How do I show that :$\sigma({p^m})$ is divisible by $4$ if $m=4k+1$ , and $k$ is an integer number?

Question

How do i show this if it's not an open problem :$\sigma({p^m})$ is divisible by $4$ if $m=4k+1$ , and $k$ is an integer number and p is prime number. and $\sigma({p^m})$ is sum divisors of $p^m$

Attempt :$σ({p}^{4l}−1) = ({p}^{4m}−1)/(p − 1)$. Write $p^m = q$ so that

this is $(q^2 + 1)(q + 1)$and both factors are even. !!!!!!

Note : I have tried to proof it using some results in number theory about divisibilty but sorry i can't succed

Thank you for any help.

Answer 1

Sum of divisors of $p^m$ is $$ \sigma(p^m) = 1 + p + p^2 + \ldots + p^m. $$ Any odd prime number have form $p=4q+1$ or $p=4q-1$. Ok, if $p=4q+1$, then $$ \sigma(p^m)\equiv 1 + 1 + \ldots + 1 = m+1 \equiv 0 \pmod 4\Longrightarrow m = 4k-1. $$ If $p=4q-1$, then $$ \sigma(p^m)\equiv 1 - 1 + 1 + \ldots + (-1)^m = \begin{cases}0,& \text{$m$ is odd}\\1,&\text{$m$ is even}\end{cases} \pmod 4, $$ and $m=2k$.