<s>[INST] Solve the following equation: $\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x= 1$ A past examination paper had the following question that I found interesting. I tried having a go at it but haven't come around with any solutions. How would one go about tackling it?
$$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x = 1$$
I'm seeing a relation between $4x$, $16x$ and $64x$ so maybe the larger can be simplified to the smaller?
I do encourage you to working within an examination environment (thus not making use of anything other than pen and paper and possibly a calculator). 

EDIT: My question had a missing $-\sqrt x$ at the end, sorry!
 [/INST] \begin{align*}
\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x  & = 1\\
\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} & = 1+\sqrt{x}\\
x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+x+2\sqrt{x}\\
\sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+2\sqrt{x}\\
4x + \sqrt {16x + \sqrt {64x + 5}} & = 1+4x+4\sqrt{x}\\
\sqrt {16x + \sqrt {64x + 5}} & = 1+4\sqrt{x}\\
16x + \sqrt {64x + 5} & = 1+16x+8\sqrt{x}\\
 \sqrt {64x + 5} & = 1+8\sqrt{x}\\
64x + 5 & = 1+64x+16\sqrt{x}\\
16\sqrt{x} & = 4\\
x & =\frac{1}{16}.
\end{align*}
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