So yeah have this question and am not sure how to even start

A positive integer n is called perfect if the sum of its divisors (excluding itself) is n. Define N(p) = (2^(p-1))(2^p-1).

Assume that 2^p-1 is a prime number. Show that N(p) is a perfect number.

You have to find all posible factors of (2^(p-1))(2^p-1) and add them up and you should get (2^(p-1))(2^p-1)

If 2^p-1 is prime that means that it is only divisible by 1 and itself

(2^(p-1)) cannot be prime because it is 2*2*2*2*.... p-1 times.

N(p) is divisible by 1,2,(2^(p-1)),(2^p-1)

Still thinking on how to proceed lol

by what brmv posted, if 2^p-1 is prime then so is p

This post was edited by zeroRooter on Mar 8 2014 08:57pm

2^(p-1) is divisible by exactly all 2^i for i=0,...,p-1 and sum(i-0,...,p-1)2^i = 2^p - 1

from here it should be easy or do you need more help?

Okay so in the end you end up with the sum of all divisors of N(p) = 2(2^p-1)

How do we show that this is a perfect number

don't forget that the two factors in 2^(p-1) * (2^p -1) are co-prime
the divisor sum D(ab) of two co-prime numbers is D(a)D(b)
now D(2^(p-1))=2^p - 1 and D(2^p-1)= 2^p -1 +1 = 2^p [since 2^p-1 is prime]

with D[2^(p-1) * (2^p-1)] = D(2^(p-1)) * D(2^p-1) = (2^p-1) * 2^p = (2^p-1) * (2^(p-1)) * 2

Makes sense! thank you for the help once again