9.7. Find all triples of prime numbers $p, q, r$ such that the fourth power of any of them, decreased by 1, is divisible by the product of the other two.

(V. Senderov)
# Answer. $2,3,5$.

Solution. It is clear that any two numbers in the triplet are distinct (if $p=q$, then $p^{4}-1$ does not divide $q$). Let $p$ be the smallest of the numbers in the triplet. We know that the number $p^{4}-1=(p-1)(p+1)\left(p^{2}+1\right)$ is divisible by $q r$. Note that $p-1$ is less than any of the prime numbers $q$ and $r$, and therefore is coprime with them. Furthermore, the number $p^{2}+1$ cannot be divisible by both $q$ and $r$, since $p^{2}+1 < p$, which is only possible if $q=p+1$. Then one of the numbers $p$ and $q$ is even, and since it is prime, $p=2, q=3$. Finally, $r$ 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.

Comment. Only the correct answer -1 point.

The idea of choosing the smallest $p$ from the three prime numbers to investigate the divisibility of $p^{4}-1$ by $q r$ is worth 1 point.

If the solution is correct but lacks the indication that the obtained triplet fits, 1 point is deducted.