Romanian Mathematical Olympiad

We shall show that only the identity function and the constant functions satisfy the conditions of the problem. It is clear that these functions verify indeed the conditions.

Because $f$ is continuous, its range $\{f(x) \mid x \in \mathbb{R}\}$ is an interval $I \subseteq \mathbb{R}$. If $I$ is degenerate at a point then $f$ is constant.

If $I$ is non-degenerate, let $a = \inf I < \sup I = b$, where $a, b \in \bar{\mathbb{R}}$. By the given condition we deduce that the restriction of $f$ to the interval $(a, b)$ is the identity:
$$f(x) = x, \quad a < x < b. \tag{1}$$
We shall show that $a = -\infty$ and $b = +\infty$, i.e. $I = \mathbb{R}$ and $f$ is the identity function. Suppose $a$ is a finite number. By the continuity of $f$ in $a$ and by (1) we get $f(a) = a$, so
$$f'(a) = f'_d(a) = \lim_{x \to a,\ x > a} \frac{f(x) - f(a)}{x - a} = \lim_{x \to a,\ x > a} \frac{x - a}{x - a} = 1. \tag{2}$$
On the other side $f$ has a minimum at $a$, because
$$f(a) = a = \inf I = \inf\{f(x) \mid x \in \mathbb{R}\},$$
so, by the Theorem of Fermat, $f'(a) = 0$, in contradiction with (2). We conclude $a = -\infty$. Analogously $b = +\infty$.

Therefore, the only differentiable functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f \circ f = f$ are the constant functions and the identity function.