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.

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.

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

The second part was achieved by Euler.