Timeline for Understanding a character calculation for arithmetic progressions of squares in finite fields
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7 at 15:37 | comment | added | orangeskid | maybe worth mentioning that Szemerédi's theorem would take care right away of the case of prime $q$... probably the character method gives much better estimates | |
| Sep 6 at 12:33 | comment | added | Jyrki Lahtonen | There is also the trivial observation that all the elements of $\Bbb{F}_q$ are squares of elements of $\Bbb{F}_{q^2}$. So if $q$ is a square, the question is trivial as long as $p>7$. Surely, that is a non-interesting case, and doesn't cover as much ground as you would like to :-) | |
| Sep 5 at 3:33 | vote | accept | Chris Wolird | ||
| Sep 4 at 15:35 | history | became hot network question | |||
| Sep 4 at 8:46 | answer | added | Jyrki Lahtonen | timeline score: 6 | |
| Sep 4 at 8:08 | comment | added | Jyrki Lahtonen | Ok @Gerry, didn't look carefully. My comment was just a knee-jerk reaction saying that the claim in boldface is trivially wrong for all the fields of characteristic $2,3,5$ or $7$, so size alone won't do. Wonder why she even mentions characters other than the Legendre character? | |
| Sep 4 at 7:58 | comment | added | Gerry Myerson | @Jyrki, $q$ disappears from the calculation early on, thereafter only $p$ appears, and the end result is given as $p>769^2$, so I think that takes care of your comment. | |
| Sep 4 at 7:51 | comment | added | Jyrki Lahtonen | Presumably you have a constraint on the characteristic :-) After all, in characteristic $p$ every arithmetic progression loops back to itself after $p$ steps. | |
| Sep 4 at 5:38 | history | asked | Chris Wolird | CC BY-SA 4.0 |