The third largest prime divisor of an odd perfect number exceeds one hundred

Let $\sigma(n)$ denote the sum of positive divisors of the natural number $n$. Such a number is said to be perfect if $\sigma(n)=2n$. 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.