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!
 A: This is a very intuitive approach based on the fact that I assume to be in the exam room with no computer and even no calculator. 
If there is a simple root $\sqrt{64x+5}$ should reduce to a whole number and $x=\frac1{16}$ is obvious (since $9$ is the closest square to $5$). From here, we can go backward (verify that every time we get another square) and check that this is the solution.
A: You get rid of the square roots by successive squarings and changes of side.
$$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} = 1,$$
$$\sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} = -x+1,$$
$$\sqrt {16x + \sqrt {64x + 5}} = x^2-6x+1,$$
$$\sqrt {64x + 5} = x^4-12x^3+38x^2-28x+1,$$
$$0=x^8-24x^7+220x^6-968x^5+2118x^4-2152x^3+860x^2-120x-4.$$
Using a polynomial solver, there are six real roots and a complex conjugate pair, with no apparent simple value.
This makes the correctness of the problem statement rather dubious.
A: \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*}
A: The best solution I can imagine is to square repeatedly and then use numerical methods to find roots of the resulting polynomial and then checking for extranneous answers from our squaring.
A: With the extra $\sqrt x$ there, you just square both side and keep doing it, miraculously some terms cancel out :) 
A: $$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}}  = 1+ \sqrt x$$
Squaring
$$ \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}  = 1+ 2\sqrt x$$
Squaring
$$ \sqrt {16x + \sqrt {64x + 5}}  = 1+ 4\sqrt x$$
Squaring
$$ \sqrt {64x + 5}  = 1+ 8\sqrt x$$
Squaring
$$ 5  = 1+ 16\sqrt x$$
