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All pythagorean triples can be generated from $(3,4,5)^T$ and $(5,4,3)^T$ by the matrices:

\[ A = \left(\begin{array}{ccc} -1 & 2 & 2 \\\\ -2 & 1 & 2 \\\\-2 & 2 & 3 \end{array} \right) \hspace{0.25in} B = \left(\begin{array}{ccc} 1 & 2 & 2 \\\\ 2 & 1 & 2 \\\\2 & 2 & 3 \end{array} \right) \hspace{0.25in} C = \left(\begin{array}{ccc} 1 & -2 & 2 \\\\ 2 & -1 & 2 \\\\2 & -2 & 3 \end{array} \right) \]

Giving pythagorean triples the structure of a ternary tree. These correspond to generators of $\mathrm{O}(\mathbb{Z}^3, x^2+y^2-z^2)$ and we're looking for integer points on the ``light cone" where the norm is zero.

I remember seeing there being 5 generators for the quadratic $x^2 + xy + y^2 = z^2$. Does anyone have the reference?

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I see, you need severa automorphs of $x^2 + x y + y^2 - z^2.$ Meanwhile, rather than primes $ \equiv 1 \pmod 4,$ For this problem you want primes $z \equiv 1 \pmod 3.$ So your first triple would be $(3,5,7)^T$ –  Will Jagy Oct 5 '11 at 6:37
    
how might I find the lattice corresponding to a quadratic form $x^2+xy+y^2-z^2$ ? What is the lattice for $x^2+y^2-z^2$ ? –  john mangual Oct 5 '11 at 14:35
    
John, as in my answer, the Gram matrix $G$ for the original Pythagorean triples would be the diagonal matrix with diagonal entries $ (1,1,-1).$ You should check, for your three matrices above, that $$A^T G A = G, \; B^T G B = G, \; C^T G C = G,$$ which is what you want if you are writing your triples as column vectors, as I do. If it turns out you are using row vectors, similar check with the transposes of your three matrices instead. –  Will Jagy Oct 5 '11 at 22:05
    
@John, there are two problems with your description of the orthogonal group of $x^2 + y^2 - z^2$. First, the vector [5,4,3] isn't even a zero of that quadratic form. Typo? Second, why do you say these three matrices generate the orthogonal group? I think you're forgetting to look at stabilizer subgroups too. For example, the diagonal matrix diag(-1,1,1) is in the orthogonal group. Can you write it in terms of your three matrices? I know five generators of that orthogonal group, not three. –  KConrad Oct 6 '11 at 3:50
    
This might be in Allcock's paper "The reflective Lorentzian lattices of rank 3". ma.utexas.edu/users/allcock –  Ian Agol Oct 7 '11 at 3:52
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1 Answer

I wouldn't call it GL, it is the orthogonal group of the lattice we are discussing. References, as i said, include Lattices and Codes by W. Ebeling, Rational Quadratic Forms by Cassels, these two being available and inexpensive.

We double the quadratic form to get an integral Gram matrix:

$$ G \; = \; \left( \begin{array}{rrr} 2 & 1 & 0 \\\ 1 & 2 & 0 \\\ 0 & 0 & -2 \end{array} \right). $$

Your solutions, the light cone, are column vectors $v$ such that $ v^T G v = 0.$ A root, since we have doubled everything to get an "even" lattice, is a vector $r$ with norm 2, $ r^T G r = 2.$ The general definition of reflection (Cassels calls this a symmetry, page 19) in any vector $w$ is that $$ x \mapsto \; \; x \; - \; \frac{2 \, x^T G w}{w^T G w} \; w.$$ As a result, when we take $w$ to be a root, the factors of 2 cancel and we are taking lattice points to other lattice points. In your original form, a root $(x,y,z)$ solves $x^2 + x y + y^2 = 1 + z^2.$ Then the reflection in the root is just a linear map, determinant $-1,$ and is therefore given by a square matrix with respect to the original basis. Finally, the reflection is an isometry, part of the orthogonal group of the quadratic form, and if we call the matix $A,$ it solves $$A^T G A = G.$$

I think it wise to include reflection in the root $(1,0,0)^T$ to get some negative values taken care of, $$ A_0 \; = \; \left( \begin{array}{rrr} -1 & -1 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \end{array} \right). $$

We also include reflection in the root $(0,1,0)^T,$ $$ A_{00} \; = \; \left( \begin{array}{rrr} 1 & 0 & 0 \\\ -1 & -1 & 0 \\\ 0 & 0 & 1 \end{array} \right). $$

Note that we have already included the automorph that interchanges the first two items, $$ (x,y,z) \mapsto (y,x,z). $$

$$ A_0 A_{00} A_0 \; = \; \left( \begin{array}{rrr} 0 & 1 & 0 \\\ 1 & 0 & 0 \\\ 0 & 0 & 1 \end{array} \right). $$

Although it may not be called a root, having norm $-2,$ we include reflection in $(0,0,1)^T,$ or $$ A_{000} \; = \; \left( \begin{array}{rrr} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & -1 \end{array} \right). $$

Also, as is usual, we explicitly include negation, $$ (x,y,z) \mapsto (-x,-y,-z), $$ so $$ A_{0000} \; = \; \left( \begin{array}{rrr} -1 & 0 & 0 \\\ 0 & -1 & 0 \\\ 0 & 0 & -1 \end{array} \right). $$

It occurred to me that we would get some reflections with quite small matrix entries by taking other vectors of norm $-2,$ for example $(1,1,2)^T.$ $$ A_{-1} \; = \; \left( \begin{array}{rrr} 4 & 3 & -4 \\\ 3 & 4 & -4 \\\ 6 & 6 & -7 \end{array} \right). $$ This still solves, as with all the $A'$s, both $A^T G A = G$ and $A^2 = I.$

Taking the root $(3,4,6)^T$ gives the reflection $$ A_1 \; = \; \left( \begin{array}{rrr} -29 & -33 & 36 \\\ -40 & -43 & 48 \\\ -60 & -66 & 73 \end{array} \right). $$

If we take the initial triple in the light cone to be $c = (3,5,7)^T,$ we get $A_1 c = (0,1,1)^T$ which is indeed another solution, and shows that care must be used in constructing the "tree."

Taking the root $(5,15,18)^T$ gives the reflection $$ A_2 \; = \; \left( \begin{array}{rrr} -124 & -175 & 180 \\\ -375 & -524 & 540 \\\ -450 & -630 & 649 \end{array} \right). $$ we get $A_2 c = (13,35,43)^T$

Taking the root $(8,19,24)^T$ gives the reflection $$ A_3 \; = \; \left( \begin{array}{rrr} -279 & -368 & 384 \\\ -665 & -873 & 912 \\\ -840 & -1104 & 1153 \end{array} \right). $$ we get $A_3 c = (11,24,31)^T$

Taking the root $(7,32,36)^T$ gives the reflection $$ A_4 \; = \; \left( \begin{array}{rrr} -321 & -497 & 504 \\\ -1472 & -2271 & 2304 \\\ -1656 & -2556 & 2593 \end{array} \right). $$ we get $A_4 c = (80,357,403)^T$

Taking the root $(12,47,54)^T$ gives the reflection $$ A_5 \; = \; \left( \begin{array}{rrr} -851 & -1272 & 1296 \\\ -3337 & -4981 & 5076 \\\ -3834 & -5724 & 5833 \end{array} \right). $$ we get $A_5 c = (159,616,709)^T$

One typically includes $\pm 1$ anyway. Ian Agol would know how many reflections are enough, but I suspect this will do. Nothing really wrong with finding too many. Note that the squarefree parts in $1+z^2$ in the five nontrivial roots I chose are $37, 13, 577, 1297, 2917.$

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