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Let $\def\ge{\geqslant} \def\le{\leqslant} b\in\Bbb N^{\ge2}$ be a Natural number and $p$ be a prime such that $p\nmid b-1$. For how many primes $q$ does $$p\mid b^q-1$$ hold?

I.e. for such $b$ and $p$ let $Q(b,p)=\{q\in\Bbb P \;/\; p\mid b^q-1\}$.

Is $\#Q(b,p)$

  1. Always finite?

  2. Always infinite?

  3. Always non-empty?

  4. It depends on $b$ and $p$?

What I have:

Let $q,r\in \Bbb P$ and $r > q$, then \begin{align} p\mid b^q-1\text{ and }p\mid b^r-1 &\quad\Rightarrow\quad p\mid b^q-b^r = b^r(b^{q-r}-1)\\ &\quad\Rightarrow\quad p\mid b^{q-r}-1 \end{align}

Then due to the Euclidean algorithm it follows that $p\mid b-1$ which was excluded. Does this mean that there is always $\#Q(b,p) \le 1$ or do I have a thinko? My intuition says that even $\#Q(b,p) = 1$, i.e. for every prime $p$ we find a sequence of ones (interpreted as number in base $b$) that is divisible by $p$. But I have no idea how to show that. Something with finite fields or so...

emacs drives me nuts
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1 Answers1

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Fermat tells us that $p\mid b^{p-1}-1$. Therefore, if $p\mid b^q-1$, then as you have noted, $p\mid b^d-1$, where $d=\gcd(q,p-1)$. Since $q$ is prime, $d=1$ or $d=q$. Since $p\nmid b-1$, we cannot have $d=1$. Therefore, $d=q$, that is, $q$ divides $p-1$. Thus, $Q(b,p)$ is always finite.

In fact, $\#Q(b,p) \le 1$ because $Q(b,p)$ is empty when the order of $b$ mod $p$ is not a prime and a singleton when $b$ mod $p$ is a prime, which must divide $p-1$.

lhf
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