Skip to main content
4 events
when toggle format what by license comment
Oct 17, 2009 at 6:28 comment added Harrison Brown Hmm, in that case I'm unconvinced that it'd be reasonable to conjecture the affirmative, and I'd look at some of the crazy stuff that's stronger than GRH. I now think that Elliott-Halberstam isn't actually enough for a negative answer, but it seems to provide evidence in that direction. (I'm not getting out pencil and paper because I plan to get some sleep tonight.)
Oct 17, 2009 at 6:20 comment added David E Speyer Nope, no heuristics as yet. It's just that this is the bound we need for the group theory problem, and it is exactly at the right place to meet the bounds that you expect from GRH.
Oct 17, 2009 at 6:20 comment added Harrison Brown I should get some sleep rather than posting further on this, but it's probably also worth thinking about the Elliott-Halberstam conjecture, which now that I think about it might be strong enough to answer the question in the negative.
Oct 17, 2009 at 6:12 history answered Harrison Brown CC BY-SA 2.5