Prove for $a, b, c \in \mathbb Q$ that $a^3 + 2b^3 + 4c^3 − 6abc \neq 0$ if at least one of $a$, $b$, and $c$ are non-zero without resorting to field theory or linear algebra.
2026-04-01 03:26:34.1775013994
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Prove that $a^3 + 2b^3 + 4c^3 − 6abc \neq 0$ if at least one of $a$, $b$, and $c$ are non-zero
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For any prime $q = 4 u^2 + 2 u v + 7 v^2$ in integers, the polynomial $ a^3 + 2b^3 + 4c^3 − 6abc $ is only divisible by $q$ when $a,b,c$ are all divisible by $q.$ The first few primes that could be used here are $$ 7,13,19,37,61,67,73,79,97,103,139,151,163,181$$
If there is any nontrivial solution, in integers, to $ a^3 + 2b^3 + 4c^3 − 6abc =0, $ we can divide out any common factor to get another solution with $\gcd(a,b,c) = 1.$
Put these two together, there is no nontrivial solution. Below are the 343 calculations of $n \equiv a^3 + 2b^3 + 4c^3 − 6abc \pmod 7.$ You can see that the only occurrence of $n \equiv 0 \pmod 7 $ is $a \equiv b \equiv c \equiv 0 \pmod 7.$
n a b c
0 0 0 0
1 0 3 3
1 0 3 5
1 0 3 6
1 0 5 3
1 0 5 5
1 0 5 6
1 0 6 3
1 0 6 5
1 0 6 6
1 1 0 0
1 1 1 1
1 1 2 4
1 1 3 4
1 1 4 2
1 1 5 1
1 1 6 2
1 2 0 0
1 2 1 4
1 2 2 2
1 2 3 2
1 2 4 1
1 2 5 4
1 2 6 1
1 3 1 0
1 3 1 1
1 3 1 6
1 3 2 0
1 3 2 3
1 3 2 4
1 3 4 0
1 3 4 2
1 3 4 5
1 4 0 0
1 4 1 2
1 4 2 1
1 4 3 1
1 4 4 4
1 4 5 2
1 4 6 4
1 5 1 0
1 5 1 2
1 5 1 5
1 5 2 0
1 5 2 1
1 5 2 6
1 5 4 0
1 5 4 3
1 5 4 4
1 6 1 0
1 6 1 3
1 6 1 4
1 6 2 0
1 6 2 2
1 6 2 5
1 6 4 0
1 6 4 1
1 6 4 6
2 0 1 0
2 0 2 0
2 0 3 1
2 0 3 2
2 0 3 4
2 0 4 0
2 0 5 1
2 0 5 2
2 0 5 4
2 0 6 1
2 0 6 2
2 0 6 4
2 1 1 2
2 1 1 3
2 1 2 1
2 1 2 5
2 1 3 2
2 1 4 4
2 1 4 6
2 1 5 4
2 1 6 1
2 2 1 1
2 2 1 5
2 2 2 4
2 2 2 6
2 2 3 1
2 2 4 2
2 2 4 3
2 2 5 2
2 2 6 4
2 3 0 3
2 3 0 5
2 3 0 6
2 3 3 4
2 3 5 1
2 3 6 2
2 4 1 4
2 4 1 6
2 4 2 2
2 4 2 3
2 4 3 4
2 4 4 1
2 4 4 5
2 4 5 1
2 4 6 2
2 5 0 3
2 5 0 5
2 5 0 6
2 5 3 1
2 5 5 2
2 5 6 4
2 6 0 3
2 6 0 5
2 6 0 6
2 6 3 2
2 6 5 4
2 6 6 1
3 0 0 3
3 0 0 5
3 0 0 6
3 1 1 0
3 1 2 0
3 1 3 5
3 1 4 0
3 1 5 3
3 1 6 6
3 2 1 0
3 2 2 0
3 2 3 6
3 2 4 0
3 2 5 5
3 2 6 3
3 3 0 1
3 3 0 2
3 3 0 4
3 3 1 4
3 3 2 2
3 3 3 1
3 3 3 5
3 3 4 1
3 3 5 2
3 3 5 3
3 3 6 4
3 3 6 6
3 4 1 0
3 4 2 0
3 4 3 3
3 4 4 0
3 4 5 6
3 4 6 5
3 5 0 1
3 5 0 2
3 5 0 4
3 5 1 1
3 5 2 4
3 5 3 2
3 5 3 3
3 5 4 2
3 5 5 4
3 5 5 6
3 5 6 1
3 5 6 5
3 6 0 1
3 6 0 2
3 6 0 4
3 6 1 2
3 6 2 1
3 6 3 4
3 6 3 6
3 6 4 4
3 6 5 1
3 6 5 5
3 6 6 2
3 6 6 3
4 0 0 1
4 0 0 2
4 0 0 4
4 1 0 3
4 1 0 5
4 1 0 6
4 1 1 4
4 1 1 5
4 1 2 2
4 1 2 6
4 1 3 3
4 1 4 1
4 1 4 3
4 1 5 6
4 1 6 5
4 2 0 3
4 2 0 5
4 2 0 6
4 2 1 2
4 2 1 6
4 2 2 1
4 2 2 3
4 2 3 5
4 2 4 4
4 2 4 5
4 2 5 3
4 2 6 6
4 3 1 2
4 3 2 1
4 3 3 0
4 3 4 4
4 3 5 0
4 3 6 0
4 4 0 3
4 4 0 5
4 4 0 6
4 4 1 1
4 4 1 3
4 4 2 4
4 4 2 5
4 4 3 6
4 4 4 2
4 4 4 6
4 4 5 5
4 4 6 3
4 5 1 4
4 5 2 2
4 5 3 0
4 5 4 1
4 5 5 0
4 5 6 0
4 6 1 1
4 6 2 4
4 6 3 0
4 6 4 2
4 6 5 0
4 6 6 0
5 0 1 3
5 0 1 5
5 0 1 6
5 0 2 3
5 0 2 5
5 0 2 6
5 0 3 0
5 0 4 3
5 0 4 5
5 0 4 6
5 0 5 0
5 0 6 0
5 1 0 1
5 1 0 2
5 1 0 4
5 1 1 6
5 1 2 3
5 1 4 5
5 2 0 1
5 2 0 2
5 2 0 4
5 2 1 3
5 2 2 5
5 2 4 6
5 3 1 5
5 3 2 6
5 3 3 2
5 3 3 6
5 3 4 3
5 3 5 4
5 3 5 5
5 3 6 1
5 3 6 3
5 4 0 1
5 4 0 2
5 4 0 4
5 4 1 5
5 4 2 6
5 4 4 3
5 5 1 3
5 5 2 5
5 5 3 4
5 5 3 5
5 5 4 6
5 5 5 1
5 5 5 3
5 5 6 2
5 5 6 6
5 6 1 6
5 6 2 3
5 6 3 1
5 6 3 3
5 6 4 5
5 6 5 2
5 6 5 6
5 6 6 4
5 6 6 5
6 0 1 1
6 0 1 2
6 0 1 4
6 0 2 1
6 0 2 2
6 0 2 4
6 0 4 1
6 0 4 2
6 0 4 4
6 1 3 0
6 1 3 1
6 1 3 6
6 1 5 0
6 1 5 2
6 1 5 5
6 1 6 0
6 1 6 3
6 1 6 4
6 2 3 0
6 2 3 3
6 2 3 4
6 2 5 0
6 2 5 1
6 2 5 6
6 2 6 0
6 2 6 2
6 2 6 5
6 3 0 0
6 3 1 3
6 3 2 5
6 3 3 3
6 3 4 6
6 3 5 6
6 3 6 5
6 4 3 0
6 4 3 2
6 4 3 5
6 4 5 0
6 4 5 3
6 4 5 4
6 4 6 0
6 4 6 1
6 4 6 6
6 5 0 0
6 5 1 6
6 5 2 3
6 5 3 6
6 5 4 5
6 5 5 5
6 5 6 3
6 6 0 0
6 6 1 5
6 6 2 6
6 6 3 5
6 6 4 3
6 6 5 3
6 6 6 6
The proof is by descent. It is obvious that if one of the variables is $0$, then they must all be. So we can assume all the variables are non-zero.
Because of the homogeneity, if there is a non-zero rational solution, then there is a non-zero integer solution. Then there is such a solution with $|a|+|b|+|c|$ minimal.
It is clear that $a$ is even. Let $a=2q$. Substituting, we get that $$2b^3+4c^3+8q^3 -12bcq=0,$$ or equivalently $$b^3+2c^3+4q^3-6bcq=0.$$ But $|b|+|c|+|q|\lt |a|+|b|+|c|$, contradicting the minimality of $|a|+|b|+|c|$.