Dependencies among roots of irreducible polynomial over GR(8,m) Let $GR\left( 8,m\right) =\left\{ a_{0}+a_{1}\zeta +\cdot \cdot \cdot
+a_{m-1}\zeta ^{m-1}:a_{0},a_{1},\ldots,a_{m-1}\in \mathbb{Z}_{8}\right\} $. Let $i,j,k,l=0,1,\ldots,2^{m}-2$, $i,j,k,l$ are distinct and $%
\zeta ^{i},\zeta ^{j},\zeta ^{k},\zeta ^{l}$ $\in GR\left( 8,m\right) $
satisfy the following equation 
\begin{equation}
\zeta^i+\zeta^j=\zeta^k+\zeta^l.
\end{equation}
Could you help me to prove or disprove that  $\zeta^i \zeta^j =\zeta^k \zeta^l$?
 A: [Edit: Adding some definitions]
The ring $GR(8,m)$ has several equivalent descriptions. It is a matter of taste, which one you would call the definition.


*

*Pick a primitive polynomial $p(x)$ of degree $m$ in $\mathbb{F}_2[x]$, i.e. the minimal polynomial of a generator of the multiplicative group of the finite field $GF(2^m)$. Hensel lift this to
a degree $m$ monic polynomial $f(x)\in \mathbb{Z}_8[x]$ such that $f(x)\equiv p(x)\pmod 2$ and $f(x)\mid x^{2^m-1}-1$ in the ring $\mathbb{Z}_8[x]$. Then 
$$
GR(8,m)=\mathbb{Z}_8[x]/\langle f(x)\rangle
$$ 
is a commutative ring of characteristic $8$ and cardinality $8^m$, $\zeta$ is the coset of $x$.

*Let $g$ be a generator of the multiplicative group of $F=GF(2^m)$. The ring of Witt vectors of length 3 is what we need
$$
GR(8,m)=W_3(F).
$$
See e.g. Jacobson, Basic Algebra II for the details. Here $\zeta=(g,0,0)$ as a Witt vector.

*Let $\zeta$ be a primitive root of unity of order $2^m-1$. Then the field
$\mathbb{Q}_2[\zeta]$ is an unramified extension of degree $m$ of the field of $2$-adic
numbers. Let $\mathcal{O}$ be the ring of integers of the extension field. Then
$$
GR(8,m)=\mathcal{O}/8\mathcal{O},
$$
and we denote the coset of $\zeta$ with $\zeta$ as well.


[/Edit]
So whichever way we look at it, $\zeta$ is a root of unity of order $2^m-1$ = a generator of the non-zero elements
of the Teichmüller set. Consequently the element $g=\zeta+2GR(8,m)$ is a generator of the multiplicative group of the finite field
$F=GR(8,m)/2GR(8,m)\cong \mathbb{F}_{2^m}.$ 
I claim that there are no solutions $i,j,k,l$ with all the exponents distinct in that range.
Assume contrariwise that both equations $\zeta^i+\zeta^j=\zeta^k+\zeta^\ell$ and
$\zeta^i\zeta^j=\zeta^k\zeta^\ell$ hold. Then by projecting to $F$ we get that the equations
$$
g^i+g^j=a=g^k+g^\ell
$$
as well as
$$
g^ig^j=b=g^kg^\ell
$$
also hold for some elements $a,b\in F$.
Consider the polynomial
$$
p(x)=x^2+ax+b\in F[x].
$$
It has a factorization
$$
(x-g^i)(x-g^j)=x^2-(g^i+g^j)x+g^ig^j=x^2+ax+b=p(x)
$$
as well as the factorization
$$
(x-g^k)(x-g^\ell)=x^2-(g^k+g^\ell)x+g^kg^\ell=x^2+ax+b=p(x).
$$
Because $F$ is a field, the polynomial $p(x)$ can have at most two zeros
in $F$. This means that the sets $\{g^i,g^j\}$ and $\{g^k,g^\ell\}$
are actually equal. But the restriction of the projection $GR(8,m)\rightarrow F$
to the Teichmüller set $\{0,1,\zeta,\zeta^2,\ldots,\zeta^{2^m-2}\}$ is a bijection,
so this implies an equality of sets $\{\zeta^i,\zeta^j\}=\{\zeta^k,\zeta^\ell\}$
and, a fortiori, of the sets $\{i,j\}=\{k,\ell\} (\subseteq\{0,1,\ldots,2^m-2\})$.
This contradicts the assumption that the exponents are distinct elements within the prescribed range.
