Giải phương trình nghiệm nguyên: x²y – xy + 2x – 1 = y² – xy² – 2y.$\left[\begin{array}{l}\left\{\begin{array}{l}x+y=1\\ xy-y+2=1\end{array}\right.\\ \left\{\begin{array}{l}x+y=-1\\ xy-y+2=-1\end{array}\right.\end{array}\right.$

x²y – xy + 2x – 1 = y² – xy² – 2y

⇔ x²y + xy² – xy – y² + 2x + 2y = 1

⇔ xy (x + y) – y(x + y) + 2 (x + y) = 1

⇔ (x + y) (xy – y + 2) = 1

Vì x, y nguyên nên:

$\left[\begin{array}{l}\left\{\begin{array}{l}x+y=1\\ xy-y+2=1\end{array}\right.\\ \left\{\begin{array}{l}x+y=-1\\ xy-y+2=-1\end{array}\right.\end{array}\right.$

⇔ $\left[\begin{array}{l}\left\{\begin{array}{l}x=1-y\\ \left(1-y\right)y-y+2=1\end{array}\right.\\ \left\{\begin{array}{l}x=-1-y\\ \left(-1-y\right)y-y+2=-1\end{array}\right.\end{array}\right.$

⇔ $\left[\begin{array}{l}\left[\begin{array}{l}\left\{\begin{array}{l}x=0\\ y=1\end{array}\right.\\ \left\{\begin{array}{l}x=2\\ y=-1\end{array}\right.\end{array}\right.\\ \left[\begin{array}{l}\left\{\begin{array}{l}x=-2\\ y=1\end{array}\right.\\ \left\{\begin{array}{l}x=2\\ y=-3\end{array}\right.\end{array}\right.\end{array}\right.$

Vậy (x; y) ∈ {(0; 1); (2; –1); (–2; 1); (2; –3)}.