SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO

Only two functions $f(x)\equiv 0$ or $f(x)\equiv 3$ are solutions to the equation. Indeed, let's put $c=f(0)$. By substituting $y=0$, we obtain a quadratic equation $$(f(x){)}^2-2f(x)-c=0,$$ whence $f(x)=1-\sqrt{1+c}$ or $1+\sqrt{1+c}$ for any $x\ne 0$. Suppose that there are two non-zero numbers $a$ and $b$ of the same sign ($ab>0$) such that $f(a)=1-\sqrt{1+c}$ and $f(b)=1+\sqrt{1+c}$. Substituting $x=a$ and $y=b$ into the original equation, we get $$(f(a+b){)}^2-f(a+b)-2=0,$$ whence $f(a+b)=-1$ or $2$. Let's consider two cases.

1) Case $f(a+b)=-1$. Since $a+b\ne 0$ (the sum of the same sign as both terms), then $f(a+b)=1\pm \sqrt{1+c}$ and then necessarily $1-\sqrt{1+c}=-1$. This means $c=3$ and $f(a)=f(a+b)=-1$. For $x=a$ and $y=a+b$ the original equation gives a quadratic equation $$(f(2a+b){)}^2=f(2a+b)-2,$$ which has no solutions for $f(2a+b)$, which is impossible.

2) Case $f(a+b)=2$. Since $a+b\ne 0$, then $f(a+b)=1\pm \sqrt{1+c}$ and so necessarily $1+\sqrt{1+c}=2$. This means that $c=0$ and $f(b)=f(a+b)=2$. For $x=a+b$ and $y=b$, the original equation gives a quadratic equation for $f(a+2b)$: $$(f(a+2b){)}^2=f(a+2b)+4,$$ which has roots $\frac{1\pm \sqrt{17}}{2}$. However, also $a+2b\ne 0$ and $f(a+2b)=1\pm \sqrt{1+0}=0$ or $2$. Contradiction.

Thus, on each of the open intervals $]-\infty ,0[$ and $]0,\infty [$ the function $f$ takes only one value $1-\sqrt{1+c}$ or $1+\sqrt{1+c}$. Therefore, if $x$ and $y$ are non-zero numbers of the same sign, then $f(x+y)=f(x)=f(y)=1\pm \sqrt{1+c}$ and $$(1\pm \sqrt{1+c}{)}^2=3(1\pm \sqrt{1+c}),$$ whence $1\pm \sqrt{1+c}=0$ or $3$. Since $1-\sqrt{1+c}\le 1\le 1+\sqrt{1+c}$, two options are possible: $$1-\sqrt{1+c}=0⟺c=0\text{or}1+\sqrt{1+c}=3⟺c=3.$$ Now let's substitute $y=-x$ into the original equation: $$(f(0){)}^2=f(0)+f(x)+f(-x)⟺f(x)+f(-x)=c^2-c\text{ for any }x\ne 0.$$ If $c=0$ then on the entire set $\mathbb{R}$ the function $f$ can take at most two values $(0$ or $1+\sqrt{1+0}=2)$. If $f(a)=2$ for some $a\ne 0$, then $f(a)+f(-a)=2+f(-a)=0^2-0=0$, which is impossible, since $f(-a)=0$ or $2$. This means $f(x)\equiv 0$. If $c=3$ then on the entire set $\mathbb{R}$ the function $f$ can take at most two values $(3$ or $1-\sqrt{1+3}=-1)$. If $f(b)=-1$ for some $b\ne 0$, then $f(b)+f(-b)=-1+f(-b)=3^2-3=6$, which is impossible, since $f(-b)=-1$ or $3$. So $f(x)\equiv 3$. We can check that the functions $f(x)\equiv 0$ or $f(x)\equiv 3$ are suitable.