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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 $\sigma(n)$ . 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^[1]

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.

↑ From Hardy and Wright, Introduction to the Theory of Numbers

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