The number $2$ has the interesting property that whenever $n>1$ is an integer, then $n \nmid (2^n-1)$. (It's a good exercise to prove this statement.)

Let's call a positive integer $b$ $2$-like if for all integers $n>1$ we have $n\nmid (b^n-1)$, and let's call it almost $2$-like if for all integers $n>1$ except finitely many we have $n\nmid (b^n-1)$.

Question. Is the collection of almost $2$-like numbers a proper superset of the collection of $2$-like numbers?


$b=2$ is the only almost 2-like number. Indeed, if $n\mid (b^n-1)$ and $p$ is a prime divisor of $(b^n-1)/n$, then $np\mid (b^{np}-1)$. That is, existence of one $n>1$ dividing $b^n-1$ implies existence of infinitely many of them. Also, for $b>2$, there exist at least one such $n$, e.g., $n=b-1$.

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    $\begingroup$ Perhaps one could modify the OP's question to $n\nmid(b^n-1)/(b-1)$ ? $\endgroup$ – Henri Cohen Aug 5 '19 at 11:59
  • $\begingroup$ @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 $(b-1)^2\mid (b^{b-1}-1)$). $\endgroup$ – Max Alekseyev Aug 5 '19 at 15:36
  • $\begingroup$ It is also worthwhile to note that $n\mid b^n-1$ implies that $n$ has a common prime divisor with $b-1$. $\endgroup$ – GH from MO Aug 5 '19 at 17:53

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