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.



On the set of divisors of $q-1$ and $q'-1$
$q=27=3^3$, $q'=13=13^1$. $3\mid13-1$, $13\mid27-1$. $$\pi(26)\cup3=\pi(12)\cup13=\{2,3,13\}$$


