If $a$ is not square then there is a class $\mod 4a$ such that $a$ is a quadratic non-residue modulo any prime in that class. I am trying to show that any non-square $a$ is a quadratic non-residue modulo an infinite number of primes, and this is my argument so far:
There is a well-defined group homomorphism $\chi : (\mathbb{Z}/4a\mathbb{Z})^{\times} \to \{\pm 1\}$ satisfying the relation $\chi(p) = \left(\frac{a}{p}\right)$ for all primes $p$ not dividing $4a$. This is because if $q = p + 4ak$ is also prime:
If $a$ is odd,
$\begin{align*}
\chi(q) &= \left(\frac{a}{q}\right) = \left(\frac{q}{a}\right) (-1)^{\frac{q-1}{2}\frac{a-1}{2}} = \left(\frac{p + 4ak}{a}\right) (-1)^{\frac{p+4ak-1}{2}\frac{a-1}{2}}\\
&=\left(\frac{p}{a}\right) (-1)^{\frac{p-1}{2}\frac{a-1}{2}}\\
&= \left(\frac{a}{q}\right) = \chi(q),
\end{align*}$
and if $a = 2b$ is even (so $q=p+8bk$),
$\begin{align*}
\chi(q) &= \left(\frac{2}{q}\right)\left(\frac{b}{q}\right) = \left(\frac{2}{p+8bk}\right)\left(\frac{b}{p}\right) \text{ (as above)}\\
&= \left(\frac{2}{p}\right)\left(\frac{b}{p}\right) = \chi(p).
\end{align*}$
Let us suppose that this homomorphism is not trivial. Then there exists a number $d \mod 4a$ such that $\chi(d) = -1$. If $d = p_1...p_k$ is a product of primes then $-1 = \chi(p_1)...\chi(p_k)$ and so there exists a class $p_i$ with $\chi(p_i) = \left(\frac{a}{p_i}\right) = -1.$
From here it is easy: the arithmetic progression $p_i + 4ak$ contains infinitely many primes by Dirichlet (clearly gcd$(p_i,4a) = 1$ by construction) and so there are infinitely many primes modulo which $a$ is a quadratic non-residue.
So, my question: Is it easy to show that $\chi$ is in fact non-trivial? If so, I would greatly appreciate a proof. Further, if I have made any mistakes in the rest of my proof I would like to know.
 A: First, we can reduce to the case where $a$ is squarefree: let $a = s^2n$ for squarefree $n>1$ (since $a$ is not a square). Then for any prime $p > a$, $\left(\frac{a}{p}\right) = \left(\frac{s^2}{p}\right)\left(\frac{n}{p}\right) = \left(\frac{n}{p}\right)$. 
So the list of primes satisfying $\left(\frac{n}{p}\right) = -1$ and the corresponding list for $\left(\frac{a}{p}\right) = -1$ are eventually identical.
So suppose $a = q_1q_2...q_k$ is squarefree. Choose $e_1,e_2,...,e_k \in \{\pm 1\}$ so that $\prod e_i = -1$, and note that for each $i$ the equation $\left(\frac{q_i}{p}\right) = e_i$ is implied by $p \equiv f_i \mod q_i$ for some $f_i$ (or if $q_i = 2$, $p \equiv f_i \mod 8$) by quadratic reciprocity.
By the Chinese Remainder Theorem these congruences have a solution $p \equiv f \mod 4a$, and hence $\left(\frac{a}{f}\right) = \prod \left(\frac{q_i}{f}\right) = -1$ as required.
Since $f$ doesn't divide any of the $q_i$ by construction, gcd$(f,a) = 1$, and so the rest of the proof in the question carries over.
If I've made any mistakes, don't hesitate to point them out. :D
