Problem 1438

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

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 $q = 3r-3 \div 3$, 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.