Numbers in Even-Even Amicable Pair are not Divisible by 3

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Theorem

Let $\tuple {m_1, m_2}$ be an amicable pair such that both $m_1$ and $m_2$ are even.

Then neither $m_1$ nor $m_2$ is divisible by $3$.


Proof

An amicable pair must be formed from a smaller abundant number and a larger deficient number.

Suppose both $m_1, m_2$ are divisible by $3$.

Since both are even, they must also be divisible by $6$.

However $6$ is a perfect number.

By Multiple of Perfect Number is Abundant, neither can be deficient.

So $m_1, m_2$ cannot form an amicable pair.

Therefore at most one of them is divisible by $3$.


Without loss of generality suppose $m_1$ is divisible by $3$.

Write:

$m_1 = 2^r 3^t a, m_2 = 2^s b$

where $a, b$ are not divisible by $2$ or $3$.


Then:

\(\displaystyle m + n\) \(=\) \(\displaystyle \map \sigma {m_2}\) Definition of Amicable Pair
\(\displaystyle \) \(=\) \(\displaystyle \map \sigma {2^s} \map \sigma b\) Sigma Function is Multiplicative
\(\displaystyle \) \(=\) \(\displaystyle \paren {2^{s + 1} - 1} \map \sigma b\) Sigma Function of Power of Prime
\(\displaystyle \) \(\equiv\) \(\displaystyle \paren {\paren {-1}^{s + 1} - 1} \map \sigma b\) \(\displaystyle \pmod 3\) Congruence of Powers

Since $m + n$ is not divisible by $3$, $s$ must be even.

Similarly, by Sigma Function is Multiplicative, $t$ must also be even.

In particular, both $s, t$ are at least $2$.


Now write:

$m_1 = 2^2 \cdot 3 k, m_2 = 2^2 \cdot l$

where $k, l$ are some integers.


By Multiple of Perfect Number is Abundant, $m_1$ is abundant number.

Therefore $m_2 > m_1$.

This leads to $l > 3 k \ge 3$.


By Abundancy Index of Product is greater than Abundancy Index of Proper Factors:

\(\displaystyle \frac {\map \sigma {m_1} } {m_1}\) \(\ge\) \(\displaystyle \frac {\map \sigma {12} } {12}\) equality occurs if and only if $m_1 = 12$
\(\displaystyle \) \(=\) \(\displaystyle \frac 7 3\)
\(\displaystyle \frac {\map \sigma {m_2} } {m_2}\) \(>\) \(\displaystyle \frac {\map \sigma 4} 4\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 7 4\)

But:

\(\displaystyle 1\) \(=\) \(\displaystyle \frac {m_1 + m_2} {m_1 + m_2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {m_1} {\map \sigma {m_1} } + \frac {m_2} {\map \sigma {m_2} }\) Definition of Amicable Pair
\(\displaystyle \) \(<\) \(\displaystyle \frac 3 7 + \frac 4 7\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)

which is a contradiction.


Therefore neither $m_1$ nor $m_2$ is divisible by $3$.

$\blacksquare$


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