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

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See this. –  J. M. Nov 30 '10 at 22:38
    
@J.M.: I found the resolution very complex. Honestly, I could not understand it. –  Paulo Argolo Nov 30 '10 at 22:43
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Actually, the link J. M. pointed to has the answer in the first paragraph — and it's the same as the two answers posted below. The rest of the page is a proof of Bertrand's postulate itself. –  ShreevatsaR Dec 1 '10 at 16:29
    
@ShreevatsaR: You're right. Thank you for participating. Thank you all. –  Paulo Argolo Dec 1 '10 at 19:51
    
Is there a proof of this fact which does not use Bertrand's postulate? –  Beni Bogosel Mar 1 '12 at 14:50
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2 Answers

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.

http://en.wikipedia.org/wiki/Bertrand_postulate

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

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Careful. You should say $n/2 < p \le n$ or else your statement is wrong when $n = 2, 3$. –  Qiaochu Yuan Nov 30 '10 at 23:04
    
@Qiaochu: Sorry. I should have added that Bertrand's postulate in this form applies for $n\geq4$. The other cases $n=2,3$ can be checked directly. –  Timothy Wagner Nov 30 '10 at 23:07
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There is a prime between n/2 and n, if I am not mistaken.

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