Find all pairs $(a,b)$ of positive integers satisfying $6a^2+a=b^2$.

Question

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.

Answer 1

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 $(a,b)=(400,980)$.

Answer 2

You may solve the equation for a:

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

$$\Delta=1+24b^2=c^2$$

and you get Pell equation:

$$c^2-24b^2=1$$

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

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

$$5a^2+(1-2t)a-t^2=0$$

solve this for a:

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

$$\Delta=24t^2-4t+1=k^2$$

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:

$\begin{cases}k-1=4t\\k+1=6t-1\end{cases}$

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

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