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The Riemann Hypothesis says that all non-trivial zeros for the Zeta function have $\Re(s)=1/2$.

What about the zeros for the following series?

(1) $\zeta(s)=1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+...$

(2) $\zeta_{\mathcal{2}}(s) =1+\frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{9^s}+...$

(3) $\zeta_{\mathcal{3}}(s) =1+\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{5^s}+\frac{1}{7^s}+...$

(4) $\zeta_{\mathcal{5}}(s) =1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{6^s}+...$

(5) ... ...

We should be able to find that they all satisfy the functional equation.

Clearly, we can see $\zeta_2(s) = (1 - 2^{-s})\zeta(s)$, etc. And we can see from $\zeta$ function has infinitely many factors $(1-p^{-s})^{-1}$, after analytically continued to all complex s, every of them contributing to zeros ($\Re(s)<1$), leading to RH.

My questions:

(1) How many zeros (non-trivial) are the same for the above series after analytically continuing to all complex s?

(2) If $\zeta$ has exact the same zeores (non-trivial, in critical strip) with $\zeta_2,\zeta_3, \zeta_4 ...$, is it a "proof" that $\zeta_q=\sum _{n =1}^{\infty}\frac{1}{n^{s}}, (q,n)=1$ has no Siegel zero?

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    $\begingroup$ Well, you know the zeroes of $1-2^{-s}$, right? So that tell you the relation between the zeroes of $\zeta$ and $\zeta_2$. $\endgroup$ – Gerry Myerson Dec 27 '13 at 1:35
  • $\begingroup$ I mean the zeros of $\zeta_2, \zeta_3$.. after analytically continued. It says all of them have similar none-trivial zeros with $\Re(s)=1/2$ $\endgroup$ – Ocean Yu Dec 27 '13 at 2:04
  • $\begingroup$ $\zeta_2$, after analytic continuation, has the same zeros as $\zeta$ has, and also has the zeros of $1-2^{-s}$. If you have formulas for $\zeta_3,\zeta_4,\dots$ similar to the one you have for $\zeta_2$, then you can find their zeros, as well. $\endgroup$ – Gerry Myerson Dec 27 '13 at 4:22
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    $\begingroup$ Trivially! The GRH, as summarized in the link you give, is a conjecture about zeros in the "critical strip" $0 < Re(s) < 1$. The zeros of the function $1 - 2^{-s}$ are all along the line $Re(s) = 0$, so the extra zeros that this factor contributes to $\zeta_2(s)$ are irrelevant to the GRH. $\endgroup$ – David Loeffler Dec 27 '13 at 9:04
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    $\begingroup$ Then maybe you could write up and post an answer incorporating what you now know. Then, later, you can accept your answer. This helps clean up the Unanswered Questions list. $\endgroup$ – Gerry Myerson Jan 2 '14 at 17:24
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$\zeta_q$ as defined in question, has exact the same zeros in critical strip $0<\Re(x)<1$.

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