TITLE: Factorial equation
QUESTION [5 upvotes]: I'm trying to find all nonnegative integer solutions to $x!^2=z!$. Intuitively, I think the solutions are the trivial ones with $x=0,1$ and $z=0,1$. I'm not sure how to show that there is no more solutions.

REPLY [2 votes]: Note that every prime factor of $x!^2$ appears to an even power.
By Bertrand's postulate we can find a prime factor $p$ of $z$ greater than $\frac z2$. Then the exponent of $p$ in $z!$ is one.