# Euclid's "perfect numbers theorem"

• Oct 1st 2010, 01:48 AM
courteous
Euclid's "perfect numbers theorem"
Quote:

Euclid’s theorem about perfect numbers* depends on the prime divisor property, which will be proved in the next exercise.
Assuming this for the moment, it follows that if $2^n-1$ is a prime $p$,
then the proper divisors of $2^{n-1}p$ are $1,2,2^2,...,2^{n-1}$ and $p,2p,2^2p...,2^{n-2}p$.

Given that the divisors of $2^{n-1}p$ are those just listed, show that $2^{n-1}p$ is perfect when $p=2^n-1$ is prime.
I've done that (just summing all the listed proper divisors and checking $\sum divisors=2^{n-1}p$).

*
That if $2^n-1$ is prime, then
$2^{n-1}(2^n-1)$ is perfect.

Quote:

We can now fill the gap in the proof of Euclid’s theorem on perfect numbers* (above exercise), using the prime divisor property.

Use the prime divisor property* to show that the proper divisors of $2^{n-1}p$, for any odd prime $p$, are $1,2,2^2,...,2^{n-1}$ and $p,2p,2^2p...,2^{n-2}p$.
Help with this one.(Worried)

* If $p$ is a prime that divides $ab$, then $p$ divides $a$ or $b$.