Consider the multiplicative cyclic group $\mathbb{Z}_p=\{1,2,\cdots,p-1\}$, where $p$ is prime. Let $a,a+1\in\mathbb{Z}_p$. Is there any relation between the orders of $a$ and $a+1$? I guess that, usually, there is none. But perhaps I am wrong; if there is some, I would like to know about it.
What really interests me is a sort of a "negative'' result. I explain:
I usually work in finite fields $\mathbb{F}_{p^n}$ (or $GF(p^n)$, if you prefer this notation; these are fields with $p^n$ elements, with $p$ an odd prime) and deal with subgroups of $\mathbb{F}_{p^n}\backslash\{0\}=\mathbb{F}^\ast_{p^n}$.
Let $\omega\in\mathbb{F}_{p^n}^*$ be an element of order $(p^{2s}-1)$ (with $0<s<n$) and consider elements in $\mathbb{F}_{p^n}$ of the form $s_j=\omega^j(1+\omega^{(p^s-1)j})$, where $j$ goes from $0$ to $p^s$.
I would like to show that there is some $j_0$, in this range, such that the order of $(1+\omega^{(p^s-1)j_0})$ does not divide $p^{2s}-1$, the order of $\omega$.
This would imply that $(\omega^{j_0})^{p^{2s}-1}=1$ but $(s_{j_0})^{p^{2s}-1}\neq 1$.
Thanks for any information.