Mongolian IMO Team Selection Tests, 2015

By the intermediate value theorem, it suffices to prove that $f$ takes both non-negative and non-positive values.

**Case $f \ge 0$:** Suppose that $f(x) \ge 0$ for all $x \in \mathbb{R}$. Then we have
$$f(x) = \sqrt{\frac{1 + f(2x)}{2}}$$
for all $x \in \mathbb{R}$. Since the function
$$\phi: [0, \infty) \to [0, \infty), \quad t \mapsto \sqrt{\frac{1+t}{2}},$$
is strictly increasing we see that $f \ge \sqrt{\frac{1}{2}}$. In fact, for the sequence defined by $t_0 = 0$ and $t_n = \phi(t_{n-1})$, $n \ge 1$, we have $f \ge t_n$ for all $n$ by induction. However, it is easy to see that $t_n \nearrow 1$. Thus we see that $f \ge 1$. Now suppose that for some $x_0 \in \mathbb{R}$, we have $f(x_0) > 1$. Then $f(2x_0) = f(x_0)^2 + f(x_0)^2 - 1 > f(x_0)^2 > 1$. By induction, we have $f(2^n x_0) > f(x_0)^{2^n}$ for all $n \in \mathbb{Z}_{\ge 1}$, which contradicts the boundedness of $f$. Hence $f \le 1$ and thus $f = 1$, which is impossible since $f$ is non-constant.

**Case $f \le 0$:** Suppose that $f(x) \le 0$ for all $x \in \mathbb{R}$. Then we have
$$f(x) = -\sqrt{\frac{1 + f(2x)}{2}}$$
for all $x \in \mathbb{R}$. Note that the function
$$\psi: [-1,0] \to [-1,0], \quad t \mapsto -\sqrt{\frac{1+t}{2}},$$
is strictly decreasing. Thus for the sequence defined by $s_0 = 0$ and $s_n = \phi(s_{n-1})$, $n \ge 1$, we have
$$s_{2n+1} \le f \le s_{2n}$$
for all $n \ge 0$ by induction. Moreover, it is clear that $s_0 \ge s_2 \ge s_4 \ge \dots$ and $s_1 \le s_3 \le s_5 \le \dots$ so we consider the limits $s_{2n} \searrow a$ and $s_{2n+1} \nearrow b$. Clearly, both $a$ and $b$ are solutions of the equation $\psi(\psi(s)) = s$ belonging to the interval $[-1,0]$. However, it is easy to check that the solutions of $\psi(\psi(s)) = s$ are roots of the polynomial $P(s) = 8s^4 - 8s^2 - s + 1 = (s-1)(2s+1)(4s^2 + 2s - 1)$. But $s = -1/2$ is the only root of $P(s)$ in the interval $[-1,0]$, thus $a = b = -1/2$. Hence $f$ is constant, a contradiction.

Combining the cases we get the solution.