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Estonian Math Competitions

Algebra Difficulty 4.9 AIME Prove it Estonia

A function f:R→Rf: \mathbb{R} \to \mathbb{R} satisfies for all real xx and yy the equation 2f(x)(f(y))2+y2f(−x∣y∣)=f(xy2)2f(x)(f(y))^2 + y^2f(-x|y|) = f(xy^2). Find all possible values of f(1)f(1).

Solution

Denote f(1)=af(1) = a and f(−1)=bf(-1) = b. Choosing x=y=1x = y = 1 and x=y=−1x = y = -1 yields the system of equations
{2a3+b=a2b3+a=b \begin{cases} 2a^3 + b = a \\ 2b^3 + a = b \end{cases}
Adding the equations and cancelling equal terms, we get 2b3=−2a32b^3 = -2a^3. Therefore b=−ab = -a. Substituting this into the first equation yields 2a3−2a=02a^3 - 2a = 0. So f(1)f(1) is one of the numbers −1-1, 00, 11. These values can be obtained by the functions f(x)=−x∣x∣f(x) = -x|x|, f(x)=0f(x) = 0 and f(x)=x∣x∣f(x) = x|x| respectively, which satisfy the equations.

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