Theorem of Even Perfect Numbers

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

Then $$a$$ is in the form $$2^{n-1} \left({2^n - 1}\right)$$ where $$2^n - 1$$ is prime.

Similarly, if $$2^n - 1$$ is prime, then $$2^{n-1} \left({2^n - 1}\right)$$ is perfect.

Proof

 * Suppose $$2^n - 1$$ is prime.

Let $$a = 2^{n-1} \left({2^n - 1}\right)$$.

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

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 $$\sigma \left({n}\right)$$ be the sigma function of $$n$$, that is, the sum of all divisors of $$n$$ (including $$n$$).

From Sigma Function is Multiplicative, it follows that $$\sigma \left({a}\right) = \sigma \left({2^{n-1}}\right)\sigma \left({2^n - 1}\right)$$.

But as $$2^n - 1$$ is prime, $$\sigma \left({2^n - 1}\right) = 2^n$$ from Sigma of Prime Number.

Then we have that $$\sigma \left({2^{n-1}}\right) = 2^n - 1$$ from Sigma of Power of Prime.

Hence it follows that $$\sigma \left({a}\right) = \left({2^n - 1}\right) 2^n = 2 a$$.

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


 * Now suppose $$a \in \N$$ is 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 $$\sigma \left({a}\right) = 2 a$$, we have:

$$ $$ $$ $$ $$

So $$\sigma \left({m}\right) = \frac {m 2^n} {2^n - 1}$$.

But $$\sigma \left({m}\right)$$ is an integer and so $$2^n-1$$ divides $$m 2^n$$.

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

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

Thus $$\frac {m} {2^n - 1}$$ divides $$m$$, and since $$2^n - 1 \ge 3$$ it follows that $$\frac {m} {2^n - 1} < m$$.

Now we can express $$\sigma \left({m}\right)$$ as:
 * $$\sigma \left({m}\right) = \frac {m 2^n} {2^n - 1} = m + \frac {m} {2^n - 1}$$

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, $$\frac {m} {2^n - 1} = 1$$.

Hence the result.

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

From Primes of the Form of a 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?

Historical Note
The first part of this proof was documented by Euclid in, book IX, Proposition 36.

The second part was achieved by Euler.