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 = $\frac12$ ?Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ 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?
