In this post I will be detailing a problem I have thought about for about a year, but has not made any progress on. The problem is one posed by my advisor, Professor Cameron Stewart, in the following paper:

His conjecture can be seen on page 816 of the above paper:

“… we conjecture that there exists an absolute constant such that for any binary form with nonzero discriminant and degree at least three there exists a number , which depends on , such that if is an integer larger than then the Thue equation [] has at most solutions in coprime integers and .”

In the same paper, Stewart proves that the equation has no more than solutions, where is a large divisor of . The crucial point is that quite often, we can pick which has few prime divisors, so that is small (see also this recent preprint due to Shabnam Akhtari: http://arxiv.org/abs/1508.03602).

However, this situation should not be typical. The basic idea is that when is large with respect to the coefficients of , that should behave `generically’, meaning that most of the points on the plane curve defined by the Thue equation should have few rational points. Indeed, generically the curve should be of general type (i.e. have genus at least 2), so one should not expect too many rational points by Faltings’ theorem.

Nevertheless, if one checks Stewart’s argument (this particular part is not necessarily new; Bombieri and Schmidt had used roughly the same idea in other papers), then one sees that a crucial ingredient is a -adic reduction lemma. In particular, for a prime and a positive integer such that , one can transform the equation into at most many equations of the form . Indeed, this is where the power of appears in Stewart’s theorem. Another crucial input of Stewart, arguably the most novel part of his paper, is that he shows that the equation has few solutions provided that is sufficiently small with respect to various invariants of ; in particular, he does not require that .

The deficit of Stewart’s argument, as with all other related results, is that it fundamentally obtains the ‘wrong’ bound: one should expect *fewer *solutions to the equation when is large, not more. In particular, all arguments involve some sort of reduction or descent argument and reducing the equation to one where the right hand side is a small integer, thereby susceptible to various diophantine approximation arguments, at a cost that is not so severe. This type of argument is unlikely to be useful in resolving Stewart’s conjecture.

However, recent groundbreaking work due to Manjul Bhargava offer hope. In this paper, Bhargava shows that most hyperelliptic curves in the sense of some natural density with respect to the coefficients of the curve written as with a binary form with integral coefficients and even degree, have no rational points. One of the most ingenious constructions in the paper is an *explicit *representation of a binary form whenever the curve

has a rational point. Indeed, if the curve above has a rational point say, then there exist integer matrices with such that

He then showed that most binary forms cannot be represented in the above way, thereby showing that most hyperelliptic curves have no rational points.

If one carries out Stewart’s -adic reduction argument all the way down to 1, we see the following. Start with a generic equation , apply a descent argument, and end up with a bunch of equations of the form . In particular, the curve defined by

has a rational point! Therefore, it must be one of the rare binary forms with a representation of the form . I call this effect ‘funneling’, because we start with a generic, typical object and end up with a special object. Therefore, all of the initial data had to pass through a ‘general to special’ funnel. Since the final object is so rare, the initial object could not be too abundant; otherwise we would ‘violate’ Bhargava’s result.

Of course, the above paragraph is too coarse and fuzzy to formulate precise statements, much less proving a rigorous theorem. However, I still believe that this `funnel’ should exist and that it will lead to a solution to Stewart’s conjecture.

If you have any insights or ideas, please feel free to contact me, I would be extremely pleased if this question gets an answer.