$x\in\mathbb{Q}$ where $12(x+1/x)+42$ is a rational square

Question

I am finding rational $x$s where $12(x+1/x)+42$ is a rational square. My attempts are as follows.
If $x=R^2$, then there is no solution by 1.
Let $x=m/n, m,n\in\mathbb{Z}$. This induces the equation $12(m^2+n^2)+42mn=mnz^2$ where $z\in\mathbb{Q}$. Letting $z=p/q, p,q\in\mathbb{Z}$ gives $12p^2(m^2+n^2)+42p^2mn=mnq^2$. This is related to the Diophantine equation $(12p^2)a+(42p^2)b=mnq^2$. Since $GCD(12p^2, 42p^2)=6p^2$, this has solutions $(a,b)$s if $6p^2|mnq^2$. I cannot proceed any further after this step.
Another approach may use 2 but this is also difficult because of the existence of $mn$ terms in front of $z^2$. Can anyone help?

Answer 1

This is a typical elliptic curve problem: firstly homogenize the equation:

$$12(x+1/x)+42=y^2\rightarrow12(x+z^2/x)+42=y^2/z\rightarrow xy^2=12z^3+42z^2+12z$$

Then let $x=1,y^{\prime}=12y,z^{\prime}=12z$, we get: $${y^{\prime}}^2 = {z^{\prime}}^3+42{z^{\prime}}^2+12z^{\prime}$$

Then with the following Sage code:

var('x,y')
E=EllipticCurve(y^2 == 12*12*x+42*x^2 + x^3)
E.gens()

The elliptic curve has rank 1 and generator $(-18 : 72 : 1)$. So you get your first solution $x=-2/3,y=-4$, other solutions are generated by adding the generator.