Not so Diophantine equation

Algebra Level 5

x 3 ( x + y ) + ( y 2 + 1 ) ( x 2 + x y + y 2 ) = 0 x^{3}(x+y)+(y^{2}+1)(x^{2}+xy+y^{2})=0

Find number of ordered pairs of ( x , y ) (x,y) .


The answer is 1.

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1 solution

Patrick Corn
Sep 4, 2014

Multiplying through by x y x-y and simplifying, we get x 5 y 5 + x 3 y 3 = 0 x^5-y^5+x^3-y^3 = 0 . But if x > y x > y , x 3 > y 3 x^3 > y^3 and x 5 > y 5 x^5 > y^5 , so the left side is > 0 > 0 . Similarly x < y x < y is impossible.

This leaves x = y x = y , which leads in the original equation to 5 x 4 + 3 x 2 = 0 5x^4 + 3x^2 = 0 , which is only possible if x = 0 x = 0 . So ( 0 , 0 ) (0,0) is the only solution.

world class sol.

Adarsh Kumar - 6 years, 8 months ago

Good solution

Akshat Sharda - 5 years, 10 months ago

Omg What a solution

Nitin Kumar - 1 year, 4 months ago

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