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
4 years, 4 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

What is Cramer's rule??

- 4 years, 1 month 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

- 3 years, 9 months ago

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.

- 4 years, 4 months ago

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

- 4 years, 4 months ago

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

- 4 years, 4 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.

- 4 years, 4 months ago

Indeed very clever and beautiful solution!

- 4 years, 4 months ago