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Question-215708




Question Number 215708 by essaad last updated on 15/Jan/25
Answered by A5T last updated on 15/Jan/25
(i)+(ii)⇒n^3 +y^3 =0⇒(n+y)(n^2 +y^2 −ny)=0...(iii)  (i)−(ii)⇒4n^3 −4y^3 −8n+8y=0  ⇒n^3 −y^3 −2n+2y=0  (n−y)(n^2 +y^2 +ny−2)=0...(iv)  (iii)&(iv) must be true implies  Case I: n+y=0  and n^2 +y^2 +ny−2=0  ⇒y^2 =2⇒y=+_− (√2)  ⇒(n,y)=((√2),−(√2));(−(√2),(√2))  Case II: n+y=0 and n−y=0⇒(n,y)=(0,0)  Case III: n^2 +y^2 −ny=0 and n−y=0⇒(n,y)=(0,0)  Case IV: n^2 +y^2 −ny=0 and n^2 +y^2 +ny−2=0  ⇒(n+y)^2 =3ny and (n+y)^2 =ny+2  ⇒ny=1 ⇒(n+y)=+_− (√3)  n,y satisfy x^2 +^− (√3)x+1=0  ⇒n,y=(((+_− )(√3)+_− (√(3−4(1))))/2)=(((+_− )(√3)+_− i)/2)  (n,y)  =((((√3)+i)/2),(((√3)−i)/2));(((−(√3)+i)/2),((−(√3)−i)/2));((√2),−(√2));(0,0)   upto permutation since equation is symmetric.
(i)+(ii)n3+y3=0(n+y)(n2+y2ny)=0(iii)(i)(ii)4n34y38n+8y=0n3y32n+2y=0(ny)(n2+y2+ny2)=0(iv)(iii)&(iv)mustbetrueimpliesCaseI:n+y=0andn2+y2+ny2=0y2=2y=+2(n,y)=(2,2);(2,2)CaseII:n+y=0andny=0(n,y)=(0,0)CaseIII:n2+y2ny=0andny=0(n,y)=(0,0)CaseIV:n2+y2ny=0andn2+y2+ny2=0(n+y)2=3nyand(n+y)2=ny+2ny=1(n+y)=+3n,ysatisfyx2+3x+1=0n,y=(+)3+34(1)2=(+)3+i2(n,y)=(3+i2,3i2);(3+i2,3i2);(2,2);(0,0)uptopermutationsinceequationissymmetric.

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