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,ca,b,c to the equation a+b23+c43=0.a+b\sqrt[3]{2}+c\sqrt[3]{4}=0.

Solution. If we multiplicate original equation by 23\sqrt[3]{2}, we'll obtain two similar equations. Lets write them all down {a+b23+c43=0a23+b43+2c=0a43+2b+c23=0\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 (2,23,43)(2,\sqrt[3]{2},\sqrt[3]{4}). Which contradicts the assumption that (a,b,c)(a,b,c) are rational numbers (keep in mind that we throw trivial (0,0,0)(0,0,0) solution away).

Can you propose any generalizations? What features of the tuple (1,23,43)(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
5 years, 11 months ago

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What is Cramer's rule??

Satvik Golechha - 5 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 - 5 years, 4 months ago

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

We note that we are considering arbitrary elements of Q(23)\mathbb{Q}(\sqrt[3]{2}). Also note that the minimal polynomial of 23\sqrt[3]{2} over Q\mathbb{Q} is x32x^3-2, which has degree 3. Thus, [Q(23):Q]=3[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3, so we have that Q(23)\mathbb{Q}(\sqrt[3]{2}) is a vector space over Q\mathbb{Q} with basis {1,23,(23)2}\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}. Thus, there cannot be a nontrivial solution to a+b23+c43=0a+b\sqrt[3]{2}+c\sqrt[3]{4}=0 because {1,23,(23)2}\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\} is linearly independent.

Jacob Erickson - 5 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 - 5 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 - 5 years, 11 months ago

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I think , there exist a very simple solution , Since a+b23+c43=0    a3+2b3+4c3=6abc 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,cZ a,b,c \in \mathbb{Z} Now we must have aa even , this will then give b b even , and further going , by FMI(Fermat method of infinite descent , we get contradiction.

Shivang Jindal - 5 years, 11 months ago

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Indeed very clever and beautiful solution!

Nicolae Sapoval - 5 years, 11 months ago

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