Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
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
1  
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
2  
Is there a proof of this fact which does not use Bertrand's postulate? –  Beni Bogosel Mar 1 '12 at 14:50

4 Answers 4

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.

share|improve this answer
1  
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 at 5:22

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

share|improve this answer
  • 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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.