Let $d(n)$ denote the number of positive divisors of the positive integer $n$. Determine those numbers $n$ for which

$$
d\left(n^{3}\right)=5 \cdot d(n)
$$
Solution. $n=1$ is not a solution to the equation, so $n \geq 2$. Let the prime factorization of $n$ be $n=$ $p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{k}^{a_{k}}$, where $k \geq 1, p_{1}, p_{2}, \ldots, p_{k}$ are distinct prime numbers, and $a_{1}, a_{2}, \ldots, a_{k}$ are positive integers. With these notations,

$$
\begin{aligned}
d(n) & =\left(a_{1}+1\right)\left(a_{2}+1\right) \ldots\left(a_{k}+1\right) \\
d\left(n^{3}\right) & =\left(3 a_{1}+1\right)\left(3 a_{2}+1\right) \ldots\left(3 a_{k}+1\right)
\end{aligned}
$$

The expression $3 a_{i}+1$ can be estimated from below:

$$
3 a_{i}+1=2 a_{i}+a_{i}+1 \geq 2 a_{i}+2=2\left(a_{i}+1\right)
$$

Using this,

$$
\begin{aligned}
5 \cdot d(n) & =5 \cdot\left(a_{1}+1\right)\left(a_{2}+1\right) \ldots\left(a_{k}+1\right)=\left(3 a_{1}+1\right)\left(3 a_{2}+1\right) \ldots\left(3 a_{k}+1\right) \geq \\
& \geq 2\left(a_{1}+1\right) 2\left(a_{2}+1\right) \ldots 2\left(a_{k}+1\right)=2^{k} \cdot d(n), \\
5 \cdot d(n) & \geq 2^{k} \cdot d(n),
\end{aligned}
$$

where $k$ is a positive integer. From this, it is already clear that only $k=1$ and $k=2$ are possible.

If $k=1$, then

$$
5\left(a_{1}+1\right)=3 a_{1}+1, \quad a_{1}=-2
$$

we do not get a solution.

Finally, let $k=2$. Then we get that

$$
\begin{aligned}
5\left(a_{1}+1\right)\left(a_{2}+1\right) & =\left(3 a_{1}+1\right)\left(3 a_{2}+1\right) \\
5 & =4 a_{1} a_{2}-2 a_{1}-2 a_{2}+1 \\
5 & =\left(2 a_{1}-1\right)\left(2 a_{2}-1\right)
\end{aligned}
$$

The number 5 can only be factored into positive integers in one way. We see that one of $a_{1}$ and $a_{2}$ is 1, and the other is 3. By symmetry, we can assume that $a_{1}=3$ and $a_{2}=1$. Therefore, $n=p^{3} \cdot q$, where $p$ and $q$ are distinct primes.

Numbers of this form are indeed solutions, because $d(n)=d\left(p^{3} q\right)=4 \cdot 2=8$ and $d\left(n^{3}\right)=d\left(p^{9} q^{3}\right)=10 \cdot 4=40$.