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# Irrationality and equations

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?

Note by Nicolae Sapoval
3 years, 3 months ago

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If 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. · 3 years, 3 months ago

Bravissimo, that is exactly what I expected -- a larger context! Any ideas how to broaden the result? Maybe some more general equations? · 3 years, 3 months ago

I don't follow what you're saying. How would you like me to generalize? · 3 years, 3 months ago

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. · 3 years, 3 months ago

Indeed very clever and beautiful solution! · 3 years, 3 months ago

What is Cramer's rule?? · 3 years ago

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 · 2 years, 8 months ago