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Does there exists a function $f:\Bbb R\to \Bbb R$ such that $f(f(x))=x^2-2$

I don't know where to start?

Arthur
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  • It would be instructive to try a few values and see where that gets you. – Fimpellizzeri Dec 07 '17 at 14:42
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    @james.nixon $F'(2) = 4$ – PersonX Dec 07 '17 at 14:52
  • @PersonX Oh my lord, missed that completely! –  Dec 07 '17 at 14:53
  • If $F(x) = x^2 - 2$, then $F(2) = 2$ and $F'(2) = 4$. There exists the Schroder function $\Psi$ such that $\Psi(F(x)) = 4\Psi(x)$. Locally in a neighborhood about $2$ we have $f(x) = \Psi^{-1}(2\Psi(x))$ will work. Not sure if it can be extended to $\mathbb{R}$. –  Dec 07 '17 at 14:55
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    For $x\in[0,2]$, let $f(x)=2\cos(\sqrt{2}\arccos(x/2))$, for $x\in[2,\infty)$, let $f(x)=2\cosh(\sqrt{2}\cosh^{-1}(x/2))$, and $f(-x)=f(x)$. –  Dec 07 '17 at 15:02
  • @ProfessorVector Ooo, this is a case where the conjugation has a nice closed form. +1 –  Dec 07 '17 at 15:05
  • @ProfessorVector That is pretty impressive actually. – Fimpellizzeri Dec 07 '17 at 16:34

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