IMC 2008 · Day 1 · P1

We prove that $f(x) = ax + b$ where $a \in \mathbb{Q}$ and $b \in \mathbb{R}$. These functions obviously satify the conditions.

Suppose that a function $f(x)$ fulfills the required properties. For an arbitrary rational $q$, consider the function $g_q(x) = f(x + q) - f(x)$. This is a continuous function which attains only rational values, therefore $g_q$ is constant.

Set $a = f(1) - f(0)$ and $b = f(0)$. Let $n$ be an arbitrary positive integer and let $r = f(1/n) - f(0)$. Since $f(x + 1/n) - f(x) = f(1/n) - f(0) = r$ for all $x$, we have $$f(k/n) - f(0) = (f(1/n) - f(0)) + (f(2/n) - f(1/n)) + \dots + (f(k/n) - f((k-1)/n) = kr$$ and $$f(-k/n) - f(0) = -(f(0) - f(-1/n)) - (f(-1/n) - f(-2/n)) - \dots - (f(-(k-1)/n) - f(-k/n) = -kr$$ for $k \ge 1$. In the case $k = n$ we get $a = f(1) - f(0) = nr$, so $r = a/n$. Hence, $f(k/n) - f(0) = kr = ak/n$ and then $f(k/n) = a \cdot k/n + b$ for all integers $k$ and $n > 0$.

So, we have $f(x) = ax + b$ for all rational $x$. Since the function $f$ is continous and the rational numbers form a dense subset of $\mathbb{R}$, the same holds for all real $x$.