Thread: 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.

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.

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!

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.

Ya, I hope so, it's a really confusing proof and I have no other idea of how to even start it.

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.

It does!! Thank you so much!! exactly what I needed!