8

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.

ulead86
  • 3,381
  • 3
    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
  • 3
    For a very beautiful geometric approach see Aubry's method. – Bill Dubuque Dec 10 '11 at 18:31
  • 3
    @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
  • Note that the result is also true for three squares, by Aubry's method. – Will Jagy Dec 10 '11 at 19:18

1 Answers1

3

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.

André Nicolas
  • 507,029
  • 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 http://mathoverflow.net/questions/69444/a-priori-proof-that-covering-radius-strictly-less-than-sqrt-2-implies-class-nu – 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
  • @Will Yes, I know about Pete's posts. In fact I believe Pete learned about these topics from some of my older posts on these topics in other forums. But, curiously, he never mentions such. Oh well, that's not the first time that's happened. – Bill Dubuque Dec 10 '11 at 19:41
  • @Bill Dubuque: I think of the Gaussian Integers argument as the "standard" one. But even if one uses a descent type of proof for dealing with sums of two squares, that would not make the argument circular. – André Nicolas Dec 10 '11 at 19:59
  • Bill, sorry to hear that. – Will Jagy Dec 10 '11 at 20:06