Proposition: Even Perfect Numbers

The number $n:=\frac{p+1}2p=2^{k-1}(2^k-1)$ is an even perfect number, if and only if $p$ is a prime number having the form $p=2^k-1$ for an $k > 1, k\in\mathbb N.$

Examples

We calculate some different cases:

$k=2$:

$p=2^2-1=3$, $n=2^1(2^2-1)=2\cdot 3=6\Longleftrightarrow 6$ is a perfect number.

$k=3$

$p=2^3-1=7$, $n=2^2(2^3-1)=4\cdot 7=28\Longleftrightarrow 28$ is a perfect number.

$k=4$

$p=2^4-1=15$ is not a prime number and $n=2^3(2^4-1)=8\cdot 15=120$ is not(!) a perfect number.

$k > 1$, $k$ composite

In general, $2^k-1$ never can be prime, if $k$ is not prime.
Since if $k$ is composite, then $k=bc$ for some $b > 1$ and $c > 1$.
But then $$2^k-1=2^{bc}-1=(2^b-1)(2^{b(c-1)}+2^{b(c-2)}+\ldots+2^{b}+1),$$ and both factors are $ > 1.$
Therefore, the search for even perfect numbers should be restricted to prime numbers $k$.

$k=5$

$p=2^5-1=31$ $n=2^4(2^5-1)=16\cdot 31=496\Longleftrightarrow 496$ is a perfect number.

Other cases and notes

Further perfect numbers can be found for the prime numbers $k=7\Rightarrow n=8128$, $k=13\Rightarrow n=33550336,$ etc.

Although perfect numbers and the above elementary result have been known for at least 2500 years (see Prop. 9.36 in Euclid's Elements), still unsolved mathematical problems are: * Are there infinitely many even perfect numbers. * Existence of odd perfect numbers.

Proofs: 1

Proposition: Numbers Being the Product of Their Divisors

Propositions: 1

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References

Bibliography

Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927

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