\section*{Problem 20}

\(\mathrm{x}\) is a real number. Define \(\mathrm{x}_{0}=1+\sqrt{ }(1+\mathrm{x}), \mathrm{x}_{1}=2+\mathrm{x} / \mathrm{x}_{0}, \mathrm{x}_{2}=2+\mathrm{x} / \mathrm{x}_{1}, \ldots, \mathrm{x}_{1985}=2+\) \(\mathrm{x} / \mathrm{x}_{1984}\). Find all solutions to \(\mathrm{x}_{1985}=\mathrm{x}\).

\section*{Answer}


\section*{Solution}

If \(\mathrm{x}=0\), then \(\mathrm{x}^{1985}=2 \neq \mathrm{x}\). Otherwise we find \(\mathrm{x}_{1}=2+\mathrm{x} /(1+\sqrt{ }(1+\mathrm{x}))=2+(\sqrt{ }(1+\mathrm{x})-1)=1+\) \(\sqrt{ }(1+x)\). Hence \(x_{1985}=1+\sqrt{ }(1+x)\). So \(x-1=\sqrt{ }(1+x)\). Squaring, \(x=0\) or 3 . We have already ruled out \(x=0\). It is easy to check that \(x=3\) is a solution.

