Timeline for Integral points on affine rational curves over $\mathbb{Q}$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 10, 2013 at 16:11 | comment | added | Felipe Voloch | @PeterMueller Not that I know of. If you convert the problem into a unit equation in the standard way, then the Brauer-Manin obstruction can be turned into congruence conditions. | |
| Oct 10, 2013 at 16:02 | comment | added | Peter Mueller | @Felipe Voloch: Is there a low-brow account of the integral Brauer-Manin obstruction which works practically in terms of the rational functions $f_1$ and $f_2$? | |
| Oct 9, 2013 at 8:09 | comment | added | Felipe Voloch | Only if there is no points, then you can step through the countably many obstructions till you prove that. But you can search for points in parallel with computing the obstructions and find a point if there is one. | |
| Oct 9, 2013 at 7:26 | comment | added | Y. Zhao | @FelipeVoloch: Is it possible that finite many computations are suffice to decide whether there is integral Brauer-Manin obstruction or not? | |
| Oct 8, 2013 at 13:07 | comment | added | Felipe Voloch | Conjecturally, if there is no integral Brauer-Manin obstruction then the curve has integral points. | |
| Oct 8, 2013 at 11:43 | history | edited | Y. Zhao |
edited tags
|
|
| Oct 8, 2013 at 11:22 | comment | added | Jame Ake | The Integral points constitute an Two-dimensional lattice on the complex plane, so the question can be transformed into this: Are all the rational function cann't intersect with the Intergral Points? | |
| Oct 8, 2013 at 11:12 | comment | added | Jame Ake | In the range of my knowledge, I suppose this question is very difficult to answer. Intuitively, the integral points must exist on the curve. | |
| Oct 8, 2013 at 10:33 | history | asked | Y. Zhao | CC BY-SA 3.0 |