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.
Find all pairs $(a,b)$ of positive integers satisfying $6a^2+a=b^2$.
diophantine equationsnumber theory
Best Answer
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 $(a,b)=(400,980)$.