Bilinear Recursion - 3

If we write $y_n = x_n - 1$ , then $y_0 = 9$ and $y_{n+1} \; = \; -\frac{1}{y_n + \sqrt{2}}$ Thus $\begin{aligned} y_1 & = \; -\frac{1}{9+\sqrt{2}} \; = \; \tfrac{1}{79}(\sqrt{2}-9) \\ y_2 & = \; -\frac{1}{\frac{1}{79}(80\sqrt{2}-9)} \; =\; \frac{79}{9 - 80\sqrt{2}} \; = \; -\tfrac{1}{161}(9 + 80\sqrt{2}) \\ y_3 & = \; -\frac{1}{\frac{1}{161}(81\sqrt{2} - 9)} \; = \; -\frac{161}{9(9\sqrt{2}-1)} \; = \; -\tfrac{1}{9}(9\sqrt{2}+1) \\ y_4 & = \; 9 = y_0\end{aligned}$ and it is now easy to see that $y_{n+4} = y_n$ , and hence $x_{n+4} = x_n$ for all $n \ge 0$ . Since $y_0,y_1,y_2,y_3$ are all distinct, we deduce that the period is $\boxed{4}$ .