Why doesn't this hold for $p=2$? I have a question about 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 $F[t, t^{-1}]$ if and only if $(\exists s \in \mathbb{Z}) m=np^s$.  

($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$) 
We assume that the characteristic of $F$ is $p>2$. Why can the characteristic not be $p=2$. Is it because then we would have $\frac{t^m-1}{t^n-1}=a^2=1$ ? 
Can we modify a little the lemma so that it stands for characteristic $p=2$ ?
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EDIT1: 
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? 
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EDIT2: 
Using te lemma:  

$t^n-1$ divides $t^m-1$ 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 $n$ divides $m$ in $\mathbb{Z}$. 

we have the following: 
If $\exists s \in \mathbb{Z}$ so that $m=2^s n$, then $2 \mid m$ and $n \mid m$. 
We have the following: 
$$2 \mid m \Leftrightarrow t^2-1 \mid t^m-1$$ 
and $$n \mid m \Leftrightarrow t^n-1 \mid t^m-1$$ 
Is this an "iff" statement? 

If $F$ is the $2$-element field, one has $\frac{t^2-1}{t-1}=t + 1$, but $1+t$ is not a square in $F[t, t^{-1}]$.
