theorem on multiples of abundant numbers

Theorem. The product $nm$ of an abundant number $n$ and any integer $m>0$ is also an abundant number, regardless of the abundance or deficiency of $m$.

Proof. Choose an abundant number $n$ with $k$ divisors $d_1,\mathrm{\dots },d_k$ (where the divisors are sorted in ascending order and $d_1=1$, $d_k=n$) that add up to $2n+a$, where $a>0$ is the abundance of $n$. For maximum flair, set $a=1$, the bare minimum for abundance (that is, a quasiperfect number). Next, for $m$ choose a spectacularly deficient number such that $\gcd (m,n)=1$, preferably some large prime number. If we choose a prime number, its divisors will only add up to $m+1$. However, the divisors of $nm$ will include each $d_{i}m$, where $d_i$ is a divisor of $n$ and . Therefore, the divisors of $nm$ will add up to

$$\sum _{i=1}^k{d}_i+\sum _{i=1}^k{d}_{i}m=2nm+a(m+1)+2n.$$

It now becomes obvious that by insisting that $m$ and $n$ be coprime we are guaranteeing that if $m$ is itself prime, it will bring at least $k$ new divisors to the table. But what if $\gcd (m,n)>1$, or in the most extreme case, $m=n$? In such a case, we just can’t use the same formula for the sum of divisors of $nm$ that we used when $m$ and $n$ were coprime, as that would count some divisors twice. However, $m=n$ still brings new divisors to the table, and those new divisors add up to

$$\sum _{i=2}^k{d}_i{d}_k=2n^2+2a^2+a.$$

Having proven these extreme cases, it is obvious that $nm$ will be abundant in other cases, such as $m$ being a composite deficient number, a perfect number, an abundant number sharing some but not all prime factors with $n$, etc.

Title	theorem on multiples of abundant numbers
Canonical name	TheoremOnMultiplesOfAbundantNumbers
Date of creation	2013-03-22 16:05:46
Last modified on	2013-03-22 16:05:46
Owner	CompositeFan (12809)
Last modified by	CompositeFan (12809)
Numerical id	15
Author	CompositeFan (12809)
Entry type	Theorem
Classification	msc 11A05
Related topic	APositiveMultipleOfAnAbundantNumberIsAbundant