$f(f(x)−y) = f(x)+f(f(y)−f(−x))+x$ find all functions

Question

Find $f:R\rightarrow R$ such that $f(f(x)−y) = f(x)+f(f(y)−f(−x))+x$.

The obvious is the substitution $y\rightarrow -x$, which gives us

$$ f(f(x)+x)=f(x)+f(f(-x)-f(-x))+x=f(x)+x+f(0).$$

Why can't we substitue now $u=f(x)+x$ and say that $f(u)=u+c$?

I'm sorry if this is stupid, I'm just beginning to learn.

Answer 1

If $f(x)+x$ is surjective (onto) then you can say that $f(u)=u+c$ but for example if $f(x)=-x$ then $f(x)+x=0$ is not surjective and is of a slightly different form $f(x)=-x+c$. Actually $f(x)=-x$ is the only solution which is proven in @Eugen Covaci's answer.

The $f(x)+x=u$ is surjective means that $u$ can take any value in $\Bbb{R}$ if you can do so then you can pick any value of $x$ that gives the corresponding $u$ and say $f(u)=u+c$

Answer 2

If $x=y=0$, then $f(f(0))=2f(0) \tag 1$

If $x=0, y=f(0)$ then $f(0)=f(0)+f(f(f(0))-f(0))$ therefore $f(f(f(0))-f(0))=0 \tag 2$

Using (1) in (2) one gets $f(f(0))=0$ and using again (1) one gets $f(0)=0$

Now let $x=0$. It follows $f(-y)=f(f(y)) \tag 3$

Finally, let $y=f(x)$. Then $0=f(x) + f(f(f(x)) -f(-x)) + x \tag 4$

Using (3) in (4) one gets $0=f(x) + x$ therefore $f(x) = -x, \forall x$

As one can see, the "solution" that OP has found is incorrect, see @kingW3 answer for further explanations.