19. (BLR) ${ }^{\mathrm{IMO} 3}$ For any positive integer $n$, let $\tau(n)$ denote the number of its positive divisors (including 1 and itself). Determine all positive integers $m$ for which there exists a positive integer $n$ such that $\frac{\tau\left(n^{2}\right)}{\tau(n)}=m$.
19. For $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{r}^{\alpha_{r}}$, where $p_{i}$ are distinct primes and $\alpha_{i}$ natural numbers, we have $\tau(n)=\left(\alpha_{1}+1\right) \cdots\left(\alpha_{r}+1\right)$ and $\tau\left(n^{2}\right)=\left(2 \alpha_{1}+1\right) \ldots\left(2 \alpha_{r}+1\right)$. Putting $k_{i}=\alpha_{i}+1$, the problem reduces to determining all natural values of $m$ that can be represented as 
$$ m=\frac{2 k_{1}-1}{k_{1}} \cdot \frac{2 k_{2}-1}{k_{2}} \cdots \frac{2 k_{r}-1}{k_{r}} $$
Since the numerator $\tau\left(n^{2}\right)$ is odd, $m$ must be odd too. We claim that every odd $m$ has a representation of the form (1). The proof will be done by induction. This is clear for $m=1$. Now for every $m=2 k-1$ with $k$ odd the result follows easily, since $m=\frac{2 k-1}{k} \cdot k$, and $k$ can be written as (1). We cannot do the same if $k$ is even; however, in the case $m=4 k-1$ with $k$ odd, we can write it as $m=\frac{12 k-3}{6 k-1} \cdot \frac{6 k-1}{3 k} \cdot k$, and this works. In general, suppose that $m=2^{t} k-1$, with $k$ odd. Following the same pattern, we can write $m$ as 
$$ m=\frac{2^{t}\left(2^{t}-1\right) k-\left(2^{t}-1\right)}{2^{t-1}\left(2^{t}-1\right) k-\left(2^{t-1}-1\right)} \cdots \frac{4\left(2^{t}-1\right) k-3}{2\left(2^{t}-1\right) k-1} \cdot \frac{2\left(2^{t}-1\right) k-1}{\left(2^{t}-1\right) k} \cdot k $$
The induction is finished. Hence $m$ can be represented as $\frac{\tau\left(n^{2}\right)}{\tau(n)}$ if and only if it is odd.