# How do you do this proof of perfect integer?

• Nov 15th 2009, 10:11 AM
mathgirl13
How do you do this proof of perfect integer?
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.
• Nov 15th 2009, 12:50 PM
RedPhoenix12
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.
• Nov 15th 2009, 04:51 PM
mathgirl13
It's look pretty good, and helps, but our professor hasn't taught us sigma, so I don't know how much I can use, it's definitely a good basis though, thank you!
• Nov 15th 2009, 05:22 PM
RedPhoenix12
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.
• Nov 15th 2009, 05:34 PM
mathgirl13
Ya, I hope so, it's a really confusing proof and I have no other idea of how to even start it.
• Nov 16th 2009, 12:07 PM
RedPhoenix12
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.
• Nov 16th 2009, 05:11 PM
mathgirl13
It does!! Thank you so much!! exactly what I needed!