# Sum of powers of divisors

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).

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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.
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. – Pavel Jun 10 '11 at 20:31
@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 $(1+p^s)\frac{p^{s(k+1)}-1}{p^s-1}=\sigma_s(p^{k+1})+d^s\sigma_s(p^{k-1})$ – Ross Millikan Jun 11 '11 at 15:23