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The question is motivated by this

As $6! = 5\cdot 12^2$, in order to prove that $6!$ is not a square, we need to know that $5$ exists between $3$ and $6$. So Chebyshev's theorem seems necessary here. However, I believe this is the only case (other than 2) where $n! = p m^2$ with $p$ prime; the reason must be also in some result of Chebyshev kind: there are at least two different primes between $n$ and $\frac {n}{2}$ or something similar. But I have no proper proof.

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zhoraster
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  • I believe that prime number theorem implies that there cannot be such factorial. – barak manos Sep 09 '15 at 05:44
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    @barakmanos, I don't believe this. It can only imply that there no infinitely many such factorials. – zhoraster Sep 09 '15 at 06:11
  • search for prime gap upper bounds in the form of $\pi(x+x^\theta)-\pi(x)$ with $\theta<1$ (I think some exponent between $0.5$ and $0.6$ is the current best result), and check the small cases by hand. – meowmeow Sep 09 '15 at 18:13

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There is another solution :

$$10!=7\times 720^2$$ This is almost surely the largest one.

This PARI/GP program

? for(n=0,2000,if(isprime(core(n!),2)==1,print(n)))
2
6
10
?

shows that the only solutions upto $n=2000$ are $2,6,10$. To prove the conjecture, it would be enough to show that for every even number $n\ge 2000$, there are at least two primes $p,q$ with $\frac{n}{2}<p<q<n$.

Peter
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