Junior Balkan Mathematical Olympiad

$(p, q, r) \in \{(3, 2, 7), (5, 3, 5), (7, 3, 2)\}$

We can rewrite in the form
$$\frac{pr + p - 4q}{q(r + 1)} = 1 \Rightarrow pr + p - 4q = qr + q \Rightarrow r(p - q) = 5q - p.$$

Hence $p \neq q$
$$r = \frac{5q-p}{p-q} = \frac{4q+q-p}{p-q} \quad \text{i.e.} \quad r = \frac{4q}{p-q} - 1.$$
So $p-q \neq q$, $p-q \neq 2q$, $p-q \neq 4q$. We have $p-q=1$ or $p-q=2$ or $p-q=4$.

i) If $p-q=1$ then $q=2, p=3, r=7$.

ii) If $p-q=2$ then $p=q+2, r=2q-1$.
If $q \equiv 1 \pmod{3}$ then $q+2 \equiv 0 \pmod{3}$, $q+2=3 \Rightarrow q=1$ contradiction.
If $q \equiv -1 \pmod{3}$ then $r \equiv -2-1 \equiv 0 \pmod{3}$, so $r=3, q=2, p=4$ contradiction.
Hence $q=3, p=5, r=5$.

iii) If $p-q=4$ then $p=q+4, r=q-1$. Hence $q=3, p=7, r=2$.