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Nov 4, 2015 at 14:42 vote accept Misha Verbitsky
S Sep 9, 2015 at 1:47 history bounty ended David E Speyer
S Sep 9, 2015 at 1:47 history notice removed David E Speyer
Sep 9, 2015 at 0:41 answer added Noam D. Elkies timeline score: 3
Sep 1, 2015 at 22:33 answer added Misha Verbitsky timeline score: 1
S Sep 1, 2015 at 15:41 history suggested GNiklasch CC BY-SA 3.0
qualifier nonzero needed in one place
Sep 1, 2015 at 15:23 review Suggested edits
S Sep 1, 2015 at 15:41
Sep 1, 2015 at 14:32 history edited David E Speyer CC BY-SA 3.0
deleted 35 characters in body
Sep 1, 2015 at 13:33 comment added Fedor Petrov Maybe, we may remove small values of (x,x) using Chinese remainder theorem?
S Sep 1, 2015 at 12:53 history bounty started David E Speyer
S Sep 1, 2015 at 12:53 history notice added David E Speyer Draw attention
Sep 1, 2015 at 12:52 history edited David E Speyer CC BY-SA 3.0
added 1297 characters in body
Aug 28, 2015 at 0:22 answer added few_reps timeline score: 3
Aug 26, 2015 at 10:59 history edited Misha Verbitsky CC BY-SA 3.0
added 10 characters in body
Aug 26, 2015 at 10:59 comment added Misha Verbitsky definitely! thanks for pointing this out, I amended the question
Aug 25, 2015 at 22:48 comment added Anthony Quas Presumably you want your lattice not to be a multiple of any other lattice?
Aug 25, 2015 at 22:16 comment added Will Jagy Oh, that probably does demand some kind of diophantine approximation argument, then; I don't quite recall what you commented earlier on the proof you found not really to your liking.
Aug 25, 2015 at 21:38 history edited Misha Verbitsky CC BY-SA 3.0
added 11 characters in body
Aug 25, 2015 at 21:38 comment added Misha Verbitsky Sorry! I misstated the question: I should add an assumption that this 2-dimensional lattice is not definite!
Aug 25, 2015 at 20:48 comment added Will Jagy seems to me there is a sublattice of dimension $\lceil \frac{n}{2} \rceil$ on which your nondegenerate form is definite. Does that suffice?
Aug 25, 2015 at 19:56 history asked Misha Verbitsky CC BY-SA 3.0