Let p be a prime, m an odd positive integer such that $ p^e||m$, where $ e>1 $ is an integer. I am trying to show that

$$ \phi(p^e)=p^e-p^{e-1}\nmid p^et-1=m-1 $$

Is it enough to just say that the left-hand side will be some multiple of p whereas the right hand side will be some multiple of p minus one so the right-hand side cannot divide the left?

  • $\begingroup$ Yes, that is enough. $\endgroup$ – lulu Nov 19 '18 at 20:14
  • $\begingroup$ how would I write that though $\endgroup$ – joseph Nov 19 '18 at 20:43
  • 1
    $\begingroup$ What you wrote is very nearly formal. If you want to say more, I'd remark that (clearly) if $a\,|\,b$ then any divisor of $a$ divides $b$. Now, $e>1$ implies that $p\,|,p^e-p^{e-1}$. Thus, if $p^e-p^{e-1}$ were a divisor of $m-1$, $p$ would also have to be a divisor of $m-1$. But $p\,|\,m \implies p\, \nmid m-1$ and we are done. $\endgroup$ – lulu Nov 19 '18 at 20:48

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