Exercise 2.13

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $\sigma_k(n) = \sum_{d\mid n} d^k$ . Show that $\sigma_k(n)$ is multiplicative and find a formula for it.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
As $n \mapsto n^k$ is multiplicative, then so is $\sigma_k$ (Ex. 2.10).

$\bullet$ Suppose that $k 
e 0$.

If $n = p^\alpha$ is a prime power ($\alpha \geq 1$),
\begin{align*}
\sigma_k(p^\alpha) &= \sum_{i=0}^{\alpha} p^{ik}\
&=\frac{ p^{(\alpha+1)k} -  1}{p^k-1}
\end{align*}

$\bullet$ if $k=0$, $\sigma_0(n)$ is the number of divisors of $n$.
\begin{align*}
\sigma_0(p^\alpha) &= \sum_{i=0}^{\alpha} 1\
&=\alpha+1
\end{align*}
Conclusion : if $n = p_1^{\alpha_1}\cdots p_t^{\alpha_t}$ is the decomposition of $n$ in prime factors, then
\begin{align*}
\sigma_0(n) &= (\alpha_1+1)\cdots(\alpha_t +1),\
\sigma_k(n) &= \prod_{i=0}^t \frac{p_i^{(\alpha_i+1)k}- 1}{p_i^k - 1}\ (k
e 0).\
\end{align*}
\end{proof}

Exercise 2.13

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Exercise 2.13

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