Exercise 4.2.15 (If $(a,b) > 1$, then $\sigma_k(ab) < \sigma_k(a) \sigma_k(b)$)

Proof. Here $k\in \mathbb{N}=\{0,1,2,\dots ,\}$. We prove first the property for $a=p^{\alpha },b=p^{\beta }$, where $p$ is a prime number.

If $a\wedge b>1$, then $\alpha >0$ and $\beta >0$.

• If $k=0$, then

  $$\begin{array}{llll}\hfill {\sigma }_0(\mathit{ab})<{\sigma }_0(a){\sigma }_0(b)& ⟺d(\mathit{ab})<d(a)d(b)& \hfill & \\ \hfill & ⟺d(p^{\alpha +\beta })<d(p^{\alpha })d(p^{\beta })& \hfill & \\ \hfill & ⟺\alpha +\beta +1<(\alpha +1)(\beta +1)& \hfill & \\ \hfill & ⟺0<\mathit{\alpha \beta }.& \hfill & \end{array}$$

  Since $\alpha >0,\beta >0$, $0<\mathit{\alpha \beta }$ is true, so

  $${\sigma }_0(\mathit{ab})<{\sigma }_0(a){\sigma }_0(b)(a=p^{\alpha },b=p^{\beta },\alpha >0,\beta >0).$$

• If $k>0$, then, using the formula of Problem 8,

  $$\begin{array}{llllll}\hfill & {\sigma }_k(\mathit{ab})<{\sigma }_k(a){\sigma }_k(b)& \hfill & & \hfill & \\ \hfill & ⟺{\sigma }_k(p^{\alpha +\beta })<{\sigma }_k(p^{\alpha }){\sigma }_k(p^{\beta })& \hfill & & \hfill & \\ \hfill & ⟺\frac{p^{k(\alpha +\beta +1)}-1}{p^k-1}<\frac{p^{k(\alpha +1)}-1}{p^k-1}\frac{p^{k(\beta +1)}-1}{p^k-1}& \hfill & & \hfill & & \hfill \\ \hfill & ⟺(p^{k(\alpha +\beta +1)}-1)(p^k-1)<(p^{k(\alpha +1)}-1)(p^{k(\beta +1)}-1)& \hfill & & \hfill & \\ \hfill & ⟺p^{k(\alpha +\beta +2)}-p^{k(a+\beta +1)}-p^k+1<p^{k(\alpha +\beta +2)}-p^{k(\alpha +1)}-p^{k(\beta +1)}+1& \hfill & & \hfill & \\ \hfill & ⟺p^{k(\alpha +1)}+p^{k(\beta +1)}<p^{k(\alpha +\beta +1)}& \hfill & & \hfill & \\ \hfill & ⟺p^{\mathit{k\alpha }}+p^{\mathit{k\beta }}<p^{\mathit{k\alpha }}{p}^{\mathit{k\beta }}& \hfill & & \hfill & \\ \hfill & ⟺(p^{\mathit{k\alpha }}-1)(p^{\mathit{k\beta }}-1)>1.& \hfill & & \hfill & \end{array}$$

  Since $p>1$, $k>0$ and $\alpha >0,\beta >0$, we have $p^{\mathit{k\alpha }}>1,p^{\mathit{k\beta }}>1$, so $(p^{\mathit{k\alpha }}-1)(p^{\mathit{k\beta }}-1)>1$. This shows that

  $${\sigma }_k(\mathit{ab})<{\sigma }_k(a){\sigma }_k(b)(a=p^{\alpha },b=p^{\beta },\alpha >0,\beta >0,k\ge 1).$$

In conclusion, for all $k\in \mathbb{N}$,

We know that if $a\wedge b=1$ (that is, if $\alpha =0$ or $\beta =0$), then ${\sigma }_k(\mathit{ab})={\sigma }_k(a){\sigma }_k(b)$, so in all cases

Now we consider positive integers $a,b$, such that $a\wedge b>1$, and $k\ge 0$. We write

where $p_1,p_1,\dots ,p_l$ are the prime divisors of $a$ or $b$, and $a_i\ge 0,b_i\ge 0$.

Since $a\wedge b>1$, there is some index $i$ such that $a_i>0$ and $b_i>0$. After renumbering, we may suppose that $i=1$, so $a_1>0,b_1>0$. Then by inequalities (1) and (2),

Since ${\sigma }_k$ is multiplicative, by (3) and (4),

In particular, for $k=0$, $d(\mathit{ab})={\sigma }_0(\mathit{ab})<{\sigma }_0(a){\sigma }_0(b)=d(a)d(b)$.

If $a\wedge b>1$, then ${\sigma }_k(\mathit{ab})<{\sigma }_k(a){\sigma }_k(b)$ for $k=1,2,\dots$ and $d(\mathit{ab})<d(a)d(b)$. □