Mathematica competitions in Croatia

The given equality is equivalent to $$\begin{array}{rl}6x^2{y}^2+3x^2-4y^2& =2012\\ 3x^2(2y^2+1)-2(2y^2+1)& =2012-2\\ (3x^2-2)(2y^2+1)& =2010\end{array}$$ Since $2010=2\cdot 3\cdot 5\cdot 67$ and the number $2y^2+1$ is odd, we conclude that $$\begin{array}{rl}2y^2+1& \in \{1,3,5,15,67,201,335,1005\},\\ 2y^2& \in \{0,2,4,14,66,200,334,1004\},\\ y^2& \in \{0,1,2,7,33,100,167,502\}.\end{array}$$ Thereby $y^2$ must be $0$, $1$ or $100$. There are three cases left: $$\{\begin{array}{l}3x^2-2=2010\\ 2y^2+1=1\end{array}\{\begin{array}{l}3x^2-2=670\\ 2y^2+1=3\end{array}\{\begin{array}{l}3x^2-2=10\\ 2y^2+1=201\end{array}$$ We see that only in the third case there are integer solutions, $$(x,y)\in \{(2,10),(-2,10),(2,-10),(-2,-10)\}.$$