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Estonian Math Competitions

Algebra Difficulty 6.4 National Olympiad Prove it Estonia

Anna, Anne and Anni seek for real solutions (x,y)(x, y) to the system of equations
{4x3y−x4−3x2y2=2021,4y3x−y4−3y2x2=2021. \begin{cases} 4x^3y - x^4 - 3x^2y^2 = 2021, \\ 4y^3x - y^4 - 3y^2x^2 = 2021. \end{cases}
Anna claims that the system of equations has a solution. Anne claims that the system of equations has no solution but at least one of the two equations has solutions. Anni claims that the system of equations has no solution and, even worse, neither of the two equations alone has a solution. Who is right?

Solution

The system does not have a solution since adding the equation gives −(x−y)4=4042-(x - y)^4 = 4042 whose l.h.s. is non-positive but r.h.s. is positive.
We show that the first equation 4x3y−x4−3x2y2=20214x^3y - x^4 - 3x^2y^2 = 2021 has solutions (the same could be done for the second equation by symmetry). Dividing the equation by x4x^4 and reordering the terms in the l.h.s. results in
−3(yx)2+4(yx)−1=2021x4.(6) -3 \left(\frac{y}{x}\right)^2 + 4 \left(\frac{y}{x}\right) - 1 = \frac{2021}{x^4}. \qquad (6)
As the discriminant of the quadratic equation −3t2+4t−1=0-3t^2 + 4t - 1 = 0 is positive, there exists a real number tt such that −3t2+4t−1-3t^2 + 4t - 1 equals a positive number ε\varepsilon. Define x=2021ε4x = \sqrt[4]{\frac{2021}{\varepsilon}} and y=xty = xt; then xx and yy satisfy (6) and also the first equation of the given system of equations. Hence Anne is right.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.