Timeline for Small residue classes with small reciprocal
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 26, 2014 at 5:53 | comment | added | Terry Tao | The paper linked to by Matt Young appears to have moved to nieuwarchief.nl/serie5/pdf/naw5-2000-01-4-380.pdf (and is entitled "Arithmetic applications of Kloosterman sums", in case it moves again). | |
| Nov 26, 2014 at 0:00 | answer | added | Dimitris Koukoulopoulos | timeline score: 14 | |
| Nov 16, 2011 at 9:30 | answer | added | Timothy Foo | timeline score: 7 | |
| Jul 11, 2011 at 7:26 | comment | added | Greg Martin | Friedlander showed ("Shifted primes without large prime factors") that a positive proportion of $p+1$ are $p^{0.3}$-friable (or so). (It's always quoted as $p-1$, but presumably it works for $p+1$ as well.) That implies that all of those $p+1$ factor as $mn$ where $p^{0.35} < m,n < p^{0.65}$. So a positive proportion of $p$ beat $p^{2/3}$. And then one could always try looking at $2p+1$, $3p+1$, ... or in general starting with the pairs $m,n$ and trying to count how many primes are divisors of some $mn-1$. | |
| Jul 11, 2011 at 2:06 | comment | added | David Hansen | A small note: For tools to estimate sums of incomplete Kloosterman sums over moduli (which may well be relevant), take a look at Duke-Friedlander-Iwaniec's "Bilinear forms with Kloosterman fractions". | |
| Jul 5, 2011 at 4:58 | comment | added | Matt Young | The Kuznetsov formula is useful for obtaining cancellation in a sum of Kloosterman sums of the form $\sum_{q < X} S(m,n;q)$ but here $q$ ranges over all integers in an interval and cannot be restricted to prime moduli. One can introduce certain congruence restrictions such as $q$ being divisible by some integer. | |
| Jul 5, 2011 at 4:19 | comment | added | Terry Tao | I'd prefer m and $\overline{m}$ to be positive, but I'll take whatever I can get at this point :-). | |
| Jul 5, 2011 at 3:17 | comment | added | Noam D. Elkies | Not that I see immediately how this would help, but do $m$ and $\overline m$ have to be positive for your application, or is it enough to have the least absolute residues be $\widetilde{O}(p^{2/3})$ (again excluding $m = \overline{m} = \pm 1$)? | |
| Jul 5, 2011 at 2:28 | comment | added | Terry Tao | Ah, thanks! A bit disappointed that it is open (at least as of 2000), but I guess I wasn't missing an utterly trivial argument at least. That paper mentions an improved Kloosterman sum estimate of Kuznetsov that averages over p and which, in principle, might let me reach the p^{2/3} mark on average (maybe even $p^{7/12+\varepsilon}$, from a back-of-the-envelope calculation) - I'll have to look into it. | |
| Jul 5, 2011 at 2:05 | comment | added | Matt Young | In a paper available at nieuwarchief.nl/serie5/deel01/dec2000/pdf/heathbrown.pdf Heath-Brown discusses this exact problem (in the section entitled "an elementary problem") and obtains $p^{3/4}$. He indicates that it is an open problem to improve on $3/4$. I haven't thought about how averaging over $p$ could improve the exponent but often that can be very helpful. | |
| Jul 5, 2011 at 1:47 | history | asked | Terry Tao | CC BY-SA 3.0 |