The post by Kevin Chung reminded me of an interesting equation, proposed to me by a friend.

**Problem.** Prove that there no non-trivial rational solutions \(a,b,c\) to the equation
\[a+b\sqrt[3]{2}+c\sqrt[3]{4}=0.\]

**Solution.** If we multiplicate original equation by \(\sqrt[3]{2}\), we'll obtain two similar equations. Lets write them all down
\[\begin{cases} a+b\sqrt[3]{2}+c\sqrt[3]{4}=0 \\
a\sqrt[3]{2}+b\sqrt[3]{4}+2c=0 \\
a\sqrt[3]{4}+2b+c\sqrt[3]{2}=0 \end{cases}
\]

Now we may try to think backwards. Could possibly a system of linear equations with rational coefficients give an irrational solution? Applying Cramer's rule we see that it's not the case. However, our system clearly has a solution \((2,\sqrt[3]{2},\sqrt[3]{4})\). Which contradicts the assumption that \((a,b,c)\) are rational numbers (keep in mind that we throw trivial \((0,0,0)\) solution away).

Can you propose any generalizations? What features of the tuple \((1,\sqrt[3]{2},\sqrt[3]{4})\) does this relation imply? Have you noticed how this relation is similar to conditions of linear independence?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestIf I may provide a "better" solution...

We note that we are considering arbitrary elements of \(\mathbb{Q}(\sqrt[3]{2})\). Also note that the minimal polynomial of \(\sqrt[3]{2}\) over \(\mathbb{Q}\) is \(x^3-2\), which has degree 3. Thus, \([\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3\), so we have that \(\mathbb{Q}(\sqrt[3]{2})\) is a vector space over \(\mathbb{Q}\) with basis \(\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}\). Thus, there cannot be a nontrivial solution to \(a+b\sqrt[3]{2}+c\sqrt[3]{4}=0\) because \(\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}\) is linearly independent.

Log in to reply

Bravissimo, that is exactly what I expected -- a larger context! Any ideas how to broaden the result? Maybe some more general equations?

Log in to reply

I don't follow what you're saying. How would you like me to generalize?

Log in to reply

I think , there exist a very simple solution , Since \[ a+b\sqrt[3]{2}+c\sqrt[3]{4}=0 \implies a^3+2b^3+4c^3=6abc \] Also , WLOG we can assume \( a,b,c \in \mathbb{Z} \) Now we must have \(a\) even , this will then give \( b \) even , and further going , by FMI(Fermat method of infinite descent , we get contradiction.

Log in to reply

Indeed very clever and beautiful solution!

Log in to reply

What is Cramer's rule??

Log in to reply

It's one of the methods for solving systems of linear equations.. If you haven't studied matrix before, that'll be touch job for you. =3=

Google for that because I don't know how to explain that easily. T__T

Log in to reply