2002 AMC 12A Problems/Problem 13

Problem

Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$ . What is $a+b$ ?

Each of the numbers $a$ and $b$ is a solution to $\left| x - \frac 1x \right| = 1$ .

Hence it is either a solution to $x - \frac 1x = 1$ , or to $\frac 1x - x = 1$ . Then it must be a solution either to $x^2 - x - 1 = 0$ , or to $x^2 + x - 1 = 0$ .

There are in total four such values of $x$ , namely $\frac{\pm 1 \pm \sqrt 5}2$ .

Out of these, two are positive: $\frac{-1+\sqrt 5}2$ and $\frac{1+\sqrt 5}2$ . We can easily check that both of them indeed have the required property, and their sum is $\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{\sqrt 5}$ .