Theorem of Even Perfect Numbers

Theorem
Let $a \in \N$ be an even perfect number.

Then $a$ is in the form:
 * $2^{n - 1} \paren {2^n - 1}$

where $2^n - 1$ is prime.

Similarly, if $2^n - 1$ is prime, then $2^{n - 1} \paren {2^n - 1}$ is perfect.

Comment
Hence, the hunt for even perfect numbers reduces to the hunt for prime numbers of the form $2^n - 1$.

From Primes of form Power Less One, we see that for $2^n - 1$ to be prime, $n$ itself must be prime.

See Mersenne prime.

Also see

 * Are All Perfect Numbers Even?