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38 Find all integers $a$ such that the equation $x^{2}+a x y+y^{2}=1$ has infinitely many integer solutions.
Find all integers $a$ such that the equation $x^{2}+a x y+y^{2}=1$ has infinitely many integer solutions. | 38. Discuss the number of integer solutions of the equation
$$x^{2}+a x y+y^{2}=1$$
for different values of $a$.
When $a=0$, the equation becomes $x^{2}+y^{2}=1$, which has only 4 sets of integer solutions.
If $a \neq 0$, then $(x, y)$ is a solution to equation (1) if and only if: $(x, -y)$ is a solution to the equatio... | |a|>1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,921 |
39 Find all positive integers $x, k, n (n \geqslant 2)$, such that
$$3^{k}-1=x^{n}$$ | 39. From the condition, we know $x \geqslant 2$.
If $n$ is even, then by the property of perfect squares $\equiv 0$ or $1(\bmod 3)$, we have $x^{n} \equiv 0$ or $1(\bmod 3)$, but $3^{k}-1 \equiv 2(\bmod 3)$, so there is no solution in this case.
If $n$ is odd, then
$$3^{k}=x^{n}-1=(x+1)\left(x^{n-1}-x^{n-2}+\cdots-x+... | (x, k, n)=(2,2,3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,922 |
Property 8 Let $a$ and $b$ be positive integers, then $[a, b]=\frac{a b}{(a, b)}$. | Let $c=\frac{a b}{(a, b)}$, then from $(a, b) \mid a$ and $(a, b) \mid b$ we know $b \mid c, a \mid c$. That is, $c$ is a common multiple of $a$ and $b$, hence $[a, b] \mid c$.
Conversely, by Bézout's theorem, there exist integers $x, y$ such that
$$a x + b y = (a, b)$$
That is,
Thus
$$\begin{array}{c}
\frac{a}{(a, ... | [a, b] = \frac{a b}{(a, b)} | Number Theory | proof | Yes | Yes | number_theory | false | 741,923 |
40 Find the smallest positive integer $n$ such that the indeterminate equation
$$n=x^{3}-x^{2} y+y^{2}+x-y$$
has no positive integer solutions. | 40. Let $F(x, y)=x^{3}-x^{2} y+y^{2}+x-y$, then $F(1,1)=1, F(1,2)=2$.
Therefore, when $n=1,2$, the equation has positive integer solutions.
Next, we prove that $F(x, y)=3$ has no positive integer solutions.
Consider the equation $F(x, y)=3$ as a quadratic equation in $y$
$$y^{2}-\left(x^{2}+1\right) y+x^{3}+x-3=0$$
I... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,924 |
Example 1 Let $a, b$ be positive integers, and $\frac{ab}{a+b}$ is also a positive integer. Prove: $(a, b)>1$.
| Prove that if $(a, b)=1$, then $(a, a+b)=1$ (this can be deduced from property 3), hence, from $a+b \mid a b$ and $(a, a+b)=1$, we get $a+b \mid b$, but $a+b>b$, so $a+b \mid b$ cannot hold. Therefore, $(a, b)>1$.
This is the translation of the provided text into English, maintaining the original formatting and line b... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,928 |
Example 2 Let positive integers $a, b, c$ satisfy $b^{2}=a c$. Prove: $(a, b)^{2}=a(a, c)$.
The text is translated while preserving the original line breaks and format. | To prove: If we can prove $(a, b)^{2}=\left(a^{2}, b^{2}\right)$, then by property 11, we know
$$(a, b)^{2}=\left(a^{2}, b^{2}\right)=\left(a^{2}, a c\right)=a(a, c)$$
The proposition is proved. $\square$
For this, let $d=(a, b)$, and assume $a=d u, b=d v$, then by property 11, $u$ and $v$ are two coprime positive int... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,929 |
Example 3 Find all positive integers $a, b (a \leqslant b)$, such that
$$a b=300+7[a, b]+5(a, b) .$$ | Let $[a, b]=x,(a, b)=y$, by property 8 we know $a b=x y$, thus, (1) becomes
$$x y=300+7 x+5 y$$
which is $(x-5)(y-7)=5 \times 67$.
Since $[a, b] \geqslant(a, b)$, hence $x \geqslant y$, thus $x-5>y-7$, there are only the following two cases.
Case one: $x-5=67$ and $y-7=5$; in this case, $x=72, y=12$, thus, we can se... | (12,72) \text{ or } (24,36) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,931 |
Example 4 Find all positive integers $a, b$ such that
$$(a, b)+9[a, b]+9(a+b)=7 a b .$$ | Let $(a, b)=d$, set $a=d x, b=d y$, then $(x, y)=1$ (by property 11), $[a, b]=d x y$ (by property 8), thus substituting into (1) we get
$$\begin{array}{c}
1+9 x y+9(x+y)=7 d x y \\
7 d=9+9\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{x y}
\end{array}$$
Therefore,
$$9<7 d \leqslant 9+9\left(\frac{1}{1}+\frac{1}{1}\righ... | (4,38), (38,4), (4,4) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,932 |
Example 5 The Fibonacci sequence is defined as follows: $F_{1}=F_{2}=1, F_{n+2}=F_{n+1}+F_{n}, n=1$, $2, \cdots$ Prove: For any positive integers $m$ and $n$, we have $\left(F_{m}, F_{n}\right)=F_{(m, n)}$. | Prove that when $m=n$, the proposition is obviously true. Now, without loss of generality, assume $m<n$, and note that
$$\begin{aligned}
F_{n} & =F_{2} F_{n-1}+F_{1} F_{n-2} \\
& =F_{2}\left(F_{n-2}+F_{n-3}\right)+F_{1} F_{n-2} \\
& =\left(F_{2}+F_{1}\right) F_{n-2}+F_{2} F_{n-3} \\
& =F_{3} F_{n-2}+F_{2} F_{n-3} \\
& ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,933 |
Example 6 Let $n$ be a positive integer greater than 1. Prove: there exist $n$ positive integers, which when arranged in ascending order form an arithmetic sequence (i.e., from the second term onwards, the difference between each term and the one before it is a constant), such that any two of them are coprime. | Consider the following $n$ numbers:
$$n!+1,2 \times(n!)+1, \cdots, n \times(n!)+1$$
These $n$ positive integers form an arithmetic sequence with a common difference of $n!$.
We prove that any two of them are coprime.
In fact, if there exist $1 \leqslant i < j \leqslant n$ such that $\gcd(i \times(n!)+1, j \times(n!)+1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,934 |
Fundamental Theorem of Arithmetic Let $n$ be a positive integer greater than 1, then $n$ can be decomposed into the product of several prime numbers, and this decomposition is unique when the order of these prime factors is not considered. That is, for any positive integer $n$ greater than 1, there exists a unique prim... | To prove the existence, we can use the previous analysis. Below, we prove the uniqueness.
If \( n \) has two prime factorization forms:
\[ n = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}} = q_{1}^{\beta_{1}} q_{2}^{\beta_{2}} \cdots q_{l}^{\beta_{l}} \]
where \( p_{1} < p_{2} < \cdots < p_{k} \) an... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,935 |
Example 1 There are 2012 lamps, numbered $1, 2, \cdots, 2012$, arranged in a row in a corridor, and initially, each lamp is on. A mischievous student performed the following 2012 operations: for $1 \leqslant k \leqslant 2012$, during the $k$-th operation, the student toggled the switch of all lamps whose numbers are mu... | Let $1 \leqslant n \leqslant 2012$, we examine the state of the $n$-th lamp. According to the problem, the switch of this lamp is pulled $d(n)$ times. An even number of pulls does not change the initial state of the lamp, while an odd number of pulls changes the state of the lamp from its initial state.
Using the prop... | 1968 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,936 |
Example 2 Find all positive integers $n$ such that $n=d(n)^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | When $n=1$, the condition is satisfied. Below, we consider the case where $n>1$.
From the condition, we know that $n$ is a perfect square, so $d(n)$ is odd. Let $d(n)=2k+1$. For any positive integer $d$, when $d \mid n$, we have $\left.\frac{n}{d} \right\rvert\, n$. Therefore, by pairing $d$ with $\frac{n}{d}$, we know... | n=1 \text{ or } 9 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,937 |
Example 3 Let $n$ be a positive integer. Prove: The number $2^{2^{n}}+2^{2^{n-1}}+1$ has at least $n$ distinct prime factors. | We make the following decomposition:
$$\begin{aligned}
& 2^{2^{n}}+2^{2^{n-1}}+1 \\
= & \left(2^{2^{n}}+1\right)^{2}-2^{2^{n-1}} \\
= & \left(2^{2^{n-1}}+2^{2^{n-2}}+1\right)\left(2^{2^{n-1}}-2^{2^{n-2}}+1\right) \\
= & \left(2^{2^{n-2}}+2^{2^{n-3}}+1\right)\left(2^{2^{n-2}}-2^{2^{n-3}}+1\right)\left(2^{2^{n-1}}-2^{2^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,938 |
Example 4 Let $m$ and $n$ be positive integers, and the product of all positive divisors of $m$ equals the product of all positive divisors of $n$. Ask: Must $m$ and $n$ be equal? | Solve $m$ and $n$ must be equal.
In fact, by pairing the positive divisor $d$ of $m$ with $\frac{m}{d}$, we know that the product of all positive divisors of $m$ is $m^{\frac{d(m)}{2}}$, thus, the condition is equivalent to
$$m^{d(m)}=n^{d(n)},$$
This equation indicates that $m$ and $n$ have the same prime factors. We... | m=n | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,939 |
Example 5 Find all positive integers $x, y$ such that
$$y^{x}=x^{50}$$ | Let $x, y$ be positive integers satisfying the conditions, and let $x=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}$ be the prime factorization of $x$. Then, since $y$ is a positive integer, for $1 \leqslant i \leqslant k$, we have $x \mid 50 \alpha_{i}$. Now, let's first discuss the prime factors of $x$.
If $x$ ha... | (x, y) = (1,1), \left(2,2^{25}\right), \left(2^{2}, 2^{25}\right), \left(5,5^{10}\right), \left(5^{2}, 5^{4}\right), \left(10,10^{5}\right), (50,50), (100,10) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,940 |
Example 2 Proof: Regardless of how many 3s are added between the two 0s in the number 12008, the resulting number is always a multiple of 19.
---
The translation maintains the original text's line breaks and format. | Let $a_{0}=12008, a_{n}=120 \underbrace{3 \cdots 3}_{n \text { 3's}} 08, n=1,2, \cdots$.
First, since
$$a_{0}=19 \times 632,$$
it follows that
$$19 \mid a_{0} .$$
Second, suppose $19 \mid a_{n}$, then from
$$\begin{array}{c}
a_{n+1}-10 a_{n}=228=19 \times 12 \\
19 \mid a_{n+1}
\end{array}$$
Therefore, for all intege... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,941 |
Example 6 Given a positive integer $n>1$, let $d_{1}, d_{2}, \cdots, d_{n}$ be positive integers satisfying: $\left(d_{1}, d_{2}, \cdots, d_{n}\right)=1$, and for $j=1,2, \cdots, n$ we have $d_{j} \mid \sum_{i=1}^{n} d_{i}\left(\right.$ here $\sum_{i=1}^{n} d_{i}=d_{1}+d_{2}+\cdots+ d_{n}$ ).
(1) Prove: $d_{1} d_{2} \c... | (1) Let $p$ be a prime factor of $d_{1} d_{2} \cdots d_{n}$, and let $k$ be the maximum power of $p$ in the prime factorization of each $d_{i}$. Then, from $d_{j} \mid \sum_{i=1}^{n} d_{i}$, it follows that $p^{k} \mid \sum_{i=1}^{n} d_{i}$, hence $p^{k(n-2)} \mid\left(\sum_{i=1}^{n} d_{i}\right)^{n-2}$.
Since $\left(... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,942 |
1 Let $n$ be a positive integer greater than 1. Prove: $n^{4}+4^{n}$ is a composite number. | 1. When $n$ is even, $n^{4}+4^{n}$ is an even number greater than 2, hence it is a composite number. When $n$ is odd, let $n=2 k+1$, then
$$n^{4}+4^{n}=n^{4}+4 \times\left(2^{k}\right)^{4}$$
Using
$$\begin{aligned}
x^{4}+4 y^{4} & =\left(x^{2}+2 y^{2}\right)^{2}-4 x^{2} y^{2} \\
& =\left(x^{2}-2 x y+2 y^{2}\right)\lef... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,943 |
2 Find all integer $x$ such that $\left|4 x^{2}-12 x-27\right|$ is a prime number.
Find all integer $x$ such that $\left|4 x^{2}-12 x-27\right|$ is a prime number. | 2. From $\left|4 x^{2}-12 x-27\right|=|(2 x+3)(2 x-9)|$, we know that only when $|2 x+3|=1$ or $|2 x-9|=1$, the number $\left|4 x^{2}-12 x-27\right|$ can possibly be a prime number. According to this, the required $x=-2,-1,4$ or 5, and the corresponding $\left|4 x^{2}-12 x-27\right|$ are 13, 11, 11 or 13, all of which ... | x=-2, -1, 4, 5 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,944 |
3 Let $m$ be a positive integer greater than 1, and $m \mid (m-1)! + 1$. Prove: $m$ is a prime number. | 3. If $m$ is a composite number, then there exists a positive integer $p$, such that $2 \leqslant p<m$, and $p \mid m$. In this case, $p \mid (m-1)!$, but $m \mid (m-1)!+1$, so $p \mid (m-1)!+1$, which leads to $p \mid 1$, a contradiction. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,945 |
Does there exist 3 different prime numbers $p, q, r$, such that the following divisibility relations all hold?
$$q r \mid p^{2}+d, r p \mid q^{2}+d, p q \mid r^{2}+d,$$
where (1) $d=10$; (2) $d=11$. | 4. Suppose $p<q<r$, then
$$q \geqslant p+1, r \geqslant q+2 \geqslant p+3$$
For the case of $d=10$, since $q r \mid p^{2}+10$, it should have
$$p^{2}+10 \geqslant(p+1)(p+3)$$
This requires $4 p \leqslant 7$, i.e., $p \leqslant 1$, which is a contradiction. Therefore, there are no $p, q, r$ that meet the requirements ... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,946 |
5 Let $p$ be a positive integer, and $2^{p}-1$ be a prime number. Prove: $p$ is a prime number. | 5. If $p$ is a composite number, let $p=q r, 2 \leqslant q \leqslant r$, then
$$2^{p}-1=\left(2^{q}\right)^{r}-1=\left(2^{q}-1\right)\left(\left(2^{q}\right)^{r-1}+\left(2^{q}\right)^{r-2}+\cdots+1\right),$$
this implies $2^{q}-1 \mid 2^{p}-1$, which contradicts that $2^{p}-1$ is a prime number. Therefore, $p$ is a pr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,947 |
6 Let $n$ be a positive integer, and $2^{n}+1$ be a prime number. Prove: there exists a non-negative integer $k$, such that $n=2^{k}$.
The text above is translated into English, preserving the original text's line breaks and format. | 6. By the Fundamental Theorem of Arithmetic, we can write \( n = 2^k \cdot q, k \geqslant 0, q \) as an odd number. If \( q > 1 \), then
\[
\begin{aligned}
2^n + 1 & = \left(2^{2^k}\right)^q + 1 \\
& = (x + 1)\left(x^{q-1} - x^{q-2} + \cdots - x + 1\right)
\end{aligned}
\]
is the product of two positive integers great... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,948 |
Find all primes of the form $n^{n}+1$ that do not exceed $10^{19}$, where $n$ is a positive integer. | 7. When $n=1$, $n^{n}+1=2$ satisfies the condition. When $n>1$, let $n=2^{k} q$, where $q$ is an odd number. If $q>1$, as shown in the previous problem, $n^{n}+1$ is not a prime number, so $n=2^{k}$, where $k$ is a positive integer. At this point,
$$n^{n}+1=2^{k \cdot 2^{k}}+1=\left(2^{2^{k}}\right)^{k}+1$$
Further an... | 2, 5, 257 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,949 |
8 Let $a, b, c, d$ be integers, and $a \neq c, a-c \mid ab+cd$. Prove: $a-c \mid ad+bc$.
The translation is provided while preserving the original text's line breaks and format. | 8. Using
$$\begin{aligned}
& (a d+b c)-(a b+c d) \\
= & d(a-c)-b(a-c) \\
= & (d-b)(a-c),
\end{aligned}$$
and $a-c \mid a b+c d$, we can get $a-c \mid a d+b c$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,950 |
9 Let $a, b, c, d$ be integers, and $ac, bc + ad, bd$ are all multiples of some integer $u$. Prove: the numbers $bc$ and $ad$ are also multiples of $u$.
| 9. From the identity
$$(b c+a d)^{2}+(b c-a d)^{2}=4 a b c d=4(a c)(b d)$$
combined with the condition, we know that $u^{2} \mid (b c-a d)^{2}$, hence $u \mid b c-a d$.
Now, let $b c+a d=u x, b c-a d=u y$, then from (1) we have,
$$x^{2}+y^{2}=4\left(\frac{a c}{u}\right)\left(\frac{b d}{u}\right)$$
Thus, $x^{2}+y^{2}$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,951 |
Example 3 Given that any 10 consecutive digits of a 1000-digit positive integer form a 10-digit number that is a multiple of $2^{10}$. Prove: the positive integer is a multiple of $2^{1000}$. | Proof Let the positive integer $x=a_{1} a_{2} \cdots a_{1000}$, where $a_{i}$ are decimal digits. From the condition, we know
$$2^{10}\left|\overline{a_{991} \cdots a_{1000}}, 2^{10}\right| \overline{a_{990} \cdots a_{999}},$$
Therefore,
$$2^{10} \mid \overline{a_{990} \cdots a_{999}} \times 10$$
Let $y=\overline{a_{... | 2^{1000} \mid x | Number Theory | proof | Yes | Yes | number_theory | false | 741,952 |
10 Let $a$, $b$, $n$ be given positive integers, and for any positive integer $k(\neq b)$, we have $b-k \mid a-k^{n}$. Prove: $a=b^{n}$. | 10. Notice that for any positive integer $k(\neq b)$, we have $b-k \mid b^{n}-k^{n}$, combined with $b-k \mid a-k^{n}$, it follows that $b-k \mid a-b^{n}$. This indicates that $a-b^{n}$ is a multiple of every positive integer, hence $a-b^{n}=0$, yielding $a=b^{n}$. | a=b^{n} | Number Theory | proof | Yes | Yes | number_theory | false | 741,953 |
11 Given a positive integer $n$, among its positive divisors, there is at least one positive integer ending in each of the digits $0,1,2, \cdots, 9$. Find the smallest $n$ that satisfies this condition. | 11. The smallest $n$ that satisfies the condition is $270$.
In fact, from the condition, we know that $10 \mid n$, and we start the discussion from the factor of the last digit of $n$ being 9. If $9 \mid n$, then $90 \mid n$, and it can be directly verified that 90 and 180 are not multiples of any number ending in 7; ... | 270 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,954 |
12 Find a 9-digit number $M$, such that the digits of $M$ are all different and non-zero, and for $m=2,3, \cdots$, 9, the left $m$ digits of $M$ are multiples of $m$.
Find a 9-digit number $M$, such that the digits of $M$ are all different and non-zero, and for $m=2,3, \cdots$, 9, the number formed by the left $m$ dig... | 12. Let $M=\overline{a_{1} a_{2} \cdots a_{9}}$ be a number that satisfies the given conditions. From the conditions, we know that $a_{5}=5$, and $a_{2}$, $a_{4}$, $a_{6}$, $a_{8}$ are a permutation of $2$, $4$, $6$, $8$. Consequently, $a_{1}$, $a_{3}$, $a_{7}$, $a_{9}$ are a permutation of $1$, $3$, $7$, $9$. Therefor... | 381654729 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,955 |
14 Let the prime numbers in ascending order be $p_{1}, p_{2}, p_{3}, \cdots$. Prove: When $n \geqslant 2$, the number $p_{n}+p_{n+1}$ can be expressed as the product of 3 integers greater than 1 (which can be the same).
保留源文本的换行和格式，翻译结果如下：
14 Let the prime numbers in ascending order be $p_{1}, p_{2}, p_{3}, \cdots$. ... | 14. When $n \geqslant 2$, $p_{n}$ and $p_{n+1}$ are both odd numbers, so $q=\frac{p_{n}+p_{n+1}}{2}$ is a positive integer. Also, $p_{n}<q<p_{n+1}$, and $p_{n}$ and $p_{n+1}$ are two consecutive primes, hence $q$ must be a composite number. Therefore, $q$ can be written as the product of two positive integers greater t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,957 |
15 Let $n$ be a positive integer greater than 1. Prove: $n$ is a composite number if and only if there exist positive integers $a, b, x, y$, such that $n=a+b, \frac{x}{a}+\frac{y}{b}=1$. | 15. If there exist $a, b, x, y$, such that
$$n=a+b \text{, and } \frac{x}{a}+\frac{y}{b}=1 \text{.}$$
We denote $d=(a, b)$, if $d=1$, then
we know
so
$$\begin{array}{l}
\frac{x}{a}+\frac{y}{b}=1 \\
b x+a y=a b \\
a|b x, b| a y
\end{array}$$
Combining $(a, b)=1$, it follows that $a|x, b| y$, thus
$$a b=b x+a y \geqsl... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,958 |
16 Prove: In the sequence $10001,100010001,1000100010001, \cdots$, every number is composite. | 16. Notice that $10001=73 \times 137$ is a composite number, and from the second term onward, we have
$$\begin{aligned}
a_{n} & =1 \underbrace{100010001 \cdots 001}_{n \uparrow 0001} \\
& =10^{4 n}+10^{4(n-1)}+\cdots+10^{4}+1 \\
& =\frac{10^{4(n+1)}-1}{10^{4}-1} \\
& =\frac{\left(10^{2(n+1)}-1\right)\left(10^{2(n+1)}+1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,959 |
17 Let $a, b, c, d$ all be prime numbers, and $a>3b>6c>12d, a^{2}-b^{2}+c^{2}-d^{2}=1749$. Find all possible values of $a^{2}+b^{2}+c^{2}+d^{2}$. | 17. From $a^{2}-b^{2}+c^{2}-d^{2}=1749$ being odd, we know that one of $a, b, c, d$ must be even, indicating that $d=2$. Then,
from
$$\begin{array}{l}
a^{2}-b^{2}+c^{2}=1753 \\
a>3 b>6 c>12 d
\end{array}$$
we know $c \geqslant 5, b \geqslant 2 c+1, a \geqslant 3 b+1$, so
$$\begin{aligned}
a^{2}-b^{2}+c^{2} & \geqslan... | 1999 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,960 |
19 The sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integers $m, n$, if $m \mid n, m<n$, then $a_{m} \mid a_{n}$, and $a_{m}<a_{n}$. Find the minimum possible value of $a_{2000}$. | 19. From the conditions, when $m \mid n$ and $m < n$, we have $a_{n} \geqslant 2 a_{m}$. Therefore, $a_{1} \geqslant 1, a_{2} \geqslant 2, a_{4} \geqslant 2 a_{2} \geqslant 2^{2}$, similarly, $a_{8} \geqslant 2^{3}, a_{16} \geqslant 2^{4}, a_{30} \geqslant 2^{5}, a_{400} \geqslant 2^{6}, a_{2000} \geqslant 2^{7}$, whic... | 128 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,962 |
Example 4 Let $m$ be a positive integer greater than 2, prove: for any positive integer $n$, we have $2^{m}-1 \nmid$
$$2^{n}+1$$ | Prove that if there exists a positive integer $n$, such that $2^{m}-1 \mid 2^{n}+1$, then take the smallest such $n$.
Since $m>2$, we know $n>1$, and further, we should have $2^{n}+1 \geqslant 2^{m}-1$, so $n \geqslant m$. If $n=m$, it would lead to $2^{m}-1 \mid 2$ (since $2=\left(2^{n}+1\right)-\left(2^{n}-1\right)$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,963 |
20 Let $p$ be an odd prime, and positive integers $m, n$ satisfy $\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{p-1}$. Prove: $p \mid m$. | 20. From the condition, we know
$$\begin{aligned}
\frac{2 m}{n} & =\left(1+\frac{1}{2}+\cdots+\frac{1}{p-1}\right)+\left(\frac{1}{p-1}+\frac{1}{p-2}+\cdots+1\right) \\
& =\left(1+\frac{1}{p-1}\right)+\left(\frac{1}{2}+\frac{1}{p-2}\right)+\cdots+\left(\frac{1}{p-1}+1\right) \\
& =\frac{p}{1 \times(p-1)}+\frac{p}{2 \tim... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,964 |
21 Let $a$, $m$, $n$ be positive integers, $a>1$, and $a^{m}+1 \mid a^{n}+1$. Prove: $m \mid n$. | 21. If $m \nmid n$, from $a^{m}+1 \mid a^{n}+1$ and $a>1$, we know that $m1$, $a^{m}+1>a^{r}+1$, which is a contradiction.
Therefore, $m \mid n$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,965 |
22 Prove: For any positive integer $n$ and positive odd integer $m$, we have $\left(2^{m}-1,2^{n}+1\right)=1$.
| 22. Let $d=\left(2^{m}-1,2^{n}+1\right)$, then
hence
that is
$$\begin{array}{c}
d \mid 2^{m}-1 \\
d \mid\left(2^{m}\right)^{n}-1^{n} \\
d \mid 2^{n}-1
\end{array}$$
Additionally, $d \mid 2^{n}+1$, and since $m$ is odd, we have
$$2^{n}+1 \mid\left(2^{n}\right)^{m}+1^{m},$$
thus
$$d \mid 2^{m}+1$$
Comparing the two de... | 1 | Number Theory | proof | Yes | Yes | number_theory | false | 741,966 |
24 Given positive integers $a, b, c, d$ whose least common multiple is $a+b+c+d$. Prove: $abcd$ is a multiple of 3 or 5. | 24. From the conditions, we know that $a, b, c, d$ are not all equal. Without loss of generality, assume $d$ is the largest number, then
$$d3$ cases.
When $z \geqslant 6$, we have
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$$
Therefore, it can only be that $x=y=z=6$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,968 |
25 Let $M_{n}$ denote the least common multiple of the positive integers $1,2, \cdots, n$. Find all positive integers $n(>1)$ such that $M_{n}=M_{n-1}$. | 25. If $n$ has at least two distinct prime factors, then we can write $n=p q$, where $2 \leqslant p<q$, $p, q$ are positive integers, and $(p, q)=1$. In this case, $2 \leqslant p<q<n-1$, thus $n \mid M_{n-1}$, so $M_{n}=M_{n-1}$.
If $n=p^{\alpha}$, where $p$ is a prime and $\alpha$ is a positive integer, then each num... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,969 |
26 Let $a, m, n$ be positive integers, $a>1$. Prove: $\left(a^{m}-1, a^{n}-1\right)=a^{(m, n)}-1$. | 26. Suppose $m>n$, then
and
$$\begin{array}{l}
\left(a^{m}-1, a^{n}-1\right)=\left(a^{m}-a^{n}, a^{n}-1\right) \\
=\left(a^{n}\left(a^{m-n}-1\right), a^{n}-1\right), \\
\left(a^{n}, a^{n}-1\right)=1, \\
\left(a^{m}-1, a^{n}-1\right)=\left(a^{m-n}-1, a^{n}-1\right),
\end{array}$$
By repeatedly applying this, performing... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,970 |
27 Let $a, n$ be positive integers, $a>1$, and $a^{n}+1$ is a prime. Prove: $d\left(a^{n}-1\right) \geqslant n$.
| 27. Given that $a^{n}+1$ is a prime number, we know that $a$ is even. Similar to problem 6, there exists a non-negative integer $k$ such that $n=2^{k}$. Therefore,
$$\begin{aligned}
& a^{n}-1 \\
= & a^{2^{k}}-1 \\
= & \left(a^{2^{k-1}}-1\right)\left(a^{2^{k-1}}+1\right) \\
= & \cdots \\
= & (a-1)(a+1)\left(a^{2}+1\righ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,971 |
For which positive integers $n(>2)$, 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 $n=3$, for any three consecutive positive integers $a-1, a, a+1$, if
then
and
so
$$\begin{array}{l}
a+1 \mid [a-1, a] \\
a+1 \mid a(a-1) \\
(a+1, a)=1 \\
a+1 \mid a-1
\end{array}$$
This is a contradiction.
When $n>3$, if $n$ is even, let $n=2m$, then among the numbers $2m-1, 2m, \cdots, 2(2m-1)$, the largest num... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,972 |
29 Let positive integers $a, b, m, n$ satisfy: $(a, b)=1, a>1$, and $a^{m}+b^{m} \mid a^{n}+b^{n}$. Prove: $m \mid n$. | 29. Utilize
$$a^{n}+b^{n}=\left(a^{n-m}+b^{n-m}\right)\left(a^{m}+b^{m}\right)-\left(a^{m} b^{n-m}+a^{n-m} b^{m}\right),$$
we know that if $n \geqslant 2 m$, then
$$a^{n}+b^{n}=\left(a^{n-m}+b^{n-m}\right)\left(a^{m}+b^{m}\right)-a^{m} b^{m}\left(a^{n-2 m}+b^{n-2 m}\right),$$
thus
$$a^{m}+b^{m} \mid a^{m} b^{m}\left(... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,973 |
30 Proof: There exist 2012 different positive integers, such that for any two different numbers $a$ and $b$, the condition $(a-b)^{2} \mid a b$ is satisfied. | 30. Generalizing the proposition, it can be proven: For any $n(\geqslant 2)$, there exist $n$ distinct positive integers such that for any two different numbers $a, b$, we have $(a-b)^{2} \mid a b$. The proof is as follows:
When $n=2$, take $a_{1}=1, a_{2}=2$, then they satisfy the condition.
Now assume $a_{1}<a_{2}<\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,975 |
31 Let $a, b$ be positive integers, and $(a, b)=1$. Prove: For any positive integer $m$, the sequence
$$a, a+b, a+2b, \cdots, a+nb, \cdots$$
contains infinitely many numbers that are coprime to $m$. | 31. For any positive integer $m$, given that $(a, b)=1$, we can write $m=m_{1} m_{2}$, such that the prime factors of $m_{1}$ are all prime factors of $a$, and
$$\left(a, m_{2}\right)=1,\left(m_{1}, b\right)=1,\left(m_{1}, m_{2}\right)=1$$
(This can be achieved by appropriately distributing the prime factorizations of ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,976 |
32 Given the pair of positive integers $(a, b)$ satisfies: the number $a^{a} \cdot b^{b}$ in decimal notation ends with exactly 98 zeros. Find the minimum value of $a b$.
| 32. Let the prime factorization of $a$ and $b$ have the powers of $2$ and $5$ as $\alpha_{1}$, $\beta_{1}$ and $\alpha_{2}$, $\beta_{2}$, respectively, then
$$\left\{\begin{array}{l}
a \cdot \alpha_{1} + b \cdot \alpha_{2} \geqslant 98 \\
a \cdot \beta_{1} + b \cdot \beta_{2} \geqslant 98
\end{array}\right.$$
and one ... | 7350 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,977 |
33 Find all positive integers $m$ such that $m=d(m)^{4}$.
Find all positive integers $m$ such that $m=d(m)^{4}$. | 33. From the conditions, we know that \( m \) is a fourth power, so we can set
$$m=2^{4 a_{2}} \cdot 3^{4 a_{3}} \cdot 5^{4 a_{5}} \cdot 7^{4 a_{7}} \cdots \cdots$$
where \(\alpha_{2}, \alpha_{3}, \alpha_{5}, \alpha_{7}, \cdots\) are non-negative integers. And
$$d(m)=\left(4 \alpha_{2}+1\right)\left(4 \alpha_{3}+1\rig... | m=1,5^{4}, 3^{8}, 3^{8} \cdot 5^{4} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,978 |
34 Prove: Every positive integer can be expressed as the difference of two positive integers, and these two positive integers have the same number of prime factors.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 34. Let $n$ be a positive integer. If $n$ is even, then the expression $n=(2 n)-n$ satisfies the requirement. If $n$ is odd, let $p$ be the smallest odd prime that does not divide $n$. Then the expression $n=p n-(p-1) n$ shows that the number of prime factors of $p n$ is one more than the number of prime factors of $n$... | null | Number Theory | proof | Yes | Yes | number_theory | false | 741,979 |
35 Find all positive integers $a$, $b$, and $c$, such that $a^{2}+1$ and $b^{2}+1$ are both prime, and satisfy
$$\left(a^{2}+1\right)\left(b^{2}+1\right)=c^{2}+1$$ | 35. Let's assume $a \leqslant b$, from the condition we have
$$a^{2}\left(b^{2}+1\right)=c^{2}+1-b^{2}-1=(c-b)(c+b)$$
Thus, $b^{2}+1 \mid c-b$ or $b^{2}+1 \mid c+b$ (here we use the fact that $b^{2}+1$ is a prime).
If
$$b^{2}+1 \mid c-b$$
then $\square$
$$c-b \geqslant b^{2}+1 \text { (note that } c>b \text { is obv... | (1,2,3) \text{ or } (2,1,3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,980 |
36 Let $p(k)$ denote the greatest odd divisor of the positive integer $k$. Prove: for any positive integer $n$, we have $\frac{2}{3} n<$
$$\sum_{k=1}^{n} \frac{p(k)}{k}<\frac{2}{3}(n+1) \text {. }$$ | 36. Let $S_{n}=\sum_{k=1}^{n} \frac{p(k)}{k}$, then by the definition of $p(k)$ we have
$$\begin{aligned}
S_{2 n} & =\sum_{k=1}^{2 n} \frac{p(k)}{k}=\sum_{k=1}^{n} \frac{p(2 k-1)}{2 k-1}+\sum_{k=1}^{n} \frac{p(2 k)}{2 k} \\
& =n+\frac{1}{2} \sum_{k=1}^{n} \frac{p(2 k)}{k}=n+\frac{1}{2} S_{n}
\end{aligned}$$
Similarly,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,981 |
37 Let $a$, $b$, $c$ all be positive integers greater than 1. Find the minimum possible value of the algebraic expression $\frac{a+b+c}{2}-\frac{[a, b]+[b, c]+[c, a]}{a+b+c}$. | 37. By symmetry, we may assume $a \geqslant b \geqslant c$. Note that when $(a, b, c)=(2,2,2),(3,2,2),(3,3,2),(4,2,2)$, the value of the given algebraic expression $A$ is $2, \frac{3}{2}, \frac{17}{8}, \frac{11}{4}$, respectively. This indicates that when $a+b+c \leqslant 8$, $A \geqslant \frac{3}{2}$.
Next, we prove ... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,982 |
38 For any given prime $p$, how many integer triples $(a, b, c)$ are there such that
(1) $1 \leqslant a, b, c \leqslant 2 p^{2}$;
(2) $\frac{[a, c]+[b, c]}{a+b}=\frac{p^{2}+1}{p^{2}+2} \cdot c$. | 38. Let $u=(a, c), v=(b, c)$, then condition (2) becomes
$$\frac{\frac{a c}{u}+\frac{b c}{v}}{a+b}=\frac{p^{2}+1}{p^{2}+2} \cdot c$$
i.e.,
$$\frac{a}{u}+\frac{b}{v}=\frac{p^{2}+1}{p^{2}+2}(a+b)$$
When $\frac{1}{2}2$, the number of arrays satisfying the condition is $p^{2}-p$. | p^{2}-p | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,983 |
39 The numbers $1,2, \cdots, 33$ are written on the blackboard. Each time, it is allowed to perform the following operation: take any two numbers $x, y$ from the blackboard that satisfy $x \mid y$, remove them from the blackboard, and write the number $\frac{y}{x}$. Continue until there are no such two numbers on the b... | 39. Consider the objective function $S=$ the product of all numbers on the blackboard.
Initially, $S=33!=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$. Each operation targets $x, y (x \mid y)$, where $y=kx$. Removing $x, y$ and replacing them with $... | 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,984 |
Property 2 If for any prime $p$ between 1 and $\sqrt{n}$, $p \nmid n$, then $n$ is a prime. Here $n(>1)$ is a positive integer. | Proof: In fact, if $n$ is a composite number, then it can be written as $n=p q, 2 \leqslant p \leqslant q$. Therefore, $p^{2} \leqslant n$, which means $p \leqslant \sqrt{n}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,985 |
40. Let $n$ be a positive integer. Prove: The number $1+5^{n}+5^{2 n}+5^{3 n}+5^{4 n}$ is a composite number. | When $n$ is even, let $n=2 m, x=5^{m}$, then
$$\begin{aligned}
A & =1+5^{n}+5^{2 n}+5^{3 n}+5^{4 n}=1+x^{2}+x^{4}+x^{6}+x^{8} \\
& =\frac{x^{10}-1}{x^{2}-1}=\frac{\left(x^{5}-1\right)\left(x^{5}+1\right)}{(x-1)(x+1)} \\
& =\left(x^{4}+x^{3}+x^{2}+x+1\right)\left(x^{4}-x^{3}+x^{2}-x+1\right)
\end{aligned}$$
Since $x=5^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,986 |
Property 2 If $a \equiv b(\bmod m), c \equiv d(\bmod m)$, then $a+c \equiv b+d(\bmod m)$, $a-c \equiv b-d(\bmod m), a c \equiv b d(\bmod m)$ | To prove: these conclusions and equations are very similar, and they are all easy to prove. We only provide the proof for the 3rd equation.
We only need to prove: $m \mid a c-b d$.
Since
$$\begin{aligned}
a c-b d & =a c-b c+b c-b d \\
& =(a-b) c+b(c-d)
\end{aligned}$$
By the condition $m|a-b, m| c-d$, we know $m \mid... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,988 |
Property 6 If $(a, m)=1$, then there exists an integer $b$, such that $a b \equiv 1(\bmod m)$. This $b$ is called the number-theoretic inverse of $a$ modulo $m$, denoted as $a^{-1}(\bmod m)$, and is often simply written as $a^{-1}$ when there is no risk of confusion. | By Bézout's Theorem, we know there exist integers $x, y$ such that
$$a x + m y = 1.$$
Thus, $m \mid a x - 1$, i.e., $a x \equiv 1 \pmod{m}$, so there exists a $b$ that satisfies the condition. | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,992 |
Example 1 Find all prime numbers $p, q, r (p \leqslant q \leqslant r)$ such that
$$p q+r, p q+r^{2}, q r+p, q r+p^{2}, r p+q, r p+q^{2}$$
are all prime numbers. | If $p>2$, then $p, q, r$ are all odd numbers, in which case $pq+r$ is an even number greater than 2, leading to a contradiction. Therefore, $p=2$. Now, the numbers
$$2q+r, 2q+r^2, qr+2, qr+4, 2r+q, 2r+q^2$$
are all prime numbers.
If $q, r$ include an even number, then $qr+2$ is an even number greater than 2, leading t... | p=2, q=3, r=5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,993 |
Example 2 Let $n$ be a positive integer greater than 1, and suppose that any two numbers from $1!, 2!, \cdots, n!$ yield different remainders when divided by $n$. Prove: $n$ is a prime number. | Notice that, $n! \equiv 0 \pmod{n}$, and when $n=4$, we have $2! \equiv 3! \pmod{4}$. Therefore, if we can prove that for any composite number $n$ greater than 4, $(n-1)! \equiv 0 \pmod{n}$, we can derive a contradiction from the conditions given in the problem. Thus, we can prove that $n$ is a prime number.
In fact, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,994 |
Example 3 Let integers $x, y, z$ satisfy
$$(x-y)(y-z)(z-x)=x+y+z .$$
Prove: $x+y+z$ is a multiple of 27. | Consider the remainders of $x$, $y$, and $z$ when divided by 3. If any two of $x$, $y$, and $z$ are not congruent modulo 3, then
$$x+y+z \equiv 0+1+2 \equiv 0(\bmod 3)$$
However, $3 \nmid (x-y)(y-z)(z-x)$, which contradicts (1).
Now, at least two of $x$, $y$, and $z$ must be congruent modulo 3. By symmetry, assume $x ... | 27 \mid x+y+z | Number Theory | proof | Yes | Yes | number_theory | false | 741,995 |
Property 3 There are infinitely many prime numbers. | Proof: If there are only a finite number of primes, let them be $p_{1}<p_{2}<\cdots<p_{n}$. Consider the number
$$x=p_{1} p_{2} \cdots p_{n}+1$$
Its smallest divisor greater than 1, $p$, is a prime, so $p$ should be one of $p_{1}, p_{2}, \cdots, p_{n}$. Let $p=p_{i}, 1 \leqslant i \leqslant n$, and $x=p_{i} y$, then $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,996 |
Example 4: Does there exist 19 different positive integers such that in their decimal representation, their digit sums are the same, and the sum of these 19 numbers is $1999$? | To solve this problem, we need to use a well-known conclusion: in decimal notation, each positive integer is congruent to the sum of its digits modulo 9. (This conclusion can be proven using $10^{k} \equiv 1(\bmod 9)$)
If there exist 19 different positive integers that satisfy the condition, then by their digit sums b... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,997 |
Example 5 Let $m, n, k$ be positive integers, $n \geqslant m+2, k$ be an odd number greater than 1, and $p=k \times 2^{n}+1$ be a prime number, $p \mid 2^{2^{m}}+1$. Prove: $k^{2^{n-1}} \equiv 1(\bmod p)$. | Given that $2^{2^{m}} \equiv-1(\bmod p)$, and $n \geqslant m+2$, hence $2^{m+1}$ is a factor of $n \cdot 2^{n-1}$, so, $2^{n \cdot 2^{n-1}} \equiv(-1)^{2 t}=1(\bmod p)$ (here $\left.t=n \cdot 2^{n m-2}\right)$.
Now, from $k \cdot 2^{n} \equiv-1(\bmod p)$, we know $k^{2^{n-1}} \cdot 2^{n \cdot 2^{n-1}} \equiv\left(-13^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,998 |
Example 6 Let $m$ be a positive integer, prove: there exist integers $a, b, k$, such that $a, b$ are both odd, and $k \geqslant 0$, and
$$2 m=a^{20}+b^{11}+k \cdot 2^{2011}.$$ | To prove that (1) is equivalent to (under the condition that the left side is not less than the right side)
$$2 m \equiv a^{20}+b^{11}\left(\bmod 2^{2011}\right)$$
We first prove: there exist odd numbers $a, b$ satisfying (2).
Notice that, for any odd numbers $x, y$, we have
$$x^{11}-y^{11}=(x-y)\left(x^{10}+x^{9} y+\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,999 |
Example 1 Proof: In decimal notation, among any 39 consecutive positive integers, there must be one number whose digit sum is a multiple of 11. | Proof: Since among any 10 consecutive positive integers, there must be one that is a multiple of 10, among any 39 consecutive positive integers, there must be 3 numbers that are multiples of 10. Among these 3 numbers, there must be one whose tens digit is no greater than 8, and this number must have at least 19 numbers... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,002 |
Example 2 Let $n$ be a positive odd number. Prove: Among the numbers
$$2-1,2^{2}-1, \cdots, 2^{n-1}-1$$
there must be a number that is a multiple of $n$. | Prove that when $n=1$, the proposition is obviously true.
Consider the case when $n>1$, at this point, in the numbers
$$1,2, \cdots, 2^{n-1}$$
there is no number that is a multiple of $n$, so the remainders when these numbers are divided by $n$ can only be $1,2, \cdots, n-1$. Therefore, there must be two numbers among... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,003 |
Example 3 Let $m$ and $n$ be positive integers, $m$ is odd, and $\left(m, 2^{n}-1\right)=1$. Prove: The number $1^{n}+$ $2^{n}+\cdots+m^{n}$ is a multiple of $m$.
| Prove that since $m$ is odd, and $1,2, \cdots, m$ is a complete residue system modulo $m$, then $2 \times 1, 2 \times 2, \cdots, 2 \times m$ is also a complete residue system modulo $m$, so,
$$1^{n}+2^{n}+\cdots+m^{n} \equiv (2 \times 1)^{n}+(2 \times 2)^{n}+\cdots+(2 \times m)^{n} \pmod{m}.$$
Thus, $m \mid \left(2^{n... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,004 |
Example 4 (1) Prove: There exist infinitely many sets of integers $(x, a, b, c)$ such that
$$x^{2}+a^{2}=(x+1)^{2}+b^{2}=(x+2)^{2}+c^{2}$$
(2) Question: Does there exist an integer set $(x, a, b, c, d)$ such that
$$x^{2}+a^{2}=(x+1)^{2}+b^{2}=(x+2)^{2}+c^{2}=(x+3)^{2}+d^{2} ?$$ | (1) For positive integers $k$ greater than 1, let $x=4 k^{3}-1, a=2 k^{2}+2 k, b=2 k^{2}+$ $1, c=2 k^{2}-2 k$, it can be known that the integer tuple $(x, a, b, c)$ meets the requirements.
The construction idea for $(x, a, b, c)$ is as follows:
From the requirements of the problem, we know $a^{2}-b^{2}=2 x+1, b^{2}-c^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,005 |
Example 5 Let $n$ be a positive integer. Prove: there exists a positive integer whose digits are all odd, which is a multiple of $5^{n}$.
The text is translated while preserving the original line breaks and format. | Prove that we can construct a qualified $n$-digit positive integer using the recursive method.
When $n=1$, take $a_{1}=5$.
Assume that when $n=m$, there exists an $m$-digit positive integer $a_{m}$ with all digits being odd, such that $5^{m} \mid a_{m}$. Let $a_{m}=5^{m} \times q$, where $q \equiv r(\bmod 5)$, and $r=0... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,006 |
Example 1
Define "Fermat number": $F_{n}=2^{2 n}+1$. When $n>1$, prove: if prime $p$ is a factor of $F_{n}$, then $p \equiv 1\left(\bmod 2^{n+1}\right)$ | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 742,026 |
Example 3
Define "Fermat number": $F_{n}=2^{2^{n}}+1$. When $n>1$, prove: if a prime $p$ is a factor of $F_{n}$, then $p \equiv 1\left(\bmod 2^{n+2}\right)$. | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 742,028 |
Example 1 Proof: $1 \underset{200 \uparrow}{10 \cdots} 1$ is divisible by 1001. | $$\begin{aligned}
\underbrace{0 \cdots 01}_{200 \uparrow 0} & =10^{201}+1=\left(10^{3}\right)^{67}+1 \\
& =\left(10^{3}+1\right)\left[\left(10^{3}\right)^{66}-\left(10^{3}\right)^{65}+\cdots-10^{3}+1\right],
\end{aligned}$$
Therefore, $10^{3}+1(=1001)$ divides $1 \underbrace{0 \cdot \cdots 0}_{200 \uparrow 0} 1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,046 |
211 girls and $n$ boys went to pick mushrooms. All these children collected a total of $n^{2}+9 n-2$ mushrooms, and each child collected the same number of mushrooms. Determine whether there are more girls or boys among the children picking mushrooms.
Translate the above text into English, please keep the original tex... | 2. Since each child collected the same number of mushrooms, the total number of children $n+11$ must divide the total number of mushrooms
$$n^{2}+9 n-2=(n+11)(n-2)+20,$$
thus $n+11$ divides 20. Since $n+11>11$, $n$ can only be 9. Therefore, the number of girls is greater than the number of boys. | null | Number Theory | proof | Yes | Yes | number_theory | false | 742,047 |
Example 3 Let $n>1,2 \nmid n$, then for any integer $m>0$, we have $n \nmid\left(m^{n-1}+1\right)$. | Proof: Suppose there exists an odd number $n > 1$ such that $n \mid (m^{n-1} + 1)$, then $(m, n) = 1$. Let $p$ be any prime divisor of $n$, and $r$ be the order of $m$ modulo $p$ (note that $p \nmid m$). Assume $n-1 = 2^k t$, where $k \geq 1$ and $2 \nmid t$. Then we have
$$m^{2^k t} \equiv -1 \pmod{p}$$
Thus, $m^{2^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,049 |
Example 4 Let $p$ be an odd prime. Prove: any positive divisor of $\frac{p^{2 p}+1}{p^{2}+1}$ is $\equiv 1(\bmod 4 p)$. | To prove that any prime divisor $q$ of $\frac{p^{2 p}+1}{p^{2}+1}$ satisfies $q \equiv 1(\bmod 4 p)$. First, note that
$$\frac{p^{2 p}+1}{p^{2}+1}=p^{2(p-1)}-p^{2(p-2)}+\cdots-p^{2}+1$$
Thus, $q \neq p$. Let $r$ be the order of $p$ modulo $q$. Since
$$p^{2 p} \equiv-1(\bmod q)$$
it follows that $p^{4 p} \equiv 1(\bmo... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,050 |
Example 5 (1) Let $p$ be an odd prime, $a \neq \pm 1, p \nmid a$. Let $r$ be the order of $a$ modulo $p$, and $k_{0}$ satisfy $p^{k_{0}} \|\left(a^{r}-1\right)$. Denote $r_{k}$ as the order of $a$ modulo $p^{k}$, then we have
$$r_{k}=\left\{\begin{array}{l}
r, \text { if } k=1, \cdots, k_{0}, \\
r p^{k-k_{0}}, \\
\text... | Prove (1) When $1 \leqslant k \leqslant k_{0}$, from $a^{r_{k}} \equiv 1\left(\bmod p^{k}\right)$ we can deduce $a^{r_{k}} \equiv 1(\bmod p)$, hence by the definition of $r$ we have $r \mid r_{k}$. On the other hand, from $a^{r} \equiv 1\left(\bmod p^{k_{0}}\right)$ we can get $a^{r} \equiv 1\left(\bmod p^{k}\right)$, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,051 |
Example 6 Let $a$ and $n$ be integers, neither of which is $\pm 1$, and $(a, n)=1$. Prove: there are at most finitely many $k$ such that $n^{k} \mid\left(a^{k}-1\right)$. | To prove that since $n \neq \pm 1$, $n$ has a prime factor. First, assume $n$ has an odd prime factor $p$, then $p \nmid a$. Let the order of $a$ modulo $p$ be $r$. Since $a \neq \pm 1$, there exists a positive integer $k_{0}$ such that $p^{k_{0}} \|\left(a^{r}-1\right)$.
If there are infinitely many $k$ such that $n^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,052 |
1 Prove: Any divisor of the Fermat number $F_{k}=2^{2^{k}}+1(k \geqslant 0)$ is $\equiv 1\left(\bmod 2^{k+1}\right)$. | 1. It suffices to prove that any prime factor $p$ of $F_{k}$ satisfies $p \equiv 1\left(\bmod 2^{k+1}\right)$. Clearly, $p \neq 2$. Let the order of 2 modulo $p$ be $r$. From $p \mid F_{k}$, we get
$$2^{2^{k}} \equiv-1(\bmod p)$$
Thus, $2^{2^{k+1}} \equiv 1(\bmod p)$, so $r \mid 2^{k+1}$, meaning $r$ is a power of 2. ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,053 |
2 (1) Let $m, n$ be coprime positive integers, $m, n>1$. Let $a$ be an integer coprime to $m n$. Suppose the orders of $a$ modulo $m$ and modulo $n$ are $d_{1}, d_{2}$, respectively, then the order of $a$ modulo $m n$ is $\left[d_{1}, d_{2}\right]$;
(2) Find the order of 3 modulo $10^{4}$. | 2. (1) Let the order of $a$ modulo $mn$ be $r$. From $a^{r} \equiv 1(\bmod m n)$, we can deduce $a^{r} \equiv 1(\bmod m)$ and $a^{r} \equiv 1(\bmod n)$. Therefore, $d_{1} \mid r$ and $d_{2} \mid r$, which implies $\left[d_{1}, d_{2}\right] \mid r$. On the other hand, from $a^{d_{1}} \equiv 1(\bmod m)$ and $a^{d_{2}} \e... | 500 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,054 |
3 Prove that for any integer $k>0$, there exists a positive integer $n$, such that $2^{k} \mid\left(3^{n}+5\right)$. | 3. Using induction. For $k=1,2$, the conclusion is obviously true. Suppose for $k \geqslant 3$ there exists $n_{0}$ such that $2^{k} \mid \left(3^{n_{0}}+5\right)$, and let $3^{n_{0}}=2^{k} u-5$. If $u$ is even, then $2^{k+1} \mid \left(3^{n_{0}}+5\right)$. Now assume $u$ is odd.
The key to the argument is to note tha... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,055 |
4 Prove that if the integer $n>1$, then $n \nmid 3^{n}-2^{n}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. Proof by contradiction, suppose there exists $n>1$, such that $n \mid 3^{n}-2^{n}$. Let $p$ be the smallest prime factor of $n$, then $3^{n} \equiv$ $2^{n}(\bmod p)$, thus $p \geqslant 5$. Hence there exists an integer $a$, such that $2 a \equiv 1(\bmod p)$. Therefore,
$$(3 a)^{n} \equiv 1(\bmod p)$$
Let $d$ be the... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,056 |
Example 1 If $n \equiv 4(\bmod 9)$, prove that the indeterminate equation
$$x^{3}+y^{3}+z^{3}=n$$
has no integer solutions $(x, y, z)$. | Prove that if equation (1) has integer solutions, then (1) modulo 9 also has integer solutions. It is well known that a perfect cube modulo 9 is congruent to $0, 1, -1$ (or equivalently $0, 1, 8$), hence
$$x^{3}+y^{3}+z^{3} \equiv 0,1,2,3,6,7,8(\bmod 9)$$
But $n \equiv 4(\bmod 9)$, so (1) modulo 9 has no solutions, wh... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,057 |
3 Let the decimal representation of the positive integer $n$ be $n=\overline{a_{k} \cdots a_{1} a_{0}}\left(0 \leqslant a_{i} \leqslant 9, a_{k} \neq 0\right)$, and let $T(n)=a_{0}-a_{1}+\cdots+(-1)^{k} a_{k}$ (the alternating sum of the digits of $n$ starting from the units digit). Prove that $n-T(n)$ is divisible by ... | 3. We have
$$n-T(n)=\left(a_{0}-a_{0}\right)+\left(10 a_{1}+a_{1}\right)+\cdots+\left(a_{k} \times 10^{k}-(-1)^{k} a_{k}\right) .$$
It is easy to see that for $i=0,1, \cdots, k$, the number $a_{i} \times 10^{i}-(-1)^{i} a_{i}$ is divisible by 11 (using decomposition (5) for even $i$ and decomposition (6) for odd $i$).... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,058 |
Example 2 Determine all non-negative integer solutions $\left(x_{1}, \cdots, x_{14}\right)$ (disregarding the order of the solutions) of the equation
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{14}^{4}=1599$$ | Solving modulo 16 can prove that the equation has no integer solutions, because the fourth power of an integer modulo 16 is congruent to 0 or 1, so the possible values of $x_{1}^{4}+x_{2}^{4}+\cdots+x_{14}^{4}$ modulo 16 are $0,1,2, \cdots, 14$, and it cannot be 15. But $1599 \equiv 15(\bmod 16)$, therefore the equatio... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,059 |
Example 3 Proof: The following numbers cannot be expressed as the sum of cubes of several consecutive integers.
(1) $385^{97}$;
(2) $366^{17}$. | Prove that using
$$1^{3}+2^{3}+\cdots+k^{3}=\left(\frac{k(k+1)}{2}\right)^{2}$$
it is easy to see that the sum of cubes of several consecutive integers can be expressed in the form
$$\left(\frac{m(m+1)}{2}\right)^{2}-\left(\frac{n(n+1)}{2}\right)^{2},$$
where \( m \) and \( n \) are integers. We need to prove that for... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,060 |
Example 4 Find all powers of 2 such that when the leading digit (in its decimal representation) is removed, the remaining number is still a power of 2. | The problem requires solving the equation
$$2^{n}=2^{k}+a \times 10^{m}$$
for all positive integer solutions $(n, k, m, a)$, where $a=1,2, \cdots, 9$. Transforming (1) yields
$$2^{k}\left(2^{n-k}-1\right)=a \times 10^{m}$$
First, we prove that $m=1$. If $m>1$, then the right-hand side of (2) is divisible by $5^{2}$, ... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,061 |
Example 5 Find all positive integers $x>1, y>1, z>1$, such that
$$1!+2!+\cdots+x!=y^{z}$$ | The key step is to prove that when $x \geqslant 8$, it must be that $z=2$. Since the left side of (1) is divisible by 3, hence $3 \mid y^{z}$, which implies $3 \mid y$, and thus the right side of (1) is divisible by $3^{z}$. On the other hand,
$$1!+2!+\cdots+8!=46233$$
is divisible by $3^{2}$, but not by $3^{3}$; and ... | x=y=3, z=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,062 |
Example 6 Proof, the indeterminate equation
$$(x+2)^{2 m}=x^{n}+2$$
has no positive integer solutions. | To prove, for the subsequent argument, we first derive some simple conclusions from equation (1).
Clearly, $n>1$. Moreover, $x$ must be odd; otherwise, taking (1) modulo 4 leads to a contradiction. Furthermore, $n$ is also odd, because if $2 \mid n$, then $x^{n}$ is the square of an odd number, making the right side of... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,063 |
Example 7 Proof: The indeterminate equation
$$8^{x}+15^{y}=17^{x}$$
has the only positive integer solution $x=y=2$. | First, we use congruences to prove that $y$ and $z$ are both even.
Equation (1) modulo 4 gives
$$(-1)^{y} \equiv 1(\bmod 4)$$
Thus, $y$ is even. Modulo 16 of equation (1) gives
$$8^{x}+(-1)^{y} \equiv 1(\bmod 16)$$
which simplifies to $8^{x} \equiv 0(\bmod 16)$, hence $x \geqslant 2$.
Note that $17^{2} \equiv 1,15^{2... | x=y=2 | Number Theory | proof | Yes | Yes | number_theory | false | 742,064 |
I prove that the indeterminate equation
$$x^{2}+3 x y-2 y^{2}=122$$
has no integer solutions. | 1. Transform the equation into
$$(2 x+3 y)^{2}=17 y^{2}+4 \times 122$$
Taking modulo 17, we get $(2 x+3 y)^{2} \equiv 12(\bmod 17)$. However, it is easy to verify that the square of an integer modulo 17 can only be one of $0, 1, 2, 4, 8, 9, 13, 15, 16$, and cannot be 12. Therefore, the original equation has no integer... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,066 |
2 Find all positive integers $m, n$, such that $\left|12^{m}-5^{n}\right|=7$.
| 2. Mod 4 reveals that the equation
$$12^{m}-5^{n}=-7$$
has no positive integer solutions. The equation
$$12^{m}-5^{n}=7$$
obviously has the solution $m=n=1$. Below we prove that when $m>1$ it has no positive integer solutions. Taking (1) mod 3 yields $-(-1)^{n} \equiv 1(\bmod 3)$, so $n$ is odd, thus $5^{n} \equiv 5(... | m=n=1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,067 |
3 Find all prime numbers $p$, such that $2^{p}+3^{p}$ is a $k$-th power of an integer (where $k \geqslant 2$).
Find all prime numbers $p$, such that $2^{p}+3^{p}$ is a $k$-th power of an integer (where $k \geqslant 2$). | 3. $p=2,5$ do not meet the requirements. Let prime $p>2$ and $p \neq 5$. By the binomial theorem, it is easy to see that
$$\begin{aligned}
2^{p}+3^{p} & =2^{p}+(5-2)^{p}=5^{p}-\mathrm{C}_{p}^{1} 5^{p-1} \times 2+\cdots+5 \mathrm{C}_{p}^{p-1} 2^{p-1} \\
& =5^{2} u+5 p \times 2^{p-1}, u \text { is an integer. }
\end{alig... | p=2,5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,068 |
4 Let $n$ integers have the following property: the product of any $n-1$ of them minus the remaining number is divisible by $n$. Prove: the sum of the squares of these $n$ numbers is also divisible by $n$.
保留源文本的换行和格式，直接输出翻译结果。 | 4. Let $a_{1}, \cdots, a_{n}$ be integers with the given property, and let $A$ be their product. For $1 \leqslant i \leqslant n$, the number $n$ divides $\frac{A}{a_{i}}-a_{i}$, and thus it also divides
$$a_{i}\left(\frac{A}{a_{i}}-a_{i}\right)=A-a_{i}^{2}$$
Therefore, $n$ divides the sum $\left(A-a_{1}^{2}\right)+\cd... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,069 |
4 Prove: The indeterminate equation
$$5^{x}-3^{y}=2$$
has only the positive integer solution $x=y=1$. | 4. The equation clearly has the solution $x=y=1$. Taking the equation modulo 4, it is easy to see that $y$ is odd. If $y>1$, taking the equation modulo 9 yields
$$5^{x} \equiv 2(\bmod 9)$$
It is not difficult to find that for $x=1,2, \cdots, 5^{x}$ modulo 9 cycles as $5,7,8,4,2,1$. Hence, by (1), $x$ must be of the fo... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,070 |
5 Prove that $x^{3}+y^{4}=7$ has no integer solutions. | 5. It is easy to verify that $x^{3} \equiv 0,1,5,8,12(\bmod 13) ; y^{4} \equiv 0,1,3,9(\bmod 13)$. From these, it is clear that $x^{3}+y^{4} \not \equiv 7(\bmod 13)$, hence the equation has no integer solutions. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,071 |
6 Let $p$ be a given odd prime, find all positive integer solutions $x, y$ of $p^{x}-y^{p}=1$.
The text has been translated while preserving the original line breaks and format. | 6. From $p^{x}=y^{p}+1=(y+1)\left(y^{p-1}-y^{p-2}+\cdots-y+1\right)$, we know that $y+1=p^{n}$, where $n$ is an integer. Since $y>0$, it follows that $n>0$. Therefore,
$$\begin{aligned}
p^{x} & =\left(p^{n}-1\right)^{p}+1 \\
& =p^{n p}-p \cdot p^{n(p-1)}+\mathrm{C}_{p}^{2} p^{n(p-2)}-\cdots-\mathrm{C}_{p}^{p-2} p^{2 n}... | x=y=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,072 |
Example 1 Let $u$ be a given positive integer, prove that the equation
$$n!=u^{x}-u^{y}$$
has at most finitely many positive integer solutions $(n, x, y)$. | Prove that we can assume $u>1$. The conclusion is equivalent to proving that the equation
$$n!=u^{r}\left(u^{s}-1\right)$$
has at most finitely many positive integer solutions $(n, r, s)$.
First, note that for a given $n$, equation (1) clearly has at most finitely many solutions $(r, s)$. We will prove that when $n$ i... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,073 |
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