Mongolian National Mathematical Olympiad

We consider the numbers $2^i 3^j$ for $1 \le i, j \le p-1$. There are $(p-1)^2 \ge p+1$ numbers, so by Dirichlet's principle, there exist different pairs $(i_1, j_1)$ and $(i_2, j_2)$ such that $1 \le i_1, i_2, j_1, j_2 \le p-1$ and $2^{i_1} 3^{j_1} \equiv 2^{i_2} 3^{j_2} \pmod p$.

By Fermat's theorem we have $2^{p-1} \equiv 3^{p-1} \equiv 1 \pmod p$. Hence we have $2^{i_1} 2^{p-1-i_2} \equiv 3^{j_2} 3^{p-1-j_1} \pmod p$ so there exist $1 \le i, j \le p-2$ such that $2^i \equiv 3^j \pmod p$.

If $i \le j$, then we can choose $m=i$ and $n=p-1-j$ and if $i > j$ then choose $m=p-1-i$ and $n=j$.