Timeline for On the Diophantine equation $m^2 - p^k = 2^r t$, where $r \geq 2$ and $\gcd(2,t)=1$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jun 29, 2021 at 9:50 | history | bounty ended | Jose Arnaldo Bebita | ||
| S Jun 29, 2021 at 9:50 | history | notice removed | Jose Arnaldo Bebita | ||
| Jun 28, 2021 at 10:45 | vote | accept | Jose Arnaldo Bebita | ||
| Jun 28, 2021 at 10:00 | answer | added | mathlove | timeline score: 1 | |
| S Jun 28, 2021 at 9:03 | history | bounty started | Jose Arnaldo Bebita | ||
| S Jun 28, 2021 at 9:03 | history | notice added | Jose Arnaldo Bebita | Draw attention | |
| Jun 25, 2021 at 13:40 | history | edited | Jose Arnaldo Bebita | CC BY-SA 4.0 |
added/corrected context
|
| Jun 25, 2021 at 4:25 | history | edited | Jose Arnaldo Bebita | CC BY-SA 4.0 |
added context, in response to Mike Bennett's comment
|
| Jun 25, 2021 at 4:24 | comment | added | Jose Arnaldo Bebita | Okay doing so now. | |
| Jun 25, 2021 at 4:23 | comment | added | Mike Bennett | Then state that in the question. | |
| Jun 25, 2021 at 3:46 | comment | added | Jose Arnaldo Bebita | @MikeBennett: For the particular problem that we are considering (the problem of determining whether $m < p^k$ holds in general for an odd perfect number $p^k m^2$ with special prime $p$), we know that $p \equiv k \equiv 1 \pmod 4$ holds. So $k = 2j$ is not possible, and your argument breaks down. | |
| Jun 25, 2021 at 3:44 | comment | added | Jose Arnaldo Bebita | @MaxAlekseyev: When $p^k m^2$ is an odd perfect number with special prime $p$ satisfying $\sigma(m^2)/p^k$ is a square, then $\sigma(m^2) \equiv 1 \pmod 4$ and $k = 1$. But $\sigma(m^2) \equiv 1 \pmod 4$ is equivalent to $p \equiv k \pmod 8$. Therefore, if $\sigma(m^2)/p^k$ is a square, then $m^2 - p^k \equiv 0 \pmod 8$. Which is the reason why we put $2^r t$ for $m^2 - p^k$ there. But we do get your point. | |
| Jun 24, 2021 at 22:33 | comment | added | Mike Bennett | If you just take $k=2j$, $m=2^r+p^j$, $t=m+p^j$ and assume that $2^r > p^{2j}$, then this is always satisfied. | |
| Jun 24, 2021 at 14:35 | comment | added | Max Alekseyev | Why not replace $2^rt$ with $4s$ having just a single variable? | |
| Jun 24, 2021 at 10:44 | history | asked | Jose Arnaldo Bebita | CC BY-SA 4.0 |