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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.)

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    $\begingroup$ See this. $\endgroup$ Commented Nov 30, 2010 at 22:38
  • $\begingroup$ @J.M.: I found the resolution very complex. Honestly, I could not understand it. $\endgroup$ Commented Nov 30, 2010 at 22:43
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    $\begingroup$ 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. $\endgroup$ Commented Dec 1, 2010 at 16:29
  • $\begingroup$ @ShreevatsaR: You're right. Thank you for participating. Thank you all. $\endgroup$ Commented Dec 1, 2010 at 19:51
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    $\begingroup$ Is there a proof of this fact which does not use Bertrand's postulate? $\endgroup$ Commented Mar 1, 2012 at 14:50

6 Answers 6

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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.

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    $\begingroup$ Careful. You should say $n/2 < p \le n$ or else your statement is wrong when $n = 2, 3$. $\endgroup$ Commented Nov 30, 2010 at 23:04
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    $\begingroup$ @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. $\endgroup$ Commented Nov 30, 2010 at 23:07
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    $\begingroup$ 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 $n=3$? $\endgroup$
    – XYZT
    Commented Feb 26, 2015 at 5:22
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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: https://projecteuclid.org/euclid.ijm/1256050816.

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  • $\begingroup$ Is there any English language source? let us know! $\endgroup$ Commented Jan 5, 2021 at 6:58
  • $\begingroup$ @SmritidipaMukherjee The old link is dead. I updated the link. $\endgroup$ Commented Jan 10, 2021 at 4:56
  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. $\endgroup$ Commented Jun 19 at 7:23
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√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.

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  • $\begingroup$ $n<p^2$ doesn't imply what is needed. You might still have $2p<n$ and then $p\cdot 2p \mid n!$ so that $n!$ is a multiple of $p^2$, and perhaps even $n! = k^2p^2$. $\endgroup$
    – MJD
    Commented Jun 19 at 7:07

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