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I recently found this problem saying:

Find all $n$ such that there exists a solution of $a^2 + b^2 = n!$ for positive integers $a, b, n$.

I first thought to show $n!$ will contain some prime of the form $4k+3$ odd number of times after some value $n_0$. Then I thought by Bertrand's postulate $n!$ (for $n >3$ ) contains at least one prime only once. If any of them is of the form $4k+3$ we would be done. But I failed to make any more progress.

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    $\begingroup$ Well, it should be easy enough to determine the answer, if you can't prove it. As to a possible proof... There are versions of Bertrand's postulate for arithmetic progressions, see, e.g., this question $\endgroup$
    – lulu
    Apr 15 '20 at 18:45
  • $\begingroup$ I think n=2 is the only solution,but i still can't prove it.Can you give me some more hint? $\endgroup$ Apr 16 '20 at 17:20
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    $\begingroup$ $2$ is not the only solution. There is one other small solution (bigger than $2$). I don't know if the variants of Bertrand I mentioned are strong enough to prove the result or not, I did not try. $\endgroup$
    – lulu
    Apr 16 '20 at 17:35
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    $\begingroup$ It is enough to prove that if $ p \equiv 3 \bmod 4$ and $p\ge 7$, then there is a prime $q \equiv 3 \bmod 4$ strictly between $p$ and $2p$. I don't know a proof. $\endgroup$
    – lhf
    Jul 6 '20 at 12:18
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    $\begingroup$ A proof is posted here that $n$ must be $\leq6$. $\endgroup$ Jul 6 '20 at 15:54

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