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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Sep 13, 2010 at 5:04 history edited Charles
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Dec 30, 2009 at 21:18 vote accept Rune
Dec 29, 2009 at 20:30 answer added Terry Tao timeline score: 89
Dec 29, 2009 at 15:15 answer added Steve Flammia timeline score: 8
Nov 6, 2009 at 5:23 answer added paarshad timeline score: 4
Nov 3, 2009 at 2:16 comment added Harrison Brown After thinking about it some more, I suspect that some variant on Pollard's rho might be able to do better for the promise problem than, say, GNFS factorization would, though still nowhere near polynomial. Still, I wonder how far you can push that idea.
Nov 2, 2009 at 23:53 comment added Rune That's a good point. It would be interesting to see if it were possible to solve the promise problem you stated.
Nov 2, 2009 at 22:28 comment added Harrison Brown I'd actually like to even see an algorithm that does the following: given a (large) integer n and promised that n has either 2 prime factors or between (ln ln n)^2 and 2(ln ln 2)^2 prime factors, it decides which in polynomial time (or even just faster than by factoring the numbers). This seems like it could be doable, since on average n would be expected to have just ln ln n prime factors, but I don't have a clue where you'd start.
Nov 2, 2009 at 19:41 comment added Ilya Nikokoshev Finding out whether a number has 2 versus >2 prime factors already seems inaccessible to me.
Nov 2, 2009 at 19:35 answer added Harrison Brown timeline score: 20
Nov 2, 2009 at 19:10 comment added Harrison Brown Determining whether the number of distinct prime factors is exactly 1 is also in P, via the easy trick of computing approximate kth roots for k < log n.
Nov 2, 2009 at 17:32 comment added Rune Both questions are interesting to me.
Nov 2, 2009 at 17:28 comment added Qiaochu Yuan Distinct or with multiplicity?
Nov 2, 2009 at 17:25 history asked Rune CC BY-SA 2.5