Timeline for Rational Points on $y^2=x^3-86069^5$
Current License: CC BY-SA 3.0
24 events
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| Sep 25, 2012 at 0:06 | history | edited | James Weigandt | CC BY-SA 3.0 |
Added point from Tom Fisher
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| Sep 24, 2012 at 22:46 | comment | added | James Weigandt | UPDATE: PointSearch(C, 10^27) for all 12-covers C took about 11 days. Nothing to report. Trying 10^30. | |
| Sep 13, 2012 at 16:28 | comment | added | James Weigandt | @Kevin: thanks. I was really hoping to go to that conference but unfortunately I have other obligations, and can't afford it. I would suggest speaking to Tom Fisher about the possibility of patching together 8-descents and 3-descents to get a 24-descent. I'm not sure if it is feasible for this curve or not, but if it were, the other generator shouldn't be too hard to find. | |
| Sep 13, 2012 at 4:31 | comment | added | Kevin Acres | @Jamie - Thank you for your continuation of this search. It's very much appreciated. Meanwhile, one of our group will be at the 'Selmer Groups, Descent and the Distribution of Ranks' conference at Warwick between 24th and 28th September. Hopefully something useful will come out of that. | |
| Sep 10, 2012 at 14:40 | comment | added | James Weigandt | UPDATE: PointSearch(C,10^24) for all 12-covers C took about 3 days and found no new points. I'll try to run something bigger on my department's servers. | |
| Sep 6, 2012 at 14:08 | comment | added | James Weigandt | @Noam: Thanks! It's nice to get one of these higher descents to work out. I've been looking at curves with bigger conductors recently and getting discouraged. @joro: My most recently search of the 12-covers was up to 10^21 and took about 8 1/2 hours. No new points to report. A big part of the descent procedure is minimizing and reducing the 12-covers so that the points on them are as small as possible. So changing the model should not help in this case. | |
| Sep 6, 2012 at 13:05 | comment | added | joro | In what time was PointSearch(C, 10^15) computed? Since E is isomporphich to E': y^2 = x^3 - 1/86069 can you please try PointSearch(E',10^15) (the coefficients are much smaller). | |
| Sep 6, 2012 at 4:45 | comment | added | Noam D. Elkies | @K.Acres: It's not hard to make examples where the height is comparable to the height of the coefficients. If $k=t^2+1$ then $(x,y)=(-1,t)$ is a rational point; here $k = 2 \cdot 86069^{50}$. | |
| Sep 6, 2012 at 4:44 | comment | added | Noam D. Elkies | Congratulations! Not a record because Heegner-point techniques go well beyond descent when they're available, but possibly a record or near-record height for a curve of rank $>1$. $$ $$ My point-search techniques aren't that abstruse -- basically all I'm using (besides LLL) is that the image of an $\epsilon$-interval under an analytic map is contained in a box of size $\ll\epsilon\times\epsilon^2\times\epsilon^3\times\cdots$. Descent technology is much more complicated. (And Jeechul Woo's $3$-descent formulas won't help here: they need a rational $3$-torsion point, not just a $3$-isogeny.) | |
| Sep 6, 2012 at 4:42 | comment | added | James Weigandt | I suppose the goal should be to maximize h(P)/log|\Delta_min(E)|, although conceivably this could still get arbitrarily large. It will not get arbitrarily small though by Szpiro's conjecture. See modular.math.washington.edu/mcs/archive/Fall2001/notes/12-10-01/… | |
| Sep 6, 2012 at 3:55 | comment | added | Kevin Acres | @Jamie - I have a script that will generate points on curves of special types. For example on y^2=x^3+k, where k=4*86069^100+1, I can construct a point of height 3033.804673. But I needed a big value of k to get there. | |
| Sep 6, 2012 at 1:42 | history | edited | James Weigandt | CC BY-SA 3.0 |
Corrected my ignorance about saturation.
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| Sep 6, 2012 at 1:21 | comment | added | James Weigandt | @Kevin - I am interested in the long term project of implementing higher descents in Sage. Jeechul Woo has some GP scripts for descent via 3-isogeny, which is applicable to Mordell curves. | |
| Sep 6, 2012 at 0:47 | history | edited | James Weigandt | CC BY-SA 3.0 |
Wrote P = (r/t^2, s/t^3)
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| Sep 6, 2012 at 0:44 | comment | added | Kevin Acres | @Jamie - It's unfortunate that I don't have Magma here, only Pari/gp. I'll have to hunt around to see if I can find any scripts for descent in that language. Other than that I'll try and understand the process and write my own, which, I'm sure, will be a big learning curve for me. | |
| Sep 5, 2012 at 23:56 | comment | added | James Weigandt | @Kevin - Thank you, but I only did the tiniest fraction of the work that went into finding this point. Tom Fisher's work on 12-Descent and Noam's ideas on point searches implemented in MAGMA by Mark Watkins are essential parts of these calculations which I only vaguely understand. | |
| Sep 5, 2012 at 23:08 | comment | added | Kevin Acres | @Jamie - Amazing job. Thank you so much for that. Surely this has to be close to, if not actually, a record height found so far! | |
| Sep 5, 2012 at 23:04 | vote | accept | Kevin Acres | ||
| Sep 5, 2012 at 23:04 | history | bounty ended | Kevin Acres | ||
| Sep 5, 2012 at 22:06 | history | edited | James Weigandt | CC BY-SA 3.0 |
Found a point of infinite order!
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| Aug 31, 2012 at 18:25 | history | edited | James Weigandt | CC BY-SA 3.0 |
I'd written "sha" was autocorrected to "she", somewhat humorous.
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| Aug 27, 2012 at 2:25 | comment | added | James Weigandt | An 8-descent took about 1500 seconds in Magma with class group bounds set to GRH. I searched up to height 10^12 and the 8-covers (this took about another 1000 seconds) and didn't find anything yet, but the coefficients of the 8-covers are very very large. | |
| Aug 27, 2012 at 1:31 | comment | added | Kevin Acres | Thanks Jamie, we'll dedicate a bit of processing time to this and see if it won't crack. | |
| Aug 27, 2012 at 1:12 | history | answered | James Weigandt | CC BY-SA 3.0 |