Timeline for On odd perfect numbers and a GCD
Current License: CC BY-SA 4.0
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| Sep 14, 2021 at 13:21 | comment | added | Jose Arnaldo Bebita | In fact, we have the implication $$G = \gcd(\sigma(q^k),\sigma(n^2)) = 1 \implies k = 1,$$ per this answer to a closely related MSE question. | |
| Mar 2, 2019 at 13:24 | history | answered | Jose Arnaldo Bebita | CC BY-SA 4.0 |