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.