unboundedness of number of integral points on elliptic curves?

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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).

Is it known, or conjectured, that $n(E)$ is unbounded as $E$ ranges over all elliptic curves?

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.

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I vaguely remember a similar question may imply some conjecture. Try searching for "integral points on minimal". Currently the 2 URLs timeout for me. May be wrong... – jerr18 Dec 29 2010 at 15:39
web.archive.org/web/20080509110233/http://… Stein: Is it conjectured that the number of integral points on minimal curves is unbounded? If so, then abc implies that ranks are unbounded. – jerr18 Dec 29 2010 at 16:07

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Not that it helps much, but this can be expressed more intrinsically in terms of the N\'eron model $\mathcal{E}$ of $E$ over $\mathbf{Z}$, avoiding the ick of minimal Weierstrass models (locally or globally). Namely, the equality $E(\mathbf{Q}) = \mathcal{E}(\mathbf{Z})$ carries the pts that are everywhere integral with respect to local (or global) minimal Weierstrass models over to exactly the pts in $\mathcal{E}(\mathbf{Z})$ disjoint from the identity section and supported in the open relative identity component $\mathcal{E}^0$. So that gives a clean defn of $n(E)$ over any number field. – BCnrd Dec 30 2010 at 5:39

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By curiosity, is the result known for elliptic curves over function fields? In this case, unboundedness of the rank was proved by Shafarevich and Tate (there are also results by Ulmer). I don't know whether these constructions yield arbitrarily many integral points. – François Brunault Jan 4 2011 at 12:54

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Over function fields you have to be careful because the number of integral points can be not just unbounded but infinite! For example, in characteristic 2 if $(x,y)$ is an integral point on the supersingular curve $y^2+y=x^3+a$ then so is $(x^2,y+x^3)$, so as long as $x \notin \overline{{\bf F}_2}$ we get an infinite sequence of integral points. For instance, let $a=t^3$ and start from $(t,0)$ to get an infinite sequence of integral points (i.e. solutions of $y^2+y=x^3+t^3$ in polynomials $x,y \in {\bf F}_2[t]$; yes, that's a Néron model). – Noam D. Elkies Jun 10 2011 at 2:52
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2 Answers

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.

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).

BTW, Kevin, don't you have some catching up to do?

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Abramovich restricts to semi-stable curves, if I understand correctly. He speculates that there is a bound on $n(E)$ depending on the number of additive places. Remark 2 after Cor 1 in springerlink.com/content/xgbvqxm4383nwhqu – Chris Wuthrich Dec 29 2010 at 18:04

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@Felipe: yes, I do. But that "catching up" involves going to work and spending a day concentrating; I can ask idle questions whilst babysitting three children! I'm at work tomorrow so I'll get things done then: I have a "no MO at work" policy, for example! – Kevin Buzzard Dec 29 2010 at 18:46

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Hi Kevin,

I proved that if $E/\mathbf{Q}$ is given using by a minimal Weiestrass equation, then

$\#E(Z) \le C^{\text{rank} E(Q) + n(j) + 1}$

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 J. Reine Angew. Math. 378 (1987), 60-100.

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 Invent. Math. 93 (1988), 419-450. It is a conjecture due to Lang.

The papers contain more general results for (quasi)-S-integral points over number fields.

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Thanks Joe! I didn't know about these results, which are pretty neat. To answer my question though we need some sort of bound the other way, right? I guess we now know that ABC implies that if there does happen to exist a universal bound for the rank then there's also a universal bound for the integral points---this presumably being the source of William Stein's quoted comment above. Can one prove "rank unbounded implies integral points on minimal models unbounded" though? – Kevin Buzzard Dec 31 2010 at 20:36

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Kevin asks: Can one prove "rank unbounded implies integral points on minimal models unbounded" though? No. In fact, it may well be that the number of integral points on minimal models is uniformly bounded (I don't have a strong feeling on that). One might guess that integral points satisfy h(P) << h(E) for an absolute constant, while Lang suggests that on "most" curves of positive rank, the smallest non-torsion point satisfies h(P) >> C^h(E). Conclusion would be that "most" curves have no integral points. So there should be lots of curves with big rank and no integral points. – Joe Silverman Jan 1 2011 at 13:50
Hi Professor Silverman! I believe the displayed formula is broken because it needs two backslashes before the pound sign. – Zev Chonoles Jan 12 2011 at 3:26
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