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.

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.


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