2. Determine all triples of composite natural numbers, each of which is divisible by the product of the superdivisors of the remaining two numbers. (A superdivisor of a number $n>1$ is its greatest divisor $d$ with the property $d<n$.)

(Jaromír Šimša)
SOLUTION. The superdivisor $d$ of a composite number $n$ is obtained when we strike out the smallest represented factor (one, if there are multiple) from the prime factorization of the number $n$. Therefore, it holds that $n = p d$, where $p$ is a prime number with the property $p \leq d$?

Let $n_{1}, n_{2}, n_{3}$ be any sought triplet. According to the introductory paragraph, for the superdivisor $d_{i}$ of the number $n_{i}$, the equality $n_{i} = p_{i} d_{i}$ holds, where $p_{i}$ is a prime number and $p_{i} \leq d_{i}$ for each $i=1,2,3$. According to the problem statement, it must simultaneously hold that $d_{2} d_{3} \mid p_{1} d_{1}$, $d_{1} d_{3} \mid p_{2} d_{2}$, and $d_{1} d_{2} \mid p_{3} d_{3}$. By multiplying these three relations, we get

$$
\left(d_{1} d_{2} d_{3}\right)^{2} \mid p_{1} p_{2} p_{3} d_{1} d_{2} d_{3}, \quad \text { or } \quad d_{1} d_{2} d_{3} \mid p_{1} p_{2} p_{3}
$$

On the other hand, by multiplying the three inequalities $p_{i} \leq d_{i}$, we obtain $p_{1} p_{2} p_{3} \leq d_{1} d_{2} d_{3}$. From the conclusions of the last two sentences, it follows that $d_{1} d_{2} d_{3} = p_{1} p_{2} p_{3}$ must hold. From the way the inequality $p_{1} p_{2} p_{3} \leq d_{1} d_{2} d_{3}$ was derived, we see that $d_{i} = p_{i}$ or $n_{i} = p_{i}^{2}$ must hold for each $i=1,2,3$. The relation $d_{2} d_{3} \mid p_{1} d_{1}$ then means that $p_{2} p_{3} \mid p_{1}^{2}$, so all three prime numbers $p_{i}$ must be equal to the same prime number $p$. Then $d_{i} = p$ and $n_{i} = p^{2}$ for each $i=1,2,3$, and the conditions $d_{i} d_{j} \mid n_{k}^{2}$ are clearly satisfied in the form $p^{2} \mid p^{2}$.

Conclusion. The solutions to the problem are precisely the triplets $\left(p^{2}, p^{2}, p^{2}\right)$, where $p$ is any prime number.