Except for a finite few outside the strip, do all complex zeros of $\zeta(a+s)\pm \zeta(a+1-s)$ reside on the critical line for all $a\lt 0$?

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.

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.  • @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 $a$. – Agno Oct 5 '14 at 9:07
• Hm, for a=-2 do you find root near $-37.999999999999999792709374591737022640535360938$. Not sure if it is numerical instability. – joro Oct 5 '14 at 9:37
• 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 "-"). – Agno Oct 5 '14 at 12:52