Prove that if a, b, c ∈ Z and a^2 + b^2 = c^2 , then 3 ∣ ab


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Question Number 44584 by Joel578 last updated on 01/Oct/18

$Prove that if a, b, c \in Z and a^{2} + b^{2} = c^{2}, then$ $3 ∣ a b$
Commented by ajfour last updated on 02/Oct/18

$t h e n 3 ∣ a b m e a n s ?! . .$
Commented by rahul 19 last updated on 02/Oct/18

$3 d i v i d e s a b . . . . .$ ${I t}^{'} s j u s t a n o t a t i o n$
Commented by ajfour last updated on 02/Oct/18

$t h a n k s r a h u l .$
Commented by Joel578 last updated on 03/Oct/18

$Yes, 3 ∣ a b means there is such number x that$ $satisfy a b = 3 x$
Commented by Joel578 last updated on 03/Oct/18

$I^{'} m having trouble how to start the proof$
	Answered by arcana last updated on 03/Oct/18

	$podemos ver solamente para el caso$ $de que a, b, c sea una s o l u c i o n p r i m i t i v a,$ $es decir (a, b, c) = 1, ya que las soluciones primitivas$ $pueden generar los demas casos .$ $A hora solo uno de los enteros a, b debe$ $ser impar, pues si los lo son e n t o n c e s a^{2} + b^{2} es$ $de la forma 4 k + 2 pero todo cuadrado$ $perfecto es de la forma 4 k o 4 k + 1 (\to\leftarrow) .$ $B ajo estas condiciones podemos usar el$ $siguiente teorema :$ $S i a, b, c \in Z e s u n a s o l u c i o n p r i m i t i v a$ $d o n d e a e s i m p a r p a r a l a e c u a c i o n$ $a^{2} + b^{2} = c^{2} s i i e x i s t e n e n t e r o s x, y$ $t a l e s q u e :$ $a = x^{2} - y^{2}$ $\Rightarrow c = x^{2} + y^{2}$ $b = 2 x y$ $c o n (x, y) = 1, x > y, x e y t i e n e n d i s t i n t a$ $p a r i d a d .$ $p o r l o a n t e r i o r t e n e m o s a b = 2 x y (x^{2} - y^{2}) .$ $y e s t o s e r e d u c e e n v e r s i 3 ∣ x \lor 3 ∣ y \lor$ $3 ∣ (x^{2} - y^{2}) .$ $s i 3 ∣ x \lor 3 ∣ y e n t o n c e a e s t a h e c h o .$ $s i 3 n o d i v i d e a x n i y, e n t o n c e s$ $x, y s o n d e l a f o r m a 3 k + 1 o 3 k + 2, k e n t e r o$ $p a r a k_{0}, k_{1} \in Z$ $c a s o 1 : s i x = 3 k_{0} + 1, y = 3 k_{1} + 1$ $l u e g o x^{2} - y^{2} = 3 (3 k_{0}^{2} + 2 k_{0} - 3 k_{1}^{2} - 2 k_{1})$ $e s t o e s 3 ∣ (x^{2} - y^{2}) \Rightarrow 3 ∣ a b$ $c a s o 2 : s i x = 3 k_{0} + 1, y = 3 k_{1} + 2$ $\Rightarrow x^{2} - y^{2} = 3 (3 k_{0}^{2} + 2 k_{0} - 3 k_{1}^{2} - 2 k_{1} - 1)$ $l u e g o 3 ∣ (x^{2} - y^{2}) \Rightarrow 3 ∣ a b$ $c a s o 3 : s i x = 3 k_{0} + 2, y = 3 k_{1} + 1$ $e s s i m i l a r a l c a s o 2$ $c a s o 4 : s i x = 3 k_{0} + 2, y = 3 k_{1} + 2$ $s i m i l a r a l c a s o 1 .$ $La demostracion del teorema se puede encontrar en el libro : teoria de numeros,$ $Rubiano$
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