# Find rational $a,b$ with $a\ne b$ such that $x²-bx-\frac{a³-b³}{3b}=0$ has rational solutions

I was working on the generalization of a geometry problem, which resulted in the quadratic equation

$$x^2 - bx - \frac{a^3 - b^3}{3b}= 0$$

Its solutions are

$$-\frac{b}{2} \pm \sqrt{\frac{4a^3 - b^3}{12b}}$$

I wanted to see if there are any rational values for a and b with a not equal to b such that the quadratic equation has rational solutions (in other words: such that $$\cfrac{4a^3-b^3}{12b}$$ is the square of a rational number). I couldn't find any, so I'm not even sure if there are any.

If there are not, I would want to prove there cannot be such a and b.

By substituting $$x=12a/b$$ (or rather, $$a=bx/12$$) and simplifying, the expression becomes $$\frac{4b^2}{12^4}(x^3-3\cdot 12^2)$$
To make this a rational square, we need to find a rational point on the elliptic curve $$y^2=x^3-3\cdot 12^2$$. The theory of rational points on elliptic curves is huge and complicated, so it is easiest to just let a computer do it.
According to Magma, the Mordell-Weil group (the group of rational points) on this curve is isomorphic to $$Z_3$$, so it has only three rational points. Those points are just the ones we already expect, namely two points with $$x=12$$ (corresponding to $$a=b$$), and a point at infinity. So there are no other rational points.
E:=EllipticCurve([0,0,0,0,-3*144]);