Timeline for Proving $k = 1 \implies q = 5$, if $q^k n^2$ is an odd perfect number with Euler prime $q$
Current License: CC BY-SA 3.0
11 events
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Sep 28, 2021 at 6:04 | comment | added | Jose Arnaldo Bebita | @PaceNielsen: I think I may have proved that $\sigma(q^k)/2$ is not squarefree, if $q^k n^2$ is an odd perfect number with special prime $q$. (I invite you to check out this answer to a closely related MO question.) | |
Sep 18, 2017 at 12:07 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
corrected a logical mistake
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Dec 29, 2016 at 22:48 | comment | added | Pace Nielsen | I think that's a wide-open problem. | |
Dec 29, 2016 at 21:49 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
minor update, question was flawed to begin with
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Dec 29, 2016 at 21:42 | vote | accept | Jose Arnaldo Bebita | ||
Dec 29, 2016 at 21:41 | comment | added | Jose Arnaldo Bebita | @PaceNielsen, I dimly recall that Hagis (there may be others, of course) proved some results similar to those in your comment. Notwithstanding, do you happen to know if $q = 5 \implies k = 1$ holds unconditionally? | |
Dec 29, 2016 at 21:40 | history | edited | Jose Arnaldo Bebita | CC BY-SA 3.0 |
edited in response to a comment from Pace Nielsen
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Dec 29, 2016 at 21:37 | comment | added | Jose Arnaldo Bebita | Thanks for pointing that out, Pace! I have noted your observation in the question. | |
Dec 29, 2016 at 21:10 | comment | added | Pace Nielsen | Jose, you are misinterpreting the result from Iannucci. His number $N$ is restricted to be an odd perfect number which additionally has all but at most two of its prime divisors less than 100. He goes on to prove that there are no such numbers, but along the way, in the lemma you cite, he shows that such numbers have restricted form. | |
Dec 29, 2016 at 20:59 | answer | added | Maxtimax | timeline score: 3 | |
Dec 29, 2016 at 20:43 | history | asked | Jose Arnaldo Bebita | CC BY-SA 3.0 |