what is the solutions of function equation

Question

For complex number $z$, equation $2^z + 3^z = 2^{1-z}+3^{1-z}$ has a simple solution $z=\frac{1}{2}$.

Are there any other solutions?

do all solutions lie on the critical line?

Answer 1

Yes, it is provable that all zeros of that expression are on the critical line, and are simple. (The existence of infinitely many zeros is a separate exercise using Hadamard's product formula results.)

There are several types of arguments to prove things of this sort, going back to P.J. Taylor 1945, Lax-Phillips 1976 ("Scattering Theory" book), L. de Branges, and more recent work of J. Lagarias and M. Suzuki 2006, W. Mueller 2008, R. C. McPhedran and C.G. Poulton 2014, K. Klinger-Logan 2017, among others.

A very simple case is simple to state: for a meromorphic function $h(s)$ not vanishing in $\Re(s)>1/2$, taking real values on $\mathbb R$, and such that $h(1-s)/h(s)$ is bounded in $\Re(s)>1/2$, $h(s)\pm h(1-s)$ has no zeros off the critical line, except for possible simultaneous zeros of $h(s)$ and $h(1-s)$.

Among the variety of proof techniques that give results similar to this, the spectral method of Lax-Phillips, Mueller, and Klinger-Logan seems to me to be the most conceptual: one constructs an unbounded, semi-bounded self-adjoint operator whose discrete spectrum, if any, is $s(1-s)\ge 1/4$ for exactly for zeros $s$ of $1-h(1-s)/h(s)$. This entails that $s$ is on the critical line.

(As it stands, naturally one wonders whether this can be adapted to prove things like RH, but by this year it does not seem that this idea alone is sufficient. Indeed, Conrey-Li proved that de Branges' version of such an approach cannot succeed for zeta. Also, whatever device might be imagined to succeed in proving that all zeros are on-the-line must somehow distinguish Epstein zetas and other linear combinations of Hecke L-functions (for example) which are known to have many off-the-line zeros...)

EDIT: in response to @ChristianRemling's request... although I will not reproduce the whole argument, I will sketch Klinger-Logan's preprint's argument for the simplest case, which includes that of the question. With $h(s)$ as just above, let $V$ be the Hilbert space of square-integrable functions on the critical line. Let $T$ be the unbounded operator of multiplication by $s(1-s)$. (Yes, just a multiplication operator, but unbounded.) The domain of $T$ can be taken to be compactly supported square-integrable functions on the critical line. Since $T$ is semi-bounded, it has a Friedrichs self-adjoint extension $T^F$. Let $\theta(s)=1-h(1-s)/h(s)$, and view integration-against $\theta$ as an unbounded functional on $V$. The restriction $T_\theta$ of $T^F$ to the kernel of $\theta$ is still self-adjoint and densely defined, so has a Friedrichs extension $S$. An alternative characterization of $S$ is that $Su=f$ if and only if $s(1-s)u=f+c\cdot \theta$ for some constant $\theta$, and $\theta(u)=0$. Some relatively straightforward computations show that for $\theta$ of this special form, $(S-s(1-s))u=0$ (and $u\in V$) implies $\theta(s)=0$, and also $\theta(u)=0$, and conversely.