Timeline for What is the rank of the Mordell equation $y^2 = x^3 - 2$?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Nov 30, 2016 at 21:01 | vote | accept | Avram Grant | ||
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| Nov 27, 2016 at 18:38 | history | edited | Michael Stoll |
Added elliptic-curves tag
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| Nov 27, 2016 at 18:34 | answer | added | Michael Stoll | timeline score: 12 | |
| Nov 27, 2016 at 16:50 | comment | added | Noam D. Elkies | Hm, good point. Often for such equations there are trivial solutions with one variable equal zero; e.g. $y^2 = x^3 + 1$ has trivial solutions $(x,y) = (-1,0)$ (2-torsion) and $(0,\pm 1)$ (3-torsion), and also the nontrivial solutions $(2,\pm 3)$ (6-torsion) and no others. But $y^2 = x^3 - 2$ has no integer solutions with $xy=0$. There is the trivial element ("point at infinity") of the group of rational points, but that point is not integral. | |
| Nov 27, 2016 at 11:40 | comment | added | John Bentin | What might be a trivial integer solution? | |
| S Nov 27, 2016 at 10:57 | history | suggested | Martin Sleziak |
added (diophantine-equations) tag
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| Nov 27, 2016 at 10:22 | review | Suggested edits | |||
| S Nov 27, 2016 at 10:57 | |||||
| Nov 27, 2016 at 8:00 | history | edited | Avram Grant | CC BY-SA 3.0 |
added 73 characters in body
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| Nov 26, 2016 at 21:35 | comment | added | Noam D. Elkies | That's about a criterion for a curve in the family $y^2 = x^3 + k$ to have rank zero. It can be much sipmler for a single curve in this family, which is what this question is asking (with $k=-2$). | |
| Nov 26, 2016 at 20:02 | answer | added | Noam D. Elkies | timeline score: 12 | |
| Nov 26, 2016 at 19:49 | comment | added | Carlo Beenakker | it seems the answer is "no" --- mathoverflow.net/a/202080/11260 | |
| Nov 26, 2016 at 19:27 | review | First posts | |||
| Nov 26, 2016 at 19:36 | |||||
| Nov 26, 2016 at 19:26 | history | asked | Avram Grant | CC BY-SA 3.0 |