An 'identitial' problem

$x^2+y^2+2xy-2008x-2008y-2009=0$ $\Rightarrow(x+y)^2=2008x+2008y+2009$ $\Rightarrow(x+y)^2=2008(x+y)+2008+1$ $\Rightarrow(x+y)^2-1^2=2008(x+y+1)$ $\Rightarrow(x+y+1)(x+y-1)=2008(x+y+1)$ $\Rightarrow\boxed{x+y=2009}$ The solutions are:

$x=1,y=2008$ $x=2,y=2007...........\space \text{and so on till}\space x=2008,y=1$

$\therefore\space$ Total number of solutions $=\large\frac{2008-1}{1}+1=\boxed{2008}.$