, 2008

Solution:
The given equation can be written as:
$$\begin{aligned} & x(x-y)=8y-7 \\ & x^{2}+7=y(x+8) \end{aligned}$$
Let $x+8=m$, $m \in \mathbb{N}$. Then we have: $x^{2}+7 \equiv 0 \pmod{m}$, and $x^{2}+8x \equiv 0 \pmod{m}$. So we obtain that $8x-7 \equiv 0 \pmod{m}$ $\quad$ (1).
Also we obtain $8x+8^{2}=8(x+8) \equiv 0 \pmod{m}$ $\quad$ (2).
From (1) and (2) we obtain $(8x+64)-(8x-7)=71 \equiv 0 \pmod{m}$, therefore $m \mid 71$, since $71$ is a prime number, we have:
$x+8=1$ or $x+8=71$. The only accepted solution is $x=63$, and from the initial equation we obtain $y=56$.
Therefore the equation has a unique solution, namely $(x, y)=(63,56)$.

Solution 2:
The given equation is $x^{2}-x y+7-8 y=0$.
Discriminant is $\Delta=y^{2}+32y-28=(y+16)^{2}-284$ and must be perfect square. So $(y+16)^{2}-284=m^{2}$, and its follow $(y+16)^{2}-m^{2}=284$, and after some casework, $y+16-m=2$ and $y+16+m=142$, hence $y=56, x=63$.