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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jan 10, 2010 at 10:48 comment added Ilya Nikokoshev @Qiaochu Yuan, you're right, sorry.
Jan 10, 2010 at 10:45 history edited Ilya Nikokoshev CC BY-SA 2.5
correction
Jan 10, 2010 at 1:21 comment added Qiaochu Yuan I'm not sure I understand. The leading homogeneous term of (x^2 - x)(y + x^{20})^2 is x^{42}, which is a sum of squares.
Jan 9, 2010 at 23:57 vote accept Ilya Nikokoshev
Jan 9, 2010 at 23:57 comment added Ilya Nikokoshev @Qiaochu Yuan: no, that's not true. You can always change $(x, y) \to (x, y + x^20)$, so you either prove there are no solutions, or lots of them. This is the primary difficulty I'm having about the problem :)
Jan 9, 2010 at 23:55 history edited Ilya Nikokoshev CC BY-SA 2.5
correction
Jan 9, 2010 at 21:52 comment added Qiaochu Yuan As far as an application to the original problem, I am reasonably sure that the "leading homogeneous term" of the polynomial must be a sum of squares (of polynomials, since it's homogeneous), but I haven't checked that yet either.
Jan 9, 2010 at 21:47 comment added Pete L. Clark @Ilya: Your "note" is somewhat confusing (I claim that every non-negative function is bounded below!). I think you are missing something like "on $\mathbb{Q} \times \mathbb{Q}$".
Jan 9, 2010 at 21:45 comment added Charles Siegel So, you've got some answers involving Hilbert 17, but is that really what you're looking for? If not, perhaps you could say something about what kinds of things you're looking for in answers.
Jan 9, 2010 at 21:45 history edited Pete L. Clark CC BY-SA 2.5
fixed grammar
Jan 9, 2010 at 21:40 answer added Pete L. Clark timeline score: 5
Jan 9, 2010 at 21:10 answer added Charles Siegel timeline score: 3
Jan 9, 2010 at 20:44 history asked Ilya Nikokoshev CC BY-SA 2.5