An Old Problem

Note first that $2 - \sqrt{3} = \dfrac{1}{2 + \sqrt{3}}$ . So letting $a = 2 + \sqrt{3}$ the given equation is

$a^{x} + \left(\dfrac{1}{a}\right)^{x} = 4 \Longrightarrow (a^{x})^{2} - 4a^{x} + 1 = 0 \Longrightarrow a^{x} = \dfrac{4 \pm \sqrt{4^{2} - 4}}{2} = 2 \pm \sqrt{3}$ .

So either $a^{x} = (2 + \sqrt{3})^{x} = 2 + \sqrt{3} \Longrightarrow x = 1$ or $(2 + \sqrt{3})^{x} = 2 - \sqrt{3} \Longrightarrow x = -1$ .

The sum of all possible real values of $x$ is thus $1 + (-1) = \boxed{0}$ .