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
$\tuple {m_1, m_2, m_3}$ are an amicable triplet if and only if:
- $\map {\sigma_1} {m_1} = \map {\sigma_1} {m_2} = \map {\sigma_1} {m_3} = m_1 + m_2 + m_3$
where $\sigma_1$ denotes the divisor sum function.
Proof
For $n \in \Z_{>0}$, 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:
\(\ds \map s {m_1}\) | \(=\) | \(\ds m_2 + m_3\) | ||||||||||||
\(\ds \map s {m_2}\) | \(=\) | \(\ds m_1 + m_3\) | ||||||||||||
\(\ds \map s {m_3}\) | \(=\) | \(\ds m_1 + m_2\) |
Then:
\(\ds \map {\sigma_1} {m_1} - m_1\) | \(=\) | \(\ds m_2 + m_3\) | Definition of Proper Divisor of Integer | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {\sigma_1} {m_1}\) | \(=\) | \(\ds m_1 + m_2 + m_3\) |
Similarly:
\(\ds \map {\sigma_1} {m_2} - m_2\) | \(=\) | \(\ds m_1 + m_3\) | Definition of Proper Divisor of Integer | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {\sigma_1} {m_2}\) | \(=\) | \(\ds m_1 + m_2 + m_3\) |
and:
\(\ds \map {\sigma_1} {m_3} - m_3\) | \(=\) | \(\ds m_1 + m_2\) | Definition of Proper Divisor of Integer | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {\sigma_1} {m_3}\) | \(=\) | \(\ds m_1 + m_2 + m_3\) |
Thus:
- $\map {\sigma_1} {m_1} = \map {\sigma_1} {m_2} = \map {\sigma_1} {m_3} = m_1 + m_2 + m_3$
The argument reverses.
$\blacksquare$