most recent 30 from http://mathoverflow.net 2013-05-23T03:43:21Z http://mathoverflow.net/feeds/question/50661 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50661/unboundedness-of-number-of-integral-points-on-elliptic-curves Kevin Buzzard 2010-12-29T14:50:10Z 2011-01-12T15:42:40Z

<p>If $E/\mathbf{Q}$ is an elliptic curve and we put it into minimal Weierstrass form, we can count how many integral points it has. A theorem of Siegel tells us that this number $n(E)$ is finite, and there are even effective versions of this result. If I'm not mistaken this number $n(E)$ is going to be a well-defined invariant of $E/\mathbf{Q}$ (because different minimal Weierstrass models will have the same number of integral points).</p> <p>Is it known, or conjectured, that $n(E)$ is unbounded as $E$ ranges over all elliptic curves?</p> <p>Note: the question is trivial if one does not put $E$ into some sort of minimal form first: e.g. take any elliptic curve of rank 1 and then keep rescaling $X$ and $Y$ to make more and more rational points integral. </p>

http://mathoverflow.net/questions/50661/unboundedness-of-number-of-integral-points-on-elliptic-curves/50670#50670 Felipe Voloch 2010-12-29T17:44:22Z 2010-12-29T17:44:22Z

<p>It is expected that the number of integral points is bounded in terms of the rank (this is known for some curves not in minimal Weierstrass form, Silverman JLMS 28 (1983), 1–7). So, if you could prove unboundedness of $n(E)$, you'd have a shot at proving unboundedness of rank which, as you know, is a hard problem.</p> <p>On the other hand, if you believe Lang's (and Vojta's) conjectures on rational points on varieties of general type, then you would conclude that $n(E)$ is uniformly bounded (Abramovich, Inv. Math. 127 (1997), 307–317).</p> <p>BTW, Kevin, don't you have some catching up to do?</p>

http://mathoverflow.net/questions/50661/unboundedness-of-number-of-integral-points-on-elliptic-curves/50766#50766 Joe Silverman 2010-12-30T21:51:52Z 2011-01-12T15:42:40Z

<p>Hi Kevin, </p> <p>I proved that if $E/\mathbf{Q}$ is given using by a minimal Weiestrass equation, then</p> <p>$ \#E(Z) \le C^{\text{rank} E(Q) + n(j) + 1} $</p> <p>where $n(j)$ is the number of distinct primes dividing the denominator of the $j$-invariant of $E$ and $C$ is an absolute constant. This is in <em>J. Reine Angew. Math.</em> <strong>378</strong> (1987), 60-100.</p> <p>Mark Hindry and I proved that if you assume the abc conjecture, then you can remove the n(j) in the above estimate. This is in <em>Invent. Math.</em> <strong>93</strong> (1988), 419-450. It is a conjecture due to Lang.</p> <p>The papers contain more general results for (quasi)-S-integral points over number fields.</p>