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Croatian Mathematical Competitions

Algebra Difficulty 5.2 AIME, harder Prove it Croatia

Let α\alpha be a real number. Find all functions f:R→Rf: \mathbb{R} \to \mathbb{R} such that
f(x+α+f(y))=f(f(x))+f(α)+y,f(x + \alpha + f(y)) = f(f(x)) + f(\alpha) + y,
for all x,y∈Rx, y \in \mathbb{R}.

Solution

Letting x=y=−αx = y = -\alpha we get
f(f(−α))=f(f(−α))+f(α)−α  ⟹  f(α)=α. f(f(-\alpha)) = f(f(-\alpha)) + f(\alpha) - \alpha \implies f(\alpha) = \alpha.
Letting x=−α,y=αx = -\alpha, y = \alpha we get
α=f(f(α))=f(f(−α))+α+α  ⟹  f(f(−α))=−α. \alpha = f(f(\alpha)) = f(f(-\alpha)) + \alpha + \alpha \implies f(f(-\alpha)) = -\alpha.
Now let us denote f(−α)=af(-\alpha) = a. We know that f(a)=−αf(a) = -\alpha. Letting x=α,y=ax = \alpha, y = a we get
α=2α+a  ⟹  a=−α. \alpha = 2\alpha + a \implies a = -\alpha.
Finally, letting y=−αy = -\alpha gives us
f(x)=f(f(x)),∀x∈R, f(x) = f(f(x)), \quad \forall x \in \mathbb{R},
and letting x=−αx = -\alpha gives us
f(f(y))=y,∀y∈R. f(f(y)) = y, \quad \forall y \in \mathbb{R}.
It follows that
f(x)=f(f(x))=x,∀x∈R. f(x) = f(f(x)) = x, \quad \forall x \in \mathbb{R}.
It is easy to check that the function f(x)=xf(x) = x really is a solution.

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