Perfect number/Advanced

Definition in terms of the sum-of-divisors function

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

$\sigma (n)=2n$.

Proof of the classification of even perfect numbers

Euclid showed that every number of the form

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

where $2^{n}-1$ 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:

• $\sigma (n)$ is a multiplicative function. In other words, if $a$ and $b$ are relatively prime positive integers, then $\sigma (ab)=\sigma (a)\sigma (b)$.

• If $p^{n}$ is a power of a prime number, then

$\sigma (p^{n})={\frac {p^{n+1}-1}{p-1}}$

The proof

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

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

Since $n$ is perfect,

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

and so

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

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

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

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

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

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

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

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

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

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