Show that $A = a^{pq} - a^{p} - a^{q} + a$ can be divided by $pq$ 
Let $p$ and $q$ be two different prime numbers, and $a$ a positive integer. Show that $$A = a^{pq} - a^{p} - a^{q} + a$$ can be divided by $pq$

I started by rewriting $A$ as $((a^{p})^{q} - a^{p}) - (a^{q} - a)$ but I'm not sure how to go from here
 A: It suffices to show that both $p$ and $q$ divide$$a^{pq}-a^p-a^q+a $$
Note that $$a^p \equiv a \mod (p)$$ therefore $$a^{pq} \equiv a^q \mod (p)$$Thus $$a^{pq}-a^p-a^q-a \equiv a^q-a^p-a^q+a \equiv 0 \mod (p)$$
Similarly we can show that  $$a^{pq}-a^p-a^q-a \equiv 0 \mod (q)$$
A: Use little Fermat, if $p$ is a prime number and $a$ a intenger then 
$p|(a^{p} - a)$
So, $p|((a^{q})^{p} - a^{q}$), and $p|(a^{p} - a)$
Then $p$ divides the difference, $p|((a^{q})^{p} - a^{q} - a^{p} + a)$
Doing the same for $q$, we have $q|((a^{p})^{q} - a^{p} - a^{q} + a)$
Then as $(p,q) = 1$, we have $pq|(a^{pq} - a^{p} - a^{q} + a)$
Well, $(p,q) = 1$, then there exists $s,t \in \mathbb{Z}$ such that $1 = ps + qt$
So if $p|c$ and $q|c$, then $ c = pk = qr$, with $k,r \in \mathbb{Z}$ then $c = c.1 = c(ps + qt) = cps + cqt = qrps + pkqt) = qp (rs + kt)$ 
Then $pq|c$
A: I suppose you mean $pq$ divides $\, a^{pq}-a^p-a^q+a$, i.e.
$$ a^{pq}-a^p-a^q+a\equiv 0\pmod{pq}.$$
Hint:
By the Chinese remainder theorem, it amounts to proving it is congruent to $0$ $\bmod p$ and $\bmod q$.
As $p$ and $q$ are prime, remember Little Fermat can be formulated as
$$\forall a,\: a^p\equiv a\pmod p$$
and similarly for $q$.
