Equivalence of Definitions of Amicable Triplet

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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.

$\blacksquare$