Find all functions $f:\mathbb{R} \to [0, \infty)$such that $f(x^2 + y^2)=f(x^2 - y^2)+ f(2xy)$. Question - 

Find all functions $f:\mathbb{R} \to [0, \infty)$ such that 
  $$f(x^2 + y^2)=f(x^2 - y^2)+ f(2xy)$$ for all $x,y\in\mathbb{R}$. 

My try - 
$f(0)=0$ and $f(x)=f(-x)$ for all $x>0$ obtained by putting $y=-x$ ...
Now I replaced $x^2 +y^2$ by $x$
And $x^2 - y^2$ by $y$ ..
I get $f(x)=f(y)+f(2xy)$
Putting $y=0$ I get $f(x)=0$ ...
Now there is another solution i.e $f(x)=cx^2$ but I am not able to apply any other substitution to reach at this...
Any hints ??
 A: Suppose that $f:\mathbb{R}\to\mathbb{R}_{\geq 0}$ is a function such that $$f(x^2+y^2)=f(x^2-y^2)+f(2xy)\,,\tag{*}$$
for all $x,y\in\mathbb{R}$.  For each $x,y\in\mathbb{R}$, let $P(x,y)$ denote the statement (*).
From $P(0,0)$, we get $f(0)=0$.  Thus, $P(0,y)$ yields $$f(+y^2)=f(-y^2)$$ for all $y\in\mathbb{R}$.  Therefore, $f$ is an even function.  Let $F:\mathbb{R}_{\geq 0}\to\mathbb{R}$ be the function defined by
$$F(t):=f(\sqrt{t})$$
for all $t\geq 0$.
Now, since $f$ is an even function,
$$\begin{align}F\big((x^2+y^2)^2\big)&=f(x^2+y^2)=f(x^2-y^2)+f(2xy)\\&=f\big(|x^2-y^2|\big)+f\big(2|xy|\big)\\&=F\big((x^2-y^2)^2\big)+F\big((2xy)^2\big)\end{align}$$
for all $x,y\in\mathbb{R}$.  This shows that $$F(u+v)=F(u)+F(v)\tag{#}$$
for all $u,v\geq 0$.  (Note that, for $u,v\geq 0$, there are $x,y\in\mathbb{R}$ such that $u=(x^2-y^2)^2$ and $v=(2xy)^2$.)  Thus, $F$ satisfies Cauchy's functional equation and $F$ is a nonnegative function.  Therefore, $F$ is a nondecreasing function.  Ergo, there exists $k\geq 0$ such that
$$F(t)=kt$$
for all $t\in\mathbb{R}_{\geq 0}$.
Since $f(x)=f\big(|x|\big)=F(x^2)$ for all $x\in\mathbb{R}$, we conclude that
$$f(x)=kx^2$$
for all $x\in\mathbb{R}$ (where $k\geq 0$ is a constant).  It is easy to see that all functions $f$ of this form obey (*).
Remark.  For a function $F:\mathbb{R}_{\geq 0}\to\mathbb{R}$ such that $F$ satisfies Cauchy's functional equation (#), we can conclude that there exists a constant $k$ such that $F(t)=kt$ for all $t\geq 0$ if at least one of the following properties are known to be true:


*

*$F$ is continuous at any point,

*$F$ is continuous at one point,

*$F$ is bounded on any bounded interval (with nonempty interior), 

*$F$ is bounded on one bounded interval (with nonempty interior), 

*$F$ is monotonic, or

*$F$ is monotonic on one interval (with nonempty interior).


Without these properties, there are noncontinuous examples of $F$ (but such examples rely on the Axiom of Choice).
