# Showing that $\dfrac{q^n -1}{(q-1)\gcd(n,q-1)}$ is an integer for $q = p^k$, $p$ prime.

Does anyone know how to prove that $\dfrac{q^n -1}{(q-1)\gcd(n,q-1)}$ is an integer for $q = p^k$, $p$ prime using only basic algebra?

It is possible to show this by showing that there is a subgroup $C$ of $\mathbb{F}_{q^n}^*$ such that $\#C = (q-1)\gcd(n,q-1)$, but is it possible to find a simpler proof?

In modulo $\gcd(n,q-1)$ we have:
• is the condition $q=p^k$ necessary ? Aug 31, 2016 at 15:14
• $\gcd(q,a)=1$ with $a=\gcd(n,q-1)$? Aug 31, 2016 at 16:01
• @Maman yes. Because of its definition, in modulo $a$ we have $q-1\equiv 0$ which implies $q \equiv 1$. Hence $\gcd(q,a)=1$. Aug 31, 2016 at 16:08
• in fact, it seems also to me that it is superfluous: $gcd(q,q-1)=1$ for all $q$ Aug 31, 2016 at 16:36