Sequence and series (2)

Given that:

$\begin{aligned} a_n & = \sqrt 2 a_{n-1} - a_{n-2} \\ \implies a_{n+2} & = \sqrt 2 a_{n+1} - a_n \\ a_{n+4} & = \sqrt 2 a_{n+3} - a_{n+2} \\ & = \sqrt 2\left(\sqrt 2 a_{n+2} - a_{n+1}\right) - a_{n+2} \\ & = a_{n+2} - \sqrt 2 a_{n+1} \\ & = \sqrt 2 a_{n+1} - a_n - \sqrt 2 a_{n+1} \\ & = - a_n \\ \implies a_{n+8} & = -a_{n+4} = a_n \end{aligned}$

Therefore, $\{a_n\}$ with $k=\sqrt 2$ has a period of $\boxed 8$ .