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
\(\ds \) \(=\) \(\ds \map {\sigma_1} m {2^n - 1}\) 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.

$\blacksquare$


Comment

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 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.


Sources

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