Exercise 2.16

Proof. $\text{∙}$ If $n=a^2$ is a square, where $a=p_1^{k_1}\cdots p_t^{k_t}$, then $\nu (n)=(2k_1+1)\cdots (2k_t+1)$ is odd.

$\text{∙}$ Conversely, if $\nu (n)=\nu (q_1^{l_1}\cdots q_r^{l_r})$ is odd, then $(l_1+1)\cdots (l_r+1)$ is odd. So each $l_i+1$ is odd, and then $l_i$ is even, for $i=1,2,\dots ,r$ : $n$ is a square. □