# Does there exist rational $a,b,c$, such that $\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$

Let $w = \sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}$.
How to prove that there are no triples $(a,b,c)$, such that

• $a,b,c \in \mathbb{Q}$;
• $a \leqslant b \leqslant c$;
• $(a,b,c)\ne (1,2,4)$;
• $w = \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$.

Or maybe there exists one?

Everything will follow from the iteration of the following lemma (whose proof is explained at the very end of this answer) :

FUNDAMENTAL LEMMA. Let ${\mathbb K}$ be a field, and let $a$ be a non-cube in ${\mathbb K}$. Then all the “new” cubes in ${\mathbb K}(\sqrt[3]a)$ are made up of $a$ and “old” cubes : they can be written as $w^3,aw^3$ or $a^2(w^3)$ for some $w\in {\mathbb K}$.

COROLLARY 1. Let ${\mathbb K}$ be a subfield of $\mathbb R$, and let $a_1,a_2, \ldots, a_n$ be a sequence of real numbers such that for any $i$, $a_i$ is a non-cube in ${\mathbb K}(\sqrt[3]{a_1},\sqrt[3]{a_2}, \ldots ,\sqrt[3]{a_{i-1}})$. Then all the cubes in ${\mathbb K}(\sqrt[3]{a_1},\sqrt[3]{a_2}, \ldots ,\sqrt[3]{a_{n}})$ can be written as $a_1^{i_1}a_2^{i_2} \ldots a_n^{i_n}(w^3)$ where $w\in {\mathbb K}$ and each $i_k$ is $0,1$ or $2$.

COROLLARY 2. Let $r_1,r_2, \ldots ,r_n$ be positive rational numbers
such that there is no-nontrivial relationship between them of the form $r_1^{i_1}r_2^{i_2} \ldots r_n^{i_n}=w^3$, where each $i_k$ is $0,1$ or $2$, and $w$ is rational (non-trivial means that not all the $i_k$ are zero). Then the $3^n$ algebraic numbers $s_1^{i_1}s_2^{i_2} \ldots s_n^{i_n}$ (where $s_i=\sqrt[3]{r_i}$) are linearly independent over the rationals.

Now, suppose that $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}$. The degree of the extension ${\mathbb Q}(\sqrt[3]{2},\sqrt[3]{4},\sqrt[3]{a},\sqrt[3]{b},\sqrt[3]{c})$ (call it $d$) is a power of three. If it were strictly greater than $3$, then $d\geq 9$ and by corollary 2, any non-trivial relation between cube roots would need at least nine terms, which is too much for the six we've got. So $d=3$, which means that $a,b,c$ must all be, up to sign, (possibly negative) powers of $2$. So there are essentially three cases :

$$\begin{eqnarray} \sqrt[3]{2^i}+\sqrt[3]{2^j}+\sqrt[3]{2^k}&=& \sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4} \\ \sqrt[3]{2^i}+\sqrt[3]{2^j}-\sqrt[3]{2^k}&=& \sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4} \\ \sqrt[3]{2^i}-\sqrt[3]{2^j}-\sqrt[3]{2^k}&=& \sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4} \\ \end{eqnarray}$$ where $i,j,k$ are integers. It is easy now to see then that $i,j,k$ must form a complete system of residues modulo $3$, and to check that $(1,2,4)$ is indeed the only solution to the original problem.

PROOF OF FUNDAMENTAL LEMMA . Let ${\mathbb L}={\mathbb K}(\sqrt[3]a)$. We have $[{\mathbb L}:{\mathbb K}]=3$. Let $c$ be a cube in $\mathbb L$. Then, there are $x_1,x_2,x_3 \in K$ such that $c=(x_1+x_2\sqrt[3]{a}+x_3\sqrt[3]{a^2})^3$. Expanding, we find $c=C_0+3C_1\sqrt[3]a+3C_2\sqrt[3]{a^2}$, with

$$\begin{eqnarray} C_0&=&\bigg(x_1^3 + 6ax_1x_2x_3 + (ax_2^3 + a^2x_3^3)\bigg), \\ C_1&=&\bigg(x_1^2x_2 + ax_1x_3^2 + ax_2^2x_3\bigg), \\ C_2&=&\bigg(x_1^2x_3 + x_1x_2^2 + ax_2x_3^2\bigg) \end{eqnarray}$$

So $C_1$ and $C_2$ must both be $0$, and hence

$$ax_2x_3(x_2^3-ax_3^3)=(x_2^2+x_1x_3)C_1-(ax_3^2+x_1x_2)C_2=0$$

Since $a$ is not a cube in $\mathbb K$, we must have $x_3=0$ or $x_2=0$. If $x_3=0$, $C_2$ simplifies to $x_1x_2^2$, and if $x_2=0$, $C_2$ simplifies to $x_1^2x_3$. So at least two of $x_1,x_2,x_3$ are zero, which means that $x_1+x_2\sqrt[3]{a}+x_3\sqrt[3]{a^2}$ is a multiple (by some element of $\mathbb K$) of $1,\sqrt[3]{a}$, or $\sqrt[3]{a^2}$, as wished.

Denote $t=\sqrt[3]2,w=e^{\frac{2\pi i}{3}}=-\frac{1}{2}+\frac{\sqrt3}{2}i,A=\sqrt[3]a,B=\sqrt[3]b,C=\sqrt[3]c.$

If $A,B,C$ are all $\notin \mathbb{Q},$ rewrite the equation as $$A+B+C-t-t^2=1.\tag1$$

Lemma: Every number $g(\theta)$ of the field $K(\theta)$ is likewise an algebraic number over $k$ of degree at most $n$. The relative conjugates of a number $a=g(\theta)$ are the distinct numbers among the numbers $g(\theta_i)\ (i=1,2,\dots, n)$. Each conjugate to $a$ appears equally often among the $g(\theta_i).$

You can find this lemma in Erich Hecke, Lectures on the theory of algebraic numbers, page 61.

Denote $K=\mathbb{Q}(A,B,C,t)=\mathbb{Q}(\theta)$ for some algebraic number $\theta.$ Assume $[K:\mathbb{Q}]=n=3^r,$ where $r$ is an integer and $1\leq r \leq4.$

Now the conjugates to $A$ with respect to $\mathbb{Q}$ are $A,Aw,Aw^2,$ so do $B,C,t.$ If $A=g_1(\theta),$ and $\theta_i\ (i=1,2,\dots, n)$ are the conjugates to $\theta$ with respect to $\mathbb{Q},$ then $g_1(\theta_i) (i=1,2,\dots, n)$ are the conjugates to $A$ with respect to $\mathbb{Q},$ and each of $A,Aw,Aw^2$ appears equally often among the $g_1(\theta_i),$ namely $3^{r-1}$ times.

Assume that $A+B+C-t-t^2=g_1(\theta)+g_2(\theta)+g_3(\theta)-g_4(\theta)-g_4(\theta)^2=G(\theta).$ If $\theta$ goes over $\theta_i\ (i=1,2,\dots, n)$ then $g_1(\theta)$ goes over $A,Aw,Aw^2,$ and each of them appears $3^{r-1}$ times. So do $B,C,t.$

Since $A,B,C,t$ are all $\notin \mathbb{Q},$we have $$\sum_{i=1}^{n}g_1(\theta_i)=\frac{n}{3}\sum_{i=0}^{2}Aw^i=0.$$ So do $B,C,t.$ Hence we get $$\sum_{i=1}^{n}G(\theta_i)=0.$$ But $$\sum_{i=1}^{n}G(\theta_i)=\sum_{i=1}^{n}1=n,$$ a contradiction. Hence $A,B,C$ are all $\notin \mathbb{Q}$ is impossible. WLOG we assume that $A \in \mathbb{Q}.$ If $B,C$ are all $\notin \mathbb{Q},$ then $$\sum_{i=1}^{n}G(\theta_i)=nA=n, A=1.$$ Now rewrite $(1)$ to $$B+C-t-t^2=0.$$

We get $$B/t+C/t-t/t-t^2/t=0.$$ $$\sqrt[3]{\frac{b}{2}}+\sqrt[3]{\frac{c}{2}}-t=1.$$ For the same reason we get $\dfrac{b}{2}=1$ or $\dfrac{c}{2}=1.$ Now we are done.

This method can easily generalize to other cases. For example, we can prove that $\sqrt[3]{3},\sqrt[5]{23},\sqrt[11]{24}$ are linearly independent in $\mathbb{Q}.$

This isn't a complete answer but I'm curious to see if this strategy could be made to work:

Suppose there existed such a triple. Note that

$$w = \sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4} \in \mathbb{Q}\left[\sqrt[3]{2}\right]=K.$$

Since $a$, $b$, and $c$ are rational, we can pull out cubes and write

$$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=q_{a}\sqrt[3]{a'}+q_{b}\sqrt[3]{b'}+q_{c}\sqrt[3]{c'}$$

where $a'$, $b'$, $c'$ are cube-free integers and $q_{a}$, $q_{b}$ $q_{c}$ $\in \mathbb{Q}$. Thus, $w \in \mathbb{Q}\left[\sqrt[3]{a'},\sqrt[3]{b'},\sqrt[3]{c'}\right]=L$. Since $\mathbb{Q}[w]=\mathbb{Q}[\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}]=\mathbb{Q}[\sqrt[3]{2}]=K$, we see that $L$ extends $K$.

$\Delta_{K}=-108=-2^2\cdot3^3$. I wrote some code to compute $\Delta_{L}$ for various $a'$, $b'$, $c'$ and noticed that the primes dividing $a'$, $b'$, $c'$ always appear in factorization of $\Delta_{L}$ along with an additional factor of at least $3^3$. For instance, if $(a',b',c')=(3,5,7)$ then $\Delta_{L}=-3^{31}\cdot 5^{18} \cdot 7^{18} \cdot 11^{18}$.

• Thank you, @JacksonWalters, for your answer. $\Delta_K$ is discriminant, right? If so, I'd like to know more about discriminants and how its factors are related to this question? Maybe you know some good literature/links? – Oleg567 Jun 17 '13 at 20:32
• Yes, $\Delta_{K}$ is the discriminant of the number field $K$. I think the question is really driving at the linear independence of radicals for the case $n=3$, from which it would follow that $(1,2,4)$ is the unique solution (see this question: math.stackexchange.com/questions/158722/…). To actually prove the linear independence you need to circle back to field theory and Galois theory, which I am only just beginning to learn about. – Jackson Jun 17 '13 at 23:50
• A tiny note (mostly because I needed to convince myself): the reason that $\mathbb{Q}[w]=\mathbb{Q}[\sqrt[3]{2}]$ is that since $w-1=\sqrt[3]{2}+\sqrt[3]{4}$, $(w-1)^2=\sqrt[3]{2}^2+2\sqrt[3]{2}\sqrt[3]{4}+\sqrt[3]{4}^2 = \sqrt[3]{4}+4+2\sqrt[3]{2}$ and so $\sqrt[3]{2} = (w-1)^2-4-(w-1)$, which is clearly polynomial in $w$. – Steven Stadnicki Jun 19 '13 at 5:24
• @StevenStadnicki Or you could say $1+\sqrt[3]{2}+\sqrt[3]{4}=\frac{2-1}{\sqrt[3]{2}-1}$ by geo sum / difference of cubes. – anon Jun 19 '13 at 5:26
• @StevenStadnicki Useful fact to always have on-hand: $K(a)=K[a]$ whenever $a$ is algebraic over a field $K$ and not a zero divisor. Proof: multiplication-by-$a$ is a linear map on the finite-dimensional $K$-vector space $K[a]$, it is injective since $a$ is not a zero divisor, hence (by finite-dimensionality) it is an isomorphism, hence $1\in K[a]$ has a preimage under this map, which must be $a^{-1}$. Alternatively, take the minimal polynomial $p(\cdot)$ for $a$ over $K$, then rearrange the relation $p(w)/w=0$ to get $1/w$ as a polynomial in $w$. – anon Jun 19 '13 at 5:42