Sieve methods for odd perfect numbers (Q2894533)

A positive integer \(N\) is perfect if \(\sigma(N)= 2N\). No odd perfect number has been found, and a well-known open problem is to show that there are none. Various conditions on the number or size of the prime factors of an odd perfect number \(N\) (if one exists) have been established. In [On the divisibility of odd perfect numbers by a high power of a prime, \url{http://arxiv.org/abs/math/0511410}] \textit{T. Yamada} obtained an upper bound for the smallest prime factor of \(N\) if certain conditions hold. The current paper establishes an improvement of this result. The prime factor \(\pi\) of \(N\) is called special if \(N=\pi^\alpha m^2\) with \(\pi\nmid m\) and \(\pi\equiv\alpha\equiv 1\pmod 4\). Let \(N\) be an odd perfect number such that if \(p^\alpha\parallel N\) with \(p\) not the special prime then either \(3|(\alpha+ 1)\) or \(5|(\alpha+ 1)\).NEWLINENEWLINE The main theorem of this paper asserts that the smallest prime factor \(p\) of \(N\) satisfies \(10^8< p< 10^{1000}\). The authors first prove a more general result that extends Yamada's first theorem and leads ultimately by applying the large sieve and results from prime number theory to the upper bound for \(p\). The proof of the lower bound involves a lot of computation the full print-outs of which are available on the second author's website. Assuming that \(N\) is not divisible by the sixth: power of a prime, the idea is to compute factor chains of prime divisors of \(N\) showing that if \(q\) is a prime dividing \(N\) with \(7\leq q< 10^8\) then a smaller prime divides \(N\) and then to show, that \(3\nmid N\), \(5\nmid N\).