$$(a^2)!\cdot (a^2-1)!=((a^2-1)!\cdot a)^2$$

Too trivial to appear in a TST

Trivial

Obvious!!

.

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.

By Erdos-Selfridge, $|m-n|\leq1$, which gives the only solutions as $(x^2-1,x^2)$, $(x^2,x^2-1)$, and $(x,x)$.

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??

What the hell was I doing