If $p,q$ are primes $(pq,a)=1$ and $p-1|q-1$ then prove $a^{q-1}\equiv 1\pmod{pq}$

Question

$p,q$ primes, $(pq,a)=1$ and $p-1|q-1$ then prove $a^{q-1}\equiv 1\pmod{pq}$

Got this in an exam today and had no idea how to do it.

Answer 1

$\color{#c00}{q\!-\!1 = (p\!-\!1)k}\ $ so $\bmod p\!:\ a^{\large\color{#c00}{q-1}} \equiv (a^{\large\color{#c00}{p-1}})^{\large\color{#c00}k}\equiv 1^k\equiv 1,\,$ & $\bmod q\!:\ a^{\large q-1}\equiv 1,\,$ by Fermat & $\,p,q\nmid a.\,$ Thus $\, p,q\mid a^{q-1}-1\,\Rightarrow\, {\rm lcm}(p,q)\mid a^{q-1}-1\,$ & $\,{\rm lcm}(p,q)=pq\,\iff \gcd(p,q) = 1$

Remark $ $ More generally see Carmichael's generalization of little Fermat / Euler.