`xy+x+y=3`
`⇒x(y+1)+y=3`
`⇒x(y+1)+(y+1)=4`
`⇒(x+1)(y+1)=4`
`⇒x+1;y+1∈Ư(4)={±1;±2;±4}`
Ta có bảng :
$\left[\begin{array}{ccc}x+1&1&2&4&-1&-2&-4\\y+1&4&2&1&-4&-2&-1\\x&0&1&3&-2&-3&-5\\y&3&1&0&-5&-3&-2\end{array}\right]$
Vậy `(x;y)=(0;3);(1;1);(3;0);(-2;-5);(-3;-3);(-5;-2)`