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4. Car 204 transports daily necessities from Town A to Village B, passing through 20 kilometers of uphill road, 14 kilometers of downhill road, and 5 kilometers of flat road. It then transports grain from Village B back to Town A. The time difference for the round trip is 10 minutes. It is known that the speed ratio of... | [Solution] (1) According to the problem, let the average speeds of the car when going uphill, downhill, and on a flat road be $3v, 6v, 5v$ (km/h) respectively.
Since $\quad\left(\frac{20}{3 v}+\frac{14}{6 v}+\frac{5}{5 v}\right)-\left(\frac{20}{6 v}+\frac{14}{3 v}+\frac{5}{5 v}\right)=\frac{1}{6}$,
that is, $\frac{6}{3... | 1 \frac{2}{3} \text{ hours, } 1 \frac{1}{2} \text{ hours} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,614 |
4-205 There is a mixture of copper and zinc weighing 400 grams, with a volume of 50 cubic centimeters. It is known that the specific gravity of copper is less than 9 grams/cubic centimeter and greater than 8.8 grams/cubic centimeter, and the specific gravity of zinc is less than 7.2 grams/cubic centimeter and greater t... | [Solution] Let the mixture contain $x$ grams of copper and $y$ grams of zinc, and denote the specific gravities of copper and zinc as $D_{1}$ and $D_{2}$, respectively. Then we have:
$$\left\{\begin{array}{l}x+y=400, \\ \frac{x}{D_{1}}+\frac{y}{D_{2}}=50 .\end{array}\right.$$
Solving this system of equations, we get:
$... | 200 \text{ grams} \leqslant x \leqslant 233 \text{ grams}, \quad 167 \text{ grams} \leqslant y \leqslant 200 \text{ grams} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,615 |
4.206 There are two forces $f_{1}$ and $f_{2}$ acting on the origin $O$ of the coordinate axis,
$$\begin{array}{l}
\vec{f}_{1}=\overrightarrow{O A}=\sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right) \\
\vec{f}_{2}=\overrightarrow{O B}=2\left[\cos \left(-30^{\circ}\right)+i \sin \left(-30^{\circ}\right)\right]
\end{... | [Solution]
$$\text { (1) Resultant force } \begin{aligned}
\vec{f}= & \overrightarrow{f_{1}}+\overrightarrow{f_{2}} \\
= & \sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right) \\
& +2\left[\cos \left(-30^{\circ}\right)+i \sin \left(-30^{\circ}\right)\right] \\
= & 1+i+\sqrt{3}-i \\
= & \sqrt{3}+1 \\
= & (\sqrt{3}+1)\... | 2.1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,616 |
4. 207 Car A is 100 meters west of Car B. Now both cars start moving east at the same time. If Car A moves forward at a constant speed of 50 meters/second, and Car B accelerates forward at 20 meters/second ${ }^{2}$, how many seconds will it take for the two cars to be closest to each other? At that time, how far apart... | ［Solution］ Let the distance between the two cars be $S$ meters after $t$ seconds.
At this time, car A is $S_{1}=v t=50 t$ meters from the starting point, and car B is
$$\begin{aligned}
S_{2} & =100+\frac{1}{2} a t^{2}=100+10 t^{2} \\
S & =S_{2}-S_{1}=100+10 t^{2}-50 t \\
& =10\left(t-\frac{5}{2}\right)^{2}+\frac{75}{2}... | \frac{75}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,617 |
4. 209 There are two small piles of bricks. If 100 bricks are taken from the first pile and placed in the second pile, then the second pile will be twice as large as the first pile. If a certain number of bricks are taken from the second pile and placed in the first pile, then the first pile will be six times the size ... | [Solution] Let $x$ represent the number of bricks in the first pile, and $y$ represent the number of bricks in the second pile. Let $z$ be the number of bricks transferred from the second pile to the first pile according to the problem's conditions. At this point, the following equations hold:
$$\begin{array}{l}
2(x-10... | 170 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,618 |
4. 210 On the same route, there are four people: the first person is in a car, the second person is on a motorcycle, the third person is on a moped, and the fourth person is on a bicycle. The speeds of the vehicles are constant. The person in the car catches up with the person on the moped at 12:00, meets the person on... | [Solution] Let the speeds of the car, motorcycle, moped, and bicycle be $v_{1}, v_{2}, v_{3}, v_{4}$, respectively, and let $x, y$ represent the distances between the bicycle rider and the car passenger, and the bicycle rider and the motorcycle driver at 12 o'clock. According to the problem, we have:
$$\begin{array}{l}... | 15:20 | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,619 |
4- 2117 elves are sitting around a round table, and in front of each elf is a glass of milk. One elf pours all his milk evenly into the other glasses. Then his first neighbor to the right does the same, followed by the next one, and so on. After the 7th elf has poured his milk into the other glasses, there is still the... | [Solution] Initially, each cup contains milk of $\frac{6}{7}, \frac{5}{7}, \frac{4}{7}, \frac{3}{7}, \frac{2}{7}, \frac{1}{7}, 0$ liters.
In fact, after each elf pours milk to the others (giving each elf $\frac{1}{7}$), they still end up with the original distribution, just shifted by one position. The total amount of ... | \frac{6}{7} | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,620 |
4. 212 The river water is flowing, entering a still lake at point $Q$, a swimmer travels downstream from $P$ to $Q$, then across the lake to $R$, taking a total of 3 hours. If he travels from $R$ to $Q$ and then to $P$, it takes a total of 6 hours. If the lake water also flows, at the same speed as the river, then trav... | [Solution] Let the swimmer's speed be 1, the water speed be $y$, $P Q=a$, $Q R=b$, then
$$\left\{\begin{array}{l}
\frac{a}{1+y}+b=3 \\
\frac{a+b}{1+y}=\frac{5}{2} \\
\frac{a}{1-y}+b=6 .
\end{array}\right.$$
(1) - (2) gives $\frac{b y}{1+y}=\frac{1}{2}$, i.e., $b=\frac{1+y}{2 y}$.
(3) - (1) gives $\frac{a \cdot 2 y}{1-y... | \frac{15}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,621 |
4. 213 Given a rectangular box that can be filled with unit cubes. If it is instead filled with as many volume-2 cubes as possible, and these cubes are aligned with the edges of the box, then the box's volume is exactly filled to $40 \%$. Try to find the volume of such a rectangular box $(\sqrt[3]{2}=1.2599 \cdots)$. | [Solution] Let $Q$ be a rectangular box with the required properties, and let $a, b, c (a \leqslant b \leqslant c)$ be the side lengths of $Q$ that meet the conditions. Since this rectangular box can be filled with unit cubes, $a, b, c$ are natural numbers. Now, placing a cube with a volume of 2, its edge length is $\s... | 60 \text{ or } 30 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,622 |
4. $214 A$, $B$, $C$ three people are playing a game: On three cards, write down the integers $p$, $q$, $r$ ($0<p<q<r$). Mix these three cards and distribute them to $A$, $B$, $C$, each getting one. Then, give each person marbles according to the number on their card, and then collect the cards, leaving the marbles wit... | [Solution] Let the game have proceeded for $n$ rounds, then
$$n(p+q+r)=20+10+9=39$$
Since $00$ is contradictory. Therefore, the number of marbles $C$ obtained in the last round must be $p$, and the number of marbles $A$ obtained in the last round is $q$, so we have
$$2 r+q=20,2 p+r=10,2 q+p=9 .$$
Solving these equatio... | p=1, q=4, r=8 | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,623 |
$4 \cdot 215$ The sports meet lasted for $n$ days, and a total of $m$ medals were issued. On the first day, 1 medal was issued, as well as $\frac{1}{7}$ of the remaining $(m-1)$ medals. On the second day, 2 medals were issued, as well as $\frac{1}{7}$ of the remaining medals after that. This pattern continued for subse... | ［Solution］ Let the sports meet last for $k$ days, and there are still $a_{k}$ medals left. Where $k=0,1$, $2, \cdots, n$, and $a_{0}=m, a_{n}=0$. Then the number of medals issued on the $k$-th day is:
$$k+\frac{1}{7}\left(a_{k-1}-k\right)=\frac{1}{7} a_{k-1}+\frac{6}{7} k$$
So $a_{k}=a_{k-1}-\left(\frac{1}{7} a_{k-1}+... | n=6, m=36 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,624 |
$4 \cdot 216$ Try to find the number of all such positive integers: in their representation in base $n$, all digits are different, and each digit, except the leftmost one, differs from some digit to its left by $\pm 1$ (express the answer as a simple explicit function of $n$), and prove your conclusion. | [Solution] Although 0 cannot be the first digit, we first do not consider whether the first digit is 0, and define $F(n)$ as the number of positive integers that meet the conditions in the $n$-ary system, including those with the first digit being 0. For example,
$$\begin{array}{l}
F(1)=1,[0] \\
F(2)=4,[0,1,01,10] \\
F... | 2^{n+1}-2n-2 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,625 |
1. 63 In the infinite decimal expansion of a real number $a$, every digit is present. Let $v_{i}$ be the number of different digit sequences of length $n$ in this expansion. Prove: if for some $n$, the condition $V_{n} \leqslant n+8$ holds, then the number $a$ is rational. | [Proof] Clearly, $v_{1}=10$ and $v_{n} \leqslant v_{n+1}$.
If for all $n$ we have $v_{n+1}>v_{n}$, then
$$\begin{aligned}
v_{n} & \geqslant v_{n-1}+1 \geqslant v_{n-2}+2 \geqslant \cdots \\
& \geqslant v_{1}+n-1 \\
& =n+9>n+8
\end{aligned}$$
This contradicts the given condition, therefore, there must exist some natura... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,626 |
$4 \cdot 217$ assembly of products A, B, and C requires three types of parts, A, B, and C. Each product A requires 2 A's and 2 B's; each product B requires 1 B and 1 C; each product C requires 2 A's and 1 C. Using the inventory of A, B, and C parts, if p pieces of product A, q pieces of product B, and r pieces of produ... | [Proof] To assemble $p$ pieces of product A, $q$ pieces of product B, and $r$ pieces of product C, a total of $(2 p+2 r)$ parts A, $(2 p+q)$ parts B, and $(q+r)$ parts C are required. Therefore, after adding the remaining parts, the inventory of parts A is $2 p+2 r+2$ pieces, parts B is $2 p+q+1$ pieces, and parts C is... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,627 |
4. 218 In a middle school mathematics competition, three problems $A$, $B$, and $C$ were given. Among the 25 students who participated in the competition, each student solved at least one problem. Among the students who did not solve $A$, the number of students who solved $B$ is twice the number of students who solved ... | [Solution]. Let the number of people who solved not only $A$ be $x$, only $B$ be $y$, and did not solve $A$ but solved $B$ and $c$ be $z$. Then (refer to the diagram on the previous page),
$$\left\{\begin{array}{l}
x+x+1+y+z+\frac{y-z}{2}=25 \\
x+1=y+\frac{y-z}{2}
\end{array}\right.$$
Solving these equations, we get
$... | 6 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,628 |
4. 219 Prove: The cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic sequence.
保持了原文的换行和格式。 | [Proof] If not, then $\sqrt[3]{p_{1}}=a, \sqrt[3]{p_{2}}=a+m d, \sqrt[3]{p_{3}}=a+n d$, where $m$ and $n$ are integers. Eliminating $a$, we get
$$\frac{\sqrt[3]{p_{2}}-\sqrt[3]{p_{1}}}{\sqrt[3]{p_{3}}-\sqrt[3]{p_{1}}}=\frac{m}{n}$$
That is, $\left(m \sqrt[3]{p_{3}}-n \sqrt[3]{p_{2}}\right)^{3}=(m-n)^{3} p_{1}$.
Expan... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,629 |
$4 \cdot 220$ There is a group of children, the sum of their ages is 50 years, the oldest is 13 years old, and one is 10 years old; except for the 10-year-old child, the ages of the remaining children form an arithmetic sequence. How many children are there? How old is each child? | [Solution] Let the age of the oldest child be $a$, i.e., $a=13$.
Suppose there are $b+1$ children excluding the 10-year-old child, and let the common difference of their ages be $d$. Then the ages of these $b+1$ children are
$$a, a-d, a-2 d, \cdots, a-b d$$
Thus, $a+(a-d)+(a-2 d)+\cdots+(a-b d)$
$$=(b+1) a-\frac{b(b+1... | 5 \text{ children, ages } 13, 11, 10, 9, 7 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,630 |
4.221 For what natural numbers $n$ and $k$, do the binomial coefficients $C_{n}^{k-1}, C_{n}^{k}, C_{n}^{k+1}$ form an arithmetic sequence? | [Solution] The necessary and sufficient condition for three binomial coefficients to form an arithmetic sequence is that the following equation holds: $\square$
$$C_{n}^{k-1}-2 C_{n}^{k}+C_{n}^{k+1}=0$$
Let $k-1 \geqslant 0$ and $k+1 \leqslant n$, i.e., $1 \leqslant k \leqslant n-1$.
Multiplying both sides of (1) by t... | n=u^2-2, k=C_u^2-1 \text{ or } k=C_{u+1}^2-1 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,631 |
4. 222 Prove: If the lengths of the three sides of a right triangle form an arithmetic sequence, then their ratio is $3: 4: 5$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | [Proof] Let the lengths of the three sides of a right-angled triangle be $a-d, a, a+d$ (where $a>d$ $>0$), then
$$(a+d)^{2}=a^{2}+(a-d)^{2},$$
Simplifying, we get $\quad a^{2}-4 a d=0$.
Since $a \neq 0$, we have $a-4 d=0$, which leads to
$$a-d=3 d, a=4 d, a+d=5 d \text {. }$$
Thus, the ratio of the three sides of thi... | null | Logic and Puzzles | other | Yes | Yes | inequalities | false | 735,632 |
$4 \cdot 224$ unit squares are divided into 9 equal parts by lines parallel to the sides, and the central part is removed. The remaining 8 smaller squares are each divided into 9 equal parts by lines parallel to the sides, and the central part is removed. Then, a similar process is applied to each of the remaining squa... | ［Solution］（1）After the first division and removing one square from the resulting 9 squares, 8 squares with a side length of $\frac{1}{3}$ remain; after the second division, $8 \times 9$ squares are obtained, from which 8 squares are removed, leaving $8^{2}$ squares with a side length of $\frac{1}{3^{2}}$; if this proce... | 1 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 735,633 |
4-225 Prove: All terms of an arithmetic sequence composed of positive integers cannot all be prime numbers (except for the degenerate case, i.e., a sequence with a common difference of zero, where all terms can equal the same prime number). | [Proof] From the condition, we know that the common difference $d$ of the sequence is a positive integer. If $a_{r}$ is its general term, then
$$a_{r+s}=a_{r}+s d$$
Since $a_{1} \geqslant 1$ and $d>0$, there always exists a term $a_{r}$ in this sequence such that $a_{r}>1$ and $s=a_{r}$. At this time,
$$a_{r+s}=a_{r}+... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,634 |
4. 226 Several residential areas are each connected to the city center by roads, but there are no roads between the residential areas. A truck loaded with all the goods that need to be delivered to each residential area is to start from the city center and deliver to each point. The cost of transportation is calculated... | [Solution] It is only necessary to prove that the total transportation cost remains unchanged when the order of any two residential points is altered.
For this, it is only necessary to prove that the total transportation cost remains unchanged when the order of any residential point and the one immediately following i... | proof | Logic and Puzzles | proof | Yes | Yes | inequalities | false | 735,635 |
1.64 Prove that the necessary and sufficient condition for the sine and cosine of any angle to be expressible as rational numbers is: the tangent of the half-angle is either a rational number or undefined.
保持了原文的换行和格式。 | [Proof] Establish a rectangular coordinate system as shown, making the center of the unit circle coincide with the origin, and construct angle $\alpha$ with $O$ as the vertex and the $O x$ axis as the initial side, then we have
$$\cos \alpha=x, \quad \sin \alpha=\dot{y}$$
where $x$ and $y$ are the coordinates of point... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,636 |
$4 \cdot 227$ A person spent one coin to buy a loaf of bread and a bottle of kvass (a drink). When prices increased by $20 \%$, the coin was only enough to buy half a loaf of bread and a bottle of kvass. Question: If prices increase by another 20%, will the coin be enough to buy just a bottle of kvass? | [Solution] Let the coin value be 1. Before the price increase, the price of a loaf of bread is $x$, and the price of a bottle of kvass is $y$. According to the problem,
$$\left\{\begin{array}{l}
x+y=1 \\
1.2(0.5 x+y)=1
\end{array}\right.$$
Solving these equations, we get
$$y=\frac{2}{3} \text {. }$$
Since $\frac{2}{3... | \frac{24}{25}<1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,637 |
4. 228 If the values $x=\sin \alpha, y=\sin \beta$ are given, then the expression
$$z=\sin (\alpha+\beta)$$
in general has four different values. Try to write the equation connecting $x, y$ and $z$ without using radicals or trigonometric functions. And find the values of $x$ and $y$ for which $z=\sin (\alpha+\beta)$ h... | [Solution] From the given information, we have
$$\cos \alpha= \pm \sqrt{1-x^{2}}, \cos \beta= \pm \sqrt{1-y^{2}}$$
Therefore, $\approx=\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$ has the following four values:
$$\begin{array}{l}
z_{1}=x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}, \\
-z_{1}=-x \sqrt{1-y^{2}... | z^{4}-2\left(x^{2}-2 x^{2} y^{2}+y^{2}\right) z^{2}+\left(x^{2}-y^{2}\right)^{2}=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,638 |
4. 229 A decimal natural number $a$ consists of $n$ identical digits $x$, and the number $b$ consists of $n$ identical digits $y$, while the number $c$ consists of $2n$ identical digits $z$. For any $n \geqslant$ 2, find the digits $x, y, z$ that make $a^{2}+b=c$ true. | Then
$$\begin{array}{l}
b=\underbrace{\frac{n \uparrow}{y w \cdots}}_{n \uparrow}=y \cdot \underbrace{11 \cdots 1}_{n \uparrow} \text {, } \\
c=\frac{n \uparrow}{z \approx \cdots}=z \cdot \frac{11 \cdots 1}{2 n \uparrow} \text {. }
\end{array}$$
That is
$$\begin{array}{l}
x^{2} \cdot \underbrace{11 \cdots 1^{2}}_{n \u... | x=3, y=2, z=1, n \geqslant 2 ; x=6, y=8, z=4, n \geqslant 2 ; x=8, y=3, z=7, n=2 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,639 |
$4 \cdot 230$ If an integer $n$ can be expressed as
$$n=a_{1}+a_{2}+\cdots+a_{k},$$
where $a_{1}, a_{2}, \cdots, a_{k}$ are positive integers (not necessarily distinct) satisfying
$$\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}}=1$$
then we call $n$ a "good number".
It is known that the integers from 33 to 73... | [Proof] If $\left(a_{1}, a_{2}, \cdots, a_{k}\right)$ is a decomposition of a "good number" $n$, then we have
$$\left\{\begin{array}{l}
n=a_{1}+a_{2}+\cdots+a_{k} \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}}=1
\end{array}\right.$$
Multiplying (2) by $\frac{1}{2}$ gives
$$\frac{1}{2 a_{1}}+\frac{1}{2 a_{2}... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,640 |
4. 231 Find all real numbers $a$ such that there exist non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfying
$$\sum_{k=1}^{5} k x_{k}=a, \sum_{k=1}^{5} k^{3} \cdot x_{k}=a^{2}, \sum_{k=1}^{5} k^{5} x_{k}=a^{3} .$$ | [Solution] Let there be non-negative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ that satisfy the conditions. Then, by the Cauchy-Schwarz inequality,
$$\begin{array}{l}
\quad a^{4}=\left(\sum_{k=1}^{5} k^{3} x_{k}\right)^{2}=\left[\sum_{k=1}^{5}\left(k^{\frac{1}{2}} x_{k}^{\frac{1}{2}}\right)\left(k^{\frac{5}{2}} ... | 0,1,4,9,16,25 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,641 |
4-232 If $a, b, c$ are positive integers, satisfying $c=(a+b i)^{3}-107 i$, find $c$ (where $i^{2}=-1$).
| [Solution] From the given information, we have
$$c=\left(a^{3}-3 a b^{2}\right)+i\left(3 a^{2} b-b^{3}-107\right) .$$
Since $c$ is a positive integer, it follows that
$$3 a^{2} b-b^{3}-107=0$$
which simplifies to $b\left(3 a^{2}-b^{2}\right)=107$.
Since $a, b$ are positive integers, and 107 is a prime number, there c... | 198 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,642 |
4. 233 Let $a, b, c, d$ be odd numbers, $0<a<b<c<d$, and $a d=b c$. Prove that if $a+d=2^{k}, b+c=2^{m}$, where $k, m$ are integers, then $a=1$.
The text is translated while preserving the original line breaks and format. | [Proof] Since $a d=b c$, we have
$$\begin{aligned}
a[(a+d)-(b+c)] & =a^{2}+a d-a b-a c \\
& =a^{2}+b c-a b-a c \\
& =(a-b)(a-c) \\
& >0
\end{aligned}$$
Thus, $a+d>b+c$,
which implies $2^{k}>2^{m}$,
hence $k>m$.
Also, $b \cdot 2^{m}-a \cdot 2^{k}=b(b+c)-a(a+d)$
$$\begin{array}{l}
=b^{2}+b c-a^{2}-a d \\
=b^{2}-a^{2},
\... | a=1 | Number Theory | proof | Yes | Yes | inequalities | false | 735,643 |
4. 237 Prove that for infinitely many primes $p$, the equation $x^{2}+x+1=p y$ has integer solutions $(x, y)$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | [Proof] When $x=4$, $4^{2}+4+1=3 \times 7$. Therefore, when $p=3$ or 7, the original equation has integer solutions.
Assume the original equation has integer solutions $(x, y)$ for only a finite number of primes $p_{1}, p_{2}, \cdots, p_{m}$. Then we can take $x=p_{1} \cdot p_{2} \cdots \cdot p_{m}$, so
$$x^{2}+x+1=\l... | null | Logic and Puzzles | other | Yes | Yes | inequalities | false | 735,644 |
4.238 Let rational numbers $x, y$ satisfy the equation
$$x^{5}+y^{5}=2 x^{2} y^{2} \text {, }$$
Prove that $\quad 1-x y$ is the square of a rational number. | [Proof] Squaring both sides of the known equation, we get
$$\left(x^{5}+y^{5}\right)^{2}=4 x^{4} y^{4}$$
Subtracting $4 x^{5} y^{5}$ from both sides of (1), we get
$$\left(x^{5}-y^{5}\right)^{2}=4 x^{4} y^{4}(1-x y)$$
If $x y=0$, then $1-x y=1$ is the square of the rational number 1.
If $x y \neq 0$, then transforming... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,645 |
4. 240 Find three natural numbers $x, y, z$ that satisfy the equation $x^{3}+y^{4}=z^{5}$, and determine whether the solution set of this equation in the set of natural numbers is finite or infinite. | [Solution] Since $2^{24}+2^{24}=2^{25}$, we have
$$\left(2^{8}\right)^{3}+\left(2^{6}\right)^{4}=\left(2^{5}\right)^{5}$$
That is, $x=2^{8}=256 ; y=2^{6}=64, z=2^{5}=32$ is a set of solutions to the given equation.
On the other hand, if $\left(x_{0}, y_{0}, z_{0}\right)$ is a set of solutions to the given equation in... | x=256, y=64, z=32 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,646 |
4. 241 Given $2 \lg (x-2 y)=\lg x+\lg y$, try to find $x: y$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | [Solution] The given equation can be transformed into $(x-2 y)^{2}=x y$,
which is $\square$
$$x^{2}-5 x y+4 y^{2}=0$$
or $\quad(x-y)(x-4 y)=0$.
From $x-y=0$, we get $\frac{x}{y}=1$. Since $x$ and $y$ should both be positive, in this case $x-2 y<0$, which contradicts the given equation. From $x-4 y=0$, we get $\frac{x}... | null | Logic and Puzzles | other | Yes | Yes | inequalities | false | 735,647 |
4・242 Let $a, b$ be integers, what is the minimum positive integer value $c$ of $\Delta=\left|\begin{array}{ll}36 & b \\ 81 & a\end{array}\right|$? For each set of positive integer solutions $(a, b)$ that satisfies $\Delta=c$, to make $a+b$ the smallest, what are $a, b$ respectively? | ［Solution］ Since $\Delta=36 a-81 b=9(4 a-9 b)$, and $a, b$ are integers, hence $4 a-9 b$ is also an integer. Since the smallest positive integer is 1, the smallest positive value $c$ of $\Delta$ is no less than 9.
Since 4 and 9 are coprime, by the theorem, there must exist integers $a, b$ such that
$$4 a-9 b=1$$
By $... | a=7, b=3, a+b=10 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,648 |
4. 243 Let $a$ be any given positive integer. Try to prove: there always exists a unique pair of positive integers $(x, y)$, such that
$$x+\frac{(x+y-1)(x+y-2)}{2}=a$$ | ［Proof］Let $x+y-1=k$, then
$$\frac{(x+y-1)(x+y-2)}{2}=\frac{k(k-1)}{2}, 0<x \leqslant k$$
For any given positive integer $a$, take the value of $k$ such that
$$\frac{k(k-1)}{2}<a \leqslant \frac{k(k-1)}{2}+k=\frac{k(k+1)}{2}$$
This inequality uniquely determines the value of $k$. Thus, it uniquely determines the valu... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,649 |
4-244 Prove: If a set of values of variables $x$ and $y$ satisfy the equation
and
$$x^{2}-3 x y+2 y^{2}+x-y=0$$
then this set of values also satisfies the equation
$$x y-12 x+15 y=0 .$$ | [Proof] Multiply both sides of equations (1) and (2) by $(x-y-9)$ and $(-x+2 y+3)$ respectively, add them, and simplify to get
$$\begin{array}{l}
\left(x^{2}-3 x y+2 y^{2}+x-y\right)(x-y-9)+\left(x^{2}-2 x y+y^{2}-5 x+\right. \\
7 y)(-x+2 y+3) \\
= 2(x y-12 x+15 y) .
\end{array}$$
Therefore, if a set of values for $x$... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,650 |
4・245 Prove that the equation
$$y_{1}+2 y_{2}+\cdots+n y_{n}=a-\frac{n(n+1)}{2}$$
has no non-negative integer solutions if and only if the equation
$$x_{1}+2 x_{2}+\cdots+n x_{n}=a$$
has no positive integer solutions (where $a$ is a positive integer). | [Proof] The numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy the equation
$$x_{1}+2 x_{2}+\cdots+n x_{n}=a$$
if and only if: the numbers $y_{1}=x_{1}-1, y_{2}=x_{2}-1, \cdots, y_{n}=x_{n}-1$ satisfy the equation
$$\left(y_{1}+1\right)+2\left(y_{2}+1\right)+\cdots+n\left(y_{n}+1\right)=a \text {. }$$
The above equation c... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,651 |
4.246 Proof: The equation $6\left(6 a^{2}+3 b^{2}+c^{2}\right)=5 n^{2}$ has no integer solutions except for $a=b=c=n=0$. | [Proof] Without loss of generality, let the greatest common divisor of the non-zero solutions $a, b, c, n$ be
$$(a, b, c, n)=1 \text {. }$$
From the equation $6\left(6 a^{2}+3 b^{2}+c^{2}\right)=5 n^{2}$,
we get $6 \mid n^{2}$, hence $6 \mid n$, and $6^{2} \mid 6\left(6 a^{2}+3 b^{2}+c^{2}\right)$, which implies $3 \m... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,652 |
1.66 Is the number $\sin \frac{\pi}{18} \sin \frac{3 \pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18} \sin \frac{9 \pi}{18}$ a rational number? | ［Solution］ From the double angle formula, we get
$$\begin{aligned}
\sin \frac{8 \pi}{18} & =2 \sin \frac{4 \pi}{18} \cos \frac{4 \pi}{18}=4 \sin \frac{2 \pi}{18} \cos \frac{2 \pi}{18} \cos \frac{4 \pi}{18} \\
& =8 \sin \frac{\pi}{18} \cos \frac{\pi}{18} \cos \frac{2 \pi}{18} \cos \frac{4 \pi}{18}
\end{aligned}$$
Notin... | \frac{1}{16} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,653 |
4. 247 Find all positive integers $n$ such that the following equation has an integer solution.
$$x^{n}+(2+x)^{n}+(2-x)^{n}=0$$ | [Solution] When $n=1$, we get $x+(2+x)+(2-x)=0, x=-4$ as its integer solution.
When $n$ is even, the original equation has integer solutions if and only if:
$$\left\{\begin{array}{l}
x=0 \\
2+x=0 \\
2-x=0
\end{array}\right.$$
But (1) is a contradictory equation, hence we know that when $n$ is even, the original equat... | n=1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,654 |
$4 \cdot 248$ To make the equation
$$\left[\frac{10^{n}}{x}\right]=1989$$
have integer solutions, what is the smallest value of the positive integer $n$? | [Solution] From the equation, we get
$$1989 \leqslant \frac{10^{n}}{x} < 1990$$. Thus, from (1), we have
$$10^{n} \cdot \frac{1}{1990}<x \leqslant 10^{n} \cdot \frac{1}{1989}$$
That is,
$$10^{n} \cdot 0.0005025 \cdots<x \leqslant 10^{n} \cdot 0.0005027 \cdots$$
When $n=1,2,3, \cdots, 6$, there is no integer $x$ that ... | 7 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,655 |
4. 249 Proof: The equation $x-y+z=1$ has infinitely many sets of positive integer solutions where $x, y, z$ are all distinct, and the product of any two of them is divisible by the third.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result ... | [Proof] Let $x, y, z$ be positive integer solutions of the original equation that meet the problem's requirements. Then $x, y, z$ are pairwise distinct, and there exist positive integers $u, v, w$ such that
$$\left\{\begin{array}{l}
x y=z u \\
y z=x v \\
z x=y w .
\end{array}\right.$$
(1) $\times$ (2) $\times$ (3) give... | null | Number Theory | proof | Yes | Yes | inequalities | false | 735,656 |
4・250 Given the equation $x^{3}-3 x y^{2}+y^{3}=n$, where $n$ is a positive integer. Prove:
(1) If the equation has one integer solution $(x, y)$, then it has at least three integer solutions;
(2) When $n=2891$, the equation has no integer solutions. | [Proof] (1) From $(y-x)^{3}=y^{3}-3 x y^{2}+3 x^{2} y-x^{3}$, we get $x^{3}-3 x y^{2}+y^{3}=(y-x)^{3}-3 x^{2}(y-x)-x^{3}$,
which means $x^{3}-3 x y^{2}+y^{3}=(y-x)^{3}-3 \cdot(y-x) \cdot(-x)^{2}+(-x)^{3}$.
Therefore, if $(x, y)$ is an integer solution to the equation
$$x^{3}-3 x y^{2}+y^{3}=n$$
then $(y-x,-x)$ is also... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,657 |
4. 251 Suppose 1987 can be written as a three-digit number $x y z$ in base $b$ and $x+y+z=1+9+8+7$, try to determine all $x, y, z$ and $b$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | [Solution] From the given, we have
$$x b^{2}+y b+z=1987(x \geqslant 1)$$
and $b^{3}>1987, b^{2}<1987$. Therefore,
$$12<b<45 \text {. }$$
Also, from the given, we know $x+y+z=25$.
Subtracting (2) from (1) gives $(b-1)(b x+x+y)=1962$.
Thus, $b-1$ can divide 1962. Considering $1962=2 \times 9 \times 109$,
and $\quad 12... | null | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,658 |
4. 252 (1) Find all positive integers whose first digit is 6, and when this 6 is removed, the resulting integer is $\frac{1}{25}$ of the original integer.
(2) Prove that there is no such positive integer whose first digit is 6, and when this 6 is removed, the resulting integer is $\frac{1}{35}$ of the original integer. | ［Solution］ Let the positive integer satisfying the given conditions be
$6 \cdot 10^{n}+m$, where $n \in N$ and $0 \leqslant m \leqslant 10^{n}-1$.
(1) From the given conditions, we have
$$m=\frac{1}{25}\left(6 \cdot 10^{n}+m\right)$$
Simplifying, we get
$$m=2^{n-2} \cdot 5^{n}$$
Therefore, the number we are looking f... | 625, 6250, 62500, \cdots | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,659 |
4. 253 Given positive integers $a$ and $b$ such that $a b+1$ divides $a^{2}+b^{2}$, prove that $\left(a^{2}+\right.$ $\left.b^{2}\right) /(a b+1)$ is the square of some positive integer. | [Proof] Let the positive integer $k=\left(a^{2}+b^{2}\right) /(a b+1)$ not be a perfect square, and consider the indeterminate equation
$$a^{2}+b^{2}-k a b=k, \text{ where } k \text{ is a constant.}$$
Obviously, the solution $(a, b)$ of this indeterminate equation will not make $a b \leqslant 0$. Let $\left(a_{0}, b_{... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,660 |
$4 \cdot 254$ Prove: There exist infinitely many numbers $B$, such that the equation
$$\left[x^{\frac{3}{2}}\right]+\left[y^{\frac{3}{2}}\right]=B$$
has at least 1980 natural number solutions $x, y$ (where $[z]$ denotes the greatest integer not exceeding $z$). | [Proof] We first prove a lemma: for any natural number $M$, there exists a number $B$ such that the original equation has at least $M+1$ natural number solutions.
Indeed, if there exists a constant $M$ such that for any number $B$, the number of natural number solutions of the original equation is no more than $M$, th... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,661 |
4. 255 Let $m$ be a given integer, prove that there exist integers $a, b$ and $k$, where $a, b$ are both not divisible by 2, $k \geqslant 0$, such that
$$2 m=a^{19}+b^{99}+k \cdot 2^{1999}$$ | [Proof] Since when $s, x, y$ are all odd,
$$x^{s}-y^{y}=(x-y)\left(x^{s-1}+x^{s-2} y+\cdots+y^{s-1}\right)$$
and $x^{s-1}+x^{s-2} y+\cdots+y^{s-1}$ is odd, the following proposition holds:
If $x, y$ are odd, $r$ is a positive integer, and
$$x \not \equiv y \quad\left(\bmod 2^{r}\right)$$
then, for any odd $s$, we hav... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,662 |
$1 \cdot 67$ Try to prove that if $n$ is a natural number, then
$$(\sqrt{2}-1)^{n}=\sqrt{m}-\sqrt{m-1}$$
where $m$ is a natural number. | [Proof] First, we prove an auxiliary proposition: For any natural number $n$, there exist natural numbers $a, b$, such that
$$\left\{\begin{array}{l}
(1-\sqrt{2})^{n}=\sqrt{a^{2}}-\sqrt{2 b^{2}}, \\
a^{2}-2 b^{2}=(-1)^{n} .
\end{array}\right.$$
When $n=1$, the conclusion is true. In fact, taking $a=b=1$ suffices. Supp... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,663 |
4・258 $p, q, r$ are pairwise distinct real numbers, satisfying the system of equations
$$\left\{\begin{array}{l}
q=p(4-p) \\
r=q(4-q) \\
p=r(4-r)
\end{array}\right.$$
Find all possible values of $p+q+r$. | [Solution] From the given,
$$p=r(4-r)=-(r-2)^{2}+4 \leqslant 4,$$
Similarly, we can get
$$q \leqslant 4, r \leqslant 4$$
Next, we prove that $p \geqslant 0$ using proof by contradiction.
If $p<0$, then from $4-p \geqslant 0$ we get
$$q \leqslant 0.$$
Again, from $4-q \geqslant 0$ we get
$$r \leqslant 0$$
Thus, we h... | 6 \text{ and } 7 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,665 |
$4 \cdot 259$ For any integer $k$, prove that the equation
$$y^{2}-k=x^{3}$$
cannot simultaneously have the following 5 sets of integer solutions $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{1}-1\right),\left(x_{3}, y_{1}-2\right)$, $\left(x_{4}, y_{1}-3\right)$ and $\left(x_{5}, y_{1}-4\right)$. If the above equation h... | [Proof] First, prove the first conclusion using proof by contradiction.
If the equation $y^{2}-k=x^{3}$, i.e.,
$$x^{3}+k=y^{2}$$
has the 5 integer solutions given in the problem, then
$$\left\{\begin{array}{l}
x_{1}^{3}+k=y_{1}^{2} \\
x_{2}^{3}+k=\left(y_{1}-1\right)^{2} \\
x_{3}^{3}+k=\left(y_{1}-2\right)^{2} \\
x_{4... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,666 |
4・260 Proof: The equation
$$x^{2}+y^{2}+z^{2}=x^{3}+y^{3}+z^{3}$$
has infinitely many integer solutions. | ［Proof］Let
$$\left\{\begin{array}{l}
x=n\left(4 n^{2}-1\right)+1, \\
y=1-n\left(4 n^{2}-1\right), \\
z=1-4 n^{2}
\end{array} \quad n=1,2, \cdots\right.$$
Then, we have
$$\begin{aligned}
x^{2}(x-1)= & {\left[n\left(4 n^{2}-1\right)+1\right]^{2} \cdot n\left(4 n^{2}-1\right) } \\
= & {\left[n\left(4 n^{2}-1\right)-1\rig... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,667 |
5.1 Given the polynomial
$$f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}$$
where the coefficients $a_{0}, a_{1}, \cdots, a_{n}$ are all integers, and $f(2)$ and $f(3)$ are divisible by 6. Prove that $f(5)$ is also divisible by 6. | [Solution] $f(2)=a_{0} 2^{n}+a_{1} 2^{n-1}+\cdots+a_{n-1} 2+a_{n}$
$$f(3)=a_{0} 3^{n}+a_{1} 3^{n-1}+\cdots+a_{n-1} 3+a_{n}$$
Since $f(2)$ is divisible by 6, it is also divisible by 2. From (1), we know that $a_{n}$ is divisible by 2. Since $f(3)$ is divisible by 6, it is also divisible by 3. From (2), we know that $a_... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,669 |
$5 \cdot 2$ Let $P(x)=a_{k} x^{k}+a_{k-1} x^{k-1}+\cdots+a_{1} x+a_{0}$ where each coefficient $a_{j}(j=0,1,2, \cdots, k)$ is an integer: Now suppose there are four distinct integers $x_{1}, x_{2}, x_{3}$, $x_{4}$ such that $p\left(x_{i}\right)(i=1,2,3,4)$ all equal 2, prove that for any integer $x, p(x)$ is never equa... | [Proof] According to the Remainder Theorem, we have
$$p(x)-2=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right) Q(x)$$
where $Q(x)$ is a polynomial with integer coefficients, or an integer. For any integer $x$, $x-x_{1}, x-x_{2}, x-x_{3}, x-x_{4}, Q(x)$ are always integers, and the first ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,670 |
5.6 If $p(x)$ is an $n$-degree polynomial, and for $k=0,1,2, \cdots, n$, we have $p(k)=\frac{k}{k+1}$, determine $p(n+1)$.
| [Proof] Let $Q(x) = (x+1) p(x) - x$. From the given conditions, we know that $Q(x)$ is a polynomial of degree no more than $n+1$, and when $x = 0, 1, 2, \cdots, n$, we have $Q(x) = 0$. Therefore, we have
$$Q(x) = a x(x-1)(x-2) \cdots (x-n)$$
where $a$ is a constant. That is,
$$(x+1) p(x) - x = a x(x-1)(x-2) \cdots (x-... | p(n+1) = \left\{\begin{array}{l}
1, \text{ if } n \text{ is odd; } \\
\frac{n}{n+2}, \text{ if } n \text{ is even. }
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,673 |
$5 \cdot 7$ Let $n$-degree polynomial $p(x)$ satisfy $p(k)=\frac{1}{C_{n}^{k}}$, where $k=0,1,2, \cdots$, $n$. Find $p(n+1)$. | ［Solution］ According to the Lagrange interpolation formula, we get
$$p(x)=\sum_{k=0}^{\prime \prime} \frac{1}{C_{n+1}^{k}} \prod_{\substack{i \neq k \\ 0 \leqslant i \leqslant n}} \frac{x-i}{k-i}$$
But $\prod_{\substack{i \neq k \\ 0 \leqslant \leqslant}}(k-i)=k(k-1) \cdots[k-(k-1)] \cdot[k-(k+1)] \cdots(k-n)$
Therefo... | p(n+1) = \begin{cases} 0 & \text{if } n \text{ is odd} \\ 1 & \text{if } n \text{ is even} \end{cases} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,674 |
5.8 Proof: For any integers \(x\) and \(y\), the algebraic expression
$$x^{5}+3 x^{4} y-5 x^{3} y^{2}-15 x^{2} y^{3}+4 x y^{4}+12 y^{5}$$
cannot equal 33. | [Proof] Factorize the given algebraic expression to get
$$(x-2 y)(x-y)(x+y)(x+2 y)(x+3 y),$$
If $y=0$, then this algebraic expression becomes $x^{5}$. It is clear that there does not exist an integer $x$ such that $x^{5}$
$$\text { = } 33$$
If $y \neq 0$, then the values of the five factors of this algebraic expressi... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,675 |
$5 \cdot 9$ Expand the expression $\left(x^{7}+x^{5}+1\right)^{20}$ and combine like terms, to get a polynomial. Find the coefficients of $x^{17}$ and $x^{18}$ in this polynomial. | [Solution] Since $17=5 \times 2+7 \times 1$, the coefficient of $x^{17}$ is
$$\frac{20!}{2!(20-2-1)!}=\frac{20!}{2!\cdot 17!}=\frac{20 \cdot 19 \cdot 18}{2}=3420 .$$
Since 18 cannot be expressed as the sum of several 5s and 7s, this polynomial does not contain a term of $x^{18}$, or in other words, the coefficient of ... | 3420, 0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,676 |
5. 10 Let $A$ and $B$ be the sums of the odd and even terms, respectively, in the expansion of $(x+a)^{n}$. Find $A^{2}-B^{2}$. | [Solution] Clearly, $A+B=(x+a)^{n}$,
$$A-B=(x-a)^{n},$$
Therefore,
$$\begin{aligned}
A^{2}-B^{2} & =(x+a)^{n}(x-a)^{n} \\
& =\left(x^{2}-a^{2}\right)^{n} .
\end{aligned}$$ | \left(x^{2}-a^{2}\right)^{n} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,677 |
5・12 Let the polynomial $1-x+x^{2}-x^{3}+\cdots+x^{16}-x^{17}$ be written as $a_{0}+a_{1} y+a_{2} y^{2}+a_{3} y^{3}+\cdots+a_{16} y^{16}+a_{17} y^{17}$, where $y=x+1$, and each $a_{i}$ is a constant, find $a_{2}$. | [Solution] $1-x+x^{2}-x^{3}+\cdots+x^{16}-x^{17}$
$$\begin{array}{l}
=\frac{1-x^{18}}{1+x} \\
\quad=\frac{1-(y-1)^{18}}{y}
\end{array}$$
The coefficient of $y^{3}$ in the numerator of the above expression is $C_{18}^{3}$, so
$$a_{2}=C_{18}^{3}=816$$ | 816 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,678 |
5- 13 Prove: There do not exist integers $a, b, c, d$, such that the expression $a x^{3} + b x^{2} + c x + d$ has a value of 1 when $x=19$, and a value of 2 when $x=62$. | [Solution] Let $p(x)=a x^{3}+b x^{2}+c x+d$, then for any integers $a, b, c$ we have
$$p(62)-p(19)=a\left(62^{3}-19^{3}\right)+b\left(62^{2}-19^{2}\right)+c(62-19)$$
can be divisible by 43, thus it cannot equal 1. Therefore, the proposition is proved. | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,679 |
5・14 Proof: The polynomial
$$p(x)=\frac{1}{630} x^{9}-\frac{1}{21} x^{7}+\frac{13}{20} x^{5}-\frac{82}{63} x^{3}+\frac{32}{35} x$$
takes integer values for all $x \in Z$. | $$\begin{array}{l}
p(x)= \frac{1}{2 \cdot 5 \cdot 7 \cdot 9}(x-4)(x-3)(x-2)(x-1) x(x+1)(x+ \\
2) (x+3)(x+4) .
\end{array}$$
Since in any 9 consecutive integers, there is always a natural number that can be divided by $2, 5, 7, 9$, and $2 \cdot 5 \cdot 7 \cdot 9$ is the product of coprime numbers, therefore for any $x... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,680 |
5. 15 Prove: If when the independent variable $x$ takes any integer value, the quadratic trinomial $a x^{2}+b x$ $+c$ always takes integer values, then $2 a, a+b, c$ are all integers. And the converse is also true. | [Proof] First, assume that when $x$ takes integer values, $f(x)=a x^{2}+b x+c$ is also an integer. Then
(1) $f(0)=c$ is an integer;
(2) $f(1)=a+b+c$ is an integer, from which it follows that $a+b=f(1)-c$ is also an integer;
(3) $f(2)=4 a+2 b+c$ is an integer, from which it follows that $2 a=f(2)-2(a+b)-c$ is also an in... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,681 |
1.69 Prove that if the sum
$$\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{4 \cdot 5 \cdot 6}+\cdots+\frac{1}{1984 \cdot 1985 \cdot 1986}$$
is expressed in its simplest fractional form, then the numerator of this fraction will be divisible by 1987. | [Solution] The sum has 662 (even number) terms. By adding terms equidistant from the first and last terms, the resulting fraction can be written as
$$\begin{aligned}
& \frac{(1987-\overline{3 m+1})(1987-\overline{3 m+2})(1987-\overline{3 m+3})+(3 m+1)(3 m+2)(3 m+3)}{(3 m+1)(3 m+2)(3 m+3)(1987-\overline{3 m+1})(1987-\ov... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,682 |
5・16 Let for any $x \in Z$, the polynomial with integer coefficients
$$a x^{3}+b x^{2}+c x+d$$
can be divisible by 5. Prove: the coefficients $a, b, c, d$ are all divisible by 5. | [Solution] Let $x=0,1,-1,2$ we get
$$\left\{\begin{array}{l}
d \equiv 0(\bmod 5) \\
a+b+c+d \equiv 0(\bmod 5) \\
-a+b-c+d \equiv 0(\bmod 5) \\
8 a+4 b+2 c+d \equiv 0(\bmod 5)
\end{array}\right.$$
Using (1) in (2), (3), (4) we get
$$\begin{array}{l}
a+b+c \equiv 0(\bmod 5) \\
-a+b-c \equiv 0(\bmod 5) \\
3 a+4 b+2 c \eq... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,683 |
5.17 Let \(a < b < c < d\). If the variables \(x, y, z, t\) are some permutation of the numbers \(a, b, c, d\), how many different values can the expression
$$n=(x-y)^{2}+(y-z)^{2}+(z-t)^{2}+(t-x)^{2}$$
take? | [Solution] Let $n(x, y, z, t) = (x-y)^{2} + (y-z)^{2} + (z-t)^{2} + (t-x)^{2}$. Since the expression
$$\begin{aligned}
& n(x, y, z, t) + (x-z)^{2} + (y-z)^{2} \\
= & n(x, y, z, t) + \left(x^{2} + y^{2} + z^{2} + t^{2}\right) - 2(x z + y t)
\end{aligned}$$
is independent of the permutation of the numbers $a, b, c, d$ c... | 3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,684 |
5. 18 Prove that (1) any quadratic trinomial with given constant coefficients $A x^{2}+$ $B x+C$ can be expressed in the form
$$k \frac{x(x-1)}{1 \cdot 2}+l x+m$$
where $k, l$, and $m$ have completely determined values.
(2) The quadratic trinomial $A x^{2}+B x+C$ takes integer values for all integer $x$ if and only if... | [Proof] (1) Since $x^{2}=2 \frac{x(x-1)}{2}+x$,
the quadratic trinomial
$$Q=A x^{2}+B x+C$$
can be transformed into
$$Q=k \frac{x(x-1)}{2}+l x+m$$
where $k=2 A, l=A+B, m=C$.
(2) If the quadratic trinomial $Q$ takes integer values for all integer $x$, then when $x$ equals 0, 1, 2, we get the corresponding values of th... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,685 |
5・19 Let the polynomial
$$P(x)=x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+1$$
have $n$ roots, and the coefficients $a_{1}, a_{2}, \cdots, a_{n-1}$ are all non-negative. Prove: $P(2) \geqslant 3^{n}$. | [Proof] Since all the coefficients of the polynomial $P(x)$ are non-negative, its roots $\alpha_{1}$, $\alpha_{2}, \cdots, \alpha_{n}$ are all negative. Therefore,
$$P(x)=\left(x+\beta_{1}\right)\left(x+\beta_{2}\right) \cdots\left(x+\beta_{n}\right)$$
where $\beta_{i}=-\alpha_{i}>0, i=1,2, \cdots, n$.
By the mean val... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,686 |
$5 \cdot 21$ Find all $x \in Z$, such that the polynomial
$$2 x^{2}-x-36$$
is the square of some prime number. | ［Solution］Let $2 x^{2}-x-36 \doteq p^{2}, p$ be a prime number, then
$$p^{2}=(x+4)(2 x-9)=a b,$$
where $a=x+4, b=2 x-9, a, b \in \mathbb{Z}, 2 a-b=17$.
Since $a$ is an integer and divides $p^{2}$, there are only the following 6 cases:
(1) $a=p^{2}, b=1$, then $2 p^{2}-1=17$, i.e., $p=3$, thus
$$x=a-4=p^{2}-4=5$$
(2) $... | x=5 \text{ and } x=13 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,688 |
$5 \cdot 22$ Given real-coefficient polynomials $p_{0}(x), p_{1}(x), \cdots, p_{n}(x)$ and real numbers $a_{1}, \cdots, a_{n}$. Prove: If the function
$$f(x)=p_{0}(x)+\sum_{K=1}^{n} a_{k}\left|p_{k}(x)\right|$$
takes different values at different points in $R$, then its range is $R$. | [Proof] Without loss of generality, assume that all polynomials $p_{1}(x), \cdots, p_{n}(x)$ are non-zero. Take such a number $x^{+}$, which is greater than all the real roots of the above polynomials. Then, when $x \geqslant x^{+}$, the function $f(x)$ is consistent with the polynomial
$$p^{+}(x)=p_{0}(x)+\sum_{k=1}^{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,689 |
5. 25 Let given integers $x_{0}<x_{1}<\cdots<x_{n}$. Prove: The polynomial $x^{n}+a_{1} x^{n-1}$ $+\cdots+a_{n}$ at the points $x_{0}, x_{1}, \cdots, x_{n}$ takes values among which there exists a number whose absolute value is not
$$\text { less than } \frac{n!}{2^{n}} \text {. }$$ | [Proof] According to the Lagrange interpolation formula, the polynomial
$$P(x)=x^{n}+a_{1} x^{n-1}+\cdots+a_{n}$$
can be expressed as \( P(x)=\sum_{j=0}^{n}\left(\prod_{\substack{i \neq j \\ 0 \leq 1 \leq n}} \frac{x-x_{i}}{x_{j}-x_{i}}\right) P\left(x_{j}\right) \).
By contradiction. Suppose the conclusion in the pro... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,690 |
5-26. Let $\left\{a_{n}\right\}$ be the Fibonacci sequence, defined as follows: $a_{1}=a_{2}=1, a_{n+2}=a_{n+1}+a_{n}, n \in N$. Prove: If the 990th degree polynomial $P(x)$ satisfies: when $k=992, \ldots, 1982$, $P(k)=a_{k}$, then $P(1983)=a_{1983}-1$. | [Proof] For $n \in N$, use induction to prove a more general conclusion: if an $n$-degree polynomial satisfies, when
$$k=n+2, n+3, \cdots, 2 n+2$$
then $P(k)=a_{k}$, then $P(2 n+3)=a_{2 n+3}-1$. When $n=1$, we have $P(3)=2$, $P(4)=3$, thus $P(x) \equiv x-1$, and
$$P(5)=4=a_{5}-1$$
Now assume the conclusion holds for ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,691 |
5.27 Proof: The necessary and sufficient condition for non-zero complex coefficient polynomials $P$ and $Q$ to have the same roots (with the same multiplicities) is that the function $f(z)=|P(z)|-|Q(z)|$ has the same sign for all non-zero $z \in \mathbb{C}$. | [Proof] Suppose the roots (including their multiplicities) of polynomials $P$ and $Q$ are the same, then we have
$$\begin{array}{l}
P(z)=a\left(z-z_{1}\right)^{n_{1}}\left(z-z_{2}\right)^{n_{2}} \cdots\left(z-z_{k}\right)^{n_{k}} \\
Q(z)=b\left(z-z_{1}\right)^{n_{1}}\left(z-z_{2}\right)^{n_{2}} \cdots\left(z-z_{k}\righ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,692 |
5. 29 Simplify the expression
$$\left(\left(\cdots\left(\left((x-2)^{2}-2\right)^{2}-2\right)^{2}-\cdots-2\right)^{2}-2\right)^{2}$$
where there are $k$ nested parentheses, and combine like terms to obtain a polynomial. Find the coefficient of $x^{2}$ in this polynomial. | [Solution] Let the obtained polynomial be $P_{k}(x)$, and set
$$P_{k}(x)=A_{k}+B_{k} x+C_{k} x^{2}+\cdots$$
Since
$$\begin{aligned}
P_{k}(0) & =\left(\left(\cdots\left(\left((0-2)^{2}-2\right)^{2}-2\right)^{2}-\cdots-2\right)^{2}-2\right)^{2} \\
& =4
\end{aligned}$$
Therefore,
$$P_{k}(x)=4+B_{k} x+C_{k} x^{2}+\cdots$... | 4^{k-1} \cdot \frac{4^{k}-1}{4-1} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,693 |
5.31 Prove: If $k$ is a natural number, then for any $x$,
$$\begin{aligned}
& (1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \cdots\left(1+x^{2 k}\right) \\
= & 1+x+x^{2}+x^{3}+\cdots+x^{m}
\end{aligned}$$
where $m$ is a natural number related to $k$, and find $m$. | [Proof] The expression
$$P_{k}=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \cdots\left(1+x^{2^{k}}\right)$$
is the product of $k+1$ binomials. After expanding the brackets, we get a sum of $2^{k+1}$ terms. By selecting one term from each bracket and then forming their product, we obtain each of the terms. Therefore,... | m=2^{k+1}-1 | Algebra | proof | Yes | Yes | inequalities | false | 735,695 |
5.32 Given a polynomial with integer coefficients \(a_{1}, a_{2}, \cdots, a_{n}\)
$$f(x)=x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+a_{n-1} x+a_{n} .$$
It is also known that there exist four distinct integers \(a, b, c, d\), such that
$$f(a)=f(b)=f(c)=f(d)=5 \text {. }$$
Prove that there is no integer \(k\), such that ... | [Proof] From the given conditions, the polynomial
$$f(x)-5$$
has four distinct integer roots $a, b, c, d$. Therefore, we can assume
$$f(x)-5=(x-a)(x-b)(x-c)(x-d) \cdot g(x)$$
where $g(x)=x^{m}+b_{2} x^{m-1}+\cdots+b_{m}$, and $b_{1}, b_{2}, \cdots, b_{m}$ are integers.
If there exists an integer $k$ such that $f(k)=8... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,696 |
5・33 If $a, b$ are integers, and $x^{2}-x-1$ is a factor of $a x^{17}+b x^{16}+1$, try to find the value of $a$.
| [Solution] The roots of the equation $x^{2}-x-1=0$ are
$$p=\frac{1+\sqrt{5}}{2}, q=\frac{1-\sqrt{5}}{2}$$
and $p+q=1, p q=-1$.
From the problem, we know that $p, q$ are also roots of the equation
$$a x^{17}+b x^{16}+1=0$$
Therefore,
$$\begin{array}{l}
a p^{17}+b p^{16}+1=0 \\
a q^{17}+b q^{16}+1=0
\end{array}$$
Multi... | 987 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,697 |
$5 \cdot 34$ Polynomial
$$(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1-z^{3}\right)^{b_{3}} \cdots\left(1-z^{32}\right)^{b_{32}}$$
In this polynomial, $b_{i}$ are positive integers $(i=1,2, \cdots, 32)$, and the polynomial has the following remarkable property: when expanded, and terms with powers of $z$ higher th... | \[ \text{Solution: Let } f(z)=(1-z)^{b_{1}}\left(1-z^{2}\right)^{b_{2}}\left(1-z^{3}\right)^{b_{3}} \cdots\left(1-z^{32}\right)^{b_{32}} = 1-2 z+g(z) \]
where every term in $g(z)$ is of degree higher than 32.
By comparing the coefficients of $z$ on both sides, we get $b_{1}=2$.
\[ \begin{aligned}
f(-z)= & (1+\approx)^{... | 2^{27}-2^{11} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,698 |
5.35 (1) Find the possible minimum value of the polynomial $P(x, y)=4+x^{2} y^{4}+x^{4} y^{2}-3 x^{2} y^{2}$.
(2) Prove that this polynomial cannot be expressed as a sum of squares of polynomials in variables $x, y$. | ［Solution］（1）By the Arithmetic Mean - Geometric Mean Inequality, we have
$$1+x^{2} y^{4}+x^{4} y^{2} \geqslant 3 x^{2} y^{2}$$
and the equality holds when $x=y=1$. Therefore, we have
$$\begin{aligned}
P(x, y) & =3+\left(1+x^{2} y^{4}+x^{4} y^{2}-3 x^{2} y^{2}\right) \\
& \geqslant 3
\end{aligned}$$
and when $x=y=1$, ... | 3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,699 |
$1 \cdot 72$ Given $2 n$ different numbers $a_{1}, a_{2}, \cdots, a_{n}, b_{1}, \cdots, b_{n}$, and fill them into a grid according to the following rule, that is, fill the number $a_{j}+b_{j}$ in the cell at the intersection of the $i$-th row and the $j$-th column, prove: If the product of the numbers in each column i... | [Proof] Let's examine the polynomial
$$f(x)=\prod_{i=1}^{n}\left(x+a_{i}\right)-\prod_{j=1}^{n}\left(x-b_{j}\right)$$
whose degree is less than $n$. If for all $j=1, \cdots, n$, we have $f\left(b_{j}\right)=\prod_{i=1}^{n}\left(a_{i}+b_{j}\right)=c$, then the polynomial $f(x)-c$ has at least $n$ distinct roots. Theref... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,700 |
$5 \cdot 36$ Try to find 3 distinct non-zero integers $a, b, c$, such that the algebraic expression
$$x(x-a)(x-b)(x-c)+1$$
can be expressed as the product of two polynomials with integer coefficients. | [Solution] (1) If $d, e, f, g$ are integers, and
$$x(x-a)(x-b)(x-c)+1=(x+d)\left(x^{3}+e x^{2}+f x+g\right)$$
Let $x=0$, we get $d g=1, d= \pm 1$. By setting $x=a, b, c$ in sequence, we obtain
$$\left\{\begin{array}{l}
a+d= \pm 1 \\
b+d= \pm 1 \\
c+d= \pm 1
\end{array}\right.$$
If $d=1$, then the non-zero integers $a... | (1,2,3),(-1,-2,-3),(1,-1,2),(1,-1,-2) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,701 |
$5 \cdot 37$ Find real numbers $a, b, p, q$, such that the equation
$$(2 x-1)^{20}-(a x+b)^{20}=\left(x^{2}+p x+q\right)^{10}$$
holds for any $x$. | [Solution] Let $x=\frac{1}{2}$, we get
$$\left(\frac{a}{2}+b\right)^{20}+\left(\frac{1}{4}+\frac{p}{2}+q\right)^{10}=0$$
From this, we obtain $\square$
$$\frac{a}{2}+b=0$$
That is
$$a=-2 b$$
Thus, the original equation can be transformed into
$$(2 x-1)^{20}-(-2 b x+b)^{20}=\left(x^{2}+p x+q\right)^{10}$$
That is $\... | a=\mp \sqrt[20]{2^{20}-1}, b= \pm \frac{1}{2} \cdot \sqrt[20]{2^{20}-1}, p=-1, q=\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,702 |
5・38 $P(x)$ is a polynomial of degree $3 n$, such that
$$\begin{array}{l}
P(0)=P(3)=\cdots=P(3 n)=2 \\
P(1)=P(4)=\cdots=P(3 n-2)=1 \\
P(2)=P(5)=\cdots=P(3 n-1)=0 \\
P(3 n+1)=730
\end{array}$$
Determine $n$. | [Solution] Notice that when $x$ takes the values $0,1,2, \cdots, 3 n$, $P(x)-1$ cyclically takes the values $1,0,-1,1,0,-1, \cdots$. Therefore, we can consider the sequence
$$\left\{w^{n}\right\}=\left\{1, w, w^{2}, 1, w, w^{2}, \cdots\right\}$$
where $w=\frac{1}{2}(-1+i \sqrt{3})$ is a cube root of 1. Using $\operato... | 4 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,703 |
5.39 It can be proven that for any given positive integer $n$, every complex number of the form $r + s i (r, s$ both integers) can be expressed as a polynomial in $(-n+i)$, and the coefficients of the polynomial all belong to $\{0,1,2, \cdots, n^{2}\}$.
That is, the equation
$$r+s i=a_{m}(-n+i)^{m}+a_{m-1}(-n+i)^{m-1}... | 【Solution】From the given conditions, we have
$$\begin{aligned}
k & =a_{3}(-3+i)^{3}+a_{2}(-3+i)^{2}+a_{1}(-3+i)+a_{0} \\
& =\left(-18 a_{3}+8 a_{2}-3 a_{1}+a_{0}\right)+\left(26 a_{3}-6 a_{2}+a_{1}\right) i
\end{aligned}$$
where $a_{1}, a_{2}, a_{3}$ are chosen from $\{0,1,2, \cdots, 9\}$.
By comparing the imaginary p... | 490 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,704 |
5. 40 Let $n$ be a positive integer, find the number of odd coefficients in the polynomial $P_{n}(x)=\left(x^{2}+x+1\right)^{n}$. | [Solution] Let $P(x), Q(x)$ be polynomials with integer coefficients. If all the coefficients of $P(x)-Q(x)$ are even, we denote this as $P \sim Q$. If $P \sim Q$, then the number of odd coefficients of $P(x)$ and $Q(x)$ is the same. Below, we discuss several cases. Let $\beta(P(x))$ denote the number of odd coefficien... | \beta\left(P_{n}(x)\right)=\prod_{i=1}^{k} \frac{1}{3}\left(2^{a_{i}+2}-(-1)^{a_{i}}\right) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,705 |
5.41 Let $p(x)$ be a polynomial function with real coefficients
$$p(x)=a x^{3}+b x^{2}+c x+d .$$
Prove that if for any $|x|<1$, we have $|p(x)| \leqslant 1$, then
$$|a|+|b|+|c|+|d| \leqslant 7$$ | [Proof] Since $p(x)$ is a continuous function, it follows from the given condition that when $|x| \leqslant 1$, we have $|p(x)| \leqslant 1$.
Let $x=\lambda$, where $\lambda= \pm 1$, we get
$$|\dot{\lambda} a+b+\lambda c+d| \leqslant 1$$
Now let $x=\frac{1}{2} \lambda$, we get
$$\left|\frac{\lambda}{8} a+\frac{1}{4} ... | 7 | Algebra | proof | Yes | Yes | inequalities | false | 735,706 |
5.42 Does there exist a real-coefficient polynomial $p(x)$: it has negative coefficients, while for $n>1, p^{n}(x)$ has all positive coefficients? | [Solution] Exists.
For example, the real-coefficient polynomial
$$\begin{aligned}
p(x) & =10\left(x^{3}+1\right)(x+1)-x^{2} \\
& =10 x^{4}+10 x^{3}-x^{2}+10 x+10
\end{aligned}$$
is a polynomial with negative coefficients. However,
$$\begin{aligned}
p^{2}(x) & =100\left(x^{3}+1\right)^{2}(x+1)^{2}-20 x^{2}\left(x^{3}+1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,707 |
$5 \cdot 43$ Let
$$\begin{array}{l}
f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0} \\
g(x)=c_{n+1} x^{n+1}+c_{n} x^{n}+\cdots+c_{0}
\end{array}$$
be two non-zero polynomials with real coefficients, and there exists a real number $r$ such that
$$g(x)=(x-r) f(x)$$
Let $a=\max \left\{\left|a_{n}\right|,\left|a_{n-1}\righ... | [Proof] Given $c_{n+1} x^{n+1}+c_{n} x^{n}+\cdots+c_{0}=(x-r)\left(a_{n} x^{n}+\right.$ $\left.a_{n-1} x^{n-1}+\cdots+a_{0}\right)$, we have
$$\left\{\begin{array}{l}
c_{n+1}=a_{n} \\
c_{n}=a_{n-1}-r a_{n} \\
c_{n-1}=a_{n-2}-r a_{n-1} \\
\cdots \cdots \\
c_{1}=a_{0}-r a_{1} \\
c_{0}=-r a_{0}
\end{array}\right.$$
Thus,... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,708 |
5.44 Let \( P(x) = x^{4} + a x^{3} + b x^{2} + c x + d \), where \( a, b, c, d \) are constants, and \( P(1) = 1993, P(2) = 3986, P(3) = 5979 \).
Try to calculate \( \frac{1}{4}[P(11) + P(-7)] \). | [Solution] Let $Q(x) = P(x) - 1993x$,
then $Q(1) = Q(2) = Q(3) = 0$,
so $Q(x) = (x-1)(x-2)(x-3)(x-r)$.
Thus, from $P(x) = Q(x) + 1993x$ we get
$$\begin{aligned}
& \frac{1}{4}[P(11) + P(-7)] \\
= & \frac{1}{4}[Q(11) + 1993 \times 11 + Q(-7) + 1993 \times (-7)] \\
= & \frac{1}{4}[Q(11) + Q(-7)] + 1993,
\end{aligned}$$
b... | 5233 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,709 |
5.45 Given non-constant polynomials $P(x)$ and $Q(x)$, whose leading coefficients are both 1. Prove: the sum of the squares of the coefficients of the polynomial $P(x) \cdot Q(x)$ is not less than the sum of the squares of the constant terms of the polynomials $P(x)$ and $Q(x)$. | [Proof] For any polynomial
$$R(x)=c_{n} x^{n}+c_{n-1} x^{n-1}+\cdots+c_{1} x+c_{0}$$
define its norm as
$$\|R\|=\sqrt{c_{n}^{2}+c_{n-1}^{2}+\cdots+c_{1}^{2}+c_{0}^{2}},$$
and its conjugate as
$$R^{*}(x)=c_{0} x^{n}+c_{1} x^{n-1}+\cdots+c_{n-1} x+c_{n} \text {. }$$
We first prove the following lemma:
Lemma For any tw... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,710 |
1. 73 In an $n \times n$ square grid, fill in 1 or $-1$ arbitrarily, where $n$ is an odd number. Below each column, write the product of all numbers in that column, and to the right of each row, write the product of all numbers in that row. Prove: the sum of the $2n$ products written is not equal to zero. | [Proof] Let $p_{1}, p_{2}, \cdots, p_{n}$ be the product of each row, and $q_{1}, q_{2}, \cdots, q_{n}$ be the product of each column, then
$$p_{1} \cdot p_{2} \cdots p_{n}=q_{1} q_{2} \cdots q_{n}$$
Therefore, the parity of the number of -1s in $p_{1}, p_{2}, \cdots, p_{n}$ and in $q_{1}, q_{2}, \cdots, q_{n}$ is the... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,711 |
5.46 Find an integer-coefficient polynomial such that $\alpha=\sqrt[3]{2}+\sqrt[3]{3}$ is its root.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
(Note: The note is not part of the translation, just a guideline for the tas... | [Solution] Since $\alpha^{3}=(\sqrt[3]{2}+\sqrt[3]{3})^{3}$
$$\begin{array}{l}
=5+3 \cdot \sqrt[3]{2} \cdot \sqrt[3]{3} \cdot(\sqrt[3]{2}+\sqrt[3]{3}) \\
=5+3 \sqrt[3]{6} \alpha
\end{array}$$
Therefore, $\left(\alpha^{3}-5\right)^{3}=(3 \sqrt[3]{6} \alpha)^{3}=162 \alpha^{3}$,
which means $\alpha^{9}-15 \alpha^{6}-87 ... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,712 |
5.47 $P(x)$ and $Q(x)$ are two polynomials, and for all real numbers $x$, they satisfy the identity $P(Q(x))=Q(P(x))$. If the equation $P(x)=Q(x)$ has no real solutions, prove that the equation
$$P(P(x))=Q(Q(x))$$
also has no real solutions. | [Proof] From the given, the polynomial $P(x)-Q(x)$ is not equal to 0 for any real number $x$, so the graph of the function $y=P(x)-Q(x)$ is always above or always below the $x$-axis. Without loss of generality, let's assume it is above the $x$-axis, i.e.,
$$P(x)-Q(x)>0 \text {. }$$
If $P(P(x))=Q(Q(x))$ has a real root... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,713 |
5.48 Determine the real number $a$ such that the polynomials $x^{2}+a x+1$ and $x^{2}+x+a$ have at least one common root. | ［Solution］ Let $x$ be a common root, then we have
$$\left\{\begin{array}{l}
x^{2} + a x + 1 = 0 \\
x^{2} + x + a = 0
\end{array}\right.$$
(1) - (2) gives $\square$
$$(a-1)(x-1)=0 .$$
If $a=1$, then equations (1) and (2) are the same, in which case the two polynomials have a common root.
If $a \neq 1$, then the common ... | a=1 \text{ or } a=-2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,714 |
$5 \cdot 49$ Let $P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}$, where the coefficients $a_{i}$ are integers. If $P(0)$ and $P(1)$ are both odd, then $P(x)$ has no integer roots. | [Proof] When $x$ is odd, the parity of $P(x)$ is the same as that of $P(1)$; when $x$ is even, the parity of $P(x)$ is the same as that of $P(0)$. Given that both $P(0)$ and $P(1)$ are odd, it follows that $P(x)$ is always odd regardless of the integer value of $n$. Therefore, $P(x) \neq 0$, and thus, $P(x)$ has no int... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,715 |
5.50 Let $f(x)=a_{0}+a_{1} x+\cdots+a_{n} x^{n}$ be an $n$-degree polynomial with integer coefficients. If $a_{n}$, $a_{0}$, and $f(1)$ are all odd, prove: $f(x)=0$ has no rational roots. | [Solution] Suppose $\frac{p}{q}$ is its rational root, where $p, q$ are integers and the greatest common divisor $(p, q)=1$, then we have
$$a_{0} q^{n}+a_{1} q^{n-1} p+\cdots+a_{n} p^{n}=0$$
From this, we know that $p$ divides $a_{0}$, and $q$ divides $a_{n}$, so $p$ and $q$ are both odd. From $a_{k} q^{n-k} p^{k}$ be... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,716 |
5.51 Proof: If the roots of the polynomial $x^{2}+p x+1$ are $\alpha$ and $\beta$, and the roots of the polynomial $x^{2}+q x+1$ are $\gamma$ and $\delta$, then we have
$$(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)=q^{2}-p^{2}$$ | ［Proof］According to Vieta's formulas, we have
$$\alpha+\beta=-p, \alpha \beta=1, \gamma+\delta=-q, \gamma \delta=1 \text {, }$$
From this, we get
$$\begin{aligned}
& (\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta) \\
= & {\left[\alpha \beta-(\alpha+\beta) \gamma+\gamma^{2}\right]\left[\alpha \beta+(\alpha+\... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,717 |
5. 52 Proof: For any non-zero $\alpha, \beta$, the polynomial
$$\alpha x^{3}-\alpha x^{2}+\beta x+\beta$$
has roots $x_{1}, x_{2}, x_{3}$ that satisfy
$$\left(x_{1}+x_{2}+x_{3}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=-1$$ | [Proof] According to Vieta's formulas, for the roots $x_{1}, x_{2}, x_{3}$ of the polynomial $\alpha x^{3}-\alpha x^{2}+\beta x+\beta$, we have
$$\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=1, \\
x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}=\frac{\beta}{\alpha} \\
x_{1} x_{2} x_{3}=-\frac{\beta}{\alpha} .
\end{array}\right.$$
Th... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,718 |
5.53 Proof: The roots \(x_{1}, x_{2}\) of the polynomial
$$x^{2}+p x-\frac{1}{2 p^{2}}, p \in R, p \neq 0$$
satisfy
$$x_{1}^{4}+x_{2}^{4} \geqslant 2+\sqrt{2} \text {. }$$ | [Proof] According to Vieta's formulas,
$$x_{1}+x_{2}=-p, x_{1} x_{2}=-\frac{1}{2 p^{2}}$$
and the inequality between the arithmetic mean and the geometric mean of two numbers, we get
$$\begin{aligned}
x_{1}^{4}+x_{2}^{4} & =\left(x_{1}+x_{2}\right)^{4}-2 x_{1} x_{2}\left[2\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,719 |
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