Theorem of Even Perfect Numbers/Necessary Condition

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.

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

We can extract the highest power of $2$ out of $a$ that we can, and write $a$ in the form:
 * $a = m 2^{n-1}$

where $n \ge 2$ and $m$ is odd.

Since $a$ is perfect and therefore $\map \sigma a = 2 a$:

So:
 * $\map \sigma m = \dfrac {m 2^n} {2^n - 1}$

But $\map \sigma m$ is an integer and so $2^n - 1$ divides $m 2^n$.

From Consecutive Integers are Coprime, $2^n$ and $2^n - 1$ are coprime.

So from Euclid's Lemma $2^n - 1$ divides $m$.

Thus $\dfrac m {2^n - 1}$ divides $m$.

Since $2^n - 1 \ge 3$ it follows that:
 * $\dfrac m {2^n - 1} < m$

Now we can express $\map \sigma m$ as:
 * $\map \sigma m = \dfrac {m 2^n} {2^n - 1} = m + \dfrac m {2^n - 1}$

This means that the sum of all the divisors of $m$ is equal to $m$ itself plus one other divisor of $m$.

Hence $m$ must have exactly two divisors, so it must be prime by definition.

This means that the other divisor of $m$, apart from $m$ itself, must be $1$.

That is:
 * $\dfrac m {2^n - 1} = 1$

Hence the result.

Historical Note
This result was achieved by.