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.)
Can n! be a perfect square when n is an integer greater than 1?
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2See [this](http://mathforum.org/library/drmath/view/54290.html). – 2010-11-30
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0@J.M.: I found the resolution very complex. Honestly, I could not understand it. – 2010-11-30
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1Actually, 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. – 2010-12-01
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0@ShreevatsaR: You're right. Thank you for participating. Thank you all. – 2010-12-01
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2Is there a proof of this fact which does not use Bertrand's postulate? – 2012-03-01
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0Another related question: http://math.stackexchange.com/questions/1812580/is-sqrtn-a-natural-number – 2016-09-19
5 Answers
Assume, $n\geq 4$. By Bertrand's postulate there is a prime, let's call it $p$ such that $\frac{n}{2}
n$. 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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1Careful. You should say $n/2 < p \le n$ or else your statement is wrong when $n = 2, 3$. – 2010-11-30
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0@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. – 2010-11-30
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2I don't get this. There is, in fact, a prime $p$ such that $3/2 < p < 3$. So, why doesn't this work for $n=3$? – 2015-02-26
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
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: http://ad.bolyai.hu/~p_erdos/1975-46.pdf.