Find the positive integer solutions of $m!=n(n+1)$
I basically have $(m,n)=(2,1)$ or $(3,2)$ and I think these are the only solutions.
I don't have a complete proof but here's what I know so far. By Bertrand's Postulate, I can find prime $p$ in the "second half" of $m!$. If $m>4$, then $p$ is odd.
Suppose $n$ be even. Then $n+1$ be odd. Also, $n=2^kq$, where $k$ is the maximum number of times $2$ can divide $m!$.
What else can be done?