Divisor Sum Function is Multiplicative/Proof 2

From ProofWiki
Jump to navigation Jump to search


The divisor sum function:

$\ds {\sigma_1}: \Z_{>0} \to \Z_{>0}: \map {\sigma_1} n = \sum_{d \mathop \divides n} d$

is multiplicative.


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 $b_j \divides b$.

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

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


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

and so their divisor sums are:

$\ds \map {\sigma_1} a = \sum_{i \mathop = 1}^r a_i$
$\ds \map {\sigma_1} b = \sum_{j \mathop = 1}^s b_j$

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

Their sum is:

\(\ds \sum_{j \mathop = 1}^s a_i b_j\) \(=\) \(\ds a_i \sum_{j \mathop = 1}^s b_j\)
\(\ds \) \(=\) \(\ds a_i \map {\sigma_1} b\)

Summing for all $a_i$:

\(\ds \map {\sigma_1} {a b}\) \(=\) \(\ds \sum_{i \mathop = 1}^r \paren {a_i \map {\sigma_1} b}\)
\(\ds \) \(=\) \(\ds \paren {\sum_{i \mathop = 1}^r a_i} \map {\sigma_1} b\)
\(\ds \) \(=\) \(\ds \map {\sigma_1} a \map {\sigma_1} b\)