Find all solutions to the functional equation $f:\mathbb{R}\rightarrow \mathbb{R}$ $$f(x+f(x))=f(x)+x$$ I have no idea how to solve this. I can substitute $x=0$ to obtain $f(f(0))=f(0)$. But other than that I can't make any progress. There are two obvious solutions, namely $f(x)=x$ and $f(x)=-x$. But are there any other solutions?
Thanks in advance.
On
I want to qualify this by saying I know almost nothing about functional equations, but here's a little gimmick I see here:
$ f : \mathbb{R} \mapsto \mathbb{R} $ defined by
$$ f( f(x) + x) = f(x) + x $$
could be thought of a lot more succinctly as a relation showing that
$ f(c) = c $, for all $ c \in \mathbb{R} $, with $ c = f(x) + x $.
If you consider the entire real line as your domain, then $ f(x) $ has to have a "fixed point" at every value $ f(x) + x $ attains. I think it shouldn't be hard to go on and show that there are only two solutions (so I rescind my comment on your original post).
That's also just my take on what the linked post says, but it's a bit more technical and general.
There are many many solutions. Let $S\subseteq\mathbb R$ be any nonempty subset satisfying $2S\subseteq S$. Pick any function $g:\mathbb R\setminus S\to S$. Let $$ f(x)=\begin{cases} x&\text{if }x\in S\\ g(x)-x&\text{otherwise.} \end{cases} $$ For each $x\in\mathbb R$ we have $f(x)+x\in S$, so $f(f(x)+x)=f(x)+x$.
Even if we require $f$ to be continuous, there are as many solutions as continuous functions on $\mathbb R$. Indeed set $S=\{x\in\mathbb R\mid x\geq0\}$ and pick any continuous function $g:\mathbb R\setminus S\to S$ satisfying $\lim_{x\to0^-}g(x)=0$; then construct $f$ as above.