How find this diophantine equation $(3x-1)^2+2=(2y^2-4y)^2+y(2y-1)^2-6y$ integer solution Find this following  Diophantine equation all integer solution
$$(3x-1)^2+2=(2y^2-4y)^2+y(2y-1)^2-6y$$
or
$$9x^2-6x+3=4y^4-12y^3+12y^2-5y$$
Maybe this equation can be solved by using Pell equation methods?
I want take right is Quadratic formula with $(ay^2+by+c)$
maybe we have
$$(3x-1)^2-A(ay^2+by+c)^2=B?$$


You can write it as
$$(3x-1)^2-(2y^2-3y+\tfrac34)^2=-\tfrac12y-\tfrac{41}{16}.$$
Factoring the LHS gives two factors at least one of which gets too large as $y$ is large, as
$$|(3x-1)-(2y^2-3y+\tfrac34)|+|(3x-1)+(2y^2-3y+\tfrac34)|\geqslant2\cdot|2y^2-3y+\tfrac34|.$$
It suffices to check the $y$'s with $-\tfrac12y-\tfrac{41}{16}=0$ (impossible) or $|-\tfrac12y-\tfrac{41}{16}|\geqslant|2y^2-3y+\tfrac34|$, that is, $y\in\{0,1,2\}$.
(Note all this makes a little more sense after denominators are cleared, but it's perfectly valid to act as if they are.)
The only solution is $(1,2)$.
