How to solve the functional equation $f(x + f(x +y ) ) = f(2x) + y$? Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ that satisfy the following equation: 
$$ f(x + f(x +y ) ) = f(2x) + y,\quad \forall x,y\in\mathbb{R}$$ 
The only function I have found is $f(x) = x$, but I think there are more. 

Given $z$, let $x=f(z)$. and $y=z-x.$ Then you get:
$$f(x+f(x+y))=f(2f(z))$$ and $$f(2x)+y=f(2f(z))+z-f(z)$$
From this you get $z=f(z).$

A cute variation of Christian's very nice answer:
$$\begin{align}
2z+f(0)&=f(f(2z))&[x=0,y=2z]\\
&=f(f(z+f(z)))&[x=z,y=0]\\
&=z+f(z)+f(0)&[x=0,y=z+f(z)]
\end{align}$$
So $f(z)=z.$
This is avoiding the reference to being an injection, by implicitly using the right inverse $g(z)=f(z)-f(0).$
