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In a paper I was reading recently, the author has made use of the following formula in his proof:

$\displaystyle\sum_{k|n}\lambda(k)=\displaystyle\sum_{k|n}2^{\nu(k)}\lambda(k)d\big(\frac{n}{k}\big)$.

Here $\nu(n)$ denotes the number of distinct prime divisors of $n$, $\lambda$ is Liouville's function and $d$ is the divisor function. I don't understand this formula and have not seen it before. Is this true? What would be a proof of it? I feel it must be a simple exercise in number theory, but am unable to prove it at first attempt. Any help will be appreciated. I think the argument for $d$ involves the Jacobi symbol. Is that so? What exactly is this identity?

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  • $\begingroup$ in general, when you have such an equality of multiplicative functions, you can look at their Dirichlet series and Euler product (as Marko Riedel did) $\endgroup$ – reuns Jun 29 '16 at 21:29
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Introduce for the RHS

$$L_1(s) = \sum_{n\ge 1} \frac{2^{\nu(n)} \lambda(n)}{n^s}$$

and $$L_2(s) = \sum_{n\ge 1} \frac{\sigma_0(n)}{n^s} = \zeta(s)^2.$$

We have $$L_1(s) = \prod_p \left(1-\frac{2}{p^s}+\frac{2}{p^{2s}}-\frac{2}{p^{3s}} + \cdots \right) \\ = \prod_p \left(-1 + 2\frac{1}{1+1/p^s}\right) = \prod_p \frac{-1-1/p^s+2}{1+1/p^s} = \prod_p \frac{1-1/p^s}{1+1/p^s} \\ = \prod_p \frac{(1-1/p^s)^2}{1-1/p^{2s}} = \frac{\zeta(2s)}{\zeta(s)^2}.$$

It follows that $$L_1(s) L_2(s) = \zeta(2s) = \sum_{n\ge 1} \frac{1}{n^{2s}}$$

which is the Dirichlet series of the indicator function of positive integers being a square.

However this is precisely the LHS as was proved at this MSE link, QED.

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  • $\begingroup$ Great answer, thank you. This was not a 'simple exercise' . $\endgroup$ – adrija Jun 29 '16 at 21:38

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