28.2 Primitive Pythagorean triple associated with ak-gonal triple 727
Proposition 28.1.The two lines lying entirely on the hyperboloidS :
x^2 +y^2 =z^2 +1and passing throughP(x 0 ,y 0 ,z 0 )have direction num-
bers
x 0 z 0 −y 0 :y 0 z 0 +x 0 :x^20 +y 02
for=± 1.
In particular, ifPis a rational point, these direction numbers are ra-
tional.
28.2 Primitive Pythagorean triple associated with ak-
gonal triple
LetPbe the rational point determined by ak-gonal triple(a, b, c),as
given by (28.5), fork ≥ 5 and (28.6) fork=3(triangular numbers).
We first note that the coordinates ofP all exceed 1. This is clear for
k =3, and fork ≥ 5 , it follows from the fact thatg= 2(kk−− 4 2) > 2.
The direction numbers of the ruling lines onSthrough the pointP,as
given in Proposition 1, are all positive. In view of (28.8), we may there-
fore choose aprimitive Pythagorean triple(p, q, r)for these direction
numbers. As is well known, every such triple is given by
p=m^2 −n^2 ,q=2mn, r=m^2 +n^2 (28.10)for relatively prime integersm>nof different parity.
We study the converse question of determiningk-gonal triples from
(primitive) Pythagorean triples.
28.3 Triples of triangular numbers ...........
Given a primitive Pythagorean triple(p, q, r)as in (28.10), we want to
determine a triangular triple(a, b, c)corresponding to it. Given anodd
integerz 0 > 1 , we obtain, from (28.7) and (28.9),
x 0 =pz 0 +q
r,y 0 =qz 0 −p
r. (28.11)
We claim that it is possible to choosez 0 > 1 so thatx 0 andy 0 are
also odd integers> 1.
By the euclidean algorithm, there areoddintegersuandvsuch that
qu+rv=1. (Note thatvmust be odd, sinceqis even. Ifuis even, we