Integer solutions.



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Integer solutions.
The equation can be transformed into
$$
\left(x^{2}+y^{2}\right)(x+y-8)=8(x y+1) .
$$

If $x+y-8 \geqslant 6$, i.e., $x+y \geqslant 14$, then $x^{2}+y^{2} \geqslant \frac{(x+y)^{2}}{2}>4$. At this time, the left side of (1) $\geqslant$ $6\left(x^{2}+y^{2}\right)=4\left(x^{2}+y^{2}\right)+2\left(x^{2}+y^{2}\right) \geqslant 8 x y+2\left(x^{2}+y^{2}\right)>8(x y+1)$, so the equation has no integer solutions. If $x+y-8 \leqslant-4$, i.e., $x+y \leqslant 4$, the left side of (1) $\leqslant-4\left(x^{2}+y^{2}\right) \leqslant-4 \times$ $2|x y|<8 x y+1$, so the equation also has no integer solutions.

Therefore, the integer solutions $(x, y)$ of the equation should satisfy $-3 \leqslant x+y-8 \leqslant 5$. On the other hand, the right side of (1) must be even, which implies that $x, y$ must have the same parity, thus $x+y-8$ must be even, so it can only be -2, $0,2,4$. When $x+y-8=0$, $x y+1=0, x= \pm 1, y=\mp 1$. This contradicts $x+y=8$, so there is no solution. When $x+y-8=4$, $(x-y)^{2}=2$, there is no solution.
When $x+y-8=-2$, $x+y=6, x^{2}+y^{2}+4 x y+4=0$, i.e., $6^{2}+2 x y+4=0$, $x y=-20$. $x, y$ have opposite signs and are the roots of the equation $u^{2}-6 u-20=0$. However, this equation has no real roots. When $x+y-8=2$, similarly, we get $x+y=10, x y=16$. Therefore, $(x, y)=(2,8)$, $(8,2)$.
The solutions to this problem are $(x, y)=(2,8),(8,2)$.