All even perfect numbers are a power of two times a Mersenne prime
Take a look at this (I'm working on the same problem btw). I'm still making sense of the sigma notation but hopefully this will get you going in the right direction.
I'm completely stuck on this proof. I know the idea of how to do it, find the factors, add them, and show its the same, but I have no idea what the factors should be. Here's the question:
Show that 2^(p-1) * ((2^p) - 1) is a perfect number when (2^p) -1 is prime.
Thanks for any help, it is really really appreciated.
All even perfect numbers are a power of two times a Mersenne prime
Take a look at this (I'm working on the same problem btw). I'm still making sense of the sigma notation but hopefully this will get you going in the right direction.
I'm actually in the same position and was trying to extrapolate from what was there... but now I'm realizing that I would have to prove the correctness of their sigma function which leaves me just as stuck. Hopefully somebody else has an answer for the both of us.
I posted the question on another forum and got this response:
"I think a sketch of a proof could go about something like this:
The sum of all the positive divisors to 2^(p-1) is:
Sum ( 2^k ), k = 0 to k = p-1 = 2^p - 1 (known power series formula)
For each term 2^k in the above series, except the last, there's another positive divisor (2^p - 1)*2^k that should be included. The sum of all of those are (2^p - 1)(2^(p-1) - 1).
Adding these to together gives us:
(2^p - 1)(2^(p-1) - 1) + 2^p - 1 = 2^(2p-1) - 2^p - 2^(p-1) + 1 + 2^p - 1 = 2^(2p-1) - 2^(p-1) = 2^(p-1)*(2^p - 1)
Note that if 2^p - 1 where not a prime there would have been more divisors."
This should do it! Hope this helps mathgirl.