# find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that : $f(f(x))=x^2-2$ [duplicate]

This is a very hard functional equation.

the problem is this :

find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that : $f(f(x))=x^2-2$

to solve it i have no idea! can we solve it with highschool olympiad education?

## marked as duplicate by M. Vinay, Jyrki Lahtonen, user147263, Michael Albanese, Will OrrickJul 7 '14 at 3:40

• Well, $f$ cannot be injective, surjective, or weakly monotone increasing, for these sets of functions are closed under composition. Also, $f$ cannot be a polynomial by degree arguments. – Robert Wolfe Jul 2 '14 at 7:06
As shown by Gottfried Helms in a linked question, a solution over $[-1,1]$ is given by a function defined over $(-2,+\infty)$: $$2\cdot T_{\sqrt{2}}(x/2)$$ where $T_n$ is a solution to the Chebyshev differential equation $$(1-x^2)\frac{d^2 y}{dx^2}-x\frac{dy}{dx}+ n^2 y = 0.$$ The first terms of the Taylor series in zero are: $$2 \cos\left(\frac{\pi }{\sqrt{2}}\right)+\sqrt{2}\,\sin\left(\frac{\pi }{\sqrt{2}}\right) x-\frac{1}{2} \cos\left(\frac{\pi }{\sqrt{2}}\right) x^2-\frac{1}{12\sqrt{2}}\sin\left(\frac{\pi }{\sqrt{2}}\right) x^3-\frac{1}{48} \cos\left(\frac{\pi }{\sqrt{2}}\right) x^4+\ldots$$
• I don't understand, on the duplicate question, the answer said that there is no such $f$, does I miss something ? – user169373 Oct 13 '15 at 14:16