New Year's Countdown Day 4: Fourth Roots

$\begin{aligned}\text{Let,}\\ m&= \sqrt[4]{1009 +\sqrt{2017}},\\ n&=\sqrt[4]{1009 -\sqrt{2017}}\\ \text{We} &\text{ need to find } m+n\\\\ m^4+n^4&=2018\\ (mn)^2&=1008\\ (m^2+n^2)^2&=(m^4+n^4+2(mn)^2)=4034\\ \implies(m^2+n^2)&=\sqrt{4034}\hspace{4mm} \color{#3D99F6}(1)\\ mn&=\sqrt{1008}\hspace{4mm} \color{#3D99F6}(2)\\ (m+n)^2&=(m^2+n^2+2mn)=\left(\sqrt{4034}+\sqrt{4032}\right)\hspace{4mm}\color{#3D99F6}\text{From } (1) \text{ and } (2)\\ (m+n)&=\sqrt{\left(\sqrt{4034}+\sqrt{4032}\right)}\\\\ \text{Let,} u&=\sqrt{4034}\\ v&=\sqrt{4032}\\ (u+v)^2&=8066+2\sqrt{4034\times4032}\\ &=8066+48\sqrt{28238}\\ (m+n)&=\sqrt{(u+v)}\\ &=\sqrt[4]{8066+48\sqrt{28238}}\\ \text{On }&\text{comparing, we get } a=8066,b=48,c=28238\\ \implies c-a&=28238-8066=\color{#EC7300}\boxed{\color{#333333}20172}\end{aligned}$

$\begin{aligned} x & = \sqrt[4]{1009+\sqrt{2017}} + \sqrt[4]{1009-\sqrt{2017}} & \small \color{#3D99F6} \text{Let }k = \sqrt{2017} \\ & = \sqrt[4]{\frac {k^2+1}2+k} + \sqrt[4]{\frac {k^2+1}2-k} \\ & = \sqrt[4]{\frac {k^2+2k+1}2} + \sqrt[4]{\frac {k^2-2k+1}2} \\ & = \sqrt[4]{\frac {(k+1)^2}2} + \sqrt[4]{\frac {(k-1)^2}2} \\ & = \frac {\sqrt{k+1}}{\sqrt[4]2} + \frac {\sqrt{k-1}}{\sqrt[4]2} \\ & = \sqrt{\left(\frac {\sqrt{k+1}}{\sqrt[4]2} + \frac {\sqrt{k-1}}{\sqrt[4]2}\right)^2} \\ & = \sqrt{\frac {k+1+2\sqrt{k^2-1}+k-1}{\sqrt 2}} \\ & = \sqrt{\frac {2k+2\sqrt{k^2-1}}{\sqrt 2}} \\ & = \sqrt[4]2 \sqrt{k+\sqrt{k^2-1}} \\ & = \sqrt[4]2 \sqrt[4]{(k+\sqrt{k^2-1})^2} \\ & = \sqrt[4]2 \sqrt[4]{k^2+2k\sqrt{k^2-1}+k^2-1} \\ & = \sqrt[4]{4k^2-2+4k\sqrt{k^2-1}} & \small \color{#3D99F6} \text{Putting back }k = \sqrt{2017} \\ & = \sqrt[4]{4(2017)-2+4\sqrt{2017(2016)}} \\ & = \sqrt[4]{8066+48\sqrt{28238}} \end{aligned}$

$\implies c - a = 28238 - 8066 = \boxed{20172}$