# Integer points of a circumference which radius in $n^{3/2}$

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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$4+4=8$. $121+4=125$. – Chris Eagle Nov 24 '11 at 16:58
Ok I feel stupid now :( What was I thinking? O.O – ugosugo Nov 24 '11 at 17:27

There is a complete description of the integers that can be written as sum of 2 squares : see http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares

(a theorem of fermat states that it is exactly the integers such that their odd prime factors all have a rest equal to 1 mod 4).

it can also be shown that any integers can be written as a sum of 4 squares.

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hum, what's wrong with my answer ? pretty sure I didn't insult anyone and gave a complete answer on the existence of solutions, with a reference... was it because I didn't mention the fact that this property is stable by product and therefore there is no loss of generality in replacing $n^3$ by a prime ? – Glougloubarbaki Nov 24 '11 at 17:41

The following theorem gives the number of representations of the positive integer $N$ as a sum of two squares. Please note that for example $5=1^2+2^2$, $5=2^2+1^2$, $5=(-1)^2+2^2$ (and so on) count as different representations. But that's exactly what you want for your geometric problem. Let $$N=2^k (p_1^{a_1} p_2^{a_2}\cdots p_s^{a_s})(q_1^{b_1} q_2^{b_2}\cdots q_t^{b_t}),$$ where the $p_i$ are distinct primes of the form $4u+1$, and the $q_i$ are distinct primes of the form $4u-1$. (Here $s$ and/or $t$ can be $0$, and $k$ may be $0$.)

If one or more of the $b_i$ is odd, there are no representations of $N$. If all the $b_i$ are even, then the number of representations of $N$ is $$4(a_1+1)(a_2+1)\cdots(a_s+1).$$ (An empty product is interpreted to be $1$.)

In your case, we have $N=n^3$. If for some prime $q$ of the form $4u-1$, the largest $b$ such that $q^b$ divides $n$ is odd, then the same will be true for $n^3$, and there are no representations. Otherwise, we can adapt the formula for the number of representations of $N$ to get a formula for the number of representations of $n^3$ in terms of the prime factorization of $n$.

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