sum of the squares of two rationals numbers

Question

Show that an irreducible fraction $x/y$, where $x, y$ are natural numbers, is the sum of the squares of two rationals numbers if and only if each of the numbers $x, y$ is the sum of the squares of two integers.

I showed $"\Leftarrow"$ and don't know how to show $"\Rightarrow"$.

Answer 1

Assume $$\frac{x}{y}=(\frac{a}{b})^2+(\frac{c}{d})^2=\frac{a^2d^2+b^2c^2}{b^2d^2}$$

If a prime $p$ divides $b^2d^2$ and $a^2d^2+b^2c^2$, then $p|b$ or $p|d$. Hence $p|ad$ or $p|bc$. This impies that $p$ divides $ad$ AND $bc$. Hence numerator and denominator are divisible by $p^2$. The squares remain squares, and we can continue until the fraction is irreducible.

Answer 2

To avoid troublesome (hence error generating) calculations, it seems better to turn to a more "functorial" approach, i.e. to view a sum of two squares of rationals as a norm from $\mathbf Q(i)/\mathbf Q$. By hypothesis, your irreducible fraction reads $x/y = N(Z), Z\in \mathbf Q(i)$. But since $\mathbf Q(i)$ is the field of fractions of the Gauss ring $\mathbf Z[i]$, we can as well write $Z=z/z'$, with co-prime $z,z'\in \mathbf Z[i]$ , and the original equation becomes (*) $xz\bar z=yz'\bar z'$, where the bar denotes complex conjugation.

We now exploit the classical property that $\mathbf Z[i]$ is a UFD. For notational convenience, we'll ignore the units of this ring (which are merely $u=\pm 1, \pm i$) when they don't play a part in the arguments. Let us show that $z\bar z$ and $z'\bar z'$ are in fact co-prime in $\mathbf Z$. When reasoning in $\mathbf Z[i]$, the only thing to show will actually be that $z, \bar z'$ are also co-prime. Let $p$ be a rational prime dividing both $z\bar z,z'\bar z'$ @. There are three cases to consider according to the prime decomposition of $p$ in $\mathbf Z[i]$ : (1) If $p$ is inert, i.e. $p$ remains a prime in $\mathbf Z[i]$, condition @ implies that $p$ divides $z$ or $\bar z$, hence $p$ divides both $z,\bar z$ because $p$ is fixed by complex conjugation; similarly $p$ divides both $z',\bar z'$: contradiction with the co-primarity of $z,z'$ ; (2) If $p$ is totally decomposed, i.e. $p=\pi\bar\pi$, where $\pi$ is a prime of $\mathbf Z[i]$ and $\bar\pi \neq u\pi$, condition @ implies, say, that $\pi$ divides $z$ (hence does not divide $z'$) and $\bar\pi$ divides $\bar z$ (hence not $\bar z'$) : contradiction ; (3) If $p$ is ramified, then necessarily $p=2=-i{\pi}^2=\pi\bar\pi$, with $\pi=1-i$, and @ implies that $\pi$ divides both $z,\bar z$, hence does not divide neither $z'$ nor $\bar z'$ : contradiction.

Conclusion. We have shown that $z\bar z$ and $z'\bar z'$ are co-prime, and since $x$ and $y$ are co-prime by hypothesis, equation (*) implies that $x=z'\bar z'$ and $y=z\bar z$ ./.