Proof: If $n \in \mathbb{N}$ is not the sum of two squares, then $n$ is also not the sum of two rational squares

I have to prove the following:

If $n \in \mathbb{N}$ is not representable by the sum of two squares, then $n$ is also not representable by the sum of two rational squares.

How do I start here? Any ideas would be fine.

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What results are you allowed to use? If you can use Fermat's characterisation of natural numbers that are sums of two squares, then this is trivial. – Srivatsan Dec 10 '11 at 18:21
I'm allowed to use Fermat's characerisation, but I don'T see how this is trivial... – ulead86 Dec 10 '11 at 18:24
For a very beautiful geometric approach see Aubry's method. – Bill Dubuque Dec 10 '11 at 18:31
@Daniel: If I have $n=r^2+s^2$ where $r,s\in\mathbb{Q}$, then I can multiply through by the common denominator and get $n a^2 =b^2+c^2$ for some integers $a,b,c$. Now ask yourself, what is the exact statement of Fermats characterization? – Eric Naslund Dec 10 '11 at 18:33
@Bill: Thanks for the link, nice approach :) and Eric: n is representable if and only if every prime divisior 4m+3 occurs with an even power. From there it was easy. Thanks for the hint. – ulead86 Dec 10 '11 at 19:11

We will show that the positive integer $n$ is representable as the sum of the squares of two integersif and only if $n$ is representable as the sum of the squares of two rationals. One direction is clear: If $n$ is representable as the sum of the squares of two integers, then $n$ is representable as the sum of the squares of two rationals. The other direction is harder.

It is well-known that if $n$ is positive, then the equation $x^2+y^2=n$ has integer solutions if and only if every prime divisor of $n$ of the form $4k+3$ occurs to an even power in the prime factorization of $n$.

If $n$ is representable as the sum of the squares of two rationals, then by bringing the rationals to a common denominator $z$, we can see that there exist integers $u$, $v$, and $w$, with $w\ne 0$, such that $$\left(\frac{x}{z}\right)^2+\left(\frac{y}{z}\right)^2=n.$$

If $(u,v,w)$ is a solution of the above equation, then $u^2+v^2=nw^2$, so $nw^2$ is representable as the sum of the squares of two integers. It follows that every prime divisor of $nw^2$ of the form $4k+3$ occurs to an even power. But then every prime divisor of $n$ of that form occurs to an even power, so $n$ is representable as the sum of the squares of two integers.

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Most likely that would be a circular proof for the OP. – Bill Dubuque Dec 10 '11 at 18:58
Ah ok, thanks. Seemed very easy :/ – ulead86 Dec 10 '11 at 19:07
@Daniel Be sure to double check that invoking this proof does not result in circularity (as it would in some textbooks). Which textbook are you using? – Bill Dubuque Dec 10 '11 at 19:16
Bill, I looked at your post on Aubry's method. Pete L. Clark on MO got very interested, part of that story culminated with mathoverflow.net/questions/69444/… – Will Jagy Dec 10 '11 at 19:25
@Bill I'm not using any textbooks, I'm using my prof's script and this was a homework question. So I guess (or better I hope :D), there's no circularity – ulead86 Dec 10 '11 at 19:32