Timeline for Are there integer solutions to $3y^2 = 4x^3-1$ other than $(1,1)$ and $(1,-1)$?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 2, 2016 at 16:35 | comment | added | Bobby Grizzard | If a CAS can do it, I guess a hand can do it, although my hands don't have much experience in the matter. Probably you do some kind of descent/ Selmer group calculation? | |
| Aug 2, 2016 at 14:02 | comment | added | Pablo | @BobbyGrizzard and could you prove by hand that the rank is zero? | |
| Aug 2, 2016 at 13:52 | comment | added | Bobby Grizzard | @pablo I just mean that the CAS can compute whole Mordell-Weil group over of the elliptic curve $\mathbb{Q}$. Apparently the group only has 3 points. To figure out what the torsion looks like by hand, you'd have to do some work, but these techniques are well-known. | |
| Aug 2, 2016 at 6:39 | comment | added | Pablo | @BobbyGrizzard how do you know that only the $3$-torsion points are defined over $\mathbb{Q}$? Why (say) there is no $5$-torsion? | |
| Aug 1, 2016 at 22:50 | comment | added | Bobby Grizzard | Maybe it's worth pointing out that, even if one did not realize it was isomorphic to the Fermat curve, one could discover quickly using a CAS that the elliptic curve has rank 0, and then that the only rational points were the obvious ones (which I guess are 3-torsion points). | |
| Aug 1, 2016 at 18:18 | comment | added | Jeremy Rouse | @Pablo and Wojowu - I used Magma, and knew off the top of my head that $y^{2} + y = x^{3} - 7$, one elliptic curve with conductor $27$, is isomorphic to the Fermat cubic. | |
| Aug 1, 2016 at 18:17 | comment | added | Wojowu | @Pablo This is a rather wild guess, but I suppose some computer algebra system (e.g. Sage) can find the isomorphisms. | |
| Aug 1, 2016 at 18:14 | vote | accept | Pablo | ||
| Aug 1, 2016 at 18:11 | comment | added | Pablo | @JeremyRouse Maybe this is an inappropriate question, but how did you know so quickly that this is going to be isomorphic to a Fermat curve? Have yous just played with the equations or is there some theory/criterion behind this? | |
| Aug 1, 2016 at 18:02 | comment | added | Jeremy Rouse | The isomorphism from $3y^2z = 4x^3 - z^3$ to $X^3 + Y^3 = Z^3$ is given by $X = (y-z)/2$, $Y = x$ and $Z = (y+z)/2$. | |
| Aug 1, 2016 at 17:55 | comment | added | Pablo | @WhatsUp Right! But how do I show that the curves are isomorphic over the rationals? | |
| Aug 1, 2016 at 17:53 | comment | added | WhatsUp | It doesn't necessarily give a bijection between integral points. But the NON-EXISTENCE of other rational points on one curve immediately implies the non-existence of other rational points on the other curve, and a fortiori the non-existence of other integral points. | |
| Aug 1, 2016 at 17:48 | comment | added | Pablo | What would an isomorphism over $\mathbb{Q}$ look like? And why does it necessarily give a bijection between integral points? | |
| Aug 1, 2016 at 17:45 | history | answered | Jeremy Rouse | CC BY-SA 3.0 |