Square of number of divisors of n equals n

Question

Find $n \in \mathbb{N}$ such that $n = \tau(n)^2$ ($\tau(n)$ being the number of positive divisors of $n$). I tried some values for $n$, it seems that besides $n = 1$ and $n = 9$ there's no other solutions. I have no idea to how to prove that nor if it's true in the first place. I could use some hints, thank you.

Answer 1

$n$ is a perfect square, so

$$n = \prod_{i=1}^k p_i^{2 \beta_i}.$$

Therefore, $$\tau(n) = \prod_{i=1}^k (2\beta_i+1),$$ so

$$n = \prod_{i=1}^k (2\beta_i+1)^2.$$

Note that $p_i^{\beta_i} > 2\beta_i + 1,$ unless $p_i=2, 3$ and $\beta_i = 1.$ For $p_i=3$ we have equality, for $p_i=2,$ the ratio between the LHS and the RHS is $2/3.$ The smallest next ratio is $5/3,$ and since $5/3 \times 2/3 = 10/9 > 1,$ the only prime we can have in the decomposition of $n$ is $3,$ so $n = 9.$