# 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 $3$.

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.

Therefore $m_2 > m_1$.

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

 $\displaystyle \frac {\map \sigma {m_1} } {m_1}$ $\ge$ $\displaystyle \frac {\map \sigma {12} } {12}$ 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 $3$.

$\blacksquare$