Prove that $3x^2 + 6y^6 + 1 = 8xy^3$ has no solutions in $\mathbb{Q}$ Prove that $3x^2 + 6y^6 + 1 = 8xy^3$ has no solutions in $\mathbb{Q}$.
I know two kind of ways of solving these kind of problems:


*

*look at $\mathbb{R}$

*look at $\mathbb{Z}$, if you can't find any solutions there and $\gcd (x,y) \neq 1$, then $(0,0,0)$ can be your only solution
The problem is that I used 2. only on homogeneous functions so I don't know for sure if one can use it here as well.
I tried using option 1., and I noted that $x$ and $y$ must be strictly positive for if not the right hand side is negative and the left hand side is positive.
Can I get a hint on how to proceed?
Kees
 A: If you let $u = y^3$ (note that cubing is bijective $\mathbb{R} \to \mathbb{R}$), then we have
$$
3x^2-8xu+6u^2 = -1
$$
But the determinant is then $64-72 = -8$ (ETA: Kees Til notes this should be $64u^2-72u^2 = -8u^2$, and so it should), so the left-hand side is always non-negative.  So there are no solutions in $\mathbb{R}$, let alone $\mathbb{Q}$.
A: Here's a way. Treat $x$ as constant, so we may solve for $y$ such that $(x, y)$ is a solution (assuming it should exist). We will show that in fact, for any given real $x$, there is no real $y$ such that $(x, y)$ is a solution. Then we get
\begin{align*}
6 y^{6} - (8x^{3}) y^{3} + (3x^{2} + 1) & = 0 \\
= 6 (y^{3})^{2} - (8x) y^{3} + (3x^{2} + 1) \\
\Rightarrow y^{3} & = \frac{ 8x \pm \sqrt{ 64x^{2} - 24(3x^{2} + 1)}}{12} \\
& = \frac{ 8x \pm \sqrt{ -8x^{2} - 24}}{12} \\
\Rightarrow y & = \left( \frac{ 8x \pm \sqrt{ -8x^{2} - 24}}{12} \right)^{1 / 3} \\
& \not \in \mathbb{R} .
\end{align*}
If $y$ is not real, then clearly it is not rational.
