# If $p$ and $q$ are distinct primes then for any integer $a, pq\mid (a^{pq}-a^p-a^q+a)$

Prove that If $p$ and $q$ are distinct primes then for any integer $a$, $$pq\mid (a^{pq}-a^p-a^q+a)$$

I start by using Fermate little theorem and I am looking for $\bmod p$ and looking for $\bmod q$ is that correct for mod p $$p\mid ((a^{q}-a)^{p}-(a^q-a_)$$ then $$p\mid (a^{pq}-a^p-a^q+a)$$

## 1 Answer

By the use of Fermat's little theorem: $$a^{p} \equiv a \mod{p}$$ $$(a^p)^{q} \equiv a^{pq} \equiv a^{q} \mod{p}$$ $$a^{pq} \equiv 0+a^{q}\equiv (a^p-a)+a^{q} \equiv a^p+a^{q}-a \mod{p}$$

Likewise, we have the same for $q$: $$a^{pq} \equiv 0+a^{p}\equiv (a^q-a)+a^{p} \equiv a^p+a^{q}-a \mod{q}$$

Therefore, $$p \mid a^{pq}-a^{p}-a^{q}+a$$ $$q \mid a^{pq}-a^{p}-a^{q}+a$$

Now, since $[p,q]=pq$, by the definition of the least common multiple, we see that:

$$pq \mid a^{pq}-a^{p}-a^{q}+a$$

Q.E.D.

Remark: Note that if $a \mid c$ and $b \mid c$, then $c$ is a common multiple of $a$ and $b$ and therefore, by definition, the least common multiple of $a,b$, i.e. $[a,b]$ divides $c$. In other words, $[a,b] \mid c$