Timeline for Status of $x^3+y^3+z^3=6xyz$
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 4, 2021 at 5:13 | answer | added | Tomita | timeline score: 2 | |
| Apr 1, 2021 at 12:21 | answer | added | Erik Dofs | timeline score: 4 | |
| Feb 27, 2021 at 16:28 | answer | added | Sam | timeline score: 1 | |
| Feb 23, 2021 at 5:40 | vote | accept | Haran | ||
| Feb 22, 2021 at 15:01 | answer | added | Timothy Chow | timeline score: 15 | |
| Feb 22, 2021 at 12:04 | comment | added | Haran | @Wojowu Thank you :) | |
| Feb 22, 2021 at 12:03 | comment | added | Wojowu |
The method to transform an arbitrary smooth cubic into a Weierstrass form is standard and is explained for instance in the lovely book by Silverman-Tate, see section 1.3 or appendix B there. It is also implemented in Sage as EllipticCurve_from_cubic
|
|
| Feb 22, 2021 at 9:46 | comment | added | Haran | @NoamD.Elkies Thank you for your response! If possible, could you let me know what substitution you made to transform the given Diophantine Equation into an Elliptic Curve? Thanks in advance. | |
| Feb 22, 2021 at 2:57 | comment | added | David Roberts♦ | @NoamD.Elkies and even more characters can be used for human-readable information that might be of use to someone down the track. :-) | |
| Feb 22, 2021 at 2:55 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
added 118 characters in body; edited tags
|
| Feb 22, 2021 at 2:45 | comment | added | Noam D. Elkies | @Gerry Myerson Fixed this now (seems that 9 of the 10 characters can be invisible whitespace ☺) | |
| Feb 22, 2021 at 2:41 | history | edited | Noam D. Elkies | CC BY-SA 4.0 |
Correct a typo signaled by **Gerry Myerson** ($z^3$, not $z^2$)
|
| Feb 21, 2021 at 22:22 | comment | added | Gerry Myerson | In the first line of the body of the question, $z^2$ should be $z^3$. (I would just edit this myself, but edits have to be at least ten characters.) | |
| Feb 21, 2021 at 20:07 | comment | added | Noam D. Elkies | @dodd as usual, each rational point on the elliptic curve corresponds to a unique primitive solution (up to multiplying $(x,y,z)$ by $-1$). $$ $$ Haran: The positive solutions constitute one of two connected components of the real locus of the elliptic curve. Since the generator is on that component (and the identity is not), its odd multiples again yield positive solutions. The next batches of positive solutions are permutations of $(1817, 3258, 5275)$ and $(4904676969, 10840875082, 15051171563)$. | |
| Feb 21, 2021 at 19:37 | comment | added | markvs | Usually in this case one is interested in co-prime solutions, @Lubin. Like in the paper linked in the OP. | |
| Feb 21, 2021 at 19:34 | comment | added | Lubin | But rational solutions to the Weierstrass equation will always correspond to integer solutions of the homogeneous equation in OP’s question, @dodd . | |
| Feb 21, 2021 at 19:34 | comment | added | Haran | @NoamD.Elkies Can we say anything specific about positive solutions alone? | |
| Feb 21, 2021 at 19:31 | comment | added | markvs | I am not sure the OP is interested in rational (not integer) solutions. | |
| Feb 21, 2021 at 19:19 | history | edited | LSpice | CC BY-SA 4.0 |
Name of paper and DOI
|
| Feb 21, 2021 at 18:57 | comment | added | Noam D. Elkies | This curve has a rational point at $(x:y:z)=(1:2:3)$ (and infinitely many others, such as $(19:21:-52)$ ). If I did this right it's isomorphic with $Y^2 + Y = X^3 - 54 X - 88$, which has torsion ${\bf Z} / 3{\bf Z}$ (as usual in this family) and rank 1, see e.g. <lmfdb.org/EllipticCurve/Q/189/b/2>. So yes, one can start with $(1:2:3)$ and its permutations and generate all rational solutions with "chords and tangents". | |
| Feb 21, 2021 at 18:29 | history | asked | Haran | CC BY-SA 4.0 |