For any $ f : \mathbb R \to \mathbb R $ and any nonnegative integer $ n $, let $ f ^ n $ denote the $ n $-th iteration of $ f $; i.e. $ f ^ 0 ( x ) = x $ and $ f ^ { n + 1 } ( x ) = f \big( f ^ n ( x ) \big) $ for all $ x \in \mathbb R $.
Find all functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f ^ 4 \left( x f ( x ) ^ 2 + y f ( x ) – x \right) = x ^ 3 + x f ^ 2 ( y ) – x \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $.
It's straightforward to check that the identity function is a solution. I suspect that it's the only solution. One witness to this is that assuming $ f $ is real analytic, some messy calculations show that identity is the only candidate. I don't know whether there are non-analytic solutions or not. I just have the following observations.
Plugging $ x = y = 0 $ in \eqref{0} we get $ f ^ 4 ( 0 ) = 0 $. Then, putting $ x = 0 $ and $ y = 1 $ in \eqref{0} we get $ f ^ 5 ( 0 ) = 0 $, and since $ f ^ 5 ( 0 ) = f \left( f ^ 4 ( 0 ) \right) $, we have $ f ( 0 ) = 0 $. Hence, setting $ y = 0 $ in \eqref{0} we get
$$ f ^ 4 \left( x f ( x ) ^ 2 – x \right) = x ^ 3 – x \tag 1 \label 1 $$
for all $ x \in \mathbb R $. In particular, \eqref{1} shows that $ f $ is surjective, since $ x ^ 3 – x $ takes every real number as value. As a special case, we must have $ f ( x _ \pm ) = \pm 1 $ for some $ x _ \pm $, and letting $ x = x _ \pm $ in \eqref{1} we get $ x _ \pm \in \{ – 1 , 0 , 1 \} $. As we know that $ f ( 0 ) = 0 \ne \pm 1 $, $ 0 $ is excluded, and because both $ – 1 $ and $ 1 $ must be covered, we get $ f ( – 1 ) = – f ( 1 ) \in \{ – 1 , 1 \} $.
Source:
Number four on the list at the end of this page.
Best Answer
You've got everything you need to finish. Here's some hidden hints, as to not spoil the solution.
From the above, see if you can deduce what $f$ must be. So all that's left...
For completeness, I'll put a full solution below (continued from your progress).