Let $a\in \mathbb{Z_m}$, $a\ne 1$ and $n \in \mathbb{N}$. Also $a \equiv_p 1$, if $p$ | $m$ where $p$ is prime and $a \equiv_4 1$ if $4$ | $m$.
Prove that the smallest number $n$, such that $\frac{a^n-1}{a-1}$ $\equiv_m$ $0$, is $m$.
Ok, my approach was to write the equality as $a^{n-1} + a^{n-2} +...+a+1$ $\equiv_m$ $0$. This irreductible over $\mathbb{Z_m}$, but what is the next step?
First, denote $\nu_p(b)=a$ if $p^a \mid b$ but $p^{a+1} \nmid b$. It suffices to prove that for any prime divisor $p$ of $m$ so that $\nu_p(m)=b$ then $m \mid \dfrac{a^n-1}{a-1}$ if and only if $p^b \mid n$.
Now, according to Lifting the Exponent Lemma, since $p \mid a-1$ so $$\nu_p(a^n-1)=\nu_p(a-1)+\nu_p(n).$$ Hence, since $p^b \mid \dfrac{a^n-1}{a-1}$ so $p^b \mid n$. Since this is true for all prime divisor of $m$, therefore $m \mid n$. Thus, $n=m$ is the smallest value so that $m \mid \dfrac{a^n-1}{a-1}$.