Continued fraction in 8th root------ any simpler approach? It seems to be a problem from the Putnam Exam. The problem asked to find the exact value of
$$x=\sqrt[8]{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{2207-\cfrac{1}{\ddots}}}}}$$
And express as $\dfrac{a+b\sqrt{c}}{d}$, in terms of some integers $a,b,c,d$.
My approach
I've tried to treat it as normal continued fractions. It is assumed by the question that it is positive and real (per the question, $x\in\mathbb Q(\sqrt c)$ for some integer $c$), it's straight-forward to have:
$$
x=\sqrt[8]{2207-\frac1{x^8}}
$$$$
x^{16}-2207x^8+1=0
$$$$
x^8=\frac{2207\pm\sqrt{2207^2-4}}{2}
$$
Let $x^4=\sqrt \alpha\pm\sqrt \beta$, as
$$\alpha+\beta=\frac{2207}{2}$$
$$4\alpha\beta=\sqrt{\frac{2207^2-4}{4}}$$
Solve and reject inappropriate (as $x^4$ is positive by our assumption) solution to get
$$x^4=\sqrt{\frac{2207}{4}+\frac12}\pm\sqrt{\frac{2207}{4}-\frac12}$$
$$=\frac{47}{2}\pm\sqrt{\frac{2205}{4}}$$
And let $x^2=\sqrt{\gamma}\pm\sqrt{\delta}$, using the same approach to get
$$x^2=\frac72\pm\sqrt{\frac{45}{4}}$$
And hence
$$x=\frac{3\pm\sqrt 5}{2}$$
But as 
$$x=\frac{3-\sqrt 5}{2}\lt\frac12\lt 1$$
When we take the fraction to the second evolution, a fallacy occurred. Thus,
$$x=\frac{3+\sqrt 5}{2}$$
I'm wondering whether my approach is valid. Also, this method seems to be a bit tedious, is there any easier one?
Thanks in advance.
 A: Let $x=a- \frac{1}{x}$ ; this satisfies $x^2-ax+1=0$ ... & $x^2$ satisfies 
\begin{eqnarray*}
(x^2+1)^2=(ax)^2 \\
(\color{blue}{x^2})^2-(a^2-2)\color{blue}{x^2}+1=0 
\end{eqnarray*}
So to "square root a continued fraction" we need to solve $a^2-2=b$ ... eighth root so we need to do this three times
\begin{eqnarray*}
a^2-2 &=&2207 \; \; \; &a&=&47 \\
b^2-2 &=&47 \; \; \; &b&=&7 \\
c^2-2 &=&7 \; \; \; &c&=&3 \\
\end{eqnarray*}
So we have $\color{red}{x=\frac{3 +\sqrt{5}}{2}}$. (Justify why the positive root has been chosen ... ?)
EDIT : There is often an ambiguity given by exactly how we define the convergents of a continued fraction ... see Continued fraction fallacy: $1=2$
A: Expanding the comment a little.
We can factor
$$
\begin{aligned}
2207^2-4&=(2207-2)(2207+2)\\
&=2205\cdot2209\\
&=5\cdot441\cdot47^2\\
&=5\cdot21^2\cdot47^2.
\end{aligned}
$$
Therefore your equation
implies that $x^8$ is one of
$$
t_1=\frac{2207+ 987\sqrt5}2\qquad\text{or}\qquad
t_2=\frac{2207-987\sqrt5}2.
$$
Those are integers of the field $\Bbb{Q}(\sqrt5)$. Because $t_1t_2=1$
(the constant term of the quadratic $T^2-2207T+1=0$), they are inversers of each other and therefore units of the ring $\mathcal{O}=\Bbb{Z}[(1+\sqrt5)/2]$.
From the basics of algebraic number theory (or from the theory of Pell equations) we know that $t_1$ is a power of the fundamental unit $u=(1+\sqrt5)/2$ of the ring $\mathcal{O}$.
The given piece of information, $x=(a+b\sqrt c)/d$, for some integers $a,b,c,d$, implies that $x$ is an element of the field $\Bbb{Q}(\sqrt c)$. We already saw that $x^8$ is an element of $\Bbb{Q}(\sqrt5)$. So unless $c=5$ and $x$ is an eighth power in $\mathcal{O}$, our results imply that the minimal polynomial of $x$ would have degree $>2$. Therefore it is strongly implied that, by happenstance, $t_1$ is an eight power in $\mathcal{O}$ as well.
Then we can just do a bit of testing to see that
$$
u^{16}=t_1.
$$
Therefore $x=\pm u^{-2}$, and you seem to know how to eliminate the wrong alternatives.
With less theory you can just repeatedly denest $\root{8}\of t_1=\sqrt{\sqrt{\sqrt{t_1}}}$. This is, again, by luck (=read problem design). Bill Dubuque has explained a general method for denesting square roots, and you can apply that thrice here (in a sense you already did).
