Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer

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 $a = b = c = 0$.

share|cite|improve this answer

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.

share|cite|improve this answer

you need at least 3 equations to solve 3 variables

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.