Continued fraction in 8th root------ any simpler approach?

Question

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.

Answer 1

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: $1=2$

Answer 2

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 inverses 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 eighth 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).