Mongolian Mathematical Olympiad

Answer: $p\ge 5$ prime, $(m,p)=(2,p),(3,p)$. Let $p\ge 5$ be a prime number.

First, consider the case where $m\ge 2$ is even. If $m$ is even, consider $a_1=2(p-1)-m$. Since $a_1$ is even, we have $q=2$. Therefore, $q\ge m$ is satisfied only for $m=2$. For $m\ge 4$, the condition is not satisfied because $q<m$.

Next, consider the case where $m=3$. For $m=3$, all terms $a_n$ are odd. Hence, $q\ge 3$. Therefore, the condition $q\ge m$ is satisfied for $m=3$.

Consider the case where $m\ge 5$ is odd. We need to show that $q<m$ for odd $m\ge 5$.

First, let $m-2=p^k$ with $k\ge 1$. If $p=5$ and $k=1$, then $m=7$. Since $3\mid a_2=27$, we have $q=3<7=m$.

For general $k$, since $3\le a_k=(p-2{)}^k-2<m$, it follows that $q<m$.

Next, consider other prime divisors $r\ne p$ dividing $m-2$. If $r\nmid p-2$, then $r\nmid a_1$. If $r\nmid p-2$, by Fermat's Little Theorem, $r\nmid a_{r-1}$. Hence, $q\le r<m$.

Therefore, for $m\ge 5$, the condition $q\ge m$ is not satisfied if $m$ is an odd number.