Equilateral triangle inscribed in parabola

Question

Count how many parabolas with integer coeficients, $y=ax^2+bx+c$ can have an equilateral triangle with vertices on the intersection of the parabola and on the x and y axis.

Answer 1

From your comments, you got $b=0$. The position of the three points are then $$\left(-\sqrt{-\frac ca},0\right),\left(+\sqrt{-\frac ca},0\right),(0,c)$$ Then the distance between the intersections with $x$ axis is $$d=2\sqrt{-\frac ca}$$ The distance between one of the intersection with $x$ axis and the intersection with $y$ axis is $$d=\sqrt{\left(\sqrt{-\frac ca}\right)^2+c^2}$$ Equating the squares of these distances we get $$c^2=-3\frac ca$$ Obviously $c=0$ is not a good solution, since the equilateral triangle will reduce to the origin. Then $$a=-\frac 3c$$ Since both $a$ and $c$ are integers, it means that the possible $c$ values $\pm1,\pm3$ and the corresponding $a$ values are $\mp3,\mp1$.