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Question Number 120812 by bramlexs22 last updated on 03/Nov/20

Find all integral solutions to the equation (x^2 +1)(y^2 +1)+2(x−y)(1−xy)=4(1+xy)

Findallintegralsolutionstotheequation(x2+1)(y2+1)+2(x−y)(1−xy)=4(1+xy)

Answered by liberty last updated on 03/Nov/20

⇔ (xy−1)^2 +(x−y)^2 −2(x−y)(xy−1)=4 ⇔ [ xy−1−(x−y) ]^2 = 4 ⇔ (x+1)(y+1)= ±2 case(1) (x+1)(y+1)=2 → { ((x+1=2)),((y+1=1)) :} , { ((x+1=1)),((y+1=2)) :} → { ((x+1=−2)),((y+1=−1 )) :}, { ((x+1=−1)),((y+1=−2)) :} we get four solutions are {(1,0),(0,1),(−3,−2),(−2,−3)} case(2) (x+1)(y+1)=−2→ { ((x+1=−2)),((y+1=1)) :}, { ((x+1=1)),((y+1=−2)) :} → { ((x+1=−1)),((y+1=2)) :} , { ((x+1=2)),((y+1=−1)) :} also we get four solutions are { (−2,1),(1,−2),(−3,0),(0,−3)} ∵ complete solutions are { (1,0),(0,1),(−3,−2),(−2,−3),(−2,1),(1,−2),(−3,0),(0,−3)}▲

⇔(xy−1)2+(x−y)2−2(x−y)(xy−1)=4⇔[xy−1−(x−y)]2=4⇔(x+1)(y+1)=±2case(1)(x+1)(y+1)=2→{x+1=2y+1=1,{x+1=1y+1=2→{x+1=−2y+1=−1,{x+1=−1y+1=−2wegetfoursolutionsare{(1,0),(0,1),(−3,−2),(−2,−3)}case(2)(x+1)(y+1)=−2→{x+1=−2y+1=1,{x+1=1y+1=−2→{x+1=−1y+1=2,{x+1=2y+1=−1alsowegetfoursolutionsare{(−2,1),(1,−2),(−3,0),(0,−3)}∵completesolutionsare{(1,0),(0,1),(−3,−2),(−2,−3),(−2,1),(1,−2),(−3,0),(0,−3)}▴

Commented by bramlexs22 last updated on 03/Nov/20

nice...

nice...

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