Let $A$ be a set of functions $f : R\to R$. For all $f_1, f_2 \in A$ there exists a $f_3 \in A$ such that $f_1(f_2(y) - x)+ 2x = f_3(x + y)$ for all $x, y \in R$. Prove that for all $f \in A$, we have $f(x - f(x))= 0$ for all $x \in R$.
Source: Dutch IMO TST 2018 day 1 p4
Tags: functional equation, function, algebra
Let $A$ be a set of functions $f : R\to R$. For all $f_1, f_2 \in A$ there exists a $f_3 \in A$ such that $f_1(f_2(y) - x)+ 2x = f_3(x + y)$ for all $x, y \in R$. Prove that for all $f \in A$, we have $f(x - f(x))= 0$ for all $x \in R$.