Estonian Math Competitions

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