MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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

share|cite|improve this answer
Many thanks for the answer – BHZ Jun 20 '13 at 6:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.