Không mất tính tổng quát, giả sử \(m\ge n\)
\(2\left(m^2+n^2\right)-1=\left(m+n\right)^2-1+\left(m-n\right)^2\)
\(=\left(m+n+1\right)\left(m+n-1\right)+\left(m-n\right)^2\)
\(\Rightarrow\left(m-n\right)^2⋮\left(m+n+1\right)\)
\(\Rightarrow\left(m-n\right)⋮\left(m+n+1\right)\) (do \(m+n+1\) nguyên tố)
\(\Rightarrow\left[{}\begin{matrix}m-n\ge m+n+1\\m-n=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2n+1\le0\left(vô-lý\right)\\m=n\end{matrix}\right.\)
\(\Rightarrow m=n\Rightarrow m.n=m^2\) là SCP
