Let $n > 1$ be a square-free natural number, which is fixed. The assertion to be proved is the following:



> 
> Let $p$ run through primes. Then, $$\left( \frac{n}{p} \right)$$ is equally distributed between $1$ and $-1$.
> 
> 
> 


The precise statement of which is to be made using the appropriate asymptotic expressions. 


J-P. Serre, "A Course in Arithmetic", outlines a proof of the above assertion just after the proof of Dirichlet's theorem on arithmetic progressions. That proof uses the zeta function of number fields. I am not able to shake off the feeling that it should be provable without using this. That is, it should be possible to prove this statement just using the properties of Dirichlet $L$-functions, and elementary arguments on quadratic residues. However I am not able to construct such a proof either, since I am not very skilled in this type of matters. So I ask here, is such a proof known?



Distribution of quadratic residues of a fixed number without using Dedekind zeta function
This answer summarizes the above discussion: Extend $p \mapsto \left( \frac{n}{p} \right)$ to a multiplicative function $\chi$ on the positive integers. By quadratic reciprocity, $\chi$ is periodic modulo $4n$, and it is multiplicative by construction, so it is a character. We know that $L(1, \chi) \neq 0$. Thus, $\lim\_{s \to 1^{+}} |\log L(s,\chi)| < \infty$.


We compute:
$$\log L(s, \chi) = - \sum \log \left( 1-\frac{\chi(p)}{p^s} \right) = \sum \frac{\chi(p)}{p^s} + O(1).$$


So $\sum \left( \frac{n}{p} \right)/p^s$ is bounded as $s \to 1^{+}$. A little more work shows that the limit as $s \to 1^{+}$ exists.


If you want to prove results like that $\sum \left( \frac{n}{p} \right)/p$ converges, or that $|\sum\_{p < N} \left( \frac{n}{p} \right)| = o(N)$, then you need Tauberian methods, as discussed in any book on analytic number theory.


