I'm trying to find a proof of the following claim, without using Dirichlet's Theorem on the arithmetic progression of primes:
Claim: If $a$ is not a square, then there are infinitely many odd primes $p$ such that $\left(\frac ap\right) = -1$.
Note $(a/p)$ represents the Legendre symbol. Here's my current attempt:
Proof: Clearly it suffices the prove the case where $a$ is squarefree. To begin with, let us prove the case where $a$ is odd.
Suppose we have already found such primes $p_1, \ldots, p_k$. Prime-factorise $a = r_1 \ldots r_s$ and note that $$\left( \frac a{p_i} \right) = -1 \implies (p_i, r_j) = 1 \ \ \forall i,j$$ Moreover, these are all odd primes, so all coprime to $8$. Thus by the Chinese Remainder Theorem, we can find $N > 1$ such that: $$ \begin{cases}N \equiv 1 \ (\text{mod} \ 8) \\ N \equiv 1 \ (\text{mod} \ p_i) \ \forall i \\ N \equiv 1 \ (\text{mod} \ r_j) \ \forall j < s \\ N \equiv c \ (\text{mod} \ r_s) \end{cases} $$
where $c$ is any quadratic non-residue modulo $r_s$.
Prime-factorise $N = q_1 \ldots q_l$. Then by construction these are all odd primes, and $q_i \ne p_j \ \forall i,j$.
Since $a$ is odd and $N \equiv 1 \ (\text{mod} \ 4)$, quadratic reciprocity gives:
$$ \prod_i \left(\frac a{q_i}\right) = \left(\frac{a}{N}\right) = \left(\frac{N}{a}\right) = \prod_{j=1}^{s-1} \left(\frac{N}{r_i}\right) \cdot \left(\frac{N}{r_s}\right) = \prod_{j=1}^{s-1} \left(\frac{1}{r_i}\right) \cdot \left(\frac{c}{r_s}\right) = -1 $$
So at least one of the terms in the product is $\left(\frac{a}{q_i}\right) = -1$, so $p_{k+1} = q_i$ is a new such prime, and the result is proven for $a$ odd.
Furthermore, we chose $N \equiv 1 \ (\text{mod} \ 8)$, so $$ \prod_j \left(\frac{2a}{q_j}\right) = \left(\frac{2}{N}\right)\left(\frac{a}{N}\right) = \left(\frac{a}{N}\right) = -1$$
and so by a similar argument the result follows for $a' = 2a$, hence for all even squarefree numbers.
Is this proof valid? If so, can it be condensed or improved in any way?
