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 $2^{i_1}{3}^{j_1}\equiv 2^{i_2}{3}^{j_2}(\mathrm{mod}p)$.

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

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 $n=j$.