Just wanted to share my latest train of thoughts and leave you with the open question if the logic makes sense or not.

The following equation can be derived from combining the Hadamard products of the non-trivial zeros ($\rho$) from the Zeta function and its reflection formula:

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

that can be rewritten as:

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

This equation must be valid for all $s$ (except 1) and to learn more about the $\rho$'s, I tried to solve the following more generic (and less infinite) equation:

$\prod_{n=1}^N \left(\frac{x_n-s}{x_n + s -1}\right) = 1$

It can be proven (see above) that when $s= \frac12 \pm yi$ the equation always produces $x_n = \frac 12 \pm w$; $x_n, w \in \mathbb{R}$. The outcome is therefore always real and this is in direct conflict with the empirical evidence that at least the first billions of non-trivial zeros lie on the (complex) critical line. The situation that $s$ = $\rho_n$ (i.e. a non-trivial zero, subset of $x_n$) can therefore never be achieved and this makes me suspicious something is wrong here.

Therefore the derived equation must be incorrect and the only decent way out of it I see is to assume the following subtle change:

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

Assumption (A):

$\rho = a + yi$ and $(1-\rho) = (1-a) - yi$ or $\rho = a - yi$ and $(1-\rho) = (1-a) + yi$.

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

that can be dramatically simplified into (note the $s$ dropping out):

$\prod_\rho \left(\frac{\rho -1}{\rho} \right) = 1$

and more generically to enable experimenting:

$\prod_{n=1}^N \left(\frac{x_n-1}{x_n}\right) = 1$

Solving this equation for all $x_n$ being equal, the outcome is always $\frac12 \pm y i$, so that's a much more promising outcome for inducing zeros of the Riemann hypothesis at $s=\rho_n$.

But we know that each $\rho_n$ is different and could lie anywhere in the critical strip $0<\Re(\rho)<1$. Obviously by simply assuming that all $\rho$'s (as a subset of $x_n$) are lying on the critical line, it is simple to proof that the equation nicely holds, since each (absolute) term in the product will be equal to 1:

$|\prod_{n=1}^N \left(\frac{\frac12 + ni -1}{\frac12 + ni}\right)| = 1$

N.B: Just briefly like to share a nice
byproduct I observed when playing with this
equation:

$k_n, a, b, c \in \mathbb{Z},
> |\prod_{n=1}^N \left(\frac{\frac12 +
> k_n i -1}{\frac12 + k_n i}\right)| =
> |\frac{a}{c} + \frac{b}{c}i| = 1
> \rightarrow a^2 + b^2 = c^2$

Only Pythagorean triples will be produced
for random values of $N$ and $k_n$.

So what will happen when some $\rho$'s are lying off the critical line? This would make at least two terms not being 1 anymore (one that causes it and one complementary to make the total product equal to 1 again). Example:

$\left(\frac{\frac13 + 6i -1}{\frac13 + 6i}\right)\left(\frac{y -1}{y}\right)=1$ gives $y=\frac23 -6i$.

But here's the trick: assumption (A) above now prohibits the $\rho$'s to switch signs in the product (only $(1-\rho)$ does that). This therefore implies that only $\rho_n = \frac12 + yi$ or $\rho_n = \frac12 - yi$ can be valid solutions for the equation.

And now I only have to proof assumption (A) is true...