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.