\(x-y=x^2+xy+y^2\)
\(\Leftrightarrow x^2+xy+y^2-x+y=0\)
\(\Leftrightarrow x^2+x\left(y-1\right)+\left(y^2+y\right)=0\)
\(\Delta=\left(y-1\right)^2-4\left(y^2+y\right)\ge0\Leftrightarrow y^2-2y+1-4y^2-4y\ge0\)\(\Leftrightarrow-3y^2-2y+1\ge0\Leftrightarrow-1\le y\le\frac{1}{3}\)
Do \(\left\{{}\begin{matrix}y\in Z\\y\ge0\end{matrix}\right.\Rightarrow y=0\)
Thay vào pt: \(x=x^2\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy pt có nghiệm \(\left(x;y\right)=\left(0;0\right);\left(0;1\right)\)