Find all the functions for which $f(x^{2})-f(y^{2})=(x+y)(f(x)-f(y)):\forall x, y\in\mathbb{R}$

Question

Hey guys I need help on solving the following problem Find all the functions for which : $$f(x^{2})-f(y^{2})=(x+y)(f(x)-f(y)):\forall x, y\in\mathbb{R}$$ I started with taking $x=y$, and there you see that the both sides are equal to $0$, therefore the solution is the function $y=x$, but after that I can't prove that that is the only solution (if it is the only). Would appreciate some help, thanks.

Answer 1

We suppose $f(0)=0$. With $y=0$ we obtain $f(x^2)=xf(x)$. Thus $$ xf(x)-yf(y)=f(x^2)-f(y^2)=(x+y)(f(x)-f(y))=xf(x)-xf(y)+yf(x)-yf(y)\implies\\ yf(x)=xf(y) $$ for all $x,y\in\mathbb{R}$. With $y=1$ this yields $f(x)=f(1)x$ thus $f$ is of the form $f(x)=cx$ for some constant $c$. We see that whatever the value of $c$ such a function satisfies the equation. If $f(0)\neq 0$, we see that $g(x)=f(x)-f(0)$ still satisfies the equation and $g(0)=0$ so $g(x)=cx$ and thus all solutions are of the form $f(x)=cx+a$ for constants $c$ and $a$.