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

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.)

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

Assume, $$n\geq 4$$. By Bertrand's postulate there is a prime, let's call it $$p$$ such that $$\frac{n}{2} . Suppose, $$p^2$$ divides $$n$$. Then, there should be another number $$m$$ such that $$p 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.

• Careful. You should say $n/2 < p \le n$ or else your statement is wrong when $n = 2, 3$. Commented Nov 30, 2010 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. Commented Nov 30, 2010 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 $n=3$?
– XYZT
Commented Feb 26, 2015 at 5:22

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

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.

• 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.

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.

• Is there any English language source? let us know! Commented Jan 5, 2021 at 6:58
• @SmritidipaMukherjee The old link is dead. I updated the link. Commented Jan 10, 2021 at 4:56
• 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. Commented Jun 19 at 7:23

√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.

• $n<p^2$ 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 $n! = k^2p^2$.
– MJD
Commented Jun 19 at 7:07