3. If $x^{2}+y^{2}+2 x-4 y+5=0$, what is $x^{2000}+2000 y$?
Solution: If $x^{2}+2 x+y^{2}-4 y+5=0$, then $x^{2}+2 x+1 y^{2}-4 y+4=0$; $(x+1)^{1}+(y-2)^{2}=0$. From here $x+1=0$ and $y-2=0$, i.e. $x=-1, y=2$. Therefore $x^{2000}+2000 y=(-1)^{2000}+2000 y=1+4000=4001$.