# [Proposed] Equation Fonctionnelle

• Trouver toutes les fonctions $f: \mathbb R \to \mathbb R$ telles que:
$f(f(x)+y)=f(x^2-y)+4yf(y) \ \ \ \ \ \forall x,y \in \mathbb R$

Proposé par Amir Hossein Parvardi

• Just some ideas :
$P(x, \frac{x^2-f( x)}{2} )$ \implies $f (x)=x^2$ and $f(\frac{x^2-f(x)}{2})=0$
First equation shows that $f (x)=x^2$ for all x in R .
Second equation needs to be used .
I thought abt showing injectivity and surjectivity hence we can find an uinR such that $f(u)=0$ and hence $x^2-f(x)=2u$ so $f (x)=x^2-2u$ .
But we all see that $f(x)=0$ is a solution , so i suppose we should use the second equation to show that , for all x .

Sauf erreur

• Ligne2: $P(x,\frac{x^2-f(x)}{2}) \rightarrow f(x) =x^2$ OU $f(\frac{x^2-f(x)}{2})=0$ non pas ET
Sinon tu es sur une bonne piste, continue.

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