how to find $n\in \Bbb N$ such that: $\tau(n)$is odd

Question

how to find $n\in \Bbb N$ such that:

$\tau(n)$is odd

$\tau(n)$ is number of positive divisor of $n$

Answer 1

We know if $n=\prod p_i^{r_i},$ where $p_i$s are distinct primes

$\tau(n)=\tau(\prod p_i^{r_i})=\prod(r_i+1)$

So, we need each $r_i+1$ to be odd $\implies r_i$ must be even for all $i$

Let $r_i=2m_i$

So, $n=\prod p_i^{r_i}=\prod p_i^{2m_i}=\left(\prod p_i^{m_i}\right)^2$ i.e., must be perfect square

Answer 2

Hint: you can pair up the divisors unless $n$ is perfect square.