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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
$$ \\ $$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
Answered by Frix last updated on 20/Aug/24
x, y ∈Q  x=y=0 ⇒ 1−xy=1=(±1)^2 ; ±1∈Q  x=y=1 ⇒ 1−xy=0=0^2 ; 0∈Q    x≠0∧y≠0∧x≠y:    x^5 −2x^2 y^2 +y^5 =0  x^(10) −2x^7 y^2 +x^5 y^5 =0  x^(10) −2x^5 (xy)^2 +(xy)^5 =0  x^(10) −2x^5 (xy)^2 +(xy)^4 =(xy)^4 −(xy)^5   (x^5 −(xy)^2 )^2 =(1−xy)(xy)^4   1−xy=(((x^5 +(xy)^2 )^2 )/((xy)^4 ))=       =(±((x^5 +x^2 y^2 )/(x^2 y^2 )))^2 ; ±((x^5 +x^2 y^2 )/(x^2 y^2 ))∈Q
$${x},\:{y}\:\in\mathbb{Q} \\ $$$${x}={y}=\mathrm{0}\:\Rightarrow\:\mathrm{1}−{xy}=\mathrm{1}=\left(\pm\mathrm{1}\right)^{\mathrm{2}} ;\:\pm\mathrm{1}\in\mathbb{Q} \\ $$$${x}={y}=\mathrm{1}\:\Rightarrow\:\mathrm{1}−{xy}=\mathrm{0}=\mathrm{0}^{\mathrm{2}} ;\:\mathrm{0}\in\mathbb{Q} \\ $$$$ \\ $$$${x}\neq\mathrm{0}\wedge{y}\neq\mathrm{0}\wedge{x}\neq{y}: \\ $$$$ \\ $$$${x}^{\mathrm{5}} −\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{5}} =\mathrm{0} \\ $$$${x}^{\mathrm{10}} −\mathrm{2}{x}^{\mathrm{7}} {y}^{\mathrm{2}} +{x}^{\mathrm{5}} {y}^{\mathrm{5}} =\mathrm{0} \\ $$$${x}^{\mathrm{10}} −\mathrm{2}{x}^{\mathrm{5}} \left({xy}\right)^{\mathrm{2}} +\left({xy}\right)^{\mathrm{5}} =\mathrm{0} \\ $$$${x}^{\mathrm{10}} −\mathrm{2}{x}^{\mathrm{5}} \left({xy}\right)^{\mathrm{2}} +\left({xy}\right)^{\mathrm{4}} =\left({xy}\right)^{\mathrm{4}} −\left({xy}\right)^{\mathrm{5}} \\ $$$$\left({x}^{\mathrm{5}} −\left({xy}\right)^{\mathrm{2}} \right)^{\mathrm{2}} =\left(\mathrm{1}−{xy}\right)\left({xy}\right)^{\mathrm{4}} \\ $$$$\mathrm{1}−{xy}=\frac{\left({x}^{\mathrm{5}} +\left({xy}\right)^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({xy}\right)^{\mathrm{4}} }= \\ $$$$\:\:\:\:\:=\left(\pm\frac{{x}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }\right)^{\mathrm{2}} ;\:\pm\frac{{x}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }\in\mathbb{Q} \\ $$
Commented by hardmath last updated on 20/Aug/24
thankyou dearprofessor
$$\mathrm{thankyou}\:\mathrm{dearprofessor} \\ $$

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