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

$$\mathrm{find}\mathrm{integral}\mathrm{solution}$$$$\mathrm{of}{\mathrm{y}}^2={\mathrm{x}}^3+1$$

Commented by bemath last updated on 15/Jun/20

$$\mathrm{oo}\mathrm{yes}\mathrm{sir}.\mathrm{thank}\mathrm{you}$$

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

$$\mathrm{Sir}(1,\pm 2){\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{satisfy}:$$$${(\pm 2)}^2\ne {\left(1\right)}^3+1$$$$4\ne 2$$$${\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{extended}\mathrm{my}\mathrm{answer}\mathrm{and}\mathrm{now}$$$$\mathrm{it}\mathrm{includes}\mathrm{all}\mathrm{solutions}.$$

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

$${\mathrm{y}}^2={\mathrm{x}}^3+1$$$${}^{\bullet }{\mathrm{x}}^3+1\mathrm{is}\mathrm{perfect}\mathrm{square}.\mathrm{It}\mathrm{may}\mathrm{be}$$$$0\mathrm{also}\mathrm{provided}\mathrm{that}\mathrm{x}\mathrm{is}\mathrm{an}\mathrm{integer}.$$$$\mathrm{So}{\mathrm{y}}^2={\mathrm{x}}^3+1=0$$$${\mathrm{x}}^3=-1$$$$\mathrm{x}=-1\Rightarrow \mathrm{y}=0$$$$(-1,0)$$$${}^{\bullet }{\mathrm{y}}^2=(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)$$$$\mathrm{x}+1={\mathrm{x}}^2-\mathrm{x}+1$$$${\mathrm{x}}^2-2\mathrm{x}=0$$$$\mathrm{x}(\mathrm{x}-2)=0$$$$\mathrm{x}=0\mid \mathrm{x}=2$$$$\mathrm{x}=0\Rightarrow {\mathrm{y}}^2=0^3+1\Rightarrow \mathrm{y}=\pm 1$$$$\mathrm{x}=2\Rightarrow {\mathrm{y}}^2=2^3+1\Rightarrow \mathrm{y}=\pm 3$$$$(-1,0),(0,\pm 1),(2,\pm 3)$$

Commented by bemath last updated on 15/Jun/20

$$\mathrm{sir}\mathrm{i}\mathrm{got}(-1,0),(1,\pm 2),(2,\pm 3)$$$$$$

Commented by 1549442205 last updated on 15/Jun/20

$$\mathbf{but}\mathbf{remain}\mathbf{the}\mathbf{case}\mathbf{that}\mathbf{x}+1={\mathbf{u}}^2,$$$${\mathbf{x}}^2-\mathbf{x}+1={\mathbf{v}}^2,{\mathbf{doesn}}^{\prime }\mathbf{t}\mathbf{consider}?$$

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

$$\mathcal{T}hankyousir!I^{\prime }lltrytocomplete$$$$myanswer.$$

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

$$y^2=(x+1)(x^2-x+1)$$$$Letx+1=u^2,x^2-x+1=v^2$$$$x=u^2-1,x^2-x+1$$$$={(u^2-1)}^2-(u^2-1)+1=v^2$$$$\Rightarrow u^4-2u^2+1-u^2+1+1=v^2$$$$\Rightarrow u^4-3u^2+3-v^2=0$$$$u^2=\frac{3\pm \sqrt{9-4(3-v^2)}}{2}$$$$u^2=\frac{3\pm \sqrt{9-12+4v^2}}{2}$$$$\Rightarrow 4v^2-3=p^2$$$$\Rightarrow 4v^2-p^2=3$$$$(2v-p)(2v+p)=3$$$$(2v-p=1\wedge 2v+p=3)\mid (2v-p=3\wedge 2v+p=1)$$$$p=1,v=1\mid p=-1,v=1$$$$u^2=\frac{3\pm \sqrt{4v^2-3}}{2}=\frac{3\pm \sqrt{4{\left(1\right)}^2-3}}{2}$$$$=\frac{3\pm 1}{2}=2,1$$$${}^{\bullet }x=u^2-1={\left(1\right)}^2-1=0\Rightarrow y^2=x^3+1$$$$\Rightarrow y^2=1\Rightarrow y=\pm 1$$$$(0,\pm 1)$$$${}^{\bullet }x=u^2-1=2-1=1\Rightarrow y^2=1^3+1$$$$\Rightarrow y=\pm \sqrt{2}\left(Notexceptable\right)$$$$Sowhenx+1\mathrm{\&}{x}^2-x+1areboth$$$$perfectsquare(x,y)=(0,\pm 1)$$

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

$$Anothermethod$$$$(y-1)(y+1)=x.x^2$$$$\Rightarrow y-1={(y+1)}^2\mid y+1={(y-1)}^2$$$${}^{\bullet }y=y^2+2y+2$$$$y^2+y+2=0$$$$y=\frac{-1\pm \sqrt{1-8}}{2}$$$$y\notin \mathbb{Z}$$$${}^{\bullet }y+1=y^2-2y+1$$$$y^2-3y=0$$$$y(y-3)=0$$$$y=0\mid y=3$$$$x^3+1=y^2$$$$x^3={\left(0\right)}^2-1=-1$$$$(-1,0)$$$$x^3={\left(3\right)}^2-1$$$$x=2$$$$(2,3)$$

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