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

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 $3$.

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:

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:

But:

which is a contradiction.

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