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class | __index_level_0__ int64 0 742k |
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Theorem 3 Let $p$ be an odd prime, $\alpha \in \mathbf{N}^{*}$, then a primitive root modulo $p^{\alpha}$ exists. | To prove a fundamental idea is "translation".
First, prove: there exists a primitive root $g$ modulo $p$ such that
$$g^{\rho-1} \not \equiv 1\left(\bmod p^{2}\right)$$
In fact, take any primitive root $g$ modulo $p$. If $g$ does not satisfy (1), we say $g+p$ is a primitive root modulo $p$ that satisfies (1).
First, u... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,474 |
Example 4 Let positive integers $a>b>c>d$ satisfy:
$$a c+b d=(b+d+a-c)(b+d-a+c) .$$
Prove: The numbers $a b+c d, a c+b d$ and $a d+b c$ are all composite. | To prove that $ab + cd$ is composite in the 42nd (2001) IMO Problem 6. In fact, $ab + cd$, $ac + bd$, and $ad + bc$ are all composite, and the number of prime factors (counting multiplicities) of each is at least 3, 3, and 2, respectively (the proof of this conclusion is left to the reader).
Let
\[
\alpha = b + d + a ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,475 |
Theorem 4 Let $p$ be an odd prime, $\alpha \in \mathbf{N}^{*}$, then a primitive root modulo $2 p^{\alpha}$ exists. | Proof: Let $g$ be a primitive root modulo $p^{a}$, then $g + p^{a}$ is also a primitive root modulo $p^{a}$. Among $g$ and $g + p^{a}$, one is odd. Let this odd number be $\widetilde{g}$, then $\left(\widetilde{g}, 2 p^{a}\right) = 1$. Therefore, by Euler's theorem, $\delta_{2 p^{a}}(\widetilde{g}) \mid \varphi\left(2 ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,476 |
Example 1 Let $p$ be an odd prime. Prove: for $j \in\{0,1,2, \cdots, p-2\}$, we have
$$\sum_{x=0}^{p-1} x^{j} \equiv 0(\bmod p)$$
Here we assume $0^{0}=1$. | When $j=0$, the proposition is obviously true.
Consider the case $j \in\{1,2, \cdots, p-2\}$. Take a primitive root $g$ modulo $p$, then when $x$ runs through a complete residue system modulo $p$, $g x$ also runs through a complete residue system modulo $p$, hence
$$\sum_{x=0}^{p-1} x^{j} \equiv \sum_{x=0}^{p-1}(g x)^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,477 |
Example 2 Let $p$ be a given odd prime. Find all functions $f$ : $\mathbf{Z} \rightarrow \mathbf{Z}$ satisfying the following conditions:
(1) For any $m, n \in \mathbf{Z}$, if $m \equiv n(\bmod p)$, then $f(m)=f(n)$;
(2) For any $m, n \in \mathbf{Z}$, $f(m n)=f(m) f(n)$. | The functions that satisfy the conditions are only the following 4:
(1) $f(x)=1$;
(2) $f(x)=0$;
(3) $f(x)=\left\{\begin{array}{ll}0, & p \mid x, \\ 1, & p \nmid x ;\end{array}\right.$
(4) $f(x)=\left\{\begin{array}{ll}0, & p \mid x, \\ 1, & x \text { is a quadratic residue modulo } p, \\ -1, & x \text { is a quadratic ... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,478 |
Example 3 Let $p$ be a given odd prime, and call a positive integer $m$ a "good number" if it satisfies the following conditions:
(1) $m \in\{1,2, \cdots, p-1\}$;
(2) there exists $n \in \mathbf{N}^{*}$, such that $m^{n} \equiv-1(\bmod p)$.
Find the number of "good numbers". | Let $g$ be a primitive root modulo $p$, then $g^{\frac{p-1}{2}} \equiv -1 \pmod{p}$. (Because $\delta_{p}(g)=p-1$, and $g^{p-1} \equiv 1 \pmod{p}$, and $g, g^{2}, \cdots, g^{p-1}$ form a reduced residue system modulo $p$.)
Now for $m \in \{g, g^{2}, \cdots, g^{p-1}\}$, let $m=g^{k}$. According to the definition of "go... | p-1-u | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,479 |
Example 4 Let $p$ be an odd prime, and $m$ be a positive multiple of $p-1$. Prove:
$$\sum_{0 \leqslant i \leqslant m} \mathrm{C}_{m}^{(p-1) i} \equiv 2+p(1+m)\left(\bmod p^{2}\right)$$ | Let $r=\frac{m}{p-1}$, then
$$S_{r}=\sum_{0 \leqslant \leq \frac{m}{p-1}} \mathrm{C}_{m}^{(p-1) t}=\sum_{i=0}^{r} \mathrm{C}_{r(p-1)}^{i(p-1)}$$
Take a primitive root $g$ modulo $p$, and let $h=g^{p}$, then by Euler's theorem we have
$$h^{p-1}=g^{p(p-1)}=g^{\alpha\left(p^{2}\right)} \equiv 1\left(\bmod p^{2}\right) .$... | 2+p(1+m)\left(\bmod p^{2}\right) | Number Theory | proof | Yes | Yes | number_theory | false | 741,480 |
1. Let $p$ be a prime, $a \in \mathbf{N}^{*}$. Prove: If $\delta_{p}(a)=3$, then
$$\delta_{p}(a+1)=6$$ | 1. From $\delta_{p}(a)=3$, we know $a \neq \pm 1(\bmod p)$, and $a^{2}+a+1 \equiv 0(\bmod p)$. Therefore, $1+a \not \equiv 1(\bmod p),(1+a)^{2}=1+2 a+a^{2} \equiv a \not \equiv 1(\bmod p),(1+a)^{3} \equiv$ $(1+a) a \equiv-1(\bmod p)$, so $\delta_{p}(a+1)=6$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,481 |
2. Let $n$ be a given positive integer, find the smallest positive integer $m$, such that
$$2^{m} \equiv 1\left(\bmod 5^{n}\right)$$ | 2. This is equivalent to finding $\delta_{5^{n}}(2)$, the answer is: the smallest positive integer $m=4 \times 5^{n-1}$.
We prove the above conclusion by induction on $n$. Direct calculation shows that $\delta_{5}(2)=4$. Now assume $\delta_{5^{n}}(2)=4 \times 5^{n-1}$, then we can set $2^{4 \times 5^{n-1}}=t \cdot 5^{... | 4 \times 5^{n-1} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,482 |
3. Let $p$ be an odd prime, and $p \equiv 2(\bmod 3)$. Consider the set
$$S=\left\{y^{2}-x^{2}-1 \mid x, y \in \mathbf{Z}, 0 \leqslant x, y \leqslant p-1\right\} .$$
Prove: $S$ contains at most $p-1$ elements that are multiples of $p$. | 3. First prove: If $1 \leqslant m < n \leqslant p-1$, then $m^{3} \not\equiv n^{3}(\bmod p)$, where $p$ is an odd prime and $p \equiv 2(\bmod 3)$.
In fact, let $t$ be the smallest positive integer such that $m^{t} \equiv n^{t}(\bmod p)$. Using the same method as proving the properties of exponents, we know that for an... | p-1 | Number Theory | proof | Yes | Yes | number_theory | false | 741,483 |
4. Let $n, b_{0} \in \mathbf{N}^{*}, n \geqslant 2, 2 \leqslant b_{0} \leqslant 2 n-1$. The sequence $\left\{b_{i}\right\}$ is defined as follows:
$$b_{i+1}=\left\{\begin{array}{ll}
2 b_{i}-1, & b_{i} \leqslant n, \\
2 b_{i}-2 n, & b_{i}>n,
\end{array} \quad i=0,1,2, \cdots\right.$$
Let $p\left(b_{0}, n\right)$ denote... | 4. Let $m=n-1, a_{i}=b_{i}-1$, then $1 \leqslant a_{0} \leqslant 2 m$, and
$$a_{i+1}=\left\{\begin{array}{ll}
2 a_{i}, & a_{i} \leqslant m, \\
2 a_{i}-(2 m+1), & a_{i}>m .
\end{array}\right.$$
This indicates: $a_{i+1} \equiv 2 a_{i}(\bmod 2 m+1)$, and for $i \in \mathbf{N}$, we have $1 \leqslant a_{i} \leqslant 2 m$.
... | p\left(2,2^{k}\right)=k+1, p\left(2,2^{k}+1\right)=2(k+1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,484 |
5. Find all two-digit numbers $n=\overline{a b}$ (where $a \geqslant 1, a, b \in\{0,1,2, \cdots, 9\}$), such that for any $x \in \mathbf{Z}$, we have $n \mid\left(x^{a}-x^{b}\right)$. | 5. First, $n=11,22, \cdots, 99$ satisfy the condition. When $a \neq b$, let $p$ be a prime factor of $n$. By the condition, for any $x \in \mathbf{Z}$, we have $p \mid (x^a - x^b)$. Therefore, for any $x \in \{1, 2, \cdots, p-1\}$, we have $x^{|a-b|} \equiv 1 \pmod{p}$. Let $g$ be a primitive root modulo $p$, then we a... | n=11,22, \cdots, 99,15,28,48 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,485 |
Example 5 Let $a, b, c, d, e, f$ all be positive integers, and the number $S(=a+b+c+d+e+f)$ is a common divisor of $a b c+d e f$ and $a b+b c+c a-d e-e f-f d$. Prove: $S$ is a composite number. | Prove that for the polynomial
$$\begin{aligned}
f(x) & =(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f) \\
& =S x^{2}+(a b+b c+c a-d e-e f-f d) x+(a b c+d e f) .
\end{aligned}$$
From the given conditions, it is known that for any \( x \in \mathbf{Z} \), \( S \mid f(x) \).
In particular, \( S \mid f(d) \), i.e.,
$$S \mid (d+a)(d+b)(d+... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,486 |
6. Let $m \in \mathbf{N}^{*}$. Prove: If
$$\left(2^{m+1}+1\right) \mid\left(3^{2^{m}}+1\right),$$
then $2^{m+1}+1$ is a prime number. | 6. Let $q=2^{m+1}+1$, by the condition we know $3^{2^{m}} \equiv-1(\bmod q)$, hence $3^{2^{m+1}} \equiv 1(\bmod$ $q)$. This indicates $\delta_{q}(3) \mid 2^{m+1}$, but $\delta_{q}(3) \times 2^{m}$. Therefore, $\delta_{q}(3)=2^{m+1}$.
On the other hand, by Euler's theorem, we know $3^{\varphi(q)} \equiv 1(\bmod q)$, so... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,487 |
7. Find all prime triples $(p, q, r)$ such that
$$p \mid \left(q^{r}+1\right), q \mid \left(r^{p}+1\right), r \mid \left(p^{q}+1\right) .$$ | 7. Let $(p, q, r)$ be an array of prime numbers, and by cyclic symmetry, assume $p = \max \{p, q, r\}$. From $p \mid (q^r + 1)$, we know $p \neq q$, similarly $q \neq r$, $r \neq p$, i.e., $p, q, r$ are pairwise distinct. We discuss the following two cases:
(1) $p > q > r$. From $q \mid (r^p + 1)$, i.e., $r^p \equiv -1... | (5, 3, 2), (3, 2, 5), (2, 5, 3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,488 |
8. Let $p$ be a prime. Prove: The set $\left\{p n+1 \mid n \in \mathbf{N}^{*}\right\}$ contains infinitely many primes. | 8. If there are only finitely many primes of the form $p n+1$, let them be $p_{1}, p_{2}, \cdots, p_{m}$, and let $t=p_{1} \cdot p_{2} \cdot \cdots \cdot p_{m} \cdot p$. Let $q$ be a prime factor of $t^{p-1}+t^{p-2}+\cdots+1$, then $q \left\lvert\, \frac{t^{p}-1}{t-1}\right.$, hence $q \mid t^{p}-1$. Therefore, $\delta... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,489 |
9. Prove: There are infinitely many prime numbers $q$ that satisfy the conditions of Example 5 in Section 3.1.
The text above is translated into English, preserving the original text's line breaks and format. | 9. If the primes satisfying the condition are finite, let them be $q_{1}, q_{2}, \cdots, q_{k}$. Let $a=$ $p \cdot\left(q_{1} \cdot q_{2} \cdot \cdots \cdot q_{k}\right)^{p}$. Notice that,
$$\frac{a^{p}-1}{a-1}=a^{p-1}+a^{p-2} \cdots+a+1 \equiv a+1 \equiv 1\left(\bmod p^{2}\right),$$
Therefore, it has a prime factor $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,490 |
10. Let $p$ be a prime. Prove: if there exists $n \in \mathbf{N}^{*}$, such that
$$p \|\left(2^{n}-1\right),$$
then $p \|\left(2^{p-1}-1\right)$. | 10. Let $d=\delta_{p}(2)$ (note that since $p \mid\left(2^{n}-1\right)$, $p$ is an odd prime, so $\delta_{p}(2)$ exists), then from $2^{n} \equiv 1(\bmod p)$, we know $d \mid n$. Furthermore, from $p \|\left(2^{n}-1\right)$, we know $p \|\left(2^{d}-1\right)$.
By Fermat's Little Theorem, we know $2^{p-1} \equiv 1(\bmo... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,491 |
11. Find all prime pairs $(p, q)$ such that $p q \mid\left(5^{p}-2^{p}\right)\left(5^{q}-2^{q}\right)$. | 11. If $p \mid\left(5^{p}-2^{p}\right)$, then $p$ is coprime with 2 and 5. Using Fermat's Little Theorem, we know $5^{p}-2^{p} \equiv 5-2(\bmod p)$, hence $p \mid 3$. Therefore, $p=3$. In this case, $5^{p}-2^{p}=3^{2} \times 13$, so $q \mid\left(5^{q}-2^{q}\right)$ or $q \mid\left(3^{2} \times 13\right)$, which gives $... | (3,3),(3,13),(13,3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,492 |
12. Find all prime pairs $(p, q)$ such that
$$p q \mid\left(2^{p}+2^{q}\right) \text {. }$$ | 12. Let $p \leqslant q$.
If $p=2$, then $q=2$ satisfies the condition. When $q>2$, $q$ is odd. From $2^{p}+2^{q}=2^{2}(1+2^{q-2})$, we know that $q \mid (2^{q-2}+1)$. Therefore,
$$2^{q-2} \equiv -1 \pmod{q}$$
By Fermat's Little Theorem, $2^{q-1} \equiv 1 \pmod{q}$, so
$$1 \equiv 2^{q-1} = 2^{q-2} \cdot 2 \equiv -2 \p... | (2, 2), (2, 3), (3, 2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,493 |
13. Let $n$ be an odd number greater than 1. Prove: for any $m \in \mathbf{N}^{*}$, we have
$$n \nmid\left(m^{n-1}+1\right) \text {. }$$ | 13. If $n$ is a prime, then when $n \mid m$, it is obvious that $n \nmid\left(m^{n-1}+1\right)$; if $n \nmid m$, then by Fermat's Little Theorem, we know $n \mid\left(m^{n-1}-1\right)$; if $n \mid\left(m^{n-1}+1\right)$, then it requires $n \mid 2$, which contradicts $n$ being an odd number greater than 1.
If $n$ is a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,494 |
14. Let $n$ be an integer greater than 1. Prove: The following statements are equivalent.
(1) For any $a \in \mathbf{N}^{*}$, we have $n \mid\left(a^{n}-a\right)$;
(2) For any prime factor $p$ of $n$, we have $p^{2} \nmid n$, and $(p-1) \mid (n-1)$. | 14. First, prove: If proposition (2) holds, then proposition (1) holds. For any $a \in \mathbf{N}^{*}$, consider any prime factor $p$ of $n$. Since $a^{p} \equiv a(\bmod p)$ (Fermat's Little Theorem), if $p \mid a$, then $a^{n} \equiv a(\bmod p)$; if $p \nmid a$, then $a^{p-1} \equiv 1(\bmod p)$, combined with $(p-1) \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,495 |
15. Consider a positive integer $n$ that has been factored into prime factors
$$n=p_{1}^{\sigma_{1}} p_{2}^{\sigma_{2}} \cdots p_{k}^{\sigma_{k}},$$
where $p_{1}<p_{2}<\cdots<p_{k}$ are all prime numbers, $\alpha_{i} \in \mathbf{N}^{*}, i=1,2, \cdots, k, m$ is a positive integer. Prove: The following statements are eq... | 15. First prove that Proposition (1) $\Rightarrow$ Proposition (2) holds.
If there exists some $\alpha_{i} \geqslant 2$, without loss of generality, let $\alpha_{1} \geqslant 2$. Take $a=p_{1}$, then by Proposition (1), $x^{m} \equiv p_{1}(\bmod n)$ has a solution, hence $p_{1}^{2} \mid (x^{m} - p_{1})$, so $p_{1} \mi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,496 |
16. Let $p$ be a prime. Prove: There exist $p-1$ integers $a_{1}, a_{2}, \cdots, a_{p-1}$, such that the numbers
$$a_{i}+a_{j}(1 \leqslant i \leqslant j \leqslant p-1)$$
are pairwise incongruent modulo $\varphi\left(p^{2}\right)$. | 16. Let $g$ be a primitive root modulo $p$. By the Chinese Remainder Theorem, there exist integers $a_{i}$ such that $a_{i} \equiv i(\bmod p-1)$ and $a_{i} \equiv g^{i}(\bmod p)$. We prove that $a_{i}(1 \leqslant i \leqslant p-1)$ has the property mentioned in the problem.
In fact, if $a_{i}+a_{j} \equiv a_{r}+a_{t}(\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,498 |
17. Let $p$ be an odd prime, and $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ be an $n$-variable integer coefficient polynomial of degree less than $n$. Prove: The number of integer solutions $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ satisfying the following conditions
(1) $0 \leqslant x_{i} \leqslant p-1, i=1,2, \cdot... | 17. Let $S=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid 0 \leqslant x_{i} \leqslant p-1, i=1,2, \cdots, n\right\}$, then $|S|=$ $p^{n}$. Consider the following sum:
$$T=\sum_{\left(y_{1}, y_{2}, \cdots, y_{n}\right) \in S} f\left(y_{1}, y_{2}, \cdots, y_{n}\right)^{p-1}$$
Notice that, the degree of $f\left(x_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,499 |
18. Let $p$ be a prime, and $p \equiv 1(\bmod 12)$. Find the number of tuples $(a, b, c, d)$ satisfying:
(1) $a, b, c, d \in\{0,1,2, \cdots, p-1\}$;
(2) $a^{2}+b^{2} \equiv c^{3}+d^{3}(\bmod p)$. | 18. Under the condition $p \equiv 1(\bmod 12)$, first prove the following lemma.
Lemma 1: The number of integer pairs $(x, y)$ that satisfy $x^{2}+y^{2} \equiv 0(\bmod p), x, y \in\{0,1,2, \cdots, p-1\}$ is $2 p-1$.
Since $p \equiv 1(\bmod 4)$, we know that -1 is a quadratic residue modulo $p$, so there exists $u$ su... | p^3 + 2p^2 - 2p | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,500 |
19. The Fibonacci sequence $\left\{F_{n}\right\}$ is defined as follows:
$$F_{1}=F_{2}=1, F_{n+2}=F_{n+1}+F_{n}, n=1,2, \cdots .$$
Prove: For any prime $p$, the number $F_{p-1}$ is the first number in the sequence $\left\{F_{n}\right\}$ that is divisible by $p$ if and only if there exists a primitive root $r$ modulo $... | 19. Direct verification shows that when $p \in\{2,3,5\}$, $F_{p-1}$ is not the first number in $\left\{F_{n}\right\}$ that is divisible by $p$; and at this time, the primitive root $r$ of $p$ does not satisfy
$$(r+1)(r+2) \equiv 1(\bmod p)$$
Now consider the case $p \geqslant 7$. Let $\alpha=\frac{1+\sqrt{5}}{2}, \bet... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,501 |
Theorem 1 The indeterminate equation
$$a x + b y = c$$
has integer solutions if and only if $(a, b) \mid c$. | The necessity is obvious. Below we prove the sufficiency.
Let $(a, b)=d, a=a_{1} d, b=b_{1} d, c=c_{1} d$, then equation (1) becomes
$$a_{1} x+b_{1} y=c_{1},\left(a_{1}, b_{1}\right)=1$$
Since $\left(a_{1}, b_{1}\right)=1$, by Bézout's theorem, there exist integers $x_{0}^{\prime}, y_{0}^{\prime}$, such that
$$a_{1} x... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,502 |
Theorem 2 Let $x_{0}, y_{0}$ be a set of integer solutions to the equation $a x+b y=c$, then all integer solutions to this equation can be expressed as
$$\left\{\begin{array}{l}
x=x_{0}+\frac{b}{(a, b)} t \\
y=y_{0}-\frac{a}{(a, b)} t
\end{array} t \in \mathbf{Z}\right.$$ | Prove that since $x_{0}, y_{0}$ is a solution, we have $a x_{0} + b y_{0} = c$. Therefore,
$$a\left(x_{0} + \frac{b}{(a, b)} t\right) + b\left(y_{0} - \frac{a}{(a, b)} t\right) = a x_{0} + b y_{0} = c$$
This shows that $x = x_{0} + \frac{b}{(a, b)} t, y = y_{0} - \frac{a}{(a, b)} t$ is a solution to the equation.
Let ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,503 |
Example 1 Arrange the simplest fractions with denominators not exceeding 99 between $[0,1]$ in ascending order, find the two numbers adjacent to $\frac{17}{76}$. | Let $x, y \in \mathbf{N}^{*}, (x, y)=1$, and $\frac{x}{y}$ is the number to the left of $\frac{17}{76}$ in the above sequence, then
$$\frac{17}{76}-\frac{x}{y}=\frac{17 y-76 x}{76 y}>0$$
Notice that $17 y-16 x$ is an integer, so $17 y-76 x \geqslant 1$.
We first solve the indeterminate equation
$$17 y-76 x=1$$
for po... | \frac{19}{85}, \frac{15}{67} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,504 |
Example 2 Find all integer solutions to the equation $3 x+7 y+16 z=40$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Since $(3,7)=1$, the equation has a solution.
Let $3 x+7 y=t$, this can be regarded as a linear Diophantine equation in two variables, its general solution is
$$x=-2 t+7 u, y=t-3 u$$
On the other hand, the general solution of $t+16 z=40$ is
$$t=40-16 v, z=v$$
Therefore, all solutions of the original equation are
$$x=... | x=-80+32 v+7 u, y=40-16 v-3 u, z=v | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,505 |
Example 3 Let $a, b, c \in \mathbf{N}^{*},(a, b)=1$. Prove: When $c>a b-a-b$, the equation
$$a x+b y=c$$
has non-negative integer solutions; while when $c=a b-a-b$, equation (4) has no non-negative integer solutions. | Prove that since $(a, b)=1$, the equation $a x+b y=c$ has integer solutions, and let these solutions be $x=x_{0}+b t, y=y_{0}-a t$ (where $t$ is an integer).
By choosing an appropriate $t$, we can ensure that $0 \leqslant x \leqslant b-1$ (by adding or subtracting multiples of $b$ from $x_{0}$, we can find such a $t$),... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,506 |
Example 4 Let $a, b \in \mathbf{N}^{*}$. Prove: In the interval $[0, a b-a-b]$, there are exactly $\frac{1}{2}(a-1)(b-1)$ integers $c$ that cannot be expressed in the form $a x+b y$, where $x, y$ are non-negative integers. | Proof
Let
$$I=\{n \mid 0 \leqslant n \leqslant a b-a-b, n \in \mathbf{Z}\}, A=a b-a-b,$$
and call a number representable if it can be written as $a x+b y(x, y \geqslant 0, x, y \in \mathbf{Z})$, otherwise it is called non-representable.
Clearly, $I$ contains
$$a b-a-b+1=(a-1)(b-1)$$
integers, so it suffices to prove... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,507 |
Example 1 Let $a, b \in \mathbf{N}^{*}$, it is known that the product of all positive divisors of $a$ and $b$ are equal. Question: Must $a=b$? | The answer is affirmative.
Notice that if $d \mid n$, then $\left.\frac{n}{d} \right\rvert\, n$, therefore, by pairing all positive divisors of $n$, we can see that the product of all positive divisors of $n$ equals $n^{\frac{d(n)}{2}}$. Thus, to prove $a=b$, we only need to prove: from $a^{d(a)}=b^{d(b)}$, we can dedu... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,508 |
Example 5 For which $n \in \mathbf{N}^{\cdot}, n \geqslant 5$, can the vertices of a regular $n$-gon be colored using no more than 6 colors, such that any 5 consecutive vertices have distinct colors? | Let the colors be $a, b, c, d, e, f$. Define the sequence $A: a, b, c, d, e$ and the sequence $B: a, b, c, d, e, f$.
If there exist non-negative integers $x, y$, such that $n=5 x+6 y$, then for a regular $n$-gon, the vertices can be colored by first coloring $y$ sequences of $B$, followed by $x$ sequences of $A$, ensu... | n \geqslant 5, \text{ except } n \in \{7,8,9,13,14,19\} | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,509 |
Example 6 Let $a, b, c$ be positive integers, $(a, b, c)=1$, and let $(a, b)=d$. Prove: when
$$n>\frac{a b}{d}+c d-a-b-c$$
then, the equation $a x+b y+c z=n$ has a non-negative integer solution $(x, y, z)$. | Given $(a, b, c)=1$, we know $(c, d)=1$, the equation
$$c z+d t=n$$
has integer solutions. Similar to the proof in Example 3, we can ensure that in the solutions of $c z+d t=n$, $0 \leqslant z < d$. When $n > \frac{a b}{d}+c d-a-b-c$,
$$t=\frac{n-c z}{d} \geqslant \frac{n-c(d-1)}{d}>\frac{a b}{d^{2}}-\frac{a}{d}-\frac... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,510 |
Example 7 Let $p_{1}, p_{2}, \cdots, p_{n}$ be $n(\geqslant 2)$ pairwise coprime positive integers, and let
$$\pi_{i}=\frac{p_{1} p_{2} \cdots p_{n}}{p_{i}}, i=1,2, \cdots, n$$
Find the largest positive integer $m$ such that the indeterminate equation
$$\pi_{1} x_{1}+\pi_{2} x_{2}+\cdots+\pi_{n} x_{n}=m$$
has no non-... | Let $M=(n-1) p_{1} p_{2} \cdots p_{n}-\sum_{i=1}^{n} \pi_{i}$, we prove that the required maximum positive integer is
$M$.
In fact, if there exist non-negative integers $x_{1}, x_{2}, \cdots, x_{n}$, such that
$$\pi_{1} x_{1}+\pi_{2} x_{2}+\cdots+\pi_{n} x_{n}=M$$
Taking modulo $p_{i}$ on both sides of (6), we get
$$... | M=p_{1} p_{2} \cdots p_{n}\left((n-1)-\sum_{i=1}^{n} \frac{1}{p_{i}}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,511 |
Example 8 System of equations
$$\left\{\begin{array}{l}
a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1 q} x_{q}=0, \\
a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2 q} x_{q}=0, \\
\cdots \\
a_{p 1} x_{1}+a_{p 2} x_{2}+\cdots+a_{p q} x_{q}=0
\end{array}\right.$$
where, $q=2 p, a_{i j} \in\{-1,0,1\}(1 \leqslant i \leqslant p, 1 \leqslant... | Proof
This is an existence problem of a solution, considering the application of the Pigeonhole Principle to prove it. Let
$$b_{i}=a_{i 1} y_{1}+a_{i 2} y_{2}+\cdots+a_{i q} y_{q}, i=1,2, \cdots, p$$
where $y_{i} \in \mathbf{Z}, 0 \leqslant y_{i} \leqslant q, i=1,2, \cdots, q$.
Considering each $y_{i}$ has $q+1$ choic... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,512 |
Example 1 Let $(x, y, z)$ be any positive integer solution of equation (1), prove: $60 \mid x y z$.
| Proof
Since $60=3 \times 4 \times 5$, and $3,4,5$ are pairwise coprime, it suffices to prove that $x y z$ can be divided by $3,4,5$ respectively.
It is easy to know that $x^{2}, y^{2}, z^{2} \equiv 0,1(\bmod 3)$. Taking $\bmod 3$ on both sides of equation $(1)$, if $x^{2}$, $y^{2}$ are both $\equiv 1(\bmod 3)$, then
$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,513 |
Theorem The integer solutions $(x, y, z)$ of the indeterminate equation (1) satisfying
$$(x, z)=1, x>0, y>0, z>0,2 \mid y$$
can be expressed as
$$x=a^{2}-b^{2}, y=2 a b, z=a^{2}+b^{2},$$
where $a, b$ are any integers satisfying $a>b>0, a, b$ one odd and one even, and $(a, b)=1$. | To prove the theorem, it consists of two parts. The first part is to show that when $a, b$ satisfy the conditions, the $(x, y, z)$ given by (3) is a solution to (1) and satisfies (2); the second part is to show that for any solution $(x, y, z)$ of (1) that satisfies (2), there must exist $a, b$ that meet the conditions... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,514 |
Example 2 Proof: For each positive integer $n$, there exist $n$ (non-congruent) right triangles, all of which have the same perimeter. | Proof: By the theorem, it is not difficult to see that there exist infinitely many non-similar right triangles (it is easy to prove that any two primitive solutions determine non-similar triangles). From these, take any \( n \) triangles, with side lengths
\[
\begin{array}{l}
a_{k}, b_{k}, c_{k}\left(0<a_{k}<b_{k}<c_{k... | proof | Geometry | proof | Yes | Yes | number_theory | false | 741,515 |
Example 3 For every positive integer $n$, how many primitive right triangles are there such that their area (numerically) equals $n$ times their perimeter? | Let the sides of a primitive right-angled triangle be $x, y, z$. Then $x, y$ are one odd and one even. Without loss of generality, assume $2 \mid y$. By the theorem, there exist $u, v \in \mathbf{N}^{*}, (u, v)=1$, and $u, v$ are one odd and one even, such that
$$x=u^{2}-v^{2}, y=2 u v, z=u^{2}+v^{2}.$$
By the given c... | 2^{k} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,516 |
Example 4 Proof: There exist infinitely many positive integer triples $(a, b, c)$ such that $a^{2}+b^{2}, a^{2}+c^{2}, b^{2}+c^{2}$ are all perfect squares. | We prove that we can construct using Pythagorean triples. Take any Pythagorean triple \((x, y, z)\) (not necessarily primitive). Let
$$a=x\left|4 y^{2}-z^{2}\right|, b=y\left|4 x^{2}-z^{2}\right|, c=4 x y z,$$
then we have
$$\begin{aligned}
a^{2}+b^{2} & =x^{2}\left(3 y^{2}-x^{2}\right)^{2}+y^{2}\left(3 x^{2}-y^{2}\ri... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,517 |
Example 5 Proof: The indeterminate equation $x^{4}+y^{4}=z^{2}$ has no integer solutions such that $x y z \neq 0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Example 5 Proof: The indeterminate equation $x^{4}+y^{4}=z^{2... | Proof:
If the equation has a solution such that $x y z \neq 0$, without loss of generality, let $x, y, z \in \mathbf{N}^{*}$, and $(x, y) = 1$. Furthermore, assume $y$ is even (it is easy to see that $x, y$ are of different parities), and let $z$ be the smallest positive integer among all positive integer solutions of ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,518 |
Example 2 Let $n$ be a positive integer greater than 1. Prove:
$$k \sqrt{n}<\sigma(n)<n \sqrt{2 k}$$
where $k=d(n)$. | Proof
Notice,
$$\sigma(n)=\sum_{d \mid n} d=\sum_{d \mid n} \frac{n}{d}$$
Here $\sum_{d \mid n}$ denotes the summation over all positive divisors of $n$.
Therefore, we have
$$\sigma(n)=\frac{1}{2} \sum_{d \mid n}\left(d+\frac{n}{d}\right) \geqslant \frac{1}{2} \sum_{d \mid n} 2 \sqrt{n}=\sqrt{n} \sum_{d \mid n} 1=k \s... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,519 |
Example 6 Let positive integers $a, b$ satisfy $a b>1$. Find all positive integer values that the algebraic expression
$$f(a, b)=\frac{a^{2}+a b+b^{2}}{a b-1}$$
can take. | Let $a, b \in \mathbf{N}^{*}$ satisfy $a b>1$, and such that
$$f(a, b)=\frac{a^{2}+a b+b^{2}}{a b-1}=k \in \mathbf{N}^{*},$$
with the conditions: $a \geqslant b$ and $b$ is the smallest positive integer.
From this, we know that the quadratic equation in $x$
i.e., $\square$
$$\begin{aligned}
x^{2}+b x+b^{2}-k(b x-1) &... | 4, 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,520 |
Example 7 Positive integers $a, b, c$ satisfy:
$$0<a^{2}+b^{2}-a b c \leqslant c+1$$
Prove: The number $a^{2}+b^{2}-a b c$ is a perfect square. | Proof
Let $a, b, c$ be positive integers satisfying the condition, and let
$$a^{2}+b^{2}-a b c=t$$
Suppose $a, b$ are the positive integers that minimize $a+b$ for fixed $c, t$.
By symmetry, assume $a \geqslant b$. Since $a$ is a positive integer solution to the quadratic equation
$$x^{2}-b c x+b^{2}-t=0$$
Let $\bar{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,521 |
Example 8 Let $c$ be a positive integer, and $c$ can be expressed as the sum of squares of 3 rational numbers. Prove: $c$ can be expressed as the sum of squares of 3 integers. | Proof
First, we transform the problem.
Let $f(x, y, z, w) = n x^{2} - y^{2} - z^{2} - w^{2}$, where $n \in \mathbf{N}^{*}, x, y, z, w \in \mathbf{Z}$. Then the proposition can be transformed into: If $f(x, y, z, w) = 0$ has an integer solution $(x, y, z, w)$ with $x \neq 0$, then $f(x, y, z, w) = 0$ has an integer solu... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,522 |
Example 1 Let $\left(x_{0}, y_{0}\right)$ be the minimal solution of equation (1). Prove: For any positive integer solution $(x, y)$ of (1), it must be that $x_{0} \leqslant x, y_{0} \leqslant y$. | Proof
First, prove $x_{0} \leqslant x$. If $x_{0}>x$, then by
$$x_{0}^{2}=D y_{0}^{2}+1, x^{2}=D y^{2}+1$$
we get
$$D y_{0}^{2}+1>D y^{2}+1$$
which means $y_{0}>y$, thus
$$x+y \sqrt{D}<x_{0}+y_{0} \sqrt{D}$$
This contradicts the fact that $\left(x_{0}, y_{0}\right)$ is the minimal solution. Similarly, we can prove $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,523 |
Theorem 1 The equation (1) has infinitely many positive integer solutions, all of which can be expressed from its fundamental solution $\left(x_{0}, y_{0}\right)$ in the following form:
$$x+y \sqrt{D}=\left(x_{0}+y_{0} \sqrt{D}\right)^{n},$$
where $n$ is any positive integer, $x, y \in \mathbf{N}^{*}$. | Below is the proof of Theorem 1.
First, we prove that equation (1) has at least one positive integer solution \((x, y)\).
By Corollary 2, there exist two distinct positive integer solutions \((x_1, y_1) \neq (x_2, y_2)\) that satisfy equation (4),
$$
x_1 \equiv x_2 \pmod{|k|}, \quad y_1 \equiv y_2 \pmod{|k|}
$$
Thus, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,524 |
Lemma 1 Let $\alpha(\alpha>1)$ be an irrational number, then for any $q>1, q \in \mathbf{N}^{*}$, there exist positive integers $x, y$, such that
$$\left|x-y_{\alpha}\right|<\frac{1}{q}(0<y \leqslant q) \text {. }$$ | Note that $0, \{\alpha\}, \{2 \alpha\}, \cdots, \{q \alpha\}$ are $q+1$ numbers all within the interval $[0,1)$ (here $\{x\}=x-[x]$, representing the fractional part of $x$). Therefore, by the pigeonhole principle, there exist $0 \leqslant i < j \leqslant q$ such that $|\{i \alpha\} - \{j \alpha\}| < \frac{1}{q}$, i.e.... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,525 |
Lemma 2 Let $D \in \mathbf{N}^{*}, D$ not a perfect square (thus, $\sqrt{D}(>1)$ is an irrational number), then there exist infinitely many pairs of positive integers $x, y$, such that
$$\left|x^{2}-D y^{2}\right|<1+2 \sqrt{D}$$ | Proof: By Corollary 1 (taking $\alpha=\sqrt{D}$), there exist infinitely many pairs of positive integers $(x, y)$ such that $|x-y \sqrt{D}|<\frac{1}{y}$. Therefore,
$$\begin{aligned}
\left|x^{2}-D y^{2}\right| & =|x-y \sqrt{D}| \cdot|x+y \sqrt{D}| \\
& <\frac{1}{y}|x+y \sqrt{D}| \leqslant \frac{1}{y}(|x-y \sqrt{D}|+2 y... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,527 |
Example 3 Let $a, b$ be two distinct positive integers, and
$$\left(a^{2}+a b+b^{2}\right) \mid a b(a+b) .$$
Prove: $|a-b|>\sqrt[3]{a b}$. | Notice,
$$a^{3}=\left(a^{2}+a b+b^{2}\right) a-a b(a+b),$$
thus, by the condition, we know $\left(a^{2}+a b+b^{2}\right) \mid a^{3}$. Similarly, we have $\left(a^{2}+a b+b^{2}\right) \mid b^{3}$.
Therefore, $a^{2}+a b+b^{2}$ is a common divisor of $a^{3}$ and $b^{3}$. Hence,
$$\left(a^{2}+a b+b^{2}\right) \mid\left(a^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,531 |
Example 2 Find all triangular numbers that are also perfect squares, i.e., find all positive integer solutions $(n, k)$ of
$$\frac{n(n+1)}{2}=k^{2}$$
determining the numbers $\frac{n(n+1)}{2}$. | Solve: Transform equation (7) into: $(2 n+1)^{2}-2(2 k)^{2}=1$, and consider the Pell's equation
$$x^{2}-2 y^{2}=1$$
for all positive integer solutions. From (8), we know that $x$ is odd. Taking both sides of (8) modulo 4, we find that $y$ is even. Let
$$n=\frac{x-1}{2}, k=\frac{y}{2}$$
When $(x, y)$ runs through all... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,532 |
Example 3 Let $k>1$ be a given integer. Prove: there are infinitely many integers $n$, such that $k n + 1$ and $(k+1) n + 1$ are both perfect squares. | Proof
Suppose $k n+1$ and $(k+1) n+1$ are both perfect squares. Let
$$k n+1=u^{2},(k+1) n+1=v^{2} .$$
Eliminating $n$, we get
$$(k+1) u^{2}-k v^{2}=1$$
Notice that, for any positive integer solution $(u, v)$ of (9), taking $n=u^{2}-v^{2}$, it is easy to see that $k n+1$ and $(k+1) n+1$ are both squares, thus the prob... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,533 |
Example 4 Find all positive integers $m(>1)$, such that the number $m^{3}$ can be expressed as the sum of squares of $m$ consecutive positive integers. | Let $m$ be a positive integer that meets the requirement, then there exists $k \in \mathbf{N}$, such that
$$m^{3}=(k+1)^{2}+(k+2)^{2}+\cdots+(k+m)^{2} .$$
Equation (11) is equivalent to
$$k^{2}+(m+1) k+\frac{1}{6}(m+1)(2 m+1)-m^{2}=0$$
Solving the above quadratic equation in $k$, we get
$$k=-\frac{1}{2}(m+1) \pm \frac... | m_{n}=x_{n}+6 y_{n} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,534 |
Example 5 Given a positive real number $\varepsilon$, if there exist $a, b \in \mathbf{N}^{*}$, such that
$$a \leqslant b<(1+\varepsilon) a \text {, and } n=a b, \text {. }$$
We call the positive integer $n$ an " $\varepsilon-$ square number". Prove: There exist infinitely many positive integers $n$, such that the con... | Proof
Our starting point is to find infinitely many $x \in \mathbf{N}^{*}$, such that
$$x^{2}, x^{2}-1, x^{2}-2, \cdots, x^{2}-5$$
are all " $\varepsilon$-square numbers".
Notice that, when $x$ is sufficiently large, $x^{2}, x^{2}-1, x^{2}-4$ are all " $\varepsilon$-square numbers", so the focus is on the three number... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,535 |
Example 6 Proof: The indeterminate equation
$$5^{a}-3^{b}=2$$
has the only positive integer solution $a=b=1$. | Proof
Obviously, when one of $a, b$ is 1, the other is also 1. Now let $(a, b)$ be a solution satisfying (18), with $a>1, b>1$.
Congruences are commonly used to handle exponential Diophantine equations. Taking (18) modulo 4, we get
$$1^{a}-(-1)^{b} \equiv 2(\bmod 4),$$
so $b$ is odd. Taking (18) modulo 3, we get
$$(-... | a=b=1 | Number Theory | proof | Yes | Yes | number_theory | false | 741,536 |
Example 7 Find all positive integer pairs $(m, n)$ that satisfy the equation
$$3^{m}=2 n^{2}+1$$ | It is easy to verify that $(m, n)=(1,1),(2,2)$ and $(5,11)$ satisfy (19). Below, we prove that these pairs are all the solutions that meet the conditions.
Case 1: $m$ is even. In this case, the problem turns into finding the positive integer solutions of the Pell equation
$$x^{2}-2 y^{2}=1$$
that make $x$ a power of ... | (m, n)=(1,1),(2,2),(5,11) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,537 |
Example 1 Find the integer solutions of the indeterminate equation $\frac{1}{2}(x+y)(y+z)(z+x)+(x+y+z)^{3}=1-$ $x y z$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Let's make the substitution, set $x+y=u, y+z=v, z+x=w$, then the original equation transforms to $4 u v w+(u+v+w)^{3}=8-(u+v-w)(u-v+w)(-u+v+w)$.
After expanding and combining like terms, we get
$$4\left(u^{2} v+v^{2} w+w^{2} u+u v^{2}+v w^{2}+w u^{2}\right)+8 u v w=8,$$
which simplifies to
$$u^{2} v+v^{2} w+w^{2} u+u ... | (x, y, z)= (1,0,0),(0,1,0),(0,0,1),(2,-1,-1),(-1,2,-1),(-1,-1,2) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,538 |
Example 2 Find the integer solutions of the equation $x^{2}+x=y^{4}+y^{3}+y^{2}+y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solving as in the previous example, multiply both sides of the equation by 4 and complete the square on the left side:
$$(2 x+1)^{2}=4\left(y^{4}+y^{3}+y^{2}+y\right)+1$$
Next, we estimate the right side of equation (1). Since
$$\begin{aligned}
4\left(y^{4}+y^{3}+y^{2}+y\right)+1 & =\left(2 y^{2}+y+1\right)^{2}-y^{2}+... | (x, y)=(0,-1),(-1,-1),(0,0),(-1,0),(-6,2),(5,2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,539 |
Example 3 Proof: The indeterminate equation
$$x^{2}+y^{2}+z^{2}+3(x+y+z)+5=0$$ has no rational solutions. | Proof:
Multiply both sides of the equation by 4, complete the square, we get
$$(2 x+3)^{2}+(2 y+3)^{2}+(2 z+3)^{2}=7$$
The necessary and sufficient condition for this equation to have rational solutions is that the following equation
$$a^{2}+b^{2}+c^{2}=7 m^{2}$$
has integer solutions $(a, b, c, m)$, and $m \in \math... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,540 |
Example 4 Find all positive integer tuples $(a, b, c, d)$ such that
$$\left\{\begin{array}{l}
b d > a d + b c, \\
(9 a c + b d)(a d + b c) = a^{2} d^{2} + 10 a b c d + b^{2} c^{2}
\end{array}\right.$$ | Let $(a, b, c, d)$ be a set of positive integers that satisfy the conditions. Then, from (3), we have $b > a$, $d > \frac{bc}{b-a}$.
Consider the following quadratic function:
\[
\begin{aligned}
f(x) & = (9ac + bx)(ax + bc) - a^2x^2 - 10abcx - b^2c^2 \\
& = a(b-a)x^2 + c(9a-b)(a-b)x + bc^2(9a-b)
\end{aligned}
\]
The ... | (a, 3a, c, 3c) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,541 |
Example 4 Let $n \in \mathbf{N}^{*}, a$ and $b$ be two distinct integers. It is known that:
$$n \mid\left(a^{n}-b^{n}\right) .$$
Prove: $n \left\lvert\, \frac{a^{n}-b^{n}}{a-b}\right.$. | Proof
For any prime factor $p$ of $n$, let $p^{a} \| n$ (i.e., $p^{a} \mid n$, but $p^{a+1} \nmid n$, this notation will be frequently used in this book), by the condition, we know that $p^{a} \mid (a^{n}-b^{n})$.
Using the Fundamental Theorem of Arithmetic, it suffices to prove: $p^{a} \left\lvert\, \frac{a^{n}-b^{n}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,542 |
Example 5 Find all integer arrays $(a, b, c, x, y, z)$, such that
$$\left\{\begin{array}{l}
a+b+c=x y z, \\
x+y+z=a b c,
\end{array}\right.$$
where $a \geqslant b \geqslant c \geqslant 1, x \geqslant y \geqslant z \geqslant 1$. | By symmetry, we only need to consider the case $x \geqslant a$. At this time, $x y z=a+b+c \leqslant 3 a \leqslant 3 x$, so $y z \leqslant 3$. Thus, $(y, z)=(1,1),(2,1),(3,1)$.
When $(y, z)=(1,1)$, $a+b+c=x$ and $x+2=a b c$, thus $a b c=a+b+c+2$. If $c \geqslant 2$, then
$$a+b+c+2 \leqslant 3 a+2 \leqslant 4 a \leqsla... | (2,2,2,6,1,1),(5,2,1,8,1,1),(3,3,1,7,1,1),(3,2,1,3,2,1),(6,1,1,2,2,2),(8,1,1,5,2,1),(7,1,1,3,3,1) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,543 |
Example 6 Find all integer pairs $(m, n)$ such that
$$\left\{\begin{array}{l}
2 m \equiv-1(\bmod n), \\
n^{2} \equiv-2(\bmod m) .
\end{array}\right.$$ | This problem is formally about divisibility, but in essence, it is about solving indeterminate equations.
Using the conditions $n|(2 m+1), m|\left(n^{2}+2\right)$, we know that $n, m$ are both odd numbers. If $(m, n)$ meets the requirements, then $(m, -n)$ also meets them, so we can assume $n$ is a positive odd number... | (27, \pm 5),(19, \pm 13),(17, \pm 7),(3, \pm 1),(3, \pm 7),(1, \pm 1),(1, \pm 3),(-1, \pm 1),(-3, \pm 1),(-3, \pm 5),(-11, \pm 3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,544 |
Example 7 Find the smallest positive integer $n$, such that the indeterminate equation
$$x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}=1599$$
has integer solutions $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$. | Note that for any $x \in \mathbf{Z}$, if $x$ is even, then $x^{4} \equiv 0(\bmod 16)$; if $x$ is odd, then $x^{2} \equiv 1(\bmod 8)$, and in this case, $x^{4} \equiv 1(\bmod 16)$.
The above discussion shows that $x_{i}^{4} \equiv 0$ or $1(\bmod 16)$, thus the remainder of $x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}$ modulo ... | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,545 |
Example 8 Find all non-negative integer arrays $(m, n, p, q)$, such that $0<p<q,(p, q)=1$, and
$$\left(1+3^{m}\right)^{p}=\left(1+3^{n}\right)^{q} .$$ | If $p>1$, by $(p, q)=1$, we can set $q=p s+r, 0<p$. Then, $\left(c^{p}-1\right) \mid\left(c^{q}-1\right)$, and
$$c^{g}-1=c^{\infty+r}-1=\left(c^{p}\right)^{s} \cdot c^{r}-1 \equiv 1^{s} \cdot c^{r}-1\left(\bmod c^{\rho}-1\right),$$
Therefore,
$$\left(c^{p}-1\right) \mid\left(c^{r}-1\right)$$
This cannot hold when $r<... | (m, n, p, q)=(1,0,1,2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,546 |
Example 9 Find the positive integer solutions $(x, y, z, t)$ of the indeterminate equation $1+5^{x}=2^{y}+2^{x} \cdot 5^{t}$. | Let $(x, y, z, t)$ be a positive integer solution to the original equation. Taking both sides of the equation modulo 5, we have $2^{y} \equiv 1(\bmod 5)$. Since 2 is a primitive root modulo 5, it follows that $4 \mid y$. At this point, taking both sides of the equation modulo 4, we get $2^{z} \equiv 2(\bmod 4)$, so $z=... | (x, y, z, t)=(2,4,1,1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,547 |
Example 10 Proof: The indeterminate equation
$$x^{2}+5=y^{3}$$
has no integer solutions. | Proof: If $(13)$ has integer solutions $(x, y)$, taking both sides of $(13)$ modulo 4, we get $y^{3} \equiv 1$ or $2(\bmod 4)$, so $y$ must be odd (since if $y$ is even, $y^{3} \equiv 0(\bmod 4)$). We can set $y=2 k+1$, substituting into (13), we get
$$x^{2}=8 k^{3}+12 k^{2}+6 k-4$$
This shows that $x$ is even. Let $x... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,548 |
Example 12 Find all integers $a$ such that the equation
$$x^{2}+a x y+y^{2}=1$$
has infinitely many integer solutions $(x, y)$, and prove your conclusion. | When $a=0$, the equation is $x^{2}+y^{2}=1$, which has only 4 solutions.
If $a \neq 0$, then $(x, y)$ is a solution to equation (16) if and only if: $(x, -y)$ is a solution to the equation $x^{2}-a x y+y^{2}=1$. Therefore, we only need to discuss $ay$, then $(x, -a x+y)$ is also a solution to (16) (this solution can be... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,550 |
Example 13 Let $(x, y, z)$ be a positive integer solution to the indeterminate equation
$$x y=z^{2}+1$$
Prove: there exist integers $a, b, c, d$, such that
$$x=a^{2}+b^{2}, y=c^{2}+d^{2}, z=a c+b d,$$
where $a, b, c, d$ satisfy: $|a d-b c|=1$. | Proof:
We prove the statement by induction on $z$.
When $z=1$, $xy=2$, so $(x, y)=(1,2)$ or $(2,1)$. Thus, $(x, y, z)=$ $(1,2,1)$ or $(2,1,1)$. By setting
$$(a, b, c, d)=(1,0,1,1),(1,1,0,1)$$
we see that the proposition holds for $z=1$.
Now suppose $(x_0, y_0, z_0)$ is a positive integer solution to (17), and that (17... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,551 |
1. Find all integer solutions to the indeterminate equation $x+y=x^{2}-x y+y^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 1. Transform the original equation to
$$x^{2}-(y+1) x+y^{2}-y=0 .$$
From $\Delta=(y+1)^{2}-4\left(y^{2}-y\right) \geqslant 0$, we know that $1-\frac{2 \sqrt{3}}{3} \leqslant y \leqslant 1+\frac{2 \sqrt{3}}{3}$, hence $y$ $\in\{0,1,2\}$. Substituting these values into the original equation, we obtain the solutions $(x,... | null | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,552 |
Example 5 Let $p$ be a prime number, for a set $M$ consisting of $p+1$ different positive integers, prove: from $M$ two numbers $a, b(a>b)$ can be taken out, such that
$$\frac{a}{(a, b)} \geqslant p+1$$ | Proof
First, we prove a lemma: Let $a, b \in \mathbf{N}^{*}, a > b$ and $p \mid (a - b)$, where $a$ and $b$ are not multiples of $p$. Then $\frac{a}{(a, b)} \geqslant p + 1$.
In fact, since $p \mid (a - b)$, we can set $a = b + px$, where $x \in \mathbf{N}^{*}$. Then we have $(a, b) \mid px$. Since $p$ is not a factor... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,553 |
2. Let $n \in \mathbf{N}^{*}, a_{n}$ be the number of positive integer solutions $(x, y)$ to the indeterminate equation $x^{2}-y^{2}=10^{2} \cdot 30^{2 n}$. Prove: $a_{n}$ is not a perfect square. | 2. Let $(x, y)$ be its positive integer solution, then $x, y$ have the same parity, hence, $\frac{x-y}{2} \cdot \frac{x+y}{2}=$ $5^{2} \times 30^{2 n}=2^{2 n} \times 3^{2 n} \times 5^{2 n+2}$.
We can obtain
$$a_{n}=\frac{1}{2}\left((2 n+1)^{2}(2 n+3)-1\right)=(n+1)\left(4 n^{2}+6 n+1\right)$$
(because $\frac{x-y}{2}<\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,554 |
3. Find all integer pairs $(a, b)$ such that the system of equations
$$\left\{\begin{array}{l}
x^{2}+2 a x-3 a-1=0 \\
y^{2}-2 b y+x=0
\end{array}\right.$$
has exactly three distinct real solutions $(x, y)$. | 3. Let $(a, b)$ be a pair of integers that satisfy the conditions, then the system of equations
$$\left\{\begin{array}{l}
x^{2}+2 a x-3 a-1=0, \\
y^{2}-2 b y+x=0
\end{array}\right.$$
has exactly 3 different real solutions, so equation (1) should have two different real roots, with the discriminant $\Delta=4 a^{2}+12 a... | (0,1), (0,-1), (-3,2), (-3,-2) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,555 |
4. Find all integer solutions to the indeterminate equation $5 x^{2}-14 y^{2}=11 z^{2}$. | 4. Taking both sides modulo 7, we know $5 x^{2} \equiv 11 z^{2} \equiv 4 z^{2}(\bmod 7)$. If $7 \times x$, then $\left(2 z x^{-1}\right)^{2} \equiv 5(\bmod 7)$, where $x^{-1}$ is the modular inverse of $x$ modulo 7. However, a square number $\equiv 0,1$ or $4(\bmod 7)$, which is a contradiction. Therefore, $7 \mid x$.... | (x, y, z)=(0, 0,0) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,556 |
5. Find all integer pairs $(x, y)$ such that $x^{3}=y^{3}+2 y^{2}+1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. Notice that, when $y>0$ or $y<-3$, we always have $y^{3}<y^{3}+2 y^{2}+1<(y+$ $1)^{3}$, at this time $y^{3}+2 y^{2}+1$ is not a cube number, so the original equation has no solution. Therefore, we only need to consider the cases $y=$ $-3,-2,-1,0$. Substituting them respectively, we get the integer solutions of the e... | (x, y)=(-2, -3), (1, -2), (1, 0) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,557 |
6. Find the integer solutions of the indeterminate equation $x^{2}(y-1)+y^{2}(x-1)=1$.
untranslated portion:
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将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
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6.... | 6. Let $(x, y)$ be an integer solution of the equation, and assume $x \leqslant y$, then $y \geqslant 2$. In this case, consider the original equation as a quadratic equation in $x$:
$$(y-1) x^{2}+y^{2} x-\left(y^{2}+1\right)=0$$
Since the equation has integer solutions, $\Delta=y^{4}+4(y-1)\left(y^{2}+1\right)$ must ... | (x, y)=(1,2),(2,1),(-5,2),(2,-5) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,558 |
7. Let $a, b, c, d$ all be prime numbers, satisfying: $a>3 b>6 c>12 d$, and
$$a^{2}-b^{2}+c^{2}-d^{2}=1749$$
Find the value of $a^{2}+b^{2}+c^{2}+d^{2}$. | 7. From the conditions, we know that $a, b, c$ are all odd numbers. If $d$ is odd, then $a^{2}-b^{2}+c^{2}-d^{2}$ is even, which is a contradiction. Therefore, $d$ is even, and thus $d=2$. Consequently, $a^{2}-b^{2}+c^{2}=1753$. From the conditions, we also know that $a \geqslant 3 b+2, b \geqslant 2 c+1, c \geqslant 5... | 1999 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,559 |
8. Positive integers $a, b, c$ and $x, y, z$ satisfy: $|x-a| \leqslant 1,|y-b| \leqslant 1$,
$$a^{2}+b^{2}=c^{2}, x^{2}+y^{2}=z^{2} .$$
Prove: The sets $\{a, b\}$ and $\{x, y\}$ are the same. | 8. By symmetry, we may assume $x \geqslant a$, then $x=a$ or $x=a+1$.
Case one: $x=a$, in this case, $b \neq a$ (otherwise, $c=\sqrt{2} a \notin \mathbf{N}^{*}$).
If $y=b$, then $\{a, b\}=\{x, y\}$.
If $y=b+1$, then by $a^{2}+b^{2}=c^{2}$, we know $bc$, so $z \geqslant c+2$, which requires
$$c^{2}+2(a-b+1) \geqslant(c... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,560 |
9. If there exists a triangle with side lengths of positive integers $a, b, c$, and the angle opposite the side of length $c$ is $120^{\circ}$, then the array $(a, b, c)$ is called a quasi-Pythagorean array. Prove: If $(a, b, c)$ is a quasi-Pythagorean array, then $c$ has a prime factor greater than 5. | 9. From the condition, we know that $c^{2}=a^{2}+a b+b^{2}$. To prove that $c$ has a prime factor greater than 5, it suffices to prove: if $c$ is a multiple of $2,3,5$, then $a, b$ are also multiples of $2,3,5$ (in this case, canceling out $2,3,5$ from both sides, we get an indeterminate equation of the same form, thus... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,561 |
10. Find all positive integer triples $(x, y, z)$ such that $y$ is a prime number, 3 and $y$ are not divisors of $z$, and $x^{3}-y^{3}=z^{2}$. | 10. From the original equation, we get $z^{2}=(x-y)\left(x^{2}+x y+y^{2}\right)$. Let
$$\begin{aligned}
\left(x-y, x^{2}+x y+y^{2}\right) & =\left(x-y,(x-y)^{2}+3 x y\right)=(x-y, 3 x y) \\
& =\left(x-y, 3 y^{2}\right)=d
\end{aligned}$$
Then $d \mid 3 y^{2}$. Since $y$ is a prime and 3 and $y$ are not factors of $z$, ... | (8,7,13) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,562 |
11. Find the integer solutions of the indeterminate equation $x^{5}+y^{5}=(x+y)^{3}$. | 11. Clearly, the integer pairs $(x, y)$ that satisfy $x+y=0$ are all solutions to the equation. Now consider the solutions to the equation when $x+y \neq 0$. In this case, dividing both sides by $x+y$ gives
$$x^{4}-x^{3} y+x^{2} y^{2}-x y^{3}+y^{4}=(x+y)^{2},$$
Rearranging and simplifying, we get
$$\left(x^{2}+y^{2}\r... | (t,-t) \mid t \in \mathbf{Z}\ | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,563 |
Example 6 Let $n$ be a positive integer of the form $a^{2}+b^{2}$, where $a, b$ are two coprime positive integers. It satisfies: if $p$ is a prime and $p \leqslant \sqrt{n}$, then $p \mid a b$. Find all positive integers $n$ that meet the requirements. | Let $n=a^{2}+b^{2}$ be a number that meets the requirement.
If $a=b$, then by $(a, b)=1$, we know $a=b=1$, in this case $n=2$ meets the requirement.
If $a \neq b$, without loss of generality, assume $a < b$. Take a prime factor $p$ of $b-a$. By $(a, b)=1$, we know $(b-a, b)=(b-a, a)=1$, thus $(b-a, ab)=1$, hence $p \nm... | 2, 5, 13 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,564 |
12. Find the integer solutions of the indeterminate equation $x^{6}+x^{3} y=y^{3}+2 y^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 12. Let $(x, y)$ be the integer solution of the equation, then when $x=0$, $y^{2}(y+2)=0$, we get $y=$ 0 or -2. When $y=0$, $x^{6}=0$, we get $x=0$. It is known that when $x, y$ have one equal to zero, the solutions are $(x, y)=(0,0)$ or $(0,-2)$.
Below, we discuss the case where $x y \neq 0$. For any prime factor $p$... | (x, y)=(0,0),(0,-2) \text{ or } (2,4) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,565 |
13. Find all pairs of positive integers $(x, y)$ such that $x^{x+y}=y^{y-x}$. | 13. Let the pair of positive integers $(x, y)$ satisfy the condition, then $x^{y}(x y)^{x}=y^{y}$, hence $x^{y} \mid y^{y}$, so $x \mid y$. Let $y=k x, k \in \mathbf{N}^{*}$, then $k^{y}=\left(k x^{2}\right)^{x}$, thus $k^{k}=k x^{2}$, which means $x^{2}=k^{k-1}$. Therefore, all pairs of positive integers $(x, y)$ that... | (2 n+1)^{n},(2 n+1)^{n+1} \text{ or } (2 m)^{4 m^{2}-1},(2 m)^{4 m^{2}+1} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,566 |
14. Prove: The indeterminate equation $x^{2}=y^{5}-4$ has no integer solutions. | 14. Suppose the original equation has integer solutions $(x, y)$. If $11 \mid y$, then $x^{2} \equiv 7(\bmod 11)$; if $11 \nmid y$, then by Fermat's Little Theorem, we know $11 \mid\left(y^{10}-1\right)$, i.e., $11 \mid\left(y^{5}-1\right)\left(y^{5}+1\right)$. Since $\left(y^{5}-1, y^{5}+1\right) \mid 2$, it follows t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,567 |
15. Prove: The indeterminate equation
$$x^{2}+y^{2}+z^{2}+u^{2}+v^{2}=x y z u v-65$$
has infinitely many sets of positive integer solutions. | 15. Notice that $(x, y, z, u, v)=(1,2,3,4,5)$ is a positive integer solution to the original equation.
Generally, let $(x, y, z, u, v)$ be a positive integer solution to the original equation, and $x<y<z<u<v$. Then, considering the original equation as a quadratic equation in $x$, we know that $(y z u v-x, y, z, u, v)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,568 |
16. Prove: There exist infinitely many $n \in \mathbf{N}^{*}$, such that the square mean of the first $n$ positive integers $\left(\right.$ i.e., $\left.\left(\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}\right)^{\frac{1}{2}}\right)$ is a positive integer. | 16. To prove that there exist infinitely many pairs of positive integers $(n, k)$, satisfying
$$k=\left(\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}\right)^{\frac{1}{2}},$$
i.e., $2 n^{2}+3 n+1=6 k^{2}$. Rearranging gives
$$(4 n+3)^{2}-48 k^{2}=1$$
It suffices to prove that the Pell equation $x^{2}-48 y^{2}=1$ has infinitely m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,569 |
17. Prove: There exist infinitely many sets of integers $(x, y, z, t)$, such that $x^{3}+y^{3}+z^{3}+t^{3}=$ 1999. | 17. It is easy to see that $10^{3}+10^{3}+0^{3}+(-1)^{3}=1999$. We seek integer solutions of the form:
$$x=10-k, y=10+k, z=m, t=-1-m$$
where $k, m$ are undetermined integers. Substituting (3) into the original equation and simplifying, we find that $k, m$ should satisfy $m(m+1)=20 k^{2}$. Multiplying both sides by 4 a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,570 |
18. Prove: There exist infinitely many arrays composed of three consecutive positive integers, each of which is the sum of the squares of two positive integers. Also, ask: Do such quadruple arrays exist? | For any $n \in \mathbf{N}^{*}$, let $x=2 n^{4}+4 n^{3}+2 n^{2}$, then
$$\begin{array}{c}
x=\left(n^{2}+n\right)^{2}+\left(n^{2}+n\right)^{2}, x+1=[n(n+2)]^{2}+\left(n^{2}-1\right)^{2} \\
x+2=\left(n^{2}+n+1\right)^{2}+\left(n^{2}+n-1\right)^{2}
\end{array}$$
Thus, there exist infinitely many sets of three consecutive ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,571 |
19. Prove: If $p$ is a prime, and $p \equiv 3(\bmod 4)$, then for any $n \in \mathbf{N}^{*}$, there does not exist a pair of positive integers $(x, y)$ such that $x^{2}+y^{2}=p^{n}$. | 19. If there exists $k \in \mathbf{N}^{*}$, and $x, y \in \mathbf{N}^{*}$, such that
$$p^{k}=x^{2}+y^{2},(x, y)=1$$
If $k$ is odd, then taking both sides of (5) modulo 4, we get
$$x^{2}+y^{2} \equiv(-1)^{k} \equiv 3(\bmod 4)$$
This is a contradiction. If $k$ is even, let $k=2 r, r \in \mathbf{N}^{*}$, then by the the... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,572 |
20. Prove: For any $n \in \mathbf{N}^{*}$, the indeterminate equation $x^{2}+y^{2}=z^{n}$ has infinitely many positive integer solutions $(x, y, z)$. | 20. An interesting construction is as follows: $x+y \mathrm{i}=(a+b \mathrm{i})^{n}$, where i is the imaginary unit. When $n$ is fixed, there are infinitely many pairs of integers $(a, b)$ such that integers $x, y$ satisfy $x y \neq 0$. In this case, $x^{2}+y^{2}=\left(a^{2}+b^{2}\right)^{n}$ (this can be obtained by t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,573 |
21. Prove: For any $m \in \mathbf{Z}$, there exist infinitely many $n \in \mathbf{N}^{*}$, such that the number $\left[n \sqrt{m^{2}+1}\right]$ is a perfect square. | 21. When $m=0$, take $n$ as a perfect square; when $m \neq 0$, the Pell equation $\left(m^{2}+1\right) x^{2}-y^{2}=1$ has infinitely many positive integer solutions $\left(x_{t}, y_{t}\right)$, which are determined by the following formula:
$$y_{t}+x_{t} \sqrt{m^{2}+1}=\left(|m|+\sqrt{m^{2}+1}\right)^{2 t+1}.$$
For th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,574 |
Example 7 Let $p_{1}, p_{2}, \cdots, p_{n}$ be $n$ distinct prime numbers greater than 3. Prove: the number $2^{p_{1} p_{2} \cdots p_{n}}+1$ has at least $4^{n}$ distinct positive divisors. | Proof
First, we prove a lemma: If $u, v$ are two coprime positive odd numbers, then $\left(2^{u}+1,2^{v}+1\right) = 3$.
Without loss of generality, assume $u > v$ (if $u = v$, then $u = v = 1$, and the proposition is obviously true), then
$$\begin{aligned}
\left(2^{u}+1,2^{v}+1\right) & =\left(2^{u}-2^{v}, 2^{v}+1\rig... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,575 |
22. Prove: There exist infinitely many $n \in \mathbf{N}^{*}$, such that there exist positive integers $a, b, c, d$, satisfying:
(1) $(a, b, c, d)=1,\{a, b\} \neq\{c, d\}$;
(2) $n=a^{3}+b^{3}=c^{3}+d^{3}$. | 22. First, we prove a lemma: there exist infinitely many pairs of positive integers \((u, v)\) such that
$$u^{2}-7 v^{2}=144$$
In fact, since \(17^{2}-7 \times 5^{2}=114\), \(\alpha=17+5 \sqrt{7}\) is a solution to (6). Note that \(\varepsilon=8+3 \sqrt{7}\) is the fundamental solution to the Pell equation \(x^{2}-7 y... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,576 |
23. Let $p$ be a given prime, $p \equiv 3(\bmod 4)$. Prove: the indefinite equation
$$(p+2) x^{2}-(p+1) y^{2}+p x+(p+2) y=1$$
has infinitely many positive integer solutions $(x, y)$, and for each positive integer solution $(x, y)$, we have $p \mid x$. | 23. Transform the equation into
$$\begin{aligned}
x^{2} & =(p+1)\left(y^{2}-x^{2}\right)-p x-p y-2 y+1 \\
& =(p+1)(y+x)(y-x)-(p+1)(x+y)+x-y+1 \\
& =((p+1)(x+y)-1)((y-x)-1) .
\end{aligned}$$
Let $z=y-1$, then the above equation becomes
$$x^{2}=(z-x)(p(z+x)+p)$$
If $(z-x,(p+1)(z+x)+p)=1$, then by (7), $z-x$ and $(p+1)(... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,577 |
24. Prove: The indeterminate equation $x^{4}-y^{4}=z^{2}$ has no integer solutions such that $x y z \neq 0$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 24. Suppose the given equation has a solution such that $x y z \neq 0$, and let $x, y, z \in \mathbf{N}^{*}$, with $x$ being the smallest among all solutions. Then $(x, y)=1$, and it is easy to see that $x$ is odd.
If $y$ is odd, then $z$ is even, and there exist $a, b \in \mathbf{N}^{*}$ such that
$$x^{2}=a^{2}+b^{2}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,578 |
25. Find the integer solutions of the indeterminate equation $8 x^{4}+1=y^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 25. Notice that $(x, y)=(0, \pm 1),( \pm 1, \pm 3)$ are solutions to the equation. If $(x, y)$ is an integer solution to the equation and $|x|>1$, we may assume $y>1$.
From $8 x^{4}=(y-1)(y+1)$, and since $y$ is odd, we can set
$$\left\{\begin{array} { l }
{ y - 1 = 2 u ^ { 4 } , } \\
{ y + 1 = 4 v ^ { 4 } , }
\end{a... | (x, y)=(0, \pm 1),( \pm 1, \pm 3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,579 |
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