Competition problem — Algebra — Maths Olympiad Prep

Library / /6 of 299

Algebra Difficulty 5.0 AIME, harder Prove it Iran

Let $f, g : \mathbb{R}^+ \to \mathbb{R}^+$ be two functions such that for all positive real numbers $x$ and $y$
$f(x + g(y))^2 = f(x^2) + y^2.$
Prove that the range of $g$ is not bounded from above.

Solution

Let $P(x, y)$ be the assertion
$f(x + g(y))^2 = f(x^2) + y^2$
If $g(f(t)) < t$ for some $t > 0$ , then $P(t - g(f(t)), f(t))$ implies
$f(t)^2 = f((t - g(t))^2) + f(t)^2 \implies f((t - g(t))^2) = 0$
which is impossible.
So $g(f(x)) \ge x$ , for all positive real numbers $x$ . Hence the solution is complete. ■

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.