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Belarusian Mathematical Olympiad

Algebra Difficulty 5.1 AIME, harder Prove it Belarus

Find all functions $f : \mathbb{R} \to \mathbb{R}$ , $g : \mathbb{R} \to \mathbb{R}$ satisfying the following equality $f(f(x+y)) = x f(y) + g(x)$ for all real $x$ and $y$ . (I. Gorodnin)

Solution

Answer: $f(x) = a$ , $g(x) = a - a x$ for all $a \in \mathbb{R}$ .

Let $a = f(0)$ , $b = g(0)$ . Note that due to the symmetry the given equation implies
$x f(y) + g(x) = y f(x) + g(y). \tag{1}$
Set $y = 0$ in (1), then we have $g(x) = b - a x$ . Now set $y = 1$ in (1), then $f(x) = (f(1) - a)x + a = a + c x$ . Now substituting $g(x)$ and $f(x)$ in the given equation we easily obtain that $c = 0$ , $b = a$ .

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