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?
 A: We can prove that if a function $f:\mathbb R\to\mathbb R$ satisfies
$$f\left(x^2-y^2\right)=(x-y)\big(f(x)+f(y)\big)\tag0\label0$$
for every real numbers $x$ and $y$, then there is a constant real number $k$ such that $f(x)=kx$ for every real number $x$.
First, let $y=x$ in \eqref{0} and get $f(0)=0$. Next, let $x=1$ and $y=-1$ in \eqref{0} and get $f(-1)=-f(1)$. Now, define $k:=f(1)$ and $g(x):=f(x)-kx$. So we have $g(1)=g(-1)=0$ and:
$$g\left(x^2-y^2\right)+k\left(x^2-y^2\right)=(x-y)\big(g(x)+g(y)+k(x+y)\big)$$
$$\therefore\quad g\left(x^2-y^2\right)=(x-y)\big(g(x)+g(y)\big)\tag1\label1$$
Thus letting $y=\pm1$ in \eqref{1} we have:
$$g\left(x^2-1\right)=(x\pm1)g(x)$$
Hence, subtracting these two equations, we conclude that $g(x)=0$ for every real number $x$, which shows that $f(x)=kx$ for every $x$.
A: We can show that $f(x)=-f(-x)$, hence $f(x)$ must be an odd function. If $f(x)$ is a polynomial function then we can write it as:$$f(x)=\sum_{i=0}^\infty a_ix^{2i+1}\tag{1}$$
We can also show that:$$f(x^2)=x\,f(x)\tag{2}$$
Using (1) we get:$$f(x^2)=\sum_{i=0}^\infty a_ix^{4i+2}=a_0x^2+a_1x^6+a_2x^{10}+\cdots\tag{3}$$
and:$$x\,f(x)=\sum_{i=0}^\infty a_ix^{2i+2}=a_0x^2+a_1x^4+a_2x^6+\cdots\tag{4}$$
Using (2) we know that (3) and (4) must be equivalent which means $a_i=0$ for all $i\gt0$. Hence:$$f(x)=a_0x$$
A: 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.
A: We can solve this one in the same manner as here.
So, it is easy to see that $f(0)=0$ and $f$ is odd.
Let $f(1)=0$. If we put $y=1$ we get $$f(x^2-1) = (x-1)f(x)$$ and if we put $-x$ instead of $x$ we get $$f(x^2-1) = -(x+1)f(-x) = (x+1)f(x)$$ So we have $$(x-1)f(x)= (x+1)f(x)\implies f(x)=0\;\;\;\forall x$$ and we are done.
