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Let $a\in \Bbb Z$, $b\in \Bbb Z$ such that $p \nmid b$, and $p$ a prime where $p \gt 2$.

If for all $x \in \Bbb Z$ such that $p \nmid x$ and $\operatorname{ord}_p(x) \ne p-1$, $p$ satisfies $\operatorname{ord}_p(a+bx) = p-1$, prove that $p$ is in the form:

$p = 2^{2^n} + 1$ for some $n$ non-negative.

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    Nothing is clear until the person asking the question writes it clearly.2012-10-29

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Remark that there are $p-1 - \phi(p-1)$ elements of order not $p-1$ modulo $p$ because there are $\phi(p-1)$ elements of order $p-1$. We can pick $p-1 - \phi(p-1)$ values of $b$ which satisfy the conditions and thus it follows there are at least $p-1 - \phi(p-1)$ distinct elements of order $p-1$. But then: $p-1 - \phi(p-1) \le \phi(p-1)$ However, this implies $p-1 \le 2\phi(p-1)$. But as $2|(p-1)$ it follows $\phi(p-1) \le \frac{p-1}{2}$ on inequality iff $p-1$ is a power of two. It immediately follows $p = 2^n + 1$ for some $n$ and thus we are done.

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    Sorry but I cannot see why by picking $p-1 - \phi(p-1)$ values of $b$ that satisfy the conditions we have at least $p-1 - \phi(p-1)$ distinct elements of order $p-1$?2012-10-31