HMMT-Feb 2022 guts, Problem 8

For any positive integer $n$, let $\tau(n)$ denote the number of positive divisors of $n$. If $n$ is a positive integer such that $\frac{\tau\left(n^{2}\right)}{\tau(n)}=3$, compute $\frac{\tau\left(n^{7}\right)}{\tau(n)}$.

Solution: Solution: Let the prime factorization of $n$ be $n=p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{k}^{e_{k}}$. Then, the problem condition is equivalent to \[ \prod_{i=1}^{k} \frac{2 e_{i}+1}{e_{i}+1}=3 \] Note that since $\frac{2 x+1}{x+1} \geq 1.5$ for $x \geq 1$, and $1.5^{3}>3$, we have $k \leq 2$. Also, $k=1$ implies $2 e_{1}+1=$ $3\left(e_{1}+1\right)$, which implies $e_{1}$ is negative. Thus, we must have $k=2$. Then, our equation becomes \[ \left(2 e_{1}+1\right)\left(2 e_{2}+1\right)=3\left(e_{1}+1\right)\left(e_{2}+1\right) \] which simplifies to $\left(e_{1}-1\right)\left(e_{2}-1\right)=3$. This gives us $e_{1}=2$ and $e_{2}=4$. Thus, we have $n=p^{2} q^{4}$ for primes $p$ and $q$, so $\frac{\tau\left(n^{7}\right)}{\tau(n)}=\frac{\tau\left(p^{14} q^{28}\right)}{\tau\left(p^{2} q^{4}\right)}=\frac{15 \cdot 29}{3 \cdot 5}=29$. $\fbox{29}$.

Answer: 29