I stumbled across the following problem in high school:$$ x^2 + y^2 = n! $$ I tested it within my laptop capabilities, watched a 3b1b video Pi in prime regularities, where he explains how to find the number of integer solutions based on prime factors. There doesn't seem to be any above $30!$. Maybe I'm wrong and there are infinitely many exceptions like $2$ and $6$, maybe the proof is too difficult for me to grasp or... I hope I'm just too blind to see the obvious.
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$\begingroup$ Hi and welcome to MO. What is your question? $\endgroup$– Amir SagivMar 30, 2019 at 12:31
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1$\begingroup$ Hi. The question is: are there any integers above 6 for which this equation has integer pairs (x,y) as solutions. $\endgroup$– BetydligMar 30, 2019 at 12:35
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24$\begingroup$ At least for sufficiently large $n$, there will be a prime $p \equiv 3 \bmod 4$ such that $p \le n < 2p$. Then $p$ divides $n!$ exactly once, hence $n!$ cannot be a sum of two squares. $\endgroup$– Michael StollMar 30, 2019 at 12:45
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For $n\geq 7$, Erdős proved in 1932 that there is a prime $n/2<p\leq n$ of the form $p=4k+3$. From this he deduces (in the same paper) that $1!$, $2!$, $6!$ are the only factorials which can be written as a sum of two squares.
answered Mar 30, 2019 at 13:42