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Feb 12, 2023 at 13:33 comment added CHUAKS Some of the observations follows from an identity: if an odd prime $p$ divides $a^n-1$, $\nu_p(a^n-1)=\nu_p(n)+\nu_p(a^{p-1}-1)$, see arxiv.org/abs/2112.04173
S Nov 24, 2013 at 11:22 history suggested Gottfried Helms CC BY-SA 3.0
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Nov 24, 2013 at 11:04 review Suggested edits
S Nov 24, 2013 at 11:22
Nov 24, 2013 at 8:54 history edited Aaron Meyerowitz CC BY-SA 3.0
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Nov 23, 2013 at 6:52 history answered Aaron Meyerowitz CC BY-SA 3.0