let p and q be distinct odd prime numbers with p-1 divides q-1. If a is with in integers with (a,pq)=1 prove that a^q-1=1mod pq
let $\displaystyle q-1=d(p-1).$ since $\displaystyle \gcd(a,q)=1,$ we have $\displaystyle a^{q-1} \equiv 1 \mod q.$ also since $\displaystyle \gcd(a,p)=1,$ we have: $\displaystyle a^{p-1} \equiv 1 \mod p,$ which gives us: $\displaystyle a^{q-1}=(a^{p-1})^d \equiv 1 \mod p.$
so we showed that both $\displaystyle p$ and $\displaystyle q$ divide $\displaystyle a^{q-1}-1,$ and thus $\displaystyle pq$ must also divide $\displaystyle a^{q-1} - 1. \ \ \ \square$