In this thread, the question was to find a $f: \mathbb{R} \to \mathbb{R}$ such that
$$f(f(x)) = f(x) + x$$
(which was revealed in the comments to be solved by $f(x) = \varphi x$ where $\varphi$ is the golden ratio $\frac{1+\sqrt{5}}{2}$).
Having read about iterated functions shortly before though, I came up with this train of thought:
$$f(f(x)) = f(x) + x$$ $$\Leftrightarrow f^2 = f^1 + f^0$$ $$f^2 - f - f^0 = 0$$
where $f^n$ denotes the $n$'th iterate of $f$.
Now I solved the resulting quadratic equation much as I did with plain numbers
$$f = \frac{1}{2} \pm \sqrt{\frac{1}{4} + 1}$$ $$f = \frac{1 \pm \sqrt{1+4}}{2} = \frac{1 \pm \sqrt{5}}{2}\cdot f^0$$
And finally the solution $$f(x) = \frac{1 \pm \sqrt{5}}{2} x .$$
Now my question is: *Is it somehow allowed to work with functions in that way?** I know that in the above, there are denotational ambiguities as $1$ is actually treated as $f^0 = id$ ... But since the result is correct, there seems to be some correct thing in this approach.
So can I actually solve certain functional equations like this? And if true, how would the correct notation of the above be?