1. ## Perfect Number Proof

Show that if 2^p - 1 is a prime number, then (2^(p - 1))(2^(p) - 1) is a perfect number.

Just so you guys know, a perfect number is a number that equals the sum of all its positive divisors (excluding itself). 1+2+4+7+14=28

2. Originally Posted by zachsch
Show that if 2^p - 1 is a prime number, then (2^(p - 1))(2^(p) - 1) is a perfect number.

Just so you guys know, a perfect number is a number that equals the sum of all its positive divisors (excluding itself). 1+2+4+7+14=28

1) If $n=p_1^{a_1}\cdot ...\cdot p_r^{a_r}\,,\,n\in\mathbb{N}\,,\,p_i\,\,primes\,,\ ,0 , then the number of different divisors of n is $\prod\limits_{i=1}^r(a_i+1)$

2) If $2^p-1$ is a prime, then $2^{p-1}(2^p-1)$ has $p\cdot 2=2p$ different divisors, and the ones that are less than the number itself are $1,2,2^2,...,2^{p-1},(2^p-1),2(2^p-1),...,2^{p-2}(2^p-1)$.

Well, now just sum up the above divisors using the well known formula for the sum of geometric series and you'll get the original number again.

Tonio

3. Originally Posted by tonio
1) If $n=p_1^{a_1}\cdot ...\cdot p_r^{a_r}\,,\,n\in\mathbb{N}\,,\,p_i\,\,primes\,,\ ,0 , then the number of different divisors of n is $\prod\limits_{i=1}^r(a_i+1)$

2) If $2^p-1$ is a prime, then $2^{p-1}(2^p-1)$ has $p\cdot 2=2p$ different divisors, and the ones that are less than the number itself are $1,2,2^2,...,2^{p-1},(2^p-1),2(2^p-1),...,2^{p-2}(2^p-1)$.

Well, now just sum up the above divisors using the well known formula for the sum of geometric series and you'll get the original number again.

Tonio
"Well, now just sum up the above divisors using the well known formula for the sum of geometric series and you'll get the original number again."
can you show how to do that?
i looked up geometric series and i can't follow how you would do it
didn't learn geometric series or slept through the class