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There was a small sign error in the original.

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edited Nov 16, 2020 at 16:01

Arseniy Sheydvasser

333
2
10

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

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

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

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.

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.

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created Nov 16, 2020 at 2:30

Arseniy Sheydvasser

333
2
10

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

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

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.