Continued fraction of $\phi^3$

Question

I found that $$\phi^3=4+\cfrac1{\small{4+\cfrac1{4+\cfrac1{4+\cfrac1{4+\ddots}}}}}$$ How should I prove this?

Attempt:

Suppose$$x= 4+\cfrac1{\small{4+\cfrac1{4+\cfrac1{4+\cfrac1{4+\ddots}}}}}$$ To solve for $x$, I did the following: $$x= 4+\frac{1}{x}$$ $$x^2-4x-1=0$$ I know one of the two solutions of the equation is $$x=\phi^3$$, but how should I show that? I know I can use the quadratic formula, but are here any easier ways of doing this?

Answer 1

You could use the property that defines $\phi$:

$$\phi = 1+\frac{1}{\phi}$$

Multiplying both sides by $\phi$ and $\phi^2$ would give

$$\phi^2 = \phi+1$$

$$\phi^3 = \phi^2+\phi = 2\phi+1$$

You know $\phi = \frac{1+\sqrt{5}}{2}$ and that $x = \frac{4+\sqrt{20}}{2} = \frac{4+2\sqrt{5}}{2} = 2+\sqrt{5} = 2\big(\frac{1+\sqrt{5}}{2}\big)+1 = 2\phi+1$, so $x = \phi^3$.

Answer 2

If you have $\phi-(1/\phi)=1$, you can cube both sides. Applying the Binomial Theorem to the cubed left side then renders

$\phi^3-3\phi+3/\phi-1/\phi^3=1$

$\phi^3-1/\phi^3=3(\phi-1/\phi)+1=4$,

from which $\phi^3$ is the positive root of $x=4+(1/x)$ and iteration of that equation gives the claimed continued fraction.