` (y+2)x^3 - y^2-2y -1=0`
` => (y+2)x^3 - (y^2+2y) -1 = 0`
`=> (y+2)x^3 - y(y+2) =1`
` => (x^3-y)(y+2) = 1`
Vì ` x ; y \inZ` nên ta có ` x^3-y ; y+2 \in Ư(1)`
Trường hợp `1`
$\begin{cases}\\ y+2=1 \\\\\\x^3-y =1 \\\\\end{cases}$
`\to`
$\begin{cases}\\ y=-1 \\\\\\ x^3 +1 = 1 \\\\\end{cases}$
`\to`
$\begin{cases}\\ y=-1 \\\\\\ x = 0 \\\\\end{cases}$
Trường hợp `2`
$\begin{cases}\\ y+2=-1 \\\\\\x^3-y =-1 \\\\\end{cases}$
`\to`
$\begin{cases}\\ y=-3 \\\\\\x^3+3 =-1 \\\\\end{cases}$
`\to`
$\begin{cases}\\ y=-3 \\\\\\ x ∉ Z \\\\\end{cases}$ (loại)
Vậy ` (x;y) \in {0;-1}`
