Finding all solutions to quartic Diophantine equation

Question

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!).

Answer 1

1. Using the change of variables $ z = \frac{x}{y}$, we have $ 6z^2 = (2y-1)(y-1)$.
2. Completing the Square on RHS gives us $ 48z^2+1 = (4y-3)^2 $.
   This is a Pell's equation with solutions $$ z = \frac{ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n } { 8 \sqrt{3} }, 4y-3 = \frac{1}{2} [ ( 7+ 4\sqrt{3} )^n - ( 7 - 4 \sqrt{3} ) ^n] .$$
3. Show that $y$ is an integer iff $n $ is even.
4. 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$.

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Previous version:

1. Using the change of variables $ z = \frac{x}{y}$, we have $ 6z^2 = (2y-1)(y-1)$.
2. 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.
3. 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] $
4. It remains to backtrack (and check for positivity / integer-ness).

Answer 2

HINT.- It seems there are infinitely many solutions.

$2y-1$ and $y-1$ are coprimes then $2y-1=au^2$ and $y-1=bv^2$ where $ab=6$. Trying with distinct possible values for $a$ and $b$ we get that the system $$2y-1=u^2\\y-1=6v^2$$ solved by Pell-Fermat equations give, in particular the solutions $$(u,v)=(7,2),(97,28),(1351,390),(18817,5432),(262087,75658),(3650401,1053780),(708158977,204427888)$$ and many other solutions.