Prove that there exist infinitely many pairs of different positive integers $(m, n)$ for which $m!n!$ is a square of an integer. Proposed by Anton Trygub
Problem
Source:
Tags: number theory
14.05.2019 22:11
$$(a^2)!\cdot (a^2-1)!=((a^2-1)!\cdot a)^2$$
05.08.2019 08:27
Too trivial to appear in a TST
04.12.2022 18:48
Trivial
27.11.2023 10:54
Obvious!!
02.04.2024 16:27
.
02.04.2024 16:40
John_Mgr wrote: let m=n=2k where k is an positive integer so m!n!=(2k!)^2 which is a square and a integer. there are infinite m=n=2K numbers thus, exists many (m,n) s.t m!n! is a square. *different* m and n.
02.04.2024 18:31
By Erdos-Selfridge, $|m-n|\leq1$, which gives the only solutions as $(x^2-1,x^2)$, $(x^2,x^2-1)$, and $(x,x)$.
06.06.2024 14:05
DottedCaculator wrote: By Erdos-Selfridge, $|m-n|\leq1$, which gives the only solutions as $(x^2-1,x^2)$, $(x^2,x^2-1)$, and $(x,x)$. woooow!! How many theorems do you know??
06.06.2024 18:35
What the hell was I doing