
NT3 Find all pairs $(x, y)$ of integers which satisfy the equation

$$
(x+y)^{2}\left(x^{2}+y^{2}\right)=2009^{2}
$$


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 $|s| \leq 41$. 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 $(40,9),(9,40),(-40,-9),(-9,-40)$.
