Show that $A = a^{pq} - a^{p} - a^{q} + a$ can be divided by $pq$

Question

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

Answer 1

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 $$a^{pq}-a^p-a^q-a \equiv 0 \mod (q)$$

Answer 2

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 $pq|c$

Answer 3

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 $q$.