Iranian Mathematical Olympiad

We shall firstly prove following lemmas;

Lemma 1. For any real number $r$, the length of the interval $f(r)$ is at most 2.
Proof. If $x, y \in f(r)$ we have $r \in f(x) \cap f(y)$ hence we have $|x - y| \le 2$.

Lemma 2. If the interval $I$ has length 2, then there is a real number $r$ such that $f(r) = I$.
Proof. If $I = [x, x+2]$ then by condition 2 there is $r \in f(x) \cap f(x+2)$ so we have $x, x+2 \in f(r)$. But the length of $f(r)$ is at most 2 hence we have $f(r) = [x, x+2]$, this completes our proof.

Lemma 3. f is an injective function and for any r the length of f(r) is 2.
Proof. Using preceding lemmas it is enough to show that if $f(x) \subset f(y)$ then $x = y$. Assume the contrary, let $I$ be an interval of length 2 such that $x \in I, y \notin I$ then by lemma 2 there exists $r$ such that $I = f(r)$. Now we have $r \in f(x), r \notin f(y)$ which contradicts $f(x) \subset f(y)$.

Define a new function $g: \mathbb{R} \to \mathbb{R}$ such that for all $r$, $g(r)$ is the midpoint of $f(r)$. Since we know the length of $f(r)$ we can forget about $f$ and work with $g$. We apt to find all bijective functions $g$ such that:
$$i. \quad |x - g(y)| \le 1 \iff |y - g(x)| \le 1.$$
$$ii. \quad |g(x) - g(y)| \le 2 \iff |x - y| \le 2.$$
$$iii. \quad g(r) = r^2 \text{ for all } 0 \le r \le 1.$$

Lemma 4. g is strictly increasing.
Proof. It suffices to prove if $x < y, y - x < 1$ then $g(x) < g(y)$. We firstly consider the case $y = 0$. For each $-2 \le x \le 0$ by condition 2 and $g(0) = 0$ we have
$$|g(x) - g(2)| > 2, |g(x)| \le 2, |g(2)| \le 2$$
so we can deduce that $g(x)$ and $g(2)$ have different signs, similarly we can deduce that $g(2), g(1) = 1$ have the same sign (they are both opposite to the sign of $g(-2)$). Hence we have $g(2) > 0, g(x) < 0$.
We prove by induction on $\lfloor x \rfloor$ (We only prove the case that $0 \le x < y$ the other case is similar). For the base if $0 \le x < y < 1$ then $x^2 < y^2$. We know that $y - x - 2 > 2$ hence by the second condition we have
$$|g(y) - g(x - 2)| > 2, |g(y) - g(x)| \le 2, |g(x) - g(x - 2)| \le 2.$$
We know by induction that $g(x) - g(x - 2) > 0$ hence we have $g(y) > g(x)$.

Lemma 5. *we have $g(x + 1) = g^{-1}(x) + 1$.*
Proof. We know that $g$ is bijective so it suffices to prove that $g(g(x) + 1) = x + 1$. Assume that $g(r) = x + 1$ then we have $r \le g(x) + 1$. Now if $r' > r$ we have $g(r') > x + 1$ hence by condition 1; $r' > g(x) + 1$ so we get $r = g(x) + 1$.

By the last lemma we obtain that there is only one function that satisfies the condition of the problem and that function is:
$$g(x) = \begin{cases} (x - \lfloor x \rfloor)^2 + \lfloor x \rfloor & \lfloor x \rfloor \equiv 0 \pmod{2}; \\ \sqrt{x - \lfloor x \rfloor} + \lfloor x \rfloor & \lfloor x \rfloor \equiv 1 \pmod{2}. \end{cases}$$
It is easy to check that this function actually works. ■