Solve-for-integers-x-2-y-2-x-4-y-4-x-y-6-

Question Number 156502 by MathSh last updated on 11/Oct/21

Solve for integers :

(x^{2} + y^{2}) (x^{4} + y^{4}) = {(x + y)}^{6}

Answered by Rasheed.Sindhi last updated on 12/Oct/21

{(x + y)}^{6} - (x^{4} + y^{4}) (x^{2} + y^{2}) = 0

6 x^{5} y + 14 x^{4} y^{2} + 20 x^{3} y^{3} + 14 x^{2} y^{4} + 6 x y^{5} = 0

xy ({6 x}^{4} + {14 x}^{3} y + {20 x}^{2} y^{2} + {14 xy}^{3} + {6 y}^{4}) = 0

x = 0 ∣ y = 0 ∣ {6 x}^{4} + {14 x}^{3} y + {20 x}^{2} y^{2} + {14 xy}^{3} + {6 y}^{4} = 0

{6 x}^{4} + {14 x}^{3} y + {20 x}^{2} y^{2} + {14 xy}^{3} + {6 y}^{4} = 0

2 (x^{2} + xy + y^{2}) ({3 x}^{2} + 4 xy + {3 y}^{2}) = 0

\underset{\underset{only integer solution}{x = 0, y = 0}}{x^{2} + xy + y^{2} = 0} ∣ \underset{\underset{only integer solution}{x = 0, y = 0}}{{3 x}^{2} + 4 xy + {3 y}^{2}} = 0

Other solutions are complex :

^{∙} x = \frac{- y \pm \sqrt{y^{2} - {4 y}^{2}}}{2} = \frac{- y \pm \sqrt{- {3 y}^{2}}}{2}

= \frac{- y \pm iy \sqrt{3}}{2} \notin Z

^{∙} x = \frac{- 4 y \pm \sqrt{{16 y}^{2} - {36 y}^{2}}}{6} \notin Z

x = 0, y = 0

Commented by MathSh last updated on 12/Oct/21

Very nice dear S er, thank you

Facebook Tweet Pin