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.

$n=2$,$m=0$ seems like the first pair to try:
$\frac{2!2!}{(2-2)!(2+0-2)!}=\frac{2!2!}{0!0!}=\frac{2*2}{1*1}=4$, which is a bit high.
Let's try $n=2$,$m=-2$ then:
$\frac{2!(2-2)!}{(2-2)!(2-2-2)!}=\frac{2!0!}{0!(-2)!}$, so that goes nowhere. sigh
