Perfect Numbers ?

Apr 2009
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Okay, I have a few questions pertaining to perfect numbers.

Q1:
The theorem in my book states that a prime power is never a perfect number. Therefore if p is a prime, then p^n cannot be a perfect number. Can anyone explain why this is?

Q2:
Suppose that p is a prime and that (2^p)-1 is also prime (in fact a Mersenne prime) . It can be prove that :

n = [2^(p-1)]*[(2^p)-1] is a perfect number. How can this be proven? I've tried like 5 times and can't get anywhere with this.
 

Drexel28

MHF Hall of Honor
Nov 2009
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Q2:
Suppose that p is a prime and that (2^p)-1 is also prime (in fact a Mersenne prime) . It can be prove that :

n = [2^(p-1)]*[(2^p)-1] is a perfect number. How can this be proven? I've tried like 5 times and can't get anywhere with this.
If \(\displaystyle 2^p-1\) is prime then \(\displaystyle n=2^{p-1}[2^{p-1}-1]\) is a prime factorization of it. So, the proper divisors o \(\displaystyle n\) are \(\displaystyle 1,2,\cdots,2^{p-1},2^p-1,2(2^p-1),\cdots2^{p-2}(2^p-1)\). So, \(\displaystyle 1+2+\cdots+2^{p-1}+2^p-1+\cdots+2^{p-2}(2^p-1)=\sum_{n=0}^{p-1}2^n+(2^p-1)\sum_{n=0}^{p-2}2^n\) which using the geometric sum formula is equal to \(\displaystyle 2^p-1+(2^p-1)(2^{p-1}-1)=(2^p-1)(1+2^{p-1}-1)=2^{p-1}(2^p-1)=n\)
 
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Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
Q1:
The theorem in my book states that a prime power is never a perfect number. Therefore if p is a prime, then p^n cannot be a perfect number. Can anyone explain why this is?
Suppose that \(\displaystyle n=p^m\) then the proper divisors are \(\displaystyle 1,\cdots,p^{m-1}\) and thus \(\displaystyle 1+\cdots+p^{m-1}=\frac{p^m-1}{p-1}\ne n\)
 
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