Timeline for Does every geometric progression contain a small remainder modulo a large prime?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Jan 18, 2021 at 2:16 | vote | accept | fedja | ||
| Jan 16, 2021 at 8:25 | answer | added | Seva | timeline score: 3 | |
| Jan 16, 2021 at 2:09 | comment | added | Mark Lewko | Also regarding the ``counting argument" part of the approach, for large $c$, an unconditional lower estimate of the form $p^{1/2} \log^c p$ is known for almost every $p$. See: citeseerx.ist.psu.edu/viewdoc/… | |
| Jan 16, 2021 at 1:43 | comment | added | Mark Lewko | In any event, what you want is certainly true and provable. | |
| Jan 16, 2021 at 1:43 | comment | added | Mark Lewko | Once the subgroup has an asymptotic size larger than $p^{1/2}$ a non-trivial estimate certainly follows from Weil's bound. But, I think you should also be able to get a bound of sufficient strength by looking at moments. I also believe there is some work due to Shparlinski which goes below the exponent $1/2$ but doesn't get down to subgroups nearly as small as Bourgain-Glibichuk-Konyagin. | |
| Jan 16, 2021 at 1:38 | comment | added | Mark Lewko | I'm not entirely sure what @Seva has in mind but what he claims should be true up to the factor of $(2\epsilon)^{-1}$ which might well also be true but I haven't checked carefully. The idea is, as our friend Don points out, that Erdos-Turan reduces the problem to showing an upperbound on an exponential sum of the form $\sum_{x} e( a x^k/ p)$ that beats the trivial estimate by more than a factor of $\log p$. Where $\{x^k : x \mod p\}$ is just another way of parameterizing our multiplicative subgroup. | |
| Jan 15, 2021 at 20:00 | comment | added | Seva | Ok, will try to post it tomorrow though, unfortunately, I cannot see a way to improve the estimates on my end. | |
| Jan 15, 2021 at 19:38 | comment | added | fedja | @Seva That may be not so easy but I'll try to push my trivial counting argument up a bit though to gain $\log p$ rather than a mere constant seems somewhat problematic. Meanwhile, if you can post your "very simple argument", we can also think if we can push it down to $\varepsilon\sqrt p$. The gap looks rather small, so, perhaps, it can be closed from one of the ends :-) | |
| Jan 15, 2021 at 19:33 | comment | added | Seva | How difficult is to show that for almost all primes $P$, the subgroup generated by $q$ has size $(2\varepsilon)^{-1}\sqrt p\log p$ at least? The rest, I think, can be done with a very simple argument. | |
| Jan 14, 2021 at 22:01 | comment | added | fedja | @katago No, what I stated was exactly what would be sufficient to figure out what I wanted to figure out. I suspected that a stronger result might have been known (as it turned out to be the case) but I tried to request the minimum I would be happy with :-) | |
| S Jan 14, 2021 at 21:10 | history | suggested | mathworker21 | CC BY-SA 4.0 |
fixed plural-ism in title
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| Jan 14, 2021 at 19:40 | review | Suggested edits | |||
| S Jan 14, 2021 at 21:10 | |||||
| Jan 14, 2021 at 19:23 | comment | added | katago | maybe the exact thing you are interested in is how fast $G(a, P)=\left\{a q^{m} \quad \bmod P: m=0,1,2, \ldots\right\}$ tend to equidistribution as $m$ grow?(fix q and let $p$ go to infinity and some uniformly discrepancy estimate on almost every prime) | |
| Jan 14, 2021 at 18:56 | comment | added | fedja | @MarkLewko Yep, for almost all primes $P$ and $a\ne 0$ we shall have $|G(a,P)|\ge P^{\frac 12-\varepsilon}$ by an elementary counting argument, so it seems to answer the question in the affirmative. That's great. Let me see if I can understand the proof in Don's reference :-) | |
| Jan 14, 2021 at 18:34 | comment | added | so-called friend Don | For almost all primes $P$, one will have that $G(a,P)$ is a coset of a subgroup of the multiplicative group mod P with size at least $P^{1/3}$. It seems an affirmative answer will follow from the result of Bourgain--Glibichuk--Konyagin appearing as Theorem 7 here, along with the Erdos-Turan inequality (Lemma 6): insa.nic.in/writereaddata/UpLoadedFiles/IJPAM/20019e3c_15.pdf | |
| Jan 14, 2021 at 18:33 | comment | added | Mark Lewko | If I understand correctly, you're almost asking if the cosets of multiplicative subgroups of the residues mod $p$ are equidistributed. This is known as long as the size of the subgroup is at most $p^{\delta}$ for some fixed $\delta$. Of course, there's still an issue of how q relates to the size of the induced multiplicative subgroup. | |
| Jan 14, 2021 at 18:16 | comment | added | markvs | $a$ is chosen after $P$? | |
| Jan 14, 2021 at 17:52 | history | edited | Glorfindel | CC BY-SA 4.0 |
edited title
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| Jan 14, 2021 at 17:50 | history | asked | fedja | CC BY-SA 4.0 |