Possible primes $p$ $q$ satisfying $a^{3pq}-a \equiv 0 \pmod {3pq}$ For any integer $a$, consider the primes $p$ and $q$ satisfying
$a^{3pq}-a \equiv 0 \pmod {3pq}$
Find all such possible $p$ and $q$.

So I tried breaking it down into 3 congruences:
$a^{3pq}-a \equiv 0 \pmod {3}$
$a^{3pq}-a \equiv 0 \pmod {p}$
$a^{3pq}-a \equiv 0 \pmod {q}$
and for $a^{3pq}-a \equiv 0 \pmod {p}$ it is equivalent to $a^{3pq-1}-1 \equiv 0 \pmod {p}$ since $gcd(a,p)=1$
and I would have $Ord_p(a) \mid (3pq-1)$
But I still cannot see by far how I can approach the final solution in this way.
 A: Let us consider $n=\prod P_i^{A_i}$ and $(a,n)=1$ where $P_i$s are distinct primes.
If $a^n\equiv a\pmod n,a^{n-1}\equiv 1\pmod n\implies  a^{n-1}\equiv 1\pmod {P_i^{A_i}}$ 
But, $P_i^{A_i}$ has primitive root, so there exists $a$ such that $ord_{(P_i^{A_i})}a=\phi(P_i^{A_i})=P_i^{A_i-1}(P_i-1)$.
So, $P_i^{A_i-1}(P_i-1)$ must divide $(n-1)=(\prod P_j^{A_j}-1)$ which is impossible if $A_i-1\ge 1$ as $(P_i^{A_i-1},\prod P_j^{A_j}-1)=1$.
This is true for $P_i\mid n$, so, $n$ reduces to $\prod P_i$ 
and $(P_i-1)$ must divide $(\prod P_j-1)$
Now $\prod P_j-1=(P_i-1)(\prod_{i\ne j} P_j-1)+(\prod_{i\ne j} P_j-1)$
$(P_i-1)$ must divide $(\prod_{i\ne j} P_j-1)$
In that case, $a^{n-1}\equiv 1\pmod {P_i}$
$\implies a^{n-1}\equiv 1\pmod {lcm(P_i)}$
But, $lcm(P_i)=\prod P_i$ as $(P_i,P_j)=1$ for $i\ne j$, so $lcm(P_i)=n$
$\implies a^{n-1}\equiv 1\pmod n$, 
which has $(P_i-1)\mid (\prod_{i\ne j} P_j-1) ∀ P_k\mid n$ as necessary and sufficient condition known as Korselt's Criterion.
Here, $P_1=3,P_2=p,P_3=q \implies p,q,3$ are all distinct primes.
So, 
(i)$(3-1)\mid(pq-1)\implies pq$ is odd, So, $p,q\ne 2$
(ii)$(p-1)\mid(3q-1)$
Now, $3$ must not divide $(p-1)$ as it does not divide $(3q-1),$
So, $p-1=3a+1,3a-1$ for some integer $a$
But $p\ne 3a$ as prime $p>3,$ so odd $p=3a+2$, so a must be odd $=2b+1$(say).
So, $p=3a+2=3(2b+1)+2=6b+5$
(iii) $(q-1)\mid (3p-1),$ so $q$ will also be $\equiv 5\pmod 6$
As we need to test for any integer $a,$ it suffices to test for $(a,3pq)=1$.
By trial, one set of values of $p,q$ is  $17,11$.
