The above is just a conjecture, there is a possibility it's wrong. The idea for it didn't come out of the blue, but I had to see some numerical evidence to see if it holds up, and none contradicted it. Moreover, even if $p | a-1$, then it still works for all $q | {a^p-1 \over p(a-1)}$.
Asked
Active
Viewed 47 times
1
-
By the way, if someone has a more concise title to describe the problem, please do because I think such a messy title will deter people from even opening the thread. – Fujoyaki Oct 08 '14 at 22:46
-
See also here – Bill Dubuque Dec 28 '23 at 18:13
1 Answers
3
If $q \mid \frac{a^p-1}{a-1}$, then in particular $q\mid a^p-1$. So the order of $a$ modulo $q$ is either $1$ or $p$. If the order of $a$ is $p$, then $p\mid q-1$, or $q \equiv 1 \pmod{p}$ and we're done.
So let's look at the case where $a \equiv 1 \pmod{q}$. Then ${a^p-1 \over a-1}=\sum_{k=0}^{p-1} a^k$. Hence
$${a^p-1 \over a-1}=\sum_{k=0}^{p-1} a^k \equiv \sum_{k=0}^{p-1} 1^k \equiv p \pmod{q}$$
and by hypothesis
$${a^p-1 \over a-1} \equiv 0 \pmod{q}$$
So, $p=q$ since they're primes, and we have $a-1 \equiv 1-1 \equiv 0 \pmod{p}$, which contradicts the hypothesis that $p \not| a-1$, which forces $ \operatorname{ord}(a)=p$.
Fujoyaki
- 2,915
Daniel Fischer
- 206,697
-
Thank you, I feel stupid now because I overestimated the problem thinking it to be complex and didn't think enough. I edited the comment to remove some of what I thought were redundancies. – Fujoyaki Oct 08 '14 at 23:33