Homework Help: Trying to show that there are no rational solutions for an equation

1. Sep 3, 2013

1. The problem statement, all variables and given/known data
I have the equation
$$a^3+2b^3+4c^3-6abc=0$$
and I want to show that for $a,b,c\in\mathbb{Q}$ and $a,b,c\not=0$ that there are no solutions.

3. The attempt at a solution
I tried many different approaches but this was the only one that seemed to work. I am not sure if this is even allowed.

Holding $a,c$ constant, I first differentiate the equation in terms of $b$. Then I have
$$6b^2=6ac\Leftrightarrow b=\sqrt{ac}.$$
Now I differentiate the original equation in terms of $a$ holding $b,c$ constant so
$$3a^2=6bc\Leftrightarrow a^2=2bc\Leftrightarrow a=\sqrt{2bc}\Leftrightarrow a=\sqrt{2c\sqrt{ac}}\Leftrightarrow a^4=4c^3a\Leftrightarrow a^3=4c^3.$$
Plugging this back into the original equation gives me
$$a^3+2b^3+4c^3=2b^3+8c^3=6abc.$$
Differentiating again in terms of $b$ holding $c$ constant gives me
$$b^2=ac\Leftrightarrow b^2=c\sqrt{2bc}\Leftrightarrow b^4=2bc^3\Leftrightarrow b^3=2c^3.$$
Plugging this back in gives
$$2b^3+8c^3=12c^3=6abc$$
so
$$12c^3=6abc\rightarrow 2c^2=ab\Leftrightarrow 2c^2=8c^6\Leftrightarrow c^4=\frac{1}{4}\Leftrightarrow c=\frac{1}{\sqrt[4]{4}}.$$
Now $\sqrt[4]{4}$ is irrational so $\frac{1}{\sqrt[4]{4}}$ is irrational and hence if $a,b,c\not=0$ there are no solutions.

I might also have to give more justification to the $\sqrt[4]{4}$ part as well but is my main idea ok? Am I allowed to differentiate like that?

2. Sep 3, 2013

Dick

No, you aren't allowed to differentiate like that. Why would you be? You are trying to show there aren't any solutions. Start it this way. Can you show that if there is a rational solution then there must also be a solution where a, b and c are all integers? Now start using divisibility arguments.

3. Sep 3, 2013

Here is an attempt.

Let
$$a=\frac{a_{1}}{a_{2}}, b=\frac{b_{1}}{b_{2}}, c=\frac{c_{1}}{c_{2}}$$
where $a_{1},a_{2},b_{1},b_{2},c_{1},c_{2}$ are integers in lowest terms. Then
$$a^3+2b^3+4c^3-6abc=0\Leftrightarrow \left(\frac{a_{1}}{a_{2}}\right)^{3}+2\left(\frac{b_{1}}{b_{2}}\right)^{3}+4\left(\frac{c_{1}}{c_{2}}\right)-6\frac{a_{1}}{a_{2}}\frac{b_{1}}{b_{2}}\frac{c_{1}}{c_{2}}=0.$$
Multiplying through by $a_{2}^{3},b_{2}^{3},c_{2}^{3}$ gives us
$$(a_{1}b_{2}c_{2})^{3}+2(b_{1}a_{2}c_{2})^{3}+4(c_{1}a_{2}b_{2})^{3}-6a_{1}a_{2}^{2}b_{1}b_{2}^{2}c_{1}c_{2}^{2}=0.$$

I feel like this is close to the part where rational solutions implies integer solutions because we have now $a=a_{1}b_{2}c_{2}$ which are integers. We have the same with $b,c$. I am not seeing how I would apply divisibility arguments here.

4. Sep 3, 2013

Dick

You are getting close to starting. Hint: if a, b and c are rational, then they can be expressed as integers over a common denominator. Nothing has to be in lowest terms here. As for divisibility, start by proving if a, b and c are integers, then a must be even. Then use that to prove more stuff.

Last edited: Sep 3, 2013
5. Sep 4, 2013

We have
$$\frac{(a_{1}b_{2}c_{2})^{3}+2(b_{1}a_{2}c_{2})^{3}+4(c_{1}a_{2}b_{2})^{3}}{(a_{2}b_{2}c_{2})^{2}}=6a_{1}b_{1}c_{1}.$$
Since $a_{1},b_{1},c_{1}$ are integers, $6a_{1}b_{1}c_{1}$ must also be an integer. Hence our left hand side must be an integer. Then
$$(a_{1}b_{2}c_{2})^{3}+2(b_{1}a_{2}c_{2})^{3}+4(c_{1}a_{2}b_{2})^{3}=6a_{1}b_{1}c_{1}(a_{2}b_{2}c_{2})^{2}$$
so
$a_{2}^{2}$ divides $(a_{1}b_{2}c_{2})^{3}+2(b_{1}a_{2}c_{2})^{3}+4(c_{1}a_{2}b_{2})^{3}$ and similarly for $b_{2},c_{2}$. I deduce that
$a_{2}^{2}$ divides $a_{1}^{3} ,b_{2}^{2}$ divides $b_{1}^{3}, c_{2}^{2}$ divides $c_{1}^{3}$.

I worked a little ahead of that but it didn't seem to get anywhere because I ended up with something like
$$\lambda_{1}b_{2}c_{2}+2\lambda_{2}a_{2}c_{2}+4\lambda_{3}a_{2}b_{2}=6a_{1}b_{1}c_{1}$$
where the $\lambda$'s are from $a_{2}^{2}\lambda=a_{1}^{3}$.

Just saw the part about divisibility. Going at it again!

Last edited: Sep 4, 2013
6. Sep 4, 2013

Dick

I'm not sure where you are going here. Maybe you are concentrating more on the TeX than the problem. I meant you can write a=d/n, b=e/n and c=f/n with d, e, f and n integers. Doesn't that mean d^3+2e^3+4f^3-6def=0? So your original equation must also have an integer solution if it has a rational solution.

Last edited: Sep 4, 2013
7. Sep 4, 2013

I end up being able to show that all three integers are even. I am not sure how to proceed from there. After putting all a,b,c in the form 2a,2b,2c, the equation doesn't seem to be changing to anything useable.

It does seem to me that no matter how I manipulate this function with the divisibility of 2, the integers still always seems to be divisible by 2. Can we use that in any way with the fact that any even integer has to become odd sooner or later after being divided by 2 over and over again?

**Edit: I answered my own question.. 0 is an integer that this doesn't work for.

8. Sep 4, 2013

Ok. Here is something else I tried to do.
We have that 2 divides a so for some $\alpha$ we have $2\alpha=a$. Hence $a^{3}=(2\alpha)^{3}$ but since $(2\alpha)^{3}$ is even we can write this as $2d$. Hence we have
$$2d+2b^{3}+4c^{3}-6\sqrt[3]{2a}bc=0.$$
I was able to show that each of a,b,c are even so repeating the same process as with a, we come to an equation like
$$2d+2e+2f-\frac{3}{2}\sqrt[3]{def}=0.$$

9. Sep 4, 2013

Dick

You've got it! If a,b,c is a solution then they are all even integers. And if a,b,c is a solution, then so is a/2,b/2,c/2. Put that together with your observation about dividing by 2 over and over again. It's going to create a contradiction unless a=b=c=0.

Last edited: Sep 4, 2013
10. Sep 4, 2013