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Let $f(x,y)\in\mathbb{Z}[x,y]$ be a binary cubic form with nonzero discriminant, and for a positive integer $m$ consider the integral representations $f(x,y)=m$. Assume that the number of representations is large, then is it true that a positive proportion of the corresponding lattice points $(x,y)$ lies in the first quadrant (that is, $x$ and $y$ are positive)? More generally, are these lattice points equidistributed by angle (asymptotically, as $m\to\infty$)?

This question is motivated by the paper of Silverman (JLMS (2) 28 (1983), 1-7), which in turn is motivated by the paper of Mahler (PLMS (2) 39 (1935), 431-466). In particular, I wonder if the main theorem in Silverman's paper holds true when we restrict $(x,y)$ to a given angular sector.

Added. Silverman's theorem states that if $f(x,y)=m_0$ defines an elliptic curve over $\mathbb{Q}$ with rank $r$ (for some $m_0$), then the number of integral representations $f(x,y)=m$ is $\gg(\log m)^\frac{r}{r+2}$. I wonder if the same is true for the number of positive integral representations.

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  • $\begingroup$ i see; finite because of Thue, chapter 22 in Mordell's book. Could you please summarize the Silverman theorem you reference? $\endgroup$
    – Will Jagy
    Jul 17, 2015 at 20:35
  • $\begingroup$ getting there, jlms.oxfordjournals.org/content/s2-28/1.toc $\endgroup$
    – Will Jagy
    Jul 17, 2015 at 20:39
  • $\begingroup$ nope, they want a prescription jlms.oxfordjournals.org/content/s2-28/1/1.full.pdf+html $\endgroup$
    – Will Jagy
    Jul 17, 2015 at 20:41
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    $\begingroup$ GH, thank you. Nice result and conjecture/wonderment, no idea on my part whatsoever. $\endgroup$
    – Will Jagy
    Jul 17, 2015 at 20:45
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    $\begingroup$ @WillJagy: I think Mahler's argument is capable of showing positive proportion, but I have not studied it yet, and hoped that someone could tell the answer right away. Equidistribution might be much subtler. $\endgroup$
    – GH from MO
    Jul 17, 2015 at 20:45

1 Answer 1

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Yes, you should be able to do this. The point is that $E(\mathbb R)$ is just a circle group (as a real Lie group), so the image of the points $n_1P_1+\cdots+n_rP_r$, say for $|n_i|\le N$ as $N\to\infty$, are equidistributed in $E(\mathbb R)$ relative to Haar measure. (This is certainly true for $r=1$, and I'm pretty sure it's okay for all $r$, but haven't actually checked.) So you'll get a positive proportion of these points sitting in any interval of the circle $E(\mathbb R)$. In particular, the points in the first quadrant form such an interval, as do the points in any angular sector. Then you can just use those points for the construction in my paper, and the only thing that will change is the $\gg$ constant.

To answer the second question. I think that the points you get from the construction are indeed equidistributed. However, that is much different from saying that the set of all integer solutions to $f(x,y)=m$ is equidistributed. The point being, of course, that the construction creates an $m$ for which one can write down a lot of integer solutions (essentially by clearing denominators), but it says nothing about the existence or size or location of any additional solutions.

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    $\begingroup$ @GHfromMO You're welcome. This also reminds me of a related question that I believe is still open, namely whether there is even a single $f$ such that there is an infinite sequence of $m$'s for which the number of solutions to $f(X,Y)=m$ is ${}\gg|m|$. (And since some people have been suggesting that there is an absolute upper bound for the rank of elliptic curves over $\mathbb Q$, it's not even clear that the elementary method in my paper will get you to ${}\gg|m|^{1-\epsilon}$ for every $\epsilon>0$.) $\endgroup$
    – Joe Silverman
    Jul 18, 2015 at 0:03
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    $\begingroup$ I think you meant $\gg\log|m|$ and $\gg(\log|m|)^{1-\epsilon}$, respectively. Indeed it occurred to me what the best growth rate could be, but I was more interested in the number of representations as a sum of two (positive) cubes (motivated by a colleague's work where this question has a particular relevance). $\endgroup$
    – GH from MO
    Jul 18, 2015 at 0:25
  • $\begingroup$ @GHfromMO Yes, of course you're right, I left off the log. Unfortunately, I can't edit the comment, but you're comment clears it up. Thanks. $\endgroup$
    – Joe Silverman
    Jul 18, 2015 at 1:17

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