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.

  • 2
    $\begingroup$ I don't know for sure. Multiple denesting rounds may be par for the course for a Putnam question! If you are familiar with the basics of elementary algebraic number theory a kind of clue is hidden in the observation: $$2207^2-4=5\cdot3^2\cdot7^2\cdot 47^2$$ implying that $x^8$ is a unit of the ring of integers of the quadratic field $\Bbb{Q}(\sqrt 5)$. So if we are lucky it is an 8th power in that group of units. Instead of algebraic number theory familiarity with Pell equations points in the same direction. This being a constructed contest question, where a trick can be expected to work... $\endgroup$ – Jyrki Lahtonen Jun 30 '17 at 6:14
  • $\begingroup$ @JyrkiLahtonen sounds nice, I think it might be what the question want. Could you add an answer to evaluate more on it, or share some learning resources (books, lecture notes or even web pages) about it? It would be greatly appreciated. Maybe something starting at a relatively not-so-high level as I'm a high school student. $\endgroup$ – BAI Jun 30 '17 at 6:21
  • $\begingroup$ @JyrkiLahtonen I mean, the part "implying that $x^8$ is a unit of the ring of integers of the quadratic field $\Bbb{Q}(\sqrt{5})$" $\endgroup$ – BAI Jun 30 '17 at 6:31
  • $\begingroup$ This is Putnam 1995 B4; see math.hawaii.edu/~dale/putnam/1995.pdf for the official solution and mks.mff.cuni.cz/kalva/putnam/psoln/psol9510.html for an unofficial one. $\endgroup$ – Michael Lugo Jun 30 '17 at 13:27

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: $1=2$

  • $\begingroup$ This is a quicker method. I like it.(+1) $\endgroup$ – BAI Jun 30 '17 at 8:29

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).

  • $\begingroup$ It's nice. Thanks for your time.(+1) $\endgroup$ – BAI Jun 30 '17 at 8:32

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