Timeline for How to find all integer points on an elliptic curve?
Current License: CC BY-SA 2.5
8 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Aug 23, 2014 at 14:25 | comment | added | user5262 | "and then use clever congruence techniques to massively cut down the search space by giving strong congruences for all the n_i. Then just do a brute force search. " Can you elaborate on this, please? | |
Apr 6, 2014 at 13:37 | comment | added | Joe Silverman | @KevinBuzzard Hi Kevin, Sorry to nit-pick (and on such an old post), but the "priveleged rational point" $O$ is definitely not an integral point. In essence, we are taking a model for $E$ over $\mathbb{Z}$, and when we say a point $Q$ is integral, we mean integral with respect to $O$, which by definition means that for every prime $p$, the points $\tilde Q \bmod p$ and $\tilde O\bmod p$ are distinct. Clearly $O$ itself fails to have that property in quite spectacular fashion! | |
Dec 6, 2009 at 16:28 | comment | added | Kevin Buzzard | My computer says it has proved those are the only ones. Whether or not you want to believe (a) my computer and (b) the program I used is up to you. If you want to try your own computer and another program (e.g. SAGE, which would also do the job) then feel free. If you want to prove it by hand then I would first buy a lot of pieces of paper, because what takes my computer 30 seconds will take a lot longer to do by hand. As for the "priviledged rational point", you can choose whether or not it's an integer point. That's not a maths question, it's a convention question. | |
Dec 6, 2009 at 15:32 | comment | added | amateur algebraist | Okay. But how can I prove those are the only one? Am I right that the priviledged rational point is not an integer point? | |
Dec 6, 2009 at 14:40 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
edited answer to reflect edits in question
|
Dec 6, 2009 at 14:27 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
added explicit solutions, now OP asked an explicit question
|
Dec 6, 2009 at 12:12 | vote | accept | amateur algebraist | ||
Dec 6, 2009 at 7:54 | history | answered | Kevin Buzzard | CC BY-SA 2.5 |