Mongolian National Mathematical Olympiad

Assume that there are positive integers $a_1,a_2,\dots ,a_{2017}$ and a prime $p$ as required. Then, for each $i$, there is a positive integer $k_i$ so that $$a_i^{2017}+a_{i+1}=p^{k_i}.(\ast )$$ Here $a_{2018}=a_1$. The sum of all $p^{k_i}$ equals to the sum of all $a_i^{2017}+a_i$ which is even and so $p=2$.

We claim that $a_i$ is odd for each $i$. Write $a_i=2^{{\alpha }_i}\cdot (2b_i+1)$ for some integer ${\alpha }_i$ and $b_i$. Since $a_i^{2017}+a_{i+1}=2^{k_i}$ we get $2017{\alpha }_i={\alpha }_{i+1}$ for each $i$ and so ${\alpha }_i=0$. We claim that $(a,m)=(1,1)$ is the only solution of $a^{2017}+1=2^m$. By contrary, there is a solution with $m>1$. Then $a>1$ and $a^{(2\cdot 2017,2^{m-1})}\equiv 1(\mathrm{mod}{2}^m)$ since $a^{2\cdot 2017}\equiv 1(\mathrm{mod}{2}^m)$ and $a^{2^{m-1}}\equiv 1(\mathrm{mod}{2}^m)$. Thus $a^2\equiv 1(\mathrm{mod}{2}^m)$ and so $a^2-1\ge 2^m=a^{2017}+1$ which is a contradiction. The claim implies that $a_i>1$ for each $i$. Indeed, if there is $i$ so that $a_i=1$ then $a_{i-1}=1$ by the claim and so $a_i=k_i=1$ for each $i$. In this case the sum of all $k_i$ is 2018. Thus $a_i^{2017}+a_{i+1}\ge 2^{2017}+a_{i+1}$ and so $k_i\ge 2018$. Let $k=\min (k_i)$. For each $1\le i\le 2017$, it follows from (*) that $$a_i^{20172017}\equiv -a_i(\mathrm{mod}{2}^k).$$ Since $a_i$ is odd we get $a_i^{(2(2017^{2017}-1),2^{k-1})}\equiv 1(\mathrm{mod}{2}^k)$ which implies $a_i^{2^6}\equiv 1(\mathrm{mod}{2}^k)$. Thus $a_i^{2^6}\ge 2^{2017}$ and so $k_i\ge 30\cdot 2017$ which means the sum of all $k_i$ is more than $2017\cdot 2023$.