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Example 4 For any positive integer $n$, prove: there exist $n$ consecutive positive integers, all of which are composite.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
However, it seems the translation request was already fu... | Proof: Let $n$ be a positive integer, then
$$(n+1)!+2,(n+1)!+3, \cdots,(n+1)!+(n+1)$$
are $n$ consecutive positive integers, and the $k$-th number is a multiple of $k+1$ (and greater than $k+1$), so they are $n$ consecutive composite numbers. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,812 |
13 Can the squares of a $2010 \times 2010$ grid be colored black or white such that the colors of squares symmetric about the center of the grid are different, and in each row and each column, the number of black and white squares are equal? | 13. It cannot be done.
In fact, if there exists a coloring method that satisfies the conditions, we write +1 in all black cells and -1 in all white cells. We divide the table into 4 parts according to the two central lines of the table, and let the sum of the numbers in the upper-left part be $A$, the upper-right part... | proof | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,813 |
14 Labeled as $1,2, \cdots, 100$, there are some matches in the matchboxes. If each question allows asking about the parity of the sum of matches in any 15 boxes, then to determine the parity of the number of matches in box 1, at least how many questions are needed? | 14. At least 3 questions are needed.
First, prove that "3 questions are sufficient." For example:
The first question is: $a_{1}, a_{2}, \cdots, a_{15}$;
The second question is: $a_{1}, a_{2}, \cdots, a_{8}, a_{16}, a_{17}, \cdots, a_{22}$;
The third question is: $a_{1}, a_{9}, a_{10}, \cdots, a_{22}$.
Here, $a_{i}$ re... | 3 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,814 |
15 Find all positive integers $n$, such that it is possible to write +1 or -1 in each cell of an $n \times n$ grid, satisfying: each cell labeled +1 has exactly one adjacent cell labeled -1, and each cell labeled -1 has exactly one adjacent cell labeled +1. | 15. Let $a_{ij}$ denote the number filled in the square at the $i$-th row and $j$-th column. If there exists a valid filling method, we can assume $a_{11}=1$ (otherwise, change the signs of all numbers in the table and then discuss), at this point, one of $a_{21}$ and $a_{12}$ must be -1, let's assume $a_{21}=-1$ (othe... | 3 \nmid n, n>1 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,815 |
16 Let $a_{1}, a_{2}, \cdots, a_{100}$ be a permutation of $1,2, \cdots, 100$, and let $b_{i}=a_{1}+a_{2}+\cdots+a_{i}$, $i=1,2, \cdots, 100$, and let $r_{i}$ be the remainder when $b_{i}$ is divided by 100. Prove that $r_{1}, r_{2}, \cdots, r_{100}$ contains at least 11 different numbers. | 16. If $r_{1}, r_{2}, \cdots, r_{100}$ contain only 10 different numbers, then for $i=1,2, \cdots, 99$, $r_{i+1}-r_{i}$ has only $10^{2}-9=91$ (here subtracting 9 is because when $r_{i+1}=r_{i}$, the resulting values are all zero) different values. However, under modulo 100, $r_{i+1}-r_{i}$ are sequentially $a_{2}, a_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,816 |
17 Find the number of all positive integers $a$ that satisfy the following condition: there exist non-negative integers $x_{0}, x_{1}, x_{2}, \cdots$, $x_{2001}$, such that $a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}}$. | 17. If $a$ is a number that satisfies the condition, then $a^{x_{0}}>1$, so $a>1$. At this point, taking
both sides modulo $a-1$, we know
so $\square$
$$\begin{array}{c}
a^{x_{0}}=a^{x_{1}}+a^{x_{2}}+\cdots+a^{x_{2001}} \\
1 \equiv \underbrace{1+\cdots+1}_{2001 \uparrow}(\bmod a-1), \\
a-1 \mid 2000 .
\end{array}$$
O... | 20 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,817 |
18 Let $m, n$ be positive integers, $m>1$, prove: $m\left(2^{m}-1\right) \mid n$ if and only if $\left(2^{m}-1\right)^{2} \mid$ $2^{n}-1$. | 18. If \( m\left(2^{m}-1\right) \mid n \), let \( n=m\left(2^{m}-1\right) k \), then
\[
\begin{aligned}
2^{n}-1 & =2^{m\left(2^{m}-1\right) k}-1 \\
& =\left(2^{m k}\right)^{\left(2^{m}-1\right)}-1 \\
& =\left(2^{m k}-1\right) A,
\end{aligned}
\]
where
\[
A=\left(2^{m k}\right)^{2^{m}-2}+\left(2^{m k}\right)^{2^{m}-3}+... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,818 |
19 Let $a, b$ be coprime positive integers, and $p$ be an odd prime. Prove: $\left(a+b, \frac{a^{p}+b^{p}}{a+b}\right)=1$ or $p$. | 19. Let $A=\frac{a^{p}+b^{p}}{a+b}=a^{p-1}-a^{p-2} b+\cdots-a b^{p-2}+b^{p-1}$, combining the fact that $p$ is odd and $b \equiv-a(\bmod a+b)$, we know
$$A \equiv \underbrace{a^{p-1}+a^{p-1}+\cdots+a^{p-1}}_{p \uparrow}=p a^{p-1}(\bmod a+b)$$
and
$$\begin{array}{c}
(a, b)=1 \\
(a, a+b)=1
\end{array}$$
Therefore,
$$\b... | 1 \text{ or } p | Number Theory | proof | Yes | Yes | number_theory | false | 741,819 |
20 Find the smallest positive integer $a$, such that for any integer $x$, we have $65 \mid\left(5 x^{13}+13 x^{5}+9 a x\right)$. | 20. From the condition, we know $65 \mid (18 + 9a)$ (take $x=1$), and since $(9,65)=1$, it follows that $65 \mid a+2$, hence $a \geq 63$.
When $a=63$, using Fermat's Little Theorem, we know that for any integer $x$, we have
$$\begin{aligned}
& 5 x^{13} + 13 x^{5} + 9 a x \\
\equiv & 13 x + 9 a x \\
\equiv & (3 + (-1) ... | 63 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,820 |
Does there exist integers $a$, $b$, $c$, such that the equations
$$a x^{2}+b x+c=0 \text { and }(a+1) x^{2}+(b+1) x+(c+1)=0$$
both have two integer roots? | 21. There do not exist such integers $a$, $b$, $c$.
In fact, if $a$, $b$, $c$ satisfy the conditions, we might as well assume $a$ is even (otherwise, use $-(a+1)$, $-(b+1)$, $-(c+1)$ to replace $a$, $b$, $c$ for discussion). By the conditions and Vieta's formulas, $-\frac{b}{a}$ and $\frac{c}{a}$ are both integers, so... | proof | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,821 |
22 Find all positive integer tuples $(x, y, z, w)$, such that $x!+y!+z!=w!$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | $$
\begin{array}{l}
22. \text{Let } x \leqslant y \leqslant z, \text{ and } w! = z! + y! + x! \\
z \leqslant 2
\end{array}
$$
If $z=1$, then $x=y=z=1$, at this time $w!=3$, there is no such $w$, so $z=2$. At this time $w \geqslant 3$, hence
$$w!\equiv 0(\bmod 3)$$
Therefore,
$$x!+y!\equiv 1(\bmod 3)$$
And
thus it ca... | (2,2,2,3) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,822 |
23 Find the number of integer pairs $(a, b)$ that satisfy the following conditions: $0 \leqslant a, b \leqslant 36$, and $a^{2}+b^{2}=$ $0(\bmod 37)$. | 23. Notice that, $a^{2}+b^{2} \equiv a^{2}-36 b^{2}(\bmod 37)$, so from the condition we have
$$37 \mid a^{2}-36 b^{2},$$
which means
$$37 \mid(a-6 b)(a+6 b),$$
thus
$37 \mid a-6 b$ or $37 \mid a+6 b$.
Therefore, for each $1 \leqslant b \leqslant 36$, there are exactly two $a(a \equiv \pm 6 b(\bmod 37))$ that satisfy... | 73 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,824 |
24 Let $m, n$ be positive integers, and $m n \mid m^{2}+n^{2}+m$. Prove: $m$ is a perfect square. | 24. From the condition, we can set \( m^{2}+n^{2}+m=k m n \), where \( k \) is a positive integer. Thus, the quadratic equation in \( n \)
\[ n^{2}-k m n+m^{2}+m=0 \]
has positive integer solutions, so
\[ \Delta=(k m)^{2}-4\left(m^{2}+m\right)=m\left(k^{2} m-4 m-4\right) \]
is a perfect square.
If \( m \) is odd, the... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,825 |
25 Prove: If a positive integer $n$ can be expressed as the sum of squares of three positive integers, then $n^{2}$ can also be expressed as the sum of squares of three positive integers. | 25. Let $n=x^{2}+y^{2}+z^{2}, x \geqslant y \geqslant z$ be positive integers, then
$$\begin{aligned}
n^{2} & =\left(x^{2}+y^{2}+z^{2}\right)^{2} \\
& =\left(x^{2}+y^{2}\right)^{2}+2\left(x^{2}+y^{2}\right) z^{2}+z^{4} \\
& =\left(x^{2}+y^{2}-z^{2}\right)^{2}+4\left(x^{2}+y^{2}\right) z^{2} \\
& =\left(x^{2}+y^{2}-z^{2... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,826 |
26 Find all positive integers $n$, such that the cube root of $n$ equals the positive integer obtained by removing the last three digits of $n$.
Find all positive integers $n$, such that the cube root of $n$ equals the positive integer obtained by removing the last three digits of $n$. | 26. Let $n=1000 x+y$, where $x$ is a positive integer, $y$ is an integer, and $0 \leqslant y \leqslant 999$. According to the problem,
$$x^{3}=1000 x+y$$
From $0 \leqslant y \leqslant 999$, we know
$$1000 x \leqslant x^{3}<1000 x+1000=1000(x+1)$$
Thus,
$$x^{2} \geqslant 1000, x^{3}+1 \leqslant 1000(x+1)$$
Therefore,... | 32768 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,827 |
27 Prove: There exist infinitely many integers $n$, such that the numbers $n$, $n+1$, and $n+2$ can all be expressed as the sum of two integers (not necessarily distinct) squared. For example: $0=0^{2}+0^{2}$, $1=0^{2}+1^{2}$, $2=1^{2}+1^{2}$, so $n=0$ is an integer that satisfies the condition. | 27. Find a positive integer $l$ such that $l^{2}-1=x^{2}+y^{2}$ has positive integer solutions. Let $x=2 m^{2}$, $y=2 m$, and $l=2 m^{2}+1$, then $l^{2}-1=x^{2}+y^{2}$. Therefore, for any positive integer $m$, take
$$n=\left(2 m^{2}+1\right)^{2}-1=4 m^{4}+4 m^{2},$$
then $\square$
$$\begin{array}{c}
n=\left(2 m^{2}\ri... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,828 |
28 Find the smallest positive integer $n$, such that in decimal notation $n^{3}$ ends with the digits 888. | 28. From the condition, we know $n^{3} \equiv 888(\bmod 1000)$, hence
$$n^{3} \equiv 888(\bmod 8), n^{3} \equiv 888(\bmod 125)$$
From the former, we know $n$ is even, let $n=2 m$, then
$$m^{3} \equiv 111(\bmod 125)$$
Therefore
$$m^{3} \equiv 111 \equiv 1(\bmod 5)$$
Noting that when $m=0,1,2,3,4(\bmod 5)$, correspond... | 192 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,829 |
29 Let $n>1$ be a positive integer, prove: the number $2^{n}-1$ is neither a perfect square nor a perfect cube. | 29. Since $n \geqslant 2$, we have $2^{n}-1 \equiv -1 \pmod{4}$, while a perfect square $\equiv 0$ or $1 \pmod{4}$. Therefore, $2^{n}-1$ is not a perfect square.
On the other hand, if there exist $n>1$ and a positive integer $x$ such that
$$2^{n}-1=x^{3}$$
then
$$2^{n}=(x+1)\left(x^{2}-x+1\right)$$
Since
$$x^{2}-x+1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,830 |
30 Let $a$, $b$, $c$ be positive integers, and $\sqrt{a}+\sqrt{b}+\sqrt{c}$ be an integer. Prove: $a$, $b$, $c$ are all perfect squares. | 30. First, we prove: For any positive integer $a$, if $\sqrt{a}$ is a rational number, then $a$ is a perfect square.
Indeed, if $\sqrt{a}=\frac{q}{p}$, where $p, q$ are positive integers and $(p, q)=1$, then $a=\frac{q^{2}}{p^{2}}$. Since $a$ is a positive integer, it follows that $p^{2} \mid q^{2}$. But $(p, q)=1$, s... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,831 |
Given that the positive integer $c$ is an odd composite number. Prove: there exists a positive integer $a$, such that $a \leqslant \frac{c}{3}-1$, and $(2 a-1)^{2}+8 c$ is a perfect square. | 31. By completing the perfect square. From the condition, we can set $c=p q, 3 \leqslant p \leqslant q, p, q$ are both odd numbers. Now we need to find $a$ such that $(2 a-1)^{2}+8 p q$ is a perfect square. A natural choice is: let $2 a-1=2 q-p$, then
$$\begin{aligned}
(2 a-1)^{2}+8 p q & =(2 q-p)^{2}+8 p q \\
& =(2 q+... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,832 |
32 Let integers $a, b$ satisfy: for any positive integer $n$, the number $2^{n} \cdot a + b$ is a perfect square. Prove $a=0$. | 32. If $a \neq 0$, note that in $a0$. Now let $2^{n} a+b=x_{n}^{2}$, where $x_{n}$ is a positive integer, then for any positive integer $n$, we have $x_{n}<x_{n+1}$.
Since $4 x_{n}^{2}-x_{n+2}^{2}=4\left(2^{n} a+b\right)-\left(2^{n+2} a+b\right)=3 b$, thus
$$3|b|=\left|2 x_{n}-x_{n+2}\right| \cdot\left|2 x_{n}+x_{n+2}... | a=0 | Number Theory | proof | Yes | Yes | number_theory | false | 741,833 |
Example 6 Let $a, b, c, d, e, f$ all be positive integers, $S=a+b+c+d+e+f$ is a factor of $abc+def$ and $ab+bc+ca-de-ef-ed$. Prove: $S$ is a composite number.
The text is translated while preserving the original line breaks and format. | Consider the polynomial
$$f(x)=(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f) .$$
After expanding, we can see that
$$f(x)=S x^{2}+(a b+b c+c a-d e-e f-f d) x+(a b c+d e f)$$
From the condition, for any $x \in \mathbf{Z}$, we have $S \mid f(x)$. In particular, taking $x=d$, we have $S \mid f(d)$, i.e., $S \mid(d+a)(d+b)(d+c)$. Since... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,834 |
33 Find the largest positive integer that cannot be expressed as the sum of a positive multiple of 42 and a composite number.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 33. For any positive integer $n$ that cannot be expressed as a positive multiple of 42 and a composite number, consider the remainder $r$ when $n$ is divided by 42. If $r=0$ or $r$ is a composite number, then $n \leqslant 42$.
Now consider the case where $r=1$ or $r$ is a prime number.
If $r \equiv 1(\bmod 5)$, then
$... | 215 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,835 |
34 Find a positive integer $n$, such that each number $n, n+1, \cdots, n+20$ is not coprime with 30030. | 34. Since $30030=2 \times 3 \times 5 \times 7 \times 11 \times 13$, if we take $N=210 k$, then $N$ and $N \pm r$ are not coprime with 30030, where $r$ is a number in $2,3, \cdots, 10$. Now consider the numbers $N \pm 1$. We choose $k$ such that
$$210 k \equiv 1(\bmod 11) \text { and } 210 k \equiv-1(\bmod 13),$$
The f... | 9440 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,836 |
35 Does there exist a sequence of 13 consecutive positive integers, each of which is a multiple of one of the numbers $2, 3, 5, 7, 11$? What about 14 consecutive integers? | 35. Notice that the 13 numbers $114, 115, \cdots, 126$ are all composite numbers, and each number is a multiple of one of $2, 3, 5, 7, 11$. Therefore, there are 13 numbers that meet the requirement.
Next, we prove that there do not exist 14 consecutive positive integers such that each number is a multiple of one of $2... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,837 |
36 Let $p$ be a prime, $a$ and $n$ be positive integers, and $2^{p}+3^{p}=a^{n}$. Prove: $n=1$.
The text has been translated while preserving the original line breaks and format. | 36. When $p=2$, $a^{n}=13$, we know $a=13, n=1$. When $p>2$, since $p$ is a prime, $p$ is an odd number, at this time
$$2^{p}+3^{p}=(2+3)\left(2^{p-1}-2^{p-2} \times 3+\cdots-2 \times 3^{p-2}+3^{p-1}\right),$$
thus $5 \mid a^{n}$, which means $5 \mid a$. If $n>1$, then $5^{2} \mid a^{n}$, in this case, we should have
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,838 |
37 points are arranged on a circle, and one of the points is labeled with the number 1. Moving clockwise, label the next point with the number 2 after counting two points, then label the next point with the number 3 after counting three points, and so on, labeling the points with the numbers $1,2, \cdots, 2000$. In thi... | 37. Equivalent to finding the smallest positive integer $n$, such that
$$1+2+\cdots+n \equiv 1+2+\cdots+2000(\bmod 2000)$$
i.e. $\square$
$$\frac{n(n+1)}{2} \equiv 1000(\bmod 2000),$$
which is equivalent to
$$n(n+1) \equiv 2000(\bmod 4000),$$
This requires
$$2000 \mid n(n+1)$$
Notice that
$$(n, n+1)=1$$
and
$$2000... | 624 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,839 |
38 There are 800 points on a circle, labeled $1,2, \cdots, 800$ in a clockwise direction, dividing the circle into 800 gaps. Now, choose one of these points and color it red, then perform the following operation: if the $k$-th point is colored red, then move $k$ gaps in a clockwise direction and color the point reached... | 38. This is equivalent to finding the maximum number of distinct numbers in the sequence $a, 2a, 2^2a, 2^3a, \cdots$ under modulo 800, where $a$ takes values in $1, 2, \cdots, 800$.
Notice that when $2^n \not\equiv 2^m \pmod{800}$, it is not necessarily true that $2^n a \not\equiv 2^m a \pmod{800}$. Conversely, when $... | 24 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,840 |
30 Let $m$ be a positive integer, and $m \equiv 2(\bmod 4)$. Prove: there exists at most one pair of positive integers $(a, b)$, such that $m=a b$, and $0<a-b<\sqrt{5+4 \sqrt{4 m+1}}$. | 39. If the value of $a+b$ can be determined (considering $m$ as a constant), then by the inverse of Vieta's theorem, there is at most one pair of positive integers $(a, b)$ that satisfies the condition.
From the condition, we know that $(a+b)^{2}=(a-b)^{2}+4 a b$ satisfies
$$\begin{aligned}
1+4 m & \leqslant(a+b)^{2} ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,841 |
40 Let $n$ be a positive integer greater than 10, and each digit of $n$ is $1$, $3$, $7$, or $9$. Prove: $n$ has a prime factor greater than 10. | 40. By contradiction, if every prime factor of $n$ is no greater than 10, using the given conditions, we know that $n$ is odd and $n$ is not a multiple of 5. Therefore, there exist non-negative integers $i, j$ such that
$$n=3^{i} \cdot 7^{j},$$
Consider the remainders of $3^{i}$ and $7^{j}$ when divided by 20. For $i=... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,842 |
Find all prime pairs $(p, q)$ such that $p q \mid p^{p}+q^{q}+1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 41. From the condition, we know $p \neq q$. Using symmetry, without loss of generality, let $p < q$, then $p$ and $q$ are both odd primes. From the condition, we have $p^{p} + 1 \equiv 0 \pmod{q}$, hence $p^{2p} \equiv 1 \pmod{q}$. Using Fermat's Little Theorem, we have $p^{q-1} \equiv 1 \pmod{q}$, thus,
$$p^{(2p, q-1)... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,843 |
42 Let $f(n)=1+n+n^{2}+\cdots+n^{2010}$. Prove: for any integer $m$, if $2 \leqslant m \leqslant 2010$, then there does not exist a positive integer $n$, such that $m \mid f(n)$. | 42. If there exists $2 \leqslant m \leqslant 2010$, such that for some positive integer $n$, we have $m \mid f(n)$. Since $f(1)=2011$ is a prime number (here 2011 is a prime number, which needs to be verified by dividing 2011 by all primes not greater than $\sqrt{2011}$), hence $n \neq 1$, and at this point we can writ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,844 |
Property 1 Let $d=(a, b)$, then there exist integers $x, y$, such that
$$a x+b y=d \text {. }$$ | We use the division algorithm to handle this, and the proof process of this conclusion is also the process of finding the greatest common divisor (GCD) of $a$ and $b$, which is called "Euclidean algorithm".
Assume $a$ and $b$ are both non-zero (when one of $a$ and $b$ is zero, the conclusion is obvious), and $|a| \leq... | d = a x + b y | Number Theory | proof | Yes | Yes | number_theory | false | 741,845 |
Does there exist integers $x, y$ such that $x^{2012}-2010=4 y^{2011}+4 y^{2010}+2011 y$? | 43. There do not exist such integers $x, y$.
Otherwise, we would have
$$x^{2012}+1=\left(4 y^{2010}+2011\right)(y+1) .$$
Notice that, $4 y^{2010}+2011 \equiv 3(\bmod 4)$, which indicates that the right-hand side of (1) has a prime factor that is congruent to 3 modulo 4. Therefore, there exists a prime $p$ such that $... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,846 |
Theorem 1 The necessary and sufficient condition for the indeterminate equation (2) to have integer solutions is $(a, b) \mid c$.
| The necessity is obvious, while the sufficiency can be obtained by Bézout's theorem.
In fact, by Bézout's theorem, we know there exist integers $x_{0}, y_{0}$ such that $a x_{0} + b y_{0} = (a, b)$. Let $c = (a, b) c_{1}$, then $\left(c_{1} x_{0}, c_{1} y_{0}\right)$ is a solution to (2), and the sufficiency is proved. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,847 |
Theorem 2 If the indeterminate equation (2) has an integer solution $\left(x_{0}, y_{0}\right)$, then all integer solutions of (2) are
$$\left\{\begin{array}{l}
x=x_{0}+\frac{b}{(a, b)} t, \\
y=y_{0}-\frac{a}{(a, b)} t .
\end{array}\right.$$ | Proof: Let $(x, y)$ be an integer solution to (2), and combine with $\left(x_{0}, y_{0}\right)$ being a solution to (2), we have
$$\left\{\begin{array}{l}
a x+b y=c, \\
a x_{0}+b y_{0}=c,
\end{array}\right.$$
Thus,
$$a\left(x-x_{0}\right)+b\left(y-y_{0}\right)=0$$
That is,
$$a\left(x-x_{0}\right)=b\left(y_{0}-y\right... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,848 |
Example 1 Find the number of positive integer solutions to the indeterminate equation
$$7 x+19 y=2012$$ | Solve: First, find a particular solution of (1).
$$x=\frac{1}{7}(2012-19 y)=287-3 y+\frac{1}{7}(3+2 y) .$$
Therefore, $\frac{1}{7}(3+2 y)$ must be an integer. Taking $y_{0}=2$, then $x_{0}=282$.
Using the conclusion of Theorem 2, the general solution of equation (1) is
$$\left\{\begin{array}{l}
x=282-19 t, \\
y=2+7 t ... | 15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,849 |
Example 2 Let $a, b$ be coprime positive integers. Prove: the indeterminate equation
$$a x+b y=a b-a-b$$
has no non-negative integer solutions. | If there exists a pair of non-negative integers $\left(x_{0}, y_{0}\right)$ satisfying (1), then
$$a\left(x_{0}+1\right)+b\left(y_{0}+1\right)=a b,$$
then, we should have
and
thus
$$\begin{array}{l}
a \mid b\left(y_{0}+1\right) \\
(a, b)=1 \\
a \mid y_{0}+1
\end{array}$$
Since $a$ and $y_{0}+1$ are both positive int... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,850 |
Example 3 Let positive integers $a, b$ be coprime, and positive integer $c$ be greater than $a b-a-b$. Prove: the indeterminate equation
$$a x+b y=c$$
has non-negative integer solutions. | Proof: Let the general solution of (1) be
$$\left\{\begin{array}{l}
x=x_{0}+b t, \\
y=y_{0}-a t,
\end{array}(\text { where } t \text { is an integer })\right.$$
where $\left(x_{0}, y_{0}\right)$ is a particular solution of (1).
Thus, by adjusting the value of $t$ (adding or subtracting $b$ from $x_{0}$), we can find a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,851 |
Example 4 Find the number of positive integer solutions to the indeterminate equation
$$x+2 y+3 z=2012$$ | Let $(x, y, z)$ be a positive integer solution to (1), then $3 z \leqslant 2009$, i.e., $1 \leqslant z \leqslant 669$, which respectively yield
$$x+2 y=2009,2006, \cdots, 5$$
Correspondingly, the range of values for $y$ are
$$\begin{array}{l}
1 \leqslant y \leqslant 1004,1 \leqslant y \leqslant 1002,1 \leqslant y \leq... | 336340 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,852 |
Example 5 Find all positive integer arrays $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, such that
$$\left\{\begin{array}{l}
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, \\
a_{1}+a_{2}+\cdots+a_{n}=26, \\
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=62, \\
a_{1}^{3}+a_{2}^{3}+\cdots+a_{n}^{3}=164 .
\end{array}\right.$... | Given that $a_{i}$ are all positive integers, and $6^{3}=216>164$, we know that each $a_{i}$ is no greater than 5. Let $x_{1}, x_{2}, \cdots, x_{5}$ be the number of 1, 2, 3, 4, 5 in $a_{1}, a_{2}, \cdots, a_{n}$, respectively. Then,
$$\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=n \\
x_{1}+2 x_{2}+3 x_{3}+4 x... | (1,1,1,1,1,2,2,2,3,3,3,3,3) \text{ and } (1,2,2,2,2,2,2,2,2,2,3,4) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,853 |
Example 6 Arrange all the simplest fractions with denominators no greater than 99 in ascending order, find the two numbers adjacent to $\frac{17}{76}$. | Let $\frac{q}{p}$ and $\frac{n}{m}$ be the two numbers adjacent to $\frac{17}{76}$, and
$$\frac{q}{p}<\frac{17}{76}<\frac{n}{m}$$
with
$$\begin{array}{l}
17 p-76 q>0 \\
17 p-76 q \geqslant 1
\end{array}$$
First, consider the equation $17 p-76 q=1$ and find the largest positive integer solution for $p \leqslant 99$. To... | \frac{19}{85} \text{ and } \frac{15}{67} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,854 |
Example 1 Find the integer solutions of the equation
$$x y-10(x+y)=1$$ | Solving by using the cross-multiplication method, we can transform (1) into
$$(x-10)(y-10)=101$$
Since 101 is a prime number, we have
$$\begin{aligned}
& (x-10, y-10) \\
= & (1,101),(101,1),(-1,-101),(-101,-1) .
\end{aligned}$$
Solving each case, we get the integer solutions of the equation as
$$(x, y)=(11,111),(111,... | (x, y)=(11,111),(111,11),(9,-91),(-91,9) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,855 |
Example 2 Does there exist integers $x, y, z$, such that
$$x^{4}+y^{4}+z^{4}=2 x^{2} y^{2}+2 y^{2} z^{2}+2 z^{2} x^{2}+24 \text{ ?}$$ | If there exist integers $x, y, z$ satisfying the condition, then
$$\begin{aligned}
-24 & =2 x^{2} y^{2}+2 y^{2} z^{2}+2 z^{2} x^{2}-\left(x^{4}+y^{4}+z^{4}\right) \\
& =-\left(x^{2}+y^{2}\right)^{2}+2\left(x^{2}+y^{2}\right) z^{2}-z^{4}+4 x^{2} y^{2} \\
& =-\left(x^{2}+y^{2}-z^{2}\right)^{2}+4 x^{2} y^{2} \\
& =\left(2... | proof | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,857 |
Example 3 Find all pairs of positive integers $(m, n)$ such that
$$n^{5}+n^{4}=7^{m}-1$$ | Solving, by rearranging and factoring (1), we get
$$\begin{aligned}
7^{m} & =n^{5}+n^{4}+1=n^{5}+n^{4}+n^{3}-\left(n^{3}-1\right) \\
& =n^{3}\left(n^{2}+n+1\right)-(n-1)\left(n^{2}+n+1\right) \\
& =\left(n^{3}-n+1\right)\left(n^{2}+n+1\right) .
\end{aligned}$$
From (1), we know $n>1$, and when $n=2$, we get $m=2$.
Nex... | (m, n)=(2,2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,858 |
Example 4 Find all integer solutions of the indefinite equation $3 x^{2}-4 x y+3 y^{2}=35$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve: Multiply both sides of the equation by 3, and complete the square to get
$$(3 x-2 y)^{2}+5 y^{2}=105$$
From equation (1), we get
$$5 y^{2} \leqslant 105$$
Therefore,
$$|y| \leqslant 4$$
When $|y|=4$, $|3 x-2 y|=5$, at this time the solutions to the original equation are
$$(x, y)=(1,4),(-1,-4)$$
When $|y|=1$,... | (x, y)=(4,1),(1,4),(-4,-1),(-1,-4) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,859 |
Example 5 Find the integer solutions of the equation $x^{2}+x=y^{4}+y^{3}+y^{2}+y$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solving as in the previous example, multiply both sides of the equation by 4 and complete the square on the left side, we get
$$(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\ri... | (x, y)=(0,-1),(-1,-1),(0,0),(-1,0),(-6,2),(5,2) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,860 |
Example 6 Find all positive integers $n \geqslant 2$, such that the system of equations
$$\left\{\begin{array}{c}
x_{1}^{2}+x_{2}^{2}+50=16 x_{1}+12 x_{2} \\
x_{2}^{2}+x_{3}^{2}+50=16 x_{2}+12 x_{3} \\
\cdots \\
x_{n-1}^{2}+x_{n}^{2}+50=16 x_{n-1}+12 x_{n} \\
x_{n}^{2}+x_{1}^{2}+50=16 x_{n}+12 x_{1}
\end{array}\right.$... | Solving by rearranging and completing the square, the system of equations transforms to
$$\left\{\begin{array}{c}
\left(x_{1}-8\right)^{2}+\left(x_{2}-6\right)^{2}=50 \\
\left(x_{2}-8\right)^{2}+\left(x_{3}-6\right)^{2}=50 \\
\cdots \\
\left(x_{n-1}-8\right)^{2}+\left(x_{n}-6\right)^{2}=50 \\
\left(x_{n}-8\right)^{2}+\... | n = 3k | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,861 |
Example 7 Find the positive integer solutions of the indeterminate equation $x^{3}-y^{3}=x y+61$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $(x, y)$ be a positive integer solution of the equation, then $x>y$. Let $x=y+d$, then $d$ is a positive integer, and
$$\begin{aligned}
(y+d) y+61 & =(y+d)^{3}-y^{3} \\
& =3 d y^{2}+3 y d^{2}+d^{3}
\end{aligned}$$
Thus we have
$$(3 d-1) y^{2}+d(3 d-1) y+d^{3}=61$$
Therefore,
$$\begin{array}{l}
d^{3}<61 \\
d \leqs... | (x, y)=(6,5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,862 |
Example 8 Find all positive integers $a, b$ such that
$$4^{a}+4 a^{2}+4=b^{2}$$ | If $(a, b)$ is a pair of positive integers satisfying (1), then $b^{2}$ is even, and $b^{2}>4^{a}$, so $b$ is even, and $b>2^{a}$, hence $b \geqslant 2^{a}+2$. Therefore,
$$4^{a}+4 a^{2}+4=b^{2} \geqslant\left(2^{a}+2\right)^{2}=4^{a}+4 \cdot 2^{a}+4$$
we have $a^{2} \geqslant 2^{a}$, which implies $a \leqslant 4$ (by... | (2,6) \text{ or } (4,18) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,863 |
Example 9 Find all positive integer tuples $(a, b, c, x, y, z)$ such that
$$\left\{\begin{array}{l}
a+b+c=x y z, \\
x+y+z=a b c,
\end{array}\right.$$
where $a \geqslant b \geqslant c, x \geqslant y \geqslant z$. | By symmetry, we only need to consider the case $x \geqslant a$. At this time,
$$x y z=a+b+c \leqslant 3 a \leqslant 3 x$$
Thus,
$$y z \leqslant 3$$
Therefore,
$$(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 \leqsl... | (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) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,864 |
Example 10 Prove: the indeterminate equation
$$x^{2}+y^{2}-8 z^{3}=6$$
has no integer solutions. | Prove that if $(x, y, z)$ is an integer solution to equation (1), taking both sides of (1) modulo 2, we can see that $x$ and $y$ have the same parity; then taking both sides of (1) modulo 4, we can see that $x$ and $y$ are both odd, thus $x^{2} \equiv y^{2} \equiv 1(\bmod 8)$, which requires
$$6=x^{2}+y^{2}-8 z^{3} \eq... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,865 |
Example 11 Find all non-negative integers $x, y, z$, such that
$$2^{x}+3^{y}=z^{2}$$ | Solve (1) When $y=0$, we have
$$2^{x}=z^{2}-1=(z-1)(z+1)$$
Thus, we can set
$$z-1=2^{\alpha}, z+1=2^{\beta}, 0 \leqslant \alpha \leqslant \beta$$
Therefore,
$$2^{\beta}-2^{\alpha}=2$$
At this point, if $\alpha \geqslant 2$, then $4 \mid 2^{\beta}-2^{\alpha}$, which contradicts $4 \nmid 2$, so $\alpha \leqslant 1$. A... | (x, y, z)=(3,0,3),(0,1,2),(4,2,5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,866 |
Example 12 Let $m, n$ be positive integers, and $n>1$. Find the minimum value of $\left|2^{m}-5^{n}\right|$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note is for you, the assistant, and should not be included... | Since $\left|2^{m}-5^{n}\right|$ is an odd number, and when $m=7, n=3$, $\left|2^{m}-5^{n}\right|=3$, if we can prove that when $n>1$, $\left|2^{m}-5^{n}\right| \neq 1$, then the minimum value sought is 3.
If there exist positive integers $m, n$, such that $n>1$, and $\left|2^{m}-5^{n}\right|=1$, then
$$2^{m}-5^{n}=1 ... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,868 |
Example 13 Proof: The equation $x^{2}+y^{5}=z^{3}$ has infinitely many integer solutions satisfying $x y z \neq 0$. | Prove that taking $x=2^{156+10}, y=2^{6 k+4}, z=2^{10 k+7}, k$ as a non-negative integer, then such $x$, $y$, $z$ satisfy $x^{2}+y^{5}=z^{3}$, so the equation has infinitely many integer solutions satisfying $x y z \neq 0$.
Another proof First find a particular solution to the equation, it is easy to see that $x=10, y... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,869 |
Example 14 Proof: For any integer $n$, the equation
$$x^{2}+y^{2}-z^{2}=n$$
has infinitely many integer solutions $(x, y, z)$. | Prove that the proposition "when $m$ is an odd number or a multiple of 4, the equation $a^{2}-b^{2}=m$ has integer solutions $(a, b)$" is useful for solving this problem. This proposition is based on the following two identities:
$$\begin{array}{c}
(k+1)^{2}-k^{2}=2 k+1 \\
(k+1)^{2}-(k-1)^{2}=4 k
\end{array}$$
For equ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,870 |
Example 15: Do there exist pairwise distinct positive integers $m, n, p, q$ such that $m+n=p+q$ and $\sqrt{m}+\sqrt[3]{n}=\sqrt{p}+\sqrt[3]{q}>2012$ both hold? | Solve for positive integers satisfying the condition.
From the structure of the equation, we look for positive integers of the form
$$m=a^{2}, n=b^{3}, p=c^{2}, q=d^{3}$$
where $a, b, c, d$ are positive integers.
At this point, the condition transforms to
$$a+b=c+d>2012, a^{2}+b^{3}=c^{2}+d^{3},$$
which simplifies to... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,871 |
Example 16 Proof: There exist infinitely many sets of positive integers $(x, y, z)$ such that $x, y, z$ are all distinct, and
$$x^{x}=y^{3}+z^{3}$$ | To prove an idea: take $x$ as a number of the form $3k+1$, at this time
$$\begin{aligned}
x^{x} & =(3 k+1)^{3 k+1} \\
& =(3 k+1)(3 k+1)^{3 k} \\
& =3 k(3 k+1)^{3 k}+(3 k+1)^{3 k}
\end{aligned}$$
Therefore, if we make $3k$ a perfect cube, then the positive integers $x, y, z$ that meet the requirements can be found.
Fo... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,872 |
Example 1 Let $(x, y, z)$ be an integer solution of the Pythagorean equation (1). Prove: $x$, $y$, $z$ must include one number that is a multiple of 3, one number that is a multiple of 4, and one number that is a multiple of 5. | Prove that using the fact that perfect squares $\equiv 0,1(\bmod 3)$, if $x$ and $y$ are not multiples of 3, then
this leads to
$$\begin{array}{c}
x^{2}+y^{2} \equiv 2(\bmod 3) \\
z^{2} \equiv 2(\bmod 3)
\end{array}$$
which is a contradiction. Hence, one of $x$ or $y$ must be a multiple of 3.
If $x$, $y$, and $z$ are... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,873 |
Example 2 Let $(x, y, z)$ be a Pythagorean triple $\left(x^{2}+y^{2}=z^{2}\right)$. Prove: $z^{2}+x y$ and $z^{2}-$ $x y$ can both be expressed as the sum of squares of two positive integers. | Notice that $z^{2} \pm x y=\frac{2 z^{2} \pm 2 x y}{2}$.
$$=\frac{z^{2}+(x \pm y)^{2}}{2}$$
And
$$2 a^{2}+2 b^{2}=(a+b)^{2}+(a-b)^{2}$$
Therefore
$$\begin{aligned}
z^{2} \pm x y & =\frac{(z+x \pm y)^{2}+(-z+x \pm y)^{2}}{4} \\
& =\left(\frac{x \pm y+z}{2}\right)^{2}+\left(\frac{x \pm y-z}{2}\right)^{2}
\end{aligned}$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,874 |
Example 3 Let $n$ be a positive integer greater than 2. Prove: there exists a right-angled triangle with integer side lengths, one of whose legs has length exactly $n$.
The above text is translated into English, preserving the original text's line breaks and format. | To prove that the indeterminate equation $x^{2}+n^{2}=z^{2}$ has positive integer solutions.
Using $(z-x)(z+x)=n^{2}$, and noting that $z-x$ and $z+x$ have the same parity, when $n$ is odd, from $(z-x, z+x)=\left(1, n^{2}\right)$, we can obtain a set of positive integer solutions
$$(x, z)=\left(\frac{n^{2}-1}{2}, \frac... | proof | Geometry | proof | Yes | Yes | number_theory | false | 741,875 |
Example 5 Let $n$ be a positive integer. Prove: there exist $n$ pairwise non-congruent Pythagorean triangles (right triangles with integer side lengths) whose perimeters are all equal.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result direc... | To prove that if we can find $n$ mutually dissimilar right-angled triangles, then by multiplying each triangle by an appropriate positive integer, we can obtain right-angled triangles with the same perimeter but not congruent to each other. This is a starting point for solving this problem.
First, we prove that any tw... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,877 |
Example 6: Do there exist positive integers $x, y$ such that $x^{2}+y^{2}=2011^{2}$ holds? | If there are such positive integers, then $x, y$ are both less than 2011. Since 2011 is a prime number (this conclusion can be directly calculated by the fact that no prime number less than or equal to $\sqrt{2011}$ can divide 2011), $x, y$ are both coprime with 2011. This indicates that $(x, y, 2011)$ is a primitive s... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,880 |
1 Given that the ages of A, B, and C are all positive integers, A's age does not exceed twice B's age, B is 7 years younger than C, the sum of the three people's ages is a prime number less than 70, and the sum of the digits of this prime number is 13. Question: What is the maximum age of A? | 1. Let Jia's age be $x$ years, and Yi's age be $y$ years, then Bing is $y+7$ years old, and $x \leqslant 2 y$.
Since the positive integers less than 70 and with a digit sum of 13 are only $49$, $58$, and $67$, the sum of the three people's ages (which is a prime number) can only be 67 years, i.e.,
$$x+y+(y+7)=67$$
Th... | 30 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,881 |
2 Given 12 bamboo poles, each 13 units long, divide them into smaller segments of lengths 3, 4, or 5, and then assemble them into 13 triangles with side lengths of 3, 4, or 5. How should they be divided? Please explain your reasoning. | 2. First, find the non-negative integer solutions of $3 a+4 b+5 c=13$. It is known that the lengths obtained after each bamboo stick is divided can only be $(3,3,3,4),(3,5,5),(4,4,5)$, a total of 3 cases. Then, let the number of times each case occurs be $x, y, z$, respectively, so we need to solve
$$\left\{\begin{arra... | (x, y, z)=(3,4,5) | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,882 |
3i Let $n$ be a positive integer, and the indeterminate equation $x+2 y+2 z=n$ has exactly 28 sets of positive integer solutions. Find the value of $n$.
| 3. When $n$ is even, let $n=2 m$, then $x$ is even. For $x=2 k, 1 \leqslant k \leqslant m-1$, we have
$$y+z=m-k$$
In this case, there are $m-k-1$ solutions, meaning $x+2 y+2 z=2 m$ has a total of $(m-2)+(m-3)+\cdots+$
$$1+0=\frac{1}{2}(m-1)(m-2)$$
sets of positive integer solutions. Therefore,
$$\frac{1}{2}(m-1)(m-2)... | n=17 \text{ or } 18 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,883 |
4. Positive integers $a, b, c, d$ satisfy: $1<a<b<c<d<1000$, and $a+d=b+c$, $bc-ad=2004$. Find the number of all such positive integer tuples $(a, b, c, d)$. | 4. Let $b=a+x, c=a+y$, then $x<y$, and $d=a+x+y$ (this is obtained from $a+d=b+c$), thus
$$b c-a d=(a+x)(a+y)-a(a+x+y)=x y$$
That is
$$x y=2004$$
Combining $a+x+y<1000$ and $2004=2^{2} \times 3 \times 167$, we know that $(x, y)=(3,668),(4,501),(6,334),(12,167)$.
Accordingly, $1<a<329,1<a<495,1<a<660,1<a<821$. Thus, ... | 2297 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,884 |
5 Find the smallest positive integer $c$, such that the indeterminate equation $x y^{2}-y^{2}-x+y=c$ has exactly three sets of positive integer solutions.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. Factorizing the left side of the equation, we get
$$(y-1)(x y+x-y)=c$$
Notice that, for any positive integer $c$, there is a solution
$$(x, y)=(1, c+1)$$
When $c$ is a prime number, there is at most one additional set of positive integer solutions. Therefore, to make the equation have exactly 3 sets of positive in... | 10 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,885 |
6 Find the positive integer solutions of the indeterminate equation $x^{2}+3 x^{2} y^{2}=30 y^{2}+517$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 6. From the conditions, we get
i.e. $\square$
$$\begin{array}{c}
3 x^{2} y^{2}+x^{2}-30 y^{2}=517 \\
\left(x^{2}-10\right)\left(3 y^{2}+1\right)=507=3 \times 13^{2}
\end{array}$$
Notice that among the positive divisors of 507, only 39 and 10 have a sum that is a perfect square, hence
$$\left(x^{2}-10,3 y^{2}+1\right)... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,886 |
7 rectangles can be divided into $n$ identical squares, and it can also be divided into $n+76$ identical squares, find the value of the positive integer $n$.
---
Note: The translation keeps the original format and line breaks as requested. However, the phrase "7-个长方形" is translated as "7 rectangles" for clarity in En... | 7. Let the side length of the squares when divided into $n$ squares be $x$, and the side length when divided into $n+76$ squares be $y$, then
$$n x^{2}=(n+76) y^{2}$$
Since both divisions are performed on the same rectangle, $\frac{x}{y}$ is a rational number (this can be seen by considering one side of the rectangle)... | 324 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,887 |
8 Positive integers $x, y, z$ satisfy $\left\{\begin{array}{l}7 x^{2}-3 y^{2}+4 z^{2}=8, \\ 16 x^{2}-7 y^{2}+9 z^{2}=-3 .\end{array}\right.$ Find the value of $x^{2}+y^{2}+z^{2}$. | 8. Let $x, y, z$ satisfy
$$\left\{\begin{array}{l}
7 x^{2}-3 y^{2}+4 z^{2}=8 \\
16 x^{2}-7 y^{2}+9 z^{2}=-3
\end{array}\right.$$
Multiplying (1) by 7 and (2) by 3, we get
Substituting back into (1) yields
$$\begin{array}{c}
x^{2}+z^{2}=65 \\
3 x^{2}-3 y^{2}=8-260
\end{array}$$
Thus,
$$\begin{aligned}
y^{2}-x^{2} & =... | 165 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,888 |
9 Find the integer solutions of the indeterminate equation $\left(x^{2}-y^{2}\right)^{2}=1+16 y$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 9. From the equation, we know $y \geqslant 0$, and $x \neq y$. Without loss of generality, let $x \geqslant 0$.
If $x>y$, then
we get
$$\begin{aligned}
1+16 y & =\left(x^{2}-y^{2}\right)^{2} \\
& \geqslant\left[(y+1)^{2}-y^{2}\right]^{2} \\
& =(2 y+1)^{2} \\
& =4 y^{2}+4 y+1 \\
0 & \leqslant y \leqslant 3 .
\end{alig... | (x, y)=( \pm 1,0),( \pm 4,3),( \pm 4,5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,889 |
Property 5 Let $a \mid b c$ , and $(a, b)=1$ , then $a \mid c$. | By property 3, we know there exist integers $x, y$ such that
$$a x+b y=1$$
Thus, $a c x+b c y=c$. By $a \mid b c$ and $a \mid a c x$, we can conclude that $a \mid c$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,890 |
10 Find all integers $x, y$ such that $x^{2} + xy + y^{2} = 1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 10. Multiply both sides by 4, then complete the square, to get
$$(2 x+y)^{2}+3 y^{2}=4$$
Thus, $4-3 y^{2}$ is a perfect square, which requires $y^{2}=0$ or 1, corresponding to $(2 x+y)^{2}=4,1$. Solving these respectively yields
$$(x, y)=( \pm 1,0),(0, \pm 1),(1,-1),(-1,1)$$ | (x, y)=( \pm 1,0),(0, \pm 1),(1,-1),(-1,1) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,891 |
111 Determine all positive integers $x, y$ that satisfy $x+y^{2}+z^{3}=x y z$, where $z$ is the greatest common divisor of $x$ and $y$. | 11. From the conditions, we can set \(x = za, y = zb\), where \(a, b\) are positive integers, and \((a, b) = 1\). Substituting into the equation, we get
$$a + z b^{2} + z^{2} = z^{2} a b$$
Thus, \(z \mid a\). Let \(a = zm\), then the equation becomes
$$m + b^{2} + z = z^{2} m b$$
Therefore, \(b \mid m + z\). Let \(m ... | (5, 2), (5, 3), (4, 2), (4, 6) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,892 |
12 Find all integers $m, n$ such that $m^{4}+(m+1)^{4}=n^{2}+(n+1)^{2}$. | 12. After expanding and rearranging, we get
$$m^{4}+2 m^{3}+3 m^{2}+2 m=n^{2}+n$$
Adding 1 to both sides and completing the square, we get
$$\left(m^{2}+m+1\right)^{2}=n^{2}+n+1$$
When $n>0$,
$$n^{2}<n^{2}+n+1<(n+1)^{2}$$
When $n<-1$, $\quad(n+1)^{2}<n^{2}+n+1<n^{2}$,
the number $n^{2}+n+1$ is never a perfect square... | (m, n)=(-1,0),(0,0),(-1,-1),(0,-1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,893 |
14 Let $x, y$ be real numbers greater than 1, and let $a=\sqrt{x-1}+\sqrt{y-1}, b=\sqrt{x+1}+$ $\sqrt{y+1}$, where $a, b$ are two non-consecutive positive integers. Find the values of $x, y$. | 14. From the conditions, we know
$$\begin{aligned}
b-a & =(\sqrt{x+1}-\sqrt{x-1})+(\sqrt{y+1}-\sqrt{y-1}) \\
& =\frac{2}{\sqrt{x+1}+\sqrt{x-1}}+\frac{2}{\sqrt{y+1}+\sqrt{y-1}} \\
& 2-\frac{2}{\sqrt{2}}=2-\sqrt{2}
\end{aligned}$$
Therefore,
$$\sqrt{x+1}+\sqrt{x-1}<\frac{2}{2-\sqrt{2}}=2+\sqrt{2}$$
Similarly,
$$\sqrt{y... | x=y=\frac{5}{4} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,895 |
15 Positive integers $a, b, c$ satisfy: $[a, b]=1000,[b, c]=2000,[c, a]=2000$. Find the number of such ordered positive integer triples $(a, b, c)$. | 15. From the conditions, we can set
$$a=2^{\alpha_{1}} \cdot 5^{\beta_{1}}, b=2^{\alpha_{2}} \cdot 5^{\beta_{2}}, c=2^{\alpha_{3}} \cdot 5^{\beta_{3}}$$
Then
$$\begin{array}{l}
\max \left\{\alpha_{1}, \alpha_{2}\right\}=3, \max \left\{\alpha_{2}, \alpha_{3}\right\}=4, \max \left\{\alpha_{3}, \alpha_{1}\right\}=4 \\
\m... | 70 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,896 |
16. Let $x, y$ be positive integers, and $y>3, x^{2}+y^{4}=2\left((x-6)^{2}+(y+1)^{2}\right)$. Prove: $x^{2}+y^{4}=1994$. | 16. Transpose and expand, we get
that is
$$\begin{array}{c}
y^{4}-2 y^{2}-4 y-2=x^{2}-24 x+72 \\
(x-12)^{2}=y^{4}-2 y^{2}-4 y+70
\end{array}$$
Notice that
$$\begin{aligned}
\left(y^{2}-2\right)^{2} & =y^{4}-4 y^{2}+4 \\
& 3 \text{ holds, and } y^{4}-2 y^{2}-4 y+70 \text{ is a perfect square, so it can only be or }
$$... | 1994 | Algebra | proof | Yes | Yes | number_theory | false | 741,897 |
17 Prove: For any positive integer $n \geqslant 3$, there exists a perfect cube that can be expressed as the sum of the cubes of $n$ distinct positive integers. | 17. Using a recursive construction method, it is based on the following two equations:
$$\begin{array}{c}
6^{3}=3^{3}+4^{3}+5^{3} \\
13^{3}=5^{3}+7^{3}+9^{3}+10^{3}
\end{array}$$
Generally, suppose there exist positive integers \(x_{1}<x_{2}<\cdots<x_{n}(n \geqslant 3)\) and a positive integer \(y\), such that
then \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,898 |
18 Let $n$ be a given positive integer. Find all positive integers $m$, such that there exist positive integers $x_{1}<x_{2}<\cdots<x_{n}$, satisfying: $\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{n}{x_{n}}=m$. | 18. Let $m$ be a positive integer that meets the requirement, i.e., there exist positive integers $x_{1}<x_{2}<\cdots<x_{n}$, such that
$$\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{n}{x_{n}}=m$$
Then
$$x_{i} \geqslant i(1 \leqslant i \leqslant n)$$
Hence
$$\frac{i}{x_{i}} \leqslant 1,1 \leqslant i \leqslant n$$
Th... | 1,2, \cdots, n | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,899 |
19 Find all positive integers $n, k$, such that $1!+2!+\cdots+n!=k^{3}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 19. When $n=1$, $k=1$;
When $n \geqslant 2$, $1!+2!+\cdots+n!\equiv 1!+2!\equiv 0(\bmod 3)$, hence $3 \mid k$, and it further requires $3^{3} \mid A$, where
$$A=1!+2!+\cdots+n!$$
Notice that, when $n \geqslant 8$, we have
$$\begin{aligned}
A & \equiv 1!+2!+\cdots+8! \\
& \equiv 1+2+6+(-3)+12+(-9)+(-9)+9 \\
& \equiv 9... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,900 |
Property 6 Let $p$ be a prime, $p \mid a b$, then $p \mid a$ or $p \mid b$.
| Proof: Since $p$ has only two positive divisors, hence $(p, a)=1$ or $(p, a)=p$. If $(p, a)=1$, then by property 5 we know $p \mid b$; if $(p, a)=p$, then $p \mid a$.
Below we introduce some concepts and properties of common multiples.
Let $a, b$ be integers not equal to zero. If the integer $c$ satisfies $a \mid c$ a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,901 |
20 Find all integer pairs $(x, y)$, such that $x^{2}+3 y^{2}=1998 x$. | 20. Let $(x, y)$ be the integer pairs that satisfy the equation, it is known that $3 \mid x^{2}$, hence $3 \mid x$, set
then
$$\begin{array}{l}
x=3 x_{1} \\
3 x_{1}^{2}+y^{2}=1998 x_{1}
\end{array}$$
Thus,
$$3 \mid y^{2},$$
hence
$$3 \mid y \text {. }$$
Set again
$$y=3 y_{1},$$
then
$$x_{1}^{2}+3 y_{1}^{2}=666 x_{... | (x, y)=(0,0),(1998,0),(1350, \pm 540),(648, \pm 540) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 741,902 |
21 Find all positive integer solutions to the indeterminate equation $7^{x}-3 \cdot 2^{y}=1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 21. When $x=1$, it is known that $y=1$. Now consider the case $x \geqslant 2$, at this time $y \geqslant 4$, taking both sides modulo 8, we should have
$$(-1)^{x} \equiv 1(\bmod 8)$$
Thus $x$ is even, let $x=2 m$, then
$$\left(7^{m}-1\right)\left(7^{m}+1\right)=3 \cdot 2^{y}$$
Since $7^{m}-1$ and $7^{m}+1$ differ by ... | (1,1),(2,4) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,903 |
Does there exist non-negative integers $a, b$ such that $\left|3^{a}-2^{b}\right|=41$? | 22. There are two cases, namely
$$3^{a}-2^{b}=41 \text {, or } 2^{b}-3^{a}=41 \text {. }$$
For the latter, we know $b>3$. Taking both sides modulo 8, we need
$$3^{a} \equiv 7(\bmod 8)$$
However, for any non-negative integer $a$, we have
$$3^{a} \equiv 1 \text { or } 3(\bmod 8),$$
which is a contradiction.
Now we onl... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,904 |
23 Let $k$ and $m$ be positive integers. Find the minimum possible value of $\left|36^{k}-5^{m}\right|$. | 23. Notice that
$$36^{k}-5^{m} \equiv \pm 1(\bmod 6)$$
and
$$36^{k}-5^{m} \equiv 1(\bmod 5)$$
Therefore, the positive integers that $36^{k}-5^{m}$ can take, in ascending order, are $1, 11, \cdots$.
If $36^{k}-5^{m}=1$, then $k>1$, so taking both sides modulo 8, we require
$$5^{m} \equiv-1(\bmod 8)$$
However, $5^{m} ... | 11 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,905 |
24 Find all positive integers $x, y, z$, such that $3^{x}+4^{y}=5^{z}$. | 24. Taking both sides of the equation modulo 3, we can see that $z$ is even, so $3^{x}+2^{2 y}$ is a perfect square. Thus, using the conclusion from Example 11 in Section 3.2, we can determine that $x=y=z=2$. | x=y=z=2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,906 |
25 Find all Pythagorean triples $(x, y, z)$ such that $x<y<z$, and $x, y, z$ form an arithmetic sequence. | 25. From the conditions, we know $x^{2}+y^{2}=z^{2}$ and $x+z=2 y$, then
$$y^{2}=(z-x)(z+x)=2 y(z-x)$$
Thus,
$$y=2(z-x)$$
Therefore, $y$ is even. Let $y=2 m$, then
we get
$$\begin{array}{c}
x+z=4 m, z-x=m \\
x=\frac{3 m}{2}, z=\frac{5 m}{2}
\end{array}$$
Hence, $m$ is even. Therefore,
$$(x, y, z)=(3 n, 4 n, 5 n)$$
... | (x, y, z) = (3n, 4n, 5n) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,907 |
27 Prove: There exists a positive integer $n$, which appears in exactly 2012 Pythagorean triples.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 27. Consider only numbers $n$ in the form of powers of prime numbers. For this, take a prime $p$ such that $p \equiv 3(\bmod 4)$ (for example, $7, 11, 19 \cdots$), and consider the number $p^{2012}$.
Using the method from Example 2 in Section 3.3, we know that the number $p^{2012}$ cannot be the hypotenuse of a Pythag... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,909 |
28 Find all right-angled triangles with integer side lengths where the perimeter equals twice the area (numerically).
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 28. Equivalent to finding the Pythagorean triple $(x, y, z)$ that satisfies the condition
$$x+y+z=xy$$
Noting that $z=\sqrt{x^{2}+y^{2}}$, (1) becomes
$$x^{2}+y^{2}=(xy-x-y)^{2}$$
i.e.,
$$x^{2}+y^{2}=x^{2} y^{2}-2 xy(x+y)+x^{2}+2 xy+y^{2},$$
Thus,
$$x^{2} y^{2}-2 xy(x+y)+2 xy=0$$
Therefore,
$$xy-2(x+y)+2=0$$
i.e.,... | null | Number Theory | proof | Yes | Yes | number_theory | false | 741,910 |
29 Find the side lengths of a right-angled triangle with integer sides and an area of 24.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 29. According to the analysis starting from Section 3.3, it is noted that the area of a Pythagorean triangle can also be expressed in the form $k^{2} m n\left(n^{2}-m^{2}\right)$, where $k, m, n$ are positive integers, $(m, n)=1, m<24$, so $k=2$. At this point, $m n\left(n^{2}-m^{2}\right)=2 \times 3$, which can only b... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,911 |
Property 7 Let $a, b$ be non-zero integers, $d, c$ be a common divisor and a common multiple of $a, b$ respectively, then $d|(a, b), [a, b] | c$.
| Prove that this property essentially reflects the attributes of the greatest common divisor (GCD) and the least common multiple (LCM). The former is the conclusion of property 2, listed here again for comparison.
For the latter, we use proof by contradiction.
If $[a, b] \nmid c$, let $c=[a, b] \cdot q+r, 0<r<[a, b]$, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,912 |
30 Given that the three sides of $\triangle A B C$ are all integers, $\angle A=2 \angle B, \angle C>90^{\circ}$. Find the minimum perimeter of $\triangle A B C$. | 30. Let the corresponding side lengths of $\triangle ABC$ be $a, b, c$. Draw the angle bisector of $\angle A$ intersecting $BC$ at point $D$, then
$$C D=\frac{a b}{b+c},$$
Using $\triangle A C D \backsim \triangle B C A$, we know
$$\begin{array}{c}
\frac{C D}{b}=\frac{b}{a} \\
a^{2}=b(b+c)
\end{array}$$
That is $\squ... | 77 | Geometry | math-word-problem | Yes | Yes | number_theory | false | 741,913 |
31 Let integers $a, b$ satisfy $5a \geqslant 7b \geqslant 0$. Prove: The system of equations for $x, y, z, w$
$$\left\{\begin{array}{l}
x+2 y+3 z+7 w=a \\
y+2 z+5 w=b
\end{array}\right.$$
has non-negative integer solutions. | 31. Let integers $a, b$ satisfy $5 a \geqslant 7 b \geqslant 0$. Let
$$w=\left[\frac{b}{5}\right] \text { (representing the greatest integer not exceeding } \frac{b}{5} \text {), } v=b-5 w \text {, }$$
then when $v=0,1,2,3,4$ (there are only 5 possible values for $v$), the original system of equations, viewed as a sys... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,914 |
33 Let $n$ be a positive integer, and let $d(n)$ denote the number of positive divisors of $n$, and $\varphi(n)$ denote the number of integers in $1,2, \cdots, n$ that are coprime to $n$. Find all $n$ such that $d(n)+\varphi(n)=n$. | 33. Let $n$ be a number that satisfies the condition, then $n>1$, and at this time, only one number (i.e., 1) is both a divisor of $n$ and coprime with $n$. Therefore, among the numbers $1,2, \cdots, n$, there is exactly one number that is neither a divisor of $n$ nor coprime with $n$.
Since $\varphi(n)$ is even when ... | n=6,8,9 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,916 |
34 Prove: There exist infinitely many triples of positive integers $(a, b, c)$, such that $a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+$ $a^{2}$ are all perfect squares. | 34. We use Pythagorean triples to construct. Take any Pythagorean triple \((x, y, z)\) (not necessarily primitive), and 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
$$\begin{aligned}
a^{2}+b^{2} & =x^{2}\left(3 y^{2}-x^{2}\right)^{2}+y^{2}\left(3 x^{2}-y^{2}\right)^{2} ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,917 |
35 Prove: The indeterminate equation $x^{2}=y^{5}-4$ has no integer solutions. | 35. Taking both sides modulo 11, by Fermat's Little Theorem, when $11 \nmid y$, we have
$$y^{10} \equiv 1(\bmod 11)$$
Thus, in this case,
$$y^{5} \equiv \pm 1(\bmod 11)$$
Therefore, we always have
$$y^{5} \equiv -1,0,1(\bmod 11)$$
That is,
$$y^{5}-4 \equiv 6,7,8(\bmod 11)$$
On the other hand, for $x^{2}$, since
$$x... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,918 |
36 Prove: There do not exist positive integers $m, n$, such that $19^{19}=m^{3}+n^{4}$. | 36. Handle both sides modulo 13. By Fermat's Little Theorem, when $13 \nmid m$, we have $m^{12} \equiv 1(\bmod 13)$, which implies $m^{6} \equiv 1$ or $25(\bmod 13)$, so $m^{3} \equiv \pm 1$ or $\pm 5(\bmod 13)$, meaning $m^{3} \equiv 0,1, 5,8,12(\bmod 13)$.
Similarly, we know that $n^{2} \equiv 0,1,4,9,3,12(\bmod 13)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,919 |
37 Does there exist positive integers $x, y, z, u, v$, such that $x, y, z, u, v$ are all greater than 2012, and satisfy $x^{2}+y^{2}+z^{2}+u^{2}+v^{2}=x y z u v-65$? | 37. 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 $x<y<z<u<v$. If we consider the original equation as a quadratic equation in $x$, using Vieta's formulas, we know that $(y... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,920 |
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