Hellenic Mathematical Olympiad

The equation can be written as
$$y = 2x^2 + 2xy + 3xy + 3y^2 \Leftrightarrow y = (x+y)(2x+3y). \quad (1)$$
By putting $x+y=z \in \mathbb{Z}$, then equation (1) is written
$$y = z(2z + y) \Leftrightarrow y = 2z^2 + yz \Leftrightarrow (z-1)y = -2z^2. \quad (2)$$
For $z=1$ equation (2) becomes: $0 \cdot y = -2$ (impossible). For $z \neq 1$ we have
$$y = -\frac{2z^2}{z-1} = -\frac{2(z^2-1)+2}{z-1} = -2(z+1) - \frac{2}{z-1}. \quad (3)$$
In order $y \in \mathbb{Z}$, $z-1$ must be a divisor of $2$, i.e.
$$z-1 \in \{-1, 1, -2, 2\} \Leftrightarrow z \in \{0, 2, -1, 3\}.$$
For $z=0$, we find $(x,y)=(0,0)$, for $z=2$, we find $(x,y)=(10,-8)$, for $z=-1$, we find $(x,y)=(-2,1)$ and for $z=3$, we find $(x,y)=(12,-9)$.

The equation can be written as
$$y = 2x^2 + 2xy + 3xy + 3y^2 \Leftrightarrow (x+y-1)(2x+3y+2) = -2 \quad (4)$$
Hence we have the cases:
$$\bullet \quad \begin{cases} x+y-1=-1 \\ 2x+3y+2=2 \end{cases} \Leftrightarrow \begin{cases} x+y=0 \\ 2x+3y=0 \end{cases} \Leftrightarrow (x,y)=(0,0)$$
$$\bullet \quad \begin{cases} x+y-1=1 \\ 2x+3y+2=-2 \end{cases} \Leftrightarrow \begin{cases} x+y=2 \\ 2x+3y=-4 \end{cases} \Leftrightarrow (x,y)=(10,-8)$$
$$\bullet \quad \begin{cases} x+y-1=2 \\ 2x+3y+2=-1 \end{cases} \Leftrightarrow \begin{cases} x+y=3 \\ 2x+3y=-3 \end{cases} \Leftrightarrow (x,y)=(12,-9)$$
$$\bullet \quad \begin{cases} x+y-1=-2 \\ 2x+3y+2=1 \end{cases} \Leftrightarrow \begin{cases} x+y=-1 \\ 2x+3y=-1 \end{cases} \Leftrightarrow (x,y)=(-2,1)$$