γ 2 ≤ c a+b. 10.47. Prove that if a + b < 3c, then tan α 2 tan β 2 < 1 2. 10.48. In an acute triangle, if α < β < γ, then sin 2α > sin 2β > sin 2γ. 10.49. Prove that cos 2α + cos 2β −cos 2γ ≤3 2. 10.50. On median BM of triangle ABC, point X is taken. Prove that if AB < BC, then ∠XAB < ∠XCB. 10.51. The inscribed circle is tangent to sides of triangle ABC at points A1, B1 and C1. Prove that triangle A1B1C1 is an acute one. 10.52. From the medians of a triangle whose angles are α, β and γ a triangle whose angles are αm, βm and γm is constructed. (Angle αm subtends median AA1, etc.) Prove that if α > β > γ, then α > αm, α > βm, γm > β > αm, βm > γ and γm > γ. See also Problems 10.90, 10.91, 10.93. §8. Inequalities for the area of a triangle 10.53. Prove that: a) 3 √ 3r2 ≤S ≤ p2 3 √ 3; b) S ≤a2+b2+c2 4 √ 3 . 10.54. Prove that a2 + b2 + c2 −(a −b)2 −(b −c)2 −(c −a)2 ≥4 √ 3S. 10.55. Prove that: a) S3 ≤ ³ √ 3 4 ´3 (abc)2; b) √hahbhc ≤ 4 √ 3 √ S ≤ 3 √rarbrc. 238 CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE 10.56. On sides BC, CA and AB of triangle ABC points A1, B1 and C1, respectively, are taken so that AA1, BB1 and CC1 meett at one point. Prove that SA1B1C1 SABC ≤1 4. 10.57. On sides BC, CA and AB of triangle ABC arbitrary points A1, B1 and C1 are taken. Let a = SAB1C1, b = SA1BC1, c = SA1B1C and u = SA1B1C1. Prove that u3 + (a + b + c)u2 ≥4abc. 10.58. On sides BC, CA and AB of triangle ABC points A1, B1 and C1 are taken. Prove that the area of one of the triangles AB1C1, A1BC1, A1B1C does not exceed: a) 1 4SABC; b) SA1B1C1. See also Problems 9.33, 9.37, 9.40, 10.9, 20.1, 20.7. §9. The greater angle subtends the longer side 10.59. In a triangle ABC, prove that ∠ABC < ∠BAC if and only if AC < BC, i.e., the longer side subtends the greater angle and the greater angle subtends the longer side. 10.60. Prove that in a triangle ABC angle ∠A is an acute one if and only if mb > 1 2a. 10.61. Let ABCD and A1B1C1D1 be two convex quadrilaterals with equal corresponding sides. Prove that if ∠A > ∠A1, then ∠B < ∠B1, ∠C < ∠C1, ∠D < ∠D1. 10.62. In an acute triangle ABC the longest height AH is equal to median BM. Prove that ∠B ≤60◦. 10.63. Prove that a convex pentagon ABCDE with equal sides whose angles satisfy inequalities ∠A ≥∠B ≥∠C ≥∠D ≥∠E is a regular one. §10. Any segment inside a triangle is shorter than the longest side 10.64. a) Segment MN is placed inside triangle ABC. Prove that the length of MN does not exceed the length of the longest side of the triangle. b) Segment MN is placed inside a convex polygon. Prove that the length of MN does not exceed that of the longest side or of the greatest diagonal of this polygon. 10.65. Segment MN lies inside sector AOB of a disk of radius R = AO = BO. Prove that either MN ≤R or MN ≤AB (we assume that ∠AOB < 180◦). 10.66. In an angle with vertex A, a circle tangent to the legs at points B and C is inscribed. In the domain bounded by segments AB, AC and the shorter arc ⌣BC a segment is placed. Prove that the length of the segment does not exceed that of AB. 10.67. A convex pentagon lies inside a circle. Prove that at least one of the sides of the pentagon is not longer than a side of the regular pentagon inscribed in the circle. 10.68. Given triangle ABC the lengths of whose sides satisfy inequalities a > b > c and an arbitrary point O inside the triangle. Let lines AO, BO, CO intersect the sides of the triangle at points P, Q, R, respectively. Prove that OP + OQ + OR < a. §11. Inequalities for right triangles In all problems of this section ABC is a right triangle with right angle ∠C. 10.69. Prove that cn > an + bn for n > 2. 10.70. Prove that a + b < c + hc. 10.71. Prove that for a right triangle 0.4 < r h < 0.5, where h is the height dropped from the vertex of the right angle. 10.72. Prove that c r ≥2(1 + √ 2). 10.73. Prove that m2 a + m2 b > 29r2. §13. INEQUALITIES IN TRIANGLES 239 §12. Inequalities for acute triangles 10.74. Prove that for an acute triangle ma ha + mb hb + mc hc ≤1 + R r . 10.75. Prove that for an acute triangle 1 la + 1 lb + 1 lc ≤ √ 2 µ1 a + 1 b + 1 c ¶ . 10.76. Prove that if a triangle is not an obtuse one, then ma + mb + mc ≥4R. 10.77. Prove that if in an acute triangle ha = lb = mc, then this triangle is an equilateral one. 10.78. In an acute triangle ABC heights AA1, BB1 and CC1 are drawn. Prove that the perimeter of triangle A1B1C1 does not exceed a semiperimeter of triangle ABC. 10.79. Let h be the longest height of a non-obtuse triangle. Prove that r + R ≤h. 10.80. On sides BC, CA and AB of an acute triangle ABC, points A1, B1 and C1, respectively, are taken. Prove that 2(B1C1 cos α + C1A1 cos β + A1B1 cos γ) ≥a cos α + b cos β + c cos γ). 10.81. Prove that a triangle is an acute one if and only if a2 + b2 + c2 > 8R2. 10.82. Prove that a triangle is an acute one if and only if p > 2R + r. 10.83. Prove that triangle ABC is an acute one if and only if on its sides BC, CA and AB interior points A1, B1 and C1, respectively, can be selected so that AA1 = BB1 = CC1. 10.84. Prove that triangle ABC is an acute one if and only if the lengths of its projections onto three distinct directions are equal. See also Problems 9.93, 10.39, 10.44, 10.48, 10.62. §13. Inequalities in triangles 10.85. A line is drawn through the intersection point O of the medians of triangle ABC. The line intersects the triangle at points M and N. Prove that NO ≤2MO. 10.86. Prove that if triangle ABC lies inside triangle A′B′C′, then rABC < rA′B′C′. 10.87. In triangle ABC side c is the longest and a is the shortest. Prove that lc ≤ha. 10.88. Medians AA1 and BB1 of triangle ABC are perpendicular. Prove that cot ∠A + cot ∠B ≥2 3. 10.89. Through vertex A of an isosceles triangle ABC with base AC a circle tangent to side BC at point M and intersecting side AB at point N is drawn. Prove that AN > CM. 10.90. In an acute triangle ABC bisector AB, median BM and height CH intersect at one point. What are the limits inside which the value of angle A can vary? 10.91. In triangle ABC, prove that 1 3π ≤πaα + bβ + cγa + b + c < 1 2π. 10.92. Inside triangle ABC point O is taken. Prove that AO sin ∠BOC + BO sin ∠AOC + CO sin ∠AOB ≤p. 10.93. On the extension of the longest side AC of triangle ABC beyond point C, point D is taken so that CD = CB. Prove that angle ∠ABD is not an acute one. 10.94. In triangle ABC bisectors AK and CM are drawn. Prove that if AB > BC, then AM > MK > KC. 10.95. On sides BC, CA, AB of triangle ABC points X, Y , Z are taken so that lines AX, BY , CZ meet at one point O. Prove that of ratios OA : OX, OB : OY , OC : OZ at least one is not greater than 2 and one is not less than 2. 240 CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE 10.96. Circle S1 is tangent to sides AC and AB of triangle ABC, circle S2 is tangent to sides BC and AB and, moreover, S1 and S2 are tangent to each other from the outside. Prove that the sum of radii of these circles is greater than the radius of the inscribed circle S. See also Problems 14.24, 17.16, 17.18. Problems for independent study 10.97. In a triangle ABC, let P = a+b+c, Q = ab+bc+ca. Prove that 3Q < P 2 < 4Q. 10.98. Prove that the product of any two sides of a triangle is greater than 4Rr. 10.99. In triangle ABC bisector AA1 is drawn. Prove that A1C < AC. 10.100. Prove that if a > b and a + ha ≤b + hb, then ∠C = 90◦. 10.101. Let O be the centre of the inscribed circle of triangle ABC. Prove that ab + bc + ca ≥(AO + BO + CO)2. 10.102. On sides of triangle ABC equilateral triangles with centers at D, E and F are constructed outwards. Prove that SDEF ≥SABC. 10.103. In plane, triangles ABC and MNK are given so that line MN passes through the midpoints of sides AB and AC and the intersection of these triangles is a hexagon of area S with pairwise parallel opposite sides. Prove that 3S < SABC + SMNK. Solutions 10.1. Let medians AA1 and BB1 meet at point M. Since BC > AC, points A and C lie on one side of the midperpendicular to segment AB and therefore, both median CC1 and its point M lie on the same side. It follows that AM < BM, i.e., ma < mb. 10.2. Suppose that, for instance, a > b. Then m < mb (Problem 10.1). Since quadrilateral A1MB1C is a circumscribed one, it follows that 1 2a + 1 3mb = 1 2b + 1 3ma, i.e., 1 2(a −b) = 1 3(ma −mb). Contradiction. 10.3. Let, for instance, BC > AC. Then MA < MB (cf. Problem 10.1); hence, BC + MB + MC > AC + MA + MC. 10.4. a) Since c ≤a + b, it follows that c2 ≤(a + b)2 = a2 + b2 + 2ab ≤2(a2 + b2). b) Let M be the intersection point of medians of triangle ABC. By heading a) MA2 + MB2 ≥1 2AB2, i.e., 4 9m2 a + 4 9m2 b ≥1 2c2. 10.5. a) Let M be the intersection point of medians, O the center of the circumscribed circle of triangle ABC. Then AO2 + BO2 + CO2 = (− − → AM + − − → MO)2 + (− − → BM + − − → MO)2 + (− − → CM + − − → MO)2 = AM 2 + BM 2 + CM 2 + 2(− − → AM + − − → BM + − − → CM, − − → MO) + 3MO2. Since − − → AM + − − → BM + − − → CM = − → 0 , it follows that AO2 + BO2 + CO2 = AM 2 + BM 2 + CM 2 + 3MO2 ≥AM 2 + BM 2 + CM 2, i.e., 3R2 ≥4 9(m2 a + m2 b + m2 c). b) It suﬃces to notice that (ma + mb + mc)2 ≤3(m2 a + m2 b + m2 c), cf. Supplement to Ch. 9. 10.6. Heron’s formula can be rewritten as 16S2 = 2a2b2 + 2a2c2 + 2b2c2 −a4 −b4 −c4. SOLUTIONS 241 Since m2 c = 1 4(2a2 + 2b2 −c2) (Problem 12.11 a)), it follows that the inequalities m2 c ≤ µa2 + b2 2c ¶2 ; m2 c ≥ µa2 −b2 2c ¶2 are equivalent to the inequalities 16S2 ≤4a2b2 and 16S2 > 0, respectively. 10.7. Let y = a2 + b2 + c2 and y1 = m2 a + m2 b + m2 c. Then 3y = 4y1 (Problem 12.11, b), y < 2x (Problem 9.7) and 2x1 + y1 < 2x + y because (ma + mb + mc)2 < (a + b + c)2 (cf. Problem 9.2). By adding 8x1 + 4y1 < 8x + 4y to 3y = 4y1 we get 8x1 < y + 8x < 10x, i.e., x1 x < 5 4. Let M be the intersection point of the medians of triangle ABC. Let us complement triangle AMB to parallelogram AMBN. Applying the above-proved statement to triangle AMN we get (x/4) (4x1/9) < 5 4, i.e., x x1 < 20 9 . 10.8. Clearly, ha ≤b, hb ≤c, hc ≤a, where at least one of these inequalities is a strict one. Hence, ha + hb + hc < a + b + c. 10.9. Let ha > 1 and hb > 1. Then a ≥hb > 1. Hence, S = 1 2aha > 1 2. 10.10. By the hypothesis BH ≥AC and since the perpendicular is shorter than a slanted line, BH ≥AC ≥AM. Similarly, AM ≥BC ≥BH. Hence, BH = AM = AC = BC. Since AC = AM, segments AC and AM coincide, i.e., ∠C = 90◦; since AC = BC, the angles of triangle ABC are equal to 45◦, 45◦, 90◦. 10.11. Clearly, 1 ha + 1 hb = a+b 2S = a+b (a+b+c)r and a + b + c < 2(a + b) < 2(a + b + c). 10.12. Since aha = r(a + b + c), it follows that ha = r ¡ 1 + b a + c a ¢ . Adding these equalities for ha, hb and hc and taking into account that x y + y x ≥2 we get the desired statement. 10.13. Since ha −hb = 2S ¡ 1 a −1 b ¢ = 2S b−a ab and 2S ≤ab, it follows that ha −hb ≤b −a. 10.14. By Problem 12.21 2 ha = 1 rb + 1 rc. Moreover, 1 rb + 1 rc ≥ 2 √rbrc. 10.15. Since 2 sin β sin γ = cos(β −γ) −cos(β + γ) ≤1 + cos α, we have ha a = sin β sin γ sin α ≤1 + cos α 2 sin α = 1 2 cot α 2 . 10.16. Since b 2R = sin β, then multiplying by 2p we get (a + b + c)(ha + hb + hc) ≤3 sin β(a2 + ac + c2). Subtracting 6S from both sides we get a(hb + hc) + b(ha + hc) + c(ha + hb) ≤3 sin β(a2 + c2). Since, for instance, ahb = a2 sin γ = a2c 2R , we obtain a(b2 + c2) −2b(a2 + c2) + c(a2 + b2) ≤0. To prove the latter inequality let us consider the quadratic exoression f(x) = x2(a + c) −2x(a2 + c2) + ac(a + c). It is easy to verify that f(a) = −a(a −c)2 ≤0 and f(c) = −c(a −c)2 ≤0. Since the coeﬃcient of x is positive and a ≤b ≤c, it follows that f(b) ≤0. 10.17. By Problem 12.35 a) l2 a = 4bcp(p−a) (b+c)2 . Moreover, 4bc ≤(b + c)2. 242 CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE 10.18. Clearly, ha la = cos 1 2(β −γ). By Problem 12.36 a) 2r R = 8 sin α 2 sin β 2 sin γ 2 = 4 sin α 2 [ cos β −γ 2 −cos β + γ 2 ] = 4x(q −x), where x = sin α 2 and q = cos β −γ 2 . It remains to notice that 4x(q −x) ≤q2. 10.19. a) By Problem 10.17 l2 a ≤p(p −a). Adding three similar inequalities we get the desired statement. b) For any numbers la, lb and lc we have (la + lb + lc)2 ≤3(l2 a + l2 b + l2 c). 10.20. It suﬃces to prove that p p(p −a) + p p(p −b) + mc ≤√3p. We may assume that p = 1; let x = 1 −a and y = 1 −b. Then m2 c = 2a2 + 2b2 −c2 4 = 1 −(x + y) + (x −y)2 4 = m(x, y). Let us consider the function f(x, y) = √x + √y + p m(x, y). We have to prove that f(x, y) ≤ √ 3 for x, y ≥0 and x + y ≤1. Let g(x) = f(x, x) = 2√x + √ 1 −2x. Since g′(x) = 1 √x − 1 √1−2x, it follows that as x grows from 0 to 1 3 and g(x) grows from 1 to √ 3 and as x grows from 1 3 to 1 2; we also see that g(x) diminishes from √ 3 to √ 2. Introduce new variables: d = x−y and q = √x+√y. It is easy to verify that (x−y)2 −2q2(x+y)+q4 = 0, i.e., x + y = d2+q4 2q2 . Hence, f(x, y) = q + s 1 −q2 2 −d2(2 −q2) 4q2 . Now, observe that q2 = (√x + √y)2 ≤2(x + y) ≤2, i.e., d2(2−q2) 4q2 ≥0. It follows that for a ﬁxed q the value of function f(x, y) is the maximal one for d = 0, i.e., x = y; the case x = y(?) is the one considered above. 10.21. Clearly, 1 a + 1 b + 1 c = ha+hb+hc 2S . Moreover, 9r ≤ha + hb + hc (Problem 10.12) and ha + hb + hc ≤ma + mb + mc ≤9 2R (Problem 10.5 b)). 10.22. First, let us prove that b + c −a < 2bc a . Let 2x = b + c −a, 2y = a + c −b and 2z = a + b −c. We have to prove that 2x < 2(x + y)(x + z) y + z , i.e., xy + xz < xy + xz + x2 + yz. The latter inequality is obvious. Since 2bc cos α = b2 + c2 −a2 = (b + c −a)(b + c + a) −2bc, it follows that 2bc cos α b + c = b + c −a + ·(b + c −a)a b + c −2bc b + c ¸ . The expression in square brackets is negative because b + c −a < 2bc a . 10.23. By Problem 12.30 we have a2 + b2 + c2 = (a + b + c)2 −2(ab + bc + ac) = 4p2 −2r2 −2p2 −8rR = 2p2 −2r2 −8rR SOLUTIONS 243 and abc = 4prR. Thus, we have to prove that 2p2 −2r2 −8rR < 2(1 −4prR), where p = 1. This inequality is obvious. 10.24. By Problem 12.30, ab + bc + ca = r2 + p2 + 4Rr. Moreover, 16Rr −5r2 ≤p2 ≤ 4R2 + 4Rr + 3r2 (Problem 10.34). 10.25. Since r(cot α + cot β) = c = rc(tan α + tan β), it follows that c2 = rrc µ 2 + tan α tan β + tan β tan β ¶ ≥4rrc. 10.26. It suﬃces to apply the results of Problems 12.36 a) and 10.45. Notice also that x(1 −x) ≤1 4, i.e., r R ≤1 2. 10.27. Since hc ≤a and hc ≤b, it follows that 4S = 2chc ≤c(a + b). Hence, 6r(a + b + c) = 12S ≤4ab + 4S ≤(a + b)2 + c(a + b) = (a + b)(a + b + c). 10.28. Since 2 ha = 1 rb + 1 rc (Problem 12.21), it follows that ra ha = 1 2 ³ ra rb + ra rc ´ . Let us write similar equalities for rb hb and rc hc and add them. Taking into account that x y + y x ≥2 we get the desired statement. 10.29. Since Rr = PS p = abc 4p (cf. Problem 12.1), we obtain 27abc ≤8p3 = (a + b + c)3. Since (a + b + c)2 ≤3(a2 + b2 + c2) for any numbers a, b and c, we have p2 ≤3 4(a2 + b2 + c2) = m2 a + m2 b + m2 c (cf. Problem 12.11 b)). It remains to notice that m2 a + m2 b + m2 c ≤27 4 R2 (Problem 10.5 a)). 10.30. Since OA = r sin A 2 , OB = r sin ∠B 2 and OC = r sin ∠X 2 and since angles 1 2∠A, 1 2∠B and 1 2∠C are acute ones, it follows that ∠A ≤∠B ≤∠C. Hence, ∠A ≤60◦and ∠B ≤90◦and, therefore, sin ∠A 2 ≤1 2 and sin ∠B 2 ≤ 1 √ 2. 10.31. If ∠C ≥120◦, then the sum of distances from any point inside the triangle to its vertices is not less than a + b (Problem 11.21); moreover, a + b ≥6r (Problem 10.27). If each angle of the triangle is less than 120◦, then at a point the sum of whose distances from the vertices of the triangle is the least one the square of this sum is equal to 1 2(a2 + b2 + c2) + 2 √ 3S (Problem 18.21 b)). Further, 1 2(a2 + b2 + c2) ≥2 √ 3S (Problem 10.53 b)) and 4 √ 3S ≥36r2 (Problem 10.53 a)). 10.32. Let α = cos ∠A 2 , β = cos ∠B 2 and γ = cos ∠C 2 . By Problem 12.17 b) a ra = α βγ, b rb = β γα and c rc = γ αβ. Therefore, multiplying by αβγ we express the inequality to be proved in the form 3(α2 + β2 + γ2) ≥4(β2γ2 + γ2α2 + α2β2). Since α2 = 1+cos ∠A 2 , β2 = 1+cos ∠B 2 and γ2 = 1+cos ∠C 2 , we obtain the inequality cos ∠A + cos ∠B + cos ∠C + 2(cos ∠A cos ∠B + cos ∠B cos ∠C + cos ∠C cos ∠A) ≤3. It remains to make use of results of Problems 10.36 and 10.43. 10.33. a) Adding equality 4R+r = ra+rb+rc (Problem 12.24) with inequality R−2r ≥0 (Problem 10.26) we get 5R −r ≥ra + rb + rc = pr((p −a)−1 + (p −b)−1 + (p −c)−1) = p(ab+bc+ca−p2) S == p(2(ab+bc+ca)−a2−b2−c2) 4S . 244 CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE It remains to observe that 2(ab + bc + ca) −a2 −b2 −c2 ≥4 √ 3S (Problem 10.54). b) It is easy to verify that 4R −ra = rb + rc −r = pr p −b + pr p −c −pr p = (p −a)(p2 −bc) S . It remains to observe that 4(p2 −bc) = a2 + b2 + c2 + 2(ab −bc + ca) = = 2(ab + bc + ca) = −a2 −b2 −c2 + 2(a2 + b2 + c2 −2bc) ≥4 √ 3S + 2(a2 + (b −c)2). 10.34. Let a, b and c be the lengths of the sides of the triangle, F = (a−b)(b−c)(c−a) = A −B, where A = ab2 + bc2 + ca2 and B = a2b + b2c + c2a. Let us prove that the required inequalities can be obtained by a transformation of an obvious inequality F 2 ≥0. Let σ1 = a + b + c = 2p, σ2 = ab + bc + ca = r2 + p2 + 4rR and σ3 = abc = 4prR, cf. Problem 12.30. It is easy to verify that F 2 = σ2 1σ2 2 −4σ3 2 −4σ3 1σ3 + 18σ1σ2σ3 −27σ2 3. Indeed, (σ1σ2)2 −F 2 = (A + B + 3abc)2 −(A −B)2 = 4AB + 6(A + B)σ3 + 9σ2 3 = 4(a3b3 + . . .) + 4(a4bc + . . .) + 6(A + B)σ3 + 21σ2 3. It is also clear that 4σ3 2 = 4(a3b3 + . . .) + 12(A + B)σ3 + 24σ2 3, 4σ3 1σ3 = 4(a4bc + . . .) + 12(A + B)σ3 + 24σ2 3, 18σ1σ2σ3 = 18(A + B)σ3 + 54σ2 3. Expressing σ1, σ2 and σ3 via p, r and R, we obtain F 2 = −4r2[(p2 −2R2 −10Rr + r2)2 −4R(R −2r)3] ≥0. Thus, we obtain p2 ≥2R2 + 10Rr −r2 −2(R −2r) p R(R −2r) = [(R −2r) − p R(R −2r)]2 + 16Rr −5r2 ≥16Rr −5r2 p2 ≤2R2 + 10Rr + r2 + 2(R −2r) p R(R −2r) = 4R2 + 4Rr + 3r2 −[(R −2r) − p R(R −2r)]2 ≤ 4R2 + 4Rr + 3r2. 10.35. Since ra +rb +rc = 4R+r and rarb +rbrc +rcra = p2 (Problems 12.24 and 12.25), it follows that r2 a + r2 b + r2 c = (4R + r)2 −2p2. By Problem 10.34 p2 ≤4R2 + 4Rr + 3r2; hence, r2 a + r2 b + r2 c = 8R2 −5r2. It remains to notice that r ≤1 2R (Problem 10.26). 10.36. a) By Problem 12.38 cos α + cos β + cos γ = R+r R . Moreover, r ≤1 2R (Problem 10.26). b) Follows from heading a), cf. Remark. 10.37. a) Clearly, sin α + sin β + sin γ = p R. Moreover, p ≤3 2 √ 3R (Problem 10.29). b) Follows from heading a), cf. Remark. 10.38. a) By Problem 12.44 a) cot α + cot β + cot γ = a2 + b2 + c2 4S . SOLUTIONS 245 Moreover, a2 + b2 + c2 ≥4 √ 3S (Problem 10.53 b)). b) Follows from heading a), cf. Remark. 10.39. a) By Problem 12.45 a) cot α 2 + cot β 2 + cot γ 2 = p r. Moreover, p ≥3 √ 3r (Problem 10.53 a)). b) Follows from heading a), cf. Remark. For an acute triangle tan α + tan β + tan γ < 0; cf., for instance, Problem 12.46. 10.40. a) By Problem 12.36 a) sin α 2 + sin β 2 + sin γ 2 = r 4R. Moreover, r ≤1 2R (Problem 10.26). b) For an acute triangle it follows from heading a), cf. Remark. For an obtuse triangle cos α cos β cos γ < 0. 10.41. a) Since sin x = 2 sin x 2 cos x 2, we see that making use of results of Problems 12.36 a) and 12.36 c) we obtain sin α sin β sin γ = pr 2R2. Moreover, p ≤3 2 √ 3R (Problem 10.29) and r ≤1 2R (Problem 10.26). b) Follows from heading a), cf. Remark. 10.42. By Problem 12.39 b) cos2 α + cos2 β + cos2 γ = 1 −2 cos α cos β cos γ. It remains to notice that cos α cos β cos γ ≤1 8 (Problem 10.40 b)) and for an obtuse triangle cos α cos β cos γ < 0. 10.43. Clearly, 2(cos α cos β + cos β cos γ + cos γ cos α) = (cos α + cos β + cos γ)2 −cos2 α −cos2 β −cos2 γ. It remains to notice that cos α + cos β + cos γ ≤3 2 (Problem 10.36 a)) and cos2 α + cos2 β + cos2 γ ≥3 4 (Problem 10.42). 10.44. Let the extensions of bisectors of acute triangle ABC with angles α, β and γ intersect the circumscribed circle at points A1, B1 and C1, respectively. Then SABC = R2(sin 2α + sin 2β + sin 2γ) 2 ; SA1B1C1 = R2(sin(α + β) + sin(β + γ) + sin(γ + α)) 2 . It remains to make use of results of Problems 12.72 and 10.26. 10.45. Clearly, 2 sin β 2 sin γ 2 = cos β −γ 2 −cos β + γ 2 ≤1 −sin α 2 . 10.46. Let us drop perpendiculars AA1 and BB1 from vertices A and B to the bisector of angle ∠ACB. Then AB ≥AA1 + BB1 = b sin γ 2 + a sin γ 2. 10.47. By Problem 12.32 tan α 2 tan β 2 = a+b−c a+b+c. Since a+b < 3c, it follows that a+b−c < 1 2(a + b + c). 246 CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE 10.48. Since π −2α > 0, π −2β > 0, π −2γ > 0 and (π −2α)+(π −2β)+(π −2γ) = π, it follows that there exists a triangle whose angles are π −2α, π −2β, π −2γ. The lengths of sides opposite to angles π −2α, π −2β, π −2γ are proportional to sin(π −2α) = sin 2α, sin 2β, sin 2γ, respectively. Since π −2α > π −2β > π −2γ and the greater angle subtends the longer side, sin 2α > sin 2β > sin 2γ. 10.49. First, notice that cos 2γ = cos 2(π −α −β) = cos 2α cos 2β −sin 2α sin 2β. Hence, cos 2α + cos 2β −cos 2γ = cos 2α + cos 2β −cos 2α cos 2β + sin 2α sin 2β. Since a cos ϕ + b sin ϕ ≤ √ a2 + b2 (cf. Supplement to Ch. 9), it follows that (1 −cos 2β) cos 2α + sin 2β sin 2α + cos 2β ≤ q (1 −cos 2β)2 + sin2 2β + cos 2β = 2\| sin β\| + 1 −2 sin2 β. It remains to notice that the greatest value of the quadratic 2t + 1 −2t2 is attained at point t = 1 2 and this value is equal to 3 2. The maximal value corresponds to angles α = β = 30◦ and γ = 120◦. 10.50. Since AB < CB, AX < CX = SABX = SBCX, it follows that sin ∠XAB > sin ∠XCB. Taking into account that angle ∠XCB is an acute one, we get the desired statement. 10.51. If the angles of triangle ABC are equal to α, β and γ, then the angles of triangle A1B1C1 are equal to 1 2(β + γ), 1 2(γ + α) and 1 2(α + β). 10.52. Let M be the intersection point of medians AA1, BB1 and CC1. Complementing triangle AMB to parallelogram AMBN we get ∠BMC1 = αm and ∠AMC1 = βm. It is easy to verify that ∠C1CB < 1 2γ and ∠B1BC < 1 2β. It follows that αm = ∠C1CB + ∠B1BC < 1 2(β + γ) < β. Similarly, γm = ∠A1AB + ∠B1BA > 1 2(α + β) > β. First, suppose that triangle ABC is an acute one. Then the heights’ intersection point H lies inside triangle AMC1. Hence, ∠AMB < ∠AHB, i.e., π −γm < π −γ and ∠CMB < ∠CHB, i.e., π −αm > π −α. Now, suppose that angle α is an obtuse one. Then angle CC1B is also an obtuse one and therefore, angle αm is an acute one, i.e., αm < α. Let us drop perpendicular MX from point M to BC. Then γm > ∠XMB > 180◦−∠HAB > γ. Since α > αm, it follows that α+(π−αm) > π, i.e., point M lies inside the circumscribed circle of triangle AB1C1. Therefore, γ = ∠AB1C1 < ∠AMC1 = βm. Similarly, α = ∠CB1A1 > ∠CMA1 = βm because γ + (π −γm) < π. 10.53 a) Clearly, S2 p = (p −a)(p −b)(p −c) ≤ µp −a + p −b + p −c 3 ¶2 = p3 27. Hence, pr = S ≤ p2 3 √ 3, i.e., r ≤ p 3 √ 3. By multiplying the latter inequality by r we get the desired statement. b) Since (a + b + c)2 ≤3(a2 + b2 + c2), it follows that S ≤ p2 3 √ 3 = (a + b + c)2 12 √ 3 ≤a2 + b2 + c2 4 √ 3 . SOLUTIONS 247 10.54. Let x = p −a, y = p −b, z = p −c. Then (a2 −(b −c)2) + (b2 −(a −c)2) + (c2 −(a −b)2) = 4(p −b)(p −c) + 4(p −a)(p −c) + 4(p −a)(p −b) = 4(yz + zx + xy) and 4 √ 3S = 4 p 3p(p −a)(p −b)(p −c) = 4 p 3(x + y + z)xyz. Thus, we have to prove that xy + yz + zx ≥ p 3(x + y + z)xyz. After squaring and simpli-ﬁcation we obtain x2y2 + y2z2 + z2x2 ≥x2yz + y2xz + z2xy. Adding inequalities x2yz ≤x2(y2 + z2) 2 , y2xz ≤y2(x2 + z2) 2 and z2xy ≤z2(x2 + y2) 2 we get the desired statement. 10.55. a) By multiplying three equalities of the form S = 1 2ab sin γ we get S3 = 1 8(abc)2 sin γ sin β sin α. It remains to make use of a result of Problem 10.41. b) Since (hahbhc)2 = (2S)6 (abc)2 and (abc)2 ≥ ³ 4 √ 3 ´3 S3, it follows that (hahbhc)2 ≤(2S)6( √ 3/4)3 S3 = ( √ 3S)3. Since (rarbrc)2 = S4 r2 (Problem 12.18, c) and r2( √ 3)3 ≤S (Problem 10.53 a), it follows that (rarbrc)2 ≥( √ 3S)3. 10.56. Let p = BA BC, q = CB1 CA and r = AC1 AC . Then SA1B1C1 SABC = 1 −p(1 −r) −q(1 −p) −r(1 −q) = 1 −(p + q + r) + (pq + qr + rp). By Cheva’s theorem (Problem 5.70) pqr = (1 −p)(1 −q)(1 −r), i.e., 2pqr = 1 −(p + q + r) + (pq + qr + rp). Moreover, (pqr)2 = p(1 −p)(1 −q)r(1 −r) ≤ µ1 4 ¶3 . Therefore, SA1B1C1 SABC = 2pqr ≤1 4. 10.57. We can assume that the area of triangle ABC is equal to 1. Then a + b + c = 1 and, therefore, the given inequality takes the form u2 ≥4abc. Let x = BA1 BC , y = CB1 CA and z = AC1 AB . Then u = 1 −(x + y + z) + xy + yz + zx and abc = xyz(1 −x)(1 −y)(1 −z) = v(u −v), where v = xyz. Therefore, we pass to inequality u2 ≥4v(u −v), i.e., (u −2v)2 ≥0 which is obvious. 10.58. a) Let x = BA1 BC , y = BC1 BA and z = AC1 AB . We may assume that the area of triangle ABC is equal to 1. Then SAB1C1 = z(1 −y), SA1BC1 = x(1 −z) and SA1B1C = y(1 −x). Since x(1 −x) ≤1 4, y(1 −y) ≤1 4 and z(1 −z) ≤1 4, it follows that the product of numbers SAB1C1, SA1BC1 and SA1B1C does not exceed ¡ 1 4 ¢3; hence, one of them does not exceed 1 4. b) Let, for deﬁniteness, x ≥ 1 2. If y ≤ 1 2, then the homothety with center C and coeﬃcient 2 sends points A1 and B1 to inner points on sides BC and AC, consequently, SA1B1C ≤SA1B1C1. Hence, we can assume that y ≥1 2 and, similarly, z ≥1 2. Let x = 1 2(1+α), y = 1 2(1 + β) and z = 1 2(1 + γ). Then SAB1C1 = 1 4(1 + γ −β −βγ), SA1B1C1 = 1 4(1 + α − γ −αγ) and SA1B1C = 1 4(1 + β −α −αβ); hence, SA1B1C1 = 1 4(1 + αβ + βγ + αγ) ≥1 4 and SAB1C1 + SA1BC1 + SA1B1C ≤3 4. 248 CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE 10.59. It suﬃces to prove that if AC < BC, then ∠ABC < ∠BAC. Since AC < BC, on side BC point A1 can be selected so that A1C = AC. Then ∠BAC < ∠AAC = ∠AA1C > ∠ABC. 10.60. Let A1 be the midpoint of side BC. If AA1 < 1 2BC = BA1 = A1C, then ∠BAA1 > ∠ABA1 and ∠CAA1 > ∠ACA1; hence, ∠A = ∠BAA1 + ∠CAA1 < ∠B + ∠C, i.e., ∠A > 90◦. Similarly, if AA1 > 1 2BC then ∠A > 90◦. 10.61. If we ﬁx two sides of the triangle, then the greater the angle between these sides the longer the third side. Therefore, inequality ∠A > ∠A1 implies that BD > B1D1, i.e., ∠C < ∠C1. Now, suppose that ∠B ≥∠B1. Then AC ≥A1C1, i.e., ∠D ≥∠D1. Hence, 360◦= ∠A + ∠B + ∠C + ∠D > ∠A1 + ∠B1 + ∠C1 + ∠D1 = 360◦. Contradiction; therefore, ∠B < ∠B1 and ∠D < ∠D1. 10.62. Let point B1 be symmetric to B through point M. Since the height dropped from point M to side BC is equal to a half of AH, i.e., to a half of BM, it follows that ∠MBC = 30◦. Since AH is the longest of heights, BC is the shortest of sides. Hence, AB1 = BC ≤AB, i.e., ∠ABB1 ≤∠AB1B = ∠MBC = 30◦. Therefore, ∠ABC = ∠ABB1 + ∠MBC ≤30◦+ 30◦= 60◦. 10.63. First, let us suppose that ∠A > ∠D. Then BE > EC and ∠EBA < ∠ECD. Since in triangle EBC side BE is longer than side EC, it follows that ∠EBC < ∠ECB. Therefore, ∠B = ∠ABE + ∠EBC < ∠ECD + ∠ECB = ∠C which contradicts the hypothesis. Thus, ∠A = ∠B = ∠C = ∠D. Similarly, the assumption ∠B > ∠E leads to inequality ∠C < ∠D. Hence, ∠B = ∠C = ∠D = ∠E. 10.64. Let us carry out the proof for the general case. Let line MN intersect the sides of the polygon at points M1 and N1. Clearly, MN ≤M1N1. Let point M1 lie on side AB and point N1 lie on PQ. Since ∠AM1N1 + ∠BM1N1 = 180◦, one of these angles is not less than 90◦. Let, for deﬁniteness, ∠AM1N1 ≥90◦. Then AN1 ≥M1N1 because the longer side subtends the greater angle. We similarly prove that either AN1 ≤AP or AN1 ≤AQ. Therefore, the length of segment MN does not exceed the length of a segment with the endpoints at vertices of the polygon. 10.65. The segment can be extended to its intersection with the boundary of the sector because this will only increase its length. Therefore, we may assume that points M and N lie on the boundary of the disk sector. The following three cases are possible: 1) Points M and N lie on an arc of the circle. Then MN = 2R sin ∠MON 2 ≤2R sin ∠AOB 2 = AB because 1 2∠MON ≤1 2∠AOB ≤90◦. 2) Points M and N lie on segments AO and BO, respectively. Then MN is not longer than the longest side of triangle AOB. 3) One of points M and N lies on an arc of the circle, the other one on one of segments AO or BO. Let, for deﬁniteness, M lie on AO and N on an arc of the circle. Then the length of MN does not exceed that of the longest side of triangle ANO. It remains to notice that AO = NO = R and AN ≤AB. 10.66. If the given segment has no common points with the circle, then a homothety with center A (and coeﬃcient greater than 1) sends it into a segment that has a common point X with arc AB and lies in our domain. Let us draw through point X tangent DE to the circle (points D and E lie on segments AB and AC). Then segments AD and AE are SOLUTIONS 249 shorter than AB and DE < 1 2(DE + AD + AE) = AB, i.e., each side of triangle ADE is shorter than AB. Since our segment lies inside triangle ADE (or on its side DE), its length does not exceed that of AB. 10.67. First, suppose that the center O of the circle lies inside the given pentagon A1A2A3A4A5. Consider angles ∠A1OA2, ∠A2OA3, . . . , ∠A5OA1. The sum of these ﬁve angles is equal to 2π; hence, one of them, say, ∠A1OA2, does not exceed 2 5π. Then segment A1A2 can be placed in disk sector OBC, where ∠BOC = 2 5π and points B and C lie on the circle. In triangle OBC, side BC is the longest one; hence, A1A2 ≤BC. If point O does not belong to the given pentagon, then the union of angles ∠A1OA2, . . . , ∠A5OA1 is less than π and each point of the angle — the union — is covered twice by these angles. Therefore, the sum of these ﬁve angles is less than 2π, i.e., one of them is less than 2 5π. The continuation of the proof is similar to the preceding case. If point O lies on a side of the pentagon, then one of the considered angles is not greater than 1 4π and if it is its vertex, then one of them is not greater than 1 3π. Clearly, 1 4π < 1 3π < 2 5π. Figure 121 (Sol. 10.68) 10.68. On sides BC, CA, AB take points A1 and A2, B1 and B2, C1 and C2, respectively, so that B1C2 ∥BC, C1A2 ∥CA, A1B2 ∥AB (Fig. 121). In triangles A1A2O, B1B2O, C1C2O sides A1A2, B1O, C2O, respectively, are the longest ones. Hence, OP < A1A2, OQ < BO, OR ≤C2O, i.e., OP + OQ + OR < A1A2 + B1O + C2O = A1A2 + CA2 + BA1 = BC. 10.69. Since c2 = a2 + b2, it follows that cn = (a2 + b2)cn−2 = a2cn−2 + b2cn−2 > an + bn. 10.70. The height of any of the triangles considered is longer than 2r. Moreover, in a right triangle 2r = a + b −c (Problem 5.15). 10.71. Since ch = 2S = r(a + b + c) and c = √ a2 + b2, it follows that r h = √ a2+b2 a+b+ √ a2+b2 = 1 x+1, where x = a+b √ a2+b2 = q 1 + 2ab a2+b2. Since 0 < 2ab a2+b2 ≤1, it follows that 1 < x ≤ √ 2. Hence, 2 5 < 1 1+ √ 2 ≤r h < 1 2. 10.72. Clearly, a + b ≥2 √ ab and c2 + a2 + b2 ≥2ab. Hence, c2 r2 = (a + b + c)2c2 a2b2 ≥(2 √ ab + √ 2ab)2 · 2ab a2b2 = 4(1 + √ 2)2. 10.73. By Problem 12.11 a) m2 a + m2 b = 1 4(4c2 + a2 + b2) = 5 4c2. Moreover, 5c2 4 ≥5(1 + √ 2)2r2 = (15 + 10 √ 2)r2 > 29r2, cf. Problem 10.72. 10.74. Let O be the center of the circumscribed circle, A1, B1, C1 the midpoints of sides BC, CA, AB, respectively. Then ma = AA1 ≤AO + OA1 = R + OA1. Similarly, 250 CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE mb ≤R + OB1 and mc ≤R + OC1. Hence, ma ha + mb hb + mc hc ≤R µ 1 ha + 1 hb + 1 hc ¶ + OA1 ha + OB1 hb + OC1 hc . It remains to make use of the result of Problem 12.22 and the solution of Problem 4.46. 10.75. By Problem 4.47 1 b + 1 c = 2 cos(α/2) la ≥ √ 2 la . Adding three analogous inequalities we get the required statement. 10.76. Denote the intersection point of medians by M and the center of the circumscribed circle by O. If triangle ABC is not an obtuse one, then point O lies inside it (or on its side); let us assume, for deﬁniteness, that it lies inside triangle AMB. Then AO+BO ≤AM+BM, i.e., 2R ≤2 3ma + 2 3mb or, which is the same, ma + mb ≥3R. It remains to notice that since angle ∠COC1 (where C1 is the midpoint of AB) is obtuse, it follows that CC1 ≥CO, i.e., mc ≥R. The equality is attained only for a degenerate triangle. 10.77. In any triangle hb ≤lb ≤mb (cf. Problem 2.67); hence, ha = lb ≥hb and mc = lb ≤mb. Therefore, a ≤b and b ≤c (cf. Problem 10.1), i.e., c is the length of the longest side and γ is the greatest angle. The equality ha = mc yields γ ≤60◦(cf. Problem 10.62). Since the greatest angle γ of triangle ABC does not exceed 60◦, all the angles of the triangle are equal to 60◦. 10.78. By Problem 1.59 the ratio of the perimeters of triangles A1B1C1 and ABC is equal to r R. Moreover, r ≤R 2 (Problem 10.26). Remark. Making use of the result of Problem 12.72 it is easy to verify that SA1B1C1 SABC = r1 2R1 ≤1 4. 10.79. Let 90◦≥α ≥β ≥γ, then CH is the longest height. Denote the centers of the inscribed and circumscribed circles by I and O, the tangent points of the inscribed circle with sides BC, CA, AB by K, L, M, respectively (Fig. 122). Figure 122 (Sol. 10.79) First, let us prove that point O lies inside triangle KCI. For this it suﬃces to prove that CK ≥KB and ∠BCO ≤∠BCI. Clearly, CK = r cot γ 2 ≥r cot β 2 = KB and 2∠BCO = 180◦−∠BOC = 180◦−2α ≤180◦−α −β = γ = 2∠BCI. Since ∠BCO = 90◦−α = ∠ACH, the symmetry through CI sends line CO to line CH. Let O′ be the image of O under this symmetry and P the intersection point of CH and IL. Then CP ≥CO′ = CO = R. It remains to prove that PH ≥IM = r. It follows from the fact that ∠MIL = 180◦−α ≥90◦. 10.80. Let B2C2 be the projection of segment B1C1 on side BC. Then BC1 ≥B2C2 = BC −BC1 cos β −CB1 cos γ. SOLUTIONS 251 Similarly, A1C1 ≥AC −AC1 cos α −CA1 cos γ; A1B1 ≥AB −AB1 cos α −BA1 cos β. Let us multiply these inequalities by cos α, cos β and cos γ, respectively, and add them; we get B1C1 cos α + C1A1 cos β + AB1 cos γ ≥a cos α + b cos β + c cos γ− −(a cos β cos γ + b cos α cos γ + c cos α cos β). Since c = a cos β + b cos α, it follows that c cos γ = a cos β cos γ + b cos α cos γ. Write three analogous inequalities and add them; we get a cos β cos γ + b cos α cos γ + c cos α cos β = a cos α + b cos β + c cos γ 2 . 10.81. Since cos2 α + cos2 β + cos2 γ + 2 cos α cos β cos γ = 1 (Problem 12.39 b)), it follows that triangle ABC is an acute one if and only if cos2 α + cos2 β + cos2 γ < 1, i.e., sin2 α + sin2 β + sin2 γ > 2. Multiplying both sides of the latter inequality by 4R2 we get the desired statement. 10.82. It suﬃces to notice that p2 −(2R + r)2 = 4R2 cos α cos β cos γ (cf. Problem 12.41 b). 10.83. Let ∠A ≤∠B ≤∠C. If triangle ABC is not an acute one, then CC1 < AC < AA1 for any points A1 and C1 on sides BC and AB, respectively. Now, let us prove that for an acute triangle we can select points A1, B1 and C1 with the required property. For this it suﬃces to verify that there exists a number x satisfying the following inequalities: ha ≤x < max(b, c) = c, hb ≤x < max(a, c) = c and hc ≤x < max(a, b) = b. It remains to notice that max(ha, hb, hc) = ha, min(b, c) = b and ha < h. 10.84. Let ∠A ≤∠B ≤∠C. First, suppose that triangle ABC is an acute one. As line l that in its initial position is parallel to AB rotates, the length of the triangle’s projection on l ﬁrst varies monotonously from c to hb, then from hb to a, then from a to hc, next from hc to b, then from b to ha and, ﬁnally, from ha to c. Since hb < a, there exists a number x such that hb < x < a. It is easy to verify that a segment of length x is encountered on any of the ﬁrst four intervals of monotonity. Now, suppose that triangle ABC is not an acute one. As line l that in its initial position is parallel to AB rotates, the length of the triangle’s projection on l monotonously decreases ﬁrst from c to hb, then from hb to hc; after that it monotonously increases, ﬁrst, from hc to ha, then from ha to c. Altogether we have two intervals of monotonity. 10.85. Let points M and N lie on sides AB and AC, respectively. Let us draw through vertex C the line parallel to side AB. Let N1 be the intersection point of this line with MN. Then N1O : MO = 2 but NO ≤N1O; hence, NO : MO ≤2. 10.86. Sircle S inscribed in triangle ABC lies inside triangle A′B′C′. Draw the tangents to this circle parallel to sides of triangle A′B′C′; we get triangle A′′B′′C′′ similar to triangle A′B′C′ and S is the inscribed circle of triangle A′′B′′C′′. Hence, rABC = rA′′B′′C′′ < rA′B′C′. 10.87. The bisector lc divides triangle ABC into two triangles whose doubled areas are equal to alc sin γ 2 and blc sin γ 2. Hence, aha = 2S = lc(a + b) sin γ 2. The conditions of the problem imply that a a+b ≤1 2 ≤sin γ 2. 252 CHAPTER 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE 10.88. Clearly, cot ∠A + cot ∠B = c hc ≥ c mc. Let M be the intersection point of the medians, N the midpoint of segment AB. Since triangle AMB is a right one, MN = 1 2AB. Therefore, c = 2MN = 2 3mc. 10.89. Since BN ·BA = BM 2 and BM < BA, it follows that BN < BM and, therefore, AN > CN. 10.90. Let us draw through point B the perpendicular to side AB. Let F be the intersection point of this perpendicular with the extension of side AC (Fig. 123). Let us prove that bisector AD, median BM and height CH intersect at one point if and only if AB = CF. Indeed, let L be the intersection point of BM and CH. Bisector AD passes through point L if and only if BA : AM = BL : LM but BL : LM = FC : CM = FC : AM. Figure 123 (Sol. 10.90) If on side AF of right triangle ABF (∠ABF = 90◦) segment CF equal to AB is marked, then angles ∠BAC and ∠ABC are acute ones. It remains to ﬁnd out when angle ∠ACB is acute. Let us drop perpendicular BP from point B to side AF. Angle ACB is an acute one if FP > FC = AB, i.e., BF sin ∠A > BF cot ∠A. Therefore, 1−cos2 ∠A = sin2 ∠A > cos ∠A, i.e., cos A < 1 2( √ 5 −1). Finally, we see that 90◦> ∠A > arccos √5 −1 2 ≈51◦50′. 10.91. Since the greater angle subtends the longer side, (a −b)(α −β) ≥0, (b −c)(β −γ) ≥0 and (a −c)(α −γ) ≥0. Adding these inequalities we get 2(aα + bβ + cγ) ≥a(β + γ) + b(α + γ) + c(α + β) = (a + b + c)π −aα −bβ −cγ, i.e., 1 3π ≤aα+bβ+cγ a+b+c . The triangle inequality implies that α(b + c −a) + β(a + c −b) + γ(a + b −c) > 0, i.e., a(β + γ −α) + b(α + γ −β) + c(α + β −γ) > 0. Since α+β +γ = π, it follows that a(π−2α)+b(π−2β) = c(π−2γ) > 0, i.e., aα+bβ+cγ a+b+c < 1 2π. 10.92. On rays OB and OC, take points C1 and B1, respectively, such that OC1 = OC and OB1 = OB. Let B2 and C2 be the projections of points B1 and C1, respectively, on a line perpendicular to AO. Then BO sin ∠AOC + CO sin ∠AOB = B2C2 ≤BC. Adding three analogous inequalities we get the desired statement. It is also easy to verify that the conditions B1C1 ⊥AO, C1A1 ⊥BO and A1B1 ⊥CO are equivalent to the fact that O is the intersection point of the bisectors. SOLUTIONS 253 10.93. Since ∠CBD = 1 2∠C and ∠B ≤∠A, it follows that ∠ABD = ∠B + ∠CBD ≥ 1 2(∠A + ∠B + ∠C) = 90◦. 10.94. By the bisector’s property, BM : MA = BC : CA and BK : KC = BA : AC. Hence, BM : MA < BK : KC, i.e., AB AM = 1 + BM MA < 1 + BK KC = CB CK . Therefore, point M is more distant from line AC than point K, i.e., ∠AKM < ∠KAC = ∠KAM and ∠KMC < ∠MCA = ∠MCK. Hence, AM > MK and MK > KC, cf. Problem 10.59. 10.95. Suppose that all the given ratios are less than 2. Then SABO + SAOC < 2SXBO + 2SXOC = 2SOBC, SABO + SOBC < 2SAOC, SAOC + SOBC < 2SABO. Adding these inequalities we come to a contradiction. We similarly prove that one of the given ratios is not greater than 2. 10.96. Denote the radii of the circles S, S1 and S2 by r, r1 and r2, respectively. Let triangles AB1C1 and A2BC2 be similar to triangle ABC with similarity coeﬃcients r1 r and r2 r , respectively. Circles S1 and S2 are the inscribed circles of triangles AB1C1 and A2BC2, respectively. Therefore, these triangles intersect because otherwise circles S1 and S2 would not have had common points. Hence, AB1 + A2B > AB, i.e., r1 + r2 > r. Chapter 11. PROBLEMS ON MAXIMUM AND MINIMUM Background 1) Geometric problems on maximum and minimum are in close connection with geometric inequalities because in order to solve these problems we always have to prove a correspond-ing geometric inequality and, moreover, to prove that sometimes it turns into an equality. Therefore, before solving problems on maximum and minimum we have to skim through Supplement to Ch. 9 once again with the special emphasis on the conditions under which strict inequalities become equalities. 2) For elements of a triangle we use the standard notations. 3) Problems on maximum and minimum are sometimes called extremal problems (from Latin extremum). Introductory problems 1. Among all triangles ABC with given sides AB and AC ﬁnd the one with the greatest area. 2. Inside triangle ABC ﬁnd the vertex of the smallest angle that subtends side AB. 3. Prove that among all triangles with given side a and height ha. an isosceles triangle is the one with the greatest value of angle α. 4. Among all triangles with given sides AB and AC (AB < AC), ﬁnd the one for which the radius of the circumscribed circle is maximal. 5. The iagonals of a convex quadrilateral are equal to d1 and d2. What the greatest value the quadrilateral’s area can attain? §1. The triangle 11.1. Prove that among all the triangles with ﬁxed angle α and area S, an isosceles triangle with base BC has the shortest length of side BC. 11.2. Prove that among all triangles with ﬁxed angle α and semiperimeter p, an isosceles triangle with base BC is of the greatest area. 11.3. Prove that among all the triangles with ﬁxed semiperimeter p, an equilateral triangle has the greatest area. 11.4. Consider all the acute triangles with given side a and angle α. What is the maximum of b2 + c2? 11.5. Among all the triangles inscribed in a given circle ﬁnd the one with the maximal sum of squared lengths of the sides. 11.6. The perimeter of triangle ABC is equal to 2p. On sides AB and AC points M and N, respectively, are taken so that MN ∥BC and MN is tangent to the inscribed circle of triangle ABC. Find the greatest value of the length of segment MN. 11.7. Into a given triangle place a centrally symmetric polygon of greatest area. 11.8. The area of triangle ABC is equal to 1. Let A1, B1, C1 be the midpoints of sides BC, CA, AB, respectively. On segments AB1, CA1, BC1, points K, L, M, respectively, are taken. What is the least area of the common part of triangles KLM and A1B1C1? 255 256 CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM 11.9. What least width From an inﬁnite strip of paper any triangle of area 1 can be cut. What is the least width of such a strip? 11.10. Prove that triangles with the lengths of sides a, b, c and a1, b1, c1, respectively, are similar if and only if √aa1 + p bb1 + √cc1 = p (a + b + c)(a1 + b1 + c1). 11.11. Prove that if α, β, γ and α1, β1, γ1 are the respective angles of two triangles, then cos α1 sin α + cos β1 sin β + cos γ1 sin γ ≤cot α + cot β + cot γ. 11.12. Let a, b and c be the lengths of the sides of a triangle of area S; let α1, β1 and γ1 be the angles of another triangle. Prove that a2 cot α1 + b2 cot β1 + c2 cot γ1 ≥4S, where the equality is attained only if the considered triangles are similar. 11.13. In a triangle a ≥b ≥c; let x, y and z be the angles of another triangle. Prove that bc + ca −ab < bc cos x + ca cos y + ab cos z ≤a2 + b2 + c2 2 . See also Problem 17.21. §2. Extremal points of a triangle 11.14. On hypothenuse AB of right triangle ABC point X is taken; M and N are the projections of X on legs AC and BC, respectively. a) What is the position of X for which the length of segment MN is the smallest one? b) What is the position of point X for which the area of quadrilateral CMXN is the greatest one? 11.15. From point M on side AB of an acute triangle ABC perpendiculars MP and MQ are dropped to sides BC and AC, respectively. What is the position of point M for which the length of segment PQ is the minimal one? 11.16. Triangle ABC is given. On line AB ﬁnd point M for which the sum of the radii of the circumscribed circles of triangles ACM and ACN is the least possible one. 11.17. From point M of the circumscribed circle of triangle ABC perpendiculars MP and MQ are dropped on lines AB and AC, respectively. What is the position of point M for which the length of segment PQ is the maximal one? 11.18. Inside triangle ABC, point O is taken. Let da, db, dc be distances from it to lines BC, CA, AB, respectively. What is the position of point O for which the product dadbdc is the greatest one? 11.19. Points A1, B1 and C1 are taken on sides BC, CA and AB, respectively, of triangle ABC so that segments AA1, BB1 and CC1 meet at one point M. For what position of point M the value of MA1 AA1 · MB1 BB1 · MC1 CC1 is the maximal one? 11.20. From point M inside given triangle ABC perpendiculars MA1, MB1, MC1 are dropped to lines BC, CA, AB, respectively. What are points M inside the given triangle ABC for which the quantity a MA1 + b MB1 + c MC1 takes the least possible value? 11.21. Triangle ABC is given. Find a point O inside of it for which the sum of lengths of segments OA, OB, OC is the minimal one. (Take a special heed to the case when one of the angles of the triangle is greater than 120◦.) §5. POLYGONS 257 11.22. Inside triangle ABC ﬁnd a point O for which the sum of squares of distances from it to the sides of the triangle is the minimal one. See also Problem 18.21 a). §3. The angle 11.23. On a leg of an acute angle points A and B are given. On the other leg construct point C the vertex of the greatest angle that subtends segment AB. 11.24. Angle ∠XAY and point O inside it are given. Through point O draw a line that cuts oﬀthe given angle a triangle of the least area. 11.25. Through given point P inside angle ∠AOB draw line MN so that the value OM + ON is minimal (points M and N lie on legs OA and OB, respectively). 11.26. Angle ∠XAY and a circle inside it are given. On the circle construct a point the sum of the distances from which to lines AX and AY is the least. 11.27. A point M inside acute angle ∠BAC is given. On legs BA and AC construct points X and Y , respectively, such that the perimeter of triangle XY M is the least. 11.28. Angle ∠XAY is given. The endpoints B and C of unit segments BO and CO move along rays AX and AY , respectively. Construct quadrilateral ABOC of the greatest area. §4. The quadrilateral 11.29. Inside a convex quadrilateral ﬁnd a point the sum of distances from which to the vertices were the least one. 11.30. The diagonals of convex quadrilateral ABCD intersect at point O. What least area can this quadrilateral have if the area of triangle AOB is equal to 4 and the area of triangle COD is equal to 9? 11.31. Trapezoid ABCD with base AD is cut by diagonal AC into two triangles. Line l parallel to the base cuts these triangles into two triangles and two quadrilaterals. What is the position of line l for which the sum of areas of the obtained triangles is the minimal one? 11.32. The area of a trapezoid is equal to 1. What is the least value the length of the longest diagonal of this trapezoid can attain? 11.33. On base AD of trapezoid ABCD point K is given. On base BC ﬁnd point M for which the area of the common part of triangles AMD and BKC is maximal. 11.34. Prove that among all quadrilaterals with ﬁxed lengths of sides an inscribed quadrilateral has the greatest area. See also Problems 9.35, 15.3 b). §5. Polygons 11.35. A polygon has a center of symmetry, O. Prove that the sum of the distances from a point to the vertices attains its minimum at point O. 11.36. Among all the polygons inscribed in a given circle ﬁnd the one for which the sum of squared lengths of its sides is minimal. 11.37. A convex polygon A1 . . . An is given. Prove that a point of the polygon for which the sum of distances from it to all the vertices is maximal is a vertex. See also Problem 6.69. 258 CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM §6. Miscellaneous problems 11.38. Inside a circle centered at O a point A is given. Find point M on the circle for which angle ∠OMA is maximal. 11.39. In plane, line l and points A and B on distinct sides of l are given. Construct a circle that passes through points A and B so that line l intercepts on the circle a shortest chord. 11.40. Line l and points P and Q lying on one side of l are given. On line l, take point M and in triangle PQM draw heights PP ′ and QQ′. What is the position of point M for which segment P ′Q′ is the shortest? 11.41. Points A, B and O do not lie on one line. Through point O draw line l so that the sum of distances from it to points A and B were: a) maximal; b) minimal. 11.42. If ﬁve points in plane are given, then considering all possible triples of these points we can form 30 angles. Denote the least of these angles by α. Find the greatest value of α. 11.43. In a town there are 10 streets parallel to each other and 10 streets that intersect them at right angles. A closed bus route passes all the road intersections. What is the least number of turns such a bus route can have? 11.44. What is the greatest number of cells on a 8 × 8 chessboard that one straight line can intersect? (An intersection should have a common inner point.) 11.45. What is the greatest number of points that can be placed on a segment of length 1 so that on any segment of length d contained in this segment not more than 1 + 1000d2 points lie? See also Problems 15.1, 17.20. §7. The extremal properties of regular polygons 11.46. a) Prove that among all n-gons circumscribed about a given circle a regular n-gon has the least area. b) Prove that among all the n-gons circumscribed about a given circle a regular n-gon has the least perimeter. 11.47. Triangles ABC1 and ABC2 have common base AB and ∠AC1B = ∠AC2B. Prove that if \|AC1 −C1B\| < \|AC2 −C2B\|, then: a) the area of triangle ABC1 is greater than the area of triangle ABC2; b) the perimeter of triangle ABC1 is greater than the perimeter of triangle ABC2. 11.48. a) Prove that among all the n-gons inscribed in a given circle a regular n-gon has the greatest area. b) Prove that among all n-gons inscribed in a given circle a regular n-gon has the greatest perimeter. Problems for independent study 11.49. On a leg of an acute angle with vertex A point B is given. On the other leg construct point X such that the radius of the circumscribed circle of triangle ABX is the least possible. 11.50. Through a given point inside a (given?) circle draw a chord of the least length. 11.51. Among all triangles with a given sum of lengths of their bisectors ﬁnd a triangle with the greatest sum of lengths of its heights. 11.52. Inside a convex quadrilateral ﬁnd a point the sum of squared distances from which to the vertices is the least possible. SOLUTIONS 259 11.53. Among all triangles inscribed in a given circle ﬁnd the one for which the value 1 a + 1 b + 1 c is the least possible. 11.54. On a chessboard with the usual coloring draw a circle of the greatest radius so that it does not intersect any white ﬁeld. 11.55. Inside a square, point O is given. Any line that passes through O cuts the square into two parts. Through point O draw a line so that the diﬀerence of areas of these parts were the greatest possible. 11.56. What is the greatest length that the shortest side of a triangle inscribed in a given square can have? 11.57. What greatest area can an equilateral triangle inscribed in a given square can have? Solutions 11.1. By the law of cosines a2 = b2 + c2 −2bc cos α = (b −c)2 + 2bc(1 −cos α) = (b −c)2 + 4S(1 −cos α) sin α . Since the last summand is constant, a is minimal if b = c. 11.2. Let an escribed circle be tangent to sides AB and AC at points K and L, respec-tively. Since AK = AL = p, the escribed circle Sa is ﬁxed. The radius r of the inscribed circle is maximal if it is tangent to circle Sa, i.e., triangle ABC is an isosceles one. It is also clear that S = pr. 11.3. By Problem 10.53 a) we have S ≤ p2 3 √ 3, where the equality is only attained for an equilateral triangle. 11.4. By the law of cosines b2 + c2 = a2 + 2bc cos α. Since 2bc ≤b2 + c2 and cos α > 0, it follows that b2 + c2 ≤a2 + (b2 + c2) cos α, i.e., b2 + c2 ≤ a2 1−cos α. The equality is attained if b = c. 11.5. Let R be the radius of the given circle, O its center; let A, B and C be the vertices of the triangle; a = − → OA, b = − − → OB, c = − → OC. Then AB2 + BC2 + CA2 = \|a −b\|2 + \|b −c\|2 + \|c −a\|2 = 2(\|a\|2 + \|b\|2 + \|c\|2) −−2(a, b) −2(b, c) −2(c, a). Since \|a + b + c\|2 = \|a\|2 + \|b\|2 + \|c\|2 + 2(a, b) + 2(b, c) + 2(c, a), it follows that AB2 + BC2 + CA2 = 3(\|a\|2 + \|b\|2 + \|c\|2) −\|a + b + c\|2 ≤ 3(\|a\|2 + \|b\|2 + \|c\|2) = 9R2, where the equality is only attained if a + b + c = 0. This equality means that triangle ABC is an equilateral one. 11.6. Denote the length of the height dropped on side BC by h. Since △AMN ∼ △ABC, it follows that MN BC = h−2r h , i.e. MN = a ¡ 1 −2r h ¢ . Since r = S p = ah 2p, we deduce that MN = a ³ 1 −a p ´ . The maximum of the quadratic expression a ³ 1 −a p ´ = a(p−a) p in a 260 CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM is attained for a = 1 2p. This maximum is equal to p 4. It remains to notice that there exists a triangle of perimeter 2p with side a = 1 2p (set b = c = 3 4p). 11.7. Let O be the center of symmetry of polygon M lying inside triangle T, let S(T) be the image of triangle T under the symmetry through point O. Then M lies both in T and in S(T). Therefore, among all centrally symmetric polygons with the given center of symmetry lying in T the one with the greatest area is the intersection of T and S(T). Point O lies inside triangle T because the intersection of T and S(T) is a convex polygon and a convex polygon always contains its center of symmetry. Figure 124 (Sol. 11.7) Let A1, B1 and C1 be the midpoints of sides BC, CA and AB, respectively, of triangle T = ABC. First, let us suppose that point O lies inside triangle A1B1C1. Then the intersection of T and S(T) is a hexagon (Fig. 124). Let side AB be divided by the sides of triangle S(T) in the ratio of x : y : z, where x + y + z = 1. Then the ratio of the sum of areas of the shaded triangles to the area of triangle ABC is equal to x2 + y2 + z2 and we have to minimize this expression. Since 1 = (x + y + z)2 = 3(x2 + y2 + z2) −(x −y)2 −(y −z)2 −(z −x)2, it follows that x2 + y2 + z2 ≥1 3, where the equality is only attained for x = y = z; the latter equality means that O is the intersection point of the medians of triangle ABC. Now, consider another case: point O lies inside one of the triangles AB1C1, A1BC1, A1B1C; for instance, inside AB1C1. In this case the intersection of T and S(T) is a paral-lelogram and if we replace point O with the intersection point of lines AO and B1C1, then the area of this parallelogram can only increase. If point O lies on side B1C1, then this is actually the case that we have already considered (set x = 0). The polygon to be found is a hexagon with vertices at the points that divide the sides of the triangles into three equal parts. Its area is equal to 2 3 of the area of the triangle. 11.8. Denote the intersection point of lines KM and BC by T and the intersection points of the sides of triangles A1B1C1 and KLM as shown on Fig. 125. Figure 125 (Sol. 11.8) SOLUTIONS 261 Then TL : RZ = KL : KZ = LC : ZB1. Since TL ≥BA1 = A1C ≥LC, it follows that RZ ≥ZB1, i.e., SRZQ ≥SZB1Q. Similarly, SQY P ≥SY A1P and SPXR ≥SXC1R. Adding all these inequalities and the inequality SPQR > 0 we see that the area of hexagon PXRZQY is not less than the area of the remaining part of triangle A1B1C1, i.e., its area is not less than SA1B1C1 2 = 1 8. The equality is attained, for instance, if point K coincides with B1 and point M with B. 11.9. Since the area of an equilateral triangle with side a is equal to a2√ 3 4 , the side of an equilateral triangle of area 1 is equal to 2 4 √ 3 and its height is equal to 4 √ 3. Let us prove that it is impossible to cut an equilateral triangle of area 1 oﬀa strip of width less than 4 √ 3. Let equilateral triangle ABC lie inside a strip of width less than 4 √ 3. Let, for deﬁniteness, the projection of vertex B on the boundary of the strip lie between the projections of vertices A and C. Then the line drawn through point B perpendicularly to the boundary of the strip intersects segment AC at a point M. The length of a height of triangle ABC does not exceed BM and BM is not greater than the width of the strip and, therefore, a height of triangle ABC is shorter than 4 √ 3, i.e., its area is less than 1. It remains to prove that any triangle of area 1 can be cut oﬀa strip of width 4 √ 3. Let us prove that any triangle of area 1 has a height that does not exceed 4 √ 3. For this it suﬃces to prove that it has a side not shorter than 2 4 √ 3. Suppose that all sides of triangle ABC are shorter than 2 4 √ 3. Let α be the smallest angle of this triangle. Then α ≤60◦and SABC = AB · AC sin α 2 < µ 2 4 √ 3 ¶2 Ã√ 3 4 ! = 1. We have obtained a contradiction. A triangle that has a height not exceeding 4 √ 3 can be placed in a strip of width 4 √ 3: place the side to which this height is dropped on a boundary of the strip. 11.10. Squaring both sides of the given equality we easily reduce the equality to the form ( p ab1 − p a1b)2 + (√ca1 −√c1a)2 + ( p bc1 − p cb1)2 = 0, i.e., a a1 = b b1 = c c1. 11.11. Fix angles α, β and γ. Let A1B1C1 be a triangle with angles α1, β1 and γ1. Consider vectors a, b and c codirected with vectors − − − → B1C1, − − − → C1A1 and − − − → A1B1 and of length sin α, sin β and sin γ, respectively. Then cos α1 sin α + cos β1 sin β + cos γ1 sin γ = −[(a, b) + (b, c) + (c, a)] sin α sin β sin γ . Since 2[(a, b) + (b, c) + (c, a)] = \|a + b + c\|2 −\|a\|2 −\|b\|2 −\|c\|2, the quantity (a, b) + (b, c) + (c, a) attains its minimum when a + b + c = 0, i.e., α1 = α, β1 = β and γ1 = γ. 11.12. Let x = cot α1 and y = cot β1. Then x + y > 0 (since α1 + β1 < π) and cot γ1 = 1 −xy x + y = x2 + 1 x + y −x. Therefore, a2 cot α1 + b2 cot β1 + c2 cot γ1 = (a2 −b2 −c2)x + b2(x + y) + c2x2 + 1 x + y . 262 CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM For a ﬁxed x this expression is minimal for a y such that b2(x + y) = c2 x2+1 x+y , i.e., c b = x + y √ 1 + x2 = sin α1(cot α1 + cot β1) = sin γ1 sin β1 . Similar arguments show that if a : b : c = sin α1 : sin β1 : sin γ1, then the considered expression is minimal. In this case triangles are similar and a2 cot α+b2 cot β+c2 cot γ = 4S, cf. Problem 12.44 b). 11.13. Let f = bc cos x + ca cos y + ab cos z. Since cos x = −cos y cos z + sin y sin z, it follows that f = c(a −b cos z) cos y + bc sin y sin z + ab cos z. Consider a triangle the lengths of whose two sides are equal to a and b and the angle between them is equal to z; let ξ and η be the angles subtending sides a and b; let t be the length of the side that subtends angle z. Then cos z = a2 + b2 −t2 2ab and cos η = t2 + a2 −b2 2at ; hence, a−b cos z t = cos η. Moreover, b t = sin η sin z. Therefore, f = ct cos(η −y) + 1 2(a2 + b2 −t2). Since cos(η −y) ≤1, it follows that f ≤1 2(a2 + b2 + c2) −1 2((c −t)2) ≤1 2(a2 + b2 + c2). Since a ≥b, it follows that ξ ≥η, consequently, −ξ ≤−η < y −η < π −z −η = ξ, i.e., cos(y −ψ) > cos ξ. Hence, f > ct cos ξ + a2 + b2 −t2 2 = c −b 2b t2 + c(b2 −a2) 2b + a2 + b2 2 = g(t). The coeﬃcient of t2 is either negative or equal to zero; moreover, t < a + b. Hence, g(t) ≥ g(a + b) = bc + ca −ab. 11.14. a) Since CMXN is a rectangle, MN = CX. Therefore, the length of segment MN is the least possible if CX is a height. b) Let SABC = S. Then SAMX = AX2·S AB2 and SBNX = BX2·S AB2 . Since AX2 + BX2 ≥1 2AB2 (where the equality is only attained if X is the midpoint of segment AB), it follows that SCMXN = S −SAMX −SBNX ≤1 2S. The area of quadrilateral CMXN is the greatest if X is the midpoint of side AB. 11.15. Points P and Q lie on the circle constructed on segment CM as on the diameter. In this circle the constant angle C intercepts chord PQ, therefore, the length of chord PQ is minimal if the diameter CM of the circle is minimal, i.e., if CM is a height of triangle ABC. 11.16. By the law of sines the radii of the circumscribed circles of triangles ACM and BCM are equal to AC 2 sin AMC and BC 2 sin BMC, respectively. It is easy to verify that sin AMC = sin BMC. Therefore, AC 2 sin AMC + BC 2 sin BMC = AC + BC 2 sin BMC . The latter expression is minimal if sin BMC = 1, i.e., CM ⊥AB. 11.17. Points P and Q lie on the circle with diameter AM, hence, PQ = AM sin PAQ = AM sin A. It follows that the length of segment PQ is maximal if AM is a diameter of the circumscribed circle. 11.18. Clearly, 2SABC = ada + bdb + cdc. Therefore, the product (ada)(bdb)(cdc) takes its greatest value if ada = bdb = cdc (cf. Supplement to Ch. 9, the inequality between the mean arithmetic and the mean geometric). Since the value abc is a constant, the product (ada)(bdb)(cdc) attains its greatest value if and only if the product dadbdc takes its greatest value. SOLUTIONS 263 Let us show that equality ada = bdb = cdc means that O is the intersection point of the medians of triangle ABC. Denote the intersection point of lines AO and BC by A1. Then BA1 : A1C = SABA1 : SACA1 = SABO : SACO = (cdc) : (bdb) = 1, i.e., AA1 is a median. We similarly prove that point O lies on medians BB1 and CC1. 11.19. Let α = MA1 AA1 , β = MB1 BB1 and γ = MC1 CC1 . Since α + β + γ = 1 (cf. Problem 4.48 a)), we have √αβγ ≤1 3(α + β + γ) = 1 3, where the equality is attained when α = β = γ = 1 3, i.e., M is the intersection point of the medians. 11.20. Let x = MA1, y = MB1 and z = MC1. Then ax + by + cz = 2SBMC + 2SAMC + 2SAMB = 2SABC. Hence, ³ a x + b y + c z ´ · 2SABC = ³ a x + b y + c z ´ (ax + by + cz) = a2 + b2 + c2 + ab ³ x y + y x ´ + bc ³ y z + z y ´ + ac ¡ z x + x z ¢ ≥ a2 + b2 + c2 + 2ab + 2bc + 2ac, where the equality is only attained if x = y = z, i.e., M is the center of the inscribed circle of triangle ABC. 11.21. First, suppose that all the angles of triangle ABC are less than 120◦. Then inside triangle ABC there exists a point O — the vertex of angles of 120◦that subtend each side. Let us draw through vertices A, B and C lines perpendicular to segments OA, OB and OC, respectively. These lines form an equilateral triangle A1B1C1 (Fig. 126). Figure 126 (Sol. 11.21) Let O′ be any point that lies inside triangle ABC and is distinct from O. Let us prove then that O′A + O′B + O′C > OA + OB + OC, i.e., O is the desired point. Let A′, B′ and C′ be the bases of the perpendiculars dropped from point O′ on sides B1C1, C1A1 and A1B1, respectively, a the length of the side of equilateral triangle A1B1C1. Then O′A′ + O′B′ + O′C′ = 2(SO′B1C1 + SO′A1C1) a = 2SA1B1C1 a = OA + OB + OC. Since a slanted line is longer than the perpendicular, O′A + O′B + O′C > O′A′ + O′B′ + O′C′ = OA + OB + OC. 264 CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM Figure 127 (Sol. 11.21) Now, let one of the angles of triangle ABC, say ∠C, be greater than 120◦. Let us draw through points A and B perpendiculars B1C1 and C1A1 to segments CA and CB and through point C line A1B1 perpendicular to the bisector of angle ∠ACB (Fig. 127). Since ∠AC1B = 180◦−∠ACB < 60◦, it follows that B1C1 > A1B1. Let O′ be any point that lies inside triangle A1B1C1. Since B1C1 · O′A′ + C1A1 · O′B′ + A1C1 · O′C′ = 2SA1B1C1, it follows that (O′A′ + O′B′ + O′C′) · B1C1 = 2SA1B1C1 + (B1C1 −A1B1) · O′C′. Since B1C1 > A1B1, the sum O′A′ +O′B′ +O′C′ is minimal for points that lie on side B1A1. It is also clear that O′A + O′B + O′C ≥O′A′ + O′B′ + O′C′. Therefore, vertex C is the point to be found. 11.22. Let the distances from point O to sides BC, CA and AB be equal to x, y and z, respectively. Then ax + by + cz = 2(SBOC + SCOA + SAOB) = 2SABC. It is also clear that x : y : z = µSBOC a ¶ : µSCOA b ¶ : µSAOB c ¶ . Equation ax + by + cz = 2S determines a plane in 3-dimensional space with coordinates x, y, z; vector (a, b, c) is perpendicular to this plane because if ax1 + by1 + cz1 = 2S and ax2 + by2 + cz2 = 2S, then a(x1 −x2) + b(y1 −y2) + c(z1 −z2) = 0. We have to ﬁnd a point (x0, y0, z0) on this plane at which the minimum of expression x2 + y2 + z2 is attained and verify that an inner point of the triangle corresponds to this point. Since x2 + y2 + z2 is the squared distance from the origin to point (x, y, z), it follows that the base of the perpendicular dropped from the origin to the plane is the desired point, i.e., x : y : z = a : b : c. It remains to verify that inside the triangle there exists point O for which x : y : z = a : b : c. This equality is equivalent to the condition µSBOC a ¶ : µSCOA b ¶ : µSAOB c ¶ = a : b : c, i.e., SBOC : SCOA : SAOB = a2 : b2 : c2. Since the equality SBOC : SAOB = a2 : c2 follows from equalities SBOC : SCOA = a2 : b2 and SCOA : SAOB = b2 : c2, the desired point is the intersection point of lines CC1 and AA1 that divide sides AB and BC, respectively, in the ratios of BC1 : C1A = a2 : b2 and CA1 : A1B = b2 : c2, respectively. SOLUTIONS 265 11.23. Let O be the vertex of the given angle. Point C is the tangent point of a leg with the circle that passes through points A and B, i.e., OC2 = OA · OB. To ﬁnd the length of segment OC, it suﬃces to draw the tangent to any circle that passes through points A and B. 11.24. Let us consider angle ∠X′A′Y ′ symmetric to angle ∠XAY through point O. Let B and C be the intersection points of the legs of these angles. Denote the intersection points of the line that passes through point O with the legs of angles ∠XAY and ∠X′A′Y ′ by B1, C1 and B′ 1, C′ 1, respectively (Fig. 128). Figure 128 (Sol. 11.24) Since SAB1C1 = SA′B′ 1C′ 1, it follows that SAB1C1 = 1 2(SABA′C +SBB1C′ 1 +SCC1B′ 1). The area of triangle AB1C1 is the least if B1 = B and C1 = C, i.e., line BC is the one to be found. 11.25. On legs OA and OB, take points K and L so that KP ∥OB and LP ∥OA. Then KM : KP = PL : LN and, therefore, KM + LN ≥2 √ KM · LN = 2 √ KP · PL = 2 √ OK · OL where the equality is attained when KM = LN = √ OK · OL. It is also clear that OM + ON = (OK + OL) + (KM + LN). 11.26. On rays AX and AY , mark equal segments AB an AC. If point M lies on segment BC, then the sum of distances from it to lines AB and AC is equal to 2(SABM+SACM) AB = 2SABC AB . Therefore, the sum of distances from a point to lines AX and AY is the lesser, the lesser is the distance between point A and the point’s projection on the bisector of angle ∠XAY . 11.27. Let points M1 and M2 be symmetric to M through lines AB and AC, respectively. Since ∠BAM1 = ∠BAM and ∠CAM2 = ∠CAM, it follows that ∠M1AM2 = 2∠BAC < 180◦. Hence, segment M1M2 intersects rays AB and AC at certain points X and Y (Fig. 129). Let us prove that X and Y are the points to be found. Figure 129 (Sol. 11.27) Indeed, if points X1 and Y1 lie on rays AB and AC, respectively, then MX1 = M1X1 and MY1 = M2Y1, i.e., the perimeter of triangle MX1Y1 is equal to the length of the broken 266 CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM line MX1Y1M2. Of all the broken lines with the endpoints at M1 and M2 segment M1M2 is the shortest one. 11.28. Quadrilateral ABOC of the greatest area is a convex one. Among all the triangles ABC with the ﬁxed angle ∠A and side BC an isosceles triangle with base BC has the greatest area. Therefore, among all the considered quadrilaterals ABOC with ﬁxed diagonal BC the quadrilateral with AB = AC, i.e., for which point O lies on the bisector of angle ∠A, is of greatest area. Further, let us consider triangle ABO in which angle ∠BAO equal to 1 2∠A and side BO are ﬁxed. The area of this triangle is maximal when AB = AO. 11.29. Let O be the intersection point of the diagonals of convex quadrilateral ABCD and O1 any other point. Then AO1 + CO1 ≥AC = AO + CO and BO1 + DO1 ≥BD = BO + DO, where at least one of the inequalities is a strict one. Therefore, O is the point to be found. 11.30. Since SAOB : SBOC = AO : OC = SAOD : SDOC, it follows that SBOC · SAOD = SAOB · SDOC = 36. Therefore, SBOC + SAOD ≥2√SBOC · SAOD = 12, where the equality takes place if SBOC = SAOD, i.e., SABC = SABD. This implies that AB ∥CD. In this case the area of the triangle is equal to 4+9+12=25. 11.31. Let S0 and S be the considered sums of areas of triangles for line l0 that passes through the intersection point of the diagonals of the trapezoid and for another line l. It is easy to verify that S = S0 + s, where s is the area of the triangle formed by diagonals AC and BD and line l. Hence, l0 is the line to be found. 11.32. Denote the lengths of the diagonals of the trapezoid by d1 and d2 and the lengths of their projections on the bottom base by p1 and p2, respectively; denote the lengths of the bases by a and b and that of the height by h. Let, for deﬁniteness, d1 ≥d2. Then p1 ≥p2. Clearly, p1 + p2 ≥a + b. Hence, p1 ≥a+b 2 = S h = 1 h. Therefore, d2 1 = p2 1 + h2 ≥ 1 h2 + h2 ≥2, where the equality is attained only if p1 = p2 = h = 1. In this case d1 = √ 2. 11.33. Let us prove that point M that divides side BC in the ratio of BM : NC = AK : KD is the desired one. Denote the intersection points of segments AM and BK, DM and CK by P and Q, respectively. Then KQ : QC = KD : MC = KA : MB = KP : PB, i.e., line PQ is parallel to the basis of the trapezoid. Let M1 be any other point on side BC. For deﬁniteness, we may assume that M1 lies on segment BM. Denote the intersection points of AM1 and BK, DM1 and CK, AM1 and PQ, DM1 and PQ, AM and DM1 by P1, Q1, P2, Q2, O, respectively (Fig. 130). Figure 130 (Sol. 11.33) We have to prove that SMPKQ > SM1P1KQ1, i.e., SMOQ1Q > SM1OPP1. Clearly, SMOQ1Q > SMOQ2Q = SM1OPP2 > SM1OPP1. 11.34. By Problem 4.45 a) we have S2 = (p −a)(p −b)(p −c)(p −d) −abcd cos2 ∠B + ∠D 2 . SOLUTIONS 267 This quantity takes its maximal value when cos ∠B+∠D 2 = 0, i.e., ∠B + ∠D = 180◦. 11.35. If A and A′ are vertices of the polygon symmetric through point O, then the sum of distances from any point of segment AA′ to points A and A′ is the same whereas for any other point it is greater. Point O belongs to all such segments. 11.36. If in triangle ABC, angle ∠B is either obtuse or right, then by the law of sines AC2 ≥AB2 + BC2. Therefore, if in a polygon the angle at vertex B is not acute, then deleting vertex B we obtain a polygon with the sum of squared lengths of the sides not less than that of the initial polygon. Since for n ≥3 any n-gon has a nonacute angle, it follows that by repeating such an operation we eventually get a triangle. Among all the triangles inscribed in the given circle an equilateral triangle has the greatest sum of squared lengths of the sides, cf. Problem 11.5. 11.37. If point X divides segment PQ in the ratio of λ : (1 −λ), then − − → AiX = (1 − λ)− − → AiP + λ− − → AiQ; hence, AiX ≤(1 −λ)AiP + λAiQ. Therefore, f(X) = X AiX ≤(1 −λ) X AiP + λ X AiQ = (1 −λ)f(P) + λf(Q). Let, for instance, f(P) ≤f(Q), then f(X) ≤f(Q); hence, on segment PQ the function f attains its maximal value at one of the endpoints; more precisely, inside the segment there can be no point of strict maximum of f. Hence, if X is any point of the polygon, then f(X) ≤f(Y ), where Y is a point on a side of the polygon and f(Y ) ≤f(Z), where Z is a vertex. 11.38. The locus of points X for which angle ∠OXA is a constant consists of two arcs of circles S1 and S2 symmetric through line OA. Consider the case when the diameter of circles S1 and S2 is equal to the radius of the initial circle, i.e., when these circles are tangent to the initial circle at points M1 and M2 for which ∠OAM1 = ∠OAM2 = 90◦. Points M1 and M2 are the desired ones because if ∠OXA > ∠OM1A = ∠OM2A, then point X lies strictly inside the ﬁgure formed by circles S1 and S2, i.e., cannot lie on the initial circle. 11.39. Let us denote the intersection point of line l and segment AB by O. Let us consider an arbitrary circle S that passes through points A and B. It intersects l at certain points M and N. Since MO · NO = AO · BO is a constant, MN = MO + NO ≥2 √ MO · NO = 2 √ AO · BO, where the equality is only attained if MO = NO. In the latter case the center of S is the intersection point of the midperpendicular to AB and the perpendicular to l that passes through point O. 11.40. Let us construct the circle with diameter PQ. If this circle intersects with l, then any of the intersection points is the desired one because in this case P ′ = Q′. If the circle does not intersect with l, then for any point M on l angle ∠PMQ is an acute one and ∠P ′PQ′ = 90◦± ∠PMQ. Now it is easy to establish that the length of chord P ′Q′ is minimal if angle ∠PMQ is maximal. To ﬁnd point M it remains to draw through points P and Q circles tangent to l (cf. Problem 8.56 a)) and select the needed point among the tangent points. 11.41. Let the sum of distances from points A and B to line l be equal to 2h. If l intersects segment AB at point X, then SAOB = h · OX and, therefore, the value of h is extremal when the value of OX is extremal, i.e., when line OX corresponds to a side or a height of triangle AOB. If line l does not intersect segment AB, then the value of h is equal to the length of the midline of the trapezoid conﬁned between the perpendiculars dropped from points A and B 268 CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM on line l. This quantity is an extremal one when l is either perpendicular to median OM of triangle AOB or corresponds to a side of triangle AOB. Now it only remains to select two of the obtained four straight lines. 11.42. First, suppose that the points are the vertices of a convex pentagon. The sum of angles of the pentagon is equal to 540◦; hence, one of its angles does not exceed 540◦ 5 = 108◦. The diagonals divide this angle into three angles, hence, one of them does not exceed 108◦ 3 = 36◦. In this case α ≤36◦. If the points are not the vertices of a convex pentagon, then one of them lies inside the triangle formed by some other three points. One of the angles of this triangle does not exceed 60◦. The segment that connects the corresponding vertex with an inner point divides this angle into two angles, hence, one of them does not exceed 30◦. In this case α ≤30◦. In all the cases α ≤36◦. Clearly, for a regular pentagon α = 36◦. 11.43. A closed route that passes through all the road crossings can have 20 turns (Fig. 131). It remains to prove that such a route cannot have less than 20 turns. After each turn a passage from a horizontal street to a vertical one or from a vertical street to a horizontal one occurs. Figure 131 (Sol. 11.43) Hence, the number of horizontal links of a closed route is equal to the number of vertical links and is equal to half the number of turns. Suppose that a closed route has less than 20 turns. Then there are streets directed horizontally, as well as streets directed vertically, along which the route does not pass. Therefore, the route does not pass through the intersection point of these streets. 11.44. A line can intersect 15 cells (Fig. 132). Let us prove now that a line cannot intersect more than 15 cells. The number of cells that the line intersects is by 1 less than the number of intersection points of the line with the segments that determine the sides of the cells. Inside a square there are 14 such segments. Hence, inside a square there are not more than 14 intersection points of the line with sides of cells. No line can intersect the boundary of the chessboard at more than 2 points; hence, the number of intersection points of the line with the segments does not exceed 16. Hence, the maximal number of cells on the chessboard of size 8 × 8 that can be intersected by one line is equal to 15. 11.45. First, let us prove that 33 points are impossible to place in the required way. Indeed, if on a segment of length 1 there are 33 points, then the distance between some two of them does not exceed 1 32. The segment with the endpoints at these points contains two points and it should contain not more than 1 + 1000 322 points, i.e., not less than two points. Now, let us prove that it is possible to place 32 points. Let us take 32 points that divide the segment into equal parts (the endpoints of the given segment should be among these 32 SOLUTIONS 269 Figure 132 (Sol. 11.44) points). Then a segment of length d contains either [31d] or [31d] + 1 points. (Recall that [x] denotes the integer part of the number x, i.e., the greatest integer that does not exceed x.) We have to prove that [31d] ≤1000d2. If 31d < 1, then [31d] = 0 < 1000d2. If 31d ≥1, then [31d] ≤31d ≤(31d)2 = 961d2 < 1000d2. 11.46. a) Let a non-regular n-gon be circumscribed about circle S. Let us circumscribe a regular n-gon about this circle and let us circumscribe circle S1 about this regular n-gon (Fig. 133). Let us prove that the area of the part of the non-regular n-gon conﬁned inside S1 is greater than the area of the regular n-gon. Figure 133 (Sol. 11.46) All the tangents to S cut oﬀS1 equal segments. Hence, the sum of areas of the segments cut oﬀS1 by the sides of the regular n-gon is equal to the sum of segments cut oﬀS1 by the sides of the non-regular n-gon or by their extensions. But for the regular n-gon these segments do not intersect (more exactly, they do not have common interior points) and for the non-regular n-gon some of them must overlap, hence, the area of the union of these segments for a regular-gon is greater than for a non-regular one. Therefore, the area of the part of the non-regular n-gon conﬁned inside S1 is greater than the area of the regular n-gon and the area of the whole non-regular n-gon is still greater than the area of the regular one. b) This heading follows from heading a) because the perimeter of the polygon circum-scribed about a circle of radius R is equal to 2S R , where S is the area of the polygon. 11.47. The sides of triangle ABC are proportional to sin α, sin β and sin γ. If angle γ is ﬁxed, then the value of \| sin α −sin β\| = 2 ¯ ¯ ¯ ¯sin α −β 2 sin γ 2 ¯ ¯ ¯ ¯ 270 CHAPTER 11. PROBLEMS ON MAXIMUM AND MINIMUM is the greater the greater is ϕ = \|α −β\|. It remains to observe that quantities S = 2R2 sin α sin β sin γ = R2 sin γ(cos α −β + cos γ) = R2 sin γ(cos ϕ + cos γ) and sin α + sin β = 2 cos γ 2 cos ϕ 2 monotonously decrease as ϕ increases. 11.48. a) Denote the length of the side of a regular n-gon inscribed in the given circle by an. Consider an arbitrary non-regular n-gon inscribed in the same circle. It will necessarily have a side shorter than an. On the other hand, it can have no side longer than an and in such a case such a polygon can be conﬁned in a segment cut oﬀa side of the regular n-gon. Since the symmetry through a side of a regular n-gon sends the segment cut oﬀthis side inside the n-gon, the area of the n-gon is greater than the area of the segment(??). Therefore, we may assume that the considered n-gon has a side shorter than an and a side longer than an. We can replace neighbouring sides of the n-gon, i.e., replace A1A2A3 . . . An with polygon A1A′ 2A3 . . . An, where point A′ 2 is symmetric to A2 through the midperpendicular to segment A1A3 (Fig. 134). Clearly, both polygons are inscribed in the same circle and their areas are equal. It is also clear that with the help of this operation we can make any two sides of the polygon neighbouring ones. Therefore, let us assume that for the n-gon considered, A1A2 > an and A2A3 < an. Figure 134 (Sol. 11.48) Let A′ 2 be the point symmetric to A2 through the midperpendicular to segment A1A3. If point A′′ 2 lies on arc ⌣A2A′ 2, then the diﬀerence of the angles at the base A1A3 of triangle A1A′′ 2A3 is less than that of triangle A1A2A3 because the values of angles ∠A1A2A′′ 2 and ∠A3A1A′′ 2 are conﬁned between the values of angles ∠A1A3A2 and ∠A3A1A2. Since A1A′ 2 < an and A1A2 > an, on arc ⌣A2A′ 2 there exists a point A′′ 2 for which A1A′′ 2 = an. The area of triangle A1A′′ 2A3 is greater than the area of triangle A1A2A3, cf. Problem 11.47 a). The area of polygon A1A′′ 2A3 . . . An is greater than the area of the initial polygon and it has at least by 1 more sides equal to an. After ﬁnitely many steps we get a regular n-gon and at each step the area increases. Therefore, the area of any non-regular n-gon inscribed in a circle is less than the area of a regular n-gon inscribed in the same circle. b) Proof is similar to the proof of heading a); one only has to make use of the result of Problem 11.47 b) instead of that of Problem 11.47 a). Chapter 12. CALCULATIONS AND METRIC RELATIONS Introductory problems 1. Prove the law of cosines: BC2 = AB2 + AC2 −2AB · AC cos ∠A. 2. Prove the law of sines: a sin α = b sin β = c sin γ = 2R. 3. Prove that the area of a triangle is equal to p p(p −a)(p −b)(p −c), where p is semiperime-ter (Heron’s formula.) 4. The sides of a parallelogram are equal to a and b and its diagonals are equal to d and e. Prove that 2(a2 + b2) = d2 + e2. 5. Prove that for convex quadrilateral ABCD with the angle ϕ between the diagonals we have SABCD = 1 2AC · BD sin ϕ. §1. The law of sines 12.1. Prove that the area S of triangle ABC is equal to abc 4R . 12.2. Point D lies on base AC of equilateral triangle ABC. Prove that the radii of the circumscribed circles of triangles ABD and CBD are equal. 12.3. Express the area of triangle ABC in terms of the length of side BC and the value of angles ∠B and ∠C. 12.4. Prove that a+b c = cos α−β 2 sin γ 2 and a−b c = sin α−β 2 cos γ 2 . 12.5. In an acute triangle ABC heights AA1 and CC1 are drawn. Points A2 and C2 are symmetric to A1 and C1 through the midpoints of sides BC and AB, respectively. Prove that the line that connects vertex B with the center O of the circumscribed circle divides segment A2C2 in halves. 12.6. Through point S lines a, b, c and d are drawn; line l intersects them at points A, B, C and D. Prove that the quantity AC·BD BC·AD does not depend on the choice of line l. 12.7. Given lines a and b that intersect at point O and an arbitrary point P. Line l that passes through point P intersects lines a and b at points A and B. Prove that the value of OA OB P A P B does not depend on the choice of line l. 12.8. Denote the vertices and the intersection points of links of a (non-regular) ﬁve-angled star as shown on Fig. 135. Prove that A1C · B1D · C1E · D1A · E1B = A1D · B1E · C1A · D1B · E1C. 12.9. Two similar isosceles triangles have a common vertex. Prove that the projections of their bases on the line that connects the midpoints of the bases are equal. 12.10. On the circle with diameter AB, points C and D are taken. Line CD and the tangent to the circle at point B intersect at point X. Express BX in terms of the radius R of the circle and angles ϕ = ∠BAC and ψ = ∠BAD. 271 272 CHAPTER 12. CALCULATIONS AND METRIC RELATIONS Figure 135 (12.8) §2. The law of cosines 12.11. Prove that: a) m2 a = 2b2+2c2−a2 4 ; b) m2 a + m2 b + m2 c = 3(a2+b2+c2) 4 . 12.12. Prove that 4S = (a2 −(b −c)2) cot α 2 . 12.13. Prove that cos2 α 2 = p(p −a) bc and sin2 α 2 = (p −b)(p −c) bc . 12.14. The lengths of sides of a parallelogram are equal to a and b; the lengths of the diagonals are equal to m and n. Prove that a4 + b4 = m2n2 if and only if the acute angle of the parallelogram is equal to 45◦. 12.15. Prove that medians AA1 and BB1 of triangle ABC are perpendicular if and only if a2 + b2 = 5c2. 12.16. Let O be the center of the circumscribed circle of scalane triangle ABC, let M be the intersection point of the medians. Prove that line OM is perpendicular to median CC1 if and only if a2 + b2 = 2c2. §3. The inscribed, the circumscribed and escribed circles; their radii 12.17. Prove that: a) a = r ¡ cot β 2 + cot γ 2 ¢ = r cos α 2 sin β 2 sin γ 2 ; b) a = ra ¡ tan β 2 + tan γ 2 ¢ = ra cos α 2 cos β 2 cos γ 2 ; c) p −b = r cot β 2 = ra tan γ 2; d) p = ra cot α 2 . 12.18. Prove that: a) rp = ra(p −a), rra = (p −b)(p −c) and rbrc = p(p −a); b) S2 = p(p −a)(p −b)(p −c); (Heron’s formula.) c) S2 = rrarbrc. 12.19. Prove that S = r2 c tan α 2 tan β 2 cot γ 2. 12.20. Prove that S = crarb ra+rb. 12.21. Prove that 2 ha = 1 rb + 1 rc. 12.22. Prove that 1 ha + 1 hb + 1 hc = 1 ra + 1 rb + 1 rc = 1r. 12.23. Prove that 1 (p −a)(p −b) + 1 (p −b)(p −c) + 1 (p −c)(p −a) = 1 r2. §5. THE SINES AND COSINES OF A TRIANGLE’S ANGLES 273 12.24. Prove that ra + rb + rc = 4R + r. 12.25. Prove that rarb + rbrc + rcra = p2. 12.26. Prove that 1 r3 −1 r3 a −1 r3 b −1 r2 c = 12R S2 . 12.27. Prove that a(b + c) = (r + ra)(4R + r −ra) and a(b −c) = (rb −rc)(4R −rb −rc). 12.28. Let O be the center of the inscribed circle of triangle ABC. Prove that OA2 bc + OB2 ac + OC2 ab = 1. 12.29. a) Prove that if for a triangle we have p = 2R + r, then this triangle is a right one. b) Prove that if p = 2R sin ϕ + r cot ϕ 2 , then ϕ is one of the angles of the triangle (we assume here that 0 < ϕ < π). §4. The lengths of the sides, heights, bisectors 12.30. Prove that abc = 4prR and ab + bc + ca = r2 + p2 + 4rR. 12.31. Prove that 1 ab + 1 bc + 1 ca = 1 2Rr. 12.32. Prove that a+b−c a+b+c = tan α 2 tan β 2. 12.33. Prove that ha = bc 2R. 12.34. Prove that ha = 2(p −a) cos β 2 cos γ 2 cos α 2 = 2(p −b) sin β 2 cos γ 2 sin α 2 . 12.35. Prove that the length of bisector la can be computed from the following formulas: a) la = q 4p(p−a)bc (b+c)2 ; b) la = 2bc cos α 2 b+c ; c) la = 2R sin β sin γ cos β−γ 2 ; d) la = 4p sin β 2 sin γ 2 sin β+sin γ . §5. The sines and cosines of a triangle’s angles Let α, β and γ be the angles of triangle ABC. In the problems of this section one should prove the relations indicated. 12.36. a) sin α 2 sin β 2 sin γ 2 = r 4R; b) tan α 2 tan β 2 tan γ 2 = r p; c) cos α 2 cos β 2 cos γ 2 = p 4R. 12.37. a) cos α 2 sin β 2 sin γ 2 = p−a 4R ; b) sin α 2 cos β 2 cos γ 2 = ra 4R. 12.38. cos α + cos β + cos γ = R+r R . 12.39. a) cos 2α + cos 2β + cos 2γ + 4 cos α cos β cos γ + 1 = 0; b) cos2 α + cos2 β + cos2 γ + 2 cos α cos β cos γ = 1. 12.40. sin 2α + cos 2β + cos 2γ = 4 sin α sin β sin γ. 12.41. a) sin2 α + sin2 β + sin2 γ = p2−r2−4rR 2R2 . b) 4R2 cos α cos β cos γ = p2 −(2R + r)2. 12.42. ab cos γ + bc cos α + ca cos β = a2+b2+c2 2 . 12.43. cos2 α 2 a + cos2 β 2 b + cos2 γ 2 c = p 4Rr. 274 CHAPTER 12. CALCULATIONS AND METRIC RELATIONS §6. The tangents and cotangents of a triangle’s angles In problems of this section one has to prove the relations indicated between the values α, β and γ of the angles of triangle ABC. 12.44. a) cot α + cot β + cot γ = a2+b2+c2 4S ; b) a2 cot α + b2 cot β + c2 cot γ = 4S. 12.45. a) cot α 2 + cot β 2 + cot γ 2 = p r; b) tan α 2 + tan β 2 + tan γ 2 = 1 2 ³ a ra + b rb + c rc ´ . 12.46. tan α + tan β + tan γ = tan σ tan β tan γ. 12.47. tan α 2 tan β 2 + tan β 2 tan γ 2 + tan γ 2 tan α 2 = 1. 12.48. a) cot α cot β + cot β cot γ + cot α cot γ = 1; b) cot α + cot β + cot γ −cot α cot β cot γ = 1 sin α sin β sin γ. 12.49. For a non-right triangle we have tan σ + tan β + tan γ = 4S a2 + b2 + c2 −8R2. §7. Calculation of angles 12.50. Two intersecting circles, each of radius R with the distance between their centers greater than R are given. Prove that β = 3α (Fig. 136). Figure 136 (12.50) 12.51. Prove that if 1 b + 1 c = 1 la, then ∠A = 120◦. 12.52. In triangle ABC height AH is equal to median BM. Find angle ∠MBC. 12.53. In triangle ABC bisectors AD and BE are drawn. Find the value of angle ∠C if it is given that AD · BC = BE · AC and AC ̸= BC. 12.54. Find angle ∠B of triangle ABC if the length of height CH is equal to a half length of side AB and ∠BAC = 75◦. 12.55. In right triangle ABC with right angle ∠A the circle is constructed with height AD of the triangle as a diameter; the circle intersects leg AB at point K and leg AC at point M. Segments AD and KM intersect at point L. Find the acute angles of triangle ABC if AK : AL = AL : AM. 12.56. In triangle ABC, angle ∠C = 2∠A and b = 2a. Find the angles of triangle ABC. 12.57. In triangle ABC bisector BE is drawn and on side BC point K is taken so that ∠AKB = 2∠AEB. Find the value of angle ∠AKE if ∠AEB = α. 12.58. In an isosceles triangle ABC with base BC angle at vertex A is equal to 80◦. Inside triangle ABC point M is taken so that ∠MBC = 30◦and ∠MCB = 10◦. Find the value of angle ∠AMC. §9. MISCELLANEOUS PROBLEMS 275 12.59. In an isosceles triangle ABC with base AC the angle at vertex B is equal to 20◦. On sides BC and AB points D and E, respectively, are taken so that ∠DAC = 60◦and ∠ECA = 50◦. Find angle ∠ADE. 12.60. In an acute triangle ABC segments BO and CO, where O is the center of the circumscribed circle, are extended to their intersection at points D and E with sides AC and AB, respectively. It turned out that ∠BDE = 50◦and ∠CED = 30◦. Find the value of the angles of triangle ABC. §8. The circles 12.61. Circle S with center O on base BC of isosceles triangle ABC is tangent to equal sides AB and AC. On sides AB and AC, points P and Q, respectively, are taken so that segment PQ is tangent to S. Prove that 4PB · CQ = BC2. 12.62. Let E be the midpoint of side AB of square ABCD and points F and G are taken on sides BC and CD, respectively, so that AG ∥EF. Prove that segment FG is tangent to the circle inscribed in square ABCD. 12.63. A chord of a circle is distant from the center by h. A square is inscribed in each of the disk segments subtended by the chord so that two neighbouring vertices of the square lie on an arc and two other vertices lie either on the chord or on its extension (Fig. 137). What is the diﬀerence of lengths of sides of these squares? Figure 137 (12.63) 12.64. Find the ratio of sides of a triangle one of whose medians is divided by the inscribed circle into three equal parts. 12.65. In a circle, a square is inscribed; in the disk segment cut oﬀthe disk by one of the sides of this square another square is inscribed. Find the ratio of the lengths of the sides of these squares. 12.66. On segment AB, point C is taken and on segments AC, BC and AB as on diameters semicircles are constructed lying on one side of line AB. Through point C the line perpendicular to AB is drawn and in the obtained curvilinear triangles ACD and BCD circles S1 and S2 are inscribed (Fig. 138). Prove that the radii of these circles are equal. 12.67. The centers of circles with radii 1, 3 and 4 are positioned on sides AD and BC of rectangle ABCD. These circles are tangent to each other and lines AB and CD as shown on Fig. 139. Prove that there exists a circle tangent to all these circles and line AB. §9. Miscellaneous problems 12.68. Find all the triangles whose angles form an arithmetic projection and sides form a) an arithmetic progression; b) a geometric progression. 276 CHAPTER 12. CALCULATIONS AND METRIC RELATIONS Figure 138 (12.66) Figure 139 (12.67) 12.69. Find the height of a trapezoid the lengths of whose bases AB and CD are equal to a and b (a < b), the angle between the diagonals is equal to 90◦, and the angle between the extensions of the lateral sides is equal to 45◦. 12.70. An inscribed circle is tangent to side BC of triangle ABC at point K. Prove that the area of the triangle is equal to BK · KC cot α 2 . 12.71. Prove that if cot α 2 = a+b a , then the triangle is a right one. 12.72. The extensions of the bisectors of triangle ABC intersect the circumscribed circle at points A1, B1 and C1. Prove that SABC SA1B1C1 = 2r R , where r and R are the radii of the inscribed and circumscribed circles, respectively, of triangle ABC. 12.73. Prove that the sum of cotangents of the angles of triangle ABC is equal to the sum of cotangents of the angles of the triangle formed by the medians of triangle ABC. 12.74. Let A4 be the orthocenter of triangle A1A2A3. Prove that there exist numbers λ1, . . . , λ4 such that AiA2 j = λi + λj and if the triangle is not a right one, then P 1 λi = 0. §10. The method of coordinates 12.75. Coordinates of the vertices of a triangle are rational numbers. Prove that then the coordinates of the center of the circumscribed circle are also rational. 12.76. Diameters AB and CD of circle S are perpendicular. Chord EA intersects diameter CD at point K, chord EC intersects diameter AB at point L. Prove that if CK : KD = 2 : 1, then AL : LB = 3 : 1. 12.77. In triangle ABC angle ∠C is a right one. Prove that under the homothety with center C and coeﬃcient 2 the inscribed circle turns into a circle tangent to the circumscribed circle. SOLUTIONS 277 12.78. A line l is ﬁxed. Square ABCD is rotated about its center. Find the locus of the midpoints of segments PQ, where P is the base of the perpendicular dropped from point D on l and Q is the midpoint of side AB. See also Problems 7.6, 7.14, 7.47, 22.15. Problems for independent study 12.79. Each of two circles is tangent to both sides of the given right angle. Find the ratio of the circles’ radii if it is known that one of the circles passes through the center of the other one. 12.80. Let the extensions of sides AB and CD, BC and AD of convex quadrilat-eral ABCD intersect at points K and M, respectively. Prove that the radii of the cir-cles circumscribed about triangles ACM, BDK, ACK, BDM are related by the formula RACM · RBDK = RACK · RBDM. 12.81. Three circles of radii 1, 2, 3 are tangent to each other from the outside. Find the radius of the circle that passes through the tangent points of these circles. 12.82. Let point K lie on side BC of triangle ABC. Prove that AC2 · BK + AB2 · CK = BC(AK2 + BK · KC). 12.83. Prove that the length of the bisector of an outer angle ∠A of triangle ABC is equal to 2bc sin α 2 \|b−c\| . 12.84. Two circles of radii R and r are placed so that their common inner tangents are perpendicular. Find the area of the triangle formed by these tangents and their common outer tangent. 12.85. Prove that the sum of angles at rays of any (nonregular) ﬁve-angled star is equal to 180◦. 12.86. Prove that in any triangle S = (p −a)2 tan α 2 cot β 2 cot γ 2. 12.87. Let a < b < c be the lengths of sides of a triangle; la, lb, lc and l′ a, l′ b, l′ c the lengths of its bisectors and the bisectors of its outer angles, respectively. Prove that 1 alal′ a + 1 clcl′ c = 1 blbl′ b. 12.88. In every angle of a triangle a circle tangent to the inscribed circle of the triangle is inscribed. Find the radius of the inscribed circle if the radii of these smaller circles are known. 12.89. The inscribed circle is tangent to sides AB, BC, CA at points K, L, M, respec-tively. Prove that: a) S = 1 2 ³ MK2 sin α + KL2 sin β + LM2 sin γ ´ ; b) S2 = 1 4(bcMK2 + caKL2 + abLM 2); c) MK2 hbhc + KL2 hcha + LM2 hahb = 1. Solutions 12.1. By the law of sines sin γ = c2R; hence, S = 1 2ab sin γ = abc 4R . 12.2. The radii of the circumscribed circles of triangles ABD and CBD are equal to AB 2 sin ∠ADB and BC 2 sin ∠BDC. It remains to notice that AB = BC and sin ∠ADB = sin ∠BDC. 12.3. By the law of sines b = a sin β sin α = a sin β sin(β+y) and, therefore, S = 1 2ab sin γ = a2 sin β sin γ 2 sin(β+γ) . 12.4. By the law of sines 1 2(a + b) = sin α+sin β sin γ . Moreover, sin α + sin β = 2 sin α + β 2 cos α −β 2 = 2 cos γ 2 cos α −β 2 and sin γ = 2 sin γ 2 cos γ 2. The second equality is similarly proved. 278 CHAPTER 12. CALCULATIONS AND METRIC RELATIONS 12.5. In triangle A2BC2, the lengths of sides A2B and BC2 are equal to b cos γ and b cos α; line BO divides angle ∠A2BC2 into angles of 90◦−γ and 90◦−α. Let line BO intersect segment A2C2 at point M. By the law of sines A2M = A2B sin ∠A2BM sin ∠A2MB = b cos γ cos α sin ∠C2MB = C2M. 12.6. Let α = ∠(a, c), β = ∠(c, d) and γ = ∠(d, b). Then AC AS BC BS = sin α sin(β + γ) and BD BS AD AS = sin γ sin(α + β). Hence (AC · BD) (BC · AD) = sin α sin γ sin(α + β) sin(β + γ). 12.7. Since OA PA = sin ∠OPA sin ∠POA and OB PB = sin ∠OPB sin ∠POB, it follows that (OA : OB) (PA : PB) = sin ∠POB sin ∠POA. 12.8. It suﬃces to multiply ﬁve equalities of the form D1A D1B = sin ∠B sin ∠A. 12.9. Let O be the common vertex of the given triangles, M and N the midpoints of the bases, k the ratio of the lengths of the bases to that of heights. The projections of the bases of given triangles on line MN are equal to k · OM sin ∠OMN and k · ON sin ∠ONM. It remains to notice that OM sin ∠ONM = ON sin ∠OMN . 12.10. By the law of sines BX sin ∠BDX = BD sin ∠BXD = 2R sin ψ sin ∠BXD. Moreover, sin ∠BDX = sin ∠BDC = sin ϕ and the value of angle ∠BXD is easy to calcu-late: if points C and D lie on one side of AB, then ∠BXD = π −ϕ −ψ and if they lie on distinct sides, then ∠BXD = \|ϕ −ψ\|. Hence, BX = 2R sin ϕ sin ψ sin \|ϕ±ψ\| . 12.11. a) Let A1 be the midpoint of segment BC. Adding equalities AB2 = AA2 1 + A1B2 −2AA1 · BA1 cos ∠BA1A and AC2 = AA2 1 + A1C2 −2AA1 · A1C cos ∠CA1A and taking into account that cos ∠BA1A = −cos ∠CA1A we get the statement desired. b) Follows in an obvious way from heading a). 12.12. By the law of cosines a2 −(b −c)2 = 2bc(1 −cos α) = 4S(1 −cos α) sin α = 4S tan α 2 . 12.13. By the law of cosines cos α = b2+c2−a2 2bc . It remains to make use of the formulas cos2 α 2 = 1 2(1 + cos α) and sin2 α 2 = 1 2(1 −cos α). 12.14. Let α be the angle at a vertex of the parallelogram. By the law of cosines m2 = a2 + b2 + 2ab cos α and n2 = a2 + b2 −2ab cos α. Hence, m2n2 = (a2 + b2)2 −(2ab cos α)2 = a4 + b4 + 2a2b2(1 −2 cos2 α). Therefore, m2n2 = a4 + b4 if and only if cos2 α = 1 2. SOLUTIONS 279 12.15. Let M be the intersection point of medians AA1 and BB1. Angle ∠AMB is a right one if and only if AM 2 + BM 2 = AB2, i.e. 4 9(m2 a + m2 b) = c2. By Problem 12.11 m2 a + m2 b = 4c2+a2+b2 4 . 12.16. Let m = C1M and ϕ = ∠C1MO. Then OC2 1 = C1M 2 + OM 2 −2OM · C1M cos ϕ and BO2 = CO2 = OM 2 + MC2 + 2OM · CM cos ϕ = OM 2 + 4C1M 2 + 4OM · C1M cos ϕ. Hence, BC2 1 = BO2 −OC2 1 = 3C1M 2 + 6OM · C1M cos ϕ, i.e., c2 = 4BC2 1 = 12m2 + 24OM · C1M cos ϕ. It is also clear that 18m2 = 2m2 c = a2 + b2 −c2 2 , cf. Problem 12.11. Therefore, equality a2 + b2 = 2c2 is equivalent to the fact that 18m2 = 3c2 2 , i.e., c2 = 12m2. Since c2 = 12m2 + 24OM · C1M cos ϕ, equality a2 + b2 = 2c2 is equivalent to the fact that ∠C1MO = ϕ = 90◦, i.e., CC1 ⊥OM. 12.17. Let the inscribed circle be tangent to side BC at point K and the escribed one at point L. Then BC = BK + KC = t cot β 2 + r cot γ 2 and BC = BL + LC = ra cot LBOa + ra cot LCOa = ra tan β 2 + ra tan γ 2. Moreover, cos α 2 = sin ¡β 2 + γ 2 ¢ . By Problem 3.2, p −b = BK = r cot β 2 and p −b = CL = ra tan γ 2. If the inscribed circle is tangent to the extensions of sides AB and AC at points P and Q, respectively, then p = AP = AQ = ra cot α 2 . 12.18. a) By Problem 12.17, p = ra cot α 2 and r cot α 2 = p −a; r cot β 2 = p −b and ra tan β 2 = p −c; rc tan β 2 = p −a and rb cot β 2 = p. By multiplying these pairs of equalities we get the desired statement. b) By multiplying equalities rp = ra(p −a) and rra = (p −b)(p −c) we get r2p = (p −a)(p −b)(p −c). It is also clear that S2 = p(r2p). c) It suﬃces to multiply rra = (p−b)(p−c) and rbrc = p(p−a) and make use of Heron’s formula. 12.19. By Problem 12.17, r = rc tan α 2 tan β 2 and p = rc cot γ 2. 12.20. By Problem 12.18 a), ra = rp p−a and rb = rp p−b. Hence, crarb = cr2p2 (p −a)(p −b) and ra + rb = rpc (p −a)(p −b) and, therefore, crarb ra+rb = rp = S. 12.21. By Problem 12.18 a), 1 rb = p−b pr and 1 rc = p−c pr , hence, 1 rb + 1 rc = a pr = a S = 2 ha. 12.22. It is easy to verify that 1 ha = a 2pr and 1 ra = p−a pr . Adding similar equalities we get the desired statement. 280 CHAPTER 12. CALCULATIONS AND METRIC RELATIONS 12.23. By Problem 12.18 a) 1 (p−b)(p−c) = 1 rra. It remains to add similar equalities and make use of the result of Problem 12.22. 12.24. By Problem 12.1, 4SR = abc. It is also clear that abc = p(p −b)(p −c) + p(p −c)(p −a) + p(p −a)(p −b) −(p −a)(p −b)(p −c) = S2 p−a + S2 p−b + S2 p−c −S2p = S(ra + rb + rc −r). 12.25. By Problem 12.18 a) rarb = p(p −c), rbrc = p(p −a) and rcra = p(p −b). Adding these equalities we get the desired statement. 12.26. Since S = rp = ra(p −a) = rb(p −b) = rc(p −c), the right-most expression is equal to p3 −(p −a)3 −(p −b)3 −(p −c)3 S3 = 3abc S3 . It remains to observe that abc S = 4R (Problem 12.1). 12.27. Let the angles of triangle ABC be equal to 2α, 2β and 2γ. Thanks to Problems 12.36 a) and 12.37 b) we have r = 4R sin α sin β sin γ and ra = 4R sin α cos β cos γ. Therefore, (r + ra)(4R + r −ra) = 16R2 sin α · (sin β sin γ + cos β cos γ)(1 + sin α(sin β sin γ −cos β cos γ)) = 16R2 sin α cos(β −γ)(1 −sin α cos(β + γ)) = 16R2 sin α cos(β −γ) cos2 α. It remains to notice that 4R sin α cos α = a and 4R sin(β + γ) cos(β −γ) = 2R(sin 2β + sin 2γ) = b + c. The second equality is similarly proved. 12.28. Since OA = r sin α 2 and bc = 2S sin α, it follows that OA2 bc = r2 cot α 2 S = r(p −a) S , cf. Problem 12.17 c). It remains to notice that r(p −a + p −b + p −c) = rp = S. 12.29. Let us solve heading b); heading a) is its particular case. Since cot ϕ 2 = sin ϕ 1−cos ϕ, it follows that p2(1 −x)2 = (1 −x2)(2R(1 −x) + r)2, where x = cos ϕ. The root x0 = 1 of this equation is of no interest to us because in this case cot ϕ 2 is undeﬁned; therefore, by dividing both parts of the equation by 1 −x we get a cubic equation. Making use of results of Problems 12.38, 12.41 b) and 12.39 b) we can verify that this equation coincides with the equation (x −cos α)(x −cos β)(x −cos γ) = 0, where α, β and γ are the angles of the triangle. Therefore the cosine of ϕ is equal to the cosine of one of the angles of the triangle; moreover, the cosine is monotonous on the interval [0, π]. 12.30. It is clear that 2pr = 2S = ab sin γ = abc 2R , i.e., 4prR = abc. To prove the second equality make use of Heron’s formula: S2 = p(p −a)(p −b)(p −c), i.e., pr2 = (p −a)(p −b)(p −c) = p3 −p2(a + b + c) + p(ab + bc + ca) −abc = = −p3 + p(ab + bc + ca) −4prR. SOLUTIONS 281 By dividing by p we get the desired equality. 12.31. Since abc = 4RS (Problem 12.1), the expression in the left-hand side is equal to c+a+b 4RS = 2p 4Rpr = 1 2Rr. 12.32. It suﬃces to observe that p−c p = r rc (Problem 12.18 a)), r = c sin α 2 sin β 2 cos γ 2 and rc = c cos α 2 cos β 2 cos γ 2 (Problem 12.17). 12.33. By Problem 12.1, S = abc 4R . On the other hand, S = aha 2 . Hence, ha = bc 2R. 12.34. Since aha = 2S = 2(p −a)ra and ra a = cos β 2 cos γ 2 cos α 2 (Problem 12.17 b)), we have ha = 2(p −a) cos β 2 cos γ 2 cos α 2 . Taking into account that (p −a) cot β 2 = rc = (p −b) cot α 2 (Problem 12.17 c)), we get ha = 2(p−b) sin β 2 cos γ 2 sin α 2 . 12.35. a) Let the extension of bisector AD intersect the circumscribed circle of triangle ABC at point M. Then AD · DM = BD · DC and since △ABC ∼△AMC, it follows that AB · AC = AD · AM = AD(AD + DM) = AD2 + BD · DC. Moreover, BD = ac b+c and DC = ab b+c. Hence, AD2 = bc − bca2 (b + c)2 = 4p(p −a)bc (b + c)2 . b) See the solution of Problem 4.47. c) Let AD be a bisector, AH a height of triangle ABC. Then AH = c sin β = 2R sin β sin γ. On the other hand, AH = AD sin ∠ADH = la sin ³ β + α 2 ´ = la sin π + β −γ 2 = la cos β −γ 2 . d) Taking into account that p = 4R cos α 2 cos β 2 cos γ 2 (Problem 12.36 c)) and sin β + sin γ = 2 sin β + γ 2 cos β −γ 2 = 2 cos α 2 cos β −γ 2 we arrive at the formula of heading c). 12.36. a) Let O be the center of the inscribed circle, K the tangent point of the inscribed circle with side AB. Then 2R sin γ = AB = AK + KB = r µ cot α 2 + cot β 2 ¶ = r sin α + β 2 sin α 2 sin β 2 . Taking into account that sin γ = 2 sin γ 2 cos γ 2 and sin α+β 2 = cos γ 2 we get the desired state-ment. b) By Problem 3.2, p−a = AK = r cot α 2 . Similarly, p−b = r cot β 2 and p−c = r cot γ 2. By multiplying these equalities and taking into account that p(p −a)(p −b)(p −c) = S2 = (pr)2 we get the desired statement. c) Obviously follows from headings a) and b). 12.37. a) By multiplying equalities r cos α 2 sin α 2 = p −a and sin α 2 sin β 2 sin γ 2 = r 4R 282 CHAPTER 12. CALCULATIONS AND METRIC RELATIONS (cf. Problems 12.17 c) and 12.36 a)) we get the desired statement. b) By Problem 12.17 c), ra tan γ 2 = p −b = r cot β 2. By multiplying this equality by r 4R = sin α 2 sin β 2 sin γ 2 we get the desired statement. 12.38. By adding equalities cos α + cos β = 2 cos α + β 2 cos α −β 2 cos γ = −cos(α + β) = −2 cos2 α + β 2 + 1 and taking into account that cos α −β 2 −cos α + β 2 = 2 sin α 2 sin β 2 we get cos α + cos β + cos γ = 4 sin α 2 sin β 2 sin γ 2 + 1 = r R + 1, cf. Problem 12.36 a). 12.39. a) Adding equalities cos 2α + cos 2β = 2 cos(α + β) cos(α −β) = −2 cos γ cos(α −β); cos 2γ = 2 cos2 γ −1 = −2 cos γ cos(α + β) −1 and taking into account that cos(α + β) + cos(α −β) = 2 cos α cos β we get the desired statement. b) It suﬃces to substitute expressions of the form cos 2α = 2 cos2 α −1 in the equality obtained in heading a). 12.40. Adding equalities sin 2α + sin 2β = 2 sin(α + β) cos(α −β) = 2 sin γ cos(α −β); sin 2γ = 2 sin γ cos γ = −2 sin γ cos(α + β) and taking into account that cos(α −β) −cos(α + β) = 2 sin α sin β we get the desired statement. 12.41. a) Clearly, sin2 α + sin2 β + sin2 γ = a2 + b2 + c2 4R and a2 + b2 + c2 = (a + b + c)2 −2(ab + bc + ca) = 4p2 −2(r2 + p2 + 4rR), cf. Problem 12.30. b) By Problem 12.39 b) 2 cos α cos β cos γ = sin2 α + sin2 β + sin2 γ −2. It remains to make use of a result of heading a). 12.42. The law of cosines can be expressed as ab cos γ = a2+b2−c2 2 . By adding three similar equalities we get the desired statement. 12.43. By Problem 12.13 cos2 α 2 a = p(p−a) abc . It remains to notice that p(p −a) + p(p −b) + p(p −c) = p2 and abc = 4SR = 4prR. SOLUTIONS 283 12.44. a) Since bc cos α = 2S cot α, it follows that a2 = b2 + c2 −4S cot α. By adding three similar equalities we get the desired statement. b) For an acute triangle a2 cot α = 2R2 sin 2α = 4SBOC, where O is the center of the circumscribed circle. It remains to add three analogous equalities. For a triangle with an obtuse angle α the quality SBOC should be taken with the minus sign. 12.45. By Problem 12.17 cot α 2 + cot β 2 = c r and tan α 2 + tan β 2 = c rc. It remains to add such equalities for all pairs of angles of the triangle. 12.46. Clearly, tan γ = −tan(α + β) = −tan α + tan β 1 −tan α tan β . By multiplying both sides of equality by 1 −tan α tan β we get the desired statement. 12.47. tan γ 2 = cot µα 2 + β 2 ¶ = · 1 −tan α 2 tan β 2 ¸ · tan α 2 + tan β 2 ¸ . It remains to multiply both sides of the equality by tan α 2 + tan β 2. 12.48. a) Let us multiply both sides of the equality by sin α sin β sin γ. Further on: cos γ(sin α cos β + sin β cos α) + sin γ(cos α cos β −sin α sin β) = cos γ sin(α + β) + sin γ cos(α + β) = cos γ sin γ −sin γ cos γ = 0. b) Let us multiply both sides of the equality by sin α sin β sin γ. Further on: cos α(sin β sin γ −cos β cos γ) + sin α(cos β sin γ + cos γ sin β) = cos2 α + sin2 α = 1. 12.49. Since sin2 α + sin2 β + sin2 γ −2 = 2 cos α cos β cos γ (see Problem 12.39 b) and S = 2R2 sin α sin β sin γ, it remains to verify that (tan α + tan β + tan γ) cos α cos β cos γ = sin γ sin β sin α. The latter equality is proved in the solution of Problem 12.48 a). 12.50. Let A and B be the vertices of angles α and β, let P be the intersection point of non-coinciding legs of these angles, Q the common point of the given circles that lies on segment PA. Triangle AQB is an isosceles one, hence, ∠PQB = 2α. Since ∠PQB + ∠QPB = β + ∠QBA, it follows that β = 3α. 12.51. By Problem 4.47, 1 b + 1 c = 2 cos α 2 la , hence, cos α 2 = 1 2, i.e., α = 120◦. 12.52. Let us drop perpendicular MD from point M to line BC. Then MD = 1 2AH = 1 2BM. In right triangle BDM, leg MD is equal to a half hypothenuse BM. Hence, ∠MBC = ∠MBD = 30◦. 12.53. The quantities AD · BC sin ADB and BE · AC sin AEB are equal because each of them is equal to the doubled area of triangle ABC. Hence, sin ADB = sin AEB. Two cases are possible: 1) ∠ADB = ∠AEB. In this case points A, E, D, B lie on one circle; hence, ∠EAD = ∠EBD, i.e., ∠A = ∠B which contradicts the hypothesis. 2) ∠ADB + ∠AEB = 180◦. In this case ∠ECD + ∠EOD = 180◦, where O is the intersection point of bisectors. Since ∠EOD = 90◦+ ∠C 2 (Problem 5.3), it follows that ∠C = 60◦. 12.54. Let B′ be the intersection point of the midperpendicular to segment AC with line AB. Then AB′ = CB′ and ∠AB′C = 180◦−2·75◦= 30◦. Hence, AB′ = CB′ = 2CH = AB, i.e., B′ = B and ∠B = 30◦. 284 CHAPTER 12. CALCULATIONS AND METRIC RELATIONS 12.55. Clearly, AKDM is a rectangle and L the intersection point of its diagonals. Since AD ⊥BC and AM ⊥BA, it follows that ∠DAM = ∠ABC. Similarly, ∠KAD = ∠ACB. Let us drop perpendicular AP from point A to line KM. Let, for deﬁniteness, ∠B < ∠C. Then point P lies on segment KL. Since △AKP ∼△MKA, it follows that AK : AP = MK : MA. Hence, AK · AM = AP · MK = AP · AD = 2AP · AL. By the hypothesis AL2 = AK · AM; hence, AL = 2AP, i.e., ∠ALP = 30◦. Clearly, ∠KMA = ∠ALP 2 = 15◦. Therefore, the acute angles of triangle ABC are equal to 15◦and 75◦. 12.56. Let CD be a bisector. Then BD = ac a+b. On the other hand, △BDC ∼△BCA, consequently, BD : BC = BC : BA, i.e., BD = a2 c . Hence c2 = a(a + b) = 3a2. The lengths of the sides of triangle ABC are equal to a, 2a and √ 3a; hence, its angles are equal to 30◦, 90◦and 60◦, respectively. 12.57. Let ∠ABC = 2x. Then the outer angle ∠A of triangle ABE is equal to ∠ABE + ∠AEB = x + α. Further, ∠KAE = ∠BAE −∠BAK = (180◦−x −α) −(180◦−2x −2α) = x + α. Therefore, AE is the bisector of the outer angle ∠A of triangle ABK. Since BE is the bisector of the inner angle ∠B of triangle ABK, it follows that E is the center of its escribed circle tangent to side AK. Hence, ∠AKE = 1 2∠AKC = 90◦−α. 12.58. Let A1 . . . A18 be a regular 18-gon. For triangle ABC we can take triangle A14A1A9. By Problem 6.59 b) the diagonals A1A12, A2A14 and A9A18 meet at one point, hence, ∠AMC = 1 2(⌣A18A2+ ⌣A9A14) = 70◦. 12.59. Let A1 . . . A18 be a regular 18-gon, O its center. For triangle ABC we can take triangle A1OA18. The diagonals A2A14 and A18A6 are symmetric through diameter A1A10; diagonal A2A14 passes through the intersection point of diagonals A1A12 and A9A18 (cf. the solution of Problem 12.58), therefore, ∠ADE = 1 2(⌣A1A2+ ⌣A12A14) = 30◦. 12.60. Since ∠BDE = 50◦and ∠CDE = 30◦, it follows that ∠BOC = ∠EOD = 180◦−50◦−30◦= 100◦. Let us assume that diameters BB′ and CC′ of the circle are ﬁxed, ∠BOC = 100◦and point A moves along arc ⌣B′C′. Let D be the intersection point of BB′ and AC, E the intersection point of CC′ and AB (Fig. 140). As point A moves from B′ to C′, segment OE increases while OD decreases, consequently, angle ∠OED decreases and angle ∠ODE increases. Therefore, there exists a unique position of point A for which ∠CED = ∠OED = 30◦and ∠BDE = ∠ODE = 50◦. Figure 140 (Sol. 12.60) Now, let us prove that triangle ABC with angles ∠A = 50◦, ∠B = 70◦, ∠C = 60◦ possesses the required property. Let A1 . . . A18 be a regular 18-gon. For triangle ABC we can take triangle A2A14A9. Diagonal A1A12 passes through point E (cf. solution of Problem 12.58). Let F be the intersection point of lines A1A12 and A5A14; line A9A16 is symmetric to line A1A12 through line A5A14 and, therefore, it passes through point F. In triangle CDF, SOLUTIONS 285 ray CE is the bisector of angle ∠C and line FE is the bisector of the outer angle at vertex F. Hence, DE is the bisector of angle ∠ADB, i.e., ∠ODE = 1 4(⌣A2A14+ ⌣A5A9) = 50◦. 12.61. Let D, E and F be the tangent points of the circle with BP, PQ and QC, respectively; ∠BOD = 90◦−∠B = 90◦−∠C = ∠COF = α, ∠DOP = ∠POE = β and ∠EOQ = ∠QOF = γ. Then 180◦= ∠BOC = 2α + 2β + 2γ, i.e., α + β + γ = 90◦. Since ∠BPO = 1 2∠DPE = 1 2(180◦−∠DOE) = 90◦−β and ∠QOC = γ + α = 90◦−β, it follows that ∠BPO = ∠COQ. It is also clear that ∠PBO = ∠OCQ. Hence, △BPO ∼△COQ, i.e., PB · CQ = BO · CO = 1 4BC2. 12.62. Let P and Q be the midpoints of sides BC and CD, respectively. Points P and Q are the tangent points of the inscribed circle with sides BC and CD. Therefore, it suﬃces to verify that PF + GQ = FG. Indeed, if F ′G′ is the segment parallel to FG and tangent to the inscribed circle, then PF ′ + G′Q = F ′G′; hence, F ′ = F and G′ = G. We may assume that the side of the square is equal to 2. Let GD = x. Since BF : EB = AD : GD, then BF = 2 x. Therefore, CG = 2 −x, GQ = x −1, CF = 2 −2 x, FP = 2 x −1, i.e., PF + GQ = x + 2 x −2 and FG2 = CG2 + CF 2 = (2 −x)2 + µ 2 −2 x ¶2 = 4 −4x + x2 + 4 −8 x + 4 x2 = µ x + 2 x −2 ¶2 = (PF + GQ)2. 12.63. Denote the vertices of the squares as shown on Fig. 141. Let O be the center of the circle, H the midpoint of the given chord, K the midpoint of segment AA1. Figure 141 (Sol. 12.63) Since tan AHB = 2 = tan A1HD1, point H lies on line AA1. Let α = ∠AHB = ∠A1HD1, then AB −A1D1 = (AH −A1H) · sin α = 2KH sin α = 2OH sin2 α. Since tan α = 2 and 1 + cot2 α = 1 sin2 α, it follows that sin2 α = 4 5. Therefore, the diﬀerence of the lengths of the squares’ sides is equal to 8 5h. 12.64. Let median BM of triangle ABC intersect the inscribed circle at points K and L, where BK = KL = LM = x. Let, for deﬁniteness, the tangent point of the inscribed circle with side AC lie on segment MC. Then since the symmetry through the midperpendicular to segment BM interchanges points B and M and ﬁxes the inscribed circle, tangent MC turns into tangent BC. Therefore, BC = MC = 1 2AC, i.e., b = 2a. Since BM 2 = 2a2+2c2−b2 4 by Problem 12.11 a), we have 9x2 = 2a2+2c2−4a2 4 = c2−a2 2 . Let P be the tangent point of the inscribed circle with side BC. Then BP = a+c−b 2 = c−a 2 . On the other hand, by a property of the tangent, BP 2 = BK · BL, i.e., BP 2 = 2x2. Hence, 286 CHAPTER 12. CALCULATIONS AND METRIC RELATIONS 2x2 = ¡ c−a 2 ¢2. Multiplying inequalities 9x2 = c2−a2 2 and ¡ c−a 2 ¢2 = 2x2 we get c+a c−a = 9 4, i.e., c : a = 13 : 5. As a result we get a : b : c = 5 : 10 : 13. 12.65. Let 2a and 2b be the length of the side of the ﬁrst and second squares, respectively. Then the distance from the center of the circle to any of the vertices of the second square that lie on the circle is equal to p (a + 2b)2 + b2. On the other hand, this distance is equal to √ 2a. Therefore, (a + 2b)2 + b2 = 2a2, i.e., a = 2b ± √ 4b2 + 5b2 = (2 ± 3)b. Only the solution a = 5b is positive. 12.66. Let P and Q be the midpoints of segments AC and AB, respectively, R the center of circle S1; a = 1 2AC, b = 1 2BC, x the radius of circle S1. It is easy to verify that PR = a + x, QR = a + b −x and PQ = b. In triangle PQR, draw height RH. The distance from point R to line CD is equal to x, hence, PH = a −x, consequently, QH = \|b −a + x\|. It follows that (a + x)2 −(a −x)2 = RH2 = (a + b −x)2 −(b −a + x)2, i.e., ax = b(a −x). As a result we get x = ab a+b. For the radius of circle S2 we get precisely the same expression. 12.67. Let x be the radius of circle S tangent to circles S1 and S2 and ray AB, let y be the radius of circle S′ tangent to circles S2 and S3 and ray BA. The position of the circle tangent to circle S1 and ray AB (resp. S3 and BA) is uniquely determined by its radius, consequently, it suﬃces to verify that x = y. By equating two expressions for the squared distance from the center of circle S to line AD we get (x + 1)2 −(x −1)2 = (3 + x)2 −(5 −x)2, i.e., x = 4 3. Considering circles S2 and S3 it is easy to verify that AB2 = (3 + 4)2 −12 = 48. On the other hand, the squared distances from the center of circle S′ to lines AD and BC are equal to (y + 3)2 −(5 −y)2 = 16(y −1) and (4 + y)2 −(4 −y)2 = 16y, respectively. Therefore, 4√y −1 + 4√y = √ 48, i.e., y = 4 3. 12.68. If the angles of a triangle form an arithmetic progression, then they are equal to α −γ, α, α + γ, where γ ≥0. Since the sum of the angles of a triangle is equal to 180◦, we deduce that α = 60◦. The sides of this triangle are equal to 2R sin(α −γ), 2R sin α, 2R sin(α + γ). Since the greater side subtends the greater angle, sin(α −γ) ≤sin α ≤ sin(α + γ). a) If the numbers sin(α −γ) ≤sin α ≤sin(α + γ) form an arithmetic progression, then sin α = 1 2(sin(α + γ) + sin(α −γ)) = sin α cos γ, i.e., either cos γ = 1 or γ = 0. Therefore, each of the triangle’s angles is equal to 60◦. b) If the numbers sin(α −γ) ≤sin α ≤sin(α + γ) form a geometric progression, then sin2 α = sin(α −γ) sin(α + γ) = sin2 α cos2 γ −sin2 γ cos2 α ≤sin2 α cos2 γ. Hence, cos γ = 1, i.e., each of the triangle’s angles is equal to 60◦. 12.69. Let us complement triangle ABC to parallelogram ABCE (Fig. 142). Let BC = x and AD = y. Then (b −a)h = 2SAED = xy sin 45◦and (b −a)2 = x2 + y2 −2xy cos 45◦= x2 + y2 −2xy sin 45◦. By Pythagoras theorem a2 + b2 = (AO2 + BO2) + (CO2 + DO2) = (BO2 + CO2) + (DO2 + AO2) = x2 + y2. Therefore, (b −a)2 = x2 + y2 −2xy sin 45◦= a2 + b2 −2(b −a)h, SOLUTIONS 287 Figure 142 (Sol. 12.69) i.e., h = ab b−a. 12.70. Since BK = 1 2(a + c −b) and KC = 1 2(a + b −c) (cf. Problem 3.2), it follows that BK · KC = a2−(b−c)2 4 = S tan α 2 , cf. Problem 12.12. 12.71. Since b+c a = cos β−γ 2 sin α 2 (Problem 12.4), it follows that cos β−γ 2 = cos α 2 , i.e., β −γ = ±α. If β = γ + α, then β = 90◦and if β + α = γ, then γ = 90◦. 12.72. It is easy to verify that SABC = 2R2 sin α sin β sin γ. Analogously, SA1B1C1 = 2R2 sin β + γ 2 sin α + γ 2 sin α + β 2 = 2R2 cos α 2 cos β 2 cos γ 2. Hence, SABC SA1B1C1 = 8 sin α 2 sin β 2 sin γ 2 = 2r R , cf. Problem 12.36 a). 12.73. The sum of cotangents of the angles of a triangle is equal to a2+b2+c2 4S , cf. Problem 12.44 a). Moreover, m2 a + m2 b + m2 c = 3(a2+b2+c2) 4 (by Problem 12.11 b)) and the area of the triangle formed by the medians of triangle ABC is equal to 3 4SABC (Problem 1.36). 12.74. One of the points Ai lies inside the triangle formed by the other three points; hence, we can assume that triangle A1A2A3 is an acute one (or a right one). Numbers λ1, λ2 and λ3 are easy to obtain from the corresponding system of equations; as a result we get λ1 = b2 + c2 −a2 2 , λ2 = a2 + c2 −b2 2 and λ3 = a2 + b2 −c2 2 , where a = A2A3, b = A1A3 and c = A1A2. By Problem 5.45 b) A1A2 4 = 4R2 −a2, where R is the radius of the circumscribed circle of triangle A1A2A3. Hence, λ4 = A1A2 4 −λ1 = 4R2 −a2 + b2 + c2 2 = A2A2 4 −λ2 = A3A2 4 −λ3. Now, let us verify that P 1 λi = 0. Since b2+c2−a2 2 = bc cos α = 2S cot α, it follows that 1 λ1 = tan α 2S. It remains to observe that 2 a2 + b2 + c2 −8R2 = tan α + tan β + tan γ 2S ?Problem 12.49. 12.75. Let (a1, b1), (a2, b2) and (a3, b3) be the coordinates of the triangle’s vertices. The coordinates of the center of the circumscribed circle of the triangle are given by the system of equations (x −a1)2 + (y −b1)2 = (x −a2)2 + (y −b2)2, (x −a1)2 + (y −b1)2 = (x −a3)2 + (y −b3)2. 288 CHAPTER 12. CALCULATIONS AND METRIC RELATIONS It is easy to verify that these equations are actually linear ones and, therefore, the solution of the considered system is a rational one. 12.76. On segments AB and CD, take points K and L that divide the segments in the ratios indicated. It suﬃces to prove that the intersection point of lines AK and CL lies on circle S. Let us take the coordinate system with the origin at the center O of circle S and axes Ox and Oy directed along rays OB and OD. The radius of circle S can be assumed to be equal to 1. Lines AK and CL are given by equations y = x+1 3 and y = 2x−1, respectively. Therefore, the coordinates of their intersection point are x0 = 4 5 and y0 = 3 5. Clearly, x2 0 + y2 0 = 1. 12.77. Let d be the distance between the center of the circumscribed circle and the image of the center of the inscribed circle under the considered homothety. It suﬃces to verify that R = d + 2r. Let (0, 0), (2a, 0) and (0, 2b) be the coordinates of the vertices of the given triangle. Then (a, b) are the coordinates of the center of the circumscribed circle, (r, r) the coordinates of the center of the inscribed circle, where r = a + b −R. Therefore, d2 = (2r −a)2 + (2r −b)2 = a2 + b2 −4r(a + b −r) + 4r2 = (R −2r)2 because a2 + b2 = R2. 12.78. Let us consider the coordinate system with the origin at the center of the square and the Ox-axis parallel to line l. Let the coordinates of the vertices of the square be (A(x, y), B(y, −x), C(−x, −y) and D(−y, x); let line l be given by the equation y = a. Then the coordinates of point Q are ¡x+y 2 , y−x 2 ¢ and those of P are (−y, a). Therefore, the locus to be found consists of points ¡ t, −t + 1 2a ¢ , where t = x−y 4 . It remains to observe that the quantity x −y varies from − p 2(x2 + y2) = −AB to AB. Chapter 13. VECTORS Background 1. We will make use of the following notations: a) − → AB and a denote vectors; b) AB and \|a\| denote the lengths of these vectors; sometimes the length of vector a will be denoted by a; a unit vector is a vector of unit length; c) (− → AB, − − → CD), (a, b) and (− → AB, a) denote the inner products of the vectors; d) (x, y) is the vector with coordinates x, y; e) − → 0 or 0 denotes the zero vector. 2. The oriented angle between the nonzero vectors a and b (notation ∠(a, b)) is the angle through which one should rotate the vector a counterclockwise to make it directed as b is. The angles that diﬀer by 360 degrees are assumed to be equal. It is easy to verify the following properties of oriented angles between vectors: a) ∠(a, b) = −∠(b, a); b) ∠(a, b) + ∠(b, c) = ∠(a, c); c) ∠(−a, b) = ∠(a, b) + 180◦. 3. The inner product of vectors a and b is the number (a, b) = \|a\| · \|b\| cos ∠(a, b) (if one of these vectors is the zero one, then by deﬁnition (a, b) = 0). The following properties of the inner product are easily veriﬁed: a) (a, b) = (b, a); b) \|(a, b)\| ≤\|a\| · \|b\|; c) (λa + µb, c) = λ(a, c) + µ(b, c); d) if a, b ̸= 0 then (a, b) = 0 if and only if a ⊥b. 4. Many of vector inequalities can be proved with the help of the following fact. Given two sets of vectors such that the sum of lengths of projections of the vectors of the ﬁrst set to any straight line does not exceed the sum of the lengths of projections of the vectors from the second set to the same line, the sum of the lengths of the vectors from the ﬁrst set does not exceed the sum of the lengths of the vectors of the second set, cf. Problem 13.39. In this way a problem on a plane reduces to a problem on a straight line which is usually easier. Introductory problems 1. Let AA1 be the median of triangle ABC. Prove that − − → AA1 = 1 2(− → AB + − → AC). 2. Prove that \|a + b\|2 + \|a −b\|2 = 2(\|a\|2 + \|b\|2). 3. Prove that if vectors a + b and a −b are perpendicular, then \|a\| = \|b\|. 4. Let − → OA + − − → OB + − → OC = − → 0 and OA = OB = OC. Prove that ABC is an equilateral triangle. 289 290 CHAPTER 13. VECTORS 5. Let M and N be the midpoints of segments AB and CD, respcetively. Prove that − − → MN = 1 2(− → AC + − − → BD). §1. Vectors formed by polygons’ (?) sides 13.1. a) Prove that from the medians of a triangle one can construct a triangle. b) From the medians of triangle ABC one constructed triangle A1B1C1 and from the medians of triangle A1B1C1 one constructed triangle A2B2C2. Prove that triangles ABC and A2B2C2 are similar with simlarity coeﬃcient 3 4. 13.2. The sides of triangle T are parallel to the respective medians of triangle T1. Prove that the medians of T are parallel to the corresponding sides of T1. 13.3. Let M1, M2, . . . , M6 be the midpoints of a convex hexagon A1A2 . . . A6. Prove that there exists a triangle whose sides are equal and parallel to the segments M1M2, M3M4, M5M6. 13.4. From a point inside a convex n-gon, the rays are drawn perpendicular to the sides and intersecting the sides (or their continuations). On these rays the vectors a1, . . . , an whose lengths are equal to the lengths of the corresponding sides are drawn. Prove that a1 + · · · + an = 0. 13.5. The sum of four unit vectors is equal to zero. Prove that the vectors can be divided into two pairs of opposite vectors. 13.6. Let E and F be the midpoints of sides AB and CD of quadrilateral ABCD and K, L, M and N are the midpoints of segments AF, CE, BF and DE, respectively. Prove that KLMN is a parallelogram. 13.7. Consider n pairwise noncodirected vectors (n ≥3) whose sum is equal to zero. Prove that there exists a convex n-gon such that the set of vectors formed by its sides coincides with the given set of vectors. 13.8. Given four pairwise nonparallel vectors whose sum is equal to zero. Prove that we can construct from them: a) a nonconvex quadrilateral; b) a self-intersecting broken line of four links. 13.9. Given four pairwise nonparallel vectors a, b, c and d whose sum is equal to zero, prove that \|a\| + \|b\| + \|c\| + \|d\| > \|a + b\| + \|a + c\| + \|a + d\|. 13.10. In a convex pentagon ABCDE side BC is parallel to diagonal AD, in addition we have CD ∥BE, DE ∥AC and AE ∥BD. Prove that AB ∥CE. §2. Inner product. Relations 13.11. Prove that if the diagonals of quadrilateral ABCD are perpendicular to each other, then the diagonals of any other quadrilateral with the same lengths of its sides are perpendicular to each other. 13.12. a) Let A, B, C and D be arbitrary points on a plane. Prove that (− → AB, − − → CD) + (− − → BC, − − → AD) + (− → CA, − − → BD) = 0. b) Prove that the hights of a triangle intersect at one point. 13.13. Let O be the center of the circle inscribed in triangle ABC and let point H satisfy OH = OA + OB + OC. Prove that H is the intersection point of heights of triangle ABC. §3. INEQUALITIES 291 13.14. Let a1, . . . , an be vectors formed by the sides of an n-gon, ϕij = ∠(ai, aj). Prove that a2 1 = a2 2 + · · · + a2 n + 2 X i>j>1 aiaj cos ϕij, where ai = \|ai\|. 13.15. Given quadrilateral ABCD and the numbers u = AD2, v = BD2, w = CD2, U = BD2 + CD2 −BC2, V = AD2 + CD2 −AC2, W = AD2 + BD2 −AB2. Prove that ((Gauss).) uU 2 + vV 2 + wW 2 = UV W + 4uvw. 13.16. Points A, B, C and D are such that for any point M the numbers (− − → MA, − − → MB) and (− − → MC, − − → MD) are distinct. Prove that − → AC = − − → DB. 13.17. Prove that in a convex k-gon the sum of distances from any inner point to the sides of the k-gon is constant if and only if the sum of vectors of unit exterior normals to the sides is equal to zero. 13.18. In a convex quadrilateral the sum of distances from a vertex to the sides is the same for all vertices. Prove that this quadrilateral is a parallelogram. §3. Inequalities 13.19. Given points A, B, C and D. Prove that AB2 + BC2 + CD2 + DA2 ≥AC2 + BD2, where the equality is attained only if ADCD is a parallelogram. 13.20. Prove that from any ﬁve vectors one can always select two so that the length of their sum does not exceed the length of the sum of the remaining three vectors. 13.21. Ten vectors are such that the length of the sum of any nine of them is smaller than the length of the sum of all the ten vectors. Prove that there exists an axis such that the projection of every of the ten vectors to the axis is positive. 13.22. Points A1, . . . , An lie on a circle with center O and − − → OA1 + · · ·+ − − → OAn = − → 0 . Prove that for any point X we have XA1 + · · · + XAn ≥nR, where R is the radius of the circle. 13.23. Given eight real numbers a, b, c, d, e, f, g, h. Prove that at least one of the six numbers ac + bd, ae + bf, ag + bh, ce + d f, cg + dh, eg + fh is nonnegative. 13.24. On the circle of radius 1 with center O there are given 2n+1 points P1, . . . , P2n+1 which lie on one side of a diameter. Prove that \|− − → OP1 + · · · + − − − − → OP2n+1\| ≥1. 13.25. Let a1, a2, . . . , an be vectors whose length does not exceed 1. Prove that in the sum c = ±a1 ± a2 ± · · · ± an we can select signs so that \|c\| ≤ √ 2. 292 CHAPTER 13. VECTORS 13.26. Point O is the beginning point of n unit vectors such that in any half plane bounded by a straight line through O there are contained not less than k vectors (we assume that the boundary line belongs to the half-plane). Prove that the length of the sum of these vectors does not exceed n −2k. §4. Sums of vectors 13.27. Prove that point X belongs to line AB if and only if − − → OX = t− → OA + (1 −t)− − → OB for some t and any point O. 13.28. We are given several points and for several pairs (A, B) of these points the vectors AB are taken in such a way that as many vectors exit from every point as terminate in it. Prove that the sum of all the selected vectors is equal to 0. 13.29. Inside triangle ABC, point O is taken. Prove that SBOC · − → OA + SAOC · − − → OB + SAOB · − → OC = − → 0 . 13.30. Points A and B move along two ﬁxed rays with common origin O so that p OA + q OB is a constant. Prove that line AB passes through a ﬁxed point. 13.31. Through the intersection point M of medians of triangle ABC a straight line is drawn intersecting BC, CA and AB at points A1, B1 and C1, respectively. Prove that ( 1 MA1 ) + ( 1 MB1 ) + ( 1 MC1 ) = 0. (Segments MA1, MB1 and MC1 are assumed to be oriented.) 13.32. On sides BC, CA and AB of triangle ABC points A1, B1 and C1, respectively, are taken. Segments BB1 and CC1, CC1 and AA1, AA1 and BB1 intersect at points A1, B2 and C2, respectively. Prove that if − − → AA2 + − − → BB2 + − − → CC2 = − → 0 , then AB1 : B1C = CA1 : A1B = BC1 : C1A. 13.33. Quadrilateral ABCD is an inscribed one. Let Ha be the orthocenter of BCD, let Ma be the midpoint of AHa; let points Mb, Mc and Md be similarly deﬁned. Prove that points Ma, Mb, Mc and Md coincide. 13.34. Quadrilateral ABCD is inscribed in a circle of radius R. a) Let Sa be the circle of radius R with center at the orthocenter of triangle BCD; let circles Sb, Sc and Sd be similarly deﬁned. Prove that these four circles intersect at one point. b) Prove that the circles of nine points of triangles ABC, BCD, CDA and DAB intersect at one point. §5. Auxiliary projections 13.35. Point X belongs to the interior of triangle ABC; let α = SBXC, β = SCXA and γ = SAXB. Let A1, B1 and C1 be the projections of points A, B and C, respectively, on an arbitrary line l. Prove that the length of vector α− − → AA1 + β− − → BB1 + γ− − → CC1 is equal to (α + β + γ)d, where d is the distance from X to l. 13.36. A convex 2n-gon A1A2 . . . A2n is inscribed into a unit circle. Prove that \|− − − → A1A2 + − − − → A3A4 + · · · + − − − − − − → A2n−1A2n\| ≤2. 13.37. Let a, b and c be the lengths of the sides of triangle ABC; let na, nb and nc be unit vectors perpendicular to the corresponding sides and directed outwards. Prove that a3na + b2nb + c2nc = 12S · − − → MO, §7. PSEUDOINNER PRODUCT 293 where S is the area, M the intersection point of the medians, O the center of the circle inscribed into triangle ABC. 13.38. Let O and R be the center and the radius, respectively, of an escribed circle of triangle ABC; let Z and r be the center and the radius of the inscribed circle, K the intersection point of the medians of the triangle with vertices at the tangent points of the inscribed circle of triangle ABC with the sides of triangle ABC. Prove that Z belongs to segment OK and OZ : ZK = 3R : r. §6. The method of averaging 13.39. Given two sets of vectors a1, . . . , an and b1, . . . , bm such that the sum of the lengths of the projections of the vectors from the ﬁrst set to any straight line does not exceed the sum of the lengths of the projections of the vectors from the second set to the same straight line. Prove that the sum of the lengths of the vectors from the ﬁrst set does not exceed the sum of the lengths of the vectors from the second set. 13.40. Prove that if one convex polygon lies inside another one, then the perimeter of the inner polygon does not exceed the perimeter of the outer one. 13.41. The sum of the length of several vectors on a plane is equal to L. Prove that from these vectors one can select several vectors (perhaps, just one) so that the length of their sum is not less than L π. 13.42. Prove that if the lengths of any side and diagonal of a convex polygon are shorter than d, then its perimeter is shorter than πd. 13.43. On the plane, there are given four vectors a, b, c and d whose sum is equal to zero. Prove that \|a\| + \|b\| + \|c\| + \|d\| ≥\|a + d\| + \|b + d\| + \|b + d\|. 13.44. Inside a convex n-gon A1A2 . . . An a point O is selected so that − − → OA1+· · ·+− − → OAn = − → 0 . Let d = OA1 + · · · + OAn. Prove that the perimeter of the polygon is not shorter than 4d n for n even and not shorter than 4dn n2−1 for n odd. 13.45. The length of the projection of a closed convex curve to any line is equal to 1. Prove that its length is equal to π. 13.46. Given several convex polygons so that it is impossible to draw a line which does not intersect any of the polygons and at least one polygon would lie on both sides of it. Prove that all the polygons are inside a polygon whose perimeter does not exceed the sum of the perimeters of the given polygons. §7. Pseudoinner product The pseudoinner product of nonzero vectors a and b is the number c = \|a\| · \|b\| sin ∠(a, b); the pseudoinner product is equal to 0 if at least one of the vectors a or b is zero. The pseudoinner product is denoted by c = a ∨b. Clearly, a ∨b = −b ∨a. The absolute value of the pseudoinner product of a and b is equal to the area of the parallelogram spanned by these vectors. In this connection the oriented area of the triple of points A, B and C is the number S(A, B, C) = 1 2(− → AB ∨− → AC). The absolute value of S(A, B, C) is equal to the area of triangle ABC. 13.47. Prove that: 294 CHAPTER 13. VECTORS a) (λa) ∨b = λ(a, b); b) a ∨(b + c) = a ∨b + a ∨c. 13.48. Let a = (a1, a2) and b = (b1, b2). Prove that a ∨b = a1b2 −a2b1. 13.49. a) Prove that S(A, B, C) = −S(B, A, C) = S(B, C, A). b) Prove that for any points A, B, C and D we have S(A, B, C) = S(D, A, B) + S(D, B, C) + S(D, C, A). 13.50. Three runners A, B and C run along the parallel lanes with constant speeds. At the initial moment the area of triangle ABC is equal to 2 in 5 seconds it is equal to 3. What might be its value after 5 more seconds? 13.51. Three pedestrians walk at constant speeds along three straight roads. At the initial moment the pedestrians were not on one straight line. Prove that the pedestrians can occure on one straight line not more than twice. 13.52. Prove Problem 4.29 b) with the help of a pseudoinner product. 13.53. Points P1, P2 and P3 not on one line are inside a convex 2n-gon A1 . . . A2n. Prove that if the sum of the areas of triangles A1A2Pi, A3A4Pi, . . . , A2n−1A2nPi is equal to the same number c for i = 1, 2, 3, then for any inner point P the sum of the areas of these triangles is equal to c. 13.54. Given triangle ABC and point P. Let point Q be such that CQ ∥AP and point R be such that AR ∥BQ and CR ∥BP. Prove that SABC = SPQR. 13.55. Let H1, H2 and H3 be the orthocenters of triangles A2A3A4, A1A3A4 and A1A2A4. Prove that the areas of triangles A1A2A3 and H1H2H3 are equal. 13.56. In a convex 5-gon ABCDE whose area is equal to S the areas of triangles ABC, BCD, CDE, DEA and EAB are equal to a, b, c, d and e, respectively. Prove that S2 −S(a + b + c + d + e) + ab + bc + cd + de + ea = 0. Problems for independent study 13.57. Let M and N be the midpoints of segments AB and AC, respectively, P the midpoint of MN and O an arbitrary point. Prove that 2− → OA + − − → OB + − → OC = 4− → OP. 13.58. Points A, B and C move uniformly with the same angle velocities along the three circles in the same direction. Prove that the intersection point of the medians of triangle ABC moves along a circle. 13.59. Let A, B, C, D and E be arbitrary points. Is there a point O such that − → OA + − − → OB + − → OC = − − → OD + − − → OE? Find all such points, if any. 13.60. Let P and Q be the midpoints of the diagonals of a convex quadrilateral ABCD. Prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2 + 4PQ2. 13.61. The midpoints of segments AB and CD are connected by a segment; so are the midpoints of segments BC and DE. The midpoints of the segments obtained are also connected by a segment. Prove that the last segment is parallel to segment AE and its length is equal to 1 4AE. 13.62. The inscribed circle is tangent to sides BC, CA and AB of triangle ABC at points A1, B1 and C1, respectively. Prove that if − − → AA1 + − − → BB1 + − − → CC1 = − → 0 , then triangle ABC is an equilateral one. SOLUTIONS 295 13.63. Quadrilaterals ABCD, AEFG, ADFH, FIJE and BIJC are parallelograms. Prove that quadrilateral AFHG is also a parallelogram. Solutions 13.1. a) Let a = − − → BC, b = − → CA and c = − → AB; let AA′, BB′ and CC′ be medians of triangle ABC. Then − − → AA′ = 1 2(c −b), − − → BB′ = 1 2(a −c) and − − → CC′ = 1 2(b −c). Therefore, − − → AA′ + − − → BB′ + − − → CC′ = − → 0 . b) Let a1 = − − → AA′, b1 = − − → BB′ and c = − − → CC′. Then 1 2(c1 −b1) = 1 4(b −a −a + c) = −3 4a is the vector of one of the sides of triangle A2B2C2. 13.2. Let a, b and c be the vectors of the sides of T. Then 1 2(b −a), 1 2(a −c) and 1 2(c −b) are the vectors of its medians. We may assume that a, b and c are the vectors directed from the intersection point of the medians of triangle T1 to its vertices. Then b−a, a −c and c −a are the vectors of its sides. 13.3. It is clear that 2− − − − → M1M2 = − − − → A1A2 + − − − → A2A3 = − − − → A1A3, 2− − − − → M3M4 = − − − → A3A5 and 2− − − − → M5M6 = − − − → A5A1. Therefore, − − − − → M1M2 + − − − − → M3M4 + − − − − → M5M6 = − → 0 . 13.4. After rotation through 90◦the vectors a1, . . . , an turn into the vectors of sides of the n-gon. 13.5. From given vectors one can construct a convex quadrilateral. The lengths of all the sides of this quadrilateral are equal to 1, therefore, this quadrilateral is a rhombus; the pairs of its opposite sides provide us with the division desired. 13.6. Let a = − → AE, b = − − → DF and v = − − → AD. Then 2− − → AK = b + v and 2− → AL = a + v + 2b and, therefore, − − → KL = − → AL −− − → AK = 1 2(a + b). Similarly, − − → NM = 1 2(a + b). 13.7. Let us draw the given vectors from one point and index them clockwise: a1, . . . , an. Consider a closed broken line A1 . . . An, where − − − − → AiAi+1 = ai. Let us prove that A1 . . . An is a convex polygon. Introduce a coordinate system and direct the Ox-axis along a1. Let the vectors a2, . . . , ak lie on one side of Ox-axis and the vectors ak+1, . . . , an lie on the other side (if there is a vector directed opposite to a1 it can be referred to either of these two groups). The projections of the vectors from the ﬁrst group on the Oy-axis are of one sign and the projections of the vectors of the other group are of the opposite sign. Therefore, the second coordinate of the points A2, A3, . . . , Ak+1 and the points Ak+1, . . . , An, A1 vary monotonously: for the ﬁrst group from 0 to a quantity d, for the second group they de-crease from d to 0. Since there are two intervals of monotonity, all the vertices of the polygon lie on one side of the line A1A2. For the other lines passing through the sides of the polygon the proof is similar. 13.8. Thanks to Problem 13.7 the given vectors form a convex quadrilateral. The rest is clear from Fig. 143. Figure 143 (Sol. 13.8) 296 CHAPTER 13. VECTORS 13.9. By Problem 13.8 b) from the given vectors we can construct a self-intersecting broken line of four links; this broken line can be viewed as the two diagonals and two opposite sides of a convex quadrilateral. Two cases are possible: the vector a can be either a side or a diagonal of this quadrilateral. But in both cases the sum in the left-hand side of the inequality is the sum of lengths of two opposite sides and two diagonals of the quadrilateral and the sum in the right-hand side is constituted by the length of the sum of vectors of the same opposite sides and the lengths of the two other opposite sides. It only remains to observe that the sum of lengths of two vectors is not shorter than the length of their sum and the sum of the length of diagonals of a convex quadrilateral is longer than the sum of lengths of the two opposite sides: cf. Problem 19.14. 13.10. Let diagonal BE intersect diagonals AD and AC at points F and G, respectively. The respective sides of triangles AFE and BCD are parallel; hence, the triangles are similar and AF : FE = BC : CD. Therefore, AD : BE = (AF + BC) : (EF + CD) = BC : CD. Similarly, AE : BD = DE : AC. From the similarity of BED and EGA we deduce that AE : DB = EG : BE = CD : BE. Thus, BC AD = CD BE = AE BD = DE AC = λ. Clearly, − − → BC + − − → CD + − − → DE + − → EA + − → AB = − → 0 , − − → AD + − − → BE + − → CA + − − → DB + − − → EC = − → 0 and − − → BC = λ− − → AD, − − → CD = λ− − → BE, − − → DE = λ− → CA, − → EA = λ− − → DB. It follows that − → 0 = λ(− − → AD + − − → BE + − → CA + − − → DB) + − → AB = −λ− − → EC + − → AB, i.e., − → AB = λ− − → EC. Hence, AB ∥EC. 13.11. Let a = − → AB, b = − − → BC, c = − − → CD and d = − − → DA. It suﬃces to verify that AC ⊥BD if and only if a2 + c2 = b2 + d2. Clearly, d2 = \|a + b + c\|2 = a2 + b2 + c2 + 2[(a, b) + (b, c) + (c, a)]. Therefore, the condition AC ⊥BD, i.e., 0 = (a + b, b + c) = b2 + (b, c) + (a, c) + (a, b) is equivalent to the fact that d2 = a2 + b2 + c2 −2b2. 13.12. a) Let us express all the vectors that enter the formula through − → AB, − − → BC and − − → CD, i.e., let us write − − → AD = − → AB + − − → BC + − − → CD, − → CA = −− → AB −− − → BC and − − → BD = − − → BC + − − → CD. After simpliﬁcation we get the statement desired. b) Let D be the intersection point of heights drawn from vertices A and C of triangle ABC. Then in the formula proved in heading a) the ﬁrst two summands are zero and, therefore, the last summand is also zero, i.e., BD ⊥AC. SOLUTIONS 297 13.13. Let us prove that AH ⊥BC. Indeed, − − → AH = − → AO+− − → OH = − → AO+− → OA+− − → OB+− → OC = − − → OB + − → OC and − − → BC = − − → BO + − → OC = −− − → OB + − → OC and, therefore, (− − → AH, − − → BC) = OC2 −OB2 = R2 −R2 = 0 because O is the center of the circumscribed circle. We similarly prove that BH ⊥AC and CH ⊥AB. 13.14. Let αi = ∠(ai, a1). Considering the projections to the straight line parallel to a1 and the straight line perpendicular to a1 we get a1 = P i>1 ai cos αi and 0 = P i>1 ai sin αi, respectively. Squaring these equalities and summing we get a2 1 = P i>1 a2 i (cos2 αi + sin2 αi) + 2 P i>j>1 aiaj(cos αi cos αj + sin αi sin αj) = a2 2 + · · · + a2 n + 2 P i>j>1 aiaj cos(αi −αj). It remains to notice that αi −αj = ∠(ai, a1) −∠(aj, a1) = ∠(ai, aj) = ϕij. 13.15. Let a = − − → AD, b = − − → BD and c = − − → CD. Since BC2 = \|b−c\|2 = BD2+CD2−2(b, c), it follows that U = 2(b, c). Similarly, V = 2(a, c) and W = 2(a, b). Let α = ∠(a, b) and β = ∠(b, c). Multiplying the equality cos2 α + cos2 β + cos2(α + β) = 2 cos α cos β cos(α + β) + 1 (cf. Problem 12.39 b)) by 4uvw = 4\|a\|2\|b\|2\|c\|2 we get the statement desired. 13.16. Fix an arbitrary point O. Let m = − − → OM, a = − → OA, . . . , d = − − → OD. Then (− − → MA, − − → MB) −(− − → MC, − − → MD) = (a −m, b −m) −(c −m, d −m) = (c + d −a −b, m) + (a, b) −(c, d). If v = c + d −a −b ̸= 0, then as the point M runs over the plane the value (v, m) attains all the real values, in particular, it takes the value (c, d) −(a, b). Hence, v = 0, i.e., − → OC + − − → OD = − → OA + − − → OB and, therefore, − → AC = − − → DB. 13.17. Let n1, . . . , nk be the unit exterior normals to the sides and let M1, . . . , Mk be arbitrary points on these sides. For any point X inside the polygon the distance from X to the i-th side is equal to (− − − → XMi, ni). Therefore, the sums of distances from the inner points A and B to the sides of the polygon are equal if and only if k X i=1 (− − → AMi, ni) = k X i=1 (− − → BMi, ni) = k X i=1 (− → BA, ni) + k X i=1 (− − → AMi, ni), i.e., (− → BA, Pk i=1 ni) = 0. Hence, the sum of distances from any inner point of the polygon to the sides is constant if and only if P ni = 0. 13.18. Let l be an arbitrary line, n the unit vector perpendicular to l. If points A and B belong to the same half-plane given by the line l the vector n belongs to, then ρ(B, l) −ρ(A, l) = (− → AB, n), where ρ(X, l) is the distance from X to l. Let n1, n2, n3 and n4 be unit vectors perpendicular to the consecutive sides of quadri-lateral ABCD and directed inwards. Denote the sum of distances from point X to the sides 298 CHAPTER 13. VECTORS of quadrilateral ABCD by P(X). Then 0 = X (B) − X (A) = (− → AB, n1 + n2 + n3 + n4). Similarly, (− − → BC, n1 + n2 + n3 + n4) = 0. Since points A, B and C do not belong to the same line, n1 + n2 + n3 + n4 = 0. It remains to make use of the result of Problem 13.5. 13.19. Let a = − → AB, b = − − → BC and c = − − → CD. Then − − → AD = a + b + c, − → AC = a + b and − − → BD = b + c. It is also clear that \|a\|2 + \|b\|2 + \|c\|2 + \|a + b + c\|2 −\|a + b\|2 −\|b + c\|2 = \|a\|2 + 2(a, c) + \|c\|2 = \|a + c\|2 ≥0. The equality is only attained if a = −c, i.e., ABCD is a parallelogram. 13.20. Consider ﬁve vectors a1, a2, a3, a4, a5 and suppose that the length of the sum of any two of them is longer than the length of the sum of the three remaining ones. Since \|a1 + a2\| > \|a3 + a4 + a5\|, it follows that \|a1\|2 + 2(a1, a2) + \|a2\|2 > \|a3\|2 + \|a4\|2 + \|a5\|2 + 2(a3, a4) + 2(a4, a5) + 2(a3, a5). Adding such inequalities for all ten pairs of vectors we get 4(\|a1\|2 + . . . ) + 2((a1, a2) + . . . ) > 6(\|a1\|2 + . . . ) + 6((a1, a2) + . . . ) i.e., \|a1 + a2 + a3 + a4 + a5\|2 < 0. Contradiction. 13.21. Denote the given vectors by e1, . . . , e10. Let − → AB = e1 + · · · + e10. Let us prove that the ray AB determines the required axis. Clearly, \|− → AB −ei\|2 = AB2 −2(− → AB, ei)+\|ei\|2, i.e., (− → AB, ei) = 1 2(AB2+\|ei\|2−\|− → AB−ei\|2). By the hypothesis AB > \|− → AB−ei\| and, therefore, (− → AB, ei) > 0, i.e., the projection of ei to AB is positive. 13.22. Let ai = − − → OAi and x = − − → OX. Then \|ai\| = R and − − → XAi = ai −x. Therefore, X XAi = X \|ai −x\| = X \|ai −x\| · \|ai\| R ≥ X ai −x, ai R = X ai, ai R −(x, P ai) R . It remains to observe that (ai, ai) = R2 and P ai = 0. 13.23. On the plane, consider four vectors (a, b), (c, d), (e, f) and (g, h). One of the angles between these vectors does not exceed 360◦ 4 = 90◦. If the angle between the vectors does not exceed 90◦, then the inner product is nonnegative. The given six numbers are inner products of all the pairs of our four vectors and, therefore, at least one of them is nonnegative. 13.24. Let us prove this statement by induction. For n = 0 the statement is obviously true. Let us assume that the statement is proved for 2n + 1 vectors. In a system of 2n + 3 vectors consider two extreme vectors (i.e., the vectors the angle between which is maximal). For deﬁniteness sake, suppose that these are vectors − − → OP1 and − − − − → OP2n+3. By the inductive hypothesis the length of − → OR = − − → OP2 + · · · + − → OP 2n+2 is not shorter than 1. The vector − → OR belongs to the interior of angle ∠P1OP2n+3 and, therefore, it forms an acute angle with the vector − → OS = − → OP 1 + − → OP 2n+3. Hence, \|− → OS + − → OR\| ≥OR ≥1. 13.25. First, let us prove that if a, b and c are vectors whose length does not exceed 1, then at least one of the vectors a ± b, a ± c, b ± c is not longer than 1. SOLUTIONS 299 Indeed, two of the vectors ±a, ±b, ±c form an angle not greater than 60◦and, therefore, the diﬀerence of these two vectors is not longer than 1 (if (??) in triangle ABC we have AB ≤1, BC ≤1 and ∠ABC ≤60◦, then AC is not the greatest side and AC ≤1). Thus, we can reduce the discussion to two vectors a and b. Then either the angle between vectors a and b or between vectors a and −b does not exceed 90◦; hence, either \|a−b\| ≤ √ 2 or \|a + b\| ≤ √ 2. 13.26. We can assume that the sum a of the given vectors is nonzero because otherwise the statement of the problem is obvious. Figure 144 (Sol. 13.26) Let us introduce a coordinate system directing Oy-axis along a. Let us enumerate the vectors of the lower half-plane clockwise: e1, e2, . . . as on Fig. 144. By the hypothesis there are not less than k of these vectors. Let us prove that among the given vectors there are also vectors v1, . . . , vk such that the second coordinate of the vector vi + ei is nonpositive for any i = 1, . . . , k. This will prove the required statement. Indeed, the length of the sum of the given vectors is equal to the sum of the second coordinates (the coordinate system was introduced just like this). The second coordinate of the sum of the vectors e1, v1, . . . , ek, vk is nonpositive and the second coordinate of any of the remaining vectors does not exceed 1. Therefore, the second coordinate of the sum of all the given vectors does not exceed n −2k. Let vectors e1, . . . , ep belong to the fourth quadrant. Let us start assigning to them the vectors v1, . . . , vp. Let us rotate the lower half plane that consists of points with nonpositive second coordinate by rotating the Ox-axis clockwise through an angle between 0◦and 90◦. If one of the two vectors that belongs to the half plane rotated this way lies in the fourth quadrant, then their sum has a nonpositive second coordinate. As the Ox-axis rotates beyond vector e1, at least one vector that belongs to the half plane should be added to the vectors e2, . . . , ek; hence, the vector which follows ek should be taken for v1. Similarly, while the Ox-axis is rotated beyond e2 we get vector v2, and so on. These arguments remain valid until the Ox-axis remains in the fourth quadrant. For the vectors ep+1, . . . , ek which belong to the third quadrant the proof is given similarly (if the ﬁrst coordinate of the vector ep+1 is zero, then we should ﬁrst disregard it; then take any of the remaining vectors for its(whose?) partner). 13.27. Point X belongs to line AB if and only if − − → AX = λ− → AB, i.e., − − → OX = − → OA + − − → AX = (1 −λ)− → OA + λ− − → OB. 13.28. Let us take an arbitrary point O and express all the selected vectors in the form − − − → AiAj = − → OAj −− → OAi. By the hypothesis every vector − → OAi enters the sum of all the chosen vectors with the “plus” sign as many times as with the “minus” sign. 300 CHAPTER 13. VECTORS 13.29. Let e1, e2 and e3 be unit vectors codirected with vectors − → OA, − − → OB and − → OC, respectively; let α = ∠BOC, β = ∠COA and γ = ∠AOB. We have to prove that e1 sin α + e2 sin β + e3 sin γ = − → 0 . Consider triangle A1B1C1 whose sides are parallel to lines OC, OA and OB. Then − → 0 = − − − → A1B1 + − − − → B1C1 + − − − → C1A1 = ±2R(e1 sin α + e2 sin β + e3 sin γ), where R is the radius of the circumscribed circle of triangle ABC. 13.30. Let a and b be unit vectors codirected with rays OA and OB, let λ = OA and µ = OB. Line AB consists of all points X such that − − → OX = t− → OA + (1 −t)− − → OB = tλa + (1 −t)µb. We have to ﬁnd numbers x0 and y0 such that x0 λ = t = 1 −y0 µ for all the considered values of λ and µ. It remains to set x0 = p c and y0 = q c. As a result we see that if p OA + q OB = c, then line AB passes through a point X such that − − → OX = pa+qb c . 13.31. Let a = − − → MA, b = − − → MB and c = − − → MC. Then e = − − → MC1 = pa + (1 −p)b and − − → MA1 = qc + (1 −q)b = −qa + (1 −2q)b. On the other hand, − − → MA1 = αe. Similarly, βe = − − → MB1 = −rb + (1 −2r)a. We have to show that 1 + 1 α + 1 β = 0. Since αpa + α(1 −p)b = αe = −qa + (1 −2q)b, it follows that αp = 1 −2r and α(1 −p) = 1 −2q and, therefore, 1 α = 1 −3p. Similarly, βp = 1 −2r and β(1 −p) = −r and, therefore, 1 β = 3p −2. 13.32. Summing up the equalities − → AA2 + − − → BB2 + − → CC2 = − → 0 and − − → A2B2 + − − → B2C2 + − − → C2A2 = − → 0 we get − → AB2 + − − → BC2 + − → CA2 = − → 0 . It follows that − → AB2 = λ− − → C2B2, − − → BC2 = λ− − → A2C2 and − → CA2 = λ− − → B2A2. Let E be a point on line BC such that A2E ∥AA1. Then − → BA1 = λ− → EA1 and − − → EC = λ− → EA1; hence, − − → A1C = − − → EC −− → EA1 = (λ −1)− → EA1. Therefore, A1C BA1 = λ−1 λ . Similarly, AB1 B1C = BC1 C1A = λ−1 λ . 13.33. Let O be the center of the inscribed circle of the given quadrilateral, a = − → OA, b = − − → OB, c = − → OC and d = − − → OD. If Ha is the orthocenter of triangle BCD, then − − → OHa = b + c + d (cf. Problem 13.13). Therefore, − − → OM a = 1 2(a + b + c + d) = − − → OM b = − − → OM c = − − → OM d. 13.34. Let O be the center of the circumscribed circle of the given quadrilateral; a = − → OA, b = − − → OB, c = − → OC and d = − − → OD. If Hd is the orthocenter of triangle ABC, then − − → OHd = a + b + c (Problem 13.13). a) Take a point K such that − − → OK = a + b + c + d. Then KHd = \|− − → OK −− − → OHd\| = \|d\| = R, i.e., K belongs to circle Sd. We similarly prove that K belongs to circles Sa, Sb and Sc. b) Let Od be the center of the circle of nine points of triangle ABC, i.e., the midpoint of OHd. Then − → OOd = − − → OHd/2 = (a+b+c)/2. Take point X such that − − → OX = (a+b+c+d)/2. Then XOd = 1 2\|d\| = 1 2R, i.e., X belongs to the circle of nine points of triangle ABC. We SOLUTIONS 301 similarly prove that X belongs to the circles of nine points of triangles BCD, CDA and DAB. 13.35. Let X1 be the projection of X on l. Vector α− → AA1+β− − → BB1+γ− → CC1 is the projection of vector α− − → AX1 + β− − → BX1 + γ− − → CX1 to a line perpendicular to l. Since α− − → AX1 + β− − → BX1 + γ− − → CX1 = α− − → AX + β− − → BX + γ− − → CX + (α + β + γ)− − → XX1 and α− − → AX + β− − → BX + γ− − → CX = − → 0 (by Problem 13.29), we get the statement required. (?)13.36. Let a = − − → A1A2 + − − → A3A4 + · · · + − − − − − − → A2n−1A2n and a ̸= 0. Introduce the coordinate system directing the Ox-axis along vector a. Since the sum of projections of vectors − − − → A1A2, − − − → A3A4, . . . , − − − − − − → A2n−1A2n on Oy is zero, it follows that the length of a is equal to the absolute value of the diﬀerence between the sum of the lengths of positive projections of these vectors to the Ox-axis and the sum of lengths of their negative projections. Therefore, the length of a does not exceed either the sum of the lengths of the positive projections or the sum of the lengths of the negative projections. It is easy to verify that the sum of the lengths of positive projections as well as the sum of the lengths of negative projections of the given vectors on any axis does not exceed the diameter of the circle, i.e., does not exceed 2. 13.37. In the proof of the equality of vectors it suﬃces to verify the equality of their projections (minding the sign) on lines BC, CA and AB. Let us carry out the proof, for example, for the projections on line BC, where the direction of ray BC will be assumed to be the positive one. Let P be the projection of point A on line BC and N the midpoint of BC. Then − − → PN = − → PC + − − → CN = b2 + a2 −c2 2a −a 2 = b2 −c2 2a (PC is found from the equation AB2 −BP 2 = AC2 −CP 2). Since NM : NA = 1 : 3, the projection of − − → MO on line BC is equal to 1 3 − − → PN = b2−c2 6a . It remains to notice that the projection of vector a3na + b3nb + c3nc on BC is equal to b3 sin γ −c3 sin β = b3c −c3b 2R = abc 2R · b2 −c2 a = 2S b2 −c2 a . 13.38. Let the inscribed circle be tangent to sides AB, BC and CA at points U, V and W, respectively. We have to prove that − → OZ = 3R r − − → ZK, i.e., − → OZ = R r (− → ZU + − → ZV + − − → ZW). Let us prove, for example, that the (oriented) projections of these vectors on line BC are equal; the direction of ray BC will be assumed to be the positive one. Let N be the projection of point O on line BC. Then the projection of vector OZ on line BC is equal to − − → NV = − − → NC + − − → CV = (a 2) −(a + b −c) 2 = (c −b) 2 . The projection of vector − → ZU + − → ZV + − − → ZW on this line is equal to the projection of vector − → ZU + − − → ZW, i.e., it is equal to −r sin V ZU + r sin V ZW = −r sin B + r sin C = r(c −b) 2R . 13.39. Introduce the coordinate system Oxy. Let lϕ be the straight line through O and constituting an angle of ϕ (0 < ϕ < π) with the Ox-axis, i.e., if point A belongs to lϕ and the second coordinate of A is positive, then ∠AOX = ϕ; in particular, l0 = lπ = Ox. 302 CHAPTER 13. VECTORS If vector a forms an angle of α with the Ox-axis (the angle is counted counterclockwise from the Ox-axis to the vector a), then the length of the projection of a on lϕ is equal to \|a\| · \| cos(ϕ −α)\|. The integral R π o \|a\| · \| cos(ϕ −α)\|dϕ = 2\|a\| does not depend on α. Let vectors a1, . . . , an; b1, . . . , bm constitute angles of α1, . . . , αn; β1, . . . , βn, respec-tively, with the Ox-axis. Then by the hypothesis \|a1\| · \| cos(ϕ −α1)\| + · · · + \|an\| · \| cos(ϕ −αn)\| ≤ \|b1\| · \| cos(ϕ −β1)\| + · · · + \|bm\| · \| cos(ϕ −βm)\| for any ϕ. Integrating these inequalities over ϕ from 0 to π we get \|a1\| + · · · + \|an\| ≤\|b1\| + · · · + \|bm\|. Remark. The value 1 b−a R b a f(x)dx is called the mean value of the function f on the segment [a, b]. The equality Z π 0 \|a\| · \| cos(ϕ −α)\|dϕ = 2\|a\| means that the mean value of the length of the projection of vector a is equal to 2 π\|a\|; more precisely, the mean value of the function f(ϕ) equal to the length of the projection of a to lϕ on the segment [0, π] is equal to 2 π\|a\|. 13.40. The sum of the lengths of the projections of a convex polygon on any line is equal to twice the length of the projection of the polygon on this line. Therefore, the sum of the lengths of the projections of vectors formed by edges on any line is not longer for the inner polygon than for the outer one. Hence, by Problem 13.39 the sum of the lengths of vectors formed by the sides, i.e., the perimeter of the inner polygon, is not longer than that of the outer one. 13.41. If the sum of the lengths of vectors is equal to L, then by Remark to Problem 13.39 the mean value of the sum of the lengths of projections of these vectors is equal to 2L/π. The value of function f on segment [a, b] cannot be always less than its mean value c because otherwise c = 1 a −b Z b a f(x)dx < (b −a)c b −a = c. Therefore, there exists a line l such that the sum of the lengths of the projections of the initial vectors on l is not shorter than 2L/π. On l, select a direction. Then either the sum of the lengths of the positive projections to this directed line or the sum of the lengths of the negative projections is not shorter than L/π. Therefore, either the length of the sum of vectors with positive projections or the length of the sum of vectors with negative porjections is not shorter than L/π. 13.42. Let AB denote the projection of the polygon on line l. Clearly, points A and B are projections of certain vertices A1 and B1 of the polygon. Therefore, A1B1 ≥AB, i.e., the length of the projection of the polygon is not longer than A1B1 and A1B1 < d by the hypothesis. Since the sum of the lengths of the projections of the sides of the polygon on l is equal to 2AB, it does not exceed 2d. The mean value of the sum of the lengths of the projections of sides is equal to 2 πP, where P is a perimeter (see Problem 13.39). The mean value does not exceed the maximal one; hence, 2 πP < 2d, i.e., P < πd. SOLUTIONS 303 13.43. By Problem 13.39 it suﬃces to prove the inequality \|a\| + \|b\| + \|c\| + \|d\| ≥\|a + d\| + \|b + d\| + \|c + d\| for the projections of the vectors on a line, i.e., we may assume that a, b, c and d are vectors parallel to one line, i.e., they are just numbers such that a + b + c + d = 0. Let us assume that d ≥0 because otherwise we can change the sign of all the numbers. We can assume that a ≤b ≤c. We have to consider three cases: 1) a, b, c ≤0; 2) a ≤0 and b, c ≥0; 3) a, b ≤0, c ≥0. All arising inequalities are quite easy to verify. In the third case we have to consider separately the subcases \|d\| ≤\|b\|, \|b\| ≤\|d\| ≤\|a\| and \|a\| ≤\|d\| (in the last subcase we have to take into account that \|d\| = \|a\| + \|b\| −\|c\| ≤\|a\| + \|b\|). 13.44. By Problem 13.39 it suﬃces to prove the inequality for the projections of vectors on any line. Let the projections of − → OA1, . . . , − → OAn on a line l be equal (up to a sign) to a1, . . . , an. Let us divide the numbers a1, . . . , an into two groups: x1 ≥x2 ≥· · · ≥xk > 0 and y′ 1 ≤y′ 2 ≤· · · ≤y′ n−k ≤0. Let yi = −y′ i. Then x1 + · · · + xk = y1 + · · · + yn−k = a and, therefore, x1 ≥a k and y1 ≥ a n−k. To the perimeter the number 2(x1 + y1) in the projection corresponds. To the sum of the vectors − → OAi the number x1 + · · · + xk + y1 + · · · + yn−k = 2a in the projection corresponds. And since 2(x1 + y1) x1 + · · · + yn−k ≥2((a/k) + (a/(n −k))) 2a = n k(n −k), it remains to notice that the quantity k(n −k) is maximal for k = n/2 if n is even and for k = (n ± 1)/2 if n is odd. 13.45. By deﬁnition the length of a curve is the limit of perimeters of the polygons inscribed in it. [Vo vvedenie] Consider an inscribed polygon with perimeter P and let the length of the projection on line l be equal to di. Let 1 −ε < di < 1 for all lines l. The polygon can be selected so that ε is however small. Since the polygon is a convex one, the sum of the lengths of the projections of its sides on l is equal to 2di. By Problem 13.39 the mean value of the quantity 2di is equal to 2 πP (cf. Problem 13.39) and, therefore, 2 −2ε < 2 πP < 2, i.e., π −πε < P < π. Tending ε to zero we see that the length of the curve is equal to π. 13.46. Let us prove that the perimeter of the convex hull of all the vertices of given polygons does not exceed the sum of their perimeters. To this end it suﬃces to notice that by the hypothesis the projections of given polygons to any line cover the projection of the convex hull. 13.47. a) If λ < 0, then (λa) ∨b = −λ\|a\| · \|b\| sin ∠(−a, b) = λ\|a\| · \|a\| sin ∠(a, b) = λ(a ∨b). For λ > 0 the proof is obvious. b) Let a = − → OA, b = − − → OB and c = − → OC. Introduce the coordinate system directing the Oy-axis along ray OA. Let A = (0, y1), B = (x2, y2) and C = (x3, y3). Then a ∨b = x2y1, a ∨c = x3y1; a ∨(b + c) = (x2 + x3)y1 = a ∨b + a ∨c. 13.48. Let e1 and e2 be unit vectors directed along the axes Ox and Oy. Then e1 ∨e2 = −e2 ∨e1 = 1 and e1 ∨e1 = e2 ∨e2 = 0; hence, a ∨b = (a1e1 + a2e2) ∨(b1e1 + b2e2) = a1b2 −a2b1. 304 CHAPTER 13. VECTORS 13.49. a) Clearly, − → AB ∨− → AC = − → AB ∨(− → AB + − − → BC) = −− → BA ∨− − → BC = − − → BC ∨− → BA. b) In the proof it suﬃces to make use of the chain of inequalities − → AB ∨− → AC = (− − → AD + − − → DB) ∨(− − → AD + − − → DC) = − − → AD ∨− − → DC + − − → DB ∨− − → AD + − − → DB ∨− − → DC = = − − → DC ∨− − → DA + − − → DA ∨− − → DB + − − → DB ∨− − → DC. 13.50. Let at the initial moment, i.e., at t = 0 we have − → AB = v and − → AC = w. Then at the moment t we get − → AB = v + t(a −b) and − → AC = w + t(c −a), where a, b and c are the velocity vectors of the runners A, B and C, respectively. Since vectors a, b and c are parallel, it follows that (b −a) ∨(c −a) = 0 and, therefore, \|S(A, B, C)\| = 1 2\|− → AB ∨− → AC\| = \|x + ty\|, where x and y are some constants. Solving the system \|x\| = 2, \|x + 5y\| = 3 we get two solutions with the help of which we express the dependence of the area of triangle ABC of time t as \|2 + t 5\| or \|2 −t\|. Therefore, at t = 10 the value of the area can be either 4 or 8. 13.51. Let v(t) and w(t) be the vectors directed from the ﬁrst pedestrian to the second and the third ones, respectively, at time t. Clearly, v(t) = ta + b and w(t) = tc + d. The pedestrians are on the same line if and only if v(t) ∥w(t), i.e., v(t)∨w(t) = 0. The function f(t) = v(t) ∨w(t) = t2a ∨c + t(a ∨d + b ∨c) + b ∨d is a quadratic and f(0) ̸= 0. We know that a quadratic not identically equal to zero has not more than 2 roots. 13.52. Let − → OC = a, − − → OB = λa, − − → OD = b and − → OA = µb. Then ±2SOPQ = − → OP ∨− → OQ = a + µb 2 ∨λa + b 2 = 1 −λµ 4 (a ∨b) and ±SABCD = ±2(SCOD −SAOB) = ±(a ∨b −λa ∨µb) = ±(1 −λµ)a ∨b. 13.53. Let aj = − − → P1Aj. Then the doubled sum of the areas of the given triangles is equal for any inner point P to (x + a1) ∨(x + a2) + (x + a3) ∨(x + a4) + · · · + (x + a2n−1) ∨(x + a2n), where x = − → PP 1 and it diﬀers from the doubled sum of the areas of these triangles for point P1 by x ∨(a1 −a2 + a3 −a4 + · · · + a2n−1 −a2n) = x ∨a. By the hypothesis x ∨a = 0 for x = − − → P1P 1 and x = − − → P3P 1 and these vectors are not parallel. Hence, a = 0, i.e., x ∨a = 0 for any x. 13.54. Let a = − → AP, b = − − → BQ and c = − → CR. Then − → QC = αa, − → RA = βb and − − → PB = γc; we additionally have (1 + α)a + (1 + β)b + (1 + γ)c = 0. It suﬃces to verify that − → AB ∨− → CA = − → PQ ∨− → RP. The diﬀerence between these quantities is equal to (a + γc) ∨(c + βb) −(γc + b) ∨(a + βb) = a ∨c + βa ∨b + a ∨b + γa ∨c = = a ∨[(1 + γ)c + (1 + β)b] = −a ∨(1 + α)a = 0. SOLUTIONS 305 13.55. Let ai = − − → A4Ai and wi = − − → A4Hi. By Problem 13.49 b) it suﬃces to verify that a1 ∨a2 + a2 ∨a3 + a3 ∨a1 = w1 ∨w2 + w2 ∨w3 + w3 ∨w1. Vectors a1 −w2 and a2 −w1 are perpendicular to vector a3 and, therefore, they are parallel to each other, i.e., (a1 −w2) ∨(a2 −w1) = 0. Adding this equality to the equalities (a2 −w3) ∨(a3 −w2) = 0 and (a3 −w1) ∨(a1 −w3) = 0 we get the statement required. 13.56. Let x = x1e1 + x2e2. Then e1 ∨x = x2(e1 ∨e2) and x ∨e2 = x1(e1 ∨e2), i.e., x = (x ∨e2)e1 + (e1 ∨x)e2 e1 ∨e2 . Multiplying this expression by (e1 ∨e2)y from the right we get (1) (x ∨e2)(e1 ∨y) + (e1 ∨x)(e2 ∨y) + (e2 ∨e1)(x ∨y) = 0. Let e1 = − → AB, e2 = − → AC, x = − − → AD and y = − → AE. Then S = a + x ∨e2 + d = c + y ∨e2 + a = d + x ∨e1 + b, i.e., x ∨e2 = S −a −d, y ∨e2 = S −c −a and x∨e1 = S −d−b. Substituting these expressions into (1) we get the statement required. Chapter 14. THE CENTER OF MASS Background 1. Consider a system of mass points on a plane, i.e., there is a set of pairs (Xi, mi), where Xi is a point on the plane and mi a positive number. The center of mass of the system of points X1, . . . , Xn with masses m1, . . . , mn, respectively, is a point, O, which satisﬁes m1 − − → OX1 + · · · + mn − − → OXn = − → 0 . The center of mass of any system of points exists and is unique (Problem 14.1). 2. A careful study of the solution of Problem 14.1 reveals that the positivity of the numbers mi is not actually used; it is only important that their sum is nonzero. Sometimes it is convenient to consider systems of points for which certain masses are positive and certain are negative (but the sum of masses is nonzero). 3. The most important property of the center of mass which lies in the base of almost all its applications is the following Theorem on mass regroupping. The center of mass of a system of points does not change if part of the points are replaced by one point situated in their center of mass and whose mass is equal to the sum of their masses (Problem 14.2). 4. The moment of inertia of a system of points X1, . . . , Xn with masses m1, . . . , mn with respect to point M is the number IM = m1MX2 1 + · · · + mnMX2 n. The applications of this notion in geometry are based on the relation IM = IO + mOM 2, where O is the center of mass of a system and m = m1 + · · · + mn (Problem 14.17). §1. Main properties of the center of mass 14.1. a) Prove that the center of mass exists and is unique for any system of points. b) Prove that if X is an arbitrary point and O the center of mass of points X1, . . . , Xn with masses m1, . . . , mn, then − − → XO = 1 m1 + · · · + mn (m1 − − → XX1 + · · · + mn − − − → XXn). 14.2. Prove that the center of mass of the system of points X1, . . . , Xn, Y1, . . . , Ym with masses a1, . . . , an, b1, . . . , bm coincides with the center of mass of two points — the center of mass X of the ﬁrst system with mass a1 + · · · + an and the center of mass Y of the second system with mass b1 + · · · + bm. 14.3. Prove that the center of mass of points A and B with masses a and b belongs to segment AB and divides it in the ratio of b : a. 307 308 CHAPTER 14. THE CENTER OF MASS §2. A theorem on mass regroupping 14.4. Prove that the medians of triangle ABC intersect at one point and are divided by it in the ratio of 2 : 1 counting from the vertices. 14.5. Let ABCD be a convex quadrilateral; let K, L, M and N be the midpoints of sides AB, BC, CD and DA, respectively. Prove that the intersection point of segments KM and LN is the midpoint of these segments and also the midpoint of the segment that connects the midpoints of the diagonals. 14.6. Let A1, B1, . . . , F1 be the midpoints of sides AB, BC, . . . , FA, respectively, of a hexagon. Prove that the intersection points of the medians of triangles A1C1E1 and B1D1F1 coincide. 14.7. Prove Ceva’s theorem (Problem 4.48 b)) with the help of mass regrouping. 14.8. On sides AB, BC, CD and DA of convex quadrilateral ABCD points K, L, M and N, respectively, are taken so that AK : KB = DM : MC = α and BL : LC = AN : ND = β. Let P be the intersection point of segments KL and LN. Prove that NP : PL = α and KP : PM = β. 14.9. Inside triangle ABC ﬁnd point O such that for any straight line through O, intersecting AB at K and intersecting BC at L the equality p AK KB + q CL LB = 1 holds, where p and q are given positive numbers. 14.10. Three ﬂies of equal mass crawl along the sides of triangle ABC so that the center of their mass is ﬁxed. Prove that the center of their mass coincides with the intersection point of medians of ABC if it is known that one ﬂy had crawled along the whole boundary of the triangle. 14.11. On sides AB, BC and CA of triangle ABC, points C1, A1 and B1, respectively, are taken so that straight lines CC1, AA1 and BB1 intersect at point O. Prove that a) CO OC1 = CA1 A1B + CB1 B1A; b) AO OA1 · BO OB1 · CO OC1 = AO OA1 + BO OB1 + CO OC1 + 2 ≥8. 14.12. On sides BC, CA and AB of triangle ABC points A1, B1 and C1, respectively, are taken so that BA1 A1C = CB1 B1A = AC1 C1B. Prove that the centers of mass of triangles ABC and A1B1C1 coincide. 14.13. On a circle, n points are given. Through the center of mass of n −2 points a straight line is drawn perpendicularly to the chord that connects the two remaining points. Prove that all such straight lines intersect at one point. 14.14. On sides BC, CA and AB of triangle ABC points A1, B1 and C1, respectively, are taken so that segments AA1, BB1 and CC1 intersect at point P. Let la, lb, lc be the lines that connect the midpoints of segments BC and B1C1, CA and C1A1, AB and A1B1, respectively. Prove that lines la, lb and lc intersect at one point and this point belongs to segment PM, where M is the center of mass of triangle ABC. 14.15. On sides BC, CA and AB of triangle ABC points A1, B1 and C1, respectively, are taken; straight lines B1C1, BB1 and CC1 intersect straight line AA1 at points M, P and Q, respectively. Prove that: a) A1M MA = A1P PA + A1Q QA ; b) if P = Q, then MC1 : MB1 = BC1 AB : CB1 AC . 14.16. On line AB points P and P1 are taken and on line AC points Q and Q1 are taken. The line that connects point A with the intersection point of lines PQ and P1Q1 §3. THE MOMENT OF INERTIA 309 intersects line BC at point D. Prove that BD CD = BP PA −BP1 P1A CQ QA −CQ1 Q1A . §3. The moment of inertia For point M and a system of mass points X1, . . . , Xn with masses m1, . . . , mn the quantity IM = m1MX2 1 + · · · + mnMX2 n is called the moment of inertia with respect to M. 14.17. Let O be the center of mass of a system of points whose sum of masses is equal to m. Prove that the moments of inertia of this system with respect to O and with respect to an arbitrary point X are related as follows: IX = IO + mXO2. 14.18. a) Prove that the moment of inertia with respect to the center of mass of a system of points of unit masses is equal to 1 n P i<j a2 ij, where n is the number of points and aij the distance between points whose indices are i and j. b) Prove that the moment of inertia with respect to the center of mass of a system of points whose masses are m1, . . . , mn is equal to 1 m P i<j mimja2 ij, where m = m1 + · · · + mn and aij is the distance between the points whose indices are i and j. 14.19. a) Triangle ABC is an equilateral one. Find the locus of points X such that AX2 = BX2 + CX2. b) Prove that for the points of the locus described in heading a) the pedal triangle with respect to the triangle ABC is a right one. 14.20. Let O be the center of the circumscribed circle of triangle ABC and H the intersection point of the heights of triangle ABC. Prove that a2 + b2 + c2 = 9R2 −OH2. 14.21. Chords AA1, BB1 and CC1 in a disc with center O intersect at point X. Prove that AX XA1 + BX XB1 + CX XC1 = 3 if and only if point X belongs to the circle with diameter OM, where M is the center of mass of triangle ABC. 14.22. On sides AB, BC, CA of triangle ABC pairs of points A1 and B2, B1 and C2, C1 and A2, respectively, are taken so that segments A1A2, B1B2 and C1C2 are parallel to the sides of triangle ABC and intersect at point P. Prove that PA1 · PA2 + PB1 · PB2 + PC1 · PC2 = R2 −OP 2, where O is the center of the circumscribed circle. 14.23. Inside a circle of radius R, consider n points. Prove that the sum of squares of the pairwise distances between the points does not exceed n2R2. 14.24. Inside triangle ABC point P is taken. Let da, db and dc be the distances from P to the sides of the triangle; Ra, Rb and Rc the distances from P to the vertices. Prove that 3(d2 a + d2 b + d2 c) ≥(Ra sin A)2 + (Rb sin B)2 + (Rc sin C)2. 14.25. Points A1, . . . , An belong to the same circle and M is their center of mass. Lines MA1, . . . , MAn intersect this circle at points B1, . . . , Bn (distinct from A1, . . . , An). Prove that MA1 + · · · + MAn ≤MB1 + · · · + MBn. 310 CHAPTER 14. THE CENTER OF MASS §4. Miscellaneous problems 14.26. Prove that if a polygon has several axes of symmetry, then all of them intersect at one point. 14.27. A centrally symmetric ﬁgure on a graph paper consists of n “corners” and k rectangles of size 1 × 4 depicted on Fig. 145. Prove that n is even. Figure 145 (14.27) 14.28. Solve Problem 13.44 making use the properties of the center of mass. 14.29. On sides BC and CD of parallelogram ABCD points K and L, respectively, are taken so that BK : KC = CL : LD. Prove that the center of mass of triangle AKL belongs to diagonal BD. §5. The barycentric coordinates Consider triangle A1A2A3 whose vertices are mass points with masses m1, m2 and m3, respectively. If point X is the center of mass of the triangle’s vertices, then the triple (m1 : m2 : m3) is called the barycentric coordinates of point X with respect to triangle A1A2A3. 14.30. Consider triangle A1A2A3. Prove that a) any point X has some barycentric coordinates with respect to △A1A2A3; b) provided m1 + m2 + m3 = 1 the barycentric coordinates of X are uniquely deﬁned. 14.31. Prove that the barycentric coordinates with respect to △ABC of point X which belongs to the interior of ABC are equal to (SBCX : SCAX : SABX). 14.32. Point X belongs to the interior of triangle ABC. The straight lines through X parallel to AC and BC intersect AB at points K and L, respectively. Prove that the barycentric coordinates of X with respect to △ABC are equal to (BL : AK : LK). 14.33. Consider △ABC. Find the barycentric coordinates with respect to △ABC of a) the center of the circumscribed circle; b) the center of the inscribed circle; c) the orthocenter of the triangle. 14.34. The baricentric coordinates of point X with respect to △ABC are (α : β : γ), where α + β + γ = 1. Prove that − − → XA = β− → BA + γ− → CA. 14.35. Let (α : β : γ) be the barycentric coordinates of point X with respect to △ABC and α + β + γ = 1 and let M be the center of mass of triangle ABC. Prove that 3− − → XM = (α −β)− → AB + (β −γ)− − → BC + (γ −α)− → CA. 14.36. Let M be the center of mass of triangle ABC and X an arbitrary point. On lines BC, CA and AB points A1, B1 and C1, respectively, are taken so that A1X ∥AM, B1X ∥BM and C1X ∥CM. Prove that the center of mass M1 of triangle A1B1C1 coincides with the midpoint of segment MX. 14.37. Find an equation of the circumscribed circle of triangle A1A2A3 (kto sut’ indexy? iz 14.36?) in the barycentric coordinates. SOLUTIONS 311 14.38. a) Prove that the points whose barycentric coordinates with respect to △ABC are (α : β : γ) and (α−1 : β−1 : γ−1) are isotomically conjugate with respect to triangle ABC. b) The lengths of the sides of triangle ABC are equal to a, b and c. Prove that the points whose barycentric coordinates with respect to △ABC are (α : β : γ) and (a2 α : b2 β : c2 γ ) are isogonally conjugate with respect to ABC. Solutions 14.1. Let X and O be arbitrary points. Then m1 − − → OX1 + · · · + mn − − → OXn = (m1 + · · · + mn)− − → OX + m1 − − → XX1 + · · · + mn − − − → XXn and, therefore, O is the center of mass of the given system of points if and only if (m1 + · · · + mn)− − → OX + m1 − − → XX1 + · · · + Mn − − − → XXn = − → 0 , i.e., − − → OX = 1 m1+···+mn(m1 − − → XX1 + · · · + mn − − − → XXn). This argument gives a solution to the problems of both headings. 14.2. Let Z be an arbitrary point; a = a1 + · · · + an and b = b1 + · · · + bm. Then − − → ZX = a1 − − → ZX1+···+an − − − → ZXn a and − → ZY = b1 − − → ZY1+···+bm − − − → ZYm b . If O is the center of mass of point X whose mass is a and of point Y whose mass is b, then − → ZO = a− − → ZX + b− → ZY a + b = a1 − − → ZX1 + · · · + an − − → ZXn + b1 − − → ZY1 + · · · + bm − − → ZYm a + b , i.e., O is the center of mass of the system of points X1, . . . , Xn and Y1, . . . , Ym with masses a1, . . . , an, b1, . . . , bm. 14.3. Let O be the center of mass of the given system. Then a− → OA + b− − → OB = − → 0 and, therefore, O belongs to segment AB and aOA = bOB, i.e., AO : OB = b : a. 14.4. Let us place unit masses at points A, B and C. Let O be the center of mass of this system of points. Point O is also the center of mass of points A of mass 1 and A1 of mass 2, where A1 is the center of mass of points B and C of unit mass, i.e., A1 is the midpoint of segment BC. Therefore, O belongs to median AA1 and divides it in the ratio AO : OA1 = 2 : 1. We similarly prove that the remaining medians pass through O and are divided by it in the ratio of 2 : 1. 14.5. Let us place unit masses in the vertices of quadrilateral ABCD. Let O be the center of mass of this system of points. It suﬃces to prove that O is the midpoint of segments KM and LN and the midpoint of the segment connecting the midpoints of the diagonals. Clearly, K is the center of mass of points A and B while M is the center of mass of points C and D. Therefore, O is the center of mass of points K and M of mass 2, i.e., O is the center of mass of segment KM. Similarly, O is the midpoint of segment LN. Considering centers of mass of pairs of points (A, C) and (B, D) (i.e., the midpoints of diagonals) we see that O is the midpoint of the segment connecting the midpoints of diagonals. 312 CHAPTER 14. THE CENTER OF MASS 14.6. Let us place unit masses in the vertices of the hexagon; let O be the center of mass of the obtained system of points. Since points A1, C1 and E1 are the centers of mass of pairs of points (A, B), (C, D) and (E, F), respectively, point O is the center of mass of the system of points A1, C1 and E1 of mass 2, i.e., O is the intersection point of the medians of triangle A1C1E1 (cf. the solution of Problem 14.4). We similarly prove that O is the intersection point of medians of triangle B1D1F1. 14.7. Let lines AA1 and CC1 intersect at O and let AC1 : C1B = p and BA1 : A1C = q. We have to prove that line BB1 passes through O if and only if CB1 : B1A = 1 : pq. Place masses 1, p and pq at points A, B and C, respectively. Then point C1 is the center of mass of points A and B and point A1 is the center of mass of points B and C. Therefore, the center of mass of points A, B and C with given masses is the intersection point O of lines CC1 and AA1. On the other hand, O belongs to the segment which connects B with the center of mass of points A and C. If B1 is the center of mass of points A and C of masses 1 and pq, respectively, then AB1 : B1C = pq : 1. It remains to notice that there is one point on segment AC which divides it in the given ratio AB1 : B1C. 14.8. Let us place masses 1, α, αβ and β at points A, B, C and D, respectively. Then points K, L, M and N are the centers of mass of the pairs of points (A, B), (B, C), (C, D) and (D, A), respectively. Let O be the center of mass of points A, B, C and D of indicated mass. Then O belongs to segment NL and NO : OL = (αβ + α) : (1 + β) = α. Point O belongs to the segment KM and KO : OM = (β + αβ) : (1 + α) = β. Therefore, O is the intersection point of segments KM and LN, i.e., O = P and NP : PL = NO : OL = α, KP : PM = β. 14.9. Let us place masses p, 1 and q in vertices A, B and C, respectively. Let O be the center of mass of this system of points. Let us consider a point of mass 1 as two coinciding points of mass xa and xc, where xa + xc = 1. Let K be the center of mass of points A and B of mass p and xa and L the center of mass of points C and B of mass q and xc, respectively. Then AK : KB = xa : p and CL : LB = xc : q, whereas point O which is the center of mass of points K and L of mass p + xa and q + xc, respectively, belongs to line KL. By varying xa from 0 to 1 we get two straight lines passing through O and intersecting sides AB and BC. Therefore, for all these lines we have pAK KB + qCL LB = xa + xc = 1. 14.10. Denote the center of mass of the ﬂies by O. Let one ﬂy be sited in vertex A and let A1 be the center of mass of the two other ﬂies. Clearly, point A1 lies inside triangle ABC and point O belongs to segment AA1 and divides it in the ratio of AO : OA1 = 2 : 1. Therefore, point O belongs to the interior of the triangle obtained from triangle ABC by a homothety with coeﬃcient 2 3 and center A. Considering such triangles for all the three vertices of triangle ABC we see that their unique common point is the intersection point of the medians of triangle ABC. Since one ﬂy visited all the three vertices of the triangle ABC and point O was ﬁxed during this, O should belong to all these three small triangles, i.e., O coincides with the intersection point of the medians of triangle ABC. 14.11. a) Let AB1 : B1C = 1 : p and BA1 : A1C = 1 : q. Let us place masses p, q, 1 at points A, B, C, respectively. Then points A1 and B1 are the centers of mass of the pairs of points (B, C) and (A, C), respectively. Therefore, the center of mass of the system of points A, B and C belongs both to segment AA1 and to segment BB1, i.e., coincides with O. It SOLUTIONS 313 follows that C1 is the center of mass of points A and B. Therefore, CO OC1 = p + q = CB1 B1A + CA1 A1B . b) By heading a) we have AO OA1 · BO OB1 · CO OC1 = 1 + q p · 1 + p q · p + q 1 = p + q + p q + q p + 1 p + 1 q + 2 = AO OA1 + BO OB1 + CO OC1 + 2. It is also clear that p + 1 p ≥2, q + 1 q ≥2 and p q + q p ≥2. 14.12. Let M be the center of mass of triangle ABC. Then − − → MA + − − → MB + − − → MC = − → 0 . Moreover, − − → AB1 + − − → BC1 + − − → CA1 = k(− → AC + − → BA + − − → CB) = − → 0 . Adding these identities we get − − − → MB1 + − − − → MC1 + − − − → MA1 = − → 0 , i.e., M is the center of mass of triangle A1B1C1. Remark. We similarly prove a similar statement for an arbitrary n-gon. 14.13. Let M1 be the center of mass of n −2 points; K the midpoint of the chord connecting the two remaining points, O the center of the circle, and M the center of mass of all the given points. If line OM intersects a(?) line drawn through M1 at point P, then OM MP = KM MM1 = n −2 2 and, therefore, the position of point P is uniquely determined by the position of points O and M (if M = O, then P = O). 14.14. Let P be the center of mass of points A, B and C of masses a, b and c, respectively, M the center of mass of points A, B and C (the mass of M is a + b + c) and Q the center of mass of the union of these two systems of points. The midpoint of segment AB is the center of mass of points A, B and C of mass a + b + c −ab c , a + b + c −ab c and 0, respectively, and the midpoint of segment A1B1 is the center of mass of points A, B and C of mass a(b+c) c , b(a+c) c and (b + c) + (a + c), respectively. Point O is the center of mass of the union of these systems of points. 14.15. a) Place masses β, γ and b + c in points B, C and A so that CA1 : BA1 = β : γ, BC1 : AC1 = b : β and AB1 : CB1 = γ : c. Then M is the center of mass of this system and, therefore, A1M AM = b+c β+γ. Point P is the center of mass of points A, B and C of masses c, β and γ and, therefore, A1P PA = c β+γ. Similarly, A1Q AQ = b b+γ. b) As in heading a), we get MC1 MB1 = c+γ b+β, BC1 AB = b b+β and AC CB1 = c+γ c . Moreover, b = c because straight lines AA1, BB1 and CC1 intersect at one point (cf. Problem 14.7). 14.16. The intersection point of lines PQ and P1Q1 is the center of mass of points A, B and C of masses a, b and c and P is the center of mass of points A and B of masses a −x 314 CHAPTER 14. THE CENTER OF MASS and b while Q is the center of mass of points A and C of masses x and c. Let p = BP PA = a−x b and q = CQ QA = x c. Then pb + qc = a. Similarly, p1b + q1c = a. It follows that BD CD = −c b = (p −p1) (q −q1). 14.17. Let us enumerate the points of the given system. Let xi be the vector with the beginning at O and the end at the point of index i and of mass mi. Then P mixi = 0. Further, let a = − − → OX. Then IO = P m2 ii, IM = P mi(xi + a)2 = P mix2 i + 2(P mixi, a) + P mia2 = IO + ma2. 14.18. a) Let xi be the vector with the beginning at the center of mass O and the end at the point of index i. Then X i,j (xi −xj)2 = X i,j (x2 i + x2 j) −2 X i,j (xi, xj), where the sum runs over all the possible pairs of indices. Clearly, X i,j (x2 i + x2 j) = 2n X i x2 i = 2nIO; X i,j (xi, xj) = X i (xi, X j xj) = 0. Therefore, 2nIO = P i,j(xi −xj)2 = 2 P i<j a2 ij. b) Let xi be the vector with the beginning at the center of mass O and the end at the point with index i. Then X i,j mimj(xi −xj)2 = X i,j mimj(x2 i + x2 j) −2 X i,j mimj(xi, xj). It is clear that X i,j mimj(x2 i + x2 j) = X i mi X j (mjx2 i + mjx2 j) = X i mi(mx2 i + IO) = 2mIO and X i,j mimj(xi, xj) = X i mi(xi, X j mjxj) = 0. Therefore, 2mIO = X i,j mimj(xi −xj)2 = 2 X i<j mimja2 ij. 14.19. a) Let M be the point symmetric to A through line BC. Then M is the center of mass of points A, B and C whose masses are −1, 1 and 1, respectively, and, therefore, −AX2 + BX2 + CX2 = IX = IM + (−1 + 1 + 1)MX2 = (−3 + 1 + 1)a2 + MX2, where a is the length of the side of triangle ABC. As a result we see that the locus to be found is the circle of radius a with the center at M. b) Let A′, B′ and C′ be the projections of point X to lines BC, CA and AB, respec-tively. Points B′ and C′ belong to the circle with diameter AX and, therefore, B′C′ = AX sin B′AC′ = √ 3 2 AX. Similarly, C′A′ = √ 3 2 BX and A′B′ = √ 3 2 CX. Therefore, if AX2 = BX2 + CX2, then ∠B′A′C′ = 90◦. SOLUTIONS 315 14.20. Let M be the center of mass of the vertices of triangle ABC with unit masses in them. Then IO = IM + 3MO2 = 1 3(a2 + b2 + c2) + 3MO2 (cf. Problems 14.17 and 14.18 a)). Since OA = OB = OC = R, it follows that IO = 3R2. It remains to notice that OH = 3OM (Problem 5.105). 14.21. It is clear that AX XA1 = AX2 AX · XA1 = AX2 R2 −OX2. Therefore, we have to verify that AX2 + BX2 + CX2 = 3(R2 −OX2) if and only if OM 2 = OX2 + MX2. To this end it suﬃces to notice that AX2 + BX2 + CX2 = IX = IM + 3MX2 = IO −3MO2 + 3MX2 = 3(R2 −MO2 + MX2). 14.22. Let P be the center of mass of points A, B and C whose masses are α, β and γ, respectively. We may assume that α + β + γ = 1. If K is the intersection point of lines CP and AB, then BC PA1 = CK PK = CP + PK PK = 1 + CP PK = 1 + α + β γ = 1 γ . Similar arguments show that the considered quantity is equal to βγa2+γαb2+αβc2 = IP (cf. Problem 14.18 b)). Since IO = αR2 +βR2 +γR2 = R2, we have IP = IO −OP 2 = R2 −OP 2. 14.23. Let us place unit masses in the given points. As follows from the result of Problem 14.18 a) the sum of squared distances between the given points is equal to nI, where I is the moment of inertia of the system of points with respect to its center of mass. Now, consider the moment of inertia of the system with respect to the center O of the circle. On the one hand, I ≤IO (see Problem 14.17). On the other hand, since the distance from O to any of the given points does not exceed R, it follows that IO ≤nR2. Therefore, nI ≤n2R2 and the equality is attained only if I = IO (i.e., when the center of mass coincides with the center of the circle) and IO = nR2 (i.e., all the points lie on the given circle). 14.24. Let A1, B1 and C1 be projections of point P to sides BC, CA and AB, respec-tively; let M be the center of mass of triangle A1B1C1. Then 3(d2 a + d2 b + d2 c) = 3IP ≥ 3IM = A1B2 1 + B1C2 1 + C1A2 1 = (Rc sin C)2 + (Ra sin A)2 + (Rb sin B)2 because, for example, segment A1B1 is a chord of the circle with diameter CP. 14.25. Let O be the center of the given circle. If chord AB passes through M, then AM · BM = R2 −d2, where d = MO. Denote by IX the moment of inertia of the system of points A1, . . . , An with respect to X. Then IO = IM + nd2 (see Problem 14.17). On the other hand, since OAi = R, we deduce that IO = nR2. Therefore, AiM · BiM = R2 −d2 = 1 n(A1M 2 + · · · + AnM 2). Set ai = AiM. Then the inequality to be proved takes the form a1 + · · · + an ≤1 n(a2 1 + · · · + a2 n)( 1 a1 + · · · + 1 an ). 316 CHAPTER 14. THE CENTER OF MASS To prove this inequality we have to make use of the inequality x + y ≤(x2 y ) + (y2 x ) which is obtained from the inequality xy ≤x2 −xy + y2 by multiplying both of its sides by x+y xy . 14.26. Let us place unit masses in the vertices of the polygon. Under the symmetry through a line this system of points turns into itself and, therefore, its center of mass also turns into itself. It follows that all the axes of symmetry pass through the center of mass of the vertices. 14.27. Let us place unit masses in the centers of the cells which form “corners” and rectangles. Let us split each initial small cell of the graph paper into four smaller cells getting as a result a new graph paper. It is easy to verify that now the center of mass of a corner belongs to the center of a new small cell and the center of mass of a rectangle is a vertex of a new small cell, cf. Fig. 146. Figure 146 (Sol. 14.27) It is clear that the center of mass of a ﬁgure coincides with its center of symmetry and the center of symmetry of the ﬁgure consisting of the initial cells can only be situated in a vertex of a new cell. Since the masses of corners and bars (rectangles) are equal, the sum of vectors with the source in the center of mass of a ﬁgure and the targets in the centers of mass of all the corners and bars is equal to zero. If the number of corners had been odd, then the sum of the vectors would have had half integer coordinates and would have been nonzero. Therefore, the number of corners is an even one. 14.28. Let us place unit masses in the vertices of the polygon A1 . . . An. Then O is the center of mass of the given system of points. Therefore, − − → AiO = 1 n(− − − → AiA1 + · · · + − − − → AiAn) and AiO ≤1 n(AiA1 + · · · + AiAn); it follows that d = A1O + · · · + AnO ≤1 n n X i,j=1 AiAj. We can express the number n either in the form n = 2m or in the form n = 2m + 1. Let P be the perimeter of the polygon. It is clear that A1A2 + · · · + AnA1 = P, A1A3 + A2A4 + · · · + AnA2 ≤2P, . . . . . . . . . . . . . . . . . . . . . . . . . . . A1Am+1 + A2Am+2 + · · · + AnAm ≤mP SOLUTIONS 317 and in the left-hand sides of these inequalities all the sides and diagonals are encountered. Since they enter the sum Pn i,j=1 AiAj twice, it is clear that d ≤1 n n X i,j=1 AiAj ≤2 n(P + 2P + · · · + mP) = m(m + 1) n P. For n even this inequality can be strengthened due to the fact that in this case every diagonal occuring in the sum A1Am+1 + · · · + AnAm+n is counted twice, i.e., instead of mP we can take m 2 P. This means that for n even we have d ≤2 n(P + 2P + · · · + (m −1)P + m 2 P) = m2 n P. Thus, we have d ≤ ½ m2 n P = n 4P if n is even m(m+1) n P = n2−1 4n P if n is odd. 14.29. Let k = BK BC = 1 −DL DC. Under the projection to a line perpendicular to diagonal BD points A, B, K and L pass into points A′, B′, K′ and L′, respectively, such that B′K′ + B′L′ = kA′B′ + (1 −k)A′B′ = A′B′. It follows that the center of mass of points A′, K′ and L′ coincides with B′. It remains to notice that under the projection a center of mass turns into a center of mass. 14.30. Introduce the following notations: e1 = − − − → A3A1, e2 = − − − → A3A2 and x = − − → XA3. Point X is the center of mass of the vertices of triangle A1A2A3 with masses m1, m2, m3 attached to them if and only if m1(x + e1) + m2(x + e2) + m3x = 0, i.e., mx = −(m1e1+m2e2), where m = m1+m2+m3. Let us assume that m = 1. Any vector x on the plane can be represented in the form x = −m1e1 −m2e2, where the numbers m1 and m2 are uniquely deﬁned. The number m3 is found from the relation m3 = 1 −m1 −m2. 14.31. This problem is a reformulation of Problem 13.29. Remark. If we assume that the areas of triangles BCX, CAX and ABX are oriented, then the statement of the problem remains true for all the points situated outside the triangle as well. 14.32. Under the projection to line AB parallel to line BC vector u = − − → XA · BL + − − → XB · AK + − − → XC · LK turns into vector − → LA · BL + − → LB · AK + − → LB · LK. The latter vector is the zero one since − → LA = − − → LK + − − → KA. Considering the projection to line AB parallel to line AC we get u = 0. 14.33. Making use of the result of Problem 14.31 it is easy to verify that the answer is as follows: a) (sin 2α : sin 2β : sin 2γ); b) (a : b : c); c) (tan α : tan β : tan γ). 14.34. Adding vector (β +γ)− − → XA to both sides of the equality α− − → XA+β− − → XB +γ− − → XC = − → 0 we get − − → XA = (β + γ)− − → XA + β− − → BX + γ− − → CX = β− → BA + γ− → CA. 14.35. By Problem 14.1 b) we have 3− − → XM = − − → XA + − − → XB + − − → XC. Moreover, − − → XA = β− → BA + γ− → CA, − − → XB = α− → AB + γ− − → CB and − − → XC = α− → AC + β− − → BC (see Problem 14.34). 14.36. Let the lines through point X parallel to AC and BC intersect the line AB at points K and L, respectively. If (α : β : γ) are the barycentric coordinates of X and α + β + γ = 1, then 2− − → XC1 = − − → XK + − − → XL = γ− → CA + γ− − → CB 318 CHAPTER 14. THE CENTER OF MASS (see the solution of Problem 14.42). Therefore, 3− − − → XM1 = − − → XA1 + − − → XB1 + − − → XC1 = 1 2(α(− → AB + − → AC) + β(− → BA + − − → BC) + γ− → CA + − − → CB) = 3 2 − − → XM (see Problem 14.35). 14.37. Let X be an arbitrary point, O the center of the circumscribed circle of the given triangle, ei = − − → OAi and a = − − → XO. If the barycentric coordinates of X are (x1 : x2 : x3), then P xi(a + ei) = P xi − − → XAi = 0 because X is the center of mass of points A1, A2, A3 with masses x1, x2, x3. Therefore, (P xi)a = −P xiei. Point X belongs to the circumscribed circle of the triangle if and only if \|a\| = XO = R, where R is the radius of this circle. Thus, the circumscribed circle of the triangle is given in the barycentric coordinates by the equation R2( X xi)2 = ( X xiei)2, i.e., R 2 X x2 i + 2R2 X i<j xixj = R2 X x2 i + 2 X i<j xixj(ei, ej) because \|ei\| = R. This equation can be rewritten in the form X i<j xixj(R2 −(ei, ej)) = 0. Now notice that 2(R2 −(ei, ej)) = a2 ij, where aij is the length of side AiAj. Indeed, a2 ij = \|ei −ej\|2 = \|ei\|2 + \|ej\|2 −2(ei, ej) = 2(R2 −(ei, ej)). As a result we see that the circumscribed circle of triangle A1A2A3 is given in the barycentric coordinates by the equation P i<j xixjaij = 0, where aij is the length of side AiAj. 14.38. a) Let X and Y be the points with barycentric coordinates (α : β : γ) and (α−1 : β−1 : γ−1) and let lines CX and CY intersect line AB at points X1 and Y1, respectively. Then AX1 : BX1 = β : α = α−1 : β−1 = BY1 : AY1. Similar arguments for lines AX and BX show that points X and Y are isotomically conjugate with respect to triangle ABC. b) Let X be the point with barycentric coordinates (α : β : γ). We may assume that α + β + γ = 1. Then by Problem 14.34 we have − − → AX = β− → AB + γ− → AC = βc( − → AB c ) + γb( − → AC b ). Let Y be the point symmetric to X through the bisector of angle ∠A and (α′ : β′ : γ′) the barycentric coordinates of Y . It suﬃces to verify that β′ : γ′ = b2 β : c2 γ . The symmetry through the bisector of angle ∠A interchanges unit vectors − → AB c and − → AC b , consequently, − → AY = βc − → AC b + γb − → AB c . It follows that β′ : γ′ = γb c : βcb = b2 β : c2 γ . Chapter 15. PARALLEL TRANSLATIONS Background 1. The parallel translation by vector − → AB is the transformation which sends point X into point X′ such that − − → XX′ = − → AB. 2. The composition (i.e., the consecutive execution) of two parallel translations is, clearly, a parallel translation. Introductory problems 1. Prove that every parallel translation turns any circle into a circle. 2. Two circles of radius R are tangent at point K. On one of them we take point A,n on the other one we take point B such that ∠AKB = 90◦. Prove that AB = 2R. 3. Two circles of radius R intersect at points M and N. Let A and B be the intersection points of these circles with the perpendicular erected at the midpoint of segment MN. It so happens that the circles lie on one side of line MN. Prove that MN 2 + AB2 = 4R2. 4. Inside rectangle ABCD, point M is taken. Prove that there exists a convex quadri-lateral with perpendicular diagonals of the same length as AB and BC whose sides are equal to AM, BM, CM, DM. §1. Solving problems with the aid of parallel translations 15.1. Where should we construct bridge MN through the river that separates villages A and B so that the path AMNB from A to B was the shortest one? (The banks of the river are assumed to be parallel lines and the bridge perpendicular to the banks.) 15.2. Consider triangle ABC. Point M inside the triangle moves parallel to side BC to its intersection with side CA, then parallel to AB to its intersection with BC, then parallel to AC to its intersection with AB, and so on. Prove that after a number of steps the trajectory of the point becomes a closed one. 15.3. Let K, L, M and N be the midpoints of sides AB, BC, CD and DA, respectively, of convex quadrilateral ABCD. a) Prove that KM ≤1 2(BC + AD) and the equality is attained only if BC ∥AD. b) For given lengths of the sides of quadrilateral ABCD ﬁnd the maximal value of the lengths of segments KM and LN. 15.4. In trapezoid ABCD, sides BC and AD are parallel, M the intersection point of the bisectors of angles ∠A and ∠B, and N the intersection point of the bisectors of angles ∠C and ∠D. Prove that 2MN = \|AB + CD −BC −AD\|. 15.5. From vertex B of parallelogram ABCD heights BK and BH are drawn. It is known that KH = a and BD = b. Find the distance from B to the intersection point of the heights of triangle BKH. 15.6. In the unit square a ﬁgure is placed such that the distance between any two of its points is not equal to 0.001. Prove that the area of this ﬁgure does not exceed a) 0.34; b) 0.287. 319 320 CHAPTER 15. PARALLEL TRANSLATIONS §2. Problems on construction and loci 15.7. Consider angle ∠ABC and straight line l. Construct a line parallel to l on which the legs of angle ∠ABC intercept a segment of given length a. 15.8. Consider two circles S1, S2 and line l. Draw line l1 parallel to l so that: a) the distance between the intersection points of l1 with circles S1 and S2 is of a given value a; b) S1 and S2 intercept on l1 equal chords; c) S1 and S2 intercept on l1 chords the sum (or diﬀerence) of whose lengths is equal to a given value. 15.9. Consider nonintersecting chords AB and CD on a circle. Construct a point X on the circle so that chords AX and BX would intercept on chord CD a segment, EF, of a given length a. 15.10. Construct quadrilateral ABCD given the quadrilateral’s angles and the lengths of sides AB = a and CD = b. 15.11. Given point A and circles S1 and S2. Through A draw line l so that S1 and S2 intercept on l equal chords. 15.12. a) Given circles S1 and S2 intersect at points A and B. Through point A draw line l so that the intercept of this line between circles S1 and S2 were of a given length. b) Consider triangle ABC and triangle PQR. In triangle ABC inscribe a triangle equal to PQR. 15.13. Construct a quadrilateral given its angles and diagonals. 15.14. Find the loci of the points for which the following value is given: a) the sum, b) the diﬀerence of the distances from these points to the two given straight lines. 15.15. An angle made of a transparent material moves so that two nonintersecting circles are tangent to its legs from the inside. Prove that on the angle a point circumscribing an arc of a circle can be marked. Problems for independent study 15.16. Consider two pairs of parallel lines and point P. Through P draw a line on which both pairs of parallel lines intercept equal segments. 15.17. Construct a parallelogram given its sides and an angle between the diagonals. 15.18. In convex quadrilateral ABCD, sides AB and CD are equal. Prove that a) lines AB and CD form equal angles with the line that connects the midpoints of sides AC and BD; b) lines AB and CD form equal angles with the line that connects the midpoints of diagonals BC and AD. 15.19. Among all the quadrilaterals with given lengths of the diagonals and an angle between them ﬁnd the one of the least perimeter. 15.20. Given a circle and two neighbouring vertices of a parallelogram. Construct the parallelogram if it is known that its other two (not given) vertices belong to the given circle. Solutions 15.1. Let A′ be the image of point A under the parallel translation by − − → MN. Then A′N = AM and, therefore, the length of path AMNB is equal to A′N + NB + MN. Since the length of segment MN is a constant, we have to ﬁnd point N for which the sum SOLUTIONS 321 A′N +NB is the least one. It is clear that the sum is minimal if N belongs to segment A′B, i.e., N is the closest to B intersection point of the bank and segment A′B. Figure 147 (Sol. 15.2) 15.2. Denote the consecutive points of the trajectory on the sides of the triangle as on Fig. 147: A1, B1, B2, C2, C3, A3, A4, B4, . . . Since A1B1 ∥AB2, B1B2 ∥CA1 and B1C ∥B2C2, it is clear that triangle AB2C2 is the image of triangle A1B1C under a parallel translation. Similarly, triangle A3BC3 is the image of triangle AB2C2 under a parallel translation and A4B4C is obtained in the same way from A3BC3. But triangle A1B1C is also the image of triangle A3BC3 under a parallel translation, hence, A1 = A4, i.e., after seven steps the trajectory becomes closed. (It is possible for the trajectory to become closed sooner. Under what conditions?) 15.3. a) Let us complement triangle CBD to parallelogram CBDE. Then 2KM = AE ≤AD + DE = AD + BC and the equality is attained only if AD ∥BC. b) Let a = AB, b = BC, c = CD and d = DA. If \|a −c\| = \|b −d\| ̸= 0 then by heading a) the maximum is attained in the degenerate case when all points A, B, C and D belong to one line. Now suppose that, for example, \|a −c\| < \|b −d\|. Let us complement triangles ABL and LCD to parallelograms ABLP and LCDQ, respectively; then PQ ≥\|b −d\| and, therefore, LN 2 = 1 4(2LP 2 + 2LQ2 −PQ2) ≤1 4(2(a2 + c2) −(b −d)2). Moreover, by heading a) KM ≤1 2(b + d). Both equalities are attained when ABCD is a trapezoid with bases AD and BC. 15.4. Let us construct circle S tangent to side AB and rays BC and AD; translate triangle CND parallelly (in the direction of bases BC and AD) until N ′ coincides with point M, i.e., side C′D′ becomes tangent to circle S (Fig. 148). Figure 148 (Sol. 15.4) For the circumscribed trapezoid ABC′D′ the equality 2MN ′ = \|AB+C′D′−BC′−AD′\| is obvious because N ′ = M. Under the passage from trapezoid ABC′D′ to trapezoid ABCD the left-hand side of this equality accrues by 2N ′N and the right-hand side accrues by CC′ + DD′ = 2NN ′. Hence, the equality is preserved. 322 CHAPTER 15. PARALLEL TRANSLATIONS 15.5. Denote the intersection point of heights of triangle BKH by H1. Since HH1 ⊥BK and KH1 ⊥BH, it follows that HH1 ∥AD and KH1 ∥DC, i.e., H1HDK is a parallelogram. Therefore, under the parallel translation by vector − − → H1H point K passes to point D and point B passes to point P (Fig. 149). Since PD ∥BK, it follows that BPDK is a rectangle and PK = BD = b. Since BH1 ⊥KH, it follows that PH ⊥KH. It is also clear that PH = BH1. Figure 149 (Sol. 15.5) In right triangle PKH, hypothenuse KP = b and the leg KH = a are known; therefore, BH1 = PH = √ b2 −a2. 15.6. a) Denote by F the ﬁgure that lies inside the unit square ABCD; let S be its area. Let us consider two vectors − − → AA1 and − − → AA2, where point A1 belongs to side AD and AA1 = 0.001 and where point A2 belongs to the interior of angle ∠BAD, ∠A2AA1 = 60◦ and AA2 = 0.001 (Fig. 150). Figure 150 (Sol. 15.6 a)) Let F1 and F2 be the images of F under the parallel translations by vectors − − → AA1 and − − → AA2, respectively. The ﬁgures F, F1 and F2 have no common points and belong to the interior of the square with side 1.001. Therefore, 2S < 1.0012, i.e., S < 0.335 < 0.34. b) Consider vector − − → AA3 = − − → AA1 + − − → AA2. Let us rotate − − → AA3 about point A through an acute angle counterclockwise so that point A3 turns into point A4 such that A3A4 = 0.001. Let us also consider vectors − − → AA5 and − − → AA6 of length 0.001 each constituting an angle of 30◦ with vector − − → AA4 and situated on both sides of it (Fig. 151). Denote by Fi the image of ﬁgure F under the parallel translation by the vector − − → AAi. Denote the area of the union of ﬁgures A and B by S(A ∪B) and by S(A ∩B) the area of their intersection. For deﬁniteness, let us assume that S(F4 ∩F) ≤S(F3 ∩F). Then S(F4 ∩F) ≤ 1 2S and, therefore, S(F4 ∪F) ≥3 2S. The ﬁgures F5 and F6 do not intersect either each other SOLUTIONS 323 Figure 151 (Sol. 15.6 b)) or ﬁgures F or F4 and, therefore, S(F ∪F4 ∪F5 ∪F6) ≥7 2S. (If it would have been that S(F3 ∩F) ≤S(F4 ∩F), then instead of ﬁgures F5 and F6 we should have taken F1 and F2.) Since the lengths of vectors − − → AAi do not exceed 0.001 √ 3, all the ﬁgures considered lie inside a square with side 1 + 0.002 √ 3. Therefore, 7S/2 ≤(1 + 0.002 √ 3)2 and S < 0.287. 15.7. Given two vectors ±a parallel to l and of given length a. Consider the images of ray BC under the parallel translations by these vectors. Their intersection point with ray BA belongs to the line to be constructed (if they do not intersect, then the problem has no solutions). 15.8. a) Let S′ 1 be the image of circle S1 under the parallel translation by a vector of length a parallel to l (there are two such vectors). The desired line passes through the intersection point of circles S′ 1 and S2. b) Let O1 and O2 be the projections of the centers of circles S1 and S2 to line l; let S′ 1 be the image of the circle S1 under the parallel translation by vector − − − → O1O2. The desired line passes through the intersection point of circles S′ 1 and S2. c) Let S′ 1 be the image of circle S1 under the parallel translation by a vector parallel to l. Then the lengths of chords cut by the line l1 on circles S1 and S′ 1 are equal. If the distance between the projections of the centers of circles S′ 1 and S2 to line l is equal to 1 2a, then the sum of diﬀerence of the lengths of chords cut by the line parallel to l and passing through the intersection point of circles S′ 1 and S2 is equal to a. Now it is easy to construct circle S′ 1. 15.9. Suppose that point X is constructed. Let us translate point A by vector − → EF, i.e., let us construct point A′ such that − → EF = − − → AA′. This construction can be performed since we know vector − → EF: its length is equal to a and it is parallel to CD. Figure 152 (Sol. 15.9) 324 CHAPTER 15. PARALLEL TRANSLATIONS Since AX ∥A′F, it follows that ∠A′FB = ∠AXB and, therefore, angle ∠A′FB is known. Thus, point F belongs to the intersection of two ﬁgures: segment CD and an arc of the circle whose points are vertices of the angles equal to ∠AXB that subtend segment A′B, see Fig. 152. 15.10. Suppose that quadrilateral ABCD is constructed. Denote by D1 the image of point D under the parallel translation by vector − − → CB. In triangle ABD1, sides AB, BD1 and angle ∠ABD1 are known. Hence, the following construction. Let us arbitrarily construct ray BC′ and then draw rays BD′ 1 and BA′ so that ∠D′ 1BC′ = 180◦−∠C, ∠A′BC′ = ∠B and these rays lie in the half plane on one side of ray BC′. On rays BA′ and BD′ 1, draw segments BA = a and BD1 = b, respectively. Let us draw ray AD′ so that ∠BAD′ = ∠A and rays BC′, AD′ lie on one side of line AB. Vertex D is the intersection point of ray AD′ and the ray drawn from D1 parallel to ray BC′. Vertex C is the intersection point of BC′ and the ray drawn from D parallel to ray D1B. 15.11. Suppose that points M and N at which line l intersects circle S2 are constructed. Let O1 and O2 be the centers of circles S1 and S2; let O′ 1 be the image of point O1 under the parallel translation along l such that O′ 1O2 ⊥MN; let S′ 1 be the image of circle S1 under the same translation. Let us draw tangents AP and AQ to circles S′ 1 and S2, respectively. Then AQ2 = AM ·AN = AP 2 and, therefore, O′ 1A2 = AP 2 +R2, where R is the radius of circle S′ 1. Since segment AP can be constructed, we can also construct segment AO′ 1. It remains to notice that point O′ 1 belongs to both the circle of radius AO′ 1 with the center at A and to the circle with diameter O1O2. 15.12. a) Let us draw through point A line PQ, where P belongs to circle S and Q belongs to circle S2. From the centers O1 and O2 of circles S1 and S2, respectively, draw perpendiculars O1M and O2N to line PQ. Let us parallelly translate segment MN by a vector − − − → MO1. Let C be the image of point N under this translation. Triangle O1CO2 is a right one and O1C = MN = 1 2PQ. It follows that in order to con-struct line PQ for which PQ = a we have to construct triangle O1CO2 of given hypothenuse O1O2 and leg O1C = 1 2a and then draw through A the line parallel to O1C. b) It suﬃces to solve the converse problem: around the given triangle PQR circumscribe a triangle equal (?) to the given triangle ABC. Suppose that we have constructed triangle ABC whose sides pass through given points P, Q and R. Let us construct the arcs of circles whose points serve as vertices for angles ∠A and ∠B that subtend segments RP and QP, respectively. Points A and B belong to these arcs and the length of segment AB is known. By heading a) we can construct line AP through P whose intercept between circles S1 and S2 is of given length. Draw lines AR and BQ; we get triangle ABC equal to the given triangle since these triangles have by construction equal sides and the angles adjacent to it. 15.13. Suppose that the desired quadrilateral ABCD is constructed. Let D1 and D2 be the images of point D under the translations by vectors − → AC and − → CA, respectively. Let us circumscribe circles S1 and S2 around triangles DCD1 and DAD2, respectively. Denote the intersection points of lines BC and BA with circles S1 and S2 by M and N, respectively, see Fig. 153. It is clear that ∠DCD1 = ∠DAD2 = ∠D, ∠DCM = 180◦−∠C and ∠DAN = 180◦−∠A. This implies the following construction. On an arbitrary line l, take a point, D, and construct points D1 and D2 on l so that DD1 = DD2 = AC. Fix one of the half planes Π determined by line l and assume that point B belongs to this half plane. Let us construct a circle S1 whose points belonging to Π serve as vertices of the angles equal to ∠D that subtend segment DD1. SOLUTIONS 325 Figure 153 (Sol. 15.13) We similarly construct circle S2. Let us construct point M on S1 so that all the points of the part of the circle that belongs to Π serve as vertices of the angles equal to 180◦−∠C that subtend segment DM. Point N is similarly constructed. Then segment MN subtends angle ∠B, i.e., B is the intersection point of the circle with center D of radius DB and the arc of the circle serve as vertices of the angles equal to ∠B that subtend segment MN (it also belongs to the half plane Π). Points C and A are the intersection points of lines BM and BN with circles S1 and S2, respectively. 15.14. From a point X draw perpendiculars XA1 and XA2 to given lines l1 and l2, respectively. On ray A1X, take point B so that A1B = a. Then if XA1 ± XA2 = a, we have XB = XA2. Let l′ 1 be the image of line l1 under the parallel translation by vector − − → A1B and M the intersection point of lines l′ 1 and l2. Then in the indicated cases ray MX is the bisector of angle ∠A2MB. As a result we get the following answer. Let the intersection points of lines l1 and l2 with the lines parallel to lines l1 and l2 and distant from them by a form rectangle M1M2M3M4. The locus to be found is either a) the sides of this rectangle; or b) the extensions of these sides. 15.15. Let leg AB of angle ∠BAC be tangent to the circle of radius r1 with center O1 and leg AC be tangent to the circle of radius r2 with center O2. Let us parallelly translate line AB inside angle ∠BAC by distance r1 and let us parallelly translate line AC inside angle ∠BAC by distance r2. Let A1 be the intersection point of the translated lines (Fig. 154). Figure 154 (Sol. 15.15) Then ∠O1A1O2 = ∠BAC. The constant(?) angle O1A1O2 subtends ﬁxed segment O1O2 and, therefore, point A1 traverses an arc of a(?) circle. Chapter 16. CENTRAL SYMMETRY Background 1. The symmetry through point A is the transformation of the plane which sends point X into point X′ such that A is the midpoint of segment XX′. The other names of such a transformation: the central symmetry with center A or just the symmetry with center A. Notice that the symmetry with center A is a particular case of two other transformations: it is the rotation through an angle of 180◦with center A and also the homothety with center A and coeﬃcient −1. 2. If a ﬁgure turns into itself under the symmetry through point A, then A is called the center of symmetry of this ﬁgure. 3. The following notations for transformations are used in this chapter: SA — the symmetry with center A; Ta — the translation by vector a. 4. We will denote the composition of symmetries through points A and B by SB◦SA; here we assume that we ﬁrst perform symmetry SA and then symmetry SB. This notation might look unnatural at ﬁrst glance, but it is, however, justiﬁed by the identity (SB ◦SA)(X) = SB(SA(X)). The composition of maps is associative: F ◦(G ◦H) = (F ◦G) ◦H. Therefore, the order of the compositions is inessential and we may simply write F ◦G ◦H. 5. The compositions of two central symmetries or of a symmetry with a parallel transla-tion are calculated according to the following formulas (Problem 16.9): a) SB ◦SA = T2− → AB; b) Ta ◦SA = SB and SB ◦Ta = SA, where a = 2− → AB. Introductory problems 1. Prove that under any central symmetry any circle turns into a circle. 2. Prove that a quadrilateral with a center of symmetry is a parallelogram. 3. The opposite sides of a convex hexagon are equal and parallel. Prove that the hexagon has a center of symmetry. 4. Consider parallelogram ABCD and point M. The lines parallel to lines MC, MD, MA and MB are drawn through points A, B, C and D, respectively. Prove that the lines drawn intersect at one point. 5. Prove that the opposite sides of a hexagon formed by the sides of a triangle and the tangents to its circumscribed circle parallel to the sides of the triangle are equal. §1. Solving problems with the help of a symmetry 16.1. Prove that if in a triangle a median and a bisector coincide, then the triangle is an isosceles one. 327 328 CHAPTER 16. CENTRAL SYMMETRY 16.2. Two players lay out nickels on a rectangular table taking turns. It is only allowed to place a coin onto an unoccupied place. The loser is the one who can not make any move. Prove that the ﬁrst player can always win in ﬁnitely many moves. 16.3. A circle intersects sides BC, CA, AB of triangle ABC at points A1 and A2, B1 and B2, C1 and C2, respectively. Prove that if the perpendiculars to the sides of the triangle drawn through points A1, B1 and C1 intersect at one point, then the perpendiculars to the sides drawn through A2, B2 and C2 also intersect at one point. 16.4. Prove that the lines drawn through the midpoints of the circumscribed quadrilat-eral perpendicularly to the opposite sides intersect at one point. 16.5. Let P be the midpoint of side AB of convex quadrilateral ABCD. Prove that if the area of triangle PCD is equal to a half area of quadrilateral ABCD, then BC ∥AD. 16.6. Unit circles S1 and S2 are tangent at point A; the center O of circle S of radius 2 belongs to S1. Circle S1 is tangent to circle S at point B. Prove that line AB passes through the intersection point of circles S2 and S. 16.7. In triangle ABC medians AF and CE are drawn. Prove that if ∠BAF = ∠BCE = 30◦, then triangle ABC is an equilateral one. 16.8. Consider a convex n-gon with pairwise nonparallel sides and point O inside it. Prove that it is impossible to draw more than n lines through O so that each line divides the area of the n-gon in halves. §2. Properties of the symmetry 16.9. a) Prove that the composition of two central symmetries is a parallel translation. b) Prove that the composition of a parallel translation with a central symmetry (in either order) is a central symmetry. 16.10. Prove that if a point is reﬂected symmetrically through points O1, O2 and O3 and then reﬂected symmetrically once again through the same points, then it assumes the initial position. 16.11. a) Prove that a bounded ﬁgure cannot have more than one center of symmetry. b) Prove that no ﬁgure can have precisely two centers of symmetry. c) Let M be a ﬁnite set of points on a plane. Point O will be called an “almost center of symmetry” of the set M if we can delete a point from M so that O becomes the center of symmetry of the remaining set. How many “almost centers of symmetry” can a set have? 16.12. On segment AB, consider n pairs of points symmetric through the midpoint; n of these 2n points are painted blue and the remaining are painted red. Prove that the sum of distances from A to the blue points is equal to the sum of distances from B to the red points. §3. Solving problems with the help of a symmetry. Constructions 16.13. Through a common point A of circles S1 and S2 draw a straight line so that these circles would intercept on it equal chords. 16.14. Given point A, a line and a circle. Through A draw a line so that A divides the segment between the intersection points of the line drawn with the given line and the given circle in halves. 16.15. Given angle ABC and point D inside it. Construct a segment with the endpoints on the legs of the given angle and with the midpoint at D. 16.16. Consider an angle and points A and B inside it. Construct a parallelogram for which points A and B are opposite vertices and the two other vertices belong to the legs of the angle. SOLUTIONS 329 16.17. Given four pairwise nonparallel straight lines and point O not belonging to these lines. Construct a parallelogram whose center is O and the vertices lie on the given lines, one on each. 16.18. Consider two concentric circles S1 and S2. Draw a line on which these circles intercept three equal segments. 16.19. Consider nonintersecting chords AB and CD of a circle and point J on chord CD. Construct point X on the circle so that chords AX and BX would intercept on chord CD segment EF which J divides in halves. 16.20. Through a common point A of circles S1 and S2 draw line l so that the diﬀerence of the lengths of the chords intercepted by circles S1 and S2 on l were of given value a. 16.21. Given m = 2n + 1 points — the midpoints of the sides of an m-gon — construct the vertices of the m-gon. Problems for independent study 16.22. Construct triangle ABC given medians ma, mb and angle ∠C. 16.23. a) Given a point inside a parallelogram; the point does not belong to the segments that connect the midpoints of the opposite sides. How many segments divided in halves by the given point are there such that their endpoints are on the sides of the parallelogram? b) A point inside the triangle formed by the midlines of a given triangle is given. How many segments divided in halves by the given point and with the endpoints on the sides of the given triangle are there? 16.24. a) Find the locus of vertices of convex quadrilaterals the midpoints of whose sides are the vertices of a given square. b) Three points are given on a plane. Find the locus of vertices of convex quadrilaterals the midpoints of three sides of each of which are the given points. 16.25. Points A, B, C, D lie in the indicated order on a line and AB = CD. Prove that for any point P on the plane we have AP + DP ≥BP + CP. Solutions 16.1. Let median BD of triangle ABC be a bisector as well. Let us consider point B1 symmetric to B through point D. Since D is the midpoint of segment AC, the quadrilateral ABCB1 is a parallelogram. Since ∠ABB1 = ∠B1BC = ∠AB1B, it follows that triangle B1AB is an isosceles one and AB = AB1 = BC. 16.2. The ﬁrst player places a nickel in the center of the table and then places nickels symmetrically to the nickels of the second player with respect to the center of the table. Using this strategy the ﬁrst player has always a possibility to make the next move. It is also clear that the play will be terminated in a ﬁnite number of moves. 16.3. Let the perpendiculars to the sides drawn through points A1, B1 and C1 intersect at point M. Denote the center of the circle by O. The perpendicular to side BC drawn through point A1 is symmetric through point O to the perpendicular to side BC drawn through A2. It follows that the perpendiculars to the sides drawn through points A2, B2 and C2 intersect at the point symmetric to M through point O. 16.4. Let P, Q, R and S be the midpoints of sides AB, BC, CD and DA, respectively, and M the intersection point of segments PR and QS (i.e., the midpoint of both of these segments, see Problem 14.5); O the center of the circumscribed circle and O′ the point symmetric to O through M. Let us prove that the lines mentioned in the formulation of the problem pass through O′. Indeed, O′POR is a parallelogram and, therefore, O′P ∥OR. Since R is the midpoint of chord CD, it follows that OR ⊥CD, i.e., O′P ⊥CD. 330 CHAPTER 16. CENTRAL SYMMETRY For lines O′Q, O′R and O′S the proof is similar. 16.5. Let point D′ be symmetric to D through P. If the area of triangle PCD is equal to a half area of quadrilateral ABCD, then it is equal toSPBC + SPAD, i.e., it is equal to SPBC + SPBD′. Since P is the midpoint of segment DD′, it follows that SPCD′ = SPCD = SPBC + SPBD′ and, therefore, point B belongs to segment D′C. It remains to notice that D′B ∥AD. 16.6. Circles S1 and S2 are symmetric through point A. Since OB is the diameter of circle S1, it follows that ∠BAO = 90◦and, therefore, under the symmetry through A point B becomes on the circle S again. It follows that under the symmetry through A point B turns into the intersection point of circles S2 and S. 16.7. Since ∠EAF = ∠ECF = 30◦, we see that points A, E, F and C belong to one circle S and if O is its center, then ∠EOF = 60◦. Point B is symmetric to A through E and, therefore, B belongs to circle S1 symmetric to circle S through E. Similarly, point B belongs to circle S2 symmetric to circle S through point F. Since triangle EOF is an equilateral one, the centers of circles S, S1 and S2 form an equilateral triangle with side 2R, where R is the radius of these circles. Therefore, circles S1 and S2 have a unique common point — B — and triangle BEF is an equilateral one. Thus, triangle ABC is also an equilateral one. 16.8. Consider a polygon symmetric to the initial one through point O. Since the sides of the polygons are pairwise nonparallel, the contours of these polygons cannot have common segments but could only have common points. Since the polygons are convex ones, each side has not more than two intersection points; therefore, there are not more than 2n intersection points of the contours (more precisely, not more than n pairs of points symmetric through O). Let l1 and l2 be the lines passing through O and dividing the area of the initial polygon in halves. Let us prove that inside each of the four parts into which these lines divide the plane there is an intersection point of the contours. Suppose that one of the parts has no such points between lines l1 and l2. Denote the intersection points of lines l1 and l2 with the sides of the polygon as indicated on Fig. 12. Figure 154 (Sol. 16.8) Let points A′, B′, C′ and D′ be symmetric trough O to points A, B, C and D, respectively. For deﬁniteness sake, assume that point A is closer to O than C′. Since segments AB and C′D′ do not intersect, point B is closer to O than D′. It follows that SABO < SC′D′O = SCDO, where ABO is a convex ﬁgure bounded by segments AO and BO and the part of the boundary of the n-gon between points A and B. On the other hand, SABO = SCDO because lines l1 and l2 divide the area of the polygon in halves. Contradiction. Therefore, between every pair of lines which divide the area of the polygon in halves there is a pair of symmetric intersection points of contours; in other words, there are not more than n such lines. 16.9. a) Let the central symmetry through O1 send point A into A1; let the central symmetry through O2 send point A1 into A2. Then O1O2 is the midline of triangle AA1A2 and, therefore, − − → AA2 = 2− − − → O1O2. SOLUTIONS 331 b) Let O2 be the image of point O1 under the translation by vector 1 2a. By heading a) we have SO1 ◦SO2 = Ta. Multiplying this equality by SO1 from the right or by SO2 from the left and taking into account that SX ◦SX is the identity transformation we get SO1 = SO2 ◦Ta and SO2 = Ta ◦SO1. 16.10. By the preceding problem SB ◦SA = T2− → AB; therefore, SO3 ◦SO2 ◦SO1 ◦SO3 ◦SO2 ◦SO1 = T2(− − − → O2O3+− − − → O3O1+− − − → O1O2) is the identity transformation. 16.11. a) Suppose that a bounded ﬁgure has two centers of symmetry: O1 and O2. Let us introduce a coordinate system whose absciss axis is directed along ray O1O2. Since SO2 ◦SO1 = T2− − − → O1O2, the ﬁgure turns into itself under the translation by vector 2− − − → O1O2. A bounded ﬁgure cannot possess such a property since the image of the point with the largest absciss does not belong to the ﬁgure. b) Let O3 = SO2(O1). It is easy to verify that SO3 = SO2 ◦SO1 ◦SO2 and, therefore, if O1 and O2 are the centers of symmetry of a ﬁgure, then O3 is also a center of symmetry, moreover, O3 ̸= O1 and O3 ̸= O2. c) Let us demonstrate that a ﬁnite set can only have 0, 1, 2 or 3 “almost centers of symmetry”. The corresponding examples are given on Fig. 13. It only remains to prove that a ﬁnite set cannot have more than three “almost centers of symmetry”. Figure 155 (Sol. 16.11) There are ﬁnitely many “almost centers of symmetry” since they are the midpoints of the segments that connect the points of the set. Therefore, we can select a line such that the projections of “almost centers of symmetry” to the line are distinct. Therefore, it suﬃces to carry out the proof for the points which belong to one line. Let n points on a line be given and x1 < x2 < · · · < xn−1 < xn be their coordinates. If we discard the point x1, then only point 1 2(x2 + xn) can serve as the center of symmetry of the remaining set; if we discard xn, then only point 1 2(x1 + xn−1) can be the center of symmetry of the remaining set and if we discard any other point, then only point 1 2(x1 + xn) can be the center of symmetry of the remaining set. Therefore, there can not be more than 3 centers of symmetry. 16.12. A pair of symmetric points is painted diﬀerent colours, therefore, it can be discarded from the consideration; let us discard all such pairs. In the remaining set of points the number of blue pairs is equal to the number of red pairs. Moreover, the sum of the distances from either of points A or B to any pair of symmetric points is equal to the length of segment AB. 332 CHAPTER 16. CENTRAL SYMMETRY 16.13. Consider circle S′ 1 symmetric to circle S1 through point A. The line to be found passes through the intersection points of S′ 1 and S2. 16.14. Let l′ be the image of line l under the symmetry through point A. The desired line passes through point A and an intersection point of line l′ with the circle S. Figure 156 (Sol. 16.15) 16.15. Let us construct the intersection points A′ and C′ of the lines symmetric to the lines BC and AB through the point D with lines AB and BC, respectively, see Fig. 14. It is clear that point D is the midpoint of segment A′C′ because points A′ and C′ are symmetric through D. 16.16. Let O be the midpoint of segment AB. We have to construct points C and D that belong to the legs of the angle so that point O is the midpoint of segment CD. This construction is described in the solution of the preceding problem. 16.17. Let us ﬁrst separate the lines into pairs. This can be done in three ways. Let the opposite vertices A and C of parallelogram ABCD belong to one pair of lines, B and D to the other pair. Consider the angle formed by the ﬁrst pair of lines and construct points A and C as described in the solution of Problem 16.15. Construct points B and D in a similar way. 16.18. On the smaller circle, S1, take an arbitrary point, X. Let S′ 1 be the image of S1 under the symmetry with respect to X, let Y be the intersection point of circles S′ 1 and S2. Then XY is the line to be found. Figure 157 (Sol. 16.19) 16.19. Suppose X is constructed. Denote the images of points A, B and X under the symmetry through point J by A′, B′ and X′, respectively, see Fig. 15. Angle ∠A′FB = 180◦−∠AXB is known and, therefore, point F is the intersection point of segment CD with the arc of the circle whose points serve as vertices of angles of value 180◦−∠AXB that subtend segment BA′. Point X is the intersection point of line BF with the given circle. 16.20. Suppose that line l is constructed. Let us consider circle S′ 1 symmetric to circle S1 through point A. Let O1, O′ 1 and O2 be the centers of circles S1, S′ 1 and S2, as shown on Fig. 16. SOLUTIONS 333 Figure 158 (Sol. 16.20) Let us draw lines l′ 1 and l2 through O′ 1 and O2 perpendicularly to line l. The distance between lines l′ 1 and l2 is equal to a half diﬀerence of the lengths of chords intercepted by l on circles S1 and S2. Therefore, in order to construct l, we have to construct the circle of radius 1 2a with center O′ 1; line l2 is tangent to this circle. Having constructed l2, drop the perpendicular from point A to l2; this perpendicular is line l. 16.21. Let B1, B2, . . . , Bm be the midpoints of sides A1A2, A2A3, . . . , AmA1 of polygon A1A2 . . . Am. Then SB1(A1) = A2, SB2(A2) = A3, . . . , SBm(Am) = A1. It follows that SBm ◦· · · ◦SB1(A1) = A1, i.e., A1 is a ﬁxed point of the composition of symmetries SBm ◦ SBm−1 ◦· · · ◦SB1. By Problem 16.9 the composition of an odd number of central symmetries is a central symmetry, i.e., has a unique ﬁxed point. This point can be constructed as the midpoint of the segment that connects points X and SBm ◦SBm−1 ◦· · · ◦SB1(X), where X is an arbitrary point. Chapter 17. THE SYMMETRY THROUGH A LINE Background 1. The symmetry through a line l (notation: Sl) is a transformation of the plane which sends point X into point X′ such that l is the midperpendicular to segment XX′. Such a transformation is also called the axial symmetry and l is called the axis of the symmetry. 2. If a ﬁgure turns into itself under the symmetry through line l, then l is called the axis of symmetry of this ﬁgure. 3. The composition of two symmetries through axes is a parallel translation, if the axes are parallel, and a rotation, if they are not parallel, cf. Problem 17.22. Axial symmetries are a sort of “bricks” all the other motions of the plane are constructed from: any motion is a composition of not more than three axial symmetries (Problem 17.35). Therefore, the composition of axial symmetries give much more powerful method for solving problems than compositions of central symmetries. Moreover, it is often convenient to de-compose a rotation into a composition of two symmetries with one of the axes of symmetry being a line passing through the center of the rotation. Introductory problems 1. Prove that any axial symmetry sends any circle into a circle. 2. A quadrilateral has an axis of symmetry. Prove that this quadrilateral is either an equilateral trapezoid or is symmetric through a diagonal. 3. An axis of symmetry of a polygon intersects its sides at points A and B. Prove that either point A is a vertex of the polygon or the midpoint of a side perpendicular to the axis of symmetry. 4. Prove that if a ﬁgure has two perpendicular axes of symmetry, it has a center of symmetry. §1. Solving problems with the help of a symmetry 17.1. Point M belongs to a diameter AB of a circle. Chord CD passes through M and intersects AB at an angle of 45◦. Prove that the sum CM 2 + DM 2 does not depend on the choice of point M. 17.2. Equal circles S1 and S2 are tangent to circle S from the inside at points A1 and A2, respectively. An arbitrary point C of circle S is connected by segments with points A1 and A2. These segments intersect S1 and S2 at points B1 and B2, respectively. Prove that A1A2 ∥B1B2. 17.3. Through point M on base AB of an isosceles triangle ABC a line is drawn. It intersects sides CA and CB (or their extensions) at points A1 and B1. Prove that A1A : A1M = B1B : B1M. 335 336 CHAPTER 17. THE SYMMETRY THROUGH A LINE §2. Constructions 17.4. Construct quadrilateral ABCD whose diagonal AC is the bisector of angle ∠A knowing the lengths of its sides. 17.5. Construct quadrilateral ABCD in which a circle can be inscribed knowing the lengths of two neighbouring sides AB and AD and the angles at vertices B and D. 17.6. Construct triangle ABC knowing a, b and the diﬀerence of angles ∠A −∠B. 17.7. Construct triangle ABC given its side c, height hc and the diﬀerence of angles ∠A −∠B. 17.8. Construct triangle ABC given a) c, a −b (a > b) and angle ∠C; b) c, a + b and angle ∠C. 17.9. Given line l and points A and B on one side of it. Construct point X on l such that AX + XB = a, where a is given. 17.10. Given acute angle ∠MON and points A and B inside it. Find point X on leg OM such that triangle XY Z, where Y and Z are the intersection points of lines XA and XB with ON, were isosceles, i.e., XY = XZ. 17.11. Given line MN and two points A and B on one side of it. Construct point X on MN such that ∠AXM = 2∠BXN. 17.12. Given three lines l1, l2 and l3 intersecting at one point and point A1 on l1. Construct triangle ABC so that A1 is the midpoint of its side BC and lines l1, l2 and l3 are the midperpendiculars to the sides. 17.13. Construct triangle ABC given points A, B and the line on which the bisector of angle ∠C lies. 17.14. Given three lines l1, l2 and l3 intersecting at one point and point A on line l1. Construct triangle ABC so that A is its vertex and the bisectors of the triangle lie on lines l1, l2 and l3. 17.15. Construct a triangle given the midpoints of two of its sides and the line that contains the bisector drawn to one of these sides. §3. Inequalities and extremals 17.16. On the bisector of the exterior angle ∠C of triangle ABC point M distinct from C is taken. Prove that MA + MB > CA + CB. 17.17. In triangle ABC median AM is drawn. Prove that 2AM ≥(b + c) cos( 1 2α). 17.18. The inscribed circle of triangle ABC is tangent to sides AC and BC at points B1 and A1. Prove that if AC > BC, then AA1 > BB1. 17.19. Prove that the area of any convex quadrilateral does not exceed a half-sum of the products of opposite sides. 17.20. Given line l and two points A and B on one side of it, ﬁnd point X on line l such that the length of segment AXB of the broken line was minimal. 17.21. Inscribe a triangle of the least perimeter in a given acute triangle. §4. Compositions of symmetries 17.22. a) Lines l1 and l2 are parallel. Prove that Sl1 ◦Sl2 = T2a, where Ta is the parallel translation that sends l1 to l2 and such that a ⊥l1. b) Lines l1 and l2 intersect at point O. Prove that Sl2 ◦Sl1 = R2α O , where Rα O is the rotation about O through the angle of α that sends l1 to l2. §6. CHASLES’S THEOREM 337 17.23. On the plane, there are given three lines a, b, c. Let T = Sa ◦Sb ◦Sc. Prove that T ◦T is a parallel translation (or the identity map). 17.24. Let l3 = Sl1(l2). Prove that Sl3 = Sl1 ◦Sl2 ◦Sl1. 17.25. The inscribed circle is tangent to the sides of triangle ABC at points A1, B1 and C1. Points A2, B2 and C2 are symmetric to these points through the bisectors of the corresponding angles of the triangle. Prove that A2B2 ∥AB and lines AA2, BB2 and CC2 intersect at one point. 17.26. Two lines intersect at an angle of γ. A grasshopper hops from one line to another one; the length of each jump is equal to 1 m and the grasshopper does not jump backwards whenever possible. Prove that the sequence of jumps is periodic if and only if γ/π is a rational number. 17.27. a) Given a circle and n lines. Inscribe into the circle an n-gon whose sides are parallel to given lines. b) n lines go through the center O of a circle. Construct an n-gon circumscribed about this circle such that the vertices of the n-gon belong to these lines. 17.28. Given n lines, construct an n-gon for which these lines are a) the midperpendic-ulars to the sides; b) the bisectors of the inner or outer angles at the vertices. 17.29. Given a circle, a point and n lines. Into the circle inscribe an n-gon one of whose sides passes through the given point and the other sides are parallel to the given lines. §5. Properties of symmetries and axes of symmetries 17.30. Point A lies at the distance of 50 cm from the center of the disk of radius 1 cm. It is allowed to reﬂect point A symmetrically through any line intersecting the disk. Prove that a) after 25 reﬂexions point A can be driven inside the given circle; b) it is impossible to perform this in 24 reﬂexions. 17.31. On a circle with center O points A1, . . . , An which divide the circle into equal archs and a point X are given. Prove that the points symmetric to X through lines OA1, . . . , OAn constitute a regular polygon. 17.32. Prove that if a planar ﬁgure has exactly two axes of symmetry, then these axes are perpendicular to each other. 17.33. Prove that if a polygon has several (more than 2) axes of symmetry, then all of them intersect at one point. 17.34. Prove that if a polygon has an even number of axes of symmetry, then it has a center of symmetry. §6. Chasles’s theorem A transformation which preserves distances between points (i.e., such that if A′ and B′ are the images of points A and B, respectively, then A′B′ = AB) is called a movement. A movement of the plane that preserves 3 points which do not belong to one line preserves all the other points. 17.35. Prove that any movement of the plane is a composition of not more than three symmetries through lines. A movement which is the composition of an even number of symmetries through lines is called a ﬁrst type movement or a movement that preserves the orientation of the plane. A movement which is the composition of an odd number of symmetries through lines is called a second type movement or a movement inversing the orientation of the plane. 338 CHAPTER 17. THE SYMMETRY THROUGH A LINE We will not prove that the composition of an odd number of symmetries through lines is impossible to represent in the form of the composition of an odd number of symmetries through lines and the other way round because this fact, though true, is beyond the scope of our book. 17.36. Prove that any ﬁrst type movement is either a rotation or a parallel translation. The composition of a symmetry through line l and the translation by a vector parallel to l (this vector might be the zero one) is called a transvection. 17.37. Prove that any second type movement is a transvection. Problems for independent study 17.38. Given a nonconvex quadrilateral of perimeter P. Prove that there exists a convex quadrilateral of the same perimeter but of greater area. 17.39. Can a bounded ﬁgure have a center of symmetry and exactly one axis of symme-try? 17.40. Point M belongs to the circumscribed circle of triangle ABC. Prove that the lines symmetric to the lines AM, BM and CM through the bisectors of angles ∠A, ∠B and ∠C are parallel to each other. 17.41. The vertices of a convex quadrilateral belong to diﬀerent sides of a square. Prove that the perimeter of this quadrilateral is not shorter than 2 √ 2a, where a is the length of the square’s side. 17.42. A ball lies on a rectangular billiard table. Construct a trajectory traversing along which the ball would return to the initial position after one reﬂexion from each side of the table. Solutions 17.1. Denote the points symmetric to points C and D through line AB by C′ and D′, respectively. Since ∠C′MD = 90◦, it follows that CM 2 + MD2 = C′M 2 + MD2 = C′D2. Since ∠C′CD = 45◦, chord C′D is of constant length. 17.2. In circle S, draw the diameter which is at the same time the axis of symmetry of circles S1 and S2. Let points C′ and B′ 2 be symmetric to points C and B2 through this diameter: see Fig. 17. Figure 159 (Sol. 17.2) Circles S1 and S are homothetic with the center of homothety at point A1; let this homothety send line B1B′ 2 into line CC′. Therefore, these lines are parallel to each other. It is also clear that B2B′ 2 ∥CC′. Therefore, points B1, B′ 2 and B2 belong to one line and this line is parallel to line CC′. SOLUTIONS 339 17.3. Let the line symmetric to line A1B1 through line AB intersect sides CA and CB (or their extensions) at points A2 and B2, respectively. Since ∠A1AM = ∠B2BM and ∠A1MA = ∠B2MB, it follows that A1AM ∼B2BM, i.e., A1A : A1M = B2B : B2M. Moreover, since MB is a bisector in triangle B1MB2, it follows that B2B : B2M = B1B : B1M. 17.4. Suppose that quadrilateral ABCD is constructed. Let, for deﬁniteness sake, AD > AB. Denote by B′ the point symmetric to B through diagonal AC. Point B′ belongs to side AD and B′D = AD −AB. In triangle B′CD, the lengths of all the sides are known: B′D = AD −AB and B′C = BC. Constructing triangle B′CD on the extension of side B′D beyond B′ let us construct point A. Further construction is obvious. 17.5. Suppose that quadrilateral ABCD is constructed. For deﬁniteness sake, assume that AD > AB. Let O be the center of the circumscribed circle; let point D′ be symmetric to D through line AO; let A′ be the intersection point of lines AO and DC; let C′ be the intersection point of lines BC and A′D′ (Fig. 18). Figure 160 (Sol. 17.5) In triangle BC′D′, side BD′ and adjacent angles are known: ∠D′BC′ = 180◦−∠B and ∠BD′C′ = ∠D. Let us construct triangle BC′D′ given these elements. Since AD′ = AD, we can construct point A. Further, let us construct O — the intersection point of bisectors of angles ABC′ and BD′C′. Knowing the position of O we can construct point D and the inscribed circle. Point C is the intersection point of line BC′ and the tangent to the circle drawn from D. 17.6. Suppose that triangle ABC is constructed. Let C′ be the point symmetric to C through the midperpendicular to segment AB. In triangle ACC′ there are known AC = b, AC′ = a and ∠CAC′ = ∠A −∠B. Therefore, the triangle can be constructed. Point B is symmetric to A through the midperpendicular to segment CC′. 17.7. Suppose that triangle ABC is constructed. Denote by C′ the point symmetric to C through the midperpendicular to side AB and by B′ the point symmetric to B through line CC′. For deﬁniteness, let us assume that AC < BC. Then ∠ACB′ = ∠ACC′ + ∠C′CB = 180◦−∠A + ∠C′CB = 180◦−(∠A −∠B) i.e., angle ∠ACB′ is known. Triangle ABB′ can be constructed because AB = c, BB′ = 2hc and ∠ABB′ = 90◦. Point C is the intersection point of the midperpendicular to segment BB′ and the arc of the circle whose points serve as vertices of angles of value 180◦−(∠A −∠B) that subtend segment AB′. 340 CHAPTER 17. THE SYMMETRY THROUGH A LINE 17.8. a) Suppose triangle ABC is constructed. Let C′ be the point symmetric to A through the bisector of angle ∠C. Then ∠BC′A = 180◦−∠AC′C = 180◦−1 2(180◦−∠C) = 90◦+ 1 2∠C and BC′ = a −b. In triangle ABC′, there are known AB = c, BC′ = a −b and ∠C′ = 90◦+ 1 2∠C. Since ∠C′ > 90◦, triangle ABC′ is uniquely constructed from these elements. Point C is the intersection point of the midperpendicular to segment AC′ with line BC′. b) The solution is similar to that of heading a). For C′ we should take the point symmetric to A through the bisector of the outer angle ∠C in triangle ABC. Since ∠AC′B = 1 2∠C < 90◦, the problem can have two solutions. 17.9. Let S be the circle of radius a centered at B, let S′ be the circle of radius AX with center X and A′ the point symmetric to A through line l. Then circle S′ is tangent to circle S and point A′ belongs to circle S′. It remains to draw circle S′ through the given points A and A′ tangent to the given circle S and ﬁnd its center X, cf. Problem 8.56 b). Figure 161 (Sol. 17.10) 17.10. Let the projection of point A to line ON be closer to point O than the projection of point B. Suppose that the isosceles triangle XY Z is constructed. Let us consider point A′ symmetric to point A through line OM. Let us drop perpendicular XH from point X to line ON (Fig. 19). Since ∠A′XB = ∠A′XO + ∠OXA + ∠Y XH + ∠HXZ = 2∠OXY + 2∠Y XH = 2∠OXH = 180◦−2∠MON, angle ∠A′XB is known. Point X is the intersection point of line OM and the arc whose points serve as vertices of angles of 180◦−2∠MON that subtend A′B. In addition, the projection of X onto ON must lie between the projections of A and B. Conversely, if ∠A′XB = 180◦−∠MON and the projection of X to line ON lies between the projections of A and B, then triangle XY Z is an isosceles one. 17.11. Suppose that point X is constructed. Let B′ be the point symmetric to point B through line MN; the circle of radius AB′ with center B′ intersects line MN at point A′. Then ray B′X is the bisector of angle ∠AB′A′. It follows that X is the intersection point of lines B′O and MN, where O is the midpoint of segment AA′. 17.12. Through point A1 draw line BC perpendicular to line l1. Vertex A of triangle ABC to be found is the intersection point of lines symmetric to line BC through lines l2 and l3. 17.13. Let point A′ be symmetric to A through the bisector of angle ∠C. Then C is the intersection point of line A′B and the line on which the bisector of angle ∠C lies. SOLUTIONS 341 17.14. Let A2 and A3 be points symmetric to A through lines l2 and l3, respectively. Then points A2 and A3 belong to line BC. Therefore, points B and C are the intersection points of line A2A3 with lines l2 and l3, respectively. 17.15. Suppose that triangle ABC is constructed and N is the midpoint of AC, M the midpoint of BC and the bisector of angle ∠A lies on the given line, l. Let us construct point N ′ symmetric to N through line l. Line BA passes through point N ′ and is parallel to MN. In this way we ﬁnd vertex A and line BA. Having drawn line AN, we get line AC. It remains to construct a segment whose endpoints belong to the legs of angle ∠BAC and whose midpoint is M, cf. the solution of Problem 16.15. 17.16. Let points A′ and B′ be symmetric to A and B, respectively, through line CM. Then AM + MB = A′M + MB > A′B = A′C + CB = AC + CB. 17.17. Let points B′, C′ and M ′ be symmetric to points B, C and M through the bisector of the outer angle at vertex A. Then AM + AM ′ = MM ′ = 1 2(BB′ + CC′) = (b + c) sin(90◦−1 2α) = (b + c) cos(1 2α). 17.18. Let point B′ be symmetric to B through the bisector of angle ∠ACB. Then B′A1 = BB1, i.e., it remains to verify that B′A1 < AA1. To this end it suﬃces to notice that ∠AB′A1 > ∠AB′B > 90◦. 17.19. Let D′ be the point symmetric to D through the midperpendicular to segment AC. Then SABCD = SABCD′ = SBAD′ + SBCD′ ≤1 2AB · AD′ + 1 2BC · CD′ = 1 2(AB · CD + BC · AD). 17.20. Let point A′ be symmetric to A through line l. Let X be a point on line l. Then AX + XB = A′X + XB ≥A′B and the equality is attained only if X belongs to segment A′B. Therefore, the point to be found is the intersection point of line l with segment A′B. 17.21. Let PQR be the triangle determined by the bases of the heights of triangle ABC and let P ′Q′R′ be any other triangle inscribed in triangle ABC. Further, let points P1 and P2 (respectively P ′ 1 and P ′ 2) be symmetric to point P (resp. P ′) through lines AB and AC, respectively (Fig. 20). Figure 162 (Sol. 17.21) Points Q and R belong to segment P1P2 (see Problem 1.57) and, therefore, the perimeter of triangle PQR is equal to the length of segment P1P2. The perimeter of triangle P ′Q′R′ is, however, equal to the length of the broken segment P ′ 1R′Q′P ′ 2, i.e., it is not shorter than 342 CHAPTER 17. THE SYMMETRY THROUGH A LINE the length of segment P ′ 1P ′ 2. It remains to notice that (P ′ 1P ′ 2)2 = P1P 2 2 + 4d2, where d is the distance from point P ′ 1 to line P1P2. 17.22. Let X be an arbitrary point, X1 = Sl1(X) and X2 = Sl2(X1). a) On line l1, select an arbitrary point O and consider a coordinate system with O as the origin and the absciss axis directed along line l1. Line l2 is given in this coordinate system by the equation y = a. Let y, y1 and y2 be ordinates of points X, X1 and X2, respectively. It is clear that y1 = −y and y2 = (a −y1) + a = y + 2a. Since points X, X1 and X2 have identical abscisses, it follows that X2 = T2a(X), where Ta is the translation that sends l1 to l2, and a ⊥l1. b) Consider a coordinate system with O as the origin and the absciss axis directed along line l1. Let the angle of rotation from line l1 to l2 in this coordinate system be equal to α and the angles of rotation from the absciss axes to rays OX, OX1 and OX2 be equal to ϕ, ϕ1 and ϕ2, respectively. Clearly, ϕ1 = −ϕ and ϕ2 = (α −ϕ1) + α = ϕ + 2α. Since OX = OX1 = OX2, it follows that X2 = R2α O (X), where Rα O is the translation that sends l1 to l2. 17.23. Let us represent T ◦T as the composition of three transformations: T ◦T = (Sa ◦Sb ◦Sc) ◦(Sa ◦Sb ◦Sc) = (Sa ◦Sb) ◦(Sc ◦Sa) ◦(Sb ◦Sc). Here Sa ◦Sb, Sc ◦Sa and Sb ◦Sc are rotations through the angles of 2∠(b, a), 2∠(a, c) and 2∠(c, b), respectively. The sum of the angles of the rotations is equal to 2(∠(b, a) + ∠(a, c) + ∠(c, b)) = 2∠(b, b) = 0◦ and this value is determined up to 2 · 180◦= 360◦. It follows that this composition of rotations is a parallel translation, cf. Problem 18.33. 17.24. If points X and Y are symmetric through line l3, then points Sl1(X) and Sl1(Y ) are symmetric through line l2, i.e., Sl1(X) = Sl2 ◦Sl1(Y ). It follows that Sl1 ◦Sl3 = Sl2 ◦Sl1 and Sl3 = Sl1 ◦Sl2 ◦Sl1. 17.25. Let O be the center of the inscribed circle; let a and b be lines OA and OB. Then Sa ◦Sb(C1) = Sa(A1) = A2 and Sb ◦Sa(C1) = Sb(B1) = B2. Points A2 and B2 are obtained from point C1 by rotations with center O through opposite angles and, therefore, A2B2 ∥AB. Similar arguments show that the sides of triangles ABC and A2B2C2 are parallel and, therefore, these triangles are homothetic. Lines AA2, BB2 and CC2 pass through the center of homothety which sends triangle ABC to A2B2C2. Notice that this homothety sends the circumscribed circle of triangle ABC into the inscribed circle, i.e., the center of homothety belongs to the line that connects the centers of these circles. 17.26. For every jump vector there are precisely two positions of a grasshopper for which the jump is given by this vector. Therefore, a sequence of jumps is periodic if and only if there exists but a ﬁnite number of distinct jump vectors. Let a1 be the jump vector of the grasshopper from line l2 to line l1; let a2, a3, a4,. . . be vectors of the successive jumps. Then a2 = Sl2(a1), a3 = Sl1(a2), a4 = Sl2(a3) , . . . Since the composition Sl1 ◦Sl2 is a rotation through an angle of 2γ (or 2π −2γ), it follows that vectors a3, a5, a7, . . . are obtained from a1 by rotations through angles of 2γ, 4γ, 6γ, . . . (or through angles of 2(π −γ), 4(π −γ), 6(π −γ), . . . ). Therefore, the set a1, a3, a5, . . . contains a ﬁnite number of distinct vectors if and only if γ/π is a rational number. The set a2, a4, a6, . . . is similarly considered. 17.27. a) Suppose polygon A1A2 . . . An is constructed. Let us draw through the center O of the circle the midperpendiculars l1, l2, . . . , ln to chords A1A2, A2A3, . . . , AnA1, respectively. Lines l1, . . . , ln are known since they pass through O and are perpendicular to SOLUTIONS 343 the given lines. Moreover, A2 = Sl1(A1), A3 = Sl2(A2), . . . , A1 = Sln(An), i.e., point A1 is a ﬁxed point of the composition of symmetries Sln ◦· · · ◦Sl1. For n odd there are precisely two ﬁxed points on the circle; for n even there are either no ﬁxed points or all the points are ﬁxed. b) Suppose the desired polygon A1 . . . An is constructed. Consider polygon B1 . . . Bn formed by the tangent points of the circumscribed polygon with the circle. The sides of polygon B1 . . . Bn are perpendicular to the given lines, i.e., they have prescribed directions and, therefore, the polygon can be constructed (see heading a)); it remains to draw the tangents to the circle at points B1, . . . , Bn. 17.28. Consider the composition of consecutive symmetries through given lines l1, . . . , ln. In heading a) for vertex A1 of the desired n-gon we have to take a ﬁxed point of this composition, and in heading b) for line A1An we have to take the(a) ﬁxed line. 17.29. The consecutive symmetries through lines l1, . . . , ln−1 perpendicular to given lines and passing through the center of the circle send vertex A1 of the desired polygon to vertex An. If n is odd, then the composition of these symmetries is a rotation through a known angle and, therefore, we have to draw through point M chord A1An of known length. If n is even, then the considered composition is a symmetry through a line and, therefore, from M we have to drop perpendicular to this line. 17.30. Let O be the center of the given disk, DR the disk of radius R with center O. Let us prove that the symmetries through the lines passing through D1 send the set of images of points of DR into disk DR+2. Indeed, the images of point O under the indicated symmetries ﬁll in disk D2 and the disks of radius R with centers in D2 ﬁll in disk DR+2. It follows that after n reﬂexions we can obtain from points of D1 any point of D2n+1 and only them. It remains to notice that point A can be “herded” inside DR after n reﬂexions if and only if we can transform any point of DR into A after n reﬂexions. 17.31. Denote symmetries through lines OA1, . . . , OAn by S1, . . . , Sn, respectively. Let Xk = Sk(X) for k = 1, . . . , n. We have to prove that under a rotation through point O the system of points X1, . . . , Xn turns into itself. Clearly, Sk+1 ◦Sk(Xk) = Sk+1 ◦Sk ◦Sk(X) = Xk+1. Transformations Sk+1 ◦Sk are rotations about O through an angle of 4π n , see Problem 17.22 b). Remark. For n even we get an n 2-gon. 17.32. Let lines l1 and l2 be axes of symmetry of a plane ﬁgure. This means that if point X belongs to the ﬁgure, then points Sl1(X) and Sl2(X) also belong to the ﬁgure. Consider line l3 = Sl1(l2). Thanks to Problem 17.24 Sl3(X) = Sl1 ◦Sl2 ◦Sl1(X) and, therefore, l3 is also an axis of symmetry. If the ﬁgure has precisely two axes of symmetry, then either l3 = l1 or l3 = l2. Clearly, l3 ̸= l1 and, therefore, l3 = l2 i.e., line l2 is perpendicular to line l1. 17.33. Suppose that the polygon has three axes of symmetry which do not intersect at one point, i.e., they form a triangle. Let X be the point of the polygon most distant from an inner point M of this triangle. Points X and M lie on one side of one of the considered axes of symmetry, l. If X′ is the point symmetric to X through l, then M ′X > MX and point X′ is distant from M further than X. The obtained contradiction implies that all the axes of symmetry of a polygon intersect at one point. 17.34. All the axes of symmetry pass through one point O (Problem 17.33). If l1 and l2 are axes of symmetry, then l3 = Sl1(l2) is also an axis of symmetry, see Problem 17.24. 344 CHAPTER 17. THE SYMMETRY THROUGH A LINE Select one of the axes of symmetry l of our polygon. The odd axes of symmetry are divided into pairs of lines symmetric through l. If line l1 perpendicular to l and passing through O is not an axis of symmetry, then there is an odd number of axes of symmetry. Therefore, l1 is an axis of symmetry. Clearly, Sl1 ◦Sl = R180◦ O is a central symmetry i.e., O is the center of symmetry. 17.35. Let F be a movement sending point A into A′ and such that A and A′ are distinct; S the symmetry through the midperpendicular l to segment AA′. Then S ◦F(A) = A, i.e., A is a ﬁxed point of S ◦F. Moreover, if X is a ﬁxed point of transformation F, then AX = A′X, i.e., point X belongs to line l; hence, X is a ﬁxed point of S ◦F. Thus, point A and all the ﬁxed points of F are ﬁxed points of the transformation S ◦F. Take points A, B and C not on one line and consider their images under the given movement G. We can construct transformations S1, S2 and S3 which are either symmetries through lines or identity transformations such that S3 ◦S2 ◦S1 ◦G preserves points A, B and C, i.e., is the identity transformation E. Multiplying the equality S3 ◦S2 ◦S1 ◦G = E from the left consecutively by S3, S2 and S1 and taking into account that Si ◦Si = E we get G = S1 ◦S2 ◦S3. 17.36. Thanks to Problem 17.35 any ﬁrst type movement is a composition of two symmetries through lines. It remains to make use of the result of Problem 17.22. 17.37. By Problem 17.35 any second type movement can be represented in the form S3 ◦S2 ◦S1, where S1, S2 and S3 are symmetries through lines l1, l2 and l3, respectively. First, suppose that the lines l2 and l3 are not parallel. Then under the rotation of the lines l2 and l3 about their intersection point through any angle the composition S3 ◦S2 does not change (see Problem 17.22 b)), consequently, we can assume that l2 ⊥l1. It remains to rotate lines l1 and l2 about their intersection point so that line l2 became parallel to line l3. Now, suppose that l2 ∥l3. If line l1 is not parallel to these lines, then it is possible to rotate l1 and l2 about their intersection point so that lines l2 and l3 become nonparallel. If l1 ∥l2, then it is possible to perform a parallel transport of l1 and l2 so that lines l2 and l3 coincide. Chapter 18. ROTATIONS Background 1. We will not give a rigorous deﬁnition of a rotation. To solve the problems it suﬃces to have the following idea on the notion of the rotation: a rotation with center O (or about the point O) through an angle of ϕ is the transformation of the plane which sends point X into point X′ such that: a) OX′ = OX; b) the angle from vector − − → OX to vector − − → OX′ is equal to ϕ. 2. In this chapter we make use of the following notations for the transformations and their compositions: Ta is a translation by vector a; SO is the symmetry through point O; Sl is the symmetry through line l; Rϕ O is the rotation with center O through an angle of ϕ; F ◦G is the composition of transformations F and G deﬁned as (F ◦G)(X) = F(G(X)). 3. The problems solvable with the help of rotations can be divided into two big classes: problems which do not use the properties of compositions of rotations and properties which make use of these properties. To solve the problems which make use of the properties of the compositions of rotations the following result of Problem 18.33 is handy: Rβ B ◦Rα A = Rγ C, where γ = α + β and ∠BAC = 1 2α, ∠ABC = 1 2β. Introductory problems 1. Prove that any rotation sends any circle into a circle. 2. Prove that a convex n-gon is a regular one if and only if it turns into itself under the rotation through an angle of 360◦ n about a point. 3. Prove that triangle ABC is an equilateral one if and only if under the rotation through 60◦(either clockwise or counterclockwise) about point A vertex B turns into vertex C. 4. Prove that the midpoints of the sides of a regular polygon determine a regular polygon. 5. Through the center of a square two perpendicular lines are drawn. Prove that their intersection points with the sides of the square determine a square. §1. Rotation by 90◦ 18.1. On sides BC and CD of square ABCD points M and K, respectively, are taken so that ∠BAM = ∠MAK. Prove that BM + KD = AK. 18.2. In triangle ABC median CM and height CH are drawn. Through an arbitrary point P of the plane in which ABC lies the lines are drawn perpendicularly to CA, CM and CB. They intersect CH at points A1, M1 and B1, respectively. Prove that A1M1 = B1M1. 18.3. Two squares BCDA and BKMN have a common vertex B. Prove that median BE of triangle ABK and height BF of triangle CBH belong to one line. The vertices of each square are counted clockwise. 345 346 CHAPTER 18. ROTATIONS 18.4. Inside square A1A2A3A4 point P is taken. From vertex A1 we drop the perpen-dicular on A2P; from A2 on A3P; from A3 on A4P and from A4 on A1P. Prove that all four perpendiculars (or their extensions) intersect at one point. 18.5. On sides CB and CD of square ABCD points M and K are taken so that the perimeter of triangle CMK is equal to the doubled length of the square’s side. Find the value of angle ∠MAK. 18.6. On the plane three squares (with same orientation) are given: ABCD, AB1C1D1 and A2B2CD2; the ﬁrst square has common vertices A and C with the two other squares. Prove that median BM of triangle BB1B2 is perpendicular to segment D1D2. 18.7. Triangle ABC is given. On its sides AB and BC squares ABMN and BCPQ are constructed outwards. Prove that the centers of these squares and the midpoints of segments MQ and AC form a square. 18.8. A parallelogram is circumscribed about a square. Prove that the perpendiculars dropped from the vertices of the parallelograms to the sides of the square form a square. §2. Rotation by 60◦ 18.9. On segment AE, on one side of it, equilateral triangles ABC and CDE are constructed; M and P are the midpoints of segments AD and BE. Prove that triangle CPM is an equilateral one. 18.10. Given three parallel lines. Construct an equilateral triangle so that its vertices belong to the given lines. 18.11. Geven a square, consider all possible equilateral triangles PKM with ﬁxed vertex P and vertex K belonging to the square. Find the locus of vertices M. 18.12. On sides BC and CD of parallelogram ABCD, equilateral triangles BCP and CDQ are constructed outwards. Prove that triangle APQ is an equilateral one. 18.13. Point M belongs to arc ⌣AB of the circle circumscribed about an equilateral triangle ABC. Prove that MC = MA + MB. 18.14. Find the locus of points M that lie inside equilateral triangle ABC and such that MA2 = MB2 + MC2. 18.15. Hexagon ABCDEF is a regular one, K and M are the midpoints of segments BD and EF, respectively. Prove that triangle AMK is an equilateral one. 18.16. Let M and N be the midpoints of sides CD and DE, respectively, of regular hexagon ABCDEF, let P be the intersection point of segments AM and BN. a) Find the value of the angle between lines AM and BN. b) Prove that SABP = SMDNP. 18.17. On sides AB and BC of an equilateral triangle ABC points M and N are taken so that MN ∥AC; let E be the midpoint of segment AN and D the center of mass of triangle BMN. Find the values of the angles of triangle CDE. 18.18. On the sides of triangle ABC equilateral triangles ABC1, AB1C and A1BC are constructed outwards. Let P and Q be the midpoints of segments A1B1 and A1C1. Prove that triangle APQ is an equilateral one. 18.19. On sides AB and AC of triangle ABC equilateral triangles ABC′ and AB′C are constructed outwards. Point M divides side BC in the ratio of BM : MC = 3 : 1; points K and L are the midpoints of sides AC′ and B′C, respectively. Prove that the angles of triangle KLM are equal to 30◦, 60◦and 90◦. 18.20. Equilateral triangles ABC, CDE, EHK (vertices are circumvent counterclock-wise) are placed on the plane so that − − → AD = − − → DK. Prove that triangle BHD is also an equilateral one. §4. COMPOSITIONS OF ROTATIONS 347 18.21. a) Inside an acute triangle ﬁnd a point the sum of distances from which to the vertices is the least one. b) Inside triangle ABC all the angles of which are smaller than 120◦a point O is taken; it serves as vertex of the angles of 120◦that subtend the sides. Prove that the sum of distances from O to the vertices is equal to 1 2(a2 + b2 + c2) + 2 √ 3S. 18.22. Hexagon ABCDEF is inscribed in a circle of radius R and AB = CD = EF = R. Prove that the midpoints of sides BC, DE and FA determine an equilateral triangle. 18.23. On sides of a convex centrally symmetric hexagon ABCDEF equilateral triangles are constructed outwards. Prove that the midpoints of the segments connecting the vertices of neighbouring triangles determine a regular hexagon. §3. Rotations through arbitrary angles 18.24. Given points A and B and circle S construct points C and D on S so that AC ∥BD and the value of arc ⌣CD is a given quantity α. 18.25. A rotation with center O transforms line l1 into line l2 and point A1 on l1 into point A2. Prove that the intersection point of lines l1 and l2 belongs to the circle circumscribed about triangle A1OA2. 18.26. Two equal letters Γ lie on the plane. Denote by A and A′ the endpoints of the shorter segments of these letters. Points A1, . . . , An−1 and A′ 1, . . . , A′ n−1 divide the longer segments into n equal parts (the division points are numbered starting from the outer endpoints of longer segments). Lines AAi and A′A′ i intersect at point Xi. Prove that points X1, . . . , Xn−1 determine a convex polygon. 18.27. Along two lines that intersect at point P two points are moving with the same speed: point A along one line and point B along the other one. They pass P not simultane-ously. Prove that at all times the circle circumscribed about triangle ABP passes through a ﬁxed point distinct from P. 18.28. Triangle A1B1C1 is obtained from triangle ABC by a rotation through an angle of α (α < 180◦) about the center of its circumscribed circle. Prove that the intersection points of sides AB and A1B1, BC and B1C1, CA and C1A1 (or their extensions) are the vertices of a triangle similar to triangle ABC. 18.29. Given triangle ABC construct a line which divides the area and perimeter of triangle ABC in halves. 18.30. On vectors − − → AiBi, where i = 1, . . . , k similarly oriented regular n-gons AiBiCiDi . . . (n ≥4) are constructed (a given vector serving as a side). Prove that k-gons C1 . . . Ck and D1 . . . Dk are regular and similarly oriented ones if and only if the k-gons A1 . . . Ak and B1 . . . Bk are regular and similarly oriented ones. 18.31. Consider a triangle. Consider three lines symmetric through the triangles sides to an arbitrary line passing through the intersection point of the triangle’s heights. Prove that the three lines intersect at one point. 18.32. A lion runs over the arena of a circus which is a disk of radius 10 m. Moving along a broken line the lion covered 30 km. Prove that the sum of all the angles of his turns is not less than 2998 radian. §4. Compositions of rotations 18.33. Prove that the composition of two rotations through angles whose sum is not proportional to 360◦is a rotation. In which point is its center and what is the angle of the rotation equal to? Investigate also the case when the sum of the angles of rotations is a multiple of 360◦. 348 CHAPTER 18. ROTATIONS 18.34. On the sides of an arbitrary convex quadrilateral squares are constructed out-wards. Prove that the segments that connect the centers of opposite squares have equal lengths and are perpendicular to each other. 18.35. On the sides of a parallelogram squares are constructed outwards. Prove that their centers form a square. 18.36. On sides of triangle ABC squares with centers P, Q and R are constructed outwards. On the sides of triangle PQR squares are constructed inwards. Prove that their centers are the midpoints of the sides of triangle ABC. 18.37. Inside a convex quadrilateral ABCD isosceles right triangles ABO1, BCO2, CDO3 and DAO4 are constructed. Prove that if O1 = O3, then O2 = O4. 18.38. a) On the sides of an arbitrary triangle equilateral triangles are constructed outwards. Prove that their centers form an equilateral triangle. b) Prove a similar statement for triangles constructed inwards. c) Prove that the diﬀerence of the areas of equilateral triangles obtained in headings a) and b) is equal to the area of the initial triangle. 18.39. On sides of triangle ABC equilateral triangles A′BC and B′AC are constructed outwards and C′AB inwards; M is the center of mass of triangle C′AB. Prove that A′B′M is an isosceles triangle such that ∠A′MB′ = 120◦. 18.40. Let angles α, β, γ be such that 0 < α, β, γ < π and α + β + γ = π. Prove that if the composition of rotations R2γ C ◦R2β B ◦R2α A is the identity transformation, then the angles of triangle ABC are equal to α, β, γ. 18.41. Construct an n-gon given n points which are the vertices of isosceles triangles constructed on the sides of this n-gon and such that the angles of these triangles at the vertices are equal to α1, . . . , αn. 18.42. On the sides of an arbitrary triangle ABC isosceles triangles A′BC, AB′C and ABC′ are constructed outwards with angles α, β and γ at vertices A′, B′ and C′, respectively, such that α + β + γ = 2π. Prove that the angles of triangle A′B′C′ are equal to 1 2α, 1 2β and 1 2γ. 18.43. Let AKL and AMN be similar isosceles triangles with vertex A and angle α at the vertex; GNK and G′LM similar isosceles triangles with angle π −α at the vertex. Prove that G = G′. (All the triangles are oriented ones.) 18.44. On sides AB, BC and CA of triangle ABC points P, Q and R, respectively, are taken. Prove that the centers of the circles circumscribed about triangles APR, BPQ and CQR constitute a triangle similar to triangle ABC. Problems for independent study 18.45. On the plane, the unit circle with center at O is drawn. Two neighbouring vertices of a square belong to this circle. What is the maximal distance from point O that the two other of the square’s vertices can have? 18.46. On the sides of convex quadrilateral ABCD, equilateral triangles ABM, CDP are constructed outwards and BCN, ADK inwards. Prove that MN = AC. 18.47. On the sides of a convex quadrilateral ABCD, squares with centers M, N, P, Q are constructed outwards. Prove that the midpoints of the diagonals of quadrilaterals ABCD and MNPQ form a square. SOLUTIONS 349 18.48. Inside an equilateral triangle ABC lies point O. It is known that ∠AOB = 113◦, ∠BOC = 123◦. Find the angles of the triangle whose sides are equal to segments OA, OB, OC. 18.49. On the plane, there are drawn n lines (n > 2) so that no two of them are parallel and no three intersect at one point. It is known that it is possible to rotate the plane about a point O through an angle of α (α < 180◦) so that each of the drawn lines coincides with some other of the drawn lines. Indicate all n for which this is possible. 18.50. Ten gears of distinct shapes are placed so that the ﬁrst gear is meshed with the second one, the second one with the third one, etc., the tenth is meshed with the ﬁrst one. Is it possible for such a system to rotate? Can a similar system of 11 gears rotate? 18.51. Given a circle and a point. a) Construct an equilateral triangle whose heights intersect at the given point and two vertices belong to the given circle. b) Construct a square two vertices of which belong to the given circle and the diagonals intersect at the given point. Solutions 18.1. Let us rotate square ABCD about point A through 90◦so that B turns into D. This rotation sends point M into point M ′ and point K into point K′. It is clear that ∠BMA = ∠DM ′A. Since ∠MAK = ∠MAB = ∠M ′AD, it follows that ∠MAD = ∠M ′AK. Therefore, ∠MA′K = ∠MAD = ∠BMA = ∠DM ′A. Hence, AK = KM ′ = KD + DM ′ = KD + BM. 18.2. Under the rotation through 90◦about point P lines PA1, PB1, PM1 and CH turn into lines parallel to CA, CB, CM and AB, respectively. It follows that under such a rotation of triangle PA1B1 segment PM1 turns into a median of the (rotated) triangle. 18.3. Consider a rotation through 90◦about point B which sends vertex K into vertex N and vertex C into A. This rotation sends point A into point A′ and point E into point E′. Since E′ and B are the midpoints of sides A′N and A′C of triangle A′NC, it follows that BE′ ∥NC. But ∠EBE′ = 90◦and, therefore, BE ⊥NC. 18.4. A rotation through an angle of 90◦about the center of the square sends point A1 to point A2. This rotation sends the perpendiculars dropped from points A1, A2, A3 and A4 into lines A2P, A3P, A4P and A1P, respectively. Therefore, the intersection point is the image of point P under the inverse rotation. 18.5. Let us turn the given square through an angle of 90◦about point A so that vertex B would coincide with D. Let M ′ be the image of M under this rotation. Since by the hypothesis MK + MC + CK = (BM + MC) + (KD + CK), it follows that MK = BM + KD = DM ′ + KD = KM ′. Moreover, AM = AM ′; hence, △AMK = △AM ′K, consequently, ∠MAK = ∠M ′AK = 1 2∠MAM ′ = 45◦. 18.6. Let R be the rotation through an angle of 90◦that sends − − → BC to − → BA. Further, let − − → BC = a, − − → CB2 = b and − − → AB1 = c. Then − → BA = Ra, − − → D2C = Rb and − − → AD1 = Rc. Hence, − − − → D2D1 = Rb −a + Ra + Rc and 2− − → BM = a + b + Ra + c. Therefore, R(2− − → BM) = − − − → D2D1 because R(Ra) = −a. 18.7. Let us introduce the following notations: a = − − → BM, b = − − → BC; let Ra and Rb be the vectors obtained from vectors a and b under a rotation through an angle of 90◦, i.e., Ra = − → BA, Rb = − − → BQ. Let O1, O2, O3 and O4 be the midpoints of segments AM, MQ, QC 350 CHAPTER 18. ROTATIONS and CA, respectively. Then − − → BO1 = (a + Ra) 2 , − − → BO2 = (a + Rb) 2 , − − → BO3 = (b −Rb) 2 , − − → BO4 = (b + Ra) 2 . Therefore, − − − → O1O2 = 1 2(Rb −Ra) = −− − − → O3O4 and − − − → O2O3 = 1 2(b −a) = −O4O1. Moreover, − − − → O1O2 = R(− − − → O2O3). 18.8. Parallelogram A1B1C1D1 is circumscribed around square ABCD so that point A belongs to side A1B1, B to side B1C1, etc. Let us drop perpendiculars l1, l2, l3 and l4 from vertices A1, B1, C1 and D1, respectively to the sides of the square. To prove that these perpendiculars form a square, it suﬃces to verify that under a rotation through an angle of 90◦about the center O of square ABCD lines l1, l2, l3 and l4 turn into each other. Under the rotation about O through an angle of 90◦points A1, B1, C1 and D1 turn into points A2, B2, C2 and D2 (Fig. 21). Figure 163 (Sol. 18.8) Since AA2 ⊥B1B and BA2 ⊥B1A, it follows that B1A2 ⊥AB. This means that line l1 turns under the rotation through an angle of 90◦about O into l2. For the other lines the proof is similar. 18.9. Let us consider a rotation through an angle of 60◦about point C that turns E into D. Under this rotation B turns into A, i.e., segment BE turns into AD. Therefore, the midpoint P of segment BE turns into the midpoint M of segment AD, i.e., triangle CPM is an equilateral one. 18.10. Suppose that we have constructed triangle ABC so that its vertices A, B and C lie on lines l1, l2 and l3, respectively. Under the rotation through an angle of 60◦with center A point B turns into point C and, therefore, C is the intersection point of l3 and the image of l2 under the rotation through an angle of 60◦about A. 18.11. The locus to be found consists of two squares obtained from the given one by rotations through angles of ±60◦about P. SOLUTIONS 351 18.12. Under the rotation through an angle of 60◦vectors − → QC and − → CP turn into − − → QD and − − → CB = − − → DA, respectively. Therefore, under this rotation vector − → QP = − → QC + − → CP turns into vector − − → QD + − − → DA = − → QA. 18.13. Let M ′ be the image of M under the rotation through an angle of 60◦about B that turns A into C. Then ∠CM ′B = ∠AMB = 120◦. Triangle MM ′B is an equilateral one and, therefore, ∠BM ′M = 60◦. Since ∠CM ′B + ∠BM ′M = 180◦, point M ′ belongs to segment MC. Therefore, MC = MM ′ + M ′C = MB + MA. 18.14. Under the rotation through an angle of 60◦about A sending B to C point M turns into point M ′ and point C into point D. The equality MA2 = MB2 + MC2 is equivalent to the equality M ′M 2 = M ′C2 + MC2, i.e., ∠MCM ′ = 90◦and, therefore, ∠MCB + ∠MBC = ∠MCB + ∠M ′CD = 120◦−90◦= 30◦ that is ∠BMC = 150◦. The locus to be found is the arc of the circle situated inside the triangle and such that the pionts of the arc serve as vertices of angles of 150◦subtending segment BC. 18.15. Let O be the center of a hexagon. Consider a rotation about A through an angle of 60◦sending B to O. This rotation sends segment OC into segment FE. Point K is the midpoint of diagonal BD of parallelogram BCDO because it is the midpoint of diagonal CO. Therefore, point K turns into M under our rotation; in other words, triangle AMK is an equilateral one. 18.16. There is a rotation through an angle of 60◦about the center of the given hexagon that sends A into B. It sends segment CD into DE and, therefore, sends M into N. Therefore, this rotation sends AM into BN, that is to say, the angle between these segments is equal to 60◦. Moreover, this rotation turns pentagon AMDEF into BNEFA; hence, the areas of the pentagons are equal. Cutting from these congruent pentagons the common part, pentagon APNEF, we get two ﬁgures of the same area: triangle ABP and quadrilateral MDNP. 18.17. Consider the rotation through an angle of 60◦about C sending B to A. It sends points M, N and D into M ′, N ′ and D′, respectively. Since AMNN ′ is a parallelogram, the midpoint E of diagonal AN is its center of symmetry. Therefore, under the symmetry through point E triangle BMN turns into M ′AN ′ and, therefore, D turns into D′. Hence, E is the midpoint of segment DD′. Since triangle CDD′ is an equilateral one, the angles of triangle CDE are equal to 30◦, 60◦and 90◦. 18.18. Consider a rotation about A sending point C1 into B. Under this rotation equilateral triangle A1BC turns into triangle A2FB1 and segment A1C1 into segment A2B. It remains to notice that BA1A2B1 is a parallelogram, i.e., the midpoint of segment A2B coincides with the midpoint of segment A1B1. 18.19. Let − → AB = 4a, − → CA = 4b. Further, let R be the rotation sending vector − → AB into − − → AC′ (and, therefore, sending − → CA into − − → CB′). Then − − → LM = (a + b) −2Rb and − − → LK = −2Rb + 4b + 2Ra. It is easy to verify that b + R2b = Rb. Hence, 2R(− − → LM) = − − → LK which implies the required statement. 18.20. Under the rotation about point C through an angle of 60◦counterclockwise point A turns into B and D into E and, therefore, vector − − → DK = − − → AD turns into − − → BE. Since the rotation about point H through an angle of 60◦counterclockwise sends K into E and − − → DK into − − → BE, it sends D into B which means that triangle BHD is an equilateral one. 18.21. a) Let O be a point inside triangle ABC. The rotation through an angle of 60◦ about A sends B, C and O into some points B′, C′ and O′, respectively, see Fig. 22. Since 352 CHAPTER 18. ROTATIONS Figure 164 (Sol. 18.21) AO = OO′ and OC = O′C′, we have: BO + AO + CO = BO + OO′ + O′C′. The length of the broken line BOO′C′ is minimal if and only if this broken line is a segment, i.e., if ∠AOB = ∠AO′C′ = ∠AOC = 120◦. To construct the desired point, we can make use of the result of Problem 2.8. b) The sum of distances from O to the vertices is equal to the length of segment BC′ obtained in heading a). It is also clear that (BC′)2 = b2 + c2 −2bc cos(α + 60◦) = b2 + c2 −bc cos α + bc √ 3 sin α = 1 2(a2 + b2 + c2) + 2 √ 3S. 18.22. Let P, Q and R be the midpoints of sides BC, DE and FA; let O be the center of the circumscribed circle. Suppose that triangle PQR is an equilateral one. Let us prove then that the midpoints of sides BC, DE′ and F ′A of hexagon ABCDE′F ′ in which vertices E′ and F ′ are obtained from vertices E and F after a rotation through an angle about point O also form an equilateral triangle. This will complete the proof since for a regular hexagon the midpoints of sides BC, DE and FA constitute an equilateral triangle and any of the considered hexagons can be obtained from a regular one with the help of rotations of triangles OCD and OEF. Figure 165 (Sol. 18.22) Let Q′ and R′ be the midpoints of sides DE′ and AF ′, see Fig. 23. Under the rotation through an angle of 60◦vector − − → EE′ turns into − − → FF ′. Since − − → QQ′ = 1 2 − − → EE′ and − − → RR′ = 1 2 − − → FF ′, this rotation sends − − → QQ′ into − − → RR′. By hypothesis, triangle PQR is an equilateral one, i.e., under the rotation through an angle of 60◦vector − → PQ turns into − → PR. Therefore, vector SOLUTIONS 353 − − → PQ′ = − → PQ + − − → QQ′ turns into vector − − → PR′ = − → PR + − − → RR′ under a rotation through an angle of 60◦. This means that triangle PQ′R′ is an equilateral one. 18.23. Let K, L, M and N be vertices of equilateral triangles constructed (wherewards?) on sides BC, AB, AF and FE, respectively; let also B1, A1 and F1 be the midpoints of segments KL, LM and MN (see Fig. 24). Figure 166 (Sol. 18.23) Further, let a = − − → BC = − → FE, b = − → AB and c = − → AF; let R be the rotation through an angle of 60◦that sends − − → BC into − − → BK. Then − − → AM = −R2c and − − → FN = −R2a. Therefore, 2− − − → A1B1 = R2c + Ra + b and 2− − − → F1A1 = R2a −c + Rb, i.e., − − − → F1A1 = R(− − − → A1B1). 18.24. Suppose a rotation through an angle of α about the center of circle S sends C into D. This rotation sends point A into point A′. Then ∠(BD, DA′) = α, i.e., point D belongs to the arc of the circle whose points serve as vertices of the angles of α that subtend segment A′B. 18.25. Let P be the intersection point of lines l1 and l2. Then ∠(OA1, A1P) = ∠(OA1, l1) = ∠(OA2, l2) = ∠(OA2, A2P). Therefore, points O, A1, A2 and P belong to one circle. 18.26. It is possible to identify similar letters Γ after a rotation about O (unless they can be identiﬁed by a parallel translation in which case AAi ∥A′A′ i). Thanks to Problem 18.25 point Xi belongs to the circle circumscribed about triangle A′OA. It is clear that the points that belong to one circle constitute a convex polygon. 18.27. Let O be the center of rotation R that sends segment A(t1)A(t2) into segment B(t1)B(t2), where t1 and t2 are certain time moments. Then this rotation sends A(t) into B(t) at any moment t. Therefore, by Problem 18.25 point O belongs to the circle circum-scribed about triangle APB. 18.28. Let A and B be points on the circle with center O; let A1 and B1 be the images of these points under the rotation through an angle of α about O. Let P and P1 be the midpoints of segments AB and A1B1; let M be the intersection point of lines AB and A1B1. The right triangles POM and P1OM have a common hypothenuse and equal legs PO = P1O, therefore, these triangles are equal and ∠MOP = ∠MOP1 = 1 2α. Point M is obtained from point P under a rotation through an angle of 1 2α and a subsequent homothety with coeﬃcient 1 cos( 1 2 α) and center O. The intersection points of lines AB and A1B1, AC and A1C1, BC and B1C1 are the vertices of a triangle which is homothetic with coeﬃcient 1 cos( 1 2 α) to the triangle determined 354 CHAPTER 18. ROTATIONS by the midpoints of the sides of triangle ABC. It is clear that the triangle determined by the midpoints of the sides of triangle ABC is similar to triangle ABC. 18.29. By Problem 5.50 the line which divides in halves both the area and the perimeter of a triangle passes through the center of its inscribed circle. It is also clear that if the line passes through the center of the inscribed circle of a triangle and divides its perimeter in halves, then it divides in halves its area as well. Therefore, we have to draw a line passing through the center of the inscribed circle of the triangle and dividing its perimeter in halves. Suppose we have constructed points M and N on sides AB and AC of triangle ABC so that line MN passes through the center O of the inscribed circle and divides the perimeter of the triangle in halves. On ray AC construct point D so that AD = p, where p is a semiperimeter of triangle ABC. Then AM = ND. Let Q be the center of rotation R that sends segment AM into segment DN (so that A goes to D and M to N). Since the angle between lines AM and CN is known, it is possible to construct Q: it is the vertex of isosceles triangle AQD, where ∠AQD = 180◦−∠A and points B and Q lie on one side of line AD. The rotation R sends segment OM into segment O′N. We can now construct point O′. Clearly, ∠ONO′ = ∠A because the angle between lines OM and O′N is equal to ∠A. Therefore, point N is the intersection point of line AC and the arc of the circle whose points serve as vertices for the angles equal to ∠A that subtend segment OO′. Constructing point N, draw line ON and ﬁnd point M. It is easy to verify that if the constructed points M and N belong to sides AB and AC, then MN is the desired line. The main point of the proof is the proof of the fact that the rotation about Q through an angle of 180◦−∠A sends M into N. To prove this fact, one has to make use of the fact that ∠ONO′ = ∠A, i.e., this rotation sends line OM into line O′N. 18.30. Suppose that the k-gons C1 . . . Ck and D1 . . . Dk are regular and similarly oriented. Let C and D be the centers of these k-gons; let ci = − − → CCi and di = − − → DDi. Then − − → CiDi = − − → CiC + − − → CD + − − → DDi = −ci + − − → CD + di. The rotation Rϕ, where ϕ is the angle at a vertex of a regular n-gon, sends − − → CiDi into − − → CiBi. Therefore, − − → XBi = − − → XC + ci + − − → CiBi = − − → XC + ci + Rϕ(−ci + − − → CD + di). Let us select point X so that − − → XC + Rϕ(− − → CD) = − → 0 . Then − − → XBi = ci + Rϕ(di −ci) = Riψu, where u = ck + Rϕ(dk −ck) and Rψ is the rotation sending ck to c1. Hence, B1 . . . Bk is a regular k-gon with center X. We similarly prove that A1 . . . Ak is a regular k-gon. The converse statement is similarly proved. 18.31. Let H be the intersection point of heights of triangle ABC; let H1, H2 and H3 be points symmetric to H through sides BC, CA and AB, respectively. Points H1, H2 and H3 belong to the circle circumscribed about triangle ABC (Problem 5.9). Let l be a line passing through H. The line symmetric to l through BC (resp. through CA and AB) intersects the circumscribed circle at point H1 (resp. H2 and H3) and at a point P1 (resp. P2 and P3). Consider another line l′ passing through H. Let ϕ be the angle between l and l′. Let us construct points P ′ 1, P ′ 2 and P ′ 3 for line l′ in the same way as points P1, P2 and P3 were constructed for line l. Then ∠PiHiP ′ i = ϕ, i.e., the value of arc ⌣PiP ′ i is equal to 2ϕ (the direction of the rotation from Pi to P ′ i is opposite to that of the rotation from l to l′). Therefore, points P ′ 1, P ′ 2 and P ′ 3 are the images of points P1, P2 and P3 under a certain rotation. It is clear that if for l′ we take the height of the triangle dropped from vertex A, then P ′ 1 = P ′ 2 = P ′ 3 = A, and, therefore, P1 = P2 = P3. SOLUTIONS 355 18.32. Suppose that the lion ran along the broken line A1A2 . . . An. Let us rectify the lion’s trajectory as follows. Let us rotate the arena of the circus and all(?) the further trajectory about point A2 so that point A3 would lie on ray A1A2. Then let us rotate the arena and the further trajectory about point A3 so that point A4 were on ray A1A2, and so on. The center O of the arena turns consecutively into points O1 = O, O2, . . . , On−1; and points A1, . . . , An into points A′ 1, . . . , A′ n all on one line (Fig. 25). Figure 167 (Sol. 18.32) Let αi−1 be the angle of through which the lion turned at point A′ i. Then ∠Oi−1A′ iOi = αi−1 and A′ iOi−1 = A′ iOi ≤10; hence, OiOi−1 ≤10αi−1. Hence, 30000 = A′ 1A′ n ≤A′ 1O1 + O1O2 + · · · + On−2On−1 + On−1A′ n ≤ 10 + 10(α1 + · · · + αn−2) + 10 i.e., α1 + · · · + αn−2 ≥2998. 18.33. Consider the composition of the rotations Rβ B ◦Rα A. If A = B, then the statement of the problem is obvious and, therefore, let us assume that A ̸= B. Let l = AB; let lines a and b pass through points A and B, respectively, so that ∠(a, l) = 1 2α and ∠(l, b) = 1 2β. Then Rβ B ◦Rα A = Sb ◦Sl ◦Sl ◦Sa = Sb ◦Sa. If a ∥b, then Sa ◦Sb = T2u, where Tu is a parallel translation sending a into b and such that u ⊥a. If lines a and b are not parallel and O is their intersection point, then Sa ◦Sb is the rotation through an angle of α + β with center O. It is also clear that a ∥b if and only if 1 2α + 1 2β = kπ, i.e., α + β = 2kπ. 18.34. Let P, Q, R and S be the centers of squares constructed outwards on sides AB, BC, CD and DA, respectively. On segments QR and SP, construct inwards isosceles right triangles with vertices O1 and O2. Then D = R90◦ R ◦R90◦ Q (B) = R180◦ O1 (B) and B = R90◦ P ◦R90◦ S (D) = R180◦ O2 (D), i.e., O1 = O2 is the midpoint of segment BD. The rotation through an angle of 90◦about point O = O1 = O2 that sends Q into R sends point S into P, i.e., it sends segment QS into RP and, therefore, these segments are equal and perpendicular to each other. 18.35. Let P, Q, R and S be the centers of squares constructed outwards on the sides AB, BC, CD and DA of parallelogram ABCD. By the previous problem PR = QS and PR ⊥QS. Moreover, the center of symmetry of parallelogram ABCD is the center of symmetry of quadrilateral PQRS. This means that PQRS is a parallelogram with equal and perpendicular diagonals, hence, a square. 18.36. Let P, Q and R be the centers of squares constructed outwards on sides AB, BC and CA. Let us consider a rotation through an angle of 90◦with center R that sends C to A. Under the rotation about P through an angle of 90◦in the same direction point A turns into B. The composition of these two rotations is a rotation through an angle of 180◦and, 356 CHAPTER 18. ROTATIONS therefore, the center of this rotation is the midpoint of segment BC. On the other hand, the center of this rotation is a vertex of an isoscceles right triangle with base PR, i.e., it is the center of a square constructed on PR. This square is constructed inwards on a side of triangle PQR. 18.37. If O1 = O3, then R90◦ D ◦R90◦ C ◦R90◦ B ◦R90◦ A = R180◦ O3 ◦R180◦ O1 = E. Therefore, E = R90◦ A ◦E ◦R−90◦ A = R90◦ A ◦R90◦ D R90◦ C ◦R90◦ B = R180◦ O4 ◦R180◦ O2 , where E is the identity transformation, i.e., O4 = O2. 18.38. a) See solution of a more general Problem 18.42 (it suﬃces to set α = β = γ = 120◦). In case b) proof is analogous. b) Let Q and R (resp. Q1 and R1) be the centers of equilateral triangles constructed outwards (resp. inwards) on sides AC and AB. Since AQ = 1 √ 3b, AR = 1 √ 3c and ∠QAR = 60◦+ α, it follows that 3QR2 = b2 + c2 −2bc cos(α + 60◦). Similarly, 3Q1R2 1 = b2 + c2 − 2bc cos(α −60◦). Therefore, the diﬀerence of areas of the obtained equilateral triangles is equal to (QR2 −Q1R2 1) √ 3 4 = bc sin α sin 60◦ √ 3 = SABC. 18.39. The combination of a rotation through an angle of 60◦about A′ that sends B to C, a rotation through an angle of 60◦about B′ that sends C to A and a rotation through an angle of 120◦about M that sends A to B has B as a ﬁxed point. Since the ﬁrst two rotations are performed in the direction opposite to the direction of the last rotation, it follows that the composition of these rotations is a parallel translation with a ﬁxed point, i.e., the identity transformation: R−120◦ M ◦R60◦ B′ ◦R60◦ A′ = E. Therefore, R60◦ B′ ◦R60◦ A′ = R120◦ M , i.e., M is the center of the rotation R60◦ B′ ◦R60◦ A′ . It follows that ∠MA′B′ = ∠MB′A′ = 30◦, i.e., A′B′M is an isosceles triangle and ∠A′MB′ = 120◦. 18.40. The conditions of the problem imply that R−2γ C = R2β B ◦R2α A , i.e., point C is the center of the composition of rotations R2β B ◦R2α A . This means that ∠BAC = α and ∠ABC = β (see Problem 18.33). Therefore, ∠ACB = π −α −β = γ. 18.41. Denote the given points by M1, . . . , Mn. Suppose that we have constructed polygon A1A2 . . . An so that triangles A1M1A2, A2M2A3, . . . , AnMnA1 are isosceles, where ∠AiMiAi+1 = αi and the sides of the polygon are bases of these isosceles triangles. Clearly, Rαn Mn ◦· · · ◦Rα1 M1(A1) = A1. If α1 + · · · + αn ̸= k · 360◦, then point A1 is the center of the rotation Rαn Mn ◦· · · ◦Rα1 M1. We can construct the center of the composition of rotations. The construction of the other vertices of the polygon is done in an obvious way. If α1 + · · · + αn = k · 360◦, then the problem is ill-posed: either an arbitrary point A1 determines a polygon with the required property or there are no solutions. 18.42. Since Rγ C′ ◦Rβ B′ ◦Rα A′(B) = Rγ C′ ◦Rβ B′(C) = Rγ C′(A) = B, it follows that B is a ﬁxed point of the composition Rγ C′ ◦Rβ B′ ◦Rα A′. Since α + β + γ = 2π, it follows that this composition is a parallel translation with a ﬁxed point, i.e., the identity transformation. It remains to make use of the result of Problem 18.40. 18.43. Since Rπ−α G′ ◦Rα A(N) = L and Rπ−α G ◦Rα A(L) = N, it follows that the transfor-mations Rπ−α G′ ◦Rα A and Rπ−α G ◦Rα A are central symmetries with respect to the midpoint of segment LN, i.e., Rπ−α G′ ◦Rα A = Rπ−α G ◦Rα A. Therefore, Rπ−α G′ = Rπ−α G and G′ = G. 18.44. Let A1, B1 and C1 be the centers of the circumscribed circles of triangles APR, BPQ and CQR. Under the successive rotations with centers A1, B1 and C1 through angles SOLUTIONS 357 2α, 2β and 2γ point R turns ﬁrst into P, then into Q, and then returns home. Since 2α+2β+2γ = 360◦, the composition of the indicated rotations is the identity transformation. It follows that the angles of triangle A1B1C1 are equal to α, β and γ (see Problem 18.40). Chapter 19. HOMOTHETY AND ROTATIONAL HOMOTHETY Background 1. A homothety is a transformation of the plane sending point X into point X′ such that − − → OX′ = k− − → OX, where point O and the number k are ﬁxed. Point O is called the center of homothety and the number k the coeﬃcient of homothety. We will denote the homothety with center O and coeﬃcient k by Hk O. 2. Two ﬁgures are called homothetic if one of them turns into the other one under a homothety. 3. A rotational homothety is the composition of a homothety and a rotation with a common center. The order of the composition is inessential since Rϕ O ◦Hk O = Hk O ◦Rϕ O. We may assume that the coeﬃcient of a rotational homothety is positive since R180◦ O ◦ Hk O = H−k O . 4. The composition of two homotheties with coeﬃcients k1 and k2, where k1k2 ̸= 1, is a homothety with coeﬃcient k1k2 and its center belongs to the line that connects the centers of these homotheties (see Problem 19.23). 5. The center of a rotational homothety that sends segment AB into segment CD is the intersection point of the circles circumscribed about triangles ACP and BDP, where P is the intersection point of lines AB and CD (see Problem 19.41). Introductory problems 1. Prove that a homothety sends a circle into a circle. 2. Two circles are tangent at point K. A line passing through K intersects these circles at points A and B. Prove that the tangents to the circles through A and B are parallel to each other. 3. Two circles are tangent at point K. Through K two lines are drawn that intersect the ﬁrst circle at points A and B and the second one at points C and D. Prove that AB ∥CD. 4. Prove that points symmetric to an arbitrary point with respect to the midpoints of a square’s sides are vertices of a square. 5. Two points A and B and a line l on the plane are given. What is the trajectory of movement of the intersection point of medians of triangle ABC when C moves along l? §1. Homothetic polygons 19.1. A quadrilateral is cut by diagonals into four triangles. Prove that the intersection points of their medians form a parallelogram. 19.2. The extensions of the lateral sides AB and CD of trapezoid ABCD intersect at point K and its diagonals intersect at point L. Prove that points K, L, M and N, where M and N are the midpoints of bases BC and AD, respectively, belong to one line. 19.3. The intersection point of diagonals of a trapezoid is equidistant from the lines to which the sides of the trapezoid belong. Prove that the trapezoid is an isosceles one. 359 360 CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY 19.4. Medians AA1, BB1 and CC1 of triangle ABC meet at point M; let P be an arbitrary point. Line la passes through point A parallel to line PA1; lines lb and lc are similarly deﬁned. Prove that: a) lines la, lb and lc meet at one point, Q; b) point M belongs to segment PQ and PM : MQ = 1 : 2. 19.5. Circle S is tangent to equal sides AB and BC of an isosceles triangle ABC at points P and K, respectively, and is also tangent from the inside to the circle circumscribed about triangle ABC. Prove that the midpoint of segment PK is the center of the circle inscribed into triangle ABC. 19.6. A convex polygon possesses the following property: if all its sides are pushed by distance 1 outwards and extended, then the obtained lines form a polygon similar to the initial one. Prove that this polygon is a circumscribed one. 19.7. Let R and r be the radii of the circumscribed and inscribed circles of a triangle. Prove that R ≥2r and the equality is only attained for an equilateral triangle. 19.8. Let M be the center of mass of an n-gon A1 . . . An; let M1, . . . , Mn be the centers of mass of the (n−1)-gons obtained from the given n-gon by discarding vertices A1, . . . , An, respectively. Prove that polygons A1 . . . An and M1 . . . Mn are homothetic to each other. 19.9. Prove that any convex polygon Φ contains two nonintersecting polygons Φ1 and Φ2 similar to Φ with coeﬃcient 1 2. See also Problem 5.87. §2. Homothetic circles 19.10. On a circle, points A and B are ﬁxed and point C moves along this circle. Find the locus of the intersection points of the medians of triangles ABC. 19.11. a) A circle inscribed into triangle ABC is tangent to side AC at point D, and DM is its diameter. Line BM intersects side AC at point K. Prove that AK = DC. b) In the circle, perpendicular diameters AB and CD are drawn. From point M outside the circle there are drawn tangents to the circle that intersect AB at points E and H and also lines MC and MD that intersect AB at points F and K, respectively. Prove that EF = KH. 19.12. Let O be the center of the circle inscribed into triangle ABC, let D be the point where the circle is tangent to side AC and B1 the midpoint of AC. Prove that line B1O divides segment BD in halves. 19.13. The circles α, β and γ are of the same radius and are tangent to the sides of angles A, B and C of triangle ABC, respectively. Circle δ is tangent from the outside to all the three circles α, β and γ. Prove that the center of δ belongs to the line passing through the centers of the circles inscribed into and circumscribed about triangle ABC. 19.14. Consider triangle ABC. Four circles of the same radius ρ are constructed so that one of them is tangent to the three other ones and each of those three is tangent to two sides of the triangle. Find ρ given the radii r and R of the circles inscribed into and circumscribed about the triangle. §3. Costructions and loci 19.15. Consider angle ∠ABC and point M inside it. Construct a circle tangent to the legs of the angle and passing through M. 19.16. Inscribe two equal circles in a triangle so that each of the circles were tangent to two sides of the triangle and the other circle. §5. ROTATIONAL HOMOTHETY 361 19.17. Consider acute triangle ABC. Construct points X and Y on sides AB and BC, respectively, so that a) AX = XY = Y C; b) BX = XY = Y C. 19.18. Construct triangle ABC given sides AB and AC and bisector AD. 19.19. Solve Problem 16.18 with the help of homothety. 19.20. On side BC of given triangle ABC, construct a point such that the line that connects the bases of perpendiculars dropped from this point to sides AB and AC is parallel to BC. 19.21. Right triangle ABC is modiﬁed so that vertex A of the right angle is ﬁxed whereas vertices B and C slide along ﬁxed circles S1 and S2 tangent to each other at A from the outside. Find the locus of bases D of heights AD of triangles ABC. See also problems 7.26–7.29, 8.15, 8.16, 8.70. §4. Composition of homotheties 19.22. A transformation f has the following property: if A′ and B′ are the images of points A and B, then − − → A′B′ = k− → AB, where k is a constant. Prove that: a) if k = 1, then f is a parallel translation; b) if k ̸= 1, then f is a homothety. 19.23. Prove that the composition of two homotheties with coeﬃcients k1 and k2, where k1k2 ̸= 1, is a homothety with coeﬃcient k1k2 and its center belongs to the line that connects the centers of these homotheties. Investigate the case k1k2 = 1. 19.24. Common outer tangents to the pairs of circles S1 and S2, S2 and S3, S3 and S1 intersect at points A, B and C, respectively. Prove that points A, B and C belong to one line. 19.25. Trapezoids ABCD and APQD have a common base AD and the length of all their bases are distinct. Prove that the intersections points of the following pairs of lines belong to one line: a) AB and CD, AP and DQ, BP and CQ; b) AB and CD, AQ and DP, BQ and CP. §5. Rotational homothety 19.26. Circles S1 and S2 intersect at points A and B. Lines p and q passing through point A intersect circle S1 at points P1 and Q1 and circle S2 at points P2 and Q2. Prove that the angle between lines P1Q1 and P2Q2 is equal to the angle between circles S1 and S2. 19.27. Circles S1 and S2 intersect at points A and B. Under the rotational homothety P with center A that sends S1 into S2 point M1 from circle S1 turns into M2. Prove that line M1M2 passes through B. 19.28. Circles S1, . . . , Sn pass through point O. A grasshopper hops from point Xi on circle Si to point Xi+1 on circle Si+1 so that line XiXi+1 passes through the intersection point of circles Si and Si+1 distinct from O. Prove that after n hops (from S1 to S2 from S2 to S3, . . . , from Sn to S1) the grasshopper returns to the initial position. 19.29. Two circles intersect at points A and B and chords AM and AN are tangent to these circles. Let us complete triangle MAN to parallelogram MANC and divide segments BN and MC by points P and Q in equal proportions. Prove then that ∠APQ = ∠ANC. 19.30. Consider two nonconcentric circles S1 and S2. Prove that there exist precisely two rotational homotheties with the angle of rotation of 90◦that send S1 into S2. 362 CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY 19.31. Consider square ABCD and points P and Q on sides AB and BC, respectively, so that BP = BQ. Let H be the base of the perpendicular dropped from B on PC. Prove that ∠DHQ = 90◦. 19.32. On the sides of triangle ABC similar triangles are constructed outwards: △A1BC ∼ △B1CA ∼△C1AB. Prove that the intersection points of medians of triangles ABC and A1B1C1 coincide. 19.33. The midpoints of sides BC and B1C1 of equilateral triangles ABC and A1B1C1 coincide (the vertices of both triangles are listed clockwise). Find the value of the angle between lines AA1 and BB1 and also the ratio of the lengths of segments AA1 and BB1. 19.34. Triangle ABC turns under a rotational homothety into triangle A1B1C1; let O be an arbitrary point. Let A2 be the vertex of parallelogram OAA1A2; let points B2 and C2 be similarly deﬁned. Prove that △A2B2C2 ∼△ABC. 19.35. On top of a rectangular map lies a map of the same locality but of lesser scale. Prove that it is possible to pierce by a needle both maps so that the points where both maps are pierced depict the same point of the locality. 19.36. Rotational homotheties P1 and P2 with centers A1 and A2 have the same angle of rotation and the product of their coeﬃcients is equal to 1. Prove that the composition P2 ◦P1 is a rotation and its center coincides with the center of another rotation that sends A1 into A2 and whose angle of rotation is equal to 2∠(− − − → MA1, − − → MN), where M is an arbitrary point and N = P1(M). 19.37. Triangles MAB and MCD are similar but have opposite orientations. Let O1 be the center of rotation through an angle of 2∠(− → AB, − − → BM) that sends A to C and O2 the center of rotation through an angle of 2∠(− → AB, − − → AM) that sends B to D. Prove that O1 = O2. 19.38. Consider a half circle with diameter AB. For every point X on this half circle a point Y is placed on ray XA so that XY = kXB. Find the locus of points Y . 19.39. Consider point P on side AB of (unknown?) triangle ABC and triangle LMN. Inscribe triangle PXY similar to LMN into triangle ABC. 19.40. Construct quadrilateral ABCD given ∠B+∠D and the lengths a = AB, b = BC, c = CD and d = DA. See also Problem 5.122. §6. The center of a rotational homothety 19.41. a) Let P be the intersection point of lines AB and A1B1. Prove that if no points among A, B, A1, B1 and P coincide, then the common point of circles circumscribed about triangles PAA1 and PBB1 is the center of a rotational homothety that sends A to A1 and B to B1 and that such a rotational homothety is unique. b) Prove that the center of a rotational homothety that sends segment AB to segment BC is the intersection point of circles passing through point A and tangent to line BC at point B and the circle passing through C and tangent to line AB at point B. 19.42. Points A and B move along two intersecting lines with constant but distinct speeds. Prove that there exists a point, P, such that at any moment AP : BP = k, where k is the ratio of the speeds. 19.43. Construct the center O of a rotational homothety with a given coeﬃcient k ̸= 1 that sends line l1 into line l2 and point A1 that belongs to l1 into point A2. (?) §7. THE SIMILARITY CIRCLE OF THREE FIGURES 363 19.44. Prove that the center of a rotational homothety that sends segment AB into segment A1B1 coincides with the center of a rotational homothety that sends segment AA1 into segment BB1. 19.45. Four intersecting lines form four triangles. Prove that the four circles circum-scribed about these triangles have one common point. 19.46. Parallelogram ABCD is not a rhombus. Lines symmetric to lines AB and CD through diagonals AC and DB, respectively, intersect at point Q. Prove that Q is the center of a rotational homothety that sends segment AO into segment OD, where O is the center of the parallelogram. 19.47. Consider wo regular pentagons with a common vertex. The vertices of each pentagon are numbered 1 to 5 clockwise so that the common vertex has number 1. Vertices with equal numbers are connected by straight lines. Prove that the four lines thus obtained intersect at one point. 19.48. On sides BC, CA and AB of triangle ABC points A1, B1 and C1 are taken so that △ABC ∼△A1B1C1. Pairs of segments BB1 and CC1, CC1 and AA1, AA1 and BB1 intersect at points A2, B2 and C2, respectively. Prove that the circles circumscribed about triangles ABC2, BCA2, CAB2, A1B1C2, B1C1A2 and C1A1B2 intersect at one point. §7. The similarity circle of three ﬁgures Let F1, F2 and F3 be three similar ﬁgures, O1 the center of a rotational homothety that sends F2 to F3. Let points O2 and O3 be similarly deﬁned. If O1, O2 and O3 do not belong to one line, then triangle O1O2O3 is called the similarity triangle of ﬁgures F1, F2 and F3 and its circumscribed circle is called the similarity circle of these ﬁgures. In case points O1, O2 and O3 coincide the similarity circle degenerates into the center of similarity and in case when not all these points coincide but belong to one line the similarity circle degenerates into the axis of similarity. In the problems of this section we assume that the similarity circle of the ﬁgures consid-ered is not degenerate. 19.49. Lines A2B2 and A3B3, A3B3 and A1B1, A1B1 and A2B2 intersect at points P1, P2, P3, respectively. a) Prove that the circumscribed circles of triangles A1A2P3, A1A3P2 and A2A3P1 intersect at one point that belongs to the similarity circle of segments A1B1, A2B2 and A3B3. b) Let O1 be the center of rotational homothety that sends segment A2B2 into segment A3B3; points O2 and O3 be similarly deﬁned. Prove that lines P1O1, P2O2 and P3O3 intersect at one point that belongs to the similarity circle of segments A1B1, A2B2 and A3B3. Points A1 and A2 are called correspondent points of similar ﬁgures F1 and F2 if the rotational symmetry that sends F1 to F2 transforms A1 into A2. Correspondent lines and correspondent segments are analogously deﬁned. 19.50. Let A1B1, A2B2 and A3B3 and also A1C1, A2C2 and A3C3 be correspondent segments of similar ﬁgures F1, F2 and F3. Prove that the triangle formed by lines A1B1, A2B2 and A3B3 is similar to the triangle formed by lines A1C1, A2C2 and A3C3 and the center of the rotational homothety that sends one of these triangles into another one belongs to the similarity circle of ﬁgures F1, F2 and F3. 19.51. Let l1, l2 and l3 be the correspondent lines of similar ﬁgures F1, F2 and F3 and let the lines intersect at point W. a) Prove that W belongs to the similarity circle of F1, F2 and F3. 364 CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY b) Let J1, J2 and J3 be distinct from W intersection points of lines l1, l2 and l3 with the similarity circle. Prove that these points only depend on ﬁgures F1, F2 and F3 and do not depend on the choice of lines l1, l2 and l3. Points J1, J2 and J3 are called constant points of similar ﬁgures F1, F2 and F3 and triangle J1J2J3 is called the constant triangle of similar ﬁgures. 19.52. Prove that the constant triangle of three similar ﬁgures is similar to the triangle formed by their correspondent lines and these triangles have opposite orientations. 19.53. Prove that constant points of three similar ﬁgures are their correspondent points. The similarity circle of triangle ABC is the similarity circle of segments AB, BC and CA (or of any three similar triangles constructed from these segments). Constant points of a triangle are the constant points of the three ﬁgures considered. 19.54. Prove that the similarity circle of triangle ABC is the circle with diameter KO, where K is Lemoin’s point and O is the center of the circumscribed circle. 19.55. Let O be the center of the circumscribed circle of triangle ABC, K Lemoin’s point, P and Q Brokar’s points, ϕ Brokar’s angle (see Problems 5.115 and 5.117). Prove that points P and Q belong to the circle of diameter KO and OP = OQ and ∠POQ = 2ϕ. Problems for independent study 19.56. Given triangles ABC and KLM. Inscribe triangle A1B1C1 into triangle ABC so that the sides of A1B1C1 wre parallel to the respective sides of triangle KLM. 19.57. On the plane, there are given points A and E. Construct a rhombus ABCD with a given height for which E is the midpoint of BC. 19.58. Consider a quadrilateral. Inscribe a rombus in it so that the sides of the rombus are parallel to the diagonals of the quandrangle. 19.59. Consider acute angle ∠AOB and point C inside it. Find point M on leg OB equidistant from leg OA and from point C. 19.60. Consider acute triangle ABC. Let O be the intersection point of its heights; ω the circle with center O situated inside the triangle. Construct triangle A1B1C1 circumscribed about ω and inscribed in triangle ABC. 19.61. Consider three lines a, b, c and three points A, B, C each on the respective line. Construct points X, Y , Z on lines a, b, c, respectively, so that BY : AX = 2, CZ : AX = 3 and so that X, Y , Z are all on one line. Solutions 19.1. A homothety with the center at the intersection point of the diagonals of the quadrilateral and with coeﬃcient 3/2 sends the intersection points of the medians of the triangles in question into the midpoints of the sides of the quadrilateral. It remains to make use of the result of Problem 1.2. 19.2. The homothety with center K that sends △KBC into △KAD sends point M into N and, therefore, K belongs to line MN. The homothety with center L that sends △LBC into △LDA sends M into N. Therefore, L belongs to line MN. 19.3. Suppose the continuations of the lateral sides AB and CD intersect at point K and the diagonals of the trapezoid intersect at point L. By the preceding problem line KL passes through the midpoint of segment AD and by the hypothesis this line divides angle ∠AKD in halves. Therefore, triangle AKD is an isosceles one (see Problem 16.1); hence, so is trapezoid ABCD. SOLUTIONS 365 19.4. The homothety with center M and coeﬃcient −2 sends lines PA1, PB1 and PC1 into lines la, lb and lc, respectively, and, therefore, the point Q to be found is the image of P under this homothety. 19.5. Consider homothety Hk B with center B that sends segment AC into segment A′C′ tangent to the circumscribed circle of triangle ABC. Denote the midpoints of segments PK and A′C′ by O1 and D, respectively, and the center of S by O. Circle S is the inscribed circle of triangle A′BC′ and, therefore, it suﬃces to show that homothety Hk B sends O1 to O. To this end it suﬃces to verify that BO1 : BO = BA : BA′. This equality follows from the fact that PO1 and DA are heights of similar right triangles BPO and BDA′. 19.6. Let k be the similarity coeﬃcient of polygons and k < 1. Shifting the sides of the initial polygon inside consecutively by k, k2, k3, . . . units of length we get a contracting system of embedded convex polygons similar to the initial one with coeﬃcients k, k2, k3, . . . . The only common point of these polygons is the center of the inscribed circle of the initial polygon. 19.7. Let A1, B1 and C1 be the midpoints of sides BC, AC and AB, respectively. The homothety with center at the intersection point of the medians of triangle ABC and with coeﬃcient −1 2 sends the circumscribed circle S of triangle ABC into the circumscribed circle S1 of triangle A1B1C1. Since S1 passes through all the vertices of triangle ABC, we can construct triangle A′B′C′ whose sides are parallel to the respective sides of triangle ABC and for which S1 is the inscribed circle, see Fig. 26. Figure 68 (Sol. 19.7) Let r and r′ be the radii of the inscribed circles of triangles ABC and A′B′C′; let R and R1 be the radii of S and S1, respectively. Clearly, r ≤r′ = R1 = R/2. The equality is attained if triangles A′B′C′ and ABC coincide, i.e., if S1 is the inscribed circle of triangle ABC. In this case AB1 = AC1 and, therefore, AB = AC. Similarly, AB = BC. 19.8. Since − − − → MMi = − − − → MA1 + · · · + − − − → MAn −− − → MAi n −1 = − − − → MAi n −1, it follows that the homothety with center M and coeﬃcient − 1 n−1 sends Ai into Mi. 19.9. Let A and B be a pair of most distant from each other points of polygon Φ. Then Φ1 = H1/2 A (Φ) and Φ2 = H1/2 B (Φ) are the required ﬁgures. Indeed, Φ1 and Φ2 do not intersect because they lie on diﬀerent sides of the midperpen-dicular to segment AB. Moreover, Φ1 and Φ2 are contained in Φ because Φ is a convex polygon. 19.10. Let M be the intersection point of the medians of triangle ABC, O the midpoints of segment AB. Clearly, 3− − → OM = − → OC and, therefore, points M ﬁll in the circle obtained from the initial circle under the homothety with coeﬃcient 1 3 and center O. 366 CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY 19.11. a) The homothety with center B that sends the inscribed circle into the escribed circle tangent to side AC sends point M into point M ′. Point M ′ is the endpoint of the diameter perpendicular to AC and, therefore, M ′ is the tangent point of the inscribed circle with AC, hence, it is the intersection point of BM with AC. Therefore, K = M ′ and K is the tangent point of the escribed circle with side AC. Now it is easy to compute that AK = 1 2(a + b −c) = CD, where a, b and c are the lengths of the sides of triangle ABC. b) Consider a homothety with center M that sends line EH into a line tangent to the given circle. This homothety sends points E, F, K and H into points E′, F ′, K′ and H′, respectively. By heading a) E′F ′ = K′H′; hence, EF = KH. 19.12. Let us make use of the solution and notations of Problem 19.11 a). Since AK = DC, then B1K = B1D and, therefore, B1O is the midline of triangle MKD. 19.13. Let Oα, Oβ, Oγ and Oδ be the centers of circles α, β, γ and δ, respectively, O1 and O2 the centers of the inscribed and circumscribed circles, respectively, of triangle ABC. A homothety with center O1 sends triangle OαOβOγ into triangle ABC. This homothety sends point O2 into the center of the circumscribed circle of triangle OαOβOγ; this latter center coincides with Oδ. Therefore, points O1, O2 and Oδ belong to one line. 19.14. Let A1, B1 and C1 be the centers of the given circles tangent to the sides of the triangle, O the center of the circle tangent to these circles, O1 and O2 the centers of the inscribed and circumscribed circles of triangle ABC. Lines AA1, BB1 and CC1 are the bisectors of triangle ABC and, therefore, they intersect at point O1. It follows that triangle A1B1C1 turns into triangle ABC under a homothety with center O1 and the coeﬃcient of the homothety is equal to the ratio of distances from O1 to the sides of triangles ABC and A1B1C1, i.e., is equal to r−ρ r . Under this homothety the circumscribed circle of triangle ABC turns into the circum-scribed circle of triangle A1B1C1. Since OA1 = OB1 = OC1 = 2ρ, the radius of the circumscribed circle of triangle A1B1C1 is equal to 2ρ. Hence, R r−ρ r = 2ρ, i.e., ρ = rR 2r+R. 19.15. On the bisector of angle ∠ABC take an arbitrary point O and construct a circle S with center O tangent to the legs of the angle. Line BM intersects circle S at points M1 and M2. The problem has two solutions: circle S turns into the circles passing through M and tangent to the legs of the angle under the homothety with center B that sends M1 into M and under the homothety with center B that sends M2 into M. 19.16. Clearly, both circles are tangent to one of the triangle’s sides. Let us show how to construct circles tangent to side AB. Let us take line c′ parallel to line AB. Let us construct circles S′ 1 and S′ 2 of the same radius tangent to each other and to line c′. Let us construct tangents a′ and b′ to these circles parallel to lines BC and AC, respectively. The sides of triangle A′B′C′ formed by lines a′, b′ and c′ are parallel to respective sides of triangle ABC. Therefore, there exists a homothety sending triangle A′B′C′ into triangle ABC. The desired circles are the images of circles S′ 1 and S′ 2 with respect to this homothety. 19.17. a) On sides AB and BC of triangle ABC ﬁx segments AX1 and CY1 of equal length a.Through point Y1 draw a line l parallel to side AC. Let Y2 be the intersection point of l and the circle of radius a with center X1 situated(who?) inside the triangle. Then point Y to be found is the intersection point of line AY2 with side BC and X is a point on ray AB such that AX = CY . b) On side AB, take an arbitrary point X1 distinct from B. The circle of radius BX1 with center X1 intersects ray BC at points B and Y1. Construct point C1 on line BC such that Y1C1 = BX1 and such that Y1 lies between B and C1. The homothety with center B that sends point C1 into C sends X1 and Y1 into points X and Y to be found. SOLUTIONS 367 19.18. Take segment AD and draw circles S1 and S2 with center A and radii AB and AC, respectively. Vertex B is the intersection point of S1 with the image of S2 under the homothety with center D and coeﬃcient −DB DC = −AB AC . 19.19. On the great circle S2 take an arbitrary point X. Let S′ 2 be the image of S2 under the homothety with center X and coeﬃcient 1 3, let Y be the intersection point of S′ 2 and S1. Then XY is the line to be found. 19.20. From points B and C draw perpendiculars to lines AB and AC and let P be their intersection point. Then the intersection point of lines AP and BC is the desired one. 19.21. Let us draw common exterior tangents l1 and l2 to circles S1 and S2, respectively. Lines l1 and l2 intersect at a point K which is the center of a homothety H that sends S1 to S2. Let A1 = H(A). Points A and K lie on a line that connects the centers of the circles and, therefore, AA1 is a diameter of S2, i.e., ∠ACA1 = 90◦and A1C ∥AB. It follows that segment AB goes into A1C under H. Therefore, line BC passes through K and ∠ADK = 90◦. Point D belongs to circle S with diameter AK. It is also clear that point D lies inside the angle formed by lines l1 and l2. Therefore, the locus of points D is the arc of S cut oﬀby l1 and l2. 19.22. The hypothesis of the problem implies that the map f is one-to-one. a) Suppose f sends point A to point A′ and B to B′. Then − − → BB′ = − → BA + − − → AA′ + − − → A′B′ = −− → AB + − − → AA′ + − → AB = − − → AA′, i.e., f is a parallel translation. b) Consider three points A, B and C not on one line. Let A′, B′ and C′ be their images under f. Lines AB, BC and CA cannot coincide with lines A′B′, B′C′ and C′A′, respectively, since in this case A = A′, B = B′ and C = C′. Let AB ̸= A′B′. Lines AA′ and BB′ are not parallel because otherwise quandrilateral ABB′A′ would have been a parallelogram and − → AB = − − → A′B′. Let O be the intersection point of AA′ and BB′. Triangles AOB and A′OB′ are similar with similarity coeﬃcient k and, therefore, − − → OA′ = k− → OA, i.e., O is a ﬁxed point of the transformation f. Therefore, − − − − → Of(X) = − − − − − − − → f(O)f(X) = k− − → OX for any X which means that f is a homothety with coeﬃcient k and center O. 19.23. Let H = H2 ◦H1, where H1 and H2 are homotheties with centers O1 and O2 and coeﬃcients k1 and k2, respectively. Denote: A′ = H1(A), B′ = H1(B), A′′ = H2(A′), B′′ = H2(B′). Then − − → A′B′ = k1 − → AB and − − − → A′′B′′ = k2 − − → A′B′, i.e., − − − → A′′B′′ = k1k2 − → AB. With the help of the preceding problem this implies that for k1k2 ̸= 1 the transformation H is a homothety with coeﬃcient k1k2 and if k1k2 = 1, then H is a parallel translation. It remains to verify that the ﬁxed point of H belongs to the line that connects the centers of homotheties H1 and H2. Since − − − → O1A′ = k1 − − → O1A and − − − → O2A′′ = k2 − − − → O2A′, it follows that − − − → O2A′′ = k2(− − − → O2O1 + − − − → O1A′) = k2(− − − → O2O1 + k1 − − → O1A) = k2 − − − → O2O1 + k1k2 − − − → O1O2 + k1k2 − − → O2A. For a ﬁxed point X we get the equation − − → O2X = (k1k2 −k2)− − − → O1O2 + k1k2 − − → O2X 368 CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY and, therefore, − − → O2X = λ− − − → O1O2, where λ = k1k2−k2 1−k1k2 . 19.24. Point A is the center of homothety that sends S1 to S2 and B is the center of homothety that sends S2 to S3. The composition of these homotheties sends S1 to S3 and its center belongs to line AB. On the other hand, the center of homothety that sends S1 to S3 is point C. Indeed, to the intersection point of the outer tangents there corresponds a homothety with any positive coeﬃcient and a composition of homotheties with positive coeﬃcients is a homothety with a positive coeﬃcient. 19.25. a) Let K, L, M be the intersection points of lines AB and CD, AP and DQ, BP and CQ, respectively. These points are the centers of homotheties HK, HL and HM with positive coeﬃcients that consequtively send segments BC to AD, AD to PQ and BC to PQ. Clearly, HL ◦HK = HM. Therefore, points K, L and M belong to one line. b) Let K, L, M be the intersection points of lines AB and CD, AQ and DP, BQ and CP, respectively. These points are the centers of homotheties, HK, HL and HM that consequtively send segments BC to AD, AD to QP, BC to QP; the coeﬃcient of the ﬁrst homothety is a positive one those of two other homotheties are negative ones. Clearly, HL ◦HK = HM. Therefore, points K, L and M belong to one line. 19.26. Since ∠(P1A, AB) = ∠(P2A, AB), the oriented angle values of arcs ⌣BP1 and ⌣BP2 are equal. Therefore, the rotational homothety with center B that sends S1 to S2 sends point P1 to P2 and line P1Q1 into line P2Q2. 19.27. Oriented angle values of arcs ⌣AM1 and ⌣AM2 are equal, consequently, ∠(M1B, BA) = ∠(M2B, BA) and, therefore, points M1, M2 and B belong to one line. 19.28. Let Pi be a rotational homothety with center O that sends circle Si to Si+1. Then Xi+1 = Pi(Xi) (see Problem 19.27). It remains to observe that the composition Pn ◦· · · ◦P2 ◦P1 is a rotational homothety with center O that sends S1 to S1, i.e., is an identity transformation. 19.29. Since ∠AMB = ∠NAB and ∠BAM = ∠BNA, we have △AMB ∼△NAB and, therefore, AN : AB = MA : MB = CN : MB. Moreover, ∠ABM = 180◦−∠MAN = ∠ANC. It follows that △AMB ∼△ACN, i.e., the rotational homothety with center A sending M to B sends C to N and, therefore, it maps Q to P. 19.30. Let O1 and O2 be the centers of given circles, r1 and r2 be their radii. The coeﬃcient k of the rotational homothety which maps S1 to S2 is equal to r1/r2 and its center O belongs to the circle with diameter O1O2. Moreover, OO1 : OO2 = k = r1/r2. It remains to verify that the circle with diameter O1O2 and the locus of points O such that OO1 : OO2 = k have precisely two common points. For k = 1 it is obvious and for k ̸= 1 the locus in question is described in the solution of Problem 7.14: it is the(A?) circle and one of its intersection points with line O1O2 is an inner point of segment O1O2 whereas the other intersection point lies outside the segment. 19.31. Consider a transformation which sends triangle BHC to triangle PHB, i.e., the composition of the rotation through an angle of 90◦about point H and the homothety with coeﬃcient BP : CB and center H. Since this transformation maps the vertices of any square into vertices of a square, it maps points C and B to points B and P, respectively. Then it maps point D to Q, i.e., ∠DHQ = 90◦. 19.32. Let P be a rotational homothety that sends − − → CB to − − → CA1. Then − − → AA1 + − − → BB1 + − − → CC1 = − → AC + P(− − → CB) + − − → CB + P(− → BA) + − → BA + P(− → AC) = − → 0 . SOLUTIONS 369 Hence, if M is the center of mass of triangle ABC, then − − − → MA1 + − − − → MB1 + − − − → MC1 = (− − → MA + − − → MB + − − → MC) + (− − → AA1 + − − → BB1 + − − → CC1) = − → 0 . 19.33. Let M be the common midpoint of sides BC and B1C1, x = − − → MB and y = − − − → MB1. Further, let P be the rotational homothety with center M, the angle of rotation 90◦and coeﬃcient √ 3 that sends B to A and B1 to A1. Then − − → BB1 = y −x and − − → AA1 = P(y) − P(x) = P(− − → BB1). Therefore, the angle between vectors − − → AA1 and − − → BB1 is equal to 90◦and AA1 : BB1 = √ 3. 19.34. Let P be the rotational homothety that sends triangle ABC to triangle A1B1C1. Then − − − → A2B2 = − − → A2O + − − → OB2 = − − → A1A + − − → BB1 = − → BA + − − − → A1B1 = −− → AB + P(− → AB). Similarly, the transformation f(a) = −a + P(a) sends the other vectors of the sides of triangle ABC to the vectors of the sides of triangle A2B2C2. 19.35. Let the initial map be rectangle K0 on the plane, the smaller map rectangle K1 contained in K0. Let us consider a rotational homothety f that maps K0 to K1. Let Ki+1 = f(Ki) for i > 1. Since the sequence Ki for i = 1, 2, . . . is a contracting sequence of embedded polygons, there exists (by Helly’s theorem) a unique ﬁxed point X that belongs to all the rectangles Ki. Let us prove that X is the required point, i.e., f(X) = X. Indeed, since X belongs to Ki, point f(X) belongs to Ki+1, i.e., point f(X) belongs also to all rectangles Ki. Since there is just one point that belongs to all rectangles, we deduce that f(X) = X. 19.36. Since the product of coeﬃcients of rotational homotheties P1 and P2 is equal to 1, their composition is a rotation (cf. Problem 17.36). Let O be the center of rotation P2◦P1 and R = P1(O). Since P2 ◦P1(O) = O, it follows that P2(R) = O. Therefore, by hypothesis A1O : A1R = A2O : A2R and ∠OA1R = ∠OA2R, i.e., △OA1R ∼△OA2R. Moreover, OR is a common side of these similar triangles; hence, △OA1R = △OA2R. Therefore, OA1 = OA2 and ∠(− − → OA1, − − → OA2) = 2∠(− − → OA1, − → OR) = 2∠(− − − → MA1, − − → MN), i.e., O is the center of rotation through an angle of 2∠(− − − → MA1, − − → MN) that maps A1 to A2. 19.37. Let P1 be the rotational homothety with center B sending A to M and P2 be rotational homothety with center D sending M to C. Since the product of coeﬃcients of these rotational homotheties is equal to (BM : BA) · (DC : DM) = 1, their composition P2 ◦P1 is a rotation (sending A to C) through an angle of ∠(− → AB, − − → BM) + ∠(− − → DM, − − → DC) = 2∠(− → AB, − − → BM). On the other hand, the center of the rotation P2 ◦P1 coincides with the center of the rotation through an angle of 2∠(− → AB, − − → AM) that sends B to D (cf. Problem 19.36). 19.38. It is easy to verify that tan ∠XBY = k and BY : BX = √ k2 + 1, i.e., Y is obtained from X under the rotational homothety with center B and coeﬃcient √ k2 + 1, the 370 CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY angle of rotation being of value arctan k. The locus to be found is the image of the given half circle under this rotational homothety. 19.39. Suppose that triangle PXY is constructed and points X and Y belong to sides AC and CB, respectively. We know a transformation that maps X to Y , namely, the rotational homothety with center P, the angle of rotation ϕ = ∠XPY = ∠MLN and the homothety coeﬃcient k = PY : PX = LN · LM. Point Y to be found is the intersection point of segment BC and the image of segment AC under this transformation. 19.40. Suppose that rectangle ABCD is constructed. Consider the rotational homothety with center A that sends B to D. Let C′ be the image of point C under this homothety. Then ∠CDC′ = ∠B + ∠D and DC′ = BC·AD AB = bd a . We can recover triangle CDC′ from CD, DC′ and ∠CDC′. Point A is the intersection point of the circle of radius d with center D and the locus of points X such that C′X : CX = d : a (this locus is a circle, see Problem 7.14). The further construction is obvious. 19.41. a) If O is the center of a rotational homothety that sends segment AB to segment A1B1, then ∠(PA, AO) = ∠(PA1, A1O) and ∠(PB, BO) = ∠(PB1, B1O) (1) and, therefore, point O is the intersection point of the inscribed circles of triangles PAA1 and PBB1. The case when these circles have only one common point P is clear: this is when segment AB turns into segment A1B1 under a homothety with center P. If P and O are two intersection points of the circles considered, then equalities (1) imply that △OAB ∼△OA1B1 and, therefore, O is the center of a rotational homothety that maps segment AB into segment A1B1. b) It suﬃces to notice that point O is the center of a rotational homothety that maps segment AB to segment BC if and only if ∠(BA, AO) = ∠(CB, BO) and ∠(AB, BO) = ∠(BC, CO). 19.42. Let A1 and B1 be the positions of the points at one moment, A2 and B2 the position of the points at another moment. Then for point P we can take the center of a rotational homothety that maps segment A1A2 to segment B1B2. 19.43. Let P be the intersection point of lines l1 and l2. By Problem 19.41 point O belongs to the circumscribed circle S1 of triangle A1A2P. On the other hand, OA2 : OA1 = k. The locus of points X such that XA2 : XA1 = k is circle S2 (by Problem 7.14). Point O is the intersection point of circles S1 and S2 (there are two such points). 19.44. Let O be the center of a rotational homothety that maps segment AB to segment A1B1. Then △ABO ∼△A1B1O, i.e., ∠AOB = ∠A1OB1 and AO : BO = A1O : B1O. Therefore, ∠AOA1 = ∠BOB1 and AO : A1O = BO : B1O, i.e., △AA1O ∼△BB1O. Hence, point O is the center of the rotational homothety that maps segment AA1 to segment BB1. 19.45. Let lines AB and DE intersect at point C and lines BD and AE intersect at point F. The center of rotational homothety that maps segment AB to segment ED is the distinct from C intersection point of the circumscribed circles of triangles AEC and BDC (see Problem 19.41) and the center of rotational homothety sending AE to BD is the intersection point of circles circumscribed about triangles ABF and EDF. By Problem 19.44 the centers of these rotational homotheties coincide, i.e., all the four circumscribed circles have a common point. 19.46. The center O of parallelogram ABCD is equidistant from the following pairs of lines: AQ and AB, AB and CD, CD and DQ and, therefore, QO is the bisector of angle ∠AQD. Let α = ∠BAO, β = ∠CDO and ϕ = ∠AQO = ∠DQO. Then α + β = ∠AOD = 360◦−α −β −2ϕ, i.e., α + β + ϕ = 180◦and, therefore, △QAO ∼△QOD. SOLUTIONS 371 19.47. Let us solve a slightly more general problem. Suppose point O is taken on circle S and H is a rotational homothety with center O. Let us prove that then all lines XX′, where X is a point from S and X′ = H(X), intersect at one point. Let P be the intersection point of lines X1X′ 1 and X2X′ 2. By Problem 19.41 points O, P, X1 and X2 lie on one circle and points O, P, X′ 1 and X′ 2 also belong to one circle. Therefore, P is an intersection point of circles S and H(S), i.e., all lines XX′ pass through the distinct from O intersection point of circles S and H(S). 19.48. Let O be the center of a rotational homothety sending triangle A1B1C1 to triangle ABC. Let us prove that, for instance, the circumscribed circles of triangles ABC2 and A1B1C2 pass through point O. Under the considered homothety segment AB goes into segment A1B1; therefore, point O coincides with the center of the rotational homothety that maps segment AA1 to segment BB1 (see Problem 19.44). By problem 19.41 the center of the latter homothety is the second intersection point of the circles circumscribed about triangles ABC2 and A1B1C2 (or is their tangent point). Figure 169 (Sol. 19.48) 19.49. Points A1, A2 and A3 belong to lines P2P3, P3P1 and P1P2 (Fig. 27). Therefore, the circles circumscribed about triangles A1A2P3, A1A3P2 and A2A3P1 have a common point V (see Problem 2.80 a)), and points O3, O2 and O1 lie on these circles (see Problem 19.41). Similarly, the circles circumscribed about triangles B1B2P3, B1B3P2 and B2B3P1 have a common point V ′. Let U be the intersection point of lines P2O2 and P3O3. Let us prove that point V belongs to the circle circumscribed about triangle O2O3U. Indeed, ∠(O2V, V O3) = ∠(V O2, O2P2) + ∠(O2P2, P3O3) + ∠(P3O3, O3V ) = ∠(V A1, A1P2) + ∠(O2U, UO3) + ∠(P3A1, A1V ) = ∠(O2U, UO3). Analogous arguments show that point V ′ belongs to the circle circumscribed about triangle O2O3U. In particular, points O2, O3, V and V ′ belong to one circle. Similarly, points O1, O2, V and V ′ belong to one circle and, therefore, points V and V ′ belong to the circle circumscribed about triangle O1O2O3; point U also belongs to this circle. We can similarly prove that lines P1O1 and P2O2 intersect at one point that belongs to the similarity circle. Line P2O2 intersects the similarity circle at points U and O2 and, therefore, line P1O1 passes through point U. 19.50. Let P1 be the intersection point of lines A2B2 and A3B3, let P ′ 1 be the intersection point of lines A2C2 and A3C3; let points P2, P3, P ′ 2 and P ′ 3 be similarly deﬁned. The rotational homothety that sends F1 to F2 sends lines A1B1 and A1C1 to lines A2B2 and A2C2, respectively, and, therefore, ∠(A1B1, A2B2) = ∠(A1C1, A2C2). Similar arguments show that △P1P2P3 ∼△P ′ 1P ′ 2P ′ 3. 372 CHAPTER 19. HOMOTHETY AND ROTATIONAL HOMOTHETY The center of the rotational homothety that maps segment P2P3 to P ′ 2P ′ 3 belongs to the circle circumscribed about triangle A1P3P ′ 3 (see Problem 19.41). Since ∠(P3A1, A1P ′ 3) = ∠(A1B1, A1C1) = ∠(A2B2, A2C2) = ∠(P3A2, A2P ′ 3), the circle circumscribed about triangle A1P3P ′ 3 coincides with the circle circumscribed about triangle A1A2P3. Similar arguments show that the center of the considered rotational homo-thety is the intersection point of the circles circumscribed about triangles A1A2P3, A1A3P2 and A2A3P1; this point belongs to the similarity circle of ﬁgures F1, F2 and F3 (see Problem 19.49 a)). 19.51. a) Let l′ 1, l′ 2 and l′ 3 be the corresponding lines of ﬁgures F1, F2 and F3 such that l′ i ∥li. These lines form triangle P1P2P3. The rotational homothety with center O3 that maps F1 to F2 sends lines l1 and l′ 1 to lines l2 and l′ 2, respectively, and, therefore, the homothety with center O3 that maps l1 to l′ 1 sends line l2 to l′ 2. Therefore, line P3O3 passes through point W. Similarly, lines P1O1 and P2O2 pass through point W; hence, W belongs to the similarity circle of ﬁgures F1, F2 and F3 (see Problem 19.49 b)). Figure 170 (Sol. 19.51 a)) b) The ratio of the distances from point O1 to lines l′ 2 and l′ 3 is equal to the coeﬃcient of the rotational homothety that maps F2 to F3 and the angle ∠P1 of triangle P1P2P3 is equal to the angle of the rotation. Therefore, ∠(O1P1, P1P2) only depends on ﬁgures F2 and F3. Since ∠(O1W, WJ3) = ∠(O1P1, P1P2), arc ⌣O1J3 is ﬁxed (see Fig. 28) and, therefore, point J3 is ﬁxed. We similarly prove that points J1 and J2 are ﬁxed. 19.52. Let us make use of notations from Problem 19.51. Clearly, ∠(J1J2, J2J3) = ∠(J1W, WJ3) = ∠(P3P2, P2P1). For the other angles of the triangle the proof is similar. 19.53. Let us prove, for instance, that under the rotational homothety with center O1 that maps F2 to F3 point J2 goes to J3. Indeed, ∠(J2O1, O1J3) = ∠(J2W, WJ3). Moreover, lines J2W and J3W are the corresponding lines of ﬁgures F2 and F3 and, therefore, the distance from lines J2W and J3W to point O1 is equal to the similarity coeﬃcient k1; hence, O1J2 O1J3 = k1. 19.54. Let Oa be the intersection point of the circle passing through point B and tangent to line AC at point A and the circle passing through point C and tangent to line AB at point A. By Problem 19.41 b) point Oa is the center of rotational homothety that sends segment BA to segment AC. Having similarly deﬁned points Ob and Oc and making use of the result of Problem 19.49 b) we see that lines AOa, BOb and COc intersect at a point that belongs SOLUTIONS 373 to the similarity circle S. On the other hand, these lines intersect at Lemoin’s point K (see Problem 5.128). The midperpendiculars to the sides of the triangle are the corresponding lines of the considered similar ﬁgures. The midperpendiculars intersect at point O; hence, O belongs to the similarity circle S (see Problem 19.51 a)). Moreover, the midperpendiculars intersect S at ﬁxed points A1, B1 and C1 of triangle ABC (see Problem 19.51 b)). On the other hand, the lines passing through point K parallel to BC, CA and AB are also corresponding lines of the considered ﬁgures (see solution to Problem 5.132), therefore, they also intersect circle S at points A1, B1 and C1. Hence, OA1 ⊥A1K, i.e., OK is a diameter of S. 19.55. If P is the ﬁrst of Brokar’s points of triangle ABC, then CP, AP and BP are the corresponding lines for similar ﬁgures constructed on segments BC, CA and AB. Therefore, point P belongs to the similarity circle S (see Problem 19.51 a)). Similarly, point Q belongs to S. Moreover, lines CP, AP and BP intersect S at ﬁxed points A1, B1 and C1 of triangle ABC (cf. Problem 19.51 b)). Since KA1 ∥BC (see the solution of Problem 19.54), it follows that ∠(PA1, A1K) = ∠(PC, CB) = ϕ, i.e., ⌣PK = 2ϕ. Similarly, ⌣KQ = 2ϕ. Therefore, PQ ⊥KO; hence, OP = OQ and ∠POQ = 1 2 ⌣PQ = 2ϕ. Chapter 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT Background 1. Solving various problems it is often convenient to consider a certain extremal or “boundary” element, i.e., an element at which a certain function takes its maximal or minimal value. For instance, the longest or the shortest side a triangle, the greatest or the smallest angle, etc. This method for solving problems is sometimes called the principle (or the rule) of an extremal element; this term, however, is not conventional. Figure 171 2. Let O be the intersection point of the diagonals of a convex quadrilateral. Its vertices can be denoted so that CO ≤AO and BO ≤DO (see Fig. ). Then under symmetries with respect to point O triangle BOC is mapped inside triangle AOD, i.e., in a certain sense triangle BOC is the smallest and triangle AOD is the greatest (see §4). 3. The vertices of the convex hull and the basic lines are also extremal elements; to an extent these notions are used in §5 where they are deﬁned and where their main properties are listed. §1. The least and the greatest angles 20.1. Prove that if the lengths of all the sides of a triangle are smaller than 1, then its area is smaller than 1 4 √ 3. 20.2. Prove that the disks constructed on the sides of a convex quadrilateral as on diameters completely cover this quadrilateral. 20.3. In a country, there are 100 airports such that all the pairwise distances between them are distinct. From each airport a plane lifts up and ﬂies to the nearest airport. Prove that there is no airport to which more than ﬁve planes can arrive. 20.4. Inside a disk of radius 1, eight points are placed. Prove that the distance between some two of them is smaller than 1. 20.5. Six disks are placed on the plane so that point O is inside each of them. Prove that one of these disks contains the center of some other disk. 20.6. Inside an acute triangle point P is taken. Prove that the greatest distance from P to the vertices of this triangle is smaller than twice the shortest of the distances from P to the sides of the triangle. 375 376 CHAPTER 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT 20.7. The lengths of a triangle’s bisectors do not exceed 1. Prove that the area of the triangle does not exceed 1 √ 3. §2. The least and the greatest distances 20.8. Given n ≥3 points on the plane not all of them on one line. Prove that there is a circle passing through three of the given points such that none of the remaining points lies inside the circle. 20.9. Several points are placed on the plane so that all the pairwise distances between them are distinct. Each of these points is connected with the nearest one by a line segment. Do some of these segments constitute a closed broken line? 20.10. Prove that at least one of the bases of perpendiculars dropped from an interior point of a convex polygon to its sides is on the side itself and not on its extension. 20.11. Prove that in any convex pentagon there are three diagonals from which one can construct a triangle. 20.12. Prove that it is impossible to cover a polygon with two polygons which are homothetic to the given one with coeﬃcient k for 0 < k < 1. 20.13. Given ﬁnitely many points on the plane such that any line passing through two of the given points contains one more of the given points. Prove that all the given points belong to one line. 20.14. In plane, there are given ﬁnitely many pairwise non-parallel lines such that through the intersection point of any two of them one more of the given lines passes. Prove that all these lines pass through one point. 20.15. In plane, there are given n points. The midpoints of all the segments with both endpoints in these points are marked, the given points are also marked. Prove that there are not less than 2n −3 marked points. See also Problems 9.17, 9.19. §3. The least and the greatest areas 20.16. In plane, there are n points. The area of any triangle with vertices in these points does not exceed 1. Prove that all these points can be placed in a triangle whose area is equal to 4. 20.17. Polygon M ′ is homothetic to a polygon M with homothety coeﬃcient equal to −1 2. Prove that there exists a parallel translation that sends M ′ inside M. §4. The greatest triangle 20.18. Let O be the intersection point of diagonals of convex quadrilateral ABCD. Prove that if the perimeters of triangles ABO, BCO, CDO and DAO are equal, then ABCD is a rhombus. 20.19. Prove that if the center of the inscribed circle of a quadrilateral coincides with the intersection point of the diagonals, then this quadrilateral is a rhombus. 20.20. Let O be the intersection point of the diagonals of convex quadrilateral ABCD. Prove that if the radii of inscribed circles of triangles ABO, BCO, CDO and DAO are equal, then ABCD is a rhombus. §5. The convex hull and the base lines While solving problems of this section we will consider convex hulls of systems of points and base lines of convex polygons. §5. THE CONVEX HULL AND THE BASE LINES 377 The convex hull of a ﬁnite set of points is the least convex polygon which contains all these points. The word “least” means that the polygon is not contained in any other such polygon. Any ﬁnite system of points possesses a unique convex hull (Fig. 29). Figure 171 A base line of a convex polygon is a line passing through its vertex and with the property that the polygon is situated on one side of it. It is easy to verify that for any convex polygon there exist precisely two base lines parallel to a given line (Fig. 30). Figure 172 20.21. Solve Problem 20.8 making use of the notion of the convex hull. 20.22. Given 2n + 3 points on a plane no three of which belong to one line and no four of which belong to one circle. Prove that one can select three points among these so that n of the remaining points lie inside the circle drawn through the selected points and n of the points lie outside the circle. 20.23. Prove that any convex polygon of area 1 can be placed inside a rectangle of area 2. 20.24. Given a ﬁnite set of points in plane prove that there always exists a point among them for which not more than three of the given points are the nearest to it. 20.25. On the table lie n cardboard and n plastic squares so that no two cardboard and no two plastic squares have common points, the boundary points included. It turned out that the set of vertices of the cardboard squares coincides with that of the plastic squares. Is it necessarily true that every cardboard square coincides with a plastic one? 20.26. Given n ≥4 points in plane so that no three of them belong to one line. Prove that if for any 3 of them there exists a fourth (among the given ones) together with which they form vertices of a parallelogram, then n = 4. 378 CHAPTER 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT §6. Miscellaneous problems 20.27. In plane, there are given a ﬁnite set of (not necessarily convex) polygons each two of which have a common point. Prove that there exists a line having a common point with all these polygons. 20.28. Is it possible to place 1000 segments on the plane so that the endpoints of every segment are interior points of certain other of these segments? 20.29. Given four points in plane not on one line. Prove that at least one of the triangles with vertices in these points is not an acute one. 20.30. Given an inﬁnite set of rectangles in plane. The vertices of each of the rectangles lie in points with coordinates (0, 0), (0, m), (n, 0), (n, m), where n and m are positive integers (each rectangle has its own numbers). Prove that among these rectangles one can select such a pair that one is contained inside the other one. 20.31. Given a convex polygon A1 . . . An, prove that the circumscribed circle of triangle AiAi+1Ai+2 contains the whole polygon. Solutions 20.1. Let α be the least angle of the triangle. Then α ≤60◦. Therefore, S = bc sin α 2 ≤ sin 60◦ 2 = √ 3 4 . 20.2. Let X be an arbitrary point inside a convex quadrilateral. Since ∠AXB + ∠BXC + ∠CXD + ∠AXD = 360◦, the maximal of these angles is not less than 90◦. Let, for deﬁniteness sake, ∠AXB ≥90◦. Then point X is inside the circle with diameter AB. 20.3. If airplanes from points A and B arrived to point O, then AB is the longest side of triangle AOB, i.e., ∠AOB > 60◦. Suppose that airplanes from points A1, . . . , An arrived to point O. Then one of the angles ∠AiOAj does not exceed 360◦ n . Therefore, 360◦ n > 60◦, i.e., n < 6. 20.4. At least seven points are distinct from the center O of the circle. Therefore, the least of the angles ∠AiOAj, where Ai and Aj are given points, does not exceed 360◦ 7 < 60◦. If A and B are points corresponding to the least angle, then AB < 1 because AO ≤1, BO ≤1 and angle ∠AOB cannot be the largest angle of triangle AOB. 20.5. One of the angles between the six segments that connect point O with the centers of the disks does not exceed 360◦ 6 = 60◦. Let ∠O1OO2 ≤60◦, where O1 and O2 are the centers of the disks of radius r1 and r2, respectively. Since ∠O1OO2 ≤60◦, this angle is not the largest angle in triangle O1OO2 and, therefore, either O1O2 ≤O1O or O1O2 ≤O2O. Let, for deﬁniteness, O1O2 ≤O1O. Since point O is inside the circles, O1O < r1. Therefore, O1O2 ≤O1O < r1, i.e., point O2 is inside the disk of radius r1 with center O1. 20.6. Let us drop perpendiculars PA1, PB1 and PC1 from point P to sides BC, CA and AB, respectively, and select the greatest of the angles formed by these perpendiculars and rays PA, PB and PC. Let, for deﬁniteness sake, this be angle ∠APC1. Then ∠APC1 ≥60◦; hence, PC1 : AP = cos APC1 ≤cos 60◦= 1 2, i.e., AP ≥2PC1. Clearly, the inequality still holds if AP is replaced with the greatest of the numbers AP, BP and CP and PC1 is replaced with the smallest of the numbers PA1, PB1 and PC1. 20.7. Let, for deﬁniteness, α be the smallest angle of triangle ABC; let AD be the bisector. One of sides AB and AC does not exceed AD/ cos(α/2) since otherwise segment BC does not pass through point D. Let, for deﬁniteness, AB ≤ AD cos(α/2) ≤ AD cos 30◦≤2 √ 3. SOLUTIONS 379 Then SABC = 1 2hcAB ≤1 2lcAB ≤ 1 √ 3. 20.8. Let A and B be those of the given points for which the distance between them is minimal. Then inside the circle with diameter AB there are no given points. Let C be the remaining point — the vertex of the greatest angle that subtends segment AB. Then inside the circle passing through points A, B and C there are no given points. 20.9. Suppose that we have obtained a closed broken line. Then AB is the longest link of this broken line and AC and BD are the links neighbouring to AB. Then AC < AB, i.e., B is not the point closest to A and BD < AB, i.e., A is not the point closest to B. Therefore, points A and B cannot be connected. Contradiction. 20.10. Let O be the given point. Let us draw lines containing the sides of the polygon and select among them the one which is the least distant from point O. Let this line contain side AB. Let us prove that the base of the perpendicular dropped from O to AB belongs to side AB itself. Suppose that the base of the perpendicular dropped from O to line AB is point P lying outside segment AB. Since O belongs to the interior of the convex polygon, segment OP intersects side CD at point Q. Clearly, OQ < OP and the distance from O to line CD is smaller than OQ. Therefore, line CD is less distant from point O than line AB. This contradicts the choice of line AB. 20.11. Let BE be the longest diagonal of pentagon ABCDE. Let us prove then that from segments BE, EC and BD one can construct a triangle. To this end, it suﬃces to verify that BE < EC + BD. Let O be the intersection point of diagonals BD and EC. Then BE < BO + OE < BD + EC. 20.12. Let O1 and O2 be the centers of homotheties, each with coeﬃcient k, sending polygon M to polygons M1 and M2, respectively. Then a point from M the most distant from line O1O2 is not covered by polygons M1 and M2. 20.13. Suppose that not all of the given points lie on one line. Through every pair of given points draw a line (there are ﬁnitely many of such lines) and select the least nonzero distance from the given points to these lines. Let the least distance be the one from point A to line BC, where points B and C are among given ones. On line BC, there lies one more of the given points, D. Drop perpendicular AQ from point A to line BC. Two of the points B, C and D lie to one side of point Q, let these be C and D. Let, for deﬁniteness, CQ < DQ (Fig. 31). Figure 173 (Sol. 20.13) Then the distance from point C to line AD is smaller than that from A to line BC which contradicts to the choice of point A and line BC. 20.14. Suppose that not all lines pass through one point. Consider the intersection points of lines and select the least nonzero distance from these points to the given lines. Let the least distance be the one from point A to line l. Through point A at least three of given lines pass. Let them intersect line l at points B, C and D. From point A drop perpendicular AQ to line l. 380 CHAPTER 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT Figure 174 (Sol. 20.14) Two of the points B, C and D lie on one side of point Q, let them be C and D. Let, for deﬁniteness, CQ < DQ (Fig. 32). Then the distance from point C to line AD is smaller than the distance from point A to line l which contradicts the choice of A and l. 20.15. Let A and B be the most distant from each other given points. The midpoints of the segments that connect point A (resp. B) with the other points are all distinct and lie inside the circle of radius 1 2AB with center A (resp. B). The two disks obtained have only one common point and, therefore, there are no less than 2(n −1) −1 = 2n −3 distinct ﬁxed points. 20.16. Among all the triangles with vertices in the given points select a triangle of the greatest area. Let this be triangle ABC. Let us draw through vertex C line lc so that lc ∥AB. If points X and A lie on diﬀerent sides of line lc, then SABX > SABC. Therefore, all the given points lie on one side of lc. Similarly, drawing lines lb and la through points B and A so that lb ∥AC and la ∥BC we see that all given points lie inside (or on the boundary of) the triangle formed by lines la, lb and lc. The area of this triangle is exactly four times that of triangle ABC and, therefore, it does not exceed 4. 20.17. Let ABC be the triangle of the greatest area among these with vertices in the vertices of polygon M. Then M is contained inside triangle A1B1C1 the midpoints of whose sides are points A, B and C. The homothety with center in the center of mass of triangle ABC and with coeﬃcient −1 2 sends triangle A1B1C1 to triangle ABC and, therefore, sends polygon M inside triangle ABC. 20.18. For deﬁniteness, we may assume that AO ≥CO and DO ≥BO. Let points B1 and C1 be symmetric to points B and C through point O (Fig. 33). Figure 175 (Sol. 20.18) Since triangle B1OC1 lies inside triangle AOD, it follows that PAOD ≥PB1OC1 = PBOC and the equality is attained only if B1 = D and C1 = A (see Problem 9.27 b)). Therefore, ABCD is a parallelogram. Therefore, AB −BC = PABO −PBCO = 0, i.e., ABCD is a rhombus. SOLUTIONS 381 20.19. Let O be the intersection point of the diagonals of quadrilateral ABCD. For deﬁniteness, we may assume that AO ≥CO and DO ≥BO. Let points B1 and C1 be symmetric to points B and C, respectively, through point O. Since O is the center of the circle inscribed into the quadrilateral, we see that segment B1C1 is tangent to this circle. Therefore, segment AD can be tangent to this circle only if B1 = D and C1 = A, i.e., if ABCD is a parallelogram. One can inscribe a circle into this parallelogram since this parallelogram is a rhombus. 20.20. For deﬁniteness, we may assume that AO ≥CO and DO ≥BO. Let points B1 and C1 be symmetric to points B and C through point O. Then triangle C1OB1 is contained inside triangle AOD and, therefore, the inscribed circle S of triangle C1OB1 is contained inside triangle AOD. Suppose that segment AD does not coincide with segment C1B1. Then circle S turns into the inscribed circle of triangle AOD under the homothety with center O and coeﬃcient greater than 1, i.e., rAOD > rC1OB1 = rCOB. We have got a contradiction; hence, A = C1 and D = B1, i.e., ABCD is a parallelogram. In parallelogram ABCD, the areas of triangles AOB and BOC are equal and, therefore, if the inscribed circles have equal radii, then they have equal perimeters since S = pr. It follows that AB = BC, i.e., ABCD is a rhombus. 20.21. Let AB be the side of the convex hull of the given points, B1 be the nearest to A of all the given points that lie on AB. Select the one of the remaining points that is the vertex of the greatest angle that subtends segment AB1. Let this be point C. Then the circumscribed circle of triangle AB1C is the one to be found. 20.22. Let AB be one of the sides of the convex hull of the set of given points. Let us enumerate the remaining points in the order of increase of the angles with vertex in these points that subtend segment AB, i.e., denote them by C1, C2, . . . , C2n+1 so that ∠AC1B < ∠AC2B ≤· · · < ∠AC2n+1B. Then points C1, . . . , Cn lie outside the circle circumscribed about triangle ABCn+1 and points Cn+2, . . . , C2n+1 lie inside it, i.e., this is the circle to be constructed. 20.23. Let AB be the greatest diagonal (or side) of the polygon. Through points A and B draw lines a and b perpendicular to line AB. If X is a vertex of the polygon, then AX ≤AB and XB ≤AB, therefore, the polygon lies inside the band formed by lines a and b. Figure 176 (Sol. 20.23) Draw the base lines of the polygon parallel to AB. Let these lines pass through vertices C and D and together with a and b form rectangle KLMN (see Fig. 34). Then SKLMN = 2SABC + 2SABD = 2SACBD. Since quadrilateral ACBD is contained in the initial polygon whose area is equal to 1, SKLMN ≤2. 382 CHAPTER 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT 20.24. Select the least of all the distances between the given points and consider points which have neighbours at this distance. Clearly, it suﬃces to prove the required statement for these points. Let P be the vertex of the convex hull of these points. If Ai and Aj are the points nearest to P, then AiAj ≥AiP and AiAj ≥AjP and, therefore, ∠AiPAj ≥60◦. It follows that P cannot have four nearest neighbours since otherwise one of the angles ∠AiPAj would have been smaller than 180◦ 3 = 60◦. Therefore, P is the point to be found. 20.25. Suppose that there are cardboard squares that do not coincide with the plastic ones. Let us discard all the coinciding squares and consider the convex hull of the vertices of the remaining squares. Let A be a vertex of this convex hull. Then A is a vertex of two distinct squares, a cardboard one and a plastic one. It is easy to verify that one of the vertices of the smaller of these squares lies inside the larger one (Fig. 35). Let, for deﬁniteness, vertex B of the cardboard square lie inside the plastic one. Then point B lies inside a plastic square and is a vertex of another plastic square, which is im-possible. This is a contradiction, hence, every cardboard square coincides with a plastic one. 20.26. Let us consider the convex hull of the given points. The two cases are possible: 1) The convex hull is a parallelogram, ABCD. If point M lies inside parallelogram ABCD, then the vertices of all three parallelograms with vertices at A, B, and M lie outside ABCD (Fig. 36). Hence, in this case there can be no other points except A, B, C and D. 2) The convex hull is not a parallelogram. Let AB and BC be edges of the convex hull. Let us draw base lines parallel to AB and BC. Let these base lines pass through vertices P and Q. Then the vertices of all the three parallelograms with vertices at B, P and Q lie outside the convex hull (Fig. 37). They even lie outside the parallelogram formed by the base lines except for the case when P and Q are vertices of this parallelogram. In this last case the fourth vertex of the parallelogram does not belong to the convex hull since the convex hull is not a parallelogram. 20.27. In plane, take an arbitrary straight line l and project all the polygons to it. We will get several segments any two of which have a common point. Let us order line l; consider left endpoints of the segments-projections and select the right-most left endpoint. The point belongs to all the segments and, therefore, the perpendicular drawn through it to l intersects all the given polygons. 20.28. Let 1000 segments lie in plane. Take an arbitrary line l not perpendicular to any of them and consider the projections of the endpoints of all these segments on l. It is clear that the endpoint of the segment whose projection is the left-most of the obtained points cannot belong to the interior of another segment. 20.29. Two variants of disposition of these four points are possible: (1) The points are vertices of a convex quadrilateral, ABCD. Take the largest of the angles of its vertices. Let this be angle ∠ABC. Then ∠ABC ≥90◦, i.e., triangle ABC is not an acute one. (2) Point D lies inside triangle ABC. Select the greatest of the angles ∠ADB, ∠BDC and ∠ADC. Let this be angle ∠ADB. Then ∠ADB ≥120◦, i.e., triangle ADB is an obtuse one. We can prove in the following way that there are no other positions of the four points. The lines that pass through three of given points divide the plane into seven parts (Fig. 38). If the fourth given point belongs to the 2nd, 4th or 6th part, then we are in situation (1); if it belongs to the 1st, 3rd, 5th or 7th part, then we are in situation (2). 20.30. The rectangle with vertices at points (0, 0), (0, m), (n, 0) and (n, m) the horizontal side is equal to n and vertical side is equal to m. From the given set select a rectangle with SOLUTIONS 383 Figure 177 (Sol. 20.25) 384 CHAPTER 20. THE PRINCIPLE OF AN EXTREMAL ELEMENT Figure 178 (Sol. 20.26) Figure 179 (Sol. 20.26) Figure 180 (Sol. 20.29) the least horizontal side. Let the length of its vertical side be equal to m1. Consider any side m1 of the remaining rectangles. The two cases are possible: 1) The vertical sides of two of these m1-rectangles are equal. Then one of them is contained in another one. 2) The vertical sides of all these rectangles are distinct. Then the vertical side of one of them is greater than m1 and, therefore, it contains the rectangle with the least horizontal side. 20.31. Consider all the circles passing through two neighbouring vertices Ai and Ai+1 and a vertex Aj such that ∠AiAjAi+1 < 90◦. At least one such circle exists. Indeed, one of the angles ∠AiAi+2Ai+1 and ∠Ai+1AiAi+2 is smaller than 90◦; in the ﬁrst case set Aj = Ai+2 and in the second case set Aj = Ai. Among all such circles (for all i and j) select a circle S of the largest radius; let, for deﬁniteness, it pass through points A1, A2 and Ak. Suppose that vertex Ap lies outside S. Then points Ap and Ak lie on one side of line A1A2 and ∠A1ApA2 < ∠A1AkA2 ≤90◦. The law of sines implies that the radius of the cir-cumscribed circle of triangle A1ApA2 is greater than that of A1AkA2. This is a contradiction and, therefore, S contains the whole polygon A1 . . . An. Let, for deﬁniteness sake, ∠A2A1Ak ≤∠A1A2Ak. Let us prove then that A2 and Ak are neighbouring vertices. If Ak ̸= A3, then 180◦−∠A2A3Ak ≤∠A2A1Ak ≤90◦ and, therefore, the radius of the circumscribed circle of triangle A2A3Ak is greater than the radius of the circumscribed circle of triangle A1A2Ak. Contradiction implies that S passes through neighbouring vertices A1, A2 and A3. Chapter 21. DIRICHLET’S PRINCIPLE Background 1. The most popular (Russian) formulation of Dirichlet’s or pigeonhole principle is the following one: “If m rabbits sit in n hatches and m > n, then at least one hatch contains at least two rabbits.” It is even unclear at ﬁrst glance why this absolutely transparent remark is a quite eﬀective method for solving problems. The point is that in every concrete problem it is sometimes diﬃcult to see what should we designate as the rabbits and the hatches and why there are more rabbits than the hatches. The choice of rabbits and hatches is often obscured; and from the formulation of the problem it is not often clear how to immediately deduce that one should apply Dirichlet’s principle. What is very important is that this method gives a nonconstructive proof (naturally, we cannot say which precisely hatch contains two rabbits and only know that such a hatch exists) and an attempt to give a constructive proof, i.e., the proof by explicitly constructing or indicating the desired object can lead to far greater diﬃculties (and more profound results). 2. Certain problems are also solved by methods in a way similar to Dirichlet’s principle. Let us formulate the corresponding statements (all of them are easily proved by the rule of contraries). a) If several segments the sum of whose lengths is greater than 1 lie on a segment of length 1, then at least two of them have a common point. b) If several arcs the sum of whose lengths is greater than 2π lie on the circle of radius 1, then at least two of them have a common point. c) If several ﬁgures the sum of whose areas is greater than 1 are inside a ﬁgure of area 1, then at least two of them have a common point. §1. The case when there are ﬁnitely many points, lines, etc. 21.1. The nodes of an inﬁnite graph paper are painted two colours. Prove that there exist two horizontal and two vertical lines on whose intersection lie points of the same colour. 21.2. Inside an equilateral triangle with side 1 ﬁve points are placed. Prove that the distance between certain two of them is shorter than 0.5. 21.3. In a 3 × 4 rectangle there are placed 6 points. Prove that among them there are two points the distance between which does not exceed √ 5. 21.4. On an 8 × 8 checkboard the centers of all the cells are marked. Is it possible to divide the board by 13 straight lines so that in each part there are not more than 1 of marked points? 21.5. Given 25 points in plane so that among any three of them there are two the distance between which is smaller than 1, prove that there exists a circle of radius 1 that contains not less than 13 of the given points. 21.6. In a unit square, there are 51 points. Prove that certain three of them can be covered by a disk of radius 1 7. 385 386 CHAPTER 21. DIRICHLET’S PRINCIPLE 21.7. Each of two equal disks is divided into 1985 equal sectors and on each of the disks some 200 sectors are painted (one colour). One of the disks was placed upon the other one and they began rotating one of the disks through multiples of 360◦ 1985. Prove that there exists at least 80 positions for which not more than 20 of the painted sectors of the disks coincide. 21.8. Each of 9 straight lines divides a square into two quadrilaterals the ratio of whose areas is 2 : 3. Prove that at least three of those nine straight lines pass through one point. 21.9. In a park, there grow 10, 000 trees planted by a so-called square-cluster method (100 rows of 100 trees each). What is the largest number of trees one has to cut down in order to satisfy the following condition: if one stands on any stump, then no other stump is seen (one may assume the trees to be suﬃciently thin). 21.10. What is the least number of points one has to mark inside a convex n-gon in order for the interior of any triangle with the vertices at vertices of the n-gon to contain at least one of the marked points? 21.11. Point P is taken inside a convex 2n-gon. Through every vertex of the polygon and P a line is drawn. Prove that there exists a side of the polygon which has no common interior points with neither of the drawn straight lines. 21.12. Prove that any convex 2n-gon has a diagonal non-parallel to either of its sides. 21.13. The nodes of an inﬁnite graph paper are painted three colours. Prove that there exists an isosceles right triangle with vertices of one colour. §2. Angles and lengths 21.14. Given n pairwise nonparallel lines in plane. Prove that the angle between certain two of them does not exceed 180◦ n . 21.15. In a circle of radius 1 several chords are drawn. Prove that if every diameter intersects not more than k chords, then the sum of the length of the chords is shorter than kπ. 21.16. In plane, point O is marked. Is it possible to place in plane a) ﬁve disks; b) four disks that do not cover O and so that any ray with the beginning in O would intersect not less than two disks? (“Intersect” means has a common point.) 21.17. Given a line l and a circle of radius n. Inside the circle lie 4n segments of length 1. Prove that it is possible to draw a line which is either parallel or perpendicular to the given line and intersects at least two of the given segments. 21.18. Inside a unit square there lie several circles the sum of their lengths being equal to 10. Prove that there exists a straight line intersecting at least four of these circles. 21.19. On a segment of length 1 several segments are marked so that the distance between any two marked points is not equal to 0.1. Prove that the sum of the lengths of the marked segments does not exceed 0.5. 21.20. Given two circles the length of each of which is equal to 100 cm. On one of them 100 points are marked, on the other one there are marked several arcs with the sum of their lengths less than 1 cm. Prove that these circles can be identiﬁed so that no one of the marked points would be on a marked arc. 21.21. Given are two identical circles; on each of them k arcs are marked, the angle value of each of the arcs is > 1 k2−k+1 · 180◦. The circles can be identiﬁed(?) so that the marked arcs of one circle would coincide with the marked arcs of the other one. Prove that these circles can be identiﬁed so that all the marked arcs would lie on unmarked arcs. SOLUTIONS 387 §3. Area 21.22. In square of side 15 there lie 20 pairwise nonintersecting unit squares. Prove that it is possible to place in the large square a unit disk so that it would not intersect any of the small squares. 21.23. Given an inﬁnite graph paper and a ﬁgure whose area is smaller than the area of a small cell prove that it is possible to place this ﬁgure on the paper without covering any of the nodes of the mesh. 21.24. Let us call the ﬁgure formed by the diagonals of a unit square (Fig. 39) a cross. Prove that it is possible to place only a ﬁnite number of nonintersecting crosses in a disk of radius 100. Figure 181 (21.24) 21.25. Pairwise distances between points A1, . . . , An is greater than 2. Prove that any ﬁgure whose area is smaller than π can be shifted by a vector not longer than 1 so that it would not contain points A1, . . . , An. 21.26. In a circle of radius 16 there are placed 650 points. Prove that there exists a ring (annulus) of inner radius 2 and outer radius 3 which contains not less than 10 of the given points. 21.27. There are given n ﬁgures in plane. Let Si1...ik be the area of the intersection of ﬁgures indexed by i1, . . . , ik and S be the area of the part of the plane covered by the given ﬁgures; Mk the sum of all the Si1...ik. Prove that: a) S = M1 −M2 + M3 −· · · + (−1)n+1Mn; b) S ≥M1 −M2 + M3 −· · · + (−1)m+1Mm for m even and S ≤M1 −M2 + M3 −· · · + (−1)m+1Mm for m odd. 21.28. a) In a square of area 6 there are three polygons of total area 3. Prove that among them there are two polygons such that the area of their intersection is not less than 1. b) In a square of area 5 there are nine polygons of total area 1. Prove that among them there are two polygons the area of whose intersection is not less than 1 9. 21.29. On a rug of area 1 there are 5 patches the area of each of them being not less than 0.5. Prove that there are two patches such that the area of their intersection is not less than 0.2. Solutions 21.1. Let us take three vertical lines and nine horizontal lines. Let us consider only intersection points of these lines. Since there are only 23 = 8 variants to paint three points two colours, there are two horizontal lines on which lie similarly coloured triples of points. Among three points painted two colours there are, by Dirichlet’s principle, two similarly 388 CHAPTER 21. DIRICHLET’S PRINCIPLE coloured points. The vertical lines passing through these points together with the two horizontal lines selected earlier are the ones to be found. 21.2. The midlines of an equilateral triangle with side 1 separate it into four equilateral triangles with side 0.5. Therefore, one of the triangles contains at least two of the given points and these points cannot be vertices of the triangle. The distance between these points is less than 0.5. 21.3. Let us cut the rectangle into ﬁve ﬁgures as indicated on Fig. 40. One of the ﬁgures contains at least two points and the distance between any two points of each of the ﬁgures does not exceed √ 5. Figure 182 (Sol. 21.3) 21.4. 28 ﬁelds are adjacent to a side of an 8 × 8 chessboard. Let us draw 28 segments that connect the centers of neighbouring end(?) ﬁelds. Every line can intersect not more than 2 such segments and, therefore, 13 lines can intersect not more than 26 segments, i.e., there are at least 2 segments that do not intersect any of 13 drawn lines. Therefore, it is impossible to split the chessboard by 13 lines so that in each part there would be not more than 1 marked point since both endpoints of the segment that does not intersect with the lines belongs to one of the parts. 21.5. Let A be one of the given points. If all the remaining points lie in disk S1 of radius 1 with center A, then we have nothing more to prove. Now, let B be a given point that lies outside S1, i.e., AB > 1. Consider disk S2 of radius 1 with center B. Among points A, B and C, where C is any of the given points, there are two at a distance less than 1 and these cannot be points A and B. Therefore, disks S1 and S2 contain all the given points, i.e., one of them contains not less than 13 points. 21.6. Let us divide a given square into 25 similar small squares with side 0.2. By Dirich-let’s principle one of them contains no less than 3 points. The radius of the circumscribed circle of the square with side 0.2 is equal to 1 5 √ 2 < 1 7 and, therefore, it can be covered by a disk of radius 1 7. 21.7. Let us take 1985 disks painted as the second of our disks and place them upon the ﬁrst disk so that they would take all possible positions. Then over every painted sector of the ﬁrst disk there lie 200 painted sectors, i.e., there are altogether2002 pairs of coinciding painted sectors. Let there be n positions of the second disk when not less 21 pairs of painted sectors coincide. Then the number of coincidences of painted sectors is not less than 21n. Therefore, 21n ≤2002, i.e., n ≤1904.8. Since n is an integer, n ≤1904. Therefore, at least for 1985 −1904 = 81 positions not more than 20 pairs of painted sectors coincide. 21.8. The given lines cannot intersect neighbouring sides of square ABCD since other-wise we would have not two quadrilaterals but a triangle and a pentagon. Let a line intersect sides BC and AD at points M and N, respectively. Trapezoids ABMN and CDNM have equal heights, and, therefore, the ratio of their areas is equal to that of their midlines, i.e., MN divides the segment that connects the midpoints of sides AB and CD in the ratio of 2 : 3. There are precisely 4 points that divide the midlines of the square in the ratio of 2 : 3. Since the given nine lines pass through these four points, then through one of the points at least three lines pass. SOLUTIONS 389 Figure 183 (Sol. 21.8) 21.9. Let us divide the trees into 2500 quadruples as shown in Fig. 41. In each such quadruple it is impossible to chop oﬀmore than 1 tree. On the other hand, one can chop oﬀall the trees that grow in the left upper corners of the squares formed by our quadruples. Therefore, the largest number of trees that can be chopped oﬀis equal to 2500. 21.10. Since any diagonal that goes out of one vertex divides an n-gon into n −2 triangles, then n −2 points are necessary. Figure 184 (Sol. 21.10) From Fig. 42 one can deduce that n −2 points are suﬃcient: it suﬃces to mark one points in each shaded triangle. Indeed, inside triangle ApAqAr, where p < q < r, there is always contained a shaded triangle adjacent to vertex Aq. 21.11. The two cases are possible. (1) Point P lies on diagonal AB. Then lines PA and PB coincide and do not intersect the sides. There remain 2n −2 lines; they intersect not more than 2n −2 sides. (2) Point P does not belong to a diagonal of polygon A1A2 . . . A2n. Let us draw diagonal A1An+1. On both sides of it there lie n sides. Let, for deﬁniteness, point P be inside polygon A1 . . . An+1 (Fig. 43). Then lines PAn+1, PAn+2, . . . , PA2n, PA1 (there are n + 1 such lines) cannot intersect sides An+1An+2, An+2An+3, . . . , A2nA1, respectively. Therefore, the remaining straight lines can intersect not more than n −1 of these n sides. 21.12. The number of diagonals of a 2n-gon is equal to 2n(2n−3) 2 = n(2n −3). It is easy to verify that there are not more than n −2 diagonals parallel to the given one. Therefore, there are not more than 2n(n−2) diagonals parallel to the sides. Since 2n(n−2) < n(2n−3), there exists a diagonal which is not parallel to any side. 390 CHAPTER 21. DIRICHLET’S PRINCIPLE Figure 185 (Sol. 21.11) 21.13. Suppose that there does not exist an equilateral right triangle whose legs are parallel to the sides of the cells and with vertices of the same colour. For convenience we may assume that it is the cells which are painted, not the nodes. Let us divide the paper into squares of side 4; then on the diagonal of each such square there are two cells of the same colour. Let n be greater than the number of distinct colorings of the square of side 4. Consider a square consisting of n2 squares of side 4. On its diagonal we can ﬁnd two similarly painted squares of side 4. Finally, take square K on whose diagonal we can ﬁnd two similarly painted squares of side 4n. Figure 186 (Sol. 21.13) Considering the square with side 4n and in it two similarly painted squares with side 4 we get four cells of the ﬁrst colour, two cells of the second colour and one cell of the third colour, see Fig. 44. Similarly, considering square K we get a cell which cannot be of the ﬁrst, or second, or third colour. 21.14. In plane, take an arbitrary point and draw through it lines parallel to the given ones. They divide the plane into 2n angles whose sum is equal to 360◦. Therefore, one of these angles does not exceed 180◦ n . 21.15. Suppose the sum of the length of the chords is not shorter than πk. Let us prove that then there exists a diameter which intersects with at least k+1 chords. Since the length of the arc corresponding to the chord is greater than the length of this chord, the sum of the lengths of the archs corresponding to given chords is longer than πk. If we add to these arcs the arcs symmetric to them through the center of the circle, then the sum of the lengths of all these arcs becomes longer than 2πk. Therefore, there exists a point covered by at least k + 1 of these arcs. The diameter drawn through this point intersects with at least k + 1 chord. SOLUTIONS 391 21.16. a) It is possible. Let O be the center of regular pentagon ABCDE. Then the disks inscribed in angles ∠AOC, ∠BOD, ∠COD, ∠DOA and ∠EOB possess the required property. b) It is impossible. For each of the four disks consider the angle formed by the tangents to the disk drawn through point O. Since each of these four angles is smaller than 180◦, their sum is less than 2 · 360◦. Therefore, there exists a point on the plane covered by not more than 1 of these angles. The ray drawn through this point intersects with not more than one disk. 21.17. Let l1 be an arbitrary line perpendicular to l. Denote the lengths of the pro-jections of the i-th segment to l and l1 by ai and bi, respectively. Since the length of each segment is equal to 1, we have ai + bi ≥1. Therefore, (a1 + · · · + a4n) + (b1 + · · · + b4n) ≥4n. Let, for deﬁniteness, a1 + · · · + a4n ≥b1 + · · · + b4n. Then a1 +· · ·+a4n ≥2n. The projection of any of the given segment is of length 2n because all of them lie inside the circle of radius n. If the projections of the given segments to l would have had no common points, then we would had a1 + · · · + a4n < 2n. Therefore, on l there exists a point which is the image under the projection of at least two of the given segments. The perpendicular to l drawn through this point intersects with at least two of given segments. 21.18. Let us project all the given circles on side AB of square ABCD. The projection of the circle of length l is a segment of length l π. Therefore, the sum of the lengths of the projections of all the given circles is equal to 10 π . Since 10 π > 3 = 3AB, on segment AB there is a point which belongs to projections of at least four circles. The perpendicular to AB drawn through this point intersects at least four circles. 21.19. Let us cut the segment into ten segments of length 0.1, stack them in a pile and consider their projection to a similar segment as shown on Fig. 45. Figure 187 (Sol. 21.19) Since the distance between any two painted points is not equal to 0.1, the painted points of neighbouring segments cannot be projected into one point. Therefore, neither of the points can be the image under the projection of painted points of more than 5 segments. It follows that the sum of the lengths of the projections of the painted segments (equal to the sum of their lengths) does not exceed 5 · 0.1 = 0.5. 21.20. Let us identify the given circles and let us place a painter in a ﬁxed point of one of them. Let us rotate this circle and let the painter paint a point of the other circle each time when it is a marked point that belongs to a marked arc. We have to prove that after a complete revolution a part of the circle would remain unpainted. The ﬁnal result of the painter’s job will be the same as if he were rotated 100 times and (s)he was asked to paint the other circle on the i-th revolution so that (s)he would have to 392 CHAPTER 21. DIRICHLET’S PRINCIPLE paint the i-th marked point that belongs to one of the marked arcs. Since in this case at each revolution less than 1 cm is being painted, it follows that after 100 revolutions there will be painted less than 100 cm. Therefore, a part of the circle will be unpainted. 21.21. Let us identify(?) our circles and place a painter into a ﬁxed point of one of them. Let us rotate this circle and let the painter paint the point of the other circle against which he moves each time when some of the marked arcs intersect. We have to prove that after a full revolution a part of the circle will be unpainted. The ﬁnal result of the painter’s job would be the same as if (s)he were rotated k times and was asked to paint the circle on the i-th revolution when the i-th marked arc on which the painter resides would intersect with a marked arc of the other circle. Let ϕ1, . . . , ϕn be the angle parameters of the marked arcs. By the hypothesis ϕ1 < α , . . . , ϕn < α, where α = 180◦ k2−k+1. During the time when the marked arcs with counters i and j intersect the painter paints an arc of length ϕi + ϕj. Therefore, the sum of the angle values of the arcs painted during the i-th revolution does not exceed kϕi(ϕ1 + · · · + ϕk) and the sum of the angle values of the arcs painted during all k revolutions does not exceed 2k(ϕ1 + · · · + ϕk). Observe that during all this we have actually counted the intersection of arcs with similar(?) counters k times. In particular, point A across which the painter moves at the moment when the marked arcs coincide has, deﬁnitely, k coats of paint. Therefore, it is desirable to disregard the arcs that the painter paints at the moment when some of the marked arcs with similar counters intersect. Since all these arcs contain point A, we actually disregard only one arc and the angle value of this arc does not exceed 2α. The sum of the angle values of the remaining part of the arcs painted during the i-th revolution does not exceed (k −1)ϕ1 + (ϕ1 + · · · + ϕk −ϕi) and the sum of the angle values of the remaining part of the arcs painted through all k revolutions does not exceed (2k −2) · (ϕ1 + · · · + ϕk) < (2k2 −2k)α. A part of the circle will be unpainted if (2k2 −2k)α ≤360◦−2α, i.e., α ≤ 180◦ k2−k+1. 21.22. Let us consider a ﬁgure consisting of all the points whose distance from the small unit square is not greater than 1 (Fig. 46). Figure 188 (Sol. 21.22) It is clear that no unit disk whose center is outside this ﬁgure intersects the small square. The area of such a ﬁgure is equal to π + 5. The center of the needed disk should also lie at a distance greater than 1 from the sides of the large square, i.e., inside the square of side 13. Obviously, 20 ﬁgures of total area π + 5 cannot cover a square of side 13 because 20(π + 5) < 132. The disk with the center in an uncovered point possesses the desired property. 21.23. Let us paint the ﬁgure to(?) the graph paper arbitrarily, cut the paper along the cells of the mesh and stack them in a pile moving them parallelly with themselves and SOLUTIONS 393 without turning. Let us consider the projection of this stack on a cell. The projections of parts of the ﬁgure cannot cover the whole cell since their area is smaller. Now, let us recall how the ﬁgure was placed on the graph paper and move the graph paper parallelly with itself so that its vertices would be in the points whose projection is an uncovered point. As a result we get the desired position of the ﬁgure. 21.24. For every cross consider a disk of radius 1 2 √ 2 with center in the center of the cross. Let us prove that if two such disks intersect, then the crosses themselves also intersect. The distance between the centers of equal intersecting disks does not exceed the doubled radius of any of them and, therefore, the distance between the centers of the corresponding crosses does not exceed 1 √ 2. Let us consider a rectangle given by bars of the ﬁrst cross and the center of the second one (Fig. 47). Figure 189 (Sol. 21.24) One of the bars of the second cross passes through this rectangle and, therefore, it intersects the ﬁrst cross since the length of the bar is equal to 1 √ 2 and the length of the diagonal of the rectangle does not exceed 1 √ 2. In the disk of a ﬁnite radius one can only place ﬁnitely many non-intersecting disks of radius 1 2 √ 2. 21.25. Let Φ be a given ﬁgure, S1, . . . , Sn unit disks with centers at points A1, . . . , An. Since disks S1, . . . , Sn do not intersect, then neither do ﬁgures Vi = Φ ∩Si, consequently, the sum of their areas does not exceed the area of ﬁgure Φ, i.e., it is smaller than π. Let O be an arbitrary point and Wi the image of Vi under the translation by vector − − → AiO. The ﬁgures Wi lie inside the unit disk S centered at O and the sum of their areas is smaller than the area of this disk. Therefore, point B of disk S does not belong to any of the ﬁgures Wi. It is clear that the translation by vector − − → BO is the desired one. 21.26. First, notice that point X belongs to the ring with center O if and only if point O belongs to a similar ring centered at X. Therefore, it suﬃces to show that if we construct rings with centers at given points, then not less than 10 rings will cover one of the points of the considered disk. The considered rings lie inside a disk of radius 16 + 3 = 19 whose area is equal to 361π. It remains to notice that 9 · 361π = 3249π and the total area of the rings is equal to 650 · 5π = 3250π. 21.27. a) Let ¡n k ¢ be the number of ways to choose k elements from n indistinguishable ones. One can verify the following Newton binomial formula (x + y)n = n X k=0 µn k ¶ xkyn−k. Denote by Wm the area of the part of the plane covered by exactly m ﬁgures. This part consists of pieces each of which is covered by certain m ﬁgures. The area of each such piece has been counted ¡n k ¢ times in calculation of Mk because from m ﬁgures we can form ¡n k ¢ 394 CHAPTER 21. DIRICHLET’S PRINCIPLE intersections of k ﬁgures. Therefore, Mk = µn k ¶ Wk + µk + 1 k ¶ Wk+1 + · · · + µn k ¶ Wn. It follows that M1 −M2 + M3 −· · · = µ1 1 ¶ W1 + ( µ2 1 ¶ − µ2 2 ¶ )W2 + · · · + ( µn 1 ¶ − µn 2 ¶ + µn 3 ¶ −. . . )Wn = W1 + · · · + Wn since µm 1 ¶ − µm 2 ¶ + µm 3 ¶ −· · · −(−1)m µm m ¶ = (−1 + µm 1 ¶ − µm 2 ¶ + . . . ) + 1) = −(1 −1)m + 1 = 1. It remains to observe that S = W1 + · · · + Wn. b) According to heading a) S −(M1 −M2 + · · · + (−1)m+1Mm) = (−1)m+2Mm+1 + (−1)m+3Mm+2 + . . . · · · + (−1)n+1Mn = n X i=1 ((−1)m+2 µ i m + 1 ¶ + · · · + (−1)n+1 µi n ¶ )Wi (it is convenient to assume that ¡n k ¢ is deﬁned for k > n so that ¡n k ¢ = 0). Therefore, it suﬃces to verify that µ i m + 1 ¶ − µ i m + 2 ¶ + µ i m + 3 ¶ −· · · + (−1)m+n+1 µi n ¶ ≥0 for i ≤n. The identity (x + y)i = (x + y)i−1(x + y) implies that ¡i j ¢ = ¡i−1 j−1 ¢ + ¡i−1 j ¢ . Hence, µ i m + 1 ¶ − µ i m + 2 ¶ + · · · + (−1)m+n+1 µi n ¶ = µi −1 m ¶ ± µi −1 n ¶ . It remains to notice that ¡i−1 n ¢ = 0 for i ≤n. 21.28. a) By Problem 21.27 a) we have 6 = 9 −(S12 + S23 + S13) + S123, i.e., S12 + S23 + S13 = 3 + S123 ≥3. Hence, one of the numbers S12, S23, S13 is not less than 1. b) By Problem 21.27 b) 5 ≥9 −M2, i.e., M2 ≥4. Since from 9 polygons one can form 9 · 8 2 = 36 pairs, the area of the common part of one of such pairs is not less than M2 36 ≥1 9. SOLUTIONS 395 21.29. Let the area of the rug be equal to M, the area of the intersection of the patches indexed by i1, . . . , ik is equal to Si1...ik and Mk = P Si1...ik. By Problem 21.27 a) M −M1 + M2 −M3 + M4 −M5 ≥0 since M ≥S. One can write similar inequalities not only for the whole rug but also for every patch: if we consider the patch S1 as the rug with patches S12, S13, S14, S15 we get S1 − X i S1i + X i<j S1ij − X i<j 13 there exists an n-gon for which the similar statement is false. 22.27. What is the largest number of acute angles in a nonconvex n-gon? 22.28. The following operations are done over a nonconvex non-selﬁntersecting polygon. If it lies on one side of line AB, where A and B are non-neighbouring vertices, then we reﬂect one of the parts into which points A and B divide the contour of the polygon through the SOLUTIONS 399 midpoint of segment AB. Prove that after several such operations the polygon becomes a convex one. 22.29. The numbers α1, . . . , αn whose sum is equal to (n −2)π satisfy inequalities 0 < αi ≤2π. Prove that there exists an n-gon A1 . . . An with anagles α1, . . . , αn at vertices A1, . . . , An, respectively. Solutions 22.1. Consider the convex hull of given points. It is a convex polygon. We have to prove that all the given points are its vertices. Suppose one of the given points (point A) is not a vertex, i.e., it lies inside or on the side of the polygon. The diagonals that go out of this vertex cut the convex hull into triangles; point A belongs to one of the triangles. The vertices of this triangle and point A cannot be vertices of a convex quadrilateral. Contradiction. 22.2. Consider the convex hull of given points. If it is a quadrilateral or a pentagon, then all is clear. Now, suppose that the convex hull is triangle ABC and points D and E lie inside it. Point E lies inside one of the triangles ABD, BCD, CAD; let for deﬁniteness sake it belong to the interior of triangle ABC. Let H be the intersection point of lines CD and AB. Point E lies inside one of the triangles ADH and BDH. If, for example, E lies inside triangle ADH, then AEDC is a convex quadrilateral (Fig. 48). Figure 190 (Sol. 22.2) 22.3. Let the convex hull of the vertices of the given n-gons be an m-gon and ϕ1, . . . , ϕm its angles. Since to every angle of the convex hull an angle of a regular n-gon is adjacent, ϕi ≥(1−( 2 n))π (in the right-hand side there stands the value of an angle of a regular n-gon). Therefore, ϕ1 + · · · + ϕm ≥m(1 −(2 n))π = (m −(2m n ))π. On the other hand, ϕ1 + · · · + ϕm = (m −2)π; hence, (m −2)π ≥(m −( 2m n ))π, i.e., m ≥n. 22.4. First, notice that it suﬃces to take 50 triangles. Indeed, let ∆k be the triangle whose sides lie on rays AkAk−1 and AkAk+1 and which contains convex polygon A1 . . . A100. Then this polygon is the intersection of the triangles ∆2, ∆4, . . . , ∆100. Figure 191 (Sol. 22.4) 400 CHAPTER 22. CONVEX AND NONCONVEX POLYGONS On the other hand, the 100-gon depicted on Fig. 49 cannot be represented as the inter-section of less than 50 triangles. Indeed, if three of its sides lie on the sides of one triangle, then one of these sides is side A1A2. All the sides of this polygon lie on the sides of n triangles and, therefore, 2n + 1 ≥100, i.e., n ≥50. 22.5. Let P be the intersection point of diagonals A1A4 and A2A5 of convex heptagon A1 . . . A7. One of the diagonals A3A7 or A3A6, let, for deﬁniteness, this be A3A6, does not pass through point P. There are ﬁnitely many intersection points of the diagonals of hexagon A1 . . . A6 and, therefore, in a vicinity of point A7 one can select a point A′ 7 such that lines A1A′ 7, . . . , A6A′ 7 do not pass through these points, i.e., heptagon A1 . . . A′ 7 is a nonsingular one. 22.6. First, let us prove that H is a convex ﬁgure. Let points A and B belong to H, i.e., A and B be the midpoints of segments C1D1 and C2D2, where C1 and C2 belong to F and D1, respectively, and D2 belong to G. We have to prove that the whole segment AB belongs to H. It is clear that segments C1C2 and D1D2 belong to F and G, respectively. The locus of the midpoints of segments with the endpoints on segments C1C2 and D1D2 is the parallelogram with diagonal AB (Fig. 50); this follows from the fact that the locus of the midpoints of segments CD, where C is ﬁxed and D moves along segment D1D2, is the midline of triangle CD1D2. Figure 192 (Sol. 22.6) In plane, take an arbitrary coordinate axis Ox. The set of all the points of the polygon whose projections to the axis have the largest value (Fig. 51) will be called the basic set of the polygon with respect to axis Ox. Figure 193 (Sol. 22.6) The convex polygon is given by its basic sets for all possible axes Ox. If basic sets F and G with respect to an axis are segments of length a and b, then the basic set of H with SOLUTIONS 401 respect to the same axis is a segment of length a+b 2 (here we assume that a point is segment of zero length). Therefore, the perimeter of H is equal to P1+P2 2 and the number of H’s sides can take any value from the largest — n1 or n2 — to n1 + n2 depending on for how many axes both basic sets of F and G are sides and not vertices simultaneously. 22.7. We will prove a more general statement. Recall that cardinality of a set is (for a ﬁnite set) the number of its element. (Ramsey’s theorem.) Let p, q and r be positive integers such that p, q ≥r. Then there exists a number N = N(p, q, r) with the following property: if r-tuples from a set S of cardinality N are divided at random into two nonintersecting families α and β, then either there exists a p-tuple of elements from S all subsets of cardinality r of which are contained in α or there exists a q-tuple all subsets of cardinality r of which are contained in β. The desired statement follows easily from Ramsey’s theorem. Indeed, let N = N(p, 5, 4) and family α consist of quadruples of elements of an N-element set of points whose convex hulls are quadrilaterals. Then there exists a subset of n elements of the given set of points the convex hulls of any its four-elements subset being quadrilaterals because there is no ﬁve-element subset such that the convex hulls of any four-element subsets of which are triangles (see Problem 22.2). It remains to make use of the result of Problem 22.1. Now, let us prove Ramsey’s theorem. It is easy to verify that for N(p, q, 1), N(r, q, r) and N(p, r, r) one can take numbers p + q −1, q and p, respectively. Now, let us prove that if p > r and q > r, then for N(p, q, r) one can take numbers N(p1, q1, r −1) + 1, where p1 = N(p −1, q, r) and q1 = N(p, q −1, r). Indeed, let us delete from the N(p, q, r)-element set S one element and divide the (r −1)-element subsets of the obtained set S′ into two families: family α′ (resp. β′) consists of subsets whose union with the deleted element enters α (resp. β). Then either (1) there exists a p1-element subset of S′ all (r −1)-element subsets of which are contained in α′ or (2) there exists a q1-element subset all whose (r −1) element subsets are contained in family β′. Consider case (1). Since p1 = N(p−1, q, r), it follows that either there exists a q-element subset of S′ all r-element subsets of which belong to β (then these q elements are the desired one) or there exists a (p −1)-element subset of S′ all the r-element subsets of which are contained in α (then these p −1 elements together with the deleted element are the desired ones). Case (2) is treated similarly. Thus, the proof of Ramsey’s theorem can be carried out by induction on r, where in the proof of the inductive step we make use of induction on p + q. 22.8. If the polygon is not a triangle or parallelogram, then it has two nonparallel non-neighbouring sides. Extending them until they intersect, we get a new polygon which contains the initial one and has fewer number of sides. After several such operations we get a triangle or a parallelogram. If we have got a triangle, then everything is proved; therefore, let us assume that we have got a parallelogram, ABCD. On each of its sides there lies a side of the initial polygon and one of its vertices, say A, does not belong to the initial polygon (Fig. 52). Let K be a vertex of the polygon nearest to A and lying on AD; let KL be the side that does not lie on AD. Then the polygon is conﬁned inside the triangle formed by lines KL, BC and CD. 22.9. The proof will be carried out by induction on n. For n = 3 the statement is obvious. Let n ≥4. By Problem 22.8 there exist lines a, b and c which are extensions of the sides of the given n-gon that constitute triangle T which contains the given n-gon. Let line l be the extension of some other side of the given n-gon. The extensions of all the sides 402 CHAPTER 22. CONVEX AND NONCONVEX POLYGONS Figure 194 (Sol. 22.8) of the n-gon except the side which lies on line l form a convex (n −1)-gon that lies inside triangle T. By the inductive hypothesis for this (n −1)-gon there exist n −3 required triangles. Moreover, line l and two of the lines a, b and c also form a required triangle. Remark. If points A2, . . . , An belong to a circle with center at A1, where ∠A2A1An < 90◦and the n-gon A1 . . . An is a convex one, then for this n-gon there exist precisely n −2 triangles required. 22.10. Proof will be carried out by induction on n. For n = 3 the proof is obvious. Now, let us consider n-gons A1 . . . An, where n ≥4. Point O lies inside triangle ApAqAr. Let Ak be a vertex of the given n-gon distinct from points Ap, Aq and Ar. Selecting vertex Ak in n-gon A1 . . . An we get a (n −1)-gon to which the inductive hypothesis is applicable. Moreover, the angles ∠AkOAp, ∠AkOAq and ∠AkOAr cannot all be acute ones because the sum of certain two of them is greater than 180◦. 22.11. Proof will be carried out by induction on n. For n = 3 the statement is obvious. Let n ≥4. Fix one acute triangle ApAqAr and let us discard vertex Ak distinct from the vertices of this triangle. The inductive hypothesis is applicable to the obtained (n −1)-gon. Moreover, if, for instance, point Ak lies on arc ApAq and ∠AkApAr ≤∠AkAqAr, then triangle AkApAr is an acute one. Indeed, ∠ApAkAr = ∠ApAqAr, ∠ApArAk < ∠ApArAq and ∠AkApAr ≤90◦; hence, ∠AkApAr < 90◦. 22.12. a) Denote the given ﬁgures by M1, M2, M3 and M4. Let Ai be the intersection point of all the ﬁgures except Mi. Two variants of arrangements of points Ai are possible. 1) One of the points, for example, A4 lies inside the triangle formed by the remaining points. Since points A1, A2, A3 belong to the convex ﬁgure M4, all points of A1A2A3 also belong to M4. Therefore, point A4 belongs to M4 and it belongs to the other ﬁgures by its deﬁnition. 2) A1A2A3A4 is a convex quadrilateral. Let C be the intersection point of diagonals A1A3 and A2A4. Let us prove that C belongs to all the given ﬁgures. Both points A1 and A3 belong to ﬁgures M2 and M4, therefore, segment A1A3 belongs to these ﬁgures. Similarly, segment A2A4 belongs to ﬁgures M1 and M3. It follows that the intersection point of segments A1A3 and A2A4 belongs to all the given ﬁgures. b) Proof will be carried out by induction on the number of ﬁgures. For n = 4 the statement is proved in the preceding problem. Let us prove that if the statement holds for n ≥4 ﬁgures, then it holds also for n + 1 ﬁgures. Given convex ﬁgures Φ1, . . . , Φn, Φn+1 every three of which have a common point, consider instead of them ﬁgures Φ1, . . . , Φn−1, Φ′ n, where Φ′ n is the intersection of Φn and Φn+1. It is clear that Φ′ n is also a convex ﬁgure. Let us prove that any three of the new ﬁgures have a common point. One can only doubt this for the triple of ﬁgures that contain Φ′ n but the preceding problem implies that ﬁgures SOLUTIONS 403 Φi, Φj, Φn and Φn+1 always have a common point. Therefore, by the inductive hypothesis Φ1, . . . , Φn−1, Φ′ n have a common point; hence, Φ1, . . . , Φn, Φn+1 have a common point. 22.13. A unit disk centered at O covers certain points if and only if unit disks centered at these points contain point O. Therefore, our problem admits the following reformulation: Given n points in plane such that any three unit disks centered at these points have a common point, prove that all these disks have a common point. This statement clearly follows from Helley’s theorem. 22.14. Consider pentagons that remain after deleting pairs of neighbouring vertices of a heptagon. It suﬃces to verify that any three of the pentagons have a common point. For three pentagons we delete not more than 6 distinct vertices, i.e., one vertex remains. If vertex A is not deleted, then the triangle shaded in Fig. 53 belongs to all three pentagons. Figure 195 (Sol. 22.14) 22.15. Let us introduce the coordinate system with Oy-axis parallel to the given seg-ments. For every segment consider the set of all points (a, b) such that the line y = ax + b intersects it. It suﬃces to verify that these sets are convex ones and apply to them Helley’s theorem. For the segment with endpoints (x0, y1) and (x0, y2) the considered set is a band between parallel lines ax0 + b = y1 and ax0 + b = y2. 22.16. Wrong. A counterexample is given on Fig. 54. Figure 196 (Sol. 22.16) 22.17. The required polygons and points are drawn on Fig. 55. Figure 197 (Sol. 22.17) 404 CHAPTER 22. CONVEX AND NONCONVEX POLYGONS 22.18. Let the whole contour of polygon A1 . . . An subtend an angle with vertex O. Then no other side of the polygon except AiAi+1 lies inside angle ∠AiOAi+1; hence, point O lies inside the polygon (Fig. 56). Any point X in plane belongs to one of the angles ∠AiOAi+1 and, therefore, side AiAi+1 subtends an angle with vertex in X. Figure 198 (Sol. 22.18) 22.19. Since all the inner angles of a convex n-gon are smaller than 180◦and their sum is equal to (n −2) · 180◦, the sum of the exterior angles is equal to 360◦, i.e., for a convex polygon we attain the equality. Figure 199 (Sol. 22.19) Now, let M be the convex hull of polygon N. Each angle of M contains an angle of N smaller than 180◦and the angle of M can be only greater than the angle of N, i.e., the exterior angle of N is not less than the exterior angle of M (Fig. 57). Therefore, even restricting to the angles of N adjacent to the angles of M we will get not less than 360◦. 22.20. a) If the polygon is a convex one, then the statement is proved. Now, suppose that the exterior angle of the polygon at vertex A is greater than 180◦. The visible part of the side subtends an angle smaller than 180◦with vertex at point A, therefore, parts of at least two sides subtend an angle with vertex at A. Therefore, there exist rays exiting point A and such that on these rays the change of (parts of) sides visible from A occurs (on Fig. 58 all such rays are depicted). Each of such rays determines a diagonal that lies entirely inside the polygon. b) On Fig. 59 it is plotted how to construct an n-gon with exactly n −3 diagonals inside it. It remains to demonstrate that any n-gon has at least n −3 diagonals. For n = 3 this statement is obvious. Suppose the statement holds for all k-gons, where k < n and let us prove it for an n-gon. By heading a) it is possible to divide an n-gon by its diagonal into two polygons: a (k + 1)-gon and an (n −k + 1)-gon, where k + 1 < n and n −k + 1 < n. These parts have at least (k + 1) −3 and (n −k + 1) −3 diagonals, respectively, that lie inside these parts. Therefore, the n-gon has at least 1 + (k −2) + (n −k −2) = n −3 diagonals that lie inside it. 22.21. First, let us prove that if A and B are neighbouring vertices of the n-gon, then either from A or from B it is possible to draw a diagonal. The case when the inner angle SOLUTIONS 405 Figure 200 (Sol. 22.20 a)) Figure 201 (Sol. 22.20 b)) of the polygon at A is greater than 180◦is considered in the solution of Problem 22.20 a). Now, suppose that the angle at vertex A is smaller than 180◦. Let B and C be vertices neighbouring A. If inside triangle ABC there are no other vertices of the polygon, then BC is the diagonal and if P is the nearest to A vertex of the polygon lying inside triangle ABC, then AP is the diagonal. Hence, the number of vertices from which it is impossible to draw the diagonal does not exceed [n 2] (the integer part of n 2). On the other hand, there exist n-gons for which this estimate is attained, see Fig. 60. Figure 202 (Sol. 22.21) 22.22. Let us prove the statement by induction on n. For n = 3 it is obvious. Let n ≥4. Suppose the statement is proved for all k-gons, where k < n; let us prove it for an n-gon. Any n-gon can be divided by a diagonal into two polygons (see Problem 22.20 a)) and the number of vertices of every of the smaller polygons is strictly less than n, i.e., they can be divided into triangles by the inductive hypothesis. 22.23. Let us prove the statement by induction. For n = 3 it is obvious. Let n ≥4. Suppose it is proved for all k-gons, where k < n, and let us prove it for an n-gon. Any n-gon can be divided by a diagonal into two polygons (see Problem 22.20 a)). If the number of sides of one of the smaller polygons is equal to k + 1, then the number of sides of the other one is equal to n −k + 1 and both numbers are smaller than n. Therefore, the sum of the 406 CHAPTER 22. CONVEX AND NONCONVEX POLYGONS angles of these polygons are equal to (k −1) · 180◦and (n −k −1) · 180◦, respectively. It is also clear that the sum of the angles of a n-gon is equal to the sum of the angles of these polygons, i.e., it is equal to (k −1 + n −k −1) · 180◦= (n −2) · 180◦. 22.24. The sum of all the angles of the obtained triangles is equal to the sum of the angles of the polygon, i.e., it is equal to (n −2) · 180◦, see Problem 22.23. Therefore, the number of triangles is equal to n −2. 22.25. Let ki be the number of triangles in the given partition for which precisely i sides are the sides of the polygon. We have to prove that k2 ≥2. The number of sides of the n-gon is equal to n and the number of the triangles of the partition is equal to n −2, see Problem 22.24. Therefore, 2k2 + k1 = n and k2 + k1 + k0 = n −2. Subtracting the second equality from the ﬁrst one we get k2 = k0 + 2 ≥2. 22.26. Suppose that there exists a 13-gon for which on any line that contains its side there lies at least one side. Let us draw lines through all the sides of this 13-gon. Since the number of sides is equal to 13, it is clear that one of the lines contains an odd number of sides, i.e., one of the lines has at least 3 sides. On these sides lie 6 vertices and through each vertex a line passes on which there lie at least 2 sides. Therefore, this 13-gon has not less than 3 + 2 · 6 = 15 sides but this is impossible. Figure 203 (Sol. 22.26) For n even, n ≥10, the required example is the contour of a “star” (Fig. 61 a)) and an idea of how to construct an example for n odd is illustrated on Fig. 61 b). 22.27. Let k be the number of acute angles of the n-gon. Then the number of its angles is smaller than k·90◦+(n−k)·360◦. On the other hand, the sum of the angles of an n-gon is equal to (n−2)·180◦(see Problem 22.23) and, therefore, k·90◦+(n−k)·360◦> (n−2)·180◦, i.e., 3k < 2n + 4. It follows that k ≤[ 2n 3 ] + 1, where [x] denotes the largest integer not exceeding x. Figure 204 (Sol. 22.27) Examples of n-gons with [2n 3 ] + 1 acute angles are given on Fig. 62. 22.28. Under these operations the vectors of the sides of a polygon remain the same only their order changes (Fig. 63). Therefore, there exists only a ﬁnite number of polygons that SOLUTIONS 407 Figure 205 (Sol. 22.28) may be obtained. Moreover, after each operation the area of the polygon strictly increases. Hence, the process terminates. 22.29. Let us carry out the proof by induction on n. For n = 3 the statement is obvious. Let n ≥4. If one of the numbers αi is equal to π, then the inductive step is obvious and, therefore, we may assume that all the numbers αi are distinct from π. If n ≥4, then 1 n n X i=1 (αi + αi+1) = 2(n −2)π n ≥π, where the equality is only attained for a quadrilateral. Hence, in any case except for a parallelogram (α1 = π −α2 = α3 = π −α4), and (?) there exist two neighbouring numbers whose sum is greater than π. Moreover, there exist numbers αi and αi+1 such that π < αi + αi+1 < 3π. Indeed, if all the given numbers are smaller than π, then we can take the above-mentioned pair of numbers; if αj > π, then we can take numbers αi and αi+1 such that αi < π and αi+1 > π. Let α∗ i = αi + αi+1 −1. Then 0 < α∗ i < 2π and, therefore, by the inductive hypothesis there exists an (n −1)-gon M with angles α1, . . . , αi−1, α∗ i , αi+2, . . . , αn. Three cases might occur: 1) α∗ i < π, 2) α∗ i = π, 3) π < α∗ i < 2π. In the ﬁrst case αi + αi+1 < 2π and, therefore, one of these numbers, say αi, is smaller than π. If αi+1 < π, then let us cut from M a triangle with angles π −αi, π −αi+1, α∗ i (Fig. 64 a)). If αi+1 > π, then let us juxtapose to M a triangle with angles αi, αi+1 −π, π −α∗ i (Fig. 64 b)). In the second case let us cut from M a trapezoid with the base that belongs to side Ai−1A∗ i Ai+2 (Fig. 64 c)). In the third case αi + αi+1 > π and, therefore, one of these numbers, say αi, is greater than π. If αi+1 > π, then let us juxtapose to M a triangle with angles αi −π, αi+1 −π, 2π −α∗ i (Fig. 64 d)), and if αi+1 < π let us cut oﬀM a triangle with angles 2π −αi, π −αi+1 and α∗ i −π (Fig. 64 e)). 408 CHAPTER 22. CONVEX AND NONCONVEX POLYGONS Figure 206 (Sol. 22.29) Chapter 23. DIVISIBILITY, INVARIANTS, COLORINGS Background 1. In a number of problems we encounter the following situation. A certain system consecutively changes its state and we have to ﬁnd out something at its ﬁnal state. It might be diﬃcult or impossible to trace the whole intermediate processes but sometimes it is possible to answer the question with the help of a quantity that characterizes the state of the system and is preserved during all the transitions (such a quantity is sometimes called an invariant of the system considered). Clearly, in the ﬁnal state the value of the invariant is the same as in the initial one, i.e., the system cannot occur in any state with another value of the invariant. 2. In practice this method reduces to the following. A quantity is calculated in two ways: ﬁrst, it is simply calculated in the initial and ﬁnal states and then its variation is studied under consecutive elementary transitions. 3. The simplest and most often encountered invariant is the parity of a number; the residue after a division not only by 2 but some other number can also be an invariant. In the construction of invariants certain auxiliary colorings are sometimes convenient, i.e., partitions of considered objects into several groups, where each group consists of the objects of the same colour. §1. Even and odd 23.1. Can a line intersect (in inner points) all the sides of a nonconvex a) (2n + 1)-gon; b) 2n-gon? 23.2. Given a closed broken plane line with a ﬁnite number of links and a line l that intersects it at 1985 points, prove that there exists a line that intersects this broken line in more than 1985 points. 23.3. In plane, there lie three pucks A, B and C. A hockey player hits one of the pucks so that it passes (along the straight line) between the other two and stands at some point. Is it possible that after 25 hits all the pucks return to the original places? 23.4. Is it possible to paint 25 small cells of the graph paper so that each of them has an odd number of painted neighbours? (Riddled cells are called neighbouring if they have a common side). 23.5. A circle is divided by points into 3k arcs so that there are k arcs of length 1, 2, and 3. Prove that there are 2 diametrically opposite division points. 23.6. In plane, there is given a non-selﬁntersecting closed broken line no three vertices of which lie on one line. A pair of non-neighbouring links of the broken will be called a singular one if the extension of one of them intersects the other one. Prove that the number of singular pairs is always even. 23.7. (Sperner’s lemma.) The vertices of a triangle are labeled by ﬁgures 0, 1 and 2. This triangle is divided into several triangles so that no vertex of one triangle lies on a side of the other one. The vertices of the initial triangle retain their old labels and the additional vertices get labels 0, 1, 2 so that any vertex on a side of the initial triangle should 409 410 CHAPTER 23. DIVISIBILITY, INVARIANTS, COLORINGS be labelled by one of the vertices of this side, see Fig. 65. Prove that there exists a triangle in the partition labelled by 0, 1, 2. Figure 207 (23.6) 23.7. The vertices of a regular 2n-gon A1 . . . A2n are divided into n pairs. Prove that if n = 4m+2 or n = 4m+3, then the two pairs of vertices are the endpoints of equal segments. §2. Divisibility 23.9. On Fig. 66 there is depicted a hexagon divided into black and white triangles so that any two triangles have either a common side (and then they are painted diﬀerent colours) or a common vertex, or they have no common points and every side of the hexagon is a side of one of the black triangles. Prove that it is impossible to ﬁnd a similar partition for a 10-gon. Figure 208 (23.9) 23.10. A square sheet of graph paper is divided into smaller squares by segments that follow the sides of the small cells. Prove that the sum of the lengths of these segments is divisible by 4. (The length of a side of a small cell is equal to 1). §3. Invariants 23.11. Given a chess board, it is allowed to simultaneously repaint into the opposite colour either all the cells of one row or those of a column. Can we obtain in this way a board with precisely one black small cell? 23.12. Given a chess board, it is allowed to simultaneously repaint into the opposite colour all the small cells situated inside a 2×2 square. Is it possible that after such repaintings there will be exactly one small black cell left? 23.13. Given a convex 2m-gon A1 . . . A2m and point P inside it not belonging to any of the diagonals, prove that P belongs to an even number of triangles with vertices at points A1, . . . , A2m. §4. AUXILIARY COLORINGS 411 23.14. In the center of every cell of a chess board stands a chip. Chips were interchanged so that the pairwise distances between them did not diminish. Prove that the pairwise distances did not actually alter at all. 23.15. A polygon is cut into several polygons so that the vertices of the obtained polygons do not belong to the sides of the initial polygon nor to the sides of the obtained polygons. Let p be the number of the obtained smaller polygons, q the number of segments which serve as the sides of the smaller polygons, r the number of points which are their vertices. Prove that p −q + r = 1. (Euler’s formula) 23.16. A square ﬁeld is divided into 100 equal square patches 9 of which are overgrown with weeds. It is known that during a year the weeds spread to those patches that have not less than two neighbouring (i.e., having a common side) patches that are already overgrown with weeds and only to them. Prove that the ﬁeld will never overgrow completely with weeds. 23.17. Prove that there exist polygons of equal size and impossible to divide into poly-gons (perhaps, nonconvex ones) which can be translated into each other by a parallel trans-lation. 23.18. Prove that it is impossible to cut a convex polygon into ﬁnitely many nonconvex quadrilaterals. 23.18. Given points A1, . . . , An. We considered a circle of radius R encircling some of them. Next, we constructed a circle of radius R with center in the center of mass of points that lie inside the ﬁrst circle, etc. Prove that this process eventually terminates, i.e., the circles will start to coincide. §4. Auxiliary colorings 23.20. In every small cell of a 5 × 5 chess board sits a bug. At certain moment all the bugs crawl to neighbouring (via a horizontal or a vertical) cells. Is it necessary that some cell to become empty at the next moment? 23.21. Is it possible to tile by 1 × 2 domino chips a 8 × 8 chess board from which two opposite corner cells are cut out? 23.22. Prove that it is impossible to cut a 10 × 10 chess board into T-shaped ﬁgures consisting of four cells. 23.23. The parts of a toy railroad’s line are of the form of a quarter of a circle of radius R. Prove that joining them consecutively so that they would smoothly turn into each other it is impossible to construct a closed path whose ﬁrst and last links form the dead end depicted on Fig. 67. Figure 209 (23.23) 412 CHAPTER 23. DIVISIBILITY, INVARIANTS, COLORINGS 23.24. At three vertices of a square sit three grasshoppers playing the leap frog as follows. If a grasshopper A jumps over a grasshopper B, then after the jump it lands at the same distance from B but, naturally, on the other side and on the same line. Is it possible that after several jumps one of the grasshoppers gets to the fourth vertex of the square? 23.25. Given a square sheet of graph paper of size 100 × 100 cells. Several nonselﬁnter-secting broken lines passing along the sides of the small cells and without common points are drawn. These broken lines are all strictly inside the square but their endpoints are invariably on the boundary. Prove that apart from the vertices of the square there will be one more node (of the graph paper inside the square or on the boundary) that does not belong to any of the broken lines. §5. More auxiliary colorings 23.26. An equilateral triangle is divided into n2 equal equilateral triangles (Fig. 68). Some of them are numbered by numbers 1, 2, . . . , m and consecutively numbered triangles have adjacent sides. Prove that m ≤n2 −n + 1. Figure 210 (23.26) 23.27. The bottom of a parallelepipedal box is tiled with tiles of size 2 × 2 and 1 × 4. The tiles had been removed from the box and in the process one tile of size 2 × 2 was lost. We replaced it with a tile of size 1 × 4. Prove that it will be impossible to tile now the bottom of the box. 23.28. Of a piece of graph paper of size 29 × 29 (of unit cells) 99 squares of size 2 × 2 were cut. Prove that it is still possible to cut oﬀone more such square. 23.29. Nonintersecting diagonals divide a convex n-gon into triangles and at each of the n-gon’s vertex an odd number of triangles meet. Prove that n is divisible by 3. 23.30. Is it possible to tile a 10 × 10 graph board by tiles of size 2 × 4? 23.31. On a graph paper some arbitrary n cells are ﬁxed. Prove that from them it is possible to select not less than n 4 cells without common points. 23.32. Prove that if the vertices of a convex n-gon lie in the nodes of graph paper and there are no other nodes inside or on the sides of the n-gon, then n ≤4. 23.33. From 16 tiles of size 1×3 and one tile of size 1×1 one constructed a 7×7 square. Prove that the 1 × 1 tile either sits in the center of the square or is adjacent to its boundary. 23.34. A picture gallery is of the form of a nonconvex n-gon. Prove that in order to overview the whole gallery [n 3] guards suﬃces. §6. Problems on colorings 23.35. A plane is painted two colours. Prove that there exist two points of the same colour the distance between which is equal to 1. SOLUTIONS 413 23.36. A plane is painted three colours. Prove that there are two points of the same colour the distance between which is equal to 1. 23.37. The plane is painted seven colours. Are there necessarily two points of the same colour the distance between which is equal to 1? (?)23.38. The points on sides of an equilateral triangle are painted two colours. Prove that there exists a right triangle with vertices of the same colour. A triangulation of a polygon is its partition into triangles with the property that these triangles have either a common side or a common vertex or have no common points (i.e., the vertex of one triangle cannot belong to a side of the other one). 23.39. Prove that it is possible to paint the triangles of a triangulation three colours so that the triangles with a common side would be of diﬀerent colours. 23.40. A polygon is cut by nonintersecting diagonals into triangles. Prove that the vertices of the polygon can be painted three colours so that all the vertices of each of the obtained triangles would be of diﬀerent colours. 23.41. Several disks of the same radius were put on the table so that no two of them overlap. Prove that it is possible to paint disks four colours so that any two tangent disks would be of diﬀerent colours. Solutions 23.1. a) Let a line intersect all the sides of the polygon. Consider all the vertices on one side of the line. To each of these vertices we can assign a pair of sides that intersect at it. Thus we get a partition of all the sides of the polygon into pairs. Therefore, if a line intersects all the sides of an m-gon, then m is even. Figure 211 (Sol. 23.1) b) It is clear from Fig. 69 how to construct 2n-gon and a line that intersects all its sides for any n. 23.2. A line l determines two half planes; one of them will be called upper the other one lower. Let n1 (resp. n2) be the number of the vertices of the broken line that lie on l for which both links that intersect at this point belong to the upper (resp. lower) half plane and m the number of all the remaining intersection points of l and the broken line. Let us circumvent the broken line starting from a point that does not lie on l (and returning to the same point). In the process we pass from one half plane to the other one only passing through any of m intersection points. Since we will have returned to the same point from which we have started, m is even. By the hypothesis n1 + n2 + m = 1985 and, therefore, n1 + n2 is odd, i.e., n1 ̸= n2. Let for deﬁniteness n1 > n2. Then let us draw in the upper halfplane a line l1 parallel to l and distant from it by a distance smaller than any nonzero distance from l to any of the vertices of the broken line (Fig. 70). The number of intersection points of the broken line with l1 is equal to 2n1 + m > n1 + n2 + m = 1985, i.e., l1 is the desired line. 414 CHAPTER 23. DIVISIBILITY, INVARIANTS, COLORINGS Figure 212 (Sol. 23.2) 23.3. No, they cannot. After each hit the orientation (i.e., the direction of the circum-venting pass) of triangle ABC changes. 23.4. Let on a graph paper several cells be painted and nk be the number of painted cells with exactly k painted neighbours. Let N be the number of common sides of painted cells. Since each of them belongs to exactly two painted cells, N = n1 + 2n2 + 3n3 + 4n4 2 = n1 + n3 2 + n2 + n3 + 2n4. Since N is an integer, n1 + n3 is even. (?) We have proved that the number of painted cells with an odd number of painted cells is always even. Therefore, it is impossible to paint 25 cells so that each of them would have had an odd number of painted neighbours. 23.5. Suppose that the circle is divided into arcs as indicated and there are no diamet-rically opposite division points. Then against the endpoints of any arc of length 1 there are no division points and, therefore, against it there lies an arc of length 3. Let us delete one of the arcs of length 1 and the opposite arc of length 3. Then the circle is divided into two arcs. If on one of them there lie m arcs of length 1 and n arcs of length 3, then on the other one there lie m arcs of length 3 and n arcs of length 1. The total number of arcs of length 1 and 3 lying on these two “great” arcs is equal to 2(k −1) and, therefore, n + m = k −1. Since beside arcs of length 1 and 3 there are only arcs of even length, the parity of the length of each of the considered arcs coincides with the parity of k −1. On the other hand, the length of each of them is equal to 6k−1−3 2 = 3k −2. We have obtained a contradiction since numbers k −1 and 3k −2 are of opposite parities. 23.6. Take neighbouring links AB and BC and call the angle symmetric to angle ∠ABC through point B a little angle (on Fig. 71 the little angle is shaded). Figure 213 (Sol. 23.6) We can consider similar little angles for all vertices of the broken line. It is clear that the number of singular pairs is equal to the number of intersection points of links with little angles. It remains to notice that the number of links of the broken line which intersect one angle is even because during the passage from A to C the broken line goes into the little angle as many times as it goes out of it. SOLUTIONS 415 23.7. Let us consider segments into which side 01 is divided. Let a be the number of segments of the form 00 and b the number of segments of the form 01. For every segment consider the number of zeros at its ends and add all these numbers. We get 2a + b. On the other hand, all the “inner” zeros enter this sum twice and there is one more zero at a vertex of the initial triangle. Consequently, the number 2a + b is odd, i.e., b is odd. Let us now divide the triangle. Let a1 be the total number of triangles of the form 001 and 011 and b1 the total number of triangles of the form 012. For every triangle consider the number of its sides of the form 01 and add all these numbers. We get 2a1 + b1. On the other hand all “inner” sides enter twice the sum and all the “boundary” sides lie on the side 01 of the initial triangle and their number is odd by above arguments. Therefore, the number 2a1 + b1 is odd in particular b1 ̸= 0. 23.8. Suppose that all the pairs of vertices determine segments of distinct lengths. Let us assign to segment ApAq the least of the numbers \|p −q\| and 2n −\|p −q\|. As a result, for the given n pairs of vertices we get numbers 1, 2, . . . , n; let among these numbers there be k even and n −k odd ones. To odd numbers segments ApAq, where numbers p and q are of opposite parity, correspond. Therefore, among vertices of the other segments there are k vertices with even numbers and k vertices with odd numbers and the segments connect vertices with numbers of the same parity. Therefore, k is even. For numbers n of the form 4m, 4m + 1, 4m + 2 and 4m + 3 the number k of even numbers is equal to 2m, 2m, 2m + 1 and 2m + 1, respectively, and therefore, either n = 4m or n = 4m + 1. 23.9. Suppose we have succeded to cut the decagon as required. Let n be the number of sides of black triangles, m the number of sides of white triangles. Since every side of an odd triangle (except the sides of a polygon) is also a side of a white triangle, then n −m = 10. On the other hand, both n and m are divisible by 3. Contradiction. 23.10. Let Q be a square sheet of paper, L(Q) the sum of lengths of the sides of the small cells that lie inside it. Then L(Q) is divisible by 4 since all the considered sides split into quadruples of sides obtained from each other by rotations through angles of ±90◦and 180◦about the center of the square. If Q is divided into squares Q1, . . . , Qn, then the sum of the lengths of the segments of the partition is equal to L(Q) −L(Q1) −· · · −L(Qn). Clearly, this number is divisible by 4 since the numbers L(Q), L(Q1), . . . , L(Qn) are divisible by 4. 23.11. Repainting the horizontal or vertical containing k black and 8 −k white cells we get 8 −k black and k white cells. Therefore, the number of black cells changes by (8 −k) −k = 8 −2k, i.e., by an even number. Since the parity of the number of black cells is preserved, we cannot get one black cell from the initial 32 black cells. 23.12. After repainting the 2 × 2 square containing k black and 4 −k white cells we get 4 −k black and k white cells. Therefore, the number of black cells changes by (4 −k) −k = 4 −2k, i.e., by an even number. Since the parity of the number of black cells is preserved, we cannot get one black cell from the initial 32 black cells. 23.13. The diagonals divide a polygon into several parts. Parts that have a common side are called neighbouring. Clearly, from any inner point of the polygon we can get into any other point passing each time only from a neighbouring part to a neighbouring part. A part of the plane that lies outside the polygon can also be considered as one of these parts. The number of the considered triangles for the points of this part is equal to zero and, therefore, it suﬃces to prove that under the passage from a neighbouring part to a neighbouring one the parity of the number of triangles is preserved. Let the common side of two neighbouring parts lie on diagonal (or side) PQ. Then for all the triangles considered, except the triangles with PQ as a side, both these parts either simultaneously belong to or do not belong to. Therefore, under the passage from one part to 416 CHAPTER 23. DIVISIBILITY, INVARIANTS, COLORINGS the other one the number of triangles changes by k1 −k2, where k1 is the number of vertices of the polygon situated on one side of PQ and k2 is the number of vertices situated on the other side of PQ. Since k1 + k2 = 2m −2, it follows that k1 −k2 is even. 23.14. If at least one of the distances between chips would increase, then the sum of the pairwise distances between chips would have also increased but the sum of all pairwise distances between chips does not vary under any permutation. 23.15. Let n be the number of vertices of the initial polygon, n1, . . . , np the number of vertices of the obtained polygons. On the one hand, the sum of angles of all the obtained polygons is equal to p X i=1 (ni −2)π = p X i=1 niπ −2pπ. On the other hand, it is equal to 2(r −n)π + (n −2)π. It remains to observe that p X i=1 ni = 2(q −n) + n. 23.16. It is easy to verify that the length of the boundary of the whole patch (of several patches) overgrown with weeds does not increase. Since in the initial moment it did not surpass 9 · 4 = 36, then at the ﬁnal moment it cannot be equal to 40. 23.17. In plane, ﬁx ray AB. To any polygon M assign a number F(M) (depending on AB) as follows. Consider all the sides of M perpendicular to AB and to each of them assign the number ±l, where l is the length of this side and the sine “plus” is taken if following this side in the direction of ray AB we get inside M and “minus” if we get outside M, see Fig. 72. Figure 214 (Sol. 23.17) Let us denote the sum of all the obtained numbers by F(M); if M has no sides perpen-dicular to AB, then F(M) = 0. It is easy to see that if polygon M is divided into the union of polygons M1 and M2, then F(M) = F(M1) + F(M2) and if M ′ is obtained from M by a parallel translation, then F(M ′) = F(M). Therefore, if M1 and M2 can be cut into parts that can be transformed into each other by a parallel translation, then F(M1) = F(M2). On Fig. 73 there are depicted congruent equilateral triangles PQR and PQS and ray AB perpendicular to side PQ. It is easy to see that F(PQR) = a and F(PQS) = −a, where a is the length of the side of these equilateral triangles. Therefore, it is impossible to divide SOLUTIONS 417 Figure 215 (Sol. 23.17) congruent triangles PQR and PQS into parts that can be translated into each other by a parallel translation. 23.18. Suppose that a convex polygon M is divided into nonconvex quadrilaterals M1, . . . , Mn. To every polygon N assign the number f(N) equal to the diﬀerence between the sum of its inner angles smaller than 180◦and the sum of the angles that complements its angles greater than 180◦to 360◦. Let us compare the numbers A = f(M) and B = f(M1) + · · · + f(Mn). To this end consider all the points that are vertices of triangles M1, . . . , Mn. These points can be divided into four types: 1) The (inner?) points of M. These points contribute equally to A and to B. 2) The points on sides of M or Mi. The contribution of each such point to B exceeds the contribution to A by 180◦. (?)3) The inner points of the polygon in which the angles of the quadrilateral smaller than 180◦in it. The contribution of every such point to B is smaller than that to A by 360◦. 4) The inner points of polygon M in which the angles of the quadrilaterals meet and one of the angles is greater than 180◦. Such points give zero contribution to both A and B. As a result we see that A ≤B. On the other hand, A > 0 and B = 0. The inequality A > 0 is obvious and to prove that B = 0 it suﬃces to verify that if N is a nonconvex quadrilateral, then f(N) = 0. Let the angles of N be equal to α, β, γ and δ, where α ≥β ≥γ ≥δ. Any nonconvex quadrilateral has exactly one angle greater than 180◦and, therefore, f(N) = β + γ + δ −(360◦−α) = α + β + γ + δ −360◦= 0◦. We have obtained a contradiction and, therefore, it is impossible to cut a convex polygon into a ﬁnite number of nonconvex quadrilaterals. 23.19. Let Sn be the circle constructed at the n-th step; On its center. Consider the quantity Fn = P(R2 −OnA2 i ), where the sum runs over points that are inside Sn only. Let us denote the points lying inside circles Sn and Sn+1 by letters B with an index; the points that lie inside Sn but outside Sn+1 by letters C with an index and points lying inside Sn+1 but outside Sn by letters D with an index. Then Fn = X (R2 −OnB2 i ) + X (R2 −OnC2 i ) and Fn+1 = X (R2 −On+1B2 i ) + X (R2 −On+1D2 i ). Since On+1 is the center of mass of the system of points B and C, it follows that X OnB2 i + X OnC2 i = qOnO2 n+1 + X On+1B2 i + X On+1C2 i , where q is the total number of points of type B and C. It follows that Fn+1 −Fn = qOnO2 n+1 + X (R2 −On+1D2 i ) − X (R2 −On+1C2 i ). 418 CHAPTER 23. DIVISIBILITY, INVARIANTS, COLORINGS All the three summands are nonnegative and, therefore, Fn+1 ≥Fn. In particular, Fn ≥ F1 > 0, i.e., q > 0. There is a ﬁnite number of centers of mass of distinct subsets of given points and, there-fore, there is also only ﬁnitely many distinct positions of circles Si. Hence, Fn+1 = Fn for some n and, therefore, qOnO2 n+1 = 0, i.e., On = On+1. 23.20. Since the total number of cells of a 5 × 5 chessboard is odd, the number of black ﬁelds cannot be equal to the number of white ﬁelds. Let, for deﬁniteness, there be more black ﬁelds than white ﬁelds. Then there are less bugs that sit on white ﬁelds than there are black ﬁelds. Therefore, at least one of black ﬁelds will be empty since only bugs that sit on white ﬁelds crawl to black ﬁelds. 23.21. Since the ﬁelds are cut of one colour only, say, of black colour, there remain 32 white and 30 black ﬁelds. Since a domino piece always covers one white and one black ﬁeld, it is impossible to tile with domino chips a 8 × 8 chessboard without two opposite corner ﬁelds. 23.22. Suppose that a 10 × 10 chessboard is divided into such tiles. Every tile contains either 1 or 3 black ﬁelds, i.e., always an odd number of them. The total number of ﬁgures themselves should be equal to 100 4 = 25. Therefore, they contain an odd number of black ﬁelds and the total of black ﬁelds is 100 2 = 50 copies. Contradiction. (?)23.23. Let us divide the plane into equal squares with side 2R and paint them in a staggered order. Let us inscribe a circle into each of them. Then the details of the railway can be considered placed on these circles and the movement of the train that follows from the beginning to the end is performed clockwise on white ﬁelds and counterclockwise on black ﬁelds (or the other way round, see Fig. 74). Figure 216 (Sol. 23.23) Therefore, a deadend cannot arise since along both links of the deadend the movement is performed in the same fashion (clockwise or counterclockwise). 23.24. Let us consider the lattice depicted on Fig. 75 and paint it two colours as indicated in Fig. (white nodes are not painted on this Fig. and the initial square is shaded so that the grasshoppers sit in its white vertices). Let us prove that the grasshoppers can only reach white nodes, i.e., under the symmetry through a white node any white node turns into a white one. To prove this, it suﬃces to prove that under a symmetry through a white node a black node turns into a black one. Let A be a black node, B a white one and A1 the image of A under the symmetry through B. Point A1 is a black node if and only if − − → AA1 = 2me1 + 2ne2, where m and n are integers. SOLUTIONS 419 Figure 217 (Sol. 23.24) It is clear that − − → AA1 = 2− → AB = 2(me1 + ne2) and, therefore, A1 is a black node. Therefore, a grasshopper cannot reach the fourth vertex of the square. Figure 218 (Sol. 23.25) 23.25. Let us paint the nodes of the graph paper in a (?)chess order (Fig. 76). Since the endpoints of any unit segment are of diﬀerent colours, the broken line with the endpoints of the same colour contains an odd number of nodes and an even number of nodes if its endpoints are of the same colour. Suppose that broken lines go out of all the nodes of the boundary (except for the vertices of the square). Let us prove then that all the broken lines together contain an even number of nodes. To this end it suﬃces to show that the number of broken lines with the endpoints of the same colour is even. Let 4m white and 4n black nodes (the vertices of the square are not counted) are placed on the boundary of the square. Let k be the number of broken lines with both endpoints white. Then there are 4m −2k broken lines with endpoints of diﬀerent colours and 4n−(4m−2k) 2 = 2(n −m) + k broken lines with black endpoints. It follows that there are k + 2(n −m) + k = 2(n −m + k) — an even number — of broken lines with the endpoints of the same colour. It remains to notice that a 100 × 100 piece of paper contains an odd number of nodes. Therefore, the broken lines with an even number of nodes cannot pass through all the nodes. Figure 219 (Sol. 23.25) 420 CHAPTER 23. DIVISIBILITY, INVARIANTS, COLORINGS 23.26. Let us paint the triangles as shown on Fig. 77. Then there are 1 + 2 + · · · + n = 1 2n(n+1) black triangles and 1+2+· · ·+(n−1) = 1 2n(n−1) white triangles. It is clear that two triangles with consecutive indices are of distinct colours. Hence, among the numbered triangles the number of black triangles is only by 1 greater than that of white ones. Therefore, the total number of numbered triangles does not exceed n(n −1) + 1. Figure 220 (Sol. 23.27) 23.27. Let us paint the bottom of the box two colours as shown on Fig. 78. Then every 2 × 2 tile covers exactly one black cell and a 1 × 4 tile covers 2 or 0 of them. Hence, the parity of the number of odd cells on the bottom of the box coincides with the parity of the number of 2 × 2 tiles. Since under the change of a 2 × 2 tile by a 1 × 4 tile the parity of the number of 2 × 2 tiles changes, we will not be able to tile the bottom of the box. Figure 221 (Sol. 23.28) 23.28. In the given square piece of graph paper, let us shade 2 × 2 squares as shown on Fig. 79. We thus get 100 shaded squares. Every cut oﬀsquare touches precisely one shaded square and therefore, at least one shaded square remains intact and can be cut oﬀ(?). 23.29. If a polygon is divided into parts by several diagonals, then these parts can be painted two colours so that parts with a common side were of distinct colours. This can be done as follows. Figure 222 (Sol. 23.29) Let us consecutively draw diagonals. Every diagonal splits the polygon into two parts. In one of them retain its painting and repaint the other one changing everywhere the white SOLUTIONS 421 colour to black and black to white. Performing this operation under all the needed diagonals, we get the desired coloring. Since in the other case at every vertex an odd number of triangles meet, then under such a coloring all the sides of the polygon would belong to triangles of the same colour, for example, black, Fig. 80. Denote the number of sides of white triangles by m. It is clear that m is divisible by 3. Since every side of a white triangle is also a side of a black triangle and all the sides of the polygon are sides of the black triangles, it follows that the number of sides of black triangles is equal to n + m. Hence, n + m is divisible by 3 and since m is divisible by 3, then n is divisible by 3. 23.30. Let us paint the chessboard four colours as shown on Fig. 81. It is easy to count the number of cells of the second colour: it is 26; that of the fourth is 24. Figure 223 (Sol. 23.30) Every 1× 4 tile covers one cell of each colour. Therefore, it is impossible to tile a 10× 10 chessboard with tiles of size 1 × 4 since otherwise there would have been an equal number of cells of every colour. 23.31. Let us paint the graph paper four colours as shown on Fig. 82. Among the given n cells there are not less than n 4 cells of the same colour and such cells do not have common points. Figure 224 (Sol. 23.32) 23.32. Let us paint the nodes of graph paper four colours in the same order as the cells on Fig. 82 are painted. If n ≥5, then there exist two vertices of an n-gon of the same colour. The midpoint of the segment with the endpoints in the nodes of the same colour is a node itself. Since the n-gon is a convex one, then the midpoint of the segment with the endpoints at its nodes lies either inside it or on its side. 422 CHAPTER 23. DIVISIBILITY, INVARIANTS, COLORINGS 23.33. Let us divide the obtained square into cells of size 1 × 1 and paint them three colours as shown on Fig. 83. It is easy to verify that it is possible to divide tiles of size 1 × 3 into two types: a tile of the ﬁrst type covers one cell of the ﬁrst colour and two cells of the second colour and a tile of the second type covers one cell of the second colour and two cells of the third colour. Figure 225 (Sol. 23.33) Suppose that all the cells of the ﬁrst colour are covered by tiles 1 × 3. Then there are 9 tiles of the ﬁrst type and 7 tiles of the second type. Hence, they cover 9 · 2 + 7 = 25 cells of the second colour and 7 · 2 = 14 cells of the third colour. We have reached a contradiction and, therefore, one of the cells of the ﬁrst colour is covered by the tile of size 1 × 1. 23.34. Let us cut the given n-gon by nonintersecting diagonals into triangles (cf. Prob-lem 22.22). The vertices of the n-gon can be painted 3 colours so that all the vertices of each of the obtained triangles are of distinct colours (see Problem 23.40). There are not more than [n 3] vertices of any colour; and it suﬃces to place guards at these points. 23.35. Let us consider an equilateral triangle with side 1. All of its three vertices cannot be of distinct colours and, therefore, two of the vertices are of the same colour; the distance between them is equal to 1. 23.36. Suppose that any two points situated at distance 1 are painted distinct colours. Consider an equilateral triangle ABC with side 1; all its vertices are of distinct colours. Let point A1 be symmetric to A through line BC. Since A1B = A1C = 1, the colour of A1 is distinct from that of B and C and A1 is painted the same colour as A. These arguments show that if AA1 = √ 3, then points A and A1 are of the same colour. Therefore, all the points on the circle of radius √ 3 with center A are of the same colour. It is clear that on this circle there are two points the distance between which is equal to 1. Contradiction. 23.37. Let us give an example of a seven-colour coloring of the plane for which the distance between any two points of the same colour is not equal to 1. Let us divide the plane into equal hexagons with side a and paint them as shown on Fig. 84 (the points belonging to two or three hexagons can be painted any of the colours of these hexagons). The greatest distance between points of the same colour that belong to one hexagon does not exceed 2a and the least distance between points of the same colour lying in distinct hexagons is not less than the length of segment AB (see Fig. 84). It is clear that AB2 = AC2 + BC2 = 4a2 + 3a2 = 7a2 > (2a)2. Therefore, if 2a < 1 < √ 7a, i.e., 1 √ 7 < a < 1 2, then the distance between points of the same colour cannot be equal to 1. 23.38. Suppose there does not exist a right triangle with vertices of the same colour. Let us divide every side of an equilateral triangle into three parts by two points. These points form a right hexagon. If two of its opposite vertices are of the same colour, then all the other SOLUTIONS 423 Figure 226 (Sol. 23.37) vertices are of the second colour and therefore, there exists a right triangle with vertices of the second colour. Hence, the opposite vertices of the hexagon must be of distinct colours. Therefore, there exist two neighbouring vertices of distinct colours; the vertices opposite to them are also of distinct colours. One of these pairs of vertices of distinct colours lies on a side of the triangle. The points of this side distinct from the vertices of the hexagon cannot be of either ﬁrst or second colour. Contradiction. 23.39. Let us prove this statement by induction on the number of triangles of the triangulation. For one triangle the needed coloring exists. Now, let us suppose that it is possible to paint in the required way any triangulation consisting of less than n triangles; let us prove that then we can paint any triangulation consisting of n triangles. Let us delete a triangle one of the sides of which lies on a side of the triangulated ﬁgure. The remaining part can be painted by the inductive hypothesis. (It is clear that this part can consist of several disjoint pieces but this does not matter.) Only two sides of the deleted triangle can be neighbouring with the other triangles. Therefore, it can be coloured the colour distinct from the colours of its two neighbouring triangles. 23.40. Proof is similar to that of Problem 23.39. The main diﬀerence is in that one must delete a triangle with two sides of the boundary of the polygon (cf. Problem 22.25). 23.41. Proof will be carried out by induction on the number of disks n. For n = 1 the statement is obvious. Let M be any point, O the most distant from M center of a(?) given disk. Then the disk centered at O is tangent to not more than 3 other given disks. Let us delete it and paint the other disks; this is possible thanks to the inductive hypothesis. Now, let us paint the deleted disk the colour distinct from the colours of the disks tangent to it. Chapter 24. INTEGER LATTICES In plane, consider a system of lines given by equations x = m and y = n, where m and n are integers. These lines form a lattice of squares or an integer lattice. The vertices of these squares, i.e., the points with integer coordinates, are called the nodes of the integer lattice. §1. Polygons with vertices in the nodes of a lattice 24.1. Is there an equilateral triangle with vertices in the nodes of an integer lattice? 24.2. Prove that for n ̸= 4 a regular n-gon is impossible to place so that its vertices would lie in the nodes of an integer lattice. 24.3. Is it possible to place a right triangle with integer sides (i.e., with sides of integer length) so that its vertices would be in nodes of an integer lattice but none of its sides would pass along the lines of the lattice? 24.4. Is there a closed broken line with an odd number of links of equal length all vertices of which lie in the nodes of an integer lattice? 24.5. The vertices of a polygon (not necessarily convex one) are in nodes of an integer lattice. Inside the polygon lie n nodes of the lattice and m nodes lie on the polygon’s boundary. Prove that the polygon’s area is equal to n + m 2 −1. (Pick’s formula.) 24.6. The vertices of triangle ABC lie in nodes of an integer lattice and there are no other nodes on its sides whereas inside it there is precisely one node, O. Prove that O is the intersection point of the medians of triangle ABC. See also Problem 23.32. §2. Miscellaneous problems 24.7. On an inﬁnite sheet of graph paper N, cells are painted black. Prove that it is possible to cut oﬀa ﬁnite number of squares from this sheet so that the following two conditions are satisﬁed: 1) all black cells belong to the cut-oﬀsquares; 2) in any cut-oﬀsquare K, the area of black cells constitutes not less than 0.2 and not more than 0.8 of the area of K. 24.8. The origin is the center of symmetry of a convex ﬁgure whose area is greater than 4. Prove that this ﬁgure contains at least one distinct from the origin point with integer coordinates. (Minkowski’s theorem.) 24.9. In all the nodes of an integer lattice except one, in which a hunter stands, trees are growing and the trunks of these trees are of radius r each. Prove that the hunter will not be able to see a hare that sits further than 1 r of the unit length from it. 24.10. Inside a convex ﬁgure of area S and semiperimeter p there are n nodes of a lattice. Prove that n > S −p. 24.11. Prove that for any n there exists a circle inside which there are exactly (not more nor less) n integer points. 24.12. Prove that for any n there exists a circle on which lies exactly (not more nor less) n integer points. 425 426 CHAPTER 24. INTEGER LATTICES Solutions 24.1. Suppose that the vertices of an equilateral triangle ABC are in nodes of an integer lattice. Then the tangents of all the angles formed by sides AB and AC with the lines of the lattice are rational. For any position of triangle ABC either the sum or the diﬀerence of certain two of such angles α and β is equal to 60◦. Hence, √ 3 = tan 60◦= tan(α ± β) = tan α ± tan β 1 ∓tan α tan β is a rational number. Contradiction. 24.2. For n = 3 and n = 6 the statement follows from the preceding problem and, therefore, in what follows we will assume that n ̸= 3, 4, 6. Suppose that there exist regular n-gons with vertices in nodes of an integer lattice (n ̸= 3, 4, 6). Among all such n-gons we can select one with the shortest side. (To prove that we can do it, it suﬃces to observe that if a is the length of a segment with the endpoints in nodes of the lattice, then a = √ n2 + m2, where n and m are integers, i.e., there is only a ﬁnite number of distinct length of segments with the endpoints in nodes of the lattice shorter than the given length.) Let − − → AiBi = − − − − − − → Ai+1Ai+2. Then B1 . . . Bn is a regular n-gon whose vertices lie in nodes of the integer lattice and its side is shorter than any side of the n-gon A1 . . . An. For n = 5 this is clear from Fig. 85 and for n ≥7 look at Fig. 86. We have arrived to a contradiction with the choice of the n-gon A1 . . . An. Figure 227 (Sol. 24.2) Figure 228 (Sol. 24.2) 24.3. It is easy to verify that the triangle with the vertices at points with coordinates (0, 0), (12, 16) and (−12, 9) possesses the required properties. 24.4. Suppose that there exists a closed broken line A1 . . . An with an odd number of links of equal length all the vertices of which lie in nodes of an integer lattice. Let ai and SOLUTIONS 427 bi be coordinates of the projections of vector − − − − → AiAi+1 to the horizontal and vertical axes, respectively. Let c be the length of the link of the broken line. Then c2 = a2 i + b2 i . Hence, the residue after the division of c2 by 4 is equal to 0, 1 or 2. If c2 is divisible by 4, then ai and bi are divisible by 4 (this is proved by a simple case-by-case checking of all possible residues after the division of ai and bi by 4). Therefore, the homothety centered at A1 with coeﬃcient 0.5 sends our broken line into a broken line with a shorter links but whose vertices are also in the nodes of the lattice. After several such operations we get a broken line for which c2 is not divisible by 4, i.e., the corresponding residue is equal to either 1 or 2. Let us consider these variants, but ﬁrst observe that a1 + · · · + am = b1 + · · · + bm = 0. 1) The residue after division of c2 by 4 is equal to 1. Then one of the numbers ai and bi is odd and the other one is even; hence, the number a1 + · · · + am is odd and cannot equal to zero. 2) The residue after division of c2 by 4 is equal to 2. Then the numbers ai and bi are both odd; hence, a1 + · · · + am + b1 + · · · + bm is odd and cannot equal to zero. 24.5. To every polygon N with vertices in nodes of an integer lattice assign the number f(N) = n + m 2 −1. Let polygon M be cut into polygons M1 and M2 with vertices in nodes of the lattice. Let us prove that if Pick’s formula holds for two of the polygons M, M1 and M2, then it is true for the third one as well. To this end it suﬃces to prove that f(M) = f(M1)+f(M2). The nodes which lie outside the line of cut contribute equally to f(M) and f(M1) + f(M2). “Nonterminal” nodes of the cut contribute 1 to f(M) and 0.5 to f(M1) and f(M2). Each of the two terminal nodes of the cut contributes 0.5 to each of f(M), f(M1) and f(M2) and, therefore, the contribution of the terminal nodes to f(M) is by 1 less than to f(M1) + f(M2). Since we have to deduct 1 from each contribution to f(M) and two from each contribution to f(M1) + f(M2), it follows that f(M) = f(M1) + f(M2). Now, let us prove the validity of Pick’s formula for an arbitrary triangle. If M is a rectangle with sides of length p and q directed along the lines of the lattice, then f(M) = (p −1)(q −1) + 2(p + q) 2 −1 = pq, i.e., Pick’s formula holds for M. Cutting triangle M into triangles M1 and M2 by a diagonal and making use of the fact that f(M) = f(M1) + f(M2) and f(M1) = f(M2) it is easy to prove the validity of Pick’s formula for any right triangle with legs directed along the lines of the lattice. Cutting several such triangles from the rectangle we can get any triangle (Fig. 87). Figure 229 (Sol. 24.5) To complete the proof of Pick’s formula, it remains to notice that any polygon can be cut by diagonals into triangles. 428 CHAPTER 24. INTEGER LATTICES 24.6. Thanks to Pick’s formula SAOB = SBOC = SCOA = 1 2; hence, O is the intersection point of medians of triangle ABC (cf. Problem 4.2). 24.7. Take a suﬃciently large square with side 2n so that all the black cells are inside it and constitute less than 0.2 of its area. Let us divide this square into four identical squares. The painted area of each of them is less than 0.8 of the total. Let us leave those of them whose painted part constitutes more than 0.2 of the total and cut the remaining ones in the same way. The painted area of the obtained 2 × 2 squares will be 1 4, 1 2 or 3 4 of the total or they will not be painted at all. Now, we have to cut oﬀthose of the obtained squares which contain painted cells. 24.8. Consider all the convex ﬁgures obtained from the given one by translations by vectors with both coordinates even. Let us prove that at least two of these ﬁgures intersect. The initial ﬁgure can be squeezed in the disk of radius R centered in the origin, where for R we can take an integer. Take those of the considered ﬁgures the coordinates of whose centers are nonnegative integers not greater than 2n. There are precisely (n + 1)2 of such ﬁgures and all of them lie inside a square with side 2(n + R). If they do not intersect, then for any n we would have had (n + 1)2S < 4(n + R)2, where S is the area of the given ﬁgure. Since S > 4, we can select n so that the inequality n+R n+1 < q S 4 holds. Figure 230 (Sol. 24.8) Let now ﬁgures with centers O1 and O2 have a common point A (Fig. 88). Let us prove that then the midpoint M of segment O1O2 belongs to both ﬁgures (it is clear that the coordinates of M are integers). Let − − → O1B = −− − → O2A. Since the given ﬁgure is centrally symmetric, point B belongs to the ﬁgure with center O1. This ﬁgure is convex and points A and B belong to it and, therefore, the midpoint of segment AB also belongs to it. Clearly, the midpoint of segment AB coincides with the midpoint of segment O1O2. 24.9. Let the hunter sit at point O and the hare at point A; let A1 be the point symmetric to A with respect to O. Consider ﬁgure Φ that contains all the points the distance from which to segment AA1 does not exceed r (Fig. 89). Figure 231 (Sol. 24.9) It suﬃces to prove that Φ contains at least one node of the lattice (if the node gets into the shaded part, then point A belongs to the trunk). The area of Φ is equal to 4rh + πr2, where h is the distance from the hunter to the hare. If h > 1 r, then 4rh + πr2 > 4. By Minkowski’s theorem Φ contains an integer point. SOLUTIONS 429 24.10. Consider the integer lattice given by equations x = k + 1 2 and y = l + 1 2, where k and l are integers. Let us prove that each small square of this lattice gives a nonnegative contribution to n −S + p. Consider two cases: 1) The ﬁgure contains the center of the square. Then n′ = 1 and S′ ≤1; hence, n′ −S′ + p′ ≥0. 2) The ﬁgure intersects the square but does not contain its center. Let us prove that in this case S′ ≤p′ and we can conﬁne ourselves with the study of the cases depicted on Fig. 90 (i.e., we may assume that the center O of the square lies on the boundary of the ﬁgure). Since the distances from the center of the square to its sides are equal to 1 2, it follows that p′ ≥1 2. Draw the base line through O to this ﬁgure; we get S′ ≤1 2. Figure 232 (Sol. 24.10) It is also clear that all the contributions of the squares cannot be zero simultaneously. 24.11. First, let us prove that on the circle with center A = ( √ 2, 1 3) there cannot lie more than one integer point. If m and n are integers, then (m − √ 2)2 + (n −1 3)2 = q −2m √ 2, where q is a rational number. Therefore, the equality (m1 − √ 2)2 + (n1 −1 3)2 = (m2 − √ 2)2 + (n2 −1 3)2 implies that m1 = m2. By Vi` eta’s theorem the sum of roots of equation (n −1 3)2 = d is equal to 2 3; hence, at least one root can be integer. Now, let us arrange the radii of the circles with center A passing through integer points in the increasing order: R1 < R2 < R3 < . . . . If Rn < R < Rn+1, then inside the circle of radius R with center A there are n integer points. 24.12. First. let us prove that the equation x2 + y2 = 5k has 4(k + 1) integer solutions. For k = 0 and k = 1 this statement is obvious. Let us prove that the equation x2 + y2 = 5k has exactly 8 solutions (x, y) such that x and y are not divisible by 5. Together with 4(k−1) solutions of the form (5a, 5b), where (a, b) is a solution of the equation a2 + b2 = 5k−2, they give the needed number of solutions. These solutions are obtained from each other by permutations of x and y and changes of signs; we will call them nontrivial solutions. Let x2 + y2 be divisible by 5. Then (x + 2y)(x −2y) = x2 + y2 −5y2 is also divisible by 5. Hence, one of the numbers x + 2y and x −2y is divisible by 5. It is also easy to verify that if x + 2y and x −2y are divisible by 5, then both x and y are divisible by 5. If (x, y) is a nontrivial solution of equation x2 + y2 = 5k, then (x + 2y, 2x −y) and (x −2y, 2x + y) are solutions of equation ξ2 + η2 = 5k+1 and precisely one of them is nontrivial. It remains to prove that a nontrivial solution is unique up to permutations of x and y and changes of signs. 430 CHAPTER 24. INTEGER LATTICES Let (x, y) be a nontrivial solution of the equation x2 + y2 = 5k. Then the pairs µ ±2x −y 5 , ±x + 2y 5 ¶ and µ ±x + 2y 5 , ±2x −y 5 ¶ (1) together with the pairs µ ±2x + y 5 , ±x −2y 5 ¶ and µ ±x −2y 5 , ±2x + y 5 ¶ (2) are solutions of the equation ξ2 + η2 = 5k−1 but the pairs of exactly one of these types will be integer since exactly one of the numbers x + 2y and x −2y is divisible by 5. Thus, we will get a nontrivial solution because (x + 2y)(x −2y) = (x2 + y2) −5y2 for k ≥2 is divisible by 5 but is not divisible by 25. Therefore, each of the 8 nontrivial solutions of the equation x2+y2 = 5k yields 8 nontrivial solutions of the equation ξ2 + η2 = 5k−1 where for one half of the solutions we have to make use of formulas (1) and for the other half of the formulas (2). Now, let us pass directly to the solution of the problem. Let n = 2k + 1. Let us prove that on the circle of radius 5k 3 with center ( 1 3, 0) there lie exactly (not more nor less) n integer points. The equation x2 + y2 = 52k has 4(2k + 1) integer solutions. Moreover, after the division of 52k by 3 we have residue 1; hence, one of the numbers x and y is divisible by 3 and the residue after the division of the other one by 3 is equal to ±1. Therefore, in precisely one of the pairs (x, y), (x, −y), (y, x) and (−y, x) the residues after the divisionof the ﬁrst and the second number by 3 are equal to −1 and 0, respectively. Hence, the equation (3z −1)2 + (3t)2 = 52k has precisely 2k + 1 integer solutions. Let n = 2k. Let us prove that on the circle of radius 5(k−1)/2 2 with center ( 1 2, 0) there lie n integer points. The equation x2 + y2 = 5k−1 has 4k integer solutions; for them one of the numbers x and y is even and the other one is odd. Hence, the equation (2z−1)2+(2t)2 = 5k−1 has 2k integer solutions. Chapter 25. CUTTINGS §1. Cuttings into parallelograms 25.1. Prove that the following properties of convex polygon F are equivalent: 1) F has a center of symmetry; 2) F can be cut into parallelograms. 25.2. Prove that if a convex polygon can be cut into centrally symmetric polygons, then it has a center of symmetry. 25.3. Prove that any regular 2n-gon can be cut into rhombuss. 25.4. A regular octagon with side 1 is cut into parallelograms. Prove that among the parallelograms there is at least two rectangles and the sum of areas of all the rectangles is equal to 2. §2. How lines cut the plane In plane, let there be drawn n pairwise nonparallel lines no three of which intersect at one point. In Problems 25.5–25.9 we will consider properties of ﬁgures into which these lines cut the plane. A ﬁgure is called an n-linked one if it is bounded by n links (i.e., a link is a line segment or a ray). 25.5. Prove that for n = 4 among the obtained parts of the plane there is a quadrilateral. 25.6. a) Find the total number of all the obtained ﬁgures. b) Find the total number of bounded ﬁgures, i.e., of polygons. 25.7. a) Prove that for n = 2k there are not more than 2k−1 angles among the obtained ﬁgures. b) Is it possible that for n = 100 there are only three angles among the obtained ﬁgures? 25.8. Prove that if among the obtained ﬁgures there is a p-linked and a q-linked ones, then p + q ≤n + 4. 25.9. Prove that for n ≥3 there are not less than 2n−2 3 triangles among the obtained parts. Now, let us abandon the assumption that no three of the considered lines intersect at one point. If P is the intersection point of two or several lines, then the number of lines of the given system passing through point P will be denoted by λ(P). 25.10. Prove that the number of segments into which the given lines are divided by their intersection points is equal to n + P λ(P). 25.11. Prove that the number of parts into which given lines divide the plane is equal to 1 + n + P(λ(P) −1) and among these parts there are 2n unbounded ones. 25.12. The parts into which the plane is cut by lines are painted red and blue so that the neighbouring parts are of distinct colours (cf. Problem 27.1). Let r be the number of red parts, b the number of blue parts. Prove that r ≤2b −2 − X (λ(P) −2) where the equality is attained if and only if the red parts are triangles and angles. 431 432 CHAPTER 25. CUTTINGS Solutions 25.1. Consider a convex polygon A1 . . . An. Prove that each of the properties 1) and 2) is equivalent to the following property: 3) For any vector − − − − → AiAi+1 there exists a vector − − − − → AjAj+1 = −− − − − → AiAi+1. Property 1) clearly implies property 3). Let us prove that property 3) implies prop-erty 1). If a convex polygon A1 . . . An possesses property 3), then n = 2m and − − − − → AiAi+1 = −− − − − − − − − → Am+iAm+i+1. Let Oi be the midpoint of segment AiAm+i. Since AiAi+1Am+iAm+i+1 is a parallelogram, we have Oi = Oi+1. Hence, all the points Oi coincide and this point is the center of symmetry of the polygon. Let us prove that property 2) implies property 3). Let a convex polygon F be divided into parallelograms. We have to prove that for any side of F there exists another side parallel and equal to it. From every side of F a chain of parallelograms departs, i.e., this side sort of moves along them parallelly so that it can be split into several parts (Fig. 91). Figure 233 (Sol. 25.1) Since a convex polygon can have only one more side parallel to the given one, all the bifurcations of the chain terminate in the same side which is not shorter than the side from which the chain starts. We can equally well begin the chain of parallelograms from the ﬁrst side to the second one or from the second one to the ﬁrst one; hence, the lengths of these sides are equal. It remains to prove that property 3) implies property 2). A way of cutting a polygon with equal and parallel opposite sides is indicated on Fig. 92. Figure 234 (Sol. 25.1) After each such operation we get a polygon with a lesser number of sides which still possesses property 3) and by applying the same process to this polygon we eventually get a parallelogram. 25.2. Let us make use of the result of the preceding problem. If a convex polygon M is cut into convex centrally symmetric polygons, then they can be cut into parallelograms. Therefore, M can be cut into parallelograms, i.e., M has a center of symmetry. SOLUTIONS 433 25.3. Let us prove by induction on n that any 2n-gon whose sides have the same length and opposite sides are parallel can be cut into rhombs. For n = 2 this is obvious and from Fig. 92 it is clear how to perform the inductive step. 25.4. Let us single out two perpendicular pairs of opposite sides in a regular octagon and, as in Problem 25.1, consider chains of parallelograms that connect the opposite sides. On the intersection of these chains rectangles stand. By considering two other pairs of opposite sides we will get at least one more rectangle. It is possible to additionally cut parallelograms from every chain so that the chain would split into several “passes” and in each pass the neighbouring parallelograms are neighboring to each other along the whole sides, not a part of a side. The union of rectangles of a new partition coincides with the union of rectangles of the initial partition and, therefore, it suﬃces to carry out the proof for the new partition. Every pass has a constant width; hence, the length of one side of each rectangle that enters a path is equal to the width of the path, and the sum of length of all the other sides is equal to the sum of the widths of the passes corresponding to the other pair of sides. Therefore, the area of all the rectangles that constitute one path is equal to the product of the width of the path by the length of the side of the polygon, i.e., its value is equal to the width of the path. Hence, the area of all the rectangles corresponding to two perpendicular pairs of opposite sides is equal to 1 and the area of the union of the rectangles is equal to 2. 25.5. Denote the intersections points of one of the given lines with the other ones by A, B and C. For deﬁniteness, let us assume that point B lies between A and C. Let D be the intersection point of lines through A and C. Then any line passing through point B and not passing through D cuts triangle ACD into a triangle and a quadrilateral. 25.6. a) Let n lines divide the plane into an parts. Let us draw one more line. This will increase the number of parts by n + 1 since the new line has n intersection points with the already drawn lines. Therefore, an+1 = an + n + 1. Since a1 = 2, it follows that an = 2 + 2 + 3 + · · · + n = n2+n+2 2 . b) Encircle all the intersection points of the given lines. It is easy to verify that the number of unbounded ﬁgures is equal to 2n. Therefore, the number of bounded ﬁgures is equal to n2 + n + 2 2 −2n = n2 −3n + 2 2 . 25.7. a) All intersection points of given lines can be encircled in a circle. Lines divide this circle into 4k arcs. Clearly, two neighbouring arcs cannot simultaneously belong to angles; hence, the number of angles does not exceed 2k, where the equality can only be attained if the arcs belonging to the angles alternate. It remains to prove that the equality cannot be attained. Suppose that the arcs belonging to angles alternate. Since on both sides from any of the given lines lie 2k arcs, the opposite arcs (i.e., the arcs determined by two lines) must belong to angles (Fig. 93) which is impossible. b) For any n there can be three angles among the obtained ﬁgures. On Fig. 94 it is shown how to construct the corresponding division of the plane. 25.8. Let us call a line which is the continuation of a segment or a ray that bounds a ﬁgure a (border?)bounding line of the ﬁgure. It suﬃces to show that two considered ﬁgures cannot have more than 4 common bounding lines. If two ﬁgures have 4 common bounding lines, then one of the ﬁgures lies in domain 1 and the other one lies in domain 2 (Fig. 95). The ﬁfth bounding line of the ﬁgure that lies in domain 1 must intersect two neighbouring sides of the quadrilateral 1; but then it cannot be bounding line for the ﬁgure that belongs to domain 2. 434 CHAPTER 25. CUTTINGS Figure 235 (Sol. 25.7 a)) Figure 236 (Sol. 25.7 b)) Figure 237 (Sol. 25.8) 25.9. Consider all the intersection points of the given lines. Let us prove that these points may lie on one side of not more than two given lines. Suppose that all the intersection points lie on one side of three given lines. These lines constitute triangle ABC. The fourth line cannot intersect the sides of this triangle only, i.e., it intersects at least one extension of a side. Let, for deﬁniteness, it intersect the continuation of side AB beyond point B; let the intersection point be M. Then points A and M lie on distinct sides of line BC. Contradiction. Hence, there exist at least n −2 lines on both sides of which there are intersection points. If in the half plane given by line l we select the nearest l intersection point, then this point is a vertex of a triangle adjacent to l. Thus, there exists not less than n −2 lines to each of which at least two triangles are adjacent and there are two lines to each of which at least one triangle is adjacent. Since every triangle is adjacent to exactly 3 lines, there are not less than 2(n−2)+2 3 triangles. SOLUTIONS 435 25.10. If P is the intersection point of given lines, then 2λ(P) segments or rays go out of P. Moreover, each of x segments have two boundary points and each of 2n rays has one boundary point. Hence, 2x + 2n = 2 P λ(P), i.e., x = −n + P λ(P). 25.11. Let us carry out the proof by induction on n. For two lines the statement is obvious. Suppose that the statement holds for n −1 line and consider a system consisting of n lines. Let f be the number of parts into which the given n lines divide the plane; g = 1 + n + P(λ(P) −1). Let us delete one line from the given system and deﬁne similarly numbers f ′ and g′ for the system obtained. If on the deleted line there lie k intersection points of lines, then f ′ = f −k −1 and g′ = 1 + (n −1) + P(λ′(P) −1). It is easy to verify that P(λ(P) −1) = −k + P(λ′(P) −1). By inductive hypothesis f ′ = g′. Therefore, f = f ′ + k + 1 = g′ + k + 1 = g. It is also clear that the number of unbounded parts is equal to 2n. 25.12. Let r′ k be the number of red k-gons, r′ the number of bounded red domains and the number of segments into which the given lines are divided by their intersection points be equal to P λ(P) −n, cf. Problem 25.10. Each segment is a side of not more than 1 red polygon, hence, 3r′ ≤P k≥3 kr′ k ≤P λ(P) −n, where the left inequality is only attained if and only if there are no red k-gons for k > 3, and the right inequality is only attained if and only if any segment is a side of a red k-gon, i.e., any unbounded red domain is an angle. The number of bounded domains is equal to 1 −n + P(λ(P) −1) = c (see Problem 25.11), hence, the number b′ of bounded blue domaions is equal to c −r′ ≥1 −n + X (λ(P) −1) − P λ(P) −n 3 = 1 −2n 3 + X (2λ(P) 3 −1). The colours of 2n unbounded domains alternate; hence, b = b′ + n ≥1 + n 3 + X (2λ(P) 3 −1) and r = r′ + n ≤2n + P λ(P) 3 and, therefore, 2b −r ≥2 + P(λ(P) −2). Chapter 26. SYSTEMS OF POINTS AND SEGMENTS. EXAMPLES AND COUNTEREXAMPLES §1. Systems of points 26.1. a) An architect wants to place four sky-scrapers so that any sightseer can see their spires in an arbitrary order. In other words, if the sky-scrapers are numbered, then for any ordered set (i, j, k, l) of sky-scrapers one can stand at an arbitrary place in the town and by turning either clockwise or counterclockwise see ﬁrst the spire of the sky-scraper i, next, that of j, k and, lastly, l. Is it possible for the architect to perform this? b) The same question for ﬁve sky-scrapers. 26.2. In plane, there are given n points so that from any foresome of these points one can delete one point so that the remaining three points lie on one line. Prove that it is possible to delete one of the given points so that all the remaining points lie on one line. 26.3. Given 400 points in plane, prove that there are not fewer than 15 distinct distances between them. 26.4. In plane, there are given n ≥3 points. Let d be the greatest distance between any two of these points. Prove that there are not more than n pairs of points with the distance between the points of any pair equal to d. 26.5. In plane, there are given 4000 points no three of which lie on one line. Prove that there are 1000 nonintersecting quadrilaterals (perhaps, nonconvex ones) with vertices at these points. 26.6. In plane, there are given 22 points no three of which lie on one line. Prove that it is possible to divide them into pairs so that the segments determined by pairs intersect at least at 5 points. 26.7. Prove that for any positive integer N there exist N points no three of which lie on one line and all the pairwise distances between them are integers. See also Problems 20.13-20.15, 22.7. §2. Systems of segments, lines and circles 26.8. Construct a closed broken line of six links that intersects each of its links precisely once. 26.9. Is it possible to draw six points in plane and to connect them with nonintersecting segments so that each point is connected with precisely four other ones? 26.10. Point O inside convex polygon A1 . . . An possesses a property that any line OAi contains one more vertex Aj. Prove that no point except O possesses such a property. 26.11. On a circle, 4n points are marked and painted alternatedly red and blue. Points of the same colour are divided into pairs and points from each pair are connected by segments of the same colour. Prove that if no three segments intersect at one point, then there exist at least n intersection points of red segments with blue segments. 26.12. In plane, n ≥5 circles are placed so that any three of them have a common point. Prove that then all the circles have a common point. 437 438 CHAPTER 26. EXAMPLES AND COUNTEREXAMPLES §3. Examples and counterexamples There are many wrong statements that at ﬁrst glance seem to be true. To refute such statements we have to construct the corresponding example; such examples are called coun-terexamples. 26.13. Is there a triangle all the hights of which are shorter than 1 cm and the area is greater than 1 m 2? 26.14. In a convex quandrilateral ABCD sides AB and CD are equal and angles A and C are equal. Must this quandrilateral be a parallelogram? 26.15. The list of sides and diagonals of a convex quandrilateral ordered with respect to length coincides with a similar list for another quandrilateral. Must these quandrilaterals be equal? 26.16. Let n ≥3. Do there exist n points that do not belong to one line and such that pairwise distances between which are irrational while the areas of all the triangles with vertices in these points are rational? 26.17. Do there exist three points A, B and C in plane such that for any point X the length of at least one of the segments XA, XB and XC is irrational? 26.18. In an acute triangle ABC, median AM, bisector BK and hight CH are drawn. Can the area of the triangle formed by the intersection points of these segments be greater than 0.499 · SABC? 26.19. On an inﬁnite list of graph paper (with small cells of size 1×1) the domino chips of size 1 × 2 are placed so that they cover all the cells. Is it possible to make it so that any line that follows the lines of the mash cuts only a ﬁnite number of chips? 26.20. Is it possible for a ﬁnite set of points to contain for every of its points precisely 100 points whose distance from the point is equal to 1? 26.21. In plane, there are several nonintersecting segments. Is it always possible to connect the endpoints of some of them by segments so that we get a closed nonselﬁntersecting broken line? 26.22. Consider a triangle. Must the triangle be an isosceles one if the center of its inscribed circle is equidistant from the midpoints of two of its sides? 26.23. The arena of a circus is illuminated by n distinct spotlights. Each spotlight illuminates a convex ﬁgure. It is known that if any of the spotlights is turned oﬀ, then the arena is still completely illuminated, but if any two spotlights are turned oﬀ, then the arena is not completely illuminated. For which n this is possible? See also problems 22.16–22.18, 22.26, 22.27, 22.29, 23.37, 24.11, 24.12. Solutions 26.1. a) It is easy to verify that constructing the fourth building inside the triangle formed by the three other buildings we get the desired position of the buildings. b) It is impossible to place in the desired way ﬁve buildings. Indeed, if we consecutively see buildings A1, A2, . . . , An, then A1A2 . . . An is a nonselﬁntersecting broken line. Therefore, if ABCD is a convex quandrilateral, then it is impossible to see its vertices in the following order: A, C, D, B. It remains to notice that of ﬁve points no three of which lie on one line it is always possible to select four points which are vertices of a convex quandrilateral (Problem 22.2). 26.2. It is possible to assume that n ≥4 and not all the points lie on one line. Then we can select four points A, B, C and D not on one line. By the hypothesis, three of them lie on one line. Let, for deﬁniteness, points A, B and C lie on line l and D does not lie on l. SOLUTIONS 439 We have to prove that all the points except for D lie on l. Suppose that a point E does not belong to l. Let us consider points A, B, D, E. Both triples A, B, D and A, B, E do not lie on one line. Therefore, on one line there lies either triple (A, D, E) or triple (B, D, E). Let, for deﬁniteness, points A, D and E lie on one line. Then no three of the points B, C, D, E lie on one line. Contradiction. 26.3. Let the number of distinct distances between points be equal to k. Fix two points. Then all the other points are intersection points of two families of concentric circles containing k circles each. Hence, the total number of points does not exceed 2k2 + 2. It remains to notice that 2 · 142 + 2 = 394 < 400. 26.4. A segment of length d connecting a pair of given points will be called a diameter. The endpoints of all the diameters that begin at point A lie on the circle centered in A and of radius d. Since the distance between any two points does not exceed d, the endpoints of all the diameters beginning in A belong to an arc whose angle value does not exceed 60◦. Therefore, if three diameters AB, AC and AD begin in point A, then one of the endpoints of these diameters lies inside the angle formed by the other two endpoints. Let, for deﬁniteness, point C lie inside angle ∠BAD. Let us prove that then not more than one diameter begins in point C. Suppose that there is another diameter, CP, and points B and P lie on diﬀerent sides of line AC (Fig. 96). Then ABCP is a convex quadrilateral; hence, AB+CP < AC +BP (see Problem 9.14), i.e., d+d < d+BP and, therefore, BP > d which is impossible. Figure 238 (Sol. 26.4) As a result we see that either from each point there goes not more than two diameters or there exists a point from which there goes not more than one diameter. Now, the required statement can be proved by induction on the number of points. For n = 3 it is obvious. Suppose the statement is proved for any system of n points; let us prove it for a system of n + 1 points. In this system either there is a point from which there goes not more than one diameter or from each point there goes not more than two diameters. In the ﬁrst case we delete this point and, making use of the fact that in the remaining system there are not more than n diameters, get the desired. The second case is obvious. 26.5. Let us draw all the lines that connect pairs of given points and select a line, l, not parallel to either of them. It is possible to divide the given points into quadruples with the help of lines parallel to l. The quadrilaterals with vertices in these quadruples of points are the desired ones (Fig. 97). 26.6. Let us divide the given points in an arbitrary way into six groups: four groups of four points each, a group of ﬁve points and a group of one point. Let us consider the group of ﬁve points. From these points we can select four points which are vertices of a convex 440 CHAPTER 26. EXAMPLES AND COUNTEREXAMPLES Figure 239 (Sol. 26.5) quadrilateral ABCD (see Problem 22.2). Let us unite points A, C and B, D into pairs. Then segments AC and BD given by pairs intersect. One of the ﬁve points is free. Let us adjoin it to the foursome of points and perform the same with the obtained 5-tuple of points, etc. After ﬁve of such operations there remain two points and we can unite them in a pair. 26.7. Since ¡ 2n n2+1 ¢2 + ³ n2−1 n2+1 ´2 = 1, there exists an angle ϕ with the property that sin ϕ = 2n n2+1 and cos ϕ = n2−1 n2+1, where 0 < 2Nϕ < π 2 for a suﬃciently large n. Let us consider the circle of radius R centered at O and points A0, A1, . . . , AN−1 on it such that ∠A0OAk = 2kϕ. Then AiAj = 2R sin(\|i −j\|ϕ). Making use of the formulas sin(m + 1)ϕ = sin mϕ cos ϕ + sin ϕ cos mϕ, cos(m + 1)ϕ = cos mϕ cos ϕ −sin mϕ sin ϕ it is easy to prove that the numbers sin mϕ and cos mϕ are rational for all positive integers m. Let us take for R the greatest common divisor of all the denominators of the rational numbers sin ϕ, . . . , sin(N −1)ϕ. Then A0, . . . , AN−1 is the required system of points. 26.8. An example is depicted on Fig. 98. Figure 240 (Sol. 26.8) 26.9. It is possible. An example is plotted on Fig. 99. 26.10. The hypothesis implies that all the vertices of the polygon are divided into pairs that determine diagonals AiAj which pass through point O. Therefore, the number of vertices is even and on both parts of each of such diagonals AiAj there are an equal number of vertices. Hence, j = i + m, where m is a half of the total number of vertices. Therefore, point O is the intersection point of diagonals that connect opposite vertices. It is clear that the intersection point of these diagonals is unique. 26.11. If AC and BD are intersecting red segments, then the number of intersection points of any line with segments AB and CD does not exceed the number of intersection SOLUTIONS 441 Figure 241 (Sol. 26.9) points of this line with segments AC and BD. Therefore, by replacing red segments AC and BD with segments AB and CD we do not increase the number of intersection points of red segments with blue ones and diminish the number of intersection points of red segments with red ones because the intersection point of AC and BD vanishes. After several such operations all red segments become nonintersecting ones and it remains to prove that in this case the number of intersection points of red segments with blue ones is not smaller than n. Let us consider an arbitrary red segment. Since the other red segments do not intersect it, we deduce that on both sides of it there lies an even number of red points or, equivalently, an odd number of blue points. Therefore, there exists a blue segment that intersects the given red segment. Therefore, the number of intersection points of red segments with blue ones is not fewer than the number of red segments i.e., is not less than n. 26.12. Let A be a common point of the ﬁrst three circles S1, S2 and S3. Denote the intersection points of S1 and S2, S2 and S3, S3 and S1 by B, C, D, respectively. Suppose there exists a circle S not passing through point A. Then S passes through points B, C and D. Let S′ be the ﬁfth circle. Each pair of points from the collection A, B, C, D is a pair of intersection points of two of the circles S1, S2, S3, S. Therefore, S′ passes through one point from each pair of points A, B, C, D. On the other hand, S′ cannot pass through three points from the set A, B, C, D because each triple of these points determines one of the circles S1, S2, S3, S. Hence, S′ does not pass through certain two of these points. Contradiction. 26.13. Let us consider rectangle ABCD with sides AB = 1 cm and BC = 500 m. Let O be the intersection point of the rectangle’s diagonals. It is easy to verify that the area of AOD is greater than 1 m2 and all its hights are shorter than 1 cm. 26.14. No, not necessarily. On Fig. 100 it is shown how to get the required quadrilateral ABCD. Figure 242 (Sol. 26.14) 26.15. Not necessarily. It is easy to verify that the list of the lengths of sides and diagonals for an isosceles trapezoid with height 1 and bases 2 and 4 coincides with the similar list for the quadrilateral with perpendicular diagonals of length 2 and 4 that are 442 CHAPTER 26. EXAMPLES AND COUNTEREXAMPLES divided by their intersection point into segments of length 1 and 1 and 1 and 3, respectively (Fig. 101). Figure 243 (Sol. 26.15) 26.16. Yes, there exist. Let us consider points Pi = (i, i2), where i = 1, . . . , n. The areas of all the triangles with vertices in the nodes of an integer lattice are rational (see Problem 24.5) and the numbers PiPj = \|i −j\| p 1 + (i + j)2 are irrational. 26.17. Yes, there exist. Let C be the midpoint of segment AB. Then XC2 = 2XA2 + 2XB2 −AB2 2 . If the number AB2 is irrational, then the numbers XA, XB and XC cannot simultaneously be rational. 26.18. It can. Consider right triangle ABC1 with legs AB = 1 and BC1 = 2n. In this triangle draw median AM1, bisector BK1 and hight C1H1. The area of the triangle formed by these segments is greater than SABM1 −SABK1. Clearly, SABK1 < 1 2 and SABM1 = n 2, i.e., SABM1 −SABK1 > ( S 2 ) −( S 2n), where S = SABC1. Hence, for a suﬃciently large n the area of the triangle formed by segments AM1, BK1 and C1H1 will be greater than 0.499 · S. Slightly moving point C1 we turn triangle ABC1 into an acute triangle ABC and the area of the triangle formed by the intersection points of segments remains greater than 0.499 · SABC. 26.19. It is possible. Let us pave, for instance, inﬁnite angles illustrated on Fig. 102. Figure 244 (Sol. 26.19) 26.20. Yes, it can. Let us prove the statement by induction replacing 100 with n. For n = 1 we can take the endpoints of a segment of length 1. Suppose that the statement is proved for n and A1, . . . , Ak is the required set of points. Let A′ 1, . . . , A′ k be the images of points A1, . . . , Ak under the parallel transport by unit vector a. To prove the inductive step it suﬃces to select the unit vector a so that a ̸= − − − → AiAj and AjA′ i ̸= 1 for i ̸= j, i.e., \|− − − → AjAi + a\| ̸= 1 for i ̸= j. Each of these restrictions excludes from the unit circle not more than 1 point. SOLUTIONS 443 Figure 245 (Sol. 26.21) 26.21. Not always. Consider the segments plotted on Fig. 103. The endpoints of each short segment can be connected with the endpoints of the nearest to it long segment only. It is clear that in this way we cannot get a closed nonselﬁntersecting broken line. 26.22. Not necessarily. Let us prove that the center O of the circle inscribed in triangle ABC with sides AB = 6, BC = 4 and CA = 8 is equidistant from the midpoints of sides AC and BC. Denote the midpoints of sides AC and BC by B1 and A1 and the bases of the perpendiculars dropped from O to AC and BC by B2 and A2, see Fig. 104. Since A1A2 = 1 = B1B2 (cf. Problem 3.2) and OA2 = OB2, it follows that △OA1A2 = △OB1B2, i.e., OA1 = OB1. Figure 246 (Sol. 26.22) 26.23. This is possible for any n ≥2. Indeed, let us inscribe into the arena a regular k-gon, where k is the number of distinct pairs that can be composed of n spotlights, i.e., k = n(n−1) 2 . Then we can establish a one-to-one correspondence between the segments cut oﬀ by the sides of the k-gon and the pairs of spotlights. Let each spotlight illuminate the whole k-gon and the segments that correspond to pairs of spotlights in which it enters. (Yeah?)It is easy to verify that this illumination possesses the required properties. Chapter 27. INDUCTION AND COMBINATORICS §1. Induction 27.1. Prove that if the plane is divided into parts (“countries”) by lines and circles, then the obtained map can be painted two colours so that the parts separated by an arc or a segment are of distinct colours. 27.2. Prove that in a convex n-gon it is impossible to select more than n diagonals so that any two of them have a common point. 27.3. Let E be the intersection point of lateral sides AD and BC of trapezoid ABCD, let Bn+1 be the intersection point of lines AnC and BD (A0 = A); let An+1 be the intersection point of lines EBn+1 and AB. Prove that AnB = 1 n+1AB. 27.4. On a line, there are given points A1, . . . , An and B1, . . . , Bn−1. Prove that n X i=1 ÃQ 1≤k≤n−1 AiBk Q j̸=i AiAj ! = 1. 27.5. Prove that if n points do not lie on one line, then among the lines that connect them there are not fewer than n distinct points. See also Problems 2.12, 5.98, 22.7, 22.9–22.12, 22.20 b, 22.22, 22.23, 22.29, 23.39–23.41, 26.20. §2. Combinatorics 27.6. Several points are marked on a circle, A is one of them. Which convex polygons with vertices in these points are more numerous: those that contain A or those that do not contain it? 27.7. On a circle, nine points are ﬁxed. How many non-closed non-selﬁntersecting broken lines of nine links with vertices in these points are there? 27.8. In a convex n-gon (n ≥4) there are drawn all the diagonals and no three of them intersect at one point. Find the number of intersection points of the diagonals. 27.9. In a convex n-gon (n ≥4) all the diagonals are drawn. Into how many parts do they divide an n-gon if no three of them intersect at one point? 27.10. Given n points in plane no three of which lie on one line, prove that there exist not fewer than (n 5) n−4 distinct convex quadrilaterals with vertices in these points. 27.11. Prove that the number of nonequal triangles with the vertices in vertices of a regular n-gon is equal to the integer nearest to n2 12. See also Problem 25.6. Solutions 27.1. Let us carry out the proof by induction on the total number of lines and circles. For one line or circle the statement is obvious. Now, suppose that it is possible to paint any 445 446 CHAPTER 27. INDUCTION AND COMBINATORICS map given by n lines and circles in the required way and show how to paint a map given by n + 1 lines and circles. Let us delete one of these lines (or circles) and paint the map given by the remaining n lines and circles thanks to the inductive hypothesis. Then retain the colours of all the parts lying on one side of the deleted line (or circle) and replace the colours of all the parts lying on the other side of the deleted line (or circle) with opposite ones. 27.2. Let us prove by induction on n that in a convex n-gon it is impossible to select more than n sides and diagonals so that any two of them have a common point. For n = 3 this is obvious. Suppose that the statement holds for any convex n-gon and prove it for an (n + 1)-gon. If from every vertex of the (n + 1)-gon there goes not more than two of the selected sides or diagonals, then the total number of selected sides or diagonals does not exceed n + 1. Hence, let us assume that from vertex A there goes three of the selected sides or diagonals AB1, AB2 and AB3, where AB2 lies between AB1 and AB3. Since a diagonal or a side coming from point B2 and distinct from AB2 cannot simultaneously intersect AB1 and AB3, it is clear that only one of the chosen diagonals can go from B2. Therefore, it is possible to delete point B2 together with diagonal AB2 and apply the inductive hypothesis. 27.3. Clearly, A0B = AB. Let Cn be the intersection point of lines EAn and DC, where DC : AB = k, AB = a, AnB = an and An+1B = x. Since CCn+1 : AnAn+1 = DCn+1 : BAn+1, it follows that kx : (an −x) = (ka −kx) : x, i.e., x = aan a+an. If an = a n+1, then x = a n+2. 27.4. First, let us prove the desired statement for n = 2. Since − − − → A1B1+− − − → B1A2+− − − → A2A1 = − → 0 , it follows that A1B1 A1A2 + A2B1 A2A1 = 1. To prove the inductive step let us do as follows. Fix points A1, . . . , An and B1, . . . , Bn−2 and consider point Bn−1 variable. Consider the function f(Bn−1) = n X i=1 ÃQ 1≤k≤n−1 AiBk Q j̸=i AiAj ! = 1. This function is a linear one and by the inductive hypothesis f(Bn−1) = 1 if Bn−1 coincides with one of the points A1, . . . , An. Therefore, this function is identically equal to 1. 27.5. Induction on n. For n = 3 the statement is obvious. Suppose we have proved it for n −1 and let us prove it then for n points. If on every line passing through two of the given points lies one more given point, then all the given points belong to one line (cf. Problem 20.13). Therefore, there exists a line on which there are exactly two given points A and B. Let us delete point A. The two cases are possible: 1) All the remaining points lie on one line l. Then there will be precisely n distinct lines: l and n −1 line passing through A. 2) The remaining points do not belong to one line. Then among the lines that connect them there are not fewer than n −1 distinct ones that connect them and all of them diﬀer from l. Together with AB they constitute not fewer than n lines. 27.6. To any polygon, P, that does not contain point A we can assign a polygon that contains A by adding A to the vertices of P. The inverse operation, however, that is deleting of the point A, can be only performed for n-gons with n ≥4. Therefore, there are more polygons that contain A than polygons without A and the diﬀerence is equal to the number of triangles with A as a vertex, i.e., (n−1)(n−2) 2 . 27.7. The ﬁrst point can be selected in 10 ways. Each of the following 8 points can be selected in two ways because it must be neighbouring to one of the points selected earlier (otherwise we get a self-intersecting broken line). Since the beginning and the end do not SOLUTIONS 447 diﬀer in this method of calculation, the result should be divided by 2. Hence, the total number of the broken lines is equal to 10·28 2 = 1280. 27.8. Any intersection point of diagonals determines two diagonals whose intersection point it serves and the endpoints of these diagonals ﬁx a convex quadrilateral. Conversely, any four vertices of a polygon determine one intersection point of diagonals. Therefore, the total number of intersection points of diagonals is equal to the number of ways to choose 4 points of n, i.e., is equal to n(n−1)(n−2)(n−3) 2·3·4 . 27.9. Let us consecutively draw diagonals. When we draw a diagonal, the number of parts into which the earlier drawn diagonals divide the polygon increases by m+1, where m is the number of intersection points of the new diagonals with the previously drawn ones, i.e., each new diagonal and each new intersection point of diagonals increase the number of parts by 1. Therefore, the total number of parts into which the diagonals divide an n-gon is equal to D + P + 1, where D is the number of diagonals, P is the number of intersection points of the diagonals. It is clear that D = n(n−3) 2 . By the above problem P = n(n−1)(n−2)(n−3) 24 . 27.10. If we choose any ﬁve points, then there exists a convex quadrilateral with vertices in these points (Problem 22.2). It remains to notice that a quadruple of points can be complemented to a 5-tuple in n −4 distinct ways. 27.11. Let there be N nonequal triangles with vertices in vertices of a regular n-gon so that among them there are N1 equailateral, N2 non-equailateral isosceles, and N3 scalane ones. Each equailateral triangle is equal to a triangle with ﬁxed vertex A, a non-equailateral isosceles is equal to three triangles with vertex A and a scalane one is equal to 6 triangles. Since the total number of triangles with vertex A is equal to (n−1)(n−2) 2 , it follows that (n−1)(n−2) 2 = N1 + 3N2 + 6N3. Clearly, the number of nonequal equailateral triangles is equal to either 0 or 1 and the number of nonequal isosceles triangles is equal to either n−1 2 or (n 2) −1, i.e., N1 = 1 −c and N1 + N2 = n−2+d 2 , where c and d are equal to either 0 or 1. Therefore, 12N = 12(N1 + N2 + N3) = 2(N1 + 3N2 + 6N3) + 6(N1 + N2) + 4N1 = (n −1)(n −2) + 3(n −2 + d) + 4(1 −c) = n2 + 3d −4c. Since \|3d −4c\| < 6, it follows that N coincides with the nearest integer to n2 12. Chapter 28. INVERSION Background 1. All the geometric transformations that we have encountered in this book so far turned lines into lines and circles into circles. The inversion is a transformation of another type which also preserves the class of lines and circles but can transform a line into a circle and a circle into a line. This and other remarkable properties of inversion serve as a foundation for its astounding eﬀectiveness in solving various geometric problems. 2. In plane, consider circle S centered at O with radius R. We call the transformation that sends an arbitrary point A distinct from O into point A∗lying on ray OA at distance OA∗= R2 OA from O the inversion relative S. The inversion relative S will be also called the inversion with center O and degree R2 and S will be called the circle of inversion. 3. It follows directly from the deﬁnition of inversion that it ﬁxes points of S, moves points from inside S outside it and points from outside S inside it. If point A turns into A∗ under the inversion, then the inversion sends A∗into A, i.e., (A∗)∗= A. The image of a line passing through the center of the inversion is this line itself. Here we should make a reservation connected with the fact that, strictly speaking, the inversion is not a transformation of the plane because O has no image. Therefore, formally speaking, we cannot speak about the “image of the line through O” and should consider instead the union of two rays obtained from the line by deleting point O. Similar is the case with the circles containing point O. Nevertheless, we will use these loose but more graphic formulations and hope that the reader will easily rectify them when necessary. 4. Everywhere in this chapter the image of point A under an inversion is denoted by A∗. 5. Let us formulate the most important properties of inversion that are constantly used in the solution of problems. Under an inversion with center O: a) a line l not containing O turns into a circle passing through O (Problem 28.2); b) a circle centered at C and passing through O turns into a line perpendicular to OC (Problem 28.3); c) a circle not passing through O turns into a circle not passing through O (Problem 28.3); d) the tangency of circles with lines is preserved only if the tangent point does not coincide with the center of inversion; otherwise, the circle and a line turn into a pair of parallel lines (Problem 28.4); e) the value of the angle between two circles (or between a circle and a line, or between two lines) is preserved (Problem 28.5). §1. Properties of inversions 28.1. Let an inversion with center O send point A to A∗and B to B∗. Prove that triangles OAB and OB∗A∗are similar. 28.2. Prove that under any inversion with center O any line l not passing through O turns into a circle passing through O. 449 450 CHAPTER 28. INVERSION 28.3. Prove that under any inversion with center O any circle passing through O turns into a line and any circle not passing through O into a circle. 28.4. Prove that tangent circles (any circle tangent to a line) turn under any inversion into tangent circles or in a circle and a line or in a pair of parallel lines. Let two circles intersect at point A. The angle between the circles is the angle between the tangents to the circles at point A. (Clearly, if the circles intersect at points A and B, then the angle between the tangents at point A is equal to the angle between the tangents at point B). The angle between a line and a circle is similarly deﬁned (as the angle between the line and the tangent to the circle at one of the intersection points). 28.5. Prove that inversion preserves the angle between circles (and also between a circle and a line, and between lines). 28.6. Prove that two nonintersecting circles S1 and S2 (or a circle and a line) can be transported under an inversion into a pair of concentric circles. 28.7. Let S be centered in O. Through point A a line l intersecting S at points M and N and not passing through O is drawn. Let M ′ and N ′ be points symmetric to M and N, respectively, through OA and let A′ be the intersection point of lines MN ′ and M ′N. Prove that A′ coincides with the image of A under the inversion with respect to S (and, therefore, does not depend on the choice of line l). §2. Construction of circles While solving problems of this section we will often say “let us perform an inversion . . . ” . Being translated into a more formal language this should sound as: “Let us construct with the help of a ruler and a compass the images of all the given points, lines and circles under the inversion relative to the given circle”. The possibility to perform such constructions follows from properties of inversion and Problem 28.8. In problems on construction we often make use of the existence of inversion that sends two nonintersecting circles into concentric circles. The solution of Problem 28.6 implies that the center and radius of such an inversion (hence, the images of the circles) can be constructed by a ruler and a compass. 28.8. Construct the image of point A under the inversion relative circle S centered in O. 28.9. Construct the circle passing through two given points and tangent to the given circle (or line). 28.10. Through a given point draw the circle tangent to two given circles (or a circle and a line). 28.11. (Apollonius’ problem.) Construct a circle tangent to the three given circles. 28.12. Through a given point draw a circle perpendicular to two given circles. 28.13. Construct a circle tangent to a given circle S and perpendicular to the two given circles (S1 and S2). 28.14. Through given points A and B draw a circle intersecting a given circle S under the angle of α. §3. Constructions with the help of a compass only According to the tradition that stems from ancient Greece, in geometry they usually consider constructions with the help of ruler and compass. But we can also make construc-tions with the help of other instruments, or we can, for instance, consider constructions with the help of one compass only, without a ruler. Clearly, with the help of a compass only one §4. LET US PERFORM AN INVERSION 451 cannot simultaneously construct all the points of a line. Therefore, let us make a convention: we will consider a line constructed if two of its points are constructed. It turns out that under such convention we can perform with the help of a compass all the constructions which can be performed with the help of a compass and a ruler. This follows from the possibility to construct using only a compass the intersection points of any line given by two points with a given circle (Problem 28.21 a)) and the intersection point of two lines (Problem 28.21 b)). Indeed, any construction with the help of ruler and compass is a sequence of determinations of the intersection points of circles and lines. In this section we will only consider constructions with a compass only, without a ruler, i.e., the word “construct” means “construct with the help of a compass only”. We will consider a segment constructed if its endpoints are constructed. 28.15. a) Construct a segment twice longer than a given segment. b) Construct a segment n times longer than a given segment. 28.16. Construct the point symmetric to point A through the line passing through given points B and C. 28.17. Construct the image of point A under the inversion relative a given circle S centered in a given point O. 28.18. Construct the midpoint of the segment with given endpoints. 28.19. Construct the circle into which the given line AB turns into under the inversion relative a given circle with given center O. 28.20. Construct the circle passing through three given points. 28.21. a) Construct the intersection points of the given circle S and the line passing through given points A and B. b) Construct the intersection point of lines A1B1 and A2B2, where A1, B1, A2 and B2 are given points. §4. Let us perform an inversion 28.22. In a disk segment, all possible pairs of tangent circles (Fig. 105) are inscribed. Find the locus of their tangent points. Figure 247 (28.22) 28.23. Find the set of tangent points of pairs of circles that are tangent to the legs of the given angle at given points A and B. 28.24. Prove that the inversion with the center at vertex A of an isosceles triangle ABC, where AB = AC, of degree AB2 sends the base BC of the triangle into the arc ⌣BC of the circumscribed circle. 28.25. In a circle segment, all the possible pairs of intersecting circles are inscribed and for each pair a line is drawn through their intersection point. Prove that all these lines pass through one point, cf. Problem 3.44. 28.26. No three of the four points A, B, C, D lie on one line. Prove that the angle between the circumscribed circles of triangles ABC and ABD is equal to the angle between the circumscribed circles of triangles ACD and BCD. 28.27. Through points A and B there are drawn circles S1 and S2 tangent to circle S and circle S3 perpendicular to S. Prove that S3 forms equal angles with circles S1 and S2. 452 CHAPTER 28. INVERSION 28.28. Two circles intersecting at point A are tangent to the circle (or line) S1 at points B1 and C1 and to the circle (or line) S2 at points B2 and C2 (and the tangency at B2 and C2 is the same as at respective points B1 and C1, i.e., either inner or outer). Prove that circles circumscribed about triangles AB1C1 and AB2C2 are tangent to each other. 28.29. Prove that the circle passing through the midpoints of triangle’s sides is tangent to its inscribed and three escribed circles. (Feuerbach’s theorem.) §5. Points that lie on one circle and circles passing through one point 28.30. Given four circles, S1, S2, S3, S4, where circles S1 and S3 intersect with both circles S2 and S4. Prove that if the intersection points of S1 with S2 and S3 with S4 lie on one circle or line, then the intersection points of S1 with S4 and S2 with S3 lie on one circle or line (Fig. 106). Figure 248 (28.30) 28.31. Given four circles S1, S2, S3, S4 such that S1 and S2 intersect at points A1 and A2, S2 and S3 at points B1 and B2, S3 and S4 at points C1 and C2, S4 and S1 at points D1 and D2 (Fig. 107). Figure 249 (28.31) Prove that if points A1, B1, C1, D1 lie on one circle (or line) S, then points A2, B2, C2, D2 lie on one circle (or line). 28.32. The sides of convex pentagon ABCDE are extended so that ﬁve-angled star AHBKCLDMEN (Fig. 108) is formed. The circles are circumscribed about triangles — §5. POINTS THAT LIE ON ONE CIRCLE 453 the rays of the star. Prove that the ﬁve intersection points of these circles distinct from A, B, C, D, E lie on one circle. Figure 250 (28.32) 28.33. In plane, six points A1, A2, A3, B1, B2, B3 are ﬁxed. Prove that if the circles circumscribed about triangles A1A2B3, A1B2A3 and B1A2A3 pass through one point, then the circles circumscribed about triangles B1B2A3, B1A2B3 and A1B2B3 intersect at one point. 28.34. In plane, six points A1, A2, B1, B2, C1, C2 are ﬁxed. Prove that if the circles cir-cumscribed about triangles A1B1C1, A1B2C2, A2B1C2, A2B2C1 pass through one point, then the circles circumscribed about triangles A2B2C2, A2B1C1, A1B2C1, A1B1C2 pass through one point. 28.35. In this problem we will consider tuples of n generic lines, i.e., sets of lines no two of which are parallel and no three pass through one point. To a tuple of two generic lines assign their intersection point and to a tuple of two generic lines assign the circle passing through the three points of their pairwise intersections. If l1, l2, l3, l4 are four generic lines, then the four circles Si corresponding to four triples of lines obtained by discarding li pass through one point (cf. Problem 2.83 a)) that we will assign to the foursome of lines. This construction can be extended: a) Let li, i = 1, . . . , 5 be ﬁve generic points. Prove that ﬁve points Ai corresponding to the foursome of lines obtained by discarding li lie on one circle. b) Prove that this construction can be continued in the following way: to every tuple of n generic points assign a point if n is even or a circle if n is odd so that n circles (points) corresponding to tuples of n −1 lines pass through this point (belong to this circle). 28.36. On two intersecting lines l1 and l2, select points M1 and M2 not coinciding with the intersection point M of these lines. Assign to this set of lines and points the circle passing through M1, M2 and M. If (l1, M1), (l2, M2), (l3, M3) are three generic lines with ﬁxed points, then by Problem 2.80 a) the three circles corresponding to pairs (l1, M1) and (l2, M2), (l2, M2), (l3, M3), (l3, M3) and (l1, M1) intersect at one point that we will assign to the triple of lines with a ﬁxed point. a) Let l1, l2, l3, l4 be four generic lines on each of which a point is ﬁxed so that these points lie on one circle. Prove that four points corresponding to the triples obtained by deleting one of the lines lie on one circle. b) Prove that to every tuple of n generic lines with a point ﬁxed on each of them so that the ﬁxed points lie on one circle one can assign a point (if n is odd) or a circle (if n is even) 454 CHAPTER 28. INVERSION so that n circles (if n is odd) or points (if n is even) corresponding to the tuples of n −1 lines pass through this point (resp. lie on this circle). §6. Chains of circles 28.37. Circles S1, S2, . . . , Sn are tangent to circles R1 and R2 and, moreover, S1 is tangent to S2 at point A1, S2 is tangent to S3 at point A2, . . . , Sn−1 is tangent to Sn at point An−1. Prove that points A1, A2, . . . , An−1 lie on one circle. 28.38. Prove that if there exists a chain of circles S1, S2, . . . , Sn each of which is tangent to two neighbouring ones (Sn is tangent to Sn−1 and S1) and two given nonintersecting circles R1 and R2, then there are inﬁnitely many such chains. (?)Namely, for any circle T1 tangent to R1 and R2 (in the same fashion if R1 and R2 do not lie inside each other, by an inner or an outer way, otherwise) there exists a similar chain of n tangent circles T1, T2, . . . , Tn. (Steiner’s porism.) 28.39. Prove that for two nonintersecting circles R1 and R2 a chain of n tangent circles (cf. the preceding problem) exists if and only if the angle between the circles T1 and T2 tangent to R1 and R2 at their intersection points with the line that connects the centers of R1 and R2 is equal to an integer multiple of 360◦ n (Fig. 109). Figure 251 (28.39) 28.40. Each of six circles is tangent to four of the remaining ﬁve circles, see Fig. 110. Figure 252 (28.40) Prove that for any pair of nonintersecting circles (of these six circles) the radii and the distance between their centers are related by the formula d2 = r2 1 + r2 2 ± 6r1r2, where “plus” is taken if the circles are not inside each other and “minus” otherwise. SOLUTIONS 455 Solutions 28.1. Let R2 be the degree of the inversion. Then OA · OA∗= OB · OB∗= R2 whence, OA : OB = OB∗: OA∗and △OAB ∼△OB∗A∗because ∠AOB = ∠B∗OA∗. 28.2. Let us drop perpendicular OC from point O to line l and take an arbitrary point M on l. Since triangles OCM and OM ∗C∗are similar (Problem 28.1), ∠OM ∗C∗= ∠OCM = 90◦, i.e., point M ∗lies on circle S with diameter OC∗. If X is a point of S distinct from O, then it is the image under the inversion of the intersection point Y of l and OX (since the image of Y lies, on the one hand, on ray OX and, on the other hand, on circle S, as is already proved). Thus, the inversion sends line l into circle S (without point O). 28.3. The case when circle S passes through O is actually considered in the preceding problem (and formally follows from it since (M ∗)∗= M). Now, suppose that O does not belong to S. Let A and B be the intersection points of circle S with the line passing through O and the center of S, let M be an arbitrary point of S. Let us prove that the circle with diameter A∗B∗is the image of S. To this end it suﬃces to show that ∠A∗M ∗B∗= 90◦. But by Problem 28.1 △OAM ∼△OM ∗A∗and △OBM ∼△OM ∗B∗; hence, ∠OMA = ∠OA∗M ∗and ∠OMB = ∠OB∗M ∗; more exactly, ∠(OM, MA) = −∠(OA∗, M ∗A∗) and ∠(OM, MB) = −∠(OB∗, M ∗B∗). (In order not to consider various cases of points’ disposition we will make use of the properties of oriented angles between lines discussed in Chapter 2.) Therefore, ∠(A∗M ∗, M ∗B∗) = ∠(A∗M ∗, OA∗) + ∠(OB∗, M ∗B∗) = ∠(OM, MA) + ∠(MB, OM) = ∠(MB, MA) = 90◦. 28.4. If the tangent point does not coincide with the center of inversion, then after the inversion these circles (the circle and the line) will still have one common point, i.e., the tangency is preserved. If the circles with centers A and B are tangent at point O, then under the inversion with center O they turn into a pair of lines perpendicular to AB. Finally, if line l is tangent to the circle centered at A at point O, then under the inversion with center O the line l turns into itself and the circle into a line perpendicular to OA. In each of these two cases we get a pair of parallel lines. 28.5. Let us draw tangents l1 and l2 through the intersection point of the circles. Since under the inversion the tangent circles or a circle and a line pass into tangent ones (cf. Problem 28.4), the angle between the images of circles is equal to the angle between the images of the tangents to them. Under the inversion centered at O line li turns into itself or into a circle the tangent to which at O is parallel to li. Therefore, the angle between the images of l1 and l2 under the inversion with center O is equal to the angle between these lines. 28.6. First solution. Let us draw the coordinate axis through the centers of the circles. Let a1 and a2 be the coordinates of the intersection points of the axes with S1, let b1 and b2 be the coordinates of the intersection points of the axes with S2. Let O be the point on the axis whose coordinate is x. Then under the inversion with center O and degree k our circles turn into the circles whose diameters lie on the axis and whose endpoints have coordinates a′ 1, a′ 2 and b′ 1, b′ 2, respectively, where a′ 1 = x + k a1 −x, a′ 2 = x + k a2 −x, b′ 1 = x + k b1 −x, b′ 2 = x + k b2 −x. 456 CHAPTER 28. INVERSION The obtained circles are concentric if a′ 1+a′ 2 2 = b′ 1+b′ 2 2 , i.e., 1 a1 −x + 1 a2 −x = 1 b1 −x + 1 b2 −x, wherefrom we have (b1 + b2 −a1 −a2)x2 + 2(a1a2 −b1b2)x + b1b2(a1 + a2) −a1a2(b + b2) = 0. The discriminant of this quadratic in x is equal to 4(b1 −a1)(b1 −a2)(b2 −a1)(b2 −a2). It is positive precisely when the circles do not intersect; this proves the existence of the required inversion. The existence of such an inversion for the case of a circle and a line is similarly proved. Another solution. On the line that connects centers O1 and O2 of the circles take point C such that the tangents drawn to the circles from C are equal. This point C can be constructed by drawing the radical axis of the circles (cf. Problem 3.53). Let l be the length of these tangents. The circle S of radius l centered in C is perpendicular to S1 and S2. Therefore, under the inversion with center O, where O is any of the intersection points of S with line O1O2, circle S turns into a line perpendicular to circles S∗ 1 and S∗ 2 and, therefore, passing through their centers. But line O1O2 also passes through centers of S∗ 1 and S∗ 2; hence, circles S∗ 1 and S∗ 2 are concentric, i.e., O is the center of the desired inversion. If S2 is not a circle but a line, the role of line O1O2 is played by the perpendicular dropped from O1 to S2, point C is its intersection point with S2, and l is the length of the tangent dropped from C to S1. 28.7. Let point A lie outside S. Then A′ lies inside S and we see that ∠MA′N = 1 2(⌣ MN+ ⌣M ′N ′) =⌣MN = ∠MON, i.e., quadrilateral MNOA′ is an inscribed one. But under the inversion with respect to S line MN turns into the circle passing through points M, N, O (Problem 28.2). Therefore, point A∗(the image of A under the inversion) lies on the circle circumscribed about quadrilateral MNOA′. By the same reason points A′ and A∗ belong to the circle passing through M ′, N ′ and O. But these two circles cannot have other common points except O and A′. Hence, A∗= A′. If A lies inside S, we can apply the already proved to line MN ′ and point A′ (which is outside S). We get A = (A′)∗. But then A′ = A∗. 28.8. Let point A lie outside S. Through A, draw a line tangent to S at point M. Let MA′ be a height of triangle OMA. Right triangles OMA and OA′M are similar, hence, A′O : OM = OM : OA and OA′ = R2 OA, i.e., point A′ is the one to be found. If A lies inside S, then we can perform the construction in the reverse order: we drop perpendicular AM to OA (point M lies on the circle). Then the tangent to S at point M intersects with ray OA at the desired point, A∗. Proof is repeated literally. 28.9. If both given points A and B lie on the given circle (or line) S, then the problem has no solutions. Let now A not lie to S. Under the inversion with center A the circle to be found turns into the line passing through B∗and tangent to S∗. This implies the following construction. Let us perform the inversion with respect to an arbitrary circle with center A. Through B∗draw the tangent l to S∗. Perform an inversion once again. Then l turns into the circle to be constructed. If point B∗lies on S∗, then the problem has a unique solution; if B∗lies outside S∗, then there are two solutions, and if B∗lies inside S∗, then there are no solutions. 28.10. The inversion with center at the given point sends circles S1 and S2 into a pair of circles S∗ 1 and S∗ 2 (or into circle S∗and line l; or into a pair of lines l1 and l2), respectively; the circle tangent to them turns into the common tangent to S∗ 1 and S∗ 2 (resp. into the SOLUTIONS 457 tangent to S∗parallel to l; or into a line parallel to l1 and l2). Therefore, to construct the desired circle we have to construct a line tangent to S∗ 1 and S∗ 2 (resp. tangent to S∗and parallel to l; or parallel to l1 and l2) and perform an inversion once again. 28.11. Let us reduce this problem to Problem 28.10. Let circle S of radius r be tangent to circles S1, S2, S3 of radii r1, r2, r3, respectively. Since the tangency of S with each of Si (i = 1, 2, 3) can be either outer or inner, there are eight possible distinct cases to consider. Let, for instance, S be tangent to S1 and S3 from the outside and to S2 from the inside (Fig. 111). Figure 253 (Sol. 28.11) Let us replace the circles S, S2, S3 with the concentric to them circles S′, S′ 2 and S′ 3, respectively, so that S′ is tangent to S′ 2 and S′ 3 and passes through the center O1 of S1. To this end it suﬃces that the radii of S′, S′ 2, S′ 3 were equal to r + r1, r2 + r1, \|r3 −r1\|, respectively. Conversely, from circle S′ passing through O1 and tangent to S′ 2 and S′ 3 (from the outside if r3 −r1 ≥0 and from the inside if r3 −r1 < 0) we can construct circle S — a solution of the problem — by diminishing the radius of S′ by r1. The construction of such a circle S′ is described in the solution of Problem 28.10 (if the type of tangency is given, then the circle is uniquely constructed). One can similarly perform the construction for the other possible types of tangency. 28.12. Under the inversion with center at the given point A the circle to be constructed turns into the line perpendicular to the images of both circles S1 and S2, i.e., into the line connecting the centers of S∗ 1 and S∗ 2. Therefore, the circle circle to be constructed is the image under this inversion of an arbitrary line passing through the centers of S∗ 1 and S∗ 2. 28.13. Let us perform an inversion that sends circles S1 and S2 into a pair of lines (if they have a common point) or in a pair of concentric circles (cf. Problem 28.6) with a common center A. In the latter case the circle perpendicular to both circles S1 and S2 turns into a line passing through A (since there are no circles perpendicular to two concentric circles); the tangent drawn from A to S∗is the image of the circle circle to be constructed under this inversion. If S∗ 1 and S∗ 2 are parallel lines, then the image of the circle circle to be constructed is any of the two lines perpendicular to S∗ 1 and S∗ 2 and tangent to S∗. Finally, if S∗ 1 and S∗ 2 are lines intersecting at a point B, then the circle circle to be constructed is the image under the inversion of any of the two circles with center B and tangent to S∗. 28.14. Under the inversion with center at point A the problem reduces to the con-struction of a line l passing through B∗and intersecting circle S∗at an angle of α, i.e., to 458 CHAPTER 28. INVERSION construction of a point X on S∗such that ∠B∗XO = 90◦−α, where O is the center of S∗. This point X lies on the intersection of S∗with the arc whose points serve as the vertices of angles of 90◦−α subtending segment B∗O. 28.15. a) Let AB be the given segment. Let us draw the circle with center B and radius AB. On this circle, mark chords AX, XY and Y Z of the same length as AB; we get equilateral triangles ABX, XBY and Y BZ. Hence, ∠ABZ = 180◦and AZ = 2AB. b) In the solution of heading a) we have described how to construct a segment BZ equal to AB on line AB. Repeating this procedure n −1 times we get segment AC such that AC = nAB. 28.16. Let us draw circles with centers B and C passing through A. Then the distinct from A intersection point of these circles is the desired one. 28.17. First, suppose that point A lies outside circle S. Let B and C be the intersection points of S and the circle of radius AO and with center A. Let us draw circles with centers B and C of radius BO = CO; let O and A′ be their intersection points. Let us prove that A′ is the desired point. Indeed, under the symmetry through line OA the circles with centers B and C turn into each other and, therefore, point A′ is ﬁxed. Hence, A′ lies on line OA. Isosceles triangles OAB and OBA′ are similar because they have equal angles at the base. Therefore, OA′ : OB = OB : OA or OA′ = OB2 OA , as required. Now, let point A lie inside S. With the help of the construction from Problem 28.15 a) let us construct on ray OA segments AA2, A2A3, . . . , An−1An, . . . of length OA until one of the points An becomes outside S. Applying to An the above-described construction we get a point A∗ n on OA such that OA∗ n = R2 n·OA = 1 nOA∗. In order to construct point A∗it only remains to enlarge segment OA∗ n n times, cf. Problem 28.15 b). 28.18. Let A and B be two given points. If point C lies on ray AB and AC = 2AB, then under the inversion with respect to the circle of radius AB centered at A point C turns into the midpoint of segment AB. The construction is reduced to Problems 28.15 a) and 28.17. 28.19. The center of this circle is the image under an inversion of point O′ symmetric to O through AB. It remains to apply Problems 28.16 and 28.17. 28.20. Let A, B, C be given points. Let us construct (Problem 28.17) the images of B and C under the inversion with center A and of arbitrary degree. Then the circle passing through A, B and C is the image of line B∗C∗under this inversion and its center can be constructed thanks to the preceding problem. 28.21. a) Making use of the preceding problem construct the center O of circle S. Next, construct points A∗and B∗— the images of A and B under the inversion with respect to S. The image of AB is circle S1 passing through points A∗, B∗and O. Making use of Problem 28.19 we construct S1. The desired points are the images of the intersection points of circles S and S1, i.e., just intersection points of S and S1. b) Let us consider an inversion with center A1. Line A2B2 turns under this inversion into the circle S passing through points A1, A∗ 2 and B∗ 2. We can construct S making use of Problem 28.19. Further, let us construct the intersection points of S and line A1B1 making use of the solution of heading a). The desired point is the image of the intersection point distinct from A1 under the inversion considered. 28.22. Under the inversion centered in the endpoint A of the segment the conﬁguration plotted on Fig. 105 turns into the pair of tangent circles inscribed into the angle at vertex B∗. Clearly, the set of the tangent points of such circles is the bisector of the angle and the desired locus is the image of the bisector under the inversion — the arc of the circle with SOLUTIONS 459 endpoints A and B that divides in halves the angle between the arc of the segment and chord AB. 28.23. Let C be the vertex of the given angle. Under the inversion with center in A line CB turns into circle S; circles S1 and S2 turn into circle S∗ 1 centered in O1 tangent to S at point B∗and line l parallel to C∗A and tangent to S∗ 1 at X, respectively (Fig. 112). Figure 254 (Sol. 28.23) In S, draw radius OD perpendicular to C∗A. Points O, B∗and O1 lie on one line and OD ∥O1X. Hence, ∠OB∗D = 90◦−∠DOB∗ 2 = 90◦−∠XO1B∗ 2 = ∠O1B∗X, therefore, point X lies on line DB∗. Applying inversion once again we see that the desired locus of tangent points is arc ⌣AB of the circle passing through points A, B and D∗. 28.24. The given inversion sends line BC into the circle passing through points A, B and C so that the image of segment BC should remain inside angle ∠BAC. 28.25. Let S1 and S2 be circles inscribed into the segment; M, N their intersection points (Fig. 113). Let us show that line MN passes through point P of the circle of the segment equidistant from its endpoints A and B. Figure 255 (Sol. 28.25) Indeed, thanks to the preceding problem the inversion with center P and of degree PA2 sends segment AB to arc ⌣AB and circles S1 and S2 to circles S∗ 1 and S∗ 2, still inscribed into a segment, respectively. But the tangents to S1 drawn from P are tangent also to S∗ 1; hence, S∗ 1 = S1 (since both these circles are similarly tangent to the three ﬁxed points). Analogously, S∗ 2 = S2; hence, points M and N change places under the inversion, i.e., M ∗= N and MN passes through the center of inversion. 28.26. Let us perform an inversion with center A. The angles of interest to us are then equal (by Problem 28.5) to the respective angles between lines B∗C∗and B∗D∗or between 460 CHAPTER 28. INVERSION line C∗D∗and the circle circumscribed about triangle B∗C∗D∗. Both these angles are equal to a half arc ⌣C∗D∗. 28.27. Performing an inversion with center A we get three lines passing through B: lines S∗ 1 and S∗ 2 are tangent to S∗and S∗ 3 is perpendicular to it. Thus, line S∗ 3 passes through the center of S∗and is the bisector of the angle formed by S∗ 1 and S∗ 2. Therefore, circle S3 divides the angle between S1 and S2 in halves. 28.28. The condition of the types of tangency implies that after an inversion with center A we get either two circles inscribed into the same angle or a pair of vertical angles. In either case a homothety with center A turns circles S∗ 1 and S∗ 2 into each other. This homothety sends one segment that connects tangent points into another one. Hence, lines B∗ 1C∗ 1 and B∗ 2C∗ 2 are parallel and their images under the inversion are tangent at point A. 28.29. Let A1, B1 and C1 be the midpoints of sides BC, CA and AB, respectively. Let us prove that, for instance, the circle circumscribed about triangle A1B1C1 is tangent to the inscribed circle S and escribed circle Sa tangent to BC. Let points B′ and C′ be symmetric to B and C, respectively, through the bisector of angle ∠A (i.e., B′C′ is the second common inner tangent to S and Sa), let P and Q be the tangent points of circles S and Sa, respectively, with side BC and let D and E be the intersection points of lines A1B1 and A1C1, respectively, with line B′C′. By Problem 3.2 BQ = CP = p −c and, therefore, A1P = A1Q = 1 2\|b −c\|. It suﬃces to prove that the inversion with center A1 and degree A1P 2 sends points B1 and C1 into D and E, respectively, (this inversion sends circles S and Sa into themselves, and the circle circumscribed about triangle A1B1C1 into line B′C′). Let K be the midpoint of segment CC′. Point K lies on line A1B1 and A1K = BC′ 2 = \|b −c\| 2 = A1P. Moreover, A1D : A1K = BC′ : BA = A1K : A1B1, i.e., A1D · A1B1 = A1K2 = A1P 2. Similarly, A1E · A1C1 = A1P 2. 28.30. After an inversion with center at the intersection point of S1 and S2 we get lines l1, l2 and l intersecting at one point. Line l1 intersects circle S∗ 4 at points A and B, line l2 intersects S∗ 3 at points C and D and line l passes through the intersection points of these circles. Hence, points A, B, C, D lie on one circle (Problem 3.9). 28.31. Let us make an inversion with center at point A1. Then circles S1, S2 and S4 turn into lines A∗ 2D∗ 1, B∗ 1A∗ 2 and D∗ 1B∗ 1; circles S3 and S4 into circles S∗ 3 and S∗ 4 circumscribed about triangles B∗ 2C∗ 1B∗ 1 and C∗ 1D∗ 1D∗ 2, respectively (Fig. 114). Figure 256 (Sol. 28.31) SOLUTIONS 461 Let us draw the circle through points B∗ 2, D∗ 2 and A∗ 2. By Problem 2.80 a) it passes through the intersection point C∗ 2 of circles S∗ 3 and S∗ 4. Thus, points A∗ 2, B∗ 2, C∗ 2, D∗ 2 lie on one circle. It follows, that points A2, B2, C2, D2 lie on one circle or line. 28.32. Let P, Q, R, S, T be the intersection points of circles S1, S2, S3, S4, S5 spoken about in the formulation of the problem (cf. Fig. 108). Let us prove, for instance, that points P, Q, R, S lie on one circle. Let us draw circle Σ circumscribed about triangle NKD. Applying the result of Problem 2.83 a) (which coincides with that of Problem 19.45) to quadrilaterals AKDE and BNDC we see that circles S4, S5 and Σ intersect at one point (namely, P) and circles S2, S3, Σ also intersect at one point (namely, S). Therefore, circle Σ passes through points P and S. Now, observe that of eight intersection points of circles Σ, S1, S2, S5 four, namely, N, A, B, K, lie on one line. It follows that by Problem 28.31 the remaining four points P, Q, R, S lie on one circle. 28.33. An inversion with center at the intersection point of circumscribed circles of triangles A1A2B3, A1B2A3 and B1A2A3 sends these circles into lines and the statement of the problem reduces to the statement that the circles circumscribed about triangles B∗ 1B∗ 2A∗ 3, B∗ 1A∗ 2B∗ 3 and A∗ 1B∗ 2B∗ 3 pass through one point, i.e., the statement of Problem 2.80 a). 28.34. Under an inversion with center at the intersection point of circles circumscribed about triangles A1B1C1, A1B2C2, A2B1C2 and A2B2C1 we get four lines and four circles circumscribed about triangles formed by these lines. By Problem 2.83 a) these circles pass through one point. 28.35. a) Denote by Mij the intersection point of lines li and lj and by Sij the circle corresponding to the three remaining lines. Then point A1 is distinct from the intersection point M34 of circles S15 and S12. Repeating this argument for each point Ai, we see that thanks to Problem 28.32 they lie on one circle. b) Let us prove the statement of the problem by induction and considere separately the cases of even and odd n. Let n be odd. Denote by Ai the point corresponding to the tuple of n −1 lines obtained by deleting line li and by Aijk the point corresponding to the tuple of n given lines without li, lj and lk. Similarly, denote by Sij and Sijkm the circles corresponding to tuples of n −2 and n −4 lines obtained by deleting li and lj or li, lj, lk and lm, respectively. In order to prove that n points A1, A2, . . . , An lie on the same circle, it suﬃces to prove that any four of them lie on one circle. Let us prove this, for instance, for points A1, A2, A3 and A4. Since points Ai and Aijk lie on Sij, it follows that circles S12 and S23 intersect at points A2 and A123; circles S23 and S34 intersect at points A3 and A234; circles S34 and S41 at points A4 and A134; circles S41 and S12 at points A1 and A124. But points A123, A234, A134 and A124 lie on one circle — circle S1234 — hence, by Problem 28.31 points A1, A2, A3 and A4 lie on one circle. Let n be even. Let Si, Aij, Sijk, Aijkm be circles and points corresponding to tuples of n −1, n −2, n −3 and n −4 lines, respectively. In order to prove that circles S1, S2, . . . , Sn intersect at one point, let us prove that this holds for any three of them. (This suﬃces for n ≥5, cf. Problem 26.12.) Let us prove, for instance, that S1, S2 and S3 intersect at one point. By deﬁnition of points Aij and circles Si and Sijk, points A12, A13 and A14 lie on circle S1; points A12, A23 and A24 on S2; points A13, A14 and A34 on S3; points A12, A14 and A24 on S124; points A13, A14, A34 on S134; points A23, A24, A34 on S234. But the three circles S124, S134 and S234 pass through point A1234; hence, by Problem 28.33 circles S1, S2 and S3 also intersect at one point. 462 CHAPTER 28. INVERSION 28.36. a) Denote by Mij the intersection point of lines li and lj. Then point A1 corre-sponding to the triple (l2, l3, l4) is the intersection point of the circles circumscribed about triangles M2M3M23 and M3M4M34. By the similar arguments applied to A2, A3 and A4 we see that points A1, A2, A3 and A4 lie on one circle thanks to Problem 28.31 because points M1, M2, M3, M4 lie on one circle. b) As in Problem 28.35 b), let us prove our statement by induction; consider the cases of even and odd n separately. Let n be even; let Ai, Sij, Aijk and Sijkm denote points and circles corresponding to tuples of n −1, n −2, n −3 and n −4 lines, respectively. Let us prove that points A1, A2, A3, A4 lie on one circle. By deﬁnition of points Ai and Aijk, circles S12 and S23 intersect at points A2 and A123; circles S23 and S34 at points A3 and A234; circles S34 and S41 at points A4 and A134; circles S41 and S12 at points A1 and A124. Points A123, A234, A134 and A124 lie on circle S1234; hence, by Problem 28.31 points A1, A2, A3, A4 lie on one circle. We similarly prove that any four of points Ai (hence, all of them) lie on one circle. Proof for n odd, n ≥5, literally repeats the proof of heading b) of Problem 28.35 for the case of n even. 28.37. If circles R1 and R2 intersect or are tangent to each other, then an inversion with the center at their intersection point sends circles S1, S2, . . . , Sn into the circles that are tangent to a pair of straight lines and to each other at points A∗ 1, A∗ 2, . . . , A∗ n−1 lying on the bisector of the angle formed by lines R∗ 1 and R∗ 2 if R∗ 1 and R∗ 2 intersect, or on the line parallel to R∗ 1 and R∗ 2 if these lines do not intersect. Applying the inversion once again we see that points A∗ 1, A∗ 2, . . . , A∗ n−1 lie on one circle. If circles R1 and R2 do not intersect, then by Problem 28.6 there is an inversion sending them into a pair of concentric circles. In this case points A∗ 1, A∗ 2, . . . , A∗ n−1 lie on a circle concentric with R∗ 1 and R∗ 2; hence, points A1, A2, . . . , An−1 lie on one circle. 28.37. Let us make an inversion sending R1 and R2 into a pair of concentric circles. Then circles S∗ 1, S∗ 2, . . . , S∗ n and T ∗ 1 are equal (Fig. 115). Figure 257 (Sol. 28.37) Turning the chain of circles S∗ 1, . . . , S∗ n about the center of the circle R∗ 1 so that S∗ 1 becomes T ∗ 1 and making an inversion once again we get the desired chain T1, T2, . . . , Tn. 28.39. The center of inversion that sends circles R1 and R2 into concentric circles lies (see the solution of Problem 28.6) on the line that connects their centers. Therefore, making SOLUTIONS 463 this inversion and taking into account that the angle between circles, as well as the type of tangency, are preserved under an inversion, we reduce the proof to the case of concentric circles R1 and R2 with center O and radii r1 and r2, respectively. Let us draw circle S with center P and of radius 1 2(r1 −r2) tangent to R1 from the inside and to R2 from the outside and let us draw circles S′ and S′′ each of radius 1 2(r1 + r2) with centers A and B, respectively, tangent to R1 and R2 at their intersection points with line OP (Fig. 116). Figure 258 (Sol. 28.39) Let OM and ON be tangent to S drawn at O. Clearly, the chain of n circles tangent to R1 and R2 exists if and only if ∠MON = m 360◦ n . (In this case the circles of the chain run m times about the circle R2.) Therefore, it remains to prove that the angle between circles S′ and S′′ is equal to ∠MON. But the angle between S′ and S′′ is equal to the angle between their radii drawn to the intersection point C. Moreover, since PO = r1 −r1 −r2 2 = r1 + r2 2 = AC, PN = r1 −r2 2 = r1 −r1 + r2 2 = OA, ∠PNO = ∠AOC = 90◦, we have △ACO = △PON. Therefore, ∠ACB = 2∠ACO = 2∠PON = ∠NOM. 28.40. Let R1 and R2 be a pair of circles without common points. The remaining four circles constitute a chain and, therefore, by the preceding problem circles S′ and S′′ tangent to R1 and R2 at the intersection points of the latter with the line connecting their centers intersect at right angle (Fig. 117). If R2 lies inside R1, then the radii r′ and r′′ of circles S′ and S′′ are equal to 1 2(r1 + r2 + d) and 1 2(r1 + r2 −d), respectively, and the distance between their centers is equal to d′ = 2r1 −r1 −r2 = r1 −r2. The angle between S′ and S′′ is equal to the angle between the radii drawn to the intersection point, hence, (d′)2 = (r′)2 + (r′′)2 or, after simpliﬁcation, d2 = r2 1 + r2 2 −6r1r2. If R1 and R2 are not inside one another, then the radii of S′ and S′′ are equal to 1 2(d + (r1 −r2)) and 1 2(d −(r1 −r2)), respectively, and the distance between their centers is d′ = r1 + r2 + d −(r′ 1 + r′ 2) = r1 + r2. As a result we get d2 = r2 1 + r2 2 + 6r1r2. 464 CHAPTER 28. INVERSION Figure 259 (Sol. 28.40) Chapter 29. AFFINE TRANSFORMATIONS §1. Aﬃne transformations A transformation of the plane is called an aﬃne one if it is continuous, one-to-one, and the image of every line is a line. Shifts and similarity transformations are particular cases of aﬃne transformations. A dilation of the plane relative axis l with coeﬃcient k is a transformation of the plane under which point M turns into point M ′ such that − − → OM ′ = k− − → OM, where O is the projection of M to l. (A dilation with coeﬃcient smaller than 1 is called a contraction.) 29.1. Prove that a dilation of the plane is an aﬃne transformation. 29.2. Prove that under an aﬃne transformation parallel lines turn into parallel ones. 29.3. Let A1, B1, C1, D1 be images of points A, B, C, D, respectively, under an aﬃne transformation. Prove that if − → AB = − − → CD, then − − − → A1B1 = − − − → C1D1. Problem 29.3 implies that we can deﬁne the image of vector − → AB under an aﬃne trans-formation L as − − − − − − − → L(A)L(B) and this deﬁnition does not depend on the choice of points A and B that determine equal vectors. 29.4. Prove that if L is an aﬃne transformation, then a) L(− → 0 ) = − → 0 ; b) L(a + b) = L(a) + L(b); c) L(ka) = kL(a). 29.5. Let A′, B′, C′ be images of points A, B, C under an aﬃne transformation L. Prove that if C divides segment AB in the ratio AC : CB = p : q, then C′ divides segment A′B′ in the same ratio. 29.6. Given two points O and O′ in plane and two bases {e1, e2} and {e′ 1, e′ 2}. a) Prove that there exists a unique aﬃne transformation that sends O into O′ and a the basis {e1, e2} into the basis {e′ 1, e′ 2}. b) Given two triangles ABC and A1B1C1 prove that there exists a unique aﬃne trans-formation that sends A into A1, B into B1 and C into C1. c) Given two parallelograms, prove that there exists a unique aﬃne transformation that sends one of them into another one. 29.7. Prove that if a non-identity aﬃne transformation L sends each point of line l into itself, then all the lines of the form ML(M), where M is an arbitrary point not on l, are parallel to each other. 29.8. Prove that any aﬃne transformation can be represented as a composition of two dilations and an aﬃne transformation that sends any triangle into a similar triangle. 29.9. Prove that any aﬃne transformation can be represented as a composition of a dilation (contraction) and an aﬃne transformation that sends any triangle into a similar triangle. 29.10. Prove that if an aﬃne transformation sends a circle into itself, then it is either a rotation or a symmetry. 465 466 CHAPTER 29. AFFINE TRANSFORMATIONS 29.11. Prove that if M ′ and N ′ are the images of polygons M and N, respectively, under an aﬃne transformation, then the ratio of areas of M and N is equal to the ratio of areas of M ′ and N ′. §2. How to solve problems with the help of aﬃne transformations 29.12. Through every vertex of a triangle two lines are drawn. The lines divide the opposite side of the triangle into three equal parts. Prove that the diagonals connecting opposite vertices of the hexagon formed by these lines intersect at one point. 29.13. On sides AB, BC and CD of parallelogram ABCD points K, L and M, respec-tively, are taken. The points divide the sides in the same ratio. Let b, c, d be lines passing through points B, C, D parallel to lines KL, KM, ML, respectively. Prove that lines b, c, d pass through one point. 29.14. Given triangle ABC, let O be the intersection point of its medians and M, N and P be points on sides AB, BC and CA, respectively, that divide these sides in the same ratio (i.e., AM : MB = BN : NC = CP : PA = p : q). Prove that: a) O is the intersection point of the medians of triangle MNP; b) O is the intersection point of the medians of the triangle formed by lines AN, BP and CM. 29.15. In trapezoid ABCD with bases AD and BC, a line is drawn through point B parallel to side CD and intersecting diagonal AC at point P; through point C a line is drawn parallel to AB and intersecting diagonal BD at Q. Prove that PQ is parallel to the bases of the trapezoid. 29.16. In parallelogram ABCD, points A1, B1, C1, D1 lie on sides AB, BC, CD, DA, respectively. On sides A1B1, B1C1, C1D1, D1A1 of quadrilateral A1B1C1D1 points A2, B2, C2, D2, respectively, are taken. It is known that AA1 BA1 = BB1 CB1 = CC1 DC1 = DD1 AD = AD2 D1D2 = D1C2 C1C2 = C1B2 B1B2 = B1A2 A1A2 . Prove that A2B2C2D2 is a parallelogram with sides parallel to the sides of ABCD. 29.17. On sides AB, BC and AC of triangle ABC, points M, N and P, respectively, are taken. Prove that: a) if points M1, N1 and P1 are symmetric to points M, N and P through the midpoints of the corresponding sides, then SMNP = SM1N1P1. b) if M1, N1 and P1 are points on sides AC, BA and CB, respectively, such that MM1 ∥ BC, NN1 ∥CA and PP1 ∥AB, then SMNP = SM1N1P1. Solutions 29.1. We have to prove that if A′, B′, C′ are images of points A, B, C under the dilation with respect to line l with coeﬃcient k and point C lies on line AB, then point C′ lies on line A′B′. Let − → AC = t− → AB. Denote by A1, B1, C1 the projections of points A, B, C, respectively, on line l and let a = − − → A1A, b = − − → B1B, c = − − → C1C, a′ = − − → A1A′, b′ = − − − → B1B′, c′ = − − → C1C′, x = − − − → A1B1, y = − − − → A1C1. SOLUTIONS 467 Since the ratio of lengths of proportional vectors under the projection on line l is preserved, then y = tx and y + (c −a) = t(y + (b −a)). By subtracting the ﬁrst equality from the second one we get (c −a) = t(b −a). By deﬁnition of a dilation a′ = ka, b′ = kb, c′ = kc; hence, − − → A′C′ = y + k(c −a) = tx + k(t(b −a)) = t(x + k(b −a)) = t− − → A′B′. 29.2. By deﬁnition, the images of lines are lines and from the property of an aﬃne transformation to be one-to-one it follows that the images of nonintersecting lines do not intersect. 29.3. Let − → AB = − − → CD. First, consider the case when points A, B, C, D do not lie on one line. Then ABCD is a parallelogram. The preceding problem implies that A1B1C1D1 is also a parallelogram; hence, − − − → A1B1 = − − − → C1D1. Now, let points A, B, C, D lie on one line. Take points E and F that do not lie on this line and such that − → EF = − → AB. Let E1 and F1 be their images. Then − − − → A1B1 = − − − → E1F1 = − − − → C1D1. 29.4. a) L(− → 0 ) = L(− → AA) = − − − − − − → L(A)L(A) = − → 0 . b) L(− → AB + − − → BC) = L(− → AC) = − − − − − − − → L(A)L(C) = − − − − − − − → L(A)L(B) + − − − − − − − → L(B)L(C) = L(− → AB) + L(− − → BC). c) First, suppose k is an integer. Then L(ka) = L(a + · · · + a) = L(a) + · · · + L(a) = kL(a). Now, let k = m n be a rational number. Then nL(ka) = L(nka) = L(ma) = mL(a); hence, L(ka) = mL(a) n = kL(a). Finally, if k is an irrational number, then there always exists a sequence kn (n ∈N) of rational numbers tending to k (for instance, the sequence of decimal approximations of k). Since L is continuous, L(ka) = L( lim n→∞kna) = lim n→∞knL(a) = kL(a). 29.5. By Problem 29.4 c) the condition q− → AC = p− − → CB implies that q− − → A′C′ = qL(− → AC) = L(q− → AC) = L(p− − → CB) = pL(− − → CB) = p− − → C′B′. 29.6. a) Deﬁne the map L as follows. Let X be an arbitrary point. Since e1, e2 is a basis, it follows that there exist the uniquely determined numbers x1 and x2 such that − − → OX = x1e1 + x2e2. Assign to X point X′ = L(X) such that − − − → O′X′ = x1e′ 1 + x2e′ 2. Since e′ 1, e′ 2 is also a basis, the obtained map is one-to-one. (The inverse map is similarly constructed.) Let us prove that the image of any line AB under L is a line. Let A′ = L(A), B′ = L(B); let a1, a2, and b1, b2 be the coordinates of points A and B, respectively, in the basis e1, e2, i.e., − → OA = a1e1 + a2e2, − − → OB = b1e1 + b2e2. Let us consider an arbitrary point C on line AB. 468 CHAPTER 29. AFFINE TRANSFORMATIONS Then − → AC = k− → AB for some k, i.e., − → OC = − → OA + k(− − → OB −− → OA) = ((1 −k)a1 + kb1)e1 + ((1 −k)a2 + kb2)e2. Hence, if C′ = L(C), then − − → O′C′ = ((1 −k)a1 + kb1)e′ 1 + ((1 −k)a2 + kb2)e′ 2 = − − → O′A′ + k(− − → O′B′ −− − → O′A′), i.e., point C′ lies on line A′B′. The uniqueness of L follows from the result of Problem 29.4. Indeed, L(− − → OX) = x1L(e1)+ x2L(e2), i.e., the image of X is uniquely determined by the images of vectors e1, e2 and point O. b) To prove it, it suﬃces to make use of the previous heading setting O = A, e1 = − → AB, e2 = − → AC, O′ = A1, e′ 1 = − − − → A1B1, e′ 2 = − − − → A1C1. c) Follows from heading b) and the fact that parallel lines turn into parallel lines. 29.7. Let M and N be arbitrary points not on line l. Denote by M0 and N0 their projections to l and by M ′ and N ′ the images of M and N under L. Lines M0M and N0N are parallel because both of them are perpendicular to l, i.e., there exists a number k such that − − − → M0M = k− − → N0N. Then by Problem 29.4 c) − − − → M0M ′ = k− − − → N0N ′. Hence, the image of triangle M0MM ′ under the parallel translation by vector − − − → M0N0 is homothetic with coeﬃcient k to triangle N0NN ′ and, therefore, lines MM ′ and NN ′ are parallel. 29.8. Since an aﬃne map is uniquely determined by the images of vertices of any ﬁxed triangle (see Problem 29.6 b)), it suﬃces to prove that with the help of two dilations one can get from any triangle an arbitrary triangle similar to any before given one, for instance, to an isosceles right triangle. Let us prove this. Let ABC be an arbitrary triangle, BN the bisector of the outer angle ∠B adjacent to side BC. Then under the dilation with respect to BN with coeﬃcient tan 45◦ tan ∠CBN we get from triangle ABC triangle A′B′C′ with right angle ∠B′. With the help of a dilation with respect to one of the legs of a right triangle one can always get from this triangle an isosceles right triangle. 29.9. Let L be a given aﬃne transformation, O an arbitrary point, T the shift by vector − − − − → L(O)O and L1 = T◦L. Then O is a ﬁxed point of L1. Among the points of the unit circle with center O, select a point A for which the vector L(− → OA) is the longest. Let H be a rotational homothety with center O that sends point L1(A) into A and let L2 = H ◦L1 = H ◦T ◦L. Then L2 is an aﬃne transformation that preserves points O and A; hence, by Problem 29.4 c) it preserves all the other points of line OA and thanks to the choice of point A for all points M we have \|− − → OM\| ≥\|L(− − → OM)\|. Let us prove (which will imply the statement of the problem) that L2 is a contraction with respect to line OA. If L2 is the identity transformation, then it is a contraction with coeﬃcient 1, so let us assume that L2 is not the identity. By Problem 29.9 all the lines of the form − − − − − − → ML2(M), where M is an arbitrary point not on OA, are parallel to each other. Let − − → OB be the unit vector perpendicular to all these lines. SOLUTIONS 469 Then B is a ﬁxed point of L2 because otherwise we would have had \|− − − − − → OL2(B)\| = p OB2 + BL2(B)2 > \|OB\|. If B does not lie on line OA, then by Problem 29.6 b) transformation L2 is the identity. If B lies on OA, then all the lines of the form ML2(M) are perpendicular to the ﬁxed line of transformation L2. With the help of Problem 29.4 c) it is not diﬃcult to show that the map with such a property is either a dilation or a contraction. 29.10. First, let us prove that an aﬃne transformation L that sends a given circle into itself sends diametrically opposite points into diametrically opposite ones. To this end let us notice that the tangent to the circle at point A turns into the line that, thanks to the property of L to be one-to-one, intersects with the circle at a (uniquely determined) point L(A), i.e., is the tangent at point L(A). Therefore, if the tangents at points A and B are parallel to each other (i.e., AB is a diameter), then the tangents at points L(A) and L(B) are also parallel, i.e., L(A)L(B) is also a diameter. Fix a diameter AB of the given circle. Since L(A)L(B) is also a diameter, there exists a movement P of the plane which is either a rotation or a symmetry that sends A and B into L(A) and L(B), respectively, and each of the arcs α and β into which points A and B divide the given circle into the image of these arcs under L. Let us prove that the map F = P −1◦L is the identity. Indeed, F(A) = A and F(B) = B; hence, all points of line AB are ﬁxed. Hence, if X is an arbitrary point of the circle, then the tangent at X intersects line AB at the same place where the tangent at point X′ = F(X) does because the intersection point is ﬁxed. Since X and X′ lie on one and the same of the two arcs α or β, it follows that X coincides with X′. Thus, P −1 ◦L = E, i.e., L = P. 29.11. Let a1 and a2 be two perpendicular lines. Since an aﬃne transformation pre-serves the ratio of the lengths of (the segments of the) parallel lines, the lengths of all the segments parallel to one line are multiplied by the same coeﬃcient. Denote by k1 and k2 these coeﬃcients for lines a1 and a2. Let ϕ be the angle between the images of these lines. Let us prove that the given aﬃne transformation multiplies the areas of all polygons by k, where k = k1k2 sin ϕ. For rectangles with sides parallel to a1 and a2 and also for a right triangle with legs parallel to a1 and a2 the statement is obvious. Any other triangle can be obtained by cuttting oﬀthe rectangle with sides parallel to a1 and a2 several right triangles with legs parallel to a1 and a2 as shown on Fig. 118 and, ﬁnally, by Problem 22.22 any polygon can be cut into triangles. Figure 260 (Sol. 29.11) 29.12. Since an aﬃne transformation sends an arbitrary triangle into an equilateral one (Problem 29.6 b)), the ratio of lengths of parallel segments are preserved (Problem 29.5). It suﬃces to prove the statement of the problem for an equilateral triangle ABC. Let points A1, A2, B1, B2, C1, C2 divide the sides of the triangle into equal parts and A′, B′, C′ be the midpoints of the sides (Fig. 119). Under the symmetry through AA′ line BB1 turns into CC2 and BB2 into CC1. Since symmetric lines intersect on the axis of symmetry, AA′ contains a diagonal of the considered hexagon. Similarly, the remaining diagonals lie on BB′ and CC′. It is clear that the medians AA′, BB′, CC′ intersect at one point. 470 CHAPTER 29. AFFINE TRANSFORMATIONS Figure 161 (Sol. 29.12) 29.13. Problem 29.6 b) implies that an aﬃne transformation sends an arbitrary parallel-ogram into a square. Since this preserves the ratio of lengths of parallel segments (Problem 29.5), it suﬃces to prove the statement of the problem for the case when ABCD is a square. Denote by P the intersection point of lines b and d. It suﬃces to prove that PC ∥MK. Seg-ment KL turns under the rotation through the angle of 90◦about the center of square ABCD into LM, hence, lines b and d which are parallel to these respective segments are perpendic-ular; hence, P lies on the circle circumscribed about ABCD. Then ∠CPD = ∠CBD = 45◦. Therefore, the angle between lines CP and b is equal to 45◦but the angle between lines MK and KL is also equal to 45◦and b ∥KL implying CP ∥MK. 29.14. a) Let us consider an aﬃne transformation that sends triangle ABC into a equilateral triangle A′B′C′. Let O′, M ′, N ′, P ′ be the images of points O, M, N, P. Under the rotation through the angle of 120◦about point O′ triangle M ′N ′P ′ turns into itself and, therefore, this triangle is a equilateral one and O′ is the intersection point of its medians. Since under an aﬃne transformation any median turns into a median, O is the intersection point of the medians of triangle MNP. b) Solution is similar to the solution of heading a). 29.15. Let us consider an aﬃne transformation that sends ABCD into an isosceles trapezoid A′B′C′D′. For such a transformation one can take the aﬃne transformation that sends triangle ADE, where E is the intersection point of AB and CD, into an isosceles triangle. Then the symmetry through the midperpendicular to A′D′ sends point P ′ into point Q′, i.e., lines P ′Q′ and A′D′ are parallel. 29.16. Any parallelogram ABCD can be translated by an aﬃne transformation into a square (for this we only have to transform triangle ABC into an isosceles right triangle). Since the problem only deals with parallel lines and ratios of segments that lie on one line, we may assume that ABCD is a square. Let us consider a rotation through an angle of 90◦sending ABCD into itself. This rotation sends quadrilaterals A1B1C1D1 and A2B2C2D2 into themselves; hence, the quadrilaterals are also squares. We also have tan ∠BA1B1 = BB1 : BA1 = A1D2 : A1A2 = tan ∠A1A2D2, i.e., AB ∥A2D2 (Fig. 120). 29.17. a) Since an aﬃne transformation sends any triangle into a equilateral one, the midpoints of the sides into the midpoints, the centrally symmetric points into centrally symmetric and triangles of the same area into triangles of the same area (Problem 29.11), it follows that we can assume that triangle ABC is an equilateral one with side a. Denote SOLUTIONS 471 Figure 262 (Sol. 29.16) the lengths of segments AM, BN, CP by p, q, r, respectively. Then SABC −SMNP = SAMP + SBMN + SCNP = 1 2 sin 60◦· (p(a −r) + q(a −p) + r(a −q)) = 1 2 sin 60◦· (a(p + q + r) −(pq + qr + rp)). Similarly, SABC −SM1N1P1 = 1 2 sin 60◦· (r(a −p) + p(a −q) + q(a −r)) = 1 2 sin 60◦· (a(p + q + r) −(pq + qr + rp)). b) By the same reasons as in heading a) let us assume that ABC is an equilateral triangle. Let M2N2P2 be the image of triangle M1N1P1 under the rotation about the center of triangle ABC through the angle of 120◦in the direction from A to B (Fig. 121). Figure 263 (Sol. 29.17) Then AM2 = CM1 = BM. Similarly, BN2 = CN and CP2 = AP, i.e., points M2, N2, P2 are symmetric to points M, N, P through the midpoints of the corresponding sides. Therefore, this heading is reduced to heading a). Chapter 30. PROJECTIVE TRANSFORMATIONS §1. Projective transformations of the line 1. Let l1 and l2 be two lines on the plane, O a point that does not lie on any of these lines. The central projection of line l1 to line l2 with center O is the map that to point A1 on line l1 assigns the intersection point of lines OA1 and l2. 2. Let l1 and l2 be two lines on the plane, l a line not parallel to either of the lines. The parallel projection of l1 to l2 along l is the map that to point A1 of line l1 assigns the intersection point of l2 with the line passing through A1 parallel to l. 3. A map P of line a to line b is called a projective one if it is the composition of central or parallel projections, i.e., if there exist lines a0 = a, a1, . . . , an = b and maps Pi of the line ai to ai+1 each of which is either a central or a parallel projection and P is the composition of the maps Pi in some order. If b coincides with a, then P is called a projective transformation of line a. 30.1. Prove that there exists a projective transformation that sends three given points on one line into three given points on another line. The cross ratio of a quadruple of points A, B, C, D lying on one line is the number (ABCD) = c −a c −b : d −a d −b, where a, b, c, d are the coordinates of points A, B, C, D, respectively. It is easy to verify that the cross ratio does not depend on the choice of the coordinate system on the line. We will also write (ABCD) = AC BC : AD BD in the sence that AC BC (resp. AD BD) denotes the ratio of the lengths of these segments, if vectors − → AC and − − → BC (resp. − − → AD and − − → BD) are similarly directed or the ratio of the lengths of these segments taken with minus sign, if these vectors are pointed in the opposite directions. The double ratio of the quadruple of lines a, b, c, d passing through one point is the number (abcd) = ±sin(a, c) sin(b, c) : sin(a, d) sin(b, d) whose sign is determined as follows: if one of the angles formed by lines a and b does not intersect with one of the lines c or d (in this case we say that the pair of lines a and b does not divide the pair of lines c and d) then (abcd) > 0; otherwise (abcd) < 0. 30.2. a) Given lines a, b, c, d passing through one point and line l that does not pass through this point. Let A, B, C, D be intersection points of l with lines a, b, c, d, respectively. Prove that (abcd) = (ABCD). b) Prove that the double ratio of the quadruple of points is preserved under projective transformations. 30.3. Prove that if (ABCX) = (ABCY ), then X = Y (all points are assumed to be pairwise distinct except, perhaps, points X and Y , and lie on one line). 473 474 CHAPTER 30. PROJECTIVE TRANSFORMATIONS 30.4. Prove that any projective transformation of the line is uniquely determined by the image of three arbitrary points. 30.5. Prove that any non-identity projective transformation of the line has not more than two ﬁxed points. 30.6. A map sends line a into line b and preserves the double ratio of any quadruple of points. Prove that this map is a projective one. 30.7. Prove that transformation P of the real line is projective if and only if it can be represented in the form P(x) = ax + b cx + d, where a, b, c, d are numbers such that ad −bc ̸= 0. (Such maps are called fractionally-linear ones.) 30.8. Points A, B, C, D lie on one line. Prove that if (ABCD) = 1, then either A = B or C = D. 30.9. Given line l, a circle and points M, N that lie on the circle and do not lie on l. Consider map P of line l to itself,; let P be the composition of the projection of l to the given circle from point M and the projection of the circle to l from point N. (If point X lies on line l, then P(X) is the intersection of line NY with line l, where Y is the distinct from M intersection point of line MX with the given circle.) Prove that P is a projective transformation. 30.10. Given line l, a circle and point M that lies on the circle and does not lie on l, let PM be the projection map of l to the given circle from point M (point X of line l is mapped into the distinct from M intersection point of line XM with the circle), R the movement of the plane that preserves the given circle (i.e., a rotation of the plane about the center of the circle or the symmetry through a diameter). Prove that the composition P −1 M ◦R ◦PM is a projective transformation. Remark. If we assume that the given circle is identiﬁed with line l via a projection map from point M, then the statement of the problem can be reformulated as follows: the map of a circle to itself with the help of a movement of the plane is a projective transformation of the line. §2. Projective transformations of the plane Let α1 and α2 be two planes in space, O a point that does not belong to any of these planes. The central projection map of α1 to α2 with center O is the map that to point A1 of plane α1 assigns the intersection point of OA1 with plane α2. 30.11. Prove that if planes α1 and α2 intersect, then the central projection map of α1 to α2 with center O determines a one-to-one correspondence of plane α1 with deleted line l1 onto plane α2 with deleted line l2, where l1 and l2 are the intersection lines of planes α1 and α2, respectively, with planes passing through O and parallel to α1 and α2. On l1, the map is not deﬁned. A line on which the central projection map is not deﬁned is called the singular line of the given projection map. 30.12. Prove that under a central projection a nonsingular line is projected to a line. In order to deﬁne a central projection everywhere it is convenient to assume that in addition to ordinary points every line has one more so-called inﬁnite point sometimes denoted by ∞. If two points are parallel, then we assume that their inﬁnite points coincide; in other words, parallel lines intersect at their inﬁnite point. §2. PROJECTIVE TRANSFORMATIONS OF THE PLANE 475 We will also assume that on every plane in addition to ordinary lines there is one more, inﬁnite line, which hosts all the inﬁnite points of the lines of the plane. The inﬁnite line intersects with every ordinary line l lying in the same plane in the inﬁnite point of l. If we introduce inﬁnite points and lines, then the central projection map of plane α1 to plane α2 with center at point O is deﬁned through(?) points of α1 and the singular line is mapped into the inﬁnite line of α2, namely, the image of point M of the singular line is the inﬁnite point of line OM; this is the point at which the lines of plane α2 parallel to OM intersect. 30.13. Prove that if together with the usual (ﬁnite) points and lines we consider inﬁnite ones, then a) through any two points only one line passes; b) any two lines lying in one plane intersect at one point; c) a central projection map of one plane to another one is a one-to-one correspondence. A map P of plane α to plane β is called a projective one if it is the composition of central projections and aﬃne transformations, i.e., if there exist planes α0 = α, α1, . . . , αn = β and maps Pi of plane αi to αi+1 each of which is either a central projection or an aﬃne transformation and P is the composition of the Pi. If plane α coincides with β, map P is called a projective transformation of α. The preimage of the inﬁnite line will be called the singular line of the given projective transformation. 30.14. a) Prove that a projective transformation P of the plane sending the inﬁnite line into the inﬁnite line is an aﬃne transformation. b) Prove that if points A, B, C, D lie on a line parallel to the singular line of a projective transformation P of plane α, then P(A)P(B) : P(C)P(D) = AB : CD. c) Prove that if a projective transformation P sends parallel lines l1 and l2 into parallel lines, then either P is aﬃne or its singular line is parallel to l1 and l2. d) Let P be a one-to-one transformation of the set of all ﬁnite and inﬁnite points of the plane, let P send every line into a line. Prove that P is a projective map. 30.15. Given points A, B, C, D no three of which lie on one line and points A1, B1, C1, D1 with the same property. a) Prove that there exists a projective transformation sending points A, B, C, D to points A1, B1, C1, D1, respectively. b) Prove that the transformation from heading a) is unique, i.e., any projective trans-formation of the plane is determined by the images of four generic points (cf. Problem 30.4). c) Prove statement of heading a) if points A, B, C lie on one line l and points A1, B1, C1, D1 on one line l1. d) Is transformation from heading c) unique? In space, consider the unit sphere with center in the origin. Let N(0, 0, 1) be the sphere’s north pole. The stereographic projection of the sphere to the plane is the map that to every point M of the sphere assigns distinct from N intersection point of line MN with plane Oxy. It is known (see, for example, Solid Problem 16.19 b)) that the stereographic projection sends a circle on the sphere into a circle in plane. Make use of this fact while solving the following two problems: 30.16. Givin a circle and a point inside it. a) Prove that there exists a projective transformation that sends the given circle into a circle and the given point into the center of the given circle’s image. 476 CHAPTER 30. PROJECTIVE TRANSFORMATIONS b) Prove that if a projective transformation sends the given circle into a circle and point M into the center of the given circle’s image, then the singular line of this transformation is perpendicular to a diameter through M. 30.17. In plane, there are given a circle and a line that does not intersect the circle. Prove that there exists a projective transformation sending the given circle into a circle and the given line into the inﬁnite line. 30.18. Given a circle and a chord in it. Prove that there exists a projective transfor-mation that sends the given circle into a circle and the given chord into the diameter of the given circle’s image. 30.19. Given circle S and point O inside it, consider all the projective maps that send S into a circle and O into the center of the image of S. Prove that all such transformations map one and the same line into the inﬁnite line. The preimage of the inﬁnite line under the above transformations is called the polar line of point O relative circle S. 30.20. A projective transformation sends a circle into itself so that its center is ﬁxed. Prove that this transformation is either a rotation or a symmetry. 30.21. Given point O and two parallel lines a and b. For every point M we perform the following construction. Through M draw a line l not passing through O and intersecting lines a and b. Denote the intersection points pf l with a and b by A and B, respectively, and let M ′ be the intersection point of OM with the line parallel to OB and passing through A. a) Prove that point M ′ does not depend on the choice of line l. b) Prove that the transformation of the plane sending M into M ′ is a projective one. 30.22. Prove that the transformation of the coordinate plane that every point with coordinates (x, y) sends into the point with coordinates ( 1 x, y x) is a projective one. (?)30.23. Let O be the center of a lens, π a plane passing through the optic axis a of the lens, a and f the intersection lines of π with the plane of the lens and the focal plane, respectively, (a ∥f). In the school course of physics it is shown that if we neglect the lens, then the image M ′ of point M that lies in plane π is constructed as follows, see Fig. 122. Figure 264 (30.23) Through point M draw an arbitrary line l; let A be the intersection point of lines a and l, let B be the intersection point of f with the line passing through O parallel to l. Then M ′ is deﬁned as the intersection point of lines AB and OM. Prove that the transformation of plane π assigning to every of its points its image is a projective one. §3. LET US TRANSFORM THE GIVEN LINE INTO THE INFINITE ONE 477 Thus, through a magnifying glass we can see the image of our world thanks to projective transformations. §3. Let us transform the given line into the inﬁnite one 30.24. Prove that the locus of the intersection points of quadrilaterals ABCD whose sides AB and CD belong to two given lines l1 and l2 and sides BC and AD intersect at a given point P is a line passing through the intersection point Q of lines l1 and l2. 30.25. Let O be the intersection point of the diagonals of quadrilateral ABCD; let E (resp. F) be the intersection point of the continuations of sides AB and CD (resp. BC and AD). Line EO intersects sides AD and BC at points K and L, respectively, and line FO intersects sides AB and CD at points M and N, respectively. Prove that the intersection point X of lines KN and LM lies on line EF. 30.26. Lines a, b, c intersect at one point O. In triangles A1B1C1 and A2B2C2, vertices A1 and A2 lie on line a; B1 and B2 lie on line b; C1 and C2 lie on line c. Let A, B, C be the intersection points of lines B1C1 and B2C2, C1A1 and C2A2, A1B1 and A2B2, respectively. Prove that points A, B, C lie on one line (Desargue’s theorem.) 30.27. Points A, B, C lie on line l and points A1, B1, C1 on line l1. Prove that the intersection points of lines AB1 and BA1, BC1 and CB1, CA1 and AC1 lie on one line (Pappus’s theorem.) 30.28. Given convex quadrilateral ABCD. Let P, Q be the intersection points of the continuations of the opposite sides AB and CD, AD and BC, respectively, R an arbitrary point inside the quadrilateral. Let K, L, M be the intersection point of lines BC and PR, AB and QR, AK and DR, respectively. Prove that points L, M and C lie on one line. 30.29. Given two triangles ABC and A1B1C1 so that lines AA1, BB1 and CC1 intersect at one point O and lines AB1, BC1 and CA1 intersect at one point O1. Prove that lines AC1, BA1 and CB1 also intersect at one point O2. (Theorem on doubly perspective triangles.) 30.30. Given two triangles ABC and A1B1C1 so that lines AA1, BB1 and CC1 intersect at one point O, lines AA1, BC1 and CB1 intersect at one point O1 and lines AC1, BB1 and CA1 intersect at one point O2, prove that lines AB1, BA1 and CC1 also intersect at one point O3. (Theorem on triply perspective triangles.) 30.31. Prove that the orthocenters of four triangles formed by four lines lie on one line. 30.32. Given quadrilateral ABCD and line l. Denote by P, Q, R the intersection points of lines AB and CD, AC and BD, BC and AD, respectively. Denote by P1, Q1, R1 the midpoints of the segments which these pairs of lines cut oﬀline l. Prove that lines PP1, QQ1 and RR1 intersect at one point. 30.33. Given triangle ABC and line l. Denote by A1, B1, C1 the midpoints of the segments cut oﬀline l by angles ∠A, ∠B, ∠C and by A2, B2, C2 the intersection points of lines AA1 and BC, BB1 and AC, CC1 and AB, respectively. Prove that points A2, B2, C2 lie on one line. 30.34. (Theorem on a complete quadrilateral.) Given four points A, B, C, D and the intersection points P, Q, R of lines AB and CD, AD and BC, AC and BD, respectively; the intersection points K and L of line QR with lines AB and CD, respectively. Prove that (QRKL) = −1. 30.35. Is it possible to paint 1991 points of the plane red and 1991 points blue so that any line passing through two points of distinct colour contains one more of coloured points? (We assume that coloured points are distinct and do not belong to one line.) 478 CHAPTER 30. PROJECTIVE TRANSFORMATIONS §4. Application of projective maps that preserve a circle The main tools in the solution of problems of this section are the results of Problems 30.16 and 30.17. 30.36. Prove that the lines that connect the opposite tangent points of a circumscribed quadrilateral pass through the intersection point of the diagonals of this quadrilateral. 30.37. Consider a triangle and the inscribed circle. Prove that the lines that connect the triangle’s vertices with the tangent points of the opposite sides intersect at one point. 30.38. a) Through point P all secants of circle S are drawn. Find the locus of the intersection points of the tangents to S drawn through the two intersection points of S with every secant. b) Through point P the secants AB and CD of circle S are drawn, where A, B, C, D are the intersection points of the secants with the circle. Find the locus of the intersection points of AC and BD. 30.39. Given circle S, line l, point M on S and not on l and point O not on S. Consider a map P of line l which is the composition of the projection map of l to S from M, of S to itself from O and S to l from M, i.e., for any point A point P(A) is the intersection point of lines l and MC, where C is the distinct from B intersection point of S with line OB and B is the distinct from A intersection point of S with line MA. Prove that P is a projective map. Remark. If we assume that a projection map from point M identiﬁes circle S with line l, then the statement of the problem can be reformulated as follows: every central projection of a circle to itself is a projective transformation. 30.40. Consider disk S, point P outside S and line l passing through P and intersecting the circle at points A and B. Denote the intersection point of the tangents to the disk at points A and B by K. a) Consider all the lines passing through P and intersecting AK and BK at points M and N, respectively. Prove that the locus of the tangents to S drawn through M and N and distinct from AK and BK is a line passing through K and having the empty intersection with the interior of S. b) Let us select various points R on the circle and draw the line that connects the distinct from R intersection points of lines RK and RP with S. Prove that all the obtained lines pass through one point and this point belongs to l. 30.41. An escribed circle of triangle ABC is tangent to side BC at point D and to the extensions of sides AB and AC at points E and F, respectively. Let T be the intersection point of lines BF and CE. Prove that points A, D and T lie on one line. 30.42. Let ABCDEF be a circumscribed hexagon. Prove that its diagonals AD, BE and CF intersect at one point. (Brianchon’s theorem.) 30.43. Hexagon ABCDEF is inscribed in circle S. Prove that the intersection points of lines AB and DE, BC and EF, CD and FA lie on one line. (Pascal’s theorem.) 30.44. Let O be the midpoint of chord AB of circle S, let MN and PQ be arbitrary chords through O such that points P and N lie on one side of AB; let E and F be the intersection points of chord AB with chords MP and NQ, respectively. Prove that O is the midpoint of segment EF. (The butterﬂy problem.) 30.45. Points A, B, C and D lie on a circle, SA and SD are tangents to this circle, P and Q are the intersection points of lines AB and CD, AC and BD, respectively. Prove that points P, Q and S lie on line line. §6. APPLICATION OF PROJECTIVE TRANSFORMATIONS: CONT. 479 §5. Application of projective transformations of the line 30.46. On side AB of quadrilateral ABCD point M1 is taken. Let M2 be the projection of M1 to line BC from D, let M3 be the projection of M2 to CD from A, M4 the projection of M3 on DA from B, M5 the projection of M4 to AB from C, etc. Prove that M13 = M1 (hence, M14 = M2, M15 = M3, etc.). 30.47. Making use of projective transformations of the line prove the theorem on a complete quadrilateral (Problem 30.34). 30.48. Making use of projective transformations of the line prove Pappus’s theorem (Problem 30.27). 30.49. Making use of projective transformations of the line prove the butterﬂy problem (Problem 30.44). 30.50. Points A, B, C, D, E, F lie on one circle. Prove that the intersection points of lines AB and DE, BC and EF, CD and FA lie on one line. (Pascal’s theorem.) 30.51. Given triangle ABC and point T, let P and Q be the bases of perpendiculars dropped from point T to lines AB and AC, respectively; let R and S be the bases of perpendiculars dropped from point A to lines TC and TB, respectively. Prove that the intersection point X of lines PR and QS lies on line BC. §6. Application of projective transformations of the line in problems on construction 30.52. Given a circle, a line, and points A, A′, B, B′, C, C′, M on this line. By Problems 30.1 and 30.3 there exists a unique projective transformation of the given line to itself that maps points A, B, C into A′, B′, C′, respectively. Denote this transformation by P. Construct with the help of a ruler only a) point P(M); b) ﬁxed points of map P. (J. Steiner’s problem.) The problem of constructing ﬁxed points of a projective transformation is the key one for this section in the sense that all the other problems can be reduced to it, cf. also remarks after Problems 30.10 and 30.39. 30.53. Given two lines l1 and l2, two points A and B not on these lines, and point E of line l2. Construct with a ruler and compass point X on l1 such that lines AX and BX intercept on line l2 a segment a) of given length a; b) divisible in halves by E. 30.54. Points A and B lie on lines a and b, respectively, and point P does not lie on any of these lines. With the help of a ruler and compass draw through P a line that intersects lines a and b at points X and Y , respectively, so that the lengths of segments AX and BY a) are of given ratio; b) have a given product. 30.55. With the help of a ruler and compass draw through a given point a line on which three given lines intercept equal segments. 30.56. Consider a circle S, two chords AB and CD on it, and point E of chord CD. Construct with a ruler and compass point X on S so that lines AX and BX intercept on CD a segment a) of given length a; b) divided in halves by E. 30.57. a) Given line l, point P outside it, a given length, and a given angle α. Construct with a ruler and compass segment XY on l of the given length and subtending an angle of value α and with vertex in P. b) Given two lines l1 and l2, points P and Q outside them, and given angles α and β. Construct with the help of a ruler and compass point X on l1 and point Y on l2 such that segment XY subtends an angle of value α with vertex in P and another angle equal to β with vertex in Q. 480 CHAPTER 30. PROJECTIVE TRANSFORMATIONS 30.58. a) Given a circle, n points and n lines. Construct with the help of a ruler only an n-gon whose sides pass through the given points and whose vertices lie on the given lines. b) With the help of ruler only inscribe in the given circle an n-gon whose sides pass through n given points. c) With the help of a ruler and compass inscribe in a given circle a polygon certain sides of which pass through the given points, certain other sides are parallel to the given lines and the remaining sides are of prescribed lengths (about each side we have an information of one of the above three types). §7. Impossibility of construction with the help of a ruler only 30.59. Prove that with the help of a ruler only it is impossible to divide a given segment in halves. 30.60. Given a circle on the plane, prove that its center is impossible to construct with the help of a ruler only. Solutions 30.1. Denote the given lines by l0 and l, the given points on l0 by A0, B0, C0 and the given points on l by A, B, C. Let l1 be an arbitrary line not passing through A. Take an arbitrary point O0 not on lines l0 and l1. Denote by P0 the central projection map of l0 to l1 with center at O0 and by A1, B1, C1 the projections of points A0, B0, C0, respectively, under P0. Let l2 be an arbitrary line through point A not coinciding with l and not passing through A1. Take point O1 on line AA1 and consider the central projection map P1 of l1 to l2 with center at O1. Denote by A2, B2, C2 the projections of points A1, B1, C1, respectively, under P1. Clearly, A2 coincides with A. Finally, let P2 be the projection map of l2 to l which in the case when lines BB2 and CC2 are not parallel is the central projection with center at the intersection point of these lines; if lines BB2 and CC2 are parallel this is the parallel projection along either of these lines. The composition P2 ◦P1 ◦P0 is the required projective transformation. 30.2. a) Denote the intersection point of the four given lines by O; let H be the projection of H on l and h = OH. Then 2SOAC = OA · OC sin(a, c) = h · AC, 2SOBC = OB · OC sin(b, c) = h · BC, 2SOAD = OA · OD sin(a, d) = h · AD, 2SOBD = OB · OD sin(b, d) = h · BD. Dividing the ﬁrst equality by the second one and the third one by the fourth one we get OA sin(a, c) OB sin(b, c) = AC BC , OA sin(a, d) OB sin(b, d) = AD BD. Dividing the ﬁrst of the obtained equalities by the second one we get \|(ABCD)\| = \|(abcd)\|. To prove that the numbers (ABCD) and (abcd) are of the same sign, we can, for example, write down all the possible ways to arrange points on the line (24 ways altogether) and verify case by case that (ABCD) is positive if and only if the pair of lines a, b does not separate the pair of lines c, d. b) follows immediately from heading a). SOLUTIONS 481 30.3. Let a, b, c, x, y be the coordinates of points A, B, C, X, Y , respectively. Then x −a x −b : c −a c −b = y −a y −b : c −a c −b. Therefore, since all the points are distinct, (x −a)(y −b) = (x −b)(y −a). By simplifying we get ax −bx = ay −by. Dividing this equality by a −b we get x = y. 30.4. Let the image of each of the three given points under one projective transformation coincide with the image of this point under another projective transformation. Let us prove then that the images of any other point under these transformations coincide. Let us denote the images of the given points by A, B, C. Take an arbitrary point and denote by X and Y its images under the given projective transformations. Then by Problem 30.2 (ABCX) = (ABCY ) and, therefore, X = Y by Problem 30.3. 30.5. This problem is a corollary of the preceding one. 30.6. On line a, ﬁx three distinct points. By Problem 30.1 there exists a projective map P which maps these points in the same way as the given map. But in the solution of Problem 30.4 we actually proved that any map that preserves the cross ratio is uniquely determined by the images of three points. Therefore, the given map coincides with P. 30.7. First, let us show that the fractionally linear transformation P(x) = ax + b cx + d, ad −bc ̸= 0 preserves the cross ratio. Indeed, let x1, x2, x3, x4 be arbitrary numbers and yi = P(xi). Then yi −yj = axi + b cxi + d −axj + b cxj + d = (ad −bc)(xi −xj) (cxi + d)(cxj + d); hence, (y1y2y3y4) = (x1x2x3x4). In the solution of Problem 30.4 we have actually proved that if a transformation of the line preserves the cross ratio, then it is uniquely determined by the images of three arbitrary distinct points. By Problem 30.2 b) projective transformations preserve the cross ratio. It remains to prove that for any two triples of pairwise distinct points x1, x2, x3 and y1, y2, y3 there exists a fractionally linear transformation P such that P(xi) = yi. For this, in turn, it suﬃces to prove that for any three pairwise distinct points there exists a fractionally linear transformation that sends them into points z1 = 0, z2 = 1, z3 = ∞. Indeed, if P1 and P2 be fractionally linear transformations such that P1(xi) = zi and P2(yi) = zi, then P −1 2 (P1(xi)) = yi. The inverse to a fractionally linear transformation is a fractionally linear transformation itself because if y = ax+c cx+d, then x = dy−b −cy+a; the veriﬁcation of the fact that the composition of fractionally linear transformations is a fractionally linear transformation is left for the reader. Thus, we have to prove that if x1, x2, x3 are arbitrary distinct numbers, then there exist numbers a, b, c, d such that ad −bc ̸= 0 and ax1 + b = 0, ax2 + b = cx2 + d, cx3 + d = 0. Find b and d from the ﬁrst and third equations and substitute the result into the third one; we get a(x2 −x1) = c(x2 −x3) wherefrom we ﬁnd the solution: a = (x2−x3), b = x1(x3−x2), c = (x2−x1), d = x3(x1−x2). We, clearly, have ad −bc = (x1 −x2)(x2 −x3)(x3 −x1) ̸= 0. 30.8. First solution. Let a, b, c, d be the coordinates of the given points. Then by the hypothesis (c −a)(d −b) = (c −b)(d −a). After simpliﬁcation we get cb + ad = ca + bd. 482 CHAPTER 30. PROJECTIVE TRANSFORMATIONS Transfer everything to the left-hand side and factorize; we get (d −c)(b −a) = 0, i.e., either a = b or c = d. Second solution. Suppose that C ̸= D, let us prove that in this case A = B. Consider the central projection map of the given line to another line, let the projection send point D into ∞. Let A′, B′, C′ be the projections of points A, B, C, respectively. By Problem 30.2 (ABCD) = (A′B′C′∞) = 1, i.e., − → AC = − − → BC. But this means that A = B. 30.9. By Problem 30.6 it suﬃces to prove that the map P preserves the cross ratio. Let A, B, C, D be arbitrary points on line l. Denote by A′, B′, C′, D′ their respective images under P and by a, b, c, d and a′, b′, c′, d′ the lines MA, MB, MC, MD and NA′, NB′, NC′, ND′, respectively. Then by Problem 30.2 a) we have (ABCD) = (abcd) and (A′B′C′D′) = (a′b′c′d′) and by the theorem on an inscribed angle ∠(a, c) = ∠(a′, c′), ∠(b, c) = ∠(b′, c′), etc.; hence, (abcd) = (a′b′c′d′). 30.10. Let N = R−1(M), m = R(l), PN be the projection map of l to the circle from point N, Q the projection map of line m to l from point M. Then P −1 M ◦R ◦PM = Q ◦R ◦P −1 N ◦PM. But by the preceding problem the map P −1 N ◦PM is a projective one. 30.11. Lines passing through O and parallel to plane α1 (resp. α2) intersect plane α2 (resp. α1) at points of line l2 (resp. l1). Therefore, if a point lies on one of the planes α1, α2 and does not lie on lines l1, l2, then its projection to another plane is well-deﬁned. Clearly, the distinct points have distinct images. 30.12. The central projection to plane α2 with center O sends line l into the intersection of the plane passing through O and l with α2. 30.13. This problem is a direct corollary of the axioms of geometry and the deﬁnition of inﬁnite lines and points. 30.14. a) Problem 30.13 c) implies that if together with the ordinary (ﬁnite) points we consider inﬁnite ones, then P is a one-to-one correspondence. Under such an assumption the inﬁnite line is mapped to the inﬁnite line. Therefore, the set of ﬁnite points is also mapped one-to-one to the set of ﬁnite points. Since P sends lines into lines, P is an aﬃne map. b) Denote by l the line on which points A, B, C, D lie and by l0 the singular line of map P. Take an arbitrary point O outside plane α and consider plane β that passes through line l and is parallel to the plane passing through line l0 and point O. Let Q be the composition of the central projection of α on β with center O with the subsequent rotation of the space about axis l that sends β into α. The singular line of map Q is l0. Therefore, the projective transformation R = P ◦Q−1 of α sends the inﬁnite line into the inﬁnite line and by heading a) is an aﬃne transformation, in particular, it preserves the ratio of segments that lie on line l. It only remains to notice that transformation Q preserves the points of line l. c) The fact that the images of parallel lines l1 and l2 are parallel lines means that the inﬁnite point A of these lines turns into an inﬁnite point, i.e., A lies on the preimage l of the inﬁnite line. Therefore, either l is the inﬁnite line and then by heading a) P is an aﬃne transformation or l is parallel to lines l1 and l2. d) Denote by l∞the inﬁnite line. If P(l∞) = l∞, then P determines a one-to-one transformation of the plane that sends every line into a line and, therefore, by deﬁnition is an aﬃne one. Otherwise denote P(l∞) by a and consider an arbitrary projective transformation Q for which a is the singular line. Denote Q ◦P by R. Then R(l∞) = l∞and, therefore, as was shown above, R is an aﬃne map. Hence, P = Q−1 ◦R is a projective map. 30.15. a) It suﬃces to prove that points A, B, C, D can be transformed by a projective transformation into vertices of a square. Let E and F be (perhaps, inﬁnite) intersection SOLUTIONS 483 points of line AB with line CD and BC with AD, respectively. If line EF is not inﬁnite, then there exists a central projection of plane ABC to a plane α for which EF is the singular line. For the center of projection one may take an arbitrary point O outside plane ABC and for plane α an arbitrary plane parallel to plane OEF and not coinciding with it. This projection maps points A, B, C, D into the vertices of a parallelogram which can be now transformed into a square with the help of an aﬃne transformation. If line EF is an inﬁnite one, then ABCD is already a parallelogram. b) Thanks to heading a) it suﬃces to consider the case when ABCD and A1B1C1D1 is one and the same parallelogram. In this case its vertices are ﬁxed and, therefore, two points on an inﬁnite line in which the extensions of the opposite sides of the parallelogram intersect are also ﬁxed. Hence, by Problem 30.14 a) the map should be an aﬃne one and, therefore, by Problem 20.6 the identity one. c) Since with the help of a projection we can send lines l and l1 into the inﬁnit line (see the solution of heading a)), it suﬃces to prove that there exists an aﬃne transformation that maps every point O into a given point O1 and lines parallel to given lines a, b, c into lines parallel to given lines a1, b1, c1, respectively. We may assume that lines a, b, c pass through O and lines a1, b1, c1 pass through O1. On c and c1, select arbitrary points C and C1, respectively, and draw through each of them two lines a′, b′ and a′ 1, b′ 1 parallel to lines a, b and a1, b1, respectively. Then the aﬃne transformation that sends the parallelogram bounded by lines a, a′, b, b′ into the parallelogram bounded by lines a1, a′ 1, b1, b′ 1 (see Problem 29.6 c)) is the desired one. d) Not necessarily. The transformation from Problem 30.21 (as well as the identity transformation) preserves point O and line a. 30.16. a) On the coordinate plane Oxz consider points O(0, 0), N(0, 1), E(1, 0). For an arbitrary point M that lies on arc ⌣NE of the unit circle (see Fig. 123), denote by P the midpoint of segment EM and by M ∗and P ∗the intersection points of lines NM and NP, respectively, with line OE. Figure 265 (Sol. 30.16) Let us prove that for an arbitrary number k > 2 we can select point M soy that M ∗E : P ∗E = k. Let A(a, b) be an arbitrary point on the plane, A∗(t, 0) the intersection point of lines NA and OE, B(0, b) the projection of point A to line ON. Then t = A∗O ON = AB BN = a 1 −b. Therefore, if (x, z) are coordinates of point M, then points P, M ∗, P ∗have coordinates P µx + 1 2 , z 2 ¶ , M ∗ µ x 1 −z, 0 ¶ , P ∗ µ(x + 1)/2 1 −(z/2), 0 ¶ , 484 CHAPTER 30. PROJECTIVE TRANSFORMATIONS respectively, and, therefore, M ∗E : P ∗E = µ x 1 −z −1 ¶ : µx + 1 2 −z −1 ¶ = x + z −1 1 −z : x + z −1 2 −z = 2 −z 1 −z. Clearly, the solution of the equation 2−z 1−z = k is z = k−2 k−1 and, if k > 2, then 0 < z < 1 and, therefore, point M( √ 1 −z2, z) is the desired one. Now, let us prove the main statement of the problem. Denote the given circle and point inside it, respectively, by S and C. If point C is the center of S, then the identity transformation is the desired projective transformation. Therefore, let us assume that C is not the center. Denote by AB the diameter that contains point C. Let, for deﬁniteness, BC > CA. Set k = BA : AC. Then k > 2 and, therefore, as was proved, we can place point M on the unit circle in plane Oxz so that M ∗E : P ∗E = k = BA : CA. Therefore, by a similarity transformation we can translate circle S into a circle S1 constructed in plane Oxy with segment EM ∗as a diameter so that the images of points A, B, C are E, M ∗, P ∗, respectively. The stereographic projection maps S1 into circle S2 on the unit sphere symmetric through plane Oxz; hence, through line EM as well. Thus, EM is a diameter of S2 and the midpoint P of EM is the center of S2. Let α be the plane containing circle S2. Clearly, the central projection of plane Oxy to plane α from the north pole of the unit sphere sends S1 into S2 and point P ∗into the center P of S2. b) The diameter AB passing through M turns into a diameter. Therefore, the tangents at points A and B turn into tangents. But if the parallel lines pass into parallel lines, then the singular line is parallel to them (see Problem 30.14 c)). 30.17. On the coordinate plane Oxz consider points O(0, 0), N(0, 1), E(1, 0). For an arbitrary point M on arc ⌣NE of the unit circle denote by P the intersection of segment EM with line z = 1. Clearly, by moving point M along arc NE we can make the ratio EM : MP equal to an arbitrary number. Therefore, a similarity transformation can send the given circle S into circle S1 constructed on segment EM as on diameter in plane α perpendicular to Oxz so that the given line l turns into the line passing through P perpendicularly to Oxz. Circle S1 lies on the unit sphere with the center at the origin and, therefore, the stereographic projection sends S1 to circle S2 in plane Oxy. Thus, the central projection of plane α to plane Oxy from N sends S1 to S2 and line l into the inﬁnite line. 30.18. Let M be an arbitrary point on the given chord. By Problem 30.16 there exists a projective transformation that sends the given circle into a circle S and point M into the center of S. Since under a projective transformation a line turns into a line, the given chord will turn into a diameter. 30.19. Let us pass through point O two arbitrary chords AC and BD. Let P and Q be the intersection points of the extensions of opposite sides of quadrilateral ABCD. Consider an arbitrary projective transformation that maps S into a circle, S1, and O into the center of S1. It is clear that this transformation sends quadrilateral ABCD into a rectangle and, therefore, it sends line PQ into the inﬁnite line. 30.20. A projective transformation sends any line into a line and since the center is ﬁxed, every diameter turns into a diameter. Therefore, every inﬁnite point — the intersection point of the lines tangent to the circle in diametrically opposite points — turns into an inﬁnite point. Therefore, by Problem 30.14 a) the given transformation is an aﬃne one and by Problem 29.12 it is either a rotation or a symmetry. 30.21. a) Point M ′ lies on line OM and, therefore, its position is uniquely determined by the ratio MO : OM ′. But since triangles MBO and MAM ′ are similar, MO : OM ′ = SOLUTIONS 485 MB : BA and the latter relation does not depend on the choice of line l due to Thales’ theorem. b) First solution. If we extend the given transformation (let us denote it by P) by deﬁning it at point O setting P(O) = O, then, as is easy to verify, P determines a one-to-one transformation of the set of all ﬁnite and inﬁnite points of the plane into itself. (In order to construct point M from point M ′ we have to take an arbitrary point A on line a and draw lines AM ′, OB so that it is parallel to AM ′, and AB.) It is clear that every line passing through O turns into itself. Every line l not passing through O turns into the line parallel to OB and passing through M. Now, it only remains to make use of Problem 30.14 c). Second solution (sketch). Denote the given plane by π and let π′ = R(π), where R is a rotation of the space about axis a. Denote R(O) by O′ and let P be the projection map of plane π to plane π′ from the intersection point of line OO′ with the plane passing through b parallel to π′. Then R−1 ◦P coincides (prove it on your own) with the transformation mentioned in the formulation of the problem. 30.22. First solution. Denote the given transformation by P. Let us extend it to points of the line x = 0 and inﬁnite points by setting P(0, k) = Mk, P(Mk) = (0, k), where Mk is an inﬁnite point on the line y = kx. It is easy to see that the map P extended in this way is a one-to-one correspondence. Let us prove that under P every line turns into a line. Indeed, the line x = 0 and the inﬁnite line turn into each other. Let ax + by + c = 0 be an arbitrary other line (i.e., either b or c is nonzero). Since P ◦P = E, the image of any line coincides with its preimage. Clearly, point P(x, y) lies on the considered line if and only if a x + by x + c = 0, i.e., cx + by + a = 0. It remains to make use of Problem 30.14 d). Second solution (sketch). Denote lines x = 1 and x = 0 by a and b, respectively, and point (−1, 0) by O. Then the given transformation coincides with the transformation from the preceding problem. 30.23. If we denote line f by b, then the transformation mentioned in this problem is the inverse to the transformation of Problem 30.21. 30.24. Consider a projective transformation for which line PQ is the singular one. The images l′ 1 and l′ 2 of lines l1 and l2 under this transformation are parallel and the images of the considered quadrilaterals are parallelograms two sides of which lie on lines l′ 1 and l′ 2 and the other two sides are parallel to a ﬁxed line (the inﬁnite point of this line is the image of point P). It is clear that the locus of the intersection points of the diagonals of such parallelograms is the line equidistant from l′ 1 and l′ 2. 30.25. Let us make a projective transformation whose singular line is EF. Then quadri-lateral ABCD turns into a parallelogram and lines KL and MN into lines parallel to the sides of the parallelogram and passing through the intersection point of its diagonals, i.e., into the midlines. Therefore, the images of points K, L, M, N are the midpoints of the parallelogram and, therefore, the images of lines KN and LM are parallel, i.e., point X turns into an inﬁnite point and, therefore, X lies on the singular line EF. 30.26. Let us make the projective transformation with singular line AB. The images of points under this transformation will be denoted by primed letters. Let us consider a homothety with center at point O′ (or a parallel translation if O′ is an inﬁnite point) that sends C′ 1 to C′ 2. Under this homothety segment B′ 1C′ 1 turns into segment B′ 2C′ 2 because B′ 1C′ 1 ∥B′ 2C′ 2. Similarly, C′ 1A′ 1 turns to C′ 2A′ 2. Therefore, the corresponding sides of triangles A′ 1B′ 1C′ 1 and A′ 2B′ 2C′ 2 are parallel, i.e., all three points A′, B′, C′ lie on the inﬁnite line. 486 CHAPTER 30. PROJECTIVE TRANSFORMATIONS 30.27. Let us consider the projective transformation whose singular line passes through the intersection points of lines AB1 and BA1, BC1 and CB1 and denote by A′, B′, . . . the images of points A, B, . . . . Then A′B′ 1 ∥B′A′ 1, B′C′ 1 ∥C′B′ 1 and we have to prove that C′A′ 1 ∥A′C′ 1 (see Problem 1.12 a)). 30.28. As a result of the projective transformation with singular line PQ the problem is reduced to Problem 4.54. 30.29. This problem is a reformulation of the preceding one. Indeed, suppose that the pair of lines OO1 and OB separates the pair of lines OA and OC and the pair of lines OO1 and O1B separates the pair of lines O1A and O1C (consider on your own in a similar way the remaining ways of disposition of these lines). Therefore, if we redenote points A1, B, B1, C1, O, O1 and the intersection point of lines AB1 and CC1 by D, R, L, K, Q, P and B, respectively, then the preceding problem implies that the needed lines pass through point M. 30.30. Let us consider the projective transformation with singular line O1O2 and denote by A′, B′, . . . the images of points A, B, . . . . Then A′C′ 1 ∥C′ 1A′ 1 ∥B′B′ 1, B′C′ 1 ∥C′B′ 1 ∥ A′A′ 1. Let us, for deﬁniteness sake, assume that point C lies inside angle ∠A′O′B′ (the remaining cases can be reduced to this one after a renotation). Making, if necessary, an aﬃne transformation we can assume that the parallelogram O′A′C′ 1B′ is a square and, therefore, O′A′ 1C′B′ 1 is also a square and the diagonals O′C′ 1 and O′C′ of these squares lie on one line. It remains to make use of the symmetry through this line. 30.31. It suﬃces to prove that the orthocenters of each triple of triangles formed by the given lines lie on one line. Select some three triangles. It is easy to see that one of the given lines (denoted by l) is such that one of the sides of each of the chosen triangles lies on l. Denote the remaining lines by a, b, c and let A, B, C, respectively, be their intersection points with l. Denote by l1 the inﬁnite line and by A1 (resp. B1, C1) the inﬁnite points of the lines perpendicular to a (resp. b, c). Then the fact that the orthocenters of the three selected triangles lie on one line is a direct corollary of Pappus’s theorem (Problem 30.27). 30.32. Perform a projective transformation with singular line parallel to l and passing through the intersection point of lines PP1 and QQ1; next, perform an aﬃne transformation that makes the images of lines l and PP1 perpendicular to each other. We may assume that lines PP1 and QQ1 are perpendicular to line l and our problem is to prove that line RR1 is also perpendicular to l (points P1, Q1, R1 are the midpoints of the corresponding segments because these segments are parallel to the singular line; see Problem 30.14 b)). Segment PP1 is both a median and a hight, hence, a bisector in the triangle formed by lines l, AB and CD. Similarly, QQ1 is a bisector in the triangle formed by lines l, AC and BD. This and the fact that PP1 ∥QQ1 imply that ∠BAC = ∠BDC. It follows that quadrilateral ABCD is an inscribed one and ∠ADB = ∠ACB. Denote the points at which l intersects lines AC and BD by M and N, respectively (Fig. 124). Then the angle between l and AD is equal to ∠ADB −∠QNM = ∠ACB −∠QMN, i.e., it is equal to the angle between l and BC. It follows that the triangle bounded by lines l, AD and BC is an isosceles one and segment RR1 which is its median is also its hight, i.e., it is perpendicular to line l, as required. 30.33. Perform a projective transformation with singular line parallel to l and passing through point A. We may assume that point A is inﬁnite, i.e., lines AB and AC are parallel. Then by Problem 30.14 b) points A1, B1, C1 are, as earlier, the midpoints of the corresponding segments because these segments lie on the line parallel to the singular one. Two triangles formed by lines l, AB, BC and l, AC, BC are homothetic and, therefore, lines BB1 and CC1, which are medians of these triangles, are parallel. Therefore, quadrilateral SOLUTIONS 487 Figure 266 (Sol. 30.32) BB2CC2 is a parallelogram because its opposite sides are parallel. It remains to notice that point A2 is the midpoint of diagonal BC of this parallelogram and, therefore, it is also the midpoint of diagonal B2C2. 30.34. Let us make the projective transformation whose singular line is line PQ. De-note by A′, B′, . . . the images of points A, B, . . . . Then A′B′C′D′ is a parallelogram, R′ the intersection point of its diagonals, Q′ is the inﬁnite point of line Q′R′, K′ and L′ the intersection points of the sides of the parallelogram on line Q′R′. Clearly, points K′ and L′ are symmetric through point R′. Hence, (Q′R′K′L′) = Q′K′ Q′L′ : R′K′ R′L′ = 1 : R′K′ R′L′ = −1. It remains to notice that (QRKL) = (Q′R′K′L′) by Problem 30.2 b). 30.35. Answer: It is possible. Indeed, consider the vertices of a regular 1991-gon (red points) and points at which the extensions of the sides of this polygon intersect the inﬁnite line (blue points). This set of points has the required properties. Indeed, for any regular n-gon, where n is odd, the line passing through its vertex and parallel to one of the sides passes through one more vertex. Any given ﬁnite set of points can be transformed by a projective transformation into a set of ﬁnite (i.e., not inﬁnite) points. 30.36. Let us make a projective transformation that sends the circle inscribed into the quadrilateral into a circle S and the intersection point of the lines connecting the opposite tangent points into the center of S, cf. Problem 30.16 a). The statement of the problem now follows from the fact that the obtained quadrilateral is symmetric with respect to the center of S. 30.37. Let us make a projective transformation that sends the inscribed circle into a circle S and the intersection point of two of the three lines under consideration into the center of S, cf. Problem 30.16 a). Then the images of these two lines are simultaneously bisectors and hights of the image of the given triangle and, therefore, this triangle is an equilateral one. For an equilateral triangle the statement of the problem is obvious. 30.38. Let us consider, separately, the following two cases. 1) Point P lies outside S. Let us make the projective transformation that sends circle S into circle S1 and point P into ∞(see Problem 30.17), i.e., the images of all lines passing through P are parallel to each other. Then in heading b) the image of the locus to be found is line l, their common perpendicular passing through the center of S1, and in heading a) the line l with the diameter of S1 deleted. 488 CHAPTER 30. PROJECTIVE TRANSFORMATIONS To prove this, we have to make use of the symmetry through line l. Therefore, the locus itself is: in heading b), the line passing through the tangent points of S with the lines drawn through point P and in heading a), the part of this line lying outside S. 2) Point P lies inside S. Let us make a projective transformation that sends circle S into circle S1 and point P into its center, cf. Problem 30.16 a). Then the image of the locus to be found in both headings is the inﬁnite line. Therefore, the locus itself is a line. The obtained line coincides for both headings with the polar line of point P relative to S, cf. Problem 30.19. 30.39. Denote by m the line which is the locus to be found in Problem 30.38 b) and by N the distinct from M intersection point of S with line OM. Denote by Q the composition of the projection of l to S from M and S to M from N. By Problem 30.9 this composition is a projective map. Let us prove that P is the composition of Q with the projection of m to l from M. Let A be an arbitrary point on l, B its projection to S from M, C the projection of B to S from O, D the intersection point of lines BN and CM. By Problem 30.38 b) point D lies on line m, i.e., D = Q(A). Clearly, P(A) is the projection of D to l from M. 30.40. Both headings of the problem become obvious after a projective transformation that sends circle S into a circle and line KP into the inﬁnite line, cf. Problem 30.17. The answer is as follows: a) The locus to be found lies on the line equidistant from the images of lines AK and BK. b) The point to be found is the center of the image of S. 30.41. Let A′, B′, . . . be the images of points A, B, . . . under the projective transforma-tion that sends an escribed circle of triangle ABC into circle S, and chord EF into a diameter of S (see Problem 30.18). Then A′ is the inﬁnite point of lines perpendicular to diameter E′F ′ and we have to prove that line D′T ′ contains this point, i.e., is also perpendicular to E′F ′. Since △T ′B′E′ ∼△T ′F ′C′, it follows that C′T ′ : T ′E′ = C′F ′ : B′E′. But C′D′ = C′F ′ and B′D′ = B′E′ as tangents drawn from one point; hence, C′T ′ : T ′E′ = C′D′ : D′B′, i.e., D′T ′ ∥B′E′. 30.42. By Problem 30.16 a) it suﬃces to consider the case when diagonals AD and BE pass through the center of the circle. It remains to make use of the result of Problem 6.83 for n = 3. 30.43. Consider the projective transformation that sends circle S into a circle and the intersection points of lines AB and DE, BC and EF into inﬁnite points (see Problem 29.17). Our problem is reduced to Problem 2.11. 30.44. Consider a projective transformation that sends circle S into circle S1 and point O into the center O′ of S1, cf. Problem 30.16 a). Let A′, B′, . . . be the images of points A, B, . . . . Then A′B′, M ′N ′ and P ′Q′ are diameters. Therefore, the central symmetry through O′ sends point E′ into F ′, i.e., O′ is the midpoint of segment E′F ′. Since chord AB is perpendicular to the diameter passing through O, Problem 30.16 b) implies that AB is parallel to the singular line. Therefore, by Problem 30.14 b) the ratio of the lengths of the segments that lie on line AB is preserved and, therefore, O is the midpoint of segment EF. 30.45. Let us consider the projective transformation that maps the given circle into circle S′ and segment AD into a diameter of S′ (see Problem 30.18). Let A′, B′, . . . be the images of A, B, . . . . Then S turns into the inﬁnite point S′ of lines perpendicular to line A′D′. But A′C′ and B′D′ are hights in △A′D′P ′ and, therefore, Q′ is the orthocenter of this triangle. Therefore, line P ′Q′ is also a hight; hence, it passes through point S′. SOLUTIONS 489 30.46. By Problem 30.15 it suﬃces to consider only the case when ABCD is a square. We have to prove that the composition of projections described in the formulation of the problem is the identity transformation. By Problem 30.4 a projective transformation is the identity if it has three distinct ﬁxed points. It is not diﬃcult to verify that points A, B and the inﬁnite point of line AB are ﬁxed for this composition. 30.47. Under the projection of line QR from point A to line CD points Q, R, K, L are mapped into points D, C, P, L, respectively. Therefore, by Problem 30.2 b) (QRKL) = (DCPL). Similarly, by projecting line CD to line QR from point B we get (DCPL) = (RQKL); hence, (QRKL) = (RQKL). On the other hand, (RQKL) = RK RL : QK QL = µQK QL : RK RL ¶−1 = (QRKL)−1. These two equalities imply that (QRKL)2 = 1, i.e., either (QRKL) = 1 or (QRKL) = −1. But by Problem 30.8 the cross ratio of distinct points cannot be equal to one. 30.48. Denote the intersection points of lines AB1 and BA1, BC1 and CB1, CA1 and AC1 by P, Q, R, respectively, and the intersection point of lines PQ and CA1 by R1. We have to prove that points R and R1 coincide. Let D be the intersection point of AB1 and CA1. Let us consider the composition of projections: of line CA1 to line l1 from point A, of l1 to CB1 from B, and of CB1 to CA1 from P. It is easy to see that the obtained projective transformation of line CA1 ﬁxes points C, D and A1 and sends R into R1. But by Problem 30.5 a projective transformation with three distinct ﬁxed points is the identity one. Hence, R1 = R. 30.49. Let F ′ be the point symmetric to F through O. We have to prove that F ′ = F. By Problem 30.9 the composition of the projection of line AB to circle S from point M followed by the projection of S back to AB from Q is a projective transformation of line AB. Consider the composition of this transformation with the symmetry through point O. This composition sends points A, B, O, E to B, A, F ′, O, respectively. Therefore, by Problem 30.2 b) (ABOE) = (BAF ′O). On the other hand, it is clear that (BAF ′O) = BF ′ AF ′ : BO AO = AO BO : AF ′ BF ′ = (ABOF ′) i.e., (ABOE) = (ABOF ′); hence, by Problem 30.3, E = F ′. (?)30.50. Denote the intersection points of lines AB and DE, BC and EF, CD and FA by P, Q, R, respectively, and the intersection point of lines PQ and CD by R′. We have to prove that points R and R′ coincide. Let G be the intersection point of AB and CD. Denote the composition of the projection of line CD on the given circle from point A with the projection of circle $ to line BC from point E. By Problem 30.9 this composition is a projective map. It is easy to see that its compo-sition with the projection of BC to CD from point P ﬁxes points C, D and G and sends point R to R′. But by Problem 30.5 a projective transformation with three ﬁxed points is the identity one. Hence, R′ = R. 30.51. Since angles ∠APT, ∠ART, ∠AST and ∠AQT are right ones, points A, P, R, T, S, Q lie on the circle constructed on segment AT as on diameter. Hence, by Pascal’s theorem (Problem 30.50) points B, C and X lie on one line. 30.52. Denote the given line and circle by l and S, respectively. Let O be an arbitrary point of the given circle and let A1, A′ 1, B1, B′ 1, C1, C′ 1 be the images of points A, A′, B, B′, 490 CHAPTER 30. PROJECTIVE TRANSFORMATIONS C, C′ under the projection map of l to S from point O, i.e., A1 (resp. A′ 1, B1, . . . ) is the distinct from O intersection point of line AO (resp. A′O, BO, . . . ) with circle S. Denote by B2 the intersection point of lines A′ 1B1 and A1B′ 1 and by C2 the intersection point of lines A′ 1C1 and A1C′ 1. Let P1 be the composition of the projection of line l to circle S from point O with the projection of S to line B2C2 from point A′ 1; let P2 be the composition of the projection of B2C2 to S from point A1 with the projection of S to l from point O. Then by Problem 30.9 transformations P1 and P2 are projective ones and their composition sends points A, B, C to A′, B′, C′, respectively. It is clear that all the considered points can be constructed with the help of a ruler (in the same order as they were introduced). a) Let M1 be the distinct from O intersection point of line MO with circle S; M2 = P1(M) the intersection point of lines A′ 1M1 and B2C2; M3 the distinct from A1 intersection point of line M2A1 with circle S; P(M) = P2(P1(M)) the intersection point of lines l and OM3. b) Let M1 and N1 be the intersection points of circle S with line B2C2. Then the ﬁxed points of transformation P are the intersection points of lines OM1 and ON1 with line l. 30.53. a) The point X to be found is the ﬁxed point of the composition of the projection of l1 to l2 from point A, the translation along line l2 at distance a and the projection of l2 to l1 from point B. The ﬁxed point of this projective map is constructed in Problem 30.52. b) Replace the shift from the solution of heading a) with the central symmetry with respect to E. 30.54. a) Denote by k the number to which the ratio AX BY should be equal to. Consider the projective transformation of line a which is the composition of the projection of a to line b from point P, the movement of the plane that sends b to a and B to A and, ﬁnally, the homothety with center A and coeﬃcient k. The required point X is the ﬁxed point of this transformation. The construction of point Y is obvious. b) Denote by k the number to which the product AX · BY should equal to and by Q the intersection point of the lines passing through points A and B parallel to lines b and a, respectively; let p = AQ · BQ. Consider the projective transformation of line a which is the composition of the projection of a to line b from point P, projection of b to a from Q and the homothety with center A and coeﬃcient k p. Let X be the ﬁxed point of this transformation, Y its image under the ﬁrst projection and X1 the image of Y under the second projection. Let us prove that line XY is the desired one. Indeed, since △AQX1 ∼△BY Q, it follows that AX1 · BY = AQ · BQ = p and, therefore, AX · BY = k p · AX1 · BY = k. 30.55. Let P be the given point; A, B, C the points of pairwise intersections of the given lines a, b, c; let X, Y , Z be the intersection points of the given lines with line l to be found (Fig. 125). By the hypothesis XZ = ZY . Let T be the intersection point of line c with the line passing through X parallel to b. Clearly, XT = AY . Since △XTB ∼△CAB, it follows that XB : XT = CB : CA which implies BX : Y A = CB : CA, i.e., the ratio BX : Y A is known. Thus, our problem is reduced to Problem 30.54 a). 30.56. a) By Problem 30.9 the composition of the projection of CD on S from A with the projection of S on CD from B is a projective transformation of line CD. Let M be a ﬁxed point of the composition of this transformation with the shift along line CD by SOLUTIONS 491 Figure 267 (Sol. 30.55) distance a. Then the projection of M on S from A is the desired point. The ﬁxed point of any projective transformation is constructed in Problem 30.52. b) In the solution of heading a) replace the shift by the central symmetry through E. 30.57. a) Let us draw an arbitrary circle S through point P. By Problem 30.10 the composition of the projection of l to S from P, the rotation about the center of S through an angle of 2α and the projection of S to l from P is a projective transformation of line l. Then (by the theorem on an escribed angle) the ﬁxed point of the composition of this transformation with the shift along line CD by given distance XY is the desired point. The ﬁxed point of any projective transformation is constructed in Problem 30.52. b) Let us construct arbitrary circles S1 and S2 passing through points P and Q, respec-tively. Consider the composition of projection of l1 to S1 from P, the rotation about the center of S1 through an angle of 2α and the projection of S1 to l2 from P. By Problem 30.10 this composition is a projective map. Similarly, the composition of the projection of l2 to S2 from Q, the rotation about the center of S2 through an angle of 2β and projection of S2 to l1 from Q is also a projective map. By the theorem on an escribed angle the ﬁxed point of the composition of these maps is the desired point X and in order to construct it we can make use of Problem 30.52. 30.58. a) Denote the given points by M1, . . . , Mn and the given lines by l1, . . . , ln. A vertex of the polygon to be found is the ﬁxed point of the projective transformation of line l1 which is the composition of projections of l1 to l2 from M1, l2 to l3 from M2, . . . , ln to l1 from Mn. The ﬁxed point of a projective transformation is constructed in Problem 30.52. b) Select an arbitrary point on a given circle and with the help of projection from the given point let us identify the given circle with line l. By Problem 30.39 the central projecting of the circle to itself is a projective transformation of line l under this identiﬁcation. Clearly, a vertex of the desired polygon is the ﬁxed point of the composition of consecutive projections of the given circle to itself from given points. The ﬁxed point of a projective transformation is constructed in Problem 30.52. c) In the solution of heading b) certain central projections should be replaced by either rotations about the center of the circle if the corresponding side is of the given length or by symmetries if the corresponding side has the prescribed direction (the axis of the symmetry should be the diameter perpendicular to the given direction). 30.59. Suppose that we managed to ﬁnd the required construction, i.e., to write an instruction the result of fulﬁlment of which is always the midpoint of the given segment. Let us perform this construction and consider the projective transformation that ﬁxes the endpoints of the given segment and sends the midpoint to some other point. We can select this transformation so that the singular line would not pass through neither of the points obtained in the course of intermediate constructions. 492 CHAPTER 30. PROJECTIVE TRANSFORMATIONS Let us perform our imaginary procedure once again but now every time that we will encounter in the instruction words “take an arbitrary point (resp. line)” we shall take the image of the point (resp. line) that was taken in the course of the ﬁrst construction. Since a projective transformation sends any line into a line and the intersection of lines into the intersection of their images and due to the choice of the projective transformation this intersection is always a ﬁnite point, it follows that at each step of the second construction we obtain the image of the result of the ﬁrst construction and, therefore, we will ﬁnally get not the midpoint of the interval but its image instead. Contradiction. Remark. We have, actually, proved the following statement: if there exists a projective transformation that sends each of the objects A1, . . . , An into themselves and does not send an object B into itself, then it is impossible to construct object B from objects A1, . . . , An with the help of a ruler only. 30.60. The statement of the problem follows directly from Remark 30.59 above and from Problem 30.16 a). Index (ABCD), see cross-ratio, 473 (abcd), see double ratio, 473 Hk O, 359 Rϕ O, 345 SA, 327 SO, 345 Sl, 345 Ta, 327, 345 ∞, 474 basic set, 400 moment of inertia, 307 principle, Dirichlet, 385 principle, pigeonhole, 385 aﬃne transformation, 465 angle between a line and a circle, 450 angle between circles, 450 angle between two intersecting circles, 57 angle, Brokar, 110 angle, oriented, 33, 289 angle, right, 188 Apollonius’ problem, 450 area, oriented, 293 axis of similarity , 363 axis of the symmetry, 335 axis, radical, 63 barycentric coordinates, 310 base line, 377 bounding line, 433 Brachmagupta, 185 Brakhmagupta, 102 Brianchon’s theorem, 64, 478 Brokar’s angle, 110 Brokar’s point, 110 butterﬂy problem, 478 cardinality, 401 Carnot’s formula, 172 center of a regular polygon, 137 center of homothety, 359 center of mass, 307 center of symmetry, 327 center, radical, 64 center, similarity , 363 central projection, 473 central projection map, 474 central symmetry, 327 Ceva’s theorem, 106 circle of inversion, 449 circle of nine points, 109 circle, circumscribed, 99 circle, escribed, 99 circle, inscribed, 99 circle, similarity , 363 circle, similarity of a triangle, 364 circumscribed circle, 99 complete quadrilateral, 477 constant point of a triangle, 364 constant point of similar ﬁgures, 364 constant triangle of similar ﬁgures, 364 contraction, 465 convex hull, 207, 377 convex polygon, 397 correspondent line, 363 correspondent point, 363 correspondent segment, 363 counterexample, 438 crescents, 57 cross ratio, 473 degree of point, 63 Desargue’s theorem, 477 Desargues’s theorem, 105 diameter, 439 dilation, 465 Dirichlet’s principle, 385 double ratio, 473 doubly perspective triangles, 477 escribed circle, 99 Euler’s formula, 411 Euler’s line, 109 Feuerbach’s theorem, 452 formula, Euler, 411 formula, Heron, 271, 272 formula, Pick, 425 fractionally-linear map, 474 Gauss line, 84 493 494 INDEX Helly’s theorem, 398 Heron’s formula, 271, 272 homothety, 359 homothety, the center of, 359 Hyppocratus, 57 Hyppocratus’ crescents, 57 inequality, Ptolmy’s, 210 inequality, triangle, 205 inertia, 307, 309 inﬁnite line, 475 inﬁnite point, 474 inner product, 289 inscribed circle, 99 invariant, 409 inversion, 449 inversion, circle of, 449 isogonally conjugate point, 107 isotomically conjugate point, 106 lattice, 425 Lemoine’s point, 111 length of a curve, 303 Lindemann, 57 line, base , 377 line, bounding, 433 line, correspondent, 363 line, Euler, 109 line, Gauss , 84 line, inﬁnite, 475 line, polar, 61, 476 line, Simson, 107 line, Simson, of the inscribed quadrilateral, 108 line, singular, 475 line, singular , 474 locus, 169 map, central projection, 474 map, fractionally-linear, 474 map, projective, 473 mean arithmetic, 212 mean geometric, 212 Menelaus’s theorem, 106 Michel’s point, 40 Minkowski’s theorem, 425 moment of inertia, 309 Morlie’s theorem, 104 movement, 337 movement, orientation inverting, 337 movement, orientation preserving, 337 node, 425 oriented angle, 33 Pappus’ theorem, 105 Pappus’s theorem, 477 parallel projection, 473 parallel translation, 319 Pascal’s theorem, 145, 478, 479 pedal triangle, 108 Pick’s formula, 425 pigeonhole principle, 385 point, Brokar, 110 point, constant of a triangle, 364 point, constant of similar ﬁgures, 364 point, correspondent, 363 point, inﬁnite, 474 point, isogonally conjugate with respect to a tri-angle, 107 point, isotomically conjugate with respect to a tri-angle, 106 point, Lemoine, 111 point, Michel, 40 polar line, 476 polygon, circumscribed, 137 polygon, convex, 137, 397 polygon, inscribed, 137 polygon, regular, 137 polygon, regular, the center of, 137 porism, Steiner, 454 problem, Apollonius, 450 problem, butterﬂy, 478 problem, J. Steiner, 479 product, inner, 289 product, pseudoinner, 293 projection, central, 473 projection, parallel, 473 projection, stereographic, 475 projective map, 473 projective transformation, 473, 475 pseudoinner product, 293 Ptolemey’s inequality, 210 Pythagorean triangle, 102 quadrature of the circle, 57 radical axis, 63 radical center, 64 Ramsey’s theorem, 401 ratio cross, 473 ratio double, 473 rotation, 345 ruler, two-sided, 187 segment, correspondent, 363 set, basic, 400 shift, 465 simedian, 111 similarity axis, 363 similarity center, 363 similarity circle, 363 similarity circle of a triangle, 364 similarity transformation, 465 similarity triangle, 363 INDEX 495 Simson’s line, 107 Simson’s line of the inscribed quadrilateral, 108 singular line, 474, 475 Sperner’s lemma, 410 Steiner’s porism, 454 Steiner’s problem, 479 stereographic projection , 475 symmetry through a line, 335 symmetry through point, 327 symmetry with center, 327 symmetry, axial, 335 tangent line, 57 theorem Brianchon, 478 theorem on a complete quadrilateral, 477 theorem on doubly perspective triangles, 477 theorem on triply perspective triangles, 477 theorem Pascal, 478, 479 theorem, Brianchon, 64 theorem, Ceva, 106 theorem, Desargue, 477 theorem, Desargues, 105 theorem, Feuerbach, 452 theorem, Helly, 398 theorem, Menelaus, 106 theorem, Minkowski, 425 theorem, Morlie, 104 theorem, Pappus, 105, 477 theorem, Pascal, 145 theorem, Ramsey, 401 transformation similarity, 465 transformation, aﬃne, 465 transformation, projective, 473, 475 translation, parallel, 319 transvection, 338 triangle inequality, 205 triangle, constant of similar ﬁgures, 364 triangle, pedal, 108 triangle, Pythagorean, 102 triangle, similarity , 363 triangulation, 413 triply perspective triangles, 477
2946	https://www.medicalnewstoday.com/articles/adenomyosis-vs-endometriosis	Health Conditions Health Products News Original Series Podcasts General Health Health Tools Quizzes About Medical News Today Find Community Follow Us Related Topics Complications Related Articles Diagnosis Related Articles Home Remedies Related Articles Pregnancy Related Articles Surgery Related Articles Causes & Risk Factors Related Articles Management Related Articles Related Conditions Related Articles Symptoms Related Articles Treatment Related Articles Types Related Articles What is the difference between adenomyosis and endometriosis? In endometriosis, endometrial-type tissue grows outside the uterus. In adenomyosis, abnormal tissue grows into the uterine muscle. They are different conditions but can both cause pelvic pain, unusual menstrual bleeding, and heavy periods. While endometriosis causes the tissue to grow outside the uterus, adenomyosis causes it to grow into the uterine muscle. A person can have both conditions at once. Because the symptoms are so similar, the conditions can be difficult to diagnose — even healthcare professionals may have difficulty distinguishing one from the other. This is especially true of endometriosis, which requires surgery to definitively diagnose. On average, people with endometriosis wait nearly 7 years for a diagnosis. Both conditions often respond well to hormonal birth control, so a doctor may recommend this as the first line of treatment if they suspect that a person has adenomyosis or endometriosis. Read on to learn about the conditions, their similarities and differences, symptoms, prevalence, causes, and more. Similarities and differences Both adenomyosis and endometriosis cause abnormal growth of tissue similar to the endometrium, the tissue that lines the inside of the uterus. In adenomyosis, the tissue overgrows, extending into the muscle of the uterus. However, in endometriosis, the tissue grows outside the uterus, often attaching to nearby structures such as the ovaries and fallopian tubes. This tissue secretes substances such as prostaglandins and inflammatory chemicals that can cause pain. Although the tissue does not line the uterus, it swells and bleeds during menstrual cycles just as the endometrium does. This can be painful and harmful. Symptoms Adenomyosis and endometriosis have some common symptoms. Both conditions can cause: Additionally, adenomyosis may cause a person’s uterus to soften and enlarge, which can make the abdomen feel swollen. Endometriosis can cause varied symptoms depending on where the endometrial tissue grows. For example, endometrial tissue can adhere to digestive organs, causing stomach pain or bowel dysfunction. Learn more about the symptoms of endometriosis and adenomyosis. Prevalence Doctors do not know the exact prevalence of endometriosis. This is because people with the condition often do not receive a diagnosis for many years, and surgery is the only way to definitively diagnose it. This means that the actual number of people with endometriosis may be much higher than current estimates. Most estimates suggest that 10–15% of reproductive-age people with uteruses have the condition. Among those with pelvic pain, the prevalence may be as high as 70%. Similar issues of under-reporting and underdiagnosis exist for adenomyosis. Adenomyosis is more common among people who have had a dilation and curettage (D&C) — a procedure for pregnancy termination — and treatment for some miscarriages. Research suggests an overall prevalence of 20–35%. Causes Researchers do not fully understand what causes either condition. One of the most widely accepted explanations for endometriosis is retrograde menstruation. This theory suggests that during a person’s period, some of the menstrual flow travels backward through the fallopian tubes, allowing the tissue a person sheds during their period to move elsewhere in their body. As for adenomyosis, researchers believe that a disruption in the boundary between the uterine muscle and the deepest layer of the endometrial tissue may cause the condition. This is why injuries and surgery can be risk factors. However, both conditions are estrogen-dependent, which means they can occur only when estrogen levels are high enough to allow the endometrial tissue to grow. Risk factors The risk factors for endometriosis include: The risk factors for adenomyosis include: Diagnosis A doctor may suspect endometriosis, adenomyosis, or both depending on a person’s symptoms. Imaging tests such as pelvic ultrasound and magnetic resonance imaging (MRI) can help doctors diagnose adenomyosis. These tests may also help rule out other potential causes of pelvic pain if a doctor suspects endometriosis. While imaging scans may suggest that a person has endometriosis, the only way to conclusively diagnose the condition is through surgery to examine the uterus. Treatment The right treatment depends on a person’s symptoms, overall health, and fertility goals. If a person does not want to become pregnant but wishes to preserve future fertility, the following medications can help with both endometriosis and adenomyosis: When a person wants to become pregnant, they can stop taking these medications. A person may also need fertility treatments to become pregnant. These may include drugs to induce ovulation and therapies such as intrauterine insemination (IUI) and in vitro fertilization (IVF). People with endometriosis who do not want to become pregnant may try taking a gonadotropin-releasing hormone (GnRh), usually for a short trial of 3 months. This can shrink the endometrial tissue and reduce symptoms. Surgery to remove endometrial adhesions or adenomyosis growths may be helpful if the medication does not work. A hysterectomy is the only way to permanently cure adenomyosis. However, in many cases, surgery and medical treatments work well to manage symptoms, so a hysterectomy is not necessary. A hysterectomy may help reduce endometriosis symptoms but will not necessarily cure endometriosis. Outlook Endometriosis has no cure. While surgery can reduce the symptoms, various studies note recurrence rates of 6–67% after surgery. In 5–59% of people who undergo medical treatment for endometriosis, pain persists. Pregnancy is possible in people with endometriosis, but rates of pregnancy complications are higher and a person may have more difficulty getting and staying pregnant. The prognosis for adenomyosis varies. More adenomyosis adhesions, especially very deep ones, usually correlate with worse symptoms. Doctors typically start with medication and, depending on the results, may then try more invasive treatments. A hysterectomy can cure adenomyosis, but without a uterus, a person cannot carry a pregnancy. People who want to have children after a hysterectomy can consider using a gestational carrier or surrogate to carry a pregnancy for them. Summary Adenomyosis and endometriosis cause similar symptoms, and a person can have both conditions at the same time. Treatments for the two conditions are similar, so if a doctor is uncertain which condition a person has, they may start by recommending birth control or other hormonal medications. If a person does not experience relief from the first-line treatments, it can be helpful for them to see a specialist for additional guidance. How we reviewed this article: Share this article Latest news Related Coverage Adenomyosis is a condition where cells from the uterus lining grow into the uterus muscle. It is related to endometriosis and has similar symptoms… Painful sex, or dyspareunia, is a common symptom of endometriosis. Learn how to make sex more comfortable. Small intestinal bacterial overgrowth (SIBO) disrupts the normal balance of bacteria in the small intestine. Research suggests the balance of… People need to avoid sexual intercourse for at least 2 weeks following endometrial ablation. However, in some cases, a doctor may suggest waiting… OUR BRANDS
2947	https://teachy.ai/en/lesson-plan/elementary-school/6th-grade/mathematics-en/or-operations-properties-or-lesson-plan-or-flipped-classroom	Students 🇺🇸 Log In Lesson plan of Operations: Properties Lara from Teachy Subject Mathematics Mathematics Source Original Teachy Original Teachy Topic Operations: Properties Operations: Properties Objectives (5 - 7 minutes) Understanding the properties of arithmetic operations - Students should be able to identify and explain the common properties of basic arithmetic operations (addition, subtraction, multiplication, and division). This includes properties such as commutativity, associativity, distributivity, and identity. Applying the properties of operations to solve problems - Once the properties are understood, students will be expected to apply them to solve a variety of mathematical problems. This includes simplifying expressions, solving equations, and verifying answers. Developing critical thinking and problem-solving skills - Through exploring and applying the properties of operations, students will develop critical thinking and problem-solving skills. They should be able to analyze a problem, identify the appropriate operation, and apply the properties correctly to reach a solution. Secondary Objectives: Encourage active student participation in the lesson through group discussions and activities Foster the development of communication and collaboration skills through teamwork Stimulate students' mathematical thinking and confidence in their math abilities Introduction (10 - 15 minutes) Review of prior knowledge (3 - 5 minutes): The teacher begins the lesson by reviewing the basic arithmetic operations with the students - addition, subtraction, multiplication, and division. Simple examples can be used to reinforce the idea of how these operations work. Problem situation 1 (3 - 5 minutes): The teacher presents the following situation: "If we have 3 apples, and we add 2 more apples to the group, we will have a total of 5 apples. But, if we have 2 apples and we add 3 more apples to the group, we still get 5 apples. Why is this?" This should lead students to think about the commutative property (the order of the numbers does not matter in addition). Problem situation 2 (3 - 5 minutes): The teacher presents the following situation: "If we have 5 groups of 2 apples, we will have a total of 10 apples. But, if we have 2 groups of 5 apples, we will still get a total of 10 apples. Why is this?" This should lead students to think about the associative property of multiplication (the order of the factors does not change the product). Contextualization (2 - 3 minutes): The teacher explains that these properties, while seeming simple when applied to apples, are fundamental in mathematics and are used to solve complex problems. For example, in algebra, the properties of operations are used to simplify expressions and solve equations. Introduction to the topic (2 - 3 minutes): Finally, the teacher introduces the topic of the lesson, explaining that they will be exploring these properties of operations in more depth and learning how to apply them to solve problems. The teacher may want to share an interesting fact about operations, such as the fact that the distributive property is used in cryptography to keep sensitive information, such as passwords and credit card numbers, secure. Development (20 - 25 minutes) "Property Circle" activity (10 - 12 minutes): Preparation: The teacher divides the class into groups of 4 or 5 and distributes colored cards to each group. On each card, the teacher has written a math problem that involves a specific operation (addition, subtraction, multiplication, or division). The problems should be challenging, yet solvable with the application of the properties of operations. The cards are then placed in a circle in the center of the room. Procedure: The teacher explains that each group must choose a card from the "Property Circle" and attempt to solve the problem. They should work together, applying the properties of operations to simplify the expression or find the unknown value. Once they are finished, they should place the card back in the circle and take a new one. This should continue until all cards have been solved. Feedback: The teacher circulates around the room, observing the groups' progress and providing guidance when needed. After the activity is completed, the teacher asks a representative from each group to share one of the solutions that they found. This allows students to see different ways to approach the problems and apply the properties of operations. "Math Challenge" activity (10 - 13 minutes): Preparation: The teacher prepares a series of math problems that involve the application of the properties of operations. The problems should be progressively more challenging, allowing students to apply what they have learned gradually. Procedure: The teacher distributes the problems to each group and gives them time to attempt to solve them. Students should work together, discussing strategies and applying the properties of operations to solve the problems. The teacher should circulate around the room, offering support and guidance as needed. Feedback: After the time is up, the teacher asks each group to share their solution to one of the problems. This allows students to see different approaches to solving the same problems and reinforces their understanding of the properties of operations. Group Discussion (3 - 5 minutes): Facilitation: The teacher brings the entire class together and leads a group discussion. The teacher should ask probing questions to ensure that students have understood the properties of operations and how to apply them to solve problems. Feedback: The teacher provides feedback to the students, praising their efforts and highlighting what they did well. If there are any recurring errors, the teacher should correct them and explain the reasoning behind the correction. Closure: The teacher wraps up the discussion by emphasizing the importance of the properties of operations in everyday life and in other areas of mathematics. Assessment (8 - 10 minutes) Group discussion (3 - 4 minutes): Solution Sharing: The teacher brings the whole class together and initiates a group discussion. Each group is invited to share their solutions or approaches to the problems they solved during the "Property Circle" and "Math Challenge" activities. This allows students to see different ways of applying the properties of operations and encourages discussion and critical thinking. Identification of Difficulties: During the discussion, the teacher should identify any difficulties or misunderstandings that students may have regarding the properties of operations. This will help to guide the teacher's feedback and explanations in the next step. Connection to Theory (2 - 3 minutes): Teacher's Explanation: The teacher explains how the students' solutions and approaches connect to the theory of the properties of operations. The teacher should highlight the effective strategies that students used and explain how these strategies reflect an understanding of the properties of operations. Addressing Misconceptions: The teacher addresses any questions or misconceptions that students may have about the theory of the properties of operations. This may involve repeating examples, demonstrating similar problems, or exploring related concepts. Final Reflection (3 - 4 minutes): Moment of Reflection: The teacher asks students to take a moment to reflect individually on the lesson. They should think about what they have learned, what questions they still have, and what they found most challenging or interesting. Reflection Questions: To guide students' reflection, the teacher may ask questions such as: "What was the most important concept you learned today?" "What questions do you still have?" "What did you find most challenging about the properties of operations?" "How can you apply what you have learned today to real-life situations or other math problems?" Sharing Reflections: The teacher may call on a few students to share their reflections with the class. This can help to reinforce learning and generate further discussion. Closure: The teacher concludes the lesson by reiterating the main points about the properties of operations and the importance of understanding them for solving mathematical problems. The teacher may also provide a preview of what will be covered in the next class. Conclusion (5 - 7 minutes) Summary and Recap (2 - 3 minutes): The teacher recaps the main points covered during the lesson. This includes the definition and explanation of the properties of arithmetic operations (commutativity, associativity, distributivity, and identity) and the importance of applying them correctly to solve mathematical problems. It is important to emphasize that the properties of operations are not just abstract rules but practical tools that can be used to simplify expressions, solve equations, and verify answers. Theory-Practice Connection (1 - 2 minutes): The teacher reinforces how the hands-on activities, such as the "Property Circle" and "Math Challenge," allowed students to apply the properties of operations in a concrete and meaningful way. Additionally, the teacher highlights how working through the practical problems reinforced their theoretical understanding of the properties of operations. Supplementary Materials (1 - 2 minutes): The teacher suggests additional study materials for students who are interested in further exploring the properties of operations. This could include textbooks, online videos, interactive math websites, and math games. It is important that the suggested materials are accessible and appropriate for students' understanding level. For example, the teacher may recommend an animated video that reviews the properties of operations in an engaging and fun way. Relevance of the Topic (1 minute): Finally, the teacher emphasizes the importance of the properties of operations in everyday life and in other areas of mathematics. For example, it can be mentioned that the ability to simplify expressions and solve equations is crucial in many professions, such as engineering, science, economics, and programming. Additionally, the teacher can stress that being able to understand and apply the properties of operations not only helps to solve math problems but also develops valuable skills such as logical thinking, problem-solving, and perseverance. Need more materials to teach this subject? I can generate slides, activities, summaries, and over 60 types of materials. That's right, no more sleepless nights here :) Users who viewed this lesson plan also liked... 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2948	https://brainly.com/question/35617910	[FREE] Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion - brainly.com 6 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +45,3k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +48,6k Ace exams faster, with practice that adapts to you Practice Worksheets +5,5k Guided help for every grade, topic or textbook Complete See more / Physics Textbook & Expert-Verified Textbook & Expert-Verified Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion coefficients in the three dimensions. 1 See answer Explain with Learning Companion NEW Asked by sukhi9031 • 08/10/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 7108328 people 7M 1.0 0 Upload your school material for a more relevant answer The volume thermal expansion coefficient (β) of a solid is equal to the sum of its linear expansion coefficients (α) in the three dimensions. When a solid is subjected to a change in temperature, it undergoes thermal expansion, resulting in changes in its dimensions. The linear expansion coefficient (α) quantifies the fractional change in length of the solid per unit change in temperature. The volume thermal expansion coefficient (β) quantifies the fractional change in volume of the solid per unit change in temperature. Consider a solid with initial dimensions: length (L₀), width (W₀), and height (H₀). When the temperature changes by ΔT, the changes in dimensions are ΔL, ΔW, and ΔH. The fractional change in volume (ΔV/V₀) can be expressed as: ΔV/V₀ = (ΔL ΔW ΔH) / (L₀ W₀ H₀) Now, using the linear expansion coefficient (α) and the change in temperature (ΔT), we can relate the changes in dimensions: ΔL = αL₀ΔT ΔW = αW₀ΔT ΔH = αH₀ΔT Substituting these expressions into the fractional change in volume equation: ΔV/V₀ = (αL₀ΔT αW₀ΔT αH₀ΔT) / (L₀ W₀ H₀) ΔV/V₀ = (αL₀ αW₀ αH₀) (ΔT)^3 / (L₀ W₀ H₀) Comparing this equation with the definition of the volume thermal expansion coefficient (β = ΔV/V₀ΔT): β = (αL₀ αW₀ αH₀) (ΔT)^2 / (L₀ W₀ H₀) From the equation, it is evident that β = αL₀ + αW₀ + αH₀, which means the volume thermal expansion coefficient of a solid is indeed equal to the sum of its linear expansion coefficients in the three dimensions. To know more about thermal expansion, refer here: brainly.com/question/14092908# SPJ11 Answered by Pawar675 •27.1K answers•7.1M people helped Thanks 0 1.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 7108328 people 7M 1.0 0 The AP Physics Collection - Gregg Wolfe, Erika Gasper, John Stoke, Julie Kretchman, David Anderson, Nathan Czuba, Sudhi Oberoi, Liza Pujji, Irina Lyublinskaya, Douglas Ingram Physics - Boundless Heat and Thermodynamics - Jeremy Tatum Upload your school material for a more relevant answer The volume thermal expansion coefficient (β) of a solid is equal to the sum of its linear expansion coefficients (α) because volume changes as a result of changes in all three dimensions. By relating the changes in length, width, and height due to temperature increase, we derive that β=3 α. This confirms that in isotropic solids, the total expansion in volume can be expressed as the sum of the linear expansions in three directions. Explanation To prove that the volume thermal expansion coefficient (β) of a solid is equal to the sum of its linear expansion coefficients (α) in three dimensions, we first need to understand what these coefficients represent. When a solid is heated, it expands in all three dimensions: length, width, and height. The linear expansion coefficient (α) measures how much one dimension (like length) changes for a unit change in temperature. For a solid with initial dimensions: Length: L 0 Width: W 0 Height: H 0 When the temperature increases by Δ T, the changes in dimensions can be represented as: Change in Length: Δ L=α L 0Δ T Change in Width: Δ W=α W 0Δ T Change in Height: Δ H=α H 0Δ T Now, the original volume V 0 of the solid is: V 0=L 0⋅W 0⋅H 0 The new volume V after expansion is given by: V=(L 0+Δ L)⋅(W 0+Δ W)⋅(H 0+Δ H) Substituting the changes, we have: V=(L 0+α L 0Δ T)⋅(W 0+α W 0Δ T)⋅(H 0+α H 0Δ T) By applying the binomial expansion and only keeping the first order terms (since Δ T is small), the new volume becomes: V≈L 0W 0H 0+(L 0W 0⋅Δ H+W 0H 0⋅Δ L+H 0L 0⋅Δ W) Calculating the fractional change in volume, we get: V 0Δ V≈3 α Δ T Hence, the volume thermal expansion coefficient can be defined as: β=V 0Δ T Δ V This gives us: β=3 α Thus, the volume thermal expansion coefficient β is equal to the sum of the linear expansion coefficients in three dimensions: β=α L+α W+α H Therefore, for isotropic materials, β effectively equals three times the linear coefficient for any given direction. Therefore, this proof shows that the volume thermal expansion coefficient of a solid is indeed equal to the sum of its linear expansion coefficients in three dimensions. Examples & Evidence For instance, consider a cube of steel with an initial side length of 1 meter. If the temperature increases, each side will expand, and thus, the change in volume can be calculated to illustrate how the expansion in length affects the total volume change. This relationship is widely accepted in thermodynamics and is supported by empirical observations and experiments demonstrating that solids expand uniformly in three dimensions when subjected to temperature variations. Thanks 0 1.0 (1 vote) Advertisement sukhi9031 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Physics solutions and answers Community Answer the coefficient of area expansion isa.double the coefficient of linear expansion.b.three halves the coefficient of volume expansion.c.half the coefficient of volume expansion.d.triple the coefficient of linear expansion. Community Answer What is the thermal expansion equation that can be used for liquids and solids? What is the coefficient of volumetric expansion and how does it compare to the coefficient of linear expansion Community Answer Which best expresses the value for the coeffecient of volume expansion, Î², for given material as a function of its corresponding coefficient of linear expansion, Î±? Community Answer 47 Whats the usefulness or inconvenience of frictional force by turning a door knob? Community Answer 5 A cart is pushed and undergoes a certain acceleration. Consider how the acceleration would compare if it were pushed with twice the net force while its mass increased by four. Then its acceleration would be? Community Answer 4.8 2 define density and give its SI unit Community Answer 9 To prevent collisions and violations at intersections that have traffic signals, use the _____ to ensure the intersection is clear before you enter it. Community Answer 4.0 29 Activity: Lab safety and Equipment Puzzle Community Answer 4.7 5 If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _______. 4% Between 4% and 48% 48% More than 48% Community Answer When a constant force acts upon an object, the acceleration of the object varies inversely with its mass 2kg. When a certain constant force acts upon an object with mass , the acceleration of the object is 26m/s^2 . If the same force acts upon another object whose mass is 13 , what is this object's acceleration New questions in Physics Which of the following best represents precision but not accuracy? The distance between the two nearest molecules that are in phase with each other is called Which of the following best represents precision but not accuracy? Halley's Comet is one of the most famous objects in the Solar System, even though we have not seen it since 1986. (a) In what year will we next be able to see Halley's Comet? (b) In 1997, people could see Comet Hale-Bopp. In what year will this comet return? Why do comets appear to have large tails flowing away? Microwaves were originally used in which equipment during the world wars? A. Sonar B. Radars C. Thermostats D. 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2949	https://www.sciencedirect.com/science/article/pii/S0021904509000811	Skip to article My account Sign in View PDF Journal of Approximation Theory Volume 162, Issue 1, January 2010, Pages 141-152 Orthogonality of Jacobi and Laguerre polynomials for general parameters via the Hadamard finite part Author links open overlay panel rights and content Under an Elsevier user license Open archive Abstract Orthogonality of the Jacobi and Laguerre polynomials, and , is established for using the Hadamard finite part of the integral which gives their orthogonality in the classical cases. RiemannHilbert problems that these polynomials satisfy are found. The results are formally similar to the ones in the classical case (when ). Keywords Jacobi polynomials Laguerre polynomials Orthogonality for general parameters RiemannHilbert problems Hadamard finite part Cited by (0) Copyright © 2009 Elsevier Inc. All rights reserved.
2950	https://newsnetwork.mayoclinic.org/discussion/mayo-clinic-minute-role-of-the-larynx-and-how-to-protect-it/	Published Time: 2024-07-18T14:30:00+00:00 Mayo Clinic Minute: Role of the larynx and how to protect it - Mayo Clinic News Network Skip to Content English Español العربية 简体中文 Português Br Journalist Pass Appointments Search Request appointment Sign in News Releases Arizona Minnesota Florida International Health Topics Overview Cancer Cardiovascular Children’s Center COVID-19 Education Gastroenterology Health & Wellness Infectious Diseases Mayo Clinic Minute Medical Innovation Neurosciences Orthopedics/Sports Research Science Saturday Sharing Mayo Clinic Transplant Medical Research Overview Aging AI and Digital Health Biotherapeutics Cancer Clinical Trials Discovery Science Healthcare Delivery Individualized Medicine Translational Science Media Contacts About Mayo Clinic Minute: Role of the larynx and how to protect it By Deb Balzer July 18, 2024 Share this: A medical milestone at Mayo Clinic, atotal larynx transplantperformed on a patient with active cancer, has generated headlines recently in the medical world. But what is the larynx and what does it do? When you talk, many parts of your body work together to make a sound. The larynx, also called the voice box, plays an important part in the process. Dr. David Lott, a Mayo Clinic otolaryngologist who led the first known clinical trial on laryngeal transplantation in the U.S., says that, along with speech production, the larynx has a number of other vital roles. Watch: The Mayo Clinic Minute Journalists: Broadcast-quality video pkg (1:02) is in the downloads at the end of the post. Please courtesy: "Mayo Clinic News Network." Read thescript. It is the primary organ that's responsible for your ability to speak, for your ability to swallow, to eat and drink, and to breathe," says Dr. Lott. The larynx also helps sense when food is swallowed. Dr. Lott says it governs those functions in two ways. "When we swallow, the vocal folds close. That movement is very important in terms of regulating how good the voice sounds, how good someone swallows and how good they can breathe," he explains. Secondly, he says it's through the movement of the larynx itself. "The entire larynx has to move forward and backward when you swallow to allow food to get into the esophagus so that it can get into your stomach," Dr. Lott says. The best way to protect your larynx is to be in tune with your voice and know if something is off. "Or if it's harder for you to speak, you can't yell or you can't sing, whatever it may be — be clued in that. Maybe there's something either functionally or structurally that may be going on," he says. If you have lingering symptoms, consider seeking medical attention. Additional resources: Mayo Clinic marks medical milestone with world's first known successful total larynx transplant performed in a patient with an active cancer as part of a clinical trial Breaking the silence: First known total larynx transplant on a patient with active cancer as part of landmark clinical trial Research prepares for Mayo's first larynx transplant What is laryngeal chondrosarcoma? Mayo Clinic recognized for disability inclusion Demystifying my diagnosis of autism Related Articles Transplant (VIDEO) Eliminating the need for lifelong immunosuppressive medications for transplant patients Cancer (VIDEO) A rare cancer. A rare weapon. Curtis Jackson’s inspiring story of survival Transplant (VIDEO) Breathing easy thanks to the gift of life and new Mayo Clinic Lung Transplant Program in Arizona About the News Network Newsbureau@mayo.edu News Releases Cancer Cardiovascular Gastroenterology Neurosciences Transplant Research Mayo Clinic Minute Podcasts Health and Wellness Orthopedics/Sports Children's Center Topics Terms and Conditions Privacy Policy Manage Cookies © . Mayo Clinic News Network. All Rights Reserved. Loading... Loading... Mayo Clinic Privacy Policy Mayo Clinic and our partners use technologies such as cookies to collect information from your browser to deliver relevant advertising on our site, in emails and across the Internet, personalize content and perform site analytics. Please visit our Privacy Policy for more information about our use of data and your rights. Click here to update your preferences I Agree x Customize Cookie Settings Trusted These technologies are used in operating the site, including remembering your preferences, ensuring security and providing certain features. They do not track your activity online. [x] Website Analytics These technologies collect information to help us understand how our websites are being used and to make improvements. [x] Advertising These technologies are used to make advertising messages more relevant to you. They perform functions like preventing the same ad from continuously reappearing, ensuring that ads are properly displayed for advertisers and selecting advertisements that are based on your interests. Save
2951	https://www.genome.gov/genetics-glossary/Intron	Intron Scan to visit Skip to main content Due to reduction in workforce efforts, the information on this website may not be up to date, transactions submitted via the website may not be processed, and the agency may not be able to respond to inquiries. Note: Securing and protecting this website will continue. Skip to navigationSkip to searchSkip to sliderSkip to aboutSkip to subscriptionSkip to footer About Genomics About Genomics Introduction to Genomics Educational Resources Policy Issues in Genomics The Human Genome Project History of Genomics Program Research Funding Research Funding Funding Opportunities Funded Programs & Projects Division and Program Directors Scientific Program Analysts Contacts by Research Area News & Events Research at NHGRI Research at NHGRI Research Areas Research Investigators Staff Clinicians Research Projects Clinical Research Data Tools & Resources News & Events About Health About Health Genomics & Medicine Family Health History For Patients & Families For Health Professionals Careers & Training Careers & Training Jobs at NHGRI Training at NHGRI Funding for Research Training Professional Development Programs NHGRI Culture News & Events News & Events News Events Social Media Broadcast Media Image Gallery Videos Press Resources About NHGRI About NHGRI Organization Mission and Vision Policies and Guidance Budget Institute Advisors Strategic Vision Partner with NHGRI Staff Search Contact Us Breadcrumb Home About Genomics Educational Resources Talking Glossary of Genomic and Genetic Terms Intron Home About Genomics Educational Resources Talking Glossary of Genomic and Genetic Terms En Español Intron updated: September 28, 2025 Definition 00:00 … An intron is a region that resides within a gene but does not remain in the final mature mRNA molecule following transcription of that gene and does not code for amino acids that make up the protein encoded by that gene. Most protein-coding genes in the human genome consist of exons and introns. Narration 00:00 … Intron. The protein coding sequences for many genes are broken into smaller pieces of coding sequences called exons separated by non-coding sequences called introns. When genes are transcribed, those exons and introns are included in the initial messenger RNA products. However, introns are removed during the process called splicing so only exons are included in the mature mRNA and used to dictate what proteins are produced. In many genes, the introns are much longer than the exons. Introns may contain sequences that regulate how genes are expressed or transcribed and how mRNA is processed. Paul P. Liu, M.D., Ph.D. Senior Investigator Translational and Functional Genomics Branch Search Back to Glossary Related Messenger RNA (mRNA) Gene Promoter Exon Get Updates Enter your email address to receive updates about the latest advances in genomics research. Subscribe! Social Media Stream Footer Links Contact Accessibility Site Map Staff Search Plug-Ins Used by HHS FOIA Privacy Copyright HHS Vulnerability Disclosure
2952	https://en.wikipedia.org/wiki/Order_(group_theory)	Jump to content Search Contents (Top) 1 Example 2 Order and structure 3 Counting by order of elements 4 In relation to homomorphisms 5 Class equation 6 See also 7 Notes 8 References Order (group theory) العربية Čeština Español Français Galego 한국어 עברית മലയാളം Nederlands 日本語 Polski Português Romnă Simple English Svenska தமிழ் Українська Tiếng Việt 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Cardinality of a mathematical group, or of the subgroup generated by an element This article is about order in group theory. For other uses in mathematics, see Order (mathematics). For other uses, see Order (disambiguation). For groups with an ordering relation, see partially ordered group and totally ordered group. \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Order" group theory – news · newspapers · books · scholar · JSTOR (May 2011) (Learn how and when to remove this message) \| \| Algebraic structure → Group theoryGroup theory \| \| Basic notions \| \| \| Subgroup Normal subgroup Group action \| \| Quotient group (Semi-)direct product Direct sum Free product Wreath product \| \| Group homomorphisms \| \| kernel image \| \| simple finite infinite continuous multiplicative additive cyclic abelian dihedral nilpotent solvable \| \| Glossary of group theory List of group theory topics \| \| \| Finite groups \| \| \| Cyclic group Zn Symmetric group Sn Alternating group An Dihedral group Dn Quaternion group Q \| \| Cauchy's theorem Lagrange's theorem Sylow theorems Hall's theorem p-group Elementary abelian group Frobenius group Schur multiplier \| \| Classification of finite simple groups \| \| cyclic alternating Lie type sporadic \| \| \| Discrete groups Lattices Integers () Free group Modular groups PSL(2, ) SL(2, ) Arithmetic group Lattice Hyperbolic group \| \| Topological and Lie groups Solenoid Circle General linear GL(n) Special linear SL(n) Orthogonal O(n) Euclidean E(n) Special orthogonal SO(n) Unitary U(n) Special unitary SU(n) Symplectic Sp(n) G2 F4 E6 E7 E8 Lorentz Poincaré Conformal Diffeomorphism Loop Infinite dimensional Lie group O(∞) SU(∞) Sp(∞) \| \| Algebraic groups Linear algebraic group Reductive group Abelian variety Elliptic curve \| \| v t e \| In mathematics, the order of a finite group is the number of its elements. If a group is not finite, one says that its order is infinite. The order of an element of a group (also called period length or period) is the order of the subgroup generated by the element. If the group operation is denoted as a multiplication, the order of an element a of a group, is thus the smallest positive integer m such that am = e, where e denotes the identity element of the group, and am denotes the product of m copies of a. If no such m exists, the order of a is infinite. The order of a group G is denoted by ord(G) or \|G\|, and the order of an element a is denoted by ord(a) or \|a\|, instead of where the brackets denote the generated group. Lagrange's theorem states that for any subgroup H of a finite group G, the order of the subgroup divides the order of the group; that is, \|H\| is a divisor of \|G\|. In particular, the order \|a\| of any element is a divisor of \|G\|. Example [edit] The symmetric group S3 has the following multiplication table. : \| • \| e \| s \| t \| u \| v \| w \| --- --- --- \| e \| e \| s \| t \| u \| v \| w \| \| s \| s \| e \| v \| w \| t \| u \| \| t \| t \| u \| e \| s \| w \| v \| \| u \| u \| t \| w \| v \| e \| s \| \| v \| v \| w \| s \| e \| u \| t \| \| w \| w \| v \| u \| t \| s \| e \| This group has six elements, so ord(S3) = 6. By definition, the order of the identity, e, is one, since e 1 = e. Each of s, t, and w squares to e, so these group elements have order two: \|s\| = \|t\| = \|w\| = 2. Finally, u and v have order 3, since u3 = vu = e, and v3 = uv = e. Order and structure [edit] The order of a group G and the orders of its elements give much information about the structure of the group. Roughly speaking, the more complicated the factorization of \|G\|, the more complicated the structure of G. For \|G\| = 1, the group is trivial. In any group, only the identity element a = e has ord(a) = 1. If every non-identity element in G is equal to its inverse (so that a2 = e), then ord(a) = 2; this implies G is abelian since . The converse is not true; for example, the (additive) cyclic group Z6 of integers modulo 6 is abelian, but the number 2 has order 3: : . The relationship between the two concepts of order is the following: if we write for the subgroup generated by a, then For any integer k, we have : ak = e if and only if ord(a) divides k. In general, the order of any subgroup of G divides the order of G. More precisely: if H is a subgroup of G, then : ord(G) / ord(H) = [G : H], where [G : H] is called the index of H in G, an integer. This is Lagrange's theorem. (This is, however, only true when G has finite order. If ord(G) = ∞, the quotient ord(G) / ord(H) does not make sense.) As an immediate consequence of the above, we see that the order of every element of a group divides the order of the group. For example, in the symmetric group shown above, where ord(S3) = 6, the possible orders of the elements are 1, 2, 3 or 6. The following partial converse is true for finite groups: if d divides the order of a group G and d is a prime number, then there exists an element of order d in G (this is sometimes called Cauchy's theorem). The statement does not hold for composite orders, e.g. the Klein four-group does not have an element of order four. This can be shown by inductive proof. The consequences of the theorem include: the order of a group G is a power of a prime p if and only if ord(a) is some power of p for every a in G. If a has infinite order, then all non-zero powers of a have infinite order as well. If a has finite order, we have the following formula for the order of the powers of a: : ord(ak) = ord(a) / gcd(ord(a), k) for every integer k. In particular, a and its inverse a−1 have the same order. In any group, There is no general formula relating the order of a product ab to the orders of a and b. In fact, it is possible that both a and b have finite order while ab has infinite order, or that both a and b have infinite order while ab has finite order. An example of the former is a(x) = 2−x, b(x) = 1−x with ab(x) = x−1 in the group . An example of the latter is a(x) = x+1, b(x) = x−1 with ab(x) = x. If ab = ba, we can at least say that ord(ab) divides lcm(ord(a), ord(b)). As a consequence, one can prove that in a finite abelian group, if m denotes the maximum of all the orders of the group's elements, then every element's order divides m. Counting by order of elements [edit] Suppose G is a finite group of order n, and d is a divisor of n. The number of order d elements in G is a multiple of φ(d) (possibly zero), where φ is Euler's totient function, giving the number of positive integers no larger than d and coprime to it. For example, in the case of S3, φ(3) = 2, and we have exactly two elements of order 3. The theorem provides no useful information about elements of order 2, because φ(2) = 1, and is only of limited utility for composite d such as d = 6, since φ(6) = 2, and there are zero elements of order 6 in S3. In relation to homomorphisms [edit] Group homomorphisms tend to reduce the orders of elements: if f: G → H is a homomorphism, and a is an element of G of finite order, then ord(f(a)) divides ord(a). If f is injective, then ord(f(a)) = ord(a). This can often be used to prove that there are no homomorphisms or no injective homomorphisms, between two explicitly given groups. (For example, there can be no nontrivial homomorphism h: S3 → Z5, because every number except zero in Z5 has order 5, which does not divide the orders 1, 2, and 3 of elements in S3.) A further consequence is that conjugate elements have the same order. Class equation [edit] An important result about orders is the class equation; it relates the order of a finite group G to the order of its center Z(G) and the sizes of its non-trivial conjugacy classes: where the di are the sizes of the non-trivial conjugacy classes; these are proper divisors of \|G\| bigger than one, and they are also equal to the indices of the centralizers in G of the representatives of the non-trivial conjugacy classes. For example, the center of S3 is just the trivial group with the single element e, and the equation reads \|S3\| = 1+2+3. See also [edit] Torsion subgroup Notes [edit] ^ Conrad, Keith. "Proof of Cauchy's Theorem" (PDF). Archived from the original (PDF) on 2018-11-23. Retrieved May 14, 2011. ^ Conrad, Keith. "Consequences of Cauchy's Theorem" (PDF). Archived from the original (PDF) on 2018-07-12. Retrieved May 14, 2011. ^ Dummit, David; Foote, Richard. Abstract Algebra, ISBN 978-0471433347, pp. 57 References [edit] Dummit, David; Foote, Richard. Abstract Algebra, ISBN 978-0471433347, pp. 20, 54–59, 90 Artin, Michael. Algebra, ISBN 0-13-004763-5, pp. 46–47 Retrieved from " Categories: Group theory Algebraic properties of elements Hidden categories: Articles with short description Short description is different from Wikidata Articles needing additional references from May 2011 All articles needing additional references Order (group theory) Add topic
2953	https://en.wikipedia.org/wiki/Calcium_oxide	Jump to content Calcium oxide Afrikaans العربية Aragonés Aymar aru Azərbaycanca تۆرکجه বাংলা Български Bosanski Català Чӑвашла Čeština Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Galego 한국어 Հայերեն हिन्दी Hrvatski Bahasa Indonesia Italiano עברית Қазақша Кыргызча Latviešu Lietuvių Magyar Македонски മലയാളം Bahasa Melayu မြန်မာဘာသာ Nederlands 日本語 Norsk bokmål Occitan Oʻzbekcha / ўзбекча Polski Português Romnă Русский Scots Shqip Simple English Slovenčina Slovenščina Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Tagalog தமிழ் ไทย Türkçe Українська اردو Tiếng Việt Winaray 吴语粵語中文 Edit links From Wikipedia, the free encyclopedia Chemical compound of calcium "Quicklime" redirects here; not to be confused with Quickline. Calcium oxide \| Ionic crystal structure of calcium oxide Ca2+ O2- \| \| \| Powder sample of white calcium oxide \| \| \| Names \| \| \| IUPAC name Calcium oxide \| \| \| Other names Lime Quicklime Burnt lime Unslaked lime Free lime (building) Caustic lime Pebble lime Calcia Oxide of calcium \| \| \| Identifiers \| \| \| CAS Number \| 1305-78-8 \| \| 3D model (JSmol) \| Interactive image \| \| ChEBI \| CHEBI:31344 \| \| ChEMBL \| ChEMBL2104397 \| \| ChemSpider \| 14095 \| \| ECHA InfoCard \| 100.013.763 \| \| EC Number \| 215-138-9 \| \| E number \| E529 (acidity regulators, ...) \| \| Gmelin Reference \| 485425 \| \| KEGG \| C13140 \| \| PubChem CID \| 14778 \| \| RTECS number \| EW3100000 \| \| UNII \| C7X2M0VVNH \| \| UN number \| 1910 \| \| CompTox Dashboard (EPA) \| DTXSID5029631 \| \| InChI InChI=1S/Ca.O Key: ODINCKMPIJJUCX-UHFFFAOYSA-N InChI=1/Ca.O/rCaO/c1-2 Key: ODINCKMPIJJUCX-BFMVISLHAU \| \| \| SMILES O=[Ca] \| \| \| Properties \| \| \| Chemical formula \| CaO \| \| Molar mass \| 56.0774 g/mol \| \| Appearance \| White to pale yellow/brown powder \| \| Odor \| Odorless \| \| Density \| 3.34 g/cm3 \| \| Melting point \| 2,613 °C (4,735 °F; 2,886 K) \| \| Boiling point \| 2,850 °C (5,160 °F; 3,120 K) (100 hPa) \| \| Solubility in water \| Reacts to form calcium hydroxide \| \| Solubility in Methanol \| Insoluble (also in diethyl ether, octanol) \| \| Acidity (pKa) \| 12.8 \| \| Magnetic susceptibility (χ) \| −15.0×10−6 cm3/mol \| \| Structure \| \| \| Crystal structure \| Cubic, cF8 \| \| Thermochemistry \| \| \| Std molar entropy (S⦵298) \| 40 J·mol−1·K−1 \| \| Std enthalpy of formation (ΔfH⦵298) \| −635 kJ·mol−1 \| \| Pharmacology \| \| \| ATCvet code \| QP53AX18 (WHO) \| \| Hazards \| \| \| GHS labelling: \| \| \| Pictograms \| \| \| Signal word \| Danger \| \| Hazard statements \| H302, H314, H315, H335 \| \| Precautionary statements \| P260, P261, P264, P270, P271, P280, P301+P312, P301+P330+P331, P302+P352, P303+P361+P353, P304+P340, P305+P351+P338, P310, P312, P321, P330, P332+P313, P362, P363, P403+P233, P405, P501 \| \| NFPA 704 (fire diamond) \| 3 0 2 W \| \| Flash point \| Non-flammable \| \| Lethal dose or concentration (LD, LC): \| \| \| LD50 (median dose) \| >2000 mg/kg oral, female rat \| \| NIOSH (US health exposure limits): \| \| \| PEL (Permissible) \| TWA 5 mg/m3 \| \| REL (Recommended) \| TWA 2 mg/m3 \| \| IDLH (Immediate danger) \| 25 mg/m3 \| \| Safety data sheet (SDS) \| ICSC 0409 \| \| Related compounds \| \| \| Other anions \| Calcium sulfide Calcium hydroxide Calcium selenide Calcium telluride \| \| Other cations \| Beryllium oxide Magnesium oxide Strontium oxide Barium oxide Radium oxide \| \| Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa). Infobox references \| \| Chemical compound Calcium oxide (formula: CaO), commonly known as quicklime or burnt lime, is a widely used chemical compound. It is a white, caustic, alkaline, crystalline solid at room temperature. The broadly used term lime connotes calcium-containing inorganic compounds, in which carbonates, oxides, and hydroxides of calcium, silicon, magnesium, aluminium, and iron predominate. By contrast, quicklime specifically applies to the single compound calcium oxide. Calcium oxide that survives processing without reacting in building products, such as cement, is called free lime. Quicklime is relatively inexpensive. Both it and the chemical derivative calcium hydroxide (of which quicklime is the base anhydride) are important commodity chemicals. Preparation [edit] Calcium oxide is usually made by the thermal decomposition of materials, such as limestone or seashells, that contain calcium carbonate (CaCO3; mineral calcite) in a lime kiln. This is accomplished by heating the material to above 825 °C (1,517 °F), a process called calcination or lime-burning, to liberate a molecule of carbon dioxide (CO2), leaving quicklime behind. This is also one of the few chemical reactions known in prehistoric times. : CaCO3(s) → CaO(s) + CO2(g) The quicklime is not stable and, when cooled, will spontaneously react with CO2 from the air until, after enough time, it will be completely converted back to calcium carbonate unless slaked with water to set as lime plaster or lime mortar. Annual worldwide production of quicklime is around 283 million tonnes. China is by far the world's largest producer, with a total of around 170 million tonnes per year. The United States is the next largest, with around 20 million tonnes per year. Hydroxyapatite's free CaO content rises with increased calcination temperatures and longer times. It also pinpoints particular temperature cutoffs and durations that impact the production of CaO, offering information on how calcination parameters impact the composition of the material. Uses [edit] Heat: Quicklime releases thermal energy by the formation of the hydrate, calcium hydroxide, by the following equation: : : CaO (s) + H2O (l) ⇌ Ca(OH)2 (aq) (ΔHr = −63.7 kJ/mol of CaO) : As it hydrates, an exothermic reaction results and the solid puffs up. The hydrate can be reconverted to quicklime by removing the water by heating it to redness to reverse the hydration reaction. One litre of water combines with approximately 3.1 kilograms (6.8 lb) of quicklime to give calcium hydroxide plus 3.54 MJ of energy. This process can be used to provide a convenient portable source of heat, as for on-the-spot food warming in a self-heating can, cooking, and heating water without open flames. Several companies sell cooking kits using this heating method. It is a food additive used as an acidity regulator, a flour treatment agent and a leavener. It has E number E529. Light: When quicklime is heated to 2,400 °C (4,350 °F), it emits an intense glow. This form of illumination is known as a limelight, and was used broadly in theatrical productions before the invention of electric lighting. Cement: Calcium oxide is a key ingredient for the process of making cement. As a cheap and widely available alkali. Petroleum industry: Water detection pastes contain a mix of calcium oxide and phenolphthalein. Should this paste come into contact with water in a fuel storage tank, the CaO reacts with the water to form calcium hydroxide. Calcium hydroxide has a high enough pH to turn the phenolphthalein a vivid purplish-pink color, thus indicating the presence of water. Chemical pulping: Calcium oxide is used to make calcium hydroxide, which is used to regenerate sodium hydroxide from sodium carbonate in the chemical recovery at kraft pulp mills. Plaster: There is archeological evidence that Pre-Pottery Neolithic B humans used limestone-based plaster for flooring and other uses. Such Lime-ash floor remained in use until the late nineteenth century. Chemical or power production: Solid sprays or slurries of calcium oxide can be used to remove sulfur dioxide from exhaust streams in a process called flue-gas desulfurization. Carbon capture and storage: Calcium oxide can be used to capture carbon dioxide from flue gases in a process called calcium looping. Mining: Compressed lime cartridges exploit the exothermic properties of quicklime to break rock. A shot hole is drilled into the rock in the usual way and a sealed cartridge of quicklime is placed within and tamped. A quantity of water is then injected into the cartridge and the resulting release of steam, together with the greater volume of the residual hydrated solid, breaks the rock apart. The method does not work if the rock is particularly hard. Disposal of corpses: Historically, it was mistakenly believed that quicklime was efficacious in accelerating the decomposition of corpses. The application of quicklime can, in fact, promote preservation. Quicklime can aid in eradicating the stench of decomposition, which may have led people to the erroneous conclusion. It has been determined that the durability of ancient Roman concrete is attributed in part to the use of quicklime as an ingredient. Combined with hot mixing, the quicklime creates macro-sized lime clasts with a characteristically brittle nano-particle architecture. As cracks form in the concrete, they preferentially pass through the structurally weaker lime clasts, fracturing them. When water enters these cracks it creates a calcium-saturated solution which can recrystallize as calcium carbonate, quickly filling the crack. The thermochemical heat storage mechanism is greatly impacted by the sintering of CaO and CaCO3. It demonstrates that the storage materials become less reactive and denser at increasing temperatures. It also pinpoints particular sintering processes and variables influencing the efficiency of these materials in heat storage. Weapon [edit] Quicklime is also thought to have been a component of Greek fire. Upon contact with water, quicklime would increase its temperature above 150 °C (302 °F) and ignite the fuel. David Hume, in his History of England, recounts that early in the reign of Henry III, the English Navy destroyed an invading French fleet by blinding the enemy fleet with quicklime. Quicklime may have been used in medieval naval warfare – up to the use of "lime-mortars" to throw it at the enemy ships. Substitutes [edit] Limestone is a substitute for lime in many applications, which include agriculture, fluxing, and sulfur removal. Limestone, which contains less reactive material, is slower to react and may have other disadvantages compared with lime, depending on the application; however, limestone is considerably less expensive than lime. Calcined gypsum is an alternative material in industrial plasters and mortars. Cement, cement kiln dust, fly ash, and lime kiln dust are potential substitutes for some construction uses of lime. Magnesium hydroxide is a substitute for lime in pH control, and magnesium oxide is a substitute for dolomitic lime as a flux in steelmaking. Safety [edit] Because of vigorous reaction of quicklime with water, quicklime causes severe irritation when inhaled or placed in contact with moist skin or eyes. Inhalation may cause coughing, sneezing, and labored breathing. It may then evolve into burns with perforation of the nasal septum, abdominal pain, nausea and vomiting. Although quicklime is not considered a fire hazard, its reaction with water can release enough heat to ignite combustible materials.[better source needed] Mineral [edit] Calcium oxide is also a separate mineral species (with the unit formula CaO), named 'Lime'. It has an isometric crystal system, and can form a solid solution series with monteponite. The crystal is brittle, pyrometamorphic, and is unstable in moist air, quickly turning into portlandite (Ca(OH)2). References [edit] ^ a b Haynes, William M., ed. (2011). CRC Handbook of Chemistry and Physics (92nd ed.). Boca Raton, Florida: CRC Press. p. 4.55. ISBN 1-4398-5511-0. ^ Calciumoxid (Archived 2013-12-30 at the Wayback Machine). GESTIS database ^ a b Zumdahl, Steven S. (2009). Chemical Principles 6th Ed. Houghton Mifflin Company. p. A21. ISBN 978-0-618-94690-7. ^ a b c d NIOSH Pocket Guide to Chemical Hazards. "#0093". National Institute for Occupational Safety and Health (NIOSH). ^ ^ "free lime". DictionaryOfConstruction.com. Archived from the original on 2017-12-09. ^ Merck Index of Chemicals and Drugs, 9th edition monograph 1650 ^ Kumar, Gupta Sudhir; Ramakrishnan, Anushuya; Hung, Yung-Tse (2007), Wang, Lawrence K.; Hung, Yung-Tse; Shammas, Nazih K. (eds.), "Lime Calcination", Advanced Physicochemical Treatment Technologies, Handbook of Environmental Engineering, vol. 5, Totowa, NJ: Humana Press, pp. 611–633, doi:10.1007/978-1-59745-173-4_14, ISBN 978-1-58829-860-7, retrieved 2022-07-26 ^ "Lime throughout history \| Lhoist - Minerals and lime producer". Lhoist.com. Retrieved 10 March 2022. ^ Miller, M. Michael (2007). "Lime". Minerals Yearbook (PDF). U.S. Geological Survey. p. 43.13. ^ Collie, Robert L. "Solar heating system" U.S. patent 3,955,554 issued May 11, 1976 ^ Gretton, Lel. "Lime power for cooking - medieval pots to 21st century cans". Old & Interesting. Retrieved 13 February 2018. ^ "Compound Summary for CID 14778 - Calcium Oxide". PubChem. ^ Gray, Theodore (September 2007). "Limelight in the Limelight". Popular Science: 84. Archived from the original on 2008-10-13. Retrieved 2009-03-31. ^ Tony Oates (2007), "Lime and Limestone", Ullmann's Encyclopedia of Industrial Chemistry (7th ed.), Wiley, pp. 1–32, doi:10.1002/14356007.a15_317, ISBN 978-3527306732 ^ Tel Aviv University (August 9, 2012). "Neolithic man: The first lumberjack?". phys.org. Retrieved 2023-02-02. ^ Karkanas, P.; Stratouli, G. (2011). "Neolithic Lime Plastered Floors in Drakaina Cave, Kephalonia Island, Western Greece: Evidence of the Significance of the Site". The Annual of the British School at Athens. 103: 27–41. doi:10.1017/S006824540000006X. S2CID 129562287. ^ Connelly, Ashley Nicole (May 2012) Analysis and Interpretation of Neolithic Near Eastern Mortuary Rituals from a Community-Based Perspective Archived 2015-03-09 at the Wayback Machine. Baylor University Thesis, Texas ^ Walker, Thomas A (1888). The Severn Tunnel Its Construction and Difficulties. London: Richard Bentley and Son. p. 92. ^ "Scientific and Industrial Notes". Manchester Times. Manchester, England: 8. 13 May 1882. ^ US Patent 255042, 14 March 1882 ^ Schotsmans, Eline M.J.; Denton, John; Dekeirsschieter, Jessica; Ivaneanu, Tatiana; Leentjes, Sarah; Janaway, Rob C.; Wilson, Andrew S. (April 2012). "Effects of hydrated lime and quicklime on the decay of buried human remains using pig cadavers as human body analogues". Forensic Science International. 217 (1–3): 50–59. doi:10.1016/j.forsciint.2011.09.025. hdl:2268/107339. PMID 22030481. ^ "Riddle solved: Why was Roman concrete so durable?", MIT News, January 6, 2023 ^ Croddy, Eric (2002). Chemical and biological warfare: a comprehensive survey for the concerned citizen. Springer. p. 128. ISBN 0-387-95076-1. ^ David Hume (1756). History of England. Vol. I. ^ Sayers, W. (2006). "The Use of Quicklime in Medieval Naval Warfare". The Mariner's Mirror. Volume 92. Issue 3. pp. 262–269. ^ "Lime" (PDF). Prd-wret.s3-us-west-2.amazonaws.com. p. 96. Archived from the original (PDF) on 2021-12-19. Retrieved 2022-03-10. ^ Mallinckrodt Baker Inc. - Strategic Services Division (December 8, 1996). "Hazards". ww25.hazard.com. Archived from the original on May 1, 2012. Retrieved 2023-02-02. ^ "List of Minerals". Ima-mineralogy.org. 21 March 2011. ^ Fiquet, G.; Richet, P.; Montagnac, G. (Dec 1999). "High-temperature thermal expansion of lime, periclase, corundum and spinel". Physics and Chemistry of Minerals. 27 (2): 103–111. Bibcode:1999PCM....27..103F. doi:10.1007/s002690050246. S2CID 93706828. Retrieved 9 February 2023. ^ Tian, Lin, Yan, X. K., S. C., J., & Zhao, C. Y. (2022). "Lime". Mindat.org. doi:10.1016/j.cej.2021.131229. Retrieved 10 March 2022.{{cite journal}}: CS1 maint: multiple names: authors list (link) External links [edit] Wikimedia Commons has media related to Calcium oxide. Lime Statistics & Information from the United States Geological Survey Factors Affecting the Quality of Quicklime American Scientist (discussion of 14C dating of mortar) Chemical of the Week – Lime Material Safety Data Sheet CDC – NIOSH Pocket Guide to Chemical Hazards \| v t e Calcium compounds \| \| --- \| \| Hydrogen & halogens \| CaH2 CaF2 CaCl2 Ca(ClO)2 Ca(ClO3)2 Ca(ClO4)2 CaClOH CaBr2 Ca(BrO3)2 CaI2 Ca(IO3)2 CaICl \| \| Chalcogens \| CaO CaO2 Ca(OH)2 CaS CaSO3 CaH2S2O6 CaSO4 CaSe \| \| Pnictogens \| Ca3N2 CaN6 Ca(NO2)2 Ca(NO3)2 Ca3P2 CaP Ca4(PO4)2O Ca3(PO4)2 CaHPO4 Ca(H2PO4)2 Ca2P2O7 CaAs Ca3(AsO4)2 \| \| Group 13 & 14 \| CaC2 Ca(CN)2 CaCN2 CaCO3 Ca(HCO3)2 CaSi CaSi2 Ca2SiO4 Ca3(BO3)2 CaAl2O4 Ca3Al2O6 \| \| Trans metals \| Ca(MnO4)2 CaCrO4 CaTiO3 \| \| Organics \| CaC2O4 Ca(HCO2)2 Ca(CH3CO2)2 Ca(C3H5O2)2 CaC4H2O4 Ca3(C6H5O7)2 C3H7CaO6P Ca(C6H5O5S)2 Ca(C6H7O6)2 C10H11CaN4O8P CaC10H12O4N5PO4 C10H16CaN2O8 C12H22CaO14 C14H26CaO16 C18H32CaO19 C36H70CaO4 C24H40B2CaO24 \| \| v t e Oxides \| \| --- \| \| Mixed oxidation states \| Antimony tetroxide (Sb2O4) Boron suboxide (B12O2) Carbon suboxide (C3O2) Chlorine perchlorate (Cl2O4) Chloryl perchlorate (Cl2O6) Cobalt(II,III) oxide (Co3O4) Dichlorine pentoxide (Cl2O5) Iron(II,III) oxide (Fe3O4) Lead(II,IV) oxide (Pb3O4) Manganese(II,III) oxide (Mn3O4) Mellitic anhydride (C12O9) Praseodymium(III,IV) oxide (Pr6O11) Silver(I,III) oxide (Ag2O2) Terbium(III,IV) oxide (Tb4O7) Tribromine octoxide (Br3O8) Triuranium octoxide (U3O8) \| \| +1 oxidation state \| Aluminium(I) oxide (Al2O) Copper(I) oxide (Cu2O) Caesium monoxide (Cs2O) Dibromine monoxide (Br2O) Dicarbon monoxide (C2O) Dichlorine monoxide (Cl2O) Gallium(I) oxide (Ga2O) Iodine(I) oxide (I2O) Lithium oxide (Li2O) Mercury(I) oxide (Hg2O) Nitrous oxide (N2O) Potassium oxide (K2O) Rubidium oxide (Rb2O) Silver oxide (Ag2O) Thallium(I) oxide (Tl2O) Sodium oxide (Na2O) Water (hydrogen oxide) (H2O) \| \| +2 oxidation state \| Aluminium(II) oxide (AlO) Barium oxide (BaO) Berkelium monoxide (BkO) Beryllium oxide (BeO) Boron monoxide (BO) Bromine monoxide (BrO) Cadmium oxide (CdO) Calcium oxide (CaO) Carbon monoxide (CO) Chlorine monoxide (ClO) Chromium(II) oxide (CrO) Cobalt(II) oxide (CoO) Copper(II) oxide (CuO) Dinitrogen dioxide (N2O2) Disulfur dioxide (S2O2) Europium(II) oxide (EuO) Germanium monoxide (GeO) Iron(II) oxide (FeO) Iodine monoxide (IO) Lead(II) oxide (PbO) Magnesium oxide (MgO) Manganese(II) oxide (MnO) Mercury(II) oxide (HgO) Nickel(II) oxide (NiO) Nitric oxide (NO) Niobium monoxide (NbO) Palladium(II) oxide (PdO) Phosphorus monoxide (PO) Polonium monoxide (PoO) Protactinium monoxide (PaO) Radium oxide (RaO) Silicon monoxide (SiO) Strontium oxide (SrO) Sulfur monoxide (SO) Thorium monoxide (ThO) Tin(II) oxide (SnO) Titanium(II) oxide (TiO) Vanadium(II) oxide (VO) Yttrium(II) oxide (YO) Zirconium monoxide (ZrO) Zinc oxide (ZnO) \| \| +3 oxidation state \| Actinium(III) oxide (Ac2O3) Aluminium oxide (Al2O3) Americium(III) oxide (Am2O3) Antimony trioxide (Sb2O3) Arsenic trioxide (As2O3) Berkelium(III) oxide (Bk2O3) Bismuth(III) oxide (Bi2O3) Boron trioxide (B2O3) Californium(III) oxide (Cf2O3) Cerium(III) oxide (Ce2O3) Chromium(III) oxide (Cr2O3) Cobalt(III) oxide (Co2O3) Curium(III) oxide (Cm2O3) Dinitrogen trioxide (N2O3) Dysprosium(III) oxide (Dy2O3) Einsteinium(III) oxide (Es2O3) Erbium(III) oxide (Er2O3) Europium(III) oxide (Eu2O3) Gadolinium(III) oxide (Gd2O3) Gallium(III) oxide (Ga2O3) Gold(III) oxide (Au2O3) Holmium(III) oxide (Ho2O3) Indium(III) oxide (In2O3) Iron(III) oxide (Fe2O3) Lanthanum oxide (La2O3) Lutetium(III) oxide (Lu2O3) Manganese(III) oxide (Mn2O3) Neodymium(III) oxide (Nd2O3) Nickel(III) oxide (Ni2O3) Phosphorus trioxide (P4O6) Praseodymium(III) oxide (Pr2O3) Promethium(III) oxide (Pm2O3) Rhodium(III) oxide (Rh2O3) Samarium(III) oxide (Sm2O3) Scandium oxide (Sc2O3) Terbium(III) oxide (Tb2O3) Thallium(III) oxide (Tl2O3) Thulium(III) oxide (Tm2O3) Titanium(III) oxide (Ti2O3) Tungsten(III) oxide (W2O3) Vanadium(III) oxide (V2O3) Ytterbium(III) oxide (Yb2O3) Yttrium(III) oxide (Y2O3) \| \| +4 oxidation state \| Americium dioxide (AmO2) Berkelium(IV) oxide (BkO2) Bromine dioxide (BrO2) Californium dioxide (CfO2) Carbon dioxide (CO2) Carbon trioxide (CO3) Cerium(IV) oxide (CeO2) Chlorine dioxide (ClO2) Chromium(IV) oxide (CrO2) Curium(IV) oxide (CmO2) Dinitrogen tetroxide (N2O4) Germanium dioxide (GeO2) Iodine dioxide (IO2) Iridium dioxide (IrO2) Hafnium(IV) oxide (HfO2) Lead dioxide (PbO2) Manganese dioxide (MnO2) Molybdenum dioxide (MoO2) Neptunium(IV) oxide (NpO2) Nitrogen dioxide (NO2) Niobium dioxide (NbO2) Osmium dioxide (OsO2) Platinum dioxide (PtO2) Plutonium(IV) oxide (PuO2) Polonium dioxide (PoO2) Praseodymium(IV) oxide (PrO2) Protactinium(IV) oxide (PaO2) Rhenium(IV) oxide (ReO2) Rhodium(IV) oxide (RhO2) Ruthenium(IV) oxide (RuO2) Selenium dioxide (SeO2) Silicon dioxide (SiO2) Sulfur dioxide (SO2) Technetium(IV) oxide (TcO2) Tellurium dioxide (TeO2) Terbium(IV) oxide (TbO2) Thorium dioxide (ThO2) Tin dioxide (SnO2) Titanium dioxide (TiO2) Tungsten(IV) oxide (WO2) Uranium dioxide (UO2) Vanadium(IV) oxide (VO2) Xenon dioxide (XeO2) Zirconium dioxide (ZrO2) \| \| +5 oxidation state \| Antimony pentoxide (Sb2O5) Arsenic pentoxide (As2O5) Bismuth pentoxide (Bi2O5) Dinitrogen pentoxide (N2O5) Diuranium pentoxide (U2O5) Neptunium(V) oxide (Np2O5) Niobium pentoxide (Nb2O5) Phosphorus pentoxide (P2O5) Protactinium(V) oxide (Pa2O5) Tantalum pentoxide (Ta2O5) Tungsten pentoxide (W2O5) Vanadium(V) oxide (V2O5) \| \| +6 oxidation state \| Chromium trioxide (CrO3) Molybdenum trioxide (MoO3) Polonium trioxide (PoO3) Rhenium trioxide (ReO3) Selenium trioxide (SeO3) Sulfur trioxide (SO3) Tellurium trioxide (TeO3) Tungsten trioxide (WO3) Uranium trioxide (UO3) Xenon trioxide (XeO3) \| \| +7 oxidation state \| Dichlorine heptoxide (Cl2O7) Manganese heptoxide (Mn2O7) Rhenium(VII) oxide (Re2O7) Technetium(VII) oxide (Tc2O7) \| \| +8 oxidation state \| Iridium tetroxide (IrO4) Osmium tetroxide (OsO4) Ruthenium tetroxide (RuO4) Xenon tetroxide (XeO4) Hassium tetroxide (HsO4) \| \| Related \| Oxocarbon Suboxide Oxyanion Ozonide Peroxide Superoxide Oxypnictide \| \| Oxides are sorted by oxidation state. Category:Oxides \| \| \| v t e Oxygen compounds \| \| --- \| \| Ag4O4 Al2O3 AmO2 Am2O3 As2O3 As2O5 Au2O3 B2O3 BaO BeO Bi2O3 BiO2 Bi2O5 BrO2 Br2O3 Br2O5 Br 3O 8 CO CO2 C3O2 CaO CaO2 CdO CeO2 Ce3O4 Ce2O3 ClO2 Cl2O Cl2O2 Cl2O3 Cl2O4 Cl2O6 Cl2O7 CoO Co2O3 Co3O4 CrO3 Cr2O3 Cr2O5 Cr5O12 CsO2 Cs2O3 CuO Dy2O3 Er2O3 Eu2O3 FeO Fe2O3 Fe3O4 Ga2O Ga2O3 GeO GeO2 H2O 2H2O 3H2O H218O H2O2 HfO2 HgO Hg2O Ho2O3 IO I2O4 I2O5 I2O6 I4O9 In2O3 IrO2 KO2 K2O2 La2O3 Li2O Li2O2 Lu2O3 MgO Mg2O3 MnO MnO2 Mn2O3 Mn2O7 MoO2 MoO3 Mo2O3 NO NO2 N2O N2O3 N2O4 N2O5 NaO2 Na2O Na2O2 NbO NbO2 Nd2O3 O2F OF OF2 O2F2 O3F2 O4F2 O5F2 O6F2 O2PtF6 more... \| \| \| Authority control databases \| \| --- \| \| National \| \| \| Other \| Yale LUX \| Retrieved from " Categories: Alchemical substances Bases (chemistry) Calcium compounds Cement Dehydrating agents Desiccants Disinfectants E-number additives Limestone Rock salt crystal structure Oxides Hidden categories: Webarchive template wayback links CS1: unfit URL CS1 maint: multiple names: authors list Articles with short description Short description is different from Wikidata ECHA InfoCard ID from Wikidata E number from Wikidata Chembox having GHS data Articles containing unverified chemical infoboxes Short description matches Wikidata All articles lacking reliable references Articles lacking reliable references from June 2023 Commons category link is on Wikidata
2954	https://www.academia.edu/35408532/MECHANICAL_ENGINEERING_PROGRAMME_INCLINED_PLANE_SAFETY_REGULATIONS_AND_GUIDELINES_FOR_STUDENTS	(PDF) MECHANICAL ENGINEERING PROGRAMME INCLINED PLANE SAFETY REGULATIONS AND GUIDELINES FOR STUDENTS close Welcome to Academia Sign up to get access to over 50 million papers By clicking Continue, you agree to our Terms of Use and Privacy Policy Continue with GoogleContinue with AppleContinue with Facebook email Continue with Email arrow_back Continue with Email Sign up or log in to continue. Email Next arrow_forward arrow_back Welcome to Academia Sign up to continue. By clicking Sign Up, you agree to our Terms of Use and Privacy Policy First name Last name Password Sign Up arrow_forward arrow_back Hi, Log in to continue. Password Reset password Log in arrow_forward arrow_back Password reset Check your email for your reset link. 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Help Center less Outline keyboard_arrow_down Title Abstract Introduction Procedure Question All Topics Engineering Civil Engineering download Download Free PDF Download Free PDF MECHANICAL ENGINEERING PROGRAMME INCLINED PLANE SAFETY REGULATIONS AND GUIDELINES FOR STUDENTS Gan Natasha description See full PDF download Download PDF format_quote Cite close Cite this paper bookmark Save to Library share Share close Sign up for access to the world's latest research Sign up for free arrow_forward check Get notified about relevant papers check Save papers to use in your research check Join the discussion with peers check Track your impact Abstract AI This document outlines safety regulations and guidelines for students engaged in mechanical engineering experimental work specifically involving inclined planes. It emphasizes the importance of adhering to safety protocols in the laboratory, such as proper dress code and maintenance of a clean workspace. 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2955	http://www.zo.utexas.edu/courses/bio373/chapters/Chapter9/Chapter9.html	\| \| \| 9 \| Population Growth and Regulation Verhulst-Pearl Logistic Equation In a finite world, no population can grow exponentially for very long (Figure 9.1). Sooner or later every population must encounter either difficult environmental conditions or shortages of its requisites for reproduction. Over a long period of time, unless the average actual rate of increase is zero, a population either decreases to extinction or increases to the extinction of other populations. So far, our populations have had fixed age-specific parameters, such as their lx and mx schedules. In this section we ignore age specificity and instead allow R0 and r to vary with population density. To do this, we define carrying capacity, K, as the density of organisms (i.e., the number per unit area) at which the net reproductive rate (R0) equals unity and the intrinsic rate of increase (r) is zero. At "zero density" (only one organism, or a perfect competitive vacuum), R0 is maximal and r becomes rmax. For any given density above zero density, both R0 and r decrease until, at K, the population ceases to grow. A population initiated at a density above K decreases until it reaches the steady state at K (Figures 9.2 and 9.3). Thus, we define ractual (or dN/dt times 1/N) as the actual instantaneous rate of increase; it is zero at K, negative above K, and positive when the population is below K. The simplest assumption we can make is that ractual decreases linearly with N and becomes zero at an N equal to K (Figure 9.2); this assumption leads to the classical Verhulst-Pearl logistic equation: dN/dt = rN - rN (N / K) = rN - {(rN2)/ K} (1) Alternatively, by factoring out an rN, equation (1) can be written dN/dt = rN {1 - (N / K)} = rN{({K - N }/ K)} (2) Or, simplifying by setting r/K in equation (1) equal to z, dN/dt = rN - zN2 (3) 1. Figure 9.1. In 1911, 25 reindeer were introduced on Saint Paul Island in the Pribolofs off Alaska. The population grew rapidly and nearly exponentially until about 1938, when there were over 2000 animals on the 41-square-mile island. The reindeer badly overgrazed their food supply (primarily lichens) and the population "crashed." Only eight animals could be found in 1950. A similar sequence of events occurred on Saint Matthew Island from 1944 through 1966. [After Krebs (1972) after V. B. Scheffer (1951). The Rise and Fall of a Reindeer Herd. Science 73: 356-362.] 1. Figure 9.2. The actual instantaneous rate of increase per individual, ra, decreases linearly with increasing population density under the assumptions of the Verhulst-Pearl logistic equation. The solid line depicts conditions in an optimal environment in which the difference between b and d is maximal. The dashed line shows how the actual rate of increase decreases with N when the death rate per head, d, is higher; equilibrium population size, N, is then less than carrying capacity, K. (Compare with Figure 9.5 which plots the same thing but separates births and deaths.) The term rN (N/K) in equation (1) and the term zN2 in equation (3) represent the density-dependent reduction in the rate of population increase. Thus, at N equal to unity (an ecologic vacuum), dN/dt is nearly exponential, whereas at N equal to K, dN/dt is zero and the population is in a steady state at its carrying capacity. Logistic equations (there are many more besides the Verhulst-Pearl one) generate so-called sigmoidal (S-shaped) population growth curves (Figure 9.3). Implicit in the Verhulst-Pearl logistic equation are three assumptions: (1) that all individuals are equivalent -- that is, the addition of every new individual reduces the actual rate of increase by the same fraction, 1/K, at every density (Figure 9.2); (2) that rmax and K are immutable constants; and (3) that there is no time lag in the response of the actual rate of increase per individual to changes in N. All three assumptions are unrealistic, so the logistic has been strongly criticized (Allee et al. 1949; Smith 1952, 1963a; Slobodkin 1962b). 1. Figure 9.3a. Left: Population growth under the Verhulst-Pearl logistic equation is sigmoidal (S-shaped), reaching an upper limit termed the carrying capacity, K. Populations initiated at densities above K decline exponentially until they reach K, which represents the only stable equilibrium. Right: dN/dt plotted against N. Also shown is the actual rate of increase at density N (red line). More plausible curvilinear relationships between the rate of increase and population density are shown in Figure 9.4. Note that density-dependent effects on birth rate and death rate are combined by the use of r (these effects are separated later in this section). Carrying capacity is also an extremely complicated and confounded quantity, for it necessarily includes both renewable and nonrenewable resources, which are variables themselves. Carrying capacity almost certainly varies a great deal from place to place and from time to time for most organisms. There is also some inevitable lag in feedback between population density and the actual instantaneous rate of increase. All these assumptions can be relaxed and more realistic equations developed, but the mathematics quickly become extremely complex and unmanageable. Nevertheless, a number of populational phenomena can be nicely illustrated using the simple Verhulst-Pearl logistic, and a thorough understanding of it is a necessary prelude to the equally simplistic Lotka-Volterra competition equations, which are taken up in Chapter 12. However, the numerous flaws of the logistic must be recognized, and it should be taken only as a first approximation for small changes in population growth, most likely to be valid near equilibrium and over short time periods (i.e., situations in which linearity should be approximated). 1. Figure 9.4. Hypothetical curvilinear relationships between instantaneous rates of increase and population density. Concave upward curves have also been postulated. [From Gadgil and Bossert (1970) and Pianka (1972).] Notice that r in the logistic equation is actually rmax. The equation can be solved for the actual rate of increase, ractual, which is a variable and a function of r, N, and K, by simply factoring out an N: ractual = ra = dN/Ndt = r {(K - N)/K} = rmax - (N/K) rmax (4) The actual instantaneous rate of increase per individual, ractual, is always less than or equal to rmax (r in the logistic). Equation (4) and Figure 9.2 show how ractual decreases linearly with increasing density under the assumptions of the Verhulst-Pearl logistic equation. The two components of the actual instantaneous rate of increase per individual, ra, are the actual instantaneous birth rate per individual, b, and the actual instantaneous death rate per individual, d. The difference between b and d (i.e., b - d) is ractual. Under theoretical ideal conditions when b is maximal and d is minimal, ractual is maximized at rmax. In the logistic equation, this is realized at a minimal density, or a perfect competitive vacuum. 1. Figure 9.5. The instantaneous birth rate per individual decreases linearly with population density under the logistic equation, whereas the instantaneous death rate per head rises linearly as population density increases. Two death rate lines are plotted, one with a high death rate (dashed line) and one with a lower death rate (solid line). Equilibrium population density, N, is lowered by either an increased death rate or by a reduced birth rate. To be more precise, we add a subscript to b and d, which are functions of density. Thus, bN - dN = rN (which is ractual at density N), and b0 - d0 = rmax. When bN = dN, ractual and dN/dt are zero and the population is at equilibrium. Figure 9.5 diagrams the way in which b and d vary linearly with N under the logistic equation. At any given density, bN and dN are given by linear equations bN = b0 - xN (5) dN = d0 + yN (6) where x and y represent, respectively, the slopes of the lines plotted in Figure 9.5 (see also Bartlett 1960, and Wilson and Bossert 1971). The instantaneous death rate, dN, clearly has both density-dependent and density-independent components; in equation (6) and Figure 9.5, yN measures the density-dependent component of dN while d0 determines the density-independent component. At equilibrium, bN must equal dN, or b0 - xN = d0 + yN (7) Substituting K for N at equilibrium, r for (b0 - d0), and rearranging terms r = (x + y) K (8) K = r/(x + y) (9) Note that the sum of the slopes of birth and death rates (x + y) is equal to z, or r/K. Clearly, z is the density-dependent constant that is analogous to the density-independent constant rmax. Derivation of the Logistic Equation The derivation of the Verhulst-Pearl logistic equation is relatively straightforward. First, write an equation for population growth using the actual rate of increase rN dN/dt = rN N = (bN - dN) N (10) Now substitute the equations for bN and dN from (5) and (6), above, into (1): dN/dt = [(b0 - xN) - (d0 + yN)] N (11) Rearranging terms, dN/dt = [(b0 - d0) - (x + y)N)] N (12) Substituting r for (b - d) and, from (9) above, r/K for (x + y), multiplying through by N, and rearranging terms, dN/dt = rN - (r/K)N2 (13) Density Dependence and Density Independence Various factors can influence populations in two fundamentally different ways. If their effects on a population do not vary with population density, but the same proportion of organisms are affected at any density, factors are said to be density independent. Climatic factors often, though by no means always, affect populations in this manner (see Table 9.1 below). If, on the other hand, a factor's effects vary with population density so that the proportion of organisms influenced actually changes with density, that factor is density dependent. Density-dependent factors and events can be either positive or negative. Death rate, which presumably often increases with increasing density, is an example of positive or direct density dependence (Figure 9.5); birth rate, which normally decreases with increasing density, is an example of negative or inverse density dependence (Figure 9.6). 1. Figure 9.6. A plot of average clutch size against the density of breeding pairs of English great tits (birds) in a particular woods in a series of years over a 17-year period. [After Perrins (1965).] Density-dependent influences on populations frequently result in an equilibrium density at which the population ceases to grow. Biotic factors, such as competition, predation, and pathogens, often (though not always) act in this way. Ecologists are divided in their opinions as to the relative importance of density dependence and density independence in natural populations (Andrewartha and Birch 1954; Lack 1954, 1966; Nicholson 1957; Orians 1962; McLaren 1971; Ehrlich et al. 1972). Detection of density dependence can be difficult. In studies of population dynamics of Thrips imaginis (a small herbivorous insect), Davidson and Andrewartha (1948) found that they could predict population sizes of these insects fairly accurately using only past population sizes and recent climatic conditions. These workers could find no evidence of any density effects; they therefore interpreted their data to mean that the populations of Thrips were controlled primarily by density-independent climatic factors. However, reanalysis of their data shows pronounced density-dependent effects at high densities (Smith 1961). Population change and population size are strongly inversely correlated, which strongly suggests density dependence. Smith also demonstrated a rapidly decreasing variance in population size during the later portion of the spring population increase. Furthermore, these patterns persisted even after partial correlation analysis, which holds constant the very climatic variables that Davidson and Andrewartha considered to be so important. This example illustrates the great difficulty ecologists frequently encounter in distinguishing cause from effect. There is now little real doubt that both density-dependent and density-independent events occur; however, their relative importance may vary by many orders of magnitude from population to population -- and even within the same population from time to time as the size of the population changes (Horn 1968a; McLaren 1971). Opportunistic versus Equilibrium Populations Periodic disturbances, including fires, floods, hurricanes, and droughts, often result in catastrophic density-independent mortality, suddenly reducing population densities well below the maximal sustainable level for a particular habitat. Populations of annual plants and insects typically grow rapidly during spring and summer but are greatly reduced at the onset of cold weather. Because populations subjected to such forces grow in erratic or regular bursts (Figure 9.7), they have been termed opportunistic populations. In contrast, populations such as those of many vertebrates may usually be closer to an equilibrium with their resources and generally exist at much more stable densities (provided that their resources do not fluctuate); such populations are called equilibrium populations. Clearly, these two sorts of populations represent endpoints of a continuum; however, the dichotomy is useful in comparing different populations. Table 9.1 Dramatic Fish Kills, Illustrating Density-Independent Mortality ________________________________________________________________________ Commercial Catch Percent --------------------------- Locality Before After Decline ________________________________________________________________________ Matagorda 16,919 1,089 93.6 Aransas 55,224 2,55295.4 Laguna Madre 12,016 149 92.6 ________________________________________________________________________ Note: These fish kills resulted from severe cold weather on the Texas Gulf Coast in the winter of 1940. Source: Gunter (1941). The significance of opportunistic versus equilibrium populations is that density-independent and density-dependent factors and events differ in their effects on natural selection and on populations. In highly variable and/or unpredictable environments, catastrophic mass mortality (such as that illustrated in Table 9.1) presumably often has relatively little to do with the genotypes and phenotypes of the organisms concerned or with the size of their populations. (Some degree of selective death and stabilizing selection has been demonstrated in winter kills of certain bird flocks.) By way of contrast, under more stable and/or predict-able environmental regimes, population densities fluctuate less and much mortality is more directed, favoring individuals that are better able to cope with high densities and strong competition. Organisms in highly rarefied environments seldom deplete their resources to levels as low as do organisms living under less rarefied situations; as a result, the former usually do not encounter such intense competition. In a "competitive vacuum" (or an extensively rarefied environment) the best reproductive strategy is often to put maximal amounts of matter and energy into reproduction and to produce as many total progeny as possible as soon as possible. Because competition is weak, these offspring often can thrive even if they are quite small and therefore energetically inexpensive to produce. 1. Figure 9.7. Population growth trajectories in an equilibrium species versus an opportunistic species subjected to irregular catastrophic mortality. However, in a "saturated" environment, where density effects are pronounced and competition is keen, the best strategy may often be to put more energy into competition and maintenance and to produce offspring with more substantial competitive abilities. This usually requires larger offspring, and because they are energetically more expensive, it means that fewer can be produced. MacArthur and Wilson (1967) designate these two opposing selective forces r- selection and K-selection, after the two terms in the logistic equation (however, one should not take these terms too literally, as the concepts are independent of the equation). Of course, things are seldom so black and white, but there are usually all shades of gray. No organism is completely r-selected or completely K-selected; rather all must reach some compromise between the two extremes. Indeed, one can think of a given organism as an "r-strategist" or a "K-strategist" only relative to some other organism; thus statements about r- and K-selection are invariably comparative. We think of an r- to K-selection continuum and an organism's position along it in a particular environment at a given instant in time (Pianka 1970, 1972). Table 9.2 lists a variety of correlates of these two kinds of selection. Table 9.2 Some of the Correlates of r- and K-Selection __________________________________________________________________ r-selection K-selection __________________________________________________________________ ClimateVariable and unpredictable;Fairly constant or pre- uncertaindictable; more certain MortalityOften catastrophic,More directed, density nondirected, densitydependent independent SurvivorshipOften Type IIIUsually Types I and II Population sizeVariable in time, nonequil-Fairly constant in time, ibrium; usually well belowequilibrium; at or near carrying capacity of envi-carrying capacity of the ronment; unsaturated com-environment; saturated munities or portions thereof; communities; no recolon- ecologic vacuums; recolon-ization necessary ization each year Intra- and inter-Variable, often lax Usually keen specific competition Selection favors1. Rapid development 1. Slower development 2. High maximal rate of 2. Greater competitive increase, rmax ability 3. Early reproduction 3. Delayed reproduction 4. Small body size 4. Larger body size 5. Single reproduction 5. Repeated reproduction 6. Many small offspring 6. Fewer, larger progeny Length of lifeShort, usually less than a yearLonger, usually more than a year Leads toProductivityEfficiency Stage in successionEarlyLate, climax __________________________________________________________________ Source: After Pianka (1970). An interesting special case of an opportunistic species is the fugitive species, envisioned as a predictably inferior competitor that is always excluded locally by interspecific competition but persists in newly disturbed regions by virtue of a high dispersal ability (Hutchinson 1951). Such a colonizing species can persist in a continually changing patchy environment in spite of pressures from competitively superior species. Hutchinson (1961) used another argument to explain the apparent "paradox of the plankton," the coexistence of many species in diverse planktonic communities under relatively homogeneous physical conditions, with limited possibilities for ecological separation. He suggested that temporally changing environments may promote diversity by periodically altering relative competitive abilities of component species, thereby allowing their coexistence. McLain (1991) suggested that the relative strength of sexual selection depends on the life history strategy, with r-strategists being less likely to be subjected to strong sexual selection than K-strategists. Winemiller (1989, 1992) points out that reproductive tactics among fishes (and probably all organisms) can be placed on a two-dimensional triangular surface in a three-dimensional space with the coordinates: juvenile survivorship, fecundity, and age of first reproduction or generation time (Figure 9.8). This two-dimensional triangular surface has three vertices corresponding to equilibrium (K-strategists), opportunistic, and seasonal species. The r- to K-selection continuum runs diagonally across this surface from the equilibrium corner to the opportunistic-seasonal edge. In fish, seasonal breeders exhibit little sexual dimorphism, whereas both opportunistic and equilibrium species display marked sexual dimorphisms (Winemiller 1992). 1. Figure 9.8. Model for a triangular life history continuum. Three-dimensional representation of reproductive tactics depicting both the r-K-selection continuum and a bet hedging axis. [After Winemiller (1992).] Under situations where survivorship of adults is high but juvenile survival is low and highly unpredictable, there is a selective disadvantage to putting all one's eggs in the same basket, and a consequent advantage to distributing reproduction out over a period of time (Murphy 1968). This sort of reproductive tactic has become known as "bet hedging" (Stearns 1976) and occurs in both r-strategists and K-strategists. Winemiller (1992) points out that a bet-hedging axis passes across his triangular surface from the opportunistic corner endpoint to the edge connecting the seasonal and equilibrium tactics (Figure 9.8). Population Regulation In the majority of real populations that have been examined, numbers are kept within certain bounds by density-dependent patterns of change. When population density is high, decreases are likely, whereas increases tend to occur when populations are low (Tanner 1966; Pimm 1982). If the proportional change in density is plotted against population density, inverse correlations usually result (Figure 9.9). 1. Figure 9.9. Increases and decreases in population size plotted against population size in the year preceding the increase or decrease for an ovenbird population in Ohio over an 18-year period. [From MacArthur and Connell (1966).] Table 9.3 summarizes such data for a variety of populations, including humans (the only species with a significant non-negative correlation). Such negative correlations are found even in cyclical and erratic populations such as those considered in the next section. Table 9.3 Frequencies of Positive and Negative Correlations Between Percentage Change in Density and Population Density for a Variety of Populations in Different Taxa _________________________________________________________________________________________________ Numbers of Populations in Various Categories PositivePositiveNegativeNegativeNegative Taxon(P<.05)(Not sig.) (Not sig.)(P<.10) (P < .05) Total _________________________________________________________________________________________________ Invertebrates (not insects)000044 Insects00 71715 Fish01 2 04 7 Birds0232164393 Mammals10411319 Totals13451871138 _________________________________________________________________________________________________ Homo sapiens Sources: Tanner (1966) and Pimm (1982) During the past half century, the human population, worldwide, has doubled from about 3 billion people to almost 6.8 billion. 6,800,000,000 is a rather large number, difficult to comprehend. Each year, the human population increases by nearly 100 million, a daily increase of more than one-quarter of a million souls. Each hour, every day, day in and day out, over 11,000 more people are born than die. Most people hold the anthropocentric opinion that planet Earth (our one and only Spaceship!) and all its resources exist primarily, or even solely, for human exploitation. Genesis prescribes: "Be fruitful, and multiply, and replenish the earth, and subdue it: and have dominion over the fish of the sea, and over the fowl of the air, and over every living thing that moveth upon the earth". We have certainly lived up to everything except "replenish the earth." The human population explosion has been fueled by habitat destruction1 -- we are usurping resources once exploited by other species. Tall grass prairies of North America have been replaced with fields of corn and wheat, native American bison have given way to cattle. In 1986, humans consumed (primarily via fisheries, agriculture, pastoral activities, and forestry) an estimated 40 percent of the planet's total production (Vitousek et al. 1986). Today we consume more than half of the solar energy trapped by plants (Vitousek et al. 1997). More atmospheric nitrogen is fixed by humanity than by all other natural terrestrial sources combined. Humans have transformed nearly one-half of the earth's land surface. More than half of all accessible surface freshwater is now used by humans. Freshwater aquatic systems everywhere are polluted and threatened. Fish and frogs are seriously threatened. All the oceans have been heavily overfished. Many species have gone extinct due to human pressures over the past century and many more are threatened and endangered. Nearly one-quarter of earth's bird species have already been driven extinct by inane human activities such as species introductions and habitat destruction. People everywhere today stand ready to rape and pillage their wildernesses ("wastelands") for whatever they can be forced to yield. Raw materials, such as ore, lumber, and even sand (used to make glass), are harvested in vast quantities. Big companies enjoy privileged status, excluding the public from extensive areas, producing great ugly clear cuts, vast strip mines, deep open pit mines, instant but permanent man-made mountains, eyesores paying testimony to the avaricious pursuit of timber, precious metals, and minerals. Deforestation is nearly complete in many parts of the world. Overgrazing is rampant. Grasses and the shrub understory have been virtually eliminated over extensive areas. Fenced graveyards sometimes protect small patches of country showing how they must have been before the land rape by the pastoral industry. Native hardwoods are wasted to make charcoal and burned for firewood. Lumberjacks will soon be out of work whether or not the remaining timber is cut. Should forest habitats be saved? Is there enough left to save? This sort of pillage continues. Virtually everywhere, often with governmental subsidies and incentives, forests, deserts, and scrublands are being leveled and turned into fields for crops. Many of these fields are marginal and will soon have to be abandoned, transformed into great man-made vegetationless deserts. Couple such activities with global warming, and more dust bowls are in the making. In some regions, replacement of the drought-adapted deep rooted native vegetation with shallow-rooted crop plants has reduced evapotranspiration, thus allowing the water table to rise, bringing deep saline waters to the surface. Such salinization reduces productivity and seems to be irreversible. Some deserts have so far been able to resist the tidal wave of advancing human exploiters, but there are people who dream of the day that technological "advances," such as water plants to move "excess" water or to distill seawater, will make it possible to develop desert regions (i.e., to replace them with vast agricultural fields, or even cities). Antonyms, such as "sustainable development," are strung together into oxymorons by biopoliticians and developers in an attempt to make all this destruction and homogenization seem less offensive. Most people consider basic biology, particularly ecology, to be a luxury that they can do without. Even many medical schools no longer require that premedical students obtain a biological major. But basic biology is not a luxury at all; rather it is an absolute necessity for living creatures such as ourselves. Despite our anthropocentric (human-centered) attitudes, other life forms are not irrelevant to our own existence. As proven products of natural selection that have adapted to natural environments over millennia, they have a right to exist, too. With human populations burgeoning and pressures on space and other limited resources intensifying, we need all the biological knowledge that we can possibly get. For example, in this day and age, a primer on "how to be a successful venereal microbe" has become essential reading for everyone! Ecological understanding is particularly vital. Basic ecological research is urgent because the worldwide press of humanity is rapidly driving other species extinct and destroying the very systems that ecologists want to understand. No natural community remains undisturbed by humans. Pathetically, many will disappear without even being adequately described, let alone remotely understood. As existing species go extinct and even entire ecosystems disappear, we lose forever the very opportunity to study them. Knowledge of their evolutionary history and adaptations vanishes with them: thus we are losing access to biological information itself. Only during the last few generations have biologists been fortunate enough to be able to travel with ease to remote wilderness areas. Panglobal comparisons have broadened our horizons immensely. This is a fleeting and unique opportunity in the history of humanity, for never before could scientists get virtually anywhere. However, all too soon, there won't be any even semipristine natural habitats left to study. More than 30 years ago, in a set piece of rational thought that deserves much more attention than it has so far received, Garrett Hardin (1968) perceived a fly in the ointment of freedom, which he explained as follows: "The tragedy of the commons develops in this way. Picture a pasture open to all. It is to be expected that each herdsman will try to keep as many cattle as possible on the commons. Such an arrangement may work reasonably satisfactorily for centuries because tribal wars, poaching, and disease keep the numbers of both man and beast well below the carrying capacity of the land. Finally, however, comes the day of reckoning, that is, the day when the long-desired goal of social stability becomes a reality. At this point, the inherent logic of the commons remorselessly generates tragedy. As a rational being, each herdsman seeks to maximize his gain. Explicitly or implicitly, more or less consciously, he asks, "What is the utility to me of adding one more animal to my herd?" This utility has one negative and one positive component. 1) The positive component is a function of the increment of one animal. Since the herdsman receives all the proceeds from the sale of the additional animal, the positive utility is nearly +1. 2) The negative component is a function of the additional overgrazing created by one more animal. Since, however, the effects of overgrazing are shared by all the herdsmen, the negative utility for any particular decision-making herdsman is only a fraction of -1. Adding together the component partial utilities, the rational herdsman concludes that the only sensible course for him to pursue is to add another animal to his herd. And another; and another . . . But this is the conclusion reached by each and every rational herdsman sharing a commons. Therein is the tragedy. Each man is locked into a system that compels him to increase his herd without limit in a world that is limited. Ruin is the destination toward which all men rush, each pursuing his own interest in a society that believes in the freedom of the commons. Freedom of the commons brings ruin to all." The tragedy of the selfish herdsman on a common grazing land is underscored by the rush to catch the last of the great whales and the ongoing destruction of earth's atmosphere (ozone depletion, acid rain, CO2 enhanced greenhouse effect, etc.). Global weather modification is a very real and an exceedingly serious threat to all of us, as well as to other species of plants and animals. Over the past few hundred years, we humans have drastically engineered our own environments, creating a modern-day urbanized indoor society that simply did not exist a mere few centuries ago. Indeed, only 500 human generations ago (about 10,000 years) we were living in caves under stone-age conditions! Our only source of light other than that of the sun, moon, and stars was firelight from campfires -- replaced now with television sets glimmering in the dark! (We're using oil to generate electricity and sending fossil sunlight back out into space!) Once "primitive" hunter-gatherers who regularly walked extensive distances and worked hard to collect enough food to stay alive, we lived in intimate small clan groups planning and plotting to somehow survive winters (greed may have been an evolutionary advantage) -- we struggled to escape all sorts of natural hazards long since eliminated, but replaced with new and markedly different dangers. We propelled ourselves into a complex, brand new, human-engineered urban world with artificial lights, electricity, air conditioning, computers, email, money, shopping markets (= ample cheap food for many), antibiotics, drugs, cars, airplanes, long-distance travel, overcrowding, regimentation, phones, and television. Humans simply haven't had time enough to adapt to all these novel environmental challenges -- today, we find ourselves misfits in our own strange new world (someone dubbed us "Stone Agers in the fast lane"). Health consequences of this self-induced mismatch between organism and environment are many and varied: anxiety, depression, schizophrenia, drug abuse (including both alcohol and tobacco), obesity, myopia, diabetes, asthma and other allergies, impotency, infertility, birth defects, bullets, new microbes and resistant strains, environmental carcinogens, and other toxins. Convincing arguments suggest that we have actually produced all these maladies as well as many more. Amazingly, we have managed to create a world full of pain and suffering unknown to our stone-age ancestors just a few hundred generations ago. What a magnificent accomplishment! People sometimes ask, "What is the carrying capacity for humans?" Nearly seven billion of us occupy roughly half of earth's land surface, consuming over half the freshwater and using about half of earth's primary productivity, but many of those persons are living in poverty and not getting adequate nutrition (somewhat of an understatement, considering famines in Africa and Asia). Certainly if our population continues to double in the next 40 years as it did during the past 40, we will finally have reached our carrying capacity at 12-14 billion in the year 2050 -- at that population density, humans will occupy all Earth's surface (there won't be any more wilderness or any wild animals), and we will be using every drop of freshwater as well as every photon that intercepts the surface of our one and only Spaceship! Population "Cycles": Cause and Effect Let's return to something more fundamental. Ecologists have long been intrigued by the regularity of certain population fluctuations, such as those of the snowshoe hare, the Canadian lynx, the ruffed grouse, and many microtine rodents (voles and lemmings) as well as their predators, including the arctic fox and the snowy owl (Elton 1942; Keith 1963). These population fluctuations (sometimes called "cycles," although they should not be) are of two types: voles, lemmings, and their predators display roughly a 4-year periodicity; hare, lynx, and grouse have approximately a 10-year cycling time. Lemming population eruptions and the fabled, but very rare, suicidal marches of these rodents into the sea have frequently been popularized (and even staged for movie production!) and are therefore all too "well known" to the lay population. The tantalizing regularity of these fluctuations in population density presents ecologists with a "natural experiment" -- hopefully one that can provide some general insights into factors influencing population densities. Many different hypotheses for the explanation of population cycles have been offered, and the literature on them is extensive (see references at end of chapter). Here, as elsewhere in ecology, it is often extremely difficult or even impossible to devise tests that separate cause from effect, and many putative causes of population cycles may in fact merely be side effects of cyclical changes in populations concerned. Several currently popular hypotheses, which are not necessarily mutually exclusive, are outlined subsequently. Descriptions and discussion of others, such as the "random peaks" hypothesis, can be found in the references. Bear in mind that two or more of these hypothetical mechanisms could act together in any given situation. Sunspot Hypothesis Earth's Sun undergoes solar cycles that exhibit a periodicity of about 10 years. Moreover, this cycle is embedded in a longer cycle with 30-50-year periods of low activity, followed by 30-50-year periods of high activity. Few sunspots occurred during the Maunder minimum (1645-1715). Three periods of high sunspot maxima have occurred (1751-1787, 1838-1870, and 1948-1993). From 1931-1948, the Canadian government compiled information on snowshoe hare populations -- the cycle was synchronized over all of Canada and Alaska, suggesting that an external forcing variable might be acting at a continental scale (Sinclair et al. 1993). These workers took tree ring cores from 368 spruce trees along a 5-km strip in the Yukon (one tree germinated in 1675!). Hares prefer fast-growing palatable shrubs such as birch and willow and avoid slow-growing unpalatable white spruce, except at high hare densities when they eat apical shoots of small spruce trees (less than 50 years old, under 150 cm high). Hares cannot reach taller older trees. As a result, during hare outbreaks, dark tree ring marks are formed. Frequency of marks correlates with the Hudson Bay Fur Company records (1751-1983). Both hare numbers and tree ring marks are correlated with sunspot numbers. Annual net snow accumulation is also in phase with hare numbers, tree marks, and sunspot activity. Sinclair et al. (1993) suggest that the sunspot solar cycle affects climate and that solar activity and climate in turn entrain hare populations (and tree ring marks). These workers suggest several tests of their hypothesis: (1) More tree ring mark data need to be gathered for very old spruce trees to study the Maunder minimum -- they predict no 10-year cycle or other periodicities would occur then. (2) Tree ring marks in old spruce that were out of reach of hares should not show the 10-year periodicity. (3) Spruce in other regions should show tree ring mark cycles like those observed in the Yukon. (4) Hare cycles in Siberia should be in phase with those in Canada. However, autocorrelations of hare population dynamics with sunspot cycles in Finland differ markedly from those in Canada (Ranta et al. 1997). Time Lags Response time (1/r) to changing density plus the time lag associated with the response can generate stable limit cycles (Gotelli 1995). This occurs when time lags are large relative to response time. In effect, the population repeatedly overshoots its carrying capacity and then crashes below K before overshooting again. Moreover, regardless of the rate of increase, the period of the cycle is always about four times the time lag -- if a population is at K, the first lag period puts it below K, the second lag period brings it back up to K, the third lag takes it above K, and the fourth brings it back down to K, completing one full cycle. Gotelli (1995) suggests that this could be a reason why microtine populations at high latitudes show peaks every four years (time lag = about one year). Stress Phenomena Hypothesis At the extremely high densities that occur during population peaks of voles, a great deal of fighting occurs among these rodents. Body sizes fluctuate with population density, such that animals are larger at peak densities. The so-called stress syndrome is manifested by the animals, their adrenal weights increase, and they become extremely aggressive -- so much so that successful reproduction is almost completely curtailed. Eventually "shock disease" may set in and large numbers of animals may die off, apparently because of the physiological stresses on them. Christian and Davis (1964) review evidence pertaining to this rather mechanistic hypothesis. Some plausible extensions to the stress hypothesis can be made using the ideas of optimal reproductive tactics. Recall that, because current offspring are "worth more" in an expanding population, there is an advantage to early and intense reproduction (high reproductive effort). However, as a population ceases to grow and enters into a decline, the opposite situation arises, favoring little or no current reproduction. Also, if, as seems highly likely, juvenile survivorship diminishes as population density increases, profits to be gained from reproduction would also decrease. Curtailment of present reproduction and total investment in aggressive survival activities could repay an individual that survived the crash with the opportunity for "sweepstakes" reproductive success! Predator-Prey Oscillation Hypothesis In simple ecological systems, predator and prey populations can oscillate because of the interaction between them. When the predator population is low, the prey increase, which then allows the predators to increase -- although this increase lags behind that of the prey. Eventually, predators overeat their prey and the prey population begins to decline; but because of time lag effects, the predator population continues to increase for a period, driving the prey to an even lower density. Finally, at low enough prey densities, many predators starve and the cycle repeats itself. However, prey populations oscillating for reasons other than predation pressures obviously constitute cyclical food supplies for their predators, which should in turn lag behind and oscillate with prey availability. Short of a predator removal experiment, it is thus extremely difficult to determine whether or not changes in prey populations are causally related to changes in predator population density. (Prey populations sometimes fluctuate regularly even in the absence of predators.) Epidemiology-Parasite Load Hypothesis Under this hypothesis, parasite epidemics spread through oscillating populations at peaks, causing massive mortality. But because parasites fail to find their hosts very efficiently at low densities, most hosts are unparasitized and the population increases again until another parasite epidemic brings it back under control. Food Quantity Hypothesis Arctic hare populations sometimes oscillate without lynx populations, perhaps due to a predator-prey "cycle" involving themselves as predators and their own food plants as prey. Under this hypothesis, dense hare populations decrease the quantity of suitable foods, which in turn causes a decline in the hare population. In time the plants recover, and after a lag period hares again increase. Clearly, lynx could entrain to such a cycle. Supplemental food was provided to a declining population of Microtus (Krebs and DeLong 1965), but this failed to reverse the decline demonstrating that the food quantity mechanism does not apply to microtine cycles in general. More unambiguous experiments like this one are needed to assess the importance of various mechanisms of population control. Nutrient Recovery Hypothesis According to this hypothesis (Pitelka 1964; Schultz 1964, 1969), one reason for the periodic decline of rodent populations (especially lemmings) is that the quality of their plant food changes in a cyclical way. During a lemming "high," the ground is blanketed with lemming fecal pellets, and many important chemical elements such as nitrogen and phosphorus are tied up and unavailable to growing plants. In the cold arctic tundra, decomposition of fecal materials takes a long time. During this period, lemmings decline due to inadequate nourishment. After a lapse of a few years, feces are decomposed and their nutrients recycled once again and taken up by plants. Because their plant food is now especially nutritious, lemmings increase in numbers and the cycle repeats. Schultz (1969) found evidence for such cyclical changes, but whether they were causes, or merely side effects, of the lemming population fluctuations was not established. Other Food Quality Hypotheses Freeland (1974) proposed a related hypothesis based on changes in the relative abundances of toxic versus palatable food plants due to preferential grazing pressures by voles. He suggested that voles graze back competitively superior palatable plants during rodent outbreaks, which allows the competitively inferior unpalatable plant species to spread. Thus, rodent diets must shift to toxic plants and voles do not fare as well at high densities as they do at low ones. A similar process, but one involving induced chemical defenses of individual plants, was suggested for snowshoe hares by Pease et al. (1979) and Bryant (1980a, 1980b). According to this hypothesis, heavily grazed plant individuals respond to intense herbivory by producing heavily protected new growth that, in turn, constitutes relatively poor fodder for snowshoe hares. Genetic Control Hypothesis This hypothesis, credited to Chitty (1960, 1967a), explains population fluctuations in terms of changing genetic composition of the population concerned. During troughs, the animals experience little competition and are relatively r-selected, whereas at peaks competition is intense and they are more K-selected. Thus, directional selection, related to population density, is always occurring; modal phenotypes are never the most fit individuals in the population, and in each generation the gene pool changes. The population always lags somewhat behind the changing selective pressures and so no stable equilibrium exists. Some evidence exists for such genetic changes in populations of Microtus (Tamarin and Krebs 1969), but here again, cause and effect are extremely difficult to disentangle. A number of interesting observations have been made on various fluctuating populations that are relevant to some of the above hypotheses. Fenced populations of voles reach much higher population densities than do unfenced ones (Krebs et al. 1969), and, as a consequence, overgraze their food supplies more seriously than unfenced natural populations. Fencing presumably does not alter predation levels, but prevents emigration of both juvenile and subordinate animals, thereby raising local density. Krebs (personal communication) suggests that fencing alters spacing behavior and dispersal, hence restricting "self-regulation" of the population. Such an argument borders on group selection (see also Wynne-Edwards 1962, and Williams 1966a), but may be compatible with classical Darwinian selection at the level of the individual if carefully rephrased. Population fluctuations are sometimes out of synchrony within fairly close proximity of one another. In one such situation, marked individuals from a population nearing its peak and about to decline were transferred to a nearby population that was just beginning to increase (Krebs, unpublished, as reported in Putnam and Wratten 1984). Transplanted animals peaked and declined as observed in their own host populations but did not adopt the behavior of the increasing alien population into which they were introduced. These observations suggest that factors intrinsic to the animals, rather than extrinsic phenomena, must be involved. According to Hanski et al. (1991), specialized mammalian predators (small mustelid weasels) maintain or enhance the cyclicity of microtine rodent populations, whereas generalist predators (larger mammals and hawks) stabilize rodent populations, particularly at lower latitudes. These authors argue that small mustelids are major predators at high latitudes in Scandinavia but that generalist predators become increasingly important as latitude decreases. Both the amplitude and the length of microtine cycles increase with latitude in Scandinavia (Hanski et al. 1991), as predicted by this theory. Some of the various factors, and how they change with lemming population size over a typical "cycle," are summarized in Figure 9.10. 1. Figure 9.10. A schematic representation of various factors involved in the population "cycle" of lemmings. Events and factors with a negative influence on the population are shown with solid arrows; those with a beneficial influence depicted with a dashed arrow. D, disease; A, aggression; M, migration; and R, recovery of health. [From Itô (1980).] One must always be wary of oversimplification and "single-factor thinking"; most or even all of the preceding hypothetical mechanisms could work in concert to produce observed population "cycles." The extreme difficulty of separating cause from effect, illustrated above, plagues much of ecology. Simple tests that actually refute hypotheses are badly needed. For scientific understanding to progress rapidly and efficiently, a logical framework of refutable hypotheses, complete with alternatives, is most useful (Platt 1964). However, while such a single-factor approach may work quite satisfactorily for systems exhibiting simple causality, it has proven to be distressingly ineffective in dealing with ecological problems where multiple causality is at work. Once again, one of the major dilemmas in ecology seems to be finding effective ways to deal with multiple causality. Selected References Verhulst-Pearl Logistic Equation Allee et al. (1949); Andrewartha and Birch (1954); Bartlett (1960); Beverton and Holt (1957); Chitty (1960, 1967a, b); Clark et al. (1967); Cole (1965); Ehrlich and Birch (1967); Errington (1946, 1956); Fretwell (1972); Gadgil and Bossert (1970); Gibb (1960); Green (1969); Grice and Hart (1962); Hairston, Smith, and Slobodkin (1960); Horn (1968a); Hutchinson (1978); Krebs (1972); McLaren (1971); Murdoch (1966a, b, 1970); Nicholson (1933, 1954, 1957); Pearl (1927, 1930); Slobodkin (1962b); F. E. Smith (1952, 1954, 1963a); Solomon (1949, 1972); Southwood (1966); Williamson (1971); Wilson and Bossert (1971). Density Dependence and Density Independence Andrewartha (1961, 1963); Andrewartha and Birch (1954); Brockelman and Fagen (1972); Davidson and Andrewartha (1948); Ehrlich et al. (1972); Gunter (1941); Horn (1968a); Hutchinson (1978); Lack (1954, 1966); McLaren (1971); Nicholson (1957); Orians (1962); Pianka (1972); F. E. Smith (1961, 1963b). Opportunistic versus Equilibrium Populations Ayala (1965); Charlesworth (1971); Clarke (1972); Dobzhansky (1950); Force (1972); Gadgil and Bossert (1970); Gadgil and Solbrig (1972); Grassle and Grassle (1974); Grime (1979); Hutchinson (1951, 1961); Itô (1980); King and Anderson (1971); Lewontin (1965); Luckinbill (1979); MacArthur (1962); MacArthur and Wilson (1967); McLain (1991); Menge (1974); Pianka (1970, 1972); Roughgarden (1971); Seger and Brockmann (1987); Wilson and Bossert (1971); Winemiller (1992). Population Regulation Krebs (1995); Lack (1954, 1966); MacArthur and Connell (1966); Neese and Wil- liams (1994); Pimentel (1968); Pimm (1982); Smith (1961, 1963); Solomon (1972); St. Amant (1970); Tanner (1966); Vitousek et al. (1986, 1997). Population "Cycles": Cause and Effect Bryant (1980a, b); Bryant, Chapin and Klein (1983); Chitty (1960, 1967a, 1996); Christian and Davis (1964); Cole (1951, 1954a); Elton (1942); Freeland (1974); Gilpin (1973); Hanski et al. (1991); Hilborn and Stearns (1982); Itô (1980); Keith (1963, 1974); Kendall et al. (1999); Krebs (1964, 1966, 1970, 1978); Krebs and DeLong (1965); Krebs and Myers (1974); Krebs, Keller, and Myers (1971); Krebs, Keller, and Tamarin (1969); Lidicker (1988); Krebs et al (1995); O'Donoghue et al (1998); Pease et al. (1979); Pitelka (1964); Platt (1964); Putman and Wratten (1984); Ranta et al. (1997); Schaeffer and Tamarin (1973); Schultz (1964, 1969); Sinclair et al (1993); Tamarin and Krebs (1969); Wellington (1960); Williams (1966a); Wynne-Edwards (1962). 1. Of course, massive consumption of fossil fuels for argriculture has also contributed greatly to overpopulation. \|
2956	https://www.collinsdictionary.com/us/dictionary/english/celerity	English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese More English Italiano Português 한국어 简体中文 Deutsch Español हिंदी 日本語 English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese Definitions Summary Synonyms Sentences Pronunciation Collocations Conjugations Grammar Credits × Definition of 'celerity' COBUILD frequency band celerity in American English (səˈlɛrɪti ) nounOrigin: Fr célérité < L celeritas < celer, swift: see hold1 swiftness in acting or moving; speed Webster’s New World College Dictionary, 5th Digital Edition. Copyright © 2025 HarperCollins Publishers. You may also like English Quiz ConfusablesSynonyms of 'celerity'Language Lover's BlogFrench Translation of 'celerity'Translate your textPronunciation PlaylistsWord of the day: 'hwyl'Spanish Translation of 'celerity'English GrammarCollins AppsEnglish Quiz ConfusablesSynonyms of 'celerity'Language Lover's BlogFrench Translation of 'celerity'Translate your textPronunciation PlaylistsWord of the day: 'hwyl'Spanish Translation of 'celerity'English GrammarCollins AppsEnglish Quiz ConfusablesSynonyms of 'celerity'Language Lover's Blog COBUILD frequency band celerity in American English (səˈlerɪti) noun swiftness; speed SYNONYMS alacrity, dispatch, briskness. See speed. Most material © 2005, 1997, 1991 by Penguin Random House LLC. Modified entries © 2019 by Penguin Random House LLC and HarperCollins Publishers Ltd Word origin [1480–90; earlier celerite ‹ MF ‹ L celeritās, equiv. to celer swift + -itās -ity] COBUILD frequency band celerity in British English (sɪˈlɛrɪtɪ ) noun rapidity; swiftness; speed Collins English Dictionary. Copyright © HarperCollins Publishers Word origin C15: from Old French celerite, from Latin celeritās, from celer swift Examples of 'celerity' in a sentence celerity These examples have been automatically selected and may contain sensitive content that does not reflect the opinions or policies of Collins, or its parent company HarperCollins. We welcome feedback: report an example sentence to the Collins team. Read more… He took his time about it: celerity had no part in the due processes of the law. Aird, Catherine A DEAD LIBERTY (2000) From the celerity with which the headwaiter approached her I decided she must be very rich or very distinguished. Elizabeth Peters LION IN THE VALLEY (2000) She moves with incredible celerity, able to move between streets and run across hallways appearing as nothing more than a blurred image. Retrieved from Wikipedia CC BY-SA 3.0 Source URL: Other aspects of the work proceeded with similar celerity, and the tower was completed and fully operational by the following summer. Retrieved from Wikipedia CC BY-SA 3.0 Source URL: Manners, civility, celerity, precision, class and clarity. Retrieved from Wikipedia CC BY-SA 3.0 Source URL: Synonyms of 'celerity' speed, expedition, dispatch, velocity More Synonyms of celerity Trends of celerity View usage over: Source: Google Books Ngram Viewer Browse alphabetically celerity celebutante celecoxib celeriac celerity celery celery cabbage celery heart All ENGLISH words that begin with 'C' ## Wordle Helper ## Scrabble Tools Quick word challenge Quiz Review Question: 1 - Score: 0 / 5 FLOWERS What is this an image of? chrysanthemumlily-of-the-valleybuttercuporchid FLOWERS What is this an image of? dahliadandelioncarnationiris FLOWERS What is this an image of? violetdandeliondaisytulip FLOWERS Drag the correct answer into the box. dahlia iris chrysanthemum daffodil FLOWERS Drag the correct answer into the box. daisy iris dandelion carnation Your score: New collocations added to dictionary Collocations are words that are often used together and are brilliant at providing natural sounding language for your speech and writing. 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2957	https://www.omnicalculator.com/math/kite-area	Kite Area Calculator If you are looking for the formula for kite area or perimeter, you're in the right place: the kite area calculator is here to help you. Whether you know the length of the diagonals or two unequal side lengths and the angle between, you can quickly calculate the area of a kite. For the kite perimeter, all you need to do is enter two kite sides. But if you are still wondering how to find the area of a kite, keep scrolling! If it's not a kite area you are looking for, check our kiteboarding calculator, which can help you choose the proper kite size. Kite area formula A kite is a quadrilateral with two pairs of equal-length sides adjacent to each other. A kite is a symmetric shape, and its diagonals are perpendicular. There are two basic kite area formulas, which you can use depending on which information you have: If you know two diagonals, you can calculate the area of a kite as: area = (e × f) / 2 , where e and f are kite diagonals. If you know two non-congruent side lengths and the size of the angle between those two sides, use the formula: area = a × b × sin(α), where α is the angle between sides a and b. Did you notice that it's a doubled formula for the triangle area, knowing side-angle-side? Yes, that's right! A kite is a symmetric quadrilateral and can be treated as two congruent triangles that are mirror images of each other. Kite perimeter To calculate the kite perimeter, you need to know two unequal sides. Then, the formula is obvious: perimeter = a + a + b + b = 2 × (a + b) You can't calculate the perimeter knowing only the diagonals – we know that one is a perpendicular bisector of the other diagonal, but we don't know where is the intersection. How to find the area of the kite? How about kite perimeter? Let's imagine we want to make a simple, traditional kite. How much paper/foil do we need? And if we're going to make an edging from a ribbon, what length is required? Think for a while and choose the formula which meets your needs. Assume we found two sticks in the forest; let's use them for our kite! Enter the diagonals of the kite. The ones we have are 12 and 22 inches long. Area of a kite appears below. It's 132 in². Calculation of the kite perimeter is a bit tricky in that case. Let's have a look: Assume you've chosen the final kite shape – you've decided where the diagonals intersect each other. For example, the shorter one will be split in the middle (6 in : 6 in) and the longer one in the 8:14 ratio, as shown in the picture. Next, the easiest way is to use our right triangle calculator (this method works only for convex kites). Type 6 and 8 as a and b – the hypotenuse is one of our kite sides, here equal to 10 in. Refresh the calculator and enter 6 and 14 – the result is 15.23 in, and that's our other side. Here you go! As we know both sides, we can calculate the perimeter. Type the a and b sides. The result for our case is 50.46 in. So buy a little bit more ribbon than that, for example, 55 inches, to make the edging. Convex and concave kites The kite can be convex – it's the typical shape we associate with the kite – or concave; such kites are sometimes called a dart or arrowheads. The area is calculated in the same way, but you need to remember that one diagonal is now "outside" the kite. The kite area calculator will work properly also for the concave kites. Is kite a rhombus? The answer is almost always no. It's working the other way around – every rhombus is a kite. Only if all four sides of a kite have the same length, it must be a rhombus (see the rhombus area calculator) – or even a square, if all the angles are right. FAQs What is the formula to find the area of a kite? You can find the area of a kite using the following two formulas: If you know the lengths of both diagonals e and f, you can use: Area = (e × f) / 2 Otherwise, if you know two non-congruent side lengths a and b and the angle α between them, you can use: Area = a × b × sin(α) How can I find the area of a kite? To calculate the area of your kite, follow the steps below: Measure the lengths of the two diagonals of your kite. Multiply the values of the two diagonals. Divide the product of diagonals by 2. The result is the area of the kite. What is a kite in geometry? A kite is a four-sided, quadrilateral shape with two pairs of equal-length sides. Each pair of opposite angles is the same in size, and the diagonals, lines from one corner to the opposite, intersect at right angles, cutting each other in half. What is the area of the kite with 5.3 ft and 6ft diagonals? The area of your kite is 15.9 ft². To calculate it, plug your given diagonals e = 5.3 ft and f = 6 ft into the following formula: Area = (e × f) / 2 Area = (5.3 ft × 6 ft) / 2 Area = 31.8 ft² / 2 Area = 15.9 ft². Perimeter Did we solve your problem today? Check out 25 similar 2d geometry calculators 📏 Area Area of a rectangle Area of crescent
2958	https://www.scribd.com/document/138931726/Result-Complement-Chinese-Grammar-Wiki	Result Complement - Chinese Grammar Wiki \| PDF \| Verb \| Languages Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 721 views 5 pages Result Complement - Chinese Grammar Wiki Result complements are used in Chinese to describe the result of a verb. Many Chinese verbs only indicate an action without specifying the outcome, so result complements are needed to provid… Full description Uploaded by warner AI-enhanced description Go to previous items Go to next items Download Save Save Result Complement - Chinese Grammar Wiki For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Result Complement - Chinese Grammar Wiki For Later You are on page 1/ 5 Search Fullscreen From Chinese Grammar Wiki Also known as: 结果补语 (jiégu ǒ b ǔ y ǔ ), complement of r esult, r esultative complement and result compound. Result complements are a kind of verbal complement that appears very frequently in Chinese. Surprisingly enough, they're used to describe the result of a verb.1 Why result complements are necessary in Chinese 2 Forming the result complement 3 Negating result complements 4 Aspect particles with result complements 5 Result complements and questions 6 Common result complement compounds 7 Result complements in 把 sentences 7.1 See also 8 Sources and further reading 8.1 Books 8.2 Websites In English, we have separate verbs to describe actions depending on their outcome. You can look, but not see, or listen, but not hear. In Chinese, verbs tend to be of the 'look' and 'listen' variety - the result isn't included like it is in 'see' and 'hear'. So if you want to indicate the result, you have to use a result complement.There are endless examples of these 'attempt' or 'procedure' verbs in Chinese which don't include an outco me. T he majority o f Chinese verbs are li ke this, in fact. Some examples: 看 means 'to look' but doesn't include the result 'to see'. 听 means 'to listen' but doesn't include 'to hear'. 写 is similar to the English verb 'to write', in that it doesn't specify what the result of writing was (e.g.whether it was correct or not). 记 describes an attempt to remember, without specifying if the information was forgotten or successfully remembered. 杀 means something like 'to try and kill', whatever the result is.All of these verbs nee d a result complement to describe a complete action with its result. R e s u l t c o m p l e m e n t - C h i n e s e G r a m m a r W i k i h t t p://r e s o u r c e s.a l l s e t l e a r n i n g.c o m/c h i n e s e/g r a m m a r/R e s u l t _ c o m p l e m e n t 1 d i 5 2 0/0 3/2 0 1 3 2 2.5 2 adDownload to read ad-free Result complements form verbal compounds that behave exactly the same as normal verbs. The compound is formed by a verb plus another verb or an adjective: Result complement examples V e r b R e s u l t c o m p o u n d E x p l a n a t i o n 看看见看 alone means 'to look'. 看见 includes the r 听听到 Again, 听 alone means 'to l isten', but 听到写写对 Write correct ⇒ to write something 学学会 Study able ⇒ to master 洗洗干净 Wash clean ⇒ to wash something and In each of the examples above, a compound verb has been created from a verb plus another verb or adjective. These verb-complement compounds behave like other verbs in terms of taking objects, being modified by adverbs and being negated.As compound verbs formed by the result complement behave much the same as other verbs, you can easily negate them as normal using 不 or 没有 .Note th at a ltho ug h th ere a re ma ny inst anc es wh ere us ing 不 with a result complement is grammatical, you don't see it as much as 没有 (remember that 不 negates verbs about the present or future, whereas 没有 is used for things in the past). This makes sense if you think about English verbs. You frequently come across things like "I didn't look" and "I'm not looking" (verbs without a result), but for verbs that include a result you tend to only come across past tense forms - "I didn't see." A sentence like "I'm not seeing" sounds unnatural in most contexts. Result complement negation S u b j e c t N e g a t i v e R C c o m p o u n d v e r b O b j e c t 我没有看见你。他不会做完他的作业。我没有找到小王。他没吃完饭。她没有考上大学。 You can also use the aspect particles 了 and 过 with result complement compound verbs, just as you would with other verbs. 了 nearly always appears with result complements, as 了 marks completed actions, and as you'd expect a result complement is usually about a completed action.Conversely, you can not use the particle 着 with result compounds, as this would be nonsensical. 着 R e s u l t c o m p l e m e n t - C h i n e s e G r a m m a r W i k i h t t p://r e s o u r c e s.a l l s e t l e a r n i n g.c o m/c h i n e s e/g r a m m a r/R e s u l t _ c o m p l e m e n t 2 d i 5 2 0/0 3/2 0 1 3 2 2.5 2 adDownload to read ad-free indicates that an action is ongoing - it wouldn't make sense to describe the result of an ongoing action. Result compleme nts with as pect par ticl es S u b j e c t R C c o m p o u n d v e r b A s p e c t p a r t i c l e O b j e c t 我看到了你。我看见过那个有名的人。我听到过这个声音。他写错了这个汉字。 The aspect particle comes after the result complement, as the compound can't be separated. It behaves as a single verb, and can't be divided into separate units.You can form questions with sentences containing result complements just as you would with any other sentence:With a question particle With a question word With positive-negative inversion Some examples: 你把作业做完了吗？你看完了那部电影吗？谁能把 ‘ 打喷嚏 ’ 的 ‘ 嚏 ’ 写对？我们刚才听到的音乐是哪个乐队的？你有没有记住我的名字？他是不是我们昨天看到的人？ Remember that a result complement can describe both intentional and unintentional results of a verb. For example, doing things correctly or incorrectly, succeeding, or breaking things can all be described with R e s u l t c o m p l e m e n t - C h i n e s e G r a m m a r W i k i h t t p://r e s o u r c e s.a l l s e t l e a r n i n g.c o m/c h i n e s e/g r a m m a r/R e s u l t _ c o m p l e m e n t 3 d i 5 2 0/0 3/2 0 1 3 2 2.5 2 adDownload to read ad-free result complements. Result compleme nt example com pounds R e s u l t E x a m p l e c o m p o u n d s E x a m p l e s e n t e n c e 见看见我没看见他。听见我没听见他说的话。到看到他没有看到我。听到你听到什么奇怪的声音了？走到怎么走到人民广场？住记住我会记住你的话。抓住记者应该抓住事情的重点。对写对他什么汉字都能写对。说对你连最复杂的那句话也能说对。错写错好像这句话写错了。做错我实习的时候总是做错事情。破打破是谁打破了这扇窗户？弄破你把我的包包弄破了。着找着你找着那本汉语书了吗？睡着宝宝很晚才睡着。上带上别忘了带上你的护照。关上请把门关上。完做完今天的工作做完了吗？吃完你能吃完这些米饭吗？开打开打开窗户吧，今天不冷了。 You might have noticed that quite a few of the example sentences in this article are 把 sentences. This is beca use 把 sentences and result complements work particularly well together, as they both deal with the result of an action or the disposal of an object.Apart from result complements involving perception and psychological verbs, most result compounds work nicely in 把 sentences. R e s u l t c o m p l e m e n t - C h i n e s e G r a m m a r W i k i h t t p://r e s o u r c e s.a l l s e t l e a r n i n g.c o m/c h i n e s e/g r a m m a r/R e s u l t _ c o m p l e m e n t 4 d i 5 2 0/0 3/2 0 1 3 2 2.5 2 adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like The Sounds of Chinese 100% (2) The Sounds of Chinese 331 pages Iconicity in Language - An Encyclopaedic Dictionary 100% (1) Iconicity in Language - An Encyclopaedic Dictionary 479 pages English Curriculum Guide 9 No ratings yet English Curriculum Guide 9 31 pages TCS Lect 22 Regular Grammar No ratings yet TCS Lect 22 Regular Grammar 39 pages A Practical Chinese Grammar For Foreigners (Revised Edition) 100% (2) A Practical Chinese Grammar For Foreigners (Revised Edition) 668 pages Cambridge IGCSE English As A Second Language 95% (40) Cambridge IGCSE English As A Second Language 153 pages English Q1 W2 Day3 No ratings yet English Q1 W2 Day3 49 pages Summarizing and Paraphrasing No ratings yet Summarizing and Paraphrasing 25 pages Ingles 1 No ratings yet Ingles 1 15 pages Conjuction & Preposition No ratings yet Conjuction & Preposition 17 pages LCT 3 No ratings yet LCT 3 19 pages Manjushri Prayer 文殊真實名經 (in Tibetan text, English and Chinese) No ratings yet Manjushri Prayer 文殊真實名經 (in Tibetan text, English and Chinese) 59 pages More Steps in Chinese Cooking PDF No ratings yet More Steps in Chinese Cooking PDF 106 pages Character Worksheet No ratings yet Character Worksheet 25 pages Intonation 2: English Phonetics and Phonology No ratings yet Intonation 2: English Phonetics and Phonology 14 pages Structural Analysis of Words No ratings yet Structural Analysis of Words 13 pages Part of Speech No ratings yet Part of Speech 15 pages Abjad Table of Arabic Letters' Heavenly Values - 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2959	https://courses.lumenlearning.com/atd-sanjac-collegealgebra/chapter/identify-vertical-and-horizontal-asymptotes/	Identify vertical and horizontal asymptotes \| College Algebra Skip to main content College Algebra Rational Functions Search for: Identify vertical and horizontal asymptotes By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location. Vertical Asymptotes The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors. How To: Given a rational function, identify any vertical asymptotes of its graph. Factor the numerator and denominator. Note any restrictions in the domain of the function. Reduce the expression by canceling common factors in the numerator and the denominator. Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur. Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities. Example 5: Identifying Vertical Asymptotes Find the vertical asymptotes of the graph of k(x)=5+2 x 2 2−x−x 2 k(x)=5+2 x 2 2−x−x 2. Solution First, factor the numerator and denominator. ⎧⎪⎨⎪⎩k(x)=5+2 x 2 2−x−x 2=5+2 x 2(2+x)(1−x){k(x)=5+2 x 2 2−x−x 2=5+2 x 2(2+x)(1−x) To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero: {(2+x)(1−x)=0 x=−2,1{(2+x)(1−x)=0 x=−2,1 Neither x=−2 x=−2 nor x=1 x=1 are zeros of the numerator, so the two values indicate two vertical asymptotes. Figure 9 confirms the location of the two vertical asymptotes. Figure 9 Removable Discontinuities Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity. For example, the function f(x)=x 2−1 x 2−2 x−3 f(x)=x 2−1 x 2−2 x−3 may be re-written by factoring the numerator and the denominator. f(x)=(x+1)(x−1)(x+1)(x−3)f(x)=(x+1)(x−1)(x+1)(x−3) Notice that x+1 x+1 is a common factor to the numerator and the denominator. The zero of this factor, x=−1 x=−1, is the location of the removable discontinuity. Notice also that x−3 x−3 is not a factor in both the numerator and denominator. The zero of this factor, x=3 x=3, is the vertical asymptote. Figure 10 A General Note: Removable Discontinuities of Rational Functions A removable discontinuity occurs in the graph of a rational function at x=a x=a if a is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value. Example 6: Identifying Vertical Asymptotes and Removable Discontinuities for a Graph Find the vertical asymptotes and removable discontinuities of the graph of k(x)=x−2 x 2−4 k(x)=x−2 x 2−4. Solution Factor the numerator and the denominator. k(x)=x−2(x−2)(x+2)k(x)=x−2(x−2)(x+2) Notice that there is a common factor in the numerator and the denominator, x−2 x−2. The zero for this factor is x=2 x=2. This is the location of the removable discontinuity. Notice that there is a factor in the denominator that is not in the numerator, x+2 x+2. The zero for this factor is x=−2 x=−2. The vertical asymptote is x=−2 x=−2. Figure 11 The graph of this function will have the vertical asymptote at x=−2 x=−2, but at x=2 x=2 the graph will have a hole. Try It 5 Find the vertical asymptotes and removable discontinuities of the graph of f(x)=x 2−25 x 3−6 x 2+5 x f(x)=x 2−25 x 3−6 x 2+5 x. Solution Horizontal asymptotes While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions. There are three distinct outcomes when checking for horizontal asymptotes: Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at y= 0. Example:f(x)=4 x+2 x 2+4 x−5 Example:f(x)=4 x+2 x 2+4 x−5 In this case, the end behavior is f(x)≈4 x x 2=4 x f(x)≈4 x x 2=4 x. This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function g(x)=4 x g(x)=4 x, and the outputs will approach zero, resulting in a horizontal asymptote at y= 0. Note that this graph crosses the horizontal asymptote. Figure 12.Horizontal Asymptote y = 0 when f(x)=p(x)q(x),q(x)≠0 where degree of p<degree of q f(x)=p(x)q(x),q(x)≠0 where degree of p<degree of q. Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote. Example:f(x)=3 x 2−2 x+1 x−1 Example:f(x)=3 x 2−2 x+1 x−1 In this case, the end behavior is f(x)≈3 x 2 x=3 x f(x)≈3 x 2 x=3 x. This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function g(x)=3 x g(x)=3 x. As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of g(x)=3 x g(x)=3 x looks like a diagonal line, and since f will behave similarly to g, it will approach a line close to y=3 x y=3 x. This line is a slant asymptote. To find the equation of the slant asymptote, divide 3 x 2−2 x+1 x−1 3 x 2−2 x+1 x−1. The quotient is 3 x+1 3 x+1, and the remainder is 2. The slant asymptote is the graph of the line g(x)=3 x+1 g(x)=3 x+1. Figure 13.Slant Asymptote when f(x)=p(x)q(x),q(x)≠0 f(x)=p(x)q(x),q(x)≠0 where degree of p>degree of q by 1 p>degree of q by 1. Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at y=a n b n y=a n b n, where a n a n and b n b n are the leading coefficients of p(x)p(x) and q(x)q(x) for f(x)=p(x)q(x),q(x)≠0 f(x)=p(x)q(x),q(x)≠0. Example:f(x)=3 x 2+2 x 2+4 x−5 Example:f(x)=3 x 2+2 x 2+4 x−5 In this case, the end behavior is f(x)≈3 x 2 x 2=3 f(x)≈3 x 2 x 2=3. This tells us that as the inputs grow large, this function will behave like the function g(x)=3 g(x)=3, which is a horizontal line. As x→±∞,f(x)→3 x→±∞,f(x)→3, resulting in a horizontal asymptote at y = 3. Note that this graph crosses the horizontal asymptote. Figure 14.Horizontal Asymptote when f(x)=p(x)q(x),q(x)≠0 where degree of p=degree of q f(x)=p(x)q(x),q(x)≠0 where degree of p=degree of q. Notice that, while the graph of a rational function will never cross a vertical asymptote, the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote. It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the end behavior of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function f(x)=3 x 5−x 2 x+3 f(x)=3 x 5−x 2 x+3 with end behavior f(x)≈3 x 5 x=3 x 4 f(x)≈3 x 5 x=3 x 4, the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient. x→±∞,f(x)→∞x→±∞,f(x)→∞ A General Note: Horizontal Asymptotes of Rational Functions The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator. Degree of numerator is less than degree of denominator: horizontal asymptote at y= 0. Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote. Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients. Example 7: Identifying Horizontal and Slant Asymptotes For the functions below, identify the horizontal or slant asymptote. g(x)=6 x 3−10 x 2 x 3+5 x 2 g(x)=6 x 3−10 x 2 x 3+5 x 2 h(x)=x 2−4 x+1 x+2 h(x)=x 2−4 x+1 x+2 k(x)=x 2+4 x x 3−8 k(x)=x 2+4 x x 3−8 Solution For these solutions, we will use f(x)=p(x)q(x),q(x)≠0 f(x)=p(x)q(x),q(x)≠0. g(x)=6 x 3−10 x 2 x 3+5 x 2 g(x)=6 x 3−10 x 2 x 3+5 x 2: The degree of p=degree of q=3 p=degree of q=3, so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at y=6 2 y=6 2 or y=3 y=3. h(x)=x 2−4 x+1 x+2 h(x)=x 2−4 x+1 x+2: The degree of p=2 p=2 and degree of q=1 q=1. Since p>q p>q by 1, there is a slant asymptote found at x 2−4 x+1 x+2 x 2−4 x+1 x+2. The quotient is x−2 x−2 and the remainder is 13. There is a slant asymptote at y=−x−2 y=−x−2. k(x)=x 2+4 x x 3−8 k(x)=x 2+4 x x 3−8: The degree of p=2<p=2< degree of q=3 q=3, so there is a horizontal asymptote y = 0. Example 8: Identifying Horizontal Asymptotes In the sugar concentration problem earlier, we created the equation C(t)=5+t 100+10 t C(t)=5+t 100+10 t. Find the horizontal asymptote and interpret it in context of the problem. Solution Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is t, with coefficient 1. In the denominator, the leading term is 10 t, with coefficient 10. The horizontal asymptote will be at the ratio of these values: t→∞,C(t)→1 10 t→∞,C(t)→1 10 This function will have a horizontal asymptote at y=1 10 y=1 10. This tells us that as the values of t increase, the values of C will approach 1 10 1 10. In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or 1 10 1 10 pounds per gallon. Example 9: Identifying Horizontal and Vertical Asymptotes Find the horizontal and vertical asymptotes of the function f(x)=(x−2)(x+3)(x−1)(x+2)(x−5)f(x)=(x−2)(x+3)(x−1)(x+2)(x−5) Solution First, note that this function has no common factors, so there are no potential removable discontinuities. The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at x=1,−2,and 5 x=1,−2,and 5, indicating vertical asymptotes at these values. The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as x→±∞,f(x)→0 x→±∞,f(x)→0. This function will have a horizontal asymptote at y=0 y=0. Figure 15 Try It 6 Find the vertical and horizontal asymptotes of the function: f(x)=(2 x−1)(2 x+1)(x−2)(x+3)f(x)=(2 x−1)(2 x+1)(x−2)(x+3) Solution A General Note: Intercepts of Rational Functions A rational function will have a y-intercept when the input is zero, if the function is defined at zero. A rational function will not have a y-intercept if the function is not defined at zero. Likewise, a rational function will have x-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, x-intercepts can only occur when the numerator of the rational function is equal to zero. Example 10: Finding the Intercepts of a Rational Function Find the intercepts of f(x)=(x−2)(x+3)(x−1)(x+2)(x−5)f(x)=(x−2)(x+3)(x−1)(x+2)(x−5). Solution We can find the y-intercept by evaluating the function at zero ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩f(0)=(0−2)(0+3)(0−1)(0+2)(0−5)=−6 10=−3 5=−0.6{f(0)=(0−2)(0+3)(0−1)(0+2)(0−5)=−6 10=−3 5=−0.6 The x-intercepts will occur when the function is equal to zero: ⎧⎪ ⎪⎨⎪ ⎪⎩0=(x−2)(x+3)(x−1)(x+2)(x−5)This is zero when the numerator is zero.0=(x−2)(x+3)x=2,−3{0=(x−2)(x+3)(x−1)(x+2)(x−5)This is zero when the numerator is zero.0=(x−2)(x+3)x=2,−3 The y-intercept is (0,−0.6)(0,−0.6), the x-intercepts are (2,0)(2,0) and (−3,0)(−3,0). Figure 16 Try It 7 Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the x– and y-intercepts and the horizontal and vertical asymptotes. Solution Candela Citations CC licensed content, Shared previously Precalculus. Authored by: Jay Abramson, et al.. Provided by: OpenStax. Located at: License: CC BY: Attribution. 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2960	https://mast.queensu.ca/~blevit/math896/896-21-04.pdf	MATH 896-2021 Lecture 4 SEP 14, 2021 Expected value of a random variable Expected values play a fundamental role in Probability and Statistics. The formal deﬁnition of the expected value, or mean value, is simple and well known. Let X ∈{xi} be a discrete r.v., with a pmf f(xi) = P(X = xi). The expected value, or expectation, EX, of the random variable X is deﬁned by EX = µ = ∑ i xiP(X = xi) = ∑ i xif(xi). (1) Sometimes, it is also called mean value of X and is denoted µ = EX. More generally, for any function g(X), its expected value is given by Eg(X) = ∑ i g(xi)f(xi). (2) Although these formulas are simple enough, they do not speak directly to our intuition. To better understand their meaning, let us consider the following simple example. Suppose we are playing a sequence of independent games, the results of which, Xi, are either 1, with a probability p, or 0, with probability q = 1 −p (independent Bernoulli trials): P(Xi = 1) = p, P(Xi = 0) = q. Let us call 1 a win, and 0 a loss. What is our average gain expected to be after n such games? We have X1 + · · · + Xn n = 1 · #{wins} + 0 · #{losses} n = 1 · #{wins} n + 0 · #{losses} n ≈ 1 · p + 0 · q = 1 · P(X = 1) + 0 · P(X = 0) = EX = expected value of X. Here we used the result we are (intuitively) familiar with, that the frequencies of wins/losses are approximately equal to their corresponding probabilities. Thus, the expected value tells us, what average gain we can expect approximately, in n independent trials! Let us generalize this example, in two ways. Assume that X1, ..., Xn are independent random variables taking on values {x1, ..., xI}, having the same pmf f(xi), i = 1, ..., I. Suppose also that we are interested not in the random variable X itself, but in a certain function g(X). Let k be the current number of the trial. Consider the average value of g(Xk), in the long run: 1 ∑n k=1 g(Xk) n = g(x1) · #{times Xk = x1} + g(x2) · #{times Xk = x2} + · · · + g(xI) · #{times Xk = xI} n = g(x1)·#{times Xk = x1} n +g(x2)·#{times Xk = x2} n +· · ·+g(xI)·#{times Xk = xI} n ≈ g(x1) · P(X = x1) + g(x2) · P(X = x2) + · · · + g(xI) · P(X = xI) = I ∑ i=1 g(xi)P(X = xi) = I ∑ i=1 g(xi)f(xi) = Eg(X). Thus, ∑n k=1 g(Xk) n ≈ E g(X), or n ∑ k=1 g(Xk) ≈ n g(X). (3) This result is called the Law of Large Numbers (LLN). It plays a fundamental role in Statistics, and will be discussed later in more detail. Its meaning is clear: although we can hardly predict the outcome of any single game, in the long run, the situation settles down, and we can easily predict the average value of any function g(X), by its expected value E g(X). When X takes on ﬁnitely many values, calculating its expectation is easy, as in the following Example 1. Let X ∼B(p) be a Bernoulli random variable. Then, EX = 0 · (1 −p) + 1 · p = p. EX2 = 02 · (1 −p) + 12 · p = p. etc. When X takes inﬁnitely many diﬀerent values, some additional caution is needed. Let us discuss this in some more detail. If g(xi) ≥0, the expectation Eg(X) = ∑ i g(xi)fX(xi) always exists, even though it may be either ﬁnite or inﬁnite. In the latter case, we say that the expected value is inﬁnite, but still well deﬁned! Similarly, if g(xi) ≤0, the expectation Eg(X) = ∑ i g(xi)fX(xi) exists, but it may again be either ﬁnite or inﬁnite. In the latter case, we say that the expected value is inﬁnite (−∞), but is still deﬁned! 2 In general case, we can write Eg(X) = ∑ g(xi)<0 g(xi)fX(xi) + ∑ g(xi)≥0 g(xi)fX(xi). If both these sums are ﬁnite, we say that the expectation is ﬁnite. If one of these sums is ﬁnite, while the other is inﬁnite, we say that the expectation is inﬁnite, but deﬁned. If both these sums are inﬁnite, we have uncertainty of the type ∞−∞. In this case, we say that the expectation is not deﬁned, or that it does not exist. To guarantee that Eg(X) exists and is ﬁnite, assume that ∑ g(xi)<0 \|g(xi)\|fX(xi) + ∑ g(xi)≥0 g(xi)fX(xi) = ∑ i \|g(xi)\|fX(xi) < ∞. In other words, when talking about expected values, we will always assume, unless stated otherwise, that the series Eg(X) = ∑ i g(xi)fX(xi) is absolute convergent, i.e., ∑ i \|g(xi)\|f(xi) < ∞. Sometimes, calculating expected values requires some special methods. Let us il-lustrate this in the case of geometric distribution. Example 2. Let X ∼G(p), fX(k) = p(1 −p)k−1, k = 1, 2, ... . By diﬀerentiating the geometric series term-by-term, we get EX = ∞ ∑ k=1 kfX(k) = p ∞ ∑ k=1 k(1 −p)k−1 = −p ∞ ∑ k=1 d dp(1 −p)k = −p d dp ∞ ∑ k=1 (1 −p)k = −p d dp ( 1 −p 1 −(1 −p) ) = −p d dp (1 −p p ) = −p d dp (1 p −1 ) = −p ( −1 p2 ) = 1 p. (4) Deﬁnition. Let X ∈{xi} and Y ∈{yj} be two discrete random variables, with a joint pmf f(xi, yj) = P(X = xi, Y = yj), and let g(X, Y ) be any given function. Then the expected value, or mean value, of the random variable g(X, Y ) is given by µ = Eg(X, Y ) = ∑ i ∑ j g(xi, yj)f(xi, yj) = ∑ j ∑ i g(xi, yj)f(xi, yj). 3 As always, we assume that this double series is absolutely convergent. Here this assumption is especially important since, by a result from Calculus, it allows to carry out summation in any particular order (ﬁrst over j, and then over i, or vice versa). Properties of expectations Expected values will be constantly used in Statistics. Some of their properties are almost obvious; some will be proved in the class; while others may be left to the assignments. Let us formulate the most important of them. 1. Uniqueness. If two random variables have the same distributions, i.e., take on the same values xi, with the same probabilities f(xi) = P(X = xi), then they have the same expected values. 2. Expected value of a constant. Constants c will be treated as a (degenerate) random variables, taking on a single value c, with probability f(c) = P(X = c) = 1. If X = c = const , then EX = Ec = c. Indeed, by the deﬁnition of expectation, EX = c · f(c) = c · 1 = c. 3. Linearity. E(aX + b) = aEX + b. This follows from the property of (absolutely) convergent series: ∑ i (axi+b)f(xi) = ∑ i axif(xi)+ ∑ i bf(xi) = a ∑ i xif(xi)+b ∑ i f(xi) = aEX+b. 4. Centered random variables. Let µ = EX. The diﬀerence X −µ is called centered random variable. Expected values of centered random variables are always 0. Indeed, by the properties 2 and 3, E(X −µ) = EX −Eµ = µ −µ = 0. 5. Additivity. If X and Y are any two discrete random variables, then E(X + Y ) = EX + EY. More generally, for any functions g(X) and h(Y ), E(g(X) + h(Y )) = Eg(X) + Eh(Y ). The proof comes in naturally. Indeed, by the elimination rule, E(g(X) + h(Y )) = ∑ i ∑ j (g(xi) + h(yj))f(xi, yj) = ∑ i g(xi) ∑ j f(xi, yj) + ∑ j h(yj) ∑ i f(xi, yj) = 4 ∑ i g(xi)fX(xi) + ∑ j h(yj)fY (yj) = Eg(X) + Eh(Y ). 6. Multiplicativity. If X and Y are independent discrete random variables, i.e., f(xi, yj) = fX(xi)fY (yj), then E(XY ) = (EX) · (EY ). Moreover, if X and Y are independent, and g(X) and h(Y ) are any two functions, then E(g(X) · h(Y )) = (Eg(X)) · (Eh(Y )) . Indeed, by the deﬁnition of expected value, E(g(X)h(Y )) = ∑ i ∑ j g(xi)h(yj)f(xi, yj) = ∑ i ∑ j g(xi)h(yj)fX(xi)fY (yj) = ∑ i g(xi)fX(xi) (∑ j h(yj)fY (yj) ) = (Eh(Y )) ∑ i g(xi)fX(xi) = (Eg(X))·(Eh(Y )) . This property can be generalized as follows. Let X1, X2, ..., Xn be independent random variables, and let g1(X1), g2(X2), ..., gn(Xn) be any given functions of them. Then E(g1(X1) · g2(X2) · · · gn(Xn)) = E(g1(X1)) · E(g2(X2)) · · · E(gn(Xn)). This can be written shorter as E ( n ∏ i=1 gi(Xi) ) = n ∏ i=1 E(gi(Xi)). 7. Positivity of expected values. Suppose a random variable X, with a pmf f(xi), takes on only nonnegative values xi ≥0. Then, obviously, EX = ∑ i xif(xi) ≥0. The equality EX = 0 is possible if and only if P(X = 0) = 1. The following terminology is universally accepted and standard in the Probability Theory. Deﬁnition. If P(X = 0) = 1, we say that random variable X equals 0 with probability 1, or that it vanishes almost surely. 8. Expectation as the best predictor. Let EX = µ. Suppose, we want to predict unknown value of the random variable X, by some constant c. For instance, 5 think of predicting the temperature X tomorrow at noon (viewing it as discrete random variable). It is customary to measure the error of any predictor c by the so called mean squared error, E(X −c)2. Obviously, we are looking for a predictor c which minimizes this mean squared error. It can be shown that min c E(X −c)2 = E(X −µ)2. (5) Before proving (5), recall the following well known Deﬁnition. Variance of a random variable X is deﬁned as σ2 = VarX = E(X −µ)2 = E(X −EX)2. Thus, we can summarize (5) as follows: the mean value µ of a discrete random variable is the best constant predictor of X, while its mean squared error equals the variance of X. To prove (5), ﬁrst expand E(X −c)2 using properties 2 and 3: E(X −c)2 = EX2 −2cEX + c2. Obviously, this is a quadratic function of c (an upward parabola). As is well known from Algebra, such a function has a unique minimum. To ﬁnd the point of its minimum, let us set its derivative to 0: d dcE(X −c)2 = −2EX + 2c = −2µ + 2c = 0. From this equation, we easily ﬁnd the value of c minimizing (5) is cmin = µ. We have discovered that the notion of expected value EX has many diﬀerent as-pects. First, its formal deﬁnition, like (1) or (2). Second, EX plays important role in predicting the average value or gain, like in the Law of large numbers (3). It is also the best constant predictor of a random variable, see (5). Finally, there is a computational aspect of calculating EX, like in the case of geometric distribution (4). Let us illustrate this last aspect of expected values in the case of binomial distribution. Example 3. Let X ∼bin(n, p). By using the pmf of X, calculation of the expected value EX reduces to ﬁnding the following sum EX = n ∑ x=0 x (n x ) px(1 −p)n−x. This may not be easy! Here is another simpler way. Recall that X has the same distribution as the sum of independent Bernoulli random variables, X1 + · · · + Xn where Xi ∼B(p). Moreover, by Example 1, EXi = p. Thus, by the additivity property 5, EX = EX1 + · · · + EXn = np. This way, the calculation of EX is much easier! 6
2961	https://leah4sci.com/cis-trans-and-e-z-geometric-isomers/	MCAT and Organic Chemistry Study Guides, Videos, Cheat Sheets, tutoring and more Cis Trans and E Z Geometric Isomers September 18, 2016 By Leah4sci Your organic chemistry course will cover many different types of isomers. Isomers have the same molecular formula but something about them is different. Geometric isomers, a type of stereoisomer, differ in their geometry or shape. This happens when substituents are LOCKED in a specific relationship to each other. I say locked because, unlike conformational isomers in Newman Projections, you can’t simply rotate the molecule to change the relationship between substituents. In this tutorial, we’ll look at alkene geometric isomers including cis trans and E Z. Cis/Trans Isomerism Cis/Trans isomerism is typically seen with substituents on either side of the alkene double bond. How does this happen? Alkene double bonds occur between sp2 hybridized carbon atoms. Recall: sp2 hybrids have a trigonal-planar or ‘flat’ geometry. (Not comfortable with this? Review sp2 Hybridization.) But it’s not the hybrid we’re looking at. Instead, it’s the un-hybridized p-orbital that forms a SECOND bond between the 2 carbon atoms. An sp3 hybridized single or simga bond is free to rotate. Sp2 pi bonds are locked in place. The only way to rotate this bond is to break the double bond, rotate, and reform the double bond –which is typically not observed. In fact, this requires high energy, as you’ll see in your Diels Alder reactions later on. Take a look at the following generic alkene and it’s 4 substituents: Carbon 1 has substituents A and B; Carbon 2 has substituents C and D. But notice specifically how A is on the same side as C, and B is on the same side as D. The only way to bring A next to D is to break the pi bond, rotate the molecule, and reform the pi bond. Otherwise A is locked in place near C, and B is locked in place near D. Cis vs Trans Alkenes Let’s take a look at 2 versions of 2-butene: 2-butene is a 4-carbon chain with a double bond between carbons 2 and 3. So, we can draw this incorrectly, as a linear molecule: Or, draw each sp2 carbon at a 120 degree bond angle. This gives me the option of placing both methyl groups up, down, or one up and one down. The first two are actually the same: both cis. You see, I can flip the molecule and make the first superimpose (overlap) the second without breaking any bonds. The third is unique. The only way to superimpose the third is to break the double bond. Cis Alkenes I like to think of cis as ‘sisters’. They are together on the same side. Cis alkenes have substituents on the same side of the double bond. Trans Alkenes I like to think of trans substituents as having ‘transferred away from each other.’ Putting them on opposite sides. Trans alkenes have their substituents on opposite sides. Naming Cis/Trans Alkenes: Once you’ve identified cis/trans alkenes, naming them is fairly simple. 1) First, name the alkene using the tutorial linked below. 2) Then, simply add ‘cis’ or ‘trans’ in front of the name. Take the 2 geometric isomers of 2-butene: Their proper names are as follows: When there is only one pi bond, you don’t have to specify which carbon is cis or trans since. It’s self-understood. When you have more than one double bond on the molecule, you must specify which is cis and which is trans. Take this molecule for example: 2,5-octadiene This molecule has 2 pi bonds. One cis and one trans. Since there is more than one pi bond, you have to specify which pi bond is cis and which is trans. Alkene stability Not all isomers have the same stability. It’s all about stability – in organic chemistry or science in general. Trans alkenes are MORE STABLE than their cis counterparts. This is more apparent with larger substituents. Trans Alkenes In a trans alkene, the substituents are facing away from each other. They don’t ‘get in each other’s faces’ and therefore, don’t mind the other groups. Cis Alkenes Cis alkene substituents are close together and will ‘get in each others faces.’ This causes ‘arguments’ when one group invades the other’s personal space. When the groups try to move away from each other, they cause strain on the molecule. All of this leads to an unhappy and higher energy cis conformation. Cis & Trans on Cyclic Compounds Ring structures or cyclic compounds can also exhibit cis/trans isomerism without the presence of a pi bond. Remember, substituents will be cis and trans if they are locked in place. Pi bonds are one way to lock them in place. Rings are another matter. For example, in 1,2-dimethylcyclohexane, I can show both substituents going into the page or both going out of the page. Since they’re pointing in the same direction, they are cis to each other. If I show one going into the page and one going out of the page. They are trans to each other. Even though the carbons are sp3 and sigma bound to each other, the molecule itself cannot rotate because of the ring structure. Locked. The only way to turn cis-1,2-dimethylcyclohexane into trans-1,2-dimethylcyclohexane, is to break open the ring, rotate, and reform the ring. What if there’s more than one substituent on the sp2 carbon? Till now, we’ve looked at molecules with just one substituent on either side of the sp2 pi bound carbon. What happens if we have a pi bond with 2 different atoms or groups on the sp2 carbon? Take a look at 3-methyl-2-pentene: Here in line structure: You can draw this molecule in 2 different ways. But will you compare the red methyl or red ethyl to the green methyl when choosing cis or trans? While some professors WILL teach you to compare the larger groups, the answer is that you CANNOT compare simply choose one for cis and trans. Introducing the E Z Notation When a pi bond has more than one substituent on each side, or contains non-carbon substituents, we’ll need a more advanced system for identifying geometric isomerism. The E Z system requires ranking the groups on either side of the pi bond. We must determine if the higher priority groups are next to each other, Z (think cis), or away from each other, E (think trans). But first, we have to learn how to rank groups using the Cahn-Ingold-Prelog notation. The video below is from my chirality series, but teaches this concept in detail. Start watching from 0:52 Cahn-Ingold-Prelog in summary: We’re ranking atoms based on their atomic number. Not mass of the group, not size of the group. The higher the atomic number of the atom directly attached, the higher the priority. Here are the 10 most common atoms you’ll come across from high to low priority: I > Br > Cl > S > P > F > O > N > C > H Here’s My Simple Approach E is for Eeposite, Z is for Ze Zame Zide If the two high priority groups are opposite to each other, think of them as being ‘eeposite’ to each other. E is for Eeposite. If the two high priority groups are on the the same side, or should I say on ‘Ze Zame Zide,’ they are Z. This applies to molecules that have more than just 1 carbon on either side of double bond. Ze Zame Zide. Let’s go back to the example above: On the left, OH outranks ethyl since oxygen has a higher atomic number when compared to carbon. OH is #1 and points down. On the right, Cl outranks methyl since chlorine has a higher atomic number when compared to carbon. Cl is #1 and points down. Since both arrows point in the same direction (down), we conclude that the priority groups are on Ze Zame Zide making it Z. 2 Equal Priority Groups Sometimes you’ll see a trick question where an sp2 carbon will have 2 of the exact same groups. Since you cannot rank one over the other, there will be NO cis/trans or E/Z isomerism. Here are 2 common examples: 1) A terminal pi bond Carbon #1 in 1-butene has 2 hydrogen atoms. Since H vs H have the same exact priority, this molecule will have no cis/trans or E/Z isomerism. 2) Same exact groups on the same sp2 pi bound carbon. Carbon #2 in 2-methyl-2-butene has 2 CH3 groups. One appears to be part of the parent chain, the second appears to be a methyl substituent. However, when CH3 is compared to CH3 they rank exactly the same. This molecule will have no cis/trans n/or E/Z isomerism. Cis and Trans vs E and Z If we go back to our cis/trans practice problems, such as cis and trans 2-butene, you’ll see that we can use the E/Z system here as well. Carbon 2 and 3 each have a methyl group outranking a hydrogen atom. When they are cis, you get Z. When they are trans you get E. A word of caution You CAN use E/Z for cis/trans isomers, but you cannot use cis/trans for complex E/Z isomers as we’ve already shown above. In Summary Cis vs trans and E vs Z isomers are geometric isomers that occur when substituents are locked in position next to or opposite each other. This is seen in both double bonds for alkenes, and substituents on ring structures. Cis alkenes are on the same size, trans alkenes are on opposite sides. When the substituents are more complicated use the more advanced E/Z notation after determining the relationship of high priority groups. 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2962	https://openstax.org/books/university-physics-volume-1/pages/5-3-newtons-second-law	Loading [MathJax]/jax/output/HTML-CSS/fonts/TeX/fontdata.js Skip to Content Go to accessibility page Keyboard shortcuts menu Log in University Physics Volume 1 5.3 Newton's Second Law University Physics Volume 1 5.3 Newton's Second Law Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Distinguish between external and internal forces Describe Newton's second law of motion Explain the dependence of acceleration on net force and mass Newton’s second law is closely related to his first law. It mathematically gives the cause-and-effect relationship between force and changes in motion. Newton’s second law is quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation that gives the exact relationship of force, mass, and acceleration, we need to sharpen some ideas we mentioned earlier. Force and Acceleration First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is acceleration. Newton’s first law says that a net external force causes a change in motion; thus, we see that a net external force causes nonzero acceleration. We defined external force in Forces as force acting on an object or system that originates outside of the object or system. Let’s consider this concept further. An intuitive notion of external is correct—it is outside the system of interest. For example, in Figure 5.10(a), the system of interest is the car plus the person within it. The two forces exerted by the two students are external forces. In contrast, an internal force acts between elements of the system. Thus, the force the person in the car exerts to hang on to the steering wheel is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton’s first law. (The internal forces cancel each other out, as explained in the next section.) Therefore, we must define the boundaries of the system before we can determine which forces are external. Sometimes, the system is obvious, whereas at other times, identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton’s laws. This concept is revisited many times in the study of physics. Figure 5.10 Different forces exerted on the same mass produce different accelerations. (a) Two students push a stalled car. All external forces acting on the car are shown. (b) The forces acting on the car are transferred to a coordinate plane (free-body diagram) for simpler analysis. (c) The tow truck can produce greater external force on the same mass, and thus greater acceleration. From this example, you can see that different forces exerted on the same mass produce different accelerations. In Figure 5.10(a), the two students push a car with a driver in it. Arrows representing all external forces are shown. The system of interest is the car and its driver. The weight w⃗ of the system and the support of the ground N⃗ are also shown for completeness and are assumed to cancel (because there was no vertical motion and no imbalance of forces in the vertical direction to create a change in motion). The vector f⃗ represents the friction acting on the car, and it acts to the left, opposing the motion of the car. (We discuss friction in more detail in the next chapter.) In Figure 5.10(b), all external forces acting on the system add together to produce the net force F⃗ net. The free-body diagram shows all of the forces acting on the system of interest. The dot represents the center of mass of the system. Each force vector extends from this dot. Because there are two forces acting to the right, the vectors are shown collinearly. Finally, in Figure 5.10(c), a larger net external force produces a larger acceleration (a′→>a⃗ ) when the tow truck pulls the car. It seems reasonable that acceleration would be directly proportional to and in the same direction as the net external force acting on a system. This assumption has been verified experimentally and is illustrated in Figure 5.10. To obtain an equation for Newton’s second law, we first write the relationship of acceleration a⃗ and net external force F⃗ net as the proportionality a⃗ ∝F⃗ net where the symbol ∝ means “proportional to.” (Recall from Forces that the net external force is the vector sum of all external forces and is sometimes indicated as ∑F⃗ .) This proportionality shows what we have said in words—acceleration is directly proportional to net external force. Once the system of interest is chosen, identify the external forces and ignore the internal ones. It is a tremendous simplification to disregard the numerous internal forces acting between objects within the system, such as muscular forces within the students’ bodies, let alone the myriad forces between the atoms in the objects. Still, this simplification helps us solve some complex problems. It also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. As illustrated in Figure 5.11, the same net external force applied to a basketball produces a much smaller acceleration when it is applied to an SUV. The proportionality is written as a∝1m, where m is the mass of the system and a is the magnitude of the acceleration. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is directly proportional to net external force. Figure 5.11 The same force exerted on systems of different masses produces different accelerations. (a) A basketball player pushes on a basketball to make a pass. (Ignore the effect of gravity on the ball.) (b) The same player exerts an identical force on a stalled SUV and produces far less acceleration. (c) The free-body diagrams are identical, permitting direct comparison of the two situations. A series of patterns for free-body diagrams will emerge as you do more problems and learn how to draw them in Drawing Free-Body Diagrams. It has been found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields Newton’s second law. Newton’s Second Law of Motion The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system and is inversely proportional to its mass. In equation form, Newton’s second law is a⃗ =F⃗ netm, where a⃗ is the acceleration, F⃗ net is the net force, and m is the mass. This is often written in the more familiar form F⃗ net=∑F⃗ =ma⃗ , 5.3 but the first equation gives more insight into what Newton’s second law means. When only the magnitude of force and acceleration are considered, this equation can be written in the simpler scalar form: Fnet=ma. 5.4 The law is a cause-and-effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is based on experimental verification. The free-body diagram, which you will learn to draw in Drawing Free-Body Diagrams, is the basis for writing Newton’s second law. Example 5.2 What Acceleration Can a Person Produce When Pushing a Lawn Mower? Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb.) parallel to the ground (Figure 5.12). The mass of the mower is 24 kg. What is its acceleration? Figure 5.12 (a) The net force on a lawn mower is 51 N to the right. At what rate does the lawn mower accelerate to the right? (b) The free-body diagram for this problem is shown. Strategy This problem involves only motion in the horizontal direction; we are also given the net force, indicated by the single vector, but we can suppress the vector nature and concentrate on applying Newton’s second law. Since Fnet and m are given, the acceleration can be calculated directly from Newton’s second law as Fnet=ma. Solution The magnitude of the acceleration a is a=Fnet/m. Entering known values gives a=51N24kg. Substituting the unit of kilograms times meters per square second for newtons yields a=51kg⋅m/s224kg=2.1m/s2. Significance The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. This is a result of the vector relationship expressed in Newton’s second law, that is, the vector representing net force is the scalar multiple of the acceleration vector. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moved forward), and the vertical forces must cancel because no acceleration occurs in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long, because the person’s top speed would soon be reached. Check Your Understanding 5.3 At the time of its launch, the HMS Titanic was the most massive mobile object ever built, with a mass of 6.0×107kg. If a force of 6 MN (6×106N) was applied to the ship, what acceleration would it experience? In the preceding example, we dealt with net force only for simplicity. However, several forces act on the lawn mower. The weight w⃗ (discussed in detail in Mass and Weight) pulls down on the mower, toward the center of Earth; this produces a contact force on the ground. The ground must exert an upward force on the lawn mower, known as the normal force N⃗ , which we define in Common Forces. These forces are balanced and therefore do not produce vertical acceleration. In the next example, we show both of these forces. As you continue to solve problems using Newton’s second law, be sure to show multiple forces. Example 5.3 Which Force Is Bigger? (a) The car shown in Figure 5.13 is moving at a constant speed. Which force is bigger, F⃗ friction or F⃗ drag? Explain. (b) The same car is now accelerating to the right. Which force is bigger, F⃗ friction or F⃗ drag? Explain. Figure 5.13 A car is shown (a) moving at constant speed and (b) accelerating. How do the forces acting on the car compare in each case? (a) What does the knowledge that the car is moving at constant velocity tell us about the net horizontal force on the car compared to the friction force? (b) What does the knowledge that the car is accelerating tell us about the horizontal force on the car compared to the friction force? Strategy We must consider Newton’s first and second laws to analyze the situation. We need to decide which law applies; this, in turn, will tell us about the relationship between the forces. Solution The forces are equal. According to Newton’s first law, if the net force is zero, the velocity is constant. In this case, F⃗ friction must be larger than F⃗ drag. According to Newton’s second law, a net force is required to cause acceleration. Significance These questions may seem trivial, but they are commonly answered incorrectly. For a car or any other object to move, it must be accelerated from rest to the desired speed; this requires that the friction force be greater than the drag force. Once the car is moving at constant velocity, the net force must be zero; otherwise, the car will accelerate (gain speed). To solve problems involving Newton’s laws, we must understand whether to apply Newton’s first law (where ∑F⃗ =0⃗ ) or Newton’s second law (where ∑F⃗ is not zero). This will be apparent as you see more examples and attempt to solve problems on your own. Example 5.4 What Rocket Thrust Accelerates This Sled? Before space flights carrying astronauts, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T, for the four-rocket propulsion system shown in Figure 5.14. The sled’s initial acceleration is 49m/s2, the mass of the system is 2100 kg, and the force of friction opposing the motion is 650 N. Figure 5.14 A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust T. The system here is the sled, its rockets, and its rider, so none of the forces between these objects are considered. The arrow representing friction (f⃗ ) is drawn larger than scale. Strategy Although forces are acting both vertically and horizontally, we assume the vertical forces cancel because there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in Figure 5.14. Solution Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. We have defined the direction of the force and acceleration as acting “to the right,” so we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with Fnet=ma where Fnet is the net force along the horizontal direction. We can see from the figure that the engine thrusts add, whereas friction opposes the thrust. In equation form, the net external force is Fnet=4T−f. Substituting this into Newton’s second law gives us Fnet=ma=4T−f. Using a little algebra, we solve for the total thrust 4T: 4T=ma+f. Substituting known values yields 4T=ma+f=(2100kg)(49m/s2)+650N. Therefore, the total thrust is 4T=1.0×105N, and the individual thrusts are T=1.0×105N4=2.5×104N. Significance The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance, and the setup was designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 g’s. (Recall that g, acceleration due to gravity, is 9.80m/s2. When we say that acceleration is 45 g’s, it is 45×9.8m/s2, which is approximately 440m/s2.) Although living subjects are not used anymore, land speeds of 10,000 km/h have been obtained with a rocket sled. In this example, as in the preceding one, the system of interest is obvious. We see in later examples that choosing the system of interest is crucial—and the choice is not always obvious. Newton’s second law is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature. Check Your Understanding 5.4 A 550-kg sports car collides with a 2200-kg truck, and during the collision, the net force on each vehicle is the force exerted by the other. If the magnitude of the truck’s acceleration is 10m/s2, what is the magnitude of the sports car’s acceleration? Component Form of Newton’s Second Law We have developed Newton’s second law and presented it as a vector equation in Equation 5.3. This vector equation can be written as three component equations: ∑Fx=max,∑Fy=may,and∑Fz=maz. 5.5 The second law is a description of how a body responds mechanically to its environment. The influence of the environment is the net force F⃗ net, the body’s response is the acceleration a⃗ , and the strength of the response is inversely proportional to the mass m. The larger the mass of an object, the smaller its response (its acceleration) to the influence of the environment (a given net force). Therefore, a body’s mass is a measure of its inertia, as we explained in Newton’s First Law. Example 5.5 Force on a Soccer Ball A 0.400-kg soccer ball is kicked across the field by a player; it undergoes acceleration given by a⃗ =3.00iˆ+7.00jˆm/s2. Find (a) the resultant force acting on the ball and (b) the magnitude and direction of the resultant force. Strategy The vectors in iˆ and jˆ format, which indicate force direction along the x-axis and the y-axis, respectively, are involved, so we apply Newton’s second law in vector form. Solution We apply Newton’s second law: F⃗ net=ma⃗ =(0.400kg)(3.00iˆ+7.00jˆm/s2)=1.20iˆ+2.80jˆN. 2. Magnitude and direction are found using the components of F⃗ net: Fnet=(1.20N)2+(2.80N)2−−−−−−−−−−−−−−−−−√=3.05Nandθ=tan−1(2.801.20)=66.8°. Significance We must remember that Newton’s second law is a vector equation. In (a), we are multiplying a vector by a scalar to determine the net force in vector form. While the vector form gives a compact representation of the force vector, it does not tell us how “big” it is, or where it goes, in intuitive terms. In (b), we are determining the actual size (magnitude) of this force and the direction in which it travels. Example 5.6 Mass of a Car Find the mass of a car if a net force of −600.0jˆN produces an acceleration of −0.2jˆm/s2. Strategy Vector division is not defined, so m=F⃗ net/a⃗ cannot be performed. However, mass m is a scalar, so we can use the scalar form of Newton’s second law, m=Fnet/a. Solution We use m=Fnet/a and substitute the magnitudes of the two vectors: Fnet=600.0N and a=0.2m/s2. Therefore, m=Fneta=600.0N0.2m/s2=3000kg. Significance Force and acceleration were given in the iˆ and jˆ format, but the answer, mass m, is a scalar and thus is not given in iˆ and jˆ form. Example 5.7 Several Forces on a Particle A particle of mass m=4.0kg is acted upon by four forces of magnitudes. F1=10.0N,F2=40.0N,F3=5.0N,andF4=2.0N, with the directions as shown in the free-body diagram in Figure 5.15. What is the acceleration of the particle? Figure 5.15 Four forces in the xy-plane are applied to a 4.0-kg particle. Strategy Because this is a two-dimensional problem, we must use a free-body diagram. First, F→1 must be resolved into x- and y-components. We can then apply the second law in each direction. Solution We draw a free-body diagram as shown in Figure 5.15. Now we apply Newton’s second law. We consider all vectors resolved into x- and y-components: ∑Fx=maxF1x−F3x=maxF1cos30°−F3x=max(10.0N)(cos30°)−5.0N=(4.0kg)axax=0.92m/s2.∑Fy=mayF1y+F4y−F2y=mayF1sin30°+F4y−F2y=may(10.0N)(sin30°)+2.0N−40.0N=(4.0kg)ayay=−8.3m/s2. Thus, the net acceleration is a⃗ =(0.92iˆ−8.3jˆ)m/s2, which is a vector of magnitude 8.4m/s2 directed at 276° to the positive x-axis. Significance Numerous examples in everyday life can be found that involve three or more forces acting on a single object, such as cables running from the Golden Gate Bridge or a football player being tackled by three defenders. We can see that the solution of this example is just an extension of what we have already done. Check Your Understanding 5.5 A car has forces acting on it, as shown below. The mass of the car is 1000.0 kg. The road is slick, so friction can be ignored. (a) What is the net force on the car? (b) What is the acceleration of the car? Newton’s Second Law and Momentum Newton actually stated his second law in terms of momentum: “The instantaneous rate at which a body’s momentum changes is equal to the net force acting on the body.” (“Instantaneous rate” implies that the derivative is involved.) This can be given by the vector equation F⃗ net=dp⃗ dt. This means that Newton’s second law addresses the central question of motion: What causes a change in motion of an object? Momentum was described by Newton as “quantity of motion,” a way of combining both the velocity of an object and its mass. We devote Linear Momentum and Collisions to the study of momentum. For now, it is sufficient to define momentum p⃗ as the product of the mass of the object m and its velocity v⃗ : p⃗ =mv⃗ . Since velocity is a vector, so is momentum. It is easy to visualize momentum. A train moving at 10 m/s has more momentum than one that moves at 2 m/s. In everyday life, we speak of one sports team as “having momentum” when they score points against the opposing team. If we substitute Equation 5.7 into Equation 5.6, we obtain F⃗ net=dp⃗ dt=d(mv⃗ )dt. When m is constant, we have F⃗ net=md(v⃗ )dt=ma⃗ . Thus, we see that the momentum form of Newton’s second law reduces to the form given earlier in this section. Interactive Explore the forces at work when pulling a cart or pushing a refrigerator, crate, or person. Put an object on a ramp and see how it affects its motion. Engage the simulation below and see how applied forces make objects move. Access multimedia content Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: William Moebs, Samuel J. Ling, Jeff Sanny Publisher/website: OpenStax Book title: University Physics Volume 1 Publication date: Sep 19, 2016 Location: Houston, Texas Book URL: Section URL: © Jul 8, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . 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2963	https://www.cobrief.app/resources/contract-definitions-library/principal-balance-definition-copy-customize-and-use-instantly/	Principal Balance definition: Copy, customize, and use instantly How it works Pricing Resources ### Business proposal templates Find free, customizable proposal templates you can edit, tailor, and implement instantly. ### Business letter templates Find professional letter templates for clear and effective communication. ### Business policy templates Browse customizable policy templates to help structure your business operations. ### Interactive business checklist templates Explore step-by-step checklists designed to streamline your business processes. ### Contract clause library Discover ready-to-use contract clauses you can adapt for your agreements. ### Contract definitions library Access plain-English definitions of key contract terms to add to your contract. ### Contract template library Essential contract templates and guides to help you navigate key business agreements. ### Legal glossary Find easy explanations for common legal terms. ### Legal tips Clear, practical articles on common contract issues facing business owners. ### Risk spotlight Everything on protecting your business from legal risks. ### AI Insights Expert insights on AI-driven contract review for your business. Blog About us DashboardLog inSign up for free ← Return to Contract definitions library Start a new document with this content. Open the editor to build from scratch — paste in what you need and keep writing. Open editor for free → Contract definitions library Principal Balance definition: Copy, customize, and use instantly Introduction The term "Principal Balance" refers to the remaining amount of a loan or debt that needs to be repaid, excluding interest, fees, and other charges. It is crucial for understanding the outstanding amount that a borrower is obligated to pay back to the lender. The principal balance is a key component in loan amortization schedules and determines the base on which interest is calculated. Below are various examples of how "Principal Balance" can be defined in different contexts. Copy the one that fits your needs, customize it, and use it in your contract. Definition of "Principal Balance" as the outstanding amount of a loan This definition ties "Principal Balance" to the remaining amount owed on a loan after payments are made. "Principal Balance" means the total amount of money that remains unpaid on a loan after subtracting any principal payments made by the borrower, excluding interest and other charges. Copy Definition of "Principal Balance" as a key figure in loan amortization This definition connects "Principal Balance" to its role in determining how much of the loan’s principal is paid off during each period. "Principal Balance" refers to the amount of the original loan that has not yet been repaid, calculated after any payments made, and is used to determine the periodic amortization of the loan. Copy Definition of "Principal Balance" in relation to a mortgage loan This definition links "Principal Balance" to its use in a mortgage loan context, where the balance is key to understanding the amount owed. "Principal Balance" means the outstanding portion of a mortgage loan, excluding interest, that is due to be repaid by the borrower over the life of the loan. Copy Definition of "Principal Balance" for credit obligations This definition ties "Principal Balance" to a credit obligation, such as a line of credit or credit card, where the balance represents the amount of credit used by the borrower. "Principal Balance" refers to the total amount of credit that has been drawn and is outstanding on a credit facility, excluding any accrued interest, fees, or charges. Copy Definition of "Principal Balance" for debt repayment calculations This definition connects "Principal Balance" to its use in calculating debt repayment amounts. "Principal Balance" means the remaining unpaid amount of the debt on a loan or credit agreement, which serves as the basis for calculating monthly repayment amounts. Copy Definition of "Principal Balance" for business loans This definition links "Principal Balance" to its use in business loans, where it indicates the remaining liability the business has toward the lender. "Principal Balance" refers to the amount still owed by a business under a loan agreement, excluding interest and fees, and is used in financial calculations related to the loan. Copy Definition of "Principal Balance" in a loan agreement This definition applies "Principal Balance" to loan agreements, where it represents the amount of the loan yet to be repaid. "Principal Balance" means the amount of the original loan principal that remains unpaid as of the date specified in the agreement, excluding any interest or fees added. Copy Definition of "Principal Balance" as a factor in loan payoff This definition connects "Principal Balance" to its role in determining how much a borrower must pay to settle a loan. "Principal Balance" refers to the remaining loan amount, excluding interest, that must be paid to fully repay the loan and terminate the financial obligation. Copy Definition of "Principal Balance" for calculating interest This definition ties "Principal Balance" to the calculation of interest on loans, as interest is often based on the remaining principal balance. "Principal Balance" means the amount of the loan that is used to calculate the interest charges, as interest is typically applied to the unpaid principal balance. Copy Definition of "Principal Balance" for personal loans This definition links "Principal Balance" to personal loans, where it represents the outstanding amount owed by the borrower to the lender. "Principal Balance" refers to the remaining sum of money owed under a personal loan agreement, excluding any accrued interest or fees. Copy Definition of "Principal Balance" in a secured loan context This definition connects "Principal Balance" to secured loans, where the outstanding amount is often secured by collateral. "Principal Balance" means the remaining amount of a secured loan that must be repaid, excluding interest, fees, and charges, and secured by collateral. Copy Definition of "Principal Balance" for a loan schedule This definition links "Principal Balance" to its use in a loan repayment schedule. "Principal Balance" refers to the balance of the loan that remains after accounting for previous payments, which is used in loan schedules to determine future payments and interest charges. Copy Definition of "Principal Balance" in relation to student loans This definition applies "Principal Balance" to student loans, where it represents the outstanding amount the borrower is obligated to repay. "Principal Balance" refers to the portion of the student loan that has not yet been paid off, excluding interest, which the borrower must pay back in the future. Copy Definition of "Principal Balance" for mortgage amortization This definition connects "Principal Balance" to mortgage amortization, a process where the loan principal is paid down over time. "Principal Balance" means the remaining amount of a mortgage loan, which is amortized over the life of the loan through regular payments of principal and interest. Copy Definition of "Principal Balance" for loan refinancing This definition links "Principal Balance" to loan refinancing, where the balance plays a role in restructuring the loan terms. "Principal Balance" refers to the remaining amount of the loan that is considered when refinancing, influencing the new terms of the refinanced loan agreement. Copy Definition of "Principal Balance" in loan prepayment scenarios This definition connects "Principal Balance" to prepayment, where the borrower may choose to pay off the remaining balance early. "Principal Balance" means the total amount of the loan that remains unpaid and can be fully or partially paid off ahead of the agreed schedule through prepayment. Copy Definition of "Principal Balance" for car loans This definition ties "Principal Balance" to car loans, where it represents the amount of money still owed after car payments. "Principal Balance" refers to the remaining balance of a car loan, which excludes any interest or fees, and is the amount the borrower must pay off to own the vehicle outright. Copy Definition of "Principal Balance" in revolving credit facilities This definition links "Principal Balance" to revolving credit facilities, such as credit lines or credit cards. "Principal Balance" means the amount of credit drawn by the borrower under a revolving credit facility, excluding interest and fees, and is the balance that must be repaid. Copy Definition of "Principal Balance" for debt consolidation loans This definition applies "Principal Balance" to debt consolidation, where it reflects the total remaining debt. "Principal Balance" refers to the total remaining balance on loans being consolidated, which is combined into a new loan for easier repayment. Copy Definition of "Principal Balance" for corporate loans This definition connects "Principal Balance" to corporate loans, where it represents the outstanding amount the company owes. "Principal Balance" refers to the remaining unpaid amount of a corporate loan, excluding any interest, fees, or other charges, that the business must repay. Copy Definition of "Principal Balance" in the context of a bridge loan This definition ties "Principal Balance" to bridge loans, where it helps assess the loan's current status. "Principal Balance" means the amount of the original loan that has not been repaid in full, typically used in the context of a short-term bridge loan to cover financial gaps. Copy Definition of "Principal Balance" for credit card debt This definition links "Principal Balance" to credit card debt, which is carried month to month. "Principal Balance" refers to the amount of outstanding credit card debt, excluding any interest or fees that the borrower must repay. Copy Definition of "Principal Balance" for home equity loans This definition connects "Principal Balance" to home equity loans, where it represents the remaining amount owed on the property. "Principal Balance" means the outstanding amount of a home equity loan, excluding interest and fees, that must be repaid by the borrower based on the terms of the loan agreement. Copy Definition of "Principal Balance" in asset-backed loans This definition ties "Principal Balance" to asset-backed loans, where the balance reflects the remaining amount secured by assets. "Principal Balance" refers to the amount remaining on a loan secured by assets, such as real estate or equipment, excluding any accrued interest or additional charges. Copy Definition of "Principal Balance" for a line of credit This definition applies "Principal Balance" to a line of credit, where it represents the outstanding drawn amount. "Principal Balance" means the total amount drawn against a line of credit, excluding interest and fees, which must be repaid by the borrower. Copy Definition of "Principal Balance" for a revolving loan This definition connects "Principal Balance" to a revolving loan, where the balance changes as the borrower borrows and repays. "Principal Balance" refers to the current amount outstanding on a revolving loan, excluding interest and fees, at any given time during the life of the loan. Copy Definition of "Principal Balance" in the context of a fixed-rate loan This definition links "Principal Balance" to fixed-rate loans, where it remains the same for the duration of the loan, excluding interest. "Principal Balance" means the fixed amount of the loan principal that remains due over the term of the loan, excluding any interest, fees, or other charges. Copy Definition of "Principal Balance" in a debt agreement This definition ties "Principal Balance" to a broader debt agreement, representing the amount owed without the added charges. "Principal Balance" refers to the portion of the debt that remains unpaid under the terms of a debt agreement, excluding interest, penalties, or other adjustments. Copy Definition of "Principal Balance" in commercial real estate loans This definition connects "Principal Balance" to commercial real estate loans, where it represents the remaining balance due on the loan secured by the property. "Principal Balance" refers to the unpaid portion of a commercial real estate loan, excluding any accrued interest or other related charges. Copy Definition of "Principal Balance" for installment loans This definition ties "Principal Balance" to installment loans, where it represents the remaining unpaid amount of the loan after periodic payments. "Principal Balance" means the remaining balance on an installment loan, excluding interest, that the borrower is obligated to pay over the life of the loan. Copy Definition of "Principal Balance" in student loan refinancing This definition links "Principal Balance" to the refinancing of student loans, where the principal is adjusted based on new terms. "Principal Balance" refers to the amount remaining on a student loan after refinancing, excluding interest, which is carried forward under the new loan agreement. Copy Definition of "Principal Balance" for an interest-only loan This definition connects "Principal Balance" to interest-only loans, where the borrower only pays interest for a specified period, and the principal balance remains unchanged. "Principal Balance" means the amount owed under an interest-only loan, which remains constant until principal payments begin after the interest-only period ends. Copy Definition of "Principal Balance" in private equity financing This definition applies "Principal Balance" to private equity financing, where the balance represents the amount of capital invested that remains outstanding. "Principal Balance" refers to the outstanding amount of capital invested in a private equity deal, excluding any profits or dividends that may have been paid out. Copy Definition of "Principal Balance" for debt restructuring This definition ties "Principal Balance" to debt restructuring, where the balance of the debt is adjusted or modified in the process. "Principal Balance" means the remaining amount of debt being restructured, which may be adjusted or renegotiated as part of the debt restructuring process. Copy This article contains general legal information and does not contain legal advice. Cobrief is not a law firm or a substitute for an attorney or law firm. The law is complex and changes often. For legal advice, please ask a lawyer. Last updated 30 Mar 2025 Product How it works Testimonials Why Cobrief? 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Sign in to start your free trial. live 00:00 00:00 1x Speed × 0.75x 1x 1.5x 2x CC Subtitles × MEDIA_ELEMENT_ERROR: Format error 13.17 : Equation of Continuity Embed Video Add to Playlist en EN - EnglishCN - 中文DE - DeutschES - EspañolKR - 한국어IT - ItalianoFR - FrançaisPT - PortuguêsTR - TürkçeJA - 日本語PL - PolskiRU - РусскийHE - עִברִיתAR - العربية Text Related Fluid motion is represented by either velocity vectors or streamlines. The volume of a fluid flowing past a given location through an area during a period of time is called the flow rate Q, or more precisely, the volume flow rate. Flow rate and velocity are related—for instance, a river has a greater flow rate if the velocity of the water in it is greater. However, the flow rate also depends on the size and shape of the river. 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This is called the equation of continuity, and it is valid for any incompressible fluid (with constant density). Since liquids are essentially incompressible, the equation of continuity is valid for all liquids. However, gases are compressible, so the equation must be applied with caution to gases if they are subjected to compression or expansion. This text is adapted fromOpenstax, University Physics Volume 1, Section 14.5: Fluid Dynamics. Transcript The volume of fluid flowing through a point in an area in unit time gives the volume flow rate. On substituting volume with area times distance, and using the relation between velocity and distance, volume flow rate equals area times velocity of the fluid. Consider an incompressible fluid with the same density at all points flowing steadily through an irregular cross-sectional pipe. For a steady flow, the velocity and density of the fluid at a point remain constant with time. 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2965	https://physics.stackexchange.com/questions/847515/difference-in-weights-between-two-submerged-objects	newtonian mechanics - Difference in weights between two submerged objects - Physics Stack Exchange Join Physics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Difference in weights between two submerged objects [duplicate] Ask Question Asked 5 months ago Modified5 months ago Viewed 1k times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. This question already has answers here: Which way does the scale tip? (7 answers) Closed 6 months ago. Both the basketball and the medicine (steel) ball are the same size, which means they're displacing the same amount of water. That should mean the buoyant force is the same on both, right? But then why does one side of the scale tip more than the other? I'm trying to figure out what extra force is acting here that the scale can "feel" — because it seems like the water should cancel things out. newtonian-mechanics free-body-diagram fluid-statics buoyancy Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Apr 12 at 14:08 Vincent Thacker 15.5k 16 16 gold badges 44 44 silver badges 63 63 bronze badges asked Apr 12 at 4:06 VishVish 123 5 5 bronze badges 2 3 Hanging from the roof or the container?Qmechanic –Qmechanic♦ 2025-04-12 10:51:10 +00:00 Commented Apr 12 at 10:51 1 Exactly this problem, basics2022.github.io/bbooks-physics-hs/ch/mechanics/…. Only in Italian here, but please rely on the automatic translator in your browser basics –basics 2025-04-12 13:33:25 +00:00 Commented Apr 12 at 13:33 Add a comment\| 8 Answers 8 Sorted by: Reset to default This answer is useful 7 Save this answer. Show activity on this post. The important point is that internal forces cancel out. Therefore, it can immediately be concluded that the weight of the system on the right is simply the sum of the container, water and basketball. Similarly, for the system on the left, without the string, its weight will be the sum of the container, water and steel ball. But the string exerts an external force equal to the difference in weight between the steel ball and an equivalent volume of water. Subtracting off this difference from the weight, it can be seen that the weight felt by the scale becomes the sum of the container, water and the equivalent volume of water i.e. that of an identical container filled to the same level with just water. Therefore, since the basketball is less dense than water as shown by the taut string, it weighs less than the equivalent volume of water. The system on the left is heavier. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 13 at 17:25 answered Apr 12 at 12:54 Vincent ThackerVincent Thacker 15.5k 16 16 gold badges 44 44 silver badges 63 63 bronze badges Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. The free body diagrams are as shown below. The steel ball forces add up to zero as do the basketball forces. The downward force on the scales is M w g+U M w g+U for the steel ball arrangement which is greater than M w g+U−t M w g+U−t for the basketball arrangement. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Apr 12 at 9:23 FarcherFarcher 106k 5 5 gold badges 88 88 silver badges 235 235 bronze badges 2 2 Same symbol for the two weights?Bob D –Bob D 2025-04-12 10:03:41 +00:00 Commented Apr 12 at 10:03 M and m? Sorry about poor handwriting and M w M w is the weight of the water.Farcher –Farcher 2025-04-12 15:52:12 +00:00 Commented Apr 12 at 15:52 Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. First consider what happens with no strings attached (no pun intended). The downward force on each side of the scale will simplify be the weight on each side. The weight on the left is greater than the right (assumes same cups and volume of water) and the scale will tilt down on the left. The buoyant forces are internal to the ball/water/cup system and therefore have no effect on the weight of each system with no strings attached. Therefore, with no strings attached, the scale will tilt down on the left. Now consider the effect of the strings. The tension T T in the string on the left is an external force on the ball/water/cup system. It therefore lessens the downward force on the scale on the left according to the following where W M B W M B is the weight of the medicine ball W l e f t=W M B+W H 2 O+W c u p−T(1)(1)W l e f t=W M B+W H 2 O+W c u p−T The tension on the left makes the net force on the medicine ball of zero. Thus the tension equals the weight of the medicine ball minus the upward buoyant force, or T=W M B−F b(2)(2)T=W M B−F b Substituting (2) into (1) W l e f t=W H 2 O+W c u p+F b(3)(3)W l e f t=W H 2 O+W c u p+F b The tension in the string on the right, which is attached to the bottom of the cup, is an internal force of the system. Therefore it does not effect the weight on the right. But the weight of the ball component of the ball/water/cup system and the buoyant force will determine the tension. The weight on the right side, where the weight of the basketball is W B B W B B, is W r i g h t=W B B+W H 2 O+W c u p(4)(4)W r i g h t=W B B+W H 2 O+W c u p The scale will be balanced if the weight on the left equals the weight on the right, or W B B=F b(5)(5)W B B=F b But equation (5) is the equivalent of stating that for the scale to be balanced, the weight of the basketball must equal the buoyant force acting on it, i.e., the density of the basketball equals the density of the water displaced by the ball. That in turn means the basketbal can can float completely submerged without the need of the string, i.e., the tension in the string would be zero. For there to be any tension in the string, the basketball would need to weigh less, which in turns would make the total weight on the right side less. From that we can conclude that if there is any tension in the string on the right, the scale will necessarily tilt down on the left. Hope this helps. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 13 at 19:25 answered Apr 12 at 12:02 Bob DBob D 83.3k 7 7 gold badges 61 61 silver badges 160 160 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Buoyancy doesn't remove weight. Throw a ball into a cup and, even if it floats, the cup will feel heavier. The two balls feel the same buoyancy, but the left-hand container is still heavier, because the density of steel > the density of the basketball. To clarify: for the steel ball, we can write the forces acting as T+ρ g V=m g T+ρ g V=m g. For the basketball, it is m′g+T′=ρ g V m′g+T′=ρ g V. Both of these must balance, since the balls aren't moving. The water exerts the full ρ g V ρ g V on the steel ball. By Newton's third law, the ball exerts the full ρ g V ρ g V on the water as well (and by extension the scales). The water also exerts the full ρ g V ρ g V on the basketball, but because this buoyancy is > the basketball's weight, there is also a tension in the string holding the basketball underwater that reduces the weight on the scales. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 12 at 13:25 answered Apr 12 at 4:13 AllureAllure 23.5k 6 6 gold badges 65 65 silver badges 108 108 bronze badges 2 1 But the steel ball is suspended by a thread. The tension in the string and mg should cancel out Vish –Vish 2025-04-12 04:18:08 +00:00 Commented Apr 12 at 4:18 @Vish not exactly since there's also a tension. I edited the answer to clarify.Allure –Allure 2025-04-12 13:26:04 +00:00 Commented Apr 12 at 13:26 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. The difference is the anchoring points, as others have said. It helps me in these cases to just remember that the principle of buoyancy is a special case of the minimization of potential energy in static equilibrium. Consider that the left cup goes up and the right cup goes down by a millimeter, how has the energy changed? Assume that the rounding of the cups doesn't matter as if there were no balls the two would balance, so idealize those as perfect cylinders with some cross section area, say, 400 cm 2 400 cm 2, filled up maybe to 30 cm 30 cm. The balls inside are size-3 mini basketballs with a radius of 8.9 cm,8.9 cm, the steel being some 23 kilos and the normal one being 280 grams. You can do this in two ways, "teleporting water" analysis or actual equations. If you analyze buoyancy by teleporting water, then when the cup holding the steel ball moves up by a millimeter, a volume of 0.1 cm×400 cm 2=40 mL 0.1 cm×400 cm 2=40 mL needs to be deleted from the bottom of the cylinder and then re-added at the top of the cylinder, it has “teleported” up by 30 cm.30 cm. But on the right a separate 40 mL 40 mL is displaced downward by 30 cm 30 cm, and these two effects cancel in terms of energy. And this is why if there were no balls, the scale would balance. However, what changes with the ball on the right is that the steel ball on the left doesn't move but the basketball on the left has also dropped 1 mm 1 mm. The ball has some mass which lowers the energy, but it also will sweep out its cross section of π r 2 π r 2 (roughly 250 cm 2 250 cm 2) for a millimeter and that water has to be teleported upwards as it sinks downwards, by an average distance 4 3 r 4 3 r (roughly 12 cm). So you have 280 grams of basketball moving down by one millimeter, energy change roughly −2.8 mJ−2.8 mJ, but balanced out by 25 grams of water teleporting up by 12 centimeters on average, energy change roughly +3.0 mJ+3.0 mJ, and we know that the second number is always greater (in absolute value) than the first number because if it weren't, the ball wouldn't float in the first place. Energy minimization therefore says that the floating basketball will pull its side up with its buoyant force, the steel ball will not push its side down with the same because it is suspended and unable to fall to reduce the potential energy. In equations, we have the left hand side sitting at vertical position +y+y, the right hand side sitting at −y−y, the potential energies on the left and right are, U L=ρ g(π R 2 h[h 2+y]−4 3 π r 3 h 2)+m Steel g h 2,U R=ρ g(π R 2 h[h 2−y]−4 3 π r 3[h 2−y])+m Basketball g[h 2−y].U L=ρ g(π R 2 h[h 2+y]−4 3 π r 3 h 2)+m Steel g h 2,U R=ρ g(π R 2 h[h 2−y]−4 3 π r 3[h 2−y])+m Basketball g[h 2−y]. It's a lot to write but you can see the above analysis poking through: when we form U=U L+U R U=U L+U R the leading π R 2 h π R 2 h terms will add together to lose their y-dependence, and then when we compute the force the d U d y d U d y terms also disappear completely, leaving just 4 3 π r 3 ρ−m Basketball,4 3 π r 3 ρ−m Basketball, which is just the criterion for whether the basketball floats or sinks. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 12 at 16:44 answered Apr 12 at 4:48 CR DrostCR Drost 39.7k 3 3 gold badges 44 44 silver badges 119 119 bronze badges 5 I also feel the way to analyze this ought to be by comparing small changes in position. However, part of your analysis confuses me: why would a shift in the scale cause the basketball to displace a different amount of water on the right? Wouldn't the water, container, string and ball all move the same direction? At least eventually?Dan Getz –Dan Getz 2025-04-12 13:45:36 +00:00 Commented Apr 12 at 13:45 Of course. But why did you write "31.4 mL of water has been displaced upward as the air pocket descends"? What water's moving upwards?Dan Getz –Dan Getz 2025-04-12 15:31:42 +00:00 Commented Apr 12 at 15:31 @DanGetz So it doesn't displace a different amount of water, but that water changes its physical location in space which changes its potential. Like, forget the steel ball and it's cup and the teeter totter: we agree that if the cup with the basketball moves down its energy decreases, yes? I added a formula at the end of the above answer to help maybe see why the derivative I am hinting at, destroys everything but the basketballs buoyant force because the basketball and the missing void of water it displaced are the only parts of U U varying with y y.CR Drost –CR Drost 2025-04-12 16:47:41 +00:00 Commented Apr 12 at 16:47 (sorry, I deleted my comment before you responded, other readers will see some of this conversation out of order as I had to run back through my paste buffer to recover it)CR Drost –CR Drost 2025-04-12 16:48:44 +00:00 Commented Apr 12 at 16:48 But yeah, as the ball comes downwards, it pushes some water that was below, out of its way, so that it can occupy the space underneath it. Then that water pushes other water around it until it occupies the space that the ball came down out of. The cross section of the ball is (10cm ball) 314 cm², times one millimeter is 31.4 mL, and that water has to travel roughly 13.3 cm upwards on average to fill the space that the basketball left behind. (I also fixed the numbers in my answer, basketballs are standardized lol)CR Drost –CR Drost 2025-04-12 16:53:03 +00:00 Commented Apr 12 at 16:53 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Both cups have the same amount of water in them. Weight of water is thus equal. The two balls have the same volume, so the buoyancy of both is the same. Weight of water displaced are the same. The wire holding the medicine ball will provide T=T=weight of medicine ball - buoyancy. This means that on the left side, the scale is holding up cup of water + displaced volume of water. This is equivalent to having no ball there but with the ball's volume replaced by water. On the right hand side, the basketball is held down by a string tied to the bottom. Clearly this string is pulling down the basketball, and it can only do that by pulling up on the cup. The cup thus weighs slightly less, hence the tilt. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Apr 12 at 4:33 naturallyInconsistentnaturallyInconsistent 16.3k 2 2 gold badges 22 22 silver badges 47 47 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. The left balance beam with the steel ball will go down! This is very easy to see. The water volume and weight in both containers is the same. Thus it is the difference in the forces on the immersed equal volume steel and basket ball that determines which side goes down. On the right balance beam, the basket ball is attached to the bottom of the container. Thus there is no effect of the internal force of buoyancy of the basket ball on the right balance beam. The total external force acting on the right beam is just the sum of weights of the container, the water and the basket ball. The total external force on the left beam is the weight of the container and the water, which is equal to the right container, plus the external force on the water exerted by the immersed hanging steel ball. According to Newton’s third law, this force must be opposite to the buoyancy exerted by the water on the steel ball, which, according to Archimedes, corresponds to the weight of the displaced water. The weight of the displaced water is obviously larger than the weight of the equal volume basket ball. Therefore the total downward force on the left balance beam with the steel ball is larger than the force on the right balance beam with the basket ball. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 13 at 17:11 answered Apr 13 at 16:46 freecharlyfreecharly 18.6k 2 2 gold badges 21 21 silver badges 44 44 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Notice the difference in how the tether that holds the ball in position is secured. The fluid container with the basket ball: the tether is secured to the fluid container. The fluid container with the steel ball: the tether is secured to something other than fluid container. So between left hand side and right hand side there are two differences that affect the outcome. The density of the ball that is used Whether the tether is secured to the container, or to something other than the container In order to make a comparison the second difference must be removed. The two options: The tether supporting the steel ball should be secured to the fluid container itself The basket ball should be held below the fluid surface by a rod that is secured to something other than the fluid container Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 12 at 6:48 answered Apr 12 at 4:33 CleonisCleonis 25.3k 1 1 gold badge 27 27 silver badges 76 76 bronze badges Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions newtonian-mechanics free-body-diagram fluid-statics buoyancy See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 86Which way does the scale tip? 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2966	https://www.youtube.com/watch?v=A3Ffwsnad0k	Discrete Math - 1.1.1 Propositions, Negations, Conjunctions and Disjunctions Kimberly Brehm 113000 subscribers 11093 likes Description 1268730 views Posted: 18 Feb 2020 This is the first video in the new Discrete Math playlist. In this video you will learn about propositions and several connectives including negations, conjunctions and disjunctions and explore their truth table values. Video Chapters: Introduction 0:00 Propositions 0:22 Negations 4:27 Truth Tables 6:34 Conjunctions 11:08 Disjunctions 15:02 Inclusive or XOR 17:20 Up Next 19:17 Textbook: Rosen, Discrete Mathematics and Its Applications, 7e Playlist: Power Point Notes: 349 comments Transcript: Introduction in this first video of the discrete mathematics playlist we are going to look at propositions what they are and then look at negations conjunctions and disjunctions and we'll also take a look at the truth tables that go along with those connectives the connect toasts are the negations conjunctions and disjunctions so our first new Propositions terminology here is a proposition and a proposition is just a declarative statement that is either true or false so notice I have five statements written down three are in pink two are in green the statements written in pink are all considered propositions because they are declarative statements that are either true or false the sky is blue is a declarative statement generally what we will do is we will take this statement and we will say let's let P represent that statement or the moon is made of cheese also a declarative statement again that could be either true or false so I'm going to say false on that one and I'm going to say that is proposition Q the moon is made of cheats again it is considered a proposition because it is an eclair ative statement that is either true or false same thing with Luke I am your father this is a declarative statement I am your father either true or false I'm gonna say that one's false and we'll say that's R now take a look at the difference of the statements that I've written in green D says sit down sit down is a statement but it is neither true nor false it I can't say true you have sat down that statement would have to be you sat down that's either true or false but sit down is just telling you what to do and that is not a proposition a lot of people struggle with E because they say well that could be true or false and while you're thinking is correct the way that it is written right now is not a proposition if I replaced X with some value or if I said where x equals 5 well now it's a proposition because I can say 5 plus 1 equals 2 is false or if I said where X equals 1 then 1 plus 1 equals 2 is true but if I don't assign a value for X and I just leave it as X then this is not a proposition because it's not true or false so propositions themselves are fairly straightforward again we're using a lower case letter PQRS etc to represent a proposition and then what we're going to do is we're going to end up making a compound proposition using these connectives so I want you to think of these connectives like operators as I would take 1 plus 2 that's an operator so these are just operators for propositions and I've put them all on one page together just to have one page that you could refer back to they're all called connectives so we're going to go each of through each of these in detail in the following slides but I did want to put them all here so let's talk about them very quickly we have the negation and again this is how we would use that negation so I would say not P and again I would say it as not but that is the symbol that I would use conjunction is and and again we'll talk about each of these in detail so obviously I would be taking P and Q a disjunction is an or so that would be P or Q an implication is an if-then so this is an if P then Q and then a by conditional notice it has an arrow on each end here says if and only if so P if and only if Q and that means they both have to share the same truth value so again we're using p q RS etc using those lowercase letters to represent each proposition so let's get into each of these connectives in further detail Negations the first connective is a negation and again the negation is not and this is the symbol that we would use so the negation of the proposition P is not P so example if I say P denotes the grass is green then not P denotes it's not the case that the grass is green now it's silly to write it that way so we don't instead we say the grass is not green so let's take a look at these few examples I have here and then I want to look at some truth tables with you to make sure that this all makes sense so the first one says my dog is the cutest dog which is a true proposition by the way so my dog is the cutest dog is my proposition P if I want to write not P then it would be my dog is not the cutest dog so that is not P again we could say it is not the case that my dog is the cutest dog but it's easier to write it the way we would normally say it in the English language the door is not open is P now again if the door is not open is P then if I'm negating something that already seems like it's negated remember in mathematics is the only place that two wrongs do make a right so the door is not not open which means the door is open again we could say it is not the case that the door is not open but in real life we would just say the door is open last one are we there yet be careful with this because this is not a proposition it's not a declarative statement that is either true or false and because of that I can't negate it so Truth Tables let's talk now about truth tables and how a truth table works so if we have a truth essentially what we have is a row for each possibility of the truth values of our propositions so I want you to think of this as having two parts so this is a very very basic truth table that we're going to do but the left side of our truth table has all of the combinations of the truth values for our propositions so in this case if I just have one proposition P that P can either be true or it can be false so let's say P again represented my dog is the cutest dog then not P represents my dog is not the cutest dog so how does the truth table work well again on the left side we're going to give all of the combinations which is very easy for one proposition and on the right side we're going to use whatever our connectives are and so we might have several columns on each side in this case we just have one column on each side but we might have several depending on how complicated our truth values or our truth table is going to get in this case let's look at our proposition P and here's how a true table works let's say P is true so here I am P is true my dog is the cutest dog is a true statement which means not P my dog is not the cutest dog would have to be false because my dog can't be the cutest dog and not the cutest dog at the same time let's say instead that my proposition was false I said my dog is the cutest dog and that is an incorrect statement then not P which says my dog is not the cutest dog would have to be a true statement so that's how a truth table works is on the left side we have all of the different combinations on the right side we have whatever connectives are going to use truth tables were going to be very important to us so that's why I wanted to introduce them to you in this video so that you had a good foundation when we get to our next video where we're going to use them in more detail before I continue on to our next proposition I just want to remind you that in this case I only gave you one proposition we're going to have several in fact our very next example is a proposition where I have to excuse me a connective that requires us to use two propositions so if I have two propositions I'm going to have 2 to the N rows which is 2 squared rows or 4 rows and really this is all about just how many combinations are there let's say I'm using P and Q and it doesn't matter what's the connective is in between there so we're going to do P and Q we're going to do P or Q and we're going to do these in just a little bit but just so we understand the left side would have to have all of the values for P and all of the values for Q all of the combinations of those two so here's the best way to go about this if I've got P and Q here then P could be true and Q could be true P could be true and Q could be false then it's also a possibility that P is false and Q is true or P is false and Q is false so notice on the left side which has two columns these are my combinations now do your professor a favor and write them in this way I have a lot of students who do true true and then false true and then false false true false and it gets very hard for me to check your work when you're all willy-nilly like that so do me a favor keep these groups together and then these will alternate and we'll continue to work on that together Conjunctions our next connective is called a conjunction and a conjunction of propositions P and Q is denoted P with the little arrow head basically Q and it's read P and Q so a conjunction is an and and one way to help you remember this is this kind of looks like a capital A if you added that little marking in the middle now the reason I bring this up is because our next one is going to be the arrow pointing down instead of pointing up so it's good to be able to keep them straight for a conjunction to be true both propositions must be true so we're going to create the truth table together but it's also important to think about the fact that p and q represent statements so let's say p is it is raining q is i am home so if i'm creating my truth table remember there are two sides to this and on the left side is where we just give all of the different possible truth table combinations and since there are true two propositions that means it's going to be 2 squared or 4 different rows so I'm going to this is not going to be a row this is just going to be where I put my values P and Q so I need 4 rows so 1 2 3 4 rows and of course a column for each each of my propositions so I've got P Q and I'm going to list all of the combinations so P could be true with Q true or true false or false true or false false so that's all of the combinations on the right side is always going to be whatever it is that you are doing a connective of so in this case we're only doing P and Q or the conjunction of P and Q and that's the only thing I will need on the right side of my truth table now before we start filling it in let's think about what this means P says it's raining Q says I'm home for this conjunction to be true both propositions must be true so it must be raining and I must be home so this says P is true Q is true it is raining and I am home the only way for P and Q to be true is for both P and Q to be true and that is the case here true true so P and Q is true for the rest of the rows this row represents it is raining I am NOT home so that's false because they're not both true because I'm not home this row represents it is not raining I am home and I am home so it's not raining and I am home of course would still be false because it's not raining and this row represents it's not raining I am NOT home again false because they both have to be true for it to be true so it doesn't matter if you put these propositions of meaning to the propositions but keep in mind that that's what you'll be doing quite often is you'll be using those propositions using those statements to write them as letters and then go from there to make a truth table Disjunctions another connective is the disjunction the disjunction of propositions P and Q is denoted P with the opposite facing arrow so basically a V and read P or Q and for a disjunction to be true I either proposition must be true so when we were talking about a conjunction both had to be true for the conjunction to be true for a disjunction either proposition must be true so again when I create my truth table I'm always going to start on that left side where I'm giving all of the different combinations and on the left side again I've got true true true false false true false false so that's just the combination side I haven't done anything yet on the right side is where I'm going to write whatever my results are in this case the connective is the disjunction so I'm going to write P or Q and again I think of a disjunction or the oar as being a cup so if you can put anything in the cup if either one is true then the result is true so if P is true and Q is true then P or Q is true because either one of them is true if P is true but Q is false P or Q is still true because P was true only one of them needs to be true they can both be true but only one of them needs to be true if P is false but Q is true P or Q is still true because Q is true so the only false value I'm going to have is where both P and Q are false because neither one is true and for the disjunction to be true either proposition has to be true so that's my solution the ORS can get a little bit Inclusive or XOR tricky most of the time in mathematics we're using that inclusive or that we just talked about that's the P or Q and that is saying for instance the prerequisite for ma 420 is either ma 315 or ma 335 and by that I mean you could have passed ma 315 you could have passed ma 335 or you could have passed both of them and you can still get into ma 420 that's the inclusive or that's the one we use most often there's also the connective or in English which is called xor and that is something like you get soup or salad with your entree now if I'm buying an entree that means I can get su or I can get salad but I can't get both I mean I can I can just pay extra but I can get one or the other but I cannot get both so while we just finished talking about the inclusive or where we said if either is true then it's true and that was true true true false the exclusive-or is only true if one is true or the other is true but not both so the difference here is because this one had two truths then the exclusive order is going to return a false value because I can't have both soup and salad I can have soup or I can have salad and then of course false false is not going is still waiting to be false so hopefully you can understand the difference between those now notice the different notation I'm going to use now this is the XOR notation which is basically just a circle with a plus sign in it up next we're going to continue Up Next our study of the connectives by setting implications and by conditionals and we're of course also going to look at the converse inverse and contrapositive of the implications I hope you can join me
2967	https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/31%3A_Radioactivity_and_Nuclear_Physics/31.04%3A_Nuclear_Decay_and_Conservation_Laws	Skip to main content 31.4: Nuclear Decay and Conservation Laws Last updated : Feb 20, 2022 Save as PDF 31.3: Substructure of the Nucleus 31.5: Half-Life and Activity Buy Print CopyView on Commons Donate Page ID : 2779 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives By the end of this section, you will be able to: Define and discuss nuclear decay . State the conservation laws . Explain parent and daughter nucleus . Calculate the energy emitted during nuclear decay . Nuclear decay has provided an amazing window into the realm of the very small. Nuclear decay gave the first indication of the connection between mass and energy , and it revealed the existence of two of the four basic forces in nature. In this section, we explore the major modes of nuclear decay ; and, like those who first explored them, we will discover evidence of previously unknown particles and conservation laws . Some nuclides are stable, apparently living forever. Unstable nuclides decay (that is, they are radioactive ), eventually producing a stable nuclide after many decays. We call the original nuclide the parent and its decay products the daughters. Some radioactive nuclides decay in a single step to a stable nucleus . For example, Co60Co60 is unstable and decays directly to Ni60Ni60, which is stable. Others, such as U238U238, decay to another unstable nuclide , resulting in a decay series in which each subsequent nuclide decays until a stable nuclide is finally produced. The decay series that starts from U238U238 is of particular interest, since it produces the radioactive isotopes Ra226Ra226 and Po210Po210, which the Curies first discovered (Figure 31.4.131.4.1). Radon gas is also produced (Rn222Rn222 in the series ), an increasingly recognized naturally occurring hazard. Since radon is a noble gas, it emanates from materials, such as soil, containing even trace amounts of U238U238 and can be inhaled. The decay of radon and its daughters produces internal damage. The U238U238 decay series ends with Pb206Pb206, a stable isotope of lead. Note that the daughters of αα decay shown in Figure 31.4.131.4.1 always have two fewer protons and two fewer neutrons than the parent . This seems reasonable, since we know that αα decay is the emission of a He4He4 nucleus , which has two protons and two neutrons. The daughters of ββ decay have one less neutron and one more proton than their parent . Beta decay is a little more subtle, as we shall see. No γγ decays are shown in the figure, because they do not produce a daughter that differs from the parent . Alpha Decay In alpha decay , a He4He4 nucleus simply breaks away from the parent nucleus , leaving a daughter with two fewer protons and two fewer neutrons than the parent (Figure 31.4.231.4.2). One example of αα decay is shown in Figure 31.4.231.4.2 for U238U238. Another nuclide that undergoes αα decay is 239Pu239Pu. The decay equations for these two nuclides are U238→Th92234+He4 U238→Th92234+He4(31.4.1) Pu239→U235+He4. Pu239→U235+He4.(31.4.2) If you examine the periodic table of the elements, you will find that Th has Z=90Z=90, two fewer than U, which has Z=92Z=92. Similarly, in the second decay equation, we see that U has two fewer protons than Pu, which has Z=94Z=94. The general rule for αα decay is best written in the format AZXNAZXN. If a certain nuclide is known to αα decay (generally this information must be looked up in a table of isotopes , such as in Appendix B), its αα decay equation is XAZN→YN−2Z−2A4−+He242(αdecay) XZAN→YN−2Z−2A−4+He224(αdecay)(31.4.3) where YY is the nuclide that has two fewer protons than XX, such as ThTh having two fewer than UU. So if you were told that 239Puα239Puα decays and were asked to write the complete decay equation , you would first look up which element has two fewer protons (an atomic number two lower) and find that this is uranium. Then since four nucleons have broken away from the original 239, its atomic mass would be 235. It is instructive to examine conservation laws related to αα decay . You can see from the equation XAZN→YN−2Z−2A4−+He242 XZAN→YN−2Z−2A−4+He224(31.4.4) that total charge is conserved. Linear and angular momentum are conserved, too. Although conserved angular momentum is not of great consequence in this type of decay , conservation of linear momentum has interesting consequences. If the nucleus is at rest when it decays, its momentum is zero. In that case, the fragments must fly in opposite directions with equal-magnitude momenta so that total momentum remains zero. This results in the αα particle carrying away most of the energy , as a bullet from a heavy rifle carries away most of the energy of the powder burned to shoot it. Total mass – energy is also conserved: the energy produced in the decay comes from conversion of a fraction of the original mass . As discussed in the modele on Atomic Physics/30%3A_Atomic_Physics "30: Atomic Physics"), the general relationship is E=(Δm)c2. E=(Δm)c2.(31.4.5) Here, EE is the nuclear reaction energy (the reaction can be nuclear decay or any other reaction), and ΔmΔm is the difference in mass between initial and final products. When the final products have less total mass , ΔmΔm is positive, and the reaction releases energy (is exothermic). When the products have greater total mass , the reaction is endothermic (ΔmΔm is negative) and must be induced with an energy input. For αα decay to be spontaneous, the decay products must have smaller mass than the parent . Example 31.4.131.4.1: Alpha Decay Energy Found from Nuclear Masses Find the energy emitted in the αα decay of Pu239Pu239. Strategy Nuclear reaction energy , such as released in α decay , can be found using the equation E=(Δm)c2E=(Δm)c2. We must first find ΔmΔm, the difference in mass between the parent nucleus and the products of the decay . This is easily done using masses given in Appendix A. Solution The decay equation was given earlier for 239Pu239Pu; it is 239Pu→235U+4He. 239Pu→235U+4He. Thus the pertinent masses are those of 239Pu239Pu, 235U235U, and the αα particle or 4He4He, all of which are listed in Appendix A. The initial mass was m(239Pu)=239.052157um(239Pu)=239.052157u. The final mass is the sum: m(235U)+m(4He)=235.043924u+4.002602u=239.046526u.m(235U)+m(4He)=235.043924u+4.002602u=239.046526u. Thus, Δm=m(239Pu)−[m(235U)+m(4He)]=239.052157u−239.046526u=0.0005631u. Δm=m(239Pu)−[m(235U)+m(4He)]=239.052157u−239.046526u=0.0005631u. Now we can find EE by entering ΔmΔm into Equation 31.4.531.4.5: E=(Δm)c2=(0.005631u)c2. E=(Δm)c2=(0.005631u)c2. We know 1u=931.5MeV/c21u=931.5MeV/c2, and so E=(0.005631)(931.5MeV/c2)(c2)=5.25MeV. E=(0.005631)(931.5MeV/c2)(c2)=5.25MeV. Discussion The energy released in this αα decay is in the MeVMeV range , about 106106 times as great as typical chemical reaction energies, consistent with many previous discussions. Most of this energy becomes kinetic energy of the αα particle (or 4He4He nucleus ), which moves away at high speed. The energy carried away by the recoil of the 235U nucleus is much smaller in order to conserve momentum. The 235U nucleus can be left in an excited state to later emit photons (γ rays). This decay is spontaneous and releases energy , because the products have less mass than the parent nucleus . The question of why the products have less mass will be discussed in Binding Energy. Note that the masses given in Appendix A are atomic masses of neutral atoms, including their electrons. The mass of the electrons is the same before and after α decay , and so their masses subtract out when finding Δm. In this case, there are 94 electrons before and after the decay . Beta Decay There are actually three types of beta decay. The first discovered was “ordinary” beta decay and is called β− decay or electron emission. The symbol β− represents an electron emitted in nuclear beta decay. Cobalt-60 is a nuclide that β− decays in the following manner: Co60→Ni60+β−+ neutrino . The neutrino is a particle emitted in beta decay that was unanticipated and is of fundamental importance. The neutrino was not even proposed in theory until more than 20 years after beta decay was known to involve electron emissions. Neutrinos are so difficult to detect that the first direct evidence of them was not obtained until 1953. Neutrinos are nearly massless, have no charge, and do not interact with nucleons via the strong nuclear force . Traveling approximately at the speed of light , they have little time to affect any nucleus they encounter. This is, owing to the fact that they have no charge (and they are not EM waves), they do not interact through the EM force . They do interact via the relatively weak and very short range weak nuclear force . Consequently, neutrinos escape almost any detector and penetrate almost any shielding . However, neutrinos do carry energy , angular momentum (they are fermions with half-integral spin), and linear momentum away from a beta decay . When accurate measurements of beta decay were made, it became apparent that energy , angular momentum , and linear momentum were not accounted for by the daughter nucleus and electron alone. Either a previously unsuspected particle was carrying them away, or three conservation laws were being violated. Wolfgang Pauli made a formal proposal for the existence of neutrinos in 1930. The Italian-born American physicist Enrico Fermi (1901–1954) gave neutrinos their name, meaning little neutral ones, when he developed a sophisticated theory of beta decay (Figure 31.4.3). Part of Fermi’s theory was the identification of the weak nuclear force as being distinct from the strong nuclear force and in fact responsible for beta decay . The neutrino also reveals a new conservation law . There are various families of particles, one of which is the electron family. We propose that the number of members of the electron family is constant in any process or any closed system . In our example of beta decay , there are no members of the electron family present before the decay , but after, there is an electron and a neutrino . So electrons are given an electron family number of +1. The neutrino in β− decay is an electron ’s antineutrino, given the symbol ¯νe, where ν is the Greek letter nu, and the subscript e means this neutrino is related to the electron . The bar indicates this is a particle of antimatter . (All particles have antimatter counterparts that are nearly identical except that they have the opposite charge. Antimatter is almost entirely absent on Earth, but it is found in nuclear decay and other nuclear and particle reactions as well as in outer space.) The electron ’s antineutrino ¯νe, being antimatter , has an electron family number of −1. The total is zero, before and after the decay . The new conservation law , obeyed in all circumstances, states that the total electron family number is constant. An electron cannot be created without also creating an antimatter family member. This law is analogous to the conservation of charge in a situation where total charge is originally zero, and equal amounts of positive and negative charge must be created in a reaction to keep the total zero. If a nuclide AZXN is known to β− decay , then its β− decay equation is AZXN→AZ+1YN−1+β−+¯νe(β−decay), where Y is the nuclide having one more proton than X (Figure 31.4.4). So if you know that a certain nuclide β− decays, you can find the daughter nucleus by first looking up Z for the parent and then determining which element has atomic number Z+1. In the example of the β− decay of 60Co given earlier, we see that Z=27 for Co and Z=28 is Ni. It is as if one of the neutrons in the parent nucleus decays into a proton , electron , and neutrino . In fact, neutrons outside of nuclei do just that—they live only an average of a few minutes and β− decay in the following manner: n→p+β−+¯νe. We see that charge is conserved in β− decay , since the total charge is Z before and after the decay . For example, in 60Co decay , total charge is 27 before decay , since cobalt has Z=27. After decay , the daughter nucleus is Ni, which has Z=28, and there is an electron , so that the total charge is also 28+(−1) or 27. Angular momentum is conserved, but not obviously (you have to examine the spins and angular momenta of the final products in detail to verify this). Linear momentum is also conserved, again imparting most of the decay energy to the electron and the antineutrino, since they are of low and zero mass , respectively. Another new conservation law is obeyed here and elsewhere in nature. The total number of nucleons A is conserved. In 60Co decay , for example, there are 60 nucleons before and after the decay . Note that total A is also conserved in α decay . Also note that the total number of protons changes, as does the total number of neutrons, so that total Z and total N are not conserved in β− decay , as they are in α decay . Energy released in β− decay can be calculated given the masses of the parent and products. Example 31.4.1: β− Decay Energy from Masses Find the energy emitted in the β− decay of 60Co. Strategy and Concept As in the preceding example, we must first find Δm, the difference in mass between the parent nucleus and the products of the decay , using masses given in Appendix A. Then the emitted energy is calculated as before, using E=(Δm)c2. The initial mass is just that of the parent nucleus , and the final mass is that of the daughter nucleus and the electron created in the decay . The neutrino is massless, or nearly so. However, since the masses given in Appendix A are for neutral atoms, the daughter nucleus has one more electron than the parent , and so the extra electron mass that corresponds to the β− is included in the atomic mass of Ni. Thus, Δm=m(60Co)−m(60Ni). Solution The β− decay equation for 60Co is 6027Co33→6028Ni32+β−+¯νe. As noticed, Δm=m(60Co)−m(60Ni). Entering the masses found in Appendix A gives Δm=59.933820u−59.930789u=0.003031u. Thus, E=(Δm)c2=(0.003031u)c2. Using 1u=031.5MeV/c2 and we obtain E=(0.003031)(931.5MeV/c2)(c2)=2.82MeV. Discussion and Implications Perhaps the most difficult thing about this example is convincing yourself that the β− mass is included in the atomic mass of 60Ni. Beyond that are other implications. Again the decay energy is in the MeV range . This energy is shared by all of the products of the decay . In many 60Co decays, the daughter nucleus 60Ni is left in an excited state and emits photons ( \gamma) rays). Most of the remaining energy goes to the electron and neutrino , since the recoil kinetic energy of the daughter nucleus is small. One final note : the electron emitted in β− decay is created in the nucleus at the time of decay . The second type of beta decay is less common than the first. It is β+ decay . Certain nuclides decay by the emission of a positive electron . This is antielectron or positron decay (Figure 31.4.5). The antielectron is often represented by the symbol e+, but in beta decay it is written as β+ to indicate the antielectron was emitted in a nuclear decay . Antielectrons are the antimatter counterpart to electrons, being nearly identical, having the same mass , spin, and so on, but having a positive charge and an electron family number of −1. When a positron encounters an electron , there is a mutual annihilation in which all the mass of the antielectron - electron pair is converted into pure photon energy . (The reaction, e++e−→γ+γ, conserves electron family number as well as all other conserved quantities.) If a nuclide AZXN is known to β+ decay , then its β+ decay equation is AZXN→AZ−1YN+1+β++νe(β+decay), where Y is the nuclide having one less proton than X (to conserve charge) and νe is the symbol for the electron ’s neutrino , which has an electron family number of +1. Since an antimatter member of the electron family (the β+) is created in the decay , a matter member of the family (here the νe must also be created. Given, for example, that 22Naβ+ decays, you can write its full decay equation by first finding that Z=11 for 22Na, so that the daughter nuclide will have Z=10, the atomic number for neon. Thus the β+ decay equation for 22Na is 2211Na11→2210Ne12+β++νe. In β+ decay , it is as if one of the protons in the parent nucleus decays into a neutron , a positron , and a neutrino . Protons do not do this outside of the nucleus , and so the decay is due to the complexities of the nuclear force . Note again that the total number of nucleons is constant in this and any other reaction. To find the energy emitted in β+ decay , you must again count the number of electrons in the neutral atoms, since atomic masses are used. The daughter has one less electron than the parent , and one electron mass is created in the decay . Thus, in β+ decay , Δm=m(parent)−[m(daughter)+2me], since we use the masses of neutral atoms. Electron capture is the third type of beta decay . Here, a nucleus captures an inner- shell electron and undergoes a nuclear reaction that has the same effect as β+ decay . Electron capture is sometimes denoted by the letters EC. We know that electrons cannot reside in the nucleus , but this is a nuclear reaction that consumes the electron and occurs spontaneously only when the products have less mass than the parent plus the electron . If a nuclide AZXN is known to undergo electron capture , then its electron capture equation is AZXN+e−→AZ−1YN+1+νe(electroncapture,orEC). nuclide that can β+ decay can also undergo electron capture (and often does both). The same conservation laws are obeyed for EC as for β+ decay . It is good practice to confirm these for yourself. All forms of beta decay occur because the parent nuclide is unstable and lies outside the region of stability in the chart of nuclides. Those nuclides that have relatively more neutrons than those in the region of stability will β− decay to produce a daughter with fewer neutrons, producing a daughter nearer the region of stability. Similarly, those nuclides having relatively more protons than those in the region of stability will β− decay or undergo electron capture to produce a daughter with fewer protons , nearer the region of stability. Gamma Decay Gamma decay is the simplest form of nuclear decay it is the emission of energetic photons by nuclei left in an excited state by some earlier process. Protons and neutrons in an excited nucleus are in higher orbitals, and they fall to lower levels by emission (analogous to electrons in excited atoms). Nuclear excited states have lifetimes typically of only about 10−14 s, an indication of the great strength of the forces pulling the nucleons to lower states. The γ decay equation is simply XN⋅AZ→XAZN+γ1+γ2+⋅⋅⋅(γdecay) where the asterisk indicates the nucleus is in an excited state. There may be one or more γs emitted, depending on how the nuclide de-excites. In radioactive decay , γ emission is common and is preceded by γ or β decay . For example, when 60Coβ− decays, it most often leaves the daughter nucleus in an excited state, written Ni60⋅. Then the nickel nucleus quickly γ decays by the emission of two penetrating γs. Ni⋅60→Ni60+γ1+γ2⋅ These are called cobalt γ rays, although they come from nickel—they are used for cancer therapy, for example. It is again constructive to verify the conservation laws for gamma decay . Finally, since γ decay does not change the nuclide to another species, it is not prominently featured in charts of decay series , such as that in Figure. There are other types of nuclear decay , but they occur less commonly than α, β, and γ decay . Spontaneous fission is the most important of the other forms of nuclear decay because of its applications in nuclear and weapons. It is covered in the next chapter. Summary When a parent nucleus decays, it produces a daughter nucleus following rules and conservation laws . There are three major types of nuclear decay , called alpha (α) beta (β) and gamma (γ). The α decay equation is AZXN→A−4Z−2YN−2+42He2. Nuclear decay releases an amount of energy E related to the mass destroyed Δm by E=(Δm)c2. There are three forms of beta decay . The β− decay equation is AZXN→AZ+1YN−1+β−+¯νe. The β+ decay equation is AZXN→AZ−1YN+1+β++νe. The electron capture equation is AZXN+e−→AZ−1YN+1+νe. β− is an electron , β+ is an antielectron or positron , νe represents an electron ’s neutrino , and ¯νe is an electron ’s antineutrino. In addition to all previously known conservation laws , two new ones arise— conservation of electron family number and conservation of the total number of nucleons . The γ decay equation is AZX∗N→AZXN+γ1+γ2+...whereγ is a high- energy photon originating in a nucleus . Glossary parent : the original state of nucleus before decay daughter : the nucleus obtained when parent nucleus decays and produces another nucleus following the rules and the conservation laws positron : the particle that results from positive beta decay ; also known as an antielectron decay : the process by which an atomic nucleus of an unstable loses mass and energy by emitting ionizing particles alpha decay : type of radioactive decay in which an atomic nucleus emits an alpha particle beta decay : type of radioactive decay in which an atomic nucleus emits a beta particle gamma decay : type of radioactive decay in which an atomic nucleus emits a gamma particle decay equation : the equation to find out how much of a radioactive material is left after a given of time nuclear reaction energy : the energy created in a nuclear reaction neutrino : an electrically neutral, weakly interacting elementary subatomic particle electron ’s antineutrino : antiparticle of electron ’s neutrino positron decay : type of beta decay in which a proton is converted to a neutron , releasing a positron and a neutrino antielectron : another term for positron decay series : process whereby subsequent nuclides decay until a stable nuclide is produced electron ’s neutrino : a subatomic elementary particle which has no net electric charge antimatter : composed of antiparticles electron capture : the process in which a proton -rich nuclide absorbs an inner atomic electron and simultaneously emits a neutrino electron capture equation : equation representing the electron capture 31.3: Substructure of the Nucleus 31.5: Half-Life and Activity
2968	https://www.writingclasses.com/toolbox/ask-writer/how-do-authors-come-up-with-stories	Writer’s Toolbox Ask The Writer All Categories Your most pressing and perplexing questions about writing answered here by Gotham teacher Brandi Reissenweber. Choose a category... Adaptation Career Character Conflict Contests Description Dialogue Fiction Figurative Language Freelance General Genre Grammar Imagery Language Nonfiction Novels Pacing Perspective Plot Poetry Point Of View Professional Publication Querying Reading Recommendations Research Revision Setting Short Stories Show Don't Tell Style Submissions Theme Tone Voice Writing Habits Writing Habits I recently heard an editor at a conference talking about how important it is for novelists to read. Why is reading so important for writers, specifically? It has been a lifelong dream of mine to write a fiction novel. I haven't written since high school and have just started writing again. I've joined a few writers’ workshops. Should I improve my writing before I really get too far into my novel to avoid major rewrites? I'd like to write a memoir with my sister, but I've never seen a book like this with two authors. Is it possible? How do you remain focused on a current work of writing without getting distracted and starting a whole new story? I often come up with many ideas that draw me in and tempt me to begin working on another story, but then I lose interest in the previous story I was working on. My current strategy is to jot down an idea I may have so that I can look back at it and flesh the rest of it out. I find feedback perplexing. I can't make everyone happy and sometimes people disagree. How do I know when advice is good? I have a lot of creative writing classes under my belt, but I feel compelled to take more. At what point are classes no longer useful? In a writing class, my teacher said we should always use present tense when talking about the action in a novel. Is this right? Do I really need to bother with a cover letter when submitting my short stories? Showing 1-8 of 52 items. « 1 2 3 4 5 6 7 » How do authors come up with stories? There’s no one source, like a wellspring of inspiration that the writer simply needs to find. Perhaps Andrea Barrett captures this best in her essay “The Sea of Information” when she writes, “Writing is mysterious; and it’s supposed to be . . . any path that gets you there is a good path in the end. But one true thing among all these paths is the need to tap a deep vein of connection between our own uncontrollable interior preoccupations and what we’re most concerned about in the world around us.” Our inspiration, then, is found in our own personal preoccupations. Andrea Barrett writes, in that same essay, that her “imagination is nourished by old books, old bones, fossils, feathers paintings, photographs, museums of every kind and size, microscopes and telescopes, plants and birds; I like to learn things and . . . all this information feeds my fiction.” Author Elizabeth McCracken is captivated by her family history. She keeps a family archive filled with diaries, letters, sketches, diplomas and poems. Stories often grow out of what she finds there. In Joyce Carol Oates’ essay, “To A Young Writer,” she advises writers look for “forbidden” passions, buried selves, and ill-understood drives. “These emotions,” she writes, “are the fuel that drives your writing and makes possible hours, days, weeks, months and years of what will appear to others, at a distance, as ‘work.’” Lauren Groff’s short story, “A Season By the Shore,” was informed by her own “anxiety and restlessness” that came with her pregnancy with her first son. While she can trace the inspiration of some aspects of this story, she, too, acknowledges the mystery inherent in the process: “In truth, my debts are endless, and I only wish I were alert enough to recognize them all.” What are your own vital interests? What details, anecdotes, concerns, images or subjects do you return to again and again? Some ideas may come to mind right away. If that’s the case, start there. Look for the intriguing character or the compelling conflict within that interest. If one doesn’t easily present itself, begin to do some exploring. This might take the form of research, conversations, looking through ephemera, or journaling. If you follow the trail of that preoccupation, you’re bound to come across the engine of a story. If a preoccupation doesn’t come to mind right away, you might find it useful to keep a journal or look through the one you already write. You might keep a folder of interesting bits—news articles, flyers or brochures from intriguing places, pictures of unexpected findings. Over time, you might return to this to see what interests have remained and which have lost their appeal. Pay attention to where your mind and imagination wander. You will find your inspiration there. Subscribe to our email list Explore Our Classes Sign up to receive writing advice, news, and special deals. Purchase Course Catalogue Gotham Shop FAQ Credit & Refund Policy Ways to Save Privacy Policy Contact [email protected] 212.974.8377 © 2025 Gotham Writers Workshop, Inc. Subscribe to our email list
2969	https://www.bilibili.com/video/BV1Jt41177fB/	代数法求解绝对值不等式，利用总结部分可以快速作答_哔哩哔哩_bilibili 首页番剧直播游戏中心会员购漫画赛事下载客户端登录大会员消息动态收藏历史创作中心投稿代数法求解绝对值不等式，利用总结部分可以快速作答 515 0 2019-07-23 03:25:20 未经作者授权，禁止转载代数法求解绝对值不等式，利用总结部分可以快速作答关注正在缓冲... 已静音开播点击恢复音量登录免费享高清视频试看30秒为TA充电后即可观看 00:02/09:00 自动 1080P 高清登录即享 720P 准高清登录即享 480P 标清登录即享 360P 流畅自动(360P) 倍速 2.0x 1.5x 1.25x 1.0x 0.75x 0.5x 字幕 AI原声翻译（测试版）关闭登录可享原声翻译体验反馈字幕添加字幕关闭关闭登录可享字幕暂无字幕主字幕副字幕字幕字幕设置字幕大小适中最小较小适中较大最大字幕颜色白色白色红色紫色深紫色靛青色蓝色亮蓝色描边方式无描边无描边重墨描边 45°投影默认位置底部居中左下角底部居中右下角左上角顶部居中右上角背景不透明度其它设置 [x] 等比缩放 [x] 淡入淡出恢复默认设置 0 [x] 镜像画面 [x] 单集循环 [x] 自动开播更多播放设置播放方式自动切集播完暂停视频比例自动 4:3 16:9 播放策略默认 AV1 HEVC AVC 音量均衡标准高动态关闭其他设置 [x] 隐藏黑边 [x] 关灯模式 -人正在看，已装填 0 条弹幕 [x] 按类型过滤滚动固定彩色高级 [x] 弹幕随屏幕缩放 [x] 防挡字幕 [x] 智能防挡弹幕弹幕观看屏蔽词同步屏蔽列表显示区域 50% 弹幕密度正常较多重叠不透明度 80% 弹幕字号 100% 弹幕速度适中高级设置全新【硬核会员弹幕模式】更多弹幕设置 [x] 弹幕速度同步播放倍数弹幕字体黑体黑体宋体新宋体仿宋微软雅黑微软雅黑 Light Noto Sans DemiLight Noto Sans Regular [x] 粗体描边类型重墨描边 45°投影恢复默认设置请先登录或注册弹幕礼仪发送优先使用播放器内置策略播放优先使用 AV1 编码视频播放优先使用 HEVC/H.265 编码视频播放优先使用 AVC/H.264 编码视频播放自动平衡不同视频间的音量大小平衡音量同时保留更多声音细节关闭音量均衡开启画中画宽屏模式网页全屏进入全屏 (f) 本视频开启原声翻译时不支持关闭字幕关闭弹幕 (d) 视频底部15%部分为空白保留区特殊颜色、运动形式的弹幕反馈 10 4 8 代数法求解绝对值不等式，利用总结部分可以快速作答 [x] 从 00:00 开始分享获取视频分享链接手机扫码观看/分享动态微信 QQ QQ空间微博贴吧嵌入代码分享稿件举报记笔记笔记记笔记还没有人发布笔记哦，快去发布一篇吧～公开发布笔记用手机看稍后再看绝对值不等式数学学习生活代数法中考知识点中考数学绝对值不等式重生之数学好南发消息尽力更新┆欢迎点赞充电关注 2882 弹幕列表弹幕列表屏蔽设定高级弹幕弹幕列表填充中... 查看历史弹幕接下来播放自动连播 03:36 中考知识点：初中生都考过的数学题目，你真的理解吗？重生之数学好南 86 0 03:42 江苏连云港2019年中考数学试卷：13圆周角圆心角重生之数学好南 147 1 04:51 中考知识点：填空题用赋值法30秒就做完了！重生之数学好南 68 0 07:06 江苏连云港中考数学试卷讲解7：二次函数解决实际问题重生之数学好南 226 0 14:01 中考必考题型，二次函数应用题，要注意自变量与因变量关系重生之数学好南 808 7 06:56 中考知识点：用替代法求解代数计算题重生之数学好南 58 0 12:36 如何证明51的51次方减1能被七整除？重生之数学好南 630 2 04:15 2019年难倒了无数考生的中考数学试题，你会做吗？重生之数学好南 165 0 09:58 高考知识点：图像法、赋值法、定义法，那种方法更适合自己呢？重生之数学好南 169 0 05:45 分析题干按照自己的思路，直接求不出，采用间接迂回的方式求解重生之数学好南 60 0 08:38 中考数学试卷讲解1-5，绝对值、二次根式、乘方等重生之数学好南 124 0 04:46 老师都讲错的概率求解题目！你会吗？重生之数学好南 93 0 04:41 10-初中的动点好难啊！重生之数学好南 165 0 04:34 中考知识点：用绝对值含义求解绝对值不等式，结合图像更简单奥重生之数学好南 626 0 13:16 中考知识点：等间隔数字和怎么求？裂项题目讲解！重生之数学好南 255 0 05:48 16 除去题干有明确要求，要多用二元一次方程解决问题重生之数学好南 222 1 14:55 中考高考：家长陪考需注意的9个问题与禁忌重生之数学好南 387 0 06:38 13-14 遇到动点问题脑海里一定要考虑清楚是不是有多种情况重生之数学好南 102 0 17:18 中考数学证明第二问无从下手，来看看老师是怎样打破困境的！重生之数学好南 216 2 11:00 江苏连云港中考数学试卷讲解8：折叠对称解直角三角形重生之数学好南 154 0 展开手办模型周边聚集地>> 小窗客服顶部赛事库课堂2021拜年纪
2970	https://askfilo.com/user-question-answers-smart-solutions/4-a-transformer-has-an-efficiency-of-it-works-at-input-of-6-3232363632353837	A transformer has an efficiency of 50 \%. It works at input of 6 kW and 1.. World's only instant tutoring platform Instant TutoringPrivate Courses Tutors Explore TutorsBecome Tutor Login StudentTutor CBSE Smart Solutions A transformer has an efficiency of 50 %. It works at input of Question Question asked by Filo student A transformer has an efficiency of 50%. It works at input of 6 kW and 100 V . If secondary current is 15 A , then voltage in secondary coil is: (1) 250 V P=II+P (3) 400 V -000 =1020±P=6∘(4)200 V Views: 5,644 students Updated on: Feb 4, 2025 Not the question you're searching for? Ask your question Ask your question Or Upload the image of your question Get Solution Text solutionVerified Concepts: Transformer, Efficiency, Voltage, Current Explanation: To find the voltage in the secondary coil of the transformer, we can use the formula for power and the efficiency of the transformer. The input power (P_in) is given as 6 kW and the efficiency (η) is 50%. The output power (P_out) can be calculated as follows: P_out = η × P_in. After calculating the output power, we can use the relationship between power, voltage, and current to find the secondary voltage (V_s). Step by Step Solution: Step 1 Convert the input power from kW to W: P_in = 6 kW = 6000 W. Step 2 Calculate the output power using the efficiency: P_out = η × P_in = 0.5 × 6000 W = 3000 W. Step 3 Use the formula for power: P_out = V_s × I_s, where I_s is the secondary current (15 A). Step 4 Rearranging the formula gives us V_s = P_out / I_s = 3000 W / 15 A = 200 V. Step 5 Thus, the voltage in the secondary coil is 200 V. Final Answer: 200 V Ask your next question Or Upload the image of your question Get Solution Get instant study help from an expert tutor 24/7 Download Filo Found 2 tutors discussing this question Levi Discussed A transformer has an efficiency of 50%. It works at input of 6 kW and 100 V . If secondary current is 15 A , then voltage in secondary coil is: (1) 250 V P=II+P (3) 400 V -000 =1020±P=6∘(4)200 V 7 mins ago Discuss this question LIVE 7 mins ago One destination to cover all your homework and assignment needs Learn Practice Revision Succeed Instant 1:1 help, 24x7 60, 000+ Expert tutors Textbook solutions Big idea maths, McGraw-Hill Education etc Essay review Get expert feedback on your essay Schedule classes High dosage tutoring from Dedicated 3 experts Download AppExplore now Trusted by 4 million+ students Students who ask this question also asked Question 1 Views: 5,162 If Young's double slit experiment, is performed in water (assume that the water is still and clear) Topic: Smart Solutions View solution Question 2 Views: 5,304 Given below are some alkyl chlorides. A. CH 3−CH 2−CH(Cl)CH=CH 2 B. CH 3−CH 2−CH(Cl)−CH 2−CH 3 C. CH 3−CH 2−CH 2−CH 2−CH 2−Cl The correct order of their reactivity towards S N1 reaction will be: Topic: Smart Solutions View solution Question 3 Views: 5,889 An inscribed circular hole is made in a triangular lamina with each side 'a'. Find the area moment of inertia of this lamina about one of the sides. a Inscribed circular hole Triangular lamina Area moment of inertia Topic: Smart Solutions View solution Question 4 Views: 5,426 Class 11 - Accounting - Cash Book Problem (Bihar Board) Enter the following transactions in Cash Book with Cash and Discount Columns: 2023 August 1: Cash in hand (₹1,500) August 2: Paid to Niraj (₹300); Discount allowed by him (₹10) August 8: Cash Purchases (₹400) August 10: Received from R. Prasad (₹980); Discount allowed to him (₹20) August 15: Sold goods for Cash (₹400) August 22: Paid to Birendra (₹295); Discount received (₹5) August 25: Paid wages (₹50) August 30: Paid to Sanjeev in full settlement of his account which showed a credit balance of ₹400 (₹390) Answer: Cash (Dr.) Balance ₹1,445; Discount (Dr.) ₹20; Discount (Cr.) ₹5 Topic: Smart Solutions View solution View more Video Player is loading. Play Video Play Skip Backward Mute Current Time 0:00 / Duration-:- Loaded: 0% Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-0:00 1x Playback Rate 2.5x 2x 1.5x 1x, selected 0.75x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected Audio Track Picture-in-Picture Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. 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2971	https://www.youtube.com/watch?v=0vQM1P5CRP4	Yield \| Actual Yield \| Theoretical yield \| percentage yield \| Efficiency of worker \|Class 11 Lect#17 Umair Khan Academy 32100 subscribers 190 likes Description 6098 views Posted: 3 May 2020 A brand new series for 1st year students by Umair Khan Academy. Plz Share to 1st year FSc students for their benefit and Free education. Chapter 1 (Chemistry) This video reveals very important points connected with yield in chemistry so it tells about what is the yield and its types such as percentage yield, theoretical yield, actual yield and many more. In this video you can also find out how a worker works its efficiency. If you are interesting to calculate how percentage yield is determined then this video is for you. This video is specially designed for the students of FSc class 11 Chemistry. Playlist for all Lectures Introduction Lecture 1(Atom) Lecture 2 (Evidence of Atom) Lecture 3 (Molecule) Lecture 4 (Ion) Lecture 5 (Relative atomic mass) Lecture 6 (Isotepes and relative abundance) Lecture 7 (Mass spectrometer) Lecture 8 (Average atomic mass and %composition of elements) Lecture 9 (How to Determine Empirical formula) lecture 10 (combustion analysis) Lecture 11 (Molecular Formula & its determination) Lecture 12 (Concept of Mole) Lecture 13 (How many Molecules in 10g ice) Lecture 14 (Molar Volume) Lecture 15 (Stoichiometry) Lecture 16 (Limiting Reactant) Lecture 17 (Yield) For Multan board, Gujranwala board, rahim yar khan board, bahawalpur board, dera ghazi khan board, lahore board, sahiwal board, Fsc all boards students, 1st year Chemistry part 1. class 11 Yield #PercentageYield #ActualYield #TheoreticalYield 21 comments Transcript: और अस्सलाम वालेकुम डेथ स्टूडेंट्स हाउ आर यू थें यू विल फाइंड हाउ टो ऑब्जर्व आज हम पढ़े जा रहे हैं जिसके बारे में छोटा सा टॉपिक है यह हमारा इस लीड की अगर बात की जाए तो यह होती क्या है तो आसान यह है जैसे इसका मतलब डरा रहा है पैदावार कि अगर हमने एक चीज बनाई है तो इसलिए केमिस्ट्री में अगर हमने पॉइंट्स बनाए हैं तो वह कितने बनाएं अगर हमें पता लग जाते हैं इसको सील कहते हैं यह एंड हम दोस्तों के फारवर्ड कर सकते हैं एक इस तरह से कि हम बैलेंस केमिकल इक्वेशन लेंगे और इस हमने लास्ट लेक्चर्स के अंदर काफी एग्जांपल की थी तो उसके मुताबिक हम क्या करेंगे अपनी मोल-मोल रिलेशनशिप मॉल मैसेज सेंसरशिप लेकर इस्तेमाल कर सकते हैं इस तरह के और रिलेशनशिप हमने पड़े थे लेकिन यहां पर एक बात की होती जाती है कि एक और तरीका काटब ही लेने का और वह यह है कि हम F1 हुए वह रिएक्शन परफॉर्म करें अपनी लाइफ में जो और यह पोर्शन हम फ्रेटिकली कर रहे थे बैलेंस केम इन क्वेश्चन लेकर मतलब हम तोते के कारण वे यानी आप रोटी के लिए भी और प्रैक्टिकली भी इज़राइल मालूम कर सकते हैं और अगर यह दोनों फ्रिल हम मालूम कर लेते हैं इनमें से एक भी काफी है बादशाह दोनों जरूरत हैं तो यह अगर दोनों पता लग जाए तो इस हम एक बहुत ही अहम की पता चला सकते हैं और वह यह है कि जो एक्सपेरिमेंट करने वाला है उसकी डेफिशियेंसी क्या है यह नहीं एडिशन सी ऑफ द वर्कर के कितना अच्छा वह काम करने वाला है कितना उसका जज्बा है वह किस तरह से रिएक्शन को डील करता है यह भी बहुत अहम बात है असल में होता यह है कि जब हम बैलेंस केमिकल इक्वेशन से जेड निकालते हैं तो उसमें बिल्कुल आइडियल एनवायरनमेंट होता है यानी यह एक्टर दिए उनको आपस में मिलाया और फिर उस बाद प्रोडक्ट बना दिए और पता लग गया कि इतने रिएक्टर नगर-ग्राम में थे या इतने मूल थे तो वह अपने प्रोडक्ट के मौर्या ग्राम बना देंगे ध्यान थिस पॉलीटिकल लीडर्स लेकिन एवं एक्सपेरिमेंट करते हैं तो वहां पर मसाइल बढ़ जा कि कैसे वहां पर बेकार होता है विक्रम में कुछ ना कुछ रेफ्रेंस होते हैं उसको आप उन्हें खेलते हैं कुछ भी कर के साथ लगे रहते हैं टेस्ट ट्यूब लेते हैं कुछ इसे उबालते हैं ट्रैक्टर्स की मिकदार कम हो जाती है Filter करते हैं ट्विटर पर में सो जाती है तो यह कुछ जगहों पर जगह-जगह पर क्या होता है कि जाया होती हैं हमारी रसिया यह कह लीजिए हमारे मटीरियल जाए होते हैं और जब पॉइंट बनता है तो वह उतना नहीं बन पाता जितना कि हमने थे टिप्पणी सोचा था यहीं फैक्ट्री के लिए एक्सपेरिमेंटली हमेशा जो हमारे रिजल्ट्स यॉवर हमेशा कम होगी और मेडिकल जी हमेशा ज्यादा होगी तो यह आपके जहन में रखेंगे एनी आफ थे आर्टिकल्स ई फील थिस वे मोड ऑन एक्सपोलियट एक्चुअली इसको आप एक्स्पेरिमेंट जल्दी कह सकते हैं और मेडिकल जेल को आप कैलकुलेटेड एंड भी बोल सकते हैं मैंने देता हूं यहां पर तो थर्ड ईयर विल कैलकुलेटेड इंटेलेक्चुअल एंड एक्सपेरिमेंटल जेल तो परिमंडल जीवन हमेशा कम होगी क्योंकि क्या होगा वहां पर जाया होगी कि यह कैसे हो सकता है कि आपके जो इस कमेंट है वह आपके लिए नुकसान का मांस बनता है समटाइम कुंवर करें वह बड़ा इन एक्सपीरियंस अच्छा काम करने वाला यह नया है अभी लैब में गया तो वह चीजें उन्हें खेलते हुए एक काम करते हुए और मुक्त करते हुए गिरा था जाता है इससे भी उसके जो इज इट होगी जितनी होनी चाहिए वह नहीं होगी मतलब प्रोडक्ट काम बनेंगे अब चाहते हो कि पता करो कि वह जो को वर्कर है उसकी एडिशन सी कैसे निकाली जाएगी कौन सा वर्कर फोन नंबर चार वर्कर्स ने काम किया था कि सरकार ने अच्छा काम किया है तो उसके लिए आपको जरूरत पड़ेगी परसेंटेज एंड की तो परसेंटेज हील्ड इज इक्वल टू एक्चुअल हेल्ड ओवर थे डिफिकल्टीज एंड मल्टीप्लाई बाय हेर ओ तो अगर आप यह कर ली है वैसे आपने गौर किया होगा कि आपके हाथों करता नंबर भट्ट आकुल नंबर ग्रुप यहीं पर सेंट एंड मास्टर्स निकालते हुआ था उसी तरह से हमेशा ज्यादा होंगे क्योंकि तुम राधा होते हैं हासिल कर दे नंबर यह ट्रेंड मा आज हम होते हैं और बड़ी यह नेट करने से परसेंटेज रहती है इसी चीज को आप डेफिशियेंसी भी बोल सकते हो डेफिशियेंसी ऑफ़ वर्कर तो आ एडिशन co-worker मालूम कर सकते हो और अपोजिट अगर आपकी अश्लील आ जाती है 90% और एक के अराध्य 90% शायद वह अच्छा काम करने वाला इसकी परसेंटेज इलायची है ठीक है वैसे इस तरह एक बंदे नंबर है 50 हजार बड़ा सोर्स आफ विजिबल टो 50 पर सेंट बेनिफिटिंग बोथ साइड और एक ने यह 70 तो उसके भीतर नंबर आए इसका मतलब यह है कि इस तरह आप फ्रेंडशिप परसेंटेज निकाल सकते हैं वर्कर के एग्जाम है रिपोर्ट में दी गई है उस गड्ढे रहेगा देख लेते हैं आपके पास लाइमस्टोन है कैल्शियम कार्बोनेट इसको आपने जला दिया को इतना एंव इसको कहते हैं सीडीओ कैल्शियम ऑक्साइड इस प्रोड्यूस अकॉर्डिंग टू द फॉलोइंग क्वेश्चन यह क्वेश्चन है यह आपने जलाया और यह प्रोड्यूस हुआ एक्सपेरिमेंट चल रहा है द एक्टिवेटिड और आपने देखा कि कैल्शियम ऑक्साइड कितना प्रिंस और 2.5 प्रोग्राम इनको ग्राम में चेंज कर लें क्योंकि हमारे पास जो यूनिट्स ज्यादा यूज होंगे क्या स्टेज प्रोग्राम में होंगे तो थायरोइड क्लिक करेंगे तो आंसर देगा टू 50 ग्राम फिटकरी वैल्यू 4.5 किलोग्राम आफ लाइम स्टोन रोस्ट यानि कि आपने कितना लाइमस्टोन जलाया था यह बता रहा है कहते 4.5 उसको भी आप चेंज करने ग्राम में यह बन जाएगा फॉर 50 ग्राम तो यह आपके पास कन्वर्जन हो गई एतिहाद पटेल प्रतिक आया ग्राम है वहां ग्राम रहने दीजिएगा बना किलोग्राम को ग्रामीण चेंज करना है अगर जाता है व्हाट इज परसेंटेज एंड ऑफ द इंडियन प्रेसिडेंट बताएं परसेंटेज एंड का फॉर्मूला है एक्चुअल लागत पिंपल अब फेस्टिवल चीन है एक्सपेरिमेंट करने के बाद वह को दे दी गई है वह स्टूडेंट क्या है एक्चुअली डैडी तो 150 ग्राम मतलब कैल्शियम ऑक्साइड मिला है 2 एफ़ 500 ग्राम जो के एक्चुअल जी अब एक्सपोलियट तुम्हारे पास आ गई थी शिरडी कल यह रह गई थ्री के मुताबिक आप क्या करना है आपको यह देखना है कि जो आपस बैलेंस केमिकल इक्वेशन है यह कहती है कि कैल्शियम कार्बोनेट आपके पास एक मॉल है एक मॉल कैल्शियम कार्बाइड का वजन इसके मोलर मैस के बराबर होता है यह कैल्शियम का फैटी कार्बन का 12 जून का 16 पति टाइम्स यह मिलाकर हो गया हंड्रेड ग्राम और इसी तरह से कैल्शियम ऑक्साइड के बारे में देखें तो कैल्शियम 4350 ऑफिस 116 यह बन गया शिफ्ट 6 ग्राम्स अब देखिए प्रिंटर में करेंगे थर्ड डिग्री अधिक रहा हूं के बैलेंस केमिकल इक्वेशन बताती है कि फंड ग्राम कैल्शियम ऑक्साइड काफी 6 ग्राम देता है अगर मैं आपसे कहूं कि अगर 1 ग्राम होगा तो कितने देगा भाई पाएंगे 1568 बाय 100 लेकिन एक बात और देगा वह कहते हैं जो कैल्शियम कार्बोनेट हमने जलाना है यूज करना है वह कितना करना है 50 ग्राम तो अगर मैं आपसे कहूं कि 15500 ग्राम अगर होता तो फिर कितना तो आप मल्टीप्लाई कर देंगे इस रकम को 156 बाय हंड्रेड्स मल्टीप्लाई बाय 14510 ट्रांसफर आपके पास आ गया कितना 2520 ग्राम्स तो आप यह देख रहे हैं कि हमने यह क्या निकाली थी कि आखिर निकाली के 100 ग्राम होगा तो इतना और पैन कार्ड इस प्रोग्राम करते थे ना आप देखेंगे एक्सपोलियट पर आकर टिक एंड बेगिन फ्रॉम लगा दें फार्मूले में लिखा है नेक्स्ट वर्ल्ड इज टू 50 ओवर थ्री डिकेड्स [संगीत] 2520 यह आता है यह कम है यह हमें ज्यादा होती है यह हमेशा कम होती है मल्टीप्लाई बाय हंड्रेड तो आंसर किराया 99.2 परसेंट यह है डीडी परसेंटेज फील्ड और सवाल यहां पर हो गया खत्म हुई रेस में आ गया होगा और चैनल को सब्सक्राइब करें लाइक करें शेयर करें और वह बता दें और कुछ भी समझे दूसरों को भी समझने का मौका दें अब तक के लिए खुदा हाफिज
2972	https://stackoverflow.com/questions/345085/how-do-trigonometric-functions-work	Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Collectives¢ on Stack Overflow Find centralized, trusted content and collaborate around the technologies you use most. Learn more about Collectives Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How do Trigonometric functions work? [closed] Ask Question Asked Modified 4 years, 8 months ago Viewed 61k times So in high school math, and probably college, we are taught how to use trig functions, what they do, and what kinds of problems they solve. But they have always been presented to me as a black box. If you need the Sine or Cosine of something, you hit the sin or cos button on your calculator and you're set. Which is fine. What I'm wondering is how trigonometric functions are typically implemented. algorithm math trigonometry Share Improve this question edited Jun 28, 2012 at 12:17 Eric Wilson 59.7k8282 gold badges208208 silver badges272272 bronze badges asked Dec 5, 2008 at 20:49 Jurassic_CJurassic_C 2,41133 gold badges2323 silver badges1919 bronze badges 3 Are you confused about what trig functions are or how they are implemented? Kyle Cronin – Kyle Cronin 2008-12-05 20:53:49 +00:00 Commented Dec 5, 2008 at 20:53 18 I know what they are. I know what they do. I know how to determine which I need for what purpose. I can tell you all about the relationship between angles and distances. What I was looking for was more along the lines of John D. Cook's answer. And everybody else who mentioned actual algorithms Jurassic_C – Jurassic_C 2008-12-05 21:48:36 +00:00 Commented Dec 5, 2008 at 21:48 This is a good question. For example, sine, cosine, and tangent are transcendental functions and those are hard to solve... On the other hand, they can be defined using a simple Taylor series expansion which will give you the correct answer up to any finite degree of accuracy required. Alex – Alex 2008-12-05 23:58:37 +00:00 Commented Dec 5, 2008 at 23:58 Add a comment \| 7 Answers 7 Reset to default 156 First, you have to do some sort of range reduction. Trig functions are periodic, so you need to reduce arguments down to a standard interval. For starters, you could reduce angles to be between 0 and 360 degrees. But by using a few identities, you realize you could get by with less. If you calculate sines and cosines for angles between 0 and 45 degrees, you can bootstrap your way to calculating all trig functions for all angles. Once you've reduced your argument, most chips use a CORDIC algorithm to compute the sines and cosines. You may hear people say that computers use Taylor series. That sounds reasonable, but it's not true. The CORDIC algorithms are much better suited to efficient hardware implementation. (Software libraries may use Taylor series, say on hardware that doesn't support trig functions.) There may be some additional processing, using the CORDIC algorithm to get fairly good answers but then doing something else to improve accuracy. There are some refinements to the above. For example, for very small angles theta (in radians), sin(theta) = theta to all the precision you have, so it's more efficient to simply return theta than to use some other algorithm. So in practice there is a lot of special case logic to squeeze out all the performance and accuracy possible. Chips with smaller markets may not go to as much optimization effort. Share Improve this answer edited Dec 5, 2008 at 21:09 answered Dec 5, 2008 at 20:59 John D. CookJohn D. Cook 30.2k1010 gold badges6969 silver badges9494 bronze badges 7 Comments Jason S Jason S Great answer -- though the CORDIC doesn't really need range reduction per se (in fact it is essentially a range reduction algorithm in its own right); it works fine for angles between -pi/2 and +pi/2, so you just have to do a 180 degree vector rotation for angles outside that range. Pascal Cuoq Pascal Cuoq Implementations that use a polynomial approximation may often use Taylor series, but they typically should use coefficients that have been determined with the Remez algorithm. lolengine.net/blog/2011/12/21/better-function-approximations Rhubbarb Rhubbarb Note that the table of values used by CORDIC must be pre-calculated. So, Taylor might still be used at "compile time". Perry Perry It seems that this answer contradicts the high-rated accepted answer to this similar question: stackoverflow.com/questions/2284860/…. This answer says that the sin() function is mostly implemented on hardware-level, while the other says in C. MattHusz MattHusz Isn't CORDIC only more efficient when you don't have access to fast hardware multiplies? Embedded devices with FPUs and FPGAs with DSP elements (which is quite a few of them) can perform multiplies very efficiently. Maybe the answer to this question has changed since this was posted, or maybe I've misunderstood the tradeoff? \| 50 edit: Jack Ganssle has a decent discussion in his book on embedded systems, "The Firmware Handbook". FYI: If you have accuracy and performance constraints, Taylor series should not be used to approximate functions for numerical purposes. (Save them for your Calculus courses.) They make use of the analyticity of a function at a single point, e.g. the fact that all its derivatives exist at that point. They don't necessarily converge in the interval of interest. Often they do a lousy job of distributing the function approximation's accuracy in order to be "perfect" right near the evaluation point; the error generally zooms upwards as you get away from it. And if you have a function with any noncontinuous derivative (e.g. square waves, triangle waves, and their integrals), a Taylor series will give you the wrong answer. The best "easy" solution, when using a polynomial of maximum degree N to approximate a given function f(x) over an interval x0 < x < x1, is from Chebyshev approximation; see Numerical Recipes for a good discussion. Note that the Tj(x) and Tk(x) in the Wolfram article I linked to used the cos and inverse cosine, these are polynomials and in practice you use a recurrence formula to get the coefficients. Again, see Numerical Recipes. edit: Wikipedia has a semi-decent article on approximation theory. One of the sources they cite (Hart, "Computer Approximations") is out of print (& used copies tend to be expensive) but goes into a lot of detail about stuff like this. (Jack Ganssle mentions this in issue 39 of his newsletter The Embedded Muse.) edit 2: Here's some tangible error metrics (see below) for Taylor vs. Chebyshev for sin(x). Some important points to note: that the maximum error of a Taylor series approximation over a given range, is much larger than the maximum error of a Chebyshev approximation of the same degree. (For about the same error, you can get away with one fewer term with Chebyshev, which means faster performance) Range reduction is a huge win. This is because the contribution of higher order polynomials shrinks down when the interval of the approximation is smaller. If you can't get away with range reduction, your coefficients need to be stored with more precision. Don't get me wrong: Taylor series will work properly for sine/cosine (with reasonable precision for the range -pi/2 to +pi/2; technically, with enough terms, you can reach any desired precision for all real inputs, but try to calculate cos(100) using Taylor series and you can't do it unless you use arbitrary-precision arithmetic). If I were stuck on a desert island with a nonscientific calculator, and I needed to calculate sine and cosine, I would probably use Taylor series since the coefficients are easy to remember. But the real world applications for having to write your own sin() or cos() functions are rare enough that you'd be best off using an efficient implementation to reach a desired accuracy -- which the Taylor series is not. Range = -pi/2 to +pi/2, degree 5 (3 terms) Taylor: max error around 4.5e-3, f(x) = x-x3/6+x5/120 Chebyshev: max error around 7e-5, f(x) = 0.9996949x-0.1656700x3+0.0075134x5 Range = -pi/2 to +pi/2, degree 7 (4 terms) Taylor: max error around 1.5e-4, f(x) = x-x3/6+x5/120-x7/5040 Chebyshev: max error around 6e-7, f(x) = 0.99999660x-0.16664824x3+0.00830629x5-0.00018363x7 Range = -pi/4 to +pi/4, degree 3 (2 terms) Taylor: max error around 2.5e-3, f(x) = x-x3/6 Chebyshev: max error around 1.5e-4, f(x) = 0.999x-0.1603x3 Range = -pi/4 to +pi/4, degree 5 (3 terms) Taylor: max error around 3.5e-5, f(x) = x-x3/6+x5 Chebyshev: max error around 6e-7, f(x) = 0.999995x-0.1666016x3+0.0081215x5 Range = -pi/4 to +pi/4, degree 7 (4 terms) Taylor: max error around 3e-7, f(x) = x-x3/6+x5/120-x7/5040 Chebyshev: max error around 1.2e-9, f(x) = 0.999999986x-0.166666367x3+0.008331584x5-0.000194621x7 Share Improve this answer edited Feb 9, 2009 at 21:08 answered Dec 27, 2008 at 0:39 Jason SJason S 190k176176 gold badges635635 silver badges1k1k bronze badges 13 Comments kquinn kquinn This comment is wrong. There is a time and a place for every approximation. If you do not know enough analysis to determine the region of convergence for ANY series approximation, you should NOT be using it. That goes for Taylor, Chebyshev, Padé, etc. series. Taylor series are often Good Enough. Jason S Jason S :shrug: I don't know about you but I've never been interested in evaluating a function in a small neighborhood around just one point. Even a quick least-squares fit over an interval is pretty damn easy to do. Anyone who's using Taylor series is just missing the point. Jason S Jason S @kquinn: the region of convergence for Chebyshev approximations isn't a useful concept since the interval over which they are calculated is an explicit input to the process. user85109 user85109 Upvoting because the responder knew Hart exists. :smile: Hart is the classic reference here, even if it was difficult to find when I bought a copy (in print) 25 years ago. It is worth every penny. Range reduction wherever possible, coupled with an appropriate approximation, be it Pade, Chebychev, even Taylor series as appropriate, is a good approach. Pade or Chebychev approximants are usually the better choice over a Taylor series though. Jason S Jason S ??? How is that any different? Taylor series out to 17th degree to calculate sin(x) from -2pi to +2pi can probably be beat by Chebyshev with a 7th or 9th degree polynomial. I wouldn't have any problem making the statement, "If you have time constraints when cutting down trees, you should not use a hand saw. Use a chainsaw." Perhaps I should reword from "should not" to something like "I would not recommend using Taylor series". Sure, you could use Taylor series in some cases, but your accuracy and performance are going to be problematic. By performance I mean CPU execution time. \| 15 I believe they're calculated using Taylor Series or CORDIC. Some applications which make heavy use of trig functions (games, graphics) construct trig tables when they start up so they can just look up values rather than recalculating them over and over. Share Improve this answer answered Dec 5, 2008 at 20:58 Jon GallowayJon Galloway 53.3k2525 gold badges128128 silver badges194194 bronze badges Comments 7 Check out the Wikipedia article on trig functions. A good place to learn about actually implementing them in code is Numerical Recipes. I'm not much of a mathematician, but my understanding of where sin, cos, and tan "come from" is that they are, in a sense, observed when you're working with right-angle triangles. If you take measurements of the lengths of sides of a bunch of different right-angle triangles and plot the points on a graph, you can get sin, cos, and tan out of that. As Harper Shelby points out, the functions are simply defined as properties of right-angle triangles. A more sophisticated understanding is achieved by understanding how these ratios relate to the geometry of circle, which leads to radians and all of that goodness. It's all there in the Wikipedia entry. Share Improve this answer answered Dec 5, 2008 at 21:03 ParappaParappa 7,67633 gold badges3737 silver badges3939 bronze badges Comments 2 Most commonly for computers, power series representation is used to calculate sines and cosines and these are used for other trig functions. Expanding these series out to about 8 terms computes the values needed to an accuracy close to the machine epsilon (smallest non-zero floating point number that can be held). The CORDIC method is faster since it is implemented on hardware, but it is primarily used for embedded systems and not standard computers. Share Improve this answer answered Feb 19, 2015 at 3:51 Joshua HowardJoshua Howard 91611 gold badge1212 silver badges2727 bronze badges Comments 1 I would like to extend the answer provided by @Jason S. Using a domain subdivision method similar to that described by @Jason S and using Maclaurin series approximations, an average (2-3)X speedup over the tan(), sin(), cos(), atan(), asin(), and acos() functions built into the gcc compiler with -O3 optimization was achieved. The best Maclaurin series approximating functions described below achieved double precision accuracy. For the tan(), sin(), and cos() functions, and for simplicity, an overlapping 0 to 2pi+pi/80 domain was divided into 81 equal intervals with "anchor points" at pi/80, 3pi/80, ..., 161pi/80. Then tan(), sin(), and cos() of these 81 anchor points were evaluated and stored. With the help of trig identities, a single Maclaurin series function was developed for each trig function. Any angle between ±infinity may be submitted to the trig approximating functions because the functions first translate the input angle to the 0 to 2pi domain. This translation overhead is included in the approximation overhead. Similar methods were developed for the atan(), asin(), and acos() functions, where an overlapping -1.0 to 1.1 domain was divided into 21 equal intervals with anchor points at -19/20, -17/20, ..., 19/20, 21/20. Then only atan() of these 21 anchor points was stored. Again, with the help of inverse trig identities, a single Maclaurin series function was developed for the atan() function. Results of the atan() function were then used to approximate asin() and acos(). Since all inverse trig approximating functions are based on the atan() approximating function, any double-precision argument input value is allowed. However the argument input to the asin() and acos() approximating functions is truncated to the ±1 domain because any value outside it is meaningless. To test the approximating functions, a billion random function evaluations were forced to be evaluated (that is, the -O3 optimizing compiler was not allowed to bypass evaluating something because some computed result would not be used.) To remove the bias of evaluating a billion random numbers and processing the results, the cost of a run without evaluating any trig or inverse trig function was performed first. This bias was then subtracted off each test to obtain a more representative approximation of actual function evaluation time. Table 2. Time spent in seconds executing the indicated function or functions one billion times. The estimates are obtained by subtracting the time cost of evaluating one billion random numbers shown in the first row of Table 1 from the remaining rows in Table 1. Time spent in tan(): 18.0515 18.2545 Time spent in TAN3(): 5.93853 6.02349 Time spent in TAN4(): 6.72216 6.99134 Time spent in sin() and cos(): 19.4052 19.4311 Time spent in SINCOS3(): 7.85564 7.92844 Time spent in SINCOS4(): 9.36672 9.57946 Time spent in atan(): 15.7160 15.6599 Time spent in ATAN1(): 6.47800 6.55230 Time spent in ATAN2(): 7.26730 7.24885 Time spent in ATAN3(): 8.15299 8.21284 Time spent in asin() and acos(): 36.8833 36.9496 Time spent in ASINCOS1(): 10.1655 9.78479 Time spent in ASINCOS2(): 10.6236 10.6000 Time spent in ASINCOS3(): 12.8430 12.0707 (In the interest of saving space, Table 1 is not shown.) Table 2 shows the results of two separate runs of a billion evaluations of each approximating function. The first column is the first run and the second column is the second run. The numbers '1', '2', '3' or '4' in the function names indicate the number of terms used in the Maclaurin series function to evaluate the particular trig or inverse trig approximation. SINCOS#() means that both sin and cos were evaluated at the same time. Likewise, ASINCOS#() means both asin and acos were evaluated at the same time. There is little extra overhead in evaluating both quantities at the same time. The results show that increasing the number of terms slightly increases execution time as would be expected. Even the smallest number of terms gave around 12-14 digit accuracy everywhere except for the tan() approximation near where its value approaches ±infinity. One would expect even the tan() function to have problems there. Similar results were obtained on a high-end MacBook Pro laptop in Unix and on a high-end desktop computer in Linux. Share Improve this answer edited Nov 6, 2018 at 20:23 answered Nov 6, 2018 at 12:38 Roger WehageRoger Wehage 10944 bronze badges Comments -6 If your asking for a more physical explanation of sin, cos, and tan consider how they relate to right-angle triangles. The actual numeric value of cos(lambda) can be found by forming a right-angle triangle with one of the angles being lambda and dividing the length of the triangles side adjacent to lambda by the length of the hypotenuse. Similarily for sin use the opposite side divided by the hypotenuse. For tangent use the opposite side divided by the adjacent side. The classic memonic to remember this is SOHCAHTOA (pronounced socatoa). Share Improve this answer answered Jan 29, 2009 at 6:09 jeffDjeffD 1,23833 gold badges1313 silver badges1818 bronze badges Comments Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithm math trigonometry See similar questions with these tags. 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2973	http://mathcentral.uregina.ca/qq/database/qq.09.04/kim1.html	Square feet and cubic feet Quandaries and Queries Name: Kim Who is asking: student Level: Middle Question: Is there a relationship between square feet and cubic feet? I tried doing some experiments and stuff to see the relationship between square feet and cubic feet, out of curiosity. What I did was take 9 sq. ft. (which equals 1 sq. yd.)exponent 3 = answer and found the square root of the answer and got 27 (27 cubic feet = 1 cubic yd). Can I use this relationship and make conversions between square feet and cubic feet? Hi Kim, Square feet and Cubic feet are units of measurement for very different concepts. Square feet are units of area and cubic feet are units of volume. Here is an example. Suppose that I want to put a fish pond in my yard. I have to dig a hole, line it with a pool liner and then fill it with water. I am considering a pool which is 4 feet by 4 feet and 4 feet deep. For the liner I need the bottom: 4 feet by 4 feet or 4 4 = 16 square feet each side: 4 feet by 4 feet or 4 4 = 16 square feet There are 4 sides so it total I need 16 + 4 16 = 80 square feet on liner. The volume of the pool is 4 4 4 = 64 cubic feet so I need 64 cubic feet of water. The other option is a pool which is 10 feet long, 5 feet wide and 1 foot deep. For the liner for this design I need the bottom: 10 feet by 5 feet or 10 5 = 50 square feet each long side: 10 feet by 1 foot or 10 1 = 10 square feet each short side: 5 feet by 1 foot or 5 1 = 5 square feet There are two long sides and two short sides and hence I need 50 + 2 10 + 2 5 = 80 square feet of liner. Hence each pool requires a purchase of 80 square feet of liner. The volume of the second design is 10 5 1 = 50 cubic feet. Hence, although the square footage of the liner required for either design is 80 square feet, the cubic feet of water required is quite different. Penny Go to Math Central
2974	https://math.stackexchange.com/questions/40864/how-do-i-find-roots-of-a-single-variate-polynomials-whose-integers-coefficients	calculus - How do I find roots of a single-variate polynomials whose integers coefficients are symmetric wrt their respective powers - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How do I find roots of a single-variate polynomials whose integers coefficients are symmetric wrt their respective powers Ask Question Asked 14 years, 4 months ago Modified8 years, 5 months ago Viewed 5k times This question shows research effort; it is useful and clear 12 Save this question. Show activity on this post. Given a polynomial such as X 4+4 X 3+6 X 2+4 X+1,X 4+4 X 3+6 X 2+4 X+1, where the coefficients are symmetrical, I know there's a trick to quickly find the zeros. Could someone please refresh my memory? calculus polynomials Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Apr 4, 2017 at 6:57 CommunityBot 1 asked May 23, 2011 at 17:05 Paul MantaPaul Manta 3,595 7 7 gold badges 37 37 silver badges 50 50 bronze badges Add a comment\| 6 Answers 6 Sorted by: Reset to default This answer is useful 16 Save this answer. Show activity on this post. Hint: This particular polynomial is very nice, and factors as (X+1)4(X+1)4. Take a look at Pascal's Triangle and the Binomial Theorem for more details. Added: Overly complicated formula The particular quartic you asked about had a nice solution, but lets find all the roots of the more general a x 4+b x 3+c x 2+b x+a.a x 4+b x 3+c x 2+b x+a. Since 0 0 is not a root, we are equivalently finding the zeros of a x 2+b x 1+c+b x−1+a x−2.a x 2+b x 1+c+b x−1+a x−2. Let z=x+1 x z=x+1 x (as suggested by Aryabhatta) Then z 2=x 2+2+x−2 z 2=x 2+2+x−2 so that a x 2+b x 1+c+b x−1+a x−2=a z 2+b z+(c−2 a).a x 2+b x 1+c+b x−1+a x−2=a z 2+b z+(c−2 a). The roots of this are given by the quadratic formula: −b+b 2−4 a(c−2 a)−−−−−−−−−−−−−√2 a,−b−b 2−4 a(c−2 a)−−−−−−−−−−−−−√2 a.−b+b 2−4 a(c−2 a)2 a,−b−b 2−4 a(c−2 a)2 a. Now, we then have x+1 x=−b±b 2−4 a(c−2 a)−−−−−−−−−−−−−√2 a x+1 x=−b±b 2−4 a(c−2 a)2 a and hence we have the two quadratics x 2+b+b 2−4 a(c−2 a)−−−−−−−−−−−−−√2 a x+1=0,x 2+b+b 2−4 a(c−2 a)2 a x+1=0, x 2+b−b 2−4 a(c−2 a)−−−−−−−−−−−−−√2 a x+1=0.x 2+b−b 2−4 a(c−2 a)2 a x+1=0. This then gives the four roots: −b+b 2−4 a(c−2 a)−−−−−−−−−−−−−√4 a±1 4(b−b 2−4 a(c−2 a)−−−−−−−−−−−−−√2 a)2−1−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷−b+b 2−4 a(c−2 a)4 a±1 4(b−b 2−4 a(c−2 a)2 a)2−1 −b−b 2−4 a(c−2 a)−−−−−−−−−−−−−√4 a±1 4(b+b 2−4 a(c−2 a)−−−−−−−−−−−−−√2 a)2−1−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷.−b−b 2−4 a(c−2 a)4 a±1 4(b+b 2−4 a(c−2 a)2 a)2−1. If we plug in a=1 a=1, b=4 b=4, c=6 c=6, we find that all four of these are exactly 1 1, so our particular case does work out. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 26, 2011 at 16:33 answered May 23, 2011 at 17:12 Eric NaslundEric Naslund 73.7k 12 12 gold badges 181 181 silver badges 272 272 bronze badges 6 You've lost a couple of x x s in the two quadratics just before the roots.Peter Taylor –Peter Taylor 2011-05-23 17:46:39 +00:00 Commented May 23, 2011 at 17:46 1 Eric, could you please supply a reference for your tantalizing allusion to Aryabhatta ?Georges Elencwajg –Georges Elencwajg 2011-05-23 17:58:55 +00:00 Commented May 23, 2011 at 17:58 2 @Elgeorges: Sorry to disappoint, but I meant Aryabhatta, the user on MSE!! (Formely Moron, Who also posted a answer here)Eric Naslund –Eric Naslund 2011-05-23 18:01:43 +00:00 Commented May 23, 2011 at 18:01 Ah, I see. I thought historians had just discovered a 1600 hundred year old Indian manuscript! +1 for making me laugh at my own misunderstanding:-)Georges Elencwajg –Georges Elencwajg 2011-05-23 23:50:46 +00:00 Commented May 23, 2011 at 23:50 3 Oops, I can't keep my promise, Eric : I had forgotten that I had already upvoted you and the software won't let me upvote you again.Sorry about that.Georges Elencwajg –Georges Elencwajg 2011-05-23 23:56:32 +00:00 Commented May 23, 2011 at 23:56 \|Show 1 more comment This answer is useful 13 Save this answer. Show activity on this post. One possibility: Divide by X 2 X 2 and write it as a polynomial in Z=X+1/X Z=X+1/X. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered May 23, 2011 at 17:12 AryabhataAryabhata 84.1k 11 11 gold badges 192 192 silver badges 281 281 bronze badges Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. I'm not sure if this is what you are thinking of, but if x x is a root, and y=1/x y=1/x, then by plugging in 1/y 1/y and multiplying the result by y n y n (where n n is the degree of the polynomial), we see that y y is also a root of the polynomial. This has two consequences. First, every root, except possibly for X=±1 X=±1 comes in pairs, and so if the polynomial is of odd degree, it must have ±1±1 as a root. Second, if it is of even degree, then for every root x x, the polynomial is divisible by (X−x)(X−1/x)=X 2−X(x+1/x)+1(X−x)(X−1/x)=X 2−X(x+1/x)+1. This indicates that we should be able to rewrite the polynomial in terms of X+1/X X+1/X. We can do this as follows. Let the polynomial be P(X)=∑n i=0 a i x i P(X)=∑i=0 n a i x i, where a i=a n−i a i=a n−i If P(X)P(X) is of odd degree, we know that 1 1 must be a root. Divide out by X−1 X−1 and X+1 X+1 until ±1±1 are no longer roots, and you will get an even degree polynomial with symmetric coefficients, so from here on we can assume that n n is even. Divide P(X)P(X) by X n/2 X n/2 to get P(X)/X n/2=a n(X n/2+1/X n/2)+a n−1(X n/2−1+1/X n/2−1)+…P(X)/X n/2=a n(X n/2+1/X n/2)+a n−1(X n/2−1+1/X n/2−1)+…. We can write X k+X−k X k+X−k as a polynomial in (X+1/X)(X+1/X) with symmetric coefficients, and subtracting off this chunk and repeating with the lower degree terms allows us to eventually rewrite P(X)/X n/2=Q(Z)P(X)/X n/2=Q(Z) where Z=X+1/X Z=X+1/X. Thus, we have reduced the problem to one of half the degree. Note that we can simplify the division step with a little calculus by noting that 1 1 is a root of P(X)P(X) of multiplicity k k if P(1)=P′(1)=P′′(1)=…P(k−1)(1)=0 P(1)=P′(1)=P″(1)=…P(k−1)(1)=0, but P(k)(1)≠0 P(k)(1)≠0, and similarly for P(−1)P(−1). Also, it is worth pointing out that this is very similar to how, when you have a real polynomial, all the complex roots have to come in pairs with their conjugates, and so once you get rid of all the real roots, you can write the polynomial as a product of quadratic factors. However, the fact that we can rewrite our polynomial here in terms of X+1/X X+1/X has no real analogue. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 23, 2011 at 17:54 answered May 23, 2011 at 17:41 AaronAaron 25.7k 2 2 gold badges 53 53 silver badges 76 76 bronze badges 2 How can X k+X−k X k+X−k be written as a polynomial in (X+1/X)(X+1/X) with symmetric coefficients? I would compute (X+1/X)k(X+1/X)k and collect terms, but the resulting coefficients are not symmetric. For instance, X 2+1/X 2=(X+1/X)2−2 X 2+1/X 2=(X+1/X)2−2.Ricardo Buring –Ricardo Buring 2014-08-11 11:36:03 +00:00 Commented Aug 11, 2014 at 11:36 It's been a while since I wrote the statement, but I believe that what I meant was that it is symmetric in X X when expanded out, which should be self evident. You are correct that it is not a symmetric polynomial of (X+1/X)(X+1/X).Aaron –Aaron 2014-08-11 11:48:21 +00:00 Commented Aug 11, 2014 at 11:48 Add a comment\| This answer is useful 4 Save this answer. Show activity on this post. You can write y=x+1 x y=x+1 x and cut the degree in half. In this case (having checked that x x cannot be 0 0) x 2+4 x+6+4 x+1 x 2=y 2+4 y+4 x 2+4 x+6+4 x+1 x 2=y 2+4 y+4 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 23, 2011 at 20:06 answered May 23, 2011 at 17:13 Ross MillikanRoss Millikan 384k 28 28 gold badges 264 264 silver badges 472 472 bronze badges 4 Is that y=x+1 x y=x+1 x? and y 2+4 y+6 y 2+4 y+6?Mitch –Mitch 2011-05-23 18:18:58 +00:00 Commented May 23, 2011 at 18:18 Yes there should be a +, but the constant term becomes 4 because you get 2 from squaring y.Ross Millikan –Ross Millikan 2011-05-23 20:05:48 +00:00 Commented May 23, 2011 at 20:05 argh..I was hoping it would be so much easier without having to recompute the constants (which would get that much worse for higher degree.Mitch –Mitch 2011-05-23 21:11:54 +00:00 Commented May 23, 2011 at 21:11 Actually as the degree goes up you need to recalculate other terms as well. If you started with an eighth degree polynomial the (x+1/x)^4 supplies quadratic terms as well. But it is a triangular system. You determine the coefficients on the outer two terms by inspection, then figure out the correction to the next two, and continue.Ross Millikan –Ross Millikan 2011-05-23 21:21:17 +00:00 Commented May 23, 2011 at 21:21 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Such a polynomial with symmetric coefficients is called a palindromic polynomial. If x 0 x 0 is a zero of a palindromic polynomial than 1 x 0 1 x 0 is a zero , too. If we divide the polynomial A x 4+B x 3+C x 2+B x+A=0 A x 4+B x 3+C x 2+B x+A=0 by A A we get x 4+a x 3+b x 2+a x+1=0(1)(1)x 4+a x 3+b x 2+a x+1=0 (x 4+1)+a(x 3+x)+b x 2=0(x 4+1)+a(x 3+x)+b x 2=0 x 2(x 2+1 x 2)+a x 2(x+1 x)+b x 2=0 x 2(x 2+1 x 2)+a x 2(x+1 x)+b x 2=0 x 2((x 2+1 x 2)+a(x+1 x)+b)=0 x 2((x 2+1 x 2)+a(x+1 x)+b)=0 because x≠0 x≠0. For the same reason we can divide the equation by x 2 x 2 to get (x 2+1 x 2)+a(x+1 x)+b=0(2)(2)(x 2+1 x 2)+a(x+1 x)+b=0 We set y=x+1 x(3)(3)y=x+1 x and by squaring we get y 2=x 2+1 x 2+2 y 2=x 2+1 x 2+2 If we substitute this in (2)(2) we get a quadratic equation of y y (y 2−2)+a y+b=0(4)(4)(y 2−2)+a y+b=0 So (1)(1) can be solved by first solving the quadratic equation (4)(4) and then the quadratic equation (3)(3) In a similar manner the equation x 4+a x 3+b x 2−a x+1=0(5)(5)x 4+a x 3+b x 2−a x+1=0 can be solved: (x 2+1 x 2)+a(x−1 x)+b=0(x 2+1 x 2)+a(x−1 x)+b=0 We set y=x−1 x(6)(6)y=x−1 x and now get y 2=x 2+1 x 2−2 y 2=x 2+1 x 2−2 and so (y 2+2)+a y+b=0(7)(7)(y 2+2)+a y+b=0 So here we have to solve the quadratic equations (7)(7) and (6)(6) to get the solution of (5)(5). An palindromic equation of degree 3 3 can be solved by a different tick: We transform x 3+a x 2+a x+1=0(8)(8)x 3+a x 2+a x+1=0 to (x 3+1)+a(x 2+x)=0(x 3+1)+a(x 2+x)=0 (x+1)((x 2−x+1)+a x)=0(x+1)((x 2−x+1)+a x)=0 So one solution is x=−1 x=−1 This can also be seen by plugging −1−1 into (8)(8). The two other solutions are the solutions of the quadratic equation (x 2−x+1)+a x=0(x 2−x+1)+a x=0 Finally the equation x 3+a x 2−a x−1=0 x 3+a x 2−a x−1=0 can be transformed to (x 3−1)+a(x 2−x)=0(x 3−1)+a(x 2−x)=0 and therefore to (x−1)((x 2+x+1)+a x)=0(x−1)((x 2+x+1)+a x)=0 An palindromic equation of 6th or 9th degree again can be solved in two steps by solving first an equation of y y of degree 3 3 or 4 4 and then solve the equation y=x+1 x y=x+1 x The palindromic equation of degree 5 5 has the solution x=−1 x=−1 and can be reduced to an equation of degree $$. The polynomial equations up to degree 4 4 can be solved by radicals, so the palindromic polynomial equations of degree 5,6,8 5,6,8 can be solved by radicals, too. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 30, 2014 at 6:56 answered Nov 29, 2014 at 22:52 miracle173miracle173 11.4k 3 3 gold badges 33 33 silver badges 65 65 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. If you use the Rational Root Theorem, you find that the only possible rational roots are ±1±1. You can check if either of those work and if so, divide by the corresponding factor. Then you get a simpler polynomial, and in your specific case you would be able to use the Rational Root Theorem again to get a quadratic. EDIT: The Rational Root Theorem says that if you have a polynomial with integer coefficients, and that polynomial has a rational root a b a b where a b a b is in lowest terms, then a a divides the constant term and b b divides the leading coefficient. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 23, 2011 at 17:44 answered May 23, 2011 at 17:33 badatmathbadatmath 4,199 6 6 gold badges 35 35 silver badges 49 49 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus polynomials See similar questions with these tags. 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2975	https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/resources/presentati_w07d2/	1P18- Class 18: Outline Hour 1: Levitation Experiment 8: Magnetic Forces Hour 2: Ampere’s Law2P18- Review: Right Hand Rules 1. Torque: Thumb = torque, fingers show rotation 2. Feel: Thumb = I, Fingers = B, Palm = F 3. Create: Thumb = I, Fingers (curl) = B 4. Moment: Fingers (curl) = I, Thumb = Moment3P18- Last Time: Dipoles4P18- Magnetic Dipole Moments Anμ GG IIA ≡≡ ˆ Generate: Feel: 1) Torque to align with external field 2) Forces as for bar magnets (seek field) -DipoleU = ⋅μ B GG5P18- Some Fun: Magnetic Levitation6P18- Put a Frog in a 16 T Magnet… For details: How does that work? First a BRIEF intro to magnetic materials8P18- Para/Ferromagnetism Applied external field B0 tends to align the atomic magnetic moments (unpaired electrons)9P18- Diamagnetism Everything is slightly diamagnetic. Why? More later. If no magnetic moments (unpaired electrons) then this effect dominates.10P18- Back to Levitation11P18- Levitating a Diamagnet 1) Create a strong field (with a field gradient!) 2) Looks like a dipole field 3) Toss in a frog (diamagnet) 4) Looks like a bar magnet pointing opposite the field 5) Seeks lower field (force up) which balances gravity S N N S Most importantly, its stable: Restoring force always towards the center S N12P18- Using ∇B to Levitate For details: •Frog •Strawberry •Water Droplets •Tomatoes •Crickets13P18- Demonstrating: Levitating Magnet over Superconductor14P18- Perfect Diamagnetism: “Magnetic Mirrors” N S N S15P18- Perfect Diamagnetism: “Magnetic Mirrors” N S N S No matter what the angle, it floats -- STABILITY16P18- Using ∇B to Levitate For details: A Sumo Wrestler17P18- Two PRS Questions Related to Experiment 8: Magnetic Forces18P18- Experiment 8: Magnetic Forces (Calculating μ0) MagneticForceRepel/16-MagForceRepel_f65_320.html19P18- Experiment Summary: Currents feel fields Currents also create fields Recall… Biot-Savart20P18- The Biot-Savart Law Current element of length ds carrying current I produces a magnetic field: 2 0 ˆ 4 r dI rs Bd × = GG π μ21P18- Today: 3 rd Maxwell Equation: Ampere’s Law Analogous (in use) to Gauss’s Law22P18- Gauss’s Law – The Idea The total “flux” of field lines penetrating any of these surfaces is the same and depends only on the amount of charge inside23P18- Ampere’s Law: The Idea In order to have a B field around a loop, there must be current punching through the loop24P18- Ampere’s Law: The Equation The line integral is around any closed contour bounding an open surface S. Ienc is current through S: ∫ =⋅ encId 0 μsB GG enc S I d= ⋅∫ J A GG25P18- PRS Question: Ampere’s Law26P18- Biot-Savart vs. Ampere Biot- Savart Law general current source ex: finite wire wire loop Ampere’s law symmetric current source ex: infinite wire infinite current sheet 0 2 ˆ 4 I d r μ π × = ∫ s r B GG ∫ =⋅ encId 0 μsB GG27P18- Applying Ampere’s Law 1. Identify regions in which to calculate B field Get B direction by right hand rule 2. Choose Amperian Loops S: Symmetry B is 0 or constant on the loop! 3. Calculate 4. Calculate current enclosed by loop S 5. Apply Ampere’s Law to solve for B ∫ ⋅ sB GG d ∫ =⋅ encId 0 μsB GG28P18- Always True, Occasionally Useful Like Gauss’s Law, Ampere’s Law is always true However, it is only useful for calculation in certain specific situations, involving highly symmetric currents. Here are examples…29P18- Example: Infinite Wire I A cylindrical conductor has radius R and a uniform current density with total current I Find B everywhere Two regions: (1) outside wire (r ≥ R) (2) inside wire (r < R)30P18- Ampere’s Law Example: Infinite Wire I I B Amperian Loop: B is Constant & Parallel I Penetrates31P18- Example: Wire of Radius R Region 1: Outside wire (r ≥ R) d⋅∫ B s G G v ckwisecounterclo 2 0 r I π μ =B G B d= ∫ s G v ( )2B r π= 0 encI μ= 0 I μ= Cylindrical symmetry Æ Amperian Circle B-field counterclockwise32P18- Example: Wire of Radius R Region 2: Inside wire (r < R) 2 0 2 r I R π μ π ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ckwisecounterclo 2 2 0 R Ir π μ =B G Could also say: ( )2 22 ; r R I JAI R I A I J encenc π π π ==== d⋅∫ B s G G v B d= ∫ s G v ( )2B r π= 0 encI μ=33P18- Example: Wire of Radius R 2 0 2 R Ir Bin π μ = r I Bout π μ 2 0 =34P18- Group Problem: Non-Uniform Cylindrical Wire I A cylindrical conductor has radius R and a non- uniform current density with total current: Find B everywhere 0 R J r =J G35P18- Applying Ampere’s Law In Choosing Amperian Loop: • Study & Follow Symmetry • Determine Field Directions First • Think About Where Field is Zero • Loop Must • Be Parallel to (Constant) Desired Field • Be Perpendicular to Unknown Fields • Or Be Located in Zero Field36P18- Other Geometries37P18- Helmholtz Coil38P18- Closer than Helmholtz Coil39P18- Multiple Wire Loops40P18- Multiple Wire Loops – Solenoid41P18- Magnetic Field of Solenoid loosely wound tightly wound For ideal solenoid, B is uniform inside & zero outside42P18- Magnetic Field of Ideal Solenoid d d d d d⋅ ⋅ + ⋅ + ⋅ + ⋅∫ ∫ ∫ ∫ ∫1 2 3 4 B s = B s B s B s B s G G G G GG G G G G v Using Ampere’s law: Think! along sides 2 and 4 0 along side 3 d⎧ ⊥⎪ ⎨ =⎪⎩ B s B G G G n: turn densityencI nlI= 0d Bl nlI μ⋅ = =∫ B s G G v 0 0 nlI B nI l μ μ= =/ : # turns/unit lengthn N L= 0 0 0Bl= + + +43P18- Demonstration: Long Solenoid44P18- Group Problem: Current Sheet A sheet of current (infinite in the y & z directions, of thickness 2d in the x direction) carries a uniform current density: Find B everywhere ˆ s J=J k G y45P18- Ampere’s Law: Infinite Current Sheet I Amperian Loops: B is Constant & Parallel OR Perpendicular OR Zero I Penetrates B B46P18- Solenoid is Two Current Sheets Field outside current sheet should be half of solenoid, with the substitution: 2nI dJ= This is current per unit length (equivalent of λ, but we don’t have a symbol for it)47P18- = 2 Current Sheets Ampere’s Law: . ∫ =⋅ encId 0 μsB GG I B B X X X X X X X X X X X X X X X X X X X X X X X X X X X X B Long Circular Symmetry (Infinite) Current Sheet Solenoid Torus48P18- Brief Review Thus Far…49P18- Maxwell’s Equations (So Far) 0Ampere's Law: Currents make curling Magnetic Fields enc C d I μ⋅ =∫ B s G G v Magnetic Gauss's Law: 0 No Magnetic Monopoles! (No diverging B Fields) S d⋅ =∫∫ B A GG w 0 Gauss's Law: Electric charges make diverging Electric Fields in S Q d ε ⋅ =∫∫ E A GG w
2976	https://math.washington.edu/~king/coursedir/m444a05/lab/lab10-napoleon.html	Math 487 Lab 10 (Napoleon Theorem) The Napoleon Figure Construct the following figure. It may go faster if you create a tool 'Eqtri with center from side' for constructing a equilateral triangle ABC with center O given points A and B (i.e., construct the triangle from the edge, but also construct the center and hide the construction lines). Use this tool to make a Napoleon figure like this one. The Napoleon Theorem by Experiment Construct the circle XYZ. drag A, B and C around and see what kind of triangle this is. Do some measuring to provide extra evidence. (This makes it pretty certain that you conjecture is correct, but it does not prove it or explain why yet.) Does the relationship still hold if you drag A across line BC so that the triangles are on the inside? What if A is on segment BC? Napoleon's theorem states that XYZ is equilateral for any choice of ABC ( Why is this Napoleon's Theorem? It is claimed (probably not entirely accurately) that this theorem was discovered by Emperor Napoleon of France. What is really true is that he did have a taste for mathematics and also surrounded himself with quite a few mathematicians in his government. He also set up some French schools like the Polytechnique, that are still very important scientific institutions today. And it makes for a very memorable name for the theorem. Proof based on transformations Part 1. Investigating the Products of 120-degree rotations In the proof we will compose three 120-degree rotatoins. So on a new, separate page, let's review what this will look like. Draw 3 points X, Y, Z on the plane and also draw a point P. Rotate P with center X by 120 degrees to get P'. Rotate P' with center Y by 120 degrees to get P''. Thus P'' = Y120X120(P). What kind of isometry does our general theory say Y120X120 is? Can you see this? Can you see the angle and the center? Rotate P'' with center Z by 120 degrees to get P'''. Thus P'' = Z120Y120X120(P). What kind of isometry does our general theory say Z120Y120X120 is? Can you see this? To visualize the relationship between P and P''', construct the segment PP'''. Leaving X, Y and Z fixed, drag P around and see how the segment PP''' changes. Does the length change? Does the direction change? Question: What kind of isometry is the transformation that takes P to P'''. This isometry is the composition T = Z120Y120X120. Answer ___________________________ Move one of X, Y or Z and again observe how PP''' changes as P moves. Is the transformation still the same kind? Is it exactly the same transformation? Now our goal is to make P''' coincide with P. It is does it for one P, then it should do it for any position of P. Leaving X and Y fixed, drag Z until P''' coincides with P. Question: Where is Z when this occurs? Answer ____________________ How is this consistent with what we have learned about computing Y120X120? Part 2. The proof of Napoleon's Theorem Let P = B in the figure, then P' = A. What is P'' in this figure if P' = A? _________________ What is P''' in this figure given the P'' above? _________________ Since in this figure P = P''', based on your conclusions from the last part, what can you say about the triangle XYZ? Answer: Triangle XYZ is __________________________________ Draw triangle XYZ in your figure and see that the conclusion appears to be correct as you drag A, B, and C around and XYZ moves in consequence. Part 3. Building a Napoleon Figure with Rotations In a new sketch, draw a point X and a point Y. Let Z be the rotation of X with center Y by 60 degrees. Then draw a point P in the sketch. Repeat the construction of P', P'', and P''' from Part 1. You should find that P''' = P, so you can construct segments PP', P'P'', P''P''' to form a triangle. (If this does not happen, ask for help!) Call the points P = B, P' = A, P'' = C so your figure matches the figure on the first page. Now construct the 3 equilateral triangles in the figure by rotating appropriate points by 120 degrees with centers X, Y, Z. Construct the equilateral triangle interiors and color the 3 triangles in 3 different colors. You can now drag point B and see how the figure changes with X, Y, and Z immobile. Part 4. Covering the plane by a Napoleon tessellation Continuing with this sketch from Part 3, begin a tessellation of the plane by rotating the whole figure by 120 degrees, first with center X. Then select the whole figure and rotate by 120 with center Y. Then select the whole figure and rotate by 120 with center Z. Then select the whole figure and rotate by 120 with center X. Continue until the pattern becomes clear. Question: What are the symmetries of this tessellation? Does this suggest another reason why the points X, Y, Z have to form an equilateral triangle?
2977	https://sites.ualberta.ca/~gingrich/courses/phys395/notes/node113.html	Input Impedance Next:Voltage and Current OffsetsUp:Analysis Using Finite Open-LoopPrevious:Output Impedance Input Impedance When calculating the output impedance we still assumed an infinite input impedance. In this section we will calculate the finite input impedance assuming a zero output impedance. We consider a model that assumes an internal resistor connecting the inverting and non-inverting input terminals of the op-amp as shown in figure6.31 Consider an inverting amplifier and remove the input resistor so that the input impedance can be calculated directly at the amplifier's input terminals. Figure 6.31:Model for calculating the input impedance of the inverting amplifier. The input impedance is defined by and the current at the summing junction is The current through the feedback resistor is and the output voltage is related to by the open-loop gain The resulting input impedance is thus For large A The closed-loop input impedance is thus small and almost independent of the large of the operational amplifier. Now consider the non-inverting amplifier shown in figure6.32. Figure 6.32:Model for calculating the input impedance of the non-inverting amplifier. The student should calculate the input impedance by recognizing that is much less than , since is much greater than or . Your result should be where is the closed-loop gain of the amplifier. Notice that in contrast to the low input impedance for the inverting amplifier, the non-inverting amplifier exhibits a closed-loop input impedance that is much larger than the open-loop value . _Doug Gingrich Tue Jul 13 16:55:15 EDT 1999_
2978	https://www.facebook.com/groups/10150112797390640/posts/10152938371230640/	Evolution and Creationism Open Debate \| Definition of a scientific theory: a well-substantiated explanation of some aspect of the natural world, based on a body of knowledge that has been re... \| Facebook Log In Log In Forgot Account? Evolution and Creationism Open Debate · Join Josh Allin · June 26, 2013 · Definition of a scientific theory: a well-substantiated explanation of some aspect of the natural world, based on a body of knowledge that has been repeatedly confirmed through observation and experiment. Other modern scientific theories: germ theory, theory of relativity, heliocentrism, atomic theory, plate tectonics, and cell theory. PS Yes, gravity is a theory. Look up "the theory of gravity" and "Newton's Law Of Gravitation". Also, a theory never becomes a law. IFLScience June 26, 2013 · Definition of a scientific theory: a well-substantiated explanation of some aspect of the natural world, based on a body of knowledge that has been repeatedly confirmed through observation and experiment. Other modern scientific theories: germ theory, theory of relativity, heliocentrism, atomic theory, plate tectonics, and cell theory. PS Yes, gravity is a theory. Look up "the theory of gravity" and "Newton's Law Of Gravitation". Also, a theory never becomes a law. All reactions: 2 11 comments Like Comment John Carrington Defintion of a self serving idiot 12y Brian Wakely Regardless of how you feel about it, that comment is purely negative for the sake of being negative. It doesn't offer any counter points and you can't take anything constructive from it. Treating people with respect regardless of differences you may ha… See more 12y 5 Carolyn Reynolds One more pointless irrelevant comment from John Carrington and he's out. 12y 3 View more comments 3 of 11 See more on Facebook See more on Facebook Email or phone number Password Log In Forgot password? or Create new account
2979	https://pubmed.ncbi.nlm.nih.gov/25646498/	Denaturation and electrophoresis of RNA with formaldehyde - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Denaturation and electrophoresis of RNA with formaldehyde Donald C Rio PMID: 25646498 DOI: 10.1101/pdb.prot080994 Item in Clipboard Denaturation and electrophoresis of RNA with formaldehyde Donald C Rio. Cold Spring Harb Protoc.2015. Show details Display options Display options Format Cold Spring Harb Protoc Actions Search in PubMed Search in NLM Catalog Add to Search . 2015 Feb 2;2015(2):219-22. doi: 10.1101/pdb.prot080994. Author Donald C Rio PMID: 25646498 DOI: 10.1101/pdb.prot080994 Item in Clipboard Full text links Cite Display options Display options Format Abstract Electrophoretic size fractionation can be used to denature and separate large mRNA molecules (0.5-10 kb) on formaldehyde-containing agarose gels. Formaldehyde contains a carbonyl group that reacts to form Schiff bases with the imino or amino groups of guanine, adenine, and cytosine. These covalent adducts prevent normal base pairing and maintain the RNA in a denatured state. Because these adducts are unstable, formaldehyde must be present in the gel to maintain the RNA in the denatured state. This protocol describes the preparation of an agarose gel with formaldehyde and its setup in a horizontal electrophoresis apparatus. RNA samples are prepared and denatured in a solution of formamide and formaldehyde and, with 0.5- to 10-kb size markers, subjected to electrophoresis through the gel. Following electrophoresis, the gel is stained to visualize RNA markers or rRNA using one of several different types of stains. © 2015 Cold Spring Harbor Laboratory Press. PubMed Disclaimer Similar articles Denaturation and electrophoresis of RNA with glyoxal.Rio DC.Rio DC.Cold Spring Harb Protoc. 2015 Feb 2;2015(2):223-6. doi: 10.1101/pdb.prot081000.Cold Spring Harb Protoc. 2015.PMID: 25646499 Denaturing RNA electrophoresis in TAE agarose gels.Masek T, Vopalensky V, Suchomelova P, Pospisek M.Masek T, et al.Anal Biochem. 2005 Jan 1;336(1):46-50. doi: 10.1016/j.ab.2004.09.010.Anal Biochem. 2005.PMID: 15582557 Agarose gel electrophoresis of denatured RNA with silver staining.Skopp RN, Lane LC.Skopp RN, et al.Anal Biochem. 1988 Feb 15;169(1):132-7. doi: 10.1016/0003-2697(88)90263-1.Anal Biochem. 1988.PMID: 2453127 Analysis of RNA by northern and slot-blot hybridization.Brown T.Brown T.Curr Protoc Immunol. 2001 May;Chapter 10:Unit 10.12. doi: 10.1002/0471142735.im1012s07.Curr Protoc Immunol. 2001.PMID: 18432675 Review. Disease proteomics of high-molecular-mass proteins by two-dimensional gel electrophoresis with agarose gels in the first dimension (Agarose 2-DE).Oh-Ishi M, Maeda T.Oh-Ishi M, et al.J Chromatogr B Analyt Technol Biomed Life Sci. 2007 Apr 15;849(1-2):211-22. doi: 10.1016/j.jchromb.2006.10.064. Epub 2006 Dec 1.J Chromatogr B Analyt Technol Biomed Life Sci. 2007.PMID: 17141588 Review. See all similar articles Cited by Concise Analysis of Single-Stranded DNA of Recombinant Adeno-Associated Virus By Automated Electrophoresis System.Yuan Y, Higashiyama K, Hashiba N, Masumi-Koizumi K, Yusa K, Uchida K.Yuan Y, et al.Hum Gene Ther. 2024 Feb;35(3-4):104-113. doi: 10.1089/hum.2023.148. Epub 2024 Jan 30.Hum Gene Ther. 2024.PMID: 38062752 Free PMC article. Viroscope: Plant viral diagnosis from high-throughput sequencing data using biologically-informed genome assembly coverage.Valenzuela SL, Norambuena T, Morgante V, García F, Jiménez JC, Núñez C, Fuentes I, Pollak B.Valenzuela SL, et al.Front Microbiol. 2022 Oct 21;13:967021. doi: 10.3389/fmicb.2022.967021. eCollection 2022.Front Microbiol. 2022.PMID: 36338106 Free PMC article. microRNA Detection via Nanostructured Biochips for Early Cancer Diagnostics.Martino S, Tammaro C, Misso G, Falco M, Scrima M, Bocchetti M, Rea I, De Stefano L, Caraglia M.Martino S, et al.Int J Mol Sci. 2023 Apr 24;24(9):7762. doi: 10.3390/ijms24097762.Int J Mol Sci. 2023.PMID: 37175469 Free PMC article.Review. Efficient circular RNA synthesis for potent rolling circle translation.Du Y, Zuber PK, Xiao H, Li X, Gordiyenko Y, Ramakrishnan V.Du Y, et al.Nat Biomed Eng. 2025 Jul;9(7):1062-1074. doi: 10.1038/s41551-024-01306-3. Epub 2024 Dec 13.Nat Biomed Eng. 2025.PMID: 39672985 Free PMC article. A Replicating Viral Vector Greatly Enhances Accumulation of Helical Virus-Like Particles in Plants.Thuenemann EC, Byrne MJ, Peyret H, Saunders K, Castells-Graells R, Ferriol I, Santoni M, Steele JFC, Ranson NA, Avesani L, Lopez-Moya JJ, Lomonossoff GP.Thuenemann EC, et al.Viruses. 2021 May 11;13(5):885. doi: 10.3390/v13050885.Viruses. 2021.PMID: 34064959 Free PMC article. See all "Cited by" articles MeSH terms Electrophoresis, Agar Gel Actions Search in PubMed Search in MeSH Add to Search Formaldehyde / chemistry Actions Search in PubMed Search in MeSH Add to Search Nucleic Acid Denaturation Actions Search in PubMed Search in MeSH Add to Search RNA / chemistry Actions Search in PubMed Search in MeSH Add to Search Substances Formaldehyde Actions Search in PubMed Search in MeSH Add to Search RNA Actions Search in PubMed Search in MeSH Add to Search Related information MedGen PubChem Compound (MeSH Keyword) LinkOut - more resources Full Text Sources HighWire Full text links[x] HighWire [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited. 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2980	https://www.youtube.com/watch?v=MBniqMKBEWY	Finding the radius of a sphere given the volume MooMooMath and Science 583000 subscribers 1055 likes Description 151725 views Posted: 30 Oct 2015 Geometry Teachers Never Spend Time Trying to Find Materials for Your Lessons Again! Join Our Geometry Teacher Community Today! The video reviews how to find the radius of a sphere given the volume. The question that is solved is, Find the radius of a sphere with a volume of 72 meters squared. The volume formula is 4/3 pi r cubed. In order to solve for the radius you have to undo the formula. Transcript The radius of a sphere with a volume of 72 m^2 This is a backwards problem. There is no easy way to do it other than just to undo it. Let’s see what we have. We have the volume and we know that volume formula is equal to 4/3 pi radius cubed. We know the volume is 72 so I'm just going to plug in 72. Now the only variable we have in this formula is the radius, but we have to get it by itself by undoing all this stuff. To undo a fraction we multiply by its reciprocal so we are going to flip it upside down and multiply both sides by 3/4 and that cancels out the 4/3 so we will multiply 72 by 3/4 so let’s multiply 3 times 72 and divide that by 4 so the left side will simplify to 54 so we have 54 = pi radius cubed. We are getting closer. Now we need to get rid of pi which is 3.14 so we can just divide by that. We will take 54 and divide both sides by pi so 54 divided by pi and I get 17.9 is equal to radius cubed. Now how do you undo a cube? You take a cube root. What we will do is take the cube root of 17.9 I will show you what to do. Go to the Math button and under Math pick the 4th choice which is the cube root. You can choose it or go down to 4. Enter 17.9 and I will type in instead of pulling down my answer. I get 2.58, so our radius is 2.58 in meters. That is how you find the radius of a sphere from the volume. A little bit of undoing and algebra that you will manipulate. At MooMooMath we provide helpful Math and Geometry videos to help you figure out how to solve Math problems or review old Math concepts you may need to refresh. If you’re a Math student or teacher, we'd love to have you subscribe and join us! We upload a new video every day Subscribe Let’s connect MooMooMath Google+ Pinterest Twitter -~-~~-~~~-~~-~- Please watch: "Study Skills Teacher's Secret Guide to your Best Grades" -~-~~-~~~-~~-~- 113 comments Transcript: The radius of a sphere with a volume of 72 m^2 This is a backwards problem. There is no easy way to do it other than just to undo it. Lets see what we have. We have the volume and we know that volume formula is equal to 4/3 pi radius cubed. We know the volume is 72 so I'm just going to plug in 72. Now the only variable we have in this formula is the radius, but we have to get it by itself by undoing all this stuff. To undo a fraction we multiply by its reciprocal so we are going to flip it upside down and multiply both sides by 3/4 and that cancels out the 4/3 so we will multiply 72 by 3/4 so lets multiply 3 times 72 and divide that by 4 so the left side will simplify to 54 so we have 54 = pi radius cubed. We are getting closer. Now we need to get rid of pi which is 3.14 so we can just divide by that. We will take 54 and divide both sides by pi so 54 divided by pi and I get 17.9 is equal to radius cubed. Now how do you undo a cube. You take a cube root. What we will do is take the cube root of 17.9 So I will show you what to do. Go to the Math button and under Math pick the 4th choice which is the cube root. You can choose it or go down to 4. Enter 17.9 and I will type in instead of pulling down my answer. I get 2.58 So our radius is 2.58 in meters. That is how you find the radius of a sphere from the volume. A little bit of undoing and algebra that you will manipulate.
2981	https://www.youtube.com/watch?v=gfBcM3uvWfs	Molecular and Empirical Formulas Khan Academy 9090000 subscribers 3822 likes Description 1041285 views Posted: 27 Aug 2009 Introduction to molecular and empirical formulas. Calculating molecular mass. More free lessons at: 331 comments Transcript: We now have a respectable understanding of the periodic table itself and the atoms in them. And now we're ready to deal with molecules themselves. And to deal with molecules, we have to have some way of representing them. And you represent them with formulas. And there's two major-- actually, three major ways to represent a molecule. One is the molecular formula. And the other is the empirical formula. And I'll do it in a different color to differentiate it. And the difference is, well, let's just talk about what the word empirical means. I remember when I first took chemistry the teacher kept using the word empirical. Well, what does empirical really mean? I clearly did not have a very deep vocabulary. I forgot what age I was. But it means achieved through observation or experiment, or based on experience. So if someone said that they empirically figured out x, y, or z, it means that they figured it out through an experiment or they observed it. The molecular formula is essentially the actual number of atoms in that molecule. Let me show you what I'm talking about. So the empirical formula tells you what people have observed. Maybe before they even knew that there was such a thing as atoms, what they would have observed is the ratio of the atoms to one another in a molecule without knowing in the exact molecule how many of that atom there are. Let me show you what I'm talking about. So if I were to give you, I don't know, a benzene, the molecular formula of benzene, you have 6 carbon atoms and you have 6 hydrogen atoms. Now, if you were some chemist in the 1800's and you didn't know about the actual atoms, but if you had a big bag of benzene and you were to measure the ratio of the carbon to the hydrogen that you had in that bag, you would find out that for every one carbon you have one hydrogen. So your empirical formula is the ratio of the two. You don't know that each atom actually has 6 of these, but you know that for every carbon, there's a hydrogen. And for every hydrogen there's a carbon. And the way to go back, you can go from the molecular formula to the empirical formula very easily. You just find the greatest common divisor of the number of atoms in the molecule. So the greatest common divisor is six, and six is obviously six, you divide both of these by six and you get the empirical formula. It's not easy. You pretty much can't go back from the empirical formula to the molecular formula. You've lost information. I don't know whether this was C6H6. Was it C2H2? You just don't know. And I mentioned right at the beginning of the video that there's a third way to represent molecules. And that's the structural formula. And we'll do that, off and on, and we've already done it a little bit. Let me show you. The structural formula for benzene would actually say how the molecular formula atoms are configured. So benzene in particular is very interesting. It looks like this. It's often drawn like this. And you'll see this a lot when you take organic chemistry. But it looks like a little hexagon, where the vertices of the hexagon are carbon atoms. So let me draw the carbon atoms in yellow. So this is carbon, carbon, carbon, carbon, carbon, carbon. They have double bonds, every other carbon. Double bonds. And then they have single bonds to hydrogen. Let me just do the hydrogen in another color. Let me do it in magenta. Hydrogen. And obviously, the structural formula gives you the most information. Right? Then you can start to think about, gee, how will this interact with other things? While the molecular formula just tells you what's in the molecule, the empirical formula really gives you the least information. It just tells you the ratio of the different items in the molecule. Structural formula. Let's do a couple more. OK, what if we're dealing with, let's say we're dealing with water. I think you know the molecular formula for water. H2O. Now what would be the empirical formula for this? Well we want to know the ratio, so for every oxygen there's two hydrogens. Or I guess you could say for every hydrogen there's a half of oxygen. So you can't reduce this. If I wrote this as H2O1, what's the greatest common divisor of 2 and 1? Well, it's 1, so you just have to divide them by 1. So in this case, the empirical and the molecular formula are the exact same thing. It's H2O. What about sulfur? And an interesting molecule, because obviously, it's just one atom. Sulfur. Sorry, no, I'm spelling it wrong. No. No, it's not a p h, it's f. Clearly, I shouldn't be making spelling videos. So sulfur. So the molecular formula S8. So it forms this neat kind of octagon-looking chain of sulfurs. And if I were to draw that, you would see that. And you could look it up on Wikipedia, if you like. But its empirical formula, if you had just a bag of sulfur, you don't know that each atom has eight sulfurs. You just have a big bag of sulfur. So, the imperative formula, there's only one atom in this molecule. You divide by eight and you get S. So you just know that all you've got there is sulfur. So let's just do one more. Glucose. I'll pick a new color. Glucose. The molecular formula is C6H12O6. So for every carbon, there are how many? For every 6 carbons, there's 12 hydrogen and 1 oxygen. So if you kind of reduce this formula to its empirical form, what do you get? Let's see, you can divide all these numbers by 6, so we get 1 carbon, 2 hydrogens, and 1 oxygen. So this just tells you the ratio that they exist in a big bag of this molecule. This tells you the exact number of atoms in that molecule. Fair enough. So now we know a little bit of the difference between molecular formula, and empirical formula, and structural formula. Now let's see if we can use what we know about the formulas and the periodic table to think a little bit about the composition, the mass composition, of some of these molecules. So the first thing to even think about is, how do you figure out the molecular mass? Right? I have my little periodic table down there. So molecular mass. So the first question is, how do you figure out-- I mean, the molecular mass is going to be the sum of all of the atoms in that molecule, right? So if you wanted to know the molecular mass of-- let's say you wanted to know how much does one molecule of benzene mass? I don't want to say weight, because it should be independent of what planet you're on. So what is the mass of one molecule of benzene? Well, all you do is you add up the masses of the different constituents. So you have 6 carbons and 6 hydrogens. So let's do benzene. You have 6 carbons and 6 hydrogens. So what's the mass of each carbon? So let's go back down to the periodic table. Just to give proper credit, I got this off of the Los Alamos National Laboratories website. So let's see, the atomic mass of carbon. The reason why I used this one instead of my previous one is my previous periodic table that I got off Wikipedia only had atomic numbers on them. But now that we're actually going to start talking about the mass composition of different atoms or different molecules. We're going to have to start looking at the atomic mass, right? Remember, the atomic mass, when you think about atomic mass units, it's just the number of protons plus the number of neutrons. So you have six protons in carbon and roughly six neutrons. And why is there this decimal? Because, as we said before, this is an average of all of the masses of the isotopes you'll find of carbon. So there's a little bit of carbon 14 on the planet, very little, but most of the carbon is carbon 12. When you proportionately average them, you get 12.01. But let's say we're dealing with carbon 12, just because that's the most common element. Carbon is 12 atomic mass units. Right? And atomic mass units is a unit of mass. And we'll talk about how small it is. It's a very, very, very, very small fraction of a gram or kilogram. And we'll talk about that, probably in the next video. So carbon is 12 atomic mass units. What about hydrogen? Let's see. We go to our periodic table. Hydrogen is here in this dark blue. And I don't know if you can read it, but this is interesting, the atomic number of hydrogen is 1. The atomic mass of hydrogen is 1.0008. So that tells us that most of the hydrogen on this planet has an atomic mass of 1. Which tells us that it essentially has no neutrons. That hydrogen is a kind of an interesting nucleus there, where there is really just a proton. Just a proton sitting in that nucleus. And so if you were to ionize hydrogen. If you were to turn into a cation and take one of its electrons away, what are you left with? You just have a proton. A proton sitting by itself, just a single proton, really is no different than a hydrogen ion. And that to me is kind of interesting. That hydrogen is that simple. It's really just a proton. So hydrogen has an atomic mass of 1. Right? If it had any neutrons in it, it would have been at least an atomic mass of 2. But hydrogen has atomic mass of 1. One atomic mass unit. So what is the mass of one molecule of benzene? Well it's 6 times the carbon mass. 6 times 12 plus 6 times the mass of hydrogen. Plus 6 times 1. So that is 6 times 12, is 72, plus 6 times 1, plus 6, is equal to 78. Now, what if someone said, what percent of benzene is carbon? Well then you say, OK, this is the piece that's carbon, right here. The carbon piece of benzene is 72 atomic mass units. Right, that's carbon. So what percentage of benzene is carbon? Well it's 72 over 78. The whole thing is 78. So it's 72 over 78. And what does that equal? Let me get a calculator going. I should've had my calculator open ahead of time. All right. So 72 divided by 78 is equal to 92.3%. So benzene is 92.3% carbon by mass. And of course, the remainder, the 7.7%, is going to be hydrogen. Let's do that for a couple of these other guys down here. So let's say we wanted to know what is the mass of a molecule of water? Fair enough. There's enough water on the planet, if you want to know what that is. Well we already know what the mass of a hydrogen is. It's 1. Right? Hydrogen is 1. One atomic mass unit. Oxygen is what? Oxygen is 16. Notice, it's exactly 16. So on most of the planet, you pretty much have, in an oxygen atom, you have 8 protons and exactly 8 neutrons. So you get an atomic mass of 16. So oxygen has an atomic mass of 16 atomic mass units. So the atomic mass of the entire molecule, you have 2 hydrogens, so you have 2 times the mass of hydrogen plus 1 oxygen-- plus 16-- so that equals 18 atomic mass units for water. And once again, if you want to say, what percent by mass of water is oxygen? Well it's 16 out of the 18, right? Is oxygen. So if we get the calculator back, you get 16 divided by 18 is water. So, let's say you round it, 88.9% water. Sorry, 88.9% oxygen. So most of water is oxygen. And this is interesting, even though you have two hydrogens here, two hydrogens for every one oxygen, oxygen's mass is so much larger-- it's 16 times larger-- that most of water is oxygen. Well, I'm probably running out of time, so the next video I'm going to talk about how do we go backwards. If someone gives you the composition, how can you get the empirical formula. Actually on a side note, slightly unrelated to what I just talked about, I was doing some research last night about metals, because they're actually interesting, about why some metals conduct more and some conduct less. Because when I first talked about, you know, these were obviously the transition metals. They're backfilling their d orbitals. And I said, hey, the periodic table that was in-- I think I was looking in a Princeton Review book that described these as metals and described these as transition metals. And I was like, hey, you know, that's kind of not fair, because I consider iron and copper and gold and silver to be as metallic as anything. Why should these be called transition metals and these be just called regular metals. And it actually turns out that a common name for these are poor metals. Poor metals. Because, to a large degree, they're softer, they have lower melting points, so the intuition was right. To a large degree, when we think of metals, these are the metals I think of. And when we think of metallic nature in a chemistry sense-- we talked a lot about that, who wants to donate their electrons the most, that's metallic nature. They're the guys down here. And as you go to the top right, these want to donate their electrons the least. These are the most electronegative. They like electrons the most, so they actually have some of the worst metallic nature, so it actually makes sense to call them poor metals. And there's some debate on whether these should even be called poor metals. If you look up a bunch of periodic tables, some will call these metals. Some will call these poor metals. But I just wanted to throw that out there just so you're exposed to it. And so, you know, for me, it is a little bit more intuitive to call these poor metals, because they have less metallic nature than the stuff, especially down here, the alkali and the alkaline earth metals. Anyway, see you in the next video.
2982	https://www.quora.com/Using-the-prime-factorization-method-what-is-the-LCM-of-3-4-and-5	Using the prime factorization method, what is the LCM of 3, 4, and 5? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics LCM and GCF Factorization Methods Least Common Multiple Prime Algorithms Arithmetic Number Theory Unique Prime Factorizatio... Arithmetic Lcm of 8 5 Using the prime factorization method, what is the LCM of 3, 4, and 5? All related (42) Sort Recommended Assistant Bot · 1y To find the Least Common Multiple (LCM) of the numbers 3, 4, and 5 using the prime factorization method, we follow these steps: Find the prime factorization of each number: [math]3[/math] is a prime number: [math]3 = 3^1[/math] [math]4[/math] can be factored into primes: [math]4 = 2^2[/math] [math]5[/math] is also a prime number: [math]5 = 5^1[/math] List the prime factors: The prime factors involved are [math]2[/math], [math]3[/math], and [math]5[/math]. Take the highest power of each prime factor: From [math]3[/math]: [math]3^1[/math] From [math]4[/math]: [math]2^2[/math] From [math]5[/math]: [math]5^1[/math] Multiply these together to get the LCM: [math]\text{LCM} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5[/math] 5. Calculate the result: [math]4 \times 3 = 12[/math] Continue Reading To find the Least Common Multiple (LCM) of the numbers 3, 4, and 5 using the prime factorization method, we follow these steps: Find the prime factorization of each number: [math]3[/math] is a prime number: [math]3 = 3^1[/math] [math]4[/math] can be factored into primes: [math]4 = 2^2[/math] [math]5[/math] is also a prime number: [math]5 = 5^1[/math] List the prime factors: The prime factors involved are [math]2[/math], [math]3[/math], and [math]5[/math]. Take the highest power of each prime factor: From [math]3[/math]: [math]3^1[/math] From [math]4[/math]: [math]2^2[/math] From [math]5[/math]: [math]5^1[/math] Multiply these together to get the LCM: [math]\text{LCM} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5[/math] 5. Calculate the result: [math]4 \times 3 = 12[/math] [math]12 \times 5 = 60[/math] Therefore, the LCM of 3, 4, and 5 is 60. Upvote · Related questions More answers below What is the LCM of 4, 5, and 65 in the prime factorization method? What is the LCM of 10 and 12 using prime factorization? How do I find LCM of three or more numbers using a prime factorization method? How do I find LCM by prime factorization? What is the LCM of the following numbers by the prime factorisation method of 10, 12, and 18? Joiyad Khan B.Tech from National Institute of Technology, Hamirpur (Graduated 2024) · Author has 120 answers and 765.8K answer views ·4y LCM stands for Least Common Multiple. using prime factorization method, first list the prime factors of each number : 3=13 , 4=122, 5=15 If the same factor occurs more than once in all given numbers then multiply the factor the greatest time it occurs. Here no factor occurs more than once in all the numbers so, LCM=345 LCM=60 I hope you will find this helpful👍👍 Upvote · William Stern Hospital Custodian (2006–present) · Author has 975 answers and 513.8K answer views ·4y The LCM (Least Common Multiple) of two or more numbers is the smallest product that all of those numbers divide into equally. The integers 3, 4, and 5 have no factors in common, so the LCM will be the product of all three, i.e. 3 [math]\times[/math] 4 [math]\times[/math] 5 = 60. Upvote · Ryan Bon Former English Teacher at Abies English Room (2017–2018) ·6y Originally Answered: What is the LCM of 3, 5 & 6? · Here’s a pretty cool trick (for me at least): Step 1. Factor out the numbers into prime numbers. 3=3x1 5=5x1 6=3x2 Step 2. Once you break the numbers into prime numbers (numbers that cannot be formed by multiplying two smaller natural numbers), multiply the factors altogether BUT consider the repeating factor as only one digit. LCM= 1 x 2 x 3 x 5 = 60 Since 1 and 3 can be found on the three numbers two times, they may only be considered as one digit. And this applies for all cases of getting the LCM. Hope this helps, goodluck! Upvote · 9 3 9 3 Related questions What is the LCM of 4, 5, and 65 in the prime factorization method? What is the LCM of 10 and 12 using prime factorization? How do I find LCM of three or more numbers using a prime factorization method? How do I find LCM by prime factorization? What is the LCM of the following numbers by the prime factorisation method of 10, 12, and 18? Can you use the prime factorization method to find the LCM of 12, 16, and 20? What is the sequence LCM of the following by prime factorization of 12, 48, and 60? What is the prime factorization to find the LCM of 16 and 36? What is the prime factorization of the LCM of 15, 18, and 24? What is the approximate LCM of 12, 15, and 18 by the prime factorisation method? What is the LCM of the following by the prime factorization method 91, 65, 75, and 39? What are the HCF and LCM of 86 and 36 by using the prime factorisation method? What is the prime factorization method LCM 15 20? What is the LCM of 10, 12, and 18 by the process of prime factorization? What are the LCM and HCF of 15, 18, and 45 by the prime factorization method? Related questions What is the LCM of 4, 5, and 65 in the prime factorization method? What is the LCM of 10 and 12 using prime factorization? How do I find LCM of three or more numbers using a prime factorization method? How do I find LCM by prime factorization? What is the LCM of the following numbers by the prime factorisation method of 10, 12, and 18? Can you use the prime factorization method to find the LCM of 12, 16, and 20? What is the sequence LCM of the following by prime factorization of 12, 48, and 60? What is the prime factorization to find the LCM of 16 and 36? What is the prime factorization of the LCM of 15, 18, and 24? What is the approximate LCM of 12, 15, and 18 by the prime factorisation method? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
2983	https://math.stackexchange.com/questions/525914/solving-equations-where-the-solution-is-an-operator	Skip to main content Solving equations where the solution is an operator Ask Question Asked Modified 7 years, 3 months ago Viewed 2k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. Ok, so here's some context. Solving regular equations we might have something like this: 2+x=5, solving for x we get 3. We might even have an equation like x+y=5 where there are multiple solutions. But what's in common with all these equations is that the process, or the algorithm, we follow to solve them is determined by the operators that show the relations between variables. Now if you think back to very early elementary school, I'm sure you solved stuff like 5_2=3 where you would fill in the blank with a minus. I'm wondering if there's a branch or mathematics that studies actual systematical ways to solve "equations" like that. It might seem trivial from this example but it obviously would grow in complexity. Perhaps the solutions would be operators AND numbers. The concrete problem I thought up that led me to this was trying to find a way to "map" any 1n to 1n+1 by just adding/multiplying/something the first fraction by a constant. The "equation" would look something like this: 1n_C=1n+1 I can solve similar problems in my head, like: n_C=n+1 where the obvious solution is + and 1. Or for example, n_C=2n where the solution is × and 2. The last one can also have + as a solution but then C would have to be n and would no longer be a constant. (I didn't know what to put as the tag) functional-analysis Share CC BY-SA 3.0 Follow this question to receive notifications edited Feb 6, 2014 at 21:33 Luka Horvat asked Oct 14, 2013 at 15:05 Luka HorvatLuka Horvat 2,67811 gold badge2222 silver badges3333 bronze badges 1 Darwin I would be interested in discussing this over emails. Please respond to barrydeer@quiknet.com respectfully your Barry Deer, Ph.D. – user126573 Commented Feb 5, 2014 at 23:51 Add a comment \| 2 Answers 2 Reset to default This answer is useful 1 Save this answer. Show activity on this post. Equations where "operations" (really just a funny notation for functions) are the unknowns are called functional equations. Solving them can range from the trivial (like the examples you give) to very complex. Some techiques are given in the Wikipedia link above. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Feb 6, 2014 at 20:59 vonbrandvonbrand 28.4k66 gold badges4545 silver badges7979 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. I'm not being funny, but I think the branch of mathematics that deals with what you're talking about is simply algebra, or algebraic manipulation. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Oct 14, 2013 at 18:17 George TomlinsonGeorge Tomlinson 1,37611 gold badge88 silver badges1111 bronze badges Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions functional-analysis See similar questions with these tags. 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2984	https://math.stackexchange.com/questions/107113/find-the-sign-of-int-02-pi-frac-sin-xx-dx	calculus - Find the sign of $\int_{0}^{2 \pi}\frac{\sin x}{x} dx$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find the sign of ∫2 π 0 sin x x d x∫0 2 π sin⁡x x d x Ask Question Asked 13 years, 7 months ago Modified4 years, 2 months ago Viewed 3k times This question shows research effort; it is useful and clear 8 Save this question. Show activity on this post. I'd love your help with finding the sign of the following integral: ∫2 π 0 sin x x d x∫0 2 π sin⁡x x d x I know that computing it is impossible. I tried to use integration by parts and maybe to learn about the sign of each part and conclude something but It didn't work for me. Any suggestions? calculus Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jul 22, 2021 at 17:52 Christian Chapman 5,046 2 2 gold badges 28 28 silver badges 46 46 bronze badges asked Feb 8, 2012 at 16:30 JozefJozef 7,238 5 5 gold badges 55 55 silver badges 91 91 bronze badges 3 Do you want an approximate?Pedro –Pedro♦ 2012-02-08 16:34:28 +00:00 Commented Feb 8, 2012 at 16:34 Wolfram alpha will give you an approximation. (by the way it's positive)Bill Cook –Bill Cook 2012-02-08 16:34:55 +00:00 Commented Feb 8, 2012 at 16:34 2 No, as I said I want to find only the sign, I want to find it by a "calculus- investigation", and not by checking it on W.A :) Thanks!Jozef –Jozef 2012-02-08 16:38:27 +00:00 Commented Feb 8, 2012 at 16:38 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 31 Save this answer. Show activity on this post. ∫2 π 0 sin x x d x=∫π 0 sin x x d x+∫2 π π sin x x d x=∫π 0 sin x x d x+∫π 0 sin(x+π)x+π d x=∫π 0(1 x−1 x+π)sin x d x=π∫π 0 sin x x(x+π)d x>0∫0 2 π sin⁡x x d x=∫0 π sin⁡x x d x+∫π 2 π sin⁡x x d x=∫0 π sin⁡x x d x+∫0 π sin⁡(x+π)x+π d x=∫0 π(1 x−1 x+π)sin⁡x d x=π∫0 π sin⁡x x(x+π)d x>0 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jul 22, 2021 at 17:51 Christian Chapman 5,046 2 2 gold badges 28 28 silver badges 46 46 bronze badges answered Feb 8, 2012 at 16:43 Julián AguirreJulián Aguirre 77.9k 2 2 gold badges 65 65 silver badges 120 120 bronze badges 1 Thanks Julian! :) very nice solution!Jozef –Jozef 2012-02-08 16:49:00 +00:00 Commented Feb 8, 2012 at 16:49 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Regarding the sign, it is easy the check that every area in each π π interval is always smaller than the preceeding one. The sign is positive. For the value, integrate in the same interval y=cos x 2 cos x 4 cos x 8 y=cos⁡x 2 cos⁡x 4 cos⁡x 8 The difference between the sinc function and that is at most ≈0.015≈0.015 in that interval. Adding cos x 16 cos⁡x 16 makes the error at most ≈0.003≈0.003 For the first one you have. I=104 105 2–√∼2–√I=104 105 2∼2 For the latter: I=1568 2145 2+2–√−−−−−−√2∼2–√I=1568 2145 2+2 2∼2 Maybe the area is 2–√2 after all. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 8, 2012 at 16:44 answered Feb 8, 2012 at 16:38 Pedro♦Pedro 126k 19 19 gold badges 239 239 silver badges 406 406 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Let f(x)=sin(x)/x f(x)=sin⁡(x)/x. So f(x)=0 f(x)=0 when sin(x)=0 sin⁡(x)=0. So the only solution in the interval (0,2 π)(0,2 π) is x=π x=π. Plugging in test points shows that f(x)f(x) is positive to the left of x=π x=π and negative to the right. Next, if you accept that 1−x/3≤sin(x)/x 1−x/3≤sin⁡(x)/x (make some argument using MacLaurin series), then ∫π 0 f(x)d x≥1 2(3)(1)=3/2∫0 π f(x)d x≥1 2(3)(1)=3/2. On the other hand \|sin(x)\|/x≤\|sin(x)\|/3\|sin⁡(x)\|/x≤\|sin⁡(x)\|/3 for x≥3 x≥3 so ∫2 π π\|f(x)\|d x≤∫2 π π\|sin(x)\|3 d x=2/3∫π 2 π\|f(x)\|d x≤∫π 2 π\|sin⁡(x)\|3 d x=2/3. So ∫2 π 0 f(x)d x≥3/2−2/3>0∫0 2 π f(x)d x≥3/2−2/3>0 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 8, 2012 at 16:45 Bill CookBill Cook 30.4k 77 77 silver badges 95 95 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 1Find the sign of the following integral: ∫2 π 0 sin x x∫0 2 π sin⁡x x. Related 9Differentiating under integral sign -- trig counterexample 9Prove: Fourier series of e cos x sin(sin x)e cos⁡x sin⁡(sin⁡x) is ∑∞n=0 sin(n x)n!∑n=0∞sin⁡(n x)n! 2f f decreasing monotonically decreasing , Prove: for every α∈(0,1)α∈(0,1) :∫α 0 f(x)d x≥α∫1 0 f(x)d x∫0 α f(x)d x≥α∫0 1 f(x)d x 2Compute ∫π 2 0 sin 2 x cos x sin x+cos x d x∫0 π 2 sin 2⁡x cos⁡x sin⁡x+cos⁡x d x 6Integrate ∫ln(sin x)sin 2 x d x.∫ln⁡(sin⁡x)sin 2⁡x d x. 4Evaluate ∫d x 1+sin x+cos x∫d x 1+sin⁡x+cos⁡x 6How do I solve ∫5−5 x 3 sin 2 x x 4+2 x 2+1 d x∫−5 5 x 3 sin 2⁡x x 4+2 x 2+1 d x? 2Calculate ∫π/2 0(sin(2 x))5 d x∫0 π/2(sin⁡(2 x))5 d x Hot Network Questions How to rsync a large file by comparing earlier versions on the sending end? Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish? Does "An Annotated Asimov Biography" exist? Alternatives to Test-Driven Grading in an LLM world How to locate a leak in an irrigation system? 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2985	https://www.youtube.com/watch?v=SDr44tqdVfA	Geometric Sequence: Sum vs. Product vs. Reciprocal Sum [Sequence Week #2] Cornerstones of Math 1790 subscribers 10 likes Description 352 views Posted: 26 Jul 2022 In my last video I have handled the arithmetic sequence, so today let us take a look at the geometric sequence - the sequence with common ratio between consecutive terms. Here, Sn is the sum of first n terms of the geometric sequence, Pn is the product of first n terms, and Hn is the sum of reciprocals of first n terms. Then, what would be the relation between Sn, Pn, and Hn? My other videos of the Sequence Week: - Express 1000 as a Sum of Consecutive Natural Numbers - A Partial Sum of 1^3, －2^3, 3^3, －4^3, 5^3, －6^3, ...... - A Simple Recurrence Relation Sequences #GeometricSequences Cornerstones Of Math features quality math problems that can strengthen both your knowledge of the fundamentals of mathematics and your ability of problem-solving. I sincerely hope that the content provided by this channel provide you some intellectual pleasure and make you appreciate the beauty of math itself. Please consider a like to this video and subscribing to my channel, it really means a lot for the creator like me, and you can be introduced into many more math videos! Transcript: hello and welcome to my video this is the second video of the sequence week series so if you haven't seen my first video of the week please check that in top right corner today we are going to solve the following problem consider a geometric sequence a n with common ratio r is not equal to 1 and all terms of a n are non-zero now define s n as simply the partial sum the sum of the first n terms of the given geometric sequence a n and p n is the product of the first n terms of the sequence a n and finally h n is sum of the reciprocals of the each term of sequence a n right so each definition is pretty clear now what is the relation between s n p n and h n how these three sums or products are related okay let's find that out we want the simplest expression so let's just let a1 as a then we already have a famous formula for sn the partial sum of the geometric sequence the first term is a the second term is a r and the third term is a r squared and the nth term is a r to the power of n minus one and this sum is given as a times r to the power of n minus 1 over r minus 1. so this a is first term and this r is common ratio of the geometric sequence and this n means number of terms in this case n this is the famous formula for the finite sum of the geometric sequence if the magnitude of the common ratio r is less than 1 and if we let n goes to infinity then we obtain the limit of sn as simply a over 1 minus r the famous formula for the geometric series right so this is the expression for s n now let's take a look at p n the finite product here the first term is a and the second term is a r and the third term is a r square and the nth term is a r to the power of n minus one therefore if you count the number of a's we have total n a's so we have a to the power of n then we have r times r square times all the way up to r to the power of n minus one so their exponents must be added so therefore we have r to the power one plus two plus data dot plus and finally n minus one now this is the sum of arithmetic sequence which is simply n times m minus 1 over 2. so therefore we have a to the power of n times r to the power of n n minus 1 over 2. so this is the expression for p n now what about h n again if we write each term of h n we have one over a and one over the second term a two is a r and one over the third term is a r square plus a dot plus and one over a n is a times r to the power of n minus one therefore the sequence one over a n is also a geometric sequence with 1 over a as the first term and 1 over r as the common ratio therefore for its partial sum h n we can apply the similar formula for sn with only different first term and a common ratio therefore h and is the first term is now 1 over a and the common ratio is now 1 over r so we have 1 over r to the power of n here minus 1 over 1 over r minus 1. now let's multiply r to the power of n at both numerator and denominator then we have for this a let's send it down to the denominator and at the numerator we simply have 1 minus r to the power of n and in the denominator we have r to the power of n minus 1 minus r to the power of n so we have 1 minus r to the power n and if we extract r to the power of n minus 1 here then we have 1 minus r here i'll notice something here the sn part appears in the expression namely in this part so this part this part n plus a consists of s n and since we have a here the denominator now must be a square and we have this part remaining so we have r to the n minus one so this is the expression for h n so if you rewrite it h n and s n are related like this and p n is given as a to the n r to the n n minus one over two so from here let's send this h n down to the denominator and move these terms up here then s n over h n is given as a square r to the n minus 1. now let's raise to the power n then we have s n over h n to the power of n is if you multiply n here we have a to the power of 2 n times r to the power of n times n minus 1. and now you can see that the expression for p n appears so this is basically a to the n r to the n n minus 1 over 2 squared which is p n squared therefore we finally have our relation which is p n square equals s n over h n to the power of n and this is the relation that we're looking for now some of you might want to take square root of this expression to write pn equals sn over hn to the power of n over 2 which is equivalent to square root of sn over hn to the power of n but this is not always true because you see if you want to write like this then it must be guaranteed that both p n and s n over h n to the power of n must be non-negative but that might not always be the case if you only take a look at p n here you can already see that p n might not always be positive more specifically if p n contains even number of negative terms then p n is positive and if p n contains all the number of negative terms then p n becomes negative so remember that a squared equals b square does not always mean a equals b and that's all for this video again if you haven't seen my previous video about sequence please check that in top right corner and also please like and subscribe to my channel to explore more math videos and i will see you in another video
2986	https://flexbooks.ck12.org/cbook/ck-12-interactive-algebra-1-for-ccss/section/1.4/related/lesson/rates-of-change-bsc-alg/	Skip to content Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? Science Grade K to 5 Earth Science Life Science Physical Science Biology Chemistry Physics Advanced Biology FlexLets Math FlexLets Science FlexLets English Writing Spelling Social Studies Economics Geography Government History World History Philosophy Sociology More Astronomy Engineering Health Photography Technology College College Algebra College Precalculus Linear Algebra College Human Biology The Universe Adult Education Basic Education High School Diploma High School Equivalency Career Technical Ed English as 2nd Language Country Bhutan Brasil Chile Georgia India Translations Spanish Korean Deutsch Chinese Greek Polski EXPLORE Flexi A FREE Digital Tutor for Every Student FlexBooks 2.0 Customizable, digital textbooks in a new, interactive platform FlexBooks Customizable, digital textbooks Schools FlexBooks from schools and districts near you Study Guides Quick review with key information for each concept Adaptive Practice Building knowledge at each student’s skill level Simulations Interactive Physics & Chemistry Simulations PLIX Play. Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up Back To Piecewise Linear FunctionsBack 1.4 Rates of Change Written by:Andrew Gloag \| Melissa Kramer \| Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Suppose a new gym opened, and it had 20 members after 1 week, 40 members after 2 weeks, and 60 members after 3 weeks. Could you calculate the rate of change in the number of gym members? How is this different than the slope? If this rate continues, how long will it take for the gym to have 300 members? Finding the Rate of Change When finding the slope of real-world situations, it is often referred to as rate of change. “Rate of change” means the same as “slope.” If you are asked to find the rate of change, use the slope formula or make a slope triangle. Let's use rate of change to solve the following problems: Andrea has a part-time job at the local grocery store. She saves for her vacation at a rate of $15 every week. What is the rate of change in the amount of money Andrea has? Begin by finding two ordered pairs. You can make a chart or use the Substitution Property to find two coordinates. Using the points (2, 30) and (10, 150). Since (2, 30) is written first, it can be considered the first point, (x1, y1). That means (10,150)=(x2, y2). (Note that it doesn't matter at all which point is considered first, as the slope will end up the same either way.) Use the formula: slope=y2−y1x2−x1=150−3010−2=1208=151 Andrea’s rate of change is $151 week A candle has a starting length of 10 inches. Thirty minutes after lighting it, the length is 7 inches. Determine the rate of change in the length of the candle as it burns. How long does it take for the candle to completely burn to nothing? Begin by finding two ordered pairs. The candle begins at 10 inches in length. So at time “zero”, the length is 10 inches. The ordered pair representing this is (0, 10). 30 minutes later, the candle is 7 inches, so the ordered pair is (30, 7). Since (0, 10) is written first, it can be called (x1, y1). That means (30,7)=(x2, y2). Use the formula: slope=y2−y1x2−x1=7−1030−0=-330=-110 The candle has a rate of change of -1 inch/10 minutes (the rate is negative because the candle is getting shorter over time). To find the length of time it will take for the candle to burn out, you can create a graph, use guess and check, or solve an equation. You can create a graph to help visualize the situation. By plotting the ordered pairs you were given and by drawing a straight line connecting them, you can estimate it will take 100 minutes for the candle to burn out. Examine the following graph. It represents a journey made by a large delivery truck on a particular day. During the day, the truck made two deliveries, each one taking one hour. The driver also took a one-hour break for lunch. Identify what is happening at each stage of the journey (stages A through E). Here is the driver's journey. A. The truck sets off and travels 80 miles in 2 hours. B. The truck covers no distance for 1 hour. C. The truck covers (120−80)=40 miles in 1 hour. D. The truck covers no distance for 2 hours. E. The truck covers 120 miles in 2 hours. To identify what is happening at each leg of the driver’s journey, you are being asked to find each rate of change. The rate of change for line segment A can be found using either the formula or the slope triangle. Using the slope triangle, vertical change=80 and the horizontal change=2. slope=riserun=80 miles2 hours=40 miles1 hour. Segments B and D are horizontal lines and each has a slope of zero. The rate of change for line segment C using the slope formula: Rate of change=△y△x=(120−80) miles(4−3) hours=40 miles per hour. The rate of change for line segment E using the slope formula: Rate of change=△y△x=(0−120) miles(8−6) hours=-120 miles2 hours=-60 miles per hour. The truck is traveling at negative 60 mph. Another way to say this is that the truck is returning home at a rate of 60 mph. Examples Example 1 Earlier, you were told that a new gym had 20 members after 1 week, 40 members after 2 weeks, and 60 members after 3 weeks. What is the rate of change in the number of gym members? If this rate continues, how long will it take for the gym to have 300 members? We have three points to use to calculate the rate of change: (1, 20), (2, 40), and (3, 60). Since only two points are needed to find the rate of change, you can pick any two points and use the slope formula. Note that this only works when the points represent a constant rate of change or a linear equation. slope=y2−y1x2−x1=40−202−1=201=20. Therefore the rate of change is 20 members per week. If you choose two other points out of the three, you will get the same result. Two determine how long it will take for the gym to have 300 members, you can use proportions: 20 members1 week=300 membersx weeks Solve for x using cross multiplication: 201=300x20×x=300×120x=30020x÷20=300÷20x=15 It will take 15 weeks for the gym to have 300 members if it continues at a rate of adding 20 members per week. Example 2 Adel spent $125 on groceries in one week. Use this to predict how much Adel will spend per month on groceries, if she keeps buying them at the same rate. Adel's weekly rate is $125 per week. Since a month is about 4 weeks, multiply $125/week times 4 weeks: $1251 week⋅4 weeks=$1251 week⋅4 weeks=$125⋅4=$500 Adel will spend about $500 on groceries per month. Review How is slope related to rate of change? In what ways is it different? The graph below is a distance-time graph for Mark’s 3.5-mile cycle ride to school. During this particular ride, he rode on cycle paths but the terrain was hilly. His speed varied depending upon the steepness of the hills. He stopped once at a traffic light and at one point he stopped to mend a tire puncture. Identify each section of the graph accordingly. Four hours after she left home, Sheila had traveled 145 miles. Three hours later she had traveled 300 miles. What was her rate of change? Jenna saves $60 every 212 weeks. What is the rate of change of her savings? Geoffrey has a rate of change of 10 feet1 second. Write a situation that could fit this slope. Quick Quiz Find the intercepts of 3x+6y=25 and graph the equation. Find the slope between (8, 5) and (–5, 6). Graph f(x)=2x+1 Name the coordinates of a point that is 4 units west and 6 units north of the origin. Using the graph below, list two “trends” about this data. A trend is something you can conclude about the given data. Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. \| Image \| Reference \| Attributions \| --- Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? No Results Found Your search did not match anything in . Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)26/ 100 This lesson has been added to your library. \|Searching in: \| \| \| Looks like this FlexBook 2.0 has changed since you visited it last time. 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This 5th Grade Fractions Adding&Subtracting Fractions with Unlike Denominators Unit includes over 80 page! Adding&subtracting mixed numbers with unlike denominators are included in this resource. This bundle makes the perfect supplement to your core curriculum or a great resource for math intervention. Best of all, this bundle requires NO PREP! Just print and teach!This 80-page bundle of 5th Grade Fractions Worksheets covers:Day 1. Estimating S 5 th - 7 th Fractions, Math, Math Test Prep CCSS 4.NF.B.3 , 5.NF.A.1 , 5.NF.A.2 Bundle (10 products) $31.50 Original Price $31.50 $18.90 Price $18.90 Rated 4.86 out of 5, based on 238 reviews 4.9(238) Add to cart Wish List Adding and Subtracting Mixed Numbers Word Problem Task Cards - Set of 28 Created by Jersey Girl Gone South This set of 28 task cards includes adding and subtracting mixed number word problem task cards. Each task card allows students to practice adding and subtracting mixed numbers with like denominators. These task cards are aligned to 4th grade Common Core standards. Includes: - set of 28 task cards - student recording sheet - answer key These task cards cover Common Core standard: CCSS.Math.Content.4.NF.B.3d Solve word problems involving addition and subtraction of fractions referring to the sam 3 rd - 5 th Basic Operations, Fractions, Math CCSS 4.NF.B.3d Also included in:4th Grade Fractions Mega Task Card Bundle - Common Core $2.50 Original Price $2.50 Rated 4.85 out of 5, based on 522 reviews 4.9(522) Add to cart Wish List Adding and Subtracting Fractions &Mixed Numbers Word Problems {No Prep} Created by Jennifer Findley Adding and Subtracting Fractions and Mixed Numbers Word Problem RTI No Prep Resource! This resource is for unlike denominators only. Click here for the bundle of all 5 of my Word Problem No Prep Intervention Resources for 5th Grade. Word Problem PracticeThis resource includes 30 word problems that focus on adding and subtracting fractions and mixed numbers with unlike denominators. (15 word problems for fractions and 15 for mixed numbers) I created this resource to use as an intervention for my 5 th - 6 th Fractions, Math, Other (Math) Also included in:5th Grade Word Problems BUNDLE with Digital Word Problems $3.00 Original Price $3.00 Rated 4.8 out of 5, based on 181 reviews 4.8(181) Add to cart Wish List Adding&Subtracting Fractions with Like Denominators Word Problems Mixed Number Created by Angela Dansie This Adding and Subtracting Fractions with Like Denominators lesson workbook walks students through addition and subtraction of fractions with the same denominator, mixed numbers, and word problems. It includes an example IEP goal, 19 step-by-step math lessons, reviews and assessments. This evidence-based math intervention is tied to fourth grade standards and is great for special education math goals and tier 2 small group math interventions (RTI). Step-by-Step Math to Mastery™ 4 th Fractions, Math CCSS 4.NF.B.3a , 4.NF.B.3c , 4.NF.B.3d Also included in:Special Education Math Curriculum K-5, Tier 3 Small Group Math Intervention $10.00 Original Price $10.00 Rated 4.91 out of 5, based on 56 reviews 4.9(56) Add to cart Wish List Adding&Subtracting Mixed Number with Unlike Denominator Word Problem Worksheet Created by Sally Witherspoon Math Are your students struggling with adding and subtracting mixed numbers word problems? Then this Adding and Subtracting Mixed Numbers with Unlike Denominators packet is foryou! 8 pages of adding and subtracting mixed number worksheets- perfect for math intervention or as a supplement to your core curriculum. Best of all, this word problem packet requires NO PREP! Just print and teach!The focus of this lesson is on word problems involving adding&subtracting mixed numbers with unlike denomi 5 th - 7 th Fractions, Math Test Prep, Other (Math) CCSS 5.NF.A.2 Also included in:Add & Subtract Fractions with Unlike Denominators Word Problems Mixed Numbers $3.50 Original Price $3.50 Rated 4.91 out of 5, based on 50 reviews 4.9(50) Add to cart Wish List Add&Subtract Mixed Numbers Word Problems - Math Scavenger Quest Created by The Math Detective Scavenger hunt activities are great ways for students to have a great time and practice critical math skills. The best part is that they are self-checking! Students that make a mistake will quickly realize that there is a problem when they either return to the original card far sooner than they should or are unable to find their answer as they search the cards. This also encourages error analysis and the very important skill of double checking answers! This Scavenger Quest allows students to p 4 th - 9 th Fractions, Math, Other (Math) CCSS 5.NF.A.1 , 5.NF.A.2 $3.00 Original Price $3.00 Rated 4.81 out of 5, based on 128 reviews 4.8(128) Add to cart Wish List Adding and Subtracting Mixed Numbers Multi-Step Word Problems 5.NF.1 & 5.NF.2 Created by Reincke's Education Store The 1st worksheet can be purchased as a Google Form at the following link:Adding and Subtracting Mixed Numbers Multi-Step Word Problems Google FormMake multi-step word problems involving the addition and subtraction of mixed numbers mastery simple. Print these 5th or 6th grade math worksheets and watch them flourish. This is a rigorous resource that can be used for homework, assessment, sub plans, review, extra practice, remediation, enrichment, or centers. This product contains 5 worksheets wit 5 th - 7 th Basic Operations, Fractions, Other (Math) CCSS 5.NF.A.1 , 5.NF.A.2 Also included in:Adding and Subtracting Mixed Numbers Bundle (14 products) SAVE 50%! $3.00 Original Price $3.00 Rated 4.86 out of 5, based on 109 reviews 4.9(109) Add to cart Wish List 5th Grade Fraction Review Word Problems Adding Subtraction and Mixed Numbers Created by Think Grow Giggle Fractions for fifth grade made simple! Help your students get a deeper understanding of solving fraction word problems with these error analysis and critical thinking math problems. Students must solve the problem and then decide if they agree or disagree with the solution presented to them. Your students will love using the agree and disagree cards to critique the work of others and defend their own mathematical thinking! Save 20% by purchasing the Agree/Disagree Math Problem Solving BundleTh 5 th Fractions, Math Test Prep, Other (Math) CCSS 5.NF.A.1 , 5.NF.A.2 , 5.NF.B.3 +14 Also included in:5th Grade Math Worksheets Packet Error Analysis Word Problems Activities Centers $4.00 Original Price $4.00 Rated 4.81 out of 5, based on 54 reviews 4.8(54) Add to cart Wish List Adding&Subtracting Fractions and Mixed Numbers Word Problems \| Solve and Snip® Created by Smith Curriculum and Consulting ✂️ Turn fraction word problem practice into an interactive, self-checking activity your students will love!With Adding&Subtracting Fractions and Mixed Numbers Solve and Snip®, students solve 10 real-world word problems involving fractions and mixed numbers, then match each problem to the correct answer from a provided set. The built-in self-checking format means they’ll know immediately if their solutions are correct — no extra grading for you! Perfect for math centers, partner work, revie 5 th - 7 th Basic Operations, Fractions, Math CCSS 5.NF.A.1 , 5.NF.A.2 , 7.NS.A.1d Also included in:5th Grade Solve and Snip Bundle - Interactive Self-Checking Word Problems $1.50 Original Price $1.50 Rated 4.81 out of 5, based on 218 reviews 4.8(218) Add to cart Wish List Adding and Subtracting Mixed Numbers Multi-Step Word Problems 5th Grade GOOGLE Created by Tanya Yero Teaching THIS RESOURCE COMES IN DIGITAL AND PRINT FORMAT. THE DIGITAL VERSION IS COMPATIBLE WITH GOOGLE CLASSROOM/GOOGLE SLIDES! Maximize the versatility of task cards to increase productivity in your classroom.What are task cards?Task cards are meant to assess procedural and conceptual understanding of the curriculum you taught to your students. Designed to challenge students, task cards will provide authentic feedback to teachers on student mastery. How can I use task cards?➥ Centers/Rotations work ➥ C 5 th - 6 th Fractions, Math, Other (Math) CCSS 5.NF.A.1 Also included in:5th Grade Math Word Problems Task Cards Multi-step Math Spiral Review $3.00 Original Price $3.00 Rated 4.82 out of 5, based on 109 reviews 4.8(109) Add to cart Wish List Adding and Subtracting Fractions &Mixed Numbers Word Problems Pixel Art Created by Pick Up and Go Resources This No-Prep Fractions and Mixed Numbers Google Sheets™ pixel art activity is what you need for paperless practice. Review Adding&Subtracting Fractions and Mixed Numbers with Unlike Denominators Word Problems with this digital self-checking pixel art. It is fun, engaging and interactive. This digital self-checking fractions activity is a great no-prep activity you could use as a review, bell ringer or in place of a traditional worksheet. Automatic grading and coloring make this activity a 4 th - 6 th Fractions, Math CCSS 5.NF.A.2 Also included in:Adding and Subtracting Fractions and Mixed Numbers Unlike Denominators Pixel Art $3.25 Original Price $3.25 Rated 4.71 out of 5, based on 21 reviews 4.7(21) Add to cart Wish List Adding and Subtracting Fractions Word Problems \| Unlike Fractions Mixed Numbers Created by Hello Learning Adding and subtracting fractions in word problems is fun when you use this set of fractions task cards. Includes word problems for adding and subtracting fractions with unlike denominators and mixed numbers. A great way to get kids moving! Click the green PREVIEW button to see what is included. This easy to use resource includes:28 task cards task cards are provided in color and black and white versionsstudent response sheetanswer keysuggestions for preparing, using and storing task cardsSkills 4 th - 6 th Fractions, Math, Other (Math) CCSS 5.NF.A.1 , 5.NF.A.2 Also included in:Adding and Subtracting Fractions BUNDLE $3.00 Original Price $3.00 Rated 4.94 out of 5, based on 17 reviews 4.9(17) Add to cart Wish List Adding and Subtracting Fractions &Mixed Numbers Word Problems (3 worksheets!) Created by Reincke's Education Store Purchase part of this product on adding and subtracting mixed numbers as a Google Form at the link below:Adding&Subtracting Mixed Numbers Word Problem Google FormsNeed something to practice real-world situations with adding and subtracting fractions and mixed numbers with unlike denominators for you 5th or 6th grade math students? This product contains 3 worksheets: one worksheet on adding and subtracting fractions (8 questions) and two worksheets with mixed numbers (6 questions per worksheet) 4 th - 7 th Basic Operations, Fractions, Other (Math) CCSS 5.NF.A.1 , 5.NF.A.2 Also included in:5th Grade Math Bundle (130 products) SAVE 50%! $2.00 Original Price $2.00 Rated 4.84 out of 5, based on 61 reviews 4.8(61) Add to cart Wish List Adding and Subtracting Fractions and Mixed Numbers Word Problems Boom Cards Created by Alyssa Teaches Word problems featuring adding and subtracting fractions and mixed numbers is fun and rigorous for students with these interactive, paperless task cards! This digital activity works well for centers, early finishers, morning work, and assessment. Use during a fractions unit, for spiral review, or for test prep! Single-step and multistep word problems include proper fractions, improper fractions, and mixed numbers with like and unlike denominators limited to 2, 3, 4, 5, 6, 8, 10, and 12. All answ 4 th - 6 th Fractions, Math, Other (Math) CCSS 5.NF.A.2 $2.20 Original Price $2.20 Rated 4.75 out of 5, based on 12 reviews 4.8(12) Add to cart Wish List Adding and Subtracting Mixed Numbers With Unlike Denominators Word Problems Sort Created by Keep On Growing Are your students struggling with adding and subtracting mixed numbers with unlike denominators? This 5th grade fractions word problem sort provides practice on adding and subtracting fractions/mixed numbers with different denominators using real life scenarios. Students will identify which operation is needed to solve the word problems: Adding or Subtracting Mixed Numbers. Adding and subtracting word problems can be difficult, but this hands-on resource will help your students master it. ⭐️ CCS 5 th - 6 th Math CCSS 5.NF.A.1 $2.25 Original Price $2.25 Rated 4.91 out of 5, based on 11 reviews 4.9(11) Add to cart Wish List Adding&Subtracting Fractions &Mixed Numbers Word Problems Scavenger Hunt Game Created by Sherri Miller - The Techie Chick Adding and subtracting fractions and mixed numbers word problems for 4th grade are more exciting and motivating with this digital math scavenger hunt activity! With a Bakery theme, your fourth graders will practice fractions skills while they search for hidden links and solve the secret message. This NO Prep, easy to use activity is both digital and print. It’s self-checking which means NO grading for you! ➡️ Click the PREVIEW above to take a closer look! Fractions skills include:✅ Adding 4 th Fractions, Math Test Prep, Other (Math) CCSS 4.NF.B.3a , 4.NF.B.3c , 4.NF.B.3d Also included in:4th Grade Math Word Problems Scavenger Hunt Bundle - Word Problem Practice Games $3.00 Original Price $3.00 Rated 5 out of 5, based on 4 reviews 5.0(4) Add to cart Wish List Adding and Subtracting Fractions &Mixed Numbers Test Prep (word problems) Created by Reincke's Education Store Make adding and subtracting fractions and mixed numbers mastery simple. Print these worksheets and watch your students flourish. This rigorous resource that can be used by Common Core and non-Common Core states. I've heard it's been used for homework, assessment, sub plans, review, extra practice, remediation, enrichment, test prep, or centers. This product contains 7 worksheets. All word problems revolve around adding and subtracting mixed numbers with unlike denominators. Worksheets 1-2: Addi 4 th - 7 th Basic Operations, Fractions, Other (Math) CCSS 5.NF.A.1 , 5.NF.A.2 Also included in:5th Grade Math Bundle (130 products) SAVE 50%! $2.50 Original Price $2.50 Rated 4.94 out of 5, based on 41 reviews 4.9(41) Add to cart Wish List 4th Grade Add&Subtract Mixed Number Fraction Word Problems\|BOOM Cards\|4.NF.B3d Created by Math MindEd Teaching Use these digital, self-grading, interactive BOOM DECK TASK CARDS to help your 4th graders solve FRACTION WORD PROBLEMS. In fourth grade students must be able to solve WORD PROBLEMS involving addition and subtraction of FRACTIONS referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem. In this activity, students will solve each word problem on the task cards. Includes:20 Total Word Problem Cards10 Fraction Addition Wo 4 th Fractions, Math CCSS 4.NF.B.3d Also included in:4th Grade Fractions BUNDLE \| 4.NF.A & 4.NF.B \| BOOM Cards $3.00 Original Price $3.00 Rated 4.17 out of 5, based on 6 reviews 4.2(6) Add to cart Wish List Addition and Subtraction of Mixed Numbers Scavenger Hunt - Word Problems Created by Branigans Beehive This fun and exciting scavenger hunt gets students up and moving around the room in search for the correct answer to addition and subtraction word problems with mixed numbers. It's a great way for students to practice and review with mixed number fractions as well as assess their own work as they rotate around the room! 3 rd - 6 th Fractions, Math CCSS 4.NF.B.3 $3.00 Original Price $3.00 Rated 4.92 out of 5, based on 12 reviews 4.9(12) Add to cart Wish List Adding and Subtracting Mixed Numbers Word Problems Google Form Created by Mrs E Teaches Math In this READY TO GO digital activity, students will practice solving word problems by adding and subtracting mixed numbers. This Google Form will do the grading for you! IT'S PERFECT FOR DISTANCE LEARNING! In this digital activity, students will use Google Forms to practice adding and subtracting mixed numbers. Students will solve problems by adding and subtracting mixed numbers. Some of the problems require students to find a common denominator. The Google Form is set as a quiz, so it will 4 th - 6 th Fractions, Math CCSS 4.NF.B.3b , 5.NF.A.1 $2.00 Original Price $2.00 Rated 5 out of 5, based on 5 reviews 5.0(5) Add to cart Wish List Adding and Subtracting Mixed Numbers Word Problems Activity: Escape Room Game Created by Escape Room EDU This breakout escape room is a fun way for students to test their skills with adding and subtracting mixed numbers through word problems. Contents: ♦ Teacher Instructions and FAQ ♦ 3 Levels to decode: Multiple Choice, Message Decoder, and a Maze ♦ Student Recording Sheet and Teacher Answer Key ♦ Link to an optional, but recommended, digital breakout roomCheck out the preview and the video preview for more details! - The video preview will show 4 th - 6 th Fractions, Other (Math) $6.00 Original Price $6.00 Rated 5 out of 5, based on 2 reviews 5.0(2) Add to cart Wish List Adding and Subtracting Mixed Numbers Word Problem Task Cards - 5.NF.2 Created by Engaged Minds Grow Would you like an engaging way to have students practice mixed number word problems? Then Adding and Subtracting Mixed Number Word Problem Task Cards are for you! Students get the opportunity to practice a variety of different 5.NF.2 problems and you get an opportunity to see their skill levels for this important fraction standard. This task card set includes a key piece to helping engagement stay high and helping your time be working with students be used effectively...the answer bank. The a 5 th Fractions, Math, Other (Math) CCSS 5.NF.A.2 FREE Rated 4.65 out of 5, based on 20 reviews 4.7(20) Log in to Download Wish List Add&Subtract Unlike Fractions &Mixed Numbers Word Problem Task Cards Activity Created by Simply Teach Engage your students with this word problems activity featuring task cards focused on adding and subtracting fractions and mixed numbers with unlike denominators. Designed to align with CCSS 5.NF.2, this resource will help your students build confidence in solving real-world fraction problems while reinforcing key math skills. You May Also Be Interested In:5th Grade Fractions 5.NF.1-7 Lessons and Notes5th Grade Fractions 5.NF.1-7 WorksheetThis set includes 16 task cards, each with a unique 4 th - 6 th Fractions, Math, Other (Math) CCSS 5.NF.A.2 Also included in:Add & Subtract Unlike Fractions Mixed Numbers \| Activity Game Center \| 5.NF.1-2 FREE Rated 4.9 out of 5, based on 10 reviews 4.9(10) Log in to Download Wish List Boom Card Deck Add&Subtract Mixed Numbers with Word Problems Created by My Little Penguins This is a Boom Deck, please scroll down if you're not a member of Boom Learning and aren't familiar with Boom Decks. Included are 22 boom deck cards in random order that will allow students to practice adding or subtracting mixed numbers with word problems. Word problems contain unlike denominators with some regrouping. Click on the "preview" button above to try out a sample of this set of boom cards. Standards Covered: CCSS.MATH.CONTENT.5.NF.A.2 Solve word problems involving addition an 4 th - 5 th Fractions, Math, Other (Math) CCSS 5.NF.A.2 $2.50 Original Price $2.50 Rated 4.67 out of 5, based on 9 reviews 4.7(9) Add to cart Wish List 1 2 3 4 5 Showing 1-24 of 4,500+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. Facebook Instagram Pinterest Twitter About Who we are We're hiring Press Blog Gift Cards Support Help & FAQ Security Privacy policy Student privacy Terms of service Tell us what you think Updates Get our weekly newsletter with free resources, updates, and special offers. 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2988	https://math.stackexchange.com/questions/1219432/prove-that-if-0x-le-y-then-1-y-le-1-x-using-ordered-field-axioms	real analysis - Prove that if $0<x\le y$ then $1/y \le 1/x$ using ordered field axioms - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Prove that if 0<x≤y 0<x≤y then 1/y≤1/x 1/y≤1/x using ordered field axioms Ask Question Asked 10 years, 6 months ago Modified5 years, 6 months ago Viewed 7k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Suppose that x,y,z∈R+x,y,z∈R+. Then, either x=y x=y or x≤y x≤y. If x=y x=y, then x−1=y−1 x−1=y−1. If x<y x<y, then via proof by contradiction let's suppose x<y x<y implies x−1<y−1 x−1<y−1. (x)(x−1)<(x)(y−1)(x)(x−1)<(x)(y−1) and hence 1<(x)y−1 1<(x)y−1 implies y<x y<x How should I take it further? real-analysis real-numbers ordered-fields Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Mar 8, 2020 at 20:18 gen-ℤ ready to perish 6,750 4 4 gold badges 32 32 silver badges 52 52 bronze badges asked Apr 4, 2015 at 2:41 MathematicingMathematicing 6,581 16 16 gold badges 87 87 silver badges 169 169 bronze badges 9 00⟹1 y≤1 x 00⟹1 y≤1 x Isn't this enough?Prasun Biswas –Prasun Biswas 2015-04-04 02:45:08 +00:00 Commented Apr 4, 2015 at 2:45 Note that it works even if x,y x,yaren't bounded below by 0 0.Prasun Biswas –Prasun Biswas 2015-04-04 02:49:07 +00:00 Commented Apr 4, 2015 at 2:49 I don't understand,Mathematicing –Mathematicing 2015-04-04 02:53:49 +00:00 Commented Apr 4, 2015 at 2:53 It uses the fact that if you keep the numerator of a fraction constant, then the fraction with the largest denominator has the smallest value.Prasun Biswas –Prasun Biswas 2015-04-04 02:55:10 +00:00 Commented Apr 4, 2015 at 2:55 2 Then, you should mention that in your question.Prasun Biswas –Prasun Biswas 2015-04-04 02:57:25 +00:00 Commented Apr 4, 2015 at 2:57 \|Show 4 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Suppose that 0<x≤y 0<x≤y. Then by the properties of inequalities, and the fact that x−1,y−1 x−1,y−1 exist in said field, 0<x y≤1 0<x y≤1. Further, 0<1 y≤1 x 0<1 y≤1 x. To elaborate further, we were able to say that x−1∗0=0 x−1∗0=0 and y−1∗0=y−1∗0= since we are in a field. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 4, 2015 at 6:02 93012939301293 2,304 16 16 silver badges 24 24 bronze badges 1 I think, and the comments agree that you may have over thought this one.9301293 –9301293 2015-04-04 06:03:43 +00:00 Commented Apr 4, 2015 at 6:03 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis real-numbers ordered-fields See similar questions with these tags. 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2989	http://www.ijmetmr.com/olmarch2017/SSreeSaiSwetha-GGaneshNaidu-46.pdf	Page 340 Analysis and Design of Gravity Dam in Finite Element Method by Using STAAD Pro S.Sree Sai Swetha M.Tech Student Department of Civil Engineering, Pace Institute of Technology & Sciences. G.Ganesh Naidu Associate Professor Department of Civil Engineering, Pace Institute of Technology & Sciences. Abstract Gravity dams are solid concrete structures that maintain their stability against design loads from the geometric shape, mass and strength of the concrete. The purposes of dam construction may include navigation, flood damage reduction, hydroelectric power generation, fish and wildlife enhancement, water quality, water supply. The design and evaluation of concrete gravity dam for earthquake loading must be based on appropriate criteria that reflect both the desired level of safety and the choice of the design and evaluation procedures. The Dam discussed in this paper is of the height 110m for which Equivalent static analysis and dynamic analysis by using time history method is carried out. Most of the organizations analyze the dams by elastic method which gives very rough results. Here Finite Element Approach is used to analyze the dam which is proved to be the realistic for such structures. Comparison is done between the equivalent static approaches of seismic analysis with dynamic analysis by using time history. Keywords: Gravity Dam, Finite element analysis, Dynamic analysis, Stress Contours, Staad-pro INTRODUCTION Any structure that is constructed will undergo many forces such as wind, seismic, self-weight or forces like ice/snow etc. Among these, seismic forces are natural and as we know earthquake is a natural calamity and is so unpredictable.inorder to prevent the structure from being collapse, it’s very important to adopt earthquake resistant design philosophy while designing the structure. Waves which arise during Seismic event carries very massive speed and when it struck with any structure it travels through foundation to the top roof resulting In-elastic deformation. there may be the possibility of collapse of whole structure or probably it will survive depending upon the design adopted but surely the structure will have some major repairing and strengthening works which will be costly. Sometimes damages caused by earthquake vibrations are very high that goes beyond repair works. Generally hydraulic structures like concrete gravity dam, canals and RCC multi-storied structures are sufficiently stiff and ductile. These structures undergo large Deformations in its inelastic region. Concrete gravity dam is massive structure having many forces acting on it. It’s very important for the dam to survive against seismic vibrations. This Paper is mainly focused on behavior of concrete gravity dam with nonlinear characteristics using seismic time history analysis. In order to study the precise behavior of structures, seismic time History or response spectrum plays an important role. These analyses methods can be adopted to study the structures having single degree of system or multi degree of freedom system possessing non-linear characteristics. Time history performs analysis which is based on Time-acceleration as an input data which is basically an already experienced acceleration w.r.t time by the ground during seismic event. Time history analysis provides the most probable shapes and directions of structure which is its dynamic structural response under loading which varies as according to specified time-acceleration function. One can predict either the structure will survive or not against these seismic vibrations by using time history analysis results. Mainly structures consist of stiffness and damping as a nonlinear parameter. Damping mostly Page 341 encountered in dynamic problems related with structural control, aerodynamics and offshore hydraulic structures. Most hydraulic structures undergo yielding under seismic vibrations. Damping orinertia, displacement and acceleration are non-linear parameters which provide the non-linear characteristics to the structure. The design lateral force shall be considered in each of two orthogonal horizontal directions of the structure. For structures which have the lateral force resisting elements in the two orthogonal directions, the design lateral force shall be considered along one direction at a time, and not in both directions simultaneously. It is known that for most world tectonic regions the ground motion can act along any horizontal direction, therefore, this implies the existence of a possible different direction of seismic incidence that would lead to an increase of structural response. Critical angles are earthquake incidence angles, producing critical responses. In this study, a four storey reinforced concrete building with moment resisting frame, of different shapes i.e., L shaped and Shaped are analyzed by Time history method of Dynamic analysis of Earthquake. A set of values from 0 to 90 degrees, with an increment of 10 degrees has been used of excitation of seismic force. The details of the study and its result are described briefly in the following section of the paper. LITERATURE REVIEW Review by the workers of analyses performed by licensees, or their consultants, ought to think about the assumptions employed in the analysis. the premise for vital assumptions appreciate allowable stresses, shear strengths, drain effectiveness, and loading conditions ought to be rigorously examined. The consultant's reports, exhibits, and supplemental data should offer justification for these assumptions appreciate foundation exploration and testing, concrete testing, instrumentation knowledge, and records maintained throughout the particular construction of the project. Also, the workers engineer's freelance information of the dam gained through website scrutinys or review of operations inspection report similarly as familiarity with previous reports and analyses, ought to be accustomed verify that the exhibits bestowed area unit representative of actual conditions. Ways of study ought to change to the standard procedures employed in the engineering profession. Conservative assumptions will scale back the number of exploration and testing needed. to Illustrate, if no cohesion or drain effectiveness is assumed in associate analysis, there would be no ought to justify those assumptions with exploration and testing. For this reason, it should generally be a lot of useful to investigate the dam with conservative assumptions instead of to undertake to justify less conservative assumptions. there's but a minimum information of the inspiration that has got to be obtained. The potential for slippy on the dam foundation is usually investigated. However, the potential for failure on a plane of weakness deep within the foundation ought to be investigated. expertise has shown that the best danger to dam stability results once vital attributes of the inspiration aren't proverbial. to Illustrate, within the case of Morris Sheep arduous Dam, 26/ P-1494, a horizontal seam beneath set the dam, providing a plane of weakness that wasn't thought-about. This oversight was solely discovered once the dam had practiced vital downstream movement Concrete gravity dam and apparent structures The basic form of a concrete gravity dam is triangular in section (Figure 1a), with the highest crest usually widened to produce a road (Figure 1b). Page 342 The increasing breadth of the section towards the bottom is logical since the water pressure additionally will increase linearly with depth as shown in Figure 1a. within the figure, h is assumed because the depth of water and γh is that the pressure at base, wherever γ is that the unit weight of water (9810 N/m³), W is that the weight of the dam body. the highest portion of the dam (Figure 1b) is widened to produce area for vehicle movement. A gravity dam ought to even have associate acceptable conduit for emotional excess flood water of the watercourse throughout monsoon months. This section appearance slightly completely different from the opposite non-overflowing sections. A typical section of a conduit is shown in Figure a pair of. FIGURE 2:Typical overflow section of a gravity dam The flood water glides over the crest associated downstream face of the conduit and meets an energy dissipating structure that helps to kill the energy of the flowing water, that otherwise would have caused erosion of the watercourse bed on the downstream. The kind of energy dissipating structure shown in Figure a pair of is named the stilling basin that dissipates energy of the quick flowing water by formation of hydraulic jump at basin location. This and different sorts of conduit and energy dissipators area unit mentioned in an exceedingly future section. Figure three shows the functioning of this kind of conduit Usually, a conduit is supplied with a gate, and a typical conduit section could have a radial gate as shown in Figure four. The axis or trunnion of the gate is command to anchorages that area unit mounted to piers. Design of concrete gravity Dam sections 4.1 Design of gravity dam Fundamentally a gravity dam ought to satisfy the subsequent criteria: 1. It shall be safe against overturning at any horizontal position among the dam at the contact with the inspiration or among the inspiration. 2. It ought to be safe against slippy at any horizontal plane among the dam, at the contact with the inspiration or on any earth science feature among the inspiration. 3. The section ought to be therefore proportional that the allowable stresses in each the concrete and therefore the foundation shouldn't exceed. Safety of the dam structure is to be checked against doable loadings, which can be classified as primary, secondary or exceptional. The classification is created in terms of the pertinence and/or for the relative importance of the load. 1. Primary hundreds area unit known as universally applicable and of prime importance of the load. Page 343 2. Secondary hundreds area unit typically discretionary and of lesser magnitude like sediment load or thermal stresses thanks to mass concreting. 3. Exceptional hundreds area unit designed on the premise of restricted general pertinence or having low likelihood of incidence like mechanical phenomenon hundreds related to unstable activity. Technically a concrete gravity dam derives its stability from the force of gravity of the materials within the section and thus the name. The gravity dam has ample weight therefore on withstand the forces and therefore the overturning moment caused by the water impounded within the reservoir behind it. It transfers the hundreds to the foundations by cantilever action and thus sensible foundations area unit pre requisite for the gravity dam. The forces that provide stability to the dam include: 1. Weight of the dam 2. Thrust of the tail water The forces that attempt to destabilize the dam include: 1. Reservoir water pressure 2. Uplift 3. Forces thanks0 to waves within the reservoir 4. Ice pressure 5. Temperature stresses 6. Silt pressure 7. unstable forces 8. Wind pressure The forces to be resisted by a gravity dam represent 2 classes as given below: 1. Forces, appreciate weight of the dam and water pressure that area unit directly calculated from the unit weight of materials and properties of fluid pressure and 2. Forces appreciate uplift, earthquake hundreds, silt pressure and ice pressure that area unit assumed solely on the premise of assumptions of variable degree of responsibility. of course to judge this class of forces, special care needs to be taken and reliance placed on the market knowledge, expertise and judgement. Figure twenty three shows the position and direction of the assorted forces expected in an exceedingly concrete gravity dam. Forces like temperature stresses and wind pressure haven't been shown. Ice pressures being uncommon in Indian context are omitted. For thought of stability of a concrete dam, the subsequent assumptions area unit made: 1. That the dam consists of individual crosswise vertical parts every of that carries its load to the inspiration while not transfer of load from or to adjacent parts. but for convenience, the steadiness analysis is often applied for the complete block. 2. That the vertical stress varies linearly from upstream face to the downstream face on any mechanical drawing. The Bureau of Indian Standards code IS 6512-1984 “Criteria for style of solid gravity dams” recommends Page 344 that a gravity dam ought to be designed for the foremost adverse load condition of the seven given kind mistreatment the security factors prescribed. Depending upon the scope and details of the assorted project parts, website conditions and construction programme one or a lot of the subsequent loading conditions could also be applicable and will would like appropriate modifications. The seven sorts of load mixtures area unit as follows: 1. Load combination A (construction condition): Dam completed however no water in reservoir or tailwater 2. Load combination B (normal in operation conditions): Full reservoir elevation, traditional dry weather tail water, traditional uplift, ice and silt (if applicable) 3. Load combination C: (Flood discharge condition) - Reservoir at most flood pool elevation ,all gates open, tailwater at flood elevation, traditional uplift, and silt (if applicable) 4. Load combination D: Combination of A and earthquake 5. Load combination E: Combination B, with earthquake however no ice 6. Load combination F: Combination C, however with extreme uplift, assumptive the emptying holes to be defunct 7. Load combination G: Combination E however with extreme uplift (drains inoperative) It would be helpful to elucidate in an exceedingly bit a lot of detail the various loadings and therefore the ways needed to calculate them. These area unit explained within the following sections. 4.2 Loadings for concrete Gravity Dams The significant loadings on a concrete gravity dam embrace the self-weight or load of the dam, the water pressure from the reservoir, and therefore the uplift pressure from the inspiration. There area unit different loadings, that either occur intermittently, like earthquake forces, or area unit smaller in magnitude, just like the pressure exerted by the waves generated within the reservoir that his the upstream of the dam face. These loadings area unit explained within the following section 4.3 load The load contains of the load of the concrete structure of the dam body additionally to pier gates and bridges, if any over the piers. The density of concrete could also be thought-about as 2400 kg/m³. Since the cross section of a dam typically wouldn't be easy, the analysis could also be applied by dividing the section into many triangles and rectangles and therefore the load (self weight) of every of those sections (considering unit breadth or the block width) computed singly so superimposed up. For locating out the instant of the load (required for calculative stresses), the moments thanks to the separate sub–parts could also be calculated singly so summed up. 4.4 Water pressure on dam The pressure thanks to water within the reservoir which of the tailwater working on vertical planes on the upstream and downstream facet of the dam severally could also be calculated by the law of hydraulics. Thus, the pressure at any depth h is given by γhkN/m² acting traditional to the surface. Once the dam contains a sloping upstream face, the water pressure will be resolved into its horizontal and vertical componenets, the vertical part being given by the load of the water prism on the upstream face and acts vertically downward through the centre of gravity of the water space supported on the dam face. In conduit section, once the gates area unit closed, the water pressure will be puzzled out within the same manner as for non–overflow sections apart from vertical load of water on the dam itself. throughout overflow, the highest portion of the pressure triangle gets truncated and a trapezium of pressure acts below fig Page 345 PREPARATION OF FOUNDATION FOR DAM CONSTRUCTION A concrete gravity dam meant to be made across a watercourse natural depression would typically be set on the arduous rock foundation below the traditional watercourse overburden that consists of sand, loose rocks and boulders. but at any foundation level the arduous rock foundation, again, might not perpetually be fully satisfactory right along the projected foundation and abutment space, since domestically there could also be cracks and joints, a number of these (called seams) being filed with poor quality rock. Thus before the concreting takes place the complete foundation space is checked and in most cases reinforced unnaturally such it's ready to sustain the hundreds that may be obligatory by the dam and therefore the reservoir water, and therefore the impact of water oozy into the foundations besieged from the reservoir. Generally the standard of foundations for a gravity dam can improve with depth of excavation. often the course of the watercourse has been determined by earth science faults or weaknesses. A plug of concrete of depth double the breadth of the seam would typically be adequate for structural support of the dam, in order that depth of excavation can, on most occasions rely upon the character of infilling material, the form of the excavated zone and therefore the depth of cutoff necessary to confirm a acceptable hydraulic gradient once the reservoir is crammed. Associate example of this kind of treatment for Bhakra dam is shown in Figure forty two. RESULTS AND DISSCUSTIONS Fig. 1: Gravity Dam II. Computer files TAKEN FOR ANALYSIS Table- 1 Input Data Taken For Analysis A. Static Load: 1) The weight of the dam: The unit weight is set from materials investigation. 2) Hydrostatic pressure of the water within the reservoir. 3) The uplift forces caused by hydrostatic pressure on the inspiration at the interface of the dam and therefore the foundation. Uplift forces area unit typically thought-about in stability and stress analysis to confirm structural adequacy and area unit assumed to be unchanged by earthquake forces. Page 346 CONCLUSION For Corner Column C1: The shear force in X direction i.e. Fx is decreasing throughout from zero to ninety degrees, it's most worth at zero degree for L structure whereas T structure additionally shows parabolic decreasing curve for Fx and attains most worth at twenty degrees. Moment concerning Y axis for corner column C1 of L structure attains most worth at eighty degrees and Moment concerning Z axis attains most at zero degrees whereas T structure attains most worth at ninety degrees for My and for Mz at twenty degrees. For aspect Column C2: Shear force Fx is constant from zero to thirty degrees so it decreases until ninety degrees for L structure whereas for T structure the curve is ceaselessly increasing i.e. minimum worth at zero degree and most at ninety degrees. L and T structure each attains most My at ninety degrees and Mz at zero degrees. For Middle Column C3: For L structure the shear force Fx at begin will increase slowly and shows a steep slope and from twenty degrees forward constant throughout. T structure shows a distinct nature as shown in figure No. 8a. Value of My i.e Moment concerning Y axis is most at ninety degrees for L structure and eighty degrees for T formed structure. Value of Mzi.e Moment concerning Z axis is most at zero degrees for L and T formed structure. From then on top of graphs and conclusions it will be ended that T formed structure needs to resist additional shear force than L formed structure. REFERENCES 1. Karuna Moy Ghosh,”Analysis and style follow of Hydraulic Concrete Structure” letter of the alphabet Learning non-public restricted. 2. US Army Corps of Engineers, “ Gravity Dam Design” EM 1110-2-2200 thirty Gregorian calendar month 1995 3. Bentley, “ STAAD.Pro” www.bentley.com/en-US/Products/STAAD.Pro 4. IS 12966 (Part 1) -1992, “Practice for Galleries and different Openings in Dam, Bureau of Indian Standards, New Delhi. 5. IS 12966 (Part 2) -1990, “Practice for Galleries and different Openings in Dam, Bureau of Indian Standards, New Delhi. 6. IS 6512 -1984, “Criteria for style of Solid Gravity Dam, Bureau of Indian Standards, New Delhi. 7. IS 1893 (Part 1) -2002, “Criteria for Earthquake Resistant style of Structures, Bureau of Indian Standards, New Delhi. 8. Design Criteria for Concrete Arch and Gravity Dams" (1977), USBR, EM No.19 9. "Gravity Dam Design", (30 June 1995), US Army Corps of Engineers, EM 1110-2- 2200 10. “IS 1893” (1984), Bureau of Indian Standards, New Delhi, India 11. IS 6512” (1984), Bureau of Indian Standards, New Delhi , India 12. Small dams - Guidelines for Design, Construction and Monitoring", (2002) French Committee On Large Dams ComitéFançais Des Grands Barrages v. Sohel Ahmed Quadri, MangulkarMadhuri N, 13. “Investigation of critical angle of incidence for the analysis of RCC Frames”, International Journal of Advances in Science Engineering and Technology, ISSN 2321-9009, Volume-2, Issue-3, July 2014 vi. Sohel Ahmed Quadri, MangulkarMadhuri N, 14. “Investigation of critical direction of seismic force for the analysis of RCC frames”, International Journal of Civil Engineering and Technology (IJCIET), ISSN 0976-6308, Volume 5, Issue 6, Hune 2014
2990	https://arxiv.org/pdf/1211.1969	arXiv:1211.1969v1 [cs.IT] 8 Nov 2012 1 Fast Converging Algorithm for Weighted Sum Rate Maximization in Multicell MISO Downlink Le-Nam Tran, Member, IEEE , Muhammad Fainan Hanif, Antti Tölli, Member, IEEE , and Markku Juntti, Senior Member, IEEE Abstract The problem of maximizing weighted sum rates in the downlink of a multicell environment is of considerable interest. Unfortunately, this problem is known to be NP-hard. For the case of multi-antenna base stations and single antenna mobile terminals, we devise a low complexity, fast and provably convergent algorithm that locally optimizes the weighted sum rate in the downlink of the system. In particular, we derive an iterative second-order cone program formulation of the weighted sum rate maximization problem. The algorithm converges to a local optimum within a few iterations. Superior performance of the proposed approach is established by numerically comparing it to other known solutions. Index Terms Weighted sum rate maximization, multicell downlink, convex approximation, beamforming. I. I NTRODUCTION For multiple-input multiple-output (MIMO) broadcast channels, dirty paper coding (DPC) is known to be the capacity-achieving scheme . However, DPC is a nonlinear interference cancellation technique Copyright (c) 2012 IEEE. Personal use of this material is permitted. However, permission to use this material for any other purposes must be obtained from the IEEE by sending a request to pubs-permissions@ieee.org. This research was supported by Tekes (the Finnish Funding Agency for Technology and Innovation), Nokia Siemens Networks, Renesas Mobile Europe, Elektrobit, Xilinx, and Academy of Finland. The authors are with the Dept. Communications Eng. and Centre for Wireless Communications, University of Oulu, Finland. Email: {ltran, mhanif, atolli, markku.juntti}@ee.oulu.fi. 2 and thus requires high complexity. Hence, linear precoding techniques are of practical interest. Herein, we consider the problem of weighted sum rate maximization (WSRM) with linear transmit precoding for multicell multiple-input single-output (MISO) downlink. Unfortunately, the WSRM problem, even for single-antenna receivers as considered in this letter, has been shown to be NP-hard in . Although optimal beamformers can be obtained using the methods presented, for instance, in –, they may not be practically useful since the complexity of finding optimal designs grows exponentially with the problem size. Hence, the need of computationally conducive suboptimal solutions to the WSRM problem still remains. Since the WSRM problem is nonconvex and NP-hard, there exists a class of beamformer designs which are based on achieving the necessary optimal conditions of the WSRM problem. In fact, this philosophy has been used, e.g., in –. Interestingly, in , the authors have numerically shown that the suboptimal designs that achieve the necessary optimal conditions of the WSRM problem perform very close to the optimal design. In , the iterative coordinated beamforming algorithm was proposed by manipulating the Karush–Kuhn–Tucker (KKT) equations. However, this algorithm is not provably convergent. In , , the WSRM problem with joint transceiver design is solved using alternating optimization between transmit and receive beamforming. As we show by numerical results, these methods have a slower convergence rate compared to our proposed design. In this letter, we propose a fast converging algorithm that locally solves the problem of WSRM for multicell MISO downlink. The idea of our iterative beamformer design is based on the framework of successive convex approximation (SCA) presented in . The numerical results show that the proposed algorithm converges within a few iterations to a locally optimal point of the WSRM problem. The general concept of the SCA method is as follows. In each step of an iterative procedure, we approximate the original nonconvex problem by an efficiently solvable convex program and then update the variables involved until convergence. We note that in the context of transmit linear precoding for multicell downlink, the SCA method has been used, for example, in . Basically, this method is based on convex relaxations of the rate function and generally arrives at more complex formulations. By proper transformations, we approximate the WSRM problem as a second-order cone program (SOCP) in each step of the SCA method. Our numerical results show that the proposed algorithm generally performs better than the known approaches, in particular, in terms of convergence rate. Notation : We use standard notations in this letter. Bold lower and upper case letters represent vectors and matrices, respectively; (.)T represents the transpose operator. Ca×b represents the space of complex matrices of dimensions given as superscripts; \|c\| represents the absolute value of a complex number. 3 Finally, ‖.‖2 represents the l2 norm. II. P ROBLEM FORMULATION Consider a system of B coordinated BSs of N transmit antennas each and K single-antenna receivers. The set of all K users is denoted by U = {1, 2 . . . , K }. We assume that data for the kth user is transmitted only from one BS, which is denoted by bk ∈ B , where B , {1, 2, . . . , B } is the set of all BSs. The set of all users served by BS b is denoted by Ub. Under flat fading channel conditions, the signal received by the kth user is yk = hbk ,k wkdk + K ∑ i=1 ,i 6=k hbi ,k widi + nk (1) where hbi ,k ∈ C1×N is the channel (row) vector from BS bi to user k, wk ∈ CN ×1 is the beam-forming vector (beamformer) from BS bk to user k, dk is the normalized complex data symbol, and nk ∼ CN (0 , σ 2) is complex circularly symmetric zero mean Gaussian noise with variance σ2. The term ∑Ki=1 ,i 6 =k hbi,k widi in (1) includes both intra- and inter-cell interference. The total power transmitted by BS b is ∑ k∈U b ∥∥wk ∥∥22. The SINR γk of user k is γk = ∣∣hbk ,k wk ∣∣2 σ2 + ∑Ki=1 ,i 6 =k ∣∣hbi,k wi ∣∣2 . (2) In this letter, we are interested in the problem of WSRM under per-BS power constraints 1, which is formulated as maximize wk ∑K k=1 αk log 2(1 + γk)subject to ∑ k∈U b ‖wk‖22 ≤ Pb, ∀b ∈ B (3) where αk’s are positive weighting factors which are typically introduced to maintain a certain degree of fairness among users. As mentioned earlier, since problem (3) is NP-hard, the globally optimal design mainly plays as a theoretical benchmark rather than a practical solution . Herein, we propose a low-complexity algorithm that solves (3) locally, i.e, satisfies the necessary optimal conditions of (3). III. P ROPOSED LOW -COMPLEXITY BEAMFORMER DESIGN To arrive at a tractable solution, we note that following monotonicity of logarithmic function, (3) is equivalent to maximize wk ∏ k (1 + γk)αk subject to ∑ k∈U b ‖wk‖22 ≤ Pb, ∀b ∈ B (4) 1It is straightforward to extend the proposed algorithm to handle per-antenna power constraints at each BS. 4 which can be equivalently recast as maximize wk,t k ∏ k tk (5a) subject to γk ≥ t1/α k k − 1, ∀k ∈ U (5b) ∑ k∈U b ‖wk‖22 ≤ Pb, ∀b ∈ B . (5c) The equivalence of (4) and (5) can be easily recognized by noting the fact that all constraints in (5b) are active at the optimum. Otherwise, we can obtain a strictly larger objective by increasing tk without violating the constraints. Next, by introducing additional slack variables βk, we can reformulate (5) as maximize wk,t k,β k ∏ k tk, (6a) subject to hbk ,k wk ≥ √ t1/α k k − 1βk, ∀k ∈ U , (6b) Im( hbk ,k wk) = 0 , ∀k ∈ U , (6c) ( σ2 + ∑ i6=k \|hbi ,k wi\|2 )1/2 ≤ βk, ∀k ∈ U , (6d) ∑ k∈U b ‖wk‖22 ≤ Pb, ∀b ∈ B . (6e) The equivalence between (5) and (6) is justified as follows. First, we note that forcing the imaginary part of hbk ,k wk to zero in (6c) does not affect the optimality of (5) since a phase rotation on wk will result in the same objective while satisfying all constraints. Second, we can show that all the constraints in (6d) hold with equality at the optimum. Suppose, to the contrary, the constraint for some k in (6d) is inactive. Let us define ˜βk , βk/η and ˜tk , {η2(t1/α k k − 1) + 1 }αk , where η is a positive scaling factor. Since the constraint (6d) is inactive, we can choose η > 1 such that the constraints in (6b) and (6d) are still met if we replace (βk, t k) by ( ˜βk , ˜tk). However, such a substitution results in a strictly larger objective because ˜tk > t k for η > 1. This contradicts the fact that we have obtained an optimal solution. As a step toward a low-complexity solution to the WSRM problem, we rewrite the constraint (6b) as hbk ,k wk ≥ √xkβk, ∀k ∈ U (7a) xk + 1 ≥ t1/α k k , ∀k ∈ U . (7b) Again, we can easily see that by replacing (6b) with (7a) and (7b), we obtain an equivalent formulation of (6). The reason of doing so becomes clear shortly. Let us define f (xk, β k ) = √xkβk for xk, β k ≥ 0 and focus on the constraint (7a) first. Note that f (xk, β k ) is nonconvex on the defined domain, and thus (7a) is not a convex constraint. To deal with nonconvex constraints, we invoke a result of which shows that if we replace f (xk, β k) by its convex upper bound and iteratively solve the resulting problem 5 by judiciously updating the variables until convergence, we can obtain a KKT point of (6). To this end, for a given φk for all k, we define the function G(xk , β k, φ k ) , φk 2 β2 k 1 2φk xk (8) which arises from the inequality of arithmetic and geometric means of φkβ2 k and φ−1 k xk. It is easy to check that G(xk , β k, φ k) is a convex overestimate of f (xk, β k ) for a fixed φk > 0, i.e., G(xk, β k, φ k ) ≥ f (xk, β k ) for all φk > 0. Moreover, when φk = √xk βk , it is plain to observe f (xk, β k ) = G(xk , β k, φ k ) (9a) ∇f (xk, β k ) = ∇G(xk , β k, φ k) (9b) where ∇f is the gradient of f . Obviously if f (xk, β k ) is replaced by G(xk, β k , φ k), (7a) can be formulated as a second-order cone (SOC) constraint as we shown in (11d). Now we turn our attention to (7b). Recall that t1/α k k is convex if 0 < α k ≤ 1 and concave if αk > 1,and that the optimal solution of (3) stays the same if we multiply all αk’s by the same positive constant. Thus, we can force (7b) to be convex by scaling down αk’s in (3) such that 0 < α k ≤ 1 for all k.However, in this case, the constraint xk + 1 ≥ t1/α k k cannot be directly written as an SOC constraint for αk ∈ R++ .2 As our goal is to arrive at an SOCP, we instead scale αk’s in (3) such that αk > 1 for all k and thus t1/α k k becomes concave. Again, in the light of , we replace the right side of the inequality in (7b) by its upper bound, which now can be obtained by the first order approximation due to the concavity of t1/α k k . Precisely, we have t 1 αk k ≤ t(n) k 1 αk 1 αk t(n) k 1 αk−1 (tk − t(n) k ) (10) where t(n) k denotes the value of variable tk in the nth iteration (i.e., the iteration corresponding to Algorithm 1 described later). In fact, we have linearized t1/α k k around the operating point t(n) k . With (10), (7b) now becomes a linear inequality. We note that the linear approximation in (10) is trivially shown to satisfy the conditions in (9a) and (9b) at t(n) k . A question naturally arises is whether the linear approximation in (10) affects the optimal sum rate. Interestingly, our numerical experiments show that the WSR obtained with the successive approximation with αk > 1 is identical to that when (7b) is forced to be convex by having 0 < α k < 1 in (3) for all k. In terms of complexity, the linear inequality in (10) is more preferable since it requires lower computational effort compared to the original nonlinear equality constraint in (7b). 2When αkis an integer or a rational number, we can transform the constraint (7b) into a number of SOC constraints . 6 maximize wk,t k,x k,β k,z ik z(0) (11a) subject to ∥∥[2z(N −1) i (t2i−1 − t2i)]T∥∥2 ≤ (t2i−1 + t2i), i = 1 , . . . , 2N −1 (11b) · · · · · · ∥∥[2z(0) (z(1) 1 − z(1) 2 )]T∥∥2 ≤ (z(1) 1 + z(1) 2 ), (11c) ∥∥[ 1 2 (hbk ,k wk − 1 2φ(n) k xk − 1) √ φ(n) k 2 βk ]T∥∥2 ≤ 1 2 (hbk ,k wk − 1 2φ(n) k xk + 1 ), ∀k ∈ U (11d) t(n) k 1/α k 1 αk t(n) k 1/α k−1 (tk − t(n) k ) ≤ xk + 1 , ∀k ∈ U , (11e) ∥∥[σ hb1 ,k w1 · · · hbk−1 ,k wk−1 hbk+1 ,k wk+1 · · · hbK ,k wK ]T∥∥2 ≤ βk, ∀k ∈ U , (11f) ∑ k∈U b ‖wk‖22 ≤ Pb, ∀b ∈ B . (11g) Replacing the right sides of (7a) and (7b) by the upper bounds in (8) and (10), respectively, we can formulate (6) as an SOCP by noting that the objective in (6), i.e., the product of tk’s admits an SOC representation , . The main ingredient in arriving at the SOCP representation is the fact that the hyperbolic constraint uv ≥ z2 where u ≥ 0, v ≥ 0 is equivalent to ‖[2 z (u − v)] T‖2 ≤ (u + v). Let us illustrate the SOCP formulation of (6) for the special case K = 2 q , where q is some positive integer. By collecting two variables at a time and incorporating the additional hyperbolic constraint corresponding to them, we rewrite (6) as the SOCP in (11), shown on the top of the page, where φ(n) k is the value of φk in the nth iteration. In the case of K 6 = 2 q , we define additional tj = 1 for j = K + 1 , . . . , 2⌈log 2 K⌉,where ⌈x⌉ is the smallest integer not less than x and the above expression still holds . Now we are in a position to present an algorithm that solves problem (3) locally. The pseudocode of the beamformer design is outlined in Algorithm 1. We now present the convergence analysis of Algorithm 1. Consider Algorithm 1 Proposed beamformer design for the WSRM problem in multicell MISO downlink. Initialization: n = 0 , (φ(n) k , t (n) k ) = random. 1: repeat 2: Solve (11) with φ(n) k and t(n) k , and denote optimal (tk, β k, x k) as (t⋆k, β ⋆k , x ⋆k). 3: Update (t(n+1) k , β (n+1) k , x (n+1) k ) = ( t⋆k, β ⋆k , x ⋆k) and φ(n+1) k = √ x(n+1) k β(n+1) k ; n := n + 1 . 4: until convergence 7 P (dBW) Average sum rate (b/s/Hz) 5 10 15 20 25 510 15 20 25 Optimal linear design Algorithm 1 SIN WMMSE Zero−forcing Fig. 1. Average sum rate comparison for single-cell scenario, N= 4 ,K= 4 . the n + 1 st iteration of Algorithm 1 that solves the optimization problem (11). If we replace (tk, β k , x k) by (t(n) k , β (n) k , x (n) k ) and wb,k by w(n) b,k , all the constraints in (11d)-(11g) are still satisfied. That is to say, the optimal solution of the nth iteration is a feasible point of the problem in the n + 1 st iteration. Thus, the objective obtained in the n + 1 st iteration is larger than or equal to that in the nth iteration. In other words, Algorithm 1 generates a nondecreasing sequence of objective values. Moreover, the problem is bounded above due to the power constraints. Hence, Algorithm 1 converges to some local optimum solution of (11). By the two properties shown in (9) and based on the arguments presented in , it can be shown that this solution also satisfies the KKT conditions of (6). Numerical results in Section IV confirm that Algorithm 1 performs very close to optimal linear design .IV. N UMERICAL RESULTS In this section, we numerically evaluate the performance of Algorithm 1 under different setups using YALMIP with SDPT3 as internal solver. In the first experiment, we consider a single-cell scenario where a BS with N = 4 transmit antennas serves K = 4 users. The entries of hb,k are CN (0 , 1) and the noise variance σ2 = 1 . In Fig. 1, we plot the average sum rate ( αk = 1 for all k) versus the total transmit power P at the BS. The achieved sum rate of Algorithm 1 is compared to those of zero-forcing 8 beamforming , the weighted sum mean-square error minimization (WMMSE) algorithm in , the soft inference nulling (SIN) scheme in , and the optimal linear design using the branch-and-bound (BB) method in , . Initial values for beamformers for the suboptimal schemes in , , and (φ(0) k , t (0) k ) in Algorithm 1 are generated randomly. The sum rate is obtained after Algorithm 1 and the iterative suboptimal schemes in , converge, i.e., the increase in the objective value between two consecutive iterations is less then 10 −2. The gap tolerance between the upper and lower bounds for the BB method is set to 10 −1 as in , , and the resulting sum rate is calculated as the average of the upper and lower bounds .3 Results reveal that the average sum rate of Algorithm 1 and other iterative beamformer designs is the same on convergence and close to that of the optimal linear approach . However, the SIN scheme, WMMSE algorithm and the optimal design have a slower convergence rate as discussed next. In the second experiment, we illustrate the convergence rate of all considered iterative suboptimal schemes. A simple two-cell scenario with each BS serving 2 users is considered. The number of transmit antennas at each BS is set to N = 8 . The weights, without loss of generality, are taken as (α1, α 2, α 3, α 4) = (0 .14 , 0.21 , 0.28 , 0.36) and the power budget of each BS is set to Pb = 12 dB for b = 1 , 2. Fig. 2 compares the weighted sum rate of the considered schemes as a function of iterations needed to obtain a stabilized output for a random channel realization. In particular, our algorithm has converged just after a few of iterations, while the WMMSE algorithm is still less than midway to convergence. In fact, for this particular case the WMMSE took hundreds of runs before converging to the local optimum solution. This observation may be attributed to the fact that optimization strategy of requires alternate updates between transmit and receive beamformers and therefore exhibits slower convergence properties. We have also noticed that for certain initial values the convergence rate of the WMMSE algorithm is greatly improved. In fact, it was reported in that the convergence rate of an alternating optimization algorithm depends on the initial values of the variables involved, and it converges quickly if initial guess is relatively close to the optimal solution. Further, we observe that Algorithm 1 with and without scaling has slightly different convergence rate (labeled in Fig. 2 as ‘approximated’ and ‘not approximated’) but same optimal value. This validates that the approximation used to arrive at an SOCP formulation has no impact on the achieved sum rate. Our numerical results reveal that for other channel realizations, the SIN scheme may have similar convergence behavior to Algorithm 1, but the average per iteration running time of Algorithm 1 is approximately four times less than that of the SIN 3The principle of the BB method to compute an optimal solution to nonconvex problems is to find provable lower and upper bounds on the globally optimal value and guarantee that the bounds converge as iterations evolve. 9 1 3 5 7 9 11 13 15 17 19 5 5.5 6 6.5 7 Iteration index Weighted sum rate (b/s/Hz) Algorithm 1 (7(b) not approximated) Algorithm 1 (7(b) approximated) SIN WMMSE Fig. 2. Convergence rate of the weighted sum rate, B= 2 ,N= 8 ,K= 4 ,Pb= 12 dB for all b. method. For the set of channel realizations considered in Fig. 2, the optimal design also converges to the same point achieved by other iterative suboptimal methods. However, it takes more than 600 iterations to reduce the gap between lower and upper bounds of the BB method to less than 10 −1. Theoretically, the faster convergence of Algorithm 1 may be attributed to solving an explicit SOCP in each of its iterations. The faster convergence of our algorithm can be much useful for distributed implementation which is left as future work. V. C ONCLUSION In the letter we have studied the problem of WSRM in the downlink of multicell MISO system. Since the problem is NP-hard, we have proposed a low-complexity approximation of the optimization problem. We show that the problem can be approximated by an iterative SOCP procedure. While the convergence of the algorithm can be proved, its global optimality cannot be established. Nonetheless, the algorithm outperforms the previously studied solutions to the WSRM problem, in particular, in terms of its convergence rate. 10 REFERENCES H. Weingarten, Y. Steinberg, and S. Shamai, “The capacity region of the Gaussian multiple-input multiple-output broadcast channel,” IEEE Trans. Inf. Theory , vol. 52, no. 9, pp. 3936–3964, Sep. 2006. Z.-Q. Luo and S. Zhang, “Dynamic spectrum management: Complexity and duality,” IEEE J. Sel. Topics Signal Process. ,vol. 2, no. 1, pp. 57–73, Feb. 2008. S. Joshi, P. Weeraddana, M. Codreanu, and M. Latva-aho, “Weighted sum-rate maximization for MISO downlink cellular networks via branch and bound,” IEEE Trans. Signal Process. , vol. 60, no. 4, pp. 2090–2095, Apr. 2012. E. Björnson, G. Zheng, M. Bengtsson, and B. Ottersten, “Robust monotonic optimization framework for multicell MISO systems,” IEEE Trans. Signal Process. , vol. 60, no. 5, pp. 2508–2523, May 2012. L. Liu, R. Zhang, and K.-C. Chua, “Achieving global optimality for weighted sum-rate maximization in the K-user Gaussian interference channel with multiple antennas,” IEEE Trans. Wireless Commun. , vol. 11, no. 5, pp. 1933–1945, May 2012. L. Venturino, N. Prasad, and X. Wang, “Coordinated linear beamforming in downlink multi-cell wireless networks,” IEEE Trans. Wireless Commun. , vol. 9, no. 4, pp. 1451–1461, Apr. 2010. Chris T. K. Ng and H. Huang, “Linear precoding in cooperative MIMO cellular networks with limited coordination clusters,” IEEE J. Sel. Areas Commun. , vol. 28, no. 9, pp. 1446–1454, Dec. 2010. S. S. Christensen, R. Agarwal, E. Carvaldho, and J. Cioffi, “Weighted sum-rate maximization using weighted MMSE for MIMO-BC beamforming design,” IEEE Trans. Wireless Commun. , vol. 7, no. 12, pp. 4792–4799, Dec. 2008. Q. Shi, M. Razaviyayn, Z.-Q. Luo, and C. He, “An iteratively weighted MMSE approach to distributed sum-utility maximization for a MIMO interfering broadcast channel,” IEEE Trans. Signal Process. , vol. 59, no. 9, pp. 4331–4340, Sep. 2011. A. Beck, A. Ben-Tal, and L. Tetruashvili, “A sequential parametric convex approximation method with applications to nonconvex truss topology design problems,” Journal of Global Optimization , vol. 47, no. 1, pp. 29–51, 2010. F. Alizadeh and D. Goldfarb, “Second-order cone programming,” Mathematical Programming , vol. 95, pp. 3–51, 2001. M. Lobo, L. Vandenberghe, S. Boyd, and H. Lebret, “Applications of second-order cone programming,” Linear Algebra and its Applications , vol. 248, pp. 193–228, Nov. 1998. J. Löfberg, “YALMIP : A toolbox for modeling and optimization in MATLAB,” in Proc. the CACSD Conference , Taipei, Taiwan., 2004. [Online]. Available: K. C. Toh, M. J. Todd, and R. Tutuncu, “SDPT3— a Matlab software package for semidefinite programming,” Optimization Methods and Software , Nov. 1999. Q. Spencer, A. Swindlehurst, and M. Haardt, “Zero-forcing methods for downlink spatial multiplexing in multiuser MIMO channels,” IEEE Trans. Signal Process. , vol. 52, no. 2, pp. 461–471, Feb. 2004. J. C. Bezdek and R. J. Hathaway, “Convergence of alternating optimization,” Neural, Parallel and Scientific Computations ,vol. 11, no. 4, pp. 351–368, Dec. 2003.
2991	https://sketchfab.com/3d-models/anatomy-of-the-inner-ear-f80bda64666c4b8aaac8f63b7b82a0a0	Anatomy of the Inner Ear 3D Model Certain aspects of this model were created from segmented MRI data, making this a highly accurate representation of the tympanic membrane, facial nerve, ossicles and vestibular system. This work “Anatomy of the Inner Ear”, is a derivative of “3D Ear” by W. Robert J. Funnell, PhD; Sam Daniel, MD, CM; and Daren Nicolson, MD, CM at McGill University, used under CC BY-NC-SA 1.0. “Anatomy of the Inner Ear” is licensed under CC BY-NC-SA 4.0. You are free to copy, reuse and remix this for non-commercial purposes but we ask that you acknowledge the University of Dundee as well as publish any remixed work under the same share-alike license as the original authors. The vestibulocochlear nerve was not derived from MRI data, however heavily referenced. You can locate the segmented MRI data from the following: Illustrations of this structure are available here: CC Attribution-NonCommercial-ShareAlikeCC Attribution-NonCommercial-ShareAlike 14 comments
2992	https://mathgeekmama.com/money-math-practice-worksheets/	Money Math Practice Worksheets (Diner Theme) Skip to content Join 165,000+ parents and teachers for new tips! Home About Terms of Use Disclosure & Privacy Policy Contact Search for: Fun teaching resources & tips to help you teach math with confidence Hands on Math Math+Technology Lessons Interactive Worksheet Fractions and Decimals 3-D Shapes Worksheets Math Art Project PI Day Easy Math Games Simple Math Card Games Literature Based Math Math Books Math Teaching Classroom strategies Common Core Help Dyscalculia Resources Math Teaching Student Help Freebies Shop Math PD Courses Course Login JOIN Math Geek Mama+ Math Geek Mama+ Login Menu Search Hands on Math Math+Technology Lessons Interactive Worksheet Fractions and Decimals 3-D Shapes Worksheets Math Art Project PI Day Easy Math Games Simple Math Card Games Literature Based Math Math Books Math Teaching Classroom strategies Common Core Help Dyscalculia Resources Math Teaching Student Help Freebies Shop Math PD Courses Course Login JOIN Math Geek Mama+ Math Geek Mama+ Login Search for: Search for: Home/Grades K-1/Silver Dollar Diner: FREE Money Math Pack Silver Dollar Diner: FREE Money Math Pack ByBethanyJuly 13, 2016 February 18, 2020 262 shares Facebook Twitter Pinterest Email One of the last things I did with my kids before we officially began our summer break was review money. This was my son’s first exposure, so I taught him each of the coins and their denomination, and began to teach him how to add coins (using some of the books in this post and play money). For my daughter, this meant putting her knowledge of money values to work. To help her practice finding the total, and work through real life examples, I created this fun diner-themed math pack. The focus of these money math practice worksheets is on adding money, and solving word problems involving addition and subtraction. Please Note: Some of the links in this post are affiliate links and help support the work of this site. Read our full disclosure here. This printable pack includes a variety of activities that can be completed over and over, and easily adapted to increase the difficulty as your kids get older. Included in this Money Math Printable Pack: Silver Dollar Diner Printable Menu (whole dollar amounts) Silver Dollar Diner Printable Menu (blank prices) 2 Adding money worksheets (using the menu prices) Draw your order and find the total worksheet 1 page of money word problems involving addition and subtraction Answer keys for each worksheet How to Use the Money Math Practice Worksheets: Each of the activities in this download is completed using a printed “Silver Dollar Diner” menu. I suggest you print and laminate the menu so you can use it with each of the activities, or as a math center in the classroom. There are two different menus so that these money worksheets can be used with kids as young as kindergarten or first grade, or adjusted for older kids. The first menu includes prices for each item on the menu. The prices are whole dollar amounts, so that kids can focus on understanding how to use money and solve real world problems, as well as work on their mental math skills. The second menu doesn’t include prices, so you could print and laminate it, then write in prices using a dry erase marker. This will allow you to customize the difficulty level depending on the level of your kids. This would also allow you to use the activities over and over, but with different prices. Once you have the menu printed and ready, simply print out the worksheets of your choice! On the first set of worksheets, kids simply find the total cost of each diner order. The next activity lets kids get creative by drawing their menu choices and then finding the total (you may need to specify how many items they have to order to make sure they have to add money to find the total). And the last worksheet includes word problems that kids solve by using the prices on the menu. At the end of the download are answer keys for each worksheet that correspond to the menu with prices included. (If you create your own menu with different prices, you will have to make your own answer keys.) I think this would make a fun math center or partner activity. Kids could each take turns ordering and adding up the totals for their partner, or compare their drawings, etc. You could also use the receipts included in this money math worksheet set with the diner-themed set as well. This would allow kids to work on adding money and making change. {Click HERE to go to my shop and grab the Silver Dollar Diner Money Pack!} Want more fun money practice? Try one of these: Counting money puzzles set Summer Math Camp: Money Resources Pigs Will Be Pigs Math Pack Books to Teach Money Math Have fun making math relevant with this set of money math practice worksheets! Never Run Out of Fun Math Ideas! If you enjoyed this post, you will love being a part of the Math Geek Mama community! Sign up to receive my weekly emails with fun and engaging math ideas, free resources and special offers. Join 163,000+ readers as we help every child succeed and thrive in math! PLUS, receive my FREE ebook, 5 Math Games You Can Play TODAY, as my gift to you! Send Me FREE Math Ideas! We won't send you spam. Unsubscribe at any time. Built with Kit Post navigation Previous The BEST Books to Teach Money Math Next Weekly Freebie Round Up Similar Posts How to Introduce and Teach Unit Rate: Practical Tips How to Tackle Math Word Problems & Make Sense of Operations Student Created Math Spinner Games for Any Grade Level Symmetry: Finish the Picture Pages {Drawing Mirror Images} Virtual 100th Day of School Math Activities {FREE Resources} {FREE} Multiply by Multiples of 10 Grid Challenges How to Introduce and Teach Unit Rate: Practical Tips How to Tackle Math Word Problems & Make Sense of Operations Student Created Math Spinner Games for Any Grade Level Symmetry: Finish the Picture Pages {Drawing Mirror Images} Virtual 100th Day of School Math Activities {FREE Resources} {FREE} Multiply by Multiples of 10 Grid Challenges How to Introduce and Teach Unit Rate: Practical Tips How to Tackle Math Word Problems & Make Sense of Operations Student Created Math Spinner Games for Any Grade Level Symmetry: Finish the Picture Pages {Drawing Mirror Images} Virtual 100th Day of School Math Activities {FREE Resources} {FREE} Multiply by Multiples of 10 Grid Challenges Welcome! I'm Bethany, a.k.a Math Geek Mama. I believe every child can succeed in math with the right help and support. So I'm here to help you provide rich and engaging math lessons without sacrificing your entire weekend. Learn more about the resources I createHERE. NEWSLETTER Math Time Doesn't Have to End in Tears Join 165,000+ parents and teachers who learn new tips and strategies, as well as receive engaging resources to make math fun. Plus, receive my guide, "5 Games You Can Play Today to Make Math Fun," as my free gift to get you started! Subscribe Find More Resources To Help Make Math Engaging! Read The BlogShop The Store Join 165k+ Parents & Teachers Who learn new tips and strategies, as well as receive engaging resources to make math fun! Subscribe Free Math Printables for K through Highschool!Math Training Classes for K-8 TeachersResources to make math fun and engaging for students Search Get Our App on iTunes! ©2025, Math Geek Mama. 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2993	https://education.ti.com/~/media/93BB05F78A2F4005A48763E40C54FFBD	Scientific Notation TEACHER NOTES ©2013 Texas Instruments Incorporated 1education.ti.com Activity Overview This activity gives students an opportunity to see where large and small numbers are used and how scientific notation offers a convenient method of writing such numbers. Topic: Numbers and Operations  Understand numbers, ways of represent ing numbers, relationships among numbers, and number systems  Develop an understanding of large numbers and recognize and appropriately use exponential, scientific, and calculator notation Teacher Preparation and Notes  To download the data list and student worksheet, go to education.ti.com/exchange/sn This activity utilizes MathPrint TM functionality and includes screen captures taken from the TI -84 Plus C Silver Edition. It is also appropriate for use with the TI -83 Plus, TI -84 Plus, and TI -84 Plus Silver Edition but slight variances may be found within the directions. Compatible Devices:  TI -84 Plus Family  TI -84 Plus C Silver Edition Associated Materials:  Scientific_Notation _Student.pdf  Scientific_No tation _Student.doc Tech Tips :  Access free tutorials at /pd/US/Online -Learning/Tutorials  Any required calculator files can be distribut ed to students via handheld -to -handheld transfer. Scientific Notation TEACHER NOTES ©2013 Texas Instruments Incorporated 2education.ti.com Part 1 – Writing Scientific Notation in Expanded Form Questions 1 -4 Have students write their answers and check them on the calculator. Help students understand the form for scientific notation. Stu dents can enter scientific notation numbers two different ways, using 10 or E. Note that the key is labeled EE but the display is E. _ 3 . 7 7 1 0 ^ 8 _ 3 . 7 7 `£ 8 To check –3.77 10 8, enter in one of the two ways shown above and see if it matc hes the ir answer. Sometimes the calculator’s limitations won’t let it write the answer in expanded form. In that case have students enter their answer s in expanded form and see if they match the original. Part 2 – Writing Numbers in Scientific Notation Questions 5 & 6 Students can change the TI -84 Plus mode to “Sci” to convert numbers to scientific notation. To change the mode, press M and ► to move to Sci, then press e. The answer will appear with an E. Explain to students that this represents 10 and the number that follows the E is the exponent. Students should not enter the commas into the calculator when converting the salaries. Discuss with students how to compare numbers with the exponent. What is different between Carson’s salary and Stacy’s? Scientific Notation TEACHER NOTES ©2013 Texas Instruments Incorporated 3education.ti.com Questions 7 & 8 Students need to change the calculator mode to Normal by pressing M e. Press ` î to return to the Home Screen. . Students can fold a piece of paper in half to understand how the thickness doubles. Ensure students understand that the  2 represents the doubled thickness. The key presses for the screen to the right are 0 . 0 0 4 e 2 e e e. Continue pressing e to solve the problem. They will need to count how many times they press enter. They do not count the first entry as a paper f old. Questions 9 & 10 Place students in pairs to discuss the probability of a head when tossing one coin. When you toss the coin twice, what is the probability? Discuss as a class if this number is getting larger or smaller. Let students guess how small the answer may be. Solve the problem with the following key presses: . 5 e . 5 e e e … They need to count the first entry because it is a flip. Students are to then write the answer in expanded form. For the first few flips you can ask the students to convert the decimal to a fraction and/or a percent. Scientific Notation TEACHER NOTES ©2013 Texas Instruments Incorporated 4education.ti.com Part 3 – Ordering Numbers in Scientific Notation Students often confuse a negative number with a negative exponent. Using a number line to look at relative location can help student understanding. Qu estions 11 & 12 If students have difficulty placing the numbers on the number line, have them convert the tick marks to expanded form for assistance. Students should ensure that the calculator is in Normal mode. Press M e ` î. Enter the first number, 1.2 5x10 -3 , or 1 . 2 5 ` £ _ 3 and then press e. Students can repeat this process for each of the numbers. Scientific Notation TEACHER NOTES ©2013 Texas Instruments Incorporated 5education.ti.com Solutions to Student Worksheet Part 1 – Writing Scientific Notation in Expanded Form -3.77 10 8 Answer: –377,000,000 1.202 10 5 Answer: 120 ,200 4.224 10 -6 Answer: 0.000004224 -5.24 10 -12 Answer: –0.00000000000524 Part 2 – Writing Numbers in Scientific Notation The following are salaries for the 5 top paid players of the Cincinnati Bengals. Write each salary in scientific notation. Answer: Player Salary Scientific Notation Carson Palmer $13,980,000 1.398 x 10 7 Stacy Andrews $7,455,000 7.455 x 10 6 Chad Johnson $6,415,370 6.41537 x 10 6 Antwan Odom $5,700,000 5.7 x 10 6 Levi Jones $5,266,666 5.266666 x 10 6 The 2007 median Ameri can household income was $5.0233 10 4. How does this compare to the salaries above? Answer: Students should notice that the exponents have a difference of 2. The football salaries are more than 100 times the median income. Imagine you could fold a piec e of paper 0.004 inches thick 50 times. About how many inches thick would the resulting paper be after the 50 th fold? Answer: 4.504x10 12 inches What is the answer in expanded form? Answer: 4,504,000,000 ,000 inches What is the percent chance of flippi ng a coin 40 times and having it come up heads each time? Answer: 9.095x10 -13 What is the answer in expanded form? Ans wer: 0.0000000000009095 Scientific Notation TEACHER NOTES ©2013 Texas Instruments Incorporated 6 education.ti.com Part 3 – Ordering Numbers in Scientific Notation 11. Place the following numbers on the number line. After p lacing them on the number line, switch with a partner to check answers. Answer: Draw an appropriate number line for the following numbers. Explain your choice . Answer: All the numbers are between –1 and 1. The larger the negative exponent, the cl oser the number is to zero. A possible number line is given.
2994	https://medium.com/@jmiller_68726/you-have-started-with-a-fundamental-assumption-that-a-fraction-such-as-1-3-is-equal-to-its-3c80c05a31aa	You have started with a fundamental assumption that a fraction, such as 1/3, is equal to its "decimal representation". This is not necessarily true, and if not true, then your proofs fall apart. - Jmiller - Medium Sitemap Open in app Sign up Sign in Write Search Sign up Sign in Take the fraction 1/3. The decimal representation of 1/3 is 0.333… repeating forever. Now multiply both sides by 3: This One Math Paradox Will Change How You See Numbers Forever 375 11 Ritvik Nayak Jmiller Follow Aug 7, 2025 31 Listen Share You have started with a fundamental assumption that a fraction, such as 1/3, is equal to its "decimal representation". This is not necessarily true, and if not true, then your proofs fall apart. 31 31 Follow Written by Jmiller ------------------ 0 followers ·1 following Follow No responses yet Write a response What are your thoughts? Cancel Respond Recommended from Medium In An Injustice! by J. Henderson 10 Charlie Kirk Quotes, Ranked from Simply Bad to Utterly Horrible ------------------------------------------------------------------ ### Everyone has the right to free speech, but what happens when it turns into radicalism and extremism? Sep 20 364 In Long. Sweet. Valuable. by Ossai Chinedum I’ll Instantly Know You Used Chat Gpt If I See This --------------------------------------------------- ### Trust me you’re not as slick as you think May 16 1441 In Bitchy by Maria Cassano Studies Show That Predators Target Women Based on One Thing ----------------------------------------------------------- ### Attackers all chose similar victims — and it wasn’t what they were wearing. 5d ago 295 Will Lockett The AI Bubble Is About To Burst, But The Next Bubble Is Already Growing ----------------------------------------------------------------------- ### Techbros are preparing their latest bandwagon. Sep 14 341 Yash Batra How Netflix Accidentally Proved Monoliths Scale Better Than Microservices ------------------------------------------------------------------------- ### For a decade, the dominant Silicon Valley mantra has been: monoliths don’t scale; microservices do. Netflix itself helped popularize this… 6d ago 40 In Coding Beauty by Tari Ibaba This Secret New Coding Model Just Shocked The World --------------------------------------------------- ### The crazy thing is nobody knows exactly who’s behind it. 6d ago 18 See more recommendations Help Status About Careers Press Blog Privacy Rules Terms Text to speech
2995	https://www.youtube.com/watch?v=pnZtSc9qaJo	The Frequency and Period of Sine and Cosine Functions Mandy's Math World 2760 subscribers Description 12990 views Posted: 21 Mar 2022 Explore the relationship between the frequency and period of the sine and cosine trigonometric functions. Access the notes and assignment to accompany the video from my TpT store at: Transcript: these are essentially your day three notes over graphing sine and cosine functions and today we're focusing on the relationship between the frequency and the period okay so you can really understand the difference between the two and how they're related so i have the sine and cosine functions and transformations shown for each the b value is the frequency whatever this value is is the frequency and that's the number of cycles how many times it repeats in a given interval which is generally 2 pi it will let me know the period because it changes the period and remember the period is the length of one cycle and i can find that period by taking 360 and dividing it by that number or 2 pi and dividing it by that number so the frequency is the value of b and i can find the period by taking 360 or 2 pi and dividing it by that number so let's look at these first two examples it says to determine the period and the frequency for each trig function so i know that the frequency is the value of b in this case the frequency is eight in number two the frequency is three-fifths now there's some other stuff going on here right like y equals four sine eight x that four is going to change my amplitude all right it has nothing to do with the period and the frequency it changes how high and low this this function is going to go in number two it says y equals negative one-half sine three-fifths x well the negative is going to reflect it across the x-axis the half changes my amplitude it has nothing to do with the frequency and the period so now let's talk about the period of each one what i'm going to do is i take to find that period i'm going to take 2 pi and i'm going to divide it by the frequency i'm going to divide it by 8. 2 pi over 8 simplifies to pi over 4. so the period for this particular function is pi over 4. that's the length of one cycle if one cycle is pi over four and generally you know one interval is generally two pi then that means this sine function is going to be pi over four there's another pi over four there's another power before there's another and i can fit eight of those in that interval of two pi so how many of those can i how many of those cycles can i fit in that interval of two pi so let's look at this next one three-fifths x that is the frequency how do i find the period i take two pi and i divide it by three over five okay well how do i divide fractions two pi over one keep change or not keep change flip or yeah change it to a multiplication and then flip the three fifths to five over three now i just multiply across two pi times five is ten pi 1 times 3 is 3 and i take 10 pi and divide it by 3 which this is kind of an ugly interval if you're going to graph it but if you want your your values for your little tick marks you have to take 10 pi and divide it by four if you're graphing it on your graph which i'm not going to make you do so so the frequency and the period because you know you can see this they are inversely related what does that mean that means as one increases the other decreases right so if the period decreases the frequency has got to increase because that's going to take more cycles to fit in the two pi right and if the period increases the frequency is going to decrease it's going to take less to fit in that 2 pi so now what about if you're given the period can you frame the frequency and if you're given the frequency can you find the period oh that looks kind of bad right there i can fix that so let's go ahead and do that um you know they are they're inversely related i've got up here if you're given the frequency you're going to do 2 pi divided by the period and if you're given the period you're going to do 2 pi divided by the frequency so i know that like let's say i wanted to show this kind of you know using some algebra kind of rearrange this equation well how can i how can i do that i could multiply both sides by the period right which what does that mean that means the period times the frequency is two pi okay so i can just use a little bit of algebra to rearrange these equations to show you you know frequency is two part divided period period is 2 pi divided by the frequency and then the frequency times the period is 2 pi so if i'm given the period what am i going to do to find the frequency i'm going to do 2 pi divided by the period which is 22 and when i simplify that i get pi over 11. the frequency for this is pi over 11 okay that means that when my peer and i know that kind of looks like weird right because normally our frequency are like whole numbers and then our periods are um you know the ones that contain pi but in this case it's um the opposite and that's okay all right just use your rules you know 2 pi divided by the period is going to be the frequency in this case we have 2 pi over 7 as the period how do i find the frequency i take 2 pi and i divided by 2 pi over 7. how do i do that 2 pi over 1 and then i multiply by the reciprocal which is 7 over two pi and i can pre-simplify that's what i call it so i can cross off my two pi's and i get seven over one which is just seven so now let's move over to the other part where what if i'm given the frequency and i need to find the period well in this case i'm given a frequency of pi over eight what do i do to find the period i take two pi and i divide it by pi over eight and how do i divide fractions i write this as a fraction and i multiply by the reciprocal which is eight over pi so what happens the pies cancel two times eight is sixteen over one which is just 16. so in this case the frequency is pi over 8 and the period is 16. looking at number 6 if the frequency is 24 what is the period well let me ask you this is the period gonna be really small or is it gonna be really big it's gonna be small because 24 cycles of this particular function are going to fit within the interval of 2 pi which means the period is going to have to be small okay so i'm going to take 2 pi and divide it by 24 and i get simplify it pi over 12. so the period is in fact pi over 12. and that concludes your notes over just extra practice with frequency and period showing how they are related how you can find one when you are given the other and remember they're inversely related so as one increases the other decreases and we just need to know one to find the other i hope i hope it was helpful
2996	https://stackoverflow.com/questions/55676436/calculate-the-eccentricity-of-ellipses	r - Calculate the eccentricity of ellipses - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Calculate the eccentricity of ellipses Ask Question Asked 6 years, 5 months ago Modified4 years ago Viewed 405 times Part of R Language Collective This question shows research effort; it is useful and clear -1 Save this question. Show activity on this post. I've like to calculate the eccentricity of ellipses x, in my example: ```r Artificial ellipse a <- 1 # semi-major axis e <- x# eccentricity b <- a sqrt(1 - e^2) # semi-minor axis c <- 1.3 # ellipse area ``` But I need to make this using the ellipse area (c) in the calculation. This is poisible? R Language Collective r geometry geometry-surface Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications asked Apr 14, 2019 at 14:35 IsabelIsabel 333 1 1 silver badge 11 11 bronze badges 7 Eccentricity is a measure of how much a shape (cones, parabola's, etc) differs from a true (i.e. original) circle. Do you link this with ellipse area?Leprechault –Leprechault 2019-04-14 14:53:27 +00:00 Commented Apr 14, 2019 at 14:53 Yes my basic idea is make this by the use of area (area= minor axis major axis pi) and eccentricity definitions.Isabel –Isabel 2019-04-14 15:01:00 +00:00 Commented Apr 14, 2019 at 15:01 1 You cannot calculate eccentricity using only area. What parameters are known?MBo –MBo 2019-04-14 16:02:20 +00:00 Commented Apr 14, 2019 at 16:02 MBo my parameters was area, minor and major axis.Isabel –Isabel 2019-04-14 16:20:01 +00:00 Commented Apr 14, 2019 at 16:20 But ....if you have both axes, you don't need area to find eccentricity. What is exact problem formulation?MBo –MBo 2019-04-14 17:09:08 +00:00 Commented Apr 14, 2019 at 17:09 \|Show 2 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. r b = c/(pia); x = (sqrt(a^2 - b^2))/a; is the formula you need. Or put togther: r x = (sqrt(a^2 - (c/(pia))^2))/a; Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered May 15, 2019 at 14:55 FuturologistFuturologist 1,924 2 2 gold badges 9 9 silver badges 11 11 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! 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2997	https://www.wordreference.com/synonyms/repel	repel - WordReference.com English Thesaurus WordReference.com \| Online Language Dictionaries English Thesaurus \| repel × Forums See Also: reparable reparation repartee repay repayment repeal repeat repeated repeatedly repeating repel repellent repelling repent repentance repentant repercussion repertoire repetition repetitious repetitive Recent searches: repel View Allrepel [links] Listen: UK:UK and possibly other pronunciations UK and possibly other pronunciations/rɪˈpɛl/US:USA pronunciation: IPA and respelling USA pronunciation: IPA/rɪˈpɛl/ ,USA pronunciation: respelling(ri pel′) ⓘ One or more forum threads is an exact match of your searched term definition \| Conjugator \| in Spanish \| in French \| in context \| images Inflections of 'repel' (v): (⇒ conjugate)repels v 3rd person singular repelling v pres p repelled v past repelled v past p repel WordReference English Thesaurus © 2025 Sense: To throw back Synonyms: rebuff, resistCollocations, deflect, withstandCollocations, opposeCollocations, repulse, drive back, beat back, hold back, push back, ward off, stave off, fight off, hold off, drive off Antonyms: fallCollocations, collapseCollocations, succumb, be beaten, be overcome, be conquered, retreatCollocations, fall back, cave in, give up, surrenderCollocations, capitulate Sense: To cause aversion Synonyms: offend, disgust Antonyms: delightCollocations, pleaseCollocations, arouse, exciteCollocations, turn on, attract, lureCollocations, drawCollocations, appealCollocations, enchant, temptCollocations, captivate, fascinate, carry away Sense: To reject Synonyms: dismissCollocations, cast aside, spurn, drive off Antonyms: acceptCollocations, agreeCollocations, tolerate, put up with, approveCollocations, sanctionCollocations, allowCollocations, aidCollocations, helpCollocations, assistCollocations Is something important missing? Report an error or suggest an improvement. 🗣️Recent forum discussions about thesaurus entries: constrain or repel - English Only forum dismiss/repel (Edit: expel) - English Only forum night repel drill - English Only forum push back / push away / repel - English Only forum repel liquids and resist wrinkles - English Only forum repel repulse - English Only forum repel someone to do something? - English Only forum surveying the operations to repel the military campaigns against Aleppo - English Only forum tested to effectively repel - English Only forum Wall was built to repel the enemy. - English Only forum Visit the English Only Forum. Help WordReference: Ask in the forums yourself. Go to Preferences page and choose from different actions for taps or mouse clicks. Translations:Spanish \| French \| Portuguese \| Italian \| German \| Dutch \| Swedish \| Polish \| Romanian \| Czech \| Greek \| Turkish \| Chinese \| Japanese \| Korean \| Arabic Links:⚙️Preferences \| Privacy Policy \| Terms of Service \| Support WR \| Forums \| SuggestionsAdvertisements Advertisements Report an inappropriate ad. WordReference.com WORD OF THE DAY travel \| rack GET THE DAILY EMAIL! Become a WordReference Supporter to view the site ad-free. Chrome users: Use search shortcuts for the fastest search of WordReference. Copyright © 2025 WordReference.com Please report any problems.
2998	https://pmc.ncbi.nlm.nih.gov/articles/PMC4923431/	Biomedical Waste Management : An Infrastructural Survey of Hospitals - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Med J Armed Forces India . 2011 Jul 21;60(4):379–382. doi: 10.1016/S0377-1237(04)80016-9 Search in PMC Search in PubMed View in NLM Catalog Add to search Biomedical Waste Management : An Infrastructural Survey of Hospitals SKM Rao SKM Rao Associate Professor, Department of Hospital Administration, Armed Forces Medical College, Pune-40 Find articles by SKM Rao , RK Ranyal RK Ranyal +Deputy Director(Hospital Services), ECHS Regional Centre, C/o Air Force Station, Begumpet, Secunderabad Find articles by RK Ranyal +, SS Bhatia SS Bhatia Research Pool Officer/Director General Medical Services-1(b), O/o Director General Medical Services(Army), New Delhi Find articles by SS Bhatia , VR Sharma VR Sharma Medical Officer (Hospital Services), Military Hospital (Cardio Thoracic Centre), Pune-40 Find articles by VR Sharma Author information Article notes Copyright and License information Associate Professor, Department of Hospital Administration, Armed Forces Medical College, Pune-40 +Deputy Director(Hospital Services), ECHS Regional Centre, C/o Air Force Station, Begumpet, Secunderabad Research Pool Officer/Director General Medical Services-1(b), O/o Director General Medical Services(Army), New Delhi Medical Officer (Hospital Services), Military Hospital (Cardio Thoracic Centre), Pune-40 Received 2002 Jul 27; Accepted 2003 Jun 20; Issue date 2004 Oct. . PMC Copyright notice PMCID: PMC4923431 PMID: 27407678 Abstract Background The Ministry of Environment & Forests notified the Biomedical Waste (management & handling) Rules, 1998” (BMW Mgt) in July 1998. In accordance with the rules, every hospital generating BMW needs to set up requisite BMW treatment facilities on site or ensure requisite treatment of waste at common treatment facility. No untreated BMW shall be kept stored beyond a period of 48 hours. The cost of construction, operation and maintenance of system for managing BMW represents a significant part of overall budget of a hospital if the BMW rules have to be implemented in their true spirit. Two types of costs are required to be incurred by hospitals for BMW Mgt, internal and external. Internal cost is the cost for segregation, mutilation, disinfection, internal storage and transportation including hidden cost of protective equipment. External costs are off site transportation, treatment and final disposal. Methods A study of hospitals was carried out from various sectors like Govt, Private, Charitable institutions etc. to assess the infrastructural requirement for BMW Mgt. Cost was worked out for a hospital where all the infrastructure as per each and every requirement of BMW rules had been implemented and then it was compared with other hospitals where hospitals have made compromises on each stage of BMW Mgt. Results Capital cost incurred by benchmarked hospital of 1047 beds was Rs.3 lakh 59 thousand excluding cost of incinerator and hospital is incurring Rs. 656/- per day as recurring expenditure. Pune city has common regional facility for BMW final disposal. Facility is charging Rs.20 per kg of infectious waste. As on Dec 2001 there were 400 institutions including nursing homes, labs and blood banks which were registered. Conclusion After analyzing the results of study it was felt that there is an urgent need to standardize the infrastructural requirement so that hospitals following BMW rules strictly do not suffer additional costs. Key Words: Biomedical waste, Cost, Hospital, Infrastructure Introduction Hospital is one of the complex institutions which is frequented by people from every walk of life in the society without any distinction between age, sex, race and religion. This is over and above the normal inhabitants of hospital i.e patients and staff. All of them produce waste which is increasing in its amount and type due to advances in scientific knowledge and is creating its impact . The hospital waste, in addition to the risk for patients and personnel who handle these wastes poses a threat to public health and environment . Keeping in view inappropriate biomedical waste management, the Ministry of Environment and Forests notified the “Biomedical Waste (management and handling) Rules, 1998” in July 1998. In accordance with these Rules (Rule 4), it is the duty of every “occupier” i.e a person who has the control over the institution and or its premises, to take all steps to ensure that waste generated is handled without any adverse effect to human health and environment. The hospitals, nursing homes, clinic, dispensary, animal house, pathological lab etc., are therefore required to set in place the biological waste treatment facilities. It is however not incumbent that every institution has to have its own waste treatment facility. The rules also envisage that common facility or any other facilities can be used for waste treatment. However it is incumbent on the occupier to ensure that the waste is treated within a period of 48 hours. Biomedical Waste Management Process Handling, segregation, mutilation, disinfection, storage, transportation and final disposal are vital steps for safe and scientific management of biomedial waste in any establishment . The key to minimisation and effective management of biomedical waste is segregation (separation) and identification of the waste. The most appropriate way of identifying the categories of biomedical waste is by sorting the waste into colour coded plastic bags or containers. Biomedical waste should be segregated into containers/ bags at the point of generation in accordance with Schedule II of Biomedical Waste (management and handling) Rules 1998 as given in Table 1. Table 1. Colour coding-biomedical waste (management and handling) rules, 1998 (schedule II) \| Colour coding \| Type of container \| Waste categories \| :--- \| Yellow \| Plastic bags \| Cat 1 human anatomical waste, \| \| \| \| Cat 2 animal waste, \| \| \| \| Cat 3 microbiology waste, \| \| \| \| Cat 6 soiled waste. \| \| Red \| Disinfected container plastic bags \| Cat 3 Microbiological Cat 6 soiled Cat 7 solid waste (Waste IV tubes catheters, etc.) \| \| Blue/White \| Plastic bag/puncture proof containers \| Cat 4 waste sharps Cat 7 plastic disposable tubings, etc. \| \| Black \| −do- \| Cat 5 discarded medicines \| \| \| \| Cat 9 incineration ash \| \| \| \| Cat 10 chemical waste \| Open in a new tab General waste like garbage, garden refuse etc. should join the stream of domestic refuse. Sharps should be collected in puncture proof containers. Bags and containers for infectious waste should be marked with Biohazard symbol. Highly infectious waste should be sterilised by autoclaving. Cytotoxic wastes are to be collected in leak proof containers clearly labelled as cytotoxic waste . Needles and syringes should be destroyed with the help of needle destroyer and syringe cutters provided at the point of generation. Infusion sets, bottles and gloves should be cut with curved scissors. Disinfection of sharps, soiled linen, plastic and rubber goods is to be achieved at point of generation by usage of sodium hypochlorite with minimum contact of 1 hour. Fresh solution should be made in each shift. On site collection requires staff to close the waste bags when they are three quarters full either by tying the neck or by sealing the bag. Kerb side storage area needs to be impermeable and hard standing with good drainage. It should provide an easy access to waste collection vehicle . Biomedical waste should be transported within the hospital by means of wheeled trolleys, containers or carts that are not used for any other purpose. The trolleys have to be cleaned daily. Off site transportation vehicle should be marked with the name and address of carrier. Biohazard symbol should be painted. Suitable system for securing the load during transport should be ensured. Such a vehicle should be easily cleanable with rounded corners. All disposable plastic should be subjected to shredding before disposing off to vendor. Final treatment of biomedical waste can be done by technologies like incineration, autoclave, hydroclave or microwave. Cost of Biomedical Waste Management The cost of construction, operation and maintenance of system for managing biomedical waste represents a significant part of overall budget of a hospital if the BMW handling rules 1998 have to be implemented in their true spirit. Govt of India in its pilot project for hospital waste management in Govt hospitals has estimated Rs.85 lakh as capital cost in 1000 bedded super speciality teaching hospital which includes on site final disposal of BMW. Two types of costs are required to be incurred by hospitals for BMW mgt, internal and external. Internal cost is the cost for segregation, mutilation, disinfection, internal storage and transportation including hidden cost of protective equipment. External cost involves off site transport of waste, treatment and final disposal . Common Regional Facility For Final Disposal of Infectious BMW Hospitals, private practitioners, emergency care centers though aware of the rules do not have the time or resources to arrange satisfactory disposal of biomedical waste. Self contained on site treatment methods may be desirable and feasible for large healthcare facilities. They will not be practical or economical for smaller institutes. An acceptable common system should be in place which will provide free supply of colour coded bags, daily collection of infectious waste, safe transportation of waste to off site treatment facility and final disposal with suitable technology. Material and Methods A study was conducted at following hospitals:- A Semi Govt teaching hospital 540 beds B Charitable trust hospital 540 beds C Govt hosp (state run hospital)1296 beds D Private hospital 223 beds E Govt hosp (service hospital)1047 beds Open in a new tab Keeping in view the infrastructural requirement for BMW management and adherence to BMW Rules 1998, total cost in terms of capital cost as well as operational cost (per month) was worked out for BMW management. Methods of storage and segregation at ward / department level, internal transportation, kerb side storage, external transportation and on site final disposal / off site disposal were studied for all 5 hospitals by direct observation and infrastructure for the same was studied. Informal discussion with various hospital functionaries was carried out. Common regional facility for final disposal of infectious waste was also studied. Discussion 1. Comparison of five hospitals studied in terms of bed strength, waste generation, method of final disposal etc. has been tabulated in Table 2. 2. Total cost of BMW management at each hospital in terms of capital cost and recurring expenditure is depicted in Table 3, Table 4. Hospital E is following BMW Rules in totality, and hence has incurred high capital costs. Recurrent expenditure was found to be less in Hospital C vis a vis the quantity of waste due to non adherence to BMW Rules. 3. It was observed that Govt service hospital where BMW management was implemented as per BMW 1998 rules had process of segregation of waste at generation level into various colour coded bags and plastic drums, internal transportation of waste through trolleys to kerb site, storage at kerb site in various colour coded metallic drums, movement of plastic disposable waste to plastic shredders and transportation of infectious waste to on site incinerator was found to be completely in place. Safai karmacharis were using complete protective equipment like gloves, masks, shoes etc while handling waste and also sodium hypochlorite was available in each and every sharp generating site in wards as well as departments. The capital and recurring cost worked out for complete process thus was found to be high as compared to other hospitals as other hospitals had some shortcomings at every step of BMW management. 4. Plastic waste receptacles of different colour codes were being used in other non Govt hospitals without any consideration of rules. However, all these hospitals were collecting infectious waste into yellow colour coded bags (biodegradable) provided by Image India, a central facility. No plastic bags were being used for lining plastic buckets / receptacles for other type of wastes and other waste was being collected in single container. All the 3 hospitals were using needle destroyers for disposable needles. Plastic iv bottles, catheters and disposables were being mutilated physically and then either were sold to contractor or donated. Hospital B, a private hospital was utilizing facility of onsite incineration for infectious waste disposal. Incinerator was found to be twin chambered oil fired with regular checks from pollution control board. Hospitals A & D were sending their infectious waste to central facility of incineration. No hospital safai karmacharis were found to be using complete protective equipment, some of them were using latex gloves. In hospital A, waste destined for incineration is physically checked by laying out the waste and manual segregation is carried out which can result in injury to health care workers and should be avoided at any cost. Sodium hypochlorite is being very sparingly used and fresh solution was not available at most of the hospitals. 5. State run hospital was having plastic buckets of assorted variety at ward and department levels for waste collection, however most of them were in broken condition and no replacement was provided. Use of hypochlorite solution for sharps was non-existent. Proper segregation was not being carried out. Plastic waste was taken by rag pickers and sold to contractors at hospital waste dump itself. Trolleys for transportation of waste were being used for other purposes like carriage of linen and stores etc. Needles were being destroyed at source, however most of the needle destroyers were non-functional and physical mutilation was being resorted to which can prove to be dangerous for health care worker (HCW). Onsite incinerator is available and is oil fired double chambered type of very old vintage, however it remains non functional most of the time resulting in all the infectious waste finding its place in main municipality dump. Local municipal body as well as state Pollution Control Board do not check existing waste disposal arrangements of this hospital. 6. This city has implemented common regional facility for final disposal of biomedical waste generated by health care establishments. It has appointed M/s Image India to, offer the services of handling BMW on pay and use basis. The services include provision of bags, collection of bags containing infectious waste from all the hospitals with more than 20 beds, their transportation to the incinerator site, its incineration and final disposal of ash. They are collecting infectious waste from 446 nursing homes / hospitals all over the city. Approximately 1000 kg per day collection of waste is made with the help of two modified Tata 407 trucks. Three incinerators of twin chamber variety with approved chimney size with a total capacity of 40 kg/hr burning capacity each are functional. They are charging Rs.20 / kg of waste. Billing is done through municipal body and facility is being monitored by Municipality / State Pollution Control Board regularly. 7. Municipal Corporation / State Pollution Control Board checks only common waste facility. Nursing homes / hospitals registered with this facility are under constant scrutiny and are punished by levying fine, if any disposable plastic / sharps are found in yellow bags leading to forced re-checks of their waste which can result in injuries to HCWs. Other nursing homes, dental practitioners, hospitals, research facilities and private practitioners continue to dump their waste into main municipal garbage. Use of central incineration facility should be made compulsory for those hospitals who have defective incinerators as they are a source of pollution. Table 2. Comparison of hospitals surveyed in Pune: method of disposal of BMW \| \| Beds \| Waste \| Infectious waste \| Plastic waste \| Final disposal \| :--- :--- :--- \| \| A Hosp Semi Govt teaching \| 540 \| 500 Kg \| 10% \| Sold without shredding \| Sent to off site facility \| \| B Hosp private \| 540 \| 510 Kg \| 10% \| Given to PMC for selling \| Onsite incineration \| \| C Hosp State Govt \| 1296 \| 735 Kg \| 9% \| Picked up by rag pickers \| Onsite incineration \| \| D Hosp charitable \| 223 \| 180 Kg \| 10% \| Mutilated and sold \| Sent to off site facility \| \| E Hosp service \| 1047 \| 821 Kg/day \| 15.9% \| Shredded & sold \| Onsite incineration \| Open in a new tab Table 3. Cost of biomedical waste management capital cost \| \| Capital cost \| A Hosp \| B Hosp \| C Hosp \| D Hosp \| E Hosp \| :--- :--- :--- \| (a) \| Plastic drums \| 54000/− \| 56400/− \| 52650/− \| 20260/− \| 45580/− \| \| (b) \| Metal drums for kerb sites \| Same are being used \| Same are being used \| Same are being used \| 9000/− \| 33600/− \| \| (c) \| Protective gear for waste handlers \| 1000/− \| 7500/− \| 1584/− \| 1000/− \| 21250/− \| \| (d) \| Syringe and needle destroyers \| 9500/− \| 13000/− \| 24000/− \| 13600/− \| 48000/− \| \| (e) \| Plastic shredder \| − \| − \| − \| − \| 180000/− \| \| \| \| \| \| \| \| (10-20 Kg/hr) \| \| (f) \| Hand carts \| 20000/− \| Not being used \| 23500/− \| 9000/− \| 40000/− \| \| (g) \| Weighing scale \| 500/− \| Not being used \| Not being used \| 1000/− \| 3250/− \| \| \| Total \| 85000/− \| 76900/− \| 101734/− \| 53860/− \| 371680/− \| Open in a new tab Table 4. Cost of consumables in BMW management per month (In rupees) \| \| A Hosp \| B Hosp \| C Hosp \| D Hosp \| E Hosp \| :--- :--- :--- \| \| Plastic bags \| 6000 \| 12000 \| 9750 \| 1800 \| 20000 \| \| Hypochlorite Rs.50/lit \| 5000 \| 6750 \| 7500 \| 5000 \| 7500 \| \| Gloves/Masks \| 2000 \| 2000 \| 2000 \| 2000 \| 7000 \| \| Outsourcing/incineration \| 20000 \| 4500 \| 9000 \| 24000 \| 2000 \| \| Total \| 33000 \| 25250 \| 28250 \| 32800 \| 36500 \| Open in a new tab Recommendations 1. After analysing the results of the study it was felt that there is an urgent need to standardise the infrastructural requirement so that hospitals following BMW handling rules meticulously do not suffer additional costs. 2. Hospitals having defunct / defective incinerators should be made to utilise central incineration facility as efforts of Govt are towards reducing the number of incinerators in cities to prevent rise in air pollution. 3. Small health care establishments in city which have still not registered with central facility should be encouraged to register thereby bringing down the operating cost of contractor and decrease the cost of incineration per kg. 4. Govt hospitals which at present are totally left on their own, should be brought into net of rigorous checking as far as BMW management is concerned and a corpus grant can be allotted to them to improve their infrastructural requirements for which provision exists in Govt of India Rules. 5. Community is utilising the services of hospitals and by “Polluter Pays” principle, it needs to contribute in building infrastructure for BMW mgt. This contribution can be in the form of assistance in sharing the cost of consumables and capital cost of BMW mgt by Municipality, State Govt, Public bodies and Voluntary bodies like Rotary Club etc. References 1.Rao SKM, Garg RK. A study of Hospital Waste Disposal System in Service Hospital. Journal of Academy of Hospital Administration July. 1994;6(2):27–31. [Google Scholar] 2.Singh IB, Sarma RK. Hospital Waste Disposal System and Technology. Journal of Academy of Hospital Administration, July. 1996;8(2):44–48. [PubMed] [Google Scholar] 3.Acharya DB, Meeta Singh. The book of Hospital Waste Management. Minerva Press; New Delhi: 2000. p. 15. [Google Scholar]; Acharya DB, Meeta Singh. The book of Hospital Waste Management. Minerva Press; New Delhi: 2000. p. 47. [Google Scholar] 4.Srivastava JN. Hospital Waste Management project at Command Hospital, Air Force, Bangalore. National Seminar on Hospital waste Management: a report 27 May 2000. 5.A study of Hospital Waste Management System in Command Hospital (Southern Command), Pune; a dissertation submitted to University of Pune. Wg Cdr RK Ranyal. Dec 2001:37. [Google Scholar] Articles from Medical Journal, Armed Forces India are provided here courtesy of Elsevier ACTIONS View on publisher site PDF (29.7 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Biomedical Waste Management Process Cost of Biomedical Waste Management Common Regional Facility For Final Disposal of Infectious BMW Material and Methods Discussion Recommendations References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
2999	https://math.stackexchange.com/questions/4476235/step-in-a-proof-that-axy-axy	real analysis - Step in a proof that $(a^x)^y = a^{xy}$. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Step in a proof that (a x)y=a x y(a x)y=a x y. Ask Question Asked 3 years, 3 months ago Modified3 years, 2 months ago Viewed 1k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I'm trying to fill in the details in this proof, which seeks to show that for any real numbers a,x,y a,x,y with a>0 a>0 we have that (a x)y=a x y(a x)y=a x y. The sketch is as follows: (a x)y=sup{(sup{a r:r∈Q<x})s:s∈Q<y}=sup{sup{(a r)s:r∈Q<x}:s∈Q<y}=sup{(a r)s:r∈Q<x∧s∈Q<y}=sup{a t:t∈Q<x y}=a x y(a x)y=sup{(sup{a r:r∈Q<x})s:s∈Q<y}=sup{sup{(a r)s:r∈Q<x}:s∈Q<y}=sup{(a r)s:r∈Q<x∧s∈Q<y}=sup{a t:t∈Q<x y}=a x y I'm having trouble figuring out how the second equality (the transition from first to second line) is arrived at. In particular, we need to show that (sup{a r:r∈Q<x})s=sup{(a r)s:r∈Q<x}.(sup{a r:r∈Q<x})s=sup{(a r)s:r∈Q<x}. I have been able to show that (a x)s(a x)s is indeed an upper bound of {(a r)s:r∈Q<x}{(a r)s:r∈Q<x}, but I'm struggling in showing that it is its least upper bound. An idea is, for a fix p p, to use the continuity of the functions t→t p t→t p and t→p t t→p t, but I would like to avoid this since the benefit of this cumbersome approach to exponentiation is precisely that no Analysis beyond the basic properties of R R is needed. I was wondering how else I could proceed. real-analysis exponentiation Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jul 3, 2022 at 2:53 2'5 9'2 57.2k 8 8 gold badges 89 89 silver badges 162 162 bronze badges asked Jun 19, 2022 at 22:25 SamSam 5,300 3 3 gold badges 19 19 silver badges 53 53 bronze badges 4 2 If you ignore all the sups and make sense of the resulting expression can you see why they're the same set?CyclotomicField –CyclotomicField 2022-06-19 22:34:04 +00:00 Commented Jun 19, 2022 at 22:34 @CyclotomicField Which sets in particular?Sam –Sam 2022-06-19 23:06:59 +00:00 Commented Jun 19, 2022 at 23:06 @CyclotomicField Careful; for increasing f f, f(sup A)=sup f(A)f(sup A)=sup f(A) is only guaranteed if f f is continuous… which is exactly what needs to be checked here.Akiva Weinberger –Akiva Weinberger 2022-06-20 00:00:51 +00:00 Commented Jun 20, 2022 at 0:00 @Momo No. My question is precisely about filling in the details in the answer to that post.Sam –Sam 2022-06-20 17:39:11 +00:00 Commented Jun 20, 2022 at 17:39 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 5 Save this answer. Show activity on this post. Note: The original problem correctly assumes a>1 a>1, so it suffices to prove the equality for a>1 a>1, x,y>0 x,y>0. You may observe that ((a x)y=sup{sup…}(a x)y=sup{sup…} does not even hold when x<0 x<0 due to a x<1)a x<1). however, negative exponents can be reduced to positive case using for example a−x=1 a x a−x=1 a x. You may use Bernoulli's inequality for rational exponent which has elementary proofs. Then for every δ>0 δ>0 you may chose r r with a r>a x−δ a r>a x−δ (we'll impose additional restrictions on δ δ later). Case 1: s≥1 s≥1. We divide by a x a x a r a x>1−δ a x a r a x>1−δ a x Then useing Bernoulli: (a r a x)s>(1−δ a x)s≥1−s δ a x(a r a x)s>(1−δ a x)s≥1−s δ a x Finally: (a r)s>(a x)s−s(a x)s δ a x>(a x)s−ϵ(a r)s>(a x)s−s(a x)s δ a x>(a x)s−ϵ The latest inequality can be proven for all ϵ>0 ϵ>0 as long as δ δ is carefully chosen, i.e. 1−δ a x>0 1−δ a x>0 and ϵ>s(a x)s δ a x ϵ>s(a x)s δ a x, conditions which can be met for δ δ "small enough". Case 2: 0≤s<1 0≤s<1. We divide by a r a r and proceed similarly (with Bernoulli's inequality reversed). Thus you said that you already proved that (a x)s(a x)s is an upper bound of {(a r)s:r∈Q<x}{(a r)s:r∈Q<x} and I proved that (a x)s−ϵ(a x)s−ϵ is not an upper bound, so (a x)s(a x)s is sup sup Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jul 3, 2022 at 3:32 answered Jun 21, 2022 at 19:15 MomoMomo 16.3k 2 2 gold badges 26 26 silver badges 40 40 bronze badges 2 Terribly sorry for the long reply, but the post regarding Bernoulli's inequality for the rationals deals with the case α≥0 α≥0, an assumption which is essential in the proofs that use AM-GM as well as in the accepted, elementary proof. In your proof, making δ δ sufficiently small may guarantee α=−δ/a x≥−1 α=−δ/a x≥−1, but to apply the inequality there requires an elementary proof of Bernoulli's inequality that includes the case α≥−1 α≥−1.Sam –Sam 2022-07-06 23:20:15 +00:00 Commented Jul 6, 2022 at 23:20 1 A proof including the case α≥−1 α≥−1 can be found here math.stackexchange.com/questions/2920237/…Sam –Sam 2022-07-06 23:50:14 +00:00 Commented Jul 6, 2022 at 23:50 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis exponentiation See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 7Bernoulli's inequality for rational exponents 4Elementary proof of Bernoulli's inequality for rational exponents 4Proving exponent law for real numbers using the supremum definition only Related 0If b>1 b>1 and B(r)B(r) is the set of all numbers b t b t, where t t is rational and t≤r t≤r, prove that b r=sup B(r)b r=sup B(r) where r r is rational. 2Let E be a set of real numbers. Show that x is not an upper bound of E if and only if there exists a number e∈ E such that e > x 1Proof that the sup of a set is the least upper bound. 2Proving that k⋅sup S=sup(k S)k⋅sup S=sup(k S) for k≥0 k≥0 2Prove that sup(c A)=c sup(A)sup(c A)=c sup(A). 1Exponential supremum proof 1Proof that sup(A 2)=(sup A)2 sup(A 2)=(sup A)2. 1Proof of alternative definition of supremum. 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