We show that the only solutions are $\boxed{f\equiv c}$ and $\boxed{f(x)=x+c\ \forall x\in\mathbb{R}}$. Here, $c\in\mathbb{R}$ is a constant. Assume that the function is non-constant. Call the assertion $p(x,y)$. Suppose the function was strictly increasing, i.e., if $x>y$ then $f(x)>f(y)$. So, $f(x_1)=f(x_2)$ and $x_1>x_2$ cannot hold simultaneously. It follows that $f(x_1)=f(x_2)\Longrightarrow x_1=x_2$. So, $f(x)+y=f(y)+x\Longrightarrow f(x)-x=f(y)-y=c$, giving the required solution. We now assume that the function is not strictly increasing. So, there do exists $a,b$ such that $a<b$ and $f(a)=f(b)$. If $x\in(a,b)$, then $f(x)\geq f(a)$ and $f(x)\leq f(b)=f(a)$ hold together. So, $f(x)=f(a)=f(b)$. Assume that for $x>b$, we have $f(x)>f(b)$. We can assume this because the function is not constant so it eventually must increase. Claim: We can find an $a$ such that if $x<a$ then $f(x)<f(a)$. Proof. We show that we can do the same with $a$. If not, then for all $x<b$, we will have $f(x)=f(b)$. For each $x>b-f(b)$ such that $f(x)\geq0$, consider $p(x,b-f(x))$. We get $f(f(x)+b-f(x))=f(b)=f(f(b-f(x))+x)=f(f(b)+x)$, a contradiction to the fact that if $x>b$ then $f(x)>f(b)$. It means that for every $x>b-f(b)$, $f(x)<0$. But if we take $x>0$ and $x>b-f(b)$, then from $p(x,b)$ we get $f(b+f(x))=f(f(b)+x)$. However, $b+f(x)<b\Longrightarrow f(b+f(x))=f(b)\Longrightarrow f(f(b)+x)=f(b)$, giving $f(b)+x<b$, a contradiction to $x>b-f(b)$. So, there exists $[a,b]$ such that for all $x<a$, $f(x)<f(a)=f(b)$ and for all $x>b$, $f(x)>f(b)=f(a)$. We have proved that the (non-constant) function can never "eventually" become constant on both sides of infinity. Claim: Let $d$ be an arbitrary real number. We show that $0<b-a\leq|f(d)|$ Proof. Note that $p(x,d)$ gives $f(d+f(x))=f(f(d)+x)$. Taking $x\in[a,b]$, we get that $f(f(d)+x)$ is constant. So, for every $x\in[a+f(d),b+f(d)]$, $f(x)=f(a+f(d))$. By induction, we get that for every $n\geq0$, if $x\in[a+nf(d),b+nf(d)]$, then $f(x)=f(a+nf(d))$. It follows that if $b-a\geq f(d)$, then the ranges $[a+nf(d),b+nf(d)]$ and $[a+(n+1)f(d),b+nf(d)]$ will overlap for each $n\geq0$. But that means that the function is eventually constant on (some) side of number line, a contradiction. So, $0<b-a\leq|f(d)|$ for all $d\in\mathbb{R}$. It follows that there exists a minimum value of $|f(d)|\ d\in\mathbb{R}$, which is positive. Let this value be $m$, and let $k$ be a real number for which $|f(k)|=m$. WLOG let $f(k)>0$. Let $k_0$ be the infimum of the set $\{x\ :\ f(x)=m\}$ We have proved a very strong claim: if $f(a)=f(b)$ then $|a-b|\leq m$. This means that for all $x,y\in\mathbb{R}$, we have $|(f(x)-x)-(f(y)-y)|\leq m$. We can take $h\rightarrow0^{+}$ such that $f(k_0+h)=m$. Then, $|f(k_0-h)|>m$, but $f(k_0-h)\leq m$. It follows that $f(k_0-h)\leq-m$. But then $m\geq|(f(k_0-h)-(k_0-h))-(f(k_0+h)-(k_0+h))|=|m+2h-f(k_0+h)|\geq2m+2h$, a contradiction. So, the only solutions to the given functional equation are as mentioned before.

First, an explanation of the title.

Second, @above: Click to reveal hidden text

Pyramix wrote: Assume that for $x>b$, we have $f(x)>f(b)$. We can assume this because the function is not constant so it eventually must increase. Probably fixable but not true

Complete_quadrilateral wrote: Pyramix wrote: Assume that for $x>b$, we have $f(x)>f(b)$. We can assume this because the function is not constant so it eventually must increase. Probably fixable but not true Why is it not true? If there does not exist such a b or an a, then the function is constant, right? EDIT: I got my mistake. Define $b$ as the supremum of the set $\{x: f(x)=f(a)\}$.

Phorphyrion wrote: First, an explanation of the title.

Second, @above: Click to reveal hidden text

[I think this argument works for floor, it may be wrong] I have made a mistake. In my second claim, the condition $f(d)\ne0$ is important. So, $f(k_0-h)=0$ is possible, and is forced as well. In that case, $f(k_0)$ is either $0$ or $m$. Now, $|f(k_0+m)-f(k_0)-m|\leq m$, so $0\leq f(k_0+m)-f(k_0)\leq2m$. However, replace $k_0$ with $k_0-h$ and $k_0+h$ to observe that $m\leq f(k_0+m+h)\leq3m$, and $0\leq f(k_0+m-h)\leq2m$. So, if $f(k_0)=m$, then function is like floor, else if $f(k_0)=0$ then it is like ceil. (This can be proven by an inductive argument on $m$, I guess). So, we get $f(x)=\lfloor \frac{x}{m}\rfloor+\beta$ for some $m,\beta$. Subs in original fe to get $m=1$. Then, $\beta=1-\lfloor k_0\rfloor$. (The same can be done with the ceil case, but we get a contradiction.)

Interesting! The answer is very surprising, it should be $$f(x)\equiv C\text{ or }f(x)=x+C\text{ or }f(x)=D\left\lfloor\frac{x+C}D\right\rfloor\text{ or }f(x)=D\left\lceil\frac{x+C}D\right\rceil,$$here $C\in\mathbb R$ and $D\in\mathbb R_+$ are arbitary constants. When $f$ is periodic we get $f(x)\equiv C$ because $f$ is weakly increasing. When $f$ is injective we get $f(x)=x+C$ because $f(x)-x=f(y)-y$ for all $x,y\in\mathbb R,$ so it is constant. Now assume not, the key step is \begin{align*}f(z+f(x)+f(y))&=f(f(z+f(x))+y)=f(f(f(z)+x)+y)\\&=f(f(y)+x+f(z))=f(f(f(y)+x)+z)\end{align*}holds for all $x,y,z\in\mathbb R.$ Therefore if there exists $f(x_0)+f(y_0)\neq f(f(y_0)+x)$ then $f$ has a period $|f(x_0)+f(y_0)-f(f(y_0)+x)|,$ contradiction, so we get $f(x+f(y))=f(x)+f(y)$ for all real numbers $x,y.$ The rest part can be done by analyzing $I:=\text{Im}f.$