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Let $\chi$ be a Dirichlet character mod 4. Show $\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s} = \frac{L(\chi,s-1)}{L(\chi,s)}$ and $\sum_{n=1}^\infty \chi(n)d(n)n^{-s}=L(\chi,s)^2$. ($\phi$ is the Euler totient function and $d(n)$ is the number of divisors of $n$.)

First, is this true just for characters mod 4 and not true in general? I'm not sure what specific properties about characters mod 4 I should use besides that $\chi(n)=0$ for $n$ even.

I took the log of both sides and tried to use the following: $$L(\chi,s)=\prod_{p \text{ prime}}\frac{1}{1-\frac{\chi(p)}{p^s}}$$

$$\log L(\chi,s)=\sum_{p \text{ prime}}\sum_{n=1}^\infty \frac{\chi(p)^n}{np^{ns}}$$

$\chi$ and $\phi$ are multiplicative, so we can express $$\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s}$$ as the Euler product $$\prod_p(1+\frac{\chi(p)\phi(p)}{p^s}+\frac{\chi(p^2)\phi(p^2)}{p^{2s}}+\cdots).$$

Manipulating things are not quite working. Any help would be appreciated.

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  • $\begingroup$ If $\chi$ is completely multiplicative and $a_\chi(n) = a(n) \chi(n)$ and $a \ast b = c$ then $a_\chi \ast b_\chi = c_\chi$. It works with analytic functions too : the twist of the logarithm is the logarithm of the twist. $\endgroup$ – reuns May 4 at 23:41
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It is easier to show $$ L(\chi, s)\sum_{ n \geq 1} \frac{\chi(n) \phi(n)}{n^{s}} = L(\chi, s-1). $$ It is enough to show for $\Re (s) >> 0$ (by analytic continuation), and actually this is true for any Dirichlet character. If you compute LHS directly (assume that the serieses converges absolutely), we get $$ \left( \sum_{n\geq 1} \frac{\chi(n)\phi(n)}{n^{s}} \right)\left( \sum_{m\geq 1} \frac{\chi(m)}{m^{s}}\right) = \sum_{n, m\geq 1} \frac{\chi(nm)\phi(n)}{(nm)^{s}} = \sum_{k\geq 1} \frac{\chi(k)}{k^{s}} \left( \sum_{d|k} \phi(d)\right) = \sum_{k\geq 1} \frac{\chi(k)}{k^{s-1}} = L(\chi, s-1) $$ where the last equality follows from $\sum_{d|k}\phi(d) = k$.

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Looking at $\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s} L(\chi,s)= \sum_{n=1}^\infty \frac{(χφ * \chi)(n)}{n^s}$, * being Dirichlet Convolution.

Now looking at $(χφ * φ)(n)=\sum_{d|n} \chi(d)\phi(d) \chi(\frac{n}{d})= \sum_{d|n} \chi(n)\phi(d)=\chi(n) \sum_{d|n}φ(d)=n\chi(n) $

(maybe it's worth mentioning that $\sum_{d|n} \phi(d)=n$.)

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