If $\text{ord}_m(c)=n$, find $\text{ord}_m(c^2)$

Question

I'm trying to solve the following problem:

if $\text{ord}_m(c)=n$, find $\text{ord}_m(c^2)$

Here's what I have so far (admittedly, not much):

1. We know that $c^n \equiv 1 \pmod{m}$, equivalently, $c^n = 1 + mp$, $p\in\mathbb{Z}$
2. We know $c^2$ = $c\cdot c$.
3. And, of course, $c^n\cdot c^n \equiv 1 \pmod{m}$

But what $p$ satisfies $(c^2)^p \equiv 1 \pmod{m}$?

Answer 1

Note that $c^k\equiv 1$ (mod $m$) if and only if $n|k$. So in particular $(c^2)^k\equiv c^{2k}\equiv 1$(mod $m$) iff $n|2k$. So we are actually looking for the smallest natural number $k$ such that $n|2k$. If $n$ is even this smallest number is clearly $k=\frac{n}{2}$ while if $n$ is odd this is $k=n$.