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Question Number 197325 by sonukgindia last updated on 13/Sep/23

Answered by witcher3 last updated on 13/Sep/23

$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)={\mathrm{x}}^2-1$$$$\mathrm{if}\mathrm{x}\in \mathbb{R}\mid {\mathrm{f}}^2\left(\mathrm{x}\right)=\mathrm{x}\Rightarrow {\mathrm{x}}^2-\mathrm{x}-1=0$$$$\mathrm{x}\in \{\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\}$$$${\mathrm{f}}^4\left(\mathrm{x}\right)=\mathrm{x}\Rightarrow {\left({\mathrm{f}}^2\left(\mathrm{x}\right)\right)}^2-1={({\mathrm{x}}^2-1)}^2-1=\mathrm{x}$$$$\iff (\mathrm{x}+1)({(\mathrm{x}-1)}^2(\mathrm{x}+1)-1)=0$$$$\mathrm{x}(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}-1)=0$$$$\mathrm{x}\in \{0,-1,\frac{1\underset{-}{+}\sqrt{5}}{2}\}$$$${\mathrm{f}}^2\left(0\right)=-1,{\mathrm{f}}^2(-1)=0$$$$\mathrm{\exists }\mathrm{a}\ne \mathrm{b},\mathrm{a},\mathrm{b}\notin \{0,-1\}\mathrm{sush}\mathrm{that}$$$$,\mathrm{f}\left(0\right)=\mathrm{a}\mathrm{and}{\mathrm{f}}^3\left(0\right)=\mathrm{f}(-1)=\mathrm{b}$$$$\mathrm{because}\mathrm{if}\mathrm{f}\left(0\right)=\mathrm{a},\mathrm{a}\in \{0;-1\}$$$$\mathrm{a}=0\Rightarrow {\mathrm{f}}^2\left(0\right)=0\mathrm{impissibl}$$$$\mathrm{f}\left(0\right)=-1$$$$\Rightarrow 0={\mathrm{f}}^4\left(0\right)={\mathrm{f}}^3(-1)=\mathrm{f}\left(0\right)\Rightarrow {\mathrm{f}}^2\left(0\right)=0\mathrm{impossible}$$$$\mathrm{f}\left(0\right)={\mathrm{f}}^3\left(0\right)\Rightarrow {\mathrm{f}}^2\left(0\right)=0\mathrm{impossible}$$$$\Rightarrow \mathrm{\exists }(\mathrm{a}\ne \mathrm{b})\notin \{-1,0\}$$$$\mathrm{such}\mathrm{f}\left(0\right)=\mathrm{a}\mathrm{and}{\mathrm{f}}^3\left(0\right)=\mathrm{f}(-1)=\mathrm{b}$$$$\Rightarrow {\mathrm{f}}^3\left(\mathrm{a}\right)=0$$$${\mathrm{f}}^4\left(\mathrm{a}\right)=\mathrm{f}\left(0\right)=\mathrm{a}$$$${\mathrm{f}}^4\left(\mathrm{b}\right)=\mathrm{b},\mathrm{a}\mathrm{and}\mathrm{b}\notin \left\{\frac{1\underset{-}{+}\sqrt{5}}{2}\right\}$$$$\mathrm{suppose}\mathrm{b}\in \Rightarrow {\mathrm{f}}^4\left(\mathrm{b}\right)=\mathrm{b}=\mathrm{f}(-1)$$$${\mathrm{f}}^2\left(\mathrm{b}\right)=\mathrm{b}={\mathrm{f}}^3(-1)={\mathrm{f}}^4\left(0\right)=0$$$$\Rightarrow \mathrm{b}=0\mathrm{impossibl}$$$$\mathrm{so}\mathrm{we}\mathrm{have}6\mathrm{fixe}\mathrm{points}\mathrm{for}{\mathrm{f}}^4$$$$\mathrm{a},\mathrm{b},0,-1,\frac{1\underset{-}{+}\sqrt{5}}{2}\mathrm{impossible}\mathrm{we}\mathrm{have}\mathrm{just}4$$$$\mathrm{f}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{exist}$$$$$$$$$$$$$$

Commented by MathematicalUser2357 last updated on 17/Sep/23

$$youmean\pm$$

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