Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Austria

Let $\alpha$ and $\beta$ be real numbers with $\beta \neq 0$ . Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$f(\alpha f(x) + f(y)) = \beta x + f(y)$
holds for all real $x$ and $y$ .

Solution

The function $f$ is injective using the variable $x$ (on the left $x$ only occurs as $f(x)$ , on the right $x$ is free with a non-vanishing factor, so substituting $x = a$ and $x = b$ with $f(a) = f(b)$ gives the desired conclusion).

We set $x = 0$ and remove the outer $f$ due to the injectivity and obtain $f(y) = y + C$ .

Substituting into the original equation shows that this is equivalent to $\alpha = \beta$ (coefficient of $x$ ) and $(1+\alpha)C = 0$ (constant coefficient).

This gives the solutions $f(x) = x$ for $\alpha = \beta$ and $f(x) = x + C$ for $\alpha = \beta = -1$ .

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