Timeline for How to construct a small coprime?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Jan 7, 2018 at 15:12 | answer | added | Dan Brumleve | timeline score: 2 | |
| Sep 14, 2017 at 23:45 | vote | accept | Igor Pak | ||
| Sep 14, 2017 at 23:45 | |||||
| Sep 14, 2017 at 5:45 | history | edited | Igor Pak | CC BY-SA 3.0 |
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| Sep 14, 2017 at 3:05 | answer | added | Timothy Chow | timeline score: 6 | |
| Sep 14, 2017 at 2:25 | comment | added | Gerry Myerson | @Tim, it's clear now. It wasn't clear before Igor edited in the bit about "One can of course, deterministically test primality of $n+1,n+2,…,n+C(\log n)^2$, but that's not yet proved to work." | |
| Sep 14, 2017 at 2:16 | comment | added | Timothy Chow | @IgorPak For this type of question, people are so used to assuming GRH and so forth when convenient, that it's become customary to say explicitly that you want an unconditional result if that's what you really want. But I think it's clear from what you wrote. | |
| Sep 13, 2017 at 22:46 | history | edited | Igor Pak | CC BY-SA 3.0 |
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| Sep 13, 2017 at 22:39 | comment | added | Igor Pak | @GerryMyerson Hoping for a result not a conjecture is not a secret condition. Maybe I am missing something about MO, but that's how questions normally work. | |
| Sep 13, 2017 at 21:59 | comment | added | Gerry Myerson | Then maybe you should say that, and any other secret conditions, in the body of the question. Please edit. | |
| Sep 13, 2017 at 16:47 | comment | added | Igor Pak | @GerryMyerson Correct. I know. But I want to have a theorem free of conjectures. | |
| Sep 13, 2017 at 15:24 | answer | added | Gerhard Paseman | timeline score: -1 | |
| Sep 13, 2017 at 13:16 | comment | added | Gerry Myerson | There is a deterministic polynomial-time algorithm to test for primality, and you only have to test about $(\log n)^2$ numbers if the standard conjectures are true. | |
| Sep 13, 2017 at 10:29 | comment | added | Igor Pak | @JasonStarr - Sure. Updated. | |
| Sep 13, 2017 at 10:28 | history | edited | Igor Pak | CC BY-SA 3.0 |
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| Sep 13, 2017 at 10:27 | comment | added | Jason Starr | Did you forget to write a condition on $q$? Why not take $q$ equal to $1$? | |
| Sep 13, 2017 at 10:14 | history | edited | Igor Pak | CC BY-SA 3.0 |
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| Sep 13, 2017 at 10:06 | comment | added | Igor Pak | Oh, no. I just want it to work for some $c$. So $c=100$ is great. | |
| Sep 13, 2017 at 10:04 | comment | added | Dirk | Is $c$ an integer or are you looking for an algorithm that works for all real $c > 1$ (i.e. an algorithm that also outputs that there is no such $q$ in some cases)? | |
| Sep 13, 2017 at 10:00 | history | asked | Igor Pak | CC BY-SA 3.0 |