jmc

Let $f(x)=\sqrt{19}+\frac{91}{x}.$ Then the given equation says $$x=f(f(f(f(f(x))))).(\ast )$$Notice that any root of $x=f(x)$ is also a root of $(\ast ),$ since if $x=f(x),$ then replacing $x$ with $f(x)$ four times gives $$x=f(x)=f(f(x))=f(f(f(x)))=f(f(f(f(x))))=f(f(f(f(f(x))))).$$In fact, the roots of $x=f(x)$ are the only roots of $(\ast ).$ This is because, upon expanding both equations, they become quadratics in $x,$ so they both have exactly two roots for $x.$

Thus, it suffices to solve $x=f(x),$ or $$x=\sqrt{19}+\frac{91}{x}⟹x^2-x\sqrt{19}-91=0.$$By the quadratic formula, we have $$x=\frac{\sqrt{19}\pm \sqrt{19+4\cdot 91}}{2}=\frac{\sqrt{19}\pm \sqrt{383}}{2}.$$The root $\frac{\sqrt{19}-\sqrt{383}}{2}$ is negative (while the other root is positive), so the sum of the absolute values of the roots is $$A=\frac{\sqrt{19}+\sqrt{383}}{2}-\frac{\sqrt{19}-\sqrt{383}}{2}=\sqrt{383}.$$The answer is $A^2=\boxed{383}.$