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Jul 27, 2022 at 1:49 vote accept Jose Arnaldo Bebita
Aug 14, 2022 at 14:25
Jan 5, 2022 at 15:00 comment added Vlad Matei If you are willing to use a different approach here is what you can do. If we assume the contrary then $\sigma(p^k)=2z^2$. You can finish for $k>1$ by using the overflow post here to conclude mathoverflow.net/questions/285507/…
Dec 25, 2021 at 16:53 history asked Jose Arnaldo Bebita CC BY-SA 4.0