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We know that $q=p^a$ and $q'=p'^b$ are odd prime powers and $p$ and $p'$ are of the form $4k+1$. Also $p\mid (q'^2+1)$ and $p'\mid (q^2+1)$. If $\pi(n)$ denotes the set of prime divisors of $n$, is it possible that the following relations are hold:

(1) $\pi(q^2-1)=\pi(q'^2-1)$

and

(2) $\pi(q^2+1)\cup \{p\}=\pi(q'^2+1)\cup \{p'\}$

We suspect that this is impossible but we can not prove it.

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  • $\begingroup$ Have you done any computations? Why do you suspect that it is impossible? $\endgroup$
    – Nick Gill
    Sep 12, 2013 at 12:23
  • $\begingroup$ Dear Nick we try to find any answer but using computer we can not get any answer for them. $\endgroup$
    – BHZ
    Sep 12, 2013 at 12:25
  • $\begingroup$ You may want to add the gt.group-theory tag - I guess this question is motivated by finding different groups $PSL_2(q)$ and $PSL_2(q')$ whose orders have the same prime divisors (?). $\endgroup$
    – Nick Gill
    Sep 12, 2013 at 12:28
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    $\begingroup$ if you've used computers, do you know, for instance, that this situation does not occur for $q$ and $q'$ both less than 1000 or something? $\endgroup$
    – Nick Gill
    Sep 12, 2013 at 12:29
  • $\begingroup$ Yes this is the same problem in Group Theory that you stated with some more conditions. For your second question we check just special cases because of the limitation of our computer. $\endgroup$
    – BHZ
    Sep 12, 2013 at 12:32

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