Square it later

The problem can be solved by Ferrari's solution as shown below:

$\begin{aligned} x^4 + \sqrt{x^2+2016} & = 2016 \\ x^4 - 2016 & = - \sqrt{x^2+2016} \\ \left(x^4 - 2016 \right)^2 & = x^2 + 2016 & \small \color{#3D99F6}{\text{Add }m \text{ both sides}} \\ \left(x^4 - 2016 + \color{#3D99F6}{m} \right)^2 & = 2\color{#3D99F6}{m} x^4 + x^2 + \color{#3D99F6}{m}^2 - 4032\color{#3D99F6}{m} + 2016 & \small \color{#3D99F6}{\text{Putting }m = \frac 12 \text{(see Note)}} \\ \left(x^4 - 2016 + \color{#3D99F6}{\frac 12} \right)^2 & = x^4 + x^2 + \frac 14 \\ \left(x^4 - \frac {4031}2 \right)^2 & = \left(x^2 + \frac 12 \right)^2 \\ \left(x^4 - \frac {4031}2 \right)^2 - \left(x^2 + \frac 12 \right)^2 & = 0 \\ \left(x^4 + x^2 - 2015 \right)\left(x^4 - x^2 - 2016 \right) & = 0 \end{aligned}$

$\implies \begin{cases} x^2 = \dfrac {\pm \sqrt{8061}-1}2 & \implies x = \pm \sqrt {\dfrac {\sqrt{8061}-1}2} \\ x^2 = \dfrac {\pm \sqrt{8065}+1}2 & \implies x = \pm \sqrt {\dfrac {\sqrt{8065}+1}2} \end{cases}$

Substituting the solutions in $x^4 + \sqrt{x^2+2016} = 2016$ , it is found that the acceptable solutions are $\pm \sqrt {\dfrac {\sqrt{8061}-1}2}$ .

$\implies \dfrac {a+b+c}{c^c} = \dfrac {8061+1+2}4 = \boxed{2016}$ .

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$\color{#3D99F6}{\text{Note:}}$ According to Ferrari's solution, for a quartic equation $y^4 + py^2 + qy + r = 0$ , $m$ is given by $8m^3 + 8pm^2 + (2p^2-8r) - q^2 = 0$ . For $x^4 - 4032x^2 - x^2 - 4062240 = 0$ , we have $8m^3-32256m^2 +16128m-1=0$ and one of its root is $m = \dfrac 12$ .