Upper bound on number of integral solutions of elliptic curves

Question

I was studying M. Bhargava Et al's seminal paper titled "Bounds on 2-torsion in class groups of number fields and integral points on elliptic curves"

And came across a very fascinating observation that they made in the paper. Theorem 1.2 states:

The number of integral points on an elliptic curve, E, denoted by $N(E)$ is at most $O_\epsilon(\left| \text{Disc}(E) \right|^{0.1117... + \epsilon}) $

However, following the proof is a little above my pay-grade right now, and I'm not entirely sure as to how they derive the proof, which is given in section 5, Theorems 5.1 and 5.2.

I have a couple of questions:

Is there a fixed value of $\epsilon$ that can act as an upper bound? Am I correct in asserting that $\epsilon$ must be quite small in nature? By quite small, I mean $0.1117... + \epsilon < 1$. The authors state that the earlier upper bound which was given by Venkatesh, Helgott etc. was $O_\epsilon(\left| \text{Disc}(E) \right|^{0.2007... + \epsilon})$. So, can I reasonably assume that $\epsilon + 0.1117... < 0.2007....$

As you can see, I am desperately trying to simply get rid of the $\epsilon$ factor, and would really appreciate a hard upper bound which is not expressed in terms of $\epsilon$ or a constant.

Edit: I know that $N(E) = C(\left| \text{Disc}(E) \right|^{0.1118})$ And would like help in computing $C$. Of course, a weak lower bound is achievable by letting $N(E) = D$, but I would like to know the least number such that $C$ satisfies the condition $\forall E$.

Any help will be much appreciated. Thank you!

Answer 1

First, the bound that you cite has been significantly improved by Alpöge and Ho. Here is Theorem 1.1 therein. Let $A,B\in\mathbb{Z}$ satisfy $\Delta_{A,B}:=-16(4A^3+16B^2)\neq 0$. If $\mathscr{E}_{A,B}$ is the affine integral model $y^2 = x^3+Ax+B$ of the associated elliptic curve $E_{A,B}$ over $\mathbb{Q}$, then the number of solutions $(x,y)\in\mathbb{Z}^2$ to $y^2 = x^3+Ax+B$ is $$O\Big(2^{\mathrm{rank}(E_{A,B})}\prod_{p^2\mid \Delta_{A,B}}\min\Big(4\Big\lfloor\frac{v_p(\Delta_{A,B})}{2}\Big\rfloor+1,7^{2^7}\Big)\Big),$$ where $v_p(n)$ is the greatest nonnegative integer such that $p^{v_p(n)}\mid n$. Considering the fact that the number of primes dividing $n$ has maximal order $O((\log n)/\log\log n)$ and normal order $O(\log\log n)$, this bound considerably improves upon the one you mentioned. Keep in mind that many people now believe that there exists an absolute constant $c>0$ such that $\mathrm{rank}(E_{A,B})<c$, so the rank contribution to this bound is widely believed to be negligible. If you don't like the big oh notation, see Theorem 1.2 therein for a completely explicit version.

Second, by "Bhargava's seminal paper," I think you really mean "the seminal paper by Bhargava, Shankar, Taniguchi, Thorne, Tsimerman, and Zhao." Don't forget the other authors, even if they are numerous!

ADDED: I should also mention the Helfgott-Venkatesh bound

$$e^{O(\omega(\Delta_{A,B}))}1.33^{\mathrm{rank}(E_{A,B})}(\log|\Delta_{A,B}|)^2.$$

Here, $\omega(n)$ is the number of distinct prime factors of $n$. This might be stronger than Alpöge-Ho, depending on the prime factorization of $\Delta_{A,B}$. We can now take their minimum:

$$\ll\min\Big\{2^{\mathrm{rank}(E_{A,B})}\prod_{p^2\mid \Delta_{A,B}}\min\Big(4\Big\lfloor\frac{v_p(\Delta_{A,B})}{2}\Big\rfloor+1,7^{2^7}\Big),e^{O(\omega(\Delta_{A,B}))}1.33^{\mathrm{rank}(E_{A,B})}(\log|\Delta_{A,B}|)^2\Big\}.$$

Answer 2

You can take any $\epsilon>0$ you like. For example, by taking $\epsilon = 0.1118 - 0.1117...$ we get a bound $C|Disc(E)|^{0.1118}$ for some constant $C$. This is an "upper bound which is not expressed in terms of $\epsilon$", as requested.