Question is not written in proper way.
Actual question is:
We call a positive integer perfect if it equals the sum of its positive divisors other than itself.
a) Prove that 6 and 28 are perfect.
b) Prove that 2p−1(2p−1) is a perfect number when 2p−1 is prime.
Solution:
(a) Let us determine all divisors of each number.
Divisors of 6=1,2,3,6
Divisors of 28=1,2,4,7,14,28
Let us determine the sum of all positive divisors other than the integer itself:
6=1+2+328=1+2+4+7+14
We then note that 6 and 28 are both perfect.
(b) Given: 2p−1 is prime
To prove: 2p−1(2p−1) is a perfect number
Poof:
Determine all divisors of 2p−1(2p−1) (using that (2p−1) is prime and thus cannot be factorized):
1,2,22,…,2p−1,(2p−1),2(2p−1),22(2p−1),….2p−2(2p−1),2p−1(2p−1)
Note: The last divisor is the number 2p−1(2p−1) itself.
Let us determine the sum of all positive divisors other than the integer itself:
1+2+22+…+2p−1+(2p−1)+2(2p−1)+22(2p−1)+…+2p−2(2p−1)=1(1+2p−1)+2(1+2p−1)+22(1+2p−1)+…+2p−2(1+2p−1)+2p−1=1(2p)+2(2p)+22(2p)+…+2p−2(2p)+2p−1=2p(1+2+22+…+2p−2)+2p−1
=2p(i=0∑p−22i)+2p−1=2p(2−12p−1−1)+2p−1=2p(2p−1−1)+2p−1=2p⋅2p−1−2p+2p−1=2p⋅2p−1−2⋅2p−1+1⋅2p−1=(2p−2+1)2p−1=(2p−1)2p−1=2p−1(2p−1)
By the definition of a perfect number, we have then shown that 2p−1(2p−1) is a perfect number.
Hence Proved
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