# Equivalence of Definitions of Amicable Triplet

## Theorem

Let $m_1, m_2, m_3 \in \Z_{>0}$ be (strictly) positive integers.

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

### Definition 1

$\tuple {m_1, m_2, m_3}$ are an amicable triplet if and only if the aliquot sum of any one of them equals the sum of the other two:

the aliquot sum of $m_1$ is equal to $m_2 + m_3$

and:

the aliquot sum of $m_2$ is equal to $m_1 + m_3$

and:

the aliquot sum of $m_3$ is equal to $m_1 + m_2$

### Definition 2

$\left({m_1, m_2, m_3}\right)$ are an amicable triplet if and only if:

$\sigma \left({m_1}\right) = \sigma \left({m_2}\right) = \sigma \left({m_3}\right) = m_1 + m_2 + m_3$

where $\sigma \left({m}\right)$ denotes the $\sigma$ function.

## Proof

For $n \in \Z_{>0}$, let $s \left({n}\right)$ denote the aliquot sum of (strictly) positive integer $n$.

The sum of all the divisors of a (strictly) positive integer $n$ is $\sigma \left({n}\right)$, where $\sigma$ is the $\sigma$ function.

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

Thus:

$s \left({n}\right) = \sigma \left({n}\right) - n$

Suppose:

 $\displaystyle s \left({m_1}\right)$ $=$ $\displaystyle m_2 + m_3$ $\displaystyle s \left({m_2}\right)$ $=$ $\displaystyle m_1 + m_3$ $\displaystyle s \left({m_3}\right)$ $=$ $\displaystyle m_1 + m_2$

Then:

 $\displaystyle \sigma \left({m_1}\right) - m_1$ $=$ $\displaystyle m_2 + m_3$ Definition of Proper Divisor of Integer $\displaystyle \leadsto \ \$ $\displaystyle \sigma \left({m_1}\right)$ $=$ $\displaystyle m_1 + m_2 + m_3$

Similarly:

 $\displaystyle \sigma \left({m_2}\right) - m_2$ $=$ $\displaystyle m_1 + m_3$ Definition of Proper Divisor of Integer $\displaystyle \leadsto \ \$ $\displaystyle \sigma \left({m_2}\right)$ $=$ $\displaystyle m_1 + m_2 + m_3$

and:

 $\displaystyle \sigma \left({m_3}\right) - m_3$ $=$ $\displaystyle m_1 + m_2$ Definition of Proper Divisor of Integer $\displaystyle \leadsto \ \$ $\displaystyle \sigma \left({m_3}\right)$ $=$ $\displaystyle m_1 + m_2 + m_3$

Thus:

$\sigma \left({m_1}\right) = \sigma \left({m_2}\right) = \sigma \left({m_3}\right) = m_1 + m_2 + m_3$

The argument reverses.

$\blacksquare$