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
3 events
| when toggle format | what | by | license | comment | |
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| Dec 29, 2016 at 21:42 | vote | accept | Jose Arnaldo Bebita | ||
| Dec 29, 2016 at 21:11 | comment | added | Gerhard Paseman | Also, reading with sufficient care the section containing Lemma 12 gives more assumptions on N than are stated above (in particular all but two of the prime factors are less than 100). We do not have either direction of the general bi-conditional. Gerhard "Sounded Too Good As Truth" Paseman, 2016.12.29. | |
| Dec 29, 2016 at 20:59 | history | answered | Maxtimax | CC BY-SA 3.0 |