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.
 A: 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.
