Timeline for Average of gcd of sum of two $k$th powers
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 5 at 6:27 | history | bounty ended | Daniel Flores | ||
| Oct 4 at 19:01 | vote | accept | Daniel Flores | ||
| Oct 5 at 6:27 | |||||
| Oct 4 at 19:01 | comment | added | Daniel Flores | Yes, you're right! thank you so much for your help. | |
| Oct 4 at 16:01 | comment | added | Stanley Yao Xiao | @DanielFlores For non-square free $q$ it suffices to note that you can use Hensel's lemma to lift roots modulo $p$ to roots modulo $p^k$ uniquely; in fact they must correspond to linear factors of $ax^k + by^k$ over $\mathbb{Q}_p$. This works always provided that a root mod $p$ exists and $p$ does not divide the discriminant of $ax^k + by^k$, which in this case is just a power of $ab$ times some constant factor depending only on $k$ I think. This should allow the argument to generalize to arbitrary $q$. | |
| Oct 3 at 21:12 | comment | added | Daniel Flores | A similar argument seems to work for squarefree q, I am unsure how the this lattice argument can be extended to q which is not squarefree since the reduction to linear congruences does not seem to be as simple. | |
| Oct 1 at 3:00 | history | answered | Stanley Yao Xiao | CC BY-SA 4.0 |