<s>[INST] how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$ Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.
My try:
If $ x=y=0$ then $f(0)=0$ and 
if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then 
$f(x)=f\left( \frac{x+1}{2}  \right)+f\left( \frac{x-1}{2}  \right)$.
But how to find all functions?
 [/INST] Let's investigate the special case that $f$ is continuous.
Let $k=f(1)$ and 
$$S=\{\,t\in\Bbb R\mid f(t)=kt\,\}. $$
Trying $x=y=0$ we find $f(0)=0$. Then with $y=-x$ we find that $f(x)=-f(-x)$, i.e., $f$ is odd. Then also $S=-S$. So far we have $\{-1,0,1\}\subseteq S$.
With $y=0$, we have
$$\tag1f(x^2)=xf(x)$$
hence
$$\begin{align}f(x^2-y^2)&=(x-y)f(x)+(x-y)f(y)\\
&=\left(1-\frac yx\right)xf(x)+\left(\frac xy-1\right)yf(y)\\
&=\left(1-\frac yx\right)f(x^2)+\left(\frac xy-1\right)f(y^2).\end{align} $$
Now if $x^2,y^2\in S$ then 
$$f(x^2-y^2)=\left(1-\frac yx\right)kx^2+\left(\frac xy-1\right)ky^2=kx^2-ky^2$$
and so also $x^2-y^2\in S$. More generally, if two of the numbers $x^2,y^2,x^2-y^2$ are $\in S$ then all three are $\in S$.
We conclude that for two nonnegative numbers $\in S$, also their sum and difference is $\in S$, which makes $S$  a subgroup of $\Bbb R$. Then at least $\Bbb Z\subseteq S$.
From $(1)$, we see that $S$ is closed under taking square roots of positive elements (i.e., $x^2\in S$ implies $x\in S$). As a subgroup, $S$ must then be a dense subset of $\Bbb R$.
By continuity of $f$, the set $S$ must be closed, hence all of $\Bbb R$, as was to be shown.
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