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Even Sum Of Two Abundant Numbers

Problem

The divisors of a positive integer, excluding the number itself, are called the proper divisors . If the sum of proper divisors is equal to the number we call the number perfect. For example, the divisors of $28$ are $1, 2, 4, 7, 14,$ and $28,$ so the sum of proper divisors is $1 + 2 + 4 + 7 + 14 = 28$.

Similarly, if the sum of the proper divisors exceeds the number we call the number abundant. For example, $12$ is abundant because the divisors of $12$ are $1, 2, 3, 4, 6, 12,$ and the sum of proper divisors, $1 + 2 + 3 + 4 + 6 = 14 \gt 12$.

Prove that every even $n \ge 48$ can be written as the sum of two abundant numbers.


Solution

Let $S(n)$ represent the sum of proper divisors of $n$.

Suppose that $S(n) = c \ge n$; in other words, $n$ is perfect or abundant.

Let us consider $mn$ where $m \ge 2$.

If $d_{1}, d_{2}, ... , d_{i}$ represents the complete list of proper divisors of $n$ then $mn$ will be divisible by $md_{1}, md_{2}, ... , md_{i}$. But this list does not represent the complete set of proper divisors of $mn$ because $m \ge 2$ and so it will be missing $1$ and possibly more factors.

Therefore $S(mn) = m(d_{1} + d_{2} + ... + d_{i}) + q = mc + q,$ where $q \ge 1$.

But if $S(n) = c \ge n$ then $mc \ge mn$.

Hence $S(mn) = mc + q \gt mn$ and we prove that $mn$ will be abundant.

In other words, the multiple of any perfect or abundant number will be abundant.

Now $S(6) = 1 + 2 + 3,$ so $6$ is perfect and consequently for $k \ge 2$ all numbers of the form $6k = \{12, 18, 24, ...\}$ will be abundant.

It follows then that all numbers of the form $6k + 12 = \{24, 30, 36, 42, 48, ...\}$ can be written as the sum of two abundant numbers.

As $S(20) = 1 + 2 + 4 + 5 + 10 = 22 \gt 20$ then $20$ is abundant.

Therefore all numbers of the form $6k + 20 = \{32, 38, 44, 50, ...\}$ will be the sum of two abundant numbers.

Similarly, because $20$ is abundant then $40$ will also be abundant, and so all numbers of the form $6k + 40 = \{52, 58, 64, ...\}$ will be the sum of two abundant numbers.

As $6k + 12 \equiv 0 \pmod{6}, 6k + 20 \equiv 2 \pmod{6},$ and $6k + 40 \equiv 4 \pmod{6}$ then we have proved that all even numbers, $n \ge 48,$ can be written as the sum of two abundant numbers.

Problem ID: 347 (08 Nov 2008)     Difficulty: 4 Star

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