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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?

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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Mar 15 '12 at 13:21
  • $\begingroup$ I am not sure, but it was a problem from AwesomeMath last year $\endgroup$ – Kirthi Raman Mar 15 '12 at 13:23
  • $\begingroup$ I'm not familiar with AwesomeMath. What tools/theory do they think the solvers should be familiar with? $\endgroup$ – Jyrki Lahtonen Mar 15 '12 at 13:26
  • $\begingroup$ Try using the converse. $\endgroup$ – Hassan Muhammad Mar 15 '12 at 13:36
  • $\begingroup$ See math.stackexchange.com/questions/158722 $\endgroup$ – Watson Nov 25 '18 at 17:35
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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.

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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 $a = b = c = 0$.

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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.

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