# Integers $b$ such that $n \nmid (b^n-1)$ for $n>1$

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