6. (i) It can be deduced from $x_{n+1}+y_{n+1} \sqrt{2}=\left(x_{n}+y_{n} \sqrt{2}\right)(1+\sqrt{2})$.
(ii) It follows from $x_{2 n+1}+y_{2 n+1} \sqrt{2}=\left(x_{n}+y_{n} \sqrt{2}\right)\left(x_{n+1}+y_{n+1} \sqrt{2}\right)$ and (i).
(iii) $1+\sqrt{2}$ is the smallest positive solution of $u^{2}-2 v^{2}=-1$, and the general solution is $(u+v \sqrt{2})^{2 n+1}$. On the other hand, from problem 4, we know that $v^{2}=((u+1) / 2)^{2}+((u-1) / 2)^{2}$, thus
$$y_{2 n+1}^{2}=\left(\left(x_{2 n+1}+1\right) / 2\right)^{2}+\left(\left(x_{2 n+1}-1\right) / 2\right)^{2}$$
(iv) Using $x_{m}^{2}-2 y_{m}^{2}=(-1)^{m}$, and $\left(x_{n}+y_{n} \sqrt{2}\right)=\left(x_{m}+y_{m} \sqrt{2}\right)\left(x_{n-m}+y_{n-m} \sqrt{2}\right)$ to deduce.
(v) It follows from $x_{2 n+1}+y_{2 n+1} \sqrt{2}=\left(x_{n+1}+y_{n+1} \sqrt{2}\right)\left(x_{n}+y_{n} \sqrt{2}\right)$ and (iv).
(vi) From $x_{2 n}+y_{2 n} \sqrt{2}=\left(x_{n}+y_{n} \sqrt{2}\right)^{2}$, deduce that $y_{2 n}=2 x_{n} y_{n}$, and then use this and (ii) to deduce the desired conclusion. This can also be seen directly from the binomial expansion, and it is always true that $2 \nmid x_{n}$.
(vii) From $x_{n}^{2}-2 y_{n}^{2}=(-1)^{n}$, it is known that to prove $u^{4}-2 v^{2}= \pm 1$, there are no other positive integer solutions except $u=v=1$. $u^{4}-2 v^{2}=1$ can be rewritten as $\left(u^{2}-1\right)\left(u^{2}+1\right)=2 v^{2}$, which clearly has no positive integer solutions. $u^{4}-2 v^{2}=-1$ can be rewritten as $u^{4}+\left(v^{2}-1\right)^{2}=v^{4}$, and from problem 15 of the second chapter, it is deduced that it has no other positive integer solutions except $u=v=1$.