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-

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

Answered by Frix last updated on 20/Aug/24

x, y \in Q

x = y = 0 \Rightarrow 1 - x y = 1 = {(\pm 1)}^{2}; \pm 1 \in Q

x = y = 1 \Rightarrow 1 - x y = 0 = 0^{2}; 0 \in Q

x \neq 0 \land y \neq 0 \land x \neq y :

x^{5} - 2 x^{2} y^{2} + y^{5} = 0

x^{10} - 2 x^{7} y^{2} + x^{5} y^{5} = 0

x^{10} - 2 x^{5} {(x y)}^{2} + {(x y)}^{5} = 0

x^{10} - 2 x^{5} {(x y)}^{2} + {(x y)}^{4} = {(x y)}^{4} - {(x y)}^{5}

{(x^{5} - {(x y)}^{2})}^{2} = (1 - x y) {(x y)}^{4}

1 - x y = \frac{{(x^{5} + {(x y)}^{2})}^{2}}{{(x y)}^{4}} =

= {(\pm \frac{x^{5} + x^{2} y^{2}}{x^{2} y^{2}})}^{2}; \pm \frac{x^{5} + x^{2} y^{2}}{x^{2} y^{2}} \in Q

Commented by hardmath last updated on 20/Aug/24

thankyou dearprofessor