Can the $\Xi(t)$ function be extended this way? In 1893 Hadamard proved that:
$$\xi(s) = \xi(0) \prod_{\rho} \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right)$$
where $\xi(z) = \frac12 z(z-1) \pi^{-\frac{z}{2}} \Gamma(\frac{z}{2}) \zeta(z)$ and $\rho = \sigma + \gamma i$ is a non-trivial zero of $\zeta(s)$ (i.e. $\gamma_n$ is the imaginary part of the n-th $\rho)$).
Riemann had already conjectured this in 1859 and since he needed an 'always real' function for his next thought steps, he assumed $s=\frac12 + t i$ and defined the function $\Xi(t)=\xi(s)$ that has been proven to be equal to:
$$\Xi(t)= \Xi(0)\prod_\gamma\left(1-\frac{t^2}{\gamma^2}\right)$$
It is known that $\xi(s)=\xi(1-s)$ and $\xi(s)=\overline{\xi(\overline{s})}$ and this implies that $\xi(s)\xi(\overline{s})$ must be real. Also known is that when there is a non-trivial zero $\rho$ lying off the critical line, then also $1-\rho, \overline{\rho}, \overline{1-\rho}$ must be zeros and their product will always be real:
$$\displaystyle\prod_\gamma\left(1-\frac{s}{\sigma+i \gamma}\right)
\left(1-\frac{s}{\sigma-i \gamma}\right)\left(1-\frac{s}{1-(\sigma+i \gamma)}\right)
\left(1-\frac{s}{\overline{1-(\sigma+i \gamma)}}\right)$$
This brought me to the following conjecture. Suppose $s=a+t i$ and $\Xi_a(t)=\xi(s)$ so $\Xi_a(0)=\xi(a)$, then the following product holds and is always real:
$$\Xi_a(t)\Xi_a(-t) = \Xi_a(0)^2\prod_\gamma \left(1-\frac{t^2}{\gamma^2}\right)\left(1-\frac{(-t)^2}{\gamma^2}\right)$$
or simpler:
$$\Xi_a(t)\Xi_a(-t) = \left(\xi(a)\prod_\gamma \left(1-\frac{t^2}{\gamma^2}\right)\right)^2$$
Numerical tests indicate that the conjecture is correct, but I am obvioulsy keen to find a proof (I do realise this is a way too big of a question to ask here!). Does anybody have a link that explains how $\Xi(t)= \Xi(0)\prod_\gamma\left(1-\frac{t^2}{\gamma^2}\right)$ has been derived? Also appreciate any other thoughts/steers on how to further progress this conjecture.
Thanks!  

For completeness' sake and to avoid confusion, I can now confirm that the conjecture (even though the numerical results are very close) must be false. It is easy to see that when: 
$$\left(1-\frac{t^2}{\gamma^2}\right)$$
would be independent of $a$, then it could be expressed as:
$$\frac{\Xi(t)}{\Xi(0)}$$
This function has the same zeros as $\zeta(s)$ and therefore its square can not be equal to $\dfrac{\Xi_a(t)\Xi_a(-t)}{\xi(a)^2}$ since the latter doesn't have zeros at those spots (otherwise the RH would also have been immediately proven false).
Back to the drawing board. 
