Roots of unity in $\Bbb{Z}[X]/(X^2+X+1)^2$ Let $f:=X^2+X+1\in\Bbb{Z}[X]$ and let $A:=\Bbb{Z}[X]/(f^2)$. I'm quite convinced that the only roots of unity in $A$ are $\pm1$, i.e. that $A^{\times}_{\mathrm{tor}}=\langle-1\rangle$. My current 'proof' is not much more than a long and cumbersome series of calculations, I'm hoping for a clear and simple argument that this is so.
The problem is greatly simplified by the fact that the map $A^{\times}_{\mathrm{tor}}\ \longrightarrow\ (A_{\mathrm{red}})^{\times}_{\mathrm{tor}}$ induced by the quotient map $A\ \longrightarrow\ A_{\mathrm{red}}=A/\sqrt{0_A}$ is injective. Here $A_{\mathrm{red}}=\Bbb{Z}[X]/(f)\cong\Bbb{Z}[\zeta_3]$, so $A^{\times}_{\mathrm{tor}}$ is cyclic and its order divides $6$. Hence to show that $A^{\times}_{\mathrm{tor}}=\langle-1\rangle$ it suffices to show that $A^{\times}$ has no $3$-torsion. My question is whether there is an easy way to see that this is so, if this is indeed so.
 A: One may use a variant of Hensel lifting to arrive at this result.
Let $ g(X) = X^2 + X + 1 $, and assume that $ P \in A $ is $ 3 $-torsion. This is equivalent to $ g(P) = 0 $ in the ring $ A $: indeed, we have that $ P^3 - 1 = (P-1)(P^2 + P + 1) = 0 $, and the factors $ P - 1 $ and $ P^2 + P + 1 $ have greatest common divisor which divides $ 3 $ in the ring $ \mathbf Z[X] $. 
Then, we see that $ P = a + (X^2 + X + 1) Q(X) $ for some $ Q \in \mathbf Z[X] $, where $ a = X $ or $ a = -X-1 $. We have that
$$ g(P) = g(a + (X^2 + X + 1) Q(X)) = g(a) + g'(a) (X^2 + X + 1) Q(X) $$
in $ A $. In either case, $ g(a) = X^2 + X + 1 $, so this simplifies, upon division by $ X^2 + X + 1 $, to the equality
$$ 1 + (2 \zeta_3 + 1) Q(\zeta_3) = 0 $$
in the ring $ \mathbf Z[\zeta_3] $. This is impossible, as it implies that the element $ 2 \zeta_3 + 1 $ is a unit in $ \mathbf Z[\zeta_3] $.
Edit: As an added perk, this method also produces the solution found by Hurkyl in the comments. Indeed, if we place ourselves in $ \mathbf Q[X]/(X^2 + X + 1)^2 $, then we obtain the result $ Q(\zeta_3) = (2\zeta_3 + 1)/3 $; so we may take $ Q(X) = (2X + 1)/3 $ to obtain
$$ P(X) = X + \frac{(X^2 + X + 1)(2X + 1)}{3} = \frac{2}{3} X^3 + X^2 + 2X + \frac{1}{3} $$
This is unsurprising, given that Newton's method for polynomials in $ p $-adic settings is another name for Hensel lifting.
