Question

Prove: If k is a positive integer and 2^k -1 is prime then 2^(k-1)(2^k -1) is perfect.
Note: An integer n is perfect if the sum of all of its positive divisors (including 1 and itself ) is 2n.
Prove: If k is a positive integer and 2^k -1 is prime then 2^(k-1)(2^k -1) is perfect.
Note: An integer n is perfect if the sum of all of its positive divisors (including 1 and itself ) is 2n.
@Mathematics

Answer 1

Hint: \[\sum_{i=0}^{k-1} 2^i = 2^k -1\]
use geometric series

Answer 2

Let me start by writing down the positive divisors of the number \(n=2^{k-1}(2^k-1)\). I will write them into two groups for simplification:
The first group contains these divisors: \(1, 2, 2^2,.., 2^{k-1}\).
The summation of these numbers is \(1+2+2^2+..+2^{k-1}=\sum_{n=0}^{k-1} 2^n=2^k-1\).
The product of each one of the divisors in the first group with the prime number \(2^k-1\) make our second group of divisors. That's \(2^k-1,2(2^k-1),.., 2^{k-1}(2^k-1)\). The summation of these numbers can be written as \((2^k-1)(1+2+..+2^{k-1})=(2^k-1)\sum_{n=0}^{k-1} 2^k=(2^k-1)(2^k-1)\).
Add the summation of the divisors in each group: \((2^k-1)(2^k-1)+(2^k-1)=(2^k-1)(2^k-1+1)=2^k(2^k-1)\)
\(=2(2^{k-1})(2^k-1)=2n.\)

Answer 3

tyvm

Answer 4

You're welcome!