Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Belarus

Find all real numbers $a$ for which there exists a function $f$ defined on the set of all real numbers which takes as its values all real numbers exactly once and satisfies the equality
$f(f(x)) = x^2 f(x) + a x^2$
for all real $x$ .

Solution

Answer: $a = 0$ .

Substituting $x$ such that $f(x) = -a$ , we get $f(-a) = 0$ .

Substituting $x = -a$ , we get $f(0) = a^3$ .

Finally, substituting $x = 0$ , we get $f(a^3) = 0$ .

Since $f$ takes all real values exactly once, $a^3 = -a$ which is equivalent to $a(a^2 + 1) = 0$ , i.e. $a = 0$ .

Clearly, for $a = 0$ the function $f(x) = x|x|$ satisfies the conditions of the problem.

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