# Given that $a+b\sqrt[3]{2} +c\sqrt[3]{4} =0$, where $a,b,c$ are integers. Show $a=b=c=0$

Given that $\displaystyle{a+b\sqrt[3]{2} +c\sqrt[3]{4} =0}$, where $a,b,c$ are integers. Show $a=b=c=0$

Do I use modular arithmetic?

-
Do you know what a basis of a field extension is? Can you think of a field extension (with a known basis) that would be useful in your problem? Modular arithmetic won't help you here. – Jyrki Lahtonen Mar 15 '12 at 13:21
I am not sure, but it was a problem from AwesomeMath last year – Kirthi Raman Mar 15 '12 at 13:23
I'm not familiar with AwesomeMath. What tools/theory do they think the solvers should be familiar with? – Jyrki Lahtonen Mar 15 '12 at 13:26
Try using the converse. – Hassan Muhammad Mar 15 '12 at 13:36

Hint $\$ If so then $\rm\:x = \sqrt[3]{2}\:$ would be a root of $\rm\:f = a+b\:x+c\:x^2\:$ and $\rm\: g = x^3-2\:$ so also a root of their gcd $\rm = e f + h g\:$ (by Bezout), contra $\rm\:x^3-2\:$ is irreducible over $\rm\:\mathbb Q\:$ by the rational root test.

Alternatively, if $\rm\:w = \sqrt[3]{2}\:$ is a nonrational root of a quadratic then there exists a conjugation automorphism $\rm\:x\mapsto x'\:$ on $\rm\mathbb Q(w)\:$ with fixed field $\rm \mathbb Q,\:$ so taking the norm $\rm\:xx'$ of $\rm\:w^3 = 2\:$ yields $\rm\:(ww')^3 = 4\:$ for $\rm\:ww'\in \mathbb Q,\:$ contradiction.

-

Start with $b \sqrt[3]{2} + c \sqrt[3]{4} = -a$. Cube this relation to find another equation of the form $B \sqrt[3]{2} + C \sqrt[3]{4} = -A$ for rationals A, B, C. Eliminating the cube root of 4 from these equations will tell you that the cube root of 2 is rational. This contradiction shows that $a = b = c = 0$.

-

By standard results in field theory, given the element $\alpha$ a root of the polynomial $x^3 - 2$, $\alpha$ by definition is algebraic over $\Bbb{Q}$ so that $[\Bbb{Q}(\alpha):\Bbb{Q}]$ is finite (in particular equal to the degree of $x^3 - 2$ that is three). Viewing $\Bbb{Q}(\alpha)$ as a vector space of dimension three over $\Bbb{Q}$, there is the usual basis for $\Bbb{Q}(\alpha)$ given by $1, \alpha, \alpha^2$. It is not hard to prove that this is a basis using the division algorithm and the fact that the kernel of the evaluation map $\Bbb{Q}[x] \longrightarrow \Bbb{Q}(\alpha)$ is the principal ideal $(x^3 - 2)$. Taking $\alpha = \sqrt[3]{2}$ shows that the numbers $1, \sqrt[3]{2}, (\sqrt[3]{2})^2$ are linearly independent over $\Bbb{Q}$, so that in particular the only solution in integers to the equation

$$a + b\sqrt[3]{2} + c(\sqrt[3]{2})^2= 0$$

is the trivial solution.

-

you need at least 3 equations to solve 3 variables

-
An example $x^y-y^x=x+y$ is a Diophantine Equation. I believe it should be using modular arithmetic – Kirthi Raman Mar 15 '12 at 13:22
Depends on the domain of the equation and the variables. A single equation of vectors of a three dimensional space may allow you to solve three scalars. – Jyrki Lahtonen Mar 15 '12 at 13:23
The equation $x^2 + y^2 + z^2 = 0$ has exactly one solution for $x, y, z \in \mathbb R$. Your statement is only true for linear equations. – Johannes Kloos Mar 15 '12 at 14:05
@user26971 No, $x^y-y^x=x+y$ is not a Diophantine equation, it is an exponential Diophantine equation. Diophantine equations always have constants in the exponent. – Thomas Andrews Mar 15 '12 at 14:45