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Question Number 210289 by Abdullahrussell last updated on 05/Aug/24

Answered by Frix last updated on 05/Aug/24

$${\mathrm{There}}^{\prime }\mathrm{s}\mathrm{one}\mathrm{number}x\mathrm{with}\mathrm{nice}\mathrm{properties}:$$$$x^2=x+1,x^3=2x+1,$$$$\frac{1}{x}=x-1,\frac{1}{x^2}=2-x,\frac{1}{x^3}=2x-1$$$$\Rightarrow$$$$\sqrt{\frac{x-1}{x^3}}=\sqrt{\frac{1}{x^4}}=\frac{1}{x^2}=2-x$$$$\sqrt{\frac{x^2-1}{x^3}}=\sqrt{\frac{1}{x^2}}=\frac{1}{x}=x-1$$$$2-x+x-1=1\mathrm{true}\Rightarrow$$$$\mathrm{The}\mathrm{number}\mathrm{is}x=\phi =\frac{1}{2}+\frac{\sqrt{5}}{2}$$

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