Finding all solutions to quartic Diophantine equation Consider the Diophantine equation $6x^2 = y^2(2y-1)(y-1)$. I am interested in finding all solutions to this such that $y$ is a positive integer -- or at the very least knowing whether there are infinitely many. Certainly there are some; the smallest being $(x,y) = (0,1)$, and then (less trivially) $(x,y) = (350,25)$. Generating more is possible.
This Diophantine equation arises in my research on certain determinants; I am not fluent enough in this area to see any immediate ways this question might be resolved. I had a look in Mordell, but could not see anything that could be shaped to give any direct answer (though perhaps I am wrong!).

*

*Using the change of variables $ z = \frac{x}{y}$, we have $ 6z^2 = (2y-1)(y-1)$.

*Completing the Square on RHS gives us $ 48z^2+1 = (4y-3)^2 $.
This is a Pell's equation with solutions $$ z = \frac{ ( 7+ 4\sqrt{3}
    )^n - ( 7 - 4 \sqrt{3} ) ^n } { 8 \sqrt{3} }, 4y-3 =
    \frac{1}{2} [ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n] .$$

*Show that $y$ is an integer iff $n $ is even.

*Hence, conclude that the solutions are .... (Just backtrack. I'm too lazy to paste the expressions.)
The first non-trivial solution is $ n = 2, 4y-3 = 97, z = 14, y = 25, x = 350$.
The next solution is $ n = 4, 4y-3 = 18817, z = 2716, y
    = 4705, x = 12778780$.


Previous version:

*

*Using the change of variables $ z = \frac{x}{y}$, we have $ 6z^2 = (2y-1)(y-1)$.

*Treating this as a quadratic in $y$, we have $y = \frac{1}{4} ( 3 \pm \sqrt{ 48z^2 + 1 } )$. We have a solution if the expression is a perfect square.

*Finally, Pell's equation on $a^2 = 48z^2 + 1$ gives us $ z = \frac{ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n } { 8 \sqrt{3} }$ and $ a = \frac{1}{2} [ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n] $

*It remains to backtrack (and check for positivity / integer-ness).

