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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!

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  • $\begingroup$ Have you tried checking this for some range of values of $p$ and $n$ and $s$? $\endgroup$
    – Gerry Myerson
    Commented Aug 10, 2016 at 22:45
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    $\begingroup$ Obviously not. If it were true, then we would have $p|j^n+1$ for some $j\le n$, which is quite difficult for large $p$. $\endgroup$
    – fedja
    Commented Aug 10, 2016 at 23:32
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    $\begingroup$ Replacing $n$ by $(p-1)/n$ in the list of congruences would make an interesting problem. The problem in here is that $n$ can be too small compared to $p$. $\endgroup$
    – Sungjin Kim
    Commented Aug 11, 2016 at 3:30
  • $\begingroup$ Better to post a new question than to completely change an old one, especially after you have already accepted an answer. But my question remains – have you tried checking the new question for some range of values? Evidently, you didn't do much of this for the original question. $\endgroup$
    – Gerry Myerson
    Commented Aug 11, 2016 at 22:54
  • $\begingroup$ Ok, I'll post a new question. The first question was not really what I needed, it would just easily (but wrong) imply what I need. Well, to discover $s$ is an annoying handwork (I cannot programming), so I confess I didn't check many cases ($p=11,n=5$ only). The second question also implies what I need, but not too easy. I checked some cases by hand ($(n,p,k,s) = (5,11,6,4), (5,31,\{6,11,16,21,26\},16), (7,29,\{8,15,22\},\{7,16\})$) and it really does not seem to be false. An obvious observation is that $2 \le s \le \min\{k-1,q-k\}$. $\endgroup$
    – Sávio
    Commented Aug 12, 2016 at 1:33

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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.

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