Algebra Problem by Kevin Tran (3)

$\begin{aligned} 16 & = \sqrt{x-\color{#3D99F6}\sqrt{x+\sqrt{x-\sqrt{x+\cdots}}}} & \small \color{#3D99F6} \text{Let }y = \sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}} \\ 16 & = \sqrt{x-\color{#3D99F6}y} & \small \color{#3D99F6} \text{Squaring both sides} \\ \implies 256 & = x - y \quad \quad ...(1) \end{aligned}$

From $y$ :

$\begin{aligned} y & = \sqrt{x+\color{#3D99F6}\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}} \\ & = \sqrt{x+\color{#3D99F6}16} & \small \color{#3D99F6} \text{Squaring both sides} \\ y^2 & = x + 16 \\ \implies x & = y^2 - 16 \quad \quad ...(2) \end{aligned}$

Then, from $(1)$ :

$\begin{aligned} 256 & = {\color{#3D99F6}x} - y & \small \color{#3D99F6} \text{Note that (2): }x = y^2 - 16 \\ & = {\color{#3D99F6}y^2 - 16} - y \\ \implies y^2 - y - 272 & = 0 \\ (y-17)(y+16) & = 0 \\ \implies y & = 17 & \small \color{#3D99F6} \text{Note that }y > 0 \\ x & = 17^2 - 16 \quad \quad ...(2) \\ & = \boxed{273} \end{aligned}$

$16\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \sqrt { x-\sqrt { x+\sqrt { x-\sqrt { x... } } } } \\ 16\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \sqrt { x-\sqrt { x+16 } } \\ 256\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad x-\sqrt { x+16 } \\ 256-x\quad \quad \quad \quad \quad \quad \quad \quad =\quad -\sqrt { x+16 } \quad \\ -256+x\quad \quad \quad \quad \quad \quad \quad =\quad \sqrt { x+16 } \\ 65536-512x+{ x }^{ 2 }\quad =\quad x+16\\ { x }^{ 2 }-513x+65520\quad =\quad 0\quad$

Therefore, $x$ = 273.