X(в квадрате) - 4ху + 3у(в квадрате) = 3 задание: решить в целых числах

$x^2- 4xy + 3y^2 = 3, \\ x^2-4xy+4y^2-y^2=3, \\ (x-2y)^2-y^2=3, \\ (x-3y)(x-y)=3, \\ 3=1\cdot3, 3=-3\cdot(-1), \\ \left \{ {{x-3y=1,} \atop {x-y=3;}} \right. \left \{ {{-2y=-2,} \atop {x=y+3;}} \right. \left \{ {{y=1,} \atop {x=4;}} \right. \\ \\ \left \{ {{x-3y=3,} \atop {x-y=1;}} \right. \left \{ {{-2y=2,} \atop {x=y+1;}} \right. \left \{ {{y=-1,} \atop {x=0;}} \right. \\$

$\left \{ {{x-3y=-3,} \atop {x-y=-1;}} \right. \left \{ {{-2y=-2,} \atop {x=y-1;}} \right. \left \{ {{y=1,} \atop {x=0;}} \right. \\ \\ \left \{ {{x-3y=-1,} \atop {x-y=-3;}} \right. \left \{ {{-2y=2,} \atop {x=y-3;}} \right. \left \{ {{y=-1,} \atop {x=-4;}} \right.$
(4;1), (-4;-1), (0;1), (0;-1).