Product factorization

Rewrite the equation as the following:

$M! = \frac{N!}{M!}$

$(M!)^2 = N!$

We are going to prove that $n!$ cannot be a perfect square for $n>1$ , so this equation has no solutions for $N>1$ . Thus the only possibility is $N=1$ and $M=1$ , but $N>M$ implies that it is not a good solution so indeed this equation cannot be satisfied.

For the proof recall Chebisev's theorem, which states that there exist a prime $p$ for every $n>0$ such that $n<p \leq 2n$ . Applying this to $n=[\frac{N}{2}]$ there must exist a prime $p$ such that

$[\frac{N}{2}] +1 \leq p \leq 2[\frac{N}{2}]$

Since $2p \geq 2([\frac{N}{2}]+1) > 2 \frac{N}{2} = N$ it follows that the highest power of $p$ dividing $N!$ is 1. This is known to imply that $N!$ can not be a perfect square.