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Lemma 3 If $a$ is a quadratic residue modulo $p$, then the congruence equation
$$x^{2} \equiv a(\bmod p)$$
has exactly two distinct solutions $(\bmod p)$. | By Definition 1 of the previous section, we know that there must be an integer $x$ such that it can make
$$x_{1}^{2} \equiv a(\bmod p)$$
hold, and thus we also have
$$\left(-x_{1}\right)^{2} \equiv a(\bmod p)$$
hold, which indicates that $-x_{1}$ is also a solution to equation (20), and it is easy to prove that $x_{1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,353 |
Let $x_{1}, \cdots, x_{n}$ be $n$ non-negative real numbers, $n \geqslant 1$ be any given natural number. If
$$x_{1} \cdot x_{2} \cdots x_{n}=1$$
then it must be true that
$$x_{1}+x_{2}+\cdots+x_{n} \geqslant n .$$ | 1. Proof: When $n=1$, the conclusion is obviously true. Now assume that the conclusion holds for $n=1, \cdots, l(l \geqslant 1)$. Let's consider the case when $n=l+1$. That is, there are $l+1$ non-negative real numbers $x_{1}, x_{2}, \cdots, x_{l+1}$ satisfying $x_{1} x_{2} \cdots x_{l+1}=1$. We will discuss the follow... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 741,355 |
Theorem 1 (Euler's Criterion)
If $a$ is a quadratic residue modulo $p$, then
$$a^{\frac{p-1}{2}} \equiv 1(\bmod p) \text {. }$$
If $b$ is a quadratic non-residue modulo $p$, then
$$b^{\frac{p-1}{2}} \equiv-1(\bmod p) .$$ | Prove that when $a$ is a quadratic residue modulo $p$, by definition 1, there exists an integer $n$ such that
$$a \equiv n^{2}(\bmod p)$$
Taking the $(p-1) / 2$ power on both sides, we get
$$a^{\frac{p-1}{2}} \equiv n^{p-1}(\bmod p)$$
From $(a, p)=1$, we get $(n, p)=1$. By Fermat's Little Theorem and the above congru... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,356 |
Lemma 5 For any prime $p$:
Any product of two quadratic residues is still a quadratic residue; the product of a quadratic residue and a non-residue is a non-residue; the product of any two non-residues is necessarily a quadratic residue. | Proof: Let $a_{1}$ and $a_{2}$ both be quadratic residues. By Definition 1, there must exist two integers $n_{1}, n_{2}$ such that respectively,
$$\begin{array}{c}
n_{1}^{2} \equiv a_{1}, \quad n_{2}^{2} \equiv a_{2}(\bmod p) \\
\cdot\left(n_{1} n_{2}\right)^{2} \equiv a_{1} a_{2}(\bmod p)
\end{array}$$
This shows th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,357 |
Lemma 6 When the prime $p \geqslant 3$ and $a_{1} \equiv a_{2}(\bmod p)$, we have
$$\left(\frac{a_{1}}{p}\right)=\left(\frac{a_{2}}{p}\right)$$ | Proof: From (30) we have
$$\left(\frac{a_{1}}{p}\right) \equiv a_{1}^{\frac{p-1}{2}} \equiv a_{2}^{\frac{p-1}{2}} \equiv\left(\frac{a_{2}}{p}\right)(\bmod p)$$
From this, we get $\left(\frac{a_{1}}{p}\right)-\left(\frac{a_{2}}{p}\right) \equiv 0(\bmod p)$, but since $\left\lvert\,\left(\frac{a_{1}}{p}\right)\right.$ $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,358 |
Lemma 7 When the prime $p \geqslant 3$ and the integer $a=a_{1} a_{2} \cdots a_{n}$, we have
$$\left(\frac{a}{p}\right)=\left(\frac{a_{1}}{p}\right)\left(\frac{a_{2}}{p}\right) \ldots\left(\frac{a_{n}}{p}\right) .$$ | Proof: Still by (30), we have
$$\begin{aligned}
\left(\frac{a}{p}\right) \equiv & a^{\frac{p-1}{2}}=a_{1} \frac{p-1}{2} a_{2} \frac{p-1}{2} \cdots a_{n} \frac{p-1}{2} \equiv\left(\frac{a_{1}}{p}\right)\left(\frac{a_{1}}{p}\right) \\
& \cdots\left(\frac{a_{n}}{p}\right)(\bmod p)
\end{aligned}$$
Thus, we get $\left(\fra... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,359 |
Lemma 9 (Gauss's Lemma) Let $p$ be an odd prime, $n$ an integer, and $p \nmid n$. Suppose that among the $(p-1) / 2$ integers
$$n, 2 n, \cdots, \frac{(p-1)}{2} n$$
the least residues modulo $p$ are
$$r_{1}, r_{2}, \cdots, r_{j},-r_{1}, \cdots,-r_{\mu},$$
where $1 \leqslant r_{i}<\frac{p}{2}(1 \leqslant i \leqslant j)... | Prove that $\lambda+\mu=(p-1) / 2$. Since the numbers in (31) are all incongruent modulo $p$, any two numbers chosen from $r_{1}, r_{2}, \cdots, r_{i}$ are not equal. Similarly, any two numbers chosen from $r_{1}, \cdots, r_{\mu}$ are not equal. Now we will prove that in the product
$$r_{1}, \cdots, r_{1}$$
if we arbi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,361 |
Lemma 10 Let $p$ be an odd prime, then
$$\left(\frac{2}{p}\right)=(-1)^{\left(p^{2}-1\right) / 8},$$
that is, when $p \equiv 1$ or $-1(\bmod 8)$, 2 is a quadratic residue modulo $p$, and when $p \equiv 3$ or $-3(\bmod 8)$, 2 is a quadratic non-residue modulo $p$. | In Lemma 9, take $n=2$. At this time, the numbers in (31) are
$$2,4, \cdots, p^{-1}$$
Solving $2 x<\frac{p}{2}$ yields $x<\frac{p}{4}$, so at this time $\lambda=\left[\frac{p}{4}\right]$, thus $\mu=\frac{p-1}{2}-\left\{\frac{p}{4}\right\rceil$, hence
$$\begin{array}{c}
\mu=\left\{\begin{array}{r}
2 n, \text { when } p... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,362 |
Example 2 Find an odd prime $p$ such that
$$x^{2} \equiv 5(\bmod p)$$
List the solution | From $5 \equiv 1(\bmod 4)$ and Theorem 2, we know
$$\left(\frac{5}{p}\right)=\left(\frac{p}{5}\right)$$
Thus, when $p \equiv 1$ or $-1(\bmod 5)$, $\left(\frac{p}{5}\right)=1$. In this case, equation (44) has a solution. When $p \equiv 2$ or $-2(\bmod 5)$, $\left(\frac{p}{5}\right)=-1$, in which case equation (44) has ... | p \equiv 1 \text{ or } -1(\bmod 5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,365 |
2. (Reverse Induction) If:
（1）Conclusion P holds for infinitely many natural numbers $n$,
（2）From P holding for a natural number $n(n \geqslant 2)$, it can be deduced that P also holds for $n-1$,
then Conclusion P holds for all natural numbers. | 2. Proof: From the given condition (1), we can assume that $P$ holds for the following infinitely many natural numbers:
$$(1 \leqslant) n_{1} \leqslant n_{2} \leqslant \cdots \leqslant n_{m} \leqslant \cdots \cdots$$
Take any natural number $r$. If there exists a natural number $i$ such that
$$r=n_{i}$$
then the conc... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,366 |
Example 3 Try to prove that there are infinitely many prime numbers of the form $4 n+1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: By contradiction, assume that there are only finitely many prime numbers of the form $4n+1$. Let these be the following $r$ primes:
$$p_{1} < p_{2} < \cdots < p_{r}$$
Consider the number
$$Q = 4(p_{1} p_{2} \cdots p_{r})^{2} + 1 > 1$$
Thus, $Q$ must have a prime factor. Let $p \mid Q$, then we have
$$-1 \equiv (... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,367 |
Example 4 Find an odd prime $p$ such that
$$x^{2}+3 \equiv 0(\bmod p)$$
has a solution. | From $p$ being a prime number and Lemma 8 and Theorem 2, we have
$$\begin{array}{l}
\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) \\
\quad=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p-1}{2}}\left(\frac{p}{3}\right)=\left(\frac{p}{3}\right) .
\end{array}$$
Thus, when $p \equiv 1(\bmod 3)$, we have $\... | p \equiv 1(\bmod 3) \text{ or } p = 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,368 |
Example 5 Prove that the number of primes of the form $6n+1$ is infinite.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof: We still use proof by contradiction. Suppose there are only $r$ prime numbers of the form $6n+1$, denoted as $p_{1}, \ldots, p_{r}$.
Thus, there must exist a prime $p \mid \mathrm{Q}$, which means
$$\left(2 p_{1} \cdots p_{r}\right)^{2}+3 \equiv 0(\bmod p)$$
This implies that the congruence equation $x^{2}+3 \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,369 |
Lemma 11 Let $n$ be an odd number greater than 1, and $(a, n)=1$, then when $a_{1} \equiv a_{2}(\bmod n)$, we have
$$\left(\frac{a_{1}}{n}\right)=\left(\frac{a_{2}}{n}\right) .$$ | Proof: Let $n=p_{1} \cdots p_{m}$, where $p_{1}, \cdots, p_{m}$ are all primes. From $a_{1} \equiv$ $a_{2}(\bmod n)$, we get
$$a_{1} \equiv a_{2}\left(\bmod p_{k}\right) \quad(1 \leqslant k \leqslant m)$$
Therefore, by Definition 3 and Lemma 6, we have
$$\left(\frac{a_{1}}{n}\right)=\left(\frac{a_{1}}{p_{1}}\right) \c... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,370 |
Lemma 12 When $a=a_{1} \cdots a_{k}$ and $n \geqslant 3$ is an odd number, we have
$$\left(\frac{a}{n}\right)=\left(\frac{a_{1}}{n}\right) \ldots\left(\frac{a_{k}}{n}\right),$$
where $a_{1}, \cdots, a_{k}$ are all integers. | Let $n=p_{1} \cdots p_{m}, p_{1}, \cdots, p_{m}$ be primes. By Definition 3 and Lemma 7, we have
$$\begin{aligned}
\left(\frac{a}{n}\right) & =\left(\frac{a}{p_{1}}\right) \ldots\left(\frac{a}{p_{m}}\right) \\
& =\left(\frac{a_{1}}{p_{1}}\right) \ldots\left(\frac{a_{k}}{p_{1}}\right) \ldots\left(\frac{a_{1}}{p_{m}}\rig... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,371 |
Lemma 13 When $n \geqslant 3$ is an odd number, we have
$$\left(\frac{-1}{n}\right)=(-1)^{\frac{n-1}{2}} \text {. }$$ | Let $n=p_{1} \cdots p_{m}$, where each $p_{i}(1 \leqslant i \leqslant m)$ is an odd prime. By definition 3, we have
$$\left(\frac{-1}{n}\right)=\left(\frac{-1}{p_{1}}\right) \cdots\left(\frac{-1}{p_{m}}\right)=(-1)^{\frac{p_{1}-1}{2}} \cdots(-1)^{\frac{p_{m}-1}{2}}$$
Thus, it remains to prove
$$\begin{aligned}
\frac{p... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,372 |
Example 6 Discuss the congruence equation
$$x^{2} \equiv -286 \pmod{4272943}$$
whether it has a solution, where 4272943 is a prime number. | Let $p=4272943$, by Lemma 7 we have
$$\left(\frac{-286}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)\left(\frac{143}{p}\right)$$
Since $4272943 \equiv 7(\bmod 8)$, we have
$$\left(\frac{-1}{p}\right)=-1,\left(\frac{2}{p}\right)=1$$
Thus,
$$\left(\frac{-286}{p}\right)=-\left(\frac{143}{p}\right)$$
Sinc... | 1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,375 |
1. Prove that any odd prime factor of $x^{2}+1$ must have the form $4k+1$.
untranslated text remains the same as requested. However, the actual content has been translated into English while maintaining the original format and structure. | 1. Proof: Let the odd prime $p \mid (x^{2}+1)$, i.e., the integer $x$ satisfies
$$x^{2}+1 \equiv 0(\bmod p)$$
This indicates that -1 is a quadratic residue modulo $p$, and by Lemma 7 of this chapter, it follows that $p \equiv 1(\bmod 4)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,376 |
3. Use reverse induction to prove the Cauchy (Cauchy) inequality
$$\sqrt[n]{a_{1} \cdots a_{n}} \leqslant \frac{a_{1}+\cdots+a_{n}}{n}$$
Here $a_{1}, \cdots, a_{n}$ are any given $n$ positive numbers. | 3. Proof: We have
Thus,
$$a_{1} a_{2}=\left(\frac{a_{1}+a_{2}}{2}\right)^{2}-\left(\frac{a_{1}-a_{2}}{2}\right)^{2} \leqslant\left(\frac{a_{1}+a_{2}}{2}\right)^{2},$$
$$\begin{aligned}
a_{1} a_{2} a_{3} a_{4} & \leqslant\left(\frac{a_{1}+a_{2}}{2}\right)^{2}\left(\frac{a_{3}+a_{4}}{2}\right)^{2}=\left\{\left(\frac{a_{... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 741,377 |
2. Prove that the odd prime factors of $x^{2}-2$ must be of the form $8k \pm 1$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. Proof: Let the odd prime $p \mid (x^{2}-2)$, i.e., the integer $x$ satisfies
$$x^{2}-2 \equiv 0(\bmod p)$$
This indicates that 2 is a quadratic residue modulo $p$. By Lemma 9 of this chapter, it must be that $p \equiv \pm 1(\bmod 8)$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,378 |
3. Let $n \geqslant 1$, and $4 n+3$ and $8 n+7$ are both prime numbers, prove that
$$\mathrm{M}_{4 n+3}=2^{4 n+3}-1$$
is not a prime number. | 3. Let $p=8n+7$, by Lemma 9 of this chapter, 2 must be a quadratic residue of $p$, so there must be an $x_0$ such that
$$x_0^2 \equiv 2 \pmod{p}$$
Thus,
$$M_{4n+3}=2^{4n+3}-1 \equiv x_0^{8n+6}-1 = x_0 p^{p-1}-1 \equiv 0 \pmod{p}.$$
Therefore,
$$p \mid M_q, \quad q=\frac{p-1}{2}.$$
Since $n=1$ gives $p=15$ which is n... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,379 |
4. Find the prime $p \geqslant 5$ for which 3 is a quadratic residue.
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。
Note: The note above should not be included in the final translation, as it is a meta-instruction. Here is the final translation:
4. Find the prime $p \geqslant 5$ for which 3 is a quadratic residue. | 4 . Solution: We have
$$\left(\frac{3}{p}\right)=(-1)^{\frac{p-1}{2}}\left(\frac{p}{3}\right)$$
It is easy to see that
$$(-1)^{\frac{p-1}{2}}=\left\{\begin{aligned}
1, & \text { when } p \equiv 1(\bmod 4), \\
-1, & \text { when } p \equiv 3(\bmod 4),
\end{aligned}\right.$$
and on the other hand,
$$\left(\frac{p}{3}\r... | p \equiv \pm 1(\bmod 12) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,380 |
5. Find the prime $p$ for which 10 is a quadratic residue. | 5. Solution: We have
$$\left(\frac{10}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{5}{p}\right)$$
By Example 2 of this chapter, we have
And by Theorem 9, we have
$$\begin{array}{l}
\left(\frac{5}{p}\right)=\left\{\begin{array}{ll}
1, & p \equiv \pm 1(\bmod 5), \\
-1, & p \equiv \pm 2(\bmod 5),
\end{array}\right. \\... | p \equiv \pm 1, \pm 3, \pm 9, \pm 13,(\bmod 40) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,381 |
6. Find the prime $p$ for which 6 is a quadratic residue.
The text has been translated while preserving the original line breaks and format. | 6. Solution: We have
$$\left(\frac{6}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{3}{p}\right)$$
By Lemma 9 and Problem 4, we have respectively
$$\begin{array}{l}
\left(\frac{2}{p}\right)=\left\{\begin{array}{l}
1, p \equiv \pm 1(\bmod 8) \\
-1, p \equiv \pm 3(\bmod 8)
\end{array}\right. \\
\left(\frac{3}{p}\right)=... | p \equiv \pm 1, \pm 5(\bmod 24) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,382 |
7. If $q$ is a natural number, $p=4q+1$ is a prime, prove that $q$ must be a quadratic residue modulo $p$.
| 7. Proof: Since $p \equiv 1(\bmod 4)$, -1 must be a quadratic residue modulo $p$, i.e., there exists an integer $x_{0}$ such that
$$x_{0}^{2} \equiv -1(\bmod p)$$
On the other hand, from $p=4 q+1$ we also have
$$4 q \equiv -1(\bmod p)$$
Therefore,
$$x_{0}^{2} \equiv 4 q(\bmod p)$$
Since $2 \nmid p$, there must be a ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,383 |
8. If $q$ is a natural number, and $p=4q+3$ is a prime, prove that 2 and $2q+1$ cannot both be quadratic residues or quadratic non-residues modulo $p$.
---
If $q$ is a natural number, and $p=4q+3$ is a prime, prove that 2 and $2q+1$ cannot both be quadratic residues or quadratic non-residues modulo $p$. | 8. Proof: By contradiction. If 2 and $2 q+1$ are both quadratic residues of $p$, or both quadratic non-residues of $p$, then by Lemma 4 of this chapter, $2(2 q+1)$ must be a quadratic residue of $p$. Therefore, there should be an integer $x_{0}$ such that
$$2(2 q+1) \equiv x_{0}^{2}(\bmod p)$$
Noting that $p=4 q+3$, w... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,384 |
9. Solve the congruence equation $x^{2} \equiv 59(\bmod 125)$ | 9. Solution: First, it is easy to see that $59 \equiv 9\left(\bmod 5^{2}\right)$, hence
$$x^{2} \equiv 59\left(\bmod 5^{2}\right)$$
has solutions $x \equiv \pm 3(\bmod 25)$. Next, we solve
$$x^{2} \equiv 59\left(\bmod 5^{3}\right)$$
(1) Let $x=25 t+3$, substituting we get
$$(25 t+3)^{2} \equiv 59\left(\bmod 5^{3}\righ... | x \equiv \pm 53\left(\bmod 5^{3}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,385 |
10. Prove: The indeterminate equation
$$p=x^{2}+2 y^{2}$$
( $p$ is an odd prime) has natural number solutions if and only if $\left(\frac{-2}{p}\right)$ $=1$.
| 10. Proof: First, we prove the necessity. Suppose the equation has a solution. Then, it must be that $(p, x) = (p, y) = 1$, for if not, say $p \mid x$, then it would also imply $p \mid y$, leading to $p = (p x_1)^2 + 2(p y_1)^2$, which is impossible. Therefore, there must be a $y_1$ such that $p \nmid y_1$. Since $y y_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,386 |
11. Prove: the indeterminate equation
$$p=x^{2}+3 y^{2}$$
( $p \geqslant 3$ is a prime) has natural number solutions if and only if $\left(\frac{-3}{p}\right)=1$. | 11. Proof: The proof of necessity is exactly the same as in the previous problem and is omitted here, left for the reader to practice. Now we prove sufficiency. Suppose $\left(\frac{-3}{p}\right)=1$, then there exists $Z\left(|Z|<\frac{p}{2}\right)$ such that
$$Z^{2} \equiv-3(\bmod p)$$
Notice that $0<Z^{2}+3<\frac{p^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,387 |
4. Let $0<x_{i} \leqslant 1 / 2, i=1, \cdots, n$, then we have
$$\frac{x_{1} \cdots x_{n}}{\left(x_{1}+\cdots+x_{n}\right)^{n}} \leqslant \frac{\left(1-x_{1}\right) \cdots\left(1-x_{n}\right)}{\left[\left(1-x_{1}\right)+\cdots+\left(1-x_{n}\right)\right]^{n}}$$ | 4. Proof: First, prove the case for $n=2$, i.e., prove
$$\frac{x_{1} x_{2}}{\left(x_{1}+x_{2}\right)^{2}} \leqslant \frac{\left(1-x_{1}\right)\left(1-x_{2}\right)}{\left[\left(1-x_{1}\right)+\left(1-x_{2}\right)\right]^{2}}$$
By simplifying and expanding the above inequality, we need to prove
$$x_{1}\left(1-x_{1}\righ... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 741,388 |
12. Define the $m$-th Fermat number as
$$\mathrm{F}_{m}=2^{2^{m}}+1$$
Prove: If $p \mid \mathrm{F}_{m}$, then there must be an integer $k$ such that $p=2^{m+2} k+1$ (for $m \geqslant 2$). | 12. Proof: Let $p \mid F_{m}$, obviously $p \neq 2$, we can set $p=2^{\prime} h_{1}+1, t \geqslant 1$, $2 \mid h_{1}$. From $p \mid F_{m}$ we have
$$2^{2^{m}} \equiv-1(\bmod p)$$
Thus,
$$\left(2^{h_{1}}\right)^{2^{m}} \equiv(-1)^{h_{1}} \equiv-1(\bmod p)$$
By Fermat's Little Theorem (since $p-1=2^{\prime} h_{1}$),
$$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,389 |
13. Let $p \geqslant 3$ be a prime number, prove the following conclusions:
(1) When $p \equiv 1(\bmod 4)$, $\sum_{r=1}^{p-1} r\left(\frac{r}{p}\right)=0$.
(2) Only when $p \equiv 1(\bmod 4)$, $\sum_{i=1}^{p-1} r=\frac{p(p-1)}{4}$.
(3) When $p \equiv 3(\bmod 4)$, $\sum_{r=1}^{p-1} r^{2}\left(\frac{r}{p}\right)=p \sum_{... | 13. Proof:
(1) Note that when $r=1,2, \cdots, p-1$, $p-r$ also passes through 1, 2, ..., p-1, thus we have
$$\begin{aligned}
\sum_{r=1}^{p-1} r\left(\frac{r}{p}\right) & =\sum_{r=1}^{p-1}(p-r)\left(\frac{p-r}{p}\right)=\sum_{r=1}^{p-1}(p-r)\left(\frac{-r}{p}\right) \\
& =\sum_{r=1}^{p-1}(p-r)\left(\frac{-1}{p}\right)\l... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,390 |
14. Let $p \geqslant 5$ and $p \equiv 3(\bmod 4)$, prove that the sum of all quadratic residues modulo $p$ is divisible by $p$. | 14. Let $p \geqslant 5$ and $p \equiv 3(\bmod 4)$, then $p$ has $1^{2}, 2^{2}, \cdots$, $\left(\frac{p-1}{2}\right)^{2}$ as all its quadratic residues. Thus, the problem reduces to proving
$$1^{2}+2^{2}+\cdots+\left(\frac{p-1}{2}\right)^{2} \equiv 0(\bmod p)$$
But it is easy to see that
$$\begin{aligned}
1^{2}+2^{2}+\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,391 |
15. Let $p \geqslant 3$, try to calculate the value of the following expression:
$$\left(\frac{1 \cdot 2}{p}\right)+\left(\frac{2 \cdot 3}{p}\right)+\cdots+\left(\frac{(p-2)(p-1)}{p}\right)$$ | 15. Solution: To solve this problem, we need to study the properties of the Legendre symbol with the general term $\left(\frac{n(n+1)}{p}\right)$, where $(n, p)=1$.
By $(n, p)$ being coprime, we know there must exist an integer $r_{n}$ such that $p \nmid r_{n}$ and $n r_{n} \equiv 1(\bmod p)$. This $r_{n}$ is called t... | -1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,392 |
16. Prove: For prime $p \equiv 1,5,17,25,37,41(\bmod 42)$, the congruence equation $x^{2} \equiv 21(\bmod p)$ has a solution. | 16. Proof: From $21 \equiv 1(\bmod 4)$ and the quadratic reciprocity law, we have
$$\left(\frac{21}{p}\right)=\left(\frac{p}{21}\right)=\left(\frac{p}{3}\right)\left(\frac{p}{7}\right),$$
We already know
$$\left(\frac{p}{3}\right)=\left\{\begin{array}{l}
1, p \equiv 1 \quad(\bmod 3) \\
-1, p \equiv-1(\bmod 3)
\end{arr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,393 |
17. Find all values of $m$ for which the congruence
$$x^{2} \equiv 6(\bmod m)$$
may have a solution. | 17. Solution: If $m$ is odd, then the necessary condition for $x^{2} \equiv 6(\bmod m)$ to have a solution is $\left(\frac{6}{m}\right)=1$. By the properties of the Legendre symbol, we have
$$\left(\frac{6}{m}\right)=\left(\frac{2}{m}\right)\left(\frac{3}{m}\right)=(-1)^{\frac{m^{2}-1}{8}} \cdot(-1)^{\frac{m-1}{2}}\lef... | m \equiv \pm 1, \pm 2, \pm 5, \pm 10(\bmod 24) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,394 |
18. Prove: there are infinitely many primes of the form $8k+7$.
| 18. Proof: Let $N$ be any given positive integer, and let $p_{1}, p_{2}, \ldots, p_{s}$ be all the primes of the form $8k+7$ that do not exceed $N$. Define
$$q=\left(p_{1} p_{2} \ldots p_{s}\right)^{2}-2$$
Since each $p_{j}$ is of the form $8k+7$, they are all odd primes, so $p_{1} p_{2} \ldots p_{s}$ is also odd. Let... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,395 |
19. Prove: There are infinitely many primes of the form $8k+3$.
untranslated part:
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。
translated part:
19. Prove: There are infinitely many primes of the form $8k+3$.
The part above was translated from Chinese to English, keeping the original text's line breaks and format. | 19. Proof: Let $N$ be any given positive integer, and let $p_{1}, \ldots, p_{s}$ be all the primes of the form $8k+3$ that do not exceed $N$. Define
$$q=\left(p_{1} p_{2} \ldots p_{s}\right)^{2}+2$$
Let $p_{j}=2 m_{j}+1$. It is easy to see that
$$p_{j}^{2}=\left(2 m_{j}+1\right)^{2}=8 \cdot \frac{m_{j}\left(m_{j}+1\ri... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,396 |
20. Prove: There are infinitely many primes of the form $8k+5$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
translated part:
20. Prove: There are infinitely many primes of the form $8k+5$.
The instruction above has been translated as requested. If you need further assistance or have additional text to... | 20. Proof: First, we prove a generalization of Exercise 1: If $(x, y)=1$, then the prime factors of $x^{2}+y^{2}$ must be of the form $4k+1$.
Let $p \mid (x^{2}+y^{2})$. Clearly, $p \nmid x$ and $p \nmid y$, because if $p \mid x$, then since $y^{2} = (x^{2}+y^{2}) - x^{2}$, it must also be true that $p \mid y^{2}$, an... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,397 |
21. Let $p$ be an odd prime, $p \nmid a$, $l$ be a positive integer, prove that the congruence $x^{2} \equiv$ $a\left(\bmod p^{l}\right)$ has a solution if and only if
$$\left(\frac{a}{p}\right)=1$$ | 21. Proof: First, we prove the necessity. Suppose $x^{2} \equiv a\left(\bmod p^{l}\right)$ has a solution, then this solution also satisfies $x^{2} \equiv a(\bmod p)$, hence $a$ must be a quadratic residue modulo $p$, thus $\left(\frac{a}{p}\right)=1$.
Next, we prove the sufficiency. Suppose $\left(\frac{a}{p}\right)=... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,398 |
22. Let $\alpha$ and $\beta$ be integers that take values $\pm 1$, and let $N(\alpha, \beta)$ denote the number of $x$ in the set 1, 2, $\cdots, p-2$ such that
$$\left(\frac{x}{p}\right)=\alpha,\left(\frac{x+1}{p}\right)=\beta$$
holds, where $p \geqslant 3$. Prove:
(1) $4 \mathrm{~N}(\alpha, \beta)=\sum_{x=1}^{p-2}\le... | 22. Proof: By definition, we have $\alpha^{2}=\beta^{2}=1$, thus
$$\begin{array}{c}
\left(1+\alpha\left(\frac{x}{p}\right)\right)\left(1+\beta\left(\frac{x+1}{p}\right)\right) \\
=\left\{\begin{array}{l}
\left(1+\alpha^{2}\right)\left(1+\beta^{2}\right)=4 \text {, when }\left(\frac{x}{p}\right)=\alpha \text { and }\lef... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,400 |
23. Prove: For each prime $p$, there exist integers $x, y$ such that
$$x^{2}+y^{2}+1 \equiv 0 \quad(\bmod p)$$ | 23. Proof: We consider two cases.
Case one. $\left(\frac{-1}{p}\right)=-1$, in this case, taking $\alpha=1$, $\beta=-1$ in the previous problem, we get $4 N(\alpha, \beta)=p+1 \geqslant 4$,
thus
$$N(\alpha, \beta) \geqslant 1$$
This indicates that there exists at least one integer $r(1 \leqslant r \leqslant p-2)$, su... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,401 |
Example 1 Try to solve the congruence equation
$$x^{2} \equiv 73(\bmod 127)$$ | First, it is easy to see that $p=127$ is a prime number. Moreover, we have $p \equiv 3(\bmod 4)$, $73 \equiv 1(\bmod 8)$. By the properties of the Legendre symbol and the Jacobi symbol, we have
$$\begin{aligned}
\left(\frac{73}{127}\right)= & \left(\frac{127}{73}\right)=\left(\frac{54}{73}\right)=\left(\frac{2}{73}\rig... | \pm 33 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,402 |
Proposition Under the assumptions of Case III, if $a^{\prime \prime} \neq \pm 1(\bmod p)$, then there must be a positive integer $\mu \leqslant \lambda$, which can make
$$\left(a^{\prime \prime}\right)^{2^{\mu}} \equiv-1(\bmod p)$$
hold. At this time, there must be an odd number $t\left(1 \leqslant t<2^{\mu}\right)$, ... | Given that $a$ is a quadratic residue modulo $p$, we know
$$a^{\frac{p-1}{2}}=a^{2 \cdot 2^{i} u}=\left(a^{u}\right)^{2^{i+1}} \equiv 1(\bmod p)$$
From this, we get $\left(a^{u}\right)^{2^{i}} \equiv +1$ or $-1(\bmod p)$. If $\left(a^{u}\right)^{2^{i}} \equiv -1(\bmod p)$, then $\mu=\lambda$. If $\left(a^{u}\right)^{2... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,404 |
Example 3 Try to solve the congruence equation
$$x^{2} \equiv 22(\bmod 29)$$ | Solving: Since 29 is a prime number,
$$\begin{aligned}
\left(\frac{22}{29}\right) & =\left(\frac{2}{29}\right)\left(\frac{11}{29}\right)=-\left(\frac{11}{29}\right)=-\left(\frac{29}{11}\right) \\
& =-\left(\frac{7}{11}\right)=\left(\frac{11}{7}\right)=\left(\frac{4}{7}\right)=1
\end{aligned}$$
Therefore, the original ... | x \equiv \pm 14(\bmod 29) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,405 |
Example 4 Try to solve the congruence equation
$$x^{2} \equiv -4(\bmod 41)$$ | Since 41 is a prime number, it is easy to see that -1 and 4 are quadratic residues modulo 41, so -4 is also a quadratic residue modulo 41, which means the original congruence equation must have a solution.
Because $41=8 \times 5+1$, we have $u=5, \frac{u+1}{2}=3$, and
$$(-4)^{3}=-64 \equiv-23 \equiv 18(\bmod 41)$$
Si... | x \equiv \pm 18(\bmod 41) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,406 |
Example 5 Try to solve the congruence equation
$$x^{2} \equiv 34(\bmod 257)$$ | Since 257 is a prime number, we have
$$\begin{aligned}
\left(\frac{34}{257}\right) & =\left(\frac{2}{257}\right)\left(\frac{17}{257}\right)=\left(\frac{17}{257}\right) \\
& =\left(\frac{257}{17}\right)=\left(\frac{2}{17}\right)=1
\end{aligned}$$
Therefore, the original congruence equation must have a solution.
Since $... | x \equiv \pm 88(\bmod 257) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,407 |
Example 8 Try to solve the congruence equation
$$x^{2} \equiv 33(\bmod 128)$$ | We have $1^{2}-33=-32$. So 1 is a root modulo 32, therefore, $1+16=17$ is a root modulo 64, because
$$17^{2}-33=289-33=256,$$
Hence 17 is also a root modulo 128, and the other root modulo 128 is
$$17+64=81 \equiv-47(\bmod 128)$$
Thus, the roots of the original congruence equation are
$$x \equiv \pm 17, \pm 47(\bmod 1... | x \equiv \pm 17, \pm 47(\bmod 128) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,413 |
Example 9 Try to solve the congruence equation
$$x^{2} \equiv 105(\bmod 256)$$ | From $1^{2}-105=-104=-8 \times 13$ we know that 1 is not a root modulo 16, so 5 is a root modulo 16. Then, by
$$5^{2}-105=-80=-16 \times 5$$
we know that 5 is not a root modulo 32, hence $5+8=13$ is a root modulo 32. And by
$$13^{2}-105=169-105=64$$
we know that 13 is not only a root modulo 32 but also a root modulo ... | x \equiv \pm 19, \pm 109(\bmod 256) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,414 |
$$x^{2} \equiv a(\bmod m),(a, m)=1$$
has solutions, then it has
$2^{n}$ solutions, when $\alpha=0$ or 1,
$2^{n+1}$ solutions, when $\alpha=2$,
$2^{n+2}$ solutions, when $\alpha \geqslant 3$. | Given that we have assumed that (27) has a solution, the $n$ congruences following in (28) must all have solutions, and each has two solutions. If $x=0$, then the solutions to (28) can be found by the system of simultaneous congruences of the form
$$\left\{\begin{array}{l}
x \equiv r_{1}\left(\bmod p_{1}^{\alpha_{1}}\r... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,415 |
Example 10 Try to find the solution of the congruence equation
$$x^{2} \equiv 19(\bmod 45)$$ | It is obvious that
$x^{2} \equiv 19 \equiv 1(\bmod 9)$ has two roots $x \equiv \pm 1(\bmod 9)$, and $x^{2} \equiv 19 \equiv 4(\bmod 5)$ has two roots $x \equiv \pm 2(\bmod 5)$. From
$$x \equiv a(\bmod 9), x \equiv b(\bmod 5)$$
and the Chinese Remainder Theorem, we get
$$x \equiv 5 \times 2 a+9 \times 4 b=10 a+36 b(\bm... | x \equiv \pm 8, \pm 17(\bmod 45) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,416 |
Theorem 2 (Division Algorithm) Let $a, b \in \mathbf{Z}, a \neq 0$, then there exists a unique pair of integers $q$ and $r$ , satisfying:
$$b=a q+r$$
here $0 \leqslant r<|a|$. | Prove the existence first.
If $a \mid b$, then $r=0, q=\frac{b}{a}$ satisfies equation (1).
If $a \nmid b$, consider the set
$$T=\{b-a q \mid q \in \mathbf{Z}\}.$$
Since $T$ contains positive integers, let $T^{*}$ be the set of all positive integers in $T$. By the well-ordering principle, $T^{*}$ has a smallest elemen... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,419 |
Example 4 Proof: There exists $k \in \mathbf{N}^{*}$, such that for any $n \in \mathbf{N}^{*}$, the number $k \cdot 2^{n}+1$ is composite. | Proof
A basic idea is similar to finding a set of congruence covers. Here, we provide a method using Fermat numbers.
Notice that $F_{m}=2^{2^{m}}+1$ is prime for $m=0,1,2,3,4$, while $F_{5}$ is a multiple of 641, and $\frac{F_{5}}{641}$ is not a multiple of 641. We let $a_{m}=2^{2^{m}}+1, m=0,1,2,3,4$, and $a_{5}=641,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,421 |
Example 5 Let $m \in \mathbf{N}^{*}, n \in \mathbf{Z}$. Prove: The number $2 n$ can be expressed as the sum of two integers that are coprime with $m$.
| Proof
First, establish a lemma: For $m \in \mathbf{N}^{*}, n \in \mathbf{Z}$, there exist $a, b \in \mathbf{Z}$, such that
$$(a, m)=(b, m)=1$$
and $2 n \equiv a+b(\bmod m)$.
In fact, consider the case where $m=p^{\alpha}, p$ is a prime, $\alpha \in \mathbf{N}^{*}$. If $p=2$, then take $a=1, b=2 n-1$; if $p>2$, when $p... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,422 |
1. Prove: For any $n \in \mathbf{N}$, the number $3^{n}+2 \times 17^{n}$ is not a perfect square. | 1. It is only necessary to notice that $3^{n}+2 \times 17^{n} \equiv 2$ or $5(\bmod 8)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,424 |
2. Find the smallest positive integer $a$, such that there exists a positive odd integer $n$, satisfying
$$2001 \mid\left(55^{n}+a \cdot 32^{n}\right)$$ | 2. From $2001=3 \times 23 \times 29$ and the conditions, we have
$$\left\{\begin{array}{ll}
a \equiv 1 & (\bmod 3) \\
a \equiv 1 & (\bmod 29), \\
a \equiv-1 & (\bmod 23)
\end{array}\right.$$
From the first two equations, we can set $a=3 \times 29 \times k+1$, substituting into the last equation gives
$$k \equiv 5(\bmo... | 436 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,425 |
3. Find the smallest prime $p$ such that there do not exist $a, b \in \mathbf{N}$, satisfying
$$\left|3^{a}-2^{b}\right|=p$$ | 3. Notice that, $2=3^{1}-2^{0}, 3=2^{2}-3^{0}, 5=2^{3}-3^{1}, 7=2^{3}-3^{0}, 11=3^{3}-$ $2^{4}, 13=2^{4}-3^{1}, 17=3^{4}-2^{6}, 19=3^{3}-2^{3}, 23=3^{3}-2^{2}, 29=2^{5}-3^{1}, 31=$ $2^{5}-3^{0}, 37=2^{6}-3^{3}$. Therefore, the required $p \geqslant 41$.
On the other hand, if $\left|3^{a}-2^{b}\right|=41$, there are tw... | 41 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,426 |
4. The sequence $\left\{p_{n}\right\}$ is defined as follows: $p_{1}=2, p_{n+1}$ is the largest prime factor of the number $1+p_{1} p_{2} \cdots p_{n}$, $n=1,2, \cdots$. Question: Does 11 appear in $\left\{p_{n}\right\}$? | 4. Suppose there exists $n$, such that $p_{n+1}=11$. Direct calculation shows $p_{1}=2, p_{2}=3, p_{3}=7$, hence $n \geqslant 3$. For $n \geqslant 3$, the number $1+p_{1} p_{2} \cdots p_{n}$ is not divisible by $2,3,7$, thus it must be that
$$1+p_{1} p_{2} \cdots p_{n}=5^{a} \times 11^{\beta}$$
Taking both sides of (1... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,427 |
5. Prove: For any $m, n \in \mathbf{N}^{*}$, there exist odd numbers $a, b$, such that
$$2 m \equiv a^{20}+b^{11}\left(\bmod 2^{n}\right)$$ | 5. Notice that, for any odd numbers $x, y$,
$$x^{11}-y^{11}=(x-y)\left(x^{10}+x^{9} y+\cdots+y^{10}\right)$$
the second term on the right side is the sum of 11 odd numbers, which is odd. Therefore,
$$x^{11} \equiv y^{11}\left(\bmod 2^{n}\right) \Leftrightarrow x \equiv y\left(\bmod 2^{n}\right)$$
This indicates
$$\le... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,428 |
6. Given integers $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ whose sum is 1, the sequence $\left\{b_{m}\right\}$ is defined as follows:
$$b_{m}=a_{m}+2 a_{m+1}+\cdots+(n-m+1) a_{n}+(n-m+2) a_{1}+\cdots+n a_{m-1},$$
where $m=1,2, \cdots, n$. Prove: $b_{1}, b_{2}, \cdots, b_{n}$ form a complete residue system modulo $... | 6. Notice that, for $1 \leqslant m \leqslant n-1$, we have
$$\begin{aligned}
b_{m+1}-b_{m}= & \left(a_{m+1}+2 a_{m+2}+\cdots+(n-m) a_{n}+(n-m+1) a_{1}\right. \\
& \left.+\cdots+n a_{m}\right)-\left(a_{m}+2 a_{m+1}+\cdots+(n-m+1) a_{n}\right. \\
& \left.+(n-m+2) a_{1}+\cdots+n a_{m-1}\right) \\
= & -a_{1}-\cdots-a_{m-1}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,429 |
7. A roundtable meeting has 2008 participants. After a break, they sit around the table again in a different order. Prove: There are at least two people such that the number of people between them is the same before and after the break. | 7. Let $n=1004$, and we label each seat. The seat numbers are marked in a clockwise direction as
$$1,2,3, \cdots, 2 n$$
Thus, each person can be represented by a pair $(i, j)$, where $i$ and $j$ are their seat numbers before and after the break, respectively. Clearly, all "x-coordinates" $i$ and "y-coordinates" $j$ co... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 741,430 |
Theorem 2 There are infinitely many prime numbers. | Prove that if there are only finitely many primes, let them be \( p_{1}, p_{2}, \cdots, p_{n} \). Consider the number \( m = p_{1} p_{2} \cdots p_{n} + 1 \). Take a prime factor \( q \) of \( m \), by the assumption, \( q \) must be in \( \{ p_{1}, p_{2}, \cdots, p_{n} \} \), which implies \( q \mid (m - p_{1} p_{2} \c... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,431 |
8. Let $n \in \mathbf{N}^{\cdot}, n \geqslant 2$. The array of positive integers $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ satisfies:
$$a_{1}+a_{2}+\cdots+a_{n}=2 n$$
If it is not possible to divide $a_{1}, a_{2}, \cdots, a_{n}$ into two groups with equal sums, then $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ is calle... | 8. Let $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ be a "good" array. For $1 \leqslant i \leqslant n$, consider the following $n+1$ numbers:
$$a_{i}, a_{i+1}, a_{i}+a_{i+1}, \cdots, a_{i}+a_{i+1}+\cdots+a_{i+n-1},$$
where $a_{n+j}=a_{j}$.
Since $a_{1}, a_{2}, \cdots, a_{n}$ cannot be divided into two groups with equal ... | (1,1, \cdots, 1, n+1) \text{ or } (2,2, \cdots, 2) \text{ when } n \text{ is odd}; (1,1, \cdots, 1, n+1) \text{ when } n \text{ is even} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,432 |
9. Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive integers, whose sum is $2 n$. Define $a_{n+i}=a_{i}$, $i=1,2, \cdots$. For $u, v \in \mathbf{N}^{*}$, let
$$S_{u, v}=a_{u}+a_{u+1}+\cdots+a_{u+v-1} .$$
Prove: For any $m \in \mathbf{N}^{*}$, there exist $u, v \in \mathbf{N}^{*}$, such that
$$S_{u, v} \in\{m, m+1\} .... | 9. From the definition of $\left\{a_{j}\right\}$ and $a_{1}+a_{2}+\cdots+a_{n}=2 n$, we know that $S_{u, v+n}=S_{u, v}+2 n$. Therefore, it suffices to prove: for $m \in\{1,2, \cdots, 2 n\}$, the proposition holds.
Let $b_{i}=a_{1}+a_{2}+\cdots+a_{i}, 1 \leqslant i \leqslant n$. If the $2 n$ numbers $b_{1}, b_{2}, \cdo... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,433 |
10. Connecting the vertices of a regular $n$-sided polygon, we obtain a closed broken line with $n$ edges (each vertex of the regular $n$-sided polygon is exactly the endpoint of two edges of this closed broken line). Prove: if $n$ is even, then there must be two parallel edges in this closed broken line; if $n$ is odd... | 10. Label the vertices of a regular $n$-sided polygon in counterclockwise order as $0,1,2, \cdots, n-1$. The closed broken line can be uniquely represented by a permutation of these $n$ numbers $a_{0}=a_{n}, a_{1}, a_{2}, \cdots, a_{n-1}$ as $a_{0} a_{1}, a_{1} a_{2}, \cdots, a_{n-1} a_{0}$. Clearly,
$$a_{i} a_{i+1} / ... | proof | Geometry | proof | Yes | Yes | number_theory | false | 741,434 |
11. Let $n \in \mathbf{N}^{*}, n \geqslant 3$, and the set $M=\{1,2, \cdots, n-1\}$. Color the numbers in $M$ according to the following rules:
(1) For $1 \leqslant i \leqslant n-1$, the numbers $i$ and $n-i$ are the same color;
(2) There exists $k \in M, (k, n)=1$, such that for any $i \in M, i \neq k$, the numbers $i... | 11. Since $0,1,2, \cdots, n-1$ form a complete residue system modulo $n$, and $(k, n)=1$, it follows that $0, k, 2k, \cdots, (n-1)k$ is also a complete residue system modulo $n$. If we denote $a_{i} \equiv i k \pmod{n}, 0 \leqslant a_{i}<n$, then $\left\{a_{0}, a_{1}, \cdots, a_{n-1}\right\}=\{0,1,2, \cdots, n-1\}$. Th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,435 |
12. Let $n$ be an odd number greater than 1. Prove: There exist $2 n$ integers $a_{1}, a_{2}, \cdots, a_{n}$; $b_{1}, b_{2}, \cdots, b_{n}$, such that for any $k \in\{1,2, \cdots, n-1\}$, the following $3 n$ numbers
$$a_{i}+a_{i+1} ; a_{i}+b_{i} ; b_{i}+b_{i+k} . \quad i=1,2, \cdots, n .$$
form a complete residue syst... | 12. Let $a_{i}=3 i-2, b_{i}=3 i-1, i=1,2, \cdots, n$. Then for any $k \in \mathbf{N}^{*}, k<n$, we have
$$a_{i}+a_{i+1} \equiv 2(\bmod 3), a_{i}+b_{i} \equiv 0(\bmod 3), b_{i}+b_{i+k} \equiv 1(\bmod 3)$$
Thus, when we denote
$$A=\left\{a_{i}+a_{i+1} \mid 1 \leqslant i \leqslant n\right\}, B=\left\{a_{i}+b_{i} \mid 1 \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,436 |
13. Let $p$ be a prime. Prove: There exist infinitely many $n \in \mathbf{N}^{*}$, such that
$$2^{n} \equiv n(\bmod p)$$ | 13. When $p=2$, take $n$ as an even number; when $p>2$, $2^{p-1} \equiv 1(\bmod p)$, at this time let $n=(k p-1)(p-1), k \in \mathbf{N}^{*}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,437 |
14. For $n \in \mathbf{N}^{*}$, if for any $a \in \mathbf{N}^{*}$, whenever $n \mid (a^{n}-1)$, it follows that
$$n^{2} \mid (a^{n}-1)$$
then $n$ is said to have property $P$.
(1) Prove that every prime number has property $P$;
(2) Does there exist infinitely many composite numbers that have property $P$? | 14. (1) For any prime $p$, from $p \mid (a^{p}-1)$, we know $(a, p)=1$, thus by Fermat's Little Theorem, we have $a^{p-1} \equiv 1 \pmod{p}$. Therefore, $a^{p} \equiv a^{p-1} \pmod{p}$, which implies $a \equiv 1 \pmod{p}$. At this point,
$$a^{p-1}+a^{p-2}+\cdots+1 \equiv 1+1+\cdots+1=p \equiv 0 \pmod{p},$$
Thus, $p^{2... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,438 |
15. Let $p$ be an odd prime, $a, n \in \mathbf{N}^{*}$, satisfying: $p^{n} \mid\left(a^{p}-1\right)$.
(1) Prove: $p^{n-1} \mid(a-1)$;
(2) When $p=2$, does the above conclusion hold? | 15. (1) From the condition, $p^{n} \mid\left(a^{p}-1\right)$, thus $p \mid\left(a^{p}-1\right)$, so $(a, p)=1$. By Fermat's Little Theorem, we know that $a^{p-1} \equiv 1(\bmod p)$, hence
$$a^{p-1} \equiv 1 \equiv a^{p}(\bmod p)$$
Therefore, $a \equiv 1(\bmod p)$.
On the other hand, let $A=a^{p-1}+a^{p-2}+\cdots+1$, t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,439 |
16. The sequence of positive integers $\left\{a_{n}\right\}$ satisfies:
$$a_{1}>1, a_{n+1} \in\left\{2 a_{n}-1,2 a_{n}+1\right\}, n=1,2, \cdots .$$
Prove: $\left\{a_{n}\right\}$ contains infinitely many composite numbers. | 16. If $\left\{a_{n}\right\}$ contains at most a finite number of composite numbers, then there exists $n_{0} \in \mathbf{N}^{*}$, such that when $n \geqslant n_{0}$, $a_{n}$ is always a prime. Suppose $a_{n_{0}}=q$ is a prime greater than 3, if $q \equiv 1(\bmod 3)$, then $2 a_{n_{0}}+1 \equiv 0(\bmod 3)$, so $a_{n_{0... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,440 |
17. If $m, n \in \mathbf{N}^{*},(m, n)=1$, prove:
$$\varphi(m n)=\varphi(m) \varphi(n)$$ | 17. For $m, n \in \mathbf{N}^{*}, (m, n)=1$, we prove: When $x$ runs through a reduced residue system modulo $m$, and $y$ runs through a reduced residue system modulo $n$, the number $n x + m y$ runs through a reduced residue system modulo $m n$. From this, we can obtain $\varphi(m n) = \varphi(m) \varphi(n)$.
First, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,441 |
Example 1 Let $n \in \mathbf{N}^{*}$, prove the following conclusions:
(1) If $2^{n}+1$ is a prime number, then there exists $k \in \mathbf{N}$, such that $n=2^{k}$;
(2) If $2^{n}-1$ is a prime number, then $n$ is a prime number. | Proof
(1) If $n$ has an odd prime factor $p$, let $n=p m, p=2 k+1, k \in \mathbf{N}^{*}$, then
$$\begin{aligned}
2^{n}+1 & =\left(2^{m}\right)^{p}+1=\left(2^{m}\right)^{2 k+1}+1 \\
& =\left(2^{m}+1\right)\left(\left(2^{m}\right)^{2 k}-\left(2^{m}\right)^{2 k-1}+\cdots-2^{m}+1\right)
\end{aligned}$$
is the product of t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,442 |
18. Prove: The set $\left\{2^{n}-3 \mid n=1,2,3, \cdots\right\}$ has an infinite subset $X$, such that any two numbers in $X$ are coprime. | 18. We provide a subsequence defined in a recursive form, where any two terms are coprime. Let the sequence $\left\{2^{n}-3\right\}$ have $k$ terms that are pairwise coprime, denoted as $u_{1}, u_{2}, \cdots, u_{k}$. Define
$$u_{k+1}=2^{q\left(u_{1}, u_{2} \cdots u_{k}\right)+1}-3,$$
where $\varphi(x)$ is the Euler's ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,443 |
19. Let $a \in \mathbf{N}^{*}, a>1$. Prove: The set $\left\{a^{n}+a^{n-1}-1 \mid n=2,3, \cdots\right\}$ has an infinite subset $X$, such that any two numbers in $X$ are coprime. | 19. Let $I=\left\{a^{n}+a^{n-1}-1 \mid n=2,3, \cdots\right\}$, take $x_{1}=a^{2}+a-1$, and assume $x_{1}, x_{2}, \cdots, x_{n}$ are already determined, where $x_{1}, x_{2}, \cdots, x_{n} \in I$ and are pairwise coprime. Let $N=$ $x_{1} x_{2} \cdots x_{n}$. Since for any $m \in \mathbf{N}^{*},\left(a, a^{m}+a^{m-1}-1\ri... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,444 |
21. Let $p$ be an odd prime. Prove:
(1) $1^{2} \cdot 3^{2} \cdot \cdots \cdot(p-2)^{2} \equiv(-1)^{\frac{p+1}{2}}(\bmod p)$;
(2) $2^{2} \cdot 4^{2} \cdot \cdots \cdot(p-1)^{2} \equiv(-1)^{\frac{p+1}{2}}(\bmod p)$. | 21. Since when $i=2,4, \cdots, p-1$, we have
$$i \equiv -(p-i) \pmod{p}$$
Combining with Wilson's theorem, we get
$$2^{2} \cdot 4^{2} \cdot \cdots \cdot (p-1)^{2} \equiv (-1)^{\frac{p-1}{2}}(p-1)! \equiv (-1)^{\frac{p+1}{2}} \pmod{p}.$$
When $i=1,3,5, \cdots, p-2$, similarly,
$$1^{2} \cdot 3^{2} \cdot \cdots \cdot (p... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,446 |
22. Let $n$ be an integer greater than 2, and let the integers $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be complete residue systems modulo $n$. Prove that $a_{1} b_{1}, a_{2} b_{2}, \cdots, a_{n} b_{n}$ is not a complete residue system modulo $n$. | 22. If $4 \mid n$, and $a_{1} b_{1}, a_{2} b_{2}, \cdots, a_{n} b_{n}$ is a complete residue system modulo $n$, then among $a_{1} b_{1}, a_{2} b_{2}, \cdots, a_{n} b_{n}$, there are exactly $\frac{n}{2}$ odd numbers. Without loss of generality, let $a_{1} b_{1}, a_{2} b_{2}, \cdots, a_{\frac{n}{2}} b_{\frac{n}{2}}$ be ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,447 |
23. Let $m, k \in \mathbf{N}^{*},(m, k)=1$. Prove: There exist integers $a_{1}, a_{2}, \cdots, a_{m}$ and $b_{1}$, $b_{2}, \cdots, b_{k}$, such that all the products $a_{i} b_{j}(1 \leqslant i \leqslant m, 1 \leqslant j \leqslant k)$ exactly form a complete residue system modulo $m k$. | 23. Let $a_{i}=i k+1, b_{j}=j m+1$, where $i=1,2, \cdots, m ; j=1,2, \cdots, k$. If $a_{i} b_{j}=a_{i} b_{j^{\prime}}(\bmod m k)$, then $i k+m j=i^{\prime} k+j^{\prime} m(\bmod m k)$, which means $k\left(i-i^{\prime}\right) \equiv m\left(j^{\prime}-j\right)(\bmod m k)$.
Since $(m, k)=1$, it follows that $k \mid\left(j^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,448 |
24. Let $p$ be a prime, $m, n \in \mathbf{N}^{*}$, satisfying: for any $k \in \mathbf{N}^{*}$, we have
$$(p k-1, m)=(p k-1, n)$$
Prove: There exists some $l \in \mathbf{Z}$, such that $m=\boldsymbol{p}^{l} \cdot n$. | 24. Let $m=p^{\alpha} \cdot x, n=p^{\beta} \cdot y$, where $\alpha, \beta \in \mathbf{N}, x, y \in \mathbf{N}^{*}$, and $x, y$ are not multiples of $p$. We only need to prove that $x=y$.
If $x>y$ (the case $x<y$ is symmetric), then by $(x, p)=1$, combined with the Chinese Remainder Theorem, there exists $a \in \mathbf... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,449 |
25. For $n \in \mathbf{N}^{*}$, let $f(n)$ denote the smallest positive integer such that: $n \mid \sum_{k=1}^{f(n)} k$. Find all $n \in \mathbf{N}^{*}$ such that $f(n)=2 n-1$. | 25. First, we prove: If \( n=2^{m}, m \in \mathbf{N} \), then \( f(n)=2 n-1 \).
In fact, on one hand,
\[
\sum_{k=1}^{2 n-1} k=(2 n-1) n=\left(2^{m+1}-1\right) 2^{m}
\]
is divisible by \( n \). On the other hand, if \( l \leqslant 2 n-2 \), then
\[
\sum_{k=1}^{l} k=\frac{1}{2} l(l+1)
\]
Since one of \( l \) and \( l+1 ... | n = 2^m, m \in \mathbf{N} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,450 |
26. Let $n \in \mathbf{N}^{*}$. Prove: there exists $m \in \mathbf{N}^{*}$, such that the congruence equation
$$x^{2} \equiv 1(\bmod m)$$
has at least $n$ roots modulo $m$. | 26. Notice that, for any odd prime $p$, the congruence equation $x^{2} \equiv 1(\bmod p)$ has exactly two distinct solutions $x \equiv 1$ or $p-1(\bmod p)$.
Now, take $s$ different odd primes, where $s$ is a positive integer satisfying $2^{s} \geqslant n$, and let $m$ $=p_{1} p_{2} \cdots p_{s}$ be the product of thes... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,451 |
27. For integer points $(x, y)$ on the plane, if $(x, y)=1$, then it is called "reduced". Prove: For any $n \in \mathbf{N}^{*}$, there exists an integer point on the plane such that its distance to any reduced integer point is greater than $n$. | 27. Let $i, j \in \mathbf{Z}, -n \leqslant i, j \leqslant n$, then $p_{ij}$ represents $(2n+1)^2$ different primes. By the Chinese Remainder Theorem, there exist $a, b$ that satisfy the congruence systems
$a \equiv i \pmod{p_{ij}}, -n \leqslant i, j \leqslant n, b \equiv j \pmod{p_{ij}}, -n \leqslant i, i \leqslant n$,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,452 |
Example 2 uses $P^{*}$ to denote the set of all odd prime numbers less than 10000. Let $p$ be a number in $P^{*}$, satisfying: for any subset $S=\left\{p_{1}, p_{2}, \cdots, p_{k}\right\}$ of $P^{*}$ that does not contain $p$ and where $k \geqslant 2$, there exists $q \in \mathbf{P}^{*} \backslash S$ such that
$$(q+1) ... | A basic idea is to minimize the number of distinct prime factors in the numerator of (1), the simplest case being to make each $p_{i}+1$ a power of 2. Therefore, we examine the Mersenne primes in $P^{*}$.
Let $T=\left\{M_{2}, M_{3}, M_{5}, M_{7}, M_{13}\right\}=\{3,7,31,127,8191\}$ (note that $M_{11}=$ $23 \times 89$ ... | \{3,7,31,127,8191\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,453 |
28. Let $a$ be an integer, and $n, r$ be integers greater than 1. $p$ is an odd prime, and $(n, p-1)=1$. Find the number of solutions to the following congruence equation:
$$x_{1}^{n}+x_{2}^{n}+\cdots+x_{r}^{n} \equiv a(\bmod p)$$
Here, the solutions $\left(x_{1}, x_{2}, \cdots, x_{r}\right)$ and $\left(x_{1}^{\prime}... | 28. First, prove: For any $b \in \mathbf{Z}$, the congruence equation $x^{n} \equiv b(\bmod p)$ has a unique solution.
In fact, by $(n, p-1)=1$ and Bézout's theorem, there exist $u, v \in \mathbf{N}^{*}$ such that $n u-(p-1) v=1$. If $b \equiv 0(\bmod p)$, then the equation $x^{n} \equiv b(\bmod p)$ has only the uniqu... | p^{r-1} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,454 |
29. Let $p$ be a prime, $a, b \in \mathbf{N}^{*}$, satisfying: $p>a>b>1$. Find the largest integer $c$, such that for all $(p, a, b)$ satisfying the conditions, we have
$$p^{c} \mid\left(\mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}\right)$$ | 29. When taking $p=5, a=3, b=2$, we should have $5^{c} \mid 3000$, so $c \leqslant 3$. Below, we prove that for any $p, a, b$ satisfying the conditions, we have $p^{3} \mid \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}$.
In fact, notice that
$$\begin{aligned}
& \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}=\frac{(a p)(a p-1) ... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,455 |
31. Find all positive integers $n$, such that
$$\frac{2^{n}-1}{3} \in \mathbf{N}^{*}$$
and there exists $m \in \mathbf{N}^{*}$, satisfying ${ }^{*}$ :
$$\left.\frac{2^{n}-1}{3} \right\rvert\,\left(4 m^{2}+1\right)$$ | 31. From $\frac{2^{n}-1}{3} \in \mathbf{N}^{\cdot}$, we know that $n$ is even. If there exists an odd number $q \geqslant 3$ such that $q \mid n$, then using factorization we know that $\left(2^{q}-1\right) \mid\left(2^{n}-1\right)$. Combining $2^{q}-1 \equiv-2(\bmod 3)$, we know that $\left(2^{q}-1\right) \left\lvert\... | n = 2^k, k \in \mathbf{N}^{*} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,457 |
32. Let $p>5, p$ be a prime number. The function
$$f_{p}(x)=\sum_{k=1}^{p-1} \frac{1}{(p x+k)^{2}}$$
Prove: For any $x, y \in \mathbf{N}^{*}$, when $f_{p}(x)-f_{p}(y)$ is written as a simplest fraction, its numerator is a multiple of $p^{3}$. | 32. Introduce the congruence notation: Use $\frac{b}{a}(\bmod n)$ to denote $b a^{-1}(\bmod n)$, where $a^{-1}$ is the modular inverse of $a$ modulo $n$ (note that this requires $(a, n)=1$).
Notice that, under the above fractional congruence notation, we only need to prove: $f_{p}(x)-f_{p}(y) \equiv 0\left(\bmod p^{3}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,458 |
If $a^{n} \equiv 1(\bmod m), n \in \mathbf{N}^{*}$, then $\delta_{m}(a) \mid n$. | Proof: Let $n=\delta_{m}(a) q+r, 0 \leqslant r < \delta_{m}(a)$, then
$$a^{r} \equiv a^{r}\left(a^{\delta_{m}(a)}\right)^{q}=a^{n} \equiv 1(\bmod m)$$
This contradicts the minimality of $\delta_{m}(a)$. Therefore, $r=0$, i.e., $\delta_{m}(a) \mid n$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,459 |
Example 1 Let $a$ be a positive integer greater than 1, $n \in \mathbf{N}^{*}$. Prove: $n \mid \varphi\left(a^{n}-1\right)$. | Notice that, $n$ is the smallest positive integer satisfying the following congruence
$$a^{n} \equiv 1\left(\bmod a^{n}-1\right)$$
because $a>1$, i.e.,
$$\delta_{a^{n}-1}(a)=n \text {. }$$
By Euler's theorem, we know
$$a^{\varphi\left(a^{n}-1\right)} \equiv 1\left(\bmod a^{n}-1\right),$$
thus, combining the properti... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,460 |
Example 2 Let $p$ be an odd prime, and $q$ a prime factor of $2^{p}-1$. Prove: $q \equiv 1(\bmod 2 p)$.
The text is translated while preserving the original line breaks and format. | Prove that from $2^{p} \equiv 1(\bmod q)$, we know $\delta_{q}(2) \mid p$, and combining with $p$ being a prime, it can only be that $\delta_{q}(2) = p$.
Then, by Fermat's Little Theorem, we know $2^{q-1} \equiv 1(\bmod q)$, thus $p \mid (q-1)$, i.e.,
$$q \equiv 1(\bmod p)$$
Using the fact that $p, q$ are both odd pr... | q \equiv 1(\bmod 2 p) | Number Theory | proof | Yes | Yes | number_theory | false | 741,461 |
Example 3 Let $q$ be a prime factor of the Fermat number $F_{n}=2^{2^{n}}+1$. Prove: when $n>1$, we have $q \equiv 1\left(\bmod 2^{n+2}\right)$. | Proof
First, we prove a lemma: Let $a$ be a positive integer greater than 1, and $q$ be an odd prime factor of the number $a^{2^{n}}+1$, then
$$q \equiv 1\left(\bmod 2^{n+1}\right)$$
In fact, from $a^{2^{n}} \equiv-1(\bmod q)$, we know
$$a^{2^{n+1}} \equiv 1(\bmod q),$$
This indicates that $\delta_{q}(a) \nmid 2^{n}$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,462 |
Example 4 Find all positive integers $n$, such that $2^{n} \equiv 1(\bmod n)$.
untranslated text remains unchanged. | When $n=1$, it is obviously true.
Below is the proof: When $n>1$, we always have
$$2^{n} \not \equiv 1(\bmod n)$$
In fact, if there exists $n>1$, such that
$$2^{n} \equiv 1(\bmod n),$$
take the smallest prime factor $p$ of $n$, then $2^{n} \equiv 1(\bmod p)$. By Fermat's Little Theorem, we know $2^{p-1} \equiv$ $1(\b... | n=1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,463 |
Example 3 If $p$ and $p+2$ are both prime numbers, then these two prime numbers are called "twin primes". Consider the following two sequences.
Fibonacci sequence: $1,1,2,3,5,8, \cdots$ (the sequence satisfying $F_{1}=1, F_{2}=1$, $F_{n+2}=F_{n+1}+F_{n}, n=1,2, \cdots$).
Twin prime sequence: $3,5,7,11,13,17,19, \cdot... | This is a problem related to the properties of the Fibonacci sequence. A certain term $F_{n}$ appears in the twin prime sequence if and only if $F_{n}-2$ and $F_{n}$ are both prime, or $F_{n}$ and $F_{n}+2$ are both prime. Therefore, to negate that $F_{n}$ appears in the twin prime sequence, we need to prove that $F_{n... | 3,5,13 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,464 |
Example 5 Let $p$ be a prime. Prove: there exists a prime $q$, such that for any $n \in \mathbf{Z}$, we have $q \nmid\left(n^{p}-p\right)$. | Proof
Notice
$$\frac{p^{p}-1}{p-1}=1+p+\cdots+p^{p-1} \equiv 1+p\left(\bmod p^{2}\right)$$
Thus, the number $\frac{p^{p}-1}{p-1}$ has a prime factor $q$, satisfying:
$$q \not \equiv 1\left(\bmod p^{2}\right)$$
We will prove that this $q$ is a prime factor that meets the requirement.
Indeed, if there exists $n \in \ma... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,465 |
Example 6 calls a positive integer $n$ a "good number" if it satisfies $\sigma(n)=3n$, where $\sigma(n)$ denotes the sum of all positive divisors of the positive integer $n$. Prove: there is no "good number" whose square root is a square-free number. | Proof:
If there exists an $n \in \mathbf{N}^{*}, n=p_{1}^{2} p_{2}^{2} \cdots p_{k}^{2}, p_{1}<p_{2}<\cdots<p_{k}$, and $p_{i} \equiv 2(\bmod 3), 1 \leq i \leq k$. Let $T=\left\{p_{1}, p_{2}, \cdots, p_{k}\right\}$. Then,
$$
\prod_{p \in T}\left(p^{2}+p+1\right)=3 \prod_{p \in T} p^{2} \quad(2)
$$
Notice that when $p ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,466 |
Example 7 Does there exist positive integers $a, b, c$ that satisfy the following three conditions?
(1) $a, b, c$ are pairwise coprime;
(2) $a, b, c$ are all greater than 1;
(3) $a\left|\left(2^{b}+1\right), b\right|\left(2^{c}+1\right), c \mid\left(2^{a}+1\right)$. | There do not exist positive integers $a, b, c$ that satisfy the conditions.
Let $\pi(n)$ denote the smallest prime factor of the positive integer $n$. First, we establish the following lemma: If $p$ is a prime, $p \mid (2^y + 1)$, and $p < \pi(y)$, then $p = 3$.
In fact, from $p \mid (2^y + 1)$, we know that $p$ is an... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,467 |
Example 8 Let $p$ be a prime, $m, n$ be positive integers greater than 1, and
$$n \mid\left(m^{\rho(n-1)}-1\right) \text {. }$$
Prove: $\left(m^{n-1}-1, n\right)>1$. | Proof by contradiction. Suppose $\left(m^{n-1}-1, n\right)=1$, and let $n=p_{1}^{Q_{1}} p_{2}^{a_{2}} \cdots p_{k}^{q_{k}}$, where $p_{1}<p_{2}<\cdots<p_{k}$ are primes, $\alpha_{i} \in \mathbf{N}^{*}, 1 \leqslant i \leqslant k$. Furthermore, let $p^{\beta} \mid(n-1)$, but $p^{\beta+1} \nmid(n-1)$, where $\beta \in \ma... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,468 |
Property 1 Let $m \in \mathbf{N}^{*}, a, b \in \mathbf{Z},(a, m)=(b, m)=1$, then $\delta_{m}(a b)=\delta_{m}(a) \delta_{m}(b)$ if and only if $\left(\delta_{m}(a), \delta_{m}(b)\right)=1$. | To prove the necessity: From $a^{\delta_{m}{ }^{(a)}} \equiv 1(\bmod m)$ and $b^{\delta}{ }_{m}^{(b)} \equiv 1(\bmod m)$, we know that $(a b)^{\left[\delta_{m}(a), \delta_{m}(b)\right]} \equiv 1(\bmod m)$,
thus, $\delta_{m}(a b) \mid\left[\delta_{m}(a), \delta_{m}(b)\right]$. Since $\delta_{m}(a b)=\delta_{m}(a) \delta... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,469 |
Theorem 1 Let $m \notin\{1,2,4\}$, and there does not exist an odd prime $p$ and $\alpha \in \mathbf{N}^{*}$, such that $m \in$ $\left\{p^{a}, 2 p^{a}\right\}$. Then for any $a \in \mathbf{Z},(a, m)=1$, we have $\delta_{m}(a)<\varphi(m)$, in this case the primitive root modulo $m$ does not exist. | Prove that if $m=2^{\alpha}, \alpha \in \mathbf{N}^{*}, \alpha \geqslant 3$, then for any odd number $a$, set $a=2 k+1$, then
$$\begin{aligned}
a^{2^{a-2}} & =(2 k+1)^{2^{a^{-2}}} \equiv 1+2^{a-2} \cdot(2 k)+\mathrm{C}_{2^{a-2}}^{2}(2 k)^{2} \\
& =1+2^{a-1} \cdot k+2^{a-1} \cdot\left(2^{a-2}-1\right) k^{2} \\
& =1+2^{a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,472 |
Theorem 2 Let $p$ be an odd prime, then a primitive root modulo $p$ exists. | Proof: First, establish a lemma: Let $a, b$ be two integers coprime with $p$, then there exists $c \in \mathbf{Z}$, such that $\delta_{p}(c)=\left[\delta_{p}(a), \delta_{p}(b)\right]$.
In fact, let $\delta_{p}(a)=r, \delta_{p}(b)=t$, and set $r=d x$, where $d=(r, t)$, then $\left[\delta_{p}(a), \delta_{p}(b)\right]=x ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,473 |
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