Timeline for Distribution of quadratic residues of a fixed number without using Dedekind zeta function
Current License: CC BY-SA 2.5
20 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2010 at 13:05 | vote | accept | Anweshi | ||
| Jul 24, 2010 at 11:45 | answer | added | David E Speyer | timeline score: 2 | |
| Jul 23, 2010 at 21:59 | comment | added | Anweshi | @Will Jagy: Will do so. Maybe I will wait for a day or so. | |
| Jul 23, 2010 at 21:46 | comment | added | Will Jagy | 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. | |
| Jul 23, 2010 at 21:10 | comment | added | Anweshi | In two comments above: I meant, I can delete this question. | |
| Jul 23, 2010 at 21:10 | comment | added | Anweshi | I might add one slight omission: Now we know that the quadratic residues of $4n$ are equally distributed between $1$ and $-1$. There remains a small checking that it indeed implies the same for $n$. | |
| Jul 23, 2010 at 21:03 | comment | added | Anweshi | Ah! I see, thanks. If you could write it down as an answer, I could accept it and forget about this question. Otherwise, if you think this question is too trivial to merit an answer, and if you don't have objections, I can delete this answer. | |
| Jul 23, 2010 at 20:57 | comment | added | David E Speyer | The character $\left( \frac{n}{} \right)$, which is a character modulo $4n$ by quadratic reciprocity. | |
| Jul 23, 2010 at 20:54 | comment | added | Anweshi | @David Speyer: What is $\chi$ here? For the statement you cite, it should be a character modulo $n$ and it escapes me which character you mean. | |
| Jul 23, 2010 at 20:34 | comment | added | David E Speyer |
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.
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| Jul 23, 2010 at 19:41 | comment | added | Anweshi | @David Speyer: It is one of the standard results. But from that how would it follow? | |
| Jul 23, 2010 at 19:34 | comment | added | David E Speyer | So is the nonvanishing of $L(1, \chi)$ not one of the standard results on Dirichlet $L$-functions for you? | |
| Jul 23, 2010 at 18:51 | comment | added | Anweshi | @Robin Chapman: I do not mean that your article does not contain a proof that $L(1, \chi) \neq 0$. I meant the implication that this statement implies the assertion I made in the statement of the question. My question is not really analytic, though I tagged it analytic-number-theory. All I need is the proof that the theorems on Dirciclet L-functions implies the equal distribution of quadratic residues. I expect that it would be rather elementary. | |
| Jul 23, 2010 at 18:39 | comment | added | Robin Chapman | For Dirichlet density one needs little more than the nonvanishing of the L-function at $1$. For natural density if one also has the non-vanishing of the $L$-function on the rest of the line $s=1+it$ (which is easier than the non-vanishing at $s=1$) then one can use standard results, for instance the Wiener-Ikehara Tauberian theorem to complete the proof. (Compare for instance Zagier's excellent Monthly paper on PNT). | |
| Jul 23, 2010 at 18:28 | comment | added | Daniel Litt | If $n$ is prime this follows easily from the quantitative form of Dirichlet's theorem and quadratic reciprocity. It might follow for general $n$ by multiplicativity, but this is not obvious, because we can't immediately rule out correlations between the primes with respect to which the factors of $n$ are residues. I suspect we can find theorems to deal with this, though. | |
| Jul 23, 2010 at 18:28 | history | edited | Anweshi | CC BY-SA 2.5 |
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| Jul 23, 2010 at 18:26 | comment | added | Anweshi | @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. | |
| Jul 23, 2010 at 18:22 | history | edited | Anweshi | CC BY-SA 2.5 |
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| Jul 23, 2010 at 18:19 | comment | added | Robin Chapman | 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. | |
| Jul 23, 2010 at 18:13 | history | asked | Anweshi | CC BY-SA 2.5 |