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This question was asked in PUTNAM 1995 B4. Infinite continuous fraction starting from 2207 with next term as negative. All terms are in the 8th root.
Solution:
Let,
\[x=\sqrt[8]{2207-\frac{1}{2207-\frac{1}{2207-\cdot\cdot\cdot}}}\]
\[\Rightarrow x^8=2207-\frac{1}{x^8}\]
\[\Rightarrow x^8+\frac{1}{x^8}+2=2207+2\]
\[\Rightarrow \left(x^4+\frac{1}{x^4}\right)^2=47^2\]
\[\Rightarrow x^4+\frac{1}{x^4}+2=47+2\]
\[\Rightarrow \left(x^2+\frac{1}{x^2}\right)^2=7^2\]
\[\Rightarrow x^2+\frac{1}{x^2}+2=7+2\]
\[\Rightarrow \left(x+\frac{1}{x}\right)^2=3^2\]
\[\Rightarrow x+\frac{1}{x}=3\]
\[\Rightarrow x=\frac{3\pm\sqrt5}{2}\]
But if we look at our first equation, it is clear that:
\[x^8>1\]
\[\Rightarrow x>1\]
Here we get only one value greater than 1:
\[\Rightarrow x=\frac{3+\sqrt5}{2}\]
Question related to Algebra, Equation.
