, 2009

Solution:
Let $x+y=s$, $x y=p$ with $s \in \mathbb{Z}^{*}$ and $p \in \mathbb{Z}$. The given equation can be written in the form
$$s^{2}\left(s^{2}-2 p\right)=2009^{2}$$
or
$$s^{2}-2 p=\left(\frac{2009}{s}\right)^{2}$$
So, $s$ divides $2009=7^{2} \times 41$ and it follows that $p \neq 0$.
If $p>0$, then $2009^{2}=s^{2}\left(s^{2}-2 p\right)=s^{4}-2 p s^{2}<s^{4}$. We obtain that $s$ divides $2009$ and $|s| \geq 49$. Thus, $s \in\{ \pm 49, \pm 287, \pm 2009\}$.
- For $s= \pm 49$, we have $p=360$, and $(x, y)=\{(40,9),(9,40),(-40,-9),(-9,-40)\}$.
- For $s \in\{ \pm 287, \pm 2009\}$ the equation has no integer solutions.
If $p<0$, then $2009^{2}=s^{4}-2 p s^{2}>s^{4}$. We obtain that $s$ divides $2009$ and $|s| \leq 41$. Thus, $s \in\{ \pm 1, \pm 7, \pm 41\}$. For these values of $s$ the equation has no integer solutions.
So, the given equation has only the solutions $(40,9),(9,40),(-40,-9),(-9,-40)$.