every even integer greater than 70 is the sum of two abundant numbers in more than one way

Theorem Every even $n>70$ can be expressed as $n=a+b$, with both $a$ and $b$ abundant numbers, in more than one way. Due to the commutative property of addition, swaps of $a$ and $b$ are not counted as separate ways.

Proof.

To prove this it is enough to find just two ways for each even $n>70$, though of course there are plenty more ways as the numbers get larger, purely for our convenience we’ll seek to choose the smallest values for $b$ possible. Since every multiple of a perfect number is an abundant number, and 6 is a perfect number, it follows that every multiple of 6 is abundant, and it is small enough a modulus that reviewing all possible cases should not prove tiresome.

If $n=6m$ and $m>5$, the two desired pairs are $a=6(m-2)$, $b=12$, and $a=6(m-3)$, $b=18$. This leaves us the cases $n=6m+2$ and $n=6m+4$ to concern ourselves with.

If $n\equiv 2mod6$ and $m>10$ then the pairs are are $a=6(m-3)$, $b=20$, and $a=6(m-9)$, $b=56$.

If $n\equiv 4mod6$ and $m>12$ then the pairs are $a=6(m-6)$, $b=40$, and $a=6(m-11)$, $b=70$.

The lower bounds of $m$ have been chosen to ensure the formulas give distinct pairs of abundant numbers and never the perfect number 6 itself, but its multiples. These values of $m$ correspond to the values of $n$ 36, 68, 82. To complete the proof we are left with the special case of $n=76$ to examine on its own. Ignoring the bounds for $m$, the formulas above give us 76 = 40 + 36, a valid pair, and 76 = 70 + 6, which is not a pair of abundant numbers. But there is one other pair, 56 + 20, of which neither $a$ nor $b$ is a multiple of 6.∎

The special case of 76 shows that there are solutions that don’t use multiples of 6. These become more readily available as the numbers get larger.