1. Zadatak

Vrijedi

$$\begin{aligned}x^2+4y^2+2x-12y+10&=x^2+2x+1+4y^2-12y+9\\&=(x+1)^2+(2y-3)^2=0.\end{aligned}$$

Zato slijedi $x=-1$, $2y-3=0$ odnosno $y=\frac32$. Na kraju

$$x^{2008}+2008y=(-1)^{2008}+2008\cdot\frac32=1+3012=3013.$$