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.



Equidistribution of representations by a binary cubic form
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.


