Mongolian Mathematical Olympiad

Since $2^n + 7^n \equiv 1 \pmod{p}$, $n$ 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 $6y \equiv 1 \pmod{p}$.

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

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

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