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, 2015

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

A function $f: \mathbb{R} \to \mathbb{R}$ satisfies
$f(f(a)) = f(a) \quad \text{and} \quad f(a+b) = f(a) + f(b)$
for all real numbers $a, b$ . Prove that, for all real $x$ , there exists a unique $y$ such that $f(y) = 0$ and $x = y + f(z)$ for some real $z$ .

Solution

Let $x$ be given. First the uniqueness is proved. Assume that $x = y + f(z)$ with $f(y) = 0$ . If $f$ is applied on both sides, then
$f(x) = f(y + f(z)) = f(y) + f(f(z)) = 0 + f(z) = f(z),$

Now we prove that $y = x - f(x)$ has the assumed property. Observe that $f(a - b) = f(a) - f(b)$ , and hence
$f(y) = f(x - f(x)) = f(x) - f(f(x)) = f(x) - f(x) = 0.$
Thus $x = y + f(x)$ as was to be proved. □

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