Main Article
Discussion
Related Articles  [?]
Bibliography  [?]
Citable Version  [?]

An advanced level version of Perfect number.

## Definition in terms of the sum-of-divisors function

Perfect numbers can be succinctly defined using the sum-of-divisors function ${\displaystyle \sigma (n)}$. If ${\displaystyle n}$ is a counting number, then ${\displaystyle \sigma (n)}$ is the sum of the divisors of ${\displaystyle n}$. A number ${\displaystyle n}$ is perfect precisely when

${\displaystyle \sigma (n)=2n}$.

## Proof of the classification of even perfect numbers

Euclid showed that every number of the form

${\displaystyle 2^{n-1}(2^{n}-1)}$

where ${\displaystyle 2^{n}-1}$ is a Mersenne prime number is perfect. A short proof that every even perfect number must have this form can be given using elementary number theory.

The main prerequisite results from elementary number theory, besides a general familiarity with divisibility, are the following:

• ${\displaystyle \sigma (n)}$ is a multiplicative function. In other words, if ${\displaystyle a}$ and ${\displaystyle b}$ are relatively prime positive integers, then ${\displaystyle \sigma (ab)=\sigma (a)\sigma (b)}$.
• If ${\displaystyle p^{n}}$ is a power of a prime number, then
${\displaystyle \sigma (p^{n})={\frac {p^{n+1}-1}{p-1}}}$

### The proof[1]

Let ${\displaystyle n}$ be an even perfect number, and write ${\displaystyle n=2^{r}b}$ where ${\displaystyle n>0}$ and ${\displaystyle b}$ is odd. As ${\displaystyle \sigma (n)}$ is multiplicative,

${\displaystyle \sigma (n)=\sigma (2^{r})\sigma (b)=(2^{r+1}-1)\sigma (b)}$.

Since ${\displaystyle n}$ is perfect,

${\displaystyle \sigma (n)=2n=2^{r+1}b}$,

and so

${\displaystyle {\frac {b}{\sigma (b)}}={\frac {2^{r+1}-1}{2^{r+1}}}}$.

The fraction on the right side is in lowest terms, and therefore there is an integer ${\displaystyle c}$ so that

${\displaystyle b=\left(2^{r+1}-1\right)c\,{\text{ and }}\,\sigma (b)=2^{r+1}c}$.

If ${\displaystyle c>1}$, then ${\displaystyle b}$ has at least the divisors ${\displaystyle b}$, ${\displaystyle c}$, and 1, so that

${\displaystyle \sigma (b)\geq b+c+1=2^{r+1}c+1>2^{r+1}c=\sigma (b)}$,

a contradiction. Hence, ${\displaystyle c=1}$, ${\displaystyle n=2^{r}\left(2^{r+1}-1\right)}$, and

${\displaystyle \sigma \left(2^{r+1}-1\right)=2^{r+1}.}$

If ${\displaystyle 2^{r+1}-1}$ is not prime, then it has divisors other than itself and 1, and

${\displaystyle \sigma \left(2^{r+1}-1\right)>2^{r+1}}$.

Hence, ${\displaystyle 2^{r+1}-1}$ is prime, and the theorem is proved.

1. From Hardy and Wright, Introduction to the Theory of Numbers