$x+y+xy=2$
$→y(x+1)+x+1-3=0$
$→(y+1)(x+1)=3$
$→(y+1);(x+1)∈Ư(3)=\{±1;±3\}$
$→$ Ta có bảng:
$\begin{array}{|c|c|c|}\hline y+1&1&-1&3&-3\\\hline x+1&3&-3&1&-1\\\hline y&0&-2&2&-4\\\hline x&2&-4&0&-2\\\hline\end{array}$
mà $(x;y)∈\mathbb Z$
$→(x;y)=\{(2;0);(-4;-2);(0;2);(-2;-4)\}$