(p÷q)-4÷(r+1)=1 then (p-q)÷q=4÷(r+1). Then p-q=4 or p-q=2. If p-q =4 then q=r+1. As q and r are both prime this is impossible. Hence p-q=2 and 2q=r+1. € 2p=5+r ,¥2p-5=r.if p is of form 6k+1 then 3 divides LHS of ¥ But r is a prime hence it is not possible. We can see that in the main equation p >q . And we get p-q=2 as p ,q both prime then p is of form =6k+1and q is of form =6k-1. But previously we showed that p is of form 6k-1 . Contradiction . The only solution is p=5,q=3and r=5 .

I got p=7,q=3,r=2 and 7/3-4/3=1 which satisfied the equation

Rearranging gives $p(r+1)=q(r+5)$. Clear that $p \neq q$. Also $q|r+1$ and $p|r+5$. Let $pk=r+5$ and $qm=r+1$. Putting back into equation we have $k=m$.$pk-qk=k(p-q)=4 \Rightarrow k=1, 2, 4$. Checking cases for $k=1$ we have $(7, 3, 2)$. For $k=2$, $p=\frac{r+5}{2}$ and $q= \frac{r+1}{2}$. We get the triples $(k+3, k+1, 2k+1)$ where all are prime numbers. For $k \leq 2$, we see only $k=2$ works and gives the triple $(5, 3, 5)$. For $k>2$ we have contradiction by using mod 3. For $k=4$, $p-q=1 \Rightarrow q=2, p=3$ and $r=7$ Thus we found all triples.