A tricky diophantine equation with factorials

Question

I am being unable to solve this diophantine equation. Does anyone have any suggestions. Let $n$ and $m$ both be non-negative integers. Find all solutions to

$$n(nm - 2)! = (n!)^m$$

How would one solve this? I suspect there is a bounding argument. Is there a classical number theory argument? What about a combinatorial argument?

Answer 1

Hint: There is a theorem that says that there is always a prime between $n$ and $2n$ (and actually, the constant $2$ can be reduced if a lower bound on $n$ is used). So if $m > 2$, you should be able to find a prime strictly in between $n$ and $nm - 2$, at least if $n > 3$, and then your equation can't hold. Also, using the lowered constant, (for $n$ large enough), you will also be able to rule out $m=2$ when $n$ is above the needed lower bound for the lowered constant.

Answer 2

If there is a prime greater than $n$ and at most $nm-2$, it divides the left side but not the right side. Bertrand's hypothesis almost gives this to you.