Which are the minimum values of $m$ and $n$ over integers such that $\frac{n!(n+m)!}{(n-2)!(n+m-2)!} = 2$ Which are the minimum values of $m$ and $n$ over integers such that $\frac{n!(n+m)!}{(n-2)!(n+m-2)!} = 2$
I am searching for $m$ and $n$ even number's.
Probably there is some software to do this, but I don't know.
Thanks a lot.

To approach this systematically, let $p = m+n$. We're then looking for $n,p \geq 2$ such that
$$\frac{n!\ p!}{(n-2)!\ (p-2)!} = 2.$$
In other words,
$$n(n-1)p(p-1) = 2.$$
But $n(n-1) \geq 2$ and $p(p-1) \geq 2$, so $n(n-1)p(p-1) \geq 4$, and there are no solutions.
