Let $p$, $q$ be prime numbers such that $n^{3pq} – n$ is a multiple of $3pq$ for all positive integers $n$. Find the least possible value of $p + q$. Recently a exam called PRMO 2017 was conducted. Question 28 went as follows,

Let $p$, $q$ be prime numbers such that $n^{3pq} – n$ is a multiple of $3pq$ for all positive integers $n$. Find the least possible value of $p + q$.

This question was considered to be quite tough. Many people are saying it was too tough to be put in a exam which is open for 14 year olds. 
How can it be solved?
I have not studied number theory, so I really couldn't attempt this.
Thanks.
 A: $\color{Green}{\text{Lemma}}$: 


*

*For every odd prime number $p$; 
and for every positive integer $\alpha$;
the multiplicative group $\mathbb{Z}_{p^{\alpha}}^*$;
is 
a cyclic group of order 
$\phi(p^{\alpha})= (p-1)p^{\alpha-1}$.
In other words:  


$$ 
\big( 
\mathbb{Z}_{p^{\alpha}}^* 
\ , \times 
\big) 
\equiv
\big( 
\mathbb{Z}_{(p-1)p^{\alpha-1}} 
\ , + 
\big) 
. 
$$


*

*For $\color{Red}{p=2}$; 
and for every positive integer $\color{Red}{3 \leq \alpha}$;
the multiplicative group $\mathbb{Z}_{2^{\alpha}}^*$;
is 
the direct sum of $\mathbb{Z}_2$ and 
a cyclic group of order 
$\color{Red}{\dfrac{1}{2}}\phi(2^{\alpha})= \color{Red}{2^{\alpha-2}}$.
In other words:  


$$ 
\big( 
\mathbb{Z}_{2^{\alpha}}^* 
\ , \times 
\big) 
\equiv
\big( 
\mathbb{Z}_2 
\oplus 
\mathbb{Z}_{\color{Red}{2^{\alpha-2}}} 
\ , + 
\big) 
. 
$$


*

*The multiplicative group $\mathbb{Z}_{2^2}^*$; 
is 
a cyclic group of order $2$.
The multiplicative group $\mathbb{Z}_{2}^*$; 
is 
the trivial group.




If $n$ is coprime with $3pq$ then we can factor $n$ and hence we have: 
$$ 
n^{3pq  }\overset{3pq}{\equiv}n 
\Longrightarrow
n^{3pq-1}\overset{3pq}{\equiv}1 
. 
$$


First case: All of $3, p, q$ are different. 
By the above lemma, 
we know that:
there is an integer $a$; with $\text{ord}_p(a)=p-1$.
On the otherhand $a^{3pq-1}\overset{p}{\equiv}1$;
which implies that $\color{Blue}{p-1 \mid 3pq-1}$. 
Similarly one can prove that
$p-1 \mid 3pq-1$ 
and 
$3-1 \mid 3pq-1$. 

But notice that 
$\color{Blue}{3pq-1=3(p-1)q}+\color{Purple}{3q-1}$;
similarly we have: 
$3pq-1=3p(q-1)+3p-1$
and 
$3pq-1=2pq+pq-1$. 
So we can conclude that: 
$$   
\color{Blue}{p-1 \mid} \color{Purple}{3q-1} 
\ \ \ \ 
\text{and} 
\ \ \ \ 
q-1 \mid 3p-1 
\ \ \ \ 
\text{and} 
\ \ \ \ 
3-1 \mid pq-1 
; 
$$ 
the last divisibility condition implies that 
both of $p,q $ must be odd;
you can check that $p=11 , q=17$ 
satisfies the above divisiblity conditions.

Why these are the least posible values? 
$ \color{Green}{
\text
{As} 
\ 
\color{Red}{\text{@Thomas Andrews}} 
\ 
\text{has been mentioned:}  
\\ 
\color{Red}{\text{"}}
\text
{we can assume} 
\ 
p \overset{6}{\equiv} 5
\ 
. 
\\
\text
{[ 
Because} 
\ 
p−1∣3q−1 
\ 
\text
{means} 
\ 
p−1 
\ 
\text
{is coprime to} 
\ 
3 
\ 
; 
\text
{so} 
\ 
p \overset{3}{\ncong} 1; 
\ 
\text
{which implies} 
\ 
p \overset{3}{\equiv} 2 
\ 
\text
{. 
]}  
\\ 
\text
{Notice that} 
\ 
p=5 
\ 
\text
{doesn't work,} 
\\ 
\text
{since} 
\ 
3p−1=14 
\ 
\text
{is not divisible 
by} 
\ 
q−1 
\ 
\text
{for any} 
\ 
q \overset{6}{\equiv} 5. 
\ 
\\ 
\text 
{So the smallest possible values for} 
\ 
p,q 
\ 
\text
{is} 
\ 
11,17 
\ 
.} 
\color{Red}{\text{"}}$


Second case: All of $3, p, q$ are not distict. 
We will show that this second case is impossible. 


*

*$p=3$ or $q=3$.
From assumtion of problem 
we know that:
$3^{3pq} \overset{9}{\equiv} 3$; 
so we have: 
$0 \overset{9}{\equiv} 3^2 \overset{9}{\equiv} 3$;
which is an obvious contradiction.  

*$p=q$.
From assumtion of problem 
we know that:
$p^{3pq} \overset{p^2}{\equiv} p$; 
so we have: 
$0 \overset{p^2}{\equiv} p^2 \overset{p^2}{\equiv} p$;
which is an obvious contradiction.  
A: This is not an answer for a beginner.  However, Part (a) of this more general result can be used, although a proof is left as an exercise.  As defined in that link, we have
$$3pq \mid g(3pq,1)\,,$$
so $g(3pq,1)>2$.  Hence, $3pq-1$ must be even and
$$g(3pq,1)=2\,\prod_{r\in D(3pq,1)}\,r\,,$$
where $D(3pq,1)$ is defined in the same link above.  It follows that $p,q\in D(3pq,1)$ are odd primes. That is, $p-1\mid 3pq-1$ and $q-1\mid 3pq-1$.  The rest goes as Famke suggests (i.e., by observing that $3$, $p$, and $q$ must be distinct odd primes with $p-1\mid 3q-1$ and $q-1\mid 3p-1$).
