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5. Find all positive integers $n$ such that $\varphi(n)=24$.
The Euler's totient function $\varphi(n)$ counts the number of positive integers up to $n$ that are relatively prime to $n$. We need to find all $n$ for which $\varphi(n) = 24$.
To solve this, we can use the formula for the Euler's totient function:
\[
\var... | 5. When $\varphi(n)=24$, $n$ can only have $2,3,5,7,13$ as its prime factors. Let $n=2^{a_{1}} 3^{a_{2}} 5^{a_{3}} 7^{a_{4}}$ $13^{a_{5}}$, we can calculate: $n=13 \cdot 3,13 \cdot 3 \cdot 2,13 \cdot 2^{2}, 7 \cdot 5,7 \cdot 5 \cdot 2,7 \cdot 3 \cdot 2^{2}, 7 \cdot 2^{3}, 5 \cdot$ $3^{2}, 5 \cdot 3^{2} \cdot 2,3^{2} \c... | 32, 27, 25, 49, 169, 39, 35, 78, 70 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,264 |
6. Find all positive integers $n$ that satisfy the following equations:
(i) $\varphi(n)=\varphi(2 n)$;
(ii) $\varphi(2 n)=\varphi(3 n)$;
(iii) $\varphi(3 n)=\varphi(4 n)$. | 6. (i) $2 \nmid n$; (ii) $2 \mid n, 3 \nmid n$; (iii) $2 \nmid n, 3 \nmid n$. In (ii), (iii) only discuss
$$n=2^{a} 3^{\beta} \cdot m, \quad(6, m)=1$$ | 2 \nmid n, 3 \nmid n | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,265 |
7. Find all positive integers $n$ such that $\varphi(n)=2^{6}$.
The text above has been translated into English, preserving the original text's line breaks and format. | 7. $n=2^{7}, 2^{6} \cdot 3, 2^{5} \cdot 5, 2^{4} \cdot 3 \cdot 5 ; 2^{3} \cdot 17, 2^{2} \cdot 3 \cdot 17, 2 \cdot 5 \cdot 17, 5 \cdot 17$. | n=2^{7}, 2^{6} \cdot 3, 2^{5} \cdot 5, 2^{4} \cdot 3 \cdot 5 ; 2^{3} \cdot 17, 2^{2} \cdot 3 \cdot 17, 2 \cdot 5 \cdot 17, 5 \cdot 17 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,266 |
9. Let $a, b$ be given positive integers. Prove: there exist infinitely many pairs of natural numbers $m, n$, such that $a \varphi(m)=b \varphi(n)$. | 9. Let $(a, b)=1 . m=a^{k} b^{k+1}, n=a^{k+1} b^{k}, k \geqslant 1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,268 |
10. Let $k$ be a given positive integer. Prove: there must exist a positive integer $n$, such that
$$\varphi(n)=\varphi(n+k) .$$ | 10. If $2 \nmid k$, then we can take $n=k$. If $2 \mid k$, let $p_{0}$ be the smallest prime such that $p_{0} \nmid k$, then we can take $n=$ $\left(p_{0}-1\right) k$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,269 |
12. Prove:
(i) $\varphi(n)>\sqrt{n} / 2$; (ii) if $n$ is composite, then
$$\varphi(n) \leqslant n-\sqrt{n} .$$ | 12. (i) When $p>2$, $\varphi\left(p^{\alpha}\right)=(p-1) p^{\alpha-1}>p^{\alpha / 2}, \varphi\left(2^{\alpha}\right)>2^{\alpha / 2} / 2$.
(ii) If $n$ is composite, it must have a prime factor $p \leqslant \sqrt{n}$. Multiples of $p$ are not coprime with $n$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,272 |
13. Find all positive integers $n$ such that $\varphi(n) \mid n$.
---
Here is the translation, maintaining the original format and line breaks. | 13. $n=1,2^{\alpha} \cdot 3^{\beta}, \alpha \geqslant 1, \beta \geqslant 0$. | n=1,2^{\alpha} \cdot 3^{\beta}, \alpha \geqslant 1, \beta \geqslant 0 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,273 |
15. Let \( m = p_{1}^{q_{1}} \cdots p_{r}^{q_{r}}, p_{j} \) be different primes,
\[ c_{j} = \varphi\left(p_{j}^{q_{j}}\right), \quad \alpha = \max \left(\alpha_{1}, \cdots, \alpha_{r}\right) \]
Prove: For any integer \( a \) we have
(i) \( a^{a + \varphi(m)} \equiv a^{\alpha} \pmod{m} \);
(ii) \( a^{m} \equiv a^{m - \... | 15. (i) If $p_{i} \mid a$, then $p_{i}^{a_{i}} \mid a^{a+q(m)}-a^{\alpha}$; if $p_{i} \nmid a$, then $p_{i}^{q_{i}} \mid a^{p\left(p_{i}^{\left.q_{i}\right)}\right.}-1$, hence $p_{i}^{q_{i}} \mid a^{\rho(m)}-1$. (ii) If $p_{i} \nmid a$, then by (i) we know $p_{i}^{g_{i}} \mid a^{\rho(m)}-1$, hence $p_{i}^{q_{i}} \mid a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,275 |
17. Let $f(x)$ be a polynomial with integer coefficients. Prove: $(f(x))^{p} \equiv f\left(x^{p}\right)(\bmod p)$, where $p$ is a prime. | 17. Let $f(x)=a_{n} x^{n}+\cdots+a_{0}$. $(f(x))^{p}=\left(a_{n} x^{n}\right)^{p}+\left(a_{n-1} x^{n-1}\right)^{p}+\cdots+a_{0}^{p}+$ $p \cdot h_{1}(x), h_{1}(x)$ is a polynomial with integer coefficients. Furthermore, by Fermat's Little Theorem, $(f(x))^{p}=a_{n}\left(x^{p}\right)^{n}+$ $a_{n-1}\left(x^{p}\right)^{n-1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,277 |
18. Let $p>2$ be a prime and $a>1$. Prove:
(i) A prime factor $q$ of $a^{p}-1$ must be a divisor of $a-1$, or $q \equiv 1(\bmod 2 p)$;
(ii) A prime factor $q$ of $a^{p}+1$ must be a divisor of $a+1$, or $q \equiv 1(\bmod 2 p)$;
(iii) There are infinitely many primes of the form $2 k p+1$. | 18. (i) There must be $(q, a)=1, q \mid a^{q-1}-1$, hence $q \mid a^{(p, q-1)}-1$. If $(p, q-1)=1$, then $q \mid a-1$; if $(p, q-1)=p$, then $q \equiv 1(\bmod 2 p)$.
(ii) $q \mid a^{2 p}-1$. By the same argument as in (i), and note that $q \nmid a-1$.
(iii) Let $s$ be a given positive integer, and take $a=2^{p^{p-1}}$.... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,278 |
19. Let $b>1, n \geqslant 1$. Prove: $n \mid \varphi\left(b^{n}-1\right)$. | 19. It is evident that we need to prove $p^{s} \mid \varphi\left(b^{p^{s}}-1\right)$, where $p$ is a prime. Taking $a=b F^{s-1}$, the conclusion we need to prove can be derived from $p^{s} \mid \varphi\left(\left(a^{p}-1\right) /(a-1)\right)$. When $b=2$, part (iii) of the 18th problem has already proven that the prime... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,279 |
20. Let $(a, 10)=1$. Prove: there must exist a positive integer $n$ whose decimal representation consists entirely of the digit 1, such that $a \mid n$. Moreover, there are infinitely many such $n$.
保留源文本的换行和格式，翻译结果如下：
20. Let $(a, 10)=1$. Prove: there must exist a positive integer $n$ whose decimal representation co... | 20. There must be a positive integer $d_{0}$, such that $9 a \mid 10^{d_{0}}-1=99 \cdots 9=9 \cdot(11 \cdots 1), d_{0}$ digits of 1. Therefore, $a$ divides $11 \cdots 1$, here with $d_{0} k$ digits of 1, $k=1,2,3, \cdots$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,280 |
22. Let $a, r$ be positive integers, $(a, r)=1$. Prove: In the arithmetic sequence $a+k r(k=0$, $1,2, \cdots)$, one can always select a geometric sequence where each term is a power of $a$. | 22. Prove that $a^{l \varphi(r)+1}$ belongs to this arithmetic sequence, $l=0,1,2, \cdots$. In fact, the condition $(a, r)=1$ is not necessary; simply take $a(1+r)^{l}(l=0,1,2, \cdots)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,283 |
Theorem 1 (Wilson's Theorem) Let $p$ be a prime, and $r_{1}, \cdots, r_{p-1}$ be a reduced residue system modulo $p$. We have
$$\prod_{r \bmod p}^{\prime} r \equiv r_{1} \cdots r_{p-1} \equiv-1(\bmod p)$$
In particular, we have
$$(p-1)!\equiv-1(\bmod p) \oplus^{\oplus}$$ | When $p=2$, the conclusion is obviously true. Therefore, we can assume $p \geqslant 3$. By Property VI of $\S 1$ and the subsequent explanation, for each $r_{i}$ in the chosen reduced residue system, there must be a unique $r_{j}$ such that
$$r_{i} r_{j} \equiv 1(\bmod p)$$
The condition for $r_{i}=r_{j}$ is
$$\text {... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,288 |
Theorem 4 Let $c=\varphi\left(2^{l}\right)=2^{l-1}, l \geqslant 1, r_{1}, \cdots, r_{c}$ be the reduced residue system modulo $2^{l}$. We have
$$r_{1} \cdots r_{c} \equiv\left\{\begin{array}{ll}
-1\left(\bmod 2^{l}\right), & l=1,2, \\
1\left(\bmod 2^{l}\right), & l \geqslant 3 .
\end{array}\right.$$ | Proof: When $l=1,2$, the conclusion can be directly verified. Now assume $l \geqslant 3$. Similarly, by property VIII of §1 and the subsequent explanation, for each $r_{i}$, there must be a unique $r_{j}$ such that
$$r_{i} r_{j} \equiv 1\left(\bmod 2^{l}\right)$$
The necessary and sufficient condition for $r_{i}=r_{j}$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,291 |
Example 1 Let $r_{0}, r_{1}, \cdots, r_{p-1}$ and $r_{0}^{\prime}, r_{1}^{\prime}, \cdots, r_{p-1}^{\prime}$ be two complete residue systems modulo $p$, where $p$ is an odd prime. Prove: $r_{0} r_{0}^{\prime}, r_{1} r_{1}^{\prime}, \cdots, r_{p-1} r_{p-1}^{\prime}$ cannot be a complete residue system modulo $p$. | Prove by contradiction. Assume that $r_{0} r_{0}^{\prime}, r_{1} r_{1}^{\prime}, \cdots, r_{p-1} r_{p-1}^{\prime}$ is a complete residue system modulo $p$, then there is exactly one that is divisible by $p$, without loss of generality, let
$$p \mid r_{0} r_{0}^{\prime}, \quad p \nmid r_{j} r_{j}^{\prime}, \quad 1 \leqs... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,293 |
6. Let the prime $p$ be an odd number. Prove:
(i) When $p=4 m+3$, for any integer $a$ we have $a^{2} \equiv -1(\bmod p)$;
(ii) When $p=4 m+1$, there exists $a$ such that $a^{2} \equiv -1(\bmod p)$;
(iii) There are infinitely many primes of the form $4 m+1$. | 6. (i) By contradiction. Consider $i j \equiv -1 \pmod{p}, 1 \leqslant i, j \leqslant p-1$. If there exists $i=j=i_{0}$ such that $i_{0}^{2} \equiv -1 \pmod{p}$, such $i_{0} (1 \leqslant i_{0} \leqslant p-1)$ exists exactly two. This implies $(p-1)! \equiv 1 \pmod{p}$, which contradicts Theorem 1;
(ii) See Example 2;
(... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,300 |
Theorem 3 (Principle of the Greatest Natural Number) Let $M$ be a non-empty subset of $N$. If $M$ has an upper bound, i.e., there exists $a \in \boldsymbol{N}$, such that for any $m \in M$ we have $m \leqslant a$, then there must be $m_{0} \in M$, such that for any $m \in M$ we have $m \leqslant m_{0}$, i.e., $m_{0}$ i... | Theorem 3 Proof: Consider the set $T$ consisting of all natural numbers $t$ such that for any $m \in M$, $m \leqslant t$. By the condition, $a \in T$, so $T$ is non-empty. By Theorem 2, the set $T$ contains a smallest natural number $t_{0}$. We will prove that $t_{0} \in M$. If not, then for any $m \in M$, we must have... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,303 |
9. Let $r_{1}, \cdots, r_{4}$ and $r_{1}^{\prime}, \cdots, r_{4}^{\prime}$ be two sets of reduced residue systems modulo 8. Are $r_{1} r_{1}^{\prime}, \cdots, r_{4} r_{4}^{\prime}$ definitely not a reduced residue system modulo 8? Provide examples to illustrate. Perform the same discussion for two sets of reduced resid... | 9. Not necessarily. For modulo 8, we can take:
$$\begin{array}{llll}
r_{1}=-3, & r_{2}=-1, & r_{3}=1, & r_{4}=3 \\
r_{1}^{\prime}=3, & r_{2}^{\prime}=-3, & r_{3}^{\prime}=1, & r_{4}^{\prime}=-1 \\
r_{1}^{\prime}=1, & r_{2}^{\prime}=-1, & r_{3}^{\prime}=3, & r_{4}^{\prime}=-3
\end{array}$$
For modulo 15, we can take: $... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,305 |
10. Let $m \geqslant 3, r_{1}, \cdots, r_{m}$ and $r_{1}^{\prime}, \cdots, r_{m}^{\prime}$ be two complete residue systems modulo $m$. Prove: $r_{1} r_{1}^{\prime}, \cdots, r_{m} r_{m}^{\prime}$ is certainly not a complete residue system modulo $m$ (Hint: Use the method in Proof 2 of Theorem 2 in §3, and Example 1 of t... | 10. There must be $\left(r_{i}, m\right)=1 \Longleftrightarrow\left(r_{i}^{\prime}, m\right)=1$. Therefore, $r_{i}$ that is not coprime with $m$ must be multiplied by $r_{i}^{\prime}$ that is also not coprime with $m$. For a given $d>1$, the number of numbers in a complete residue system that have the greatest common d... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,306 |
Example 1 Find the solution to the congruence equation $4 x^{2}+27 x-12 \equiv 0(\bmod 15)$. | Solve the absolute minimal complete residue system modulo 15: $-7,-6, \cdots,-1,0,1$, $2, \cdots, 7$. Direct calculation shows that $x=-6,3$ are solutions. Therefore, the solutions to this congruence equation are
$$x \equiv-6,3(\bmod 15),$$
The number of solutions is 2. | x \equiv -6, 3 (\bmod 15) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,308 |
Example 2 Find the solution to the congruence equation $4 x^{2}+27 x-7 \equiv 0(\bmod 15)$. | Similarly, direct calculation shows that $x=-7,-2,-1,4$ are solutions. Therefore, the solutions are
$$x \equiv-7,-2,-1,4(\bmod 15),$$
The number of solutions is 4. | x \equiv -7, -2, -1, 4 \pmod{15} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,309 |
Example 5 Solve the congruence equation
$$f(x)=2 x^{7}-x^{5}-3 x^{3}+6 x+1 \equiv 0(\bmod 5) .$$ | Solve using the constant congruence identity (3). By polynomial division, we get
$$\begin{aligned}
f(x) & =\left(2 x^{2}-1\right)\left(x^{5}-x\right)-x^{3}+5 x+1 \\
& \equiv\left(2 x^{2}-1\right)\left(x^{5}-x\right)-x^{3}+1(\bmod 5) .
\end{aligned}$$
Therefore, the original congruence equation is equivalent to
$$x^{3}... | x \equiv 1(\bmod 5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,313 |
1. Find the solutions and the number of solutions to the following congruence equations by direct computation:
(i) $x^{5}-3 x^{2}+2 \equiv 0(\bmod 7)$;
(ii) $3 x^{4}-x^{3}+2 x^{2}-26 x+1 \equiv 0(\bmod 11)$;
(iii) $3 x^{2}-12 x-19 \equiv 0(\bmod 28)$;
(iv) $3 x^{2}+18 x-25 \equiv 0(\bmod 28)$;
(v) $x^{2}+8 x-13 \equiv ... | (i) $x \equiv 1,5(\bmod 7)$.
(ii) $x \equiv -1(\bmod 11)$.
(iii) $x \equiv 1,3,15,17(\bmod 28)$.
(iv) $x \equiv 3,5,17,19(\bmod 28)$.
(v) $x \equiv -5,-3,9,11(\bmod 28)$.
(vi) No solution. $141=3 \cdot 47.4 x^{2}+21 x-32 \equiv x^{2}+1 \equiv 0(\bmod 3)$ has no solution.
(vii) Using $x^{5} \equiv x(\bmod 5)$ to simplif... | x \equiv 1,5(\bmod 7) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,316 |
2. Let $(2 a, m)=1$. Prove: The congruence equation $a x^{2}+b x+c \equiv 0(\bmod m)$ can always be transformed into $(d x+e)^{2} \equiv f(\bmod m)$. Use this method to solve the congruence equations in Example 1, Example 2, and Example 5 of §1. | 2. The original equation is equivalent to
$$4 a\left(a x^{2}+b x+c\right) \equiv(2 a x+b)^{2}+4 a c-b^{2} \equiv 0(\bmod m)$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,317 |
6. For which values of $a$, does the congruence equation $x^{3} \equiv a(\bmod 9)$ have a solution? | 6. $a \equiv 0, \pm 1(\bmod 9)$. | a \equiv 0, \pm 1(\bmod 9) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,321 |
7. Find the solution to $2^{x} \equiv x^{2}(\bmod 3)$. | 7. Substitute $x=2 k, 2 k+1$. $1 \equiv 2^{2 k} \equiv (2 k)^{2} \equiv k^{2}(\bmod 3), k \equiv \pm 1(\bmod 3)$, thus the positive integer $x \equiv \pm 2(\bmod 6)$ is a solution; $2 \equiv 2^{2 k+1} \equiv (2 k+1)^{2} \equiv k^{2}+k+1(\bmod 3)$, no solution. | x \equiv \pm 2(\bmod 6) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,322 |
8. 求 $x^{4}+y^{4} \equiv 1(\bmod 5)$ 的全部解 $\{x, y\}$. | 8. $\begin{array}{ll}\{0 \bmod 5, \pm 1 \bmod 5\}, & \{0 \bmod 5, \pm 2 \bmod 5\} \\ \{ \pm 1 \bmod 5,0 \bmod 5\}, & \{ \pm 2 \bmod 5,0 \bmod 5\} .\end{array}$ | \begin{array}{ll}\{0 \bmod 5, \pm 1 \bmod 5\}, & \{0 \bmod 5, \pm 2 \bmod 5\} \\ \{ \pm 1 \bmod 5,0 \bmod 5\}, & \{ \pm 2 \bmod 5,0 \bmod 5\} .\end{array} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,323 |
9. Prove: $x^{3}+y^{3}+z^{3} \equiv 0(\bmod 9)$ has no solution where $3 \nmid x y z$. | 9. For any integer
$$a, 3 \nmid a, a^{3} \equiv \pm 1(\bmod 9) .$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,324 |
10. Prove: A congruence equation (2) can always be transformed into a congruence equation of a polynomial of degree $<m$ (including the case where all coefficients are multiples of $m$). How can this result be improved for composite moduli using Problem 15 of Exercise 3 in Chapter 3, Section 3? | 10. Using the result from Exercise 3, Chapter 3, Question 15 (ii): $x^{m} \equiv x^{m-\varphi(m)}(\bmod m)$, we can reduce the degree of the congruence equation to less than $m$. In the case of a composite modulus, we can use the result from (iii).
利用第三章习题三第 15 题 (ii) 的结果: $x^{m} \equiv x^{m-\varphi(m)}(\bmod m)$, 就可把... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,325 |
22. Let the integer-coefficient polynomial $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}, a_{n} \neq 0$. Prove: There must be infinitely many integer values of $x$, such that $P(x)$ is composite. | 22. (i) For any $x, P(x)$ is a composite number; (ii) $P\left(x_{0}\right)= \pm p, p$ is a prime number, then
$$p \mid P\left(x_{0}+j p\right), \quad j \in \boldsymbol{Z}$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,326 |
11. 证明: 同余方程 (2) 的解数
$$T=\frac{1}{m} \sum_{l=0}^{m-1} \sum_{x=0}^{m-1} \mathrm{e}^{2 \pi \mathrm{i} l f(x) / m}$$
由此推出, 当 $f(x)=a x-b$ 时,
$$T=\left\{\begin{array}{ll}
(a, m), & \text { 当 }(a, m) \mid b ; \\
0, & \text { 当 }(a, m) \nmid b .
\end{array}\right.$$ | 11. 当 $f(x)=a x-b$ 时, 令 $d=m /(a, m)$. 我们有
$$\begin{aligned}
\sum_{l=0}^{m-1} \sum_{x=0}^{m-1} \mathrm{e}^{2 \pi i l(a x-b) / m} & =\sum_{l=0}^{m-1} \mathrm{e}^{-2 x i l b / m} \sum_{x=0}^{m-1} \mathrm{e}^{2 x i l a x / m}=m \sum_{\substack{l=0 \\
m \mid h}}^{m-1} \mathrm{e}^{-2 x i l b / m}=m \sum_{k=0}^{m / d-1} \mat... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,327 |
Example 1 Solve the congruence equation $589 x \equiv 1026(\bmod 817)$. | Solve $589 x \equiv 1026(\bmod 817) \stackrel{(\mathrm{i})}{\Longleftrightarrow}-228 x \equiv 209(\bmod 81)$
This indicates that the last congruence equation about $u$ has 19 solutions modulo 19:
$$u \equiv 0,1,2, \cdots, 18(\bmod 19)$$
According to (iii), i.e., equation (13), tracing back step by step, we get: the c... | x \equiv 43 u + 17 \pmod{817}, \quad u = 0, 1, \cdots, 18 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,332 |
Example $221 x \equiv 38(\bmod 117)$. | Solve $21 x \equiv 38(\bmod 117)$
$$\begin{array}{l}
\stackrel{(\text { ii) }}{\Longleftrightarrow} 117 y \equiv -38(\bmod 21) \stackrel{(\text { i) }}{\Longleftrightarrow} -9 y \equiv 4(\bmod 21) \\
(\text { ii) } \\
\Longleftrightarrow 21 z \equiv -4(\bmod 9) \stackrel{(\text { i) }}{\Longleftrightarrow} 3 z \equiv -... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,333 |
1. Solve the following linear congruences.
(i) $3 x \equiv 2(\bmod 7)$;
(ii) $9 x \equiv 12(\bmod 15)$;
(iii) $7 x \equiv 1(\bmod 31)$;
(iv) $20 x \equiv 4(\bmod 30)$;
(v) $17 x \equiv 14(\bmod 21)$;
(vi) $64 x \equiv 83(\bmod 105)$;
(vii) $128 x \equiv 833(\bmod 1001)$;
(viii) $987 x \equiv 610(\bmod 1597)$;
(ix) $57 ... | 1. (i) $x \equiv 3(\bmod 7)$;
(ii) $x \equiv 3,8,13(\bmod 15)$;
(iii) $x \equiv 9(\bmod 31)$;
(iv) $(20,30)=10 \nmid 4$, no solution;
(v) $x \equiv 7(\bmod 21)$;
(vi) $x \equiv 62(\bmod 105)$;
(vii) $1001=7 \cdot 11 \cdot 13 . x \equiv-189(\bmod 1001)$;
(viii) $x \equiv-1(\bmod 1597)$;
(ix) $x \equiv-4,31,66(\bmod 105)... | \text{no solution} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,334 |
2. Using the following two properties of congruences:
(a) $a \equiv b(\bmod m) \Longleftrightarrow a \equiv b+m t(\bmod m)$;
(b) When $(c, m)=1$, $c a \equiv c b(\bmod m) \Longleftrightarrow a \equiv b(\bmod m)$, propose a simple method to solve the following congruence equations:
(i) $2^{k} x \equiv b(\bmod m),(2, m)=... | 2. (i) $m=2 l-1 . x \equiv l^{k} b(\bmod m)$, then find the remainder of $l^{k} b$ modulo $m=2^{l}-1$;
(ii) $m=3 l \pm 1, x \equiv( \pm 1)^{k} l^{k} b(\bmod m)$, then find the remainder of $l^{k} b$ modulo $m=3 l \pm 1$. | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,335 |
3. Use the method proposed in question 2 to solve:
(i) question 1 (vi) and (vii);
(ii) $256 x \equiv 179(\bmod 337)$;
(iii) $1215 x \equiv 560(\bmod 2755)$;
(iv) $1296 x \equiv 1105(\bmod 2413)$. | 3. (i) As an example from Question 1 (vi). $2^{6} x \equiv 83(\bmod 105) . x \equiv 52^{6} \cdot 83 \equiv 1+52^{6} \cdot 19 \equiv$ $1+13 \cdot 13 \cdot 19 \equiv 1+64 \cdot 19 \equiv 62(\bmod 105)$.
(ii) $x \equiv 179 \cdot 168^{8} \equiv 179 \cdot 21 \cdot 21 \equiv 179 \cdot 104 \equiv 81(\bmod 337)$.
(iii) $243 x ... | 62, 81, 200, -783 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,336 |
4. Let $a$ be a positive integer, $a \nmid m$, and $a_{1}$ be the smallest positive remainder of $m$ modulo $a$. Prove that the solution to the congruence equation $a x \equiv b(\bmod m)$ is also a solution to the congruence equation
$$a_{1} x \equiv -b[m / a](\bmod m)$$
Is the converse true? Provide an example to ill... | 4. $m=[m / a] a+a_{1}, 1 \leqslant a_{1}<a . a x \equiv b(\bmod m)$ is always a solution to $a[m / a] x \equiv b[m / a](\bmod m)$, which is also a solution to $a_{1} x \equiv -b[m / a](\bmod m)$. The converse is not necessarily true. The converse is always true only when $([m / a], m)=1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,338 |
5. Can you use the previous question to propose a method for solving a linear congruence equation? Use the method you propose to solve (i) $6 x \equiv 7(\bmod 23)$; (ii) $5 x \equiv 1(\bmod 12)$. Point out what to pay attention to when applying this method. | 5. (i) $m=23, a=6$. $[23 / 6]=3,(3,23)=1$. So the original equation is equivalent to $5 x \equiv -21 \equiv 2(\bmod 23) .[23 / 5]=4,(4,23)=1, 3 x \equiv -8(\bmod 23) .[23 / 3]=7,(7,23)=1$, $2 x \equiv 10(\bmod 23), x \equiv 5(\bmod 23)$. (ii) $[12 / 5]=2,(2,12)=2$, so this method should not be used. | x \equiv 5(\bmod 23) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,339 |
6. Let $(a, m)=1, x_{1}$ be the solution to $a x \equiv 1(\bmod m)$. Suppose $k$ is a positive integer, and $y_{k}=1-\left(1-a x_{1}\right)^{k}$. Prove: $a \mid y_{k}$, and $x_{k}=y_{k} / a$ is a solution to the congruence equation
$$a x \equiv 1\left(\bmod m^{k}\right)$$ | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 740,340 |
7. Use the previous problem to solve (i) $3 x \equiv 1(\bmod 125)$; (ii) $5 x \equiv 1(\bmod 243)$. | 7. (i) $3 x \equiv 1\left(\bmod 5^{3}\right), 3 \cdot 2 \equiv 1(\bmod 5), 1-(1-3 \cdot 2)^{3}=126$, so $x=42$ is a solution to the original congruence equation. (ii) $5 x \equiv 1\left(\bmod 3^{5}\right) .5 \cdot(-1) \equiv 1(\bmod 3) .1-(1-5 \cdot(-1))^{5}=$ $-7775 . x=-1555$ is a solution. | -1555 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,341 |
8. Let $(a, m)=1, b$ be an integer; and let $f(x)$ be a polynomial with integer coefficients,
$$g(y)=f(a y+b)$$
Prove: The congruence equation $f(x) \equiv 0(\bmod m)$ has the same number of solutions as $g(y) \equiv 0(\bmod m)$. Indicate how to find the solutions of $g(y) \equiv 0(\bmod m)$ from the solutions of $f(x... | 8. $a y+b \equiv x(\bmod m)$ determines a one-to-one correspondence between the solutions $y \bmod m$ of $g(y) \equiv 0(\bmod m)$ and the solutions $x \bmod m$ of $f(x) \equiv$ $0(\bmod m)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,342 |
9. Use the previous problem to solve Example 1, Example 2, and Example 3 in §1.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
However, it seems you might want the translation of the Chinese text provided, not the instructi... | 9. If $x_{0}$ is a solution to $a x \equiv b(\bmod m)$. Take $y_{0}=\left(b-a x_{0}\right) / m$. $x=x_{0}+m t /(a, m)$, $y=y_{0}-m t /(a, m), t=0, \pm 1, \pm 2, \cdots$, then this gives the solutions to the indeterminate equation. | null | Algebra | math-word-problem | Yes | Yes | number_theory | false | 740,343 |
Example 1 Solve the system of congruences
$$\left\{\begin{array}{l}
x \equiv 1(\bmod 3), \\
x \equiv-1(\bmod 5), \\
x \equiv 2(\bmod 7), \\
x \equiv-2(\bmod 11)
\end{array}\right.$$ | Let $m_{1}=3, m_{2}=5, m_{3}=7, m_{4}=11$, satisfying the conditions of Theorem 1. In this case, $M_{1}=5 \cdot 7 \cdot 11, M_{2}=3 \cdot 7 \cdot 11, M_{3}=3 \cdot 5 \cdot 11, M_{4}=3 \cdot 5 \cdot 7$. We will find $M_{j}^{-1}$. Since $M_{1} \equiv(-1) \cdot(1) \cdot(-1) \equiv 1(\bmod 3)$, we have
$$1 \equiv M_{1} M_{... | 394 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,347 |
24. Let $k \geqslant 3$. Find all sets of positive integers $\left\{a_{1}, \cdots, a_{k}\right\}$ such that
(i) $a_{1}, \cdots, a_{k}$ are distinct positive integers;
(ii) the sum of any three of these numbers is divisible by each of the three numbers. | 24. Let $1 \leqslant a_{1}<a_{2}<a_{3}$. If $a_{3}\left|a_{1}+a_{2}, a_{2}\right| a_{3}+a_{1}, a_{1} \mid a_{2}+a_{3}$, then $a_{3}=a_{1}+a_{2}, a_{2}=2 a_{1}$. All sets are $k=3,\left\{a_{1}, 2 a_{1}, 3 a_{1}\right\}$. | \{a_{1}, 2a_{1}, 3a_{1}\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,348 |
Example 2 Find four consecutive integers, each of which is divisible by $2^{2}, 3^{2}, 5^{2}$, and $7^{2}$ respectively. | Let the four consecutive integers be $x-1, x, x+1, x+2$. According to the requirements, they should satisfy:
$$\begin{array}{ll}
x-1 \equiv 0\left(\bmod 2^{2}\right), & x \equiv 0\left(\bmod 3^{2}\right) \\
x+1 \equiv 0\left(\bmod 5^{2}\right), & x+2 \equiv 0\left(\bmod 7^{2}\right)
\end{array}$$
Thus, this is a probl... | 29348,29349,29350,29351 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,349 |
Example 3 Find a complete residue system modulo 11, such that each number leaves remainders of $1, -1, 1, -1$ when divided by $2, 3, 5, 7$ respectively. | Solve: In Theorem 2, take $m_{1}=2, m_{2}=3, m_{3}=5, m_{4}=7, m_{5}=11$, and $x^{(1)}=1, x^{(2)}=-1, x^{(3)}=1, x^{(4)}=-1$. Thus, by Theorem 2, when $x^{(5)}$ runs through a complete residue system modulo 11,
$$x=M_{1} M_{1}^{-1}-M_{2} M_{2}^{-1}+M_{3} M_{3}^{-1}-M_{4} M_{4}^{-1}+M_{5} M_{5}^{-1} x^{(5)}$$
gives the... | 41, 251, 461, 671, 881, 1091, 1301, 1511, 1721, 1931, 2141 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,350 |
Example 4 Solve the system of congruences
$$x \equiv 3(\bmod 8), \quad x \equiv 11(\bmod 20), \quad x \equiv 1(\bmod 15)$$ | Here $m_{1}=8, m_{2}=20, m_{3}=15$ are not pairwise coprime, so we cannot directly apply Theorem 1. It is easy to see that the solution to this system of congruences is the same as the solution to the system of congruences
$$\left\{\begin{array}{l}
x \equiv 3(\bmod 8), \\
x \equiv 11(\bmod 4), \\
x \equiv 11(\bmod 5), ... | x \equiv -29(\bmod 120) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,351 |
Example 5 Solve the congruence equation $19 x \equiv 556(\bmod 1155)$. | This is a linear congruence equation, which can of course be solved using the method in §2. Here, we transform it into a system of linear congruence equations with smaller moduli, which is sometimes more convenient. Since \(1155 = 3 \cdot 5 \cdot 7 \cdot 11\), by the property [X] in Chapter 3 §1, this congruence equati... | x \equiv 394(\bmod 1155) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,352 |
Example 6 Solve the system of congruences
$$x \equiv 3(\bmod 7), \quad 6 x \equiv 10(\bmod 8) .$$ | This is not in the form of the system of congruences in Theorem 1. It is easy to see that the second congruence has solutions and the number of solutions is 2 (the specific solution is left to the reader):
$$x \equiv -1, 3 \pmod{8}$$
Therefore, the solutions to the original system of congruences are the solutions to t... | x \equiv 3, 31 \pmod{56} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,353 |
Example 7 Solve the system of congruences
$$\left\{\begin{array}{l}
3 x \equiv 1(\bmod 10) \\
4 x \equiv 7(\bmod 15)
\end{array}\right.$$ | Solve by using the method of Example 4. The solution to this system of congruences is the same as that of the system of congruences
$$\left\{\begin{array}{l}
3 x \equiv 1(\bmod 2), \\
3 x \equiv 1(\bmod 5), \\
4 x \equiv 7(\bmod 3), \\
4 x \equiv 7(\bmod 5)
\end{array}\right.$$
The second congruence $3 x \equiv 1(\bmo... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,354 |
1. Solve the following system of linear congruences:
(i) $x \equiv 1(\bmod 4), x \equiv 2(\bmod 3), x \equiv 3(\bmod 5)$;
(ii) $x \equiv 4(\bmod 11), x \equiv 3(\bmod 17)$;
(iii) $x \equiv 2(\bmod 5), x \equiv 1(\bmod 6), x \equiv 3(\bmod 7), x \equiv 0(\bmod 11)$;
(iv) $3 x \equiv 1(\bmod 11), 5 x \equiv 7(\bmod 13)$;... | 1. (i) $x=1+4 y, 4 y \equiv 1(\bmod 3), 4 y \equiv 2(\bmod 5)$. Thus, $y \equiv 1(\bmod 3), y \equiv -2(\bmod 5) . y=1+3 z, 3 z \equiv -3(\bmod 5), z \equiv -1(\bmod 5) . y \equiv -2(\bmod 15), x \equiv -7(\bmod 60)$.
(ii) $x=4+11 y, 11 y \equiv -1(\bmod 17), y \equiv 3(\bmod 17), x \equiv 37(\bmod 187)$.
(iii) $m_{1}=... | x \equiv -7(\bmod 60) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,355 |
2. Convert the congruence equation into a system of congruence equations to solve:
(i) $23 x \equiv 1(\bmod 140)$;
(ii) $17 x \equiv 229(\bmod 1540)$. | 2. (i) $x \equiv -1(\bmod 4), 2x \equiv -1(\bmod 5), 2x \equiv 1(\bmod 7)$, the solution is $x \equiv 67(\bmod 140)$;
(ii) $x \equiv 1(\bmod 4), 2x \equiv -1(\bmod 5), 3x \equiv 5(\bmod 7), 2x \equiv 3(\bmod 11)$, the solution is $x \equiv 557(\bmod 1540)$ | x \equiv 557(\bmod 1540) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,356 |
3. Find all integers that leave remainders of $1, 2, 3$ when divided by $3, 4, 5$ respectively. | 3. $x \equiv 1(\mod 3), x \equiv 2(\mod 4), x \equiv 3(\mod 5)$. The solution is $x \equiv -2(\mod 60)$. | x \equiv -2(\mod 60) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,357 |
4. There is a person who rests for two days after working for eight days. Once, he rested on Saturday and Sunday. How many weeks at least will it take for him to rest on Sunday again?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 4. The smallest positive integer that satisfies $7 x+1 \equiv 0(\bmod 10)$, and the smallest positive integer that satisfies $7 x \equiv 0 \equiv(\bmod 10)$, the smaller one of these two is the required number of weeks, i.e., after 7 weeks, one can rest on Sunday. | null | Logic and Puzzles | other | Yes | Yes | number_theory | false | 740,358 |
5. Let $k$ be any given positive integer. Prove: there must exist $k$ consecutive integers, each of which can be divided by a cube greater than 1. | 5. Let $p_{1}, \cdots, p_{k}$ be pairwise distinct positive integers. Consider the system of congruences: $x \equiv -j+1(\bmod$ $\left.p_{j}^{3}\right), j=1, \cdots, k$. If $x_{0}$ is a solution, then $x_{0}, x_{0}+1, \cdots, x_{0}+k-1$ satisfy the requirements. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,360 |
7. Let $(a, b)=1, c \neq 0$. Prove: there must exist an integer $n$ such that
$$(a+b n, c)=1$$ | 7. Let the prime factors that divide $c$ but do not divide $b$ be $p_{1}, \cdots, p_{r}$. The $x$ that satisfies the system of congruences $b x+a \equiv 1\left(\bmod p_{j}\right), 1 \leqslant j \leqslant r$ meets the requirement. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,362 |
11. Let $m=\left[m_{1}, \cdots, m_{k}\right]$. Prove:
(i) There must exist a set of positive integers $m_{1}^{\prime}, \cdots, m_{k}^{\prime}$ satisfying: $m_{j}^{\prime} \mid m_{j}(1 \leqslant j \leqslant k)$, $m_{1}^{\prime}, \cdots, m_{k}^{\prime}$ are pairwise coprime, and $m=m_{1}^{\prime} \cdots m_{k}^{\prime}$;
... | 11. (i) Let $m_{j}=p_{1}^{q_{1} j} \cdots p_{r^{p}}^{p}, 1 \leqslant j \leqslant k . m=p_{1}^{f_{1}} \cdots p_{r}^{R}, a_{i}=\max _{1 \leqslant j \leqslant k}\left(a_{i j}\right)$. First, in each $m_{j}$, retain the prime powers that have the same exponent as in $m$, and delete the others, to get $m_{j}^{\prime \prime}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,366 |
26. Let $a, b, n$ satisfy $a \mid b n, a x+b y=1, x, y$ are two integers. Prove: $a \mid n$. | 26. $a \mid b y n=(1-a x) n$, so $a \mid n$.
переведено как:
26. $a \mid b y n=(1-a x) n$, so $a \mid n$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,370 |
Theorem 1 Let $m=m_{1} \cdots m_{k}, m_{1}, \cdots, m_{k}$ be pairwise coprime, then the congruence equation (1) and the system of congruence equations
$$f(x) \equiv 0\left(\bmod m_{j}\right), \quad 1 \leqslant j \leqslant k$$
have the same solutions and the same number of solutions, and
$$T(m ; f)=T\left(m_{1} ; f\ri... | Proof: By Property IX of §1 in Chapter 3, the solutions of the congruence equation (1) and the system of congruence equations (2) (i.e., the values of \( x \) that satisfy both) are the same. Let \( t = T(m; f) \) and \( t_j = T(m_j; f) \) for \( 1 \leqslant j \leqslant k \). If one of the equations in (2) (let's assum... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,373 |
Theorem 3 Let $p$ be a prime, and let the polynomial with integer coefficients
$$f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}, \quad n \geqslant 2 ;$$
Suppose the integer $\alpha \geqslant 2, c$ is a solution of
$$f(x) \equiv 0\left(\bmod p^{\alpha-1}\right)$$
Then, the congruence equation
$$f(x) \equiv 0\le... | To prove that this is actually solving the system of congruences given by equations (17) and (18). Any $x$ satisfying (18) must be
$$x=c+p^{\sigma-1} y .$$
Substituting the above expression into the congruence equation (17), equation (17) becomes
$$\begin{aligned}
a_{n}(c+ & \left.p^{\sigma-1} y\right)^{n}+a_{n-1}\lef... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,375 |
Example 1 Solve the congruence equation $x^{3}+5 x^{2}+9 \equiv 0\left(\bmod 3^{4}\right)$. | Solve (1) the congruence equation
$$x^{3}+5 x^{2}+9 \equiv 0(\bmod 3)$$
It has two solutions: $\square$
$$x \equiv 0,1(\bmod 3)$$
Now $f(x)=x^{3}+5 x^{2}+9, f^{\prime}(x)=3 x^{2}+10 x, f^{\prime}(0)=0, f^{\prime}(1)=13$. Therefore,
$$3 \mid f^{\prime}(0), \quad 3 \nmid f^{\prime}(1) \text {. }$$
Next, solve the cong... | x \equiv -21, 6, 33, -24, 3, 30, 40 \left(\bmod 3^{4}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,377 |
Example 2 Solve the congruence equation $x^{3}+5 x^{2}+9 \equiv 0\left(\bmod 7 \cdot 3^{4}\right)$. | Solve: By Theorem 1, we need to solve the system of congruences
$$\left\{\begin{array}{l}
x^{3}+5 x^{2}+9 \equiv 0(\bmod 7), \\
x^{3}+5 x^{2}+9 \equiv 0\left(\bmod 3^{4}\right)
\end{array}\right.$$
By direct calculation, the solution to the first congruence is
$$x \equiv-2(\bmod 7)$$
By Example 1, the solution to the... | x \equiv 222, -156, 33, 138, -240, -51, 40 (\bmod 567) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,378 |
Example 3 Solve the congruence equation $x^{2}+x+7 \equiv 0\left(\bmod 3^{3}\right)$. | From the equivalent transformation III of $\S 1$, we know that the solution to this congruence equation is the same as the solution to the congruence equation
$$4\left(x^{2}+x+7\right) \equiv 0\left(\bmod 3^{3}\right)$$
This congruence equation is
$$(2 x+1)^{2}+27 \equiv(2 x+1)^{2} \equiv 0\left(\bmod 3^{3}\right)$$
... | x \equiv -5, 4, 13 \left(\bmod 3^3\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,379 |
Example 4 Find the solutions to the congruence equation $x^{2} \equiv 1\left(\bmod 2^{l}\right)$. | When $l=1$, the number of solutions is 1,
$$x \equiv 1(\bmod 2)$$
When $l=2$, the number of solutions is 2,
$$x \equiv-1,1\left(\bmod 2^{2}\right)$$
When $l \geqslant 3$, the congruence equation can be written as
$$(x-1)(x+1) \equiv 0\left(\bmod 2^{l}\right)$$
Since $x$ is a solution, it must be expressible as $x=2 ... | x \equiv 1,1+2^{l-1},-1,-1+2^{l-1}\left(\bmod 2^{l}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,380 |
Example 5 Let $p>2$ be a prime. Find the solutions to the congruence equation $x^{2} \equiv 1\left(\bmod p^{l}\right)$. | The congruence equation can be written as
$$(x-1)(x+1) \equiv 0\left(\bmod p^{l}\right)$$
Since $(x-1)+(x+1)=2$, the above equation is equivalent to
$$x-1 \equiv 0\left(\bmod p^{l}\right) \quad \text { or } \quad x+1 \equiv 0\left(\bmod p^{l}\right) \text {. }$$
Therefore, for any $l \geqslant 1$ the solutions are
$$... | x \equiv -1, 1 \left(\bmod p^{l}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,382 |
Example 6 Solve the congruence equation $x^{2} \equiv 2\left(\bmod 7^{4}\right)$. | The complete residue system of $7^{4}$ can be represented as
$$\begin{array}{c}
x=x_{0}+x_{1} \cdot 7+x_{2} \cdot 7^{2}+x_{3} \cdot 7^{3} \\
-3 \leqslant x_{j} \leqslant 3, \quad 0 \leqslant j \leqslant 3
\end{array}$$
We solve the congruence equations sequentially
$$\left(x_{0}+x_{1} \cdot 7+\cdots+x_{j} \cdot 7^{j}\... | x_{1} \equiv -235\left(\bmod 7^{4}\right), \quad x_{2} \equiv 235\left(\bmod 7^{4}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,383 |
2. 求下列模为素数幂的同余方程的解:
(i) $x^{3}+x^{2}+10 x+1 \equiv 0\left(\bmod 3^{3}\right)$;
(ii) $x^{3}+25 x+3 \equiv 0\left(\bmod 3^{3}\right)$;
(iii) $x^{3}-5 x^{2}+3 \equiv 0\left(\bmod 3^{4}\right)$;
(iv) $x^{5}+x^{4}+1 \equiv 0\left(\bmod 3^{4}\right)$;
(v) $x^{3}-2 x+4 \equiv 0\left(\bmod 5^{3}\right)$;
(vi) $x^{3}+x+57 \equi... | 2. (i) $x \equiv-10\left(\bmod 3^{3}\right)$;
(ii) $x \equiv-12\left(\bmod 3^{3}\right)$;
(iii) $x \equiv 11\left(\bmod 3^{4}\right)$;
(iv) 无解;
(v) $x \equiv-56,-2+25 \cdot j, 8+25 j(\bmod$
$\left.5^{3}\right), j=0, \pm 1, \pm 2$;
(vi) $x \equiv 4\left(\bmod 5^{3}\right)$;
(vii) 无解;
(viii) $x \equiv 23\left(\bmod 7^{3}... | x \equiv -10 \pmod{3^3} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,385 |
3. Solve the congruence equation
$$(x+1)^{7}-x^{7}-1 \equiv 0\left(\bmod 7^{7}\right)$$
for solutions satisfying $7 \nmid x(1+x)$. | 3. The original equation is equivalent to $\left(x^{2}+x+1\right)^{2} \equiv 0\left(\bmod 7^{6}\right)$, and further equivalent to $x^{2}+x+1 \equiv$ $0\left(\bmod 7^{3}\right), x \equiv-19,18\left(\bmod 7^{3}\right)$. (These are all the solutions, not the number of solutions). | x \equiv -19, 18 \left(\bmod 7^3\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,386 |
5. Let $d \mid m$, and the coefficients of the integer polynomial $f(x)$ are all divisible by $d$, $d \geqslant 1$.
Prove: $T(m ; f)=d T(m / d ; f / d)$, where $T(m ; f)$ denotes the number of solutions to congruence equation $(1)$ in $\S 4$, and $f / d$ represents the polynomial $f(x) / d$. | 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". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 740,388 |
13. (i) Continue to solve the congruence equation $x^{2} \equiv 2\left(\bmod 7^{l}\right), l=5,6,7$ using the method from Example 5;
(ii) $x^{2} \equiv -1\left(\bmod 5^{6}\right) ;$
(iii) $x^{2} \equiv 4\left(\bmod 7^{4}\right)$. | 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". ... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,397 |
14. Let $f(x)$ be a non-constant polynomial with integer coefficients. Prove that there exist infinitely many prime numbers $p$ such that the congruence equation $f(x) \equiv 0(\bmod p)$ has a solution. | 14. Let $f(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0} \cdot n \geqslant 1, a_{n} \neq 0$. When $a_{0}=0$, the conclusion holds. When $a_{0} \neq 0$, $f\left(a_{0} x\right)=a_{0}\left(b_{n} x^{n}+\cdots+b_{1} x+1\right)=a_{0} g(x), b_{n} \neq 0$. If $g(x) \equiv 0(\bmod p)$ is solvable only for a finite number of primes $p=p_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,398 |
15. Let $f(x)$ be a non-constant polynomial with integer coefficients, and let $r, s$ be any given positive integers. Prove: there must exist an integer $a$, such that each of $f(a), f(a+1), \cdots, f(a+r-1)$ has at least $s$ distinct prime factors. | 15. From the previous problem and the Chinese Remainder Theorem, there must be an $x_{0}$ such that $f\left(x_{0}\right)$ has $s$ distinct prime factors, denoted as $p_{1}, \cdots, p_{s}$. This proves the conclusion for $r=1$. Let $P=p_{1} \cdots p_{s}$, for any $t$, $f\left(x_{0}+P t\right)$ must also have at least $s... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,399 |
Theorem 1 In a reduced residue system modulo $p$, there are exactly $(p-1) / 2$ quadratic residues modulo $p$, and $(p-1) / 2$ quadratic non-residues modulo $p$. Moreover, if $d$ is a quadratic residue modulo $p$, then the number of solutions to the congruence equation (5) is 2. | To prove the evident, we only need to consider the absolute minimal reduced residue system modulo $p$:
$$-\frac{p-1}{2},-\frac{p-1}{2}+1, \cdots,-1,1, \cdots, \frac{p-1}{2}-1, \frac{p-1}{2}$$
$d$ is a quadratic residue modulo $p$ if and only if
$$\begin{aligned}
d \equiv & \left(-\frac{p-1}{2}\right)^{2},\left(-\frac{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,400 |
Example 1 Find the quadratic residues and non-residues for $p=11,17,19,29$.
保留源文本的换行和格式，直接输出翻译结果。
(注意：最后一句为说明，不需翻译，仅作为指导。) | \begin{tabular}{|c|rrrrl|}
\hline$j$ & 1 & 2 & 3 & 4 & 5 \\
\hline$d \equiv j^{2}(\bmod 11)$ & 1 & 4 & -2 & 5 & 3 \\
\hline
\end{tabular}
Quadratic residues modulo 11 are: $1,-2,3,4,5$; quadratic non-residues are: $-1,2,-3$, $-4,-5$
\begin{tabular}{|c|rrrrrrrr|}
\hline$j$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline$d \e... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,401 |
Theorem 2 Let $p>2$ be a prime, $p \nmid d$. Then, $d$ is a quadratic residue modulo $p$ if and only if
$$d^{(p-1) / 2} \equiv 1(\bmod p);$$
$d$ is a quadratic non-residue modulo $p$ if and only if
$$d^{(p-1) / 2} \equiv-1(\bmod p)$$ | First, we prove that for any $d, p \nmid d$, either equation (9) or (10) holds, but not both. By Theorem 3 in Chapter 3, §3, we know
$$d^{p-1} \equiv 1 \pmod{p}$$
Thus, we have
$$\left(d^{(p-1) / 2}-1\right)\left(d^{(p-1) / 2}+1\right) \equiv 0 \pmod{p}$$
Since the prime $p > 2$ and
$$\left(d^{(p-1) / 2}-1\right) + \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,402 |
1. Find the common divisors of the following arrays, and thereby find their greatest common divisor:
(i) $72, -60$;
(ii) $-120, 28$;
(iii) $168, -180, 495$. | 1. (i) $\mathscr{D}(72,-60)=\{ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\},(72,-60)=12$;
(ii) $\mathscr{T}(-120,28)=\{ \pm 1, \pm 2, \pm 4\},(-120,28)=4$;
(iii) $\mathscr{D}(168,-180,495)=\{ \pm 1, \pm 3\},(168,-180,495)=3$. | 12, 4, 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,403 |
Proposition 4 Let $p>2$ be a prime, $p \nmid d_{1}, p \nmid d_{2}$, then
(i) If $d_{1}, d_{2}$ are both quadratic residues modulo $p$, then $d_{1} d_{2}$ is also a quadratic residue modulo $p$; $\square$
(ii) If $d_{1}, d_{2}$ are both quadratic non-residues modulo $p$, then $d_{1} d_{2}$ is a quadratic residue modulo ... | This follows immediately from Theorem 2 and
$$\left(d_{1} d_{2}\right)^{(p-1) / 2}=d_{1}^{(p-1) / 2} \cdot d_{2}^{(p-1) / 2}$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,405 |
Example 2 Use Theorem 2 to determine:
(i) whether 3 is a quadratic residue modulo 17;
(ii) whether 7 is a quadratic residue modulo 29. | From $3^{3} \equiv 10(\bmod 17), 3^{4} \equiv 30 \equiv-4(\bmod 17), 3^{8} \equiv-1(\bmod 17)$ we know that 3 is a quadratic non-residue modulo 17.
From $7^{2} \equiv-9(\bmod 29), 7^{3} \equiv-5(\bmod 29), 7^{6} \equiv-4(\bmod 29), 7^{7} \equiv$ $1(\bmod 29), 7^{14} \equiv 1(\bmod 29)$ we know that 7 is a quadratic re... | 3 \text{ is a quadratic non-residue modulo 17, and 7 is a quadratic residue modulo 29} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,406 |
Example 3 Determine the number of solutions to the following congruence equations:
(i) $x^{2} \equiv -1(\bmod 61)$;
(ii) $x^{2} \equiv 16(\bmod 51)$;
(iii) $x^{2} \equiv -2(\bmod 209)$;
(iv) $x^{2} \equiv -63(\bmod 187)$. | From Corollary 3, we know that the number of solutions to equation (i) is 2. The congruence equation (ii) is equivalent to the system of equations: \(x^{2} \equiv 1(\bmod 3), x^{2} \equiv-1(\bmod 17)\). Both of these equations have 2 solutions each, and by Theorem 1 of § 4, the number of solutions to the congruence equ... | 2, 4, 4, 0 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,407 |
1. Find the quadratic residues and non-residues modulo $p=13,23,37,41$. | 1. $p=13$. Quadratic residues: $1,3,4,9,10,12$; $p=23$. Quadratic residues: $1,2,3,4,6,8,9$, $12,13,16,18 ; p=37$. Quadratic residues: $1,3,4,7,9,10,11,12,16,21,25,26,27,28,30$, $33,34,36 ; p=41$. Quadratic residues: $1,2,4,5,8,9,10,16,18,20,21,23,25,31,32,33$, $36,37,39,40$ | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,408 |
3. Use Theorem 2 to determine:
(i) whether -8 is a quadratic residue modulo 53;
(ii) whether 8 is a quadratic residue modulo 67. | 3. (i) $(-8)^{26} \equiv 11^{13} \equiv 11 \cdot 15^{6} \equiv 11 \cdot 13^{3} \equiv -16 \cdot 10 \equiv -1(\bmod 53), -8$ is not a quadratic residue modulo 53; (ii) $8^{33} \equiv 8 \cdot (3)^{16} \equiv 8 \cdot 14^{4} \equiv 8 \cdot 5^{2} \equiv -1(\bmod 67), 8$ is not a quadratic residue modulo 67. | -8 \text{ is not a quadratic residue modulo 53}; 8 \text{ is not a quadratic residue modulo 67} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,410 |
5. Let $p$ be an odd prime, $p \nmid a$. Prove: There exist integers $u, v, (u, v)=1$, such that $u^{2}+a v^{2} \equiv 0(\bmod p)$ if and only if $-a$ is a quadratic residue modulo $p$. | 5. If $u^{2}+a v^{2} \equiv 0(\bmod p)$ holds, then it must be that $p \nmid u v$, hence we have
$$v v^{\prime} \equiv 1(\bmod p), \quad\left(v^{\prime} u\right)^{2} \equiv-a(\bmod p)$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,412 |
Theorem 4 (Second Principle of Mathematical Induction) Let $P(n)$ be a property or proposition concerning the natural number $n$. If
(i) when $n=1$, $P(1)$ holds;
(ii) for $n>1$, if $P(m)$ holds for all natural numbers $m<n$, then $P(n)$ can certainly be deduced,
then $P(n)$ holds for all natural numbers $n$. | Proof by contradiction. If the theorem does not hold, let $T$ be the set of all natural numbers for which $P(n)$ does not hold, and $T$ is non-empty. By Theorem 2, the set $T$ must have a smallest natural number $t_{0}$. Since $P(1)$ holds, $t_{0} > 1$. By condition (ii) (taking $n=t_{0}$), there must be a natural numb... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,414 |
7. Let $p$ be an odd prime. Prove:
(i) The product of all quadratic residues modulo $p$ is congruent to $(-1)^{(p+1) / 2}$ modulo $p$.
(ii) The product of all quadratic non-residues modulo $p$ is congruent to $(-1)^{(p-1) / 2}$ modulo $p$.
(iii) The sum of all quadratic residues modulo $p$ is congruent to: 1, when $p=3... | 7. (i), (ii) Use Wilson's Theorem; (iii) Find the residue of $1^{2}+2^{2}+\cdots+((p-1) / 2)^{2}$ modulo $p$, using the sum of squares formula; (iv) Determine the result from (iii). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,416 |
9. Let $(a, m)=1$. If $x^{2} \equiv a(\bmod m)$ has a solution, then $a$ is called a quadratic residue modulo $m$.
(i) Prove: When $m>2$, a necessary condition for $a$ to be a quadratic residue modulo $m$ is $a^{\phi(m) / 2} \equiv 1(\bmod m)$. Is this condition sufficient? Provide an example to illustrate.
(ii) Prove:... | 9. (i) Derive the necessity from Euler's theorem. However, it is not sufficient. For example, when $m=15$, the quadratic residues are only 1, 4. But $2^{4} \equiv 1(\bmod 15)$, and 2 is not a quadratic residue.
(ii) $x^{2} \equiv a(\bmod m)$ is equivalent to $\left(x^{-1}\right)^{2} \equiv b(\bmod m)$, where $x^{-1}$ d... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,418 |
11. Let $p$ be an odd prime. Prove: the sum of all quadratic residues modulo $p$ in $1,2, \cdots, p-1$ is
$$S=p\left(p^{2}-1\right) / 24-p \sum_{j=1}^{(p-1) / 2}\left[\frac{j^{2}}{p}\right] .$$
From this, deduce that when $p \equiv 1(\bmod 4)$,
$$p \sum_{j=1}^{(p-1) / 2}\left[\frac{j^{2}}{p}\right]=\frac{p\left(p^{2}-... | 11. Let $j^{2}=p q_{j}+r_{j}, 1 \leqslant r_{j}<p, 1 \leqslant j \leqslant(p-1) / 2$. From this and $q_{j}=\left[j^{2} / p\right]$ we get
$$S=\Sigma r_{j}=\Sigma j^{2}-p \Sigma\left[j^{2} / p\right]$$
Then use part (ii) of problem 10. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,420 |
12. (i) Prove: When $0 \leqslant\{x\}1 / 2$, i.e., $\left[2 j^{2} / p\right]-2\left[j^{2} / p\right]=1$.
(iii) Prove: The number of quadratic non-residues modulo $p$ in $1,2, \cdots,(p-1) / 2$ is
$$N_{1}=\sum_{j=1}^{(p-1) / 2}\left[\left[\frac{2 j^{2}}{p}\right]-2\left[\frac{j^{2}}{p}\right]\right]$$
(iv) Prove: When $... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 740,421 |
13. Let $m>1,(a, m)=1$. Prove: The binary linear congruence equation
$$a x+y \equiv 0(\bmod m)$$
must have a solution $x=x_{0}, y=y_{0}$, satisfying $0<\left|x_{0}\right| \leqslant \sqrt{m}, 0<\left|y_{0}\right| \leqslant \sqrt{m}$ (Hint: Use the Pigeonhole Principle). | 13. Consider the set $\{a x+y: 0 \leqslant x \leqslant[\sqrt{m}], 0 \leqslant y \leqslant[\sqrt{m}]\}$, the number of its elements $=([\sqrt{m}]+1)^{2}>m$.
Using the pigeonhole principle and $(a, m)=1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,422 |
14. Improve the conclusion of the previous problem to $x_{0}, y_{0}$ satisfying
$$0<\left|x_{0}\right| \leqslant \sqrt{m}, \quad 0<\left|y_{0}\right|<\sqrt{m} .$$ | 14. When $m$ is not a square number, it follows from the previous problem; when $m$ is a square number, consider the set
$$\{a x+y: 0 \leqslant x \leqslant \sqrt{m}, 0 \leqslant y \leqslant \sqrt{m}-1\},$$
the number of its elements is $(\sqrt{m}+1) \sqrt{m}>m$. Use the pigeonhole principle and $(a, m)=1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,423 |
16. Let prime $p \equiv 1(\bmod 8)$. Prove: There must be an odd prime $q, 0<q<\sqrt{p}$, which is a quadratic non-residue modulo $p$ (Hint: Take any quadratic non-residue $a$ modulo $p$, and consider the linear congruence equation $a x+y \equiv 0(\bmod p))$. | 16. Using the hint, from question 14 we know that the congruence equation must have a solution $00$. If $x_{0}$ is not a quadratic non-residue, then $\pm y_{0}$ must both be quadratic non-residues (since $p \equiv 1(\bmod 4)$). Therefore, there must be a positive integer $c, 0<c<\sqrt{p}$, that is a quadratic non-resid... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,425 |
3. Prove: (i) $(a, b, c) \leqslant (a, b), [a, b, c] \geqslant [a, b]$;
(ii) If $a \mid b$, then $[a, c] \leqslant [b, c], (a, c) \leqslant (b, c)$;
(iii) $(a, b) \leqslant (a+b, a-b)$;
(iv) $(a, b) \leqslant (a x+b y, a u+b v)$, where $x, y, u, v$ are any integers. | 3. (i) $\mathscr{D}(a, b, c) \subseteq \mathscr{D}(a, b), \mathscr{L}(a, b, c) \supseteq \mathscr{L}(a, b)$;
(ii) $\mathscr{L}(a, c) \subseteq \mathscr{L}(b, c), \mathscr{D}(a, c) \subseteq \mathscr{L}(b, c)$;
(iii) and (iv) $\mathscr{D}(a, b) \subseteq \mathscr{D}(a x+b y, a u+b v)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,426 |
Lemma 2 (Gauss Lemma) Let $p>2$ be a prime, $p \nmid d$; and let $1 \leqslant j<p / 2$,
$$t_{j} \equiv j d(\bmod p), \quad 0<t_{j}<p$$
Let $n$ denote the number of the $(p-1) / 2$ numbers $t_{j}(1 \leqslant j<p / 2)$ that are greater than $p / 2$. Then, we have
$$\left(\frac{d}{p}\right)=(-1)^{n} .$$ | For any $1 \leqslant j<i<p / 2$,
$$t_{j} \pm t_{i} \equiv(j \pm i) d \not \equiv 0(\bmod p)$$
i.e., $\square$
$$t_{j} \not \equiv \pm t_{i}(\bmod p)$$
We denote by $r_{1}, \cdots, r_{n}$ the numbers in $t_{j}(1 \leqslant j<p / 2)$ that are greater than $p / 2$, and by $s_{1}, \cdots, s_{k}$ the numbers in $t_{j}(1 \l... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,429 |
Theorem 3 We have
$$\left(\frac{2}{p}\right)=(-1)^{\left(p^{2}-1\right) / 8} \text {. }$$ | Prove: Using the notation in Lemma 2, take $d=2$. It is easy to see that,
$$\begin{array}{c}
1 \leqslant t_{j}=2 j<p / 2, \quad 1 \leqslant j<p / 4 \\
p / 2<t_{j}=2 j<p, \quad p / 4<j<p / 2
\end{array}$$
From the second equation, we know
$$n=\frac{p-1}{2}-\left[\frac{p}{4}\right]$$
Therefore, we have
$$n=\left\{\begi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,430 |
Example 1 Calculate $\left(\frac{137}{227}\right)$. | To determine if 227 is a prime number, by Theorem 1 we get
$$\begin{aligned}
\left(\frac{137}{227}\right) & =\left(\frac{-90}{227}\right)=\left(\frac{-1}{227}\right)\left(\frac{2 \cdot 3^{2} \cdot 5}{227}\right) \\
& =(-1)\left(\frac{2}{227}\right)\left(\frac{3^{2}}{227}\right)\left(\frac{5}{227}\right) \\
& =(-1)\left... | -1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,433 |
Example 2 Determine whether the following congruence equations have solutions, and find the number of solutions if they do:
(i) $x^{2} \equiv -1(\bmod 365)$;
(ii) $x^{2} \equiv 2(\bmod 3599)$. | (i) 365 is not a prime, $365=5 \cdot 73$. Therefore, the congruence equation and the system of congruence equations
$$x^{2} \equiv -1(\bmod 5), \quad x^{2} \equiv -1(\bmod 73)$$
are equivalent. By Theorem 1(v), we know
$$\left(\frac{-1}{5}\right)=\left(\frac{-1}{73}\right)=1,$$
so the system of congruence equations h... | 0 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,434 |
Example 3 Find all odd prime numbers $p$ for which 3 is a quadratic residue. | Solve: This is to find all odd primes $p$ such that $\left(\frac{3}{p}\right)=1$. By Theorem 5, we know
$$\left(\frac{3}{p}\right)=(-1)^{(p-1) / 2}\left(\frac{p}{3}\right)$$
It is evident that $p$ is an odd prime greater than 3. By
$$(-1)^{(p-1) / 2}=\left\{\begin{array}{ll}
1, & p \equiv 1(\bmod 4) \\
-1, & p \equiv-... | p \equiv \pm 1(\bmod 12) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,435 |
Example 4 Find all odd prime numbers $p$ for which 11 is a quadratic residue.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | By Theorem 5, we know
$$\left(\frac{11}{p}\right)=(-1)^{(p-1) / 2}\left(\frac{p}{11}\right)$$
By direct calculation, we have
$$\begin{array}{l}
\left(\frac{p}{11}\right)=\left\{\begin{array}{ll}
1, & p \equiv 1,-2,3,4,5(\bmod 11) \\
-1, & p \equiv-1,2,-3,-4,-5(\bmod 11)
\end{array}\right. \\
(-1)^{(p-1) / 2}=\left\{\b... | p \equiv \pm 1, \pm 5, \pm 7, \pm 9, \pm 19(\bmod 44) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 740,436 |
4. If $(a, b)=1, c \mid a+b$, then $(c, a)=(c, b)=1$.
| 4. $d|c, d| a \Longrightarrow d|a+b, d| a \Longrightarrow d|b, d| a \Longrightarrow d= \pm 1$, so $(c, a)=1$. Similarly, we can prove $(c, b)=1$. Or use the result from question 3 (ii) and $(a, b)=(a, a+b)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,437 |
Example 5 Proof: If $\left(\frac{d}{p}\right)=-1$, then $p$ cannot be expressed in the form $x^{2}-d y^{2}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
However, the translation you requested is already in English. Here ... | Prove by contradiction. If $p=x^{2}-d y^{2}$, since $p$ is a prime, it must be that $(p, x)=(p, y)=1$. Therefore, by Theorem 1, we have
$$1=\left(\frac{x^{2}}{p}\right)=\left(\frac{d y^{2}}{p}\right)=\left(\frac{d}{p}\right)\left(\frac{y^{2}}{p}\right)=\left(\frac{d}{p}\right)$$
This is a contradiction. Thus, by Examp... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 740,438 |
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