Show that $x$, $y$, $z$ are integers when $3x$, $3x^2-6yz$, $x^3+2y^3+4z^3-6xyz$ are integers.

Question

I was trying to show that $\{1, \alpha, \alpha^2\}$ is a integral basis of $\mathbb{Q}(\alpha)$ where $\alpha= \sqrt[3]{2}$.

And after some steps it remains to prove that if $$3x, \quad 3x^2-6yz, \quad x^3+2y^3+4z^3-6xyz \quad $$ are integers and $x$, $y$, $z$ $\in \mathbb{Q}$ then $x$, $y$, $z$ are integers. And I know that the denominators of $x$, $y$, $z$ are divisors of $6$. I managed to prove some facts but nothing that really helps to conclude.

Any easy ways to solve this one?

Yeah exactly, my mistake. But it is kinda obvious if you think of the original task. — Harto Saarinen, Nov 17 '14 at 15:50
See here or here for more discussion. And this is in Murty & Esmonde, which you can borrow from a teacher near you. — Jyrki Lahtonen, Nov 17 '14 at 16:07

score 0 · Answer 1 · answered Nov 17 '14 at 15:25

0

Your assertion seems untrue.

Let $x=1$, $yz=1$.

Let $t+\frac{2}{t}=4$ and $y^3=t, z^3=\frac{1}{t}$.

A counter-example.

answered Nov 17 '14 at 15:25

Ma Ming

7,482

I think when OP said "the denominators of $x,y,z$ are divisors of $6$, it implied that $x, y,$ and $z$ are supposed to be rational, which your value of $t$ is not. (But I really like your example!) – John Hughes Nov 17 '14 at 15:28
Yeah, the numbers should be rational. – Harto Saarinen Nov 17 '14 at 15:56

Show that $x$, $y$, $z$ are integers when $3x$, $3x^2-6yz$, $x^3+2y^3+4z^3-6xyz$ are integers.

1 Answers1