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Math Help - Perfect Numbers ?

  1. #1
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    Perfect Numbers ?

    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.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by 1337h4x View Post
    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 2^p-1 is prime then n=2^{p-1}[2^{p-1}-1] is a prime factorization of it. So, the proper divisors o n are 1,2,\cdots,2^{p-1},2^p-1,2(2^p-1),\cdots2^{p-2}(2^p-1). So, 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 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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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by 1337h4x View Post
    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 n=p^m then the proper divisors are 1,\cdots,p^{m-1} and thus 1+\cdots+p^{m-1}=\frac{p^m-1}{p-1}\ne n
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  4. #4
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    Thank you! That makes good sense!
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