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The question is: with a fixed integer $n$, what are the points with integer coordinates $(a,b)$ so that $a^2 + b^2 = n^3$?

The equation is symmetric in $a$ and $b$, so if $(x,y)$ is a solution, then $(y,x)$ is a solution as well.

Obviously if $n$ is a perfect square, so we always have the solution $a=n^{3/2}$; b=0.

I think there is a solution only if $n$ is a perfect square and that this is the only solution.

I tried to prove it this way:

I can always write

$\begin{align*}a&=n^{3/2} \sin(t)\\ b&=n^{3/2} \cos(t)\end{align*}$

if $n$ is not a perfect square I would like to say that $a$ and $b$ can't both be integers, but I really can't :(

if $n$ is a perfect square I need that if $\sin(t)$ is rational then $\cos(t)$ can't be (except for the case $\sin(t)=1$; $\cos(t)=0$). I tried to use the equality $\sin^2 (t) + \cos^2 (t) = 1$ but I know I'm missing something.

Thank you for the help.

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    $4+4=8$. $121+4=125$.2011-11-24
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    Ok I feel stupid now :( What was I thinking? O.O2011-11-24

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