Prove Theorem (8.7).
Let's remind ourselves of Converse of Sum of Two Squares Theorem (8.7).
Let $m = p_1 \times p_2 \times \ldots \times p_r × N^2$ where $p$’s are distinct primes. If $m$ can be expressed as sum of two squares then none of these primes $p_j$ satisfies $p_j \equiv 3 \pmod 4$ for $j= 1, \ldots , r$.
Suppose for the purpose of contradiction that there exists a prime $p$ amongst the $p_j$ such that $p\equiv3 \pmod 4$.
We are assuming $m$ can be expressed as a sum of two squares, so for some natural numbers $a,b$ we have
$$ m= a^2 +b^2 = p_1 \times p_2 \times \ldots \times p_r \times N^2 $$
This means $p \mid a^2 + b^2$. That is
$$ a^2 \equiv -b^2 \pmod p $$
Let's now consider the quadratic residue $-b^2 \pmod p$
$$ (\frac{-b^2}{p}) = (\frac{-1}{p}) \times (\frac{b^2}{p}) = (\frac{-1}{p}) \times (1) = (\frac{-1}{p}) $$
For $ a^2 \equiv -b^2 \pmod p $ to have a solution, this requires $p \equiv 1 \pmod 4$ by Proposition (7.11).
This contradicts $p \equiv 3 \pmod 4$, and so proves the Theorem.