Let $p\ge 11$ be a prime number, $n \ge 5$ be an odd positive divisor of $p-1$ and $s \in \mathbb Z\_p$ such that $ord\_p(s) = n$.


Is it true that the geometric progression $\{s^k\}\_{k \in \mathbb Z\_n}$ intersects some of the classes $\overline{p-n}, \;\; \overline{p-n+1}, \;\; \dots, \;\; \overline{p-1} \pmod p$?


Remark: since $n$ is odd, the class $\overline{p-1}$ is actually never achieved, so I could have written until the class $\overline{p-2}$.


**Solved until here**


**EDIT:** From now on, under the same assumptions for $p,n,s$, we define $k$ as a positive integer with $k \equiv 1 \pmod n$ and $p > k > n$. Is it possible that the sets $A = \{1, 2, 3, \dots, k−1\}$ and $B = \{k, k+1, k+2, \dots, p−1\}$ of classes modulo $p$ satisfy $sA = A$ and $sB = B$, simultaneously? I mean, are $A$ and $B$ invariants by multiplication by $s$ for some $n,p,k,s$?


Thanks!



Geometric progression modulo p
I don't see a reason why this should hold.


Counterexample: $p=31$, $n=5$, $s=16$.
The powers of $s$ give 16, 8, 4, 2, 1 modulo 31.


