Exercise 8. Let a be a strictly positive integer, such that $a^{3}$ has 5 times more positive divisors than a.

How many positive divisors does $a$ have?
Solution to Exercise 8

Since the divisors of $1^{3}$ are exactly the same as those of 1, we must have $a \geqslant 2$. Let then $a=p_{1}^{\alpha_{1}} \cdots p_{k}^{\alpha_{k}}$ be the prime factorization of $a$, with $\alpha_{i} \geqslant 1$ for all $i$. The number of positive divisors of $a$ is therefore $\left(\alpha_{1}+1\right) \times \cdots \times\left(\alpha_{k}+1\right)$. Moreover, the prime factorization of $a^{3}$ is then $a^{3}=p_{1}^{3 \alpha_{1}} \cdots p_{k}^{3 \alpha_{k}}$, and the number of positive divisors of $a^{3}$ is therefore $\left(3 \alpha_{1}+1\right) \times \cdots \times\left(3 \alpha_{k}+1\right)$. Thus, the condition in the statement translates to

$$
\left(3 \alpha_{1}+1\right) \times \cdots \times\left(3 \alpha_{k}+1\right)=5\left(\alpha_{1}+1\right) \times \cdots \times\left(\alpha_{k}+1\right)
$$

It is easy to verify that if $x \geqslant 1$, then $3 x+1 \geqslant 2(x+1)$. Therefore, $\left(3 \alpha_{1}+1\right) \times \cdots \times\left(3 \alpha_{k}+1\right) \geqslant 2^{k}\left(\alpha_{1}+1\right) \times \cdots \times\left(\alpha_{k}+1\right)$, and to achieve (1), we need $2^{k} \leqslant 5$. Since $k$ is a strictly positive integer, this means $k=1$ or $k=2$.

For $k=1$, equation (1) reduces to $3 \alpha_{1}+1=5 \alpha_{1}+5$, which is clearly impossible.

For $k=2$, equation (1) becomes $\left(3 \alpha_{1}+1\right)\left(3 \alpha_{2}+1\right)=5\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)$, which, after expansion, leads to $2 \alpha_{1}+2 \alpha_{2}-4 \alpha_{1} \alpha_{2}+4=0$, or $\left(2 \alpha_{1}-1\right)\left(2 \alpha_{2}-1\right)=5$. Each of the two factors on the left-hand side are strictly positive integers and 5 is prime. Without loss of generality, we can assume that $\alpha_{1} \leqslant \alpha_{2}$, and it follows that $2 \alpha_{1}-1=1$ and $2 \alpha_{2}-1=5$, or $\alpha_{1}=1$ and $\alpha_{2}=3$.

Finally, $a$ has exactly 8 positive divisors.