Timeline for Are the 'semi' trivial zeros of $\zeta(s) \pm \zeta(1-s)$ all on the critical line?

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Feb 29, 2012 at 13:36	comment	added	GH from MO		I proved unconditionally that if $s$ is not real and not a zero of $\zeta(s)$, but a zero of $\zeta(s)\pm\zeta(1-s)$, then the real part of $s$ is bounded. One still needs to prove that the imaginary part of $s$ is bounded, in order to conclude that the number of such values $s$ is finite. I expect that this can be established rather easily via some horizontal monotonicity property of the gamma function, similar to the mentioned treatment of the zeros of $\Gamma(s)\pm\Gamma(1-s)$, but I had not time and mood to work out the details.
Feb 29, 2012 at 9:36	comment	added	Agno		@joro: I don't think the finiteness of the exceptions assumes RH since it only applies to $h(s)-h(2a-s)$ and not to the terms individually. I believe the most promising route is to use $h(s)=\eta(s)$. E.g. swapping items between both terms could force the non trivial zeros for each term to disappear, whilst potentially still retaining a form of $h(s)-h(2a-s)$ with all zeros (now including the non trivial ones) on a critical line $a$ (but no longer at $a = \frac12$). A mixing and matching approach could f.i. be to swap all even numbers or better: swap out all the prime numbers.
Feb 29, 2012 at 6:41	comment	added	joro		Does the finiteness of the exceptions assumes RH or is unconditional?
Feb 28, 2012 at 22:53	history	answered	Agno	CC BY-SA 3.0