Competition problem — Algebra — Maths Olympiad Prep

Library / /65 of 196

Algebra Difficulty 4.8 AIME Prove it Soviet Union

Problem:

$x$ is a real number. Define $x_0 = 1 + \sqrt{1 + x}$ , $x_1 = 2 + x / x_0$ , $x_2 = 2 + x / x_1$ , ..., $x_{1985} = 2 + x / x_{1984}$ . Find all solutions to $x_{1985} = x$ .

Solution

Solution:

If $x = 0$ , then $x_{1985} = 2 \neq x$ .

Otherwise we find
$x_1 = 2 + \frac{x}{1 + \sqrt{1 + x}} = 2 + (\sqrt{1 + x} - 1) = 1 + \sqrt{1 + x}.$
Hence $x_{1985} = 1 + \sqrt{1 + x}$ .

So $x - 1 = \sqrt{1 + x}$ .

Squaring, $x = 0$ or $3$ .

We have already ruled out $x = 0$ . It is easy to check that $x = 3$ is a solution.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.