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Question Number 212790 by RojaTaniya last updated on 24/Oct/24

Answered by A5T last updated on 24/Oct/24

x−y=(y−1+x−1)(y−1−x+1) ⇒x=y or −1=x+y−2⇒x+y=1 ⇒x+1=x^2 −2x+1 or x+1=x^2 ⇒x^2 +y^2 =x+y+2=3 ⇒x=0 or 3; x=0⇒y=0; x=3⇒y=3 ⇒x^2 +y^2 =0 or 18 or 3⇒(((x^2 +y^2 )/3))^(2020) =0 or 1 or 6^(2020) when x≠y; (((x^2 +y^2 )/3))^(2020) =1

x−y=(y−1+x−1)(y−1−x+1)⇒x=yor−1=x+y−2⇒x+y=1⇒x+1=x2−2x+1orx+1=x2⇒x2+y2=x+y+2=3⇒x=0or3;x=0⇒y=0;x=3⇒y=3⇒x2+y2=0or18or3⇒(x2+y23)2020=0or1or62020whenx≠y;(x2+y23)2020=1

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