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  • How do I show that $\sigma (2k)=4k$ if and only if $k=2^{p-2}(2^p-1)$ where $2^p-1$ is a prime number.

  • I want to show that if $k$ is odd and $\sigma(k) = 2k$ then $k=p^am^2$ for some p with $(p,m)=1$ and $p=a=1$(mod 4).

I know that $\sigma (2k)=\sum \limits_{d|2k} d$, where I can see that d|2...

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for (1) (it should help you with ideas for (2) as well)

$\sigma(n)$ is the sum of divisors of $n$. In your case $n=2k=2^{p-1}q$ where $q=2^p-1$ itself being a prime. So all possible divisors of $n$ will be of the form $2^aq^b$, where $a \in \{0,1,2, \ldots , p-1\}$ and $b \in \{0,1\}$. So \begin{align*} \sigma(2k) & =\sum_{a=0}^{p-1}2^a+\sum_{a=0}^{p-1}2^aq\\ & = \frac{2^p-1}{2-1}+q\left(\frac{2^p-1}{2-1}\right)\\ &=(2^p-1)(q+1)\\ &=2^p(2^p-1))\\ & =2^22^{p-2}(2^p-1)\\ & = 4k \end{align*}

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  • $\begingroup$ Thank you @AnuragA. But I could not get b). Here at a. we had 2 as a base and it was easy simplifying but at b. I got stuck $\endgroup$ – gmath May 21 '15 at 20:06
  • $\begingroup$ This only proves half of (1) (indeed, the half that was known to Euclid...). Sorry, but this line of argument does nothing to help with (2), which is in the opposite direction (deducing properties of $n$ from $\sigma(n) = 2n$). $\endgroup$ – Erick Wong Jun 9 '15 at 16:02

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