problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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26. Find all integer solutions $(x, y, z)$ that satisfy the following system of equations:
$$\left\{\begin{array}{l}
x+1=8 y^{2}, \\
x^{2}+1=2 z^{2} .
\end{array}\right.$$ | 26. First, $(x, y, z)=(-1,0, \pm 1)$ is a solution to the system of equations. Next, consider the solutions to the system of equations when $y \neq 0$. By eliminating $x$ from the system, we get
$$1+\left(2(2 y)^{2}-1\right)^{2}=2 z^{2},$$
which simplifies to $(2 y)^{4}+\left((2 y)^{2}-1\right)^{2}=z^{2}$. Therefore, ... | (x, y, z)=(-1,0, \pm 1),(7, \pm 1, \pm 5) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,580 |
27. Positive integers $a, b, c$ satisfy: $a^{2}+b^{2}=c^{2}$. Prove: $c-a$ and $c+a$ cannot both be the lengths of two sides of a right triangle. | 27. If $c-a$ and $c+a$ are the lengths of the two legs of a right triangle, and the length of the hypotenuse is $d$, then
$$\begin{aligned}
(c+a)^{4}+b^{4} & =(c+a)^{4}+\left(c^{2}-a^{2}\right)^{2} \\
& =(c+a)^{2}\left((c+a)^{2}+(c-a)^{2}\right) \\
& =(c+a)^{2} d^{2}
\end{aligned}$$
This contradicts the conclusion of ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,581 |
28. Prove: The indeterminate equation
$$x^{2} y^{2}=z^{2}\left(z^{2}-x^{2}-y^{2}\right)$$
has no positive integer solutions. | 28. Let $(x, y, z)$ be a positive integer solution to the original equation, then $z \mid x y$. Furthermore, we assume $\frac{x y}{z}$ is the smallest. At this point, it must be that $(x, y, z)=1$ (otherwise $\left(\frac{x}{d}, \frac{y}{d}, \frac{z}{d}\right)$ is also a solution to the equation, but when $d=(x, y, z)>1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,582 |
29. Does there exist a positive integer $m$ such that the indeterminate equation
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y z}=\frac{m}{x+y+z}$$
has infinitely many positive integer solutions $(x, y, z)$? | 29. There exists a positive integer $m$. For example, when $m=12$, the indeterminate equation
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y z}=\frac{12}{x+y+z}$$
has infinitely many positive integer solutions.
After simplifying and rearranging equation (8), we get
$$x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)+x+y+z-9 x y z=... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,583 |
30. Find all $n \in \mathbf{N}^{*}$, such that there exist $x, y, z \in \mathbf{N}^{*}$, satisfying:
$$n=\frac{(x+y+z)^{2}}{x y z}$$ | 30. Let the set of $n$ that meets the requirements be $S$. By taking $(x, y, z)=(9,9,9)$, $(4,4,8)$, $(3,3,3)$, $(2,2,4)$, $(1,4,5)$, $(1,2,3)$, $(1,1,2)$, $(1,1,1)$, we know $\{1,2,3,4,5,6,8,9\} \subseteq S$.
Now, let $n \in S$, and $x, y, z$ be the positive integers that satisfy $n=\frac{(x+y+z)^{2}}{x y z}$, such t... | S=\{1,2,3,4,5,6,8,9\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,584 |
31. Let $m, n$ be two positive integers of the same parity, and
$$\left(m^{2}-n^{2}+1\right) \mid\left(n^{2}-1\right)$$
Prove: $m^{2}-n^{2}+1$ is a perfect square. | 31. Let $\frac{n^{2}-1}{m^{2}+1-n^{2}}=k$, then $k \in \mathbf{N}$, and
$$(k+1)\left(n^{2}-1\right)=k m^{2}$$
Thus, $m^{2}+1-n^{2}=\frac{m^{2}}{k+1}$. Therefore, to prove the proposition, it suffices to show that $k+1$ is a perfect square.
For convenience, let $k+1=s, x=\frac{m+n}{2}, y=\frac{m-n}{2}$. Using the fact... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,585 |
Example 8 defines $\operatorname{rad}(n)$ as follows: $\operatorname{rad}(1)=1$, and for any positive integer $n$ greater than 1, $\operatorname{rad}(n)$ is the product of all distinct prime factors of $n$.
The sequence $a_{1}, a_{2}, \cdots$ is defined as follows:
$$a_{1} \in \mathbf{N}^{*}, a_{n+1}=a_{n}+\operatorna... | Proof
From the recursive formula of the sequence and the definition of $\operatorname{rad}(n)$, we know that for any $n \in \mathbf{N}^{*}$, we have $\operatorname{rad}\left(a_{n}\right) \mid \operatorname{rad}\left(a_{n+1}\right)$. Furthermore, we can obtain the following property: for any $m \in \mathbf{N}^{*}, m \ge... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,586 |
32. Let $\alpha, \beta, \gamma \in \mathbf{N}^{*}$, and $(1+\alpha \beta)(1+\beta \gamma)(1+\gamma \alpha)$ is a perfect square. Prove: $1+\alpha \beta, 1+\beta \gamma, 1+\gamma \alpha$ are all perfect squares. | 32. First, prove a lemma: If $\alpha, \beta, \gamma \in \mathbf{N}^{*}$, such that $\alpha \beta+1, \beta \gamma+1$, and $\gamma \alpha+1$ are all perfect squares, then there exists $\delta \in \mathbf{N}^{*}$, such that the product of any two of $\alpha, \beta, \gamma, \delta$ plus 1 is a perfect square.
First, consi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,587 |
33. Let $p$ be a prime, $p \equiv 1(\bmod 4)$. Prove: The indeterminate equation
$$x^{2}-p y^{2} \equiv-1(\bmod 4)$$
has integer solutions. | 33. Let $\left(x_{0}, y_{0}\right)$ be the fundamental solution of the Pell equation $x^{2}-p y^{2}=1$. Then, from
$$x_{0}^{2}-p y_{0}^{2} \equiv x_{0}^{2}-y_{0}^{2} \equiv 1(\bmod 4),$$
we know that $x_{0}$ is odd and $y_{0}$ is even. Therefore,
$$\frac{x_{0}-1}{2} \cdot \frac{x_{0}+1}{2}=\frac{1}{4}\left(x_{0}^{2}-1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,588 |
34. Prove: The indeterminate equation $x^{2}-34 y^{2}=-1$ has no integer solutions. | 34. Direct calculation shows that the fundamental solution of the Pell's equation $x^{2}-34 y^{2}=1$ is $\left(x_{0}, y_{0}\right)=$ $(35,6)$. If the equation $x^{2}-34 y^{2}=-1$ has integer solutions, let its fundamental solution be $(a, b)$, then it should be that $(a+b \sqrt{34})^{2}=x_{0}+y_{0} \sqrt{34}$ (this con... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,589 |
35. Find the positive integer tuples $(a, b, x, y)$ that satisfy the following condition:
$$(a+b)^{x}=a^{y}+b^{y}$$ | 35. When $a=b=1$, it is easy to see that the solution is $(a, b, x, y)=(1,1,1, y)$, where $y$ is any positive integer.
When $y=1$, the solution is $(a, b, 1,1)$, where $a, b$ are any positive integers.
When $y=2$, it is known that $x=1$, at this time $a+b=a^{2}+b^{2}$, so $a(a-1)=b(1-b) \geqslant 0$, which gives $a=b=... | (a, b, x, y)=(1,1,1, y),(a, b, 1,1),(2,1,2,3),(1,2,2,3),(2,2,2,3),\left(2^{t}, 2^{t}, t s+1,(t+1) s+1\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,590 |
36. Find all positive integer triples $(a, m, n)$ such that
$$\left(a^{m}+1\right) \mid(a+1)^{n}$$ | 36. Notice that when $a, m$ is equal to 1, it is easy to get the positive integer solutions $(a, m, n)=(1, m, n)$ or $(a, 1, n)$, where $a, m, n \in \mathbf{N}^{*}$.
Below, we discuss the case where $a, m$ are both greater than 1.
First, we prove a lemma: Let $u, v, l \in \mathbf{N}^{*}$, and $u \mid v^{l}$, then $u \... | (a, m, n)=(1, m, n), (a, 1, n), (2,3, r) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,591 |
Example 1 Find all positive integers $n$ such that $2^{n}-1$ is divisible by 7. | Since $2^{n}=(\underbrace{100 \cdots 0}_{n \uparrow})_{2}$, then $2^{n}-1=(\underbrace{11 \cdots 1}_{n \uparrow})_{2}$. And $7=(111)_{2}$, therefore, for $7 \mid\left(2^{n}-1\right)$, i.e., $(111)_{2} \mid(\underbrace{11 \cdots 1}_{n \uparrow})_{2}$, $n$ must be a multiple of 3. | n \text{ must be a multiple of 3} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,592 |
Example 2 Let $x>1$, and denote $a_{n}=\left[x^{n}\right], n=1,2, \cdots$, where $[y]$ represents the greatest integer not exceeding the real number $y$. Denote
$$S=0 . a_{1} a_{2} a_{3} \cdots,$$
which is a decimal fraction, indicating that the decimal representations of the positive integers $a_{1}, a_{2}, \cdots$ a... | Prove that there does not exist a real number $x (>1)$ such that $S$ is a rational number.
Our starting point is to prove: for any $x (>1)$ and $m \in \mathbf{N}^{*}$, there exists $n \in \mathbf{N}^{*}$, such that $\left[x^{n}\right]$ has a sequence of $m$ consecutive zeros in its decimal representation.
If the above... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,593 |
Example 3 Proof: At most only a finite number of positive integers cannot be represented as the sum of squares of several different positive integers.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Example 3 Proof: At most ... | Proof:
Take a positive integer $x$ such that both $x$ and $2x$ can be expressed as the sum of squares of two different positive integers (the reason for finding such a positive integer can be seen from the subsequent representation). Then for a sufficiently large positive integer $p$, set
$$p \equiv r(\bmod 4x), \quad ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,594 |
Example 4 Let the decimal representations of real numbers $A, B$ be:
$A=0 . a_{1} a_{2} \cdots a_{k}>0, B=0 . b_{1} b_{2} \cdots b_{k}>0$ (here $a_{k}, b_{k}$ can be 0).
Let $S$ be the number of numbers $0 . c_{1} c_{2} \cdots c_{k}$ satisfying
$$\text { 0. } c_{1} c_{2} \cdots c_{k}<A, 0 . c_{k} c_{k-1} \cdots c_{1}<B... | To prove that the number $0 . c_{1} c_{2} \cdots c_{k}$ satisfies (6) if and only if: there exists $t \in\{1,2, \cdots, k\}$, such that
$$\begin{array}{c}
0 . c_{1} c_{2} \cdots c_{t-1}=0 . a_{1} a_{2} \cdots a_{t-1}, c_{t}<a_{t} \\
\text { 0. } c_{k} c_{k-1} \cdots c_{t+1}<0 . b_{1} b_{2} \cdots b_{k-t}
\end{array}$$
... | 10^{k} A B-9 k \leqslant S \leqslant 10^{k} A B+9 k | Combinatorics | proof | Yes | Yes | number_theory | false | 741,595 |
Example 5 The sequence $\left\{a_{n}\right\}$ is defined as follows: if the number of digit 1s in the binary representation of $n$ is even, then $a_{n}=0$; otherwise, $a_{n}=1$. Prove: there do not exist $k, m \in \mathbf{N}^{\cdot}$, such that for any $0 \leqslant j \leqslant$ $m-1$, we have
$$a_{k+j}=a_{k+m+j}=a_{k+2... | Proof
From the conditions, we know that
\[
\left\{\begin{array}{l}
a_{2 n} \equiv a_{n}(\bmod 2), \\
a_{2 n+1} \equiv a_{2 n}+1 \equiv a_{n}+1(\bmod 2).
\end{array}\right.
\]
Assume there exist \( k, m \in \mathbf{N}^{*} \) such that for any \( 0 \leqslant j \leqslant m-1 \), (7) holds, and set \( k+m \) to be the sma... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,596 |
1. Let $a, b \in \mathbf{N}^{*}$, and $\frac{a b}{a+b} \in \mathbf{N}^{*}$. Prove: $(a, b)>1$. | 1. If $(a, b)=1$, then it leads to $(a+b) \mid b$, which is a contradiction. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,597 |
Property 3 If $x, y \in \mathbf{R}$, then
$$[x]+[y] \leqslant[x+y] \leqslant[x]+[y]+1$$ | Proof: From conclusion (2), we know $[x] \leqslant x, [y] \leqslant y$, hence
$$[x]+[y] \leqslant x+y,$$
Therefore, $[x]+[y] \leqslant[x+y]$ (here we use the definition of the floor function).
From conclusion (2), we also know
$$x<[x]+1, y<[y]+1,$$
Thus, $x+y<[x]+[y]+2$, and consequently
$$[x+y]<[x]+[y]+2,$$
Hence, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,602 |
Example 2 Let $n \in \mathbf{N}^{*}$, consider the region in the coordinate plane bounded by the inequalities $y>0, 0<x \leqslant n, y \leqslant$ $\sqrt{x}$. Try to express the number of integer points within this region using $n$ and $a(=[\sqrt{n}])$. | From property 6, we know that the number of integer points in the region sought in the problem is
$$T=\sum_{k=1}^{n}[\sqrt{k}]$$
We use the following method to calculate $T$.
For a proposition $P$, we introduce the notation $[P]$: if $P$ holds, then $[P]=1$; if $P$ does not hold, then $[P]=0$. Thus, we have
$$\begin{a... | (n+1) a-\frac{1}{6} a(a+1)(2 a+1) | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,607 |
2. Let $m, n \in \mathbf{N}^{*}, m$ be an odd number. Prove: $\left(2^{m}-1,2^{n}+1\right)=1$.
| 2. Let $\left(2^{m}-1,2^{n}+1\right)=d$, then
$$1 \equiv\left(2^{m}\right)^{n}=\left(2^{n}\right)^{m} \equiv(-1)^{m}=-1(\bmod d)$$
This leads to $d \mid 2$. Combining this with $d$ being odd, we conclude $d=1$. | 1 | Number Theory | proof | Yes | Yes | number_theory | false | 741,608 |
Example 3 Let the function $f: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*}$ be defined as follows: Let
$$\frac{(2 n)!}{n!(n+1000)!}=\frac{A(n)}{B(n)}$$
where $n \in \mathbf{N}^{*}, A(n), B(n)$ are coprime positive integers. If $B(n)=1$, then $f(n)=1$; if $B(n)>1$, then $f(n)$ is the largest prime factor of $B(n)$.
Prov... | Proof:
The idea of the proof is to find a constant such that for any $n \in \mathbf{N}^{*}$, the product of $\frac{A(n)}{B(n)}$ and this constant is a positive integer, thereby deducing that $f$ is a bounded function.
First, we prove a lemma: For any non-negative real numbers $x, y$,
$$[2 x]+[2 y] \geqslant[x]+[x+y]$$... | 1999 | Number Theory | proof | Yes | Yes | number_theory | false | 741,609 |
Example 4 Find all pairs of positive real numbers $(a, b)$ such that for any $n \in \mathbf{N}^{*}$, we have
$$[a[b n]]=n-1$$ | From (5), we know that for any $n \in \mathbf{N}^{*}$, we have
$$\begin{aligned}
n-1 & =[a[b n]] \leqslant a[b n] \leqslant a b n1$, then for the irrational number $b$, there exists $n \in \mathbf{N}^{*}$ such that $\{n b\}>\frac{1}{a}$, which contradicts (6). Therefore, $a<1$.
In summary, the positive real numbers $(... | (a, b) \mid a b=1,0<a<1, a, b \text{ are irrational numbers} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,610 |
Example 5 Let the real number $r \geqslant 1$, satisfy: for any $m, n \in \mathbf{N}^{*}$, if $m \mid n$, then $[m r] \mid [n r]$. Prove: $r$ is an integer.
保留源文本的换行和格式，翻译结果如下：
Example 5 Let the real number $r \geqslant 1$, satisfy: for any $m, n \in \mathbf{N}^{*}$, if $m \mid n$, then $[m r] \mid [n r]$. Prove: $r$... | Proof:
If $r \notin \mathbf{Z}$, combined with $r \geqslant 1$, it follows that there exists $a \in \mathbf{N}^{*}$ such that $a r \notin \mathbf{Z}$, and $[a r]>1$. At this point, there exists a unique $k \in \mathbf{N}^{*}$ such that
$$\frac{1}{k+1} \leqslant\{a r\}<\frac{1}{k} \text {. }$$
Thus,
$$[(k+1) a r]=(k+1)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,611 |
Example 6 Let $m, n$ be positive integers, find the necessary and sufficient condition for the equation
$$\sum_{i=0}^{m n-1}(-1)\left[\frac{i}{m}\right]+\left[\frac{i}{n}\right]=0$$
to hold. | Let
$$f_{m, n}(i)=\left[\frac{i}{m}\right]+\left[\frac{i}{n}\right], S(m, n)=\sum_{i=0}^{m n-1}(-1)^{f_{m, n} n^{(i)}},$$
The problem is equivalent to finding the necessary and sufficient condition for $S(m, n)=0$.
Case 1: Both $m$ and $n$ are odd. In this case, $S(m, n)$ is the sum of an odd number $(m n)$ of odd num... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,612 |
Example 7 Find the largest real number $c$ such that for any $n \in \mathbf{N}^{*}$, we have $\{\sqrt{2} n\}>\frac{c}{n}$.
| Notice that, for any $n \in \mathbf{N}^{*}$, let $k=[\sqrt{2} n]$, then
$$\begin{aligned}
\{\sqrt{2} n\} & =\sqrt{2} n-[\sqrt{2} n]=\sqrt{2} n-k=\frac{2 n^{2}-k^{2}}{\sqrt{2} n+k} \\
& \geqslant \frac{1}{\sqrt{2} n+k}>\frac{1}{\sqrt{2} n+\sqrt{2} n}=\frac{1}{2 \sqrt{2} n}
\end{aligned}$$
Here we use the fact that $\sq... | \frac{1}{2 \sqrt{2}} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,613 |
Example 8 Let the sequence of non-negative integers $a_{1}, a_{2}, \cdots, a_{2000}$ satisfy
$$a_{i}+a_{j} \leqslant a_{i+j} \leqslant a_{i}+a_{j}+1,1 \leqslant i<j \leqslant 2000, i+j \leqslant 2000 .$$
Prove: There exists a real number $x$, such that for any $n \in\{1,2, \cdots, 2000\}$, we have $a_{n}=[n x]$. | Proof
Let $I_{n}=\left[\frac{a_{n}}{n}, \frac{a_{n}+1}{n}\right), n=1,2, \cdots, 2000$. If there exists a real number $x \in \bigcap_{n=1}^{2000} I_{n}$, then the proposition is proved.
For this, let $L=\max _{1 \leqslant n \leqslant 2000}\left\{\frac{a_{n}}{n}\right\}, U=\min _{1 \leqslant n \leqslant 2000}\left\{\fr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,614 |
Example 9 Given three columns of numbers $A, B, C$ as follows: Column $A$ consists of numbers of the form $10^{k}$ in decimal, where $k \in \mathbf{N}^{*}$, and columns $B$ and $C$ are the representations of the numbers in column $A$ in binary and quinary, respectively.
\begin{tabular}{ccc}
$A$ & $B$ & $C$ \\
10 & 1010... | Notice that the number $10^{k}$ in binary representation has exactly $\left[k \log _{2} 10\right]+1$ digits, and in quinary representation, it has exactly $\left[k \log _{5} 10\right]+1$ digits. Therefore, we only need to prove that the sequences $\left\{\left[k \log _{2} 10\right]\right\}$ and $\left\{\left[k \log _{5... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,616 |
Example 10 Find all pairs of positive integers $(m, n)$ such that
$$[n \sqrt{2}]=[2+m \sqrt{2}] .$$ | Let $(m, n)$ be a pair of positive integers satisfying (14), then $n>m$.
Notice
$$\begin{aligned}
{[(m+3) \sqrt{2}] } & =[m \sqrt{2}+3 \sqrt{2}] \geqslant[m \sqrt{2}]+[3 \sqrt{2}] \\
& =[m \sqrt{2}+4]
\end{aligned}$$
Therefore, $n=m+1$ or $n=m+2$.
If $n=m+1$, since $\frac{1}{\sqrt{2}}+\frac{1}{2+\sqrt{2}}=1$, by the B... | (m, n)=([(\sqrt{2}+1) h],[(\sqrt{2}+1) h]+1) \text{ or } \left\{\begin{array}{l} m=3 k+2[\sqrt{2} k], \\ n=3 k+2+2[\sqrt{2} k] \end{array}\right.} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,617 |
3. Let $n, k \in \mathbf{N}^{*}, n>k$. Prove: $\mathrm{C}_{n}^{k}, \mathrm{C}_{n+1}^{k}, \cdots, \mathrm{C}_{n+k}^{k}$ have a greatest common divisor equal to 1. | 3. Let the greatest common divisor be $d$, then $d \mid\left(\mathrm{C}_{n+j}^{k}-\mathrm{C}_{n+j-1}^{k}\right), j=1,2, \cdots, k$, i.e., $d \mid \mathrm{C}_{n+j-1}^{k-1}$, thus $d \mid\left(\mathrm{C}_{n-1}^{k-1}, \mathrm{C}_{n}^{k-1}, \cdots, \mathrm{C}_{n+k-1}^{k-1}\right)$. By tracing back step by step, we get $d \... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 741,619 |
Example 3 Let $n \in \mathbf{N}^{*}$. Prove: $n$ can be expressed as the sum of squares of 2 integers if and only if: if $q$ is a prime factor of $n$, and $q \equiv 3(\bmod 4)$, then in the prime factorization of $n$, the exponent of $q$ is even. | Proof
First, we prove the sufficiency. Consider the following fact: If $a, b$ can both be expressed as the sum of squares of two integers, then $ab$ can also be expressed as the sum of squares of two integers. Indeed, if $a=x^{2}+y^{2}, b=s^{2}+t^{2}$, then
$$a b=\left(x^{2}+y^{2}\right)\left(s^{2}+t^{2}\right)=(x s+y ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,621 |
Example 4 Let $p \equiv 1(\bmod 4), p$ be a prime. Prove: There exists a prime $q<\sqrt{p}$, such that $q^{\rho-1} \not \equiv 1\left(\bmod p^{2}\right)$ | To prove that using the conclusion of Example 1, there exist \( x, y \in \mathbf{N}^{*} \) such that \( p = x^{2} + y^{2} \). To prove the proposition, we first prove the following:
\[ x^{\rho-1} \not\equiv 1 \pmod{p^{2}} \text{ or } y^{\rho-1} \not\equiv 1 \pmod{p^{2}} \]
must hold for at least one of them.
In fact... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,622 |
Example 5 Find all positive integers $n$ such that the indeterminate equation
$$n=x_{1}^{2}+x_{2}^{2}+\cdots+x_{5}^{2}$$
has a unique solution in integers satisfying $0 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{5}$. | Using Lagrange's four-square theorem, we know that when $n \geqslant 17$, there exists the following representation:
$$\begin{aligned}
n & =x_{0}^{2}+y_{0}^{2}+z_{0}^{2}+w_{0}^{2}, \\
n-1^{2} & =x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+w_{1}^{2}, \\
n-2^{2} & =x_{2}^{2}+y_{2}^{2}+z_{2}^{2}+w_{2}^{2}, \\
n-3^{2} & =x_{3}^{2}+y_{3}... | 1,2,3,6,7,15 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,623 |
Example 7 Proof: For any $n \in \mathbf{N}^{*}$, there exist $n$ consecutive positive integers, each of which cannot be expressed as the sum of squares of two integers. | Proof
For any $n \in \mathbf{N}^{*}$, we take $n$ distinct prime numbers $p_{1}, p_{2}, \cdots, p_{n}$ that are congruent to 3 modulo 4. By the Chinese Remainder Theorem, there exists $x \in \mathbf{N}^{*}$ such that
$$x \equiv p_{i}-i\left(\bmod p_{i}^{2}\right), i=1,2, \cdots, n$$
simultaneously.
Notice that, for $1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,625 |
Example 1 Proof: There exist infinitely many "deficient numbers," and there also exist infinitely many "abundant numbers." | Prove that in fact, all prime numbers are "deficient numbers," and thus "deficient numbers" are infinite in number.
Notice that, $\sigma(12)=28$ is an "abundant number," and if $a$ is an "abundant number," $b \in \mathbf{N}^{*}$, then
$$\sigma(a b) \geqslant b \sum_{d \mid a} a=b \sigma(a)>2 a b,$$
hence $a b$ is an "... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,626 |
The Even Perfect Number Theorem (Euler) states that a positive even number $n$ is a perfect number if and only if
$$n=2^{p-1}\left(2^{p}-1\right),$$
where $p$ and $2^{p}-1$ are both prime numbers. | Prove that if $n=2^{p-1}\left(2^{p}-1\right)$, where $p$ and $2^{p}-1$ are both primes, then
$$\begin{aligned}
\sigma(n) & =\sigma\left(2^{p}-1\right) \sigma\left(2^{p-1}\right)=\left(1+2^{p}-1\right)\left(1+2+\cdots+2^{p-1}\right) \\
& =2^{p}\left(2^{p}-1\right)=2 n
\end{aligned}$$
Thus, $n$ is an even perfect number... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,627 |
Example 2 The two smallest perfect numbers are 6 and 28, which are multiples of 3 and 7, respectively. Let $n$ be a perfect number, prove:
(1) If $3 \mid n$, and $n>6$, then $9 \mid n$;
(2) If $7 \mid n$, and $n>28$, then $7^{2} \mid n$. | Proof
(1) If $9 \gamma_{n}$, then $\left(3, \frac{n}{3}\right)=1$, hence,
$$2 n=\sigma(n)=\sigma(3) \sigma\left(\frac{n}{3}\right)=4 \sigma\left(\frac{n}{3}\right)$$
This indicates that $n$ is even, thus, $6 \mid n$. Combining with $n>6$, we get
$$\sigma(n) \geqslant n+\frac{n}{2}+\frac{n}{3}+\frac{n}{6}+1=2 n+1,$$
w... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,628 |
4. (1) Find all positive integers $n$, such that there exist $a, b \in \mathbf{N}^{*}$, satisfying: $[a, b]=n!$, $(a, b)=1998 ;$
(2) Under the condition that (1) holds, to ensure that the number of pairs of positive integers $(a, b)$ with $a \leqslant b$ does not exceed 1998, what condition should $n$ satisfy? | 4. (1) From $(a, b)=2 \times 3^{3} \times 37$, we know $37 \mid[a, b]$, which means $37 \mid n$, so $n \geqslant 37$. When $n \geqslant 37$, $1998 \mid n!$, at this time, taking $a=1998, b=n!$ will suffice. Therefore, the positive integers $n$ that satisfy the condition are all positive integers not less than 37.
(2) L... | 37 \leqslant n \leqslant 40 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,629 |
Example 3 Proof: If $n$ is an odd perfect number; then
$$n \equiv 1(\bmod 4) \text {. }$$ | Proof
If $n \equiv 3(\bmod 4)$, and $\sigma(n)=2 n$, then $n$ is not a perfect square. At this time,
$$2 n=\sum_{d \mid n} d=\sum_{d \mid n, d<\sqrt{n}}\left(d+\frac{n}{d}\right)$$
In the right-hand side of (2), the product of each term $d$ and $\frac{n}{d}$ is $n \equiv 3(\bmod 4)$, so among these two odd numbers, one... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,630 |
Example 4 Proof: There does not exist a perfect number of the form $p^{a} \cdot q^{b} \cdot r^{c}$, where $p, q, r$ are pairwise distinct odd primes, and $a, b, c \in \mathbf{N}^{*}$. | Proof
If there exists a perfect number of the form $p^{a} \cdot q^{b} \cdot r^{c}$, without loss of generality, let $3 \leqslant p < q < r$.
This requires $p=3, q=5, r \in\{7,11,13\}$.
From (3), we have
$$\frac{3^{a+1}-1}{2} \cdot \frac{5^{b+1}-1}{4} \cdot \frac{r^{c+1}-1}{r-1}=2 \cdot 3^{a} \cdot 5^{b} \cdot r^{c}$$
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,631 |
Example 5 Proof: Any two consecutive positive integers cannot both be perfect numbers. | Proof
Let $n$ be an even perfect number. By the theorem of even perfect numbers, we know that
$$n=2^{p-1}\left(2^{p}-1\right),$$
where $p$ and $2^{p}-1$ are both primes. We will prove that $n-1$ and $n+1$ are not perfect numbers.
When $p=2$, $n=6$, and it can be directly verified that 5 and 7 are not perfect numbers.
... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,632 |
Example 6 Proof: If there exists $n \in \mathbf{N}^{*}$, such that $\frac{\sigma(n)}{n}=\frac{5}{3}$, then $5 n$ is an odd perfect number. | Notice, if $m \mid n$, then by
$$\frac{\sigma(n)}{n}=\frac{1}{n} \sum_{d \mid n} \frac{n}{d}=\sum_{d \mid n} \frac{1}{d} \geqslant \sum_{d \mid m} \frac{1}{d}$$
we know that $\frac{\sigma(n)}{n} \geqslant \frac{\sigma(m)}{m}$.
Now suppose $n$ satisfies: $3 \sigma(n)=5 n$, then $3 \mid n$. If $n$ is even, then $6 \mid ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,633 |
Example 1 Let $m$ be a given positive integer, and each term of the sequence $\left\{a_{n}\right\}$ is a positive integer, and for any $n \in \mathbf{N}^{*}$, we have $0<a_{n+1}-a_{n} \leqslant m$. Prove: There exist infinitely many pairs of positive integers $(p, q)$, such that $p<q$, and $a_{p} \mid a_{q}$. | Proof
First, we prove that there exists a pair $(p, q)$ that satisfies the condition.
Consider the following number table:
$$\begin{array}{l}
x_{0,1}, x_{0,2}, \cdots, x_{0, m} \\
x_{1,1}, x_{1,2}, \cdots, x_{1, m} \\
\cdots \\
x_{m, 1}, x_{m, 2}, \cdots, x_{m, m}
\end{array}$$
where $x_{0,1}=a_{1}, x_{0, j} \rightlef... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,634 |
Example 3 Let $k, n \in \mathbf{N}^{*}$, and let $\sigma_{k}(n)$ denote the sum of the $k$-th powers of all positive divisors of $n$. Prove: When $k \geqslant 2$, there always exist infinitely many $n \in \mathbf{N}^{*}$ such that $n \mid \sigma_{k}(n)$. | Proof Consider the case when $k=1$, which requires $n \mid \sigma(n)$, an unsolved problem. In particular, whether there are infinitely many perfect numbers $(\sigma(n)=2n)$ involves some famous conjectures. By comparing this case, we arrive at the conclusion of this problem.
We use a recursive method to construct.
Ta... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,636 |
Example 4 We call a positive integer of the form $n^{k}$ a military number, where $n, k \in \mathbf{N}^{*}$, and $k>1$. Question: Does there exist a set $S$ consisting of 1000 positive integers, such that the sum of the elements of any non-empty subset of $S$ is a military number? | The set $S$ that meets the requirements exists.
Although $\{1,2, \cdots, 1000\}$ does not meet the requirements, there exists $d \in \mathbf{N}^{*}$ such that $S=\{d$, $2 d, \cdots, 1000 d\}$ meets the requirements. To prove this conclusion, we establish the following lemma:
For any $m \in \mathbf{N}^{*}$, there exist... | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,637 |
Example 5 Let the set of all integer points (points with integer coordinates) in the plane be denoted as $S$. It is known that for any $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ in $S$, there exists another point $P$ in $S$ such that the segments $A_{i} P(i=1,2, \cdots, n)$ do not contain any points from $S$ internally. ... | The maximum value of the required $n$ is 3.
On one hand, when $n \geqslant 4$, take 4 points $A(0,0), B(1,0), C(0,1)$, $D(1,1)$ in $S$ (the remaining $n-4$ points are chosen arbitrarily), then for any other point $P$ in $S$, analyzing the parity of the coordinates of $P$, there are only 4 cases: (odd, odd), (odd, even)... | 3 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,638 |
Example 6 Let $p$ be an odd prime. Prove: from any $2 p-1$ integers, one can select $p$ numbers, the sum of which is a multiple of $p$.
The text is translated while preserving the original line breaks and format. | Let the set of the given $2 p-1$ numbers be $S=\left\{x_{1}, x_{2}, \cdots, x_{2 p-1}\right\}$. If the proposition does not hold, then for any $p$-element subset $\left\{x_{i_{1}}, x_{i_{2}}, \cdots, x_{i_{p}}\right\}$ of $S$, we have $x_{i_{1}}+x_{i_{2}}+\cdots+x_{i_{p}} \not \equiv 0(\bmod p)$. Thus, using Fermat's L... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,639 |
Example 1 Let $a, b, n$ be given positive integers, and it is known that for any $k \in \mathbf{N}^{*}(k \neq b)$, we have $(b-k) \mid\left(a-k^{n}\right)$. Prove: $a=b^{n}$. | Notice that, for any $k \in \mathbf{N}^{*} (k \neq b)$, we have
$$b^{n}-k^{n}=(b-k)\left(b^{n-1}+b^{n-2} k+\cdots+k^{n-1}\right),$$
thus $(b-k) \mid\left(b^{n}-k^{n}\right)$. Combining this with $(b-k) \mid\left(a-k^{n}\right)$, we get
$$(b-k) \mid\left(\left(a-k^{n}\right)-\left(b^{n}-k^{n}\right)\right),$$
which me... | a=b^{n} | Number Theory | proof | Yes | Yes | number_theory | false | 741,640 |
5. Let $a, b, m, n \in \mathbf{N}^{*}$, satisfying $(a, b)=1$ and $a>1$. Prove: If
$$\left(a^{m}+b^{m}\right) \mid\left(a^{n}+b^{n}\right)$$
then $m \mid n$. | 5. Using $a^{n}+b^{n}=\left(a^{m}+b^{m}\right)\left(a^{n-m}+b^{n-m}\right)-a^{m} b^{m}\left(a^{n-2 m}+b^{n-2 m}\right)$, we know that when $n \geqslant 2 m$, we have $\left(a^{n}+b^{n}\right) \mid\left(a^{n-2 m}+b^{n-2 m}\right)$ (using $\left.(a, b)=1\right)$, by reducing the degree to $n<2 m$, we can prove that $m \m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,641 |
1. Let $S(n)$ denote the sum of the digits of the positive integer $n$ in its decimal representation.
(1) Prove that for any $n \in \mathbf{N}^{*}$, we have $S(n) \leqslant 2 S(n) \leqslant 10 S(2 n)$;
(2) Prove that for any $m \in \mathbf{N}^{*}$, there exists $n \in \mathbf{N}^{*}$ such that $S(n)=m S(3 n)$. | 1. (1) Using $S(a+b) \leqslant S(a)+S(b)$ for any $a, b \in \mathbf{N}^{*}$, we know that $S(2 n) \leqslant 2 S(n)$.
On the other hand, direct verification shows that for $n \in\{0,1,2, \cdots, 9\}$, we have $S(n) \leqslant 5 n$. Now assume that for any $n<m\left(m \in \mathbf{N}^{*}, m \geqslant 10\right)$, we have $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,643 |
2. In decimal notation, how many $m \in\{1,2, \cdots, 2009\}$ are there such that there exists $n \in \mathbf{N}^{*}$, satisfying: $S\left(n^{2}\right)=m$? Here $S(x)$ denotes the sum of the digits of the positive integer $x$. | 2. Notice that, perfect squares $\equiv 0,1,4,7(\bmod 9)$, and for $k \in \mathbf{N}^{*}$, let $n=$ $\underbrace{11 \cdots 1} \underbrace{22 \cdots 25}$, then $S(n)=3 k+4$, and
$k-1$ 个 $\underbrace{2+0}_{k \text { 个 }}$
$$\begin{aligned}
n & =\frac{10^{k-1}-1}{9} \times 10^{k+1}+\frac{2 \times\left(10^{k}-1\right)}{9} ... | 893 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,644 |
3. Do there exist 19 different positive integers $a_{1}, a_{2}, \cdots, a_{19}$, such that $a_{1}+a_{2}+\cdots+$ $a_{19}=1999$, and $S\left(a_{1}\right)=S\left(a_{2}\right)=\cdots=S\left(a_{19}\right)$? Here $S(n)$ denotes the sum of the digits of $n$ in its decimal representation. | 3. If it exists, then from $\frac{1999}{19}<106$, it follows that among $a_{1}, a_{2}, \cdots, a_{19}$, there must be one that is no greater than 105, so $S\left(a_{1}\right)=S\left(a_{2}\right)=\cdots=S\left(a_{19}\right)<18$.
Using $a_{i} \equiv S\left(a_{i}\right)(\bmod 9)$, we know that $19 S\left(a_{i}\right) \eq... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,645 |
4. Let the set $T$ represent the collection of all non-negative integers that have at most $n$ digits in their decimal representation, and let $T_{k}$ represent the set of all elements in $T$ whose digit sum is less than $k$. For which $n \in \mathbf{N}^{*}$ does there exist a $k \in \mathbf{N}^{*}$ such that $|T|=2\le... | 4. When $n$ is odd, take $k=\frac{1}{2}(9 n+1)$, and make the following correspondence:
$$x=\overline{a_{1} a_{2} \cdots a_{n}} \leftrightarrow y=\overline{\left(9-a_{1}\right)\left(9-a_{2}\right) \cdots\left(9-a_{n}\right)},$$
It can be known that $S(x)+S(y)=9 n$, so $S(x)k$. In this case, $|T|=2\left|T_{k}\right|$
... | proof | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,646 |
5. Let $m$ be a given positive integer, for any $n \in \mathbf{N}^{*}$, use $S_{m}(n)$ to denote the sum of the $m$-th powers of the digits of $n$ in its decimal representation, for example $S_{3}(172)=1^{3}+7^{3}+2^{3}=352$. For any $n_{0} \in \mathbf{N}^{*}$, define $n_{k}=S_{m}\left(n_{k-1}\right), k=1,2, \cdots$.
(... | 5. If $n \geqslant 10^{m+1}$, then there exists $p \in \mathbf{N}^{*}, p \geqslant m+1$, such that $10^{p} \leqslant n<10^{p+1}$, at this time
$$S_{m}(n) \leqslant(p+1) \times 9^{m}<9^{p}+C_{p}^{1} \cdot 9^{p-1}<(9+1)^{p}=10^{p} \leqslant n,$$
Therefore, if $n_{k} \geqslant 10^{m+1}$, then $n_{k+1}<n_{k}$.
On the othe... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,647 |
6. Let \( x=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{100000}} \). Find the value of \([x]\). | 6. Notice that, when $k \geqslant 2$, we have
$$\begin{array}{l}
\frac{1}{\sqrt{k}}=\frac{2}{2 \sqrt{k}}\frac{2}{\sqrt{k}+\sqrt{k+1}}=2(\sqrt{k+1}-\sqrt{k}) .
\end{array}$$
Therefore,
$$\begin{aligned}
x & =1+\sum_{k=2}^{10^{6}} \frac{1}{\sqrt{k}}2 \sum_{k=1}^{10^{6}}(\sqrt{k+1}-\sqrt{k}) \\
& =2\left(\sqrt{10^{6}+1}-... | 1998 | Calculus | math-word-problem | Yes | Yes | number_theory | false | 741,648 |
7. Let $n$ be a positive integer greater than 1. Prove: $8 \mid\left(\left[(\sqrt[3]{n}+\sqrt[3]{n+2})^{3}\right]+1\right)$. | 7. To prove: When $n \geqslant 2$, we have $\left[(\sqrt[3]{n}+\sqrt[3]{n+2})^{3}\right]=8 n+7$.
In fact, when $n \geqslant 2$, we have
$$\begin{aligned}
x & =(\sqrt[3]{n}+\sqrt[3]{n+2})^{3}=n+3 \sqrt[3]{n^{2}(n+2)}+3 \sqrt[3]{n(n+2)^{2}}+n+2 \\
& =2 n+2+3 \sqrt[3]{n^{2}(n+2)}+3 \sqrt[3]{n(n+2)^{2}} \\
& n+(n+1)=2 n+1... | proof | Algebra | proof | Yes | Yes | number_theory | false | 741,649 |
8. Prove: For any $n \in \mathbf{N}^{*}$, we have $[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4 n+2}]$. | 8. Notice that,
$$(\sqrt{n}+\sqrt{n+1})^{2}=n+2 \sqrt{n(n+1)}+n+1=2 n+1+2 \sqrt{n(n+1)},$$
and $n<\sqrt{n(n+1)}<n+\frac{1}{2}$, so for any $n \in \mathbf{N}^{\cdot}$, we have
$$\sqrt{4 n+1}<\sqrt{n}+\sqrt{n+1}<\sqrt{4 n+2}$$
If $[\sqrt{n}+\sqrt{n+1}] \neq[\sqrt{4 n+2}]$, then there exists $k \in \mathbf{N}^{*}$, such... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,650 |
9. Find all positive integers $a, b$, such that
$$\left[\frac{a^{2}}{b}\right]+\left[\frac{b^{2}}{a}\right]=\left[\frac{a^{2}+b^{2}}{a b}\right]+a b$$ | 9. When $a=b$, the left side $=a+b$, and the right side $=ab+2$. Also, $ab+2-a-b=(a-1)(b-1)+1>0$, so the left and right sides are not equal, hence $a \neq b$.
Assume without loss of generality that $ab>a$, thus $a^2-b2a$, which is a contradiction. Therefore, $b<a^2+2$.
In summary, $b=a^2+1$. Direct verification shows... | (a, b)=\left(m, m^2+1\right) \text{ or } \left(m^2+1, m\right), m \in \mathbf{N}^{*} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,651 |
6 . Find all positive integers $a, b$ such that
$$(a, b)+9[a, b]+9(a+b)=7 a b .$$ | 6. Let $(a, b)=d, a=d x, b=d y$, then $(x, y)=1$. Substituting into the condition, we get
$$1+9 x y+9(x+y)=7 d x y$$
Thus, $\square$
$$9<7 d=9+9\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{x y} \leqslant 9+9 \times 2+1=28 \text {. }$$
From this, we know that $d=2,3$ or 4. Substituting these values into the indetermi... | (a, b)=(4,4),(4,38),(38,4) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,652 |
10. For each $k \in \mathbf{N}^{*}$, let $e(k)$ denote the number of positive even divisors of $k$, and $o(k)$ denote the number of positive odd divisors of $k$. Prove that for any $n \in \mathbf{N}^{*}$, the numbers $\sum_{k=1}^{n} e(k)$ and $\sum_{k=1}^{n} o(k)$ differ by at most $n$. | 10. Notice that, for any $d \in \mathbf{N}^{*}$, the number of multiples of $d$ among $1,2, \cdots, n$ is exactly $\left[\frac{n}{d}\right]$, therefore
$$\sum_{k=1}^{n} o(k)-\sum_{k=1}^{n} e(k)=\sum_{i=1}^{+\infty}\left(\left[\frac{n}{2 i-1}\right]-\left[\frac{n}{2 i}\right]\right) \geqslant 0$$
On the other hand, we ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,653 |
11. Let positive integers $a, d$ be coprime. Prove: In the arithmetic sequence $\{a+k d \mid k=0,1,2, \cdots\}$, there are infinitely many terms that have the same prime factor. | 11. It is only necessary to prove: there exist infinitely many $n \in \mathbf{N}^{*}, n \geqslant 2$, such that $a^{n} \equiv a(\bmod d)$. Given that $(a, d)=1$, we know $a^{\varphi(d)} \equiv 1(\bmod d)$ (Euler's theorem), so let $n=m \varphi(d)+1, m \in \mathbf{N}^{*}$, then we can find infinitely many terms, all of ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,654 |
12. The geometric mean of any $n$ non-negative real numbers is the $n$-th root of their product.
(1) For any $n \in \mathbf{N}^{*}$, do there exist $n$ distinct positive integers such that the geometric mean of any $k$ of them is an integer? Here $1 \leqslant k \leqslant n$;
(2) Do there exist infinitely many distinct ... | 12. (1) For any $n \in \mathbf{N}^{*}$, the following $n$ numbers satisfy the requirement:
$$1^{n!}, 2^{n!}, \cdots, n^{n!}$$
(2) There do not exist infinitely many different positive integers that satisfy the requirement.
In fact, if such numbers exist, take two of them, $a$ and $b$. By the problem's condition, for an... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,655 |
13. Prove: There exist infinitely many $n \in \mathbf{N}^{*}$, such that $n \mid\left(2^{n}+1\right)$. | 13. First, $n=1$ satisfies the condition. Next, let $n \in \mathbf{N}^{*}, n \mid\left(2^{n}+1\right)$, then $n$ is an odd number, and we can set $2^{n}+1=n q$, where $q$ is an odd number greater than 1. Let $m=2^{n}+1$, then $2^{m}+1=\left(2^{n}\right)^{q}+1=\left(2^{n}+1\right)\left(\left(2^{n}\right)^{q-1}-\left(2^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,656 |
14. Prove: There exist infinitely many $n \in \mathbf{N}^{*}$, such that $n \mid\left(2^{n}+2\right)$. | 14. The smallest two $n$ that satisfy the condition are $n=2,6$. Based on this analysis, we prove the following strengthened proposition: there exist infinitely many positive even numbers $n_{1}<n_{2}<\cdots$ such that for any $k \in \mathbf{N}^{*}$, we have $n_{k} \mid\left(2^{n_{k}}+2\right)$, and $\left(n_{k}-1\righ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,657 |
15. Prove: For any $n \in \mathbf{N}^{*}, n \geqslant 2$, there exist $n$ distinct positive integers $a_{1}, a_{2}, \cdots$, $a_{n}$, such that for $1 \leqslant i<j \leqslant n$, we have $\left(a_{i}-a_{j}\right) \mid\left(a_{i}+a_{j}\right)$. | 15. When $n=2$, take $\left\{a_{1}, a_{2}\right\}=\{1,2\}$. Suppose the proposition holds for $n$, and let $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ satisfy the requirements of the problem. For the case of $n+1$, let $A=\left(a_{1}+a_{2}+\cdots+\right.$ $\left.a_{n}\right)$ !, and take $b_{1}=A, b_{2}=A+a_{1}, \cdot... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,658 |
16. Prove: For any $n \in \mathbf{N}^{*}, n \geqslant 2$, there exist $n$ distinct positive integers, such that for any two different numbers $a, b$, we have $(a-b)^{2} \mid a b$. | 16. When $n=2$, take $\left\{a_{1}, a_{2}\right\}=\{1,2\}$. Suppose $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ satisfies that for any two different numbers $a_{i}, a_{j}$, $\left(a_{i}-a_{j}\right)^{2} \mid a_{i} a_{j}$. Let $A=a_{1} a_{2} \cdots a_{k}$. Consider $k+1$ numbers: $A, A+a_{1}, \cdots, A+a_{k}$. Take any... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,659 |
17. Prove: For any $n \in \mathbf{N}^{*}$, there exist $n$ consecutive positive integers, none of which is a prime power.
Proof: For any $n \in \mathbf{N}^{*}$, consider the sequence of integers from $(n+1)! + 2$ to $(n+1)! + (n+1)$. Each of these integers can be expressed as $(n+1)! + k$ for $k = 2, 3, \ldots, n+1$. ... | 17. Consider the following $n$ numbers:
$$((n+1)!)^{2}+2,((n+1)!)^{2}+3, \cdots,((n+1)!)^{2}+(n+1)$$
These numbers are multiples of $2,3, \cdots, n+1$, and thus none of them are prime. If there exists $2 \leqslant j \leqslant n+1$, such that $((n+1)!)^{2}+j=p^{\alpha}, p$ is a prime, $\alpha \in \mathbf{N}^{*}$, then ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,660 |
18 , Proof: For any $n \in \mathbf{N}^{*}$, there exist $n$ distinct pairwise coprime positive integers, such that the sum of any $k$ of them is a composite number, where $2 \leqslant k \leqslant n$. | 18. Take the following $n$ numbers:
$$n!+1,2(n!)+1, \cdots, n(n!)+1$$
It is clear that the sum of any $k$ of these numbers is a multiple of $k$, so when $2 \leqslant k \leqslant n$, the sum of any $k$ of these numbers is composite.
Now let $d=(i(n!)+1, j(n!)+1), 1 \leqslant i < j \leqslant n$. Clearly, $d>1$, let $p$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,661 |
19. Prove: There exist infinitely many $n \in \mathbf{N}^{*}$, such that $\sigma(n)<\sigma(n-1)$. | 19. Let $n$ be a prime number $p$ greater than 3, then $\sigma(n)=1+p$, and $\sigma(n-1) \geqslant 1+2+(p-1)=p+2$. Therefore, the proposition holds. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,662 |
7. Let $S$ be a finite set composed of integers, and each number in $S$ is greater than 1. It is known that for any $n \in \mathbf{Z}$, there exists $s \in S$, such that $(n, s)=1$ or $s$. Prove: There exist $s, t \in S(s, t$ can be the same), such that $(s, t)$ is a prime. | 7. Take the smallest positive integer $n$ that satisfies the following condition: For any $s \in S$, we have $(s, n)>1$. According to the condition, there exists $s \in S$ such that $(s, n)=1$ or $s$, and since $(s, n)>1$, it follows that $(s, n)=s$. Now take $p$ to be a prime factor of this $s$. By the definition of $... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,663 |
20. Prove: There exist infinitely many $n \in \mathbf{N}^{*}$, such that for $k \in\{1,2, \cdots, n-1\}$, we have
$$\frac{\sigma(n)}{n}>\frac{\sigma(k)}{k}$$ | 20. Suppose the conclusion does not hold, i.e., there are only finitely many $n \in \mathbf{N}^{*}$ such that for $k=1,2, \cdots, n-1$, $a_{n}>a_{k}$, where $a_{n}=\frac{\sigma(n)}{n}$. Let $N$ be the largest of these $n$, then the maximum value of the sequence $\left\{A_{n}\right\}$ is $A_{N}$, where $A_{n}=\max _{1 \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,664 |
21. Let $n \in \mathbf{N}^{*}, n \geqslant 3$. Prove: there exists a positive integer, the cube of which can be expressed as the sum of the cubes of $n$ distinct positive integers. | 21. When $n=3$, $3^{3}+4^{3}+5^{3}=6^{3}$; when $n=4$, $5^{3}+7^{3}+9^{3}+10^{3}=$ $13^{3}$. Therefore, the proposition holds for $n=3,4$.
Now assume that the proposition holds for $n=k, k+1(k \geqslant 3)$, and consider the case for $n=k+2$. By the induction hypothesis, for $y^{3}=x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,665 |
22. Prove: For any positive integer $n$ not less than 4, the number $n^{3}$ can be expressed as the sum of the cubes of 5 integers, and the absolute value of each addend is less than $n$.
| 22. Notice,
$$\begin{array}{c}
4^{3}=3^{3}+3^{3}+2^{3}+1^{3}+1^{3} \\
5^{3}=4^{3}+4^{3}+(-1)^{3}+(-1)^{3}+(-1)^{3} \\
6^{3}=5^{3}+4^{3}+3^{3}+0^{3}+0^{3}, 7^{3}=6^{3}+5^{3}+1^{3}+1^{3}+0^{3}
\end{array}$$
For odd numbers \( n \geq 9 \), let \( n=2 m+1, m \geq 4 \), we have
$$n^{3}=(2 m+1)^{3}=(2 m-1)^{3}+(m+4)^{3}+(4-... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,666 |
23. Let $p$ be a prime, $p \equiv 1(\bmod 4)$. Prove: There exists only one pair of positive integers $(x, y)$, $x \leqslant y$, such that $p=x^{2}+y^{2}$. | 23. From Example 1 in Section 5.3, we know that such positive integers exist. If there exist pairs of positive integers $(x, y)$ and $(a, b)$, with $x \leqslant y, a \leqslant b$, such that $p=x^{2}+y^{2}=a^{2}+b^{2}$, then
$$\begin{aligned}
p^{2} & =\left(x^{2}+y^{2}\right)\left(a^{2}+b^{2}\right)=(a x-b y)^{2}+(a y+b... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,667 |
25. Prove: There exist infinitely many $n \in \mathbf{N}^{*}$, such that the equation
$$d(n x)=x$$
has no positive integer solution $x$. | 25. Let $p$ be a prime number no less than 5. We prove: when $n=p^{p-1}$, the equation $d(n x)=x$ has no positive integer solutions.
In fact, if there exists $x \in \mathbf{N}^{*}$ such that $d\left(p^{p-1} x\right)=x$, if $p \nmid x$, then $x=d\left(p^{p-1} x\right)=d\left(p^{p-1}\right) d(x)=p d(x)$, leading to $p \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,669 |
26. Let $p$ be a prime number, and $m$ be a given positive integer. Prove: There exists a positive integer $n$, such that in the decimal representation of $p^{n}$, there are at least $m$ consecutive zeros. | 26. If $p \neq 2,5$, then $\left(p, 10^{m+1}\right)=1$. Using the following $10^{m+1}+1$ numbers
$$1, p, p^{2}, \cdots, p^{10^{m+1}}$$
there must be two numbers that are congruent modulo $10^{m+1}$, which means there exist $0 \leqslant r<s \leqslant 10^{m+1}$, such that
$$p^{s} \equiv p^{r}\left(\bmod 10^{m+1}\right) ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,670 |
27. Prove: For any positive integer $k$ greater than 1, there exists $n \in \mathbf{N}^{*}$, such that in the decimal representation, the last $k$ digits of $2^{n}$ contain at least half 9s (for example: when $k=2$, there is $2^{12}=$ $4096$; when $k=3$, there is $2^{53}=\cdots 992$ and so on). | 27. First, we prove: for any $k \in \mathbf{N}^{*}$, there exists $m \in \mathbf{N}^{*}$ such that $2^{m} \equiv -1 \pmod{5^{k}}$.
We use induction on $k$. When $k=1$, taking $m=2$ suffices. Now assume the proposition holds for $k$, i.e., there exists $m \in \mathbf{N}^{*}$ such that $2^{m} \equiv -1 \pmod{5^{k}}$. If... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,671 |
28. Let $a_{1}, a_{2}, \cdots, a_{1000000} \in\{1,2, \cdots, 9\}$. Prove: there are at most 40 positive integers $k$, such that: $1 \leqslant k \leqslant 1000000$ and the decimal number $\overline{1_{1} a_{2} \cdots a_{k}}$ is a perfect square. | 28. Let $\alpha(x)$ denote the number of digits in the decimal representation of the positive integer $x$. First, we prove a lemma: If $x, y \in \mathbf{N}^{*}, y > x, x, y$ are perfect squares, and $y = 10^{2b} x + c, b, c \in \mathbf{N}^{*}$, $c < 10^{2b} x$, then $\sqrt{y} > 10^{b} \sqrt{x}$. Since $\sqrt{y}$ and $1... | 40 | Number Theory | proof | Yes | Yes | number_theory | false | 741,672 |
29. Let $X$ be the set of positive integers of the form:
$$a_{2 k} \cdot 10^{2 k}+a_{2 k-2} \cdot 10^{2 k-2}+\cdots+a_{2} \cdot 10^{2}+a_{0},$$
where $k \in \mathbf{N}, a_{2 i} \in\{1,2, \cdots, 9\}, 0 \leqslant i \leqslant k$. Prove: For any $p, q \in \mathbf{N}, X$ contains a number that is a multiple of $2^{p} \cdo... | 29. First, prove a lemma: For any $p \in \mathbf{N}^{*}$, there exists a $2 p-1$ digit positive integer in $X$ that is a multiple of $4^{p}$.
Prove by induction on $p$. When $p=1$, it is obviously true. Assume the lemma holds for $p=k$, i.e., there exists a $2 k-1$ digit number $x \in X$ such that $4^{k} \mid x$. We c... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,673 |
8. Does there exist 100 different positive integers such that their sum equals their least common multiple? | 8. Existence. In fact, when $n \geqslant 3$, take the numbers $3, 2 \times 3, 2 \times 3^{2}, \cdots, 2 \times 3^{n-2}$, $3^{n-1}$, then their least common multiple is $2 \times 3^{n-1}$, and the sum is also $2 \times 3^{n-1}$. Therefore, the conclusion holds for $n \geqslant 3$. Taking $n=100$ gives the solution. | proof | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,674 |
30. A positive integer $k$ has the following property: in decimal representation, for any $n \in \mathbf{N}^{*}$, if $n$ is a multiple of $k$, then the reverse of $n$ is also a multiple of $k$. Prove: $k$ is a divisor of 99. | 30. Notice that for any $k \in \mathbf{N}^{*}$, there exists a number starting with the digit 1 that is a multiple of $k$ (in fact, among the numbers $1, 11, \cdots$, there must be two numbers that have the same remainder when divided by $k$, and their difference is a number starting with 1 that is divisible by $k$). B... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,675 |
31. There are three piles of stones on the ground. Each time, the following operation is allowed: take some stones from one pile and put them into another pile, so that the number of stones in the pile that receives the stones becomes twice its original amount.
(1) Prove: It is always possible to merge all the stones i... | 31. (1) Let the initial number of stones in the three piles be $a, b, c$. Without loss of generality, assume $a \leqslant b \leqslant c$.
If $a=0$, then no operation is needed.
Consider the case where $a>0$, and let $b=a q+r, 0 \leqslant r<a$, and express $q$ in binary as
$$q=m_{0}+2 m_{1}+2^{2} m_{2}+\cdots+2^{k} m_{... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 741,676 |
32. Let $a, b, n \in \mathbf{N}^{*}$, and $\left(b^{n}-1\right) \mid a$. Prove: The number $a$ in base $b$ representation has at least $n$ non-zero digits. | 32. Discuss in the $b$-ary representation. Let the minimum number of non-zero digits in the $b$-ary representation of all numbers divisible by $b^{n}-1$ be $s$, and among all these numbers with $s$ non-zero digits, take the number $A$ with the smallest digit sum. Let $A=a_{1} b^{n_{1}}+a_{2} b^{n_{2}}+\cdots+a_{s} b^{n... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,677 |
33. Prove: The set $\{[\sqrt{2} n] \mid n=1,2, \cdots\}$ contains infinitely many numbers that are powers of 2. | 33. Since $\sqrt{2}$ is an irrational number, in its binary representation, there are infinitely many 1s after the decimal point, i.e., there exist infinitely many $k \in \mathbf{N}^{*}$ such that $\left\{2^{k} \sqrt{2}\right\}>\frac{1}{2}$. For these $k$, we take $n=\left[2^{k} \sqrt{2}\right]+1$, then
$$\begin{aligne... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,678 |
34. If a sequence of positive integers is such that each term from the third onward is equal to the sum of the two preceding terms, we call this sequence an "F-sequence." Can the set of positive integers be partitioned into:
(1) a finite number of
(2) an infinite number of
pairwise disjoint "F-sequences" whose union i... | 34. (1) If $\mathbf{N}^{*}$ can be partitioned into a finite union of "F-sequences," let it be $m$ sequences. Consider the following $2m+1$ numbers:
$$2m, 2m+1, \cdots, 4m,$$
Among these, there must be three numbers belonging to the same "F-sequence," but any two of these $2m+1$ numbers have a sum greater than the thi... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,679 |
35. Let $q$ be a given real number, satisfying: $\frac{1+\sqrt{5}}{2}<q<2$. For any $n \in \mathbf{N}^{*}$, let the binary representation of $n$ be:
$$n=2^{k}+a_{k-1} \cdot 2^{k-1}+\cdots+a_{1} \cdot 2+a_{0}$$
where $a_{i} \in\{0,1\}, 0 \leqslant i \leqslant k-1$. Define the sequence $\left\{p_{n}\right\}$ as follows:... | 35. We prove: when $n$ in binary representation has the form $n=(\underbrace{1010 \cdots 10}_{m \uparrow 10})_{2}$, i.e., it meets the requirement (at this time, $n=\frac{2(4^{m}-1)}{3}$), then $n$ is even, denoted as $n=2k$.
In fact, for such $n \in \mathbf{N}^{*}$, we have
$$p_{2k}=q^{2m-1}+q^{2m-3}+\cdots+q, p_{2k+... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,680 |
36. For any $a, b \in \mathbf{N}^{*}$, prove:
$$a+(-1)^{b} \sum_{m=0}^{a}(-1)^{\left[\frac{[m}{a}\right]} \equiv b+(-1)^{a} \sum_{n=0}^{b}(-1)^{\left[\frac{m}{b}\right]}(\bmod 4),$$
where $\lceil x\rceil$ denotes the smallest integer not less than $x$. | 36. Let $S_{a, b}$ denote the number of $m$ in $\{0,1,2, \cdots, a\}$ such that $\left\lceil\frac{b m}{a}\right\rceil$ is odd. The proposition is equivalent to proving:
$$a+(-1)^{b}\left(a+1-2 S_{a, b}\right) \equiv b+(-1)^{a}\left(b+1-2 S_{b, a}\right)(\bmod 4) .$$
If $S_{a, b} \equiv S_{b, a}(\bmod 2)$, then $2 S_{a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,681 |
37. Let $n \in \mathbf{N}^{*}, n \geqslant 4$. Prove: $\left|\sum_{k=1}^{n}(-1)^{k}\left\{\frac{n}{k}\right\}\right| \leqslant 3 \sqrt{n}$. | 37. Let $A \in \mathbf{N}^{*}, A$ to be determined, we have
$$\begin{aligned}
\left|\sum_{k=1}^{n}(-1)^{k}\left\{\frac{n}{k}\right\}\right|= & \left|\sum_{k=1}^{A-1}(-1)^{k}\left\{\frac{n}{k}\right\}+\sum_{k=A}^{n}(-1)^{k}\left(\frac{n}{k}-\left[\frac{n}{k}\right]\right)\right| \\
\leqslant & \left|\sum_{k=1}^{A-1}(-1)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,682 |
38. Prove: In the sequence $\left\{a_{n}\right\}$, there are infinitely many terms that are even, and infinitely many terms that are odd. Here $a_{n}=[\sqrt{2} n]+[\sqrt{3} n], n=1,2, \cdots$. | 38. Using
$$\begin{array}{l}
{[\sqrt{2} n]+[\sqrt{2}] \leqslant[\sqrt{2}(n+1)] \leqslant[\sqrt{2} n]+[\sqrt{2}]+1} \\
{[\sqrt{3} n]+[\sqrt{3}] \leqslant[\sqrt{3}(n+1)] \leqslant[\sqrt{3} n]+[\sqrt{3}]+1}
\end{array}$$
we know that $a_{n}+2 \leqslant a_{n+1} \leqslant 4$, i.e., $a_{n+1}-a_{n} \in\{2,3,4\}$.
If $\left\{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,683 |
39. The sequence $\left\{a_{n}\right\}$ is defined as follows:
$$a_{1}=0, a_{n}=a\left[\frac{n}{2}\right]+(-1)^{\frac{n(n+1)}{2}}, n=2,3, \cdots$$
For each $k \in \mathbf{N}$, find the number of indices $n$ that satisfy $2^{k} \leqslant n<2^{k+1}$ and $a_{n}=0$. | For each $n \in \mathbf{N}^{*}$, let $x_{n}$ be the number of times the adjacent pairs 00 and 11 appear in the binary representation of $n$, and $y_{n}$ be the number of times the adjacent pairs 01 and 10 appear. We prove:
$$a_{n}=x_{n}-y_{n}$$
In fact, when $n=1$, $x_{1}=y_{1}=0$, so (7) holds for $n=1$.
Now assume (... | \mathrm{C}_{k}^{\frac{k}{2}} | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 741,684 |
9. For any given $k \in \mathbf{N}^{*}, k>1$, let the least common multiple of $n, n+1, \cdots, n+k$ be denoted as $Q(n)$. Prove: There exist infinitely many $n \in \mathbf{N}^{*}$, such that
$$Q(n)>Q(n+1) .$$ | 9. For any $r \in \mathbf{N}^{*}, r \geqslant 3$, take $n=r \cdot k!-1$, we prove: $Q(n)>Q(n+$ 1), thus the proposition holds.
In fact, let $m=[n+1, n+2, \cdots, n+k]$. Since for $1 \leqslant j \leqslant k$, we have $(n, j) \equiv (-1, j)=1$, hence $(n, n+j)=1$, so $Q(n)=m n$.
On the other hand, $n+k+1=r \cdot k!+k$ ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,685 |
40. The function $f: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*}$ is defined as follows: $f(1)=1$, and for any $n \in \mathbf{N}^{*}$, we have
$$f(n+1)=\left\{\begin{array}{ll}
f(n)+2, & n=f(f(n)-n+1), \\
f(n)-1, & \text { otherwise } n .
\end{array}\right.$$
(1) Prove that for any $n \in \mathbf{N}^{*}$, $f(f(n)-n+1) \i... | 40. From the recurrence relation, we know
$$f(n) \leqslant f(n-1)+2 \leqslant \cdots \leqslant f(1)+2(n-1)=2 n-1,$$
Thus, $f(n)-n+1 \leqslant n$. Therefore, if the values of $f(1), f(2), \cdots, f(n)$ are determined, then the value of $f(n+1)$ is uniquely determined. Hence, there exists a unique function $f$ that sati... | f(n) = \left[\frac{1+\sqrt{5}}{2} n\right] | Algebra | proof | Yes | Yes | number_theory | false | 741,686 |
41. Let $a \bmod n$ denote the remainder of $a$ divided by $n$ (i.e., $\left.a-\left[\frac{a}{n}\right] \cdot n\right)$. For given $m, n \in \mathbf{N}^{*}$, find $\max _{0 \leqslant a_{1}, a_{2}, \cdots, a_{m}<n} \min _{0 \leqslant k<n} \sum_{j=1}^{m}\left(\left(a_{j}+k\right) \bmod n\right)$, where $a_{1}$, $a_{2}, \... | 41. The answer is: $\frac{1}{2}((m-1)(n-1)+(m, n)-1)$.
Because
$$a_{j}+k(\bmod n)=\left\{\begin{array}{l}
a_{j}+k, a_{j}<n-k \\
a_{j}+k-n, a_{j} \geqslant n-k
\end{array}\right.$$
Hence
$$\begin{aligned}
S & =\sum_{j=1}^{m}\left(\left(a_{j}+k\right) \bmod n\right) \\
& =\sum_{j=1}^{m} a_{j}+m k-\left(\mu_{n-1}+\mu_{n... | \frac{1}{2}((m-1)(n-1)+(m, n)-1) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,687 |
42. If positive integers $m, n$ satisfy: $\sigma(m)=\sigma(n)=m+n$, then $m, n$ are called "amicable".
(1) Prove: For any $a \in \mathbf{N}^{*}, a$ and $\varphi(a)$ are not "amicable";
(2) Prove: If $a, b \in \mathbf{N}^{*}, b>1$, then $a$ and $\left(2^{b}-1\right) a+1$ are not "amicable". | 42. First, look at two lemmas.
Lemma 1: When $m \geqslant 3$, $\varphi(m)$ is even. (This is a well-known conclusion.)
Lemma 2: If $\sigma(n)$ is odd, then $n$ is a perfect square or twice a perfect square. $\square$
In fact, let $n=2^{s} \cdot t, s \geqslant 0, t$ be an odd number. By $\sigma(n)$ being odd, we know ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,688 |
43. Let $r$ be a given positive integer. Prove: there exist infinitely many positive integers $k>r$!, such that for any positive integer $j$ satisfying $r!<j<k$, we have
$$j(j-1) \cdots(j-r+1) \nmid_{k}(k-1) \cdots(k-r+1) .$$ | 43. For any $m \in \mathbf{N}^{*}, m>r$ !, let $n=m!+r$, and denote $j(r)$ as $\frac{j!}{(j-r)!} (= j(j-1) \cdots(j-(r-1)))$. Then for any $j \in \mathbf{N}^{*}$, if $r!r$ !, so, $j(r) \times n(r)$. This indicates: there exists $n \in \mathbf{N}^{*}, n>$ $m$, such that for any $r!m$, and for $r!<j \leqslant m$, we have... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,689 |
44. Let $n, k \in \mathbf{N}^{*}$, positive integers $a_{1}, a_{2}, \cdots, a_{k}$ satisfy: $1 \leqslant a_{1}<a_{2}<\cdots<a_{k} \leqslant$ $n$, and for $1 \leqslant i<j \leqslant k$, we have $\left[a_{i}, a_{j}\right] \leqslant n$. Prove: $k \leqslant 2 \sqrt{n}+1$. | 44. Let $t=[\sqrt{n}]$, and suppose $1 \leqslant a_{1}<a_{2}<\cdots<a_{j} \leqslant t<a_{j+1}<\cdots<a_{k} \leqslant n$. Using $\left(a_{i}, a_{j}\right) \leqslant a_{i}-a_{j}(1 \leqslant j<i \leqslant k)$, we know $\left[a_{i}, a_{j}\right]=\frac{a_{i} a_{j}}{\left(a_{i}, a_{j}\right)} \geqslant \frac{a_{i} a_{j}}{a_{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,690 |
45. Find the largest real number $\alpha$, such that there exists an infinite sequence of positive integers $\left\{a_{n}\right\}_{n=1}^{+\infty}$, satisfying:
(1) For any $n \in \mathbf{N}^{*}$, we have $a_{n}>2008^{n}$;
(2) For any $n \in \mathbf{N}^{*}, n \geqslant 2$, the number $a_{n}^{\alpha}$ does not exceed the... | 45. First, prove $\alpha \leqslant \frac{1}{2}$.
In fact, let $\left\{a_{n}\right\}$ be a sequence of positive integers satisfying conditions (1) and (2). We have: for any $0 < \varepsilon < 1$, there are infinitely many positive integers $n$ such that
$$a_{n}^{2-\varepsilon} \leqslant a_{2 n}.$$
If (11) does not hold... | \frac{1}{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,691 |
46. Prove: In any right-angled triangle with integer side lengths, the lengths of the two legs cannot both be perfect numbers. | 46. If not, let $(x, y, z)$ be a Pythagorean triplet, satisfying $x^{2}+y^{2}=z^{2}$, and both $x, y$ are perfect numbers. Note that $x, y$ cannot both be odd (otherwise, $z^{2} \equiv 2(\bmod 4)$, which contradicts the fact that a square number $\equiv 0$, $1(\bmod 4)$), hence at least one of $x, y$ is even.
Euler's ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,692 |
47. Find all positive integers $p(\geqslant 2)$, such that there exists $k \in \mathbf{N}^{*}$, satisfying: the number $4 k^{2}$ in base $p$ representation consists only of the digit 1. | 47. It can be proven that the positive integers $p \in \{3,7\}$ satisfy the condition.
When $p=3$, taking $k=1$ is sufficient. When $p=7$, taking $k=10$, we have $4 k^{2}=400=(1111)_7$. Therefore, when $p \in \{3,7\}$, the condition is satisfied.
On the other hand, let $p$ be a positive integer that satisfies the con... | p \in \{3,7\} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 741,693 |
48. Define the sequence $\left\{a_{n}\right\}$ as follows: $a_{1}=1, a_{2}=2, a_{3}=3$, when $n \geqslant 4$, $a_{n}$ is the smallest positive integer not in $\left\{a_{1}, a_{2}, \cdots, a_{n-1}\right\}$ such that $\left(a_{n-1}, a_{n}\right) \geqslant 3$ (the first few terms of this sequence are $1,2,3,6,9,12,4,8, \c... | 48. Prove a more general proposition: Let $1 \leqslant n \leqslant d$ when $a_{n}=n$, and for $n \geqslant d+1$, the number $a_{n}$ is the smallest positive integer not in $\left\{a_{1}, a_{2}, \cdots, a_{n-1}\right\}$ such that $\left(a_{n-1}, a_{n}\right) \geqslant d$. Then the conclusion also holds. Clearly, to prov... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 741,694 |
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