I have a function $f : R → R$ such that $(f \circ f \circ f)(x) = (f \circ f)(x) + x \hspace{4mm} \forall x \in R$
how would i go about proving its injectivity? I have tried finding $f(x)$ but to no avail, would this be possible? Would it also be possible given this function, to compute something out of $f(x)$? Something such as $f(0)$?
By contradiction (not necessary, but this makes the argument a tiny bit cleaner, in my opinion). Suppose there exist reals $x\neq y$ such that $f(x)=f(y)$.
This implies $(f\circ f)(x)=(f\circ f)(y)$ and $(f\circ f\circ f)(x)=(f\circ f\circ f)(y)$, the latter from which by definition $$ (f\circ f)(x) + x = (f\circ f)(y) +y $$ and using the fact that $(f\circ f)(x)=(f\circ f)(y)$, this implies in turn $$ x=y, $$ contradiction.
Let us write $y\stackrel{\rm def}{=} f(0)$. From our equation, we know that $$f^{(2)}(y) = f^{(3)}(0)= f^{(2)}(0)+0 = f^{(2)}(0) = f(y)$$ which implies, applying $f$ to both sides, $ f^{(3)}(y) = f^{(2)}(y) $ so that, using again our initial relation which states that $f^{(3)}(y)=f^{(2)}(y)+y$, we get $$ f^{(2)}(y) = f^{(2)}(y)+y $$ and we conclude that $y=0$.