Showing that the Diophantine equation $3x^2 + 6y^6 + 1 = 8xy^3$ has no solutions $x,y \in \mathbb{Q}$

Question

I want to show that the Diophantine equation $3x^2 + 6y^6 + 1 = 8xy^3$ has no solutions $x,y \in \mathbb{Q}$.

I tried factoring, but didn't manage (but I'm not good at factoring). Then I tried reducing $\mod{7}$, but this didn't gave decisive results.

Answer 1

Rewriting it as $6y^6-8xy^3+3x^2+1=0$, we have a quadratic equation in $y^3$, so that

$$y^3={4x\pm\sqrt{(4x)^2-6(3x^2+1)}\over6}={4x\pm\sqrt{-(2x^2+6)}\over6}$$

The square root is imaginary for all real $x$, so there are not only no rational solutions, there aren't any real ones either.

Alternatively, it's a quadratic in $x$, with solution

$$x={4y^3\pm\sqrt{(4y^3)^2-3(6y^6+1)}\over3}={4y^3\pm\sqrt{-(2y^6+3)}\over3}$$

for which the square root is imaginary for real $y$. (For some reason I noticed the equation as a quadratic in $y^3$ first!)

Answer 2

$$ 3 u^2 - 8uv + 6 v^2 = \frac{1}{3} (3u-4v)^2 + \frac{2}{3} v^2 $$ is positive definite for real $u,v$

$$ Q^T D Q = H $$ $$\left( \begin{array}{rr} 1 & 0 \\ - \frac{ 4 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rr} 3 & 0 \\ 0 & \frac{ 2 }{ 3 } \\ \end{array} \right) \left( \begin{array}{rr} 1 & - \frac{ 4 }{ 3 } \\ 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rr} 3 & - 4 \\ - 4 & 6 \\ \end{array} \right) $$

Answer 3

Let the pair $(a, b)$ soluation for that equation $$ \\6\sqrt{2}>8=> \\8ab^3=3a^2+6b^6+1\ge2\sqrt{18a^2b^6}+1>6\sqrt{2}|ab^3|\ge8|ab^3|\ge8ab^3=> \\8ab^3>8ab^3 $$ A contradiction.