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Thread: 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; Jan 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 $\displaystyle \tfrac{\sigma(n)}{n}=\tfrac{
    \sum\limits_{\left. d \right|n} d
    }{n}=
    \sum\limits_{\left. d \right|n} {\left( {\tfrac{d}
    {n}} \right)} =
    \sum\limits_{\left. d \right|n} {\left( {\tfrac{1}
    {d}} \right)}
    $

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

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

    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: $\displaystyle
    2p^k = \sum\limits_{\left. d \right|p^k } d = 1+p+...+p^k=1+p\cdot (1+p+...+p^{k-1})
    $ which is a contradiction because the LHS is a multiple of p while the RHS is not.
    Last edited by PaulRS; Jan 24th 2009 at 04:31 AM. Reason: Add an optional proof
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  3. #3
    MHF Contributor

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