# 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!).

• Are you familiar with Pell's equation? You get a solution iff $48x^2 + 1$ is a perfect square. Sep 14, 2021 at 23:01
• @CalvinLin Yes, I know the basics of it. I do not see the reduction immediately, could you clarify how it reduces? Sep 14, 2021 at 23:03
• You stated that you wanted solutions in the positive integers. Then you should exclude $(0,1)$. If you allow $0$, then the smallest is not $(0,1)$, it is $(0,0)$. Sep 14, 2021 at 23:30
• @jjagmath Sure, edited accordingly (my main interest is in $y$ being a positive integer). Sep 14, 2021 at 23:33

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 $$n = 4, 4y-3 = 18817, z = 2716, y = 4705, x = 12778780$$.

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).
• Just to be clear regarding your original comment, with $x=14$ we have that $48x^2+1$ is a perfect square (being $97^2$). But I see no solution to my original equation with $x=14$. I am also unsure about your final step ("check for integer-ness"), and why this should be trivial. Sep 14, 2021 at 23:10
• @jjagmath As $y$ always divides $x$, we always have that $z$ is an integer (so we lose no generality). Sep 14, 2021 at 23:50
• It's not $x=14$, it's $z=14$. This gives $y=25$ and then $x=yz=350$. Sep 15, 2021 at 4:11
• @GerryMyerson I meant in CalvinLin's original comment. Sep 15, 2021 at 15:28
• @JeanCharles A) In my original comment, I did the "obvious" (to me) change of variables to simplify the equation. There was an abuse of notation, where I set $X = y/x$. So yes, $z = 14$ gives $x = 350$. B) The "check for integerness" is looking at $y = 1/4 \times \ldots$, and we just had to ensure the numerator was a multiple of 4 (and had to exclude some cases). $\quad$ EG As an alternative to step 2, we could have written it as $48 z^2 +1 = (4y-3)^2$ directly. I've done this as a separate writeup, and also added the case checking. Sep 15, 2021 at 15:50

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.