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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.

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    @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})$2011-06-11