An integer n is called perfect if it equals the sum of all its divisors d 1<d<n
Let a be a positive integer. Prove that if 2^a -1 is prime, the n= 2^(a-1) (2^a - 1) is perfect.
The smallest perfect number is a product of three integers. n obviously has two, but the third is 1, so you can try
$\displaystyle 2^{a-1} \cdot (2^a - 1) \cdot 1= 2^{a-1} + (2^a - 1) + 1 $
and solve for $\displaystyle a$. Substitute the number into 2^{a -1} and see if it's prime, and see if n is perfect.
All factors of $\displaystyle n $ are $\displaystyle \{2^i,\; 2^i\cdot(2^a-1) | 0\leq i<a\} $
So $\displaystyle \sigma(n)=\sum_{d\mid n} d = \sum_{i=0}^{a-1} (2^i+2^i\cdot(2^a-1)) = \sum_{i=0}^{a-1} 2^{i+a} = 2^a\sum_{i=0}^{a-1} 2^i = 2^a\cdot(2^a-1) = 2n $, where the last sum is just a geometric sum.
I have mistaken the vertical bar as a divisor. It took me a while before I realized that it's $\displaystyle \{2^i,\; 2^i\cdot(2^a-1): 0\leq i<a\} $
I did not know what $\displaystyle \sigma (n)$ meant until I read up on the Mersenne prime and Euclid's proof.
My book has devoted only one chapter on number theory, and I was expected to this proof without much instruction. I am glad you showed the proof.
I am an engineering student. I learn pure math on my own. Could you recommend me a book on number theory that's easy for my independent study.
Let $\displaystyle p=2^k-1 $ a prime number.
Let $\displaystyle n=2^{k-1}p$.
$\displaystyle gcd(2^{k-1},p)=1$
Now we compute $\displaystyle \sigma(n)$.
$\displaystyle \sigma(n)=\sigma(2^{k-1}p)=\sigma(2^{k-1})\sigma(p)=(2^k-1)(p+1)=(2^k-1)2^k=2n$
An exercise:
Prove the opposite way.