`x+4 = y(x+1)`
`-> x +4 =xy +y`
`-> x - xy = -4 + y =-1 + y - 3`
`-> x(1-y) + (1-y) =-3`
`-> (1-y)(x+1) =-3`
`-> 1-y ; x+1 \in Ư(-3) = { ±1 ; ±3}`
Ta có bảng sau :
$\begin{array}{|c|c|} \hline x+1&1&-1&3&-3 \\\hline 1-y&-3&3&-1&1 \\\hline x&0&-2&2&-4 \\\hline y&4&-2&2&0 \\\hline\end{array}$
Vậy `(x ; y) = (0 ;4) ; (-2 ; -2) ; (2 ; 2) ; (-4 ; 0)`