1. ## Perfect Integers

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.

2. Originally Posted by teramaries
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

$2^{a-1} \cdot (2^a - 1) \cdot 1= 2^{a-1} + (2^a - 1) + 1$

and solve for $a$. Substitute the number into 2^{a -1} and see if it's prime, and see if n is perfect.

3. Originally Posted by teramaries
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.
All factors of $n$ are $\{2^i,\; 2^i\cdot(2^a-1) | 0\leq i

So $\sigma(n)=\sum_{d\mid n} d = \sum_{i=0}^{a-1} (2^i+2^i\cdot(2^a-1)) = \sum_{i=0}^{a-1} 2^{i+a} = 2^a\sum_{i=0}^{a-1} 2^i = 2^a\cdot(2^a-1) = 2n$, where the last sum is just a geometric sum.

4. Originally Posted by chiph588@
All factors of $n$ are $\{2^i,\; 2^i\cdot(2^a-1) | 0\leq i
I have mistaken the vertical bar as a divisor. It took me a while before I realized that it's $\{2^i,\; 2^i\cdot(2^a-1): 0\leq i
I did not know what $\sigma (n)$ 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.

5. Let $p=2^k-1$ a prime number.

Let $n=2^{k-1}p$.

$gcd(2^{k-1},p)=1$

Now we compute $\sigma(n)$.

$\sigma(n)=\sigma(2^{k-1}p)=\sigma(2^{k-1})\sigma(p)=(2^k-1)(p+1)=(2^k-1)2^k=2n$

An exercise:

Prove the opposite way.

6. Originally Posted by Also sprach Zarathustra
Let $p=2^k-1$ a prime number.

Let $n=2^{k-1}p$.

$gcd(2^{k-1},p)=1$

Now we compute $\sigma(n)$.

$\sigma(n)=\sigma(2^{k-1}p)=\sigma(2^{k-1})\sigma(p)=(2^k-1)(p+1)=(2^k-1)2^k=2n$
Beautiful proof.

An exercise:

Prove the opposite way.
That's a good idea. Euler did it.

7. Indeed.