728 Polygonal triples
replace(u, v)by(u−r, v+q), in which both entries are odd). Clearly,
the integerz 0 =puis such thatqz 0 −p=p(qu−1)is divisible by
r. This makesy 0 an integer. The correspondingx 0 is also an integer.
Replacingz 0 byz 0 +rtfor a positive integertif necessary, the integers
z 0 ,x 0 , andy 0 can be chosen greater than 1. From (28.11), the integers
x 0 andy 0 are both odd, sincepandqare of different parity andz 0 is odd.
We summarize this in the following theorem.
Theorem 28.2.Let(p, q, r)be a primitive Pythagorean triple. There
are two infinite families of triangular triples(a(t),b(t),c(t)),=± 1 ,
such that one of the lines(P),P=P′(3;a(t),b(t),c(t)), has direc-
tion numbersp:q:r.
Triangular triples from primitive Pythagorean triples
(m, n) (p, q, r) (a+(0),b+(0),c+(0)) (a−(0),b−(0),c−(0))
(2,1) (3, 4 ,5) (2, 2 ,3) (3, 5 ,6)
(4,1) (15, 8 ,17) (9, 4 ,10) (5, 3 ,6)
(3,2) (5, 12 ,13) (4, 9 ,10) (5, 14 ,15)
(6,1) (35, 12 ,37) (20, 6 ,21) (14, 5 ,15)
(5,2) (21, 20 ,29) (6, 5 ,8) (14, 14 ,20)
(4,3) (7, 24 ,25) (6, 20 ,21) (7, 27 ,28)
(8,1) (63, 16 ,65) (35, 8 ,36) (27, 7 ,28)
(7,2) (45, 28 ,53) (35, 21 ,41) (9, 6 ,11)
(5,4) (9, 40 ,41) (8, 35 ,36) (9, 44 ,45)
28.4 k-gonal triples determined by a Pythagorean triple
Now, we considerk≥ 5. We shall adopt the notation
h′:=
{
h ifhis odd,
h
2 ifhis even,
for an integerh.
Theorem 28.3.Letk≥ 5 andg= 2(kk−− 2 4). The primitive Pythagorean
triple(p, q, r)defined in (28.10) by relatively prime integersm>nwith
different parity corresponds to ak-gonal triple if and only if one of^2 gn
and2(mg−n)is an integer.
Sincemandnare relatively prime, the integer(k−2)′> 1 cannot
divide bothnandm−n. This means that a primitive Pythagorean triple
