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Jul 13, 2012 at 12:21 answer added Eduardo Jose timeline score: -2
Oct 1, 2011 at 17:09 answer added Noam D. Elkies timeline score: 4
Oct 1, 2011 at 11:05 answer added Eduardo José timeline score: 0
Mar 26, 2011 at 19:30 history edited Charles
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Mar 26, 2011 at 18:09 history edited Manuel Silva CC BY-SA 2.5
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Mar 25, 2011 at 20:05 comment added Manuel Silva @Mark: The points do not have to be on a straight line. Take for example (1,1), (5,25) and (7, 49) which correspond to the 3-AP of squares 1, 25, 49.
Mar 25, 2011 at 20:02 comment added Thomas Bloom This doesn't help with your problem, but it is interesting to note that as soon as we allow two variables, we get arbitrarily long arithmetic progressions from only a quadratic example: the polynomial $x^2+y^2$, for example. This is an immediate consequence of the Green-Tao theorem.
Mar 25, 2011 at 19:08 comment added Mark Bennet Don't terms in arithmetic progression lie on a straight line, and we are talking about the intersection of a polynomial with a straight line?
Mar 25, 2011 at 17:35 history edited Manuel Silva CC BY-SA 2.5
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Mar 25, 2011 at 0:40 history edited Manuel Silva CC BY-SA 2.5
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Mar 24, 2011 at 23:39 answer added Gerry Myerson timeline score: 1
Mar 24, 2011 at 22:23 history edited Manuel Silva CC BY-SA 2.5
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Mar 24, 2011 at 21:50 comment added Kevin Buzzard If there was no uniform bound for $f(A_p)$ for degree three polynomials, then (after scaling etc) I could find a cubic $f(x)$ and rationals $x_1$, $x_2$, $x_3$...$x_{1000000}$ such that $f(x_n)=n$. Now because 1000 of these $n$ are squares I have now got 1000 points on $y^2=f(x)$ and now perhaps I am pretty close to constructing elliptic curves with arbitrarily large rank. It's unclear to me what the concensus is about these things existing. Similarly I can construct curves of genus 3, say, over Q with as many points as you like and again it's not clear that such things should exist.
Mar 24, 2011 at 20:33 history asked Manuel Silva CC BY-SA 2.5