Senderovv B.A.

Find all such triples of prime numbers $p, q, r$, such that the fourth power of each of them, decreased by 1, is divisible by the product of the other two.
Clearly, any two numbers of the triplet are distinct (if $p=q$, then $p^{4}-1$ is not divisible by $q$). Let $p$ be the smallest number of the triplet. The number

$p^{4}-1=(p-1)(p+1)\left(p^{2}+1\right)$ is divisible by $q r$. But $p-1$ is less than any of the prime numbers $q$ and $r$, and thus is coprime with them. The number $p^{2}+1$ cannot be divisible by both numbers $q$ and $r$, since $p^{2}+1<p$, this is only possible if $q=p+1$.

Therefore, $p=2, q=3$. $r$ is a prime divisor of the number $p^{4}-1=15$, different from $q=3$, so $r=$ 5.

It remains to check that the triplet $\{2,3,5\}$ satisfies the conditions of the problem.

## Answer

$\{2,3,5\}$.