Sum of Unitary Divisors is Multiplicative

Theorem

Let $\map {\sigma^*} n$ denote the sum of unitary divisors of $n$.

Then the function:

$\ds \sigma^*: \Z_{>0} \to \Z_{>0}: \map {\sigma^*} n = \sum_{\substack d \mathop \divides n \\ d \mathop \perp \frac n d} d$

is multiplicative.

Proof

Let $a, b$ be coprime integers.

Because $a$ and $b$ have no common divisor, the divisors of $a b$ are integers of the form $a_i b_j$, where $a_i$ is a divisor of $a$ and $b_j$ is a divisor of $b$.

That is, any divisor $d$ of $a b$ is in the form:

$d = a_i b_j$

in a unique way, where $a_i \divides a$ and $b_j \divides b$.

First we show that:

$d$ is an unitary divisor of $a b$ if and only if $a_i, b_j$ are unitary divisors of $a, b$ respectively

In the forward implication we are given $d \perp \dfrac {a b} d$.

By Divisor of One of Coprime Numbers is Coprime to Other:

$a_i, b_j \perp \dfrac {a b} d$

By Divisor of One of Coprime Numbers is Coprime to Other again:

$a_i \perp \dfrac a {a_i} \land b_j \perp \dfrac b {b_j}$

In the backward implication we are given $a_i \perp \dfrac a {a_i} \land b_j \perp \dfrac b {b_j}$.

By Divisor of One of Coprime Numbers is Coprime to Other:

$a \perp b \implies \paren {a_i \perp b_j \land \dfrac a {a_i} \perp \dfrac b {b_j} }$

By Product of Coprime Pairs is Coprime:

$d = a_i b_j \perp \dfrac a {a_i} \dfrac b {b_j} = \dfrac {a b} d$

$\Box$

We can list the unitary divisors of $a$ and $b$ as

$1, a_1, a_2, \ldots, a$

and:

$1, b_1, b_2, \ldots, b$

and thus the sum of their unitary divisors are:

$\ds \map {\sigma^*} a = \sum_{i \mathop = 1}^r a_i$

$\ds \map {\sigma^*} b = \sum_{j \mathop = 1}^s b_j$

Consider all unitary divisors of $a b$ with the same $a_i$.

Their sum is:

\(\ds \sum_{j \mathop = 1}^s a_i b_j\)

\(=\)

\(\ds a_i \sum_{j \mathop = 1}^s b_j\)

\(\ds \)

\(=\)

\(\ds a_i \map {\sigma^*} b\)

Summing over all $a_i$:

\(\ds \map {\sigma^*} {a b}\)

\(=\)

\(\ds \sum_{i \mathop = 1}^r \paren {a_i \map {\sigma^*} b}\)

\(\ds \)

\(=\)

\(\ds \paren {\sum_{i \mathop = 1}^r a_i} \map {\sigma^*} b\)

\(\ds \)

\(=\)

\(\ds \map {\sigma^*} a \map {\sigma^*} b\)

$\blacksquare$