Theorem

Let $b\in\mathbb{Z},n\in\mathbb{N}$ 1. $d|n\implies(b^{d}-1)|(b^{n}-1)$ 2. \begin{array} &&(b,m)=1 \\ &b^{a}\equiv 1\pmod{m} \\ &b^{c}\equiv 1\pmod{m} \end{array}\implies b^{d}\equiv 1\pmod{m}where $d=(a,c)$.