# Perfect Numbers

• January 24th 2009, 02:31 AM
clairette
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?

• January 24th 2009, 03:25 AM
PaulRS
Hi! (Hi) Welcome to the forum.

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

Assume $p^k$ is perfect, since $p \geqslant 2$ we have: $
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: $
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.
• January 24th 2009, 03:47 AM
Laurent
Quote:

Originally Posted by clairette
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?

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.