For a natural number $n$ we denote by $\pi(n)$, for the set of prime divisors of $n$.
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 $r\mid (q-1)$. I need to prove that this is impossible that $\pi(q-1)\cup \pi(p)=\pi(q'-1)\cup \pi(r)$?
Is it true?
Thanks for your helps.