Timeline for Covering the primes by arithmetic progressions
Current License: CC BY-SA 2.5
10 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 21, 2010 at 20:46 | vote | accept | Joseph O'Rourke | ||
| Nov 21, 2010 at 6:58 | comment | added | Fedor Petrov | No no, I meant what I said, but I've been mistaken, sorry. The natural conjecture is that $p$ lies in arithmetic progression of length $p$, but of course not more, by trivial reasons (modulo $p$). So, most probably we may cover all primes but finite number of them by arithmetic progressiond of arbitrary length. | |
| Nov 21, 2010 at 6:34 | comment | added | Idoneal | In such problems, it is interesting to ignore the small primes and ask what happens for all sufficiently large primes. Perhaps that is what Fedor had meant. | |
| Nov 21, 2010 at 0:28 | comment | added | George Lowther | I can't claim credit for that phrase, as I was just quoting Ben Green's linked answer. | |
| Nov 21, 2010 at 0:16 | comment | added | Joseph O'Rourke | @George: Clever proof that 3 is not in a progression of length 4! Similarly 5 is not in a progression of length 6. So the question could be modified to ask for coverage of all but the beginning of the sequence. But I won't attempt to reformulate the question in light of your recognizing that deciding between 2 and 3 is already "beyond the current technology" (a useful [and optimistic!] phrase). | |
| Nov 20, 2010 at 23:57 | history | edited | George Lowther | CC BY-SA 2.5 |
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| Nov 20, 2010 at 23:38 | history | edited | George Lowther | CC BY-SA 2.5 |
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| Nov 20, 2010 at 23:37 | comment | added | George Lowther | I think maybe, in his comment, Fedor was suggesting that it can be done using arithmetic progressions with arbitrarily large step size, rather than arbitrarily large length. | |
| Nov 20, 2010 at 23:31 | history | answered | George Lowther | CC BY-SA 2.5 |