Functional equation $f(x^2+y^2)=f(x^2-y^2)+2xy \quad \forall\; x,y\in\Bbb R$

A double differentiable function satisfies $$f(x^2+y^2)=f(x^2-y^2)+2xy \quad \forall\; x,y\in\Bbb R$$ Given that $$f(0)=0$$ and $$f''(0)=2$$, determine $$f(x)$$.

I tried to convert the equation into some form of Cauchy's additive function by observing that $$f(x)$$ is an even function. However, I am unable to make any significant progress. Any hints will be appreciated.

Edit: My apologies for posting the incorrect question. Correct question:

$$f(x^2+y^2)=f(x^2-y^2)+f(2xy) \quad \forall\; x,y\in\Bbb R$$ Given that $$f(0)=0$$ and $$f''(0)=2$$, determine $$f(x)$$.

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

There is no such function. Putting $$x=y$$ we get $$f(2x^{2})=2x^{2}$$ which implies that $$f(t)=t$$ for all $$t \geq 0$$. Since $$f$$ is also even (as seen by putting $$x=0$$) we get $$f(t)=|t|$$ which is not even differentiable.
• Even if the additional assumptions of (double) differentiability, $f ( 0 ) = 0$ and $f '' ( 0 ) = 2$ were dropped, the functional equation itself would give us $f ( x ) = f ( 0 ) + | x |$ by your method. Putting this back into the functional equation gives $$f ( 0 ) + x ^ 2 + y ^ 2 = f ( 0 ) + \left | x ^ 2 - y ^ 2 \right | + 2 x y \text ,$$ or equivalently $$( x - y ) ^ 2 = \left | x ^ 2 - y ^ 2 \right |$$ for all real number $x$ and $y$, which doesn't hold for cases as simple as $x = 2$ and $y = 1$. So, there is no function even only satisfying the functional equation. Commented Jan 7, 2022 at 21:18