TITLE: Is the continued fraction of the square root of a base $\phi$ (golden ratio) number periodic when the continued fraction is expressed in base $\phi$?
QUESTION [5 upvotes]: I have been looking at concise ways to represent irrational numbers using only integers.
I was thinking about base $\phi$ (golden ratio base) and how it can represent the quadratic extension of the rationals with $\sqrt 5$ (i.e. $\mathbb Q[\sqrt 5]$) in a finite (potentially infinite but periodic) representation.  For example, $5 + \sqrt 5 = 10000.01$ exactly.
I also noticed that $\sqrt a$ always results in a periodic continued fraction (CF) when $a$ is a square-free integer.  The golden ratio $\phi$ is the least accurate irrational number to approximate by truncating its continued fraction, as it is [1, 1, 1, ...] (see Hurwitz's theorem and the Wikipedia Continued Fractions page).
Since truncating $\phi$'s CF is not as accurate a rational approximation as for other irrationals, and given that base $\phi$ allows us to represent $\phi$ concisely ($\phi = 10$) and allows a representation of integers (e.g. $2 = 10.01$), which is what CF's already use,

Are there any potential problems with using base $\phi$ as the base for a CF representation considering it gets around the accuracy issue for $\phi$ and allows continued fractions involving $\sqrt 5$ to be finite (I think)?
Is the continued fraction of the square root of a base $\phi$ number periodic when the CF is expressed in base $\phi$, in much the same manner as a square free integer?
Are there any other advantages/disadvantages to representing continued fractions in this base instead of another more typical base?

This is my first post, so please let me know if I need to correct anything.

REPLY [2 votes]: Steven has given an excellent answer, but I fear many people will not read all the comments and decide that the answer is basically no instead of yes due to the loss of a canonical form of continued fractions.
I will now try to demonstrate how to go about finding a good representation for base-$\phi$ square roots and why this might be useful.
It turns out that the sine or cosine of any integer multiple of $\pi/60$ can be represented as a linear combination of the following 6 factors (see Exact Trigonometric Constants):
$1$
$\sqrt 2$
$\sqrt 3$
$\sqrt 5$
$\sqrt{5+\sqrt 5}$
$\sqrt{5-\sqrt 5}$
Except for 1, these are basically quadratic extensions of $\mathbb Q$.
I want to represent a subset of the constructable numbers that use only sines and cosines of angles
$\dfrac{n\pi}{60}$ where $n \in \mathbb Z$.
Generalized continued fractions (GCF) allow me to represent square roots using the following formula (see Methods of Computing Square Roots):
$\sqrt z = \sqrt{x^2 + y} = x + \dfrac{y}{2x + \dfrac{y}{2x + \dfrac {y}{2x + \dfrac{y}{\ddots}}}}$
where $x$ and $y$ are positive real numbers, thus guaranteeing convergence as long as the product of successive $2x$'s grows faster than the product of successive $y$'s (see the Convergence Problem).  Generally we want $x$ to be as large as possible and $y$ to be as small as possible for faster convergence.
For $\sqrt 2$, $x = 1$ and $y = 1$.
For $\sqrt 3$, x = 1 and $y = 2$ does not converge quickly. Let's take a look in base-$\phi$.
$3 = 100.01 = \phi^2 + \phi^{-2}$ so $x = \phi$ and $y = \phi^{-2}$ will give much faster convergence.  You might wonder if $\sqrt 2$ has a similar faster convergence, but as $2 = 10.01 = \phi + \phi^{-2}$ we don't have an easy way to represent $x$ as $\sqrt \phi$.  If we re-write it as $2 = 1.11 = 1 + \phi^{-1} + \phi^{-2}$ we can take the $1$ for $x$ leaving $y = 0.11 = 1.00 = 1$ for $y$, which simply recovers $x = 1$ and $y = 1$.
$\sqrt 5$ is already directly representable in base-$\phi$, but if we wanted to use integers, $x = 2$ and $y = 1$ works just fine, and base-$\phi$ for $5$ doesn't give us a better convergence.
Now for the fun part.
$5 + \sqrt 5 = 10000.01 = \phi^4 + \phi^{-2}$, so $x = \phi^2$ and $y = \phi^{-2}$ will give good convergence, and
$5 - \sqrt 5 = 100.0001 = \phi^2 + \phi^{-4}$, so $x = \phi$ and $y = \phi^{-4}$ will give good convergence, both in a finitely representable periodic form.
Combinations of these factors like $\sqrt 6 = \sqrt 2\sqrt 3$ (for $15^\circ$) and $\sqrt 2\sqrt{5 + \sqrt 5}$ (for $12^\circ$) are also periodic GCF's.
This gives us a nice finite way to represent vectors made from sines and cosines of integer multiples of $3^\circ$.
So to answer my original questions:

Yes. One problem is that uniqueness is lost for CF representations.  There are infinitely many ways to represent the same CF.  This is related to the problem base-$\phi$ has in that there are infinitely many finite ways to represent the same number (using the 100 = 011 conversion).  This is not a problem as long as we can find a consistent representation for numbers we are interested in.
Yes, at the cost of uniqueness.
One advantage is potentially faster convergence of a GCF than with only integers.  Another is the ability to represent quadratic extensions (square roots) of $\mathbb Z[\phi]$ in a periodic form.