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

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.

• I think your problem statement is missing something... what do you want to show about the Diophantine equation? – RandomMathGuy Dec 7 '18 at 20:36
• I fixed the question with what I thought would be the correct question. Please immediately edit it if my guess was wrong. – Batominovski Dec 7 '18 at 20:46
• You were right in the modification, thanks! Indeed, I wanted to show that there are no solutions. – Jens Wagemaker Dec 7 '18 at 20:47

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!)

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

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

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.

• What is $AM$ and $GM$? – Jens Wagemaker Dec 7 '18 at 22:05
• Arifmetic and Geometric mean. – Samvel Safaryan Dec 7 '18 at 22:05
• How did you come up with these estimates? – Jens Wagemaker Dec 7 '18 at 22:07
• if $a, b\ge 0$ then $a+b\ge2\sqrt{ab}$ – Samvel Safaryan Dec 7 '18 at 22:09