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Question Number 63945 by mathmax by abdo last updated on 11/Jul/19

solve at Z^2 x^2 −2y^2 +xy +2 =0

$${solve}\:{at}\:{Z}^{\mathrm{2}} \:\:{x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \:+{xy}\:+\mathrm{2}\:=\mathrm{0} \\ $$

Commented by hknkrc46 last updated on 11/Jul/19

x^2 −2y^2 +xy+2=(x^2 −y^2 )+(xy−y^2 )+2=0 (x−y)(x+y)+y(x−y)+2=0 (x−y)(x+y+y)+2=0 (x−y)(x+2y)+2=0 1) x−y=1∧x+2y=−2⇒(x,y)=(0,−1) 2) x−y=−1∧x+2y=2⇒(x,y)=(0,1) 3) x−y=2∧x+2y=−1⇒(x,y)=(1,−1) 4) x−y=−2∧x+2y=1⇒(x,y)=(−1,1)

$${x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} +{xy}+\mathrm{2}=\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)+\left({xy}−{y}^{\mathrm{2}} \right)+\mathrm{2}=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+{y}\right)+{y}\left({x}−{y}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+{y}+{y}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+\mathrm{2}{y}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{x}−{y}=\mathrm{1}\wedge{x}+\mathrm{2}{y}=−\mathrm{2}\Rightarrow\left({x},{y}\right)=\left(\mathrm{0},−\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:{x}−{y}=−\mathrm{1}\wedge{x}+\mathrm{2}{y}=\mathrm{2}\Rightarrow\left({x},{y}\right)=\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right)\:{x}−{y}=\mathrm{2}\wedge{x}+\mathrm{2}{y}=−\mathrm{1}\Rightarrow\left({x},{y}\right)=\left(\mathrm{1},−\mathrm{1}\right) \\ $$$$\left.\mathrm{4}\right)\:{x}−{y}=−\mathrm{2}\wedge{x}+\mathrm{2}{y}=\mathrm{1}\Rightarrow\left({x},{y}\right)=\left(−\mathrm{1},\mathrm{1}\right) \\ $$

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