3 Fixed the punctuation.

In the early 90s, I took a class from Victor Kac and in it, he explained a very beautiful way of generating all the primitive solutions for the Pythagorean equation for a sum of $n-1$ perfect squares is equal to a perfect square where $n$ can be 3, 4, 5, ..., 10. Unfortunately, I do not know where in the literature this is described in detail. It might be in his Infinite Dimensional Lie Algebras book, but I don't know.

The idea is to realize the solutions as the isotropic roots for a certain root system.

Consider the lattice ${\Bbb Z}^n$ with bilinear form $-x_0y_0 + x_1y_1 + \cdots + x_{n-1}y_{n-1}$ and standard basis $v_0$, $v_1$, $v_2, v_2$, \dots, v_{n-1}$.$v_{n-1}$. Change basis to:$\alpha_1 = v_1 - v_2$,$\alpha_2 = v_2 - v_3$, \dots,$\alpha_{n-2} = v_{n-2} - v_{n-1}$, and$\alpha_{n-1} = v_{n-1}$. If$n \geq 4$, let$\alpha_n = -v_0 - v_1 - v_2 - v_3$. If$n=3$, let$\alpha_n = -v_0 - v_1 - v_2$. The corresponding Cartan matrix$a_{ij} = \frac{2(\alpha_i, \alpha_j)}{(\alpha_i, \alpha_i)}$is represented by the diagram: Then the set of primitive solutions to the equation$x_0^2 = x_1^2 + x_2^2 + \cdots + x_{n-1}^2$is the orbit under the corresponding Coxeter group of$(1, 1, 0, \dots, 0)$if$n < 10$. If$n=10$, then you have to add the orbit$(3, 1, 1, 1, \dots, 1)$to get them all. One doesn't need the theoretical machinery to prove the result. One can just construct the matrices and use a descent argument to show that it works. For the Pythagorean triple case, for instance, you take the orbit of the vector$(1, 1, 0)$under the action of the group generated by the matrices: $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix} end{pmatrix},$$ $$\begin{pmatrix} \pm 1 & 0 & 0 \\ 0 & \pm 1 & 0 \\ 0 & 0 & \pm 1 \end{pmatrix} end{pmatrix},$$ and $$\begin{pmatrix} 3 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} end{pmatrix}.$$. For$n=4$, you can use the matrices that permute the appropriate variables, change the sign of any variable, and the following: $$\begin{pmatrix} 2 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix} end{pmatrix}.$$. I'm sorry I cannot give a reference to the literature...I hope someone else is able to. 2 improved formatting In the early 90s, I took a class from Victor Kac and in it, he explained a very beautiful way of generating all the primitive solutions for the Pythagorean equation for a sum of$n-1$perfect squares is equal to a perfect square where$n$can be 3, 4, 5, ..., 10. Unfortunately, I do not know where in the literature this is described in detail. It might be in his Infinite Dimensional Lie Algebras book, but I don't know. The idea is to realize the solutions as the isotropic roots for a certain root system. Consider the lattice${\Bbb Z}^n$with bilinear form$-x_0y_0 + x_1y_1 + \cdots + x_{n-1}y_{n-1}$and standard basis$v_0$,$v_1$,$v_2$, v_2, \dots,$v_{n-1}$.v_{n-1}$.

Change basis to:

$\alpha_1 = v_1 - v_2$, $\alpha_2 = v_2 - v_3$, \dots, $\alpha_{n-2} = v_{n-2} - v_{n-1)$v_{n-1}$, and$\alpha_{n-1} = v_{n-1}$. If$n \geq 4$, let$\alpha_n = -v_0 - v_1 - v_2 - v_3$. If$n=3$, let$\alpha_n = -v_0 - v_1 - v_2$. The corresponding Cartan matrix$a_{ij} = \frac{2(\alpha_i, \alpha_j)}{(\alpha_i, \alpha_i)}$is represented by the diagram: Then the set of primitive solutions to the equation$x_0^2 = x_1^2 + x_2^2 + \cdots + x_{n-1}^2$is the orbit under the corresponding Coxeter group of$(1, 1, 0, \dots, 0)$if$n < 10$. If$n=10$, then you have to add the orbit$(3, 1, 1, 1, \dots, 1)$to get them all. One doesn't need the theoretical machinery to prove the result. One can just construct the matrices and use a descent argument to show that it works. For the Pythagorean triple case, for instance, you take the orbit of the vector$(1, 1, 0)$under the action of the group generated by the matrices: $\left( $\begin{array}{ccc} begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 1\\ 0 & 1 & 0 \end{array} \right)$,end{pmatrix} $\left($

$$\begin{array}{ccc} begin{pmatrix} \pm 1 & 0 & 0 \\ 0 & \pm 1 & 0 \\ 0 & 0 & \pm 1 \end{array} \right), end{pmatrix}$$ and

$\left($ \begin{array}{ccc} begin{pmatrix} 3 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array} \right)$.end{pmatrix} $$. For n=4, you can use the matrices that permute the appropriate variables, change the sign of any variable, and the following: \left( \begin{array}{cccc} begin{pmatrix} 2 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right).end{pmatrix}$$. I'm sorry I cannot give a reference to the literature...I hope someone else is able to. 1 In the early 90s, I took a class from Victor Kac and in it, he explained a very beautiful way of generating all the primitive solutions for the Pythagorean equation for a sum of$n-1$perfect squares is equal to a perfect square where$n$can be 3, 4, 5, ..., 10. Unfortunately, I do not know where in the literature this is described in detail. It might be in his Infinite Dimensional Lie Algebras book, but I don't know. The idea is to realize the solutions as the isotropic roots for a certain root system. Consider the lattice${\Bbb Z}^n$with bilinear form$-x_0y_0 + x_1y_1 + \cdots + x_{n-1}y_{n-1}$and standard basis$v_0$,$v_1$,$v_2$, \dots,$v_{n-1}$. Change basis to:$\alpha_1 = v_1 - v_2$,$\alpha_2 = v_2 - v_3$, \dots,$\alpha_{n-2} = v_{n-2} - v_{n-1)$, and$\alpha_{n-1} = v_{n-1}$. If$n \geq 4$, let$\alpha_n = -v_0 - v_1 - v_2 - v_3$. If$n=3$, let$\alpha_n = -v_0 - v_1 - v_2$. The corresponding Cartan matrix$a_{ij} = \frac{2(\alpha_i, \alpha_j)}{(\alpha_i, \alpha_i)}$is represented by the diagram: Then the set of primitive solutions to the equation$x_0^2 = x_1^2 + x_2^2 + \cdots + x_{n-1}^2$is the orbit under the corresponding Coxeter group of$(1, 1, 0, \dots, 0)$if$n < 10$. If$n=10$, then you have to add the orbit$(3, 1, 1, 1, \dots, 1)$to get them all. One doesn't need the theoretical machinery to prove the result. One can just construct the matrices and use a descent argument to show that it works. For the Pythagorean triple case, for instance, you take the orbit of the vector$(1, 1, 0)$under the action of the group generated by the matrices:$\left( \begin{array}{ccc} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{array} \right)$,$\left( \begin{array}{ccc} \pm 1 & 0 & 0 \ 0 & \pm 1 & 0 \ 0 & 0 & \pm 1 \end{array} \right)$, and$\left( \begin{array}{ccc} 3 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{array} \right)$. For$n=4$, you can use the matrices that permute the appropriate variables, change the sign of any variable, and the following:$\left( \begin{array}{cccc} 2 & 1 & 1 & 1 \ 1 & 0 & 1 & 1 \ 1 & 1 & 0 & 1 \ 1 & 1 & 1 & 0 \end{array} \right)\$.

I'm sorry I cannot give a reference to the literature...I hope someone else is able to.