problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
Example 2 Find all integers $n>1$, such that $\frac{2^{n}+1}{n^{2}}$ is an integer. | It is easy to conjecture that $n=3$ is the only solution that meets the requirements. Below, we prove that this is indeed the case.
The proof requires several steps. The first point is to consider the smallest prime divisor $p$ of $n$, and from $n \mid (2^n + 1)$, derive that $p=3$, as seen in Example 1 of Unit 8. The... | n=3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,074 |
Example 3 Proof: For each $n>1$, the equation
$$\frac{x^{n}}{n!}+\frac{x^{n-1}}{(n-1)!}+\cdots+\frac{x^{2}}{2!}+\frac{x}{1!}+1=0$$
has no rational roots. | Proof Let $a$ be a rational root of the given equation, then it is easy to see that
$$a^{n}+\frac{n!}{(n-1)!} a^{n-1}+\cdots+\frac{n!}{k!} a^{k}+\cdots+\frac{n!}{1!} a+n!=0,$$
Thus, $a$ is a rational root of an integral polynomial with leading coefficient 1, so $a$ must be an integer (see Exercise 2, Question 4).
Sin... | proof | Algebra | proof | Yes | Yes | number_theory | false | 742,075 |
Example 4 Let $n>1, x_{1}, \cdots, x_{n}$ be $n$ real numbers, and their product be denoted as $A$. If for $i=1, \cdots, n$, the number $A-x_{i}$ is an odd integer. Prove: Each $x_{i}$ is an irrational number. | Proof by contradiction: If there exists an $i$ such that $x_{i}$ is a rational number, then because $A-x_{i}$ is an odd integer, $A$ must be a rational number. Let $A-x_{i}=a_{i} (i=1, \cdots, n)$. Then from $x_{1} \cdots x_{n}=A$, we get
$$\left(A-a_{1}\right) \cdots\left(A-a_{n}\right)=A$$
Since $a_{i}$ are all (odd... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,076 |
Example 5 Let $a, b, c$ be integers, $f(x)=x^{3}+a x^{2}+b x+c$. Prove: there are infinitely many positive integers $n$, such that $f(n)$ is not a perfect square. | We prove that for any positive integer $n \equiv 1(\bmod 4)$, among the four integers $f(n), f(n+1), f(n+2), f(n+3)$, at least one is not a perfect square, thereby proving the conclusion of the problem.
It is easy to see that
$$\begin{array}{l}
f(n) \equiv 1+a+b+c(\bmod 4) \\
f(n+1) \equiv 2 b+c(\bmod 4) \\
f(n+2) \eq... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,077 |
Example 6 Let $p(x)$ be a polynomial with integer coefficients, and for any $n \geqslant 1$ we have $p(n)>n$. Define $x_{1}=$ $1, x_{2}=p\left(x_{1}\right), \cdots, x_{n}=p\left(x_{n-1}\right)(n \geqslant 2)$. If for any positive integer $N$, the sequence $\left\{x_{n}\right\}$ $(n \geqslant 1)$ contains a term divisib... | Proof We proceed in two steps. First, we prove that for any fixed $m>1$, the sequence $\left\{x_{n}\right\}$ modulo $x_{m}-1$ is a periodic sequence. Clearly, $x_{m} \equiv 1=x_{1}\left(\bmod x_{m}-1\right)$. Since $p(x)$ is a polynomial with integer coefficients, for any integers $u, v (u \neq v)$, we have $(u-v) \mid... | p(x)=x+1 | Algebra | proof | Yes | Yes | number_theory | false | 742,078 |
Example 7 Let $f(x)$ be a quadratic polynomial with real coefficients. If for all positive integers $n$, $f(n)$ is the square of an integer. Prove that $f(x)$ is the square of a linear polynomial with integer coefficients. | Proving this problem is not easy, but there are several completely different solutions. The method introduced here is based on the knowledge of the limit of sequences, which is relatively simple.
Let $f(x)=a x^{2}+b x+c, a_{n}=f(n)(n \geqslant 1)$, then it is easy to know
$$\begin{aligned}
\sqrt{a_{n}}-\sqrt{a_{n-1}} ... | proof | Algebra | proof | Yes | Yes | number_theory | false | 742,079 |
5 Let integers $a, b, c, d$ satisfy $ad - bc > 1$, prove: at least one of $a, b, c, d$ is not divisible by $ad - bc$.
保留源文本的换行和格式，直接输出翻译结果。 | 5. If $a, b, c, d$ are all divisible by $ad-bc$, then $(ad-bc)^2$ divides $ad$ and $bc$, hence divides $ad-bc$, from which it follows that $|ad-bc|=1$, which contradicts the given $ad-bc>1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,080 |
Example 8 Let $n>1, n$ positive integers sum to $2 n$. Prove that, among them, we can always select some numbers whose sum equals $n$, unless the given numbers satisfy one of the following conditions:
(1) There is one number that is $n+1$, and the rest are 1;
(2) When $n$ is odd, all numbers are equal to 2. | Proof - Let the given positive integer be $01$, then we have
$$\ddot{a_{2}}+\cdots+a_{k}+a_{n} \equiv 0(\bmod n)$$
The left side of the above equation is clearly a positive integer less than $a_{1}+\cdots+a_{n}=2 n$, hence
$$a_{2}+\cdots+a_{k}+a_{n}=n$$
(iv) Suppose $a_{1}-a_{n} \equiv 0(\bmod n)$. We have already pro... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,081 |
Example 9 Let $p$ be a prime, and given $p+1$ distinct positive integers. Prove that one can select a pair of these numbers such that when the larger of the two is divided by their greatest common divisor, the quotient is at least $p+1$.
| Prove that dividing the given $p+1$ numbers by their greatest common divisor obviously does not affect the conclusion of this problem, so we can assume that these $p+1$ numbers are coprime. In particular, there must be a number that is not divisible by $p$. Let these $p+1$ numbers be
$$x_{1}, \cdots, x_{k}, x_{k+1}=p^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,082 |
Example 10 Let $S$ be a subset of $\left\{1,2, \cdots, 2^{m} n\right\}$, and the number of elements in $S$ is $|S| \geqslant\left(2^{m}-\right.$ 1) $n+1$. Prove that there are $m+1$ distinct numbers $a_{0}, \cdots, a_{m}$ in $S$, such that $a_{i-1} \mid a_{i}(i=1, \cdots, m)$. | Prove that each positive integer $a$ can be uniquely represented in the form $2 u k$, where $u \geqslant 0$ and $k$ is odd. We call $k$ the odd part of $a$, and if the odd part of $a$ does not exceed $n$, then $n$ is called a good number. This proof is based on a lower bound estimate of the number of good numbers in $S... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 742,083 |
Example 11 Let $A$ be an $n$-element set of positive integers $(n \geqslant 2)$. Prove that $A$ has a subset $B$, satisfying $|B|>\frac{n}{3}$, and for any $x, y \in B$, $x+y \notin B$.
| Let the numbers in $A$ be $a_{1}, \cdots, a_{n}$. From Exercise 3, part 2, we know that there are infinitely many primes that are $-1$ modulo 3, so we can take such a prime $p > a_{i} (1 \leqslant i \leqslant n)$, and set $p = 3k - 1$. Consider the following ($p$ rows and $n$ columns) $pn$ numbers:
$$\begin{array}{l}
a... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 742,084 |
Example 12 Given $n \geqslant 2$. Prove that there exist $n$ distinct positive integers with the following properties:
(1) These numbers are pairwise coprime;
(2) The sum of any $k$ of these numbers $(2 \leqslant k \leqslant n)$ is a composite number. | Proof. When $n=2$, the conclusion is obviously true. Suppose we already have $n$ positive integers $a_{1}, \cdots, a_{n}$ that meet the requirements. We will construct $n+1$ numbers based on this.
Since there are infinitely many primes, we can take $2^{n}-1$ distinct primes $p_{i}\left(1 \leqslant i \leqslant 2^{n}-1\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,085 |
Example 13 Find all positive integers $k$ such that there exists a positive integer $n$ satisfying
$$\frac{\tau\left(n^{2}\right)}{\tau(n)}=k$$
where $\tau(n)$ denotes the number of positive divisors of $n$. | From the formula for calculating $\tau(n)$ in Unit 3 (6), we know that $\tau\left(n^{2}\right)$ must be an odd number, so $k$ that satisfies (1) must be an odd number. We will now prove that every positive odd number $k$ meets the requirement.
$k=1$ clearly meets the requirement. For $k>1$, by the formula for $\tau(n)$... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,086 |
1 Prove that for any integer $a \geqslant 3$, there are infinitely many positive integers $n$ such that $a^{n}-1$ is divisible by $n$. (Please compare Example 2 of Unit 8.) | 1. Since $a \geqslant 3$, it follows that $a-1$ has a prime factor $p$. By Fermat's Little Theorem, we know that $a^{p} \equiv a \equiv 1(\bmod p)$. Using induction, it is easy to prove that $n=p^{k}(k=1,2, \cdots)$ all meet the requirements. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,087 |
2 Let $n_{1}, \cdots, n_{k}$ be positive integers, with the following property:
$$n_{1}\left|\left(2^{n_{2}}-1\right), n_{2}\right|\left(2^{n_{3}}-1\right), \cdots, n_{k} \mid\left(2^{n_{1}}-1\right) .$$
Prove: $n_{1}=\cdots=n_{k}=1$. | 2. The given conditions can be restated as
$$2^{n_{2}} \equiv 1\left(\bmod n_{1}\right), 2^{n_{3}} \equiv 1\left(\bmod n_{2}\right), \cdots, 2^{n_{1}} \equiv 1\left(\bmod n_{k}\right)$$
Let $D=\left[n_{1}, \cdots, n_{k}\right]$. Then from the above equations, we have
$$2^{D} \equiv 1\left(\bmod n_{i}\right)(i=1, \cdot... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,088 |
3 Let positive integers $a, b$ satisfy $a^{2} b \mid\left(a^{3}+b^{3}\right)$, prove $a=b$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. Let $a^{3}+b^{3}=m a^{2} b$, then $\left(\frac{a}{b}\right)^{3}-m\left(\frac{a}{b}\right)^{2}+1=0$, which means the rational number $\frac{a}{b}$ is a root of the integral coefficient equation with leading coefficient 1:
$$x^{3}-m x^{2}+1=0$$
Therefore, $\frac{a}{b}$ must be an integer (Exercise 2, Question 4). On ... | a=b | Combinatorics | MCQ | Yes | Yes | number_theory | false | 742,089 |
4 Prove: The indeterminate equation $x^{n}+1=y^{n+1}$ has no positive integer solutions $(x, y, n)$, where $(x, n+1)=1$, $n>1$. | 4. Clearly $y>1$. The original equation can be factored as
$$(y-1)\left(y^{n}+y^{n-1}+\cdots+y+1\right)=x^{n}$$
The key is to prove that $y-1$ and $y^{n}+y^{n-1}+\cdots+y+1$ are coprime. If their greatest common divisor $d>1$, then $d$ has a prime factor $p$. From $y \equiv 1(\bmod p)$, we know that $y^{i} \equiv 1(\b... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,090 |
5 (1) Prove: For any given positive integer $n$, there exist positive rational numbers $a, b, a \neq b$, that are not integers, such that $a-b, a^{2}-b^{2}, \cdots, a^{n}-b^{n}$ are all integers.
(2) Let $a, b$ be positive rational numbers, $a \neq b$. If there are infinitely many positive integers $n$ such that $a^{n}... | 5. (1) For example, take $a=2^{n}+\frac{1}{2}, b=\frac{1}{2}$, then for $k=1, \cdots, n$,
$$\begin{aligned}
a^{k}-b^{k} & =(a-b)\left(a^{k-1}+a^{k-2} b+\cdots+a b^{k-2}+b^{k-1}\right) \\
& =2^{n} \cdot a^{k-1}+2^{n} \cdot a^{k-2} b+\cdots+2^{n} \cdot a b^{k-2}+2^{n} \cdot b^{k-1}
\end{aligned}$$
Since $k \leqslant n$,... | proof | Algebra | proof | Yes | Yes | number_theory | false | 742,092 |
6 Let $n \geqslant 4$ be an integer, and $a_{1}, \cdots, a_{n}$ be distinct positive integers less than $2 n$. Prove: from these numbers, one can select several such that their sum is divisible by $2 n$.
untranslated text remains the same as requested. However, if you need any further assistance or a different format... | 6. If each $a_{i}$ is not equal to $n$, the conclusion is easy to prove. Because the $2n$ numbers
$$a_{1}, a_{2}, \cdots, a_{n}, 2n-a_{1}, 2n-a_{2}, \cdots, 2n-a_{n}$$
are all positive integers and less than $2n$, hence there must be two that are equal, i.e., there exist $i, j$ such that $a_{i}=2n-a_{j}$. If $i=j$, it... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,093 |
Example 1 For any integer $n$, prove that the fraction $\frac{21 n+4}{14 n+3}$ is in its simplest form. | To prove that $21n+4$ and $14n+3$ are coprime, it is easy to see that these two numbers satisfy Bézout's identity:
$$3(14n+3)-2(21n+4)=1$$
Therefore, the conclusion holds.
In general, it is not easy to derive Bézout's identity for coprime integers $a$ and $b$. Therefore, we often use the following workaround: create a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,095 |
Example 2 Let $n$ be a positive integer, prove that $(n!+1,(n+1)!+1)=1$.
The text is translated while preserving the original line breaks and format. | Prove we have the equation
$$(n!+1)(n+1)-((n+1)!+1)=n$$
Let $d=(n!+1,(n+1)!+1)$, then by (1) we know $d \mid n$.
Furthermore, since $d \mid n$, hence $d \mid n!$, combining with $d \mid(n!+1)$ we know $d \mid 1$, so $d=1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,096 |
Example 3 Let $F_{k}=2^{2^{k}}+1, k \geqslant 0$. Prove: If $m \neq n$, then $\left(F_{m}, F_{n}\right)=1$.
---
The translation maintains the original text's line breaks and format. | Assume without loss of generality that $m>n$. The key to the argument is to use $F_{n} \mid\left(F_{m}-2\right)$ (see Example 2 in Unit 1), which means there is an integer $x$ such that
$$F_{m}+x F_{n}=2 \text {. }$$
Let $d=\left(F_{m}, F_{n}\right)$, then from the above equation, we deduce $d \mid 2$, so $d=1$ or 2. ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,097 |
Example 4 Let $a>1, m, n>0$, prove:
$$\left(a^{m}-1, a^{n}-1\right)=a^{(m, n)}-1$$ | Let $D=\left(a^{m}-1, a^{n}-1\right)$. We prove that $\left(a^{(m, n)}-1\right) \mid D$ and $D \mid \left(a^{(m, n)}-1\right)$ to derive $D=a^{(m, n)}-1$, which is a common method in number theory to prove the equality of two numbers.
Since $(m, n)|m, (m, n)| n$, by the factorization formula (5) in the first unit, we ... | D=a^{(m, n)}-1 | Number Theory | proof | Yes | Yes | number_theory | false | 742,098 |
Example 2 Let $m>n \geqslant 0$, prove: $\left(2^{2^{n}}+1\right) \mid\left(2^{2^{m}}-1\right)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Prove that in the factorization (5) by taking $x=2^{2^{n+1}}, y=1$, and substituting $2^{m-n-1}$ for $n$ there, we get
$$2^{2^{m}}-1=\left(2^{2^{n+1}}-1\right)\left[\left(2^{2^{n+1}}\right)^{2^{m-n-1}-1}+\cdots+2^{2^{n+1}}+1\right]$$
Hence
$$\left(2^{2^{2+1}}-1\right) \mid\left(2^{2^{m}}-1\right)$$
Also, $2^{2^{n+1}}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,099 |
Example 5 Let $m, n>0, m n \mid\left(m^{2}+n^{2}\right)$, then $m=n$. | Proof: Let $(m, n)=d$, then $m=m_{1} d, n=n_{1} d$, where $\left(m_{1}, n_{1}\right)=1$. Thus, the given condition becomes $m_{1} n_{1} \mid\left(m_{1}^{2}+n_{1}^{2}\right)$, hence we also have $m_{1} \mid\left(m_{1}^{2}+n_{1}^{2}\right)$, which implies $m_{1} \mid n_{1}^{2}$. But $\left(m_{1}, n_{1}\right)=1$, so $\le... | m=n | Number Theory | proof | Yes | Yes | number_theory | false | 742,100 |
Example 6 Let positive integers $a, b, c$ have a greatest common divisor of 1, and
$$\frac{ab}{a-b}=c$$
Prove: $a-b$ is a perfect square. | Proof: Let $(a, b)=d$, then $a=d a_{1}, b=d b_{1}$, where $\left(a_{1}, b_{1}\right)=1$. Since $(a, b, c)=1$, it follows that $(d, c)=1$.
Now, the equation in the problem can be transformed into
$$d a_{1} b_{1}=c a_{1}-c b_{1}$$
From this, it is clear that $a_{1}$ divides $c b_{1}$. Since $\left(a_{1}, b_{1}\right)=1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,101 |
Example 7 Let $k$ be a positive odd number, prove: $1+2+\cdots+n$ divides $1^{k}+2^{k}+\cdots+n^{k}$. | To prove that since $1+2+\cdots+n=\frac{n(n+1)}{2}$, the problem is equivalent to proving: $n(n+1)$ divides $2\left(1^{k}+2^{k}+\cdots+n^{k}\right)$. Because $n$ and $n+1$ are coprime, this is equivalent to proving
$$n \mid 2\left(1^{k}+2^{k}+\cdots+n^{k}\right)$$
and
$$(n+1) \mid 2\left(1^{k}+2^{k}+\cdots+n^{k}\right... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,102 |
I Let $n$ be an integer, prove: $(12 n+5,9 n+4)=1$.
| 1. We have $4(9 n+4)-3(12 n+5)=1$.
| proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,103 |
2 Let $m, n$ be positive integers, $m$ is odd, prove: $\left(2^{m}-1,2^{n}+1\right)=1$.
| 2. Let $d=\left(2^{m}-1,2^{n}+1\right)$. Then $2^{m}-1=d u, 2^{n}+1=d v$, where $u, v$ are integers. It is easy to see that $(d u+1)^{n}=(d v-1)^{m}$, expanding both sides (note that $m$ is odd), we get $d A+1=d B-1$ ($A, B$ are some two integers), from which it follows that $d \mid 2$, i.e., $d=1$ or 2. But clearly, $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,104 |
3 Let $(a, b)=1$, prove: $\left(a^{2}+b^{2}, a b\right)=1$.
untranslated text remains the same as requested. However, if you need any further assistance or a different format, please let me know! | 3. Since $(a, b)=1$, we have $\left(a^{2}, b\right)=1$, thus $\left(a^{2}+b^{2}, b\right)=1$. Similarly, $\left(a^{2}+b^{2}\right.$, $a)=1$. Therefore, $\left(a^{2}+b^{2}, a b\right)=1$ (using (6) of this unit). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,105 |
4 If the $k$-th power of a rational number is an integer $(k \geqslant 1)$, then this rational number must be an integer. More generally, prove: a rational root of an integer-coefficient polynomial with a leading coefficient of $\pm 1$ must be an integer.
| 4. Let the reduced rational number $\frac{p}{q}$ be a root of the integer-coefficient polynomial $f(x)=x^{n}+$ $a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}$ (with the leading coefficient 1). From $f\left(\frac{p}{q}\right)=0$ we easily get
$$p^{n}+a_{1} p^{n-1} q+\cdots+a_{n-1} p q^{n-1}+a_{n} q^{n}=0$$
Since $a_{1} p^{n-1} ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,106 |
Let $m$, $n$, $k$ be positive integers, satisfying $[m+k, m]=[n+k, n]$, prove: $m=n$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. From this unit (10), we know that the given condition is
$$\frac{(m+k) m}{(m+k, m)}=\frac{(n+k) n}{(n+k, n)}$$
Since $(m+k, m)=(m, k),(n+k, n)=(n, k)$, we get from the above equation
$$\frac{(m+k) m}{(m, k)}=\frac{(n+k) n}{(n, k)}$$
Let $(m, k)=d_{1}$, then $m=m_{1} d_{1}, k=k_{1} d_{1}$, where $\left(m_{1}, k_{1}... | null | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,107 |
Example 1 Proof: The infinite sequence $10001,100010001, \cdots$ contains no prime numbers. | Let $a_{n}=\underbrace{10001 \cdots 10001}_{n \text { ones}}(n \geqslant 2)$, then
$$a_{n}=1+10^{4}+10^{8}+\cdots+10^{4(n-1)}=\frac{10^{4 n}-1}{10^{4}-1} .$$
To decompose the number on the right-hand side of the above equation into the product of two (greater than 1) integers, we distinguish two cases:
- $n$ is even. ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,109 |
Example 3 For a positive integer $n$, let $S(n)$ denote the sum of the digits in the decimal representation of $n$. Prove: $9 \mid n$ if and only if $9 \mid S(n)$. | Proof Let $n=a_{k} \times 10^{k}+\cdots+a_{1} \times 10+a_{0}$ (here $0 \leqslant a_{i} \leqslant 9$, and $a_{k} \neq 0$), then $S(n)=a_{0}+a_{1}+\cdots+a_{k}$. We have
$$n-S(n)=a_{k}\left(10^{k}-1\right)+\cdots+a_{1}(10-1)$$
For $1 \leqslant i \leqslant k$, by the factorization formula (5) we know $9 \mid\left(10^{i}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,110 |
Example 2 Proof: For any integer $n>1$, the number $n^{4}+4^{n}$ is not a prime.
untranslated text remains the same as requested. However, if you need the entire text to be translated, please let me know! | Prove that if $n$ is even, then $n^{4}+4^{n}$ is greater than 2 and divisible by 2, so it is not a prime. However, for odd $n$, it is easy to see that $n^{4}+4^{n}$ does not have a fixed divisor (greater than 1). We use a different approach:
Let the odd number $n=2 k+1, k \geqslant 1$, then
$$\begin{aligned}
n^{4}+4^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,111 |
Example 4 Proof: If integers $a, b$ satisfy $2a^2 + a = 3b^2 + b$, then $a - b$ and $2a + 2b + 1$ are both perfect squares. | Prove that the given relation is
$$(a-b)(2a+2b+1)=b^2$$
The first point of the argument is to prove that the integers $a-b$ and $2a+2b+1$ are coprime. Let $d=(a-b, 2a+2b+1)$. If $d>1$, then $d$ has a prime factor $p$, and thus from (1) we know $p \mid b^2$. Since $p$ is a prime, it follows that $p \mid b$. Combining t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,113 |
Example 5 Let $n, a, b$ be integers, $n>0$ and $a \neq b$. Prove: If $n \mid\left(a^{n}-b^{n}\right)$, then $n \left\lvert\, \frac{a^{n}-b^{n}}{a-b}\right.$. | Proof: Let $p$ be a prime, and $p^{a} \| n$. We will prove that $p^{a} \left\lvert\, \frac{a^{n}-b^{n}}{a-b}\right.$, which will lead to the conclusion of this problem (see the note below).
Let $t=a-b$. If $p \nmid t$, then $\left(p^{a}, t\right)=1$. Since $n \mid\left(a^{n}-b^{n}\right)$, it follows that $p^{a} \mid\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,114 |
Example 6 Let $m, n$ be non-negative integers, prove that $\frac{(2 m)!(2 n)!}{m!n!(m+n)!}$ is an integer.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
However, it seems the translation request was already fulfilled in the ... | To prove, we only need to show that for each prime $p$, the power of $p$ in the standard factorization of the denominator $m!n!(m+n)!$ does not exceed the power of $p$ in the numerator $(2m)!(2n)!$. By the formula in (7), this is equivalent to proving
$$\sum_{l=1}^{\infty}\left(\left[\frac{2 m}{p^{l}}\right]+\left[\fra... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,115 |
I prove: For any given positive integer $n>1$, there exist $n$ consecutive composite numbers. | 1. It is easy to verify that $(n+1)!+2, (n+1)!+3, \cdots, (n+1)!+(n+1)$ are $n$ consecutive composite numbers. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,117 |
2 Prove: there are infinitely many primes of the form $4k-1$, and there are also infinitely many primes of the form $6k-1$ ($k$ being a positive integer). | 2. The method used by Euclid to prove that there are infinitely many primes can be slightly modified to argue: Assume that there are only finitely many primes of the form $4k-1$, and let them all be $p_{1}, \cdots, p_{m}$. Consider the number $N = 4 p_{1} \cdots p_{m} - 1$. Clearly $m > 1$, so $N > 1$, and thus $N$ has... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,118 |
3 Prove: There are infinitely many odd numbers $m$, such that $8^{m}+9 m^{2}$ is a composite number. | 3. Take $m=9 k^{3}(k=1,3, \cdots)$, then $8^{m}+9 m^{2}=\left(2^{m}\right)^{3}+\left(9 k^{2}\right)^{3}$. It is easy to see that it has a proper divisor $2^{m}+9 k^{2}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,119 |
4 Let integers $a, b, c, d$ satisfy $a>b>c>d>0$, and
$$a^{2}+a c-c^{2}=b^{2}+b d-d^{2},$$
Prove: $a b+c d$ is not a prime. | 4. Proof by contradiction, suppose there exists a set of $a, b, c, d$ satisfying the conditions such that $ab + cd$ is a prime number, denoted as $p$. Substituting $a = \frac{p - cd}{b}$ into the given equation, we get
$$p(p - 2cd + bc) = (b^2 + c^2)(b^2 + bd - d^2)$$
Since $p$ is a prime number, $p$ must divide $b^2 ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,120 |
Example 1 A positive integer, when added to 100, becomes a perfect square. If 168 is added to it, it becomes another perfect square. Find this number.
| Let the number to be found be $x$. By the problem, there exist positive integers $y, z$, such that
$$\left\{\begin{array}{l}
x+100=y^{2} \\
x+168=z^{2}
\end{array}\right.$$
Eliminating $x$ from the above two equations, we get
$$z^{2}-y^{2}=68$$
Factoring the left side of this binary quadratic equation and standardizi... | 156 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,122 |
Example 2 Find all integer solutions of the indefinite equation:
$$x^{4}+y^{4}+z^{4}=2 x^{2} y^{2}+2 y^{2} z^{2}+2 z^{2} x^{2}+24$$ | The key step (and the main difficulty of the problem) is to recognize that the equation can be factored as
$$(x+y+z)(x+y-z)(y+z-x)(z+x-y)=-2^{3} \times 3 .$$
Since the four factors on the left side of the equation are all integers, by the fundamental theorem of arithmetic, we can decompose (1) into several systems of ... | not found | Algebra | math-word-problem | Yes | Yes | number_theory | false | 742,123 |
Example 3 Proof: The product of two consecutive positive integers cannot be a perfect square, nor can it be a perfect cube. | Proof by contradiction, we assume there exist positive integers $x, y$, such that
$$x(x+1)=y^{2} .$$
Multiplying both sides of the equation by 4, we transform it into $(2 x+1)^{2}=4 y^{2}+1$, which can be factored as
$$(2 x+1+2 y)(2 x+1-2 y)=1$$
Since the two factors on the left are positive integers, we have
$$\left... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,124 |
Example 4 Prove: The equation
$$y+y^{2}=x+x^{2}+x^{3}$$
has no integer solutions for $x \neq 0$. | Proof: Suppose the equation has an integer solution $x \neq 0$. We decompose it as
$$(y-x)(y+x+1)=x^{3}$$
We first prove that $(y-x, y+x+1)=1$. If this is not true, then there exists a prime $p$ that is a common divisor of $y-x$ and $y+x+1$. From (1), we know $p \mid x^{3}$, so the prime $p$ divides $x$. Combining $p ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,125 |
Example 5 Let $k$ be a given positive integer, $k \geqslant 2$, prove: the product of three consecutive positive integers cannot be an integer's $k$-th power.
保留了源文本的换行和格式。 | Prove that if there are positive integers $x \geqslant 2$ and $y$, such that
$$(x-1) x(x+1)=y^{k}$$
Note that the three factors $x-1, x, x+1$ are not always pairwise coprime, so we cannot conclude from (1) that they are all $k$-th powers. One way to overcome this difficulty is to transform (1) into
$$\left(x^{2}-1\rig... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,126 |
Example 6 Find all integer solutions of $\left(x^{2}-y^{2}\right)^{2}=1+16 y$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve: Since the left side of the equation $\geqslant 0$, the right side $\geqslant 0$, thus $y \geqslant 0$. It is also clear that $x^{2}-y^{2} \neq 0$, and since $x_{1} y$ is an integer, we have $|x| \geqslant y+1$, or $|x| \leqslant y-1$.
When $|x| \geqslant y+1$, the left side of the equation $\geqslant\left((y+1)... | ( \pm 1,0),( \pm 4,3), ( \pm 4,5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,127 |
Example 7 Let positive integers $x, y, z$ satisfy $2 x^{x}=y^{y}+z^{z}$, then $x=y=z$. | Proof First, expand $(x+1)^{x+1}$ to know
$$(x+1)^{x+1}>x^{x+1}+(x+1) x^{x}>2 x^{x}$$
From this, we know that $y$ and $z$ must both be $\leqslant x$: for if $y$ or $z$ is greater than $x$, without loss of generality, assume $y>x$. Since $y$ and $x$ are integers, we have $y \geqslant x+1$, thus
$$y^{y}+z^{z}>y^{y} \geq... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,128 |
1 Prove: The product of four consecutive positive integers cannot be a perfect square.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the request is to translate the text and not to restate the instruction, h... | 1. Let $x(x+1)(x+2)(x+3)=y^{2}, x, y$ are positive integers. Then we have
$$\left(x^{2}+3 x+1\right)^{2}-y^{2}=1$$
It is evident that this is impossible. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,129 |
2 Find all integers that can be expressed as the difference of squares of two integers.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 2. Let the integer $n$ be expressible as the difference of squares of two integers: $n=x^{2}-y^{2}$, i.e., $n=(x-y)(x+y)$. Since $x-y$ and $x+y$ have the same parity, either $n$ is odd, or $n$ is divisible by 4. Conversely, if $n$ is odd, we can take $x-y=1, x+y=n$, i.e., $x=\frac{n+1}{2}, y=\frac{n-1}{2}$; if $4 \mid ... | null | Number Theory | proof | Yes | Yes | number_theory | false | 742,130 |
Find all integer solutions of the indefinite system of equations
$$\left\{\begin{array}{l}
x+y+z=3 \\
x^{3}+y^{3}+z^{3}=3
\end{array}\right.$$ | 3. Eliminate $z$ from the system of equations, to get
$$8-9 x-9 z+3 x^{2}+6 x y+3 y^{2}-x^{2} y-x y^{2}=0$$
Transform it into
$$8-3 x(3-x)-3 y(3-x)+x y(3-x)+y^{2}(3-x)=0,$$
which is $(3-x)\left(3 x+3 y-x y-y^{2}\right)=8$. Hence $(3-x) \mid 8$, so $3-x= \pm 1, \pm 2, \pm 4, \pm 8$, which means $x=-5,-1,1,2,4,5,7,11$.... | (x, y, z)=(1,1,1),(-5,4,4),(4,-5,4),(4,4,-5) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 742,131 |
Example 4 Let $k \geqslant 1$ be an odd number, prove: for any positive integer $n$, the number $1^{k}+2^{k}+\cdots+n^{k}$ cannot be divisible by $n+2$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Prove that the conclusion is obviously true when $n=1$. Let $n \geqslant 2$, and denote the said sum as $A$, then
$$2 A=2+\left(2^{k}+n^{k}\right)+\left(3^{k}+(n-1)^{k}\right)+\cdots+\left(n^{k}+2^{k}\right) .$$
Since $k$ is a positive odd number, by the factorization formula (6), for each $i \geqslant 2$, the number ... | null | Number Theory | proof | Yes | Yes | number_theory | false | 742,132 |
4 Find all integer solutions to $x^{3}=y^{3}+2 y^{2}+1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. First note that if $y^{2}+3 y>0$, then from the original equation we deduce that $(y+1)^{3}>x^{3}>y^{3}$, which means $x^{3}$ lies between two consecutive perfect cubes, which is impossible. Therefore, it must be that $y^{2}+3 y \leqslant 0$, yielding the integer solutions $y=$ $-3,-2,-1,0$. Substituting these value... | (x, y)=(1,0), (1,-2), (-2,-3) | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false | 742,133 |
Find all positive integers $x, y$ such that $x^{2}+3 y, y^{2}+3 x$ are both perfect squares.
The text above has been translated into English, preserving the original text's line breaks and format. | 5. Let $\left\{\begin{array}{l}x^{2}+3 y=u^{2} \\ y^{2}+3 x=v^{2}\end{array}\right.$ Since $x, y$ are positive integers, we have $u>x, v>y$. We set $u=x+a, v=y+b$, where $a, b$ are positive integers. From
$$\left\{\begin{array}{l}
x^{2}+3 y=(x+a)^{2} \\
y^{2}+3 x=(y+b)^{2}
\end{array}\right.$$
we can derive
$$\left\{\... | (1,1),(16,11),(11,16) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,134 |
Example 1 Let $m \geqslant n \geqslant 1$, prove: $\frac{(m, n)}{m} \mathrm{C}_{m}^{n}$ is an integer. | Prove that $\frac{x}{m} \mathrm{C}_{m}^{n}$ is an integer when $x=m$, which is $\mathrm{C}_{m}^{n}$; when $x=n$, it is $\frac{n}{m}$. $\frac{m}{n} \mathrm{C}_{m-1}^{n-1}=\mathrm{C}_{m-1}^{n-1}$, which is also an integer. Furthermore, by Bézout's identity, there exist integers $u$ and $v$ such that
$$(m, n)=m u+n v,$$
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,135 |
Example 2 Let $a, b$ be two distinct positive integers, and $ab(a+b)$ is a multiple of $a^{2}+ab+b^{2}$. Prove: $|a-b|>\sqrt[3]{ab}$. | To prove that since $a b(a+b)$ is divisible by $a^{2}+a b+b^{2}$, we first divide $a b(a+b)$ by $a^{2}+a b+b^{2}$, obtaining
$$a b(a+b)=\left(a^{2}+a b+b^{2}\right) a-a^{3}$$
Thus, $\left(a^{2}+a b+b^{2}\right) \mid a^{3}$. Similarly, $\left(a^{2}+a b+b^{2}\right) \mid b^{3}$, meaning $a^{2}+a b+b^{2}$ is a common div... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,136 |
Example 3 Take several different integers between two consecutive perfect squares $n^{2}$ and $(n+1)^{2}$, prove that their pairwise products are all different. | Proof: Let integers $a, b, c, d$ satisfy $n^{2}a$ and $c > a$, leading to $q > p$ and $v > u$. Since $p, q, u, v$ are all integers, we have $q \geqslant p+1, v \geqslant u+1$. Therefore, we get (note $a = p u > n^{2}$)
$$\begin{array}{l}
d = q v \geqslant (p+1)(u+1) = p u + (p+u) + 1 \\
> n^{2} + 2 \sqrt{p u} + 1 > n^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,137 |
Example 4 Find all positive integer solutions to the indeterminate equation
$$(n-1)!=n^{k}-1$$ | When $n=2$, from (1) we get the solution $(n, k)=(2,1)$. When $n>2$, the left side of (1) is even, so its right side must also be even, hence $n$ is odd. When $n=3,5$, from (1) we solve $(n, k)=(3,1), (5,2)$.
Below, let $n>5$ and $n$ be odd. In this case, $\frac{n-1}{2}$ is an integer and $\frac{n-1}{2}(n-1)!$
This i... | (n, k)=(2,1),(3,1),(5,2) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,138 |
Example 7 Proof: From $1,2, \cdots, 100$, if any 51 numbers are taken, there must be two numbers that are coprime. | The problem becomes extremely simple when broken down: We take consecutive pairs of numbers from $1,2, \cdots, 100$, forming the following 50 pairs
$$\{1,2\},\{3,4\}, \cdots,\{99,100\},$$
then any selection of 51 numbers must include one of the above pairs. Since these two numbers are consecutive, they are of course c... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,141 |
Example 8 Proof: There exist 1000 consecutive positive integers, among which exactly 10 are prime. | The basis of this proof is Exercise 3, Question 1, which concludes that there exist 1000 consecutive positive integers
$$a, a+1, \cdots, a+999$$
where each number is not a prime.
Now, perform the following operation on (1): delete the rightmost number $a+999$ and add $a-1$ to the left. Clearly, the resulting sequence
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,142 |
Example 5 Let $m, n$ be positive integers, $m>2$, prove: $\left(2^{m}-1\right) \nmid\left(2^{n}+1\right)$. | First, when $n \leqslant m$, it is easy to see that the conclusion holds. In fact, when $m=n$, the conclusion is trivial; when $n < m$, the conclusion is also trivial (and can be seen from the divisibility property (3)).
Finally, the case $n > m$ can be reduced to the above special case: by the division algorithm, $n ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,143 |
Example 9 If in the standard factorization of a positive integer, the exponent of each prime factor is greater than 1, then it is called a power number. Prove: There exist infinitely many distinct positive integers, such that neither they nor the sum of any different numbers among them are power numbers. | Proof Let $2=p_{1}<p_{2}<\cdots<p_{n}<\cdots$ be all the prime numbers, then
$$p_{1}, p_{1}^{2} p_{2}, p_{1}^{2} p_{2}^{2} p_{3}, \cdots, p_{1}^{2} p_{2}^{2} \cdots p_{n-1}^{2} p_{n}, \cdots$$
satisfies the requirement.
To verify this claim, we denote the $n$-th number in the sequence as $a_{n}$. First, each $a_{n}$ i... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,144 |
Example 11 Proof: There are infinitely many positive integers $n$, such that $n \mid\left(2^{n}+2\right)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Example 11 Proof: There are infinitely many positive integers $n$, su... | At first glance, this problem seems very similar to Example 10, but it is actually much more challenging. We still use the method of inductive construction, with the key move being to strengthen the inductive hypothesis. Below is the proof: If $n$ satisfies
$$2|n, n|\left(2^{n}+2\right),(n-1) \mid\left(2^{n}+1\right),$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,146 |
1 Let $m$ be an integer greater than 4, and not a prime. Prove $m \mid (m-1)$!. | 1. Since $m$ is not a prime number, $m$ can be expressed as $m=a b$, where $14$, so $a>2$, and thus $a^{2}>2 a$, which means $m>2 a$. Therefore, $a$ and $2 a$ are two distinct terms in the sequence $1,2, \cdots, m-1$, and hence $(m-1)!$ is divisible by $a \cdot 2 a=2 m$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,147 |
2 Prove: A positive integer $n$ can be expressed as the sum of several consecutive positive integers (at least two) if and only if $n$ is not a power of 2. | 2. Let $n=x+(x+1)+\cdots+(x+k-1), x$ be a positive integer, $k \geqslant 2$. That is,
$$(2 x+k-1) k=2 n$$
If $n$ is a power of 2, then $k$ and $2 x-1+k$ are both powers of 2, but $2 x-1$ is odd, so it must be $k=1$, which contradicts the problem's condition.
Conversely, if $n$ is not a power of 2, let $n=2^{m-1}(2 t+... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,148 |
3 Prove: Any positive integer $n$ can be expressed as $a-b$, where $a$ and $b$ are positive integers, and the number of distinct prime factors of $a$ and $b$ is the same. | 3. When $n$ is even, we can take $a=2n, b=n$. If $n$ is odd, let $p$ be the smallest odd prime that does not divide $n$, then $p-1$ either has no odd prime factors (i.e., is a power of 2), or its odd prime factors all divide $n$. Therefore, $a=pn, b=(p-1)n$ have the number of distinct prime factors equal to the number ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,149 |
For any given integer $n \geqslant 3$, prove that there exists an $n$-term arithmetic sequence composed of positive integers (with a non-zero common difference), such that any two terms are coprime. | 4. The sequence $\{k \cdot n!+1\}(k=1, \cdots, n)$ meets the requirement. Suppose there exist $s, t(1 \leqslant s<t \leqslant n)$ such that $s \cdot n!+1$ and $t \cdot n!+1$ are not coprime, then there is a prime $p$ that divides both numbers, and thus divides their difference, i.e., $p \mid (t-s) n!$. Since $p$ is a p... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,150 |
5 Proof: For each $n \geqslant 2$, there exist $n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}$, such that $\left(a_{i}-\right.$ $\left.a_{j}\right) \mid\left(a_{i}+a_{j}\right)(1 \leqslant i, j \leqslant n, i \neq j)$. | 5. Using the inductive construction method. When $n=2$, we can take $a_{1}=1, a_{2}=2$. Assuming that for $n=k$, we already have $a_{1}, \cdots, a_{k}$ that meet the requirements, let $b_{0}$ be the least common multiple of $a_{1}, \cdots, a_{k}, a_{i}-a_{j}(1 \leqslant i, j \leqslant k, i \neq j)$, then the $k+1$ numb... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,151 |
Example 6 Given any $n>1$, prove: there exists a positive integer $a$, such that all numbers in $a^{a}+1, a^{a^{a}}+1, \cdots$ are divisible by $n$.
---
The translation maintains the original text's line breaks and format. | We note that if $a$ is odd, then $a^{a}, a^{a^{a}}, \cdots$ are all odd, and thus from (6) we know that $a^{a}+1, a^{a^{a}}+1=a^{\left(a^{a}\right)}+1, \cdots$ all have the factor $a+1$. Therefore, taking $a=2n-1$ satisfies the requirement in the problem. | a=2n-1 | Number Theory | proof | Yes | Yes | number_theory | false | 742,154 |
Example 1 Let $a, b, c, d$ be positive integers, prove: $a^{4b+d}-a^{4c+d}$ is divisible by 240. | To prove that since $240=2^{4} \times 3 \times 5$, we will separately prove that $a^{4 b+d}-a^{4 c+d}$ is divisible by 3, 5, and 16, thereby proving the conclusion (see the note in Example 5 of Unit 3).
First, we prove $3 \mid\left(a^{4 b+d}-a^{4 c+d}\right)$. By the result in (12) that $a^{2} \equiv 0,1(\bmod 3)$, it... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,155 |
Example 2 Let integers $a, b, c$ satisfy $a+b+c=0$, and let $d=a^{1999}+b^{1999}+c^{1999}$. Prove: $|d|$ is not a prime number. | To prove that $|d|$ has a non-trivial fixed divisor, we will use congruences here:
First, for any integer $u$, the number $u^{1999}$ has the same parity as $u$, i.e., $u^{1999} \equiv u(\bmod 2)$, hence $d \equiv a+b+c \equiv 0(\bmod 2)$, which means $2 \mid d$.
Furthermore, for any integer $u$, it is easy to verify ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,156 |
Example 3 Let integers $x, y, z$ satisfy
$$(x-y)(y-z)(z-x)=x+y+z,$$
Prove: $x+y+z$ is divisible by 27. | We will deduce from (1) that $x$, $y$, and $z$ must be pairwise congruent modulo 3, thus $27 \mid (x-y)(y-z)(z-x)$, and hence by (1) we know $27 \mid (x+y+z)$.
Proof by contradiction: First, assume that exactly two of $x$, $y$, and $z$ are congruent modulo 3. Without loss of generality, let $x \equiv y \pmod{3}$, but ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,158 |
Example 4 Let $n>1$, prove: $\underbrace{11 \cdots 1}_{n \uparrow 1}$ is not a perfect square. | Proof by contradiction, suppose there exists some $n>1$ and integer $x$, such that
$$\underbrace{11 \cdots 1}_{n \text { ones }}=x^{2}.$$
From (1), we know that $x$ is odd (this is obtained by taking (1) modulo 2, noting that $x^{2} \equiv x(\bmod 2)$). Furthermore, since $2 \nmid x$, it follows that $x^{2} \equiv 1(\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,159 |
Example 5 Use the digits $1, 2, 3, 4, 5, 6, 7$ to form seven-digit numbers, each digit being used exactly once. Prove: None of these seven-digit numbers is a multiple of another. | Prove that if there are two seven-digit numbers $a, b (a \neq b)$ such that
$$a = bc,$$
where $c$ is an integer greater than 1. Since the sum of the digits of $a$ and $b$ is $1+2+3+4+5+6+7 \equiv 1(\bmod 9)$, we have $a \equiv b \equiv 1(\bmod 9)$ (see Note 2 of Example 3 in Unit 1). Now, taking (1) modulo 9, we get $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,160 |
Example 6 The sequence $\left\{x_{n}\right\}$ is $1,3,5,11, \cdots$ and satisfies the recurrence relation
$$x_{n+1}=x_{n}+2 x_{n-1}, n \geqslant 2$$
The sequence $\left\{y_{n}\right\}$ is $7,17,55,161, \cdots$ and satisfies the recurrence relation
$$y_{n+1}=2 y_{n}+3 y_{n-1}, n \geqslant 2$$
Prove: These two sequence... | Consider modulo 8. First, we prove that the sequence $\left\{x_{n}\right\}$ modulo 8 is a periodic sequence
$$1,3,5,3,5, \cdots$$
Since $x_{2} \equiv 3, x_{3} \equiv 5(\bmod 8)$. If we already have
$$x_{n-1} \equiv 3, x_{n} \equiv 5(\bmod 8)$$
then by the recurrence formula (1), we get
$$\begin{array}{l}
x_{n+1}=x_{n... | proof | Algebra | proof | Yes | Yes | number_theory | false | 742,161 |
Example 7 Let $p$ be a given positive integer, try to determine the minimum positive value of $(2 p)^{2 m}-(2 p-1)^{n}$, where $m, n$ are any positive integers. | The smallest positive value sought is $(2 p)^{2}-(2 p-1)^{2}=4 p-1$. To prove this, we first note that by
and
$$\begin{array}{c}
(2 p)^{2}=(4 p-2) p+2 p \\
(2 p-1)^{2}=(4 p-2)(p-1)+(2 p-1)
\end{array}$$
it is easy to deduce that
$$(2 p)^{2 m}-(2 p-1)^{n} \equiv(2 p)-(2 p-1) \equiv 1(\bmod 4 p-2)$$
Furthermore, we pr... | 4p-1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,162 |
Example 8 Connect the vertices of a regular $n$-sided polygon to form a closed $n$-broken line. Prove: if $n$ is even, there are two parallel lines in the connections; if $n$ is odd, it is impossible to have exactly two parallel lines in the connections. | Prove that this is a geometric problem not suitable for a geometric solution, as it is related to the complete residue system modulo $n$. $\square$
Label the vertices in a counterclockwise order with the numbers $0,1, \cdots, n-1$. Let the closed broken line in the problem be $a_{0} \rightarrow a_{1} \rightarrow \cdot... | proof | Geometry | proof | Yes | Yes | number_theory | false | 742,163 |
Example 9 Let $n>3$ be an odd number, prove: after removing any one element from the $n$-element set $S=\{0,1, \cdots, n-1\}$, the remaining elements can always be divided into two groups, each with $\frac{n-1}{2}$ numbers, such that the sums of the two groups are congruent modulo $n$. | To prove a key point of the argument, for any $x \in S, x \neq 0$, the set $S \backslash\{x\}$ can be obtained from $T=$ $\{1,2, \cdots, n-1\}$ by the transformation
$$T+x(\bmod n)=\{a+x(\bmod n), a \in T\}$$
This reduces the problem to proving a special case: $T=S \backslash\{0\}$ can be divided into two groups, each... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,164 |
Example 7 Given any $n \geqslant 2$, prove: there exist $n$ distinct positive integers, such that the sum of any two of them divides the product of these $n$ numbers. | We arbitrarily take $n$ distinct positive integers $a_{1}, \cdots, a_{n}$, and select a (positive integer) parameter $K$, hoping that the product $K^{n} a_{1} \cdots a_{n}$ of $K a_{1}, \cdots, K a_{n}$ is divisible by the sum of any two terms $K a_{i}+K a_{j}$ $(1 \leqslant i, j \leqslant n, i \neq j)$. Since $n \geqs... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,165 |
Example 10 Proof: For any integer $n \geqslant 4$, there exists an $n$-degree polynomial
$$f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$$
with the following properties:
(1) $a_{0}, a_{1}, \cdots, a_{n-1}$ are all positive integers;
(2) For any positive integer $m$, and any $k(k \geqslant 2)$ distinct positive integ... | To prove the basic idea of this problem is to require that two integers cannot be equal, and congruence is precisely useful for this (see the note below Example 3).
We wish to construct a polynomial of degree \( n \) with positive integer coefficients and leading coefficient 1, such that for any integer \( a \), we ha... | proof | Algebra | proof | Yes | Yes | number_theory | false | 742,166 |
1 Label the vertices of a cube with the numbers +1 or -1, and label a face with a number that equals the product of the numbers at the four vertices of that face. Prove: the sum of the 14 numbers thus labeled cannot be 0. | 1. Let $S$ be the sum in question. If we change any -1 at a vertex to +1, then four numbers in $S$, denoted as $a, b, c, d$, will have their signs changed. Let $S^{\prime}$ represent the sum of the 14 numbers after the change. Since $a+b+c+d \equiv 0(\bmod 2)$, we have
$$S-S^{\prime}=2(a+b+c+d) \equiv 0(\bmod 4)$$
Rep... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 742,168 |
2 Find all positive integers $n$, such that the decimal integer composed of $n-1$ digits 1 and one digit 7 is a prime number.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 2. A positive integer $N$ composed of $n-1$ digits 1 and one digit 7 can be expressed in the form $N = A_{n} + 6 \times 10^{k}$, where $0 \leqslant k \leqslant n-1$, and $A_{n}$ is an integer composed of $n$ ones.
When $3 \mid n$, the sum of the digits of $A_{n}$ is divisible by 3, so $3 \mid A_{n}$, and thus $3 \mid ... | n = 1,2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 742,169 |
3 Let $p$ be a prime, $a \geqslant 2, m \geqslant 1, a^{m} \equiv 1(\bmod p), a^{p-1} \equiv 1\left(\bmod p^{2}\right)$. Prove: $a^{m} \equiv 1\left(\bmod p^{2}\right)$. | 3. From $a^{m} \equiv 1(\bmod p)$ we get $a^{m}=1+p x$. Therefore,
$$a^{p m}=(1+p x)^{p}=1+p^{2} x+C_{p}^{2} p^{2} x^{2}+\cdots \equiv 1\left(\bmod p^{2}\right) .$$
Also, $a^{p-1} \equiv 1\left(\bmod p^{2}\right)$, so $a^{(p-1) m} \equiv 1\left(\bmod p^{2}\right)$, thus $a^{m m} \equiv a^{m}\left(\bmod p^{2}\right)$. ... | a^{m} \equiv 1\left(\bmod p^{2}\right) | Number Theory | proof | Yes | Yes | number_theory | false | 742,170 |
4. Let $m$ be a given positive integer. Prove: the sequence $\left\{x_{n}\right\}$ defined by
$$x_{1}=x_{2}=1, x_{k+2}=x_{k+1}+x_{k}(k=1,2, \cdots)$$
contains at least one term among its first $m^{2}$ terms that is divisible by $m$. | 4. Without loss of generality, let $m>1$. We use $\bar{x}_{k}$ to denote the remainder when $x_{k}$ is divided by $m$. Consider the ordered pairs
$$\left\langle\bar{x}_{1}, \bar{x}_{2}\right\rangle,\left\langle\bar{x}_{2}, \bar{x}_{3}\right\rangle, \cdots,\left\langle\bar{x}_{n}, \bar{x}_{n+1}\right\rangle, \cdots$$
S... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,171 |
Fermat's Little Theorem: Let $p$ be a prime, and $a$ be any integer coprime to $p$, then
$$a^{p-1} \equiv 1(\bmod p)$$
Fermat's Little Theorem has a variant form, which is sometimes more applicable:
For any integer $a$, $a^{p} \equiv a(\bmod p)$.
(When $p \nmid a$, the two propositions are equivalent; when $p \mid a$,... | It is not difficult to give a proof of Fermat's Little Theorem by induction: It is easy to see that we only need to prove the proposition for $a=0$, $1, \cdots, p-1$. When $a=0$, the conclusion is obviously true. If we already have $a^{p}=a(\bmod p)$, then since $p \mid C_{p}^{i}(i=1,2, \cdots, p-1)$, we have
$$(a+1)^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,172 |
(2) Euler's Theorem Let $m>1$ be an integer, $a$ any integer coprime to $m$, and $\varphi(m)$ the Euler's function (see Unit 6), then
$$a^{\phi(m)} \equiv 1(\bmod m)$$ | Euler's theorem can be proved as follows: Take $r_{1}, r_{2}, \cdots, r_{\phi(m)}$ as a reduced residue system modulo $m$. Since $(a, m)=1$, it follows that $a r_{1}, a r_{2}, \cdots, a r_{\phi(m)}$ is also a reduced residue system modulo $m$ (see Unit 6). Because two complete (reduced) residue systems modulo $m$ are p... | a^{\phi(m)} \equiv 1(\bmod m) | Number Theory | proof | Yes | Yes | number_theory | false | 742,173 |
Example 1 Let $p$ be a given prime. Prove: The sequence $\left\{2^{n}-n\right\}(n \geqslant 1)$ has infinitely many terms divisible by $p$.
| Prove that the conclusion is obviously true when $p=2$. Let $p>2$, then by Fermat's Little Theorem, we have $2^{p-1} \equiv$ $1(\bmod p)$, hence for any positive integer $m$,
$$2^{m(p-1)} \equiv 1(\bmod p)$$
We take $m \equiv-1(\bmod p)$, then by (1), we get
$$2^{m(p-1)}-m(p-1) \equiv 1+m \equiv 0(\bmod p)$$
Therefor... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,175 |
1 Let $n$ and $k$ be positive integers, then there are exactly $\left[\frac{n}{k}\right]$ numbers in $1,2, \cdots, n$ that are divisible by $k$. | 1. Among $1,2, \cdots, n$, the numbers divisible by $k$ are $k, 2 k, \cdots, d k$, where the positive integer $d$ satisfies $d k \leqslant n$ but $(d+1) k>n$, thus $\frac{n}{k}-1<d \leqslant \frac{n}{k}$, i.e., $d=\left[\frac{n}{k}\right]$, hence there are $\left[\frac{n}{k}\right]$ numbers divisible by $k$. | \left[\frac{n}{k}\right] | Number Theory | proof | Yes | Yes | number_theory | false | 742,176 |
Example 2 Proof: The sequence $1,31,331,3331, \cdots$ contains infinitely many composite numbers. | Prove that since 31 is a prime, by Fermat's Little Theorem, we know that $10^{30} \equiv 1(\bmod 31)$, hence for any positive integer $k$, we have $10^{30 k} \equiv 1(\bmod 31)$, thus
$$\frac{.1}{3}\left(10^{30 k}-1\right) \equiv 0(\bmod 31)$$
This indicates that the number composed of $30 k$ threes is divisible by 31... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,177 |
Example 3 Proof: For any given positive integer $n$, there exist $n$ consecutive positive integers, each of which has a square factor greater than 1. | Proof: Since there are infinitely many primes, we can take $n$ distinct primes $p_{1}$, $p_{2}$, ..., $p_{n}$, and consider the system of congruences
$$x \equiv -i \left(\bmod p_{i}^{2}\right), i=1,2, \cdots, n$$
Since $p_{1}^{2}, p_{2}^{2}, \cdots, p_{n}^{2}$ are clearly pairwise coprime, by the Chinese Remainder The... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,178 |
Example 4 (1) Prove: For any positive integer $n$, there exist $n$ consecutive positive integers, none of which is a power number; $\square$
(2) Prove that there exist infinitely many distinct positive integers, such that neither they nor the sum of any distinct subset of them is a power number.
(Power number is define... | Proof (1) We prove that there exist $n$ consecutive positive integers, each of which has at least one prime factor that appears exactly once in its prime factorization, making the number not a power.
Since there are infinitely many primes, we can take $n$ distinct primes $p_{1}, \cdots, p_{n}$. Consider the system of ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,179 |
Example 5 Given a positive integer $n$, let $f(n)$ be the smallest positive integer such that $\sum_{k=1}^{f(n)} k$ is divisible by $n$. Prove that $f(n)=2n-1$ if and only if $n$ is a power of 2. | The first half of the problem is quite easy to prove. If $n=2^{m}$, then on the one hand,
$$\sum_{k=1}^{2 n-1} k=\frac{(2 n-1) \times 2 n}{2}=\left(2^{m+1}-1\right) \cdot 2^{m}$$
is divisible by $2^{m}=n$. On the other hand, if $r \leqslant 2 n-2$, then
$$\sum_{k=1}^{r} k=\frac{r(r+1)}{2}$$
is not divisible by $2^{m}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,180 |
Example 6 Let $f(x)$ be an integer-coefficient polynomial, and $a_{1}, \cdots, a_{m}$ be given non-zero integers, with the following property: for any integer $n$, the number $f(n)$ is divisible by one of the integers $a_{1}, \cdots, a_{m}$. Prove: there exists an $a_{i}(1 \leqslant i \leqslant m)$, such that for all i... | If one of $a_{1}, \cdots, a_{m}$ is $\pm 1$, then the conclusion is obviously true. Hereafter, assume each $a_{i} \neq \pm 1$. Suppose the conclusion is not true, then for each $a_{i}$, there is an integer $x_{i}$ such that $a_{i} \nmid f\left(x_{i}\right)$. $i=1, \cdots, m$. We will construct an integer $n$ such that ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,181 |
1 Let $p$ be an odd prime, $n=\frac{2^{2 p}-1}{3}$. Prove: $2^{n-1} \equiv 1(\bmod n)$. | 1. From the condition, we have
$$3(n-1)=4\left(2^{n-1}+1\right)\left(2^{n-1}-1\right) .$$
Given a prime $p>3$, by Fermat's Little Theorem, we have $p \mid\left(2^{p-1}-1\right)$. Combining this with (1), we deduce $2 p \mid(n-1)$, thus $\left(2^{2 p}-1\right) \mid\left(2^{n-1}-1\right)$. From the condition, we know $n... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,182 |
2 Let $m \geqslant 2, a_{1}, a_{2}, \cdots, a_{m}$ be given positive integers. Prove: there are infinitely many positive integers $n$, such that the number $a_{1} \cdot 1^{n}+a_{2} \cdot 2^{n}+\cdots+a_{m} \cdot m^{n}$ is composite. | 2. Clearly $a_{1}+2 a_{2}+\cdots+m a_{m}>1$, so there exists a prime $p$ that divides $a_{1}+2 a_{2}+\cdots+m a_{m}$. Take $n=k(p-1)+1, k=1,2, \cdots$, then for $1 \leqslant i \leqslant m$, if $p \nmid i$, by Fermat's Little Theorem we have
$$i^{n}=i \cdot\left(i^{k}\right)^{p-1} \equiv i(\bmod p)$$
If $p \mid i$, the... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,183 |
3 Let $m, n$ be positive integers, and $m \geqslant n$, with the property: the equation
$$(11 k-1, m)=(11 k-1, n)$$
holds for all positive integers $k$. Prove: $m=11^{r} n, r$ is a non-negative integer. | 3. Let $m=11^{i} u, n=11^{j} v$, where $i, j$ are non-negative integers, and $u, v$ are positive integers not divisible by 11. We prove that there must be $u=v$, from which it follows that $m=11^{i-j} n$. If $u \neq v$, without loss of generality, assume $u>v$. Since $(u, 11)=1$, by the Chinese Remainder Theorem, there... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,184 |
The smallest positive integer $r$ that satisfies $a^{r} \equiv 1(\bmod n)$ is called the order of $a$ modulo $n$. From the above argument, we know that $1 \leqslant r \leqslant n-1$. The following (1) indicates that the order of $a$ modulo $n$ has a very sharp property:
(1) Suppose $(a, n)=1$, and the order of $a$ modu... | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | not found | Number Theory | proof | Yes | Yes | number_theory | false | 742,185 |
Example 1 Let $n>1, n \mid\left(2^{n}+1\right)$, prove $3 \mid n$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Prove that $n$ is obviously odd. Let $p$ be the smallest prime divisor of $n$. We will prove that $p=3$, hence $3 \mid n$. Let the order of 2 modulo $p$ be $r$. From $2^{n} \equiv -1 \pmod{n}$, we know
$$2^{2n} \equiv 1 \pmod{p}$$
Since $p \geqslant 3$, Fermat's Little Theorem gives
$$2^{p-1} \equiv 1 \pmod{p}$$
From... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,186 |
4. Let $a, b$ be two coprime positive integers. Prove: There must exist positive integers $m, n$, such that $a b \mid a^{m}+b^{n}-1$. | 4. Hint: Take $m=\varphi(b), n=\varphi(a)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 742,189 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.