55rd Ukrainian National Mathematical Olympiad - Fourth Round

Let's rewrite the given equality as follows: $$\frac{q}{p-1}-\frac{q}{p}=\frac{r}{p}-\frac{r}{p+1}+\frac{1}{p}\iff \frac{q}{p(p-1)}=\frac{r}{p(p+1)}+\frac{1}{p}\iff \frac{q}{p-1}=\frac{r}{p+1}+1\iff q=\frac{(p-1)r}{p+1}+p-1\iff q=p+r-1-\frac{2r}{p+1}(1)$$

Since $p$, $q$, $r$ are prime, then the number $\frac{2r}{p+1}$ is a positive integer. The number $2r$ has only four divisors: $1$, $2$, $r$ and $2r$. Since $p+1\ge 3$, then two cases are possible: $p+1=r$ or $p+1=2r$.

1) Suppose that $p+1=r$ which means $p=r-1$. $p+1\ge 3$, so the only pair of consecutive prime numbers is $2$ and $3$, and thus, $p=2$ and $r=3$. Then from (1) we find that $q=2$. After checking we make certain that $p=q=2$, $r=3$ is an answer.

2) Suppose that $p+1=2r$ which means $p=2r-1$. Then from (1) we find that $q=3r-3÷3$, and since $q$ is prime then $q=3$. Further, consistently find that $r=2$, $p=3$. Checking shows that $p=q=3$, $r=2$ is an answer as well.