From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Pythagorean Quadruplets (=?iso-8859-1?Q?a=B2?= + =?iso-8859-1?Q?b=B2?= + 
Date: 1 Aug 2000 18:41:05 -0400
Newsgroups: sci.math
Summary: [missing]

In article <39868BFB.1061BFD2@yamashita.dontspamme>,
Sophus Lefouque (M) <vng@letterbox.com> wrote:
:[1 4 8; 9]   1² + 4² + 8² = 9²
:[4 4 7; 9]   4² + 4² + 7² = 9²
:[1 2 2; 3]   1² + 2² + 2² = 3²

:How to obtain other primitive Pythagorean Quadruplets such as these?
:
 Relate it to the following problem: - in every finite dimension:

(1) Given a unit vector v with only rational entries, find an orthogonal
matrix Q with only rational entris, whose first column is v, and such that
Q+I is invertible. (That is, name an orthonormal basis of rational vectors
that starts with v.)

(2) These matrices Q are easy to describe: take any skew-symmetric H with
all entries rational, and compute Q=(I+H)*inv(I-H) (no problem with
singularity!). Conversely, if Q is given as above, then H=(Q-I)*inv(Q+I)
comes out as skew-symmetric with all entries rational.

So, picking H skew-symmetric, we get not only one rational unit vector but
a basis. And if you play around with switching signs, you can show that
every unit rational vector can be obtained this way.

To get just any orthogonal rational matrix from first column v: define
(assuming v is not parallel to e_1 = [1, 0,...,0]')

w = e_1 + v

and then (Householder reflector)

R = I - 2 * w * w' / (w' * w)

maps e_1 onto (-v) (check it), but, alas, has (-1) as an eigenvalue. This
can be fixed if you make Q from R by switching the sign of the first
column.

One of your examples:
v =
[1/9]
[4/9]
[8/9]

gives w = 

[10/9]
[ 4/9]
[ 8/9]

and a reflector

 [  -1/9         -4/9         -8/9     ]
 [  -4/9         37/45       -16/45    ]
 [  -8/9        -16/45        13/45    ]

leading to

Q =  

[    1/9         -4/9         -8/9     ]
[    4/9         37/45       -16/45    ]
[    8/9        -16/45        13/45    ]

and (Q-I)*inv(Q+I) will be

H =

[      0          -2/5         -4/5     ]
[     2/5           0            0      ]
[     4/5           0            0      ]

General exercise: Show that if you switch the signs of the first column of
a reflector (whose first column entries are non-zero, for ease of
proving), you obtain an orthogonal matrix Q such that I+Q is invertible.

Have fun, ZVK(Slavek).