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If $n\geq2$, then $n!$ is not a perfect square. The proof of this follows easily from Chebyshev's theorem, which states that for any positive integer $n$ there exists a prime strictly between $n$ and $2n-2$. A proof can be found here.

Two weeks and four days ago, one of my classmates told me that it's possible to prove that $n!$ is never a perfect square for $n\geq2$ without using Chebyshev's theorem. I've been trying since that day to prove it in that manner, but the closest I've gotten is, through the use of the prime number theorem, showing that there exists a natural number $N$ such that if $n\geq N$, $n!$ is not a perfect square. This isn't very close at all. I've tried numerous strategies over the past weeks and am now trying to use the Sylow theorems on $S_n$ to somehow show that $|S_n|$ can't be square (I haven't made any progress).

Was my classmate messing with me, or is there really a way to prove this result without Chebyshev's Theorem? If it is possible, can someone point me in the right direction for a proof?


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"Isn't close at all"? All but finitely many cases is almost the entire problem: if you can get effective bounds on $N$ you only need to check finitely many cases. But the PNT is much stronger than Bertrand's postulate. –  Qiaochu Yuan Apr 9 '11 at 20:12
I'll take your suggestion and work on finding a bound tonight. However, the strategy of my prime number theorem proof was to prove what is essentially Chebyshev's theorem just for primes. I'm not sure if this would count as not using Chebyshev's theorem, since my (eventual?) proof could be extended to the all natural numbers if the $N$ for both of them is the same. I'll check, and maybe they'll be different, but hopefully there's another way to prove this that uses more elementary methods. –  Lincoln Blackham Apr 9 '11 at 21:42
@Qiaochu: Using a bound on $\pi(x)$ that I found, I was able to get an upper bound of 91 for $N$. However, as I said in my previous comment, my solution still petty much uses Chebyshev's theorem. I still want to see if there's a solution that doesn't use it in any form. –  Lincoln Blackham Apr 12 '11 at 12:40
* You would like a prime $p$ such that $p\equiv -1 \mod 4$ and $n!\equiv -1 \mod p$, but unfortunately the residue of $n!+1$ is not $-1$ modulo 4 and all simple variations do not seem to work, either. –  Phira Apr 18 '11 at 20:17
*Another possible approach: Can you find an argument that, say, $\binom {2k}{k}$ is not a square? –  Phira Apr 18 '11 at 20:17

1 Answer 1

Yes, use Bertrand's postulate. This is a well known way to prove your question. The postulate states that for n>3, there exists a prime number such that n>p>(n/2). Let's assume for now that there exists one value of p such that p^2 divides n. Now, if p^2 divides n, then there must be some n-1>K>p such that p divides K. So, K>(or)=2p>n (from Bertrand's postulate mentioned earlier. Multiply entire inequality by 2). But this is a contradiction to our assumption that n>K. So, for n>3, n! cannot be divisible by p^2 and hence cannot be a perfect square. You can see for a fact that 3! and 2! aren't perfect squares. Hence, proved.

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Bertrand's postulate is exactly the result that the author was trying to avoid. –  Charles May 27 at 16:10
Well, one could always use induction. –  ShankRam May 27 at 16:25
@ShankRam ok, how could one use induction? –  supinf Jun 26 at 15:58
@supinf refer this : mathforum.org/library/drmath/view/54290.html –  ShankRam Jul 30 at 4:40

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