Number of triplets for primes for the equations

From the first equation we may immediately conclude that $r^2 + 1 \equiv 0 \mod 2 \implies r \equiv 1 \mod 2$ . This and the second equation imply $pq \equiv 0 \mod 2$ which implies $p$ or $q$ are the prime $2$ . By the symmetry of the equation we notice that $p$ and $q$ are interchangeable so for now let's say $p = 2$ .

Plugging this into the second equation we get $q = \frac{r+1}{2}$ . Now plug this expression for $q$ into the first equation and then simplify to reach the polynomial equation $-\frac{1}{2}r^2 + r + \frac{15}{2} = 0$ . It has two solutions, $r=5,-3$ of which only the first is valid as primes are assumed to be positive. And if $r=5$ then $q=3$ so we have a valid solution. Remembering to count symmetry, the only solutions are $p=2,q=3,r=5$ and $p=3,q=2,r=5$ .