Iranian Mathematical Olympiad

Let $P$ denote the assertion that: $$f(x^3+xf(xy))=f(xy)+x^{2}f(x+y)$$ By $P(x,\frac{y}{x})$ we have $$A(x,y):f(x^3+xf(y))=f(y)+x^{2}f\left(x+\frac{y}{x}\right)$$ If there exists a pair $(x,y)$ such that $f(x^3+xf(y))=0$. Then $$A(x,y)\to f(y)+x^{2}f\left(x+\frac{y}{x}\right)=0(1)$$ From the definition of function we know that $\forall x\ge 0:f(x)\ge 0$ so (1) infers that $f(y)=0$. Thus it suffices to prove for each $t\ge 0$ there exists $x\ge 0$ such that $f(x^3+xf(t))=0$.

Consider the polynomial $P(x)=x^3+xf(1)-1$. It has at least one positive real root, namely $x_0$ since $P(0)<0$ and the leading coefficient of polynomial is positive. From $A(x_0,1)$ we have $$f(x_0^3+x_{0}f(1))=f(1)⟹f\left(x_0+\frac{1}{x_0}\right)=0$$ Thus there is a positive number $c$ such that $f(c)=0$. For any non-negative number $t$ consider the polynomial $Q_t(x)=x^3+xf(t)-c$. Again $Q_t$ has at least one positive root like $x_1$ because $Q_t(0)<0$ and the leading coefficient is positive. Then $$f(x_1^3+x_{1}f(t))=f(c)=0$$ And $x_1$ is what we supposed to find for each $t$ to conclude $f(t)=0$. ■