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.

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 Sigma 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 Sigma Function of Prime Number.

Then we have that $\map {\sigma_1} {2^{n - 1} } = 2^n - 1$ from Sigma Function 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.