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Algebra Difficulty 4.9 AIME Prove it Belarus

Find all triples $(a; b; c)$ of real numbers for which there exists a non-zero function $f$ , $f: \mathbb{R} \to \mathbb{R}$ , such that
$af(xy + f(z)) + bf(yz + f(x)) + cf(zx + f(y)) = 0$
for all real $x, y, z$ .

Solution

First, note that if $a + b + c = 0$ , then the function $f \equiv \lambda \neq 0$ satisfies the condition.

Now suppose that $a + b + c \neq 0$ . In particular, at least one of $a, b, c$ is not zero. Say, $a \neq 0$ . Let $(*)$ denote the given functional equation.

Set $x = y = 0$ in $(*)$ , then $af(f(z)) = 0$ , or $f(f(z)) = 0$ for all $z \in \mathbb{R}$ .

Now set $z = 0$ in $(*)$ , then $af(xy + f(0)) = 0$ , or $f(xy + f(0)) = 0$ . But $xy + f(0)$ takes all real values, hence $f$ is a zero constant. This contradiction does prove that $a + b + c = 0$ .

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