Formulas For Generating Pythagorean Triples - Pythagorean Triples By Use of Matrices and Linear Transformations

Pythagorean Triples By Use of Matrices and Linear Transformations

Let be a primitive triple with odd. Then 3 new triples may be produced from using matrix multiplication and Berggren’s three matrices A, B, C. Triple is termed the "parent" of the three new triples (the "children"). Each child is itself the parent of 3 more children, and so on. If one begins with primitive triple, all primitive triples will eventually be produced by application of these matrices. The result can be graphically represented as an infinite ternary tree with at the root node. An equivalent result may be obtained using Berggrens's three linear transformations shown below.

\overset{A}{\mathop{\left[ \begin{matrix} -1 & 2 & 2 \\ -2 & 1 & 2 \\ -2 & 2 & 3 \\ \end{matrix} \right]}} \left[ \begin{matrix} a \\ b \\ c \\ \end{matrix} \right]=\left[ \begin{matrix} a_1 \\ b_1 \\ c_1 \\ \end{matrix} \right],\quad \text{ }\overset{B}{\mathop{\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \\ \end{matrix} \right]}} \left[ \begin{matrix} a \\ b \\ c \\ \end{matrix} \right]=\left[ \begin{matrix} a_2 \\ b_2 \\ c_2 \end{matrix} \right],\quad \text{ }\overset{C}{\mathop{\left[ \begin{matrix} 1 & -2 & 2 \\ 2 & -1 & 2 \\ 2 & -2 & 3 \end{matrix} \right]}} \left[ \begin{matrix} a \\ b \\ c \end{matrix} \right]=\left[ \begin{matrix} a_3 \\ b_3 \\ c_3 \end{matrix} \right]

Berggren's three linear transformations are:

\begin{align} & \begin{matrix} -a+2b+2c=a_1 \quad & -2a+b+2c=b_1 \quad & -2a+2b+3c=c_1 & \quad\to \left \\ \end{matrix} \\ & \begin{matrix} +a+2b+2c={{a}_{2}} \quad & +2a+b+2c={{b}_{2}} \quad & +2a+2b+3c={{c}_{2}} & \quad\to \left \\ \end{matrix} \\ & \begin{matrix} +a-2b+2c={{a}_{3}} \quad & +2a-b+2c={{b}_{3}} \quad & +2a-2b+3c={{c}_{3}} & \quad\to \left \\ \end{matrix} \\ & \end{align}

Alternatively, one may also use 3 different matrices found by Price. These matrices A', B', C' and their corresponding linear transformations are shown below.

\overset{{{A}'}}{\mathop{\left[ \begin{matrix} 2 & 1 & -1 \\ -2 & 2 & 2 \\ -2 & 1 & 3 \end{matrix} \right]}} \left[ \begin{matrix} a \\ b \\ c \end{matrix} \right]=\left[ \begin{matrix} a_1 \\ b_1 \\ c_1 \end{matrix} \right],\quad \text{ }\overset{{{B}'}}{\mathop{\left[ \begin{matrix} 2 & 1 & 1 \\ 2 & -2 & 2 \\ 2 & -1 & 3 \end{matrix} \right]}} \left[ \begin{matrix} a \\ b \\ c \\ \end{matrix} \right]=\left[ \begin{matrix} a_2 \\ b_2 \\ c_2 \end{matrix} \right],\quad \text{ }\overset{{{C}'}}{\mathop{\left[ \begin{matrix} 2 & -1 & 1 \\ 2 & 2 & 2 \\ 2 & 1 & 3 \\ \end{matrix} \right]}} \left[ \begin{matrix} a \\ b \\ c \\ \end{matrix} \right]=\left[ \begin{matrix} a_3 \\ b_3 \\ c_3 \end{matrix} \right]

Price's three linear transformations are:

\begin{align} & \begin{matrix} +2a+b-c=a_1 \quad & -2a+2b+2c=b_1 \quad & -2a+b+3c=c_1 & \quad \to \left \end{matrix} \\ & \begin{matrix} +2a+b+c=a_2 \quad & +2a-2b+2c=b_2 \quad & +2a-b+3c=c_2 & \quad \to \left \end{matrix} \\ & \begin{matrix} +2a-b+c=a_3 \quad & +2a+2b+2c=b_3 \quad & +2a+b+3c=c_3 & \quad \to \left \end{matrix} \\ & \end{align}

The "3 children" produced by each of the two sets of matrices are not the same, but each set separately produces all primitive triples.

EXAMPLE: Using as the parent, we get two sets of three children:

\begin{matrix} {} & \left & {} \\ A & B & C \\ \left & \left & \left \end{matrix}\quad \quad \quad \quad \quad \quad \begin{matrix} {} & \left & {} \\ {{A}'} & {{B}'} & {{C}'} \\ \left & \left & \left \end{matrix}

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