Timeline for Find all possible rational pairs of a parametric sextic and all cases where it is reducible for either parameter
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 2, 2019 at 1:25 | vote | accept | Lorenz H Menke | ||
| Jun 3, 2015 at 1:37 | answer | added | Jeremy Rouse | timeline score: 2 | |
| Jun 2, 2015 at 18:14 | comment | added | Lorenz H Menke | How do I map this sextic to an elliptic curve. I do not see any of the possible steps. It is really puzzling. Thanks for the help. | |
| Jun 2, 2015 at 1:21 | comment | added | Jeremy Rouse | Yes, you are correct. Sorry for the typo. | |
| Jun 2, 2015 at 1:02 | comment | added | Lorenz H Menke | It looks like the elliptic curve is $E: {y}^{2} = x^{3} - 159{,}600\, x + 31{,}482{,}500$. | |
| May 29, 2015 at 17:02 | comment | added | Lorenz H Menke | Yes, I have been working on the derivation with no success. In the problems that I am studying I usually can derive a condition that establishes an elliptic curve in which case I sometimes get a rank 3 curve and I have to test the solutions like you did up to some $|a|, |b|, |c| \le 20-50$. Other times I obtain an hyperelliptic curve in which case MAGMA can find the solutions up to a given height. | |
| May 29, 2015 at 12:00 | comment | added | Jeremy Rouse | The elliptic curve is $E : y^{2} = x^{3} - 159600x + 314825000$. Generators are $(-2600/9,-197450/27)$, $(-1015/4,59675/8)$, and $(700/9,-119350/27)$. I'm guessing you probably also need the map from your curve to the elliptic curve. | |
| May 28, 2015 at 17:26 | comment | added | Lorenz H Menke | Yes, that is what I need, the elliptic curve. I misunderstood what you said and thought there was any hyper elliptic curve. | |
| May 28, 2015 at 15:28 | comment | added | Jeremy Rouse | There's no genus 3 hyperelliptic curve that I found. Do you want the elliptic curve $E$ and the generators $P_{1}$, $P_{2}$ and $P_{3}$? | |
| May 28, 2015 at 4:39 | comment | added | Lorenz H Menke | Can you show the genus 3 hyperellipic cure that you deriived for this problem. | |
| May 27, 2015 at 1:28 | comment | added | Jeremy Rouse | Your equation defines a curve of genus $7$. Setting $b = a^{2}$ gives a curve (in terms of $b$ and $z$) of genus $4$, which has an involution. The quotient by this involution is a rank $3$ elliptic curve $E$. This does not give a good method of proving your set of rational solutions is complete, but it does give an efficient way to search for points. I looked for preimages on your original curve of points on $E$ of the form $aP_{1} + bP_{2} + cP_{3}$, where $P_{1}$, $P_{2}$ and $P_{3}$ are generators of the MW-group and $|a|, |b|, |c| \leq 7$ and only found the 19 points you know about. | |
| May 27, 2015 at 0:45 | history | asked | Lorenz H Menke | CC BY-SA 3.0 |