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, 2012

Algebra Difficulty 4.9 AIME Prove it Belarus

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

Solution

Answer: such function does not exist.

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

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