Let $x_0 = 1$ and $x_1 = 2003$ and define the sequence $(x_n)_{n \ge 0}$ by: $x_{n+1} =\frac{x^2_n + 1}{x_{n-1}}$ , $\forall n \ge 1$ Prove that for every $n \ge 2$ the denominator of the fraction $x_n$, when $x_n$ is expressed in lowest terms is a power of $2003$.