Tintarn wrote: Consider all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying $f(1-f(x))=x$ for all $x \in \mathbb{R}$. a) By giving a concrete example, show that such a function exists. Define $h(x)=\frac 12-f(x+\frac 12)$ so that equation is $h(h((x))=-x$ Example of such $h(x)$ are easy to give : Split $\mathbb R^+$ in two idempotent sets $A,B$ Let $k(x)$ any bijection from $A\to B$

Concrete example

Define $h(x)$ as : $h(0)=0$ $\forall x>0$ : If $x\in A$ : $h(x)=k(x)$ If $x\in B$ : $h(x)=-k^{-1}(x)$ $\forall x<0$ : $h(x)=-h(-x)$ Tintarn wrote: Consider all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying $f(1-f(x))=x$ for all $x \in \mathbb{R}$. b) For each such function define the sum \[S_f=f(-2017)+f(-2016)+\dots+f(-1)+f(0)+f(1)+\dots+f(2017)+f(2018).\]Determine all possible values of $S_f$. Replacing in original equation $x$ by $1-f(x)$, we have easily $f(x)+f(1-x)=1$ And so required sum is $\boxed{2018}$

Let $1-f(x)=g(x)$ the function becomes $g(g(x))=1-x $ and put $x=g(x)$ we have that $g(x)+g(1-x)=1$ . b) $g(x)+g(1-x)=1$ implies that $g(-2017)+\dots g(0)+\dots g(2018)=2018$ and $g(x)=1-f(x) $ we have that $2\cdot 2018 -S_f=2018 \Longleftrightarrow S_f=2018$

XbenX wrote: a) We can see that if $g$ is a polynomial then it must be of odd degree , and since there is no linear such function , we can check for third degree polynomials , wich indeed has infinitely many solutions for example $g(x)=2x^3-3x^2+x+\frac{1}{2}$ Wrong. No continuous function can be an example : $g(g(x))=1-x$ implies $g(x)$ bijective Continuous and bijective implies monotonous monotonous implies $g(g(x))$ increasing and so can not be $1-x$ So examples can only be found in non continuous functions (so for example no polynomial can be an example)