Why do the cyclotomic cosets have the same size? So I have this problem: when $p=2$ and $n=2^m-1$ show that $|C_1|=|C_3|=m$ where $m\geq 3$.
$C_s=\{s,ps,p^2s,\dots p^{m_s-1}\}$ is the cyclotomic coset.
I know that the size of the coset is $m_s$ which is the smallest positive integer such that $p^{m_s}*s\equiv s$ modulo $p^m-1$.
I get that for $C_1$ that $2^{m_{s_1}}\equiv 1$ mod $n$ and for $C_3$ that $3*2^{m_{s3}}\equiv 3$ mod $n$. How do I prove that $m_{s1}=m{s_3}=m$?

The number of elements in the cyclotomic coset $C_s$ can be found by expressing
$s$ in base-$2$ notation as $\mathbf x = (x_{m-1}, x_{m-2}, \ldots, x_1, x_0), x_i \in \{0,1\}$ and then finding the period of $\mathbf x$ under cyclic shifts.
It is straightforward to figure out that $00\cdots 01$ and $00\cdots 11$ both
have period $m$; it takes $m$ cyclic shifts to get back to where we started from.
Hence, $C_1$ and $C_3$ each have $m$ elements.
As noted by David Wheeler, when $m=2$, the argument above breaks down.
Note that
$11$ has period $1$ while $01$ has period $2$.
