Prove Lemma (8.2).
Let's remind ourselves of Lemma (8.2).
If prime $p \equiv 1 \pmod 4$ then there exist positive integers $x$ and $y$ such that $x^2 + y^2 = kp$ where $k < p$ and it is a positive integer.
We can write $x^2 + y^2 = kp$ as
$$ x^2 \equiv -y^2 \pmod p $$
Since $p \equiv 1 \pmod 4$, by Proposition (7.11) we have
$$ (\frac{-1}{p}) = 1$$
By definition of the Legendre symbol
$$ (\frac{y^2}{p}) = 1 $$
Using the multiplicity of the Legendre symbol
$$ (\frac{-y^2}{p}) = (\frac{-1}{p}) \times (\frac{y^2}{p}) = 1 \times 1 = 1 $$
But $(\frac{-y^2}{p})=1$ means that the following has integer solutions for $x$,
$$ x^2 \equiv - y^2 \pmod p $$
That is, $x^2 + y^2 = kp$ has solutions.
Because $(-x)^2 = x^2$ and $(-y)^2 = y^2$, we can additionally state that positive integer solutions exist.
We now need to show that $k < p$ still permits solutions.
If $x \pmod p$ is a solution, so is $-x \equiv p-x \pmod p$. This means that if solutions exist, they include solutions in the range $0 \le x < \frac{p}{2}$.
This means
$$ 0 \le x^2 + y^2 < (\frac{p}{2})^2 + (\frac{p}{2})^2 = \frac{p^2}{2} = \frac{p}{2} \times p $$
That is, solutions to $x^2 + y^2 =kp$ exist when $k < \frac{p}{2} < p$.