Timeline for Can a multi-perfect number be a perfect square?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2020 at 10:38 | history | edited | CommunityBot | Commonmark migration | |
| Sep 25, 2016 at 2:16 | history | edited | Alexis Olson | CC BY-SA 3.0 | Final comment. |
| Sep 25, 2016 at 2:05 | history | edited | Alexis Olson | CC BY-SA 3.0 | Minor formatting |
| Sep 25, 2016 at 1:43 | comment | added | Alexis Olson | @user11235813 I see. I ignored that case as OP claimed to have already covered it in the posted question. | |
| Sep 25, 2016 at 1:39 | comment | added | Jose Arnaldo Bebita Dris | @AlexisOlson, what I'm saying is that your answer does not cover the possibility of odd multiperfect / $k$-perfect numbers with $k$ even. | |
| Sep 25, 2016 at 1:28 | comment | added | Alexis Olson | @user11235813 I'm not sure I understand your comment as I've not claimed anything except for when $sn$ is odd. | |
| Sep 25, 2016 at 0:57 | comment | added | Jose Arnaldo Bebita Dris | @AlexisOlson, there is an inherent flaw in your reasoning. If $sn=k$, where $k$ corresponds to "the" $k$ in odd $k$-perfect/multiperfect number, $k$ could be even. Hence, $sn$ is not always odd, as you claim. It is the number $N$ satisfying $\sigma(N)=kN$ that is supposed to be odd, per the OP's original question. In short, the argument so presented seems only to cover the case when $k=sn \equiv 1 \pmod 2$. | |
| Sep 25, 2016 at 0:51 | history | bounty awarded | CommunityBot | ||
| Sep 24, 2016 at 3:07 | vote | accept | CommunityBot | ||
| Sep 24, 2016 at 3:07 | comment | added | user141854 | Perfect. Thank you! I have to wait 17 more hours before it will let me award the bounty. I'll be sure to do so tomorrow. | |
| Sep 24, 2016 at 3:06 | comment | added | Alexis Olson | I believe so, for odd $n$ anyway, since $s=3$ was not used at all as far as I can see. | |
| Sep 24, 2016 at 3:05 | comment | added | user141854 | I see. Does this also generalize to any $s$-fold (or ($s-1$)-fold in Kanold's case) where $s$ is odd? That is, if $s$ is odd then all $Q$ in $\sigma(Q)=sQ$ must be perfect squares? | |
| Sep 24, 2016 at 3:00 | comment | added | Alexis Olson | They have a rather odd definition. In the opening paragraph of the paper, they define $(s-1)$-fold perfect to mean $\sigma(n) = sn$ rather than what you assumed. | |
| Sep 24, 2016 at 2:51 | comment | added | user141854 | How is the relationship in the proof established? The author states $n$ is an $(s-1)$-fold perfect number, which implies $\sigma(n)=\prod_{i=1}^k\sigma(p_i^{a_i})=n(s-1)\neq ns$... | |
| Sep 24, 2016 at 2:39 | comment | added | Alexis Olson | @fruitegg I've got it. It was rather simpler than I anticipated. | |
| Sep 24, 2016 at 2:34 | history | edited | Alexis Olson | CC BY-SA 3.0 | Minor edits |
| Sep 24, 2016 at 1:55 | comment | added | user141854 | Interesting articles for sure! I'm curious to see your extraction from the German manuscript. | |
| Sep 24, 2016 at 1:02 | history | answered | Alexis Olson | CC BY-SA 3.0 |