# If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$, is $a=b=c=0$?

If $a,b,c$ are integers such that $4a^3+2b^3+c^3=6abc$ , then is it true that $a=b=c=0$ ? I was thinking of infinite descent but can't actually proceed , please help. Thanks in advance

The number $c$ would have to be even, $c = 2k$. Divide the equation by $2$ to get

$$4k^3 + 2 a^3 + b^3 = 6kab$$

This gives the infinite descent, since you can always deduce that the number with the coefficent $1$ is even.

We may assume that $a$, $b$, and $c$ are relatively prime. If not, we can divide out by the cube of the common factor. Observe that $$c^3=2(3abc-2a^3-b^3),$$ so $c$ is even. Let $c=2n$.

Then, the equation becomes $4a^3+2b^3+8n^3=12abn$. Therefore, $$2b^3=4(3abn-a^3-2n^3).$$ Hence, $b$ is even. Let $b=2m$.

Then, the equation becomes $4a^3+16m^3+8n^3=24amn$. Therefore, $$4a^3=8(3amn-2m^3-n^3).$$ Hence $a$ is even, a contradiction.

Infinite descent seems to work for me. Assume that there exists a non-zero solution, then I consider the minimal solution, in terms of absolute value, $\{a_0,b_0,c_0\}$.

Since $$c_0^3 = 2(3a_0b_0c_0 -b_0^3 -2a_0^3)$$ we deduce that $2/c_0$. Then we proceed with $b_0$: $$2b_0^3 = 4( 3a_0b_0(c_0/2) -2(c_0/2)^3 -a_0^3).$$ Therefore also $b_0$ can be divided by $2$. Finally, with the same technique, we also conclude that $a_0$ must be even.

Therefore $\{a_0/2, b_0/2, c_0/2\}$ is another solution with strictly smaller absolute value. Contradiction.

What's problem with infinite decent ? Note $2|C \implies 2|b \implies 2|a$ and similarly going on for all $n\in \mathbb {N}$ you'll get $2^n$ divides all of $a,b,c$. That's all.

$a=1$, $b=1$ and $c=1$ is a solution of $4a^3+2b^3+c^3=6abc$, so $a=b=c=0$ is not the only solution.