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 deficiencyMathworldPlanetmath of m.

Proof. Choose an abundant number n with k divisorsMathworldPlanetmathPlanetmath d1,,dk (where the divisors are sorted in ascending order and d1=1, dk=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 numberMathworldPlanetmath, its divisors will only add up to m+1. However, the divisors of nm will include each dim, where di is a divisor of n and 0<ik. Therefore, the divisors of nm will add up to


It now becomes obvious that by insisting that m and n be coprimeMathworldPlanetmath 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


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 factorsMathworldPlanetmath 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