Timeline for Integers $b$ such that $n \nmid (b^n-1)$ for $n>1$
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 5, 2019 at 17:53 | comment | added | GH from MO | It is also worthwhile to note that $n\mid b^n-1$ implies that $n$ has a common prime divisor with $b-1$. | |
| Aug 5, 2019 at 15:36 | comment | added | Max Alekseyev | @HenriCohen: This would not help as $n=b-1$ still starts an infinite series of exceptional values in this case for any $b>2$ (notice that $(b-1)^2\mid (b^{b-1}-1)$). | |
| Aug 5, 2019 at 11:59 | comment | added | Henri Cohen | Perhaps one could modify the OP's question to $n\nmid(b^n-1)/(b-1)$ ? | |
| Aug 5, 2019 at 10:58 | vote | accept | Dominic van der Zypen | ||
| Aug 5, 2019 at 10:46 | history | answered | Max Alekseyev | CC BY-SA 4.0 |