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 $2x_0{y}_0=a$. Hence $$P(x_0,y_0)\to g(b)-g(a)=g(g(x)-y^2)\ge 0$$ So $g$ is an increasing function on non-negative numbers, which mixed with the non-negativity of $g$ results in $g(0)\ge 0⟹g(g(0))\ge g(0)\ge 0$. $$P(0,0)\to g(0)+g(g(0))=g(0)⟹0\le g(0)\le (g(0))=0⟹g(0)=0$$ And by $P(\frac{1}{2},-y)$ and $P(\frac{1}{2},y)$ we have $$\begin{gathered} g(-y)+g\left(g\left(\frac{1}{2}\right)-y^2\right)=g\left(\frac{1}{4}+y^2\right)=g(y)+g\left(g\left(\frac{1}{2}\right)-y^2\right) \\ ⟹g(y)=g(-y) \end{gathered}$$ 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 $2xy=x^2+y^2⟹x=y⟹x^2=g(x)$ which proves our point, so it's left to prove that $g$ is injective. For this purpose, first we prove that $g(a)=0⟺a=0$. 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 $$\{\begin{array}{l}x\le a⟹g(x)=0\\ y^2=a⟹g(-y^2)=g(y^2)=0\\ 2xy\le 2\frac{\sqrt{a}}{2}y=a⟹g(2xy)=0\end{array}$$ 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 $x=\frac{\sqrt{a}}{2}⟺a\ge \frac{\sqrt{a}}{2}⟺a\ge \frac{1}{4}$ and $x=a⟺a\le \frac{1}{4}$. In other words, if $a\ge \frac{1}{4}$ we would have $Q(a)⟹Q(\frac{5a}{4})$ and if $a\le \frac{1}{4}$, we would have $Q(a)⟹Q(a^2+a)$. Note that $a\ge \frac{1}{4}$, $Q(a)⟹Q(\frac{5a}{4})⟹Q(\frac{25a}{16})$ 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 $Q(a)⟹Q(a(na+1))$ for all $n\in {\mathbb{Z}}^{+}$; The basis is already proven: $Q(a)⟹Q(a^2+a)$. For the inductive step, suppose that $Q(a)⟹Q(n{a}^2+a)$. If $n{a}^2+a\ge \frac{1}{4}$, it's already proven that $g(b)=0$ for all $b$, therefore we suppose the contrary, so $$Q(n{a}^2+a)⟹Q\left((n{a}^2+a{)}^2+(n{a}^2+a)\right).$$ But it's easy to see that $$(n{a}^2+a{)}^2\ge a^2⟹(n{a}^2+a{)}^2+(n{a}^2+a)\ge a^2+n{a}^2+a=a((n+1)a+1),$$ therefore by $g$ being increasing, $$Q(n{a}^2+a)⟹Q\left((n{a}^2+a{)}^2+(n{a}^2+a)\right)⟹Q(a((n+1)a+1))$$ which completes the induction. With this proven, it's concluded that $g(x)=0⟺x=0$. 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 $P(y,x)$ $$\begin{array}{rlrlr}g(g(x)-y^2)& =g(g(y)-x^2)=g(a)-g(b)=0& ⟹g(x)& =y^2,g(y)=x^2& \\ ⟹x\ge y& ⟹g(x)\ge g(y)& ⟹y^2\ge x^2& ⟹y\ge x& ⟹y=x\\ ⟹2xy& =x^2+y^2& ⟹a& =b& \end{array}$$ Which is a clear contradiction. Hence $f(x)=\pm x$ is the only possible option for non-constant $f$, which is indeed a solution. ■