Can $n!$ be a perfect square when $n$ is an integer greater than $1$?

Question

Can $n!$ be a perfect square when $n$ is an integer greater than $1$?

Clearly, when $n$ is prime, $n!$ is not a perfect square because the exponent of $n$ in $n!$ is $1$. The same goes when $n-1$ is prime, by considering the exponent of $n-1$.

What is the answer for a general value of $n$? (And is it possible, to prove without Bertrand's postulate. Because Bertrands postulate is quite a strong result.)

Answer 1

There is a prime between n/2 and n, if I am not mistaken.

Answer 2

Assume, $n\geq 4$. By Bertrand's postulate there is a prime, let's call it $p$ such that $\frac{n}{2}<p<n$ . Suppose, $p^2$ divides $n$. Then, there should be another number $m$ such that $p<m\leq n$ such that $p$ divides $m$. So, $\frac{m}{p}\geq 2$, then, $m\geq 2p > n$. This is a contradiction. So, $p$ divides $n!$ but $p^2$ does not. So, $n!$ is not a perfect square.

Bertrand's postulate

That leaves two more cases. We check directly that, $2!=2$ and $3!=6$ are not perfect squares.

Answer 3

√n ≤ n/2 for n ≥ 4. Thus if p is a prime such that n/2 < p ≤ n, we have √n < p → n < p² so p² cannot be a factor of n if n ≥ 4.