$$x+f(y+z)=f(y+z+f(x))=f(f(f(z)+x)+y)=x+f(z)+f(y)$$Thus $f(x+y)=f(x)+f(y)$ and hence $f(x)=cx$ for all $x$ and constant $c$. Thus $f(x)=x$ for all $x$.

I have this idea but I am not 100% sure if it is correct: $f(f(x)+y)=x+f(y)$, so letting $z$ equal $f(x)+y$, $f(f(z))=z$, implying $f(z)=f'(z)$, and the only function satisfying this is $f(x)=x$. My problem comes from the fact that we assume that any $z\in R_+$ can be expressed in the form $f(x)+y$. Would this be a false assumption, and is there anyway to show that it must be the case that $f(x)+y$ covers all $z$?

Lemma: $f$ is injective. Proof: Assume $f(a)=f(b)$. Give $x=a, y=c$ and $x=b, y=c$. Then $a=b$. So $f$ is injective. $f(f(f(x))+y)=f(x)+f(y)=f(y)+f(x)=f(f(f(y))+x)$ and $f$ is injective, thus $f(f(x))-x=f(f(y))-y$ for all $x,y\in R_+$ . Then for all $x \in R_+$ we have $f(f(x))-x=k\Rightarrow f(f(x))=x+k$. $f(f(x)+y)=x+f(y)\Rightarrow f(f(f(x)+y))=f(x+f(y))\Rightarrow f(x)+y+k=f(x+f(y))=f(x)+y\Rightarrow k=0\Rightarrow f(f(x))=x$. $x=f(x), y=y$ gives us $f(x+y)=f(x)+f(y)$. Assume $f(a)>f(b)$. $x=b, y=f(a)-f(b)$ gives us $a=b+f(f(a)-f(b))>b$ so $f$ is increasing. With $f(x+y)=f(x)+f(y)$ and $f$ is increasing, we can say $f(x)=cx$ for all $x\in R_+$ and $c$ is constant. So $f(x)=x$ for all $x\in R_+$.

parmenides51 wrote: Find all functions$ f : R_+ \to R_+$ such that $f(f(x)+y)=x+f(y)$ , for all $x, y \in R_+$ (Folklore)

Solved with complex_afi $\color{blue}\boxed{\textbf{Answer: }f\equiv x}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Cute $$f(f(x)+y)=x+f(y)...(\alpha)$$In $(\alpha) x\to y+f(x):$ $$\Rightarrow f(f(f(x)+y)+y)=y+f(x)+f(y)$$By $(\alpha):$ $$\Rightarrow f(x+y+f(y))=y+f(x)+f(y)$$By $(\alpha):$ $$\Rightarrow f(x+y)+y=y+f(x)+f(y)$$$$\Rightarrow f(x)+f(y)=f(x+y)$$And since it is in $\mathbb{R}^+$ we have that it is strictly increasing $$\Rightarrow f\equiv cx$$Replacing in $(\alpha):$ $$\Rightarrow c(cx+y)=x+cy$$$$\Rightarrow c=1$$$$\Rightarrow \boxed{f\equiv x}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

Here is my solution for this Problem: Let $P(x,y)$ be the assertion: $f(f(x)+y)=x+f(y)$ $P(x+f(z),y)$ and with $P(z,x)$ and $P(x,y+z)$ we have $f(y+z)+x=f(f(x)+y+z)=f(f(x+f(z))+y)=x+f(y)+f(z)$ Then we will have $f(y+z)=f(y)+f(z)$ so that $f(x)=ax \quad \forall x>0$ Plugging this back in original equation, we get $a=1$ so $f(x)=x \quad \forall x>0$