Can a product of Cohn matrices over the Eisenstein integers with non-zero, non-unit coefficients be a Cohn matrix?

Question

For $k > 1$, is it possible that $\begin{pmatrix} a_1 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} a_2 & 1 \\ -1 & 0 \end{pmatrix}\ldots \begin{pmatrix} a_k & 1 \\ -1 & 0 \end{pmatrix} = \pm \begin{pmatrix} b & 1 \\ -1 & 0 \end{pmatrix}$ if $a_1,a_2,\ldots a_k,b$ are Eisenstein integers and $|a_i| > 1$ for $i=1,2,\ldots k$?

If the Eisenstein integers are replaced with the Gaussian integers, this is possible. $\begin{pmatrix} 3 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 - i & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 + i & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 - i & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} -2 & 1 \\ -1 & 0 \end{pmatrix} = -\begin{pmatrix} i & 1 \\ -1 & 0 \end{pmatrix}$

This problem came up in my research; I am primarily interested in the case where $b = (\pm 1 \pm \sqrt{-3})/2$, but I suspect that there might not be a solution for any choice of $b$.

I originally asked this question on math.stackexchange (https://math.stackexchange.com/q/3903567/202799), but it seems to be more difficult than I first thought.

Answer 1

To my surprise, not only is there a solution for some $b$, there is actually a very simple infinite family of solutions for every $b$. Let $\omega = \frac{1 + \sqrt{-3}}{2}$. Then

$\begin{pmatrix} a_0 + a_1\omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 -\omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 + \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 - \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 + \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 - \omega & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} b_0 + b_1\omega & 1 \\ -1 & 0 \end{pmatrix} = -\begin{pmatrix} a_0 + b_0 - 1 + (a_1 + b_1 - 1)\omega & 1 \\ -1 & 0 \end{pmatrix}$

I wish I could say that there was some clever trick to finding this solution, but I cannot: it was found by a brute force search through $\approx 5$ million possibilities using Python after I realized that the bottom right coordinate of the product of $n$ Cohn matrices only depends on the inner $n - 2$ matrices.

I do not know if there are shorter solutions. One can prove that there are no solutions with $k < 5$, but there could be solutions with $k = 5$ or $k = 6$. If there are, however, they have to involve some elements with squared norm at least $4$, as I checked everything smaller.