While doing some old INMO (Indian National Mathematical Olympiad) problems I am stuck on a question which is as follow:
Find all functions $f:\mathbf{R} \to \mathbf{R}$ satisfying the relation $f(x^2+y(f(x)))=x(f(x+y))$.
Though I have worked on many problems related to functions, but still I m clueless at this one. I shall be highly thankful if you can give me some hints/suggetions. Thanks.
Claim: $f(x)=0$ or $f(x)=x$ for all $x \in \mathbf{R}$.
Set $x=0$, then $f(yf(0))=0$ hence $f\equiv 0$ is a solution. Otherwise $f$ is not constant and $f(0)=0$.
Set $y=0$, then $f(x^2)=xf(x)$ for all $x$. In particular $x^2=(-x)^2$ hence $xf(x)=-xf(-x)$. Therefore $f$ is a odd function.
Suppose there exists $x_0\neq 0$ such that $f(x_0)=0$. Setting $x=x_0$ we have $f(x_0^2)=x_0f(x_0+y)$; but $f$ is not constant, hence it is a contradiction.
Set $x+y=0$, then $f(x^2-xf(x))=0$. Using (2) and (3), we have $f(x^2-f(x^2))=0$ hence $x^2-f(x^2)=0$. Therefore $f(z)=z$ for all $z\ge 0$. Using that $f$ is odd by (2), then $f$ is the identity.