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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, 11 months ago

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What is Cramer's rule??

Satvik Golechha - 3 years, 8 months ago

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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

Samuraiwarm Tsunayoshi - 3 years, 4 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.

Jacob Erickson - 3 years, 11 months ago

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

Nicolae Sapoval - 3 years, 11 months ago

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

Jacob Erickson - 3 years, 11 months ago

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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.

Shivang Jindal - 3 years, 11 months ago

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Indeed very clever and beautiful solution!

Nicolae Sapoval - 3 years, 11 months ago

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