Solve $a^3 + b^3 + c^3 = 6abc$

Find solutions for $a^3 + b^3 + c^3 = 6abc$ in $\mathbb{N}$, such that $gcd(a,b,c) = 1$, except for $(1,2,3)$ and its permutations.


Using trial and error I found out that if $a,b,c$ are solution of the equation, then they are in arithmetic progression. I’ve managed to prove that conjecture, assuming that $c>b>a$ and let $k$ be their common difference in the arithmetic progression. Then WLOG we have:

$$b = c-k \quad \quad a = c-2k$$

Now the equation looks like:

$$(c-2k)^3 + (c-k)^3 + c^3 = 6(c-2)(c-1)c$$

After expanding we have:

$$c^3 – 6kc^2 + 12ck^2 – 8k^3 + c^3 – 3kc^2 + 3ck^2 -k^3 + c^3 = 6c^3 – 18kc^2 + 12ck^2$$
$$3c^3 – 9kc^2 + 15ck^2 – 9k^3 = 6c^3 – 18kc^2 + 12ck^2$$
$$c^3 – 3kc^2 + 5ck^2 – 3k^3 = 2c^3 – 6kc^2 + 4ck^2$$
$$-c^3 + 3kc^2 + ck^2 – 3k^3 = 0$$

Now it’s easy to see that if $k=c$, then the LHS will be zero, so one of the zeroes of the polynomial is $c_1 = k$, now factorizing we have:

$$(c-k)(3a^2 + 2ax – x^2) = 0$$
$$(c-k)(c+k)(c-3k) = 0$$

Now we have three distinct cases:

Case 1: $c = k$

This implies that $b = 0$ and $a = -k$. But because $k \in \mathbb{N}$, both $a,b \not\in \mathbb{N}$, violating the initial conditions.

Case 2: $c = -k$

Obviously the initial condition is already violated, becasue $k \in \mathbb{N}$, so from the relation $c \not\in \mathbb{N}$

Case 2: $c = 3k$

This implies that $b = 2k$ and $a = k$. Now we have one 3-tuple $(3k,2k,k)$ and it’s permutation as solution, where $k \in \mathbb{N}$. But it’s easy to note that $k$ is a common factor for $a,b,c$ so we have:

$$gcd(a,b,c) = k$$

But because we want $gcd(a,b,c) = 1$, this implies that $k=1$, which means we have only one solution for $a^3 + b^3 + c^3 = 6abc$ in $\mathbb{N}$, such that $gcd(a,b,c) = 1$ and it $(1,2,3)$, solution that is already given.

Now my question is what I’m missing. Is there really no other solutions such that $gcd(a,b,c) = 1$? Or maybe there is a different way to obtain solution except for my method using arithmetic progression?

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It’s easy to prove that at least one of the variables needs to be an even number. We know that:

$$6|(n-1)n(n+1)$$

Because in three consecutive numbers, one is divisible with three and at least one is divisible with 2. So we have:

$$6|n^3 – n$$
$$n^3 = n \pmod 6$$

Now we have:

$$a^3 + b^3 + c^3 \equiv a + b + c \equiv 0 \pmod 6$$

Beacuse the modulo is an even number that means that the sum $a+b+c$ is an even number also. We know that the sum of 3 odd numbers will be odd number, so it’s impossible $a,b,c$ to be odd number, because there won’t be solution. So it means that at least one of the variables is an even number.

WLOG we can set $b=2k$. Now we can continue:

$$b-2k = 0$$

Now we can multiply both sides with $b(b+2k)$. Note that won’t give another solution, because it’ll imply that b is $0$ or a negative number, which violate the condition. Now we have:

$$b(b+2k)(b-2k) = 0$$
$$b(b^2 – 4k^2) = 0$$
$$b^3 – 4bk^2 = 0$$
$$3b^3 – 12bk^2 = 0$$
$$6b^3 – 6bk^2 = 3b^3 + 6bk^2 = 0$$
$$6b(b^2 – k^2) = (b^2 – 3kb^2 + 3bk^2 – k^3) + b^3 + (b^3 + 3kb^2 + 3bk^2 + k^3) = 0$$
$$6b(b-k)(b+k) = (b-k)^3 + b^3 + (b+k)^3$$

Now if we substitute WLOG:

$$b+k=c \quad \quad b-k=a$$

$$6abc = a^3 + b^3 + c^3$$

Because $b,k \in \mathbb{N}$ it means that also $a,c \in \mathbb{N}$. So this proves that for any $b=2k$, there are integer solutions, such $a=k$ and $c=3k$.

But because $k$ is a factor of all of them it’s easy to see that:

$$gcd(a,b,c) = gcd(k,2k,3k) = k$$

Because we want $gcd(a,b,c) = 1$, that implies that $k=1$ and that the only primitive solution of this equation is $(1,2,3)$ and its permutation.