Mathematical Reflections 2017, Issue 3 - Problem S411

Problem:
Solve in real numbers the system of equations
$$\left\{\begin{array}{rcl} \sqrt{x}-\sqrt{y}&=&45 \\ \sqrt[3]{x-2017}-\sqrt[3]{y}&=&2. \end{array} \right.$$

Proposed by Adrian Andreescu, Dallas, Texas, USA

Solution:
Squaring both sides of the first equation, cubing both sides of the second equation and using $\sqrt[3]{x-2017}-\sqrt[3]{y}=2$, we obtain the system of equations
$$\begin{array}{rcl} x+y-2\sqrt{xy}&=&2025 \\ x-y-6\sqrt[3]{(x-2017)y}&=&2025. \end{array}$$
Subtracting side by side these two equations, we get
$$2\sqrt[3]{y}(\sqrt[3]{y^2}-\sqrt{x}\cdot \sqrt[6]{y}+3\sqrt[3]{x-2017})=0.$$
If $y=0$, we get $x=2025$. If $y \neq 0$, then $\sqrt[3]{y^2}-\sqrt{x}\cdot \sqrt[6]{y}+3\sqrt[3]{x-2017}=0$, i.e.
$$\sqrt[6]{y}(\sqrt{y}-\sqrt{x})+3\sqrt[3]{x-2017}=0 \iff -45\sqrt[6]{y}+3\sqrt[3]{x-2017}=0,$$ which gives $x=3375\sqrt{y}+2017$.
Substituting this equation into the first and the second equation of the system, we get
$$y=\left(\dfrac{16}{3285+223\sqrt{217}}\right)^2, \qquad y=\left(\dfrac{3285+223\sqrt{217}}{2}\right)^2.$$
So, $$(x,y)=(2025,0),$$ $$(x,y)=\left(2017+\dfrac{54000}{3285+223\sqrt{217}},\left(\dfrac{16}{3285+223\sqrt{217}}\right)^2\right),$$
$$(x,y)=\left(2017+\dfrac{3375}{2}(3285+223\sqrt{217}),\left(\dfrac{3285+223\sqrt{217}}{2}\right)^2\right).$$