A few proofs of the ubiquitous $\small \bf CCRT = \bf{Constant\!\!-\!\!case \ CRT}\,$ [Chinese Remainder Theorem].
Theorem (CCRT) $\rm\ \ \ \begin{align}&\rm x\equiv a\!\!\pmod{\!p}\\ &\rm x\equiv a\!\!\pmod{\!q}\end{align}\!\iff x\equiv a\pmod{\!pq}\,$ if $\rm\,p,q\,$ are coprime,
$\rm\qquad said\ equivalently\!:\quad p,q\mid x-a\iff pq\mid x-a,\ $ if $\ \rm p,q\,$ are coprime
$\rm\qquad said\ equivalently\!:\qquad\ \ \ lcm(p,q)\, =\, pq,\ \ if\ \ p,q\,$ are coprime
Proof $\ $ For variety we give a few different proofs.
$\rm(1)\ \ \ x \equiv a\pmod {\!pq}\:$ is clearly a solution, and the solution is $\color{#C00}{unique}$ mod $\rm\,pq\,$ by CRT. $ $ [Note: rotely applying the common CRT formula also yields this obvious solution].
$\rm(2)\ \ \ p,q\:|\:x\!-\!a\iff lcm(p,q)\:|\:x\!-\!a\:$ by the Universal Property of $\rm lcm$.
$\qquad\! $ Further $\rm\:(p,q)=1^{\phantom{|^|}}\!\!\!\!\iff lcm(p,q) = pq,\,$ by this answer.
$(3)\ \, $ By Euclid's Lemma: $\rm\:(p,q)=1,\ p\:|\:qn =\:x\!-\!a\:\Rightarrow\:p\:|\:n\:\Rightarrow\:pq\:|\:nq = x\!-\!a.$
$\rm(4)\ \, $ The list of prime factors of $\rm\,p\,$ occurs in one factorization of $\,\rm x-a\,$, and the disjoint list of prime factors of $\rm\,q\,$ occurs in another. By $\color{#C00}{uniqueness}$, the prime factorizations are the same up to order, so the concatenation of these disjoint lists of primes occurs in $\rm\,x-a,\,$ hence $\rm\,pq\mid x-a$.
$\rm(5)\ \, $ Applying the mod Distributive Law, a handy operational form of CRT
$$\rm p\mid x\!-\!a\,\Rightarrow\, x\!-\!a\bmod pq = p\left[\dfrac{\color{#c00}{x\!-\!a}}p\bmod q\right] = p[\color{#c00}0]=0\ \ {\rm by}\ \ \color{#c00}{x\equiv a}\!\!\!\pmod{\!q}\qquad$$
Remark $\ $ This constant-case optimization of CRT arises frequently in practice so is well-worth memorizing, e.g. see some prior posts for many examples.
Quite frequently $\color{#C00}{uniqueness}$ theorems provide powerful tools for proving equalities.
More generally this idea works for values & moduli in A.P. $\ $ if $\,(a,b) = 1\,$ then
APCRT $\ \ \left\{\,x\equiv d\!-\!ck\!\pmod{b\!-\!ak}\,\right\}_{k=0}^{n}\!\!\iff\! x\equiv \dfrac{ad\!-\!bc}a\!\!\!\pmod{{\rm lcm}\{b\!-\!ak\}_{k=0}^n}$
$ $ e.g. here $\ \underbrace{\left\{\,x \equiv 3-k\pmod{7-k}\,\right\}_{k=0}^2}_{\!\!\!\!\!\!\textstyle{x\equiv 3,2,1\pmod{\!7,6,5}}}\!\iff x\equiv \dfrac{1(3)\!-\!7(1)^{\phantom{|^{|^|}}}}1\equiv -4\pmod{\!210}$
$a^p \equiv a \pmod{p}$ by Fermat's little theorem but by hypothesis $a^q \equiv a\pmod{p}$, therefore
$$a^{pq} \equiv (a^p)^q \equiv a^p \equiv a \pmod{p}$$.
Do the same with the other one, and then use the CRT.
EDIT:
By doing the same to the other one you'll get:
$$a^{pq} \equiv (a^q)^p \equiv a^q \equiv a \pmod{q} $$
Now the CRT is useful here only when you want to conclude that $a^{pq} \equiv a \pmod{pq}$.
This is what the Chinese Remainder theorem says for the case of two congruence systems:
If we're looking for $x$ such that $x \equiv a \pmod{p}$ and $x \equiv b \pmod{q}$ for two distinct primes (or two coprime moduli) then the CRT asserts that such an $x$ exists and it is unique $\mod {pq}$.
Now in our case, you can use the CRT by thinking that both $a^{pq}$ and $a$ satisfy $x \equiv a \pmod{p}$ and $x \equiv a \pmod{q}$. So, by applying the uniqueness part of the CRT, we must have $a^{pq} \equiv a \pmod{pq}$.
I don't like this approach though. All you need to know is that the least common multiple of two relatively prime numbers is their product, then by using the definition of the least common multiple, you can say that if $p \mid a^{pq}-a$ and $q \mid a^{pq}-a$ then $pq \mid a^{pq}-a$ because $\operatorname{LCM}(p,q)=pq$. This one is a better approach I think.
Best Answer
Contrapositive is: $\bmod m\!=\!pq\!:\,\ a\equiv\color{#c09} b\,\Rightarrow\, 0\equiv a(a^{\large \phi(m)+1}\!-\color{#c00}b)\equiv \color{#0a0}{a^{\large 2}}(a^{\large \phi(m)}\!-1),\, $ which is true more generally for all integers $m$ whose prime factors occur to power at most $\rm\color{#0a0}{two},$ by
Theorem $ $ Suppose that $\ m\in \mathbb N\ $ has the prime factorization $\:m = p_1^{\large e_{1}}\cdots\:p_k^{\large e_k}\ $ and suppose also that for all $\,i,\,$ $\ \color{#0a0}{e_i\le e}\ $ and $\ \phi(p_i^{\large e_{i}})\mid f.\ $ Then $\ m\mid \color{#0a0}{a^{\large e}}(a^{\large f}-1)\ $ for all $\: a\in \mathbb Z.$
Proof $\ $ Notice that if $\ p_i\mid a\ $ then $\:p_i^{\large e_{i}}\mid \color{#0a0}{a^{\large e}}\ $ by $\ \color{#0a0}{e_i \le e}.\: $ Else $\:a\:$ is coprime to $\: p_i\:$ so by Euler's phi theorem, $\!\bmod q = p_i^{\large e_{i}}:\, \ a^{\large \phi(q)}\equiv 1 \Rightarrow\ a^{\large f}\equiv 1\, $ by $\: \phi(q)\mid f\, $ and modular order reduction. Thus since all prime powers $\ p_i^{\large e_{i}}$ divide $\, a^{\large e} (a^{\large f} - 1)\ $ so too does their lcm = product = $m$.
Examples $\ $ You can find many illuminating examples in prior questions, e.g. below
$\qquad\qquad\quad$ $24\mid a^3(a^2-1)$
$\qquad\qquad\quad$ $40\mid a^3(a^4-1)$
$\qquad\qquad\quad$ $88\mid a^5(a^{20}\!-1)$
$\qquad\qquad\quad$ $6p\mid a\,b^p - b\,a^p$