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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: $$x^2+y^2=2005^{2004}$$

Surfer on the fall
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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 $x^2+y^2 = 2004^{2005}$.

Daniel Fischer
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  • and to solve $x^2+y^2=2005^{2004}$? – Surfer on the fall Feb 19 '14 at 12:13
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    $2005 = 5\cdot 401$. We have $5 = 2^2+1^2$ and $401 = 20^2 + 1^2$. We can get a solution from $(2+i)^{2004}\cdot (20+i)^{2004}$ for example then. There are many more. Or, to make it "simple", $x = 4\cdot 5^{1001}\cdot 401^{1002},; y = 3\cdot 5^{1001}\cdot 401^{1002}$. – Daniel Fischer Feb 19 '14 at 12:18
  • How did you find the "simple" solution? thanks a lot! – Surfer on the fall Feb 19 '14 at 12:22
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    We know that $5^2 = 4^2+3^2$, and that leaves a factor $5^{2002}\cdot 401^{2004}$, which is a square. Multiply the decomposition $5^2 = 4^2+3^2$ with that, $$(5\cdot\sqrt{5^{2002}401^{2004}})^2 = (4\cdot 5^{1001}401^{1002})^2 + (3\cdot 5^{1001}401^{1002})^2.$$ – Daniel Fischer Feb 19 '14 at 12:26