problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
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10. Let $a, n \in \mathbf{N}^{*}, a>1$ and $a^{n}+1$ be a prime. Prove: $d\left(a^{n}-1\right) \geqslant n$. | 10. If $a^{n}+1$ is a prime number and $a>1$, then $a$ is even, and similarly to Example 1 in Section 1.4, we can prove: $n=2^{k}, k \in \mathbf{N}$. Therefore,
$$a^{n}-1=a^{2^{k}}-1=(a-1)(a+1)\left(a^{2}+1\right) \cdot \cdots \cdot\left(a^{2^{2-1}}+1\right) \text {. }$$
Furthermore, since $\left(a^{2^{m}}+1, a^{2^{m}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,695 |
11. Let $n \in \mathbf{N}^{*}, n>1$, and suppose that the numbers $1!, 2!, \cdots, n!$ yield distinct remainders when divided by $n$. Prove: $n$ is a prime number. | 11. If $n$ is a composite number, then we can set $n=p^{2}$ or $n=qr$, where $p$ is a prime number, and $2 \leqslant q<r$, $q, r \in \mathbf{N}^{*}$. For the former, if $p=2$, then by $2!\equiv 3!(\bmod 4)$, we get a contradiction. If $p \geqslant 3$, then by $(3 p)!\equiv(2 p)!\equiv 0\left(\bmod p^{2}\right)$, we get... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,696 |
12. Let $a_{n}$ denote the sum of the first $n$ prime numbers, i.e., $a_{1}=2, a_{2}=2+3, \cdots$. Prove: there is at least one perfect square in the interval $\left[a_{n}, a_{n+1}\right], n \in \mathbf{N}^{*}$. | 12. When $n \leqslant 4$, it can be directly verified that the proposition holds. When $n \geqslant 5$, let $p_{1}, p_{2}$, $p_{3}, \cdots$ represent the sequence of all prime numbers in ascending order, and set $p_{n}=2 m+1$. It is known that
$$p_{1}+p_{2}+\cdots+p_{n}<1+3+5+7+\cdots+(2 m+1)=(m+1)^{2},$$
i.e., $a_{n}<... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,697 |
13. Let $n \in \mathbf{N}^{*}, n>1, p$ be a prime number. Given
$$n|(p-1), p|\left(n^{3}-1\right)$$
Prove: $4 p-3$ is a perfect square. | 13. From $n \mid(p-1)$ we know $nn$, and from $p \mid n(n+1-$ $m)$, we can get $n+1-m=0$. Therefore, $p=n^{2}+n+1$, and consequently $4 p-3=(2 n+1)^{2}$. | 4 p-3=(2 n+1)^{2} | Number Theory | proof | Yes | Yes | number_theory | false | 741,698 |
14. Let $A$ be a 1000-digit number, and it is known that any 10 consecutive digits of $A$ form a number that is a multiple of $2^{10}$. Prove: $2^{1000} \mid A$. | 14. Let $A=\overline{a_{1} a_{2} \cdots a_{1000}}$. From the conditions, we know that $2^{10} \mid \overline{a_{990} a_{991} \cdots a_{999}}$ and $2^{10} \mid$ $\overline{a_{991} a_{992} \cdots a_{1000}}$. Let $x=\overline{a_{991} a_{992} \cdots a_{999}}$, then $2^{10}\left|\left(a_{990} \times 10^{9}+x\right), 2^{10}\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,699 |
Example 2 Let $k \in \mathbf{N}^{*}, k \geqslant 2$, and $n$ is a positive integer no less than $2 k$.
(1) Prove: There exists an integer $i \in\{0,1,2, \cdots, k-1\}$, such that $(n-i) \times C_{n}^{k}$;
(2) Prove: For each $k \geqslant 2$, there exists a positive integer $n_{k} \geqslant 2 k$, such that there is exac... | Proof (1) Proof by contradiction, if not, suppose there exists $k \geqslant 2$ and $n \geqslant 2k$, such that $n, n-1, \cdots, n-(k-1)$ are all divisors of the combination number $\mathrm{C}_{n}^{k}$.
From $n(n-1) \cdot \cdots \cdot(n-(k-1))=k!C_{n}^{k}$, we have
$$\left\{\begin{array}{l}
(n-1)(n-2) \cdot \cdots \cdo... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,700 |
15. Prove: In decimal representation, the number of positive divisors of each positive integer that end in the digit 1 or 9 is not less than the number of positive divisors that end in the digit 3 or 7. | 15. Let $d_{1}(m), d_{3}(m), d_{7}(m), d_{9}(m)$ denote the number of positive divisors of $m$ whose last digit is $1, 3, 7, 9$ respectively. We need to prove that for any $m \in \mathbf{N}^{*}$, we have
$$d_{1}(m)-d_{3}(m)-d_{7}(m)+d_{9}(m) \geqslant 0 .$$
When $m=1$, (1) is obviously true.
Assume (1) holds for all n... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,701 |
16. Given $a, b \in \mathbf{N}^{*}$, and in decimal notation, the number $a^{a} \cdot b^{b}$ ends with exactly 98 zeros. Find all pairs $(a, b)$ that minimize $ab$.
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。
Note: The last sentence is a note and not part of the problem statement, so it is not translated. | 16. Let the powers of 2 and 5 in the prime factorizations of $a$ and $b$ be $\alpha_{1}, \beta_{1}$ and $\alpha_{2}, \beta_{2}$, respectively. Then, by the given condition, one of the equations $a \cdot \alpha_{1} + b \cdot \alpha_{2} = 98$ or $a \cdot \beta_{1} + b \cdot \beta_{2} = 98$ must hold. If $a \cdot \beta_{1... | (98, 75) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,702 |
17. Prove: Any positive integer can be expressed as the difference of two positive integers $a, b$ and the number of distinct prime factors of $a$ and $b$ is the same. | 17. For $n \in \mathbf{N}^{*}$, if $n$ is even, then take $a=2n, b=n$; if $n$ is odd, take the smallest odd prime $p$ that does not divide $n$, and let $a=p n, b=(p-1) n$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,703 |
18. Let $m, n \in \mathbf{N}^{*}$, and $m \leqslant \frac{n^{2}}{4}, m$ every prime factor of $m$ is not greater than $n$. Prove: $m \mid n!$. | 18. It suffices to prove: For any prime $p$ and positive integer $k$, if $p^{k} \mid m$, then $p^{k} \mid n!$.
When $k=1$, since the prime factor $p$ of $m \leqslant n$, it follows that $p \mid n!$.
When $k>1$, by $m \leqslant \frac{n^{2}}{4}$, we have $p^{k} \leqslant \frac{n^{2}}{4}$, i.e., $n \geqslant 2 \sqrt{p^{k... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,704 |
19. Given a positive integer $n \geqslant 2$. Let $d_{1}, d_{2}, \cdots, d_{n} \in \mathbf{N}^{*}$, and $\left(d_{1}, d_{2}, \cdots, d_{n}\right)=$ $1, d_{i} \mid\left(d_{1}+d_{2}+\cdots+d_{n}\right), i=1,2, \cdots, n$.
(1) Prove: $d_{1} d_{2} \cdots d_{n} \mid\left(d_{1}+d_{2}+\cdots+d_{n}\right)^{n-2}$;
(2) For each ... | 19. (1) Let $p$ be a prime factor of $d_{1} d_{2} \cdots d_{n}$, and let $k=\max \left\{v_{p}\left(d_{i}\right) \mid\right.$ $i=1,2, \cdots, n\}$. Then, by the condition, $p^{k} \mid \sum_{i=1}^{n} d_{i}$. Consequently, $p^{k(n-2)} \mid\left(\sum_{i=1}^{n} d_{i}\right)^{n-2}$.
On the other hand, since $\left(d_{1}, d_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,705 |
20. Let $n \in \mathbf{N}^{*}, n \geqslant 2$. Prove: there exists $m \in \mathbf{N}^{*}$, such that
$$3^{n} \|\left(m^{3}+17\right)$$ | 20. When $n=2$, taking $m=1$ suffices. Suppose the proposition holds for $n(\geqslant 2)$, i.e., there exists $m \in \mathbf{N}^{*}$ such that $m^{3}+17=3^{n}(3 q+r)$, where $r \in\{1,2\}$ and $q \in \mathbf{N}$. In this case, $3 \nmid m$, hence $m^{2}=1(\bmod 3)$, and thus
$$3^{n} \cdot m^{2} \equiv 3^{n}\left(\bmod 3... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,706 |
21. Let $m$ be a given positive integer, and a sequence of primes $p_{1}, p_{2}, \cdots$ satisfies the following condition: when $n \geqslant 3$, $p_{n}$ is the largest prime factor of $p_{n-1}+p_{n-2}+m$. Prove: this sequence is a bounded sequence. | 21. Take $k \in \mathbf{N}^{*}$, such that $\max \left\{p_{1}, p_{2}\right\} \leqslant k \cdot((m+3)!)+1$. We will prove that for any $n \in \mathbf{N}^{*}$, we have
$$p_{n} \leqslant k \cdot((m+3)!)+1,$$
thus the proposition holds.
In fact, assume the proposition holds for $n-1, n-2 (n \geqslant 3)$. If $p_{n-1}, p_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,707 |
22. Let $A=\left\{x \mid x \in \mathbf{N}^{*}, x\right.$ in decimal representation does not contain the digit zero, and the sum of the digits of $x$ is a divisor of $x$ $\}$. Prove: for any $k \in \mathbf{N}^{*}, A$ contains an element that is a $k$-digit number. | 22. It is easy to prove by mathematical induction: for any $n \in \mathbf{N}^{*}$, there exists an $n$-digit positive integer $x_{n}$ consisting only of the digits 1 and 2, such that $2^{n} \mid x_{n}$.
Returning to the original problem, for $k=1,2,3,4,5$, we can take the numbers $1,12,112,4112,42112$ respectively, an... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,708 |
23. The sequence $a_{1}, a_{2}, \cdots$ consists of positive integers in increasing order. If $a_{k}$ can be expressed as the sum of several terms (which can be the same) from this sequence, then $a_{k}$ is called "good".
(1) Prove: There are at most finitely many terms in the sequence that are not "good";
(2) If the s... | 23. (1) Let $d$ be the largest positive integer such that for any $k \in \mathbf{N}^{*}$, we have $d \mid a_{k}$. If $d>1$, then we can replace each $a_{k}$ with $\frac{a_{k}}{d}$ and continue the discussion, so we may assume $d=1$.
Now let $p_{1}, p_{2}, \cdots, p_{r}$ be all the prime factors of $a_{1}$. By $d=1$, f... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,709 |
24. Let $p_{1}, p_{2}, \cdots$ be the sequence of all prime numbers in ascending order. Prove: when $n \geqslant 4$, we have $p_{n+1} p_{n+2}<p_{1} p_{2} \cdots p_{n}$ | 24. For $4 \leqslant n \leqslant 9$, it can be directly verified. When $n \geqslant 10$, if there exists an $n$ for which the proposition does not hold, i.e., $p_{1} p_{2} \cdots p_{n} \leqslant p_{n+1} p_{n+2}<p_{n+2}^{2}$, then $p_{1} p_{2} \cdots p_{i}<p_{n+2}$, where $i=\left[\frac{n}{2}\right]$.
Consider the foll... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,710 |
Bezout's Theorem: Let $a, b$ be integers, not both zero, then there exist integers $x, y$ such that
$$a x + b y = (a, b) \text{.}$$ | Let $d=(a, b)$, in the Euclidean algorithm of Theorem 3 from the previous section, take $u_{0}=a, u_{1}=b$ (here we assume $b \neq 0$), then by the properties of divisibility, it can be known that $d\left|u_{2}, d\right| u_{3}, \cdots$, $d \mid u_{k+1}$. Therefore, $d \leqslant u_{k+1}$.
Conversely, by the properties ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,711 |
25. Find the largest positive integer $m$, such that for $k \in \mathbf{N}^{*}$, if $1<k<m$ and $(k, m)=1$, then $k$ is a power of some prime.
---
The translation maintains the original format and line breaks as requested. | 25. The required maximum positive integer $m=60$.
On the one hand, it can be directly verified that $m=60$ meets the requirement (it is only necessary to note that the product of the smallest two primes coprime with 60, 7 and 11, is greater than 60).
On the other hand, let $p_{1}, p_{2}, \cdots$ represent the sequenc... | 60 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,712 |
26. (1) Find all prime sequences $p_{1}<p_{2}<\cdots<p_{n}$, such that
$$\left(1+\frac{1}{p_{1}}\right)\left(1+\frac{1}{p_{2}}\right) \cdot \cdots \cdot\left(1+\frac{1}{p_{n}}\right)$$
is an integer; $\square$
(2) Does there exist $n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}, n \in \mathbf{N}^{\vee}$, g... | 26. (1) When $n \geqslant 3$, since $p_{n}>p_{i}+1, 1 \leqslant i \leqslant n-1$, thus $p_{n}$ is coprime with the number $\prod_{i=1}^{n-1}\left(1+p_{i}\right)$, hence it requires $p_{n} \mid\left(1+p_{n}\right)$, leading to $p_{n}=1$, which is a contradiction. When $n=1$, $1+\frac{1}{p_{1}} \notin \mathbf{N}^{*}$. Wh... | (2,3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,713 |
27. (1) Find all pairs of positive integers $(a, b), a \neq b$, such that $b^{2}+a$ is a prime power and $\left(b^{2}+a\right) \mid\left(a^{2}+b\right)$;
(2) Let $a, b$ be two different positive integers greater than 1, and
$$\left(b^{2}+a-1\right) \mid\left(a^{2}+b-1\right) .$$
Prove: $b^{2}+a-1$ has at least two dis... | 27. (1) Let $(a, b)$ be a pair of positive integers satisfying the condition, then we can set $b^{2}+a=p^{m}$, where $p$ is a prime number, and $m \in \mathbf{N}^{*}$. Note that, $a \equiv -b^{2} \pmod{b^{2}+a}$, hence
$$0 \equiv a^{2}+b \equiv \left(-b^{2}\right)^{2}+b=b\left(b^{3}+1\right) \pmod{b^{2}+a}$$
This mean... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 741,714 |
28. Let $n \in \mathbf{N}^{\cdot}, I$ be an open interval on the number line with length $\frac{1}{n}$. Find the maximum number of simplest fractions $\frac{a}{b}$ such that $1 \leqslant b \leqslant n$ and $\frac{a}{b} \in I$. | 28. Let $\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \cdots, \frac{a_{r}}{b_{r}} \in I$ be all the simplest fractions in $I^{\prime}$ satisfying: $1 \leqslant b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{r} \leqslant n$. Then for $1 \leqslant i\left|\frac{a_{j}}{b_{j}}-\frac{a_{i}}{b_{i}}\right|=\frac{M}{\left[b_{... | \left[\frac{n+1}{2}\right] | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,715 |
29. Find all prime numbers $p$ such that there exist $m, n \in \mathbf{N}^{*}$, satisfying: $p=m^{2}+n^{2}$, and $p \mid\left(m^{3}+n^{3}-4\right)$. | 29. From the given conditions, we know that \( p \mid (2m^3 + 2n^3 - 8) \), and \( p = m^2 + n^2 \), thus,
\[ p \mid (2m^3 + 2n^3 - 8 - 3(m^2 + n^2)(m + n)) \]
which means \( p \mid (-(m+n)^3 - 8) \). Therefore,
\[ p \mid (m+n+2)((m+n)^2 - 2(m+n) + 4) \]
Since \( p \) is a prime, we have \( p \mid (m+n+2) \) or \( p ... | p = 2 \text{ or } 5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,716 |
30. Find all integers $n > 1$ such that any of its divisors greater than 1 have the form $a^{r}+1$, where $a \in \mathbf{N}^{*}, r \geqslant 2, r \in \mathbf{N}^{*}$. | 30. Let all numbers that meet the conditions form a set $S$. For $x \in S(x>2)$, it is known that there exist $a, r \in \mathbf{N}^{\cdot}$, such that $x=a^{r}+1$, where $a, r \geqslant 2$. For this $x$, we set $a$ to be the smallest positive integer in this representation of $x$, then $r$ must be even (otherwise $(a+1... | S=\left\{10\right. \text{ or primes of the form } a^{2}+1, \text{ where } a \in \mathbf{N}^{*}\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,717 |
31. For a positive integer $n>1$, let
$$a_{n}=\frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots+\frac{1}{p_{m}}$$
where $p_{1}, p_{2}, \cdots, p_{m}$ are all distinct prime factors of $n$. Prove that for any positive integer $N>1$, we have
$$a_{2}+a_{2} a_{3}+\cdots+a_{2} a_{3} \cdots a_{N}<1 .$$ | 31. Direct calculation shows that $a_{2}=\frac{1}{2}, a_{3}=\frac{1}{3}, a_{4}=\frac{1}{2}, a_{5}=\frac{1}{5}$. To prove the proposition, we first prove that when $N \geqslant 6$, we have
$$a_{2} a_{3} \cdots a_{N}<\frac{3}{2^{N-1}}$$
In fact, by the arithmetic mean inequality, we know
$$a_{2} a_{3} \cdots a_{N}<\left... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,718 |
34. Let the number of distinct prime factors of a positive integer $n$ be denoted by $w(n)$. Prove: There exist infinitely many $n \in \mathbf{N}^{*}$, such that
$$w(n)<w(n+1)<w(n+2)$$ | 34. Comparing with the construction in problem 32, we prove: there exist infinitely many $k \in \mathbf{N}^{*}$ such that $n=2^{k}$ satisfies the requirement. For this, we first prove the following lemma:
If $k \in \mathbf{N}^{*}, k \neq 3$ and $k$ is not a power of 2, then $w\left(2^{k}+1\right)>1$.
Indeed, if $w\lef... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,721 |
Example 1 Let $n \geqslant m>0$. Prove: $\frac{(m, n)}{n} \mathrm{C}_{n}^{m}$ is a positive integer. | Proof
By Bézout's Theorem, there exist $x, y \in \mathbf{Z}^{*}$, such that
$$(m, n)=m x+n y \text{, }$$
Therefore,
$$\frac{(m, n)}{n} \mathrm{C}_{n}^{m}=\left(\frac{m}{n} \mathrm{C}_{n}^{m}\right) x+\left(\mathrm{C}_{n}^{m}\right) y=\mathrm{C}_{n-1}^{m-1} \cdot x+\mathrm{C}_{n}^{m} \cdot y \in \mathbf{Z},$$
Combined... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,722 |
36. Let $x, y \in \mathbf{R}, x \neq y$. It is known that there exist 4 consecutive positive integers $n$, such that
$$\frac{x^{n}-y^{n}}{x-y} \in \mathbf{Z}$$
Prove: For any $n \in \mathbf{N}^{*}$, the number $\frac{x^{n}-y^{n}}{x-y}$ is an integer. | 36. Let $a=x+y, b=x y, t_{n}=\frac{x^{n}-y^{n}}{x-y}$, then
$$\begin{aligned}
t_{n+2} & =\frac{x^{n+2}-y^{n+2}}{x-y}=\frac{1}{x-y}\left((x+y)\left(x^{n+1}-y^{n+1}\right)-x y\left(x^{n}-y^{n}\right)\right) \\
& =a t_{n+1}-b t_{n}
\end{aligned}$$
From this, combined with $t_{1}=1, t_{2}=a$, we only need to prove that $a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,724 |
37. Let $n$ be an integer greater than 1. Prove: if for integers $k$ satisfying $0 \leqslant k \leqslant \sqrt{\frac{n}{3}}$, the number $k^{2}+k+n$ is prime, then for integers $k$ satisfying $0 \leqslant k \leqslant n-2$, the number $k^{2}+k+n$ is prime. | 37. Use the second mathematical induction. Suppose that for $0 \leqslant k \leqslant t-1(k \in \mathbf{Z})$, the number $k^{2}+k+n$ is a prime, where $\left[\sqrt{\frac{n}{3}}\right]+1 \leqslant t \leqslant n-2$. Prove: $t^{2}+t+n$ is a prime.
In fact, if $t^{2}+t+n$ is not a prime, let $p$ be its smallest prime facto... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,725 |
38. Prove: The number $\sum_{n=1}^{+\infty} \frac{\sigma(n)}{n!}$ is irrational, where $\sigma(n)$ denotes the sum of all positive divisors of $n$. | 38. If $h=\sum_{n=1}^{+\infty} \frac{\sigma(n)}{n!}=\frac{r}{s}$ is a rational number, where $(r, s)=1$, and let $p>\max \{s, 6\}$ be a prime number. From $h=\sum_{n=1}^{p-1} \frac{\sigma(n)}{n!}+\sum_{n=p}^{+\infty} \frac{\sigma(n)}{n!}$, we get
$(p-1)!h=(p-1)!\sum_{n=1}^{p-1} \frac{\sigma(n)}{n!}+\sum_{k=0}^{+\infty}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,726 |
39. Let $x, a, b \in \mathbf{N}^{*}$, satisfying
$$x^{a+b}=a^{b} \cdot b$$
Prove: $a=x, b=x^{x}$. | 39. Let $x, a, b$ be positive integers satisfying
$$x^{a+b}=a^{b} b$$
If $x=1$, then $a=b=1$, and the proposition holds.
Now consider the case where $x>1$. Suppose the standard factorization of $x$ is
$$x=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{n}^{r_{n}},$$
At this point, since $a, b$ are positive divisors of $x^{a+b}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,727 |
Example 1 Find the number of positive integers $a$ that satisfy the following condition: there exist non-negative integers $x_{0}$, $x_{1}, \cdots, x_{2009}$, such that
$$a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2009}} .$$ | Solution: Clearly, $a$ that satisfies the condition is greater than 1. Taking both sides of (1) modulo $(a-1)$, we get a necessary condition as
$$1^{x_{0}} \equiv 1^{x_{1}}+1^{x_{2}}+\cdots+1^{x_{2009}}(\bmod a-1)$$
This indicates: $(a-1) \mid 2008$, the number of such $a$ is $=d(2008)=8$.
On the other hand, let $a \i... | 8 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,728 |
Example 2 Let $p$ be a prime, $m, n, k \in \mathbf{N}^{*}, n \geqslant m+2, k$ be an odd number greater than 1, and $p=k \cdot 2^{n}+1, p \mid\left(2^{2^{m}}+1\right)$. Prove:
$$k^{2^{n-1}} \equiv 1(\bmod p)$$ | Proof
From the condition, we know that $k \cdot 2^{n} \equiv -1 \pmod{p}$. Raising both sides to the $2^{n-1}$ power, we get
$$k^{2^{n-1}} \cdot 2^{n \cdot 2^{n-1}} \equiv 1 \pmod{p}$$
On the other hand, from $2^{2^{m}} \equiv -1 \pmod{p}$ and $n \geqslant m+2$, we know that $2^{n \cdot 2^{n-1}} \equiv 1 \pmod{p}$ (si... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,729 |
Example 2 Find all sequences of positive integers $\left\{a_{n}\right\}$ that satisfy the following conditions:
(1) The sequence $\left\{a_{n}\right\}$ is bounded above, i.e., there exists $M>0$, such that for any $n \in \mathbf{N}^{\cdot}$, we have $a_{n} \leqslant M ;$
(2) For any $n \in \mathbf{N}^{*}$, we have $a_{... | Let $g_{n}=\left(a_{n}, a_{n+1}\right)$, then $g_{n} a_{n+2}=a_{n+1}+a_{n}$. Combining this with $g_{n+1}\left|a_{n+1}, g_{n+1}\right| a_{n+2}$, we know that $g_{n+1} \mid a_{n}$, which indicates that $g_{n+1}$ is a common divisor of $a_{n}$ and $a_{n+1}$, hence $g_{n+1} \leqslant g_{n}$. This means that the sequence $... | a_{n}=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,733 |
Example 1 The sequence $a_{0}, a_{1}, \cdots$ is defined as follows: $a_{0}=1, a_{n}=a_{n-1}+a\left[\frac{n}{3}\right], n=1$, $2, \cdots$, where $[x]$ denotes the greatest integer not exceeding $x$. Prove: for every prime $p \in\{2, 3,5,7,11,13\}$, there exist infinitely many $n \in \mathbf{N}^{*}$, such that $p \mid a... | Notice that, $2\left|a_{6}(=12), 3\right| a_{2}(=3), 5\left|a_{3}(=5), 7\right| a_{4}(=7), 11 \mid a_{11}(= 22)$, $13 \mid a_{20}(=117)$, therefore, for $p \in\{2,3,5,7,11,13\}$, there exists $m(\geqslant 2)$, such that $p \mid a_{m}$.
We will use proof by contradiction to handle this.
If there exists some $p \in\{2,3... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,734 |
Example 2 Let $n$ be an odd number greater than 1, and $k_{1}, k_{2}, \cdots, k_{n}$ be $n$ given integers. For each permutation $a=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ of $1,2, \cdots, n$, let
$$S(a)=\sum_{i=1}^{n} k_{i} a_{i}$$
Prove: There exist two permutations $b$ and $c(b \neq c)$ of $1,2, \cdots, n$ such t... | Proof
If for any two different permutations $b$ and $c$ of $1,2, \cdots, n$, we have $n!Y(S(b) - S(c))$, then when $a$ takes all permutations of $1,2, \cdots, n$ (a total of $n!$ permutations), $S(a)$ will traverse a complete residue system modulo $n!$. Therefore,
$$\sum_{a} S(a) \equiv 1+2+3+\cdots+n!=\frac{n!}{2}(n!+... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 741,735 |
Example 3 Let $m, n \in \mathbf{N}^{*}, m$ be an odd number, and $\left(m, 2^{n}-1\right)=1$. Prove: The number $1^{n}+$ $2^{n}+\cdots+m^{n}$ is a multiple of $m$. | Proof
Since $m$ is odd, and $1,2, \cdots, m$ is a complete residue system modulo $m$, hence $2 \times 1,2 \times 2, \cdots, 2 \times m$ is also a complete residue system modulo $m$, so,
$$\sum_{k=1}^{m} k^{n} \equiv \sum_{k=1}^{m}(2 k)^{n}(\bmod m)$$
which means $m \mid\left(2^{n}-1\right) \sum_{k=1}^{m} k^{n}$. Combi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,736 |
Example 4 A quirky mathematician, he climbs up and down a ladder with a total of $n$ steps, each time ascending $a$ steps or descending $b$ steps. Here $a, b$ are given positive integers. If he can start from the ground, climb to the very top of the ladder, and then return to the ground. Find the minimum value of $n$ (... | The minimum value of $n$ is $a+b-(a, b)$.
In fact, we only need to consider the case where $(a, b)=1$ (otherwise, if $(a, b)=d$, we can treat $d$ steps as 1 step, thus reducing the problem to the case where $(a, b)=1$).
When $(a, b)=1$, we prove it in two steps.
(1) When $n=a+b-1$, the mathematician can complete the r... | a+b-(a, b) | Number Theory | proof | Yes | Yes | number_theory | false | 741,737 |
Example 5 Let $a_{1}, a_{2}, \cdots$ be a sequence of integers, in which there are infinitely many positive integers and infinitely many negative integers. Moreover, for any $n \in \mathbf{N}^{*}$, the remainders of the numbers $a_{1}$, $a_{2}, \cdots, a_{n}$ when divided by $n$ are all distinct. Prove: Each integer ap... | Proof
First, we prove that any two terms in the sequence are distinct.
Indeed, if there exist $i, j \in \mathbf{N}^{*}, i<j$, such that $a_{i}=a_{j}$, then among $a_{1}, a_{2}, \cdots, a_{j}$, there are two numbers $\left(a_{i}\right.$ and $\left.a_{j}\right)$ that have the same remainder when divided by $j$.
Next, we... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,738 |
Fermat's Little Theorem: Let $p$ be a prime, $a \in \mathbf{Z}$, then $a^{p} \equiv a(\bmod p)$. In particular, if $p \nmid a$, then $a^{p-1} \equiv 1(\bmod p)$. | Here we provide a direct proof using the binomial theorem. It suffices to prove the case where $a$ is a positive integer.
When $a=1$, the proposition is obviously true.
Now assume that the proposition holds for $a=n$, i.e., $p \mid\left(n^{p}-n\right)$, then
$$\begin{aligned}
(n+1)^{p}-(n+1) & =n^{p}+\mathrm{C}_{p}^{1}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,741 |
Euler's Theorem: If $(a, m)=1$, then $a^{\varphi(m)} \equiv 1(\bmod m)$, where $m \in \mathbf{N}^{*}, a \in \mathbf{Z}$ | Proof: Let $a_{1}, a_{2}, \cdots, a_{\phi(m)}$ be a reduced residue system modulo $m$. Then, by $(a, m)=1$, it follows that $a a_{1}, a a_{2}, \cdots, a a_{\phi(m)}$ is also a reduced residue system modulo $m$. Therefore,
$$a_{1} a_{2} \cdots a_{\varphi(m)} \equiv\left(a a_{1}\right)\left(a a_{2}\right) \cdot \cdots \c... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,742 |
Example 1 Let $n$ be an odd number greater than 1, and let the numbers $a_{1}, a_{2}, \cdots, a_{q(n)}$ be all positive integers in $1,2, \cdots, n$ that are coprime with $n$. Prove:
$$\left|\prod_{k=1}^{\varphi(n)} \cos \frac{a_{k} \pi}{n}\right|=\frac{1}{2^{\alpha(n)}}$$ | $$\begin{array}{l}
\text { Let } A=\left|\prod_{k=1}^{q(n)} \cos \frac{a_{k} \pi}{n}\right|, B=\left|\prod_{k=1}^{\phi(n)} \sin \frac{a_{k} \pi}{n}\right|, \text { then } \\
\begin{aligned}
2^{q(n)} \cdot A \cdot B & =\left|\prod_{k=1}^{q(n)}\left(2 \cos \frac{a_{k} \pi}{n} \sin \frac{a_{k} \pi}{n}\right)\right| \\
& =... | \left|\prod_{k=1}^{q(n)} \cos \frac{a_{k} \pi}{n}\right|=\frac{1}{2^{q(n)}} | Number Theory | proof | Yes | Yes | number_theory | false | 741,743 |
Example 3 Let $m, n \in \mathbf{N}^{*}, m>n$. Prove:
$$[m, n]+[m+1, n+1]>\frac{2 m n}{\sqrt{m-n}}$$ | Proof
Let $m-n=k$, then by conclusion (5) we have
$$\begin{aligned}
{[m, n]+[m+1, n+1] } & =\frac{m n}{(m, n)}+\frac{(m+1)(n+1)}{(m+1, n+1)} \\
& >\frac{m n}{(n+k, n)}+\frac{m n}{(n+k+1, n+1)} \\
& =\frac{m n}{(k, n)}+\frac{m n}{(k, n+1)}
\end{aligned}$$
Now let $(k, n)=d_{1},(k, n+1)=d_{2}$, then $\left(d_{1}, d_{2}\... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 741,744 |
Example 2 Let $a, m \in \mathbf{N}^{*}$. Prove:
$$a^{m} \equiv a^{m-\varphi(m)}(\bmod m) .$$ | Notice, when $(a, m)=1$, this conclusion is precisely Euler's theorem, hence, this conclusion can be regarded as a generalization of Euler's theorem.
If one of $a, m$ equals 1, the proposition is obvious, so we assume $a, m$ are both greater than 1.
Let $m=m_{1} \cdot m_{2}$, where the prime factors of $m_{1}$ are al... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,745 |
Example 3 Let $p$ be a prime number greater than 3. Prove: there exists $a \in\{1,2, \cdots, p-2\}$, such that $a^{p-1}-1$ and $(a+1)^{p-1}-1$ are not multiples of $p^{2}$.
---
Note: The original text had a minor typo, where it said $(a+1)^{p^{-1}}-1$ which is likely a mistake and should be $(a+1)^{p-1}-1$ for the st... | Proof
By Fermat's Little Theorem, the two given numbers are multiples of $p$, and this problem is also a discussion around Fermat's Little Theorem.
Let $S=\{1,2, \cdots, p-1\}, A=\left\{a \mid a \in S\right.$ and $\left.a^{p-1} \equiv 1\left(\bmod p^{2}\right)\right\}$. First, we prove that for any $a \in S$, the numb... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,746 |
Example 4 Let $p$ be an odd prime. Prove:
$$\sum_{i=1}^{p-1} 2^{i} \cdot j^{p-2} \equiv \sum_{i=1}^{\frac{p-1}{2}} i^{p-2}(\bmod p)$$ | Proof
Since $p$ is an odd prime, for any $1 \leqslant i \leqslant p-1$, we have
$$\begin{aligned}
& (p-1)(p-2) \cdot \cdots \cdot(p-(i-1)) \\
\equiv & (-1)(-2) \cdot \cdots \cdot(-(i-1)) \\
\equiv & (-1)^{i-1} \cdot(i-1)!(\bmod p)
\end{aligned}$$
Therefore, combining with $(p,(i-1)!)=1$, we know $\mathrm{C}_{p-1}^{i-1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,747 |
Example 5 The Fibonacci sequence $\left\{F_{n}\right\}$ is defined as follows: $F_{1}=1, F_{2}=1, F_{n+2}=F_{n+1}+F_{n}, n=1,2, \cdots$.
Let $p$ be a prime number greater than 5. Prove:
$$p \mid F_{p+1}\left(F_{p+1}-1\right)$$ | To prove the characteristic method of finding the general term of a recursive sequence, we know that
$$F_{n}=\frac{1}{\sqrt{5}}\left(\left(\frac{\sqrt{5}+1}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right)$$
For a prime $p$ greater than 5, by Fermat's Little Theorem, we know: $5^{p-1} \equiv 1(\bmod p)$, thu... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,748 |
Example 6 Let positive integers $n, k$ satisfy: $3 \times n$ and $k \geqslant n$. Prove: there exists a multiple $m$ of $n$, such that the sum of the digits of $m$ in decimal representation equals $k$. | Proof
First, consider the case where $(n, 10)=1$. In this case, by Euler's theorem, we have: $10^{\rho(n)} \equiv 1 \pmod{n}$. For convenience, let $d=\varphi(n)$. Thus, for any $i, j \in \mathbf{N}^{*}$, we have
$$10^{j d+1} \equiv 10 \pmod{n}, \quad 10^{i d} \equiv 1 \pmod{n}$$
Next, we need to find $u, v \in \mathb... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,749 |
Example 7 Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ rational numbers. It is known that for any $m \in \mathbf{N}^{*}$, the number
$$a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}$$
is an integer.
Prove: $a_{1}, a_{2}, \cdots, a_{n}$ are all integers. | Let $S_{m}=a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}$, and suppose $a_{i}=\frac{r_{i}}{t_{i}}, r_{i}, t_{i} \in \mathbf{Z}, t_{i}>0$, and $\left(r_{i}, t_{i}\right)=1, i=1,2, \cdots, n$.
If $a_{1}, a_{2}, \cdots, a_{n}$ are not all integers, then there must be a $t_{i}>1$. Let $p$ be a prime factor of $t_{1} t_{2} \cdots t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,750 |
Let $p$ be a prime, and consider an $n$-degree polynomial with integer coefficients under modulo $p$:
$$f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}\left(p \nmid a_{n}\right),$$
then the congruence equation $f(x) \equiv 0(\bmod p)$ has at most $n$ distinct solutions modulo $p$.
This conclusion is known as Lagrange's ... | Below is the proof of Lagrange's Theorem.
We use induction on \( n \). When \( n=0 \), since \( p \nmid a_{0} \), the congruence equation \( f(x) \equiv 0(\bmod p) \) has no solutions, so the theorem holds for all polynomials \( f(x) \) of degree \( n=0 \).
Now assume the proposition holds for all polynomials of degre... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,751 |
Example 1 Let $p$ be a prime number greater than 3, and
$$1+\frac{1}{2}+\cdots+\frac{1}{p}=\frac{r}{p s}$$
where $r, s \in \mathbf{N}^{*}$, and $(r, s)=1$. Prove: $p^{3} \mid(r-s)$. | Proof
Consider the polynomial
$$f(x)=(x-1)(x-2) \cdot \cdots \cdot(x-(p-1))-\left(x^{p-1}-1\right) .$$
Let \( f(x)=\alpha_{p-2} x^{p-2}+\alpha_{p-3} x^{p-3}+\cdots+\alpha_{1} x+\alpha_{0} \). Using Lagrange's theorem (see the previous discussion), we know that
$$\alpha_{p-2} \equiv \alpha_{p-3} \equiv \cdots \equiv \a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,752 |
Example 2 (Euler's Criterion) Let $p$ be an odd prime, $a \in \mathbf{Z},(a, p)=1$. Prove the following conclusions:
If $a^{\frac{p-1}{2}} \equiv 1(\bmod p)$, then $a$ is a quadratic residue modulo $p$;
If $a^{\frac{p-1}{2}} \equiv -1(\bmod p)$, then $a$ is not a quadratic residue modulo $p$. | Proof
This conclusion was given in the previous section, and here we prove it using Lagrange's theorem.
Notice that, by Fermat's Little Theorem, we have \( p \mid \left(a^{p-1}-1\right) \), i.e.,
\[ p \left\lvert \left(a^{\frac{p-1}{2}}-1\right)\left(a^{\frac{p-1}{2}}+1\right) \right., \]
and
\[ \left(a^{\frac{p-1}{2}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,753 |
Example 3 Find all pairs of positive integers $(m, n), m, n \geqslant 2$, such that for any $a \in\{1$, $2, \cdots, n\}$, we have $a^{n} \equiv 1(\bmod m)$. | When $m$ is an odd prime and $n=m-1$, by Fermat's Little Theorem, the pair $(m, n)$ satisfies the condition. Conversely, must $(m, n)=(p, p-1)$, where $p$ is an odd prime?
Now, let $(m, n)$ be a pair of positive integers that satisfy the condition, and let $p$ be a prime factor of $m$. Then $p>n$ (otherwise, if $p \le... | (m, n)=(p, p-1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,754 |
For which $n \in \mathbf{N}^{*}(n \geqslant 3)$, do there exist $n$ consecutive positive integers such that the largest number is a divisor of the least common multiple of the other $n-1$ numbers? | When $n=3$, if there exist positive integers $a, a+1, a+2$ such that $a+2$ is a divisor of $[a, a+1]$, then by $(a, a+1)=1$, we know $(a+2) \mid a(a+1)$, but $(a+2, a+1)=1$, hence leading to $(a+2) \mid a$, a contradiction. Therefore, $a=3$ does not meet the condition.
When $n \geqslant 4$, we prove that there exist $... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,755 |
Example 4 Proof: There do not exist $m, n \in \mathbf{N}^{*}$, such that the following equation holds:
$$m^{3}+n^{4}=19^{19} .$$ | Proof:
If there exist positive integers $m, n$ that satisfy (3), then for any positive integer $k$, taking the modulus $k$ of both sides of (3) should yield equal remainders. In particular, taking the modulus 13 of both sides of (3) should also be equal.
By Fermat's Little Theorem, for any $x \in \mathbf{N}^{*}$, we h... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,756 |
Wilson's Theorem: If $p$ is a prime, then $(p-1)!\equiv-1(\bmod p)$. | In the previous section, we provided a proof using Lagrange's theorem. Below, we give a proof based on the idea of "pairing."
For this purpose, we introduce the concept of "number-theoretic inverse."
Let $m \in \mathbf{N}^{*}, a \in \mathbf{Z}$, and $(a, m)=1$. Using the basic properties of complete residue systems, w... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,757 |
Example 1 Prove the following conclusions:
(1) There exist infinitely many pairs of positive integers $(n, k)$, such that $n, k \geqslant 2, n \neq k$, and
$$(n!+1, k!+1)>1$$
(2) There exist infinitely many pairs of positive integers $(n, k)$, such that $n, k \geqslant 2, n \neq k$, and
$$(n!-1, k!-1)>1$$ | Proof
(1) Take $k \in \mathbf{N}^{*}$, such that $k+1$ is composite (there are infinitely many such $k$), and then take a prime factor $p$ of $k!+1$, then $p \neq k+1$. By Wilson's theorem, we know that,
$$(p-1)!+1 \equiv 0(\bmod p),$$
Thus,
$$(k!+1,(p-1)!+1) \geqslant p>1$$
Therefore, $(n, k)=(p-1, k)$ satisfies the... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,758 |
Example 2 Let $n$ be an integer greater than 4. Prove that the following two statements are equivalent:
(1) $n$ and $n+1$ are both composite;
(2) The integer closest to $\frac{(n-1)!}{n(n+1)}$ is even. | Proof
First, we prove a lemma: Let $m$ be a composite number greater than 4, then $m \mid (m-2)!$, and $\frac{(m-2)!}{m}$ is even.
In fact, if $m = pq$, and $16$ when, in the set of numbers $\{1, 2, \cdots, m-2\}$, there are at least 3 even numbers. Comparing the above two cases, we can see that $\frac{(m-2)!}{m}$ is... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,759 |
Example 3 Proof: Any set of 18 consecutive positive integers cannot be partitioned into two subsets such that the product of the numbers in each subset is equal. | Prove that if there exists a set $A=\{a+1, a+2, \cdots, a+18\}$ (where $a \in \mathbf{N}$), which can be divided into two sets $B$ and $C$ (i.e., $B \cap C=\varnothing, B \cup C=A$), such that the product of the numbers in $B$ and $C$, $\pi(B)$ and $\pi(C)$, are equal, then from the fact that there is at most one numbe... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,760 |
Example 4 Proof: There do not exist non-negative integers $k$ and $m$, such that
$$k! + 48 = 48(k + 1)^{m}.$$ | Proof
Obviously, when $k=0$ or $m=0$, there does not exist a pair $(k, m)$ that satisfies the condition. Therefore, if there exist non-negative integers $k$ and $m$ that satisfy (1), then $k, m$ must both be positive integers.
We will discuss the cases where $k+1$ is composite and where $k+1$ is prime.
$1^{\circ}$ If ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,761 |
Example 1 Proof: For any positive integer $n$, there exist $n$ consecutive positive integers, none of which is a power of a prime (and thus, none of them is a prime). | Proof:
A basic idea is: to find $n$ consecutive positive integers, each of which has two distinct prime factors. For this, for any $n \in \mathbf{N}^{*}$, take $2n$ different primes $p_{1}, p_{2}, \cdots, p_{n}; q_{1}, q_{2}, \cdots, q_{n}$. By the Chinese Remainder Theorem, there exists $m \in \mathbf{N}^{*}$, such th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,763 |
Example 2 Proof: There exists an increasing sequence of positive integers $\left\{a_{n}\right\}$, such that for any $k \in \mathbf{N}^{*}$, the sequence $\left\{k+a_{n}\right\}$ contains at most a finite number of prime numbers. | Let $p_{1}, p_{2}, \cdots$ denote all prime numbers in ascending order.
Now, we construct the sequence $\left\{a_{n}\right\}$ that meets the requirements.
Let $a_{1}=2$, and suppose $a_{1}, a_{2}, \cdots, a_{n}$ have been determined. Take $a_{n+1}$ as the smallest positive integer greater than $a_{n}$ that satisfies th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,764 |
Example 3: Does there exist a sequence composed of positive integers such that each positive integer appears exactly once in the sequence, and for any positive integer $k$, the sum of the first $k$ terms of the sequence is a multiple of $k$? | The required sequence does exist, and we still construct it using recursion.
Take $a_{1}=1$, and suppose $a_{1}, a_{2}, \cdots, a_{m}$ have all been determined. Let $t$ be the smallest positive integer not appearing in $a_{1}, a_{2}, \cdots, a_{m}$.
By the Chinese Remainder Theorem, there exist infinitely many $r \in ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,765 |
Example 1 If $a|n, b| n$, and there exist integers $x, y$, such that $a x+b y=1$, prove: $a b \mid n$.
| Proof: From the conditions, we can set \( n = a u, n = b v, \) where \( u \) and \( v \) are integers. Thus,
\[
\begin{aligned}
n & = n(a x + b y) \\
& = n a x + n b y \\
& = a b v x + a b u y \\
& = a b(v x + u y),
\end{aligned}
\]
Therefore,
\[
a b \mid n.
\] | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,766 |
Example 6 A roundtable conference has 2012 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. | Let $n=1006$, we label each seat, and 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 $i, j$ are the seat numbers before and after the break, respectively. Clearly, all "x-coordinates" $i$ and "y-coordinates" $j$ take on all ... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 741,769 |
Fermat's Little Theorem: Let $p$ be a prime, $a$ an integer, then $a^{p} \equiv a(\bmod p)$. In particular, if $p \nmid a$, then $a^{p-1} \equiv 1(\bmod p)$. | Prove that when $p \mid a$, the conclusion is obviously true.
When $p \nmid a$, let $x_{1}, x_{2}, \cdots, x_{p-1}$ be a permutation of $1,2, \cdots, p-1$. We first prove that $a x_{1}, a x_{2}, \cdots, a x_{p-1}$ are pairwise incongruent modulo $p$.
In fact, if there exist $1 \leqslant i<j \leqslant p-1$ such that $a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,770 |
Example 1 Let $n$ be a positive integer. Prove: $7 \mid 3^{n}+n^{3}$ if and only if $7 \mid 3^{n} n^{3}+1$. | \begin{tabular}{lr}
& If $7 \mid 3^{n}+n^{3}$, then $7 \nmid n$, \\
thus, by Fermat's Little Theorem, we know & $n^{6} \equiv 1(\bmod 7)$, \\
therefore, by & $7 \mid 3^{n}+n^{3}$, \\
we know & $7 \mid\left(3^{n}+n^{3}\right) n^{3}$, \\
hence & $7 \mid 3^{n} n^{3}+1$. \\
\multicolumn{1}{c}{ Conversely, if } & $7 \mid 3... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,771 |
Example 2 Let $x$ be an integer, $p$ is an odd prime factor of $x^{2}+1$, prove: $p \equiv 1(\bmod 4)$. | Proof that since $p$ is an odd prime, if $p \neq 1(\bmod 4)$, then $p \equiv 3(\bmod 4)$, we can set $p=$ $4k+3$. In this case, from $x^{2} \equiv -1(\bmod p)$, we get
$$x^{p-1}=x^{4k+2}=\left(x^{2}\right)^{2k+1} \equiv (-1)^{2k+1} \equiv -1(\bmod p)$$
By Fermat's Little Theorem, we should have
$$x^{p-1} \equiv 1(\bmo... | p \equiv 1(\bmod 4) | Number Theory | proof | Yes | Yes | number_theory | false | 741,772 |
Example 3 Let $x$ be an integer, and $p$ a prime factor of the number $x^{6}+x^{5}+\cdots+1$. Prove: $p=7$ or $p \equiv$ $1(\bmod 7)$. | Prove that when $x=1$, $p=7$; when $x \neq 1$, $p$ is a prime factor of $\frac{x^{7}-1}{x-1}$, hence, $x^{7} \equiv 1(\bmod p)$, which implies $p \nmid x$. Therefore, by Fermat's Little Theorem, we have $x^{p-1} \equiv 1(\bmod p)$, and thus $x^{(7, p-1)} \equiv 1(\bmod p)$.
If $7 \nmid p-1$, i.e., $p \neq 1(\bmod 7)$,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,773 |
Example 4 Let $p$ be a prime. Prove: There exist infinitely many positive integers $n$, such that $p \mid 2^{n}-n$.
untranslated text remains the same as requested. | Prove that if $p=2$, then taking $n$ as an even number, we have $p \mid 2^{n}-n$, and the proposition holds. Suppose $p>2$, then by Fermat's Little Theorem, we know
$$2^{p-1} \equiv 1(\bmod p)$$
Therefore, for any positive integer $k$, we have
$$2^{k(p-1)} \equiv 1(\bmod p)$$
So, it suffices to prove that there are i... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,774 |
Example 5 By Fermat's Little Theorem, for any odd prime $p$, we have $2^{p-1} \equiv 1(\bmod p)$. Question: Does there exist a composite number $n$ such that $2^{n-1} \equiv 1(\bmod n)$ holds? | $$\begin{array}{l}
\text { Hence } \\
\text { Therefore }
\end{array}$$
$$\begin{array}{l}
2^{10}-1=1023=341 \times 3 \\
2^{10} \equiv 1(\bmod 341) \\
2^{340} \equiv 1^{34} \equiv 1(\bmod 341)
\end{array}$$
Hence 341 meets the requirement.
Furthermore, let $a$ be an odd composite number that meets the requirement, the... | 341 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,775 |
Example 6 Find all prime numbers $p$ such that $\frac{2^{p-1}-1}{p}$ is a perfect square.
Find all prime numbers $p$ such that $\frac{2^{p-1}-1}{p}$ is a perfect square. | Let $p$ be a prime number that satisfies the condition, then obviously $p$ is an odd prime. By Fermat's Little Theorem, we have
Thus,
$$\begin{array}{c}
p \mid 2^{p-1}-1 \\
2^{p-1}-1=\left(2^{\frac{p-1}{2}}-1\right)\left(2^{\frac{p-1}{2}}+1\right) \\
p \left\lvert\, 2^{\frac{p-1}{2}}-1\right. \text { or } p \left\lver... | p=3 \text{ or } 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,776 |
Example 1 Let $n$ be a positive integer greater than 1. Prove: The number $n^{5}+n^{4}+1$ is not a prime.
| Notice that
$$\begin{aligned}
& n^{5}+n^{4}+1 \\
= & n^{5}+n^{4}+n^{3}-\left(n^{3}-1\right) \\
= & n^{3}\left(n^{2}+n+1\right)-(n-1)\left(n^{2}+n+1\right) \\
= & \left(n^{3}-n+1\right)\left(n^{2}+n+1\right)
\end{aligned}$$
Therefore, if $n^{5}+n^{4}+1$ is a prime number, then $n^{3}-n+1=1$, which requires $n=0$ or $\p... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,779 |
Example 1 Given that $p$ is a prime number, find all integer pairs $(x, y)$ such that $|x+y|+(x-y)^{2}=p$. | Notice that, $x+y$ and $x-y$ are either both odd or both even, so $|x+y|+(x-y)^{2}$ is even, which implies $p=2$. This indicates that $|x+y|+(x-y)^{2}=2$.
Since $|x+y|$ and $(x-y)^{2}$ have the same parity, and $(x-y)^{2}$ is a perfect square, we have $\left(|x+y|,(x-y)^{2}\right)=(2,0),(1,1)$. Solving these respectiv... | (x, y)=(1,1),(-1,-1),(0,1),(0,-1),(1,0),(-1,0) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,781 |
Example 2 Fill the numbers $1,2, \cdots, 49$ into a $7 \times 7$ table (one number per cell), and calculate the sum of the numbers in each row and each column, resulting in 14 sums. Let $A$ represent the sum of the odd numbers among these 14 sums, and $B$ represent the sum of the even numbers among these 14 sums. Is th... | If there is a way to fill the table such that $A=B$, then
$$A=B=\frac{1}{2}(A+B)=\frac{1}{2} \times 2 \times(1+2+\cdots+49)=25 \times 49$$
This requires $B$ to be an odd number, but $B$ is the sum of several even numbers, which cannot be odd, leading to a contradiction.
Therefore, there is no way to fill the table suc... | proof | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,782 |
Example 3 In decimal notation, add a 17-digit number to its reverse. Prove: the sum must contain an even digit.
Also, if 17 is replaced by a general positive integer $n$, does the proposition hold? For which $n$ does it hold? | Prove that if there exists a 17-digit number $\overline{a_{1} a_{2} \cdots a_{17}}$, such that $M=\overline{a_{1} \cdots a_{17}}+\overline{a_{17} a_{16} \cdots a_{1}}$ has all digits odd, then by examining the units digit, we can see that $a_{1}+a_{17}$ is odd. Now, examining the sum of the leading digit, if $a_{2}+a_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,783 |
Example 4 (1) Given that there are $n$ integers, their sum is zero, and their product is $n$. Prove: $4 \mid n$;
(2) Let the positive integer $n$ be a multiple of 4. Prove: there exist $n$ integers, whose sum is zero, and whose product is $n$. | (1) Let integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy:
$$\left\{\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=0 \\
a_{1} a_{2} \cdots a_{n}=n
\end{array}\right.$$
If $n$ is odd, then by (2) we know that $a_{1}, a_{2}, \cdots, a_{n}$ are odd, so $a_{1}+a_{2}+\cdots+a_{n}$ is the sum of an odd number ($n$) of odd numbers, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,784 |
Example 5 It is known that among 4 coins, there may be counterfeit coins, where genuine coins each weigh 10 grams, and counterfeit coins each weigh 9 grams. Now there is a balance scale, which can weigh the total weight of the objects on the tray. Question: What is the minimum number of weighings needed to ensure that ... | At least 3 weighings can achieve this.
In fact, let the 4 coins be $a, b, c, d$. Weigh $a+b+c, a+b+d$, and $a+c+d$ three times. The sum of these three weights is $3a + 2(b+c+d)$, so if the sum of these three weights is odd, then $a$ is the counterfeit coin; otherwise, $a$ is genuine. Once $a$ is determined, solving the... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false | 741,785 |
Example 6 A cube with a side length of 3 is divided into 27 unit cubes. The numbers $1, 2, \cdots$, 27 are randomly placed into the unit cubes, one number in each. Calculate the sum of the 3 numbers in each row (horizontal, vertical, and column), resulting in 27 sum numbers. Question: What is the maximum number of odd ... | Solve: To calculate the sum $S$ of these 27 sums, since each number appears in exactly 3 rows, we have
$$S=3 \times(1+2+\cdots+27)=3 \times 27 \times 14$$
Thus, $S$ is an even number, so the number of odd numbers among these 27 sums must be even.
If 26 of these 27 sums are odd, let's assume that the even number is the... | 24 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,786 |
Example 2 Consider the following sequence:
$$101,10101,1010101, \cdots$$
Question: How many prime numbers are there in this sequence? | It is easy to know that 101 is a prime number. Next, we prove that this is the only prime number in the sequence.
Let $a_{n}=\underbrace{10101 \cdots 01}_{n \uparrow 01}$, then when $n \geqslant 2$, we have
$$\begin{aligned}
a_{n} & =10^{2 n}+10^{2(n-1)}+\cdots+1 \\
& =\frac{10^{2(n+1)}-1}{10^{2}-1} \\
& =\frac{\left(1... | 1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,790 |
Example 1 Let the prime numbers be arranged in ascending order as $p_{1}, p_{2}, \cdots$. Prove: For any positive integer $n$ greater than 1, the numbers $p_{1} p_{2} \cdots p_{n}-1$ and $p_{1} p_{2} \cdots p_{n}+1$ are not perfect squares. | Notice that, for $n \geqslant 2$, $3 \mid p_{1} p_{2} \cdots p_{n}$, hence
$$p_{1} p_{2} \cdots p_{n}-1 \equiv 2(\bmod 3)$$
Therefore, $p_{1} p_{2} \cdots p_{n}-1$ is not a perfect square.
Also, for $n \geqslant 2$, $p_{2} \cdots p_{n}$ is odd. Let $p_{2} \cdots p_{n}=2 k+1$, then we have
$$p_{1} p_{2} \cdots p_{n}+1=... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,791 |
Example 2 Given positive integers $a, b$ satisfy the equation
$$2 a^{2}+a=3 b^{2}+b$$
Prove: $a-b$ and $2 a+2 b+1$ are both perfect squares. | Proof: From the condition, we know
$$b^{2}=2 a^{2}+a-\left(2 b^{2}+b\right)=(a-b)(2 a+2 b+1) .$$
The left side of the above equation is greater than zero, and on the right side, \(2 a+2 b+1\) is greater than zero, so \(a-b\) is greater than zero.
From (1), to prove that \(a-b\) and \(2 a+2 b+1\) are both perfect squar... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,792 |
Example 3 Let positive integers $x, y, z$ satisfy $(x, y, z)=1$, and $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$. Prove: $x+y, x-z, y-z$ are all perfect squares. | Proof: Let $(x, y)=m$, and set $x=m n, y=m l$, where $m, l, n$ are all positive integers, and $(l, n)=1$. Thus, from the condition we have
$$(l+n) z=m l n .$$
Using $(x, y, z)=1$, we know $(m, z)=1$, hence, from (1) we know $z \mid ln$. Since $(l, n)=1$, it follows that $(l, l+n)=1, (n, l+n)=1$, therefore, from (1) we... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,793 |
Example 4 Find all prime numbers $p$ such that $p^{3}-4 p+9$ is a perfect square.
Find all prime numbers $p$ such that $p^{3}-4 p+9$ is a perfect square. | Let $p^{3}-4 p+9=x^{2}, x$ be a non-negative integer, then $p \mid x^{2}-9$, i.e., $p \mid (x-3)(x+3)$. Combining with $p$ being a prime, we can set $x=k p \pm 3, k$ as a non-negative integer. Thus,
$$p^{3}-4 p=x^{2}-9=k^{2} p^{2} \pm 6 k p$$
We get $p^{2}-4=k^{2} p \pm 6 k$, which indicates: $p \mid 6 k \pm 4$.
When ... | p=2,7,11 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,794 |
Example 5 Given that $n$ is a positive integer, and $2n+1$ and $3n+1$ are both perfect squares. Prove: $40 \mid n$.
---
The translation maintains the original text's line breaks and format. | Proof: Let $2n+1=x^2, 3n+1=y^2$, where $x, y$ are positive integers.
By Property 1, we know $x^2 \equiv 1 \pmod{8}$ (since $x^2$ is odd, hence $x$ is odd), thus
$$n \equiv 0 \pmod{4}$$
Furthermore, $3n+1$ is odd, hence
i.e. $\square$
$$\begin{array}{c}
y^2 \equiv 1 \pmod{8} \\
3n+1 \equiv 1 \pmod{8}
\end{array}$$
Th... | 40 \mid n | Number Theory | proof | Yes | Yes | number_theory | false | 741,795 |
Example 6 If $a, b$ are positive integers such that $ab+1$ is a perfect square, then we denote $a \sim b$. Prove: If $a \sim b$, then there exists a positive integer $c$, such that $a \sim c, b \sim c$.
untranslated text remains the same as requested. | Proof: Given $a \sim b$, we can set $ab + 1 = x^2$, where $x$ is a positive integer. The next perfect square related to $a$, $b$, and $x$ is $(a + x)^2$ or $(b + x)^2$. Therefore, we take $c = 2x + a + b$, then
$$\begin{aligned}
ac + 1 & = a(2x + a + b) + 1 \\
& = 2ax + a^2 + ab + 1 \\
& = 2ax + a^2 + x^2 = (x + a)^2, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,796 |
Example 7 Find all pairs of positive integers $(a, b)$ such that
$$a^{3}+6 a b+1, b^{3}+6 a b+1,$$
are both perfect cubes. | Let's assume $a \leqslant b$, then
$$b^{3}<b^{3}+6 a b+1 \leqslant b^{3}+6 b^{2}+1<(b+2)^{3}$$
Since $b^{3}+6 a b+1$ is a perfect cube, we have
$$b^{3}+6 a b+1=(b+1)^{3},$$
which implies $\square$
$$\begin{aligned}
6 a b & =3 b^{2}+3 b \\
b & =2 a-1
\end{aligned}$$
Thus, $\square$
$$a^{3}+6 a b+1=a^{3}+12 a^{2}-6 a+... | (1,1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,797 |
Example 8 Find the smallest positive integer $n$, such that there exist integers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}=1599$$ | From property 1, for any integer $a$, we know that
$$a^{2} \equiv 0(\bmod 4) \text { or } a^{2} \equiv 1(\bmod 8),$$
From this, we can deduce that
$$a^{4} \equiv 0 \text { or } 1(\bmod 16).$$
Using this conclusion, if $n<15$, let
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4} \equiv m(\bmod 16),$$
then
and
$$\begin{array}{c... | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,798 |
1 Let $p$ and $q$ both be prime numbers, and $7p + q$, $pq + 11$ are also prime numbers. Find the value of $\left(p^{2} + q^{p}\right)\left(q^{2} + p^{q}\right)$. | 1. If $p$ and $q$ are both odd, then $7p + q$ is even, and it is not a prime number. Therefore, one of $p$ and $q$ must be even.
Case one: Suppose $p$ is even, then $p = 2$. In this case, since $7p + q$ is a prime number, $q$ must be an odd prime. If $q \neq 3$, then $q \equiv 1$ or $2 \pmod{3}$.
If $q \equiv 1 \pmod... | 221 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,799 |
2 Let $p_{1}<p_{2}<p_{3}<p_{4}<p_{5}$ be 5 prime numbers, and $p_{1}, p_{2}, p_{3}, p_{4}, p_{5}$ form an arithmetic sequence. Find the minimum value of $p_{5}$. | 2. Let $d$ be the common difference, then $p_{1}, p_{1}+d, p_{1}+2 d, p_{1}+3 d, p_{1}+4 d$ are all primes. If $2 \nmid d$, i.e., $d$ is odd, then one of $p_{1}+d$ and $p_{1}+2 d$ is even, and it is not a prime. If $3 \nmid d$, then one of $p_{1}+d, p_{1}+2 d, p_{1}+3 d$ is a multiple of 3 (they form a complete residue... | 29 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,800 |
Example 3 Find all positive integers $n$, such that $\frac{n(n+1)}{2}-1$ is a prime number.
Find all positive integers $n$, such that $\frac{n(n+1)}{2}-1$ is a prime number. | Let $a_{n}=\frac{n(n+1)}{2}-1$, then $a_{1}=0$ is not a prime number, so we only need to discuss the case when $n>1$. We will use the fact that $n$ can only be of the form $4k$, $4k+1$, $4k+2$, or $4k+3$ and discuss each case separately.
When $n$ is of the form $4k+2$ or $4k+1$, $a_{n}$ is always even. For $a_{n}$ to ... | n=2 \text{ or } 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,801 |
For each positive integer $n$, let $S(n)$ denote the sum of the digits of $n$ in its decimal representation. Prove: for any positive integer $m$, there exists a positive integer $n$ such that $S(n) = m S(3 n)$. | 3. Notice that, for any positive integer $k, S(\underbrace{1}_{k \uparrow} \underbrace{0 \cdots 0})=9$, thus, let $1 \underbrace{0 \cdots 08}_{k \uparrow}=3 n$, then
$$n=\underbrace{3 \cdots 36}_{k \uparrow},$$
so $S(n)=3 k+6$. Therefore, for any positive integer $m$, take $k=3 m-2$, then
$$S(n)=m S(3 n) \text {. }$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,802 |
4 Find the largest positive integer $k$, such that there exists a positive integer $n$, satisfying $2^{k} \mid 3^{n}+1$.
| 4. Notice that, when $n$ is even, let $n=2 m$, we have
$$3^{n}=9^{m} \equiv 1(\bmod 8)$$
When $n=2 m+1$,
$$3^{n}=9^{m} \times 3 \equiv 3(\bmod 8)$$
Therefore, for any positive integer $n$, we have
$$3^{n}+1 \equiv 2 \text { or } 4(\bmod 8),$$
so $k \leqslant 2$. Also, $2^{2} \mid 3^{1}+1$, hence, the maximum value o... | 2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,803 |
5I Let $n$ be a positive integer. Prove: there exists a positive integer $m$ whose decimal representation contains only the digits 0 and 1, such that $n \mid m$.
| 5. Consider the sequence
$$1,11,111, \cdots, \underbrace{1 \cdots 1}_{n+1 \uparrow},$$
where there must be two numbers that are congruent modulo $n$ (because the remainder of any integer divided by $n$ can only be 0, 1, 2, ..., $n-1$, a total of $n$ cases), their difference (the larger minus the smaller) is the requir... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,804 |
6. Let $n$ be a positive odd number. Prove: There exists a positive integer $m$ whose decimal representation consists entirely of odd digits, such that $n \mid m$. | 6. If $(5, n)=1$, then by the conclusion of the previous problem, there exists $m=\underbrace{1 \cdots 1}_{i \uparrow} \underbrace{0 \cdots 0}_{i \uparrow}$, such that $n \mid m$, and $n$ is odd. Combining this with $5 \nmid n$, we know $(n, 10)=1$, hence $n \mid \underbrace{1 \cdots 1}_{i \uparrow}$. The proposition i... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,805 |
7 Prove: For each positive integer $n$, the number $19 \times 8^{n}+17$ is composite. | 7. If $n$ is even, then
$$19 \times 8^{n}+17 \equiv 1 \times(-1)^{n}+2 \equiv 0(\bmod 3)$$
If $n \equiv 1(\bmod 4)$, write $n=4 k+1$, then
$$\begin{aligned}
19 \times 8^{n}+17 & =19 \times 64^{2 k} \times 8+17 \\
& \equiv 6 \times(-1)^{2 k} \times 8+4 \\
& \equiv 0(\bmod 13)
\end{aligned}$$
If $n \equiv 3(\bmod 4)$, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,806 |
8 Fibonacci sequence $\left\{F_{n}\right\}$ is defined as follows: $F_{1}=F_{2}=1, F_{n+2}=F_{n+1}+F_{n}, n=1$, $2, \cdots$.
(1) Prove: The sum of any 10 consecutive terms of this sequence is a multiple of 11;
(2) Find the smallest positive integer $k$, such that the sum of any $k$ consecutive terms of this sequence is... | 8. Consider the sequence $\left\{F_{n}\right\}$ where each term is taken modulo 11 (or 12).
(1) $\left\{F_{n}(\bmod 11)\right\}: 1,1,2,3,5,-3,2,-1,1,0,1,1, \cdots$, so $\left\{F_{n}(\bmod 11)\right\}$ is a purely periodic sequence with a period of 10. Therefore,
the sum of any 10 consecutive terms in $\left\{F_{n}\righ... | 36 | Number Theory | proof | Yes | Yes | number_theory | false | 741,807 |
9 Let integers $a, b$ satisfy: $21 \mid a^{2}+b^{2}$. Prove: $441 \mid a^{2}+b^{2}$. | 9. First prove separately:
(1) If $a^{2}+b^{2} \equiv 0(\bmod 3)$, then $a \equiv b \equiv 0(\bmod 3)$;
(2) If $a^{2}+b^{2} \equiv 0(\bmod 7)$, then $a \equiv b \equiv 0(\bmod 7)$.
This only requires noting that, for any integer $x$, we have
$$x^{2} \equiv 0 \text { or } 1(\bmod 3),$$
and
$$x^{2} \equiv 0,1,2 \text { ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,808 |
10 positive integers $a$, $b$, $c$ satisfy: $c^{2}=a^{2}+b^{2}+ab$. Prove: $c$ has a prime factor greater than 5. | 10. We prove respectively:
(1) If $2 \mid c$, then $2|a, 2| b$;
(2) If $3 \mid c$, then $3|a, 3| b$;
(3) If $5 \mid c$, then $5|a, 5| b$.
The proof of (1) is trivial.
For (2), it suffices to note that
$$c^{2}=a^{2}+a b+b^{2}=(a-b)^{2}+3 a b,$$
which makes the proof straightforward.
For (3), from the condition, we know... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,809 |
I1 Fill the integers $1,2, \cdots, 9$ into a $3 \times 3$ table, one number per cell, such that the sum of the numbers in each row, each column, and each diagonal is a multiple of 9.
(1) Prove that the number in the center cell of the table is a multiple of 3;
(2) Provide a placement method that satisfies the condition... | 11. (1) Let the number filled in the cell at the $i$-th row and $j$-th column be $a_{ij}, 1 \leqslant i \leqslant 3, 1 \leqslant j \leqslant 3$, then
$$\begin{aligned}
& a_{11}+a_{22}+a_{33} \\
\equiv & a_{13}+a_{22}+a_{31} \\
\equiv & a_{12}+a_{22}+a_{32} \\
\equiv & a_{21}+a_{22}+a_{23} \\
\equiv & 0(\bmod 9)
\end{al... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,810 |
12 The following formula provides a method to determine whether a number is a multiple of 19: each time, remove the last digit of the number, double it, and add it to the remaining number. Continue this process until the number is reduced to a number less than 20. If the final number is 19, then the original number is ... | 12. Generally, let the number $\overline{a_{n} a_{n-1} \cdots a_{0}}$ be an $n+1$-digit number in decimal notation. Then, if it is a multiple of 19, we have
$$10 \overline{a_{n} a_{n-1} \cdots a_{1}}+a_{0}=\overline{a_{n} a_{n-1} \cdots a_{0}} \equiv 0(\bmod 19)$$
Thus,
$$20 \overline{a_{n} a_{n-1} \cdots a_{1}}+2 a_{... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,811 |
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