Timeline for Find the least prime so that p-1 has two factors greater than $m$ and $n$
Current License: CC BY-SA 3.0
12 events
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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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| Oct 4, 2011 at 18:43 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Oct 4, 2011 at 16:00 | comment | added | Gerhard Paseman | Further, standard heuristics predict, assuming n > m, an upper bound of nmm, while recent progress on Linnik's theorem provide some bound like O(nm^6), I suspect. Gerhard "Ask Me About System Design" Paseman, 2011.10.04 | |
| Oct 4, 2011 at 15:59 | comment | added | Aaron Sterling | @Gerhard: I see what you are saying. I will have to think about this for a bit, to make sure I am asking for what I really need. Thank you. | |
| Oct 4, 2011 at 15:55 | comment | added | Gerhard Paseman | I think you want (p-1) to be the product of two numbers, each sufficiently large. Otherwise any prime p > max(n,m) will have a factor of p-1 larger than both m and n, and any p > 2max(n,m) will have one factor bigger than twice n and another bigger than m. You might also borrow with attribution my and quid's and Igor's comments about the product version of your problem, which is what I suspect you really want. Gerhard "Ask Me About System Design" Paseman, 2011.10.04 | |
| Oct 4, 2011 at 15:51 | comment | added | Aaron Sterling | @Someone: hahaha!!! Yes, you have a point. Better now? | |
| Oct 4, 2011 at 15:50 | history | edited | Aaron Sterling | CC BY-SA 3.0 |
fixed title
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| Oct 4, 2011 at 15:49 | comment | added | Someone | You should improve the title of your question. | |
| Oct 4, 2011 at 15:41 | history | edited | Aaron Sterling | CC BY-SA 3.0 |
added example
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| Oct 4, 2011 at 15:38 | comment | added | Aaron Sterling | Hmm... I wish to find the smallest $p$ for which this is true of any $i$ and $j$. So $i$ and $j$ "don't matter." I will add an example to the question. | |
| Oct 4, 2011 at 15:34 | comment | added | Igor Rivin | Do you really mean the question as you pose it? I thought you wanted $p-1 = (m+i)(n+j)?!$ | |
| Oct 4, 2011 at 15:30 | history | asked | Aaron Sterling | CC BY-SA 3.0 |