India_2017

Let $t$ be an integer such that $p$ divides $t^2+1$. There exists such an integer since $p-1$ is divisible by $4$. Let $m=1$ and $n=t+2$. Then $$n^2=t^2+4t+4\equiv 4t+3\equiv 4n-5m(\mathrm{mod}p).$$ Therefore, if $a_0=1$ and $a_1=n$ then $a_2\equiv n^2(\mathrm{mod}p)$. By induction, it is easy to see that $a_k\equiv n_k(\mathrm{mod}p)$. Since $p>5$, it follows that $p$ does not divide $n$. Therefore $p$ does not divide $a_k$ for any $k\ge 0$.