Question Number 210840 by hardmath last updated on 19/Aug/24
$$ \\ $$Prove that if x, y are rational numbers satisfying the equation
x^5 + y^5 = 2(x^2)(y^2)
then 1 – xy is the square of rational number
x^5 + y^5 = 2(x^2)(y^2)
then 1 – xy is the square of rational number
Answered by Frix last updated on 20/Aug/24
$${x},\:{y}\:\in\mathbb{Q} \\ $$$${x}={y}=\mathrm{0}\:\Rightarrow\:\mathrm{1}−{xy}=\mathrm{1}=\left(\pm\mathrm{1}\right)^{\mathrm{2}} ;\:\pm\mathrm{1}\in\mathbb{Q} \\ $$$${x}={y}=\mathrm{1}\:\Rightarrow\:\mathrm{1}−{xy}=\mathrm{0}=\mathrm{0}^{\mathrm{2}} ;\:\mathrm{0}\in\mathbb{Q} \\ $$$$ \\ $$$${x}\neq\mathrm{0}\wedge{y}\neq\mathrm{0}\wedge{x}\neq{y}: \\ $$$$ \\ $$$${x}^{\mathrm{5}} −\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{5}} =\mathrm{0} \\ $$$${x}^{\mathrm{10}} −\mathrm{2}{x}^{\mathrm{7}} {y}^{\mathrm{2}} +{x}^{\mathrm{5}} {y}^{\mathrm{5}} =\mathrm{0} \\ $$$${x}^{\mathrm{10}} −\mathrm{2}{x}^{\mathrm{5}} \left({xy}\right)^{\mathrm{2}} +\left({xy}\right)^{\mathrm{5}} =\mathrm{0} \\ $$$${x}^{\mathrm{10}} −\mathrm{2}{x}^{\mathrm{5}} \left({xy}\right)^{\mathrm{2}} +\left({xy}\right)^{\mathrm{4}} =\left({xy}\right)^{\mathrm{4}} −\left({xy}\right)^{\mathrm{5}} \\ $$$$\left({x}^{\mathrm{5}} −\left({xy}\right)^{\mathrm{2}} \right)^{\mathrm{2}} =\left(\mathrm{1}−{xy}\right)\left({xy}\right)^{\mathrm{4}} \\ $$$$\mathrm{1}−{xy}=\frac{\left({x}^{\mathrm{5}} +\left({xy}\right)^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({xy}\right)^{\mathrm{4}} }= \\ $$$$\:\:\:\:\:=\left(\pm\frac{{x}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }\right)^{\mathrm{2}} ;\:\pm\frac{{x}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }\in\mathbb{Q} \\ $$
Commented by hardmath last updated on 20/Aug/24
$$\mathrm{thankyou}\:\mathrm{dearprofessor} \\ $$