A 120 degree triangle preserving matrix

Let

$$\mathbf{M}=\begin{pmatrix} m_{1,1}&m_{1,2}&m_{1,3}\\ m_{2,1}&m_{2,2}&m_{2,3}\\ m_{3,1}&m_{3,2}&m_{3,3} \end{pmatrix}.$$

From table (7), set

$$\mathbf{A}=\begin{pmatrix} 8&7&13\\ 105&104&181\\ 1456&1455&2521 \end{pmatrix}$$   and$$\quad \mathbf{B}=\begin{pmatrix} 105&104&181\\ 1456&1455&2521\\ 20273&20272&35113 \end{pmatrix}.$$

So solving for $\mathbf{M}$ in the matrix equation

$$\mathbf{A} \mathbf{M}=\mathbf{B}$$

we find that

$$\boxed{\mathbf{M}=\begin{pmatrix}4 & 3 & 6\ 3 & 4 & 6\ 4 & 4 & 7 \end{pmatrix}.}$$ (39)

If $a^2+b^2+ab=c^2$ then

$$\begin{pmatrix} a & b & c \end{pmatrix}\begin{pmatrix} 4 & 3 & 6\\ 3 & 4 & 6\\ 4 & 4 & 7 \end{pmatrix} = \begin{pmatrix} u &v & w \end{pmatrix}$$

where $u^2+v^2+uv=w^2.$ And

$$\begin{pmatrix}m^2-n^2&2mn+n^2&m^2+n^2+mn\end{pmatrix}\mathbf{M}= \begin{pmatrix}s^2-t^2&2st+t^2&s^2+t^2+st\end{pmatrix}$$

where $s=3m+2n$ and $t=m+n.$ So, if $\left(m^2-n^2,2mn+n^2,m^2+n^2+mn\right)$ is a primitive triple, $s$ and $t$ are, relatively prime, positive integers, $s>t,$ where, $3\nmid s-t=2m+n.$ That is, $\mathbf{M}$ takes a primitive triple and outputs a primitive triple.

Note that $u-v=a-b.$ For example.

$$\begin{pmatrix} 3& 5& 7 \end{pmatrix}\mathbf{M} = \begin{pmatrix} 55& 57& 97 \end{pmatrix}.$$

If $n$ is a positive integer then

$$\begin{pmatrix} 1 & 0 & 1 \end{pmatrix} \mathbf{M}^n = \begin{pmatrix} a_n & b_n & c_n \end{pmatrix},$$

where $(a_n,b_n,c_n)$ is the $n^{th}$ solution to the primitive 120 degree triangle $a^2+b^2+ab=c^2$ such that $a_n-b_n=1.$

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