Timeline for What are the rational solutions of $y^2 = 4x^n + z^{n-1}$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 30, 2020 at 17:43 | comment | added | Maciej Ulas | @joro Exactly. That is what I have done. | |
| Aug 29, 2020 at 11:20 | comment | added | joro | So you gave rational parametrization of two families of rational surfaces? | |
| Aug 28, 2020 at 13:39 | comment | added | Maciej Ulas | Some solutions are omitted. If $n$ is odd and $x=0$ then we need to take $v=\pm 1$. However, then we get $y=0$ from our parametrization. This is not a real problem because we can parametrize $y^2=z^{n-1}$ easily. The same situation occurs in the case $n$ even and $v=0$. In both cases the reason for this is simple: the maps constructed are rational and not defined everywhere. | |
| Aug 28, 2020 at 13:01 | comment | added | joro | Are there other solutions besides yours? | |
| Aug 28, 2020 at 9:57 | vote | accept | Q_p | ||
| Aug 28, 2020 at 9:38 | history | answered | Maciej Ulas | CC BY-SA 4.0 |