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Pythagorean triple preserving matrices

From Table (1), let

$\displaystyle \mathbf{C}= \begin{pmatrix}
3&4&5\\
21&20&29\\
119&120&169
\end{pmatrix}$   and$\displaystyle \quad \mathbf{D}=
\begin{pmatrix}
21&20&29\\
119&120&169\\
697&696&985
\end{pmatrix}$

So, solving for $ \mathbf{A}$ in the matrix equation $ \mathbf{C}\mathbf{A}=\mathbf{D}$ , we find that

$\displaystyle \mathbf{A}=
\begin{pmatrix}
1&2&2\\
2&1&2\\
2&2&3
\end{pmatrix}.
$

Let $ a^2+b^2=c^2$ . Then $ \begin{pmatrix}a&b&c\end{pmatrix}\mathbf{A}=\begin{pmatrix}u&v&w\end{pmatrix}$ where $ u^2+v^2=w^2$ , and $ u-v=b-a$ . Furthermore,

$\displaystyle \begin{pmatrix}
m^2-n^2&2mn&m^2+n^2
\end{pmatrix}\begin{pmatrix}
...
...&3
\end{pmatrix}=
\begin{pmatrix}
\mu^2-\nu^2&2\mu\nu&\mu^2+\nu^2
\end{pmatrix}$

where $ \mu=2m+n$ and $ \nu=m$ . Therefore, if $ (a,b,c)$ is a primitive Pythagorean triple then so is $ (u,v,w)$ . Thus, if $ n\in
\mathbb{Z}^+$ , then $ \begin{pmatrix}1&0&1\end{pmatrix}\mathbf{A}^n=\begin{pmatrix}u_n&v_n&w_n\end{pmatrix}$ gives the $ n^{th}$ primitive Pythagorean triple such that $ u_n-v_n=\pm 1$ .


\boxed{\mbox{c-a or c-b a constant by matrix
action}}


We have, using matrix $ \mathbf{A}$ , if $ (a,b,c)$ is a primitive Pythagorean triple then

$\displaystyle \begin{pmatrix}-a&b&c
\end{pmatrix}\begin{pmatrix}
1&2&2\\
2&1&2...
...ix}
-1&-2&-2\\
2&1&2\\
2&2&3
\end{pmatrix}=\begin{pmatrix}r&s&t
\end{pmatrix}$

where $ r, s$ and $ t$ are pairwise relatively prime, $ r^2+s^2=t^2$ , and

$\displaystyle t-r=(-2a+2b+3c)-(-a+2b+2c)=c-a.$

Also, $ r, s$ , and $ t$ are positive integers, since $ c>a$ . Hence $ (r,s,t)$ is a primitive Pythagorean triple.

Similarly, if $ (a,b,c)$ is a primitive Pythagorean triple then

$\displaystyle \begin{pmatrix}a&b&c
\end{pmatrix}\begin{pmatrix}
1&2&2\\
-2&-1&-2\\
2&2&3
\end{pmatrix}=\begin{pmatrix}x&y&z
\end{pmatrix}$

where $ (x,y,z)$ is a primitive Pythagorean triple such that $ z-y=c-b$ .

Let

$\displaystyle \mathbf{D}=\begin{pmatrix}
-1&-2&-2\\
2&1&2\\
2&2&3
\end{pmatrix}$   and$\displaystyle \quad
\mathbf{U}=\begin{pmatrix}
1&2&2\\
-2&-1&-2\\
2&2&3
\end{pmatrix}.
$

Joe Roberts2 has shown that $ (a,b,c)$ is a Primitive Pythagorean triple if and only if

$\displaystyle \begin{pmatrix}a&b&c
\end{pmatrix}=\begin{pmatrix}3&4&5
\end{pmatrix}\mathbf{\mathfrak{M}}
$

where $ \mathbf{\mathfrak{M}}$ is a finite product of the matrices $ \mathbf{U}, \mathbf{A}$ , and $ \mathbf{D}$ .


next up previous contents
Next: Primitive Pythagorean triangles where Up: Primitive Pythagorean triangles where Previous: Primitive Pythagorean triangles where   Contents
f. barnes 2008-04-29
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