IMO Team Selection Test 3

By Fermat's Little Theorem, we have $a^{p-1}\equiv 1\mathrm{mod}p$ for all $a$ such that $p\nmid a$. As $p>10$, $p$ is odd, so $p-1$ is even. We have $$(a^{\frac{p-1}{2}}-1)(a^{\frac{p-1}{2}}+1)=a^{p-1}-1\equiv 0\mathrm{mod}p.$$ So $p\mid (a^{\frac{p-1}{2}}-1)(a^{\frac{p-1}{2}}+1)$, so $p$ is a divisor of at least one of the factors. Hence $a^{\frac{p-1}{2}}$ is congruent to $1$ or $-1$ modulo $p$. We apply this to $a=5$ and $a=7$. Note that $p>10$, so $p\ne 5,7$. If $5^{\frac{p-1}{2}}\equiv 1\mathrm{mod}p$ and $7^{\frac{p-1}{2}}\equiv 1\mathrm{mod}p$, we choose $m=n=\frac{p-1}{2}$, which satisfy the condition. The same holds if $5^{\frac{p-1}{2}}\equiv -1\mathrm{mod}p$ and $7^{\frac{p-1}{2}}\equiv -1\mathrm{mod}p$. The remaining case is that one is congruent to $1$ and the other is congruent to $-1$. Assume that $5^{\frac{p-1}{2}}\equiv 1\mathrm{mod}p$ and $7^{\frac{p-1}{2}}\equiv -1\mathrm{mod}p$. The case in which it is the other way around is analogous. If there exists an $n$ such that $0<n<\frac{p-1}{2}$ and $7^n\equiv 1\mathrm{mod}p$, then we choose this $n$ and $m=\frac{p-1}{2}$, which satisfy the condition. If not, then no $n$ such that $\frac{p-1}{2}<n<p-1$ and $7^n\equiv 1\mathrm{mod}p$ can exist either, since otherwise $7^{p-1-n}\equiv 7^{p-1}(7^n{)}^{-1}\equiv 1\mathrm{mod}p$, while $0<p-1-n<\frac{p-1}{2}$, which is a contradiction. Moreover, no $i,j$ such that $1\le i<j\le p-1$ and $7^i\equiv 7^j\mathrm{mod}p$ can exist, since otherwise $7^{j-i}\equiv 1\mathrm{mod}p$ with $1\le j-i<p-1$. Hence $7^i$ assumes distinct values for $1\le i\le p-1$ modulo $p$, which are all non-zero. So $7^i$ assumes all non-zero values modulo $p$ for $1\le i\le p-1$. In particular, there exists an $n$ such that $7^n\equiv 5^{-1}\mathrm{mod}p$. Then $n\le p-2$, since $7^{p-1}\equiv 1\ne 5^{-1}\mathrm{mod}p$. Choosing this $n$ and $m=1$ gives $7^n\cdot 5^m\equiv 5^{-1}\cdot 5\equiv 1\mathrm{mod}p$. Therefore it is always possible to find $m$ and $n$ satisfying the conditions. $\square$