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, 2012

Algebra Difficulty 4.6 AIME Prove it Slovenia

Prove that there is no injective function $f: \mathbb{R} \to \mathbb{R}$ such that
$f(f(x) + y) = f(x + y) + f(2012) \quad \text{for all } x, y \in \mathbb{R}.$

Solution

If we substitute $y = -x$ in the equation, we get $f(f(x)-x) = f(0)+f(2012)$ . Since the right side of the equation is a constant, and $f$ is an injective function, $f(x)-x$ must also be a constant. Hence $f(x) = x + c$ for a real number $c$ . If we substitute this in the initial equation, we get $x + y + 2c = x + y + 2c + 2012$ , which gives us the contradiction $0 = 2012$ .

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