Let $p$ and $q$ be distinct primes and $n \in \mathbb{N}$ such that:
$q\vert \sigma(p^n)=\frac{p^{n+1}-1}{p-1}$ and $q \vert p-1$
How would one go about proving the equivalence to $q \vert n+1$?
It would be pretty useful to have $\frac{p^{n+1}-1}{p-1}\equiv n+1\,(\text{mod }q)$ in the case of $q \vert p-1$, since the result would follow immediately. Unfortunately I do not know how to prove this (or even know if it holds true). Since $$\sigma(p^n)=\frac{p^{n+1}-p+p-1}{p-1}=\frac{p^{n+1}-p}{p-1}+1=p\,\sigma(p^{n-1})+1$$ This should be equivalent to $p\,\sigma(p^{n-1})\equiv n\,(\text{mod } q)$ but that doesn't seem helpful. Any suggestions?
Thanks in advance!
$$q\mid p^n+...+p^2+p+1\implies p^n+...+p^2+p+1 \equiv 0 \pmod {q}$$
Since $$q\mid p-1\implies p \equiv 1 \pmod {q}$$
we have $$\underbrace{1+1+...+1+1+1}_{n+1} \equiv 0 \pmod {q}$$ so $q\mid n+1$.