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, 2011

Algebra Difficulty 7.7 National Olympiad, round 2 Prove it Baltic Way

Let $f: \mathbb{Z} \to \mathbb{Z}$ be a function such that, for all integers $x$ and $y$ , the following holds:
$f(f(x) - y) = f(y) - f(f(x)).$
Show that $f$ is bounded, ie. that there is a $C$ such that
$-C < f(x) < C$
for all $x$ .

Solution

First, setting $y = f(x)$ one obtains $f(0) = 0$ . Secondly $y = 0$ yields $f(f(x)) = 0$ for all $x$ , thus
$f(f(x) - y) = f(y).$
Setting $x = 0$ yields $f(-y) = f(y)$ , and finally $y := -z$ yields
$f(f(x) + z) = f(-z) = f(z).$
If $f(x) = 0$ for all $x$ , then $f$ is obviously bounded. If on the other hand there exists an $x_0$ such that $f(x_0) \neq 0$ , then, with $x = x_0$ , the last equality gives that $f$ is periodic with period $|f(x_0)|$ and thus $f$ must be bounded.

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