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Nov 14, 2023 at 3:46 comment added Navvye I'm trying to solve a question that actually needs the most explicit bound possible, the best I've found is $10h_{3}(-108d)$ where $h_3(-108d)$ is the class number for cubic form with a discriminant $d$ — and since we don't have an estimate on the bound of the rank of elliptic curves, I don't know what to do. In case you're interested, the actual problem I'm solving is how many elliptic curves exist such that $y^2 = x^3 + D$ has exactly $D$ solutions
Nov 14, 2023 at 3:31 comment added 2734364041 @Navvye Theorem 1.2 in Alpöge-Ho is completely explicit, no big-oh's at all. If you want something completely explicit without rank dependence, you'll need to work incredibly hard to make Bhargava-Shankar-Taniguchi-Thorne-Tsimerman-Zhao explicit. That is probably not worth the effort.
Nov 14, 2023 at 3:23 comment added Navvye Thank you so much. I have one last question: Do you know of any papers where an explicit bound for the number of solutions is given in terms of the discriminant (apart from the one I mentioned)? Either way, thank you very much for your response!
Nov 14, 2023 at 3:15 comment added 2734364041 @Navvye Theorem 1.1 follows as a consequence of Theorem 1.2. You are welcome to read their paper. There are lots of reasons to believe that the rank is absolutely bounded, just like there are lots of reasons to believe otherwise. There is a lot of literature out there for you to read on the matter; I do not know of a good summarizing paper.
Nov 14, 2023 at 2:52 comment added Navvye @2734364041 Thank you for your response. Two questions: Firstly, isn't theorem 1.2 related to the number of $S$ integral points over an elliptic curve instead of just the integral points? Secondly, why do people believe $\mathrm{rank}(E_{A,B})<c$? Didn't Prof. Elkies discover a curve with rank $28$?
Nov 14, 2023 at 2:24 comment added 2734364041 @GHfromMO Indeed! It's now fixed, thanks.
Nov 14, 2023 at 2:24 history edited 2734364041 CC BY-SA 4.0
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Nov 14, 2023 at 1:57 comment added GH from MO The number of primes dividing $n$ has maximal order $O(\log n/\log\log n)$ instead of $O(\log n)$.
Nov 14, 2023 at 0:39 history edited Peter Humphries CC BY-SA 4.0
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Nov 14, 2023 at 0:39 comment added Joe Silverman And if you believe the $abc$-conjecture (or alternatively Lang's height lower bound conjecture), then on a quasi-minimal Weierstrass equation (meaning $\gcd(A^3,B^2)$ is 12th power free), there is a bound $$\#E_{A,B}(\mathbb Z)=O(C^{1+\text{rank }E(\mathbb Q)}),$$ where $C$ is an absolute constant. So if you also believe the rank upper bound conjecture that you alluded to, there should be an absolute bound for $\#E_{A,B}(\mathbb Z)$.
Nov 14, 2023 at 0:38 history edited 2734364041 CC BY-SA 4.0
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Nov 14, 2023 at 0:38 history edited Peter Humphries CC BY-SA 4.0
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Nov 14, 2023 at 0:37 history edited 2734364041 CC BY-SA 4.0
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Nov 14, 2023 at 0:01 history answered 2734364041 CC BY-SA 4.0