It is easy to prove, by induction $ a_{n+1}=2ka_n-a_{n-1}$ and $ a_{n+1}a_{n-1}+1=a_n^2$. Then it is easy to verify the modulo of $ 2k$ is $ 0,1,0,-1,0,1,0,-1,...$.

This is standard...If you rearrange and square it, you get $ a_{n+1}^2-2ka_na_{n+1}+k^2a_n^2=(k^2-1)a_n^2+1$ Then $ a_{n+1}^2-2ka_n(a_{n+1})+a_n^2-1=0$ Note that $ x^2-2ka_n x+a_n^2-1=0$ has roots $ a_{n+1}$, $ a_{n-1}$ the function is increasing so $ a_{n+1}>a_n>a_{n-1}$ so $ a_{n+1}\ne a_{n-1}$ so they are the distinct roots of that quadratic and so $ x_1+x_2=2ka_n$ which gives that nice recursion. The rest of the problem is then trivial by induction.

So rearrange the condition a little bit and square it we get $a_{n+1}^2-2ka_na_{n+1}+a_n^2-1=0$ We express $ a_{n+1}^2 = 2ka_na_{n+1}-a_n^2+1$ So now let $ n \rightarrow n+1$, we have the following: $$ a_{n+2}^2-2ka_{n+1}a_{n+2}+a_{n+1}^2-1=0$$Plugging in the definition of $a_{n+1}^2$ we get $$(a_{n+2}-a_n)(a_{n+2}+a_n-2ka_{n+1})=0$$Because the sequence $\{a_n\}$ is increasing we have the following: $a_{n+2}+a_{n}=2ka_{n+1}$ We can calculate $a_1=1$ So now by induction we see that $\{a_n\}$ is a subsequnce of $\mathbb{Z}$ Since $2k | a_n + a_{n+2}$ and since $2k|a_0$.By induction with a step of 2 we get that $2k|a_{2n}$, $\forall n \in \mathbb{N}_0$