$2009^2 \geq \frac{(x+y)^2 (x+y)^2}{2} \to |x+y|\leq 53$ $|x+y| |2009=7^2*41$ So $|x+y|=1,7,41,49$ Soving all cases we will find $(x,y)=(-9,-40),(9,40),(-40,-9),(40,9)$

RagvaloD wrote: $2009^2 \geq \frac{(x+y)^2 (x+y)^2}{2}$ How did you come to this inequality?

programjames1 wrote: RagvaloD wrote: $2009^2 \geq \frac{(x+y)^2 (x+y)^2}{2}$ How did you come to this inequality? For$x\in \mathbb{R}^+$, we have $$(x-y)^2 \geq 0$$$$\Rightarrow x^2+y^2 \geq 2xy$$$$\Rightarrow 2(x^2+y^2) \geq x^2+2xy+y^2$$$$\Rightarrow x^2+y^2 \geq \dfrac {(x+y)^2}{2}$$$$\Rightarrow (x+y)^2 (x^2+y^2) \geq \dfrac{(x+y)^2 \cdot (x+y)^2}{2}$$$$\Rightarrow 2009^2 \geq \dfrac{(x+y)^2 \cdot (x+y)^2}{2}$$

Another solution: Let $s=x+y$ and $p=xy$ The equation is equivalent with $$s^2-2p^2=(2009/s)^2$$We have that $s|2009$. Now we consider cases $p>0$ and $p<0$ and find restrictions for $|s|$...