Odd Perfect Number has at least Three Distinct Prime Factors

Theorem

Let $n \in \N$ be an odd perfect number.

Then $n$ has at least $3$ distinct prime factors.

Aiming for a contradiction, suppose the contrary: that $n$ is a perfect number with at most $2$ distinct prime factors.

Hence let $n = p^a q^b$ where $p$ and $q$ are distinct primes and where $a, b \in \N_{>0}$.

As $n$ is odd, both $p$ and $q$ are also odd.

By definition of perfect number:

$\map {\sigma_1} n = 2 n$

where $\map {\sigma_1} n$ denotes the divisor sum of $n$.

Hence:

$\ds \map {\sigma_1} n$

$=$

$\ds 2 n$

Definition of Perfect Number

$\ds \leadsto \ \ $

$\ds \dfrac {\map {\sigma_1} n} n$

$=$

$\ds 2$

$\ds \leadsto \ \ $

$\ds \map h n$

$=$

$\ds 2$

where $\map h n$ denotes the abundancy index of $n$

$\ds $

$<$

$\ds \dfrac p {p - 1} \dfrac q {q - 1}$

$\ds $

$<$

$\ds \dfrac 3 2 \dfrac 5 4$

as $\dfrac x {x - 1}$ is decreasing on $x > 1$, and $p$ and $q$ are distinct odd primes

$\ds \leadsto \ \ $

$\ds 2$

$<$

$\ds \dfrac {15} 8$

contradicting $2 > \dfrac {15} 8$

$\blacksquare$