x² + 2y² + 2xy + 3y − 4 = 0

⇔ 4x² + 8y² + 8xy + 12y − 16 = 0

⇔ (4x² + 8xy + 4y²) + (4y² + 12y + 9) = 25

⇔ (2x + 2y)² + (2y + 3)² = 25 = 0 + 5² = 3² + 4²

Do x; y là số nguyên và 2y + 3 là số lẻ ⇒ (2y + 3)² thuộc {5²; 3²}.

Xét các TH xảy ra:

TH1) $\left\{\begin{array}{l}2x+2y=0\\ 2y+3=5\end{array}\right.\iff \left\{\begin{array}{l}x+y=0\\ y=1\end{array}\right.\iff \left\{\begin{array}{l}x=-1\\ y=1\end{array}\right.$

TH2) $\left\{\begin{array}{l}2x+2y=0\\ 2y+3=-5\end{array}\right.\iff \left\{\begin{array}{l}x+y=0\\ y=-4\end{array}\right.\iff \left\{\begin{array}{l}x=4\\ y=-4\end{array}\right.$

TH3) $\left\{\begin{array}{l}2x+2y=4\\ 2y+3=3\end{array}\right.\iff \left\{\begin{array}{l}x+y=2\\ y=0\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=0\end{array}\right.$

TH4) $\left\{\begin{array}{l}2x+2y=-4\\ 2y+3=-3\end{array}\right.\iff \left\{\begin{array}{l}x+y=-2\\ y=-3\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ y=-3\end{array}\right.$

TH5) $\left\{\begin{array}{l}2x+2y=4\\ 2y+3=-3\end{array}\right.\iff \left\{\begin{array}{l}x+y=2\\ y=-3\end{array}\right.\iff \left\{\begin{array}{l}x=5\\ y=-3\end{array}\right.$

TH6) $\left\{\begin{array}{l}2x+2y=-4\\ 2y+3=3\end{array}\right.\iff \left\{\begin{array}{l}x+y=-2\\ y=0\end{array}\right.\iff \left\{\begin{array}{l}x=-2\\ y=0\end{array}\right.$

Vậy các cặp số nguyên (x, y) ∈ {(–1, 1); (4, –4); (2, 0); (1, –3); (5, –3); (–2, 0)} thỏa mãn đề bài.