This question expands on this one and seems to have a stronger result.

Take the Riemann $\xi$-function $\xi(s) =\frac12 s\,(s-1) \,\pi^{-\frac{s}{2}}\, \Gamma\left(\frac{s}{2}\right)\, \zeta(s)$. We know that $\xi(s)$ has the same non-trivial zeros as $\zeta(s)$ and also that $\xi(s) = \xi(1-s)$.

With $a \in \mathbb{R}/0$, "stretch" $\xi(s)$ as follows:

$$f(s,a):= \xi(a\,s) \pm \xi\left(a\,(1-s)\right)$$

I like to conjecture that for each $a$, all zeros (in and outside the strip) reside on the critical line.

The graph below shows the first zeros ($\pm = +$), all at $\Re\left(\frac12\right)$, for $a$ from $-2$ to $2$ in steps of $0.01$.

enter image description here

At first sight this "curtain" graph looks quite symmetrical, but clearly isn't so.


1) Are there any exceptions that contradict this conjecture?

2) If not, is this just a consequence of the RH or would the RH be a consequence of the conjecture?

3) Since $\xi(s)$ is an entire function that can be expressed as a Hadamard product involving its zeros as factors, $f(s,a)$ should have a unique one too. Appreciate any ideas around what it could look like.


I believe the same conjecture could be applied to:

$$\xi(a+s) \pm \xi\left(a + 1-s\right), \text{ all zeros on } \Re(s)=\frac12$$

enter image description here


$$\xi(a+s) \pm \xi\left(a - s\right), \text{ all zeros on } \Re(s)=0$$ enter image description here

Both are symmetrical around $a=0$ and $a=\frac12$ respectively, also the values where the $\rho$'s reside.



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