SELECTION EXAMINATION

a. From the given recurrence relation we get $$(2x_{n+1}-3x_n{)}^2=5x_n^2-4\Rightarrow 4x_{n+1}^2-12x_{n+1}{x}_n+4x_n^2=-4\Rightarrow x_{n+1}^2-3x_{n+1}{x}_n+x_n^2=-1(1)$$ which can be written also in the form $$x_{n+2}^2-3x_{n+2}{x}_{n+1}+x_{n+1}^2=-1(2).$$ 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: $$x_n+x_{n+2}=3x_{n+1}(3)\text{and}{x}_n{x}_{n+2}=x_{n+1}^2+1(4).$$ 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: $$\begin{array}{rl}2011\mid x_{s+1}^2+1& \Rightarrow x_{s+1}^2\equiv -1(\mathrm{mod}2011)\\ & \Rightarrow (x_{s+1}^2{)}^{1005}\equiv (-1{)}^{1005}\equiv -1(\mathrm{mod}2011)\\ & \Rightarrow x_{s+1}^{2010}\equiv -1(\mathrm{mod}2011)\end{array}(5)$$ 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: $$x_{s+1}^{2010}\equiv 1(\mathrm{mod}2011),(6)$$ which contradicts relation (5).