# Show the following including $\sigma$ function

• 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...

$\sigma(n)$ 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*}
• 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 $\sigma(n) = 2n$). – Erick Wong Jun 9 '15 at 16:02