Let $d_1,d_2>0$ be the smallest positive integers such that \[ 5^{d_1}\equiv 7^{d_2}\equiv 1\pmod{p}. \]Clearly $d_1,d_2\mid p-1$. Now, if $d_1,d_2<p-1$ then we in fact have $d_1,d_2\le (p-1)/2$. So taking $m=d_1$ and $n=d_2$ we get the result. Now assume one of them, say $d_1$, is equal to $p-1$. Then $5^i,1\le i\le p-1$ sweeps all non-zero residues modulo $p$. So taking $n=1$, we find that there is an $m$ such that $5^m\equiv 7^{-1}\pmod{p}$. Clearly $m\ne p-1$ as $p>10$, so we must have $m\le p-2$. The case $d_2=p-1$ is analogous.

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