Timeline for Every prime number > 19 divides one plus the product of two smaller primes?
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27 events
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| Jul 28, 2017 at 3:52 | vote | accept | CommunityBot | moved from User.Id=6976 by developer User.Id=69903 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| S Oct 17, 2015 at 15:47 | history | bounty ended | Gil Kalai | ||
| S Oct 17, 2015 at 15:47 | history | notice removed | Gil Kalai | ||
| Oct 13, 2015 at 22:49 | comment | added | JMP | i can prove kq-1 is squarefree for some k<q. will this do? | |
| S Oct 11, 2015 at 9:04 | history | bounty started | Gil Kalai | ||
| S Oct 11, 2015 at 9:04 | history | notice added | Gil Kalai | Draw attention | |
| Jan 4, 2012 at 12:52 | history | edited | user6976 | CC BY-SA 3.0 |
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| Jan 3, 2012 at 6:21 | history | edited | Eric Naslund |
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| Dec 30, 2011 at 8:11 | answer | added | Timothy Foo | timeline score: 11 | |
| Dec 29, 2011 at 10:28 | answer | added | Eric Naslund | timeline score: 30 | |
| Dec 28, 2011 at 23:34 | comment | added | user6976 | @Kevin: The trig. sum method might work or give similar weaker results like: every prime $\gg 1$ divides one plus a product of at most 3 (4, 5,...) smaller primes. It is somewhat similar to the Goldbach conjecture and various versions of it that have been proved. Of course since I do not have a proof I can only hope that it exists. | |
| Dec 28, 2011 at 23:07 | comment | added | Kevin Ventullo | The problem is equivalent to asking whether $A^{-1}\cap -A=\emptyset$. Any subgroup of $Z/pZ^*$ which doesn't contain $-1$ satisfies this, so density arguments alone won't work (as Ben Green indicated above). | |
| Dec 28, 2011 at 8:40 | answer | added | joro | timeline score: 21 | |
| Dec 27, 2011 at 20:35 | comment | added | Ben Green | Denis, I don't understand your comment. I only know two things that count as good reasons in the theory of prime numbers. First, a proof. Second, a decent heuristic or "statistical" argument. Here, the set of $1 + rs$ ought to look like a fairly random (give or take some irregularities mod small primes) subset of $[1,p^2]$ of density $1/\log^2 p$. The probability of $x \leq p^2$ being divisible by $p$ is $1/p$. I'd expect subsets of $[1,X]$ of densities $\alpha$ and $\beta$ to intersect as soon as $\alpha \beta \gg 1/X$ unless there is some good reason why not. | |
| Dec 27, 2011 at 19:21 | comment | added | Geoff Robinson | @Denis,Mark: Also somewhat related to Denis's question is a 1953 paper of John Thompson "A method for finding primes", see MR052448 | |
| Dec 27, 2011 at 18:49 | comment | added | user6976 | @Denis: The Maple program that has been running since yesterday is currently considering numbers close to $10^6$, so it is not just statistics. I feel that this problem may have a trig. sum (= Fourier analysis) solutions. It looks like people who know the subject much better than I feel the same. It does not seem trivial, though, which may be good. Your problem does look similar. I did not see it before. | |
| Dec 27, 2011 at 17:38 | comment | added | Denis Serre | Related to the question (Constructing prime numbers, mathoverflow.net/questions/38794) I posed a year ago. Your construction is more flexible. It has however the flaw that there is no good reason, other than statistical, why at least one $1+rs$ should have a prime factor greater than $p$. | |
| Dec 27, 2011 at 16:39 | comment | added | Ben Green | Geoff- there certainly is work on this. It's known that the smallest prime congruent to $a$ mod $q$ is $\ll q^{5.4}$ (or so). It's conjectured that it's $\ll q^{1 + \epsilon}$. The GRH would give $\ll q^{2 + \epsilon}$. | |
| Dec 27, 2011 at 15:37 | history | edited | user6976 | CC BY-SA 3.0 |
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| Dec 27, 2011 at 14:35 | comment | added | Ben Green | Mark - yes, there are "sum-product" results of this type. The one most applicable here is that due to Bourgain-Katz-Tao: it says that if $A \subseteq Z/pZ$ and $|A + A|, |A \cdot A| \leq K|A|$ then either $|A| \leq K^C$ or $|A| \geq K^{-C} p$. When $A$ has size about $p/\log p$, it tells you rather little (namely either $|A + A|$ or $|A \cdot A|$ has size at least $p/(\log p)^{1 - c}$) and I think exponential sum techniques are likely to be better. | |
| Dec 27, 2011 at 13:17 | comment | added | user6976 | @Gjergji: You are right. I remember seeing a result that says that for every finite set $A$ of natural numbers either $A\cdot A$ or $A+A$ is large (contains more than $|A|^{1_\epsilon}$ elements). I do not know if a similar result holds for finite cyclic groups. @Ben: Thanks for your comments. | |
| Dec 27, 2011 at 12:49 | comment | added | Ben Green | I might add that I would expect trigonometric sums to allow one to show that $A \cdot A$ is almost all of $Z/pZ$ as $p \rightarrow \infty$. | |
| Dec 27, 2011 at 12:48 | comment | added | Ben Green | Interesting problem. I don't think either the sum-product idea or the trig sums idea will work immediately. In the latter case because one has a binary problem instead of one with three variables, and in the former because I think you can have both $A.A \subsetneq Z/pZ^*$ and $A + A^{-1} \subsetneq Z/pZ$ with $A$ reasonably large, even positive density. But both of these are good ideas on which to base further thought. | |
| Dec 27, 2011 at 10:59 | comment | added | Gjergji Zaimi | Of course, $A\cdot A$ contains $-1$ iff $A+A^{-1}$ contains zero. I get a feeling that even for arbitrary A, at least one of these two sets has to be large. For primes, it's likely both are. | |
| Dec 27, 2011 at 10:58 | history | edited | user6976 | CC BY-SA 3.0 |
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| Dec 27, 2011 at 10:34 | history | asked | user6976 | CC BY-SA 3.0 |