3
$\begingroup$

How do I prove the following:
$$ \sigma_k(u)\sigma_k(v) = \sum_{d|gcd(u,v)} d^k\sigma_k\left(\frac{uv}{d^2}\right) $$ when $$ \sigma_k(n) = \sum_{d|n} d^k $$

Can someone give me a clue on that one? (even for the special case when u prime).

$\endgroup$
4
$\begingroup$

If $u$ and $v$ are coprime, the only $d$ to consider is $1$. If you consider a prime dividing $u$, you should be able to convince yourself that the function $\sigma_k$ is multiplicative.

Then it suffices to consider the case $u=p^a, v=p^b$ for $p$ prime and $a \ge b$. You have $\sigma_k(p^a)=\frac{p^{k(a+1)}-1}{p^k-1}$. If you plug that into the sum I think you can get there.

$\endgroup$
  • $\begingroup$ Thanks. Just one more thing - let's say I proved $\sigma_s(p)\sigma_s(p^k) = \sum_{d|gcd(p,p^k)} d^s\sigma_s\left(\frac{pp^k}{d^2}\right)$. How do I make it $\sigma_s(n)\sigma_s(p)$ (if $p^k$ is the highest degree of $p$ dividing $n$)? Multiplying what I proved by $\sigma_s(\frac{n}{p^k})$ makes two sums one inside another. $\endgroup$ – Pavel Jun 10 '11 at 20:31
  • $\begingroup$ @Pavel: Note that gcd$(p,p^k)=p$, so your sum is just over $1$ and $p$. Note you changed from $k$ to $s$ in the comment. So it is $(1+p^s)\frac{p^{s(k+1)}-1}{p^s-1}=\sigma_s(p^{k+1})+d^s\sigma_s(p^{k-1})$ $\endgroup$ – Ross Millikan Jun 11 '11 at 15:23

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.