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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 $f(x^2 + y^2)=f(x^2 - y^2)+ f(2xy)$.

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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Jan 7, 2022 at 21:18

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