Solving a high school conjecture

Solution 1:

This is what's called a continued fraction. Let's denote your continued fraction as $f(x)$. Notice that if we subtract $x$ from $f(x)$ and take its reciprocal, we get $f(x)$ back. Therefore, $$\frac{1}{f(x)-x} = f(x)$$ or $$f(x)^2-xf(x)-1 = 0$$ Solving for $f(x)$ using the quadratic formula, we get $$f(x) = \frac{x\pm\sqrt{x^2+4}}{2}$$ It's not hard to see that if $x$ is positive, then $f(x)$ must be positive (as the reciprocal of a positive number is always positive, as is the sum of two positive numbers). Therefore, for $x > 0$, we have $$f(x) = \frac{x+\sqrt{x^2+4}}{2}$$ This is the same as $\frac{x}{2}+\sqrt{\left(\frac{x}{2}\right)^2+1}$ (where we have just distributed the $\frac{1}{2}$). Notice that $f(1) = \frac{1+\sqrt{5}}{2} = \phi$.