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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.

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Do you know of two numbers $a\gt b\gt1$ such that $\pi_3(a)=\pi_3(b)$ and $\pi_4(a)=\pi_4(b)$? –  Gerry Myerson Jun 20 '13 at 23:49
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Just the fact that$\pi_3(2)=\pi_3(18)$ is intriguing to me. I wonder if there is a result similar to Stormers Theorem on finitely many consecutive numbers whose prime factors all belong to a finite set $S$ that would say for a given finite set $S$ of primes that $\pi_n(m) = S$ has only finitely many positive integer solutions $m$ and $n> 1$. Gerhard "Also Relates To Odd Perfects" Paseman, 2013.06.20 –  Gerhard Paseman Jun 21 '13 at 2:03
    
Thank you very much for your answers. First let me explain that this questions arose from a problem in the theory of finite groups. In fact I have no example for the question of Gerry and for the same reason I suspect it is impossible, but I can not prove it. –  BHZ Jun 22 '13 at 2:19

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