Timeline for On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part II
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 30, 2021 at 10:39 | comment | added | Jose Arnaldo Bebita | We have decided to award the 100 bounty to your answer, @mathlove, as it helped to clarify some things for us. | |
| Oct 30, 2021 at 10:35 | history | bounty ended | Jose Arnaldo Bebita | ||
| Oct 30, 2021 at 6:54 | comment | added | Jose Arnaldo Bebita | Okay, thank you for your feedback, @mathlove! =) | |
| Oct 30, 2021 at 6:46 | comment | added | mathlove | @Jose Arnaldo Bebita Dris : No, I don't. Someone might be able to prove in another way that $p^k\lt m\implies |2^r-t|=1$. | |
| Oct 30, 2021 at 6:32 | comment | added | Jose Arnaldo Bebita | Do you think that this rules out any proof for the implication $$p^k < m \implies \left|2^r - t\right| = 1,$$ @mathlove? | |
| Oct 23, 2021 at 10:30 | history | answered | mathlove | CC BY-SA 4.0 |