Exercise 3.6.8 (If $n = r^2 + s^2,\ r,s \in \mathbb{Q}$ then $n = a^2 + b^2,\ a,b \in \mathbb{Z}$)

Proof. Let $n$ be a positive integer, such that $n=r^2+s^2$, where $r,s$ are rational numbers. We can write $r=a∕d,s=b∕d$ with a common denominator $d$, where $a,b\in \mathbb{Z},d\in {\mathbb{N}}^{\text{∗}}$. Then

Let $q$ be a prime factor of $n$ of the form $q=4k+3$, and $\gamma ={\nu }_q(n)$ the exponent of $q$ in the decomposition of $n$ in prime number. Put $N=d^{2}n=a^2+b^2$. Since $N$ is a sum of two squares, ${\nu }_q(N)$ is even byTheorem 3.22. Then ${\nu }_q(N)=2{\nu }_q(d)+{\nu }_q(n)$, therefore $\gamma ={\nu }_q(n)$ is even. Since this is true for every prime factor of $n$ of the form $4k+3$, $n$ is the sum of the squares of two integers by Theorem 3.22.

If a positive integer can be expressed as a sum of the squares of two rational numbers then it can be expressed as the sum of the squares of two integers □