2.5. (Yugoslavia, 74). Solve the equation

$$
x^{2}+x y+y^{2}=x^{2} y^{2}
$$

in integers.
2.5. Let the pair of numbers $x, y \in \mathbf{Z}$ satisfy the equation, then $x^{2}+2 x y+y^{2}=x^{2} y^{2}+x y$, i.e., $(x+y)^{2}=x y(x y+1)$.

If $x y>0$, then

$$
x y+1>\sqrt{x y(x y+1)}>x y
$$

$$
|x+y|=\sqrt{x y(x y+1)}
$$

lies between two consecutive integers $x y$ and $x y+1$, and thus cannot be an integer. Similarly, if $x y<-1$, then

$$
-x y-1<\sqrt{x y(x y+1)}<-x y
$$

and thus the number $|x+y|$ cannot be an integer again. Therefore, we have either $x y=0$ or $x y=-1$. In both cases, the equation is equivalent to the equality $x+y=0$, so either $x=y=0$, or $x=-y= \pm 1$ respectively. Thus, we get the answer: $(0 ; 0),(1 ;-1),(-1 ; 1)$.