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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.

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Do you have the source of the problem? – Ragib Zaman Oct 16 '11 at 13:32

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.

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It can be shown that the only function satisfying $$f(x+f(x+y))=f(x-y)+f(x)^2\qquad(1)$$ is the constant zero function. To show this, we let $x=0$ in (1) and get: $$f(f(y))=f(-y)+f(0)^2\qquad(2)$$ Also, substituting $-x$ for $y$ in (1) we have: $$f(x+f(0))=f(2x)+f(x)^2\qquad(3)$$ Letting $y=0$ and $y=f(0)$ in (3), we respectively have: $$f(f(0))=f(0)+f(0)^2$$ $$f(2f(0))=f(2f(0))+f(f(0))^2$$ So we conclude that $f(f(0))=0$ and $f(0)\in\{0,-1\}$. Suppose that $f(0)=-1$. letting $y=-1$ in (2) we get $f(1)=-2$. So if we let $x=1$ in (3) we'll have $f(2)=-5$. Now applying $f$ to both sides of (2) we have $f(f(f(y)))=f(f(-y)+1)$ which by (2) yields $f(-f(y))+1=f(f(-y)+1)$. By letting $y=1$ in the last equation, we have $f(2)=-3$ which leads to a contradiction. So $f(0)$ can't be equal to $-1$ and we must have $f(0)=0$.

Now substituting $f(y)-x$ for $y$ in (1) we have $$f(x+f(f(y)))=f(2x-f(y))+f(x^2)$$ which by (2) yields $$f(x+f(-y))=f(2x-f(y))+f(x)^2$$ Again, substituting $-x-y$ for $y$ in (1) we'll have $$f(x+f(-y))=f(2x+y)+f(x)^2$$ Compairing the last two equations above, we get $f(2x-f(y))=f(2x+y)$. Substituting $\frac{f(y)}2$ for $x$, we'll have $f(y+f(y))=0$. Also by letting $y=0$ in (1) we know that $f(x+f(x))=f(x)+f(x)^2$. So for every $x$ we find out that $f(x)\in\{0,-1\}$. Hence, adding $f(x)$ to both sides of (3) we'll get $2f(x)=f(2x)$ which shows that $f(x)$ can not be equal to $-1$. So $f$ is the constant zero function.

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