$ f(x+f(x+y))=f(x-y)+f(x)^2 \quad \forall x,y\in \mathbb R$

Question

We have to find all functions $f\colon \mathbb R\to\mathbb R$ such that f:

$$\forall x,y\in \mathbb R \quad f(x+f(x+y))=f(x-y)+f(x)^2.$$

Could somebody help me solve this problem?

Thank you.

Answer 1

For a starter:

We denote $a=f(0)$, $b=f(a)$. Substituting $y=-x$ into the original equation gives $$f(x+a)=f(2x)+f(x)^2\ \ \ \ (1).$$

Substituting $x=a$ into (1), we have $b=0$. Again, substituting $x=0$ into (1), we have $b=f(0)+f(0)^2$. This shows $f(0)+f(0)^2=0$, hence $f(0)=0$ or $-1$.

You can proceed from here.