parmenides51 19.09.2024 01:19 Find all pairs $(m, n)$ of positive integers such that $m^2 + n^2 =(m + 1)(n + 1).$
navier3072 03.10.2024 11:40 $m^2-(n+1)m+n^2-n-1=0$ Thus, $(n+1)^2-4(n^2-n-1)=-3n^2+6n+5$ is a perfect square. But $-3n^2+6n+5 \equiv 2 \pmod{3}$, contradiction.