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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
43
votes
3
answers
2k
views
Proving $\sum_{i=1}^{n}\sum_{j=1}^{n}\left\{\frac{x_{i}}{x_{j}}\right\}\le \frac{9}{14}n^2$?
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 …
-1
votes
1
answer
287
views
Show $p\nmid k!+1$
Question: let $k$ be a positive integer, $p$ a prime number, such that $p=3k+1$, $r<p$ be a positive integer, such that $2^{k+1}\equiv r\pmod p$, and
$r\not\equiv 4,5\pmod 6$. Show that
$$p\nmid k!+1. …
0
votes
1
answer
197
views
Sums of Legendre symbols with integer-valued polynomials
Let $f(x)$ be an integer-valued polynomial (when $x\in \mathbb{Z}$, then $f(x)\in \mathbb{Z}$), and $a,b$ be positive integers, and $p$ be a prime number with $(a,p)=1$.
Show that
$$\sum_{x=0}^{p-1}\l …
8
votes
1
answer
682
views
Prove an inequality related to sums of Legendre symbols
$\newcommand\Legendre{\genfrac(){}{}}$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
$$\sum_{1\le i<j\le p}\Legendre{i-j}{p}x_{i}x_{ …
6
votes
0
answers
232
views
Are there infinitely many positive integers $\ n\ $ satisfying $\ \varphi(n)\mid \sigma(n)$?
Are there infinitely many postive integers $\ n\ $ satisfying
$$\varphi(n)\mid \sigma(n)$$
where $\varphi(n)$ is Euler’s totient function, and $\sigma(n)$ is the sum of divisors of $n$? If yes, can …
3
votes
1
answer
900
views
Prove that $\frac{2^n-2}{n}$ is composite number
If positive integer $n$ such $n\mid2^n-2$,where $n>1$, we called $n$ is Poulet number, see: https://en.wikipedia.org/wiki/Super-Poulet_number
I found if $n>2$ is Poulet number, then $\dfrac{2^n-2}{n}$ …
1
vote
0
answers
140
views
On the number of asymptotic solutions of the linear Diophantine equation
Given positive integers $a,b,c$ which are pairwise coprime. Let $f(n)$ denotes the number of the non-negative integer solution $(x,y,z)$ to the equation
$$ax+by+cz=n.$$
we have Prove that there exists …
6
votes
3
answers
846
views
Is there a simple proof that $Ax^3+By^3=C$ has only finitely many integer solutions
One can use Thue's 1909 result to show that the Diophantine equation $Ax^3 + By^3 = C$ ($A,B$ not perfect cubes, $C\neq 0$) has finitely many integer solutions $(x,y)$.
But does there exist a simple w …
4
votes
1
answer
670
views
How to solve this equation $a^2+3b^2c^2=7^c$
Let $a,b,c$ be poistive integers,and such $\gcd(a,b)=\gcd(b,c)=\gcd(a,c)=1$,fine the all $a,b,c$ such
$$a^2+3b^2c^2=7^c$$
I'm not sure that this question has been studied, but I've been trying for a l …
1
vote
1
answer
183
views
inequality for sum $\sum_{j_{i}=1,i=1,\cdots,k}^{n}\gcd(j_{1},j_{2},\cdots,j_{k})$
if $n>k>1$ be postive integer,show that
$$S_{k}(n)=\dfrac{1}{n^k}\sum_{j_{1}=1}^{n}\sum_{j_{2}=1}^{n}\cdots\sum_{j_{k}=1}^{n}\gcd(j_{1},j_{2},\cdots,j_{k})\le\dfrac{\zeta(k-1)}{\zeta(k)} \tag{1}$$
…
8
votes
2
answers
594
views
On the conjecture : $|m^2-n^3|>\frac{1}{5}\sqrt[6]{m^2+n^3}$
Let $m,n$ be positive integers, such that $m^2\neq n^3$. I conjecture the inequality
$$|m^2-n^3|>\dfrac{1}{5}\sqrt[6]{m^2+n^3}.$$
I've tried a lot of numbers, and they all seem to work, but how do I …
10
votes
2
answers
752
views
Find all $m$ such $2^m+1\mid5^m-1$
The problem comes from a problem I encountered when I wrote the article
Find all positive integer $m$ such
$$2^{m}+1\mid5^m-1$$
it seem there no solution. I think it might be necessary to use qu …
0
votes
1
answer
342
views
Find the positive integers $x^3+y^3=3z^3$ [closed]
By Fermat Last theorem, I don't know if that's been discussed.
Find all positive integers $x,y,z$ such
$$x^3+y^3=3z^3$$
13
votes
2
answers
688
views
Show that there exist $k\in\{1,2,\cdots,n\}$ such that $\frac{1}{n}\sum_{i=1}^{n}\left(\{kx_...
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 …
7
votes
3
answers
940
views
Prove $ n!$ is divisible by the number of its positive divisors
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! …