# 2006 Indonesia MO Problems/Problem 1

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## Problem

Find all pairs $(x,y)$ of real numbers which satisfy $x^3-y^3=4(x-y)$ and $x^3+y^3=2(x+y)$.

## Solution

Factoring and rearranging terms for both equations results in $(x-y)(x^2 + xy + y^2 - 4) = 0$ and $(x+y)(x^2 - xy + y^2 - 2) = 0$. By the Zero Product Property, $x = y$ or $x^2 + xy + y^2 = 4$ in the first equation, and $x = -y$ and $x^2 - xy + y^2 = 2$ in the second equation. In order for the solution to satisfy both equations, it must satisfy at least one of the conditions in each equation. Now there are four cases to consider.

Case 1: $x=y$ and $x=-y$

Solving this system results in $(0,0)$.

Case 2: $x=y$ and $x^2 - xy + y^2 = 2$

Substitution results in $x^2 = 2$, so solving this system results in $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$.

Case 3: $x^2 + xy + y^2 = 4$ and $x=-y$

Substitution results in $y^2 = 4$, so solving this system results in $(2,-2)$ and $(-2,2)$.

Case 4: $x^2 + xy + y^2 = 4$ and $x^2 - xy + y^2 = 2$

Subtracting the second equation from the first equation results in $2xy = 2$, so $xy = 1$. Thus, $x^2 + y^2 = 3$ and $y = \tfrac{1}{x}$, so $x^2 + \tfrac{1}{x^2} = 3$.

Multiplying both sides by $x^2$ and bringing all terms to one side results in $x^4 - 3x^2 + 1 = 0$. By using the Quadratic Formula, $x^2 = \tfrac{3 \pm \sqrt{5}}{2}$, and both quantities are positive. If $x^2 = \tfrac{3 + \sqrt{5}}{2}$, then $y^2 = \tfrac{3 - \sqrt{5}}{2}$ (and vice versa). Since $xy = 1$, $x$ and $y$ must have the same sign.

Therefore, there are four ordered pairs — $(\sqrt{\frac{3 + \sqrt{5}}{2}},\sqrt{\frac{3 - \sqrt{5}}{2}}), (-\sqrt{\frac{3 + \sqrt{5}}{2}}, -\sqrt{\frac{3 - \sqrt{5}}{2}}),$ $(\sqrt{\frac{3 - \sqrt{5}}{2}}, \sqrt{\frac{3 + \sqrt{5}}{2}}), (-\sqrt{\frac{3 - \sqrt{5}}{2}}, -\sqrt{\frac{3 + \sqrt{5}}{2}}).$

In total, there are nine solutions, which are listed in each case.