How find the $f(x+f(y))+f(y+f(z))+f(z+f(x))=0,$ :$f:R\longrightarrow R$ ,and is continuous
 such that  $$f(x+f(y))+f(y+f(z))+f(z+f(x))=0,$$
find all  $f$
 A: There are two solutions: either $f(x) = c-x$ for some constant $c$, or $f(x) = 0$ for all $x$.


*

*We claim that $f(x+f(y)) = f(x) - f(y)$.


*

*Put $x = y = z = 0$, we get $f(f(0)) = 0$.

*Put $x = y = f(0)$, we get $f(f(0) + f(z)) = -f(z)$.

*Put $x = f(0)$, we get $-f(y) + f(y+f(z)) + f(z) = 0$, hence the claim.


*Note that the image of $f$ is an additive subgroup of $\mathbb{R}$. There are two cases - either this subgroup is dense at 0 (hence in $\mathbb{R}$), or is not dense at 0 (hence discrete in $\mathbb{R}$)


*

*For the first case, let $g(x) = x + f(x)$. Note that
$$f(x+f(y)) = f(x) - f(y) \Rightarrow x + f(y) + f(x+f(y)) = x + f(x) \Rightarrow g(x+f(y)) = g(x)$$
for any $y$. Density of image of $f$ and continuity of $f$ then implies that $g$ is a constant, say $c$, then $f(x) = g(x) - x = c-x$.

*For the second case, $f$ is a continuous map on a connected set. If the image is discrete, it must be a constant. It's easy to check that the only possible constant is 0.
A: Let me just point out that after you obtained $f(x+f(y))=f(x)-f(y)$ $(1)$ the solution can be finished as follows: as Vadim pointed out $f(x)=k-x$ on the image of $f.$ So if $f$ is not identical zero $f$ takes both negative and positive values (follows from the original equation). So $(1)$ implies that the image of $f$ is unbounded from below and from above. By the intermediate value theorem it is $\mathbb{R},$ and so $f(x)=k-x$ for all $x\in\mathbb{R}.$
A: *

*Take $x=y=z=0$, we have $3f(f(0))=0$ so $f(f(0))=0$.  Set $k=f(0)$; we have $f(k)=0$.

*Take $y=0, x=z=k$, we have $f(k+k)+f(0)+f(f(0))=0$ so $f(2k)=-k$.

*Take $y=k, z=0$, we have $f(x)+f(2k)+f(f(x))=0$.  Hence $f(x)+f(f(x))=k$, for all $x$.  

*In particular, for all $t$ in the range of $f$, we have $f(t)=k-t$.  The range, as a subset of $\mathbb{R}$, must therefore be symmetric about the point $\frac{k}{2}$.  
If the range is $\mathbb{R}$, then $f(x)=k-x$ is the only family of solutions (one for each $k\in \mathbb{R}$).  By inspection all of these work.  There is likely some continuity argument to prove that this must be the case, but right now I can't come up with it.
