Greece — Selection Examination

a.
From the given recurrence relation we get
$$(2x_{n+1} - 3x_n)^2 = 5x_n^2 - 4 \Rightarrow 4x_{n+1}^2 - 12x_{n+1}x_n + 4x_n^2 = -4 \Rightarrow x_{n+1}^2 - 3x_{n+1}x_n + x_n^2 = -1 \quad (1)$$
which can be written also in the form
$$x_{n+2}^2 - 3x_{n+2}x_{n+1} + x_{n+1}^2 = -1 \quad (2).$$
We consider now the second degree equation: $x^2 - 3x \cdot x_{n+1} + x_{n+1}^2 + 1 = 0$.
From (1), (2), we observe that two solutions of this equation are $x_n, x_{n+2}$, and hence by using Vieta's formulas we get:
$$x_n + x_{n+2} = 3x_{n+1} \quad (3) \quad \text{and} \quad x_n x_{n+2} = x_{n+1}^2 + 1 \quad (4).$$
Writing relation (3) in the form $x_{n+2} = 3x_{n+1} - x_n$ and taking in mind that $x_1 = 1$ and $x_2 = 2$ we conclude by induction that all terms of the sequence are integers

β.
We suppose that there exists a term $x_s$ of the sequence such that: $2011|x_s$. Then from (4) for $n=s$, we get $x_s x_{s+2} = x_{s+1}^2 + 1$. Since all terms of the sequence are integers and $2011|x_s$, we have:
$$\begin{aligned} 2011 \mid x_{s+1}^2 + 1 &\Rightarrow x_{s+1}^2 \equiv -1 \pmod{2011} \\ &\Rightarrow (x_{s+1}^2)^{1005} \equiv (-1)^{1005} \equiv -1 \pmod{2011} \\ &\Rightarrow x_{s+1}^{2010} \equiv -1 \pmod{2011} \end{aligned} \quad (5)$$
We know that $2011$ is prime and it is easy to see that $(x_{s+1}, 2011) = 1$.
In fact, if $(x_{s+1}, 2011) = d > 1$, then $d|x_{s+1}$, $d|2011 \Rightarrow d|x_{s+1}$, $d|x_s$ and then from relation $x_s x_{s+2} = x_{s+1}^2 + 1$ we conclude that $d|1$, absurd.
Therefore from Fermat's theorem we have:
$$x_{s+1}^{2010} \equiv 1 \pmod{2011}, \quad (6)$$
which contradicts relation (5).