Mongolian Mathematical Olympiad

Let us define two sets $D=\{(x+y+z,xy+yz+zx)\mid x,y,z\in \mathbb{R}\}$ and $E=\{(a,b)\in {\mathbb{R}}^2\mid a^2\ge 3b\}$. It is obvious that $D\subseteq E$. Now let us show that $E\subseteq D$. Indeed, if $a^2\ge 3b$ there exists $c$ such that the polynomial $P(x)=x^3-a{x}^2+bx-c$ has three real roots. Thus, $D=E$ Since the function $h$ can take any value on ${\mathbb{R}}^2\setminus D$, we only need to determine the values of $h$ on $D$. Answer: $f(x)=c$, $h(a,b)=c^2-c$ and $f(x)=x/4$, $h(a,b)=(a^2-4b)/16$, $(a,b)\in D$. These functions obviously satisfy the equation, so let us show that there is no other solution. $(x,y,z)\mapsto (x,z,y)$ implies that $$f(x-y+z{)}^2-f(xz)=h(x+y+z,xy+yz+zx)=f(x+y-z{)}^2-f(xy)(\ast )$$ Now if $x=1$, then $f(y)-f(z)=f(1-y+z{)}^2-f(1+y-z{)}^2$ and $(y,z)\mapsto (y-z,0)$ implies that $f(y-z)-f(0)=f(1-y+z{)}^2-f(1+y-z{)}^2=f(y)-f(z)$. And if $(y,z)\mapsto (x+y,x)$ then $f(x+y)=f(x)+f(y)-f(0)$. Therefore, the function $g(x)=f(x)-f(0)$ is a Cauchy's function. If $z=0$ in $(\star )$, then $f(x+y{)}^2-f(x-y{)}^2=f(xy)-f(0)$ and $g(xy)=(g(x+y)-g(x-y))(g(x+y)+g(x-y)+2f(0))=4g(y)(g(x)+f(0))$. Hence, $f(0)(g(x)-g(y))=0$. If the function $g(x)$ is not constant, $f(0)=0$ and $g(xy)=4g(x)g(y)$. Thus, $g(x)$ is an increasing function, so $g(x)=ax$ and $g(1)=4g(1{)}^2$, $a=1/4$ and $f(x)=x/4$. It is not difficult to see that $h(a,b)=(a^2-4b)/16$ on $D$.