Nested Impossibilities

$\color{#D61F06}{x}=\frac{1}{2+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{2+\frac{1}{8+\color{#D61F06}{x}}} } } } }$ $=\frac{1}{2+\frac{1}{1+\frac{1}{3+\frac{1}{1+\color{#3D99F6}{\frac{8+x}{17+2x} } } } } }$ $=\frac{1}{2+\frac{1}{1+\frac{1}{3+\color{#3D99F6}{\frac{17+2x}{25+3x} } }} }$ $=\frac{1}{2+\frac{1}{1+\color{#3D99F6}{\frac{25+3x}{92+11x} } }}$ $={\frac{1}{2+\color{#3D99F6}{\frac{92+11x}{117+14x} } }}$ $\color{#3D99F6}{=\frac{117+14x}{ 326+39x}}$ $\Rightarrow 39x^2+312x-117=0$ $\Rightarrow \color{#D61F06}{x}=-4 \pm \sqrt{19}$ Hence, $\sqrt{A}=4+x=4+\color{#D61F06}{(-4+\sqrt{19})}=\sqrt{19}$ $\Rightarrow \color{magenta}{A=19}$

Trivially, $18 < \left(4+\frac1{2+\frac11}\right)^2 < k^2 = A < \left(4+\frac1{2+\frac1{1 + \frac13}}\right)^2 < 20$

Hence, the answer is $\boxed{19}$ . (After computing the simple continued fraction expansion of $\sqrt{19}$ to prove that $19$ indeed works.)

$\sqrt{19}=[4; \overline{2,1,3,1,2,8}]$

The sequence of the continued fraction can be found on OEIS( A010124 )

This python code repeats $2, 1, 3, 1, 2, 8$ only 3 times. You can see that it converges to $\sqrt{19}$ .

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def continuedFraction(rep, times):
    new = 0.0
    for i in range(0, times):
        for j in rep[::-1]:
            new = 1 / (new + j)
    return new

sqrtValue = 4 + continuedFraction([2, 1, 3, 1, 2, 8], 3)
print(sqrtValue * sqrtValue)
# prints 18.999999999999996