There is another way of generating the primitive Pythagorean triples using matrices which was discovered by Berggren in 1934 and again by Barning in 1963.

The matrices are :

\(

\mathbf{A} =

\begin{pmatrix}

1&-2&2\\

2&-1&2\\

2&-2&3

\end{pmatrix}

\)

\(

\mathbf{B} =

\begin{pmatrix}

1&2&2\\

2&1&2\\

2&2&3

\end{pmatrix}

\)

\(

\mathbf{C} =

\begin{pmatrix}

-1&2&2\\

-2&1&2\\

-2&2&3

\end{pmatrix}

\)

\((3, 4, 5)\) is the root of this ternary tree and has three children using these matrices.

We showed previously that there is an infinite set of Pythagorean triples where \(m\; – n = 1\) and another infinite set where \((m\; – n)/n \approx \sqrt{2}\).

We show another infinite set where \(n = 1\).

We chose this time to use \((3, 4, 5) = (x, y, z)\) where \(x = m^2\;- n^2, y = 2mn\) and \(z = m^2 + n^2\).

We observe that \(x + 2 = z\). That is :

\(

\begin{align} m^2 – n^2 + 2 &= m^2 + n^2\\

2n^2 &= 2\\

n^2 &= 1\\

n &= 1

\end{align}

\)

The set begins :

\(

\begin{array} {c c c c c}

m&n&x&y&z\\

\hline

2&1&3&4&5\\

4&1&15&8&17\\

6&1&35&12&37\\

8&1&63&16&65\\

10&1&99&20&101

\end{array}

\)

All three of the above properties hold for the first primitive Pythagorean triple \((3, 4, 5)\).

We express \((3, 4, 5)\) as a column vector

\(

\mathbf{P}_0 =

\begin{pmatrix}3\\4\\5\end{pmatrix} \)

Matrix multiplication is not commutative. ** ** \(\mathbf{AB}\neq\mathbf{BA}\) in general.

We calculate as follows :

\(\begin{pmatrix}

a&b&c\\

d&e&f\\

g&h&i

\end{pmatrix}

\begin{pmatrix}

j\\

k\\

l

\end{pmatrix}

=

\begin{pmatrix}

aj\;\; + &bk\;\; + &cl\\

dj\;\; + &ek\;\; + &fl\\

gj\;\; + &hk\;\;+ &il

\end{pmatrix}

=

\begin{pmatrix}

m\\

n\\

o

\end{pmatrix}

\)

We apply each matrix:

\(\mathbf{AP}_0 =

\begin{pmatrix}

1&-2&2\\

2&-1&2\\

2&-2&3

\end{pmatrix}

\begin{pmatrix}3\\4\\5\end{pmatrix} = \begin{pmatrix}5\\12\\13\end{pmatrix} = \mathbf{P}_1\)

\(\mathbf{BP}_0 =\;

\begin{pmatrix}

1&2&2\\

2&1&2\\

2&2&3

\end{pmatrix}

\begin{pmatrix}3\\4\\5\end{pmatrix} = \begin{pmatrix}21\\20\\29\end{pmatrix} = \mathbf{P}_2

\)

\(\mathbf{CP}_0 =

\begin{pmatrix}

-1&2&2\\-2&1&2\\-2&2&3\end{pmatrix}

\begin{pmatrix}3\\4\\5\end{pmatrix} = \begin{pmatrix}15\\8\\17\end{pmatrix} = \mathbf{P}_3\)

The property \(m\, – n = 1\) holds for \(\mathbf{P}_1\).

The property \((m\, – n)/n \approx \sqrt 2\) holds for \(\mathbf{P}_2\).

The property \(n = 1\) holds for \(\mathbf{P}_3\).

The three children each have three children ad infinitum.

By inspection of the second generation, we see that the first child of \(\mathbf{P}_1\) inherits the property \(m\,-n = 1\).

The middle child of \(\mathbf{P}_2\) inherits the property \((m\,-n)/n \approx \sqrt 2\).

The third child of \(\mathbf{P}_3\)inherits the property \(n = 1\).

This rooted ternary tree produces every primitive Pythagorean triple exactly once.

© OldTrout 2017