Timeline for Difficult elliptic curves for $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28 at 0:06 | comment | added | Tomita | I appreciate your help so much. I've made a step forward. | |
| Feb 27 at 15:50 | comment | added | Jeremy Rouse | Yes - the computation you did in Magma is sufficient to show there is no rational solution. It's not exactly equivalent to what I did but it's enough. | |
| Feb 27 at 14:39 | comment | added | Tomita | Thank you for the detailed explanation. I applied $D^2$=quartic in $v=97/72$ to TwoCoverDescent function on Magma, Magma says 2-Selmer set is empty. Does this mean that $D^2$=quartic in $v=97/72$ has no rational solution? | |
| Feb 26 at 17:22 | comment | added | Jeremy Rouse | Regrading the OP's comment, these $D^{2} = \text{quartic in } v$ curves are $2$-covers of the Jacobian. (The Jacobian is $E_{u}$.) If a $2$-cover has a rational point on it, it is isomorphic to its Jacobian, but it can have no rational points on it at all. Roughly speaking a $2$-cover corresponds to an element of the 2-Selmer group, and the group $E(\mathbb{Q})/2E(\mathbb{Q})$ is the subset of the 2-Selmer group consisting of the $2$-covers that have rational points. (So knowing the rank is equivalent to knowing how many $2$-covers have rational points.) | |
| Feb 26 at 17:18 | comment | added | Jeremy Rouse | Regrading Tomita's question, Magma has the capability of doing a second two-descent on an elliptic curve over $\mathbb{Q}$: the input takes the form $y^{2} = g(x)$ where $\deg(g) = 4$ corresponds to a 2-cover of its Jacobian. The output is a list of genus $1$ intersections of two quadric in $\mathbb{P}^{3}$ which, when doubled in the 4-Selmer group equal the input. For $u = 97/72$ and $u = 103/78$, this output is empty, showing that $D^{2} = \text{ quartic in } v$ is not twice an element in Sha. (There's also a way to do this using the Cassels-Tate pairing.) | |
| Feb 26 at 3:14 | comment | added | Tomita | Why does the $D^2$= quartic in $v$ equation for $u=97/72$ and $u=103/78$ represent a 2-torsion element of the Shafarevich-Tate group of its Jacobian? Could you please explain how you did that? | |
| Feb 21 at 8:26 | vote | accept | Tito Piezas III | ||
| Feb 21 at 2:41 | comment | added | Tito Piezas III | Thanks. I am surprised at the result though. In p. 10 of MacLeod's paper, he gives an elliptic curve, $$E_u : M^2 = N^3 - 3K N^2 + 576u(u + 1)(u - 1)^3N$$ where $K = u^4 - 8u^3 - 6u^2 + 24u - 7,$ and we changed notation for consistency. He states the curve always has rank at least one and gives Table 4.1 \begin{array}{|c|c|}\hline u & \text{Estimated rank}\\ \hline 67/42 & 1-3\\ \hline 97/72 & 2-4\\ \hline 103/78 & 2-4\\ \hline \end{array} I am uncertain why he is considering $E_u$ if the estimated positive rank does not imply rational solutions $(x,y,z)$ to the original problem. | |
| Feb 20 at 21:56 | history | answered | Jeremy Rouse | CC BY-SA 4.0 |