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Can n! be a perfect square when n is an integer greater than 1? (But is it possible, to prove without Bertrand's postulate. Because bertrands postulate is quite a strong result.)

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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
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

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.

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
I don't get this. There is, in fact, a prime $p$ such that $3/2 < p < 3$. So, why doesn't this work for $n=3$? – Nikhil Mahajan Feb 26 '15 at 5:22

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

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Hopefully this is a little more intuitive (although quite a bit longer) than the other answers up here.

Let's begin by stating a simple fact : (1) when factored into its prime factorization, any perfect square will have an even number of each prime factor.

If $n$ is a prime number, then $n$ will not repeat in any of the other factors of $n!$, meaning that $n!$ cannot be a perfect square (1). Consider if $n$ is composite. $n!$ will contain at least two prime factors ($n=4$ is the smallest composite number that qualifies the restraints), so let's call $p$ the largest prime factor of $n!$

The only way that $n!$ can be a perfect square is if $n!$ contains $p$ and a second multiple of $p$ (1). Obviously, this multiple must be greater than $p$ and less than $n.$

Using Bertrand's postulate, we know that there exists an additional prime number, let's say $p'$, such that $p < p' < 2p$. Because $p$ is the largest prime factor of $n!$, we know that $p' > n$ (If it were the opposite, then we would reach a contradiction).

Thus it follows that $2p > p' > n$. Because $2p$ is the smallest multiple of $p$ and $2p > n$, then $n!$ only contains one factor of $p$. Therefore it is impossible for $n!$ to be a perfect square.

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  • If $n$ is prime, then for $n!$ to be a perfect square, one of $n-1, n-2, ... , 2$ must contain n as a factor. But this means one of $n-1, n-2, ... , 2 \geq n$, which is impossible.

  • If $n$ is not prime, then the first prime less than $n$ will be $p = n-k$, $0<k<n-1, 2\leq p<n$. No number less than $p$ will contain $p$ as a factor, so for $n!$ to be a perfect square there must exist a multiple of $p$, I'll call it $bp$, $1<b<n,$ such that$ p<bp\leq n$. Now according to chebyshev's theorem for any no. $p$ there exists a prime number between $p$and $2p.$ so if $r< n < 2r$ and also $p<n$ , so such an $n!$ would never be a perfect square. Hope this helps.

You can refer this.

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Your statement has a generalization. There is a work by Erdos and Selfridge stating that the product of at least two consecutive natural numbers is never a power. Here is it:

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