Theorem of Even Perfect Numbers

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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.


Proof

Sufficient Condition

Suppose $2^n - 1$ is prime.

Let $a = 2^{n - 1} \paren {2^n - 1}$.

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

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 $\map {\sigma_1} n$ be the divisor sum of $n$, that is, the sum of all divisors of $n$ (including $n$).

From Divisor Sum Function is Multiplicative, it follows that $\map {\sigma_1} a = \map {\sigma_1} {2^{n - 1} } \map {\sigma_1} {2^n - 1}$.

But as $2^n - 1$ is prime, $\map {\sigma_1} {2^n - 1} = 2^n$ from Divisor Sum of Prime Number.

Then we have that $\map {\sigma_1} {2^{n - 1} } = 2^n - 1$ from Divisor Sum of Power of Prime.

Hence it follows that $\map {\sigma_1} a = \paren {2^n - 1} 2^n = 2 a$.

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

$\Box$


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 $\map {\sigma_1} a = 2 a$:

\(\ds m 2^n\) \(=\) \(\ds 2 a\)
\(\ds \) \(=\) \(\ds \map {\sigma_1} a\)
\(\ds \) \(=\) \(\ds \map {\sigma_1} {m 2^{n - 1} }\)
\(\ds \) \(=\) \(\ds \map {\sigma_1} m \map {\sigma_1} {2^{n - 1} }\) Divisor Sum Function is Multiplicative
\(\ds \) \(=\) \(\ds \map {\sigma_1} m \paren {2^n - 1}\) Divisor Sum of Power of Prime

So:

$\map {\sigma_1} m = \dfrac {m 2^n} {2^n - 1}$

But $\map {\sigma_1} m$ 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 $\map {\sigma_1} m$ as:

$\map {\sigma_1} m = \dfrac {m 2^n} {2^n - 1} = m + \dfrac m {2^n - 1}$

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.

$\blacksquare$


Also see


Historical Note

The first part of this proof of the Theorem of Even Perfect Numbers was documented by Euclid in The Elements: Proposition $36$ of Book $\text{IX} $: Theorem of Even Perfect Numbers/Sufficient Condition.

The second part was achieved by Euler.


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.


Sources

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