Does $\zeta(s)^2 \pm \zeta(1-s)^2$ have roots at the $\rho$s?

Question

Maybe a strange (or stupid) question, but does

$$\zeta(s)^2 \pm \zeta(1-s)^2$$

also have roots equal to the non-trivial zeros ($\rho$) ?

At first sight you would expect so, however when I tried to 'root find' the zeros, I don't get any results at the $\rho$s for accuracies $>$ 14 digits. If this is indeed the case, could there exist any exponent (other than $1$) that does generate roots at the $\rho$s?

Answer 1

Yes, the pair of functions $\zeta(s)^2 \pm \zeta(1-s)^2$ has zeroes at all the nontrivial zeroes of $\zeta(s)$. In general, $$\pi^{-s/2} \Gamma \left( \frac{s}{2} \right) \zeta(s) := \Lambda(s) = \Lambda(1-s)$$ and neither powers of $\pi$ nor the Gamma function have zeroes in the critical strip. So if $\zeta(\rho) = 0$, then $\zeta(1 - \rho) = 0$. This indicates that it is your root-finding algorithm that is encountering trouble.

Roughly speaking you should also expect very many extra zeroes to be present as well.