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 $\displaystyle 2^n-1$ is a __prime__ $\displaystyle p$,

then the proper divisors of $\displaystyle 2^{n-1}p$ are $\displaystyle 1,2,2^2,...,2^{n-1}$ and $\displaystyle p,2p,2^2p...,2^{n-2}p$.

**Given that the divisors of **$\displaystyle 2^{n-1}p$** are those just listed, show that **$\displaystyle 2^{n-1}p$** is **__perfect__ when $\displaystyle p=2^n-1$** is **__prime__.

I've done that (just summing all the listed proper divisors and checking $\displaystyle \sum divisors=2^{n-1}p$).

* That if $\displaystyle 2^n-1$ is *prime*, then $\displaystyle 2^{n-1}(2^n-1)$ 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 $\displaystyle 2^{n-1}p$, for any odd prime $\displaystyle p$, are $\displaystyle 1,2,2^2,...,2^{n-1}$ and $\displaystyle p,2p,2^2p...,2^{n-2}p$.**

Help with this one.(Worried)

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