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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 $$\nu_2(\sigma_k(n))=\nu_2(\sigma_{\ell}(n)).$$

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    $\begingroup$ As (I think) Wojowu said in a comment on the deleted version, it's enough to check this on prime powers. In fact, if I'm not mistaken, it reduces to: for any prime power $q^m$, if $k,l$ have the same parity, then $q^{k(m+1)}-1$ has the same $2$-adic valuation as $q^{l(m+1)}-1$. Proving this shouldn't be too hard, although it will have a few cases ($q=2, q^{m+1} = 1(mod 4), q^{m+1}=3(mod 4)$). $\endgroup$ – user44191 Sep 1 '18 at 18:44
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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 $2^{n+1}$).

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