Let $q=p^\alpha$ be a prime power. We call $r$ a primitive prime divisor of $q^n-1$ where $r\mid (q^n-1)$ but $r\nmid (q^i-1)$ for each $1\leq i\leq n-1$. The set of all primitive prime divisors of $q^n-1$ is denoted by $\pi_n(q)$.
Let $q=p^\alpha$ and $q'=r^\beta$ where $r$ and $p$ are odd prime numbers and $\alpha,\beta$ are natural numbers.
Let $p\mid (q'-1)$ and $r\mid (q-1)$. Is it possible that $\pi_m(q)=\pi_m(q')$ for each $3\leq m\leq 10$? I suspect it's impossible.
Thanks for your helps.