$c>2$ $4\not | a^2+b^2+1 \to c<4 \to c=3$ $a^2+b^2+1=6 \to (a,b,c)=(2,1,3),(1,2,3)$

liliput wrote: Determine all positive integers $a$, $b$ and $c$ which satisfy the equation $$a^2+b^2+1=c!.$$ Proposed by Nikola Velov By two square sum theorem we have $a^2+b^2\not\equiv 3 \pmod{4}$ so $c<4$. $c=3$ gives us $(2,1,3)$ and $(1,2,3)$. $c<2$ gives us no solutions.

From mod 4 , we get that if c>2 then $a^2+b^2$ is odd , from there we get that this number isn't dividable with 4 c<4, abd the solutions is (2,1,3 ) , (1,2,3)