find-integral-solution-of-y-2-x-3-1-

Question Number 98602 by bemath last updated on 15/Jun/20

find integral solution

of y^{2} = x^{3} + 1

Commented by bemath last updated on 15/Jun/20

oo yes sir . thank you

Commented by Rasheed.Sindhi last updated on 15/Jun/20

Sir (1, \pm 2) {doesn}^{'} t satisfy :

{(\pm 2)}^{2} \neq {(1)}^{3} + 1

4 \neq 2

I^{'} ve extended my answer and now

it includes all solutions .

Commented by Rasheed.Sindhi last updated on 15/Jun/20

y^{2} = x^{3} + 1

^{∙} x^{3} + 1 is perfect square . It may be

0 also provided that x is an integer .

So y^{2} = x^{3} + 1 = 0

x^{3} = - 1

x = - 1 \Rightarrow y = 0

(- 1, 0)

^{∙} y^{2} = (x + 1) (x^{2} - x + 1)

x + 1 = x^{2} - x + 1

x^{2} - 2 x = 0

x (x - 2) = 0

x = 0 ∣ x = 2

x = 0 \Rightarrow y^{2} = 0^{3} + 1 \Rightarrow y = \pm 1

x = 2 \Rightarrow y^{2} = 2^{3} + 1 \Rightarrow y = \pm 3

(- 1, 0), (0, \pm 1), (2, \pm 3)

Commented by bemath last updated on 15/Jun/20

sir i got (- 1, 0), (1, \pm 2), (2, \pm 3)

Commented by 1549442205 last updated on 15/Jun/20

but remain the case that x + 1 = u^{2},

x^{2} - x + 1 = v^{2}, {doesn}^{'} t consider ?

Commented by Rasheed.Sindhi last updated on 15/Jun/20

T h a n k y o u s i r! I^{'} l l t r y t o c o m p l e t e

m y a n s w e r .

Answered by Rasheed.Sindhi last updated on 17/Jun/20

y^{2} = (x + 1) (x^{2} - x + 1)

L e t x + 1 = u^{2}, x^{2} - x + 1 = v^{2}

x = u^{2} - 1, x^{2} - x + 1

= {(u^{2} - 1)}^{2} - (u^{2} - 1) + 1 = v^{2}

\Rightarrow u^{4} - 2 u^{2} + 1 - u^{2} + 1 + 1 = v^{2}

\Rightarrow u^{4} - 3 u^{2} + 3 - v^{2} = 0

u^{2} = \frac{3 \pm \sqrt{9 - 4 (3 - v^{2})}}{2}

u^{2} = \frac{3 \pm \sqrt{9 - 12 + 4 v^{2}}}{2}

\Rightarrow 4 v^{2} - 3 = p^{2}

\Rightarrow 4 v^{2} - p^{2} = 3

(2 v - p) (2 v + p) = 3

(2 v - p = 1 \land 2 v + p = 3) ∣ (2 v - p = 3 \land 2 v + p = 1)

p = 1, v = 1 ∣ p = - 1, v = 1

u^{2} = \frac{3 \pm \sqrt{4 v^{2} - 3}}{2} = \frac{3 \pm \sqrt{4 {(1)}^{2} - 3}}{2}

= \frac{3 \pm 1}{2} = 2, 1

^{∙} x = u^{2} - 1 = {(1)}^{2} - 1 = 0 \Rightarrow y^{2} = x^{3} + 1

\Rightarrow y^{2} = 1 \Rightarrow y = \pm 1

(0, \pm 1)

^{∙} x = u^{2} - 1 = 2 - 1 = 1 \Rightarrow y^{2} = 1^{3} + 1

\Rightarrow y = \pm \sqrt{2} (N o t e x c e p t a b l e)

S o w h e n x + 1 & x^{2} - x + 1 a r e b o t h

p e r f e c t s q u a r e (x, y) = (0, \pm 1)

Answered by Rasheed.Sindhi last updated on 18/Jun/20

A n o t h e r m e t h o d

(y - 1) (y + 1) = x . x^{2}

\Rightarrow y - 1 = {(y + 1)}^{2} ∣ y + 1 = {(y - 1)}^{2}

^{∙} y = y^{2} + 2 y + 2

y^{2} + y + 2 = 0

y = \frac{- 1 \pm \sqrt{1 - 8}}{2}

y \notin Z

^{∙} y + 1 = y^{2} - 2 y + 1

y^{2} - 3 y = 0

y (y - 3) = 0

y = 0 ∣ y = 3

x^{3} + 1 = y^{2}

x^{3} = {(0)}^{2} - 1 = - 1

(- 1, 0)

x^{3} = {(3)}^{2} - 1

x = 2

(2, 3)