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Is there an analytic function with zeroes only at:

  • every $-2n$,
  • $\frac12\pm it$, and
  • at least one at $\frac12\pm\epsilon\pm it$ where $0<\epsilon<\frac12, t\neq0$ (and these zeroes observing the known reflective symmetries within the critical strip)?

A answer assuming the Riemann hypothesis true would be fine.

To my mind the uniqueness of any analytic continuation of $\zeta(s)$ is suggestive of the existence of such a function being incompatible with the Riemann Hypothesis.

If not, uniqueness of $\zeta$ is of course a nice simple sufficiency for the Riemann hypothesis.

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Take $\zeta(z)\left(z-\frac14-i\right)\left(z-\frac14+i\right)\left(z-\frac34-i\right)\left(z-\frac34+i\right)$.

Every condition that you mentioned holds.

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  • $\begingroup$ @RobertFrost Can you please explain that? I have no doubt that the zeros of my function are the zeros of the $\zeta$ functions plus $\frac14$. $\endgroup$ – José Carlos Santos Sep 5 '18 at 9:34
  • $\begingroup$ @RobertFrost I've edited my answer. What do you think now? $\endgroup$ – José Carlos Santos Sep 5 '18 at 10:28
  • $\begingroup$ @RobertFrost What about$$\zeta(z)\left(z-\frac14-i\right)\left(z-\frac14+i\right)\left(z-\frac34-i\right)\left(z-\frac34+i\right)?$$But note that the extra zeros of my previous answer ($\frac14$ and $\frac34$) were already symmetric with respect to the real axis. $\endgroup$ – José Carlos Santos Sep 5 '18 at 10:36
  • $\begingroup$ Yes, I saw they were already symmetric; I should have required they have a distinct reflection or $t\neq0$. If you put this in your answer I'll accept, thank-you for your help... I may inquire in another question to what degree we can generalise these examples. If your example is $\zeta(z)g(a+bi)$ then I am immediately struck by the question; suppose $a+bi$ is a counterexample to RH then what do $\zeta(z)g(a+bi)$ and $\zeta(z)/g(a+bi)$ look like? $\endgroup$ – user334732 Sep 5 '18 at 11:05
  • $\begingroup$ @RobertFrost Perhaps that you could post that question as a separate question. And if my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ – José Carlos Santos Sep 5 '18 at 11:24
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I think this is an open question. As long as your function is automorphic, the Grand Riemann Hypothesis asks this exact question.

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    $\begingroup$ Thanks, I didn't know that. Does the Grand Riemann Hypothesis ask if this is the case, and permit one to assume the Riemann hypothesis? Because if not, it would seem a potentially important result, if it were true, that to prove GRH assuming RH, would prove RH. $\endgroup$ – user334732 Sep 5 '18 at 9:39
  • $\begingroup$ That doesn't sound right. $\endgroup$ – Uri George Peterzil Sep 5 '18 at 15:34

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