Suppose $a>1$ is an integer, and $p$ is an odd prime number. Prove that each odd prime factor of $(a^p)-1$ which does not divide $a-1$ should be in the form $2pt+1$.

My Approaching: ($a^p)-1$ is divisible by q which is any odd prime factor of $(a^p)-1$. then I rewrite the function $(a^p)-1 as (a-1)*((a^p-1)+...+1)$ So whether $q$ divides $(a-1)$ or "the rest" Then I stuck because I do not know how make "the rest" equal to $2pt+1$.



Suppose that $q$ is prime, $q\ne2$ and $q\mid a^p-1$ but $q\not\mid a-1$. Writing the last two conditions in congruence form, $$a^p\equiv1\pmod q\quad\hbox{but}\quad a\not\equiv1\pmod q\ .$$ This means <please supply reasons> that the order of $a$ modulo $q$ is $p$, and hence $p$ is a factor of $q-1$. See if you can complete the proof.

  • $\begingroup$ Thank you. It's very helpful $\endgroup$ – Jestem Jul 9 '14 at 1:09

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