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Find all triplets of prime numbers $(p, q, r)$ such that

$$
\frac{p}{q}-\frac{4}{r+1}=1
$$
We multiply everything by $q(r+1)$ to obtain $p(r+1)-4 q=q(r+1)$, i.e., $p(r+1)=q(r+5)$. Then we look modulo $q$: $q$ must divide $p(r+1)$, so either $q=p$, or $q$ divides $r+1$. In the first case, dividing by $q$ gives $r+1=r+5$, which is impossible. We write $r+1=k q$. Then similarly, we get $r+5=k' p$. Necessarily, $k=k'$ and $k \mid 4$, so $k=1, 2$ or 4. If $k=1$, we get $r=2, q=3, p=7$.

If $k=2$, we write $p+q=r+3$ and $p-q=2$. By looking at the possibilities for the remainder of $r$ modulo 6, we find $q=3, p=5, r=7$.

Finally, if $k=4$, we get $p-q=1$ so $p=3, q=2, r=5$.

In the end, the solutions are $(7,3,2),(5,3,5)$ and $(5,3,7)$.