# Solve in positive integers $n!=m^2$

Is anybody know a solution of this problem?

(Sorry, correct question is here.)

• m=1, n=0 or 1. For larger n, there is a prime between n/2 and n, which guarantees an unsquared prime factor in the factorial. – Hugo van der Sanden Sep 18 '10 at 10:17

From Bertrand's postulate it follows swiftly that there are no solutions with $n>1$.
• Just to be more detailed for those who may need it: If $p$ be is the largest prime less than $n$ then $p$ divides $n!$, hence $p$ divides $m^2$ so $p^2$ divides $n!$. So $n>=2p$ which contradicts Bertrand's postulate (and the only solutions are the trivial ones). – danseetea Sep 18 '10 at 10:29