modular arithmetic - Fermat's Little Theorem and Prime Moduli

I am given two distinct primes $p$ and $q$, where $$m = p*q$$ Also,

$$\begin{cases} r\equiv 1\mod p-1\\ r\equiv 1\mod q-1 \end{cases}$$

I have to show that given an integer a, show that $$a^r \equiv a \mod (m)$$

I'm not sure how to get started. I know I can tie this in to Fermat's Little Theorem and I've found something here CRT + Fermat's Little Theorem that was somewhat related to my question but I had a hard time seeing where to go. Just need a little hint to get me in the right direction! Thanks.

Answer

Hint #1. It's enough to show that $a^r\equiv a\pmod p$ and $a^r\equiv a\pmod q$; since $p$ and $q$ are relatively prime, it will follow from that that $a^r\equiv a\pmod{pq}$. By symmetry, it's enough to show that $a^r\equiv a\pmod p$.

Hint #2. What does $\ r\equiv1\pmod{p-1}\ \ $ mean? It means that $r-1$ is divisible by $p-1$; in other words, that $\ r=1+(p-1)m\ \ $ for some integer $m$.