Find the positive integer solutions of $m!=n(n+1)$

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.

$(n,n+1)=1$

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?

Solution 1:

It is very unlikely that this problem can be resolved with current techniques. See BROCARD and section D25 in Unsolved Problems in Number Theory by Richard K. Guy.

Berend and Osgood showed in 1992 that the set of solutions to $P(x) = n!$ has an integer solution has density zero if $P(x)$ is a fixed polynomial of degree at least two with integer coefficients. Among conditional results, F. Luca showed that the ABC conjecture implies that the number of solutions to $P(x) = n!$ is finite. Luca can be downloaded from the wikipedia page.

In case there has been a misunderstanding in stating the problem, there is an infinite family of solutions to $$ m! = n! \; (n+1)!, $$ in that $$ (x!)! = (x! -1)! \; x!, $$ so $m=x! \; , \; n = x! -1.$