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Algebra Difficulty 4.9 AIME Prove it Belarus

Find all possible values of real number $a$ such that there exist a function $f : \mathbb{R} \to \mathbb{R}$ , and real number $\alpha$ satisfying the equalities $f(\alpha) = 0$ and $f(f(x)) = x f(x) + a$ for all real $x$ .

Solution

Answer: $a = 0$ .
Indeed, if $a = 0$ , then the function $f \equiv 0$ satisfies the condition. Now let $a \neq 0$ . Suppose that $f(\alpha) = 0$ for some $\alpha$ . We have $f(0) = f(f(\alpha)) = \alpha \cdot f(\alpha) + a = a$ . Then $f(a) = f(f(0)) = 0 \cdot f(0) + a = a$ . Therefore, $a = f(a) = f(f(a)) = a \cdot f(a) + a = a \cdot a + a = a^2 + a$ , i.e. $a = a^2 + a$ , and then $a = 0$ , a contradiction.

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