IMC 1999 · Day 1 · P4

Let $g(x) = \dfrac{f(x)}{x}$. We have $g \left( \dfrac{x}{g(x)} \right) = g(x)$. By induction it follows that $g \left( \dfrac{x}{g^n(x)} \right) = g(x)$, i.e. $$\tag{1} f \left( \frac{x}{g^n(x)} \right) = \frac{x}{g^{n-1}(x)}, \quad n \in \mathbb{N}.$$ On the other hand, let substitute $x$ by $f(x)$ in $f \left( \dfrac{x^2}{f(x)} \right) = x$. From the injectivity of $f$ we get $\dfrac{f^2(x)}{f(f(x))} = x$, i.e. $g(x g(x)) = g(x)$. Again by induction we deduce that $g(x g^n(x)) = g(x)$ which can be written in the form $$\tag{2} f(x g^n(x)) = x g^{n-1}(x), \quad n \in \mathbb{N}.$$ Set $f^{(m)} = \underbrace{f \circ f \circ \dots \circ f}_{m \text{ times}}$. It follows from (1) and (2) that $$\tag{3} f^{(m)}(x g^n(x)) = x g^{n-m}(x), \quad m, n \in \mathbb{N}.$$ Now, we shall prove that $g$ is a constant. Assume $g(x_1) < g(x_2)$. Then we may find $n \in \mathbb{N}$ such that $x_1 g^n(x_1) \le x_2 g^n(x_2)$. On the other hand, if $m$ is even then $f^{(m)}$ is strictly increasing and from (3) it follows that $x_1^m g^{n-m}(x_1) \le x_2^m g^{n-m}(x_2)$. But when $n$ is fixed the opposite inequality holds $\forall m \gg 1$. This contradiction shows that $g$ is a constant, i.e. $f(x) = Cx$, $C > 0$.

Conversely, it is easy to check that the functions of this type verify the conditions of the problem.