$(1+(\theta-1)t) = \theta^n(1+(\theta-1)^2t')$ for $t,t'\in \mathbb Z[\theta], n\in \mathbb Z$ and $\theta$ is a primitive $p^s$-th root of unity. I found the following while reading and computing problems in algebraic number theory.
Suppose $\theta$ is a primitive $9$-th root of unity over rationals. Consider the elements $a = 1+(\theta-1)\theta,  b = 1+(\theta-1)\theta^2$. 
By some calculations we can see that $a=\theta(1+\theta^8(\theta-1)^2), b = \theta(1+(\theta^8+1)(\theta-1)^2)$ 
So my question (in general )is

Let $s\ge 1$ and $p$ be a prime. If $\theta$ is a $p^s$-th root of unity over rationals then is it true that for every $t\in \mathbb Z[\theta]$, there is some $t' \in \mathbb Z[\theta]$ and $n\in \mathbb Z$ such that $(1+(\theta-1)t)= \theta^n(1+(\theta-1)^2t')$?

I have checked few cases in GAP (computer) which seems true, but I am not sure whether it is true and how t prove it?
Sorry for not showing much work from my end, but I am quite stuck.
Any help will be really appreciated. Thanks in advance.
 A: Your property is true, and one can show it as follows. Let $q=p^s$. There is a polynomial $P$ with integer coefficients such that $t=P(\theta)$. There is a unique $m\in[|0,q-1|]$ such that $m$ is congruent to $-P(1)$ modulo $q$. Then, $P(1)=-m+bq$ for some $b\in{\mathbb Z}$. we can write $P(X)=-m+bq+(X-1)T$ where $T$ is a polynomial with integer coefficients. Since 
$$q=q-\sum_{j=0}^{q-1}\theta^j=(1-\theta)(\sum_{j=0}^{q-2} (q-1-j)\theta^j),$$
we have 
$$t=-m+(\theta-1)Q(\theta)$$ 
where $Q=T-b\sum_{j=0}^{q-2} (p-1-j)X^j$. We deduce : 
$$
\begin{array}{lcl}
\theta^m(1 + (\theta-1)t) 
&=& 
\theta^m(1 -m(\theta-1)+(\theta-1)^2Q(\theta)) \\
&=& 
\theta^m- m\theta^m(\theta-1)+(\theta-1)^2Q(\theta)\theta^m \\
&=&
1+(\theta-1)(\sum_{j=0}^{m-1}\theta^j)-m\theta^m(\theta-1)+(\theta-1)^2Q(\theta)\theta^m \\
&=&
1+(\theta-1)\Bigg(\sum_{j=0}^{m-1}\theta^j-m\theta^m\Bigg)+(\theta-1)^2Q(\theta)\theta^m \\
\end{array}
$$
since the polynomial $R=\sum_{j=0}^{m-1}X^j-mX^m$ is zero at $1$, we can write  $R(X)=(X-1)S(X)$ where $S$ has integer coefficients. Then
$$
\begin{array}{lcl}
\theta^m(1 + (\theta-1)t) 
&=& 
1+(\theta-1)^2\Bigg(S(\theta)+Q(\theta)\theta^m\Bigg) \\
\end{array}
$$
So putting $t'=S(\theta)+Q(\theta)\theta^m$ we are done.
