19th Turkish Mathematical Olympiad

Observe that $a_{n+1}-2=a_n^2(a_n-2)$ for all $n\ge 1$. By induction on $n$ we obtain $a_{n+1}-2=3a_n^2{a}_{n-1}^2\cdots a_1^2$ for all $n\ge 1$. Therefore $a_{2011}+1=3(a_{2010}^2{a}_{2009}^2\cdots a_1^2+1)=(a_{2010}{a}_{2009}\cdots a_1{)}^2+1$.

Let $p\equiv 3(\mathrm{mod}4)$ be a prime divisor of $a_{2011}+1$. It is well known that if $q$ is a prime divisor of $(a_{2010}{a}_{2009}\cdots a_1{)}^2+1$, then $q\equiv 1(\mathrm{mod}4)$ or $q=2$. Thus $p|3$. That is $p=3$.