$ 1 + 2006x_1x_2$ is supposed to be square. This inspire us to choose $ x_1x_2 = 2006 + 2$, in other words, $ a = 1$ and $ b = 2008$. So $ x_0 = 0$, $ x_1 = 1$ and $ x_2 = 2008$, with recurence relation $ x_{n + 1} = 2008x_n - x_{n - 1}$. We we'll prove by induction that $ 1 + 2008x_nx_n + 1 = x_n^2 + x_{n + 1}^2 \Leftrightarrow 1 + 2006x_nx_{n + 1} = (x_{n + 1} - x_n)^2$. For $ n = 0$ this is obvious. $ 1 + 2008x_{n + 2}x_{n + 1} = 1 + 2008(2008x_{n + 1} - x_n)x_{n + 1} = 1 + 2008^2x_{n + 1}^2 - 2008x_nx_{n + 1} = x_n^2 + x_{n + 1}^2 - 2008x_nx_{n + 1} + 2008^2x_{n + 1}^2 - 2008x_nx_{n + 1} = (2008x_{n + 1} - x_n)^2 + x_{n + 1}^2 = x_{n + 2}^2 + x_{n + 1}^2$ and that's it Bye PS I don't know why does it format like this, if you can fix it, feel free to do it

We choose $ a = 1$ and $ b = 2008$ . Then by induction or other method can easily prove that $ (x_n - x_{n - 1})^2 = 1 + 2006x_nx_{n - 1}$ For all these the basic idea was to find a,b such that $ (b - a)^2 = 2006ab + 1$ Edit : Sponge Bob was faster

In the same method I found that : Prove that for $ m\in N$ and $ m\geq 3$ . Consider the sequence : $ x_1 = a,x_2 = b$ $ x_{n + 2} = mx_{n + 1} - x_n$ Prove that for every k there exist $ a,b\in N$ such that $ k^2 + (m - 2)x_{n}x_{n + 1}$ is a perfect square for all n.