## Even Perfect Numbers

#### Problem

The divisors of a positive integer, excluding the number itself, are called the proper divisors . If the sum of proper divisors is equal to the number we call the number perfect. For example, the divisors of 28 are 1, 2, 4, 7, 14, and 28, so the sum of proper divisors is 1 + 2 + 4 + 7 + 14 = 28.

The first eight perfect numbers are 6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128.

Prove that P is an even perfect number iff it is of the form $2^{n-1}(2^n - 1)$ where $2^n - 1$ is prime.

#### Solution

Throughout this proof we shall make use of the sum of divisor function, $\sigma(n)$. As this function sums ALL the divisors, including $P$ itself, then $P$ will be perfect iff $\sigma(P) = 2P$.

Given that $P = 2^{n-1}(2^n - 1)$, we shall begin by showing that if $2^n - 1$ is prime then $P$ will be perfect.

$\sigma(P) = \sigma \left(2^{n-1}(2^n - 1)\right) = \sigma(2^{n-1}) \times \sigma(2^n - 1)$ due to the multiplicative property of the function.

Now $\sigma(2^{n-1}) = 1 + 2 + 4 + 8 + ... + 2^{n-1} = 2^n - 1$.

As $2^n - 1$ is prime, $\sigma(2^n - 1) = 1 + 2^n - 1 = 2^n$.

Therefore $\sigma(P) = 2^n(2^n - 1) = 2 \times 2^{n-1}(2^n - 1) = 2P$, and we prove that $P$ is perfect.

We shall now prove the converse; that is, if $P$ is an even perfect number then it must be of the form $P = 2^{n-1}(2^n - 1)$.

Suppose that $P$ is an even perfect number. All evens can be written in the form, $q \times 2^{n-1}$, where $q$ is odd and $n \ge 2$.

Therefore, $\sigma(P) = \sigma(q \times 2^{n-1}) = \sigma(q) \times \sigma(2^{n-1}) = \sigma(q) \times (2^n - 1)$.

But as $P$ is perfect, $\sigma(P) = 2P = q \times 2^n = \sigma(q) \times (2^n - 1) \implies \sigma(q) = \dfrac{q \times 2^n}{2^n - 1}$.

Because $\sigma(q)$ is integer, $q$ must be divisible by $2^n - 1$, so let $q = r(2^n - 1)$.

$\therefore \sigma(q) = \dfrac{q \times 2^n}{2^n - 1} = r \times 2^n$.

As $q$ is divisible by $r$, we know that its sum of divisors will at least be $q + r$.

Therefore, $\sigma(q) = r \times 2^n \ge q + r = r(2^n - 1) + r = r(2^n - 1 + 1) = r \times 2^n \implies \sigma(q) = q + r$.

As $q$ and $r$ are the only divisors, $q$ must be prime and $r$ must be 1.

And as $q + r = r \times 2^n$, $q + 1 = 2^n$, hence $q$ will be a prime of the form $2^n - 1$.

And so, if $P$ is an even perfect number then it will be of the form $2^{n-1}(2^n - 1)$. **Q.E.D.**

Note that no one has yet proved that an odd perfect number exists, but Euler proved that if any do exist they will be of the form $q^a b^2$, where $q$ is a prime of the form $4k + 1$. See Odd Perfect Numbers.