System of 3rd degree equations.

Algebra Level 3

Given the system of equations x 3 − 3 x y 2 + p x 2 − p y 2 + q x + r = 0 − y 3 + 3 x 2 y + 2 p x y + q y = 0 \begin{aligned} x^3-3xy^2+px^2-py^2+qx+r &=0\\-y^3+3x^2y+2pxy+qy=0\end{aligned} where p , p, q q and r r are real numbers. Find the minimum number of pairs ( x , y ) (x, y) of real numbers that are solutions of the system.


The answer is 1.00.

Arturo Presa by Arturo Presa , 62, USA

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2 solutions

Otto Bretscher
Dec 14, 2018

There is always at least one solution: We can find a real solution x x of x 3 + p x 2 + q x + r = 0 x^3+px^2+qx+r=0 and let y = 0 y=0 . If we let p = q = r = 0 p=q=r=0 , we find the unique real solution x = y = 0. x=y=0. Thus the answer is 1 \boxed{1} .

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Arturo Presa
Dec 14, 2018

The given system is equivalent to the equation ( x + i y ) 3 + p ( x + i y ) 2 + q ( x + i y ) + r = 0 (x+iy)^3+p(x+iy)^2+q(x+iy)+r=0 . So, any pair ( x , y ) (x,y) that is a solution of the system corresponds to a solution x+iy of the polynomial equation z 3 + p z 2 + q z + r = 0 z^3+pz^2+qz+r=0 and, therefore the minimum number of solutions is 1.

1 Helpful 1 Interesting 0 Brilliant 0 Confused

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