The Zeta-function can be written as the following infinite Hadamard product of its non-trivial zeroes: 


$\zeta(s) = \pi^{\frac{s}{2}} \dfrac{\prod\_\rho \left(1- \frac{s}{\rho} \right)}{2(s-1)\Gamma(1+\frac{s}{2})}$


this also implies that:


$\zeta(1-s) = \pi^{\frac{(1-s)}{2}} \dfrac{\prod\_\rho \left(1- \frac{(1-s)}{\rho} \right)}{2((1-s)-1)\Gamma(1+\frac{(1-s)}{2})}$


Take the reflection formula: 


$\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)$


and substitute the Hadamard products for $\zeta(s)$ and $\zeta(1-s)$. The result is that:


$\prod\_\rho \left(1- \frac{s}{\rho} \right) = \prod\_\rho \left(1- \frac{(1-s)}{\rho} \right)$


that can be rewritten as:


$\prod\_\rho \left(\frac{\rho -s}{\rho + s -1} \right) = 1$


This equation has zeros for $\rho = s$ as long as $2\rho-1 \ne 0$.


The ρ's could obviously lie anywhere in the already proven strip between $0<\Re(\rho)<1$ and I just take them as 'givens' wherever they might be located. 


I started to experiment with solving s for different numbers of terms as follows:


$x=\frac13$ (e.g.)


$\prod\_{n=1}^y \left(\frac{(x + ni) -s}{(x + ni) + s -1} \right) = 1$


Whatever value I pick for x between 0 and 1, the solution is always a complex number (ignoring $s=\frac12$, that is always a solution). However the exception occurs when $x = \frac12$ that always seems to only produce real numbers as solution(s). 


Does anybody see why that is (or must be) the case?



Non trivial zeros of the Zeta function
Using the notation $s=u+1/2$ your conjecture can be reformulated and generalized as follows. 


**Proposition.** Let $v\_1,v\_2,\dots,v\_N$ be arbitrary positive numbers, then all solutions of the equation
$$ \prod\_{n=1}^N \frac{v\_ni-u}{v\_ni+u} = 1 $$
are real.


**Proof.** The degree of the polynomial $\prod\_{n=1}^N(v\_ni-u)-\prod\_{n=1}^N(v\_ni+u)$ is $N$ or $N-1$ depending on whether $N$ is odd or even (the polynomial is always odd). Therefore it suffices to show that there are the same number of real solutions to the displayed equation. As $u$ grows from $-\infty$ to $\infty$, each fraction under the product traverses the unit circle continuously in the positive direction, starting from and arriving back to $-1$. Using ideas similar to how one proves that the fundamental group of the unit circle is $\mathbb{Z}$, we see that the product traverses the unit circle $N$ times in the positive direction, starting from and arriving back to $(-1)^N$. In particular, the product passes $1$ exactly $N$ or $N-1$ times depending on whether $N$ is odd or even. QED


