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单词 TheoremOnMultiplesOfAbundantNumbers
释义

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

i=1kdi+i=1kdim=2nm+a(m+1)+2n.

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

i=2kdidk=2n2+2a2+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 factorsMathworldPlanetmath with n, etc.

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更新时间:2025/5/4 22:33:02