Team Selection Test for IMO

We prove that at $a=1$, $b=2008$ all terms of the sequence are perfect squares. Let us prove by induction that for all $n\ge 1$ $$x_n^2+x_{n+1}^2-1=2008x_n{x}_{n+1}.(1)$$ 1. $n=1:1^2+2008^2-1=2008\cdot 1\cdot 2008$. 2. Suppose (1) is held for $n=k:x_k^2+x_{k+1}^2-1=2008x_k{x}_{k+1}$. Then $x_k^2+x_{k+1}^2-1=x_k\cdot 2008x_{k+1}=x_k\cdot (x_k+x_{k+2})=x_k^2+x_k{x}_{k+2}$. Then $x_{k+1}^2-1=x_k{x}_{k+2}=(2008x_{k+1}-x_{k+2})x_{k+2}=2008x_{k+1}{x}_{k+2}-x_{k+2}^2$. Therefore, $x_{k+1}^2+x_{k+2}^2-1=2008x_{k+1}{x}_{k+2}$ and (1) is held for $n=k+1$.

Now we get $1+2006x_n{x}_{n+1}=1+2008x_n{x}_{n+1}-2x_n{x}_{n+1}=x_n^2+x_{n+1}^2-2x_n{x}_{n+1}=(x_{n+1}-x_n{)}^2$. Done.