Rational solutions of the equation $Y^3=X^3-5 X^2+8 X-4$

Question

Source: Exercise from algebraic structures class.

The question is:

a) Find all the rational solutions of the equation $$ Y^3=X^3-5 X^2+8 X-4 $$ b) Find all the integer solutions of that equation.

I've tryied: To find a particular solution $(x_0,y_0)$ and then, trying to find a parametrization of the conic obtained by intersecting it with a "pencil" of lines going through the point $(x_0,y_0)$. But this seems difficult to me and also it is a geometrical approach instead of a purely algebraic approach expected from an algebra class.

Thank you in advance :)

Answer 1

Let $W=\frac Y{X-2}$, then since the cubic on the right factorises as $(X-2)^2(X-1)$ we have $$W^2\cdot W(X-2)=X-1$$ $$W^3=\frac{X-1}{X-2}$$ $$X=\frac1{W^3-1}+2$$ Thus (almost) all rational points can be obtained by choosing an arbitrary rational $W$, computing $X$ using the last line above and then computing $Y=W(X-2)$. The parametrisation is $$(X,Y)=\left(\frac1{W^3-1}+2,\frac W{W^3-1}\right)$$ In fact, with rational $W$ the only point not covered is the "point at infinity" – at $W=\infty$ the cusp at $(2,0)$ is obtained.

For integer points we require $W^3-1=1/k$ for some integer $k$ from $X$, whence $W$ must necessarily be rational or at infinity from $Y$ – but then $W^3=\frac{k+1}k$, so $k$ and $k+1$ must be perfect integer cubes. This is only true for $k=-1$ and $k=0$, giving the only integer points on the curve as $(1,0)$ for $W=0$ and $(2,0)$ for $W=\infty$.