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

  • 3
    $\begingroup$ 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

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 $2^{n+1}$).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.