Powers of Jumbled Square Roots

Inverse of (3-√5) is (3+√5)/4. Repeatedly multiplying r.h.s. 8 by this three times gives 9+4√5, which results in two equations: x+3y=9 and -√5y=4. Solving for x and y gives y=-4 and x=21.

The trick is to apply the properties of golden ratio $\phi = \frac {1+\sqrt5}{2}$ , so $\phi^2 = \phi + 1 = \frac {3 + \sqrt5}{2}$ and $\frac {1}{\phi^2} = \frac {2}{3 - \sqrt5}$

$\begin{aligned} (3 - \sqrt5)^3 x + (3 -\sqrt5)^4 y & = & 8 \\ 8 \left ( \frac {3 - \sqrt5}{2} \right)^3 x + 16 \left ( \frac {3 - \sqrt5}{2} \right)^4 y & = & 8 \\ \left ( \frac {3 - \sqrt5}{2} \right)^3 x + 2 \left ( \frac {3 - \sqrt5}{2} \right)^4 y & = & 1 \\ \left ( \frac {1}{\phi^2} \right )^3 x + 2 \left ( \frac {1}{\phi^2} \right )^4 y & = & 1 \\ \frac {x}{\phi^6} + \frac {2y}{\phi^8} & = & 1 \\ \phi^2 x + 2y & = & \phi^8 \\ (1+\phi) x + 2y & = & (\phi^2)^4 \\ \phi x + (x+2y) & = & (\phi+ 1)^4 \\ & = & \phi^4 + 4\phi^3 + 6\phi^2 + 4\phi + 1 \\ & = & (\phi^2)^2 + 4\phi (\phi^2 + 1) + 6(\phi^2) + 1 \\ & = & (\phi + 1)^2 + 4\phi (\phi + 2) + 6(\phi + 1) + 1 \\ & = & 5\phi^2 + 16 \phi + 8 \\ & = & 5(\phi + 1) + 16 \phi + 8 \\ & = & 21 \phi + 13 \\ \end{aligned}$

Because $x$ and $y$ are integers, we just need to compare coefficients:

$x = 21, x + 2y = 13 \Rightarrow y = -4 \Rightarrow 10x - y = \boxed{214}$