Vijetnam 2011

Consider the sequence of integers $(b_n)$ determined by $$b_0=1,b_1=-1\text{ and }{b}_n=6b_{n-1}+2016b_{n-2}\text{ for all }n\ge 2.$$ Clearly, for all $n\ge 0$, we have $a_n\equiv b_n(\mathrm{mod}2011)$. () The characteristic equation of sequence $(b_n)$: $x^2-6x-2016=0$, or $(x-48)(x+42)=0$. Consequently, the general term of $(b_n)$ has the form: $b_n=C_1(-42{)}^n+C_{2}48^n$. By the initial conditions for sequence $(b_n)$, we obtain $$\{\begin{array}{l}{C}_1+C_2=1\\ 42C_1-48C_2=1.\end{array}$$ Hence $C_1=\frac{49}{90}$ and $C_2=\frac{41}{90}$. Thus $b_n=\frac{49(-42{)}^n+41\cdot 48^n}{90}\forall n\ge 0$. Since $2011$ is a prime, according to the little Fermat theorem we have: $$(-42{)}^{2010}\equiv 48^{2010}\equiv 1(\mathrm{mod}2011).$$ Hence $90b_{2012}=49\cdot (-42{)}^{2012}+41\cdot 48^{2012}=49\cdot (-42{)}^2+41\cdot 48^2=90b_2(\mathrm{mod}2011)$. Consequently $b_{2012}\equiv b_2(\mathrm{mod}2011)$ (since $(90,2011)=1$). But $b_2=6b_1+2016b_0=2010$, hence $b_{2012}=2010(\mathrm{mod}2011)$. Thus $a_{2012}=2010(\mathrm{mod}2011)$ (by means of ()).