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So far i've tried looking at an equation which is true for perfect numbers:

$\sigma(n) = 2n$. where $\sigma(n) = \sum_{d|n} d$

i started looking at a odd perfect number which has 1 prime divisor say $p$, then:

$\sigma(n) = \sum_{i \in \mathbb{N}} p_i$, where all $p_i$'s are the same except for $p_1 = 1$. But then $\frac{\sigma(n)}{n} = 2$ right...?

Any helpfull tips here?

Kees

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Hints:

(1) $\sigma(p^a)=1+p+p^2+...+p^a=\frac{p^{a+1}-1}{p-1}$. And so $\frac{\sigma(p^a)}{p^a}<\frac{p}{p-1}$.

(2) So $\frac{\sigma(p^a\cdot q^b)}{p^a\cdot q^b}<\frac{p}{p-1}\cdot\frac{q}{q-1}$

(3) $\frac{p}{p-1}$ is decreasing (i.e., it is larger for small primes).

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  • $\begingroup$ nice, i think i get it but try to understand why the last inequality of (1) is true if $p^\alpha$ is odd... $\endgroup$ – Kees Til Sep 8 '15 at 16:10
  • $\begingroup$ @KeesTil The inequality in (2)? Or... $\endgroup$ – paw88789 Sep 8 '15 at 16:12
  • $\begingroup$ i especially mean is this true for even perfect numbers as well? $\endgroup$ – Kees Til Sep 8 '15 at 16:12
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    $\begingroup$ The inequalities hold also for $p=2$. $\endgroup$ – paw88789 Sep 8 '15 at 16:13
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    $\begingroup$ $\frac{p^{a+1}-1}{p^a (p-1)}=\frac{p}{p-1}-\frac{1}{p^a (p-1)}<\frac{p}{p-1}$ $\endgroup$ – paw88789 Sep 8 '15 at 16:27

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