Exercise 4.2.12 (When $d(n)$ and $\sigma_k(n)$ are odd?)

beginproof

(a)

Let $n$ be a positive integer.

• Suppose that $d(n)$ is odd, where $n=p_1^{a_1}{p}_2^{a_2}\cdots p_l^{a_l}$ is the canonical prime decomposition of $n$. Then

  $$d(n)=(a_1+1)(a_2+1)\cdots (a_l+1)$$

  is odd, therefore every factor $a_i+1$ is odd, thus $a_i$ is even for all $i$. This shows that $n$ is a perfect square.

• Conversely, suppose that $n$ is a perfect square, where $n=p_1^{a_1}{p}_2^{a_2}\cdots p_l^{a_l}$. By the unique factorization theorem, every exponent $a_i$ is even, so $a_i=2b_i$, where $b_i$ is an integer, and $n=p_1^{2b_1}{p}_2^{2b_2}\cdots p_l^{2b_l}$. Then

  $$d(n)=(2b_1+1)(2a_2+1)\cdots (2a_l+1)$$

  is a product of odd numbers, so $d(n)$ is odd.

The number of divisors of $n$ is odd if and only if $n$ is a perfect square

(b)

Let $k$ be a positive integer.

• Suppose that ${\sigma }_k(n)$ is odd. We write $n=2^{\alpha }{p}_1^{a_1}\cdots p_l^{a_l}$ the canonical prime decomposition of $n$, where $\alpha \ge 0,a_i>0$ and the $p_i$ are odd prime numbers. By Problem 8,

  $${\sigma }_k(n)=\frac{2^{k(\alpha +1)}-1}{2^k-1}\underset{i=1}{\overset{l}{\prod }}\frac{p_i^{k(a_i+1)}-1}{p_i^k-1}.$$

  Since ${\sigma }_k(n)$ is odd, each factor is odd, so $\frac{p_i^{k(a_i+1)}-1}{p_i^k-1}$ is odd for every $i\in [[1,l]]$. Moreover

  $$\frac{p_i^{k(a_i+1)}-1}{p_i^k-1}=\underset{j=0}{\overset{a_i}{\sum }}{p}_i^{kj}=1+p_i^k+p_i^{2k}+\cdots +p_i^{k{a}_i}$$

  is the sum of $a_i+1$ odd numbers, therefore $a_i+1$ is odd, so $a_i=2b_i$ for some integers $b_i$. Then

  $$n=\{\begin{array}{cc}{(2^{\beta }{p}_1^{b_1}\cdots p_l^{b_l})}^2\hfill & \text{if }\alpha =2\beta \text{ is even,}\hfill \\ 2{(2^{\beta }{p}_1^{b_1}\cdots p_l^{b_l})}^2\hfill & \text{if }\alpha =2\beta +1\text{ is odd.}\hfill \end{array}$$

  So $n$ is a square or double a square.

• Conversely, suppose that $n=m^2$ or $n=2m^2$ for some integer $m$. We write $m=2^{\beta }{p}_1^{b_1}\cdots p_l^{b_l}$ the canonical prime decomposition of $m$, where the $p_i$ are odd primes. Then

  $$n=2^{\alpha }{p}_1^{2b_1}{p}_2^{2b_2}\cdots p_l^{2b_l},$$

  where $\alpha =2\beta +\eta$ and $\eta =0$ or $\eta =1$. The formula of Problem 8 gives

  $${\sigma }_k(n)=\frac{2^{k(\alpha +1)}-1}{2^k-1}\underset{i=1}{\overset{l}{\prod }}\frac{p_i^{k(2b_i+1)}-1}{p_i^k-1}$$

  Since $\alpha +1>0$ and $k>0$, $2^{k(\alpha +1)}-1$ is odd, and $2^k-1$ is odd, where $2^k-1\mid 2^{k(\alpha +1)}-1$, thus $\frac{2^{k(\alpha +1)}-1}{2^k-1}$ is an odd integer. Moreover for every $i\in [[1,l]]$,

  $$\frac{p_i^{k(2b_i+1)}-1}{p_i^k-1}=1+p_i^k+p_i^{2k}+\cdots +p_i^{2k{b}_i}$$

  is the sum of $2b_i+1$ odd integers, thus $\frac{p_i^{k(2b_i+1)}-1}{p_i^k-1}$ is odd for every $i$. This shows that ${\sigma }_k(n)$ is odd.

In conclusion, ${\sigma }_k(n)$ is odd if and only $n$ is a square or double a square.