Mongolian Mathematical Olympiad

Since $2^n+7^n\equiv 1(\mathrm{mod}p)$, $n$ must be odd. Let $p$ be the least prime divisor of $n$. Hence $(p,6)=1$ and there exists $x,y\in \mathbb{Z}$ such that $px+6y=1$. Therefore $6y\equiv 1(\mathrm{mod}p)$.

Let us consider $a$ such that $a\equiv 7y(\mathrm{mod}p)$. Then $a^n+1\equiv (7y{)}^n+(6y{)}^n(\mathrm{mod}p)$ and $7^n+6^n\equiv (7y{)}^n+(6y{)}^n(\mathrm{mod}p)$. Consequently $a^n+1\equiv 0(\mathrm{mod}p)$ and $a^n\equiv -1(\mathrm{mod}p)$. Since $n$ is odd $(-a{)}^n\equiv 1(\mathrm{mod}p)$.

On the other hand $(-a{)}^{p-1}\equiv 1(\mathrm{mod}p)$ by Fermat's theorem. Therefore $(-a{)}^{(n,p-1)}\equiv 1(\mathrm{mod}p)$. Since $p$ is least prime divisor of $n$ we get $(n,p-1)=1$ and $-a\equiv 1(\mathrm{mod}p)$. From here we deduce $7y\equiv 1(\mathrm{mod}p)$.

Finally $-7\cdot 6y\equiv (\mathrm{mod}\Rightarrow )-7\equiv 6(\mathrm{mod}p)\Rightarrow 13\equiv 0(\mathrm{mod}p)$ and $13|n$.