Question Number 63945 by mathmax by abdo last updated on 11/Jul/19
$${solve}\:{at}\:{Z}^{\mathrm{2}} \:\:{x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \:+{xy}\:+\mathrm{2}\:=\mathrm{0} \\ $$
Commented by hknkrc46 last updated on 11/Jul/19
$${x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} +{xy}+\mathrm{2}=\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)+\left({xy}−{y}^{\mathrm{2}} \right)+\mathrm{2}=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+{y}\right)+{y}\left({x}−{y}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+{y}+{y}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+\mathrm{2}{y}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{x}−{y}=\mathrm{1}\wedge{x}+\mathrm{2}{y}=−\mathrm{2}\Rightarrow\left({x},{y}\right)=\left(\mathrm{0},−\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:{x}−{y}=−\mathrm{1}\wedge{x}+\mathrm{2}{y}=\mathrm{2}\Rightarrow\left({x},{y}\right)=\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right)\:{x}−{y}=\mathrm{2}\wedge{x}+\mathrm{2}{y}=−\mathrm{1}\Rightarrow\left({x},{y}\right)=\left(\mathrm{1},−\mathrm{1}\right) \\ $$$$\left.\mathrm{4}\right)\:{x}−{y}=−\mathrm{2}\wedge{x}+\mathrm{2}{y}=\mathrm{1}\Rightarrow\left({x},{y}\right)=\left(−\mathrm{1},\mathrm{1}\right) \\ $$