find x ,y from Z wich verify y^2 =x(x+1)(x+7)(x+8)


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Question Number 50369 by prof Abdo imad last updated on 16/Dec/18

$f i n d x, y f r o m Z w i c h v e r i f y$ $y^{2} = x (x + 1) (x + 7) (x + 8)$
	Answered by mr W last updated on 16/Dec/18

	$l e t t = x + 4$ $y^{2} = (t - 4) (t - 3) (t + 3) (t + 4)$ $y^{2} = (t^{2} - 16) (t^{2} - 9)$ $l e t s = t^{2}$ $y^{2} = (s - 16) (s - 9)$ $s^{2} - 25 s + 144 - y^{2} = 0$ $s = \frac{25 \pm \sqrt{25^{2} - 4 (144 - y^{2})}}{2}$ $s = \frac{25 \pm \sqrt{4 y^{2} + 49}}{2}$ $4 y^{2} + 49 = {(2 n + 1)}^{2}$ $49 = {(2 n + 1)}^{2} - {(2 y)}^{2} = (2 n + 1 - 2 y) (2 n + 1 + 2 y)$ $2 n + 1 - 2 y = 1 \Rightarrow n = y$ $2 n + 1 + 2 y = 49 \Rightarrow 4 y = 48 \Rightarrow y = 12$ $w i t h y = 12 :$ $\Rightarrow s = 25 o r 0$ $\Rightarrow t = 5 o r 0$ $\Rightarrow x = 1 o r - 4$ $s o l u t i o n s a r e :$ $x = - 4, y = 12$ $x = 1, y = 12$
	Commented by maxmathsup by imad last updated on 16/Dec/18

	$t h a n k y o u s i r f o r t h i s h a r d w o r k .$
	Commented by Necxx last updated on 17/Dec/18

	$p l e a s e w h y d i d y o u e q u a t e 4 y^{2} + 49 = {(2 n + 1)}^{2} ? ? ?$
	Commented by mr W last updated on 17/Dec/18

	$s = \frac{25 \pm \sqrt{4 y^{2} + 49}}{2}$ $s u c h t h a t s i s i n t e g e r,$ $\sqrt{4 y^{2} + 49} m u s t b e a n o d d i n t e g e r, s a y$ $2 n + 1, t h e r e f o r e$ $4 y^{2} + 49 = {(2 n + 1)}^{2}$
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