Proving Perfect Number Proof: 2p-1(2p-1)

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Homework Help Overview

The discussion revolves around proving that if \(2^p - 1\) is prime, then \(2^{p-1}(2^p - 1)\) is a perfect number. Participants are examining a specific aspect of the proof related to the function \(\sigma(2^p - 1)\) and its value.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning why \(\sigma(2^p - 1) = 2^p - 1\) and discussing the implications of the divisors of \(2^{p-1}\) forming a geometric series.

Discussion Status

Some participants have acknowledged the existence of relevant information in the referenced paper, specifically in Theorem 4, which may provide clarity on the topic. There appears to be an ongoing exploration of the proof's details without a clear consensus on the understanding of the specific function.

Contextual Notes

Participants are working from a specific proof found in an external document, which may impose constraints on their discussion and understanding of the problem.

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Homework Statement



Prove: If 2p-1 is prime, then 2p-1(2p-1) is a perfect number.

Homework Equations



I am simply having trouble understanding one part of the end of the proof, namely:

Why does [itex]\sigma[/itex](2p-1)=2p-1 ?


The proof I'm working off of (trying to understand is on page 4 of this link: http://www.math.dartmouth.edu/~jvoight/notes/perfelem.pdf


The Attempt at a Solution


 
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srfriggen said:

Homework Statement



Prove: If 2p-1 is prime, then 2p-1(2p-1) is a perfect number.

Homework Equations



I am simply having trouble understanding one part of the end of the proof, namely:

Why does [itex]\sigma[/itex](2p-1)=2p-1 ?


The proof I'm working off of (trying to understand is on page 4 of this link: http://www.math.dartmouth.edu/~jvoight/notes/perfelem.pdf


The Attempt at a Solution


It's pretty easy. The divisors of 2^(p-1) are 1,2,2^2,2^3,...2^(p-1). The divisors form a geometric series. What's its sum?
 
aha, thanks!
 
srfriggen said:
aha, thanks!

Yeah, and actually this is already mentioned in the paper in Theorem 4.
 
Dick said:
Yeah, and actually this is already mentioned in the paper in Theorem 4.

oh man I can't believe I missed that :rolleyes:
 

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