\(x^2+y^2-4\left(x-y-1\right)=0\)
\(\Leftrightarrow x^2+y^2-4x+4y+4=0\)
\(\Leftrightarrow x^2-4x+4+y^2+4y+4-4=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+2\right)^2-4=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+2\right)^2=4\)
Ta có: x,y nguyên \(\Leftrightarrow\) x - 2 nguyên, y + 2 nguyên
Từ 2 điều trên \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=0\\y+2=2\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=2\\y+2=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy (x,y) = (2;0) hoặc (x,y) = (4;-2).
