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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 \boldsymbol 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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