Exercise 2.17

Proof. $\text{∙}$ Note that for all $r\ge 0$, $\sigma (2^r)=1+2+2^2+\cdots +2^r=2^{r+1}-1$ is always odd.

If $p\ne 2$, $\sigma (p^{2k})=1+p+p^2+\cdots +p^{2k}$ is a sum of $2k+1$ odd numbers, so is odd.

Therefore, if $n=a^2$, or $n=2a^2,a\in \mathbb{Z}$, $\sigma (n)$ is odd.

$\text{∙}$ Conversely, suppose that $\sigma (n)$ is odd, where $n=p_1^{k_1}{p}_2^{k_2}\cdots p_t^{k_t}$, with $p_1=2<p_2<\cdots <p_t$. Then

is odd. Then each $\frac{p_i^{k_i+1}-1}{p_i-1}=1+p_i+\cdots +p_i^{k_i}(i=2,\cdots ,t)$ is odd. As each $p_i^j,j=0,\dots ,k_i$ is odd, the number of terms $k_i+1$ is odd, so $k_i$ is even ( $i=2,\dots ,t$). Moreover, if $k_1$ is odd, $2^{k_1}$ is twice a square. Thus $n$ is a square, or twice a square. □