Equivalence of Definitions of Amicable Pair

Theorem

Let $m \in \Z_{>0}$ and $n \in \Z_{>0}$ be (strictly) positive integers.

The following definitions of the concept of Amicable Pair are equivalent:

Definition 1

$m$ and $n$ are an amicable pair if and only if:

the aliquot sum of $m$ is equal to $n$

and:

the aliquot sum of $n$ is equal to $m$.

Definition 2

$m$ and $n$ are an amicable pair if and only if:

$\map {\sigma_1} m = \map {\sigma_1} n = m + n$

where $\sigma_1$ denotes the divisor sum function.

Definition 3

$m$ and $n$ are an amicable pair if and only if they form a sociable chain of order $2$.

Proof

Definition 1 is equivalent to Definition 2

Let $\map s n$ denote the aliquot sum of (strictly) positive integer $n$.

The sum of all the divisors of a (strictly) positive integer $n$ is $\map {\sigma_1} n$, where $\sigma_1$ is the divisor sum function.

The aliquot sum of $n$ is the sum of the divisors of $n$ with $n$ excluded.

Thus:

$\map s n = \map {\sigma_1} n - n$

Suppose:

$\map s n = m$

and:

$\map s m = n$

Then:

 $\ds \map {\sigma_1} n - n$ $=$ $\ds m$ Definition of Proper Divisor of Integer $\ds \leadsto \ \$ $\ds \map {\sigma_1} n$ $=$ $\ds m + n$

Similarly:

 $\ds \map {\sigma_1} m - m$ $=$ $\ds n$ Definition of Proper Divisor of Integer $\ds \leadsto \ \$ $\ds \map {\sigma_1} m$ $=$ $\ds m + n$

Thus:

$\map s n = \map s m = m + n$

The argument reverses.

$\Box$

Definition 1 is equivalent to Definition 3

From the definition of definition 1 of an amicable pair:

$m$ and $n$ are an amicable pair if and only if:

the aliquot sum of $m$ is equal to $n$

and:

the aliquot sum of $n$ is equal to $m$.

From the definition of a sociable chain:

Let $m$ be a positive integer.

Let $\map s m$ be the aliquot sum of $m$.

Define the sequence $\sequence {a_k}$ recursively as:

$a_{k + 1} = \begin{cases} m & : k = 0 \\ \map s {a_k} & : k > 0 \end{cases}$

A sociable chain is such a sequence $\sequence {a_k}$ where:

$a_r = a_0$

for some $r > 0$.

Here it is seen that setting $r = 2$ gives that:

$\map s {a_0} = a_1$
$\map s {a_1} = a_0$

and the equivalence follows.

$\blacksquare$