Putnam Math competition

We are being asked to calculate the eight root of the continued fraction $\beta = \sqrt[8]{2207 - \frac{1}{2207 - \frac{1}{2207 - \frac{1}{2207 - \cdots}}}}$ Consider the sequence $a_n \; = \; 2207 - \frac{1}{a_{n-1}}$ with $a_0 = 2207$ , and let $\alpha$ be the larger root of the equation $x^2 - 2207x + 1 \; = \; 0$ namely $\alpha \; = \; \tfrac12\big(2207 + 987\sqrt{5}\big)$ Note that $\alpha < a_0 \le 2207$ . Thus $2207 > 2207 - a_0^{-1} > 2207- \alpha^{-1} = \alpha$ and hence $\alpha < a_1 < a_0 \le 2207$ . If $n \in \mathbb{N}$ and $\alpha < a_n < a_{n-1} \le 2207$ then $a_{n+1} \; = \; 2207 - a_n^{-1} > 2207 - \alpha^{-1} \; = \; \alpha$ while $a_n - a_{n+1} \; = \; a_n^{-1} - a_{n-1}^{-1} \; = \; \frac{a_{n-1} - a_n}{a_na_{n-1}} \; > \; 0$ so that $\alpha < a_{n+1} < a_n \le 2207$ . Hence, by induction, we have shown that $(a_n)$ is a monotonic decreasing sequence with $a_n > \alpha$ for all $n$ , and hence $(a_n)$ converges to some limit $a \ge \alpha$ . Since $a_{n+1}a_n - 2207a_n + 1 = 0$ for all $n$ , we deduce that $a^2 - 2207a + 1 = 0$ . From this we deduce that $a = \alpha$ . Thus the continued fraction is equal to $\alpha = \tfrac12(2207 + 987\sqrt{5})$ , and we need to calculate the eighth root $\beta = \sqrt[8]{\alpha}$ of $\alpha$ .

Now $4414 = 47^2 + 5\times21^2$ and $987 = 21 \times 47$ , and hence $\sqrt{\alpha} = \tfrac12(47 + 21\sqrt{5})$ . Since $94 = 7^2 + 5\times3^2$ and $21 = 7\times3$ we see that $\sqrt[4]{\alpha} = \tfrac12(7 + 3\sqrt{5})$ . Finally we see that $\beta = \sqrt[8]{\alpha} = \tfrac12(3 + \sqrt{5})$ , making the answer $3 + 1 + 5 + 2 = \boxed{11}$

My solution is very similar to Mark's, with minor differences in terminology. I suspect that there is essentially only one (correct) way to do these kinds of problems.

Consider the iteration function $f(x)=2207-\frac{1}{x}$ , an increasing function for positive $x$ . We are asked to find $\lim_{n \to \infty} f^n(2207)$ , if this limit does indeed exist. Solving the quadratic equation $f(x)=x$ produces two fixed points, the larger one being $a=\frac{1}{2}(2207+987\sqrt{5})<2207$ . Note that $f(x)>f(a)=a$ for $x>a$ since $f$ is increasing; thus $f^n(2207)>a$ for all $n$ . Now $f(x)<x$ for $x>a$ (consider $x=2207$ as a sample point). Thus $f^n(2207)$ is a decreasing sequence, bounded by $a$ , and as such it has a limit. Since this limit is a fixed point of $f$ , by continuity, we can conclude that $\lim_{n \to \infty} f^n(2207)=a$ .

As Mark lucidly explains, $\sqrt[8]{a}=\frac{1}{2}(3+\sqrt{5})$ , and the answer is $\boxed{11}$ .