1
$\begingroup$

Suppose that $n\in\mathbb{N}$, $x\in\mathbb{Z}$ and that $M$ is the order of $x$ modulo $n$. Prove that if $m\in\mathbb{N}$ and $x^m\equiv 1\;(\mathrm{mod}\;n)$, then $M\,\vert\,m$.

We know that $M$ is the order of $x$ modulo $n$, $M$ is the smallest integer such that $x^M\equiv 1\;(\mathrm{mod}\,n)$. Also $x^m\equiv1\;(\mathrm{mod}\;n)$. We have $M<m$. I got stuck at here. Can someone give me a hint or suggestion to continue or start a new proof?

$\endgroup$
1
$\begingroup$

Let $m = qM + r$ for some $q\in \mathbb{Z}_+$, $r \in \{0,1,\dots, M-1\}$. Then $x^m = x^{qM+r}$ is congruent to $1$ modulo $n$. As $x^{qM}$ is $1$ mod $n$, we must have $x^r \equiv 1 \textrm{ mod } n$. But since $r<M$, $r=0$, i.e. $M$ divides $m$.

$\endgroup$
  • $\begingroup$ Why $x^{qM}\equiv1\;(\mathrm {mod}n)$? $\endgroup$ – Simple Oct 13 '16 at 5:41
  • $\begingroup$ $x^M \equiv 1 \textrm{ mod } n$, so $x^{qM} \equiv (1)^q \equiv 1 \textrm{ mod } n$ $\endgroup$ – GiantTortoise1729 Oct 13 '16 at 14:43

Your Answer

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

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