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For any postive integer $n$ and for any postive real numbers $x_{1},x_{2},\cdots,x_{n}$, show that $$\sum_{i=1}^{n}\sum_{j=1}^{n}\left\{\dfrac{x_{i}}{x_{j}}\right\}\le \dfrac{9}{14}n^2$$ Let \begin{al …

I've been thinking about such an interesting question these last few days. I wonder if anyone has studied it. Question: Find all postive integers $n$ such that $$S(n)=1^n+2^n+3^n+4^n+\cdots+n^n …

Find the positive integers $(2^x-1)(3^y-1)=2z^2$ have three solutions $$(1,1,1),(1,2,2),(1,5,11)$$I already know $(2^x-1)(3^y-1)=z^2$ has no solution. See: P.G.Walsh December 2006 [On Diophantine equ …

Two years ago, I made a conjecture on stackexchange: Today, I tried to find all solutions in integers $a,b,c$ to $$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}.$$ I have found some sol …

Cross-Posted from Math Stackexchange Two positive integers $p,q$ and a prime $r$ are given, such that $r>p>q>1$. I have to show that there exist $n$ such that $$r|\binom{p^n}{q^n}$$ Should I use Luc …

The following question has post mathsatck :Nice problem, perhaps from a question of expectation.The problem seemed so elementary, but I had tried to solve it for a long time and eventually failed.I 'd …

The problem comes from a problem I encountered when I wrote the article Find all positive integer $m$ such $$2^{m}+1\mid5^m-1$$ it seem there no solution. I think it might be necessary to use qu …

Conjecture Let $m$ be give positive integer numbers,a sequence $p_{1},p_{2},\cdots $of primes satisfies the following condition:for $n\ge 3$,$p_{n}$ is the greatest prime divisor $p_{n-1}+p_{n-2}+ …

Let $m,n$ be positive integers, such that $m^2\neq n^3$. I conjecture the inequality $$|m^2-n^3|>\dfrac{1}{5}\sqrt[6]{m^2+n^3}.$$ I've tried a lot of numbers, and they all seem to work, but how do I …

$\newcommand\Legendre{\genfrac(){}{}}$Let $p\equiv 1\pmod 4$ be a prime number, and $x_{i}\ge 0$ be such that $$x_{1}+x_{2}+\dotsb+x_{p}=1.$$ Show that $$\sum_{1\le i<j\le p}\Legendre{i-j}{p}x_{i}x_{ …

@Allan methods it's nice! here is my answer: since $$\left(\dfrac{1-a^2}{2a}\right)\left(\dfrac{1-b^2}{2b}\right)\left(\dfrac{1-c^2}{2c}\right)=1$$ so let $$\dfrac{x}{y}=\dfrac{1-a^2}{2a},\;\dfrac{y}{ …

Prove that for all positive integers $ n$ different from $ 3$ and $ 5$, $ n!$ is divisible by the number of its positive divisors. I tried some things,such as the number of divisors of $ n! …

Conjecture: Let $m$ be a positive integer. Then $$f(m)=(2m)^{2m+1}+m^{2m+1}\cdot (2m+1)^m+(2m+1)^{2m}$$ is not a prime number. One can prove it when $m$ is odd number, it is clear that $f(m) …

One can use Thue's 1909 result to show that the Diophantine equation $Ax^3 + By^3 = C$ ($A,B$ not perfect cubes, $C\neq 0$) has finitely many integer solutions $(x,y)$. But does there exist a simple w …

Are there infinitely many postive integers $\ n\ $ satisfying $$\varphi(n)\mid \sigma(n)$$ where $\varphi(n)$ is Euler’s totient function, and $\sigma(n)$ is the sum of divisors of $n$? If yes, can …