Danube Mathematical Competition, 2014

We will show that the only solution is $a = 6$ (for $p = q = r$).

If exactly two of the three prime numbers are equal; e.g. $p = q \neq r$, then
$$a = 2\left(\frac{p}{r} + \frac{r}{p}\right) + 2 \in \mathbb{N},$$
so there exists $n \in \mathbb{N}$ so that
$$\frac{n}{2} = \frac{p}{r} + \frac{r}{p},$$
which is
$$n = \frac{2(p^2 + r^2)}{pr}.$$
Hence $p \mid 2(p^2 + r^2)$ and, since $(p, r) = 1$, we have $p \mid 2$, so $p = 2$. It follows that
$$n = \frac{r^2 + 4}{r} = r + \frac{4}{r}$$
and, as $n \in \mathbb{N}$, $r$ is a prime divisor of $4$, so $r = 2 = p$, a contradiction.

If $p, q, r$ are pairwise distinct, then
$$apqr = pq(p+q) + qr(q+r) + rp(r+p),$$
so $p \mid qr(q+r)$, which leads to $p \mid q+r$ and, furthermore, $p \mid p+q+r$. Analogously,
$q \mid p+q+r$ and $r \mid p+q+r$, so $pqr \mid p+q+r$, hence $pqr \le p+q+r$.

As $p, q, r$ are distinct and at least $2$, then $pqr \ge 2qr > 4r$, and, similarly, $pqr > 4q$ and $pqr > 4p$. It follows that $3pqr > 4(p+q+r) \ge 4qr$, which is impossible.