IRL_ABooklet

We claim that, for all $n\ge 1$, $$x_{n-1}{x}_{n+1}=x_n^2+(-1{)}^n.(10)$$ Based on this, if $k>1$ we then have $x_n=k{x}_{n-1}+x_{n-2}\ge k{x}_{n-2}\ge k\ge 2$ for $n\ge 2$, so $x_n^2\pm 1$ cannot be a square number. If $k=1$ then $x_2=2$ and $x_n=x_{n-1}+x_{n-2}\ge x_{n-2}\ge 2$ for $n\ge 3$ and the same conclusion holds. With $k=1$ and $n=2$ we have $x_{n-1}{x}_{n+1}=x_1{x}_3=2$ which is not a square number.

To prove (10) by induction, we first settle the base case $n=1$: $$x_0{x}_2-x_1^2=-1=(-1{)}^1.$$ We now suppose that $x_{n-1}{x}_{n+1}=x_n^2+(-1{)}^n$ for some $n\ge 1$ and use the recurrence relation $x_{n+2}=k{x}_{n+1}+x_n$ and $k{x}_n=x_{n+1}-x_{n-1}$ to find $$\begin{array}{rl}{x}_n{x}_{n+2}& =x_n(k{x}_{n+1}+x_n)\\ & =k{x}_n{x}_{n+1}+x_n^2\\ & =(x_{n+1}-x_{n-1})x_{n+1}+x_n^2\\ & =x_{n+1}^2-x_{n-1}{x}_{n+1}+x_n^2\\ & =x_{n+1}^2-(x_n^2+(-1{)}^n)+x_n^2\\ & =x_{n+1}^2+(-1{)}^{n+1}.\end{array}$$ This completes the proof by induction.