Timeline for Integer solutions of an algebraic equation
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 15, 2023 at 12:56 | comment | added | Michael Stoll | The six rational points on the original curve are exactly its singularities. | |
| Jan 15, 2023 at 12:54 | comment | added | Michael Stoll | @JoachimKönig It looks like you are right and Imessed it up somewhere. The curve is 14a5 in the LMFDB (Cremona label 14a4) and has indeed only six rational points. Sorry about the confusion! | |
| Jan 15, 2023 at 12:26 | comment | added | Joachim König | @MichaelStoll Strange, I would have thought that's exactly the curve I'm mentioning in my answer, but I get rank 0... Did I mess it up? | |
| Jan 15, 2023 at 11:53 | comment | added | Michael Stoll | Dividing by the involution swapping $a$ and $b$ gives an elliptic curve, which unfortunately has Mordell-Weil rank 1... | |
| Jan 15, 2023 at 11:50 | comment | added | Michael Stoll | It is a curve of genus 4, so will have only finitely many rational points. Very likely, $(0:\pm 1;1)$, $(\pm 1:0:1)$ and $(\pm 1:1:0)$ are the only ones (this would imply that the integer solutions listed by @YCor are indeed all). To show this may be fairly hard, however, but perhaps not impossible. | |
| Jan 15, 2023 at 10:38 | comment | added | Alm | Thus, if the curve is parametrizable, one can generate the solutions under the given inequality constraints | |
| Jan 15, 2023 at 10:31 | comment | added | YCor | There are obvious integer solutions $(n,\pm n,0)$, $(n,0,\pm n)$, $(0,n,\pm n)$. | |
| Jan 15, 2023 at 10:12 | history | answered | Alm | CC BY-SA 4.0 |