Determine all pairs of integers $(x, y)$ such that $9 x y - x^{2} - 8 y^{2} = 2005$.
We have:

$$
\begin{aligned}
9 x y - x^{2} - 8 y^{2} = 2005 & \Leftrightarrow x y - x^{2} + 8 x y - 8 y^{2} = 2005 \\
& \Leftrightarrow x(y - x) + 8 y(x - y) = 2005 \\
& \Leftrightarrow (x - y)(8 y - x) = 2005(*)
\end{aligned}
$$

Observe that the prime factorization of 2005 is $5 \cdot 401$.

Moreover, the sum of the factors $x - y$ and $8 y - x$ is $7 y$, which is a multiple of 7. We must then write 2005 as the product of two factors, whose sum is a multiple of 7. For this, the factors must be $\pm 5$ and $\pm 401$. The sum of the factors is $\pm 406$.

Thus, by $(*)$ we have:

$$
\left\{\begin{array} { l l l } 
{ x - y = 5 } & { \text { and } 8 y - x = 401 } \\
{ x - y = 401 } & { \text { and } 8 y - x = 5 } \\
{ } & { \text { or } } \\
{ x - y = -5 } & { \text { and } 8 y - x = -401 } \\
{ } & { \text { or } } & { \text { or } y = 58 } \\
{ x - y = -401 } & { \text { and } 8 y - x = -5 }
\end{array} \quad \Leftrightarrow \left\{\begin{array}{ll}
x = 63 & \text { and } y = 58 \\
x = 459 & \text { or } \\
x = -63 & \text { and } y = -58 \\
x = -459 & \text { and } y = -58
\end{array}\right.\right.
$$

Therefore, the solutions are $(63, 58)$, $(459, 58)$, $(-63, -58)$, and $(-459, -58)$.