[tex]x(y^{2}+1)+2y(x-2)=0\Rightarrow x(y^2+2y+1)-4y=0 \Rightarrow x= \frac{4y}{y^2+2y+1}[/tex]
Để [tex]x\in \mathbb{Z}\Leftrightarrow 4y\vdots y^2+2y+1\Rightarrow 4y(y+2)\vdots y^2+2y+1\Rightarrow 4(y^2+2y+1)-4\vdots (y+1)^2\Rightarrow 4\vdots (y+1)^2\Rightarrow (y+1)^2\in \left \{ 1;4 \right \}\Rightarrow y\in \left \{ 0;-2;1;-3 \right \}[/tex]
Thế vào ta có các cặp (x,y) thỏa mãn: (0,0);(-8,-2);(1,1);(-3,-3)