Giải phương trình nghiệm nguyên: x2y – xy + 2x – 1 = y2 – xy2 – 2y.

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:

x+y=1xy−y+2=1x+y=−1xy−y+2=−1

⇔ x=1−y1−yy−y+2=1x=−1−y−1−yy−y+2=−1

⇔ x=0y=1x=2y=−1x=−2y=1x=2y=−3

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