Abstract. Let σ(n) denote the sum of positive divisors of the natural number n. Such a number is said to be perfect if σ(n) = 2n. It is well known that a number is even and perfect if and only if it has the form 2 p−1 (2 p − 1) where 2 p − 1 is prime.No odd perfect numbers are known, nor has any proof of their nonexistence ever been given. In the meantime, much work has been done in establishing conditions necessary for their existence. One class of necessary conditions would be lower bounds for the distinct prime divisors of an odd perfect number.For example, Cohen and Hagis have shown that the largest prime divisor of an odd perfect number must exceed 10 6 , and Hagis showed that the second largest must exceed 10 3 . In this paper, we improve the latter bound. In particular, we prove the statement in the title of this paper.
Abstract. Let σ(n) denote the sum of positive divisors of the natural number n. Such a number is said to be perfect if σ(n) = 2n. It is well known that a number is even and perfect if and only if it has the form 2 p−1 (2 p − 1) where 2 p − 1 is prime.It is unknown whether or not odd perfect numbers exist, although many conditions necessary for their existence have been found. For example, Cohen and Hagis have shown that the largest prime divisor of an odd perfect number must exceed 10 6 , and Iannucci showed that the second largest must exceed 10 4 . In this paper, we prove that the third largest prime divisor of an odd perfect number must exceed 100.
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