I have already tried treating the equation as a quadratic on a and b, but it doesn't work. I also have plugged in some values. $(6,10)$ is a solution, but I didn't manage to find any other. Are there any general methods to solve equations like this one? Namely $Ax^2+Bx+Cy^2+Dy+E=0$ for integers x and y.

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    $\begingroup$ Use the fact that $a$ and $6a+1$ are coprime. Hence $a=n^2, \ 6a+1=m^2$ and $nm=b$ $\endgroup$
    – RFZ
    Commented Apr 16, 2022 at 19:45
  • $\begingroup$ For example, $a=400$, $b=980$ is a solution. $\endgroup$
    – rogerl
    Commented Apr 16, 2022 at 19:45
  • $\begingroup$ @ZFR that nice idea leads to the Pell equation $m^2-6n^2=1$ which ends up similar to my posted solution. $\endgroup$ Commented Apr 16, 2022 at 19:48
  • $\begingroup$ @GregMartin, I did not notice your answer. Sorry! I'll take a look. $\endgroup$
    – RFZ
    Commented Apr 16, 2022 at 19:48
  • $\begingroup$ no worries, your comment won by 1 minute :) $\endgroup$ Commented Apr 16, 2022 at 19:50

2 Answers 2


The equation is equivalent to $144a^2+24a=24b^2$, and thus to $(12a+1)^2-24b^2=1$, which is a special case of a Pell equation $x^2-24b^2=1$ where we want only the solutions with $x\equiv1\pmod{12}$. The trivial solution $(x,b)=(1,0)$ corresponds to $(a,b)=(1,0)$; the fundamental solution $(x,b)=(49,10)$ corresponds to the solution $(a,b)=(4,10)$ (not $(6,10)$); and infinitely many solutions can be found by calculating $(49+10\sqrt{24})^n = x_n + y_n\sqrt{24}$. For example, with $n=2$, we get the solution $(x_2,y_2) = (4801,980)$, corresponding to the solution $(a,b)=(400,980)$.

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    $\begingroup$ I like to throw in $b_{n+2} = 98 b_{n+1} - b_n$ and $a_{n+2} = 98 a_{n+1} - a_n + 8$ $\endgroup$
    – Will Jagy
    Commented Apr 16, 2022 at 22:08

You may solve the equation for a:

$$a=\frac{-1\pm\sqrt{1+24 b^2}}{12}$$


and you get Pell equation:


Which its solution is well expressed in other answer. You can also do this:

Let $b=a+t$, putting in equation you get:


solve this for a:

$$a=\frac{-(1-2t)\pm \sqrt{24t^2-4t+1}}{10}$$


Or: $$k^2-24t^2=1-4t$$

which is a pell like equation and has infinite solutions and gives rational and integer solution for t, a and b, for example:


One solution is $(t, a, b)=(\frac 32, \frac 9{10}, \frac {24}{10} )$

another solution is $(t, a, b)=(580, 400, 980)$.


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