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 and Answers Forum

All Questions Topic List
Number Theory Questions
Previous in All Question Next in All Question
Previous in Number Theory Next in Number Theory
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
	$x=2ab y=a^2 −b^2 z=a^2 +b^2 x,y,z primitive ⇒gcd(a,b)=1 and either a= 2k &b=2w+1 because if gcd(a,b)=a≥2 ⇒a∣x,y,z abvious and if a=b[2]⇒2∣x,y,z abvious x=2(2k)(2w+1)=8kw+4k=4k(8) y=(4k^2 )−4w^2 −4w−1 y=4k^2 −4w(w+1)−1=4k^2 −1(8) cause 2∣w(w+1),∀w∈IN just use either w=2s or w=2s+1 to see that if kor k=0(2)⇒x=0(8),y=−1(8)⇒ { ((x−y=1(8))),((x+y=−1(8)=7(8))) :} k=1(2)⇒x=4(8),y=3(8)⇒ { ((x−y=1(8))),((x+y=7(8))) :} ⇒x+y ,x−y congurent modulo 8 to either 1 or 7$
	$x = 2 a b$ $y = a^{2} - b^{2}$ $z = a^{2} + b^{2}$ $x, y, z p r i m i t i v e \Rightarrow g c d (a, b) = 1 a n d e i t h e r a = 2 k & b = 2 w + 1$ $b e c a u s e i f g c d (a, b) = a ⩾ 2$ $\Rightarrow a ∣ x, y, z a b v i o u s$ $a n d i f a = b [2] \Rightarrow 2 ∣ x, y, z a b v i o u s$ $x = 2 (2 k) (2 w + 1) = 8 k w + 4 k = 4 k (8)$ $y = (4 k^{2}) - 4 w^{2} - 4 w - 1$ $y = 4 k^{2} - 4 w (w + 1) - 1 = 4 k^{2} - 1 (8)$ $c a u s e 2 ∣ w (w + 1), \forall w \in I N j u s t u s e e i t h e r w = 2 s o r w = 2 s + 1 t o s e e t h a t$ $i f k o r k = 0 (2) \Rightarrow x = 0 (8), y = - 1 (8) \Rightarrow {\begin{cases} x - y = 1 (8) \\ x + y = - 1 (8) = 7 (8) \end{cases}$ $k = 1 (2) \Rightarrow x = 4 (8), y = 3 (8) \Rightarrow {\begin{cases} x - y = 1 (8) \\ x + y = 7 (8) \end{cases}$ $\Rightarrow x + y, x - y c o n g u r e n t m o d u l o 8 t o e i t h e r 1 o r 7$
	Commented by Rasheed.Sindhi last updated on 08/Oct/19

	$N o d o u b t, m i n d i s p o w e r!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum