Nested fight back!

Let $f(x) = \sqrt{19} + \frac{91}{x}$ . Then $x = f(f(f(f(f(x)))))$ , from which we realize that $f(x) = x$ . This is because if we expand the entire expression, we will get a fraction of the form $\frac{ax + b}{cx + d}$ on the right hand side, which makes the equation simplify to a quadratic. As this quadratic will have two roots, they must be the same roots as the quadratic $f(x)=x.$ The given finite expansion can then be easily seen to reduce to the quadratic equation $x_{}^{2}-\sqrt{19}x-91=0$ . The solutions are $x_{\pm}^{}=\frac{\sqrt{19}\pm\sqrt{383}}{2}$ . Therefore, $A_{}^{}=\vert x_{+}\vert+\vert x_{-}\vert=\sqrt{383}$ . We conclude that $\boxed {A_{}^{2}=383.}$

Highly overrated problem! Still,

$x=\sqrt{19} + 91/x$ . Solving this quadratic yields,

$x_1= \sqrt{19} +\sqrt{383}/2$ And $x_2= \sqrt{19} -\sqrt{383}/2$

Hence $(|x_1| + |x_2|)^{2}$ = $\boxed{383}$

We can convert problem in infinite loop problem by repeated substitution of x and then use infinite loop method to solve it