How to solve equations like $x^2+y^2=2004^{2005}$?

Question

I've found this kind of equation but I think I haven't enough mathematical tools to solve it. What would you do? $$x^2+y^2=2004^{2005}$$ Another kind: $$x^2+y^2=2005^{2004}$$

Answer 1

Look at the prime factorisation of $2004 = 2^2\cdot 3\cdot 167$. Note that if $3$ divides the sum of two squares, it divides both squares themselves (the same for $167$, but we need only one such prime). Then, since $2004^{2005}$ has an odd number of factors $3$, conclude that there are no integers $x,y$ with $x^2+y^2 = 2004^{2005}$.