Timeline for Composite pairs of the form n!-1 and n!+1
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 31, 2010 at 15:03 | answer | added | Franz Lemmermeyer | timeline score: 3 | |
| Aug 31, 2010 at 5:55 | answer | added | Noah Schweber | timeline score: 2 | |
| Jul 4, 2010 at 0:21 | history | edited | François G. Dorais | CC BY-SA 2.5 |
redaction & addendum
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| Jul 2, 2010 at 15:42 | history | edited | François G. Dorais | CC BY-SA 2.5 |
redaction
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| Jul 2, 2010 at 15:23 | history | edited | François G. Dorais | CC BY-SA 2.5 |
addendum
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| Jul 1, 2010 at 12:50 | answer | added | Andrey Rekalo | timeline score: 5 | |
| Jun 30, 2010 at 23:03 | comment | added | Kevin Buzzard | @Dror: I don't think your assertion is correct. Try $q=13$ for a counterexample. Let me conjecture the slip you made: if $p$ (your $2q-3$) is a prime which is 3 mod 4 and $n=(p-1)/2$ then $(n!)^2$ is 1 mod $p$ but the problem is that $n!$ could be either $+1$ or $-1$ mod $p$. | |
| Jun 30, 2010 at 22:34 | history | edited | François G. Dorais | CC BY-SA 2.5 |
clarification
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| Jun 30, 2010 at 22:33 | comment | added | Joel David Hamkins | François, if you considered models of TA rather than PA, then any kind of proof would suffice, and you could discard that requirement. | |
| Jun 30, 2010 at 22:25 | comment | added | François G. Dorais | @Pete: For the intended application to work, the proof has to be formalizable in PA. I'm not that picky, any proof will do for now. | |
| Jun 30, 2010 at 22:19 | comment | added | Pete L. Clark | It seems like "elementary proof" has a specific technical meaning here; what is it? Also, just to be clear: we do not as yet know of any proof, right? | |
| Jun 30, 2010 at 21:07 | comment | added | Dror Speiser | If for a prime $q$, $2q−3$ is also prime, then $n=q-2$ makes for a composite pair. Simply put, this doesn't help | |
| Jun 30, 2010 at 19:59 | comment | added | Kevin Buzzard | I don't know of an elementary proof, but in practice it's very rare that either of $n!\pm1$ are prime, so the result is surely true. For example, both $n!+1$ and $n!-1$ are composite for all $4000\leq n\leq 6000$. Caldwell and Gallot found that $6380!+1$ and $6917!-1$ were prime, but it gets harder and harder to find examples up there. NB I discovered this by computing the first few n for which $n!+1$ was prime and looking it up in Sloane and chasing up the references. | |
| Jun 30, 2010 at 19:43 | history | asked | François G. Dorais | CC BY-SA 2.5 |