Romanian Mathematical Olympiad

Let $(f, g)$ be a pair of functions satisfying the given condition. We shall show that $f''$ is constant. Suppose that $f''$ is not constant. Then, because $(f(x) - g(0)) \cdot (f'(x) - g'(0)) \cdot (f''(x) - g''(0)) = 0$, for all $x \in \mathbb{R}$, and $f''$ is continuous, there is $a \in \mathbb{R}$ and $r > 0$ such that $f''(x) \notin \{0, g''(0)\}$, for any $x \in (a - r, a + r)$. It follows that $(f(x) - g(0)) \cdot (f'(x) - g'(0)) = 0$, for $x \in (a - r, a + r)$.

Two cases are possible.
Case 1. There exists $b \in (a - r, a + r)$ such that $f'(b) \ne g'(0)$. By continuity of $f'$, there is $s > 0$ such that $(b - s, b + s) \subset (a - r, a + r)$ and $f'(x) \ne g'(0)$, for all $x \in (b - s, b + s)$, implying $f(x) = g(0)$ for $x \in (b - s, b + s)$. We get $f''(x) = 0$ for $x \in (b - s, b + s)$, a contradiction.

Case 2. $f'(x) = g'(0)$, for any $x \in (a-r, a+r)$. We obtain $f''(x) = 0$, for all $x \in (a-r, a+r)$, a contradiction again.

Thus $f''$ is constant on $\mathbb{R}$.
Let $m \in \mathbb{R}$ such that $f''(x) = 2m$, for any $x \in \mathbb{R}$. As $(f'(x) - 2mx)' = 0$, for all $x \in \mathbb{R}$, there is $n \in \mathbb{R}$ such that $f'(x) - 2mx - n = 0$, for $x \in \mathbb{R}$. As a consequence there exists $p \in \mathbb{R}$ such that $f(x) = mx^2 + nx + p$, for all $x \in \mathbb{R}$.
In the same way, $g''$ is constant on $\mathbb{R}$, implying the existence of $m', n', p' \in \mathbb{R}$ such that $g(x) = m'x^2 + n'x + p'$, for $x \in \mathbb{R}$. In the case $m \neq m'$, as the equation $(f(x) - g(x)) \cdot (f'(x) - g'(x)) = 0$ has at most three real solution, the condition of the problem is not possible. Thus $m = m'$, so the pairs of functions satisfying the given conditions are of the form $f(x) = mx^2 + nx + p$ and $g(x) = mx^2 + n'x + p'$, for any $x \in \mathbb{R}$.