2
$\begingroup$

Let g be a primitive root modulo $p$($p$ is an odd prime) with $g^{p-1}\ncong{1}\ (mod\ p^{2})$. I am interested in proving that

$$g^{(p-1)p^{m-2}}\ncong{1}\ (mod\ p^{m})$$ for every $m\geq{2}$

So far, I have proven that $a^{p^{m}-p^{m-1}}\equiv{1}\ (mod\ p^{m})$ given that $gcd(a,p)=1$. If that might be of help...

$\endgroup$
  • 1
    $\begingroup$ math.stackexchange.com/questions/332760/… $\endgroup$ – lab bhattacharjee Nov 2 '17 at 3:35
  • $\begingroup$ @labbhattacharjee I'm sorry, but it is not immediately clear to me, how this is relevant $\endgroup$ – Gwen Di Nov 2 '17 at 3:48
  • $\begingroup$ From the given condition, $g$ is a primitive root $\pmod{p^2}$ . It is sufficient to establish $g$ is a primitive root $\pmod{p^m}$ $\endgroup$ – lab bhattacharjee Nov 2 '17 at 3:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.