The proof that $\Gamma(z)\pm \Gamma(1-z)$ only has zeros for $z \in \mathbb{R}$ or $z= \frac12 +i \mathbb{R}$ has been given here:


[Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?](https://mathoverflow.net/questions/89324/are-all-zeros-of-gammas-pm-gamma1-s-on-a-line-with-real-part-frac12)


An obvious follow up question is whether $\zeta(s) \pm \zeta(1-s)$ also has zeros (other than its non-trivial ones that would induce 0+0 or 0-0). 


This is indeed the case and $\zeta(s)^2 - \zeta(1-s)^2$ has the following zeros:


$\frac12 \pm 0.819545329 i$


$\frac12 \pm 3.436218226 i$


$\frac12 \pm 9.666908056 i$


$\frac12 \pm 14.13472514 i$ (the first non trivial)


$\frac12 \pm 14.51791963 i$


$\frac12 \pm 17.84559954 i$


$\dots$


These 'semi' trivial zeros appear to all lie on the critical line. I wonder if anything is known or proven about their location (I guess not, since a proof that they must have real part of $\frac12$ would automatically imply RH, right?).


**EDIT:**
Two counterexamples found by Joro in the answers below. Both have real parts outside the critical strip, so I would like to rephrase my question as:


*Are the 'semi' trivial zeros that are located within the critical strip all on the critical line?*



Are the 'semi' trivial zeros of $\zeta(s) \pm \zeta(1-s)$ all on the critical line?
If $\zeta(s)$ is nonzero, but $\zeta(s)\pm\zeta(1-s)=0$, then by the functional equation of the Riemann zeta function we have
$$ \pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\pm \pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)=0.$$
That is, your question is just the Riemann Hypothesis plus a more elementary one similar to your earlier question about the zeros of $\Gamma(s)\pm\Gamma(1-s)$. I would expect that the exact same techniques work here, i.e. one can show by known estimates for the gamma function that all nonreal solutions of the displayed equation lie on $\Re s=1/2$.


**EDIT 1.** To keep up with new developments I now expect that within the critical strip all nonreal solutions of the displayed equation lie on $\Re s=1/2$. Moreover, it seems reasonable to believe that there are no nonreal solutions with $|\Re s|$ sufficiently large.


**EDIT 2.** It follows from a generalized Rouché's theorem and Stirling's approximation that there are no nonreal solutions with $|\Re s|$ sufficiently large. More precisely, consider the rectangular contour $C\_n$ with vertices $2n\pm it$ and $2n+2\pm it$, where $n>0$ is a large integer and $t>0$ is sufficiently large in terms of $n$. It suffices to show that along $C\_n$ we have 
$$ \left|\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\right|<\left|\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\right|,$$
because this implies that inside $C\_n$ there is precisely one solution of the above displayed equation (which must be real by the reflection principle). One can show that the right hand side divided by the left hand side is
$ \gg n^{2n-\frac{1}{2}}(\pi e)^{-2n}$ on the vertical sides of $C\_n$, while it is $\gg\_n t^{2n-\frac{1}{2}}$ on the horizontal sides of $C\_n$. The claim follows.


