# Circle $x^2 + y^2 = n!$ doesn't hit any lattice points for any $n$ except for $0$, $1$, $2$ and $6$ or does it?

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.

• Hi and welcome to MO. What is your question? – Amir Sagiv Mar 30 '19 at 12:31
• Hi. The question is: are there any integers above 6 for which this equation has integer pairs (x,y) as solutions. – Betydlig Mar 30 '19 at 12:35
• 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. – Michael Stoll Mar 30 '19 at 12:45

For $$n\geq 7$$, Erdős proved in 1932 that there is a prime $$n/2 of the form $$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.