Timeline for If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$ [closed]
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Jun 17 at 10:12 | history | closed |
Dave Benson Daniele Tampieri Andy Putman Yemon Choi Mikhail Katz |
Not suitable for this site | |
| Jun 15 at 14:27 | review | Close votes | |||
| Jun 17 at 10:12 | |||||
| S Jun 15 at 13:33 | history | bounty ended | Jose Arnaldo Bebita | ||
| S Jun 15 at 13:33 | history | notice removed | Jose Arnaldo Bebita | ||
| Jun 15 at 13:33 | vote | accept | Jose Arnaldo Bebita | ||
| Jun 15 at 11:23 | comment | added | Alex Ravsky | Done. I am also happy to make my humble contribution to solve the odd perfect number problem. :-D | |
| Jun 15 at 11:22 | answer | added | Alex Ravsky | timeline score: 0 | |
| Jun 15 at 11:12 | comment | added | Jose Arnaldo Bebita | Ohh, yes! But then that does not require proving that $m^2 - p^k \neq \square$! Ohh my goodness! If you could just write that out as an answer, I will be more than happy to accept it (and award the bounty to you), @AlexRavsky =) | |
| Jun 15 at 8:43 | comment | added | Alex Ravsky | If $p^k<m^2$ always holds then $p^k<2(m-1)m$ and so we can put $a=m-1$, which provides an affirmative answer to your question, right? | |
| Jun 15 at 8:34 | comment | added | Jose Arnaldo Bebita | Thank you for your comment, @AlexRavsky! I can see that your claimed inequality $p^k < 2(m - 1)m$ is equivalent to $m^2 - p^k > 0 > 2m - m^2$, which is always true. (We do know, however, that $p^k < m^2$ always holds. So does that take care of your concern about the failure of my "finishing arguments"?) | |
| Jun 15 at 7:09 | comment | added | Alex Ravsky | I do not understand why you devoted so many efforts to show that $m^2-p^k$ is not a square. Indeed, if $m>2$ and $m^2-p^k\ge 0$ then $p^k<2(m-1)m$ and so we can put $a=m-1$. On the other hand, if $m^2-p^k<0$ then it cannot be between two squares, so your finishing arguments fail. | |
| Jun 11 at 1:42 | comment | added | Jose Arnaldo Bebita | I appreciate your comment, @ToddTrimble! | |
| Jun 10 at 18:52 | comment | added | Todd Trimble | I have trouble following the reasoning pretty much by the first display line. However, the answer to the question "Does the following statement necessarily hold? 'Since $m^2 - p^k$ is not a square, then it is between two (consecutive) squares.' " is yes, and the proof is very easy. Since the natural numbers are well-ordered, there is a least square $M^2$ greater than this $N := m^2 - p^k$. Then $(M-1)^2 \leq N$; otherwise, $M^2$ wouldn't have been the least square. $(M-1)^2 < N$ since $N$ is assumed not to be a square. So $(M-1)^2 < N < M^2$. | |
| Jun 10 at 15:55 | history | edited | Jose Arnaldo Bebita | CC BY-SA 4.0 |
trimmed down the original question, deleted the request to check completeness of proof to conform to MO policy
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| Jun 10 at 15:54 | comment | added | Andy Putman | It's not a matter of how it is worded: as far as I can tell, this question is fundamentally a request for someone to either find the flaw in your proof or fill in some missing details, and no amount of rephrasing will make that on-topic. | |
| Jun 10 at 15:40 | comment | added | Jose Arnaldo Bebita | @AndyPutman: In that case, please do allow me to edit my question, to conform to MO policy, so that it does not look like a "request to check completeness of proofs". | |
| Jun 10 at 15:35 | comment | added | Andy Putman | This question is off-topic. Requests to check completeness of proofs are not allowed. The bounty prevents me from voting to close at the moment, but this question absolutely should be closed. | |
| S Jun 10 at 15:16 | history | bounty started | Jose Arnaldo Bebita | ||
| S Jun 10 at 15:16 | history | notice added | Jose Arnaldo Bebita | Draw attention | |
| Jun 10 at 13:45 | history | undeleted | Jose Arnaldo Bebita | ||
| Apr 27 at 10:31 | history | deleted | Jose Arnaldo Bebita | via Vote | |
| Apr 27 at 9:36 | answer | added | Jose Arnaldo Bebita | timeline score: -3 | |
| Apr 27 at 9:30 | review | Close votes | |||
| Apr 27 at 10:36 | |||||
| Apr 27 at 7:59 | history | edited | Jose Arnaldo Bebita | CC BY-SA 4.0 |
added context
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| Apr 27 at 7:30 | history | asked | Jose Arnaldo Bebita | CC BY-SA 4.0 |