AoPS Wiki talk:Problem of the Day/August 2, 2011

Problem: Find all $(x,y)\in\mathbb{R}^2$ such that $x^2-y^2=24$ and $x^3-y^3=218$.

Solution: Let $X=x^3$ and $Y=y^3$. Then $X=218+Y$ and $(218+Y)^{2/3} = 24 + Y^{2/3}$. Hence, \[218+y^3=(24+y^2)^3\] \[(y-5)(6740+1348y-76y^2+72y^3)=0\] This quartic has two real roots; $y=5$ and \[y = \frac{19}{54} - \frac{17837}{54(-7881974+25353\sqrt{105481})^{1/3}}+\frac{1}{54}(-7881974+25353\sqrt{105481})^{1/3}\] by the cubic formula. Call this second root (which evaluates to about -3.01) $\psi$. The two real solutions are therefore: \[(x_1,y_1) = (7,5)\] and \[(x_2,y_2) = (\sqrt[3]{218 + \psi^3},\psi)\]

There are more solutions in $\mathbb{C}^2$, corresponding to the complex roots of $\frac{6740}{72}+\frac{1348}{72}y-\frac{19}{18}y^2+y^3$.