\(\Leftrightarrow y^2-1=\left(y+2\right)x^2\)
Nhận thấy \(y=-2\) không phải nghiệm
\(\Rightarrow x^2=\frac{y^2-1}{y+2}=y-2+\frac{3}{y+2}\)
Do \(x\) nguyên \(\Rightarrow y+2=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow y=\left\{-5;-3;-1;1\right\}\) \(\Rightarrow x=\left\{-8;-8;0;0\right\}\)