## 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, $1 + 2 + 3 + 4 + 6 = 14 \gt 12$.

By first showing that $315p$ is abundant for all primes, $p \le 103,$ prove that all integers greater than $28123$ can be written as the sum of two abundant numbers.

#### Solution

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

$S(315) = S(3^2 \times 5 \times 7) = 1 + 3 + 5 + 7 + 9 + 15 + 21 + 35 + 45 + 63 + 105 = 309$.

When considering $S(315p)$ we must deal with two cases.

Case 1: $p$ is coprime with $315$ ($p \ne 3, 5, 7$)

$\therefore S(315p) = p(1 + 3 + ... + 105) + (1 + 3 + ... + 105) + 315 = 309p + 624$

For $315p$ to be abundant, $315p \lt 309p + 624 \Rightarrow p \lt 104$.

Case 2: $p = 3, 5, 7$

$S(315p) = p(1 + 3 + ... + 105) + q,$ where $q$ is the sum of divisors not containing a factor of $3, 5,$ or $7$ respectively.

So for $315p$ to be abundant it is sufficient to show $315p \lt 309p + q \Rightarrow 6p \lt q$.

When $p = 3, 6p = 18, q = 1 + 5 + 7 + 35 = 48 \Rightarrow 6p \lt 48$
When $p = 5, 6p = 30, q = 1 + 3 + 7 + 9 + 21 + 63 = 104 \Rightarrow 6p \lt 104$
When $p = 7, 6p = 42, q = 1 + 3 + 5 + 9 + 15 + 45 = 48 \Rightarrow 6p \lt 78$

Hence $315p$ is abundant for all primes, $p \le 103$.

It can be shown that multiples of abundant numbers are also abundant (see Even Sum Of Two Abundant Numbers). Thus all values of $m$ from $2$ to $103$ will either be prime or contain a prime in that domain. So although $315$ is deficient, $315m$ is guaranteed to be abundant for $2 \le m \le 103$.

We now search for the smallest abundant number that is coprime with $315$. Considering numbers of the form $2^k \times 11$ we find that $88$ is the smallest such example. Therefore the expression $315a + 88b$ will be the sum of two abundant numbers for $2 \le a \le 103$ and $b \ge 1$.

Clearly the expression produces integers congruent with $315a \pmod{88},$ and although $0 le a le 87$ will produce all possible congruences we require $a \ge 2$ to ensure $315a$ is abundant; that is, $2 \le a \le 89$ will produce all possible congruences.

It is necessary to have at least one multiple of $88$, so $315 \times 89 + 88$ represents that last integer before the congruences repeat. Hence every integer $n \gt 315 \times 89 + 88 = 28123$ can be written in the form $315a + 88b,$ which will be the sum of two abundant numbers. Q.E.D.

Problem ID: 348 (08 Nov 2008)     Difficulty: 4 Star

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