Greece — Selection Examination

The given equation can be written as
$$\begin{aligned} 8x^3 + y^3 - 6xy &= 4 \Leftrightarrow (2x)^3 + y^3 + 1^3 - 3 \cdot 2x \cdot y \cdot 1 = 5 \\ &\Leftrightarrow (2x + y + 1)(4x^2 + y^2 + 1 - 2xy - 2x - y) = 5 \end{aligned} \quad (1)$$
$$\Leftrightarrow \frac{1}{2}(2x+y+1)[(2x-y)^2+(2x-1)^2+(y-1)^2]=5 \quad (2)$$
From (2), since $(2x-y)^2 + (2x-1)^2 + (y-1)^2 > 0$, we have $2x + y + 1 > 0$, and hence from (1) we get
$$2x + y + 1 = 1 \text{ or } 2x + y + 1 = 5 \Leftrightarrow 2x + y = 0 \text{ or } 2x + y = 4.$$
From (1), for $2x + y = 4$, we have
$$4x^2 + y^2 + 1 - 2xy - 2x - y = 1 \Leftrightarrow (2x + y)^2 - 6xy - (2x + y) = 0 \Leftrightarrow xy = 2$$
From the system $2x + y = 4$, $xy = 2$ we get the solution $(x, y) = (1, 2)$.
Also, from (1) for $2x + y = 0$ we have
$$4x^2 + y^2 + 1 - 2xy - 2x - y = 5 \Leftrightarrow (2x + y)^2 - 6xy - (2x + y) = 4 \Leftrightarrow xy = -\frac{2}{3}.$$
However, from the system $2x + y = 0$, $xy = -\frac{2}{3}$ we get no solutions.
We also can work in the following way: The equation can be written as
$$8x^3 + y^3 - 6xy = 4 \Leftrightarrow (2x + y)^3 - 3 \cdot 2x \cdot y \cdot (2x + y) - 6xy = 4,$$ and so by putting $2x + y = s$, $2xy = p$, we get
$$s^3 - 3ps - 3p = 4 \Rightarrow p = \frac{s^3 - 4}{3(s + 1)}.$$
Since $3p \in \mathbb{Z}$, we have that
$$\frac{s^3 - 4}{s + 1} = \frac{s^3 + 1 - 5}{s + 1} = s^2 - s + 1 - \frac{5}{s + 1} \in \mathbb{Z}.$$
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Thus $s+1$ must be a divisor of 5, that is
$$s+1 \in \{-1, 1, -5, 5\} \text{ or } s \in \{-2, 0, -6, 4\},$$
And hence we find the pairs
$$(s, p) = (-2, 4), (s, p) = \left(0, -\frac{4}{3}\right), (s, p) = \left(-6, \frac{44}{3}\right), (s, p) = (4, 4),$$
From which only the last gives integer values for $x, y$, i. e. $x = 1, y = 2$.