\(x^2-2x-xy=y-2\Leftrightarrow\left(x^2-2x+1\right)-y\left(x-1\right)=2y-1\)
\(z^2-yz=2y-1\)
\(4z^2-4yz+y^2=y^2+8y-4\)
\(\left(2z-y\right)^2=\left(y+4\right)^2-20\)
\(\Leftrightarrow\left(y+4\right)^2-\left(2z-y\right)^2=20\)
Chỉ có cặp nghiệm nguyên (2.10)
\(\left\{{}\begin{matrix}\left(y+4\right)=-6\\\left(y+4\right)=6\end{matrix}\right.\Rightarrow\)\(\left\{{}\begin{matrix}y=-10\\y=2\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}y=-10\\x^2-8x+12=0\end{matrix}\right.\)\(\left\{{}\begin{matrix}y=-10\\\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}y=2\\x^2-4x=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}y=2\\\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\end{matrix}\right.\)
Kết luận: (x,y)=(2,-10);(6,-10);(0,2);(4,2)