[ Number of divisors and their sum of a number] $[\quad \underline{\text { equations in integers }}]$

A certain natural number $n$ has two prime divisors. Its square has a) 15; b) 81 divisors. How many divisors does the cube of this number have?
b) Let $n=p^{\alpha} q^{\beta}$, where $p$ and $q$ are prime numbers. Then $n^{2}=p^{2 \alpha} q^{2 \beta}, 81=\tau\left(n^{2}\right)=(2 \alpha+1)(2 \beta+1)$ (see problem $\left.\underline{60537}\right)$, from which $\alpha=\beta=4$ or $\{\alpha, \beta\}=\{2,13\}$. Therefore, $\tau\left(n^{3}\right)=(3 \alpha+1)(3 \beta+1)=13^{2}$ or $5 \cdot 40$.

## Answer

a) 28 ; b) 160 or 169.