Theorem of Even Perfect Numbers

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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.


Sufficient Condition

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.


Necessary Condition

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 $\sigma \left({a}\right) = 2 a$:

\(\displaystyle m 2^n\) \(=\) \(\displaystyle 2 a\)                    
\(\displaystyle \) \(=\) \(\displaystyle \sigma \left({a}\right)\)                    
\(\displaystyle \) \(=\) \(\displaystyle \sigma \left({m 2^{n-1} }\right)\)                    
\(\displaystyle \) \(=\) \(\displaystyle \sigma \left({m}\right) \sigma \left({2^{n-1} }\right)\)          Sigma Function is Multiplicative          
\(\displaystyle \) \(=\) \(\displaystyle \sigma \left({m}\right) \left({2^n - 1}\right)\)          Sigma of Power of Prime          


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

But $\sigma \left({m}\right)$ 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 $\sigma \left({m}\right)$ as:

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

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:

$\dfrac m {2^n - 1} = 1$

Hence the result.



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

Historical Note

The first part of this proof was documented by Euclid in The Elements: Book $\text{IX}$, Proposition $36$.

The second part was achieved by Euler.