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This question already has an answer here:

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?

please help : )

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ math.stackexchange.com/questions/832792/… $\endgroup$ – Dario Jul 2 '14 at 6:34
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    $\begingroup$ 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. $\endgroup$ – Robert Wolfe Jul 2 '14 at 7:06
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    $\begingroup$ @Dr.AKA You should look at the answers there. $\endgroup$ – M. Vinay Jul 2 '14 at 8:58
  • $\begingroup$ Yes, there is the sesquicentennial solution of Schroeder. $\endgroup$ – Cosmas Zachos Jun 2 '17 at 15:14
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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: $$2 \cos\left(\frac{\pi }{\sqrt{2}}\right)+\sqrt{2}\,\sin\left(\frac{\pi }{\sqrt{2}}\right) x-\frac{1}{2} \cos\left(\frac{\pi }{\sqrt{2}}\right) x^2-\frac{1}{12\sqrt{2}}\sin\left(\frac{\pi }{\sqrt{2}}\right) x^3-\frac{1}{48} \cos\left(\frac{\pi }{\sqrt{2}}\right) x^4+\ldots$$

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  • $\begingroup$ I don't understand, on the duplicate question, the answer said that there is no such $f$, does I miss something ? $\endgroup$ – user169373 Oct 13 '15 at 14:16
  • $\begingroup$ Nah, at the high-school level you can explicitly verify 362193 works. It was discovered in 1870 by Ernst Schroeder. $\endgroup$ – Cosmas Zachos Jun 2 '17 at 15:11

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