Iranian Mathematical Olympiad

Obviously, the constant function $f(x) = 0$ is an answer. Let $f$ be a non-constant function satisfying the problem.
Define $g$ to be $g(x) = f(x)^2$ for all $x$. Since $f(x)^2 \ge 0$, $g$ would always be non-negative.
Let $P$ denote the assertion that
$$g(2xy) + g(g(x) - y^2) = g(x^2 + y^2)$$
For every $a \ge b \ge 0$, there exists $x_0, y_0$ where $x_0^2 + y_0^2 = b$ and $2x_0y_0 = a$. Hence
$$P(x_0, y_0) \rightarrow g(b) - g(a) = g(g(x) - y^2) \ge 0$$
So $g$ is an increasing function on non-negative numbers, which mixed with the non-negativity of $g$ results in $g(0) \ge 0 \implies g(g(0)) \ge g(0) \ge 0$.
$$P(0, 0) \rightarrow g(0) + g(g(0)) = g(0) \implies 0 \le g(0) \le (g(0)) = 0 \implies g(0) = 0$$
And by $P(\frac{1}{2}, -y)$ and $P(\frac{1}{2}, y)$ we have
$$g(-y) + g\left(g\left(\frac{1}{2}\right) - y^2\right) = g\left(\frac{1}{4} + y^2\right) = g(y) + g\left(g\left(\frac{1}{2}\right) - y^2\right) \\ \implies g(y) = g(-y)$$
Take $x$ to be a real number, it suffices to prove that $g(x) = x^2$ and then $f(x) = \pm x$ would satisfy the problem. Since $g$ is an even function, by proving that $g(x) = x^2$ for $x \ge 0$, the same would be proven for $x < 0$; hence it suffices to prove that $g(x) = x^2$ for $x > 0$.
Since $g(x) \ge 0$, there exists $y$ in which $y^2 = g(x)$. $P(x, y)$ yields in
$$g(2xy) = g(2xy) + g(g(x) - g(x)) = g(x^2 + y^2)$$
If we prove $g$ to be an injective function, this will result in $2xy = x^2 + y^2 \implies x = y \implies x^2 = g(x)$ which proves our point, so it's left to prove that $g$ is injective.
For this purpose, first we prove that $g(a) = 0 \iff a = 0$. Suppose the contrary: There exists $a > 0$ in which $g(a) = 0$ and let $Q(a)$ denote this proposition, also there exists $b > 0$ satisfying $g(b) > 0$. $g$ is increasing, hence $g(x) = 0$ for all $0 \le x \le a$. Let $y = \sqrt{a}$ and $x = \min\left(a, \frac{\sqrt{a}}{2}\right)$; we clearly have
$$\begin{cases} x \le a \implies g(x) = 0 \\ y^2 = a \implies g(-y^2) = g(y^2) = 0 \\ 2xy \le 2\frac{\sqrt{a}}{2}y = a \implies g(2xy) = 0 \end{cases}$$
By $P(x, y)$ we conclude
$$g(2xy) + g(g(x) - y^2) = g(0 - y^2) = 0 = g(x^2 + y^2) = g(a + x^2)$$
So $x = \frac{\sqrt{a}}{2} \iff a \ge \frac{\sqrt{a}}{2} \iff a \ge \frac{1}{4}$ and $x = a \iff a \le \frac{1}{4}$. In other words, if $a \ge \frac{1}{4}$ we would have $Q(a) \implies Q(\frac{5a}{4})$ and if $a \le \frac{1}{4}$, we would have $Q(a) \implies Q(a^2+a)$. Note that $a \ge \frac{1}{4}$, $Q(a) \implies Q(\frac{5a}{4}) \implies Q(\frac{25a}{16})$ which inductively proves that $Q(\frac{5^n a}{4^n})$ for all $n \in \mathbb{Z}^+$, and it's easy to see that there exists $k \in \mathbb{Z}^+$ satisfying $b \le \frac{5^k a}{4^k}$ which is a clear contradiction. On the other hand, we shall prove that $a \le \frac{1}{4}$ results in $Q(a) \implies Q(a(na+1))$ for all $n \in \mathbb{Z}^+$; The basis is already proven: $Q(a) \implies Q(a^2+a)$. For the inductive step, suppose that $Q(a) \implies Q(na^2+a)$. If $na^2+a \ge \frac{1}{4}$, it's already proven that $g(b) = 0$ for all $b$, therefore we suppose the contrary, so
$$Q(na^2 + a) \implies Q\left((na^2 + a)^2 + (na^2 + a)\right).$$
But it's easy to see that
$$(na^2 + a)^2 \ge a^2 \implies (na^2 + a)^2 + (na^2 + a) \ge a^2 + na^2 + a = a((n+1)a + 1),$$
therefore by $g$ being increasing,
$$Q(na^2 + a) \implies Q\left((na^2 + a)^2 + (na^2 + a)\right) \implies Q(a((n+1)a + 1))$$
which completes the induction. With this proven, it's concluded that $g(x) = 0 \iff x = 0$.
Now we prove that $g$ is injective. Suppose on the contrary that $g(a) = g(b)$ for some $a > b$. There exists some $x, y \ge 0$ satisfying $x^2+y^2 = a$ and $2xy = b$. Without loss of generalization assume that $x \ge y$. Therefore by $P(x, y)$ and $P(y, x)$
$$\begin{align*} g(g(x) - y^2) &= g(g(y) - x^2) = g(a) - g(b) = 0 & \implies g(x) &= y^2, g(y) = x^2 \\ \implies x \ge y &\implies g(x) \ge g(y) &\implies y^2 \ge x^2 &\implies y \ge x &\implies y = x \\ \implies 2xy &= x^2 + y^2 &\implies a &= b \end{align*}$$
Which is a clear contradiction. Hence $f(x) = \pm x$ is the only possible option for non-constant $f$, which is indeed a solution. ■