Consider the sum of $k^{th}$-power of divisors of $n$, denoted 
$$\sigma\_k(n)=\sum\_{d\vert n}d^k.$$
Let $\nu\_p(x)$ stand for the $p$-adic valuation of the integer $x$.


The following appears to be true but is it?



> 
> **Question:** Fix $k, \ell\in\mathbb{N}$. If $k$ and $\ell$ have the same parity then 
>  $$\nu\_2(\sigma\_k(n))=\nu\_2(\sigma\_{\ell}(n)).$$
> 
> 
> 



$2$-adic valuations and sum of divisor function
As noted in the comment by user44191, one needs only check this for prime powers. Note that this is trivial for $q=2$ and so we may assume that we have an odd prime $q$. Then the claim is that when $k \equiv \ell$ one has that $$v\_2(\sigma\_k(q^m)) = v\_2\sigma\_\ell (q^m).$$


This is the same as asserting that $$v\_2(1+q^{k} + q^{2k} \cdots q^{mk}) = v\_2(1+q^{\ell} + q^{2\ell} \cdots q^{m\ell}) $$


or equivalently that $$v\_2\left( \frac{(q^k)^m-1}{q^k-1}\right) = v\_2\left( \frac{(q^\ell)^m-1}{q^\ell-1}\right). $$


And that's the same as $$v\_2\left( \frac{(q^m)^k-1}{q^k-1}\right) = v\_2\left( \frac{(q^m)^\ell-1}{q^\ell-1}\right). $$ 


But this is true not just for prime $q$ but in fact for all odd $q$ because $\phi(2^n)$ is always a power of 2, and if $x \equiv 1$ (mod $2^n$) then $x^2 \equiv 1$ (mod $2^{n+1}$). 


