Two real sequences $(x_n)$ and $(y_n)$ are defined by
$$x_{n+1}=x_n-(x_ny_n+x_{n+1}y_{n+1}-2)(y_n+y_{n+1})$$
$$y_{n+1}=y_n-(x_ny_n+x_{n+1}y_{n+1}-2)(x_n+x_{n+1})$$ with $x_0=1$ and $y_0=2007.$ I need to show that $|x_n|\lt \sqrt{2007}$ for all $n\in\mathbb{N}.$
I proved that $$x_{n+1}^2-x_n^2=y_{n+1}^2-y_n^2\,\,\,\,\,\,\,\,\,\,\forall n\in\mathbb{N},$$ which implies $|x_n|\lt|y_n|$ and $$x_n^2=y_n^2-2007^2+1.$$
Also I would like to know that, Is $x_n$ convergent? Any Idea?
Some idea:
let $A=x_{n}y_{n}+x_{n+1}y_{n+1}-2$,since $(2)^2-(1)^2$ then we have $$y^2_{n+1}-x^2_{n+1}=y^2_{n}-2Ay_{n}(x_{n+1}+x_{n})+A^2(x_{n}+x_{n+1})^2-x^2_{n}+2(y_{n}+y_{n+1})x_{n}A-A^2(y_{n}+y_{n+1})^2$$ since $$y^2_{n+1}-x^2_{n+1}=y^2_{n}-x^2_{n}$$then we have $$A(y_{n}y_{n+1}-x_{n}x_{n+1})=y_{n+1}x_{n}-x_{n+1}y_{n}$$ or $$y_{n+1}x_{n}-x_{n+1}y_{n}=(x_{n}y_{n}+x_{n+1}y_{n+1}-2)(y_{n}y_{n+1}-x_{n}x_{n+1})\tag{2}$$
as your comment $$(2)x_{n}:y_{n+1}x_{n}=x_{n}y_{n}-A(x_{n}+x_{n+1})x_{n}\tag{3}$$ $$(1)y_{n}:x_{n+1}y_{n}=x_{n}y_{n}-A(y_{n}+y_{n+1})y_{n}\tag{4}$$ so $(3)-(4)$ $$x_{n}y_{n+1}-x_{n+1}y_{n}=A[y^2_{n}-x^2_{n}+y_{n}y_{n+1}-x_{n}x_{n+1})\tag{6}$$ for $(2),(6)$ you can have ?