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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!).

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  • $\begingroup$ Are you familiar with Pell's equation? You get a solution iff $48x^2 + 1 $ is a perfect square. $\endgroup$
    – Calvin Lin
    Sep 14, 2021 at 23:01
  • $\begingroup$ @CalvinLin Yes, I know the basics of it. I do not see the reduction immediately, could you clarify how it reduces? $\endgroup$
    – Jean Charles
    Sep 14, 2021 at 23:03
  • $\begingroup$ 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 $(0,0)$. $\endgroup$
    – jjagmath
    Sep 14, 2021 at 23:30
  • $\begingroup$ @jjagmath Sure, edited accordingly (my main interest is in $y$ being a positive integer). $\endgroup$
    – Jean Charles
    Sep 14, 2021 at 23:33

2 Answers 2

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  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).
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  • $\begingroup$ 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. $\endgroup$
    – Jean Charles
    Sep 14, 2021 at 23:10
  • $\begingroup$ @jjagmath As $y$ always divides $x$, we always have that $z$ is an integer (so we lose no generality). $\endgroup$
    – Jean Charles
    Sep 14, 2021 at 23:50
  • $\begingroup$ It's not $x=14$, it's $z=14$. This gives $y=25$ and then $x=yz=350$. $\endgroup$
    – Gerry Myerson
    Sep 15, 2021 at 4:11
  • $\begingroup$ @GerryMyerson I meant in CalvinLin's original comment. $\endgroup$
    – Jean Charles
    Sep 15, 2021 at 15:28
  • $\begingroup$ @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. $\endgroup$
    – Calvin Lin
    Sep 15, 2021 at 15:50
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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.

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