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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?

Kirthi Raman
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5 Answers5

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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.

Bill Dubuque
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  • Alternatively, if $x$ is a root of both $f$ and $g$, then $x$ is a root of $g \bmod f$, a polynomial of degree at most $1$. But then $x$ would be rational. – lhf Jul 05 '22 at 18:44
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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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A decade late but here's a nice elementary proof:

Take $x = a$, $y = \sqrt[3]{2}b$ and $z = \sqrt[3]{4}c$

Note the identity $$x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x + y + z)((x - y)^2 + (y - z)^2 + (x - z)^2)$$ If $x = y = z$, we have $a = b = c = 0$. Otherwise $x + y + z = 0 \iff x^3 + y^3 + z^3 - 3xyz = 0$

Substitute back to obtain $$\iff a^3 + 2b^3 + 4c^3 - 6abc = 0$$

Consider a solution $(a_0, b_0, c_0) \not = (0, 0, 0)$, take modulo $2$ and note $a_0 = 2a_1$ for some integer $a_1$.

Then, $$8a_1^3 + 2b_0^3 + 4c_0^3 - 12a_1b_0c_0 = 0$$

Divide by 2 to obtain $$b_0^3 + 2c_0^3 + 4a_1^3 - 6b_0c_0a_1 = 0$$

So $(b_0, c_0, a_1)$ is also a solution! We can keep applying this transformation 2 more times to get $(a_1, b_1, c_1)$ is also a solution, where similarly $(b_1, c_1) = (2b_0, 2c_0)$. However this means that we can divide factors of 2 out of our original solution an arbitrary amount of times, contradiction.

cranberry128
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  • This is lovely, worth the 10 year wait! Seems to be working with any $d\ne 1$, square free, instead of $d=2$ ? – orangeskid Jul 05 '22 at 18:46
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    @orangeskid Yup! – cranberry128 Jul 05 '22 at 22:29
  • Your idea can do it for $d$ not a cube. Write $d = d_1 \cdot d_2^2 \cdot d_3^3$ ( extract the cubic part, and some remainders). Works for any other coefficients relatively prime to $d$ instead of $1$ ( and any coefficient instead of $-3$). Or maybe simpler. Why not write that up? You could come up with a criterion for a homogenous diophantine to have no non-zero solutions. – orangeskid Jul 05 '22 at 23:31
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Let $\alpha=\sqrt[3]{2}$

Then $m_{\alpha}({x})=x^3-2$ is the minimal polynomial of $\alpha$ over $\Bbb{Q}$.

$f(x)=a+bx+cx^2$ is an anhilating polynomial of $\sqrt[3]{2}$ as $f(\alpha) =0$

Hence $m_{\alpha}(x) \mid f(x) $

$x^3-2\mid a+bx+cx^2$ implies $a=b=c=0$

SoG
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