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Assume $a \in \mathbb{R}$ and $s \in \mathbb{C}$.

Numerical evidence suggests that all complex zeros, except for a finite few outside the strip, of:

$$\zeta(a+s)\pm \zeta(a+1-s)$$

lie on the line $\Re(s)=\frac12$ for $a \le 0$. When $a=0$ the problem reduces to this question (including the RH).

This question shows that it can be proven unconditionally that all complex zeros, except a finite few, of:

$$\zeta(a+s) \pm \zeta(a-s)$$

lie on the line $\Re(s)=0$ for $a\le 0$.

These two appear to be quite similar despite their difference only being $1$. I also did experiment with $\zeta(a+s)\pm \zeta(a+x-s)$ and the conjecture seems to hold for all $x \le 1$ with zeros lying on the line $\frac{x}{2}$.

Questions:

1) Is there any (known) counter example for the conjecture about the critical line?

2) I guess that the unconditional proof for the line $\Re(s)=0$ can not be easily applied to the conjecture about the critical line $\Re(s)=\frac12$, however I would be keen to understand why it would fail.

Thanks.

Addition:

Below are two graphs of the zeros at $s=\frac12 + y\,i$ for respectively $\zeta(a+s) + \zeta(a+1-s)$ and $\zeta(a+s) - \zeta(a+1-s)$. The $a$ varies in steps of $0.01$ from $-2$ till $+0.2$. The imaginary parts of any zeros lying off the critical line for $a>0$ are suppressed. Note that the distribution of the imaginary parts tends to becomes more and more regular when we move further towards the left.

enter image description here

enter image description here

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  • $\begingroup$ Didn't find counterexamples with after short searching. Have you found a single counterexample? $\endgroup$
    – joro
    Oct 5, 2014 at 6:57
  • $\begingroup$ @Joro. In the category a 'finite few' I found a couple e.g. $-8.249081527+3.530547997*I$ for $a=-2$ or $-6.250001448+3.459776750*I$ when $a=-4$. There appear to be a finite few only outside the strip for each $a$. $\endgroup$
    – Agno
    Oct 5, 2014 at 9:07
  • $\begingroup$ Hm, for a=-2 do you find root near $-37.999999999999999792709374591737022640535360938$. Not sure if it is numerical instability. $\endgroup$
    – joro
    Oct 5, 2014 at 9:37
  • $\begingroup$ You don't count real zeros? $\endgroup$
    – joro
    Oct 5, 2014 at 9:40
  • $\begingroup$ I focus on the complex zeros, but there are real zeros outside the strip as well. I find exactly the same real root as you for $a=-2$ (assuming we take the "-"). $\endgroup$
    – Agno
    Oct 5, 2014 at 12:52

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