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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?

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  • $\begingroup$ This is fairly straightforward to prove if one knows that $L(1,\chi)\ne0$ where $\chi$ is the Dirichlet character extending $p\mapsto (n/p)$. See mathoverflow.net/questions/25794/… for my thoughts on this. $\endgroup$
    – Robin Chapman
    Commented Jul 23, 2010 at 18:19
  • $\begingroup$ @Robin Chapman: I had indeed read your "real variable" proof of Dirichlet's theorem, if that is what you mean. But that article does not contain a proof of this assertion. I also feel that the proof should not be difficult; but I am not able to make the connection myself. $\endgroup$
    – Anweshi
    Commented Jul 23, 2010 at 18:26
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    $\begingroup$ If you accept $\lim_{s \to 1^{+}} \left| \sum \left( \frac{n}{p} \right) p^{-s} \right| < \infty$ as the way that you make your density statement precise, then this follows easily from $L(1, \chi) \neq 0$, as $\log L(s, \chi) = \sum \chi(p)/p^s + O(1)$. If you want to use some other measure of density, then you need the usual Tauberian arguments. $\endgroup$
    – David E Speyer
    Commented Jul 23, 2010 at 20:34
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    $\begingroup$ The character $\left( \frac{n}{} \right)$, which is a character modulo $4n$ by quadratic reciprocity. $\endgroup$
    – David E Speyer
    Commented Jul 23, 2010 at 20:57
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    $\begingroup$ If David does not post an answer within a few hours, you can also just collect together the best observations and put those as your answer. If it would bother you about some votes going to you instead of David you can make your answer, or the whole question, community wiki. I do not know how to do that myself but it cannot be too difficult. That way people who might be interested but did not follow the first few hours of comments can see an organized answer in one place. $\endgroup$
    – Will Jagy
    Commented Jul 23, 2010 at 21:46

1 Answer 1

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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.

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