$xy+x-y=4$
$⇒x(y+1)=4+y$
Nếu $y+1=0⇒y=-1⇒0x=3$ (loại)
Chia 2 vế cho $y+1$$\neq0$, ta được:
$x=\frac{y+4}{y+1}$
$⇒x=1+\frac{3}{y+1}$
$⇒y+1∈Ư(3)=${$±1;±3$}
$y+1=1⇒y=0⇒x=4$
$y+1=-1⇒y=-2⇒x=-2$
$y+1=3⇒y=2⇒x=2$
$y+1=-3⇒y=-4⇒x=0$
Vậy $(x,y)∈${$(4;0);(-2;-2);(2;2);(0;-4)$}