I am working on a finite field theory/number theory question, and my problem has been simplified down, via a chain of "if and only if"s, to the above question. The precise problem and its suppositions are as follows:
Suppose $n\geq2$, $k>1$ such that $k|n$, and $p$ is an odd prime such that $p^{k} \equiv 1 \pmod{2n}$. Then $n|p^{k-1} + \cdots + p + 1$.
The supposition is a bit weird, and some of them may be unnecessary for the problem. I have checked a few numbers, and it seems to be true, but I am unsure of how to approach proving it. I have looked at using the Euler Totient function, but I wasn't able to get far. Is there some kind of sleek number theory/group theory argument or trick that I am forgetting?
Thank you so much for the help!
Edit: Thank you so much for your helpful counterexamples. They helped prove my original conjecture false. However, I now wonder what conditions on $n$, $k$, and $p$ would be necessary to satisfy $n|p^{k-1}+\cdots+p+1$. If you have any extra tips, I would love to hear them!
As an extension to the answer Dietrich provided. It's possible to come with many, many counterexamples. Because $p-1$ is always even, take $p,n$ such that $p-1=2n$, $n$-not prime and $k<n$. With this, we have $$p-1=2n \Rightarrow \\ p\equiv 1 \pmod{2n} \Rightarrow \\ p\equiv 1 \pmod{n}$$ Then $$1\equiv 1 \pmod{n}$$ $$p\equiv 1 \pmod{n}$$ $$p^2\equiv 1 \pmod{n}$$ $$...$$ $$p^{k-1}\equiv 1 \pmod{n}$$ or $$p^{k-1}+...+p^2+p+1 \equiv k \pmod{n} \tag{1}$$ Because $k<n$, then $k \not\equiv 0 \pmod{n}$ always and from $(1)$ $$n \nmid p^{k-1}+...+p^2+p+1$$