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Aug 12, 2016 at 7:17 history closed Franz Lemmermeyer
Wolfgang
Michael Zieve
Jan-Christoph Schlage-Puchta
Felipe Voloch
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Aug 12, 2016 at 1:48 comment added Sávio mathoverflow.net/questions/247338/…
Aug 12, 2016 at 1:33 comment added Sávio 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 $2 \le s \le \min\{k-1,q-k\}$.
Aug 11, 2016 at 22:54 comment added Gerry Myerson 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.
Aug 11, 2016 at 20:46 history edited Sávio CC BY-SA 3.0
Adding a new related question
Aug 11, 2016 at 6:28 review Close votes
Aug 12, 2016 at 7:24
Aug 11, 2016 at 3:30 comment added Sungjin Kim 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 $p$.
Aug 11, 2016 at 0:58 vote accept Sávio
Aug 10, 2016 at 23:34 answer added Max Alekseyev timeline score: 2
Aug 10, 2016 at 23:32 comment added fedja 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 $p$.
Aug 10, 2016 at 22:45 comment added Gerry Myerson Have you tried checking this for some range of values of $p$ and $n$ and $s$?
Aug 10, 2016 at 22:01 history asked Sávio CC BY-SA 3.0