This is part (2) of the question that I asked here. Inspired by the answer in this question, I want to see if I am on the right track with this question.
I was given this exercise, based on a theorem from Cox's book, which can be found here.
Theorem. Let $D=4n$ for $n\in\Bbb Z$. Then, there is a unique homomorphism $\chi: (\Bbb Z/D\Bbb Z)^{\times}\rightarrow\{\pm 1\}$ such that $\chi([p])=(D/p)$ for all primes $p$ that do not divide $D$.
Show that this function is a homomorphism.
Here is my attempt:
Let $x$ and $y$ be primes with $m$ being chosen so that $xy+mD$ is prime. Then, we observe that $xy+mD\equiv 3\bmod 4$ if and only if $x$ and $y$ are not in the same congruence class modulo $4$. We define $n$ to be the number of indices $i$ where $xy\equiv 3\bmod 4$
\begin{align*} \chi([xy])=\Big(\frac{D}{xy+mD}\Big)&=\prod_{i=1}^k\Big(\frac{p_i}{xy+mD}\Big)\\ &=(-1)^n\prod_{i=1}^k\Big(\frac{xy+mD}{p_i}\Big)\\ &=(-1)^n\prod_{i=1}^k\Big(\frac{xy}{p_i}\Big)\\ &=(-1)^{2n}\prod_{i=1}^k\Big(\frac{p_i}{xy}\Big)\\ &=\prod_{i=1}^k\Big(\frac{p_i}{x}\Big)\Big(\frac{p_i}{y}\Big)\\ &=\Big(\frac{D}{x}\Big)\Big(\frac{D}{y}\Big)\\ &=\chi([x])\chi([y]) \end{align*}
It's the bolded sentence that I'm not too confident in right now. Is this the way to do it?