Iest * Find the integer solutions $x, y$ of the equation $(x y-1)^{2}=(x+1)^{2}+(y+1)^{2}$.
Substitute. Let $u=x+1, v=y+1$ into the equation, and after rearrangement, we get $u v(u v-2 u-2 v+2) = 0$.
(1) If $u=0$, then $x=-1, u \in \mathbf{Z}$ is a solution to the equation.
(2) If $v=0$, then $y=-1, x \in \mathbf{Z}$ is a solution to the equation.
(3) If $u v-2 u-2 v+2=0$, then $v=\frac{2 u-2}{u-2}=2+\frac{2}{u-2}$. Since $v$ is an integer, we know that $u-2 = \pm 1, \pm 2$. Thus, $u=3,1,4,0$.
When $u=3$, $v=4$, so $x=2, y=3$; when $u=4$, $v=3$, so $x=3, y=2$; when $u=1$, $v=0$, which is the case in (2); when $u=0$, it is the case in (1).
Therefore, the solutions to the equation are $\left\{\begin{array}{l}x=-1 \\ y \in \mathbf{Z}\end{array},\left\{\begin{array}{l}x \in \mathbf{Z} \\ y \in -1\end{array},\left\{\begin{array}{l}x=2 \\ y=3\end{array},\left\{\begin{array}{l}x=3 \\ y=2 .\end{array}\right.\right.\right.\right.$