I'll finish up my solution later because i have to go right now..or someone else can finish it up for me if they wish.

$3|pq(p-q)(p+q)$ for every $p,q$ If $3|p-q \to 3|p+q$-contradiction, so $3|pq$ Case 1: $p=3$ $3+q=r(3-q)^n$ $(3-q) | 6$ $3-q =(-6,-3,-2,-1,1,2) \to q=2,5$ $q=2$: $5=r 1^n \to r=5$ $q=5$: $8=r(-2)^n \to (r,n)=(2,2)$ Case 2: $q=3$ $3+p=r(p-3)^n$ $(p-3)|6$ $p-3=(-1,1,2,3,6) \to p=2,5$ $p=2$: $5=r(-1)^n \to (r,n)=(5,2k)$ $p=5$: $8=r 2^n \to (r,n)=(1,3),(2,2),(4,1)$ Answer: $(p,q,r,n) = (3,2,5,k),(3,5,2,2),(2,3,5,2k),(5,3,1,3),(5,3,2,2),(5,3,4,1)$

As $n$ is a positive integer, then $p-q$ divides $r{ (p-q) }^{ n }=p+q$. Now, see that the $gcd(p+q,p-q)=gcd(p+q,2p)=gcd(2q,2p)$ which is $1$ or $2$; but, as $p-q$ divides $p+q$ then $p-q$ divides $2$. If $p-q=\pm 1$ then is clear than $p=2$, $q=3$, $r=5$ and $n$ is even, or $p=3$, $q=2$, $r=5$ and $n$ positive integer. If $p-q=\pm 2$, then $p$ or $q=3$; otherwise, $p$ and $q$ are $1$ and $-1$ $mod$ $6$ and $3$ divides $p+q$. So, if $q=3$ and $5=5$, sp $p+q=8$. So $n=3$ and $r=1$, $n=2$ and $r=2$ or $n=1$ and $r=8$. If $p=3$ and $q=5$ then $n=2$ and $r=2$.