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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.

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    what is the co-domain of the function? – 2017-01-22
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    The problem only gives information that the formula is true for any x and y – 2017-01-22
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    This seems to be the same question as [Solve functional equation $(x+y)(f(x)-f(y))=f(x^2)-f(y^2)$](http://math.stackexchange.com/q/1079188) and in somewhat modified version [Find all functions $f$ such that $(x+y)f(x)+f(y^2)=(x+y)f(y)+f(x^2)$](http://math.stackexchange.com/q/379978). You can try [internal search](http://math.stackexchange.com/search?q=%22f%28x%5E%7B2%7D%29-f%28y%5E%7B2%7D%29%3D%28x%2By%29%28f%28x%29-f%28y%29%29%22) or [Google](https://www.google.com/search?q=%22f(x%5E%7B2%7D)-f(y%5E%7B2%7D)=(x%2By)(f(x)-f(y))%22%20site:math.stackexchange.com) or - as already suggested - – 2017-01-22
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    you can also try [Approach0](https://approach0.xyz/search/?q=%24f(x%5E%7B2%7D)-f(y%5E%7B2%7D)%3D(x%2By)(f(x)-f(y))%24&p=1). – 2017-01-22

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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 $a$.

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    Plugging $x=y=0$ yields $0=0$. We can't conclude anything. – 2017-01-22
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    Sorry, you're right; I fixed it. – 2017-01-22