Hellenic Mathematical Olympiad

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

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Alternative solution.

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