If-x-y-z-is-a-primitive-Pythagorean-triple-prove-that-x-y-and-x-y-are-congruent-modulo-8-to-either-1-or-7-

Question Number 70525 by Rasheed.Sindhi last updated on 05/Oct/19

I f x, y, z i s a p r i m i t i v e P y t h a g o r e a n

t r i p l e, p r o v e t h a t x + y a n d x - y a r e

c o n g r u e n t m o d u l o 8 t o e i t h e r 1 o r 7 .

Answered by mind is power last updated on 07/Oct/19

x = 2 a b

y = a^{2} - b^{2}

z = a^{2} + b^{2}

x, y, z p r i m i t i v e \Rightarrow g c d (a, b) = 1 a n d e i t h e r a = 2 k & b = 2 w + 1

b e c a u s e i f g c d (a, b) = a ⩾ 2

\Rightarrow a ∣ x, y, z a b v i o u s

a n d i f a = b [2] \Rightarrow 2 ∣ x, y, z a b v i o u s

x = 2 (2 k) (2 w + 1) = 8 k w + 4 k = 4 k (8)

y = (4 k^{2}) - 4 w^{2} - 4 w - 1

y = 4 k^{2} - 4 w (w + 1) - 1 = 4 k^{2} - 1 (8)

c a u s e 2 ∣ w (w + 1), \forall w \in I N j u s t u s e e i t h e r w = 2 s o r w = 2 s + 1 t o s e e t h a t

i f k o r k = 0 (2) \Rightarrow x = 0 (8), y = - 1 (8) \Rightarrow {\begin{cases} x - y = 1 (8) \\ x + y = - 1 (8) = 7 (8) \end{cases}

k = 1 (2) \Rightarrow x = 4 (8), y = 3 (8) \Rightarrow {\begin{cases} x - y = 1 (8) \\ x + y = 7 (8) \end{cases}

\Rightarrow x + y, x - y c o n g u r e n t m o d u l o 8 t o e i t h e r 1 o r 7

Commented by Rasheed.Sindhi last updated on 08/Oct/19

N o d o u b t, m i n d i s p o w e r!