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Generaliztion of 1999 USAMO
EthanWYX2009   5
N an hour ago by kingu
Let $p > 2$ be a prime and let $a_1,a_2,\ldots,a_{2n}$ be integers not divisible by $p,$ such that
\[ \left\{ \dfrac{ra_1}{p} \right\} + \left\{ \dfrac{ra_2}{p} \right\} + \cdots+ \left\{ \dfrac{ra_{2n}}{p} \right\} = n  \]for any integer $r$ not divisible by $p$. Prove that there exists $2\le j\le 2n$ such that $a_1+a_j$ is divisible by $p$.

Created by Haojia Shi
5 replies
EthanWYX2009
Nov 14, 2024
kingu
an hour ago
No more topics!
Diophantine equation
Kiewi   2
N Yesterday at 9:41 AM by tobiSALT
Does the diophantine equation
$x^3-y^3=z!$
has any positive integer solutions.

2 replies
Kiewi
Sep 7, 2024
tobiSALT
Yesterday at 9:41 AM
Diophantine equation
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Kiewi
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Y by Mhuyyy
Does the diophantine equation
$x^3-y^3=z!$
has any positive integer solutions.
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Kiewi
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tobiSALT
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Let $x, y, z$ be positive integers such that $x^3 - y^3 = z!$. If $x = y$, then $0 = z!$, which implies $z = 0$, a contradiction since $z$ must be positive. Therefore, we must have $x > y$.

Let $x = y + k$ for some positive integer $k$. Then
$$ (y+k)^3 - y^3 = z! $$$$ y^3 + 3y^2k + 3yk^2 + k^3 - y^3 = z! $$$$ k(3y^2 + 3yk + k^2) = z! $$We know that $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$. Let $x - y = a$. Then $x = y+a$ and $x^3 - y^3 = a(x^2 + xy + y^2) = z!$.

Let $p$ be a prime such that $z/2 < p \le z$. Then $p$ divides $z!$ but $p^2$ does not divide $z!$.
Since $x^3 - y^3 = (x-y)(x^2 + xy + y^2) = z!$, we must have $p \mid (x-y)$ or $p \mid (x^2 + xy + y^2)$.

Suppose $p \mid x-y$. Then $x \equiv y \pmod{p}$. Then $x^2 + xy + y^2 \equiv 3y^2 \pmod{p}$.
Since $z! = (x-y)(x^2+xy+y^2)$ and $p \mid (x-y)$, then $x-y = mp$ for some integer $m$. Then $z! = mp(x^2+xy+y^2)$.
If $p \mid (x^2+xy+y^2)$, then $3y^2 \equiv 0 \pmod{p}$, which implies $p \mid y$ since $p > 3$. Thus $y = np$ for some integer $n$, and $x = (m+n)p$.
Then $x^3 - y^3 = (m+n)^3 p^3 - n^3 p^3 = p^3 ((m+n)^3 - n^3) = z!$.
However, $p^2$ does not divide $z!$.

$\boxed{\text{Therefore, there are no positive integer solutions to } x^3 - y^3 = z!.}$
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