## Task A-4.1.

Determine all natural numbers $n$ such that the product of all positive divisors of $n$ is equal to $n^{3}$. Present them in canonical form, i.e., using the prime factorization.
## First Solution.

Let $1=d_{1}<d_{2}<\ldots<d_{k}=n$ be all the positive divisors of the number $n$. Notice that

$$
d_{1} \cdot d_{k}=d_{2} \cdot d_{k-1}=d_{3} \cdot d_{k-2}=n
$$

From this, we conclude that the product of the divisors of the number $n$ is equal to $n^{3}$ if and only if $n$ has exactly six divisors.

For a prime number $p$, the divisors of the number $p^{\alpha}$ are $1, p, p^{2}, \ldots, p^{\alpha}$; there are $\alpha+1$ of them. Therefore, numbers of the form $p^{5}$, where $p$ is a prime number, have six divisors.

If $n$ is a number divisible by three different prime numbers $p, q$, and $r$, then $1, p, q, r, p q, q r, r p, p q r$ are distinct divisors of $n$, so a number divisible by at least three different prime numbers has more than six divisors.

Now consider numbers of the form $n=p^{\alpha} q^{\beta}$, where $p$ and $q$ are different prime numbers, and $\alpha$ and $\beta$ are natural numbers. Without loss of generality, let $\alpha \geqslant \beta$.

If $\alpha=\beta=1$, the only divisors of the number $n=p q$ are $1, p, q$, and $p q$.

If $\alpha=2$, $\beta=1$, then both $p^{2}$ and $p^{2} q$ divide $n$, so $n=p^{2} q$ has exactly six divisors.

If $\alpha \geqslant 3$ or $\beta \geqslant 2$, then $n$ has more than six divisors (in addition to the six mentioned, $n$ itself is also a divisor).

The desired numbers are numbers of the form $n=p^{5}$ for some prime number $p$ and $n=p^{2} q$ for different prime numbers $p$ and $q$.