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11. A, B, and C together have 100 yuan. If A's money becomes 6 times the original, B's money becomes $1 / 3$ of the original, and C's money remains unchanged, then the three of them still have a total of 100 yuan. C's money does not exceed 30 yuan. Question: How much money do A, B, and C each have? | 11. $x, y, z$ represent the amounts of money that A, B, and C have, respectively. $x+y+z=100, 18x+y+3z=300$. $x=10-2t, y=75-15t, z=15+17t, t=0,1, \cdots, 5. t=0$ gives the required solution. | x=10, y=75, z=15 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,084 |
12. A person bought a total of 12 packs of black and white sunflower seeds, spending 9 yuan and 9 jiao. Each pack of white sunflower seeds is 3 jiao more expensive than each pack of black sunflower seeds, and the number of packs of white sunflower seeds is more than that of black sunflower seeds. Question: How many pac... | 12. Let $x, y$ represent the number of packs of black and white sunflower seeds, respectively, and let the price per pack of black sunflower seeds be $z$ cents. $x+y=12, x z+y(z+3)$ $=99, y>x$. First solve $12 z+3 y=99$. Finally, get $x=3, y=9$, and the price per pack of black sunflower seeds is 6 cents. | x=3, y=9 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,085 |
13. Person A and Person B each took 40 and 30 eggs to the market to sell. Initially, they both sold them at 50 cents each. After selling some, they lowered the price but still sold them at the same price (each for several cents). When all the eggs were sold, they found that the money they earned was the same. How much ... | 13. Let the reduced price be $u$ cents, the number of eggs sold by A at 5 cents and $u$ cents be $x_{1}, x_{2}$, and by B be $y_{1}, y_{2} . x_{1}+x_{2}=40, y_{1}+y_{2}=30.5 x_{1}+u x_{2}=5 y_{1}+u y_{2} . u=3$, and the total amount is at most 15 yuan, at least 12 yuan. | 15 \text{ yuan, } 12 \text{ yuan} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,086 |
14. Class A has 7 children, Class B has 10 children. Now there are 100 apples to be distributed between Class A and Class B. How many apples should each class get so that each child in Class A gets the same number of apples, and each child in Class B also gets the same number of apples. | 14. $7 x+10 y=100$. Class A gets 70, Class B gets 30. | 7 x+10 y=100 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,087 |
15. (i) Express the fraction $23 / 30$ as the sum of three irreducible fractions, whose denominators are pairwise coprime;
(ii) Express $23 / 30$ as the sum of two irreducible fractions, whose denominators are coprime. | 15. (i) Let $23 / 30=b_{1} / a_{1}+b_{2} / a_{2}+b_{3} / a_{3}$. $\left(a_{1}, a_{2}\right)=\left(a_{2}, a_{3}\right)=\left(a_{3}, a_{1}\right)=1$, $a_{j} \geqslant 2,\left(a_{j}, b_{j}\right)=1$. Since $30=2 \cdot 3 \cdot 5$, we can set $a_{1}=2, a_{2}=3, a_{3}=5.15 b_{1}+10 b_{2}+$ $6 b_{3}=23.23 / 30=1 / 2-1 / 3+3 /... | \frac{23}{30} = \frac{3}{5} + \frac{1}{6} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,088 |
16. There are five sailors and a monkey on a small island. During the day, they collected some coconuts as food. At night, one of the sailors woke up and decided to take his share of the coconuts. He divided the coconuts into five equal parts, with one left over, so he gave the extra one to the monkey and hid his share... | 16. This pile of coconuts has at least 3121. The number of coconuts each of the five people received in turn is $828, 703, 603$, 523, 459, and the monkey ate 5. | 3121 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,089 |
17. Find all positive solutions of the following system of indeterminate equations:
(i) $2 x_{1}+x_{2}+x_{3}=100, 3 x_{1}+5 x_{2}+15 x_{3}=270$;
(ii) $x_{1}+x_{2}+x_{3}=31, x_{1}+2 x_{2}+3 x_{3}=41$. | 17. (i) $x_{1}=40, x_{2}=15, x_{3}=5$;
(ii) $\{22,8,1\},\{23,6,2\},\{24,4,3\},\{25,2,4\}$. | (i) x_{1}=40, x_{2}=15, x_{3}=5; (ii) \{22,8,1\},\{23,6,2\},\{24,4,3\},\{25,2,4\} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,090 |
20. $63 x_{1}+110 x_{2}=6893$ Does it have a positive solution? | 20. $x_{1}=71, x_{2}=22$. Although $6893<63 \cdot 1100$, there is still a positive solution. | x_{1}=71, x_{2}=22 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,093 |
22. Let $a_{1}, a_{2}, a_{3}$ be pairwise coprime positive integers. Prove: the indeterminate equation $a_{2} a_{3} x_{1} + a_{3} a_{1} x_{2} + a_{1} a_{2} x_{3} = c$, when $c > 2 a_{1} a_{2} a_{3} - a_{1} a_{2} - a_{2} a_{3} - a_{3} a_{1}$, must have non-negative solutions; when $c = 2 a_{1} a_{2} a_{3} - a_{1} a_{2} ... | 22. Let $x_{1}^{0}, x_{2}^{0}, x_{3}^{0}$ be a particular solution, and any solution $x_{1}, x_{2}, x_{3}$ must satisfy $a_{2} a_{3}\left(x_{1}-x_{1}^{0}\right)+$ $a_{3} a_{1}\left(x_{2}-x_{2}^{0}\right)+a_{1} a_{2}\left(x_{3}-x_{3}^{0}\right)=0$. Therefore, $a_{1}\left|x_{1}-x_{1}^{0}, a_{2}\right| x_{2}-x_{2}^{0}, a_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,096 |
Lemma 1 The primitive solutions $x, y, z$ of the indeterminate equation (1) must satisfy the conditions:
$$\begin{aligned}
(x, y)= & (y, z)=(z, x)=1, \\
& 2 \nmid x+y .
\end{aligned}$$ | Prove that if $x, y$ are not coprime, then there exists a prime $p$ such that $p \mid x, p \mid y$. By (1), we know $p \mid z^{2}$. From this and Theorem 1 of Chapter 1, §5, we conclude $p \mid z$. However, this contradicts $(x, y, z)=1$. Similarly, we can prove $(y, z)=1$ and $(z, x)=1$. By $(x, y)=1$, we know that $x... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,099 |
Theorem 2 The complete set of primitive solutions for which $y$ is even in the indeterminate equation (1) is given by the following formulas:
$$x=r^{2}-s^{2}, \quad y=2 r s, \quad z=r^{2}+s^{2},$$
where $r, s$ are any integers satisfying the following conditions:
$$r>s>0, \quad(s, r)=1, \quad 2 \nmid r+s .$$ | First, we prove that $x, y, z$ given by equations (6) and (7) are certainly primitive solutions of (1) and $2 \mid y$. It is easy to verify that for any $r, s$ (not necessarily satisfying (7)), $x, y, z$ given by equation (6) are certainly solutions of (1) and $2 \mid y$. From $r>s>0$, we know that these are positive s... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,100 |
Example 1 Find all primitive solutions given by equations (6) and (7) when $r \leqslant 7$.
| Table 1 gives all the required primitive solutions.
Table1
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$r$ & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline 1 & $3,4,5$ & & $15,8,17$ & & $35,12,37$ & \\
\hline 2 & & $5,12,13$ & & $21,20,29$ & & $45,28,53$ \\
\hline 3 & & & $7,24,25$ & & & \\
\hline 4 & & & & $40,9,41$ & & $33,56,65$ \\
\hl... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,102 |
Example 3 Let $x, y, z$ be solutions of (1). Prove: (i) $3|x, 3| y$ at least one holds;
(ii) $5|x, 5| y, 5 \mid z$ at least one holds. | To prove that for $x, y, z$ being primitive solutions (why), without loss of generality, assume the solution is given by equation (6). If $3 \nmid y$ then $3 \nmid r, 3 \nmid s$. Therefore, it must be that
$$r=3 k \pm 1, \quad s=3 h \pm 1$$
In any case, $3 \mid x=r^{2}-s^{2}$. This proves (i). If $5 \nmid y$, then $5 ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,104 |
2. If $x^{2}+a x+b=0$ has an integer root $x_{0} \neq 0$, then $x_{0} \mid b$. Generally, if
$$x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}=0$$
has an integer root $x_{0} \neq 0$, then $x_{0} \mid a_{0}$. | 2. $a_{0}=-x_{0}\left(x_{0}^{n-1}+a_{n-1} x_{0}^{n-2}+\cdots+a_{1}\right)$. | a_{0}=-x_{0}\left(x_{0}^{n-1}+a_{n-1} x_{0}^{n-2}+\cdots+a_{1}\right) | Number Theory | proof | Yes | Yes | number_theory | false | 740,105 |
Theorem 4 The indeterminate equation
$$x^{4}+y^{4}=z^{2}$$
has no solution for $x y z \neq 0$. | To prove the evident, we only need to prove that equation (13) has no positive integer solutions. We use proof by contradiction. Suppose equation (13) has positive integer solutions, then among all positive integer solutions, there must be a solution \(x_{0}, y_{0}, z_{0}\) such that \(z_{0}\) takes the minimum value. ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,106 |
Theorem 6 The positive integer solutions of the indeterminate equation
$$x^{2}+y^{2}=z^{4}$$
satisfying the condition $(x, y)=1$ are
$$x=\left|6 a^{2} b^{2}-a^{4}-b^{4}\right|, y=4 a b\left(a^{2}-b^{2}\right), z=a^{2}+b^{2}$$
and
$$x=4 a b\left(a^{2}-b^{2}\right), y=\left|6 a^{2} b^{2}-a^{4}-b^{4}\right|, z=a^{2}+b^{... | Let $x, y, z$ be positive integer solutions of (16), satisfying $(x, y)=1$. Therefore, $x, y, z^{2}$ are primitive solutions of equation (1). By Lemma 1, $x, y$ are one odd and one even, without loss of generality, assume $2 \mid y$. By Theorem 2, we must have
$$x=r^{2}-s^{2}, \quad y=2 r s, \quad z^{2}=r^{2}+s^{2},$$
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,109 |
2. Find all Shang Gao triangles, and all primitive Shang Gao triangles, with one side length of (i) 15, (ii) 22, (iii) 50. | 2. (i) Primitive: $\{15,8,17\},\{15,112,113\}$, Non-primitive: $\{15,36,39\}$, $\{15,20,25\},\{9,12,15\}$; (ii) No primitive, Non-primitive: $\{22,120,122\}$, (iii) No primitive, Non-primitive: $\{14,48,50\},\{30,40,50\},\{50,120,130\},\{50,624,626\}$. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,111 |
3. Discuss for which positive integers $n$, the indeterminate equation $x^{2}-y^{2}=n$ (i) has solutions, (ii) has solutions satisfying $(x, y)=1$. For $n=30,60,120$, determine whether this equation has solutions. If it has solutions, find all of them and all solutions satisfying $(x, y)=1$. Furthermore, propose a meth... | 3. (i) There is a solution when $2 \nmid n$ or $4 \mid n$; (ii) There is a primitive solution when $2 \nmid n$ or $8 \mid n$. By discussing
$$n=n_{1} n_{2}, \quad x-y=n_{1}, \quad x+y=n_{2}, \quad 2 \mid n_{1}+n_{2}$$
we obtain the solutions of the equation. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,112 |
5. Find all Pythagorean triangles with hypotenuse (i) 1105, (ii) 5525, (iii) 117, (iv) 351, including all primitive Pythagorean triangles. | 5. (i) $1105=5 \cdot 13 \cdot 17 \cdot 5=2^{2}+1^{2}, 13=3^{2}+2^{2}, 17^{2}=4^{2}+1^{2}$, and thus using the result from question 4 (i): $5 \cdot 13=8^{2}+1^{2}=7^{2}+4^{2}, 5 \cdot 17=9^{2}+2^{2}=7^{2}+6^{2}, 13 \cdot 17=14^{2}+5^{2}=$ $11^{2}+10^{2}, 5 \cdot 13 \cdot 17=33^{2}+4^{2}=32^{2}+9^{2}=31^{2}+12^{2}=24^{2}... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,114 |
7. Find all Pythagorean triangles with an area equal to (i) 78 ; (ii) 360. | 7. Let $x, y, z$ be given by equation (6) and equation (7). The area $A=d^{2} r s(r-s)(r+s)$ of a right-angled triangle with sides $d x, d y, d z$. Trying to solve with $A=78,360$, it is found that no such triangle exists. | no such triangle exists | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,117 |
9. Prove: All positive solutions of the indeterminate equation $x^{2}+2 y^{2}=z^{2}$ satisfying $(x, y, z)=1$ are $x=\left|u^{2}-2 v^{2}\right|, y=2 u v, z=u^{2}+2 v^{2}$, where $u, v$ are any positive integers satisfying $(u, v)=1,2 \nmid u$. | 9. There must be $2 \nmid x, 2 \nmid z$ and $2 \mid y$. So the equation becomes $2(y / 2)^{2}=(z-x) / 2 \cdot(z+x) / 2$. Then, discuss as in Theorem 2. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,119 |
11. Find all solutions to $x^{2}+3 y^{2}=z^{2}$ satisfying $(x, y)=1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 11. $3 \mid z+x$ or $3 \mid z-x$ holds and only one of them is true, $(z-x, z+x)=1$ or 2. Write the equation as $y^{2}=(z+x) / 3 \cdot(z-x)$ or $y^{2}=(z+x) \cdot(z-x) / 3$, and then discuss similarly to Theorem 2. | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,121 |
14. Prove: $x^{4}+4 y^{4}=z^{2}$ has no solution for $x y z \neq 0$.
| 14. (i) If there is a solution $x_{0}, y_{0}, z_{0}$, then there must be a set of pairwise coprime positive solutions $x_{1}, y_{1}, z_{1}$;
(ii) Let $x_{1}, y_{1}, z_{1}$ be the set of all pairwise coprime positive solutions that makes $y_{1}$ the smallest. Using $\left(\left(z_{0}-x_{0}^{2}\right) /\right.$ $\left.2,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,124 |
21. Prove: In any solution of the indeterminate equation $w^{2}+x^{2}+y^{2}=z^{2}$, at least two of $w, x, y$ are even. Furthermore, prove: the general solution of this equation with $x, y$ being even is
$$w=\left(l^{2}+m^{2}-n^{2}\right) / n, x=2 l, y=2 m, z=\left(l^{2}+m^{2}+n^{2}\right) / n,$$
where $n, m, l$ are a... | 21. Let $x=2 l, y=2 m .(z+w)(z-w)=2^{2}\left(l^{2}+m^{2}\right)$. Therefore, it must be that
$$z+w=2 n, \quad z-w=2\left(l^{2}+m^{2}\right) / n, \quad n \mid l^{2}+m^{2} .$$
From this, it follows that $x, y, z, w$ have the stated form. Conversely, direct verification shows that the given forms are solutions. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,132 |
22. Prove: The indeterminate equation $x^{n}+y^{n}=z^{n+1}$ has infinitely many solutions. | 22. $x=\left(1+k^{n}\right), y=k\left(1+k^{n}\right), z=\left(1+k^{n}\right), k$ any integer. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,133 |
23. Prove: The indeterminate equation $x^{n}+y^{n}=z^{n-1}$ has infinitely many solutions. | 23. $x=\left(1+k^{n}\right)^{n-2}, y=k\left(1+k^{n}\right)^{n-2}, z=\left(1+k^{n}\right)^{n-1}, k$ any integer. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,134 |
24. Prove: The indeterminate equation $x^{2}-2 y^{4}=1$ has no positive integer solutions. | 24. Use proof by contradiction. If there is a solution, then $2 \nmid x$, let $x=2k+1$, we get $2k(k+1)=y^{4}$. Therefore, it must be that (i) $2k=u^{4}, k+1=v^{4}$, or (ii) $k=u^{4}, 2(k+1)=v^{4}, 2 \nmid uv, (u, v)=1$. If (i) holds, let $u=2^{r}u_1$, we get $v^{4}=1+2^{4r-1}u_1^{4}, v^{4}+\left(2^{2r-1}u_1^{2}\right)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,135 |
25. Prove: The indeterminate equation $x^{4}-2 y^{2}=-1$ has no other positive integer solutions except $x=y=1$. | 25. The original equation is equivalent to $y^{4}=x^{4}+\left(y^{2}-1\right)^{2}$. From this and problem 15, we can derive the required conclusion. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,136 |
26. Prove: The indeterminate equation $x^{2}-8 y^{4}=1$ has no other positive integer solutions except $x=3, y=1$. | 26. $(x-1)(x+1)=8 y^{4}$, there must be (i) $x-1=2 u^{4}, x+1=2^{4 r+2} v^{4}$; or (ii) $x-1=$ $2^{4 r+2} u^{4}, x+1=2 v^{4}, 2 \nmid u v,(u, v)=1$. Similar to the discussion in problem 24, and using problem 25. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,137 |
5. If $5 \mid n$ and $17 \mid n$, then $85 \mid n$. | 5. $7 \cdot 5 + (-2) \cdot 17 = 1$. Using $\S 2$ Example 3. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,138 |
Theorem $1 a$ is congruent to $b$ modulo $m$ if and only if $a$ and $b$ yield the same smallest non-negative remainder when divided by $m$, that is, if
$$\begin{array}{ll}
a=q_{1} m+r_{1}, & 0 \leqslant r_{1}<m, \\
b=q_{2} m+r_{2}, & 0 \leqslant r_{2}<m,
\end{array}$$
then $r_{1}=r_{2}$. | Prove that we have $a-b=\left(q_{1}-q_{2}\right) m+\left(r_{1}-r_{2}\right)$. Therefore, the necessary and sufficient condition for $m \mid a-b$ is $m \mid r_{1}-r_{2}$. From this and $0 \leqslant\left|r_{1}-r_{2}\right|<m$, we get $r_{1}=r_{2}$. Proof completed. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,139 |
Property II Congruences can be added, that is, if
$$a \equiv b(\bmod m), \quad c \equiv d(\bmod m),$$
$$a+c \equiv b+d(\bmod m)$$ | To prove: $m|a-b, m| c-d \Rightarrow m \mid(a-b)+(c-d)=(a+c)-(b+d)$, which proves the desired conclusion. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,142 |
Property UI Congruence
$$c a \equiv c(\bmod m)$$
is equivalent to
$$a \equiv b(\bmod m /(c, m))$$
In particular, when $(c, m)=1$, the congruence (7) is equivalent to
$$a \equiv b(\bmod m),$$
that is, the factor $c$ can be canceled from both sides of the congruence (7). | To prove the congruence (7), i.e., \( m \mid c(a-b) \), this is equivalent to
$$\frac{m}{(c, m)} \left\lvert\, \frac{c}{(c, m)}(a-b) .\right.$$
By \((m /(c, m), c /(c, m))=1\) and Theorem 6 of Chapter 1, §4, we know that this is equivalent to
$$\left.\frac{m}{(c, m)} \right\rvert\, a-b$$
This completes the proof of t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,147 |
Property VII If $m \geqslant 1,(a, m)=1$, then there exists $c$ such that
$$c a \equiv 1(\bmod m)$$
We call $c$ the inverse of $a$ modulo $m$, denoted by $a^{-1}(\bmod m)$ or $a^{-1}$. | Proof: By Theorem 8 (k=2) of Chapter 1, §4, we know that there exist $x_{0}, y_{0}$ such that $a x_{0} + m y_{0} = 1$. Taking $c = x_{0}$ satisfies the requirement.
For example:
\begin{tabular}{|c|cccccc|}
\hline$a$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline$a^{-1}$ & 1 & 4 & 5 & 2 & 3 & 6 \\
\hline
\end{tabular}
$(\bmod 7)$
\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,148 |
Property X system of congruences
$$a \equiv b\left(\bmod m_{j}\right), \quad j=1,2, \cdots, k$$
The sufficient and necessary condition for the simultaneous establishment is
$$a \equiv b\left(\bmod \left[m_{1}, \cdots, m_{k}\right]\right) .$$ | Proof: By Theorem 1 of Chapter 1, §4, we know that the necessary and sufficient condition for $m_{j} \mid a-b \ (j=1, \cdots, k)$ to hold simultaneously is $\left[m_{1}, \cdots, m_{k}\right] \mid a-b$. This is the conclusion to be proved. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,150 |
Example 1 Find the units digit of $3^{406}$ when it is written in decimal form. | Solve: According to the problem, we are required to find the smallest non-negative remainder $a$ when $3^{406}$ is divided by 10, i.e., $a$ satisfies
$$3^{406} \equiv a(\bmod 10), \quad 0 \leqslant a \leqslant 9$$
Obviously, we have $3^{2} \equiv 9 \equiv-1(\bmod 10), 3^{4} \equiv 1(\bmod 10)$, and thus
$$3^{404} \equ... | 9 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,151 |
Example 2 Find the last two digits of $3^{406}$ when written as a decimal number. | Solve: This is to find the smallest non-negative remainder $b$ when $3^{106}$ is divided by 100, i.e., $b$ satisfies
$$3^{406} \equiv b(\bmod 100), \quad 0 \leqslant b \leqslant 99$$
Notice that $100=4 \cdot 25, (4,25)=1$. Clearly, $3^{2} \equiv 1(\bmod 4), 3^{4} \equiv 1(\bmod 5)$. Note that 4 is the smallest power, ... | 29 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,152 |
Example 3 Prove that $641 \mid 2^{2^{5}}+1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | To prove that due to the large numbers, direct division is very cumbersome. Using congruences, we need to prove that
$$2^{32} \equiv -1 \pmod{641}.$$
641 is a prime number. By step-by-step calculations, we get:
$$\begin{array}{ll}
2^{9} \equiv 512 \equiv -129 \pmod{641}, & 2^{11} \equiv -516 \equiv 125 \pmod{641}, \\
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,153 |
Example 4 Prove that the indeterminate equation $x^{2}+2 y^{2}=203$ has no solution.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove by contradiction. From $203=7 \cdot 29$, we know that if there are solutions $x_{0}, y_{0}$, then $\left(x_{0} y_{0}, 203\right) = 1$ (why). It is evident that $x_{0}^{2} \equiv -2 y_{0}^{2} (\bmod 7)$. Since $\left(y_{0}, 7\right) = 1$, by Property VIII, $y_{0}$ has an inverse $y_{0}^{-1}$ modulo 7. Multiplying ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,154 |
Example 5 Let $n \geqslant 1, b$ whose prime factors are all greater than $n$. Prove: for any positive integer $a$ there must be $n! \mid a(a+b)(a+2 b) \cdots(a+(n-1) b)$ | Given that $(b, n!)=1$. By Property VII, $b$ has an inverse $b^{-1}$ modulo $n!$. We have
$$\begin{aligned}
\left(b^{-1}\right)^{n} & \cdot a(a+b) \cdots(a+(n-1) b) \\
& \equiv a b^{-1}\left(a b^{-1}+b b^{-1}\right) \cdots\left(a b^{-1}+(n-1) b b^{-1}\right) \\
& \equiv a b^{-1}\left(a b^{-1}+1\right) \cdots\left(a b^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,155 |
Example 6 Let $m>n \geqslant 1$. Find the smallest $m+n$ such that
$$\text {1000| } 1978^{m}-1978^{n} \text {. }$$ | Solve: Using the congruence symbol, the problem is to find the smallest $m+n$ such that
$$1978^{m}-1978^{n} \equiv 0(\bmod 1000)$$
is satisfied. First, let's discuss what conditions $m, n$ must meet for the above equation to hold. Let $k=m-n$. Equation (9) becomes
$$2^{n} \cdot 989^{n}\left(1978^{k}-1\right) \equiv 0\... | 106 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,156 |
3. For which moduli $m$ do the congruences $32 \equiv 11(\bmod m)$ and $1000 \equiv -1(\bmod m)$ both hold? In general, what condition must the modulus $m$ satisfy for the congruences $a \equiv b(\bmod m)$ and $c \equiv d(\bmod m)$ to both hold? | 3. $m \mid (a-b, c-d)$.
The translation is provided as requested, maintaining the original text's format and line breaks. | m \mid (a-b, c-d) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,159 |
7. Let $n \neq 1$. Prove: $(n-1)^{2} \mid n^{k}-1$ if and only if $(n-1) \mid k$.
The translation is complete as requested, maintaining the original text's line breaks and format. | 7. $n^{k}-1=((n-1)+1)^{k}-1=A(n-1)^{2}+k(n-1), A$ is an integer. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,160 |
4. (i) For a prime $p>2$, what are the possible values of the smallest non-negative residue, the smallest positive residue, and the absolute smallest residue modulo $m=4$?
(ii) Change (i) to $p>3, m=6$.
(iii) Change (i) to $p>5, m=30$. Provide specific examples of primes $p$ that achieve the residues mentioned in (i), ... | 4. (i) Least non-negative residue: 1,3 ; Least positive residue: 1,3 ; Absolute least residue: $-1,1$. $p=5,3$. (ii) Least non-negative residue and least positive residue system: 1,5 ; Absolute least residue: $-1,1$. $p=7$, 5. (iii) Least non-negative residue and least positive residue: $1,7,11,13,17,19,23,29$; Absolut... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,161 |
5. Prove: (i) $a \equiv b(\bmod m)$ is equivalent to $a-b \equiv 0(\bmod m)$;
(ii) If $a \equiv b(\bmod m), c \equiv d(\bmod m)$, then $a-c \equiv b-d(\bmod m)$. Explain the significance of these results from the perspective of congruence operations. | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,162 |
7. Determine whether the following statements are true or false. Provide a proof for true statements and a counterexample for false ones:
(i) If $a^{2} \equiv b^{2}(\bmod m)$ holds, then $a \equiv b(\bmod m)$;
(ii) If $a^{2} \equiv b^{2}(\bmod m)$, then $a \equiv b(\bmod m)$ or $a \equiv -b(\bmod m)$ at least one holds... | 7. (i) Not true. $5^{2} \equiv 7^{2}(\bmod 8), 5 \not \equiv 7(\bmod 8)$. (ii) Not true. For example (i) $5 \not \equiv \pm 7$ $(\bmod 8)$ . (iii) Not true. $5 \equiv-3(\bmod 8), 25 \not \equiv 9(\bmod 64)$ . (iv) True. Because $2 \mid a-b$ $\Rightarrow 2 \mid a+b$. (v) True. Because $p|(a-b)(a+b), p \nmid(a-b, a+b)| 2... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,164 |
8. When the positive integer $m$ satisfies what condition,
$$1+2+\cdots+(m-1)+m \equiv 0(\bmod m)$$
is always true (without calculating the left-hand sum)? | 8. $1+2+\cdots+(m-1)+m \equiv m+(1+(m-1))+(2+(m-2))+\cdots \equiv$ $0(\bmod m)$, when $2 \nmid m ; \equiv m / 2 \not \equiv 0(\bmod m)$, when $2 \mid m$. | 2 \nmid m | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,165 |
9. When the positive integer $m$ satisfies what condition,
$$1^{3}+2^{3}+\cdots+(m-1)^{3}+m^{3} \equiv 0(\bmod m)$$
is always true (without calculating the left-hand side sum)? | 9. $1^{3}+2^{3}+\cdots+(m-1)^{3}+m^{3} \equiv m^{3}+\left(1+(m-1)^{3}\right)+\cdots \equiv 0(\bmod m)$, when $2 \nmid$ $m ; \equiv(m / 2)^{3}(\bmod m)$, when $2 \mid m$. It holds when $2 \nmid m$ or $4 \mid m$, and does not hold when $2 \mid m, 4 \nmid m$. | 2 \nmid m \text{ or } 4 \mid m | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,166 |
10. For any positive integer $n \geqslant 1, 1+2+\cdots+n$ when expressed as a decimal, the last digit of the number can take which of the following values? | 10. $0,1,3,5,6,8$.
The text above has been translated into English, maintaining the original text's line breaks and format. | 0,1,3,5,6,8 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,167 |
12. Let $n>4$. Prove: $n$ is a composite number if and only if
$$(n-2)!\equiv 0(\bmod n)$$ | 12. $n=d_{1} d_{2}, 2 \leqslant d_{1}, d_{2} \leqslant n / 2$. When $d_{1} \neq d_{2}$, it is evident that $(n-2)!\equiv 0(\bmod n)$. When $d_{1} = d_{2}$, since $n>4$, we have $d_{1} \geqslant 3, 2 \leqslant d_{1}<2 d_{1} \leqslant n-2$. Therefore, $(n-2)!\equiv 0(\bmod n)$ also holds. This proves the necessity. The s... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,169 |
14. (i) Find the least non-negative residue of $2^{400}$ modulo 10;
(ii) Find the last two digits in the decimal representation of $2^{1000}$;
(iii) Find the last two digits in the decimal representation of $9^{9^{9}}$ and $9^{9^{9^{9}}}$;
(iv) Find the least non-negative residue of $\left(13481^{56}-77\right)^{28}$ wh... | 14. (i) 6 ; (ii) $2^{22} \equiv 4(\bmod 100), 2^{1000} \equiv 2^{100} \equiv 2^{20} \equiv 76(\bmod 100), 76$;
(iii) $9^{10}$ $\equiv(10-1)^{10} \equiv 1(\bmod 100), 9^{9} \equiv(10-1)^{9} \equiv 9(\bmod 10)$, so $9^{9^{9}} \equiv 9^{9} \equiv -11 \equiv$ $89(\bmod 100)$, the last two digits are $89, 9^{9^{9}} \equiv 9... | 6, 76, 89, 70, 6 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,172 |
18. Find the integer $n$ that satisfies $n \equiv 1(\bmod 4), n \equiv 2(\bmod 3)$. | 18. $n=4 k+1 \equiv 2(\bmod 3), k=1, n=5$. | 5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,176 |
19. Prove: For any integer $n$, at least one of the following five congruences holds:
$$\begin{array}{ll}
n \equiv 0(\bmod 2), & n \equiv 0(\bmod 3), \quad n \equiv 1(\bmod 4) \\
n \equiv 5(\bmod 6), & n \equiv 7(\bmod 12)
\end{array}$$ | 19. $n \equiv 0(\bmod 2) \Longleftrightarrow n \equiv 0,2,4,6,8$, or $10(\bmod 12) ; n \equiv 0(\bmod 3)$ $\Longleftrightarrow n \equiv 0,3,6$, or $9(\bmod 12) ; n \equiv 1(\bmod 4) \Longleftrightarrow n \equiv 1,5$, or $9(\bmod 12) ; n \equiv$ $5(\bmod 6) \Longleftrightarrow n \equiv 5$ or $11(\bmod 12) ;$ and $n \equ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,177 |
22. Find all positive integer triples $\{a, b, c\}$, satisfying the conditions: $a \equiv b(\bmod c), \quad b \equiv c(\bmod a), \quad c \equiv a(\bmod b)$. | 22. If $a, b, c$ satisfy the conditions, then for any $k \geqslant 1, k a, k b, k c$ also satisfy them, so we can first assume $(a, b, c)=1$, and $1 \leqslant a \leqslant b \leqslant c$. From this and $c \mid a-b$ we deduce $a=b$, and then by $a \mid c$ and $(a, b, c)=1$ we deduce $a=b=1$. Therefore, all solutions are ... | \{1,1, c\}, c \text{ being any positive integer} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,180 |
23. Find all non-zero integer triples $\{a, b, c\}$, satisfying the conditions:
$$a \equiv b(\bmod |c|), \quad b \equiv c(\bmod |a|), \quad c \equiv a(\bmod |b|)$$ | 23. Similar to the previous problem, we can first assume $(a, b, c)=1, c>0$ and $|a| \leqslant|b| \leqslant c$. It follows that we must have (i) $a=b$; (ii) $a-b= \pm c$; (iii) $a-b= \pm 2 c$. These three cases can respectively yield (i) $\{1,1, c\}$, $\{-1,-1, c\}$; (ii) $\{1,-n, n+1\},\{2,-(2 n+1), 2 n+3\}, n$ being ... | \{1,1, c\}, \{-1,-1, c\}, \{1,-n, n+1\}, \{2,-(2 n+1), 2 n+3\}, \{-1,1,2\}, \{-1,2,3\}, \{1,-1,1\}, \{-1,1,1\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,181 |
9. Prove:
(i) A necessary condition for $n \geqslant 1.2^{n}+1$ to be a prime is that $n=2^{k}$.
(ii) A necessary condition for $2^{n}-1$ to be a prime is that $n$ is a prime.
Give a few examples of primes of these two forms. | 9. (i) If $n \neq 2^{k}$, then $n=a \cdot m, 2 \nmid m>1$.
$$2^{n}+1=\left(2^{a}\right)^{m}+1=\left(2^{a}+1\right)\left(\left(2^{a}\right)^{m-1}-\left(2^{a}\right)^{m-2}+\cdots+1\right) .$$
(ii) If $n$ is composite, then $n=a \cdot m, a>1, m>1.2^{n}-1=\left(2^{a}\right)^{m}-1=\left(2^{a}-1\right) \cdot$
$$\begin{array}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,182 |
25. Let $p$ be a prime, and $f(x)$ be a polynomial with integer coefficients; also let $a_{1}, \cdots, a_{k}$ be pairwise incongruent modulo $p$, satisfying $f\left(a_{j}\right) \equiv 0(\bmod p), 1 \leqslant j \leqslant k$. Prove: There exists a polynomial $q(x)$ with integer coefficients, such that
$$f(x) \equiv q(x)... | 25. The first part uses polynomial division, and proves by induction. From this, derive the conclusions in the second part, and the last conclusion (d) should be derived by comparing the coefficients of $x^{p-3}$ on both sides of equation (a). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,184 |
26. Let $p>3$ be a prime. Prove:
$$\frac{(p-1)!}{1}+\frac{(p-1)!}{2}+\cdots+\frac{(p-1)!}{(p-1)} \equiv 0\left(\bmod p^{2}\right)$$
You can do IMO problems (see Appendix Four): [17.4], [20.1], [25.2], [34.6], $[37.4]$. | 26. Let $(x-1) \cdots(x-p+1)=x^{p-1}+s_{1} x^{p-2}+\cdots+s_{p-2} x+(p-1)$!. From this and parts (a), (c) of question 25, deduce that $p \mid\left(s_{1}, \cdots, s_{p-2}\right)$. By substituting $x=p$ in the above equation, it follows that $p^{2} \mid s_{p-2}$, which is the conclusion to be proven. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,185 |
Theorem 2 For a given modulus $m$, there are exactly $m$ distinct congruence classes modulo $m$, which are
$$0 \bmod m, 1 \bmod m, \cdots,(m-1) \bmod m$$
We denote the set composed of these congruence classes as
$$\boldsymbol{Z} / m \boldsymbol{Z}=\boldsymbol{Z}_{m}=\{j \bmod m: 0 \leqslant j \leqslant m-1\}$$ | Proof: By Theorem 1 (ii), we know that these are $m$ pairwise distinct congruence classes. For each integer $a$, by Theorem 1 in Chapter 1 § 3, we know that
$$a=q m+r, \quad 0 \leqslant r<m.$$
Therefore, by Theorem 1 (i), we know that $a \in j \bmod m$, i.e., $a$ must belong to one of the congruence classes in (1). Pr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,187 |
Theorem 3 (i) Among any given $m+1$ integers, there must be two numbers that are congruent modulo $m$;
(ii) There exist $m$ numbers that are pairwise incongruent modulo $m$. | Proof: Since there are $m$ congruence classes modulo $m$ given by (1), there must be two numbers among the $m+1$ numbers that belong to the same congruence class modulo $m$. These two numbers are congruent modulo $m$. This proves (i). By selecting a number $x_{r}$ as a representative in each congruence class $r \bmod m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,188 |
Theorem 2 (Principle of the Least Natural Number) Let $T$ be a non-empty subset of $\boldsymbol{N}$. Then, there exists $t_{0} \in T$ such that for any $t \in T$, $t_{0} \leqslant t$, i.e., $t_{0}$ is the least natural number in $T$. | The proof of Theorem 2 considers the set $S$ consisting of all natural numbers $s$ such that for any $t \in T$, we have $s \leqslant t$. Since 1 satisfies this condition, $1 \in S$, so $S$ is non-empty. Furthermore, if $t_{1} \in T$ (since $T$ is non-empty, there must be a $t_{1}$), then $t_{1}+1 > t_{1}$, so $t_{1}+1 ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,192 |
1. Prove: For any given positive integer $K$, there must be $K$ consecutive positive integers that are all composite. | 10. $(K+1)!+2,(K+1)!+3, \cdots,(K+1)!+(K+1)$. | (K+1)!+2,(K+1)!+3, \cdots,(K+1)!+(K+1) | Number Theory | proof | Yes | Yes | number_theory | false | 740,193 |
Theorem 5 Any two integers $a_{1}, a_{2}$ in a congruence class modulo $m$ have the same greatest common divisor with $m$, i.e., $\left(a_{1}, m\right)=\left(a_{2}, m\right)$. | Proof: Let $a_{1} \in r \bmod m, a_{2} \in r \bmod m$. By Theorem 1(i), we know $a_{j}=r+k_{j} m$, $j=1,2$. Then, by Theorem 8(iv) in Chapter 1 § 2, we get
$$\left(a_{j}, m\right)=\left(r+k_{j} m, m\right)=(r, m), \quad j=1,2 .$$
This completes the proof of the desired conclusion. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,194 |
Theorem 8 Let $p$ be a prime, $k \geqslant 1$, then
$$\varphi\left(p^{k}\right)=p^{k-1}(p-1),$$
and the reduced residue classes modulo $p^{k}$ are
$$(a+b p) \bmod p^{k}, \quad 1 \leqslant a \leqslant p-1, \quad 0 \leqslant b \leqslant p^{k-1}-1 .$$ | Proof: By Theorem 6, $\varphi\left(p^{k}\right)$ is equal to the number of $r$ satisfying:
$$1 \leqslant r \leqslant p^{k}, \quad\left(r, p^{k}\right)=1$$
Since $p$ is a prime, we have
$$(r, p)=\left\{\begin{array}{ll}
1, & p \nmid r, \\
p, & p \mid r .
\end{array}\right.$$
From this and Theorem 5 in Chapter 1, §4, w... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,197 |
Theorem 10 If $(a, m)=1$, then the necessary and sufficient condition for $x$ to traverse a complete (reduced) residue system modulo $m$ is that $a x$ traverses a complete (reduced) residue system modulo $m$. That is, $x_{1}, \cdots, x_{s}$ is a complete (reduced) residue system modulo $m$ if and only if $a x_{1}, \cdo... | Proof: By property $\mathbb{I I}$ of $\S 1$, when $(a, m)=1$, for any $i, j$ we have $x_{i} \not \equiv x_{j}(\bmod m) \Longleftrightarrow a x_{i} \not \equiv a x_{j}(\bmod m) ;$
By Theorem 5 of $\S 4$ in Chapter 1, when $(a, m)=1$, for any $i$ we have
$$\left(a x_{i}, m\right)=\left(x_{i}, m\right)$$
For a complete r... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,199 |
Theorem 12 Let $m=m_{1} m_{2} \cdots m_{k}$ and
$$x=x^{(1)}+m_{1} x^{(2)}+m_{1} m_{2} x^{(3)}+\cdots+m_{1} \cdots m_{k-1} x^{(k)},$$
Then, when $x^{(j)}(1 \leqslant j \leqslant k)$ respectively traverse the complete residue system modulo $m_{j}$, $x$ traverses the complete residue system modulo $m$. That is, when $x_{... | Proof: By Theorem 11, the conclusion holds for $k=2$. Assume the conclusion holds for $k=n (\geqslant 2)$. When $k=n+1$, let $\bar{m}_{n}=m_{1} \cdots m_{n}$ and
$$\bar{x}^{(n)}=x^{(1)}+m_{1} x^{(2)}+\cdots+m_{1} \cdots m_{n-1} x^{(n)}$$
Then we have
$$x=\bar{x}^{(n)}+\bar{m}_{n} x^{(n+1)} .$$
By the induction hypoth... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,201 |
11. Prove: An odd number can always be expressed as the difference of two square numbers. | 11. $2 n+1=(n+1)^{2}-n^{2}$.
Translating the above text into English, while preserving the original text's line breaks and format, results in:
11. $2 n+1=(n+1)^{2}-n^{2}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,204 |
Example 2 Use the complete residue systems modulo 10 and modulo 199 to represent the complete residue system modulo 1990. | (i) $x=x^{(1)}+10 x^{(2)}$, when $x^{(1)}$ traverses the complete residue system modulo 10 and when $x^{(2)}$ traverses the complete residue system modulo 199, $x$ traverses the complete residue system modulo 1990. Specifically, when $1 \leqslant x^{(1)} \leqslant 10,0 \leqslant x^{(2)} \leqslant 198$, $x$ takes values... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,205 |
Example 3 Use the residue system modulo 3 to represent the residue system modulo $3^{n}(n \geqslant 2)$. | From Theorem 13, we know that when $x^{(1)}$ runs through the complete (reduced) residue system modulo 3, and $x^{(j)} (2 \leqslant j \leqslant n)$ runs through the complete residue system modulo 3,
$$x=x^{(1)}+3 x^{(2)}+\cdots+3^{n-1} x^{(n)}$$
runs through the complete (reduced) residue system modulo $3^{n}$. Specif... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,206 |
Theorem 14 Let $m=m_{1} m_{2},\left(m_{1}, m_{2}\right)=1$,
$$x=m_{2} x^{(1)}+m_{1} x^{(2)},$$
Then, the necessary and sufficient condition for $x$ to traverse the complete (reduced) residue system modulo $m$ is that $x^{(1)}, x^{(2)}$ simultaneously traverse the complete (reduced) residue systems modulo $m_{1}, m_{2}... | Proof: First, it should be pointed out that the form of Theorem 14 is different from the previous theorems; the condition here is both sufficient and necessary, making the conclusion stronger. We will first prove it for a complete residue system. We start by proving the sufficiency (which can be derived using Theorem 1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,207 |
Theorem 15 Let $m=m_{1} \cdots m_{k}, m_{1}, \cdots, m_{k}$ be pairwise coprime; and let $m=m_{j} M_{j}$ $(1 \leqslant j \leqslant k)$ and
$$x=M_{1} x^{(1)}+\cdots+M_{k} x^{(k)}$$
Then, $x$ runs through a complete (reduced) residue system modulo $m$ if and only if $x^{(1)}, \cdots, x^{(k)}$ run through complete (reduc... | Prove that for $k=2$, Theorem 14 holds, so it is true. Assume that for $k=n (\geqslant 2)$, the theorem holds. When $k=n+1$, $m=m_{1} \cdots m_{n} m_{n+1}$. Let $x$ be given by equation (20) for $k=n+1$,
$$\bar{x}^{(n)}=\frac{m}{m_{1} m_{n+1}} x^{(1)}+\cdots+\frac{m}{m_{n} m_{n+1}} x^{(n)}$$
We have
$$x=m_{n+1} \bar{x... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,208 |
Example 6 Specifically gives a complete residue system and a reduced residue system modulo 30 in the form of equations (20) and (21).
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve (i) In equation (20), take $k=2$ (i.e., equation (18)), $m_{1}=6, m_{2}=5$. At this time, $M_{1}=5, M_{2}=6$,
$$x=5 x^{(1)}+6 x^{(2)} .$$
Take the complete residue system modulo 6 for $x^{(1)}$:
$$-2,-1,0,1,2,3,$$
Thus, $5 x^{(1)}$ also traverses the complete residue system modulo 6, with values
$$-10,-5,0,5,10... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,212 |
Example 8 Let the prime factorization of $m$ be $p_{1}^{q_{1}} \cdots p_{r}^{\sigma_{r}}$. Find the value of the sum
$$S(m)=\sum_{x \bmod m}^{\prime} \mathrm{e}^{2 \pi \mathrm{ix} / m}$$
where the summation $\sum_{x \bmod m}^{\prime}$ indicates the sum over any complete set of reduced residues modulo $m$. | Since for any integer $a, \mathrm{e}^{2 \times \mathrm{x} a}=1$, the value of the exponential sum (38) is independent of the specific choice of the reduced residue system. Let $\sum_{x \bmod m}$ denote the sum over any complete residue system modulo $m$. It is evident that for any integer $c$ we have
$$\sum_{x \bmod m}... | S(m)=\left\{\begin{array}{ll}
(-1)^{r}, & \alpha_{1}=\cdots=\alpha_{r}=1, \\
0, & \text { otherwise. }
\end{array}\right.} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,214 |
12. Let $n>1$ be an odd number. Prove: $n$ is a prime number if and only if $n$ cannot be expressed as the sum of three or more consecutive positive integers. | 12. The sum of $k+1$ consecutive positive integers $m, m+1, \cdots, m+k$ is $(k+1)(2 m+k) / 2$. Let $n=m+(m+1)+\cdots+(m+k)=(k+1)(2 m+k) / 2$. If $k \geqslant 2$, it is evident that $n$ is a composite number. This proves the necessity. If an odd number $n>1$ is a composite number, then $n=n_{1} n_{2}, n_{1} \geqslant n... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,215 |
3. Using the method of determining $k_{r}$ in formula $(7)(m=7)$ of $\S 2$, write out a complete residue system modulo 7 such that its elements belong to: (i) the residue class $0 \bmod 3$; (ii) the residue class $1 \bmod 3$; (iii) the residue class $2 \bmod 3$. | 3. (i) $21,15,3,-3,12,6$;
(ii) each number in (i) plus 1; (iii) each number in (i) minus 1. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,218 |
4. Let $(a, m)=1, s$ be any integer. Using formula (7) of § 2, prove: there definitely exists a complete residue system modulo $m$, all of whose elements belong to the residue class $s \bmod a$. | 4. For any $r$, there exist $h_{r}, k_{r}$ such that $h_{r} a+k_{r} m=s-r$.
| proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,219 |
5. Prove: When $m>2$, $0^{2}, 1^{2}, \cdots,(m-1)^{2}$ cannot be a complete residue system modulo $m$. | 5. Since $j^{2} \equiv (m-j)^{2} \pmod{m}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,220 |
7. Let there be $m$ integers, none of which belong to the residue class $0 \bmod m$, then there must be two numbers among them whose difference belongs to the residue class $0 \bmod m$. | 7. By the Pigeonhole Principle, there must be two numbers belonging to the same residue class. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,222 |
8. Among any arbitrarily chosen $[m / 2]+1$ numbers that are pairwise incongruent modulo $m$, there must be two numbers whose difference belongs to the residue class $1 \bmod m$. How can this problem be generalized? | 8. When $m$ is even, group the $m$ residue classes into pairs as follows: $1 \bmod m, 2 \bmod m$ as one group, $3 \bmod m, 4 \bmod m$ as another group, $\cdots$, $(m-1) \bmod m, m \bmod m$ as the last group. This way, by taking $[m / 2]+1$ numbers, there must be two numbers in the same group, so the conclusion holds.
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,223 |
9. (i) Write the residue class $1 \bmod 5$ as a sum of residue classes modulo 15;
(ii) Write the residue class $6 \bmod 10$ as a sum of residue classes modulo 120;
(iii) Write the residue class $6 \bmod 10$ as a sum of residue classes modulo 80;
(iv) Find the intersection of the residues $1 \bmod 5$ and $2 \bmod 3$;
(v... | 9. (i) $1 \bmod 5=\bigcup_{0 \leqslant r \leqslant 2}(1+5 r) \bmod 15$;
(ii) $6 \bmod 10=\bigcup_{0 \leqslant r \leqslant 11}(6+10 r) \bmod 120$;
(iii) $6 \bmod 10=\bigcup_{0 \leqslant r \leqslant 7}(6+10 r) \bmod 80$. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,224 |
10. (i) The necessary and sufficient condition for $n \equiv 1(\bmod 2)$ is that the absolute minimal residue of $n$ modulo 10 is which numbers?
(ii) The necessary and sufficient condition for $n \equiv-1(\bmod 5)$ is that the minimal positive residue of $n$ modulo 45 is which numbers? | 10. (i) $\pm 1, \pm 3,5$;
(ii) $4,9,14,19,24,29,34,39,44$. | 4, 9, 14, 19, 24, 29, 34, 39, 44 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,225 |
13. Let $p$ be the smallest prime factor of the positive integer $n$. Prove: if $p>n^{1 / 3}$, then $n / p$ is a prime. | 13. If $n / p$ is not a prime, then there must be a prime factor $p^{\prime}$ satisfying: $p^{\prime} \leqslant (n / p)^{1 / 2}, p^{\prime} \geqslant p$. This implies $p \leqslant n^{1 / 3}$. This contradicts the assumption. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,226 |
11. Let $n>2$ be a given integer. How many residue classes modulo $n-2$ must a complete residue system modulo $2n-1$ at least belong to? Generally, for $K>m \geqslant 1$, how many residue classes modulo $m$ must a complete residue system modulo $K$ at least belong to? | 11. $(2 n-1, n-2)=1$, when $3 \nmid n-2$; $=3$, when $3 \mid n-2$. When $(2 n-1, n-2)=1$, it belongs to at least 1 residue class modulo $n-2$; when $(2 n-1, n-2)=3$, it belongs to at least 3 residue classes modulo $n-2$. Generally, it belongs to at least $(K, m)$ residue classes modulo $m$. | (K, m) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,227 |
18. Let $d \geqslant 1, d \mid n$. Prove: $n-\varphi(n) \geqslant d-\varphi(d)$, equality holds only when $d=n$. | 18. If $(a, d)>1$, then it must be that $(a, n)>1$. Moreover, among $d$ consecutive integers, the number of integers not coprime to $d$ is $d-\varphi(d)$. Therefore, $n-\varphi(n) \geqslant(n / d)(d-\varphi(d))$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,234 |
19. Prove: There exist infinitely many positive integers $n$ such that $\varphi(n)>\varphi(n+1)$. Moreover, provide examples of positive integers $n$ such that
(i) $\varphi(n)=\varphi(n+1)$;
(ii) $\varphi(n)=\varphi(n+2)$;
(iii) $\varphi(n)=\varphi(n+3)$. | 19. For a prime $p>3$, it must be that $\varphi(p)>\varphi(p+1)$. $\varphi(3)=\varphi(4)=\varphi(6)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,235 |
14. Let $p_{1} \leqslant p_{2} \leqslant p_{3}$ be prime numbers, and $n$ be a positive integer. If $p_{1} p_{2} p_{3} \mid n$, then
$$p_{1} \leqslant n^{1 / 3}, \quad p_{2} \leqslant(n / 2)^{1 / 2}$$ | 14. $p_{1}^{3} \leqslant p_{1} p_{2} p_{3} \leqslant n . \quad 2 p_{2}^{2} \leqslant p_{1} p_{2} p_{3} \leqslant n$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,237 |
22. Using the method of problem 20, prove:
(i) Let $p_{1}, p_{2}, p_{3}$ be three distinct primes, then
$$\varphi\left(p_{1} p_{2} p_{3}\right)=\left(p_{1}-1\right)\left(p_{2}-1\right)\left(p_{3}-1\right) ;$$
(ii) Let $p_{1}, \cdots, p_{k}$ be $k$ distinct primes, then
$$\varphi\left(p_{1} \cdots p_{k}\right)=\left(p_{... | 22. (i) $\varphi\left(p_{1} p_{2} p_{3}\right)$ is the number of integers in $1,2, \cdots, p_{1} p_{2} p_{3}$ that are not divisible by $p_{1}, p_{2}$, or $p_{3}$. The numbers divisible by $p_{1}, p_{2}, p_{3}$ are $p_{2} p_{3}, p_{3} p_{1}, p_{1} p_{2}$ respectively, where the numbers divisible by two primes are count... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,239 |
3. Let $m>1,(a, m)=1$. Prove:
(i) For any integer $b, \sum_{x \bmod m}\left\{\frac{a x+b}{m}\right\}=\frac{1}{2}(m-1)$;
(ii) $\sum_{x \bmod m}^{\prime}\left\{\frac{a x}{m}\right\}=\frac{1}{2} \varphi(m)$. | 3. (i) Sum $=\sum_{r=0}^{m-1}\left\{\frac{r}{m}\right\}=\sum_{r=0}^{m-1} \frac{r}{m}$;
(ii) Sum $=\sum_{\substack{r=1 \\(r, m)=1}}^{m} \frac{r}{m}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,244 |
5. Write out a complete residue system modulo 23 $r_{1}, \cdots, r_{23}$ that satisfies the following two conditions: $r_{j} \equiv 0(\bmod 7), r_{j} \equiv j(\bmod 5), 1 \leqslant j \leqslant 23$. | 5. $r_{j}=7 k_{j}, 7 k_{j} \equiv j(\bmod 5), k_{j} \equiv 3_{j}(\bmod 5)$. Just take $k_{j}$ to be a complete residue system modulo 23 and satisfy $k_{j} \equiv 3 j(\bmod 5)$. Set $k_{j}=j+23 h_{j}$, then $h_{j} \equiv -j(\bmod 5)$. Choose any $h_{j}$ that satisfies this condition, determine $k_{j}$, and finally get $... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,246 |
15. Use the Sieve of Eratosthenes to find all prime numbers within 300. | 15. See Appendix 1. Use prime numbers not exceeding $\sqrt{300}$: $2,3,5,7,11,13,17$ to sieve. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,248 |
7. Find a complete residue system $r_{1}, \cdots, r_{4}$ modulo 4 and a complete residue system $s_{1}, \cdots, s_{5}$ modulo 5, such that $\square$
(i) $r_{i} s_{j}(1 \leqslant i \leqslant 4,1 \leqslant j \leqslant 5)$ is a complete residue system modulo 20;
(ii) $r_{i}+s_{j}(1 \leqslant i \leqslant 4,1 \leqslant j \l... | 7. Take $r_{i}=5 i+4,1 \leqslant i \leqslant 4, s_{j}=5+4 j, 1 \leqslant j \leqslant 5$ so that (i) and (ii) both hold. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,249 |
8. Can the two conclusions of the previous question hold for a reduced residue system?
---
The translation maintains the original text's line breaks and format as requested. | 8. (i) It holds. Take the reduced residue system from the complete residue system given in question 7.
(ii) It does not hold. Because in this case, there must be $\left(r_{i} s_{j}, 20\right)=1$. When $\left\{r_{i}\right\}$ is a reduced residue system modulo 4, for any fixed $s_{j_{0}}$, from $\left(s_{j_{0}}, 4\right)... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,250 |
Theorem 1 Let $m=m_{1} m_{2}$. (i) If $m_{1}$ and $m$ have the same prime factors, then
$$\varphi(m)=m_{2} \varphi\left(m_{1}\right)$$
In particular, if $m>1$,
$$m=p_{1}^{\alpha_{1}} \cdots p_{r}^{\alpha}, \quad \alpha_{j} \geqslant 1$$
then
(ii) If $\left(m_{1}, m_{2}\right)=1$, then
$$\varphi(m)=\varphi\left(m_{1}\... | Proof In Theorem 13 of §2, taking $k=2$ yields equation (1), because by definition, the number of elements in a reduced residue system modulo $m$ is $\varphi(m)$, the number of elements in a reduced residue system modulo $m_{1}$ is $\varphi\left(m_{1}\right)$, and the number of elements in a complete residue system mod... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,253 |
Theorem 3 (Fermat-Euler Theorem) Let $(a, m)=1$, then we have
$$a^{\varphi(m)} \equiv 1(\bmod m)$$
In particular, when $p$ is a prime, for any $a$ we have
$$a^{p} \equiv a(\bmod p)$$
The equation (14) is usually referred to as Fermat's Little Theorem, while equation (13) is called Euler's Theorem. | Let $r_{1}, \cdots, r_{\varphi(m)}$ be a set of reduced residues modulo $m$. By Theorem 10 in §2, when $(a, m)=1$, $a r_{1}, \cdots, a r_{\varphi(m)}$ is also a set of reduced residues modulo $m$. Therefore, from equation (12) we get
$$\prod_{j=1}^{\varphi(m)} r_{j} \equiv \prod_{j=1}^{\varphi(m)}\left(a r_{j}\right)=a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,255 |
Theorem 4 If $(a, m)=1$, then $d_{0}=\delta_{m}(a)$ if and only if
$$a^{d_{0}} \equiv 1(\bmod m)$$
and
$$a^{0}=1, a, \cdots, a^{d_{0}-1}$$
are pairwise incongruent modulo $m$. In particular, $d_{0}=\varphi(m)$ if and only if equation $(19)\left(d_{0}=\right.$ $\varphi(m))$ provides a complete set of reduced residues ... | First, we prove the first part of the theorem. If $d_{0}=\delta_{m}(a)$, then equation (18) naturally holds. If there exist $0 \leqslant i \leqslant j<d_{0}$ such that
$$a^{j} \equiv a^{i}(\bmod m),$$
then by property $\mathbb{I I}$ in § 1, we get
$$a^{j-i} \equiv 1(\bmod m)$$
but $1 \leqslant j-i<d_{0}$, which contr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,256 |
Example 1 Let $m=2^{l}(l \geqslant 3), a=5$. Find $d_{0}=\delta_{m}(a)$. | Given $\varphi\left(2^{l}\right)=2^{l-1}$ and $d_{0} \mid \varphi\left(2^{l}\right)$, we know that $d_{0}=2^{k}, 0 \leqslant k \leqslant l-1$. First, we prove that for any $a, 2 \nmid a$, it must be true that
$$a^{2^{2-2}} \equiv 1\left(\bmod 2^{l}\right)$$
We use induction on $l$ to prove equation (20). Let $a=2 t+1$... | d_{0}=2^{l-2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,257 |
16. Using the result of problem 14, propose a method similar to the Sieve of Eratosthenes to find all positive integers not exceeding 100 that are the product of at most two prime numbers. | 16. Let $N$ be a given positive integer. $p_{11}, p_{12}, \cdots, p_{1 r}$ denote all primes not exceeding $N^{1 / 3}$, and $p_{21}, p_{22}, \cdots, p_{2 s}$ denote all primes not exceeding $(N / 2)^{1 / 2}$. Thus, by the 14th problem, any integer $n$: $N^{1 / 3}(N / 2)^{1 / 2}<n \leqslant N$, is a product of three or ... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,259 |
1. Prove: (i) there must be infinitely many positive integers $n$ such that $3 \nmid \varphi(n)$;
(ii) for any positive integer $d \geqslant 3$, there must be infinitely many positive integers $n$ such that $d \nmid \varphi(n)$. | (i) As long as $3^{2} \nmid n$ and $n$ has no prime factor of the form $3 k+1$.
(ii) Let $p$ be the largest prime factor of $d$. If $p \geqslant 3$, then as long as $p^{2} \nmid n$ and $n$ has no prime factor of the form $p k+1$, it follows that $p \nmid(n)$, and thus $d \nmid \varphi(n)$; if $p=2$, then $4 \mid d$, so... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,260 |
2. For a given positive integer $k$, there are only finitely many $n$ such that $\varphi(n)=k$. | 2. Because when $n \rightarrow \infty$, $\varphi(n) \rightarrow \infty$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,261 |
4. Find the smallest positive integer $k$, such that $\varphi(n)=k$ has no solution; exactly two solutions; exactly three solutions; exactly four solutions (an unsolved conjecture is: there does not exist a positive integer $k$, such that $\varphi(n)=$ $k$ has exactly one solution).
| 4. $k=3$ has no solution; $k=1$ has exactly two solutions: $\varphi(2)=\varphi(1)=1$, because $2 \mid \varphi(n), n \geqslant 3 ; k=2$ has exactly three solutions: $\varphi(6)=\varphi(4)=\varphi(3)=2$, because at this time $n$ can only have prime factors less than 5; $k=4$ has exactly four solutions: $\varphi(12)=\varp... | 3, 1, 2, 4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,263 |
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