theorem on multiples of abundant numbers
Proof. Choose an abundant number with divisors (where the divisors are sorted in ascending order and , ) that add up to , where is the abundance of . For maximum flair, set , the bare minimum for abundance (that is, a quasiperfect number). Next, for choose a spectacularly deficient number such that , preferably some large prime number. If we choose a prime number, its divisors will only add up to . However, the divisors of will include each , where is a divisor of and . Therefore, the divisors of will add up to
It now becomes obvious that by insisting that and be coprime we are guaranteeing that if is itself prime, it will bring at least new divisors to the table. But what if , or in the most extreme case, ? In such a case, we just can’t use the same formula for the sum of divisors of that we used when and were coprime, as that would count some divisors twice. However, still brings new divisors to the table, and those new divisors add up to
Having proven these extreme cases, it is obvious that will be abundant in other cases, such as being a composite deficient number, a perfect number, an abundant number sharing some but not all prime factors with , etc.
|Title||theorem on multiples of abundant numbers|
|Date of creation||2013-03-22 16:05:46|
|Last modified on||2013-03-22 16:05:46|
|Last modified by||CompositeFan (12809)|