# Theorem of Even Perfect Numbers/Sufficient Condition

## Theorem

Let $n \in \N$ be such that $2^n - 1$ is prime.

Then $2^{n - 1} \paren {2^n - 1}$ is perfect.

In the words of Euclid:

If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect.

## Proof

Suppose $2^n - 1$ is prime.

Let $a = 2^{n - 1} \paren {2^n - 1}$.

Then $n \ge 2$ which means $2^{n - 1}$ is even and hence so is $a = 2^{n - 1} \paren {2^n - 1}$.

Note that $2^n - 1$ is odd.

Since all divisors (except $1$) of $2^{n - 1}$ are even it follows that $2^{n - 1}$ and $2^n - 1$ are coprime.

Let $\map {\sigma_1} n$ be the divisor sum of $n$, that is, the sum of all divisors of $n$ (including $n$).

From Divisor Sum Function is Multiplicative, it follows that $\map {\sigma_1} a = \map {\sigma_1} {2^{n - 1} } \map {\sigma_1} {2^n - 1}$.

But as $2^n - 1$ is prime, $\map {\sigma_1} {2^n - 1} = 2^n$ from Divisor Sum of Prime Number.

Then we have that $\map {\sigma_1} {2^{n - 1} } = 2^n - 1$ from Divisor Sum of Power of Prime.

Hence it follows that $\map {\sigma_1} a = \paren {2^n - 1} 2^n = 2 a$.

Hence from the definition of perfect number it follows that $2^{n - 1} \paren {2^n - 1}$ is perfect.

$\blacksquare$

## Historical Note

This proof is Proposition $36$ of Book $\text{IX}$ of Euclid's The Elements.