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Math Help - Perfect Numbers

  1. #1
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    Question Perfect Numbers

    Hi, I'm new here and so I first want to say "Hi, everybody"!!
    I've got a little problem with Euclid's method to find perfect numbers.
    The question is, how to prove that any number like p^n, where p is a prime and n is a positive integer, it cant be perfect?

    Actually 3^4 isnt perfect at all...and this should be enough...or not?

    Thanks in advance
    Last edited by clairette; January 24th 2009 at 02:43 AM.
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  2. #2
    Super Member PaulRS's Avatar
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    Hi! Welcome to the forum.

    First, note that \tfrac{\sigma(n)}{n}=\tfrac{<br />
\sum\limits_{\left. d \right|n} d <br />
}{n}=<br />
\sum\limits_{\left. d \right|n} {\left( {\tfrac{d}<br />
{n}} \right)} = <br />
\sum\limits_{\left. d \right|n} {\left( {\tfrac{1}<br />
{d}} \right)} <br />

    A number <br />
n \in \mathbb{Z}^ +  <br />
is perfect by definition iff: <br />
\tfrac{{\sigma \left( n \right)}}<br />
{n} = \sum\limits_{\left. d \right|n} {\left( {\tfrac{1}<br />
{d}} \right)}  = 2<br />

    Assume p^k is perfect, since p \geqslant 2 we have: <br />
2 = \sum\limits_{j = 0}^\infty  {\left( {\tfrac{1}<br />
{{2^j }}} \right)}  \geqslant \sum\limits_{j = 0}^\infty  {\left( {\tfrac{1}<br />
{{p^j }}} \right)}  > 1 + \tfrac{1}<br />
{p} + ... + \tfrac{1}<br />
{{p^k }} = \sum\limits_{\left. d \right|p^k } {\left( {\tfrac{1}<br />
{d}} \right)}  = 2<br />

    This is a contradiction, thus the power of a prime can't be perfect.

    We could also prove it as follows, if it was true: <br />
2p^k  = \sum\limits_{\left. d \right|p^k } d  = 1+p+...+p^k=1+p\cdot (1+p+...+p^{k-1})<br />
which is a contradiction because the LHS is a multiple of p while the RHS is not.
    Last edited by PaulRS; January 24th 2009 at 04:31 AM. Reason: Add an optional proof
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  3. #3
    MHF Contributor

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    Quote Originally Posted by clairette View Post
    Hi, I'm new here and so I first want to say "Hi, everybody"!!
    I've got a little problem with Euclid's method to find perfect numbers.
    The question is, how to prove that any number like p^n, where p is a prime and n is a positive integer, it cant be perfect?

    Actually 3^4 isnt perfect at all...and this should be enough...or not?

    Thanks in advance
    The following is just another way of expressing PaulRS's answer, but you may find it easier to understand:

    If p is prime, the divisors of p^n are 1, p, p^2,\ldots, p^n, so that the sum of the divisors is 1+p+\cdots+p^n. It remains to justify why this sum is not equal to 2p^n.
    This is because 1+p+\cdots+p^n=p^n\left(\frac{1}{p^n}+\frac{1}{p^{  n-1}}+\cdots+1\right)\leq p^n\left(\frac{1}{2^n}+\frac{1}{2^{n-1}}+\cdots+1\right) and \frac{1}{2^n}+\frac{1}{2^{n-1}}+\cdots+1<1+\frac{1}{2}+\frac{1}{2^2}+\cdots = 2 (the right-hand side is the series \sum_{k=0}^\infty \frac{1}{2^k}). Or because of the other argument PaulRS gives at the end.
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