Let $a$ be a constant, let $q=p^n$ be a prime power, and define $$\zeta(s) = \frac{q^{2s}-aq^s+q}{(q^s-1)(q^s-q)}.$$
Question. To show that if $\zeta(s)=0$, then $\Re(s) = \frac12$.
Here is my attempt so far:
Noice that for all $s$, $\zeta(s) = \zeta(1-s)$, thus a zero of $\zeta$ satisfies $$(q^s)^2 - a(q^s)+q=0 \quad \text{and} \quad (q^{1-s})^2 - a(q^{1-s})+q=0.$$ Thus $q^s$ and $q^{1-s}$ are the zeros of the quadratic $z^2 - az + q$. For any quadratic, $\alpha$ is a root if and only if the conjugate $\bar\alpha$ is also a root. Moreover, $\overline {q^{1-s}} = q^{1-\bar s}$ and $\overline{q^s} = q^{\bar s}$. Therefore one of the following must be true: \begin{align*} q^s &= q^{1-s}, \\ q^s = q^{\bar s} \quad &\text{and} \quad q^{1-s} = q^{1-\bar s},\\ q^s = q^{1-\bar s} \quad &\text{and} \quad q^{1-s} = q^{\bar s}. \end{align*} The first and third cases clearly imply that $\Re(s)=\frac12$. But in the second case, all we can deduce is that $s$ is real.
How can I show that $\Re(s) = \frac12$ in this case also?
$a$ is real. The roots of $X^2-aX+q$ are $\frac{a\pm \sqrt{a^2-4q}}{2}$, iff $a^2-4q\le 0$ then the two roots are complex conjugate so they have the same absolute value $q^{1/2}$. If $a^2-4q> 0$ then the result isn't true. That's why the Hasse bound is important for elliptic curves (its generalization to higher genus curves is https://en.wikipedia.org/wiki/Weil_conjectures)