[tex]xy^2+2xy+x=32y\Rightarrow x(y+1)^2=32y\Rightarrow x=\frac{32y}{(y+1)^2}[/tex]
Để x nguyên thì [tex]\frac{32y}{(y+1)^2}\in \mathbb{Z}\Rightarrow \frac{32y(y+1)}{(y+1)^2}\in \mathbb{Z}\Rightarrow \frac{32(y+1)^2-32}{(y+1)^2}\in \mathbb{Z}\Rightarrow \frac{32}{(y+1)^2}\in \mathbb{Z}\Rightarrow 32\vdots (y+1)^2\Rightarrow (y+1)^2\in \left \{ 1;4;16 \right \}\Rightarrow y\in \left \{ 0;-2;1;-3;3;-5 \right \}[/tex]
Từ đó thế lại tìm x.