Đề thiếu $x,y \in Z$
$x^2+y^2-x-y=8$
$\to 4x^2+4y^2-4x-4y=32$
$\to 4x^2-4x+1+4y^2-4y+1=34$
$\to (2x-1)^2+(2y-1)^2=34=25+9=9+25$(do (2x-1)^2 và (2y-1)^2 lẻ)
$TH1:$
$\begin{cases}(2x-1)^2=25\\(2y-1)^2=9\\\end{cases}$
$\to \begin{cases}\left[ \begin{array}{l}2x-1=-5\\2x-1=5\end{array} \right.\\\left[ \begin{array}{l}2y-1=-3\\2y-1=3\end{array} \right.\\\end{cases}$
$\to \begin{cases}\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\\\left[ \begin{array}{l}y=-1\\y=2\end{array} \right.\\\end{cases}$
$TH2:$
$\begin{cases}(2y-1)^2=25\\(2x-1)^2=9\\\end{cases}$
$\to \begin{cases}\left[ \begin{array}{l}2y-1=-5\\2y-1=5\end{array} \right.\\\left[ \begin{array}{l}2x-1=-3\\2x-1=3\end{array} \right.\\\end{cases}$
$\to \begin{cases}\left[ \begin{array}{l}y=-2\\y=3\end{array} \right.\\\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.\\\end{cases}$
