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Question Number 98602 by bemath last updated on 15/Jun/20
find integral solution of y^2 = x^3 +1
$$\mathrm{find}\:\mathrm{integral}\:\mathrm{solution} \\ $$$$\mathrm{of}\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{x}^{\mathrm{3}} +\mathrm{1}\: \\ $$
Commented by bemath last updated on 15/Jun/20
oo yes sir. thank you
$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Jun/20
Sir (1,±2) doesn′t satisfy: (±2)^2 ≠(1)^3 +1 4≠2 I′ve extended my answer and now it includes all solutions.
$$\mathrm{Sir}\:\left(\mathrm{1},\pm\mathrm{2}\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{satisfy}: \\ $$$$\:\:\:\:\:\left(\pm\mathrm{2}\right)^{\mathrm{2}} \neq\left(\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4}\neq\mathrm{2} \\ $$$$\mathrm{I}'\mathrm{ve}\:\mathrm{extended}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{and}\:\mathrm{now} \\ $$$$\mathrm{it}\:\mathrm{includes}\:\mathrm{all}\:\mathrm{solutions}. \\ $$
Commented by Rasheed.Sindhi last updated on 15/Jun/20
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⇒y=0 (−1,0) ^• y^2 =(x+1)(x^2 −x+1) x+1=x^2 −x+1 x^2 −2x=0 x(x−2)=0 x=0 ∣ x=2 x=0⇒y^2 =0^3 +1⇒y=±1 x=2⇒y^2 =2^3 +1⇒y=±3 (−1,0),(0,±1),(2,±3)
$$\mathrm{y}^{\mathrm{2}} =\mathrm{x}^{\mathrm{3}} +\mathrm{1} \\ $$$$\:^{\bullet} \mathrm{x}^{\mathrm{3}} +\mathrm{1}\:\mathrm{is}\:\mathrm{perfect}\:\mathrm{square}.\mathrm{It}\:\mathrm{may}\:\mathrm{be} \\ $$$$\mathrm{0}\:\mathrm{also}\:\mathrm{provided}\:\mathrm{that}\:\mathrm{x}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}. \\ $$$$\mathrm{So}\:\mathrm{y}^{\mathrm{2}} =\mathrm{x}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{3}} =−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=−\mathrm{1}\Rightarrow\mathrm{y}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(−\mathrm{1},\mathrm{0}\right) \\ $$$$\:^{\bullet} \mathrm{y}^{\mathrm{2}} =\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right) \\ $$$$\mathrm{x}+\mathrm{1}=\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{2x}=\mathrm{0} \\ $$$$\mathrm{x}\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{0}\:\:\mid\:\mathrm{x}=\mathrm{2} \\ $$$$\mathrm{x}=\mathrm{0}\Rightarrow\mathrm{y}^{\mathrm{2}} =\mathrm{0}^{\mathrm{3}} +\mathrm{1}\Rightarrow\mathrm{y}=\pm\mathrm{1} \\ $$$$\mathrm{x}=\mathrm{2}\Rightarrow\mathrm{y}^{\mathrm{2}} =\mathrm{2}^{\mathrm{3}} +\mathrm{1}\Rightarrow\mathrm{y}=\pm\mathrm{3} \\ $$$$\left(−\mathrm{1},\mathrm{0}\right),\left(\mathrm{0},\pm\mathrm{1}\right),\left(\mathrm{2},\pm\mathrm{3}\right) \\ $$
Commented by bemath last updated on 15/Jun/20
sir i got (−1,0), (1,±2), (2,±3)
$$\mathrm{sir}\:\mathrm{i}\:\mathrm{got}\:\left(−\mathrm{1},\mathrm{0}\right),\:\left(\mathrm{1},\pm\mathrm{2}\right),\:\left(\mathrm{2},\pm\mathrm{3}\right)\: \\ $$$$ \\ $$
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?
$$\boldsymbol{\mathrm{but}}\:\boldsymbol{\mathrm{remain}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{case}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{x}}+\mathrm{1}=\boldsymbol{\mathrm{u}}^{\mathrm{2}} , \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}+\mathrm{1}=\boldsymbol{\mathrm{v}}^{\mathrm{2}} ,\boldsymbol{\mathrm{doesn}}'\boldsymbol{\mathrm{t}}\:\boldsymbol{\mathrm{consider}}? \\ $$
Commented by Rasheed.Sindhi last updated on 15/Jun/20
Thank you sir! I′ll try to complete my answer.
$$\mathcal{T}{hank}\:{you}\:{sir}!\:{I}'{ll}\:{try}\:{to}\:{complete} \\ $$$${my}\:{answer}. \\ $$
Answered by Rasheed.Sindhi last updated on 17/Jun/20
y^2 =(x+1)(x^2 −x+1) Let 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 ⇒u^4 −2u^2 +1−u^2 +1+1=v^2 ⇒u^4 −3u^2 +3−v^2 =0 u^2 =((3±(√(9−4(3−v^2 ))))/2) u^2 =((3±(√(9−12+4v^2 )))/2) ⇒4v^2 −3=p^2 ⇒4v^2 −p^2 =3 (2v−p)(2v+p)=3 ( 2v−p=1∧2v+p=3 ) ∣ ( 2v−p=3∧2v+p=1) p=1,v=1 ∣ p=−1,v=1 u^2 =((3±(√(4v^2 −3)))/2)=((3±(√(4(1)^2 −3)))/2) =((3±1)/2)=2,1 ^• x=u^2 −1=(1)^2 −1=0⇒y^2 =x^3 +1 ⇒y^2 =1⇒y=±1 (0,±1) ^• x=u^2 −1=2−1=1⇒y^2 =1^3 +1 ⇒y=±(√2) (Not exceptable) So when x+1 & x^2 −x+1 are both perfect square (x,y)=(0,±1)
$${y}^{\mathrm{2}} =\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$${Let}\:{x}+\mathrm{1}={u}^{\mathrm{2}} ,\:{x}^{\mathrm{2}} −{x}+\mathrm{1}={v}^{\mathrm{2}} \\ $$$${x}={u}^{\mathrm{2}} −\mathrm{1},{x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\left({u}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}={v}^{\mathrm{2}} \\ $$$$\:\Rightarrow{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} +\mathrm{1}−{u}^{\mathrm{2}} +\mathrm{1}+\mathrm{1}={v}^{\mathrm{2}} \\ $$$$\:\Rightarrow{u}^{\mathrm{4}} −\mathrm{3}{u}^{\mathrm{2}} +\mathrm{3}−{v}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:{u}^{\mathrm{2}} =\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}\left(\mathrm{3}−{v}^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$$\:\:\:{u}^{\mathrm{2}} =\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{12}+\mathrm{4}{v}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{v}^{\mathrm{2}} −\mathrm{3}={p}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}{v}^{\mathrm{2}} −{p}^{\mathrm{2}} =\mathrm{3} \\ $$$$\:\:\:\:\:\left(\mathrm{2}{v}−{p}\right)\left(\mathrm{2}{v}+{p}\right)=\mathrm{3} \\ $$$$\:\:\:\:\left(\:\mathrm{2}{v}−{p}=\mathrm{1}\wedge\mathrm{2}{v}+{p}=\mathrm{3}\:\right)\:\mid\:\left(\:\mathrm{2}{v}−{p}=\mathrm{3}\wedge\mathrm{2}{v}+{p}=\mathrm{1}\right) \\ $$$$\:\:\:{p}=\mathrm{1},{v}=\mathrm{1}\:\:\mid\:\:{p}=−\mathrm{1},{v}=\mathrm{1} \\ $$$${u}^{\mathrm{2}} =\frac{\mathrm{3}\pm\sqrt{\mathrm{4}{v}^{\mathrm{2}} −\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{3}\pm\sqrt{\mathrm{4}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{3}\pm\mathrm{1}}{\mathrm{2}}=\mathrm{2},\mathrm{1} \\ $$$$\:^{\bullet} {x}={u}^{\mathrm{2}} −\mathrm{1}=\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\Rightarrow{y}^{\mathrm{2}} ={x}^{\mathrm{3}} +\mathrm{1} \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\mathrm{1}\Rightarrow{y}=\pm\mathrm{1} \\ $$$$\left(\mathrm{0},\pm\mathrm{1}\right) \\ $$$$\:^{\bullet} {x}={u}^{\mathrm{2}} −\mathrm{1}=\mathrm{2}−\mathrm{1}=\mathrm{1}\Rightarrow{y}^{\mathrm{2}} =\mathrm{1}^{\mathrm{3}} +\mathrm{1} \\ $$$$\Rightarrow{y}=\pm\sqrt{\mathrm{2}}\:\:\:\:\left({Not}\:{exceptable}\right) \\ $$$${So}\:{when}\:{x}+\mathrm{1}\:\&\:{x}^{\mathrm{2}} −{x}+\mathrm{1}\:{are}\:{both} \\ $$$${perfect}\:{square}\:\left({x},{y}\right)=\left(\mathrm{0},\pm\mathrm{1}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 18/Jun/20
Another method (y−1)(y+1)=x.x^2 ⇒y−1=(y+1)^2 ∣ y+1=(y−1)^2 ^• y=y^2 +2y+2 y^2 +y+2=0 y=((−1±(√(1−8)))/2) y∉Z ^• y+1=y^2 −2y+1 y^2 −3y=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)
$${Another}\:{method} \\ $$$$\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)={x}.{x}^{\mathrm{2}} \\ $$$$\Rightarrow{y}−\mathrm{1}=\left({y}+\mathrm{1}\right)^{\mathrm{2}} \:\mid\:{y}+\mathrm{1}=\left({y}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:^{\bullet} {y}={y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{2} \\ $$$$\:\:\:{y}^{\mathrm{2}} +{y}+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:{y}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{8}}}{\mathrm{2}} \\ $$$$\:\:\:\:{y}\notin\mathbb{Z} \\ $$$$\:\:^{\bullet} {y}+\mathrm{1}={y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1} \\ $$$$\:\:\:{y}^{\mathrm{2}} −\mathrm{3}{y}=\mathrm{0} \\ $$$$\:{y}\left({y}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\:{y}=\mathrm{0}\:\mid\:{y}=\mathrm{3} \\ $$$${x}^{\mathrm{3}} +\mathrm{1}={y}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} =\left(\mathrm{0}\right)^{\mathrm{2}} −\mathrm{1}=−\mathrm{1} \\ $$$$\:\:\:\:\:\left(−\mathrm{1},\mathrm{0}\right) \\ $$$${x}^{\mathrm{3}} =\left(\mathrm{3}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$${x}=\mathrm{2} \\ $$$$\:\:\:\left(\mathrm{2},\mathrm{3}\right) \\ $$

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