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Let $p$ be a prime and $N$ be an integer.

Then $p$ is called semi-primitive modulo $N$ if there exists a positive integer $j$ such that $p^j \equiv -1 \pmod{N}$.

Now let $m$ be a positive integer and let $N = \operatorname{rad}\left(\dfrac{p^m-1}{p-1}\right)$, where $\operatorname{rad}(c)$ denotes the product of all the (distinct) prime divisors of an integer $c$.

Question: What are the solutions $(p,m)$ to the equation: $$ p^j \equiv -1 \pmod{\operatorname{rad}\left(\frac{p^m-1}{p-1}\right)}? $$ That is, for which $(p,m)$ is $p$ semi-primitive modulo $\operatorname{rad}\left(\frac{p^m-1}{p-1}\right)$?

One particular example is the case when $p$ is a Mersenne prime and $m = 4$.

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The proof is a bit tedious but it turns out that this only works when $m = 2$ or when $p$ is a Mersenne prime and $m = 4$...

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  • $\begingroup$ But could you at least show us a sketch of the proof? $\endgroup$ – punctured dusk Oct 27 '14 at 9:14

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