Tìm x, y, z nguyên dương thỏa mãn x + 2y + 2z = xyz
Xét 2 TH
TH1: $x=1$
$\Leftrightarrow (y - 2)(z - 2) = 5$
$\Leftrightarrow (y - 2, z - 2) = (1, 5) = (5, 1) \Leftrightarrow (y, z) = (3, 7) = (7, 3)$
$\Rightarrow (x, y, z) = (1, 3, 7) = (1, 7, 3)$
TH2: $x \geq 2$
$\Leftrightarrow 2(y + z) = x(yz - 1) \geq 2(yz - 1) \Leftrightarrow y + z \geq yz - 1$
$(y - 1)(z - 1) \leq 2 \Leftrightarrow (y - 1, z - 1) = (1, 1)$ hoặc $(y - 1, z - 1) = (1, 2) = (2, 1)$
Nghiệm $(y, z) = (2, 2)$ bị loại, $(y, z) = (2, 3) = (3, 2) \Rightarrow (x, y, z) = (2, 2, 3) = (2, 3, 2)$