Particular diophantine in three variables

Question

I want solve this diophantine in a elementary way with $m,k, n $ positive integer and $k| n $ $$n^2\cdot k =m! $$ I haven't ideas to solve this diophantine.

Answer 1

Let's say $m\geq 2$. This means that $2\mid n$, so $4\mid m!$, which means $m \geq 4$. This again, means that $3\mid m!$, so $3\mid n$, and therefore $9\mid n$. This again means that $m\geq 6$, so $5\mid n$, which means that $25\mid m!$, which again implies $m \geq 10$. By Bertrand's postulate, we can just keep going, always finding a new prime $p$ that divides $m!$, which means that $p\mid n$, which again means $m \geq 2p$, which means we can find a new prime between $p$ and $2p$, and so on.

That means that we must have $m = 0$ or $1$. These may be checked by hand and verified to work with $n = 1$ in both cases.