- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey!
We have the following lemma:
Assume that the characteristic of $F$ is $p$ and $p>2$.
Then $(t^m-1)/(t^n-1)$ is a square in $F[t, t^{-1}]$ ($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$) if and only if $(\exists s \in \mathbb{Z}) m=np^s$.
Can we say something about $p=2$ ?
$$(\exists s \in \mathbb{Z})m=2^sn \Leftrightarrow \dots$$
If $(\exists s \in \mathbb{Z})m=2^sn$ then we have that:
$$t^m=t^{2^sn}=\left (t^n\right )^{2^s} \Rightarrow t^m-1=\left (t^n\right )^{2^s}-1=\left (t^n-1\right )^{2^s}$$
Can we write with that a $\Leftrightarrow$ relation?
We have the following lemma:
Assume that the characteristic of $F$ is $p$ and $p>2$.
Then $(t^m-1)/(t^n-1)$ is a square in $F[t, t^{-1}]$ ($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$) if and only if $(\exists s \in \mathbb{Z}) m=np^s$.
Can we say something about $p=2$ ?
$$(\exists s \in \mathbb{Z})m=2^sn \Leftrightarrow \dots$$
If $(\exists s \in \mathbb{Z})m=2^sn$ then we have that:
$$t^m=t^{2^sn}=\left (t^n\right )^{2^s} \Rightarrow t^m-1=\left (t^n\right )^{2^s}-1=\left (t^n-1\right )^{2^s}$$
Can we write with that a $\Leftrightarrow$ relation?