Finding all rational solutions of $~y^2 = -x(4x^2 - 9x + 4)~$

Question

Finding all rational solutions of $$y^2 = -x(4x^2 - 9x + 4)$$

I need to all of these points. It's easy to notice that $~(0,0),~ (1,1),~ (1,-1)~$ are rational points, though I can't find any others and have a feeling that these are the only rational points.

How would I prove this? I am not very familiar with elliptic curves but would like to know how to solve such an equation.

Answer 1

These are the only ones. If you define $x=-u/4, y=-v/4$, the curve becomes $v^2=u^3+9u^2+16u$. Using elltors in Pari-gp gives 3 finite torsion points which lead to the points you have found. Using Denis Simon's ellrank code, also in Pari, gives rank $0$ which means there are no other points.

Answer 2

Here is an answer using sage code. (Sage uses pari/gp for issues related to elliptic curves.) The equation $y^2=-4x^3+9x^2-4x$ can be rewritten in the form $$ \begin{aligned} \left(\frac y2\right)^2 &= (-x)^3 + \frac 94(-x)^2 + (-x)\ ,\\ &\qquad\text{ so we are looking for rational points on} \\ Y^2 &= X^3+\frac 94X^2+X\ . \end{aligned} $$

sage: E = EllipticCurve( [ 0, 9/4, 0, 1, 0 ] )
sage: E
Elliptic Curve defined by y^2 = x^3 + 9/4*x^2 + x over Rational Field
sage: E.rank()
0
sage: E.torsion_points()
[(-1 : -1/2 : 1), (-1 : 1/2 : 1), (0 : 0 : 1), (0 : 1 : 0)]

The values

$(-1,-1/2)$, $(-1,1/2)$, $(0,0)$

obtained for $(X,Y)$ correspond then to the values

$(1,-1)$, $(1,1)$, $(0,0)$

for the variable $(x,y)$ from the OP.