# Does $a^3 + 2b^3 + 4c^3 = 6abc$ have solutions in $\mathbb{Q}$

Does $a^3 + 2b^3 + 4c^3 = 6abc$ have solutions in $\mathbb{Q}$?

This is not a homework problem. Indeed, I have no prior experience in number theory and would like to see a showcase of common techniques used to solve problems such as this. Thanks

Edit Apart from $a=b=c=0$.

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$a=b=c=0$ works. – vadim123 Jun 17 '13 at 3:13
What makes you think this is solvable? Is it from a contest? Finding the rational solutions of a general diophantine equation is a very hard problem (cf. Fermat's lat theorem). – Potato Jun 17 '13 at 3:13
@Potato (i) I have no idea whether or not it is solvable, which is why I asked; (ii) This is not from a contest, but from a textbook; (iii) Of course, but this is not a general diophantine equation – user71815 Jun 17 '13 at 3:23
@user71815 Generally textbook problems are solvable. :). – Potato Jun 17 '13 at 3:27
– Martin Sleziak Jan 22 at 17:01

First, note that $(a,b,c)$ is a solution if and only if $(ka,kb,kc)$ is; hence we may assume $a,b,c$ are integers, with no common factor (divide by that common factor if necessary).

Because $6abc, 2b^3+4c^3$ are even, so is $a^3$ and hence $a$. Write $a=2a'$ and we have $$8(a')^3+2b^3+4c^3=12a'bc$$ and hence $$4(a')^3+b^3+2c^3=6a'bc$$

By similar logic, $b$ is even, so write $b=2b'$ and we have $$4(a')^3+8(b')^3+2c^3=12a'b'c$$ But now $$2(a')^2+4(b')^3+c^3=6a'b'c$$ and hence $c$ is even. Hence $a,b,c$ are all even; this contradicts $a,b,c$ having no common factor.

Followup: The same proof works if the coefficients $\{1,2,4,6\}$ are replaced by $\{\alpha_1, \alpha_2, \alpha_3,\alpha_4\}$ so long as there is some prime $p$ with $\nu_p(\alpha_1)=0, \nu_p(\alpha_2)=1, \nu_p(\alpha_3)=2, \nu_p(\alpha_4)\ge 1$. (Here $\nu_p(\cdot)$ denotes the p-adic valuation). For example, apart from $(0,0,0)$, there are no rational solutions to $$7a^3+15b^3+18c^3=45abc$$ where here $p=3$.

$~$

Double followup: The same proof works with $n$ variables $$\alpha_0a_0^n+\alpha_1a_1^n+\cdots+\alpha_{n-1}a_{n-1}^n=\alpha_n(a_0a_1\cdots a_{n-1})$$ provided that $\nu_p(\alpha_i)=i$ (for $0\le i\le n-1$) and $\nu_p(\alpha_n)\ge 1$.

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Nice. Reminds one of a traditional proof about $\sqrt{2}$ :) +1 – coffeemath Jun 17 '13 at 3:25
@user71815, alas Diophantine equations have many many techniques to solve them. Finding a divisor and making a substitution (as I did 3 times) is a common one. – vadim123 Jun 17 '13 at 3:25
Yes indeed! Thanks – user71815 Jun 17 '13 at 3:26

As it happens, the cubic form $C(a,b,c)=a^3+2b^3+4c^3-6abc$ is the norm form for the extension $K=\mathbb Q(\root 3 \of 2)$ over $\mathbb Q$. That is, if you look at a general element of $K$, say $a + b\root3\of2+c\root3\of2^2$, and take its field-theoretic norm, what you get is exactly $C(a,b,c)$. Now, the norm doesn’t vanish on an algebraic extension, except at zero. So, just because $C$ happens to be a norm form, you can say immediately that the trivial zero is the only one.

How did I spot this? By having done lots of examples, first by hand over many years, then, more recently, with symbolic algebra programs.

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Is it true that, for prime $q = 4 u^2 + 2 u v + 7 v^2,$ so that 2 is not a cubic residue, and $C(a,b,c) \equiv 0 \pmod q,$ then each $a,b,c \equiv 0 \pmod q?$ See math.stackexchange.com/questions/329936/… Note that Noam Elkies gave me a proof for a very similar problem at mathoverflow.net/questions/127160/… – Will Jagy Jun 17 '13 at 4:33
Oh man, @Will Jagy, I don’t know nothin’ about things like that. Sorry. – Lubin Jun 17 '13 at 15:56

HINT Make use of $$x^3 + y^3 + z^3 -3xyz = \left(x+y+z \right) \left(x^2+y^2+z^2-xy-yz-zx \right)$$where $x=a$, $y=b\sqrt[3]{2}$ and $z=c \sqrt[3]{4}$.

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Does this help if one seeks rational $a,b,c$ as in the posted question? – coffeemath Jun 17 '13 at 3:19
Oh, I think I see where this was going. The left side is zero iff the equation of the OP holds, for the integers $a,b,c$. So a solution would imply one of the two factors on the right is zero. The first one clearly isn't, if we assume $1,\sqrt[3]{2},\sqrt[3]{4}$ are rationally independent, and the second more complicated factor might be clearly nonzero for similar reasons, though I didn't check. So I'm giving this an upvote, but would advise inclusion of some of the explanation just indicated. – coffeemath Jun 17 '13 at 4:13
I downvoted this answer, and would happily remove it after seeing coffeemath's explanation above. The second term can't vanish unless $x=y=z$ (by writing it as sum of squares), and it's not hard to check that there's no solution in that case. – user27126 Jun 17 '13 at 6:18
You can show rational independence directly (no Galois) by assuming $a+b\sqrt[3]{2}+c\sqrt[3]{4}=0$ for rational $a,b,c$, and then start by isolating $\sqrt[3]{4}$ and cube both sides, then plug in to eliminate $\sqrt[3]{4}$ from the first formula into the cubed out formula. The result if zero for nonzero $c$ would give that $\sqrt[3]{2}$ is rational. – coffeemath Jun 18 '13 at 8:06

if negative numbers are accepted, (57,63,-156) is OK

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Those numbers satisfy the equation $a^3+b^3+c^3=6abc$, not the OP's equation. – Barry Cipra Apr 9 '14 at 13:31

(Just to give to user $71815$ a distinct way to solve).

We use $(a+b+c)^3=a^3+b^3+c^3+3ab(a+b)+3ac(a+c)+3bc(b+c)+6abc$.

Taking in account that $$\left(\frac{a}{\sqrt 6}\right)^3+\left(\frac{b}{\sqrt3}\right)^3+\left(\frac{c\sqrt2}{\sqrt3}\right)^3=abc$$ we get after calculation and equating to zero the coefficients of $\sqrt[3]2$ and $\sqrt[3]4$ the equations $$c(a^2-2ac-2b^2c)=0$$ $$abc=0$$

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