# Task A-1.1.

Determine all pairs $(x, y)$ of integers for which

$$
6 x^{2} y^{2}-4 y^{2}=2012-3 x^{2}
$$
## Solution.

The given equation is equivalent to $s$

$$
\begin{gathered}
6 x^{2} y^{2}-4 y^{2}=2012-3 x^{2} \\
6 x^{2} y^{2}+3 x^{2}-4 y^{2}=2012 \\
3 x^{2}\left(2 y^{2}+1\right)-4 y^{2}=2012 \\
3 x^{2}\left(2 y^{2}+1\right)-2\left(2 y^{2}+1\right)=2012-2 \\
\left(3 x^{2}-2\right)\left(2 y^{2}+1\right)=2010
\end{gathered}
$$

Since $2010=2 \cdot 3 \cdot 5 \cdot 67$, and $2 y^{2}+1$ is an odd natural number, we conclude

$$
\begin{aligned}
2 y^{2}+1 & \in\{1,3,5,15,67,201,335,1005\} \\
2 y^{2} & \in\{0,2,4,14,66,200,334,1004\} \\
y^{2} & \in\{0,1,2,7,33,100,167,502\}
\end{aligned}
$$

Therefore, $y^{2} \in\{0,1,100\}$.

We are left with three cases:

$$
\begin{aligned}
& \left\{\begin{array} { c } 
{ 3 x ^ { 2 } - 2 = 2 0 1 0 } \\
{ 2 y ^ { 2 } + 1 = 1 }
\end{array} \quad \left\{\begin{array} { c } 
{ 3 x ^ { 2 } - 2 = 6 7 0 } \\
{ 2 y ^ { 2 } + 1 = 3 }
\end{array} \quad \left\{\begin{array}{c}
3 x^{2}-2=10 \\
2 y^{2}+1=201
\end{array}\right.\right.\right. \\
& \left\{\begin{array} { c } 
{ 3 x ^ { 2 } = 2 0 1 2 } \\
{ y ^ { 2 } = 0 }
\end{array} \quad \left\{\begin{array} { c } 
{ 3 x ^ { 2 } = 6 7 2 } \\
{ y ^ { 2 } = 1 }
\end{array} \quad \left\{\begin{array}{l}
3 x^{2}=12 \\
y^{2}=100
\end{array}\right.\right.\right. \\
& \left\{\begin{array} { c } 
{ x ^ { 2 } = \frac { 2 0 1 2 } { 3 } } \\
{ y ^ { 2 } = 0 }
\end{array} \quad \left\{\begin{array} { c } 
{ x ^ { 2 } = 2 2 4 } \\
{ y ^ { 2 } = 1 }
\end{array} \quad \left\{\begin{array}{c}
x^{2}=4 \\
y^{2}=100
\end{array}\right.\right.\right.
\end{aligned}
$$

We see that only in the third case there are integer solutions.

The given equation has four solutions:

$$
(x, y) \in\{(2,10),(-2,10),(2,-10),(-2,-10)\}
$$