55rd Ukrainian National Mathematical Olympiad - Fourth Round

Let $p,q,r$ be the prime numbers that satisfy the conditions of the problem. If $p>2$ then all $p,q,r$ are odd thus the number $A=3p+5q$ is even and the number $B=(r-p)(r-q)(q-p)+1$ is odd, which contradicts the conditions of the problem. Therefore, $p=2$, thus:

$$(r-2)(r-q)(q-2)+1=6+5q.$$

$p<q<r$, that means $r-2>q-2$ and $r-q\ge 2$. Therefore, $6+5q>2(q-2{)}^2+1$. After solving the last inequality, we will have that $q<7$. Prime numbers that satisfy this inequality are $q=3$ and $q=5$.

If $q=3$, then $B=3p+5q=21$ is not prime.

If $q=5$, then $B=3p+5q=31$ is prime, that implies $A=(r-2)(r-5)\cdot 3+1=31$, and $r^2-7r=0$, that means $r=7$.