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Algebra Difficulty 4.9 AIME Prove it Belarus

Do there exist a function $f : \mathbb{R} \to \mathbb{R}$ , and real number $\alpha$ such that $f(\alpha) = -2$ and $f(f(x)) = x f(x) + 2x$ for all real $x$ ?

Solution

Answer: such function does not exist.

Suppose, contrary to our claim, that there exists a function satisfying the equality $f(f(x)) = x f(x) + 2x$ for all real $x$ , and $f(\alpha) = -2$ for some $\alpha$ . We have $f(-2) = f(f(\alpha)) = \alpha f(\alpha) + 2\alpha = -2\alpha + 2\alpha = 0$ . Then $f(0) = f(f(-2)) = -2 f(-2) - 4 = -4$ . Further, $f(-4) = f(f(0)) = 0 \cdot f(0) + 2 \cdot 0 = 0$ . Now we have $f(0) = f(f(-4)) = -4 f(-4) - 8 = -4 \cdot 0 - 8 = -8$ . But $f(0) = -4$ , a contradiction.

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