An Old Favorite Putnam Problem

The infinite continued fraction is defined as the limit of the sequence $L_{0} = 2207, L_{n+1} = 2207-1/L_{n}$ .

Notice that the sequence is strictly decreasing (by induction) and thus indeed has a limit $L$ , which satisfies $L = 2207 - 1/L$ , or rewriting, $L^{2} - 2207L + 1 = 0$ .

Moreover, we want the greater of the two roots.

Now how to compute the eighth root of $L$ ? Notice that if $x$ satisfies the quadratic $x^{2} - ax + 1 = 0$ , then we have $(x^{2} - ax + 1)(x^{2} + ax + 1) = 0 \\ \\ x^{4} - (a^{2} - 2)x^{2} + 1 = 0.$ Clearly, then, the positive square roots of the quadratic $x^{2} - bx + 1$ satisfy the quadratic $x^{2} - (b^{2}+2)^{1/2}x + 1 = 0$ .

Thus we compute that $L^{1/2}$ is the greater root of $x^{2} - 47x + 1 = 0$ , $L^{1/4}$ is the greater root of $x^{2} - 7x+ 1 =0$ , and $L^{1/8}$ is the greater root of $x^{2} - 3x + 1 = 0$ , otherwise known as $(3 + \sqrt{5})/2$ .

Thus, $a= 3, b=1, c= 5, \text{and} \ d= 2 \implies a+b+c+d= 11$ .