Two Primitive Pythagorean triples

$(a, \; b, \; c)$   and   $(x, y, z)$   are two primitive Pythagorean triples

The primitive Pythagorean triples   $(x, \; y, \; z)$   are characterized by

$gcd \, (x,y,z) \; = \; 1$    and    $x^2 \; + \; y^2 \; = \; z^2$

$(x, y) \; = \; (2 r s, \; r^2 - s^2)$

for some positive integers   $r$   and   $s$,   and that then   $z \; = \; r^2 \; + \; s^2$

For the given Pythagorean triples   $(a, \; b, \; c)$   and   $(x, \; y, \; z)$,

where    $a \; > \; b \; > \; c$
and    $x \; > \; y \; > \; z$

we must have, for some positive integers   $r, \; s, \; u, \; v$

(1)

$a \; = \; r^2 \; + \; s^2$
$b \; = \; 2 \, r \,s$
$c \; = \; r^2 \; - \; s^2$

or

(2)

$a \; = \; r^2 \; + \; s^2$
$b \; = \; r^2 \; - \; s^2$
$c \; = \; 2 \, r \,s$

(3)

$x \; = \; u^2 \; + \; v^2$
$y \; = \; 2 \, u \,v$
$z \; = \; u^2 \; - \; v^2$

or

(4)

$x \; = \; u^2 \; + \; v^2$
$y \; = \; u^2 \; - \; v^2$
$z \; = \; 2 \, u \,v$

$a \,x \; + \; b \,z \; + \; c \,y$
$a \,x \; + \; b \,z \; - \; c \,y$
$a \,x \; - \; b \,z \; + \; c \,y$
$a \,x \; - \; b \,z \; - \; c \,y$
$a \,x \; + \; b \,y \; + \; c \,z$
$a \,x \; + \; b \,y \; - \; c \,z$
$a \,x \; - \; b \,y \; + \; c \,z$
$a \,x \; - \; b \,y \; - \; c \,z$

are either a square or twice a square in (1) and (3) ,   (2) and (4),    (1) and (4),    (2) and (3)

math grad - Interest: Number theory
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3 Responses to Two Primitive Pythagorean triples

1. paul says:

Substituting for {a,b,c,x,y,z} the resulting formulae reduces to terms with squares for the latter examples, namely :-

a x + (b y – c z) = 2(n u + m v)^2 and
a x – (b y – c z) = 2(m u – n v)^2

Paul.