TITLE: Solve $a^3 + b^3 + c^3 = 6abc$
QUESTION [8 upvotes]: 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?

REPLY [4 votes]: Since we are looking for non-trivial solutions, we will assume $c \ne 0$.  
Divide both side by $c^3$ and let $(u,v) = (\frac{a}{c},\frac{b}{c})$, the equation becomes
$$E(u,v) \stackrel{def}{=} u^3 + v^3 + 1 - 6uv  = 0$$
The problem of finding non-trivial primitive positive integer solutions of $a,b,c$ becomes one finding positive rational solutions $u,v$ for $E(u,v) = 0$.
With following change of variables
$$(u,v) = \left(\frac{6(6-x)+y}{6(x+36)},\frac{6(6-x)-y}{6(x+36)} \right)$$
Equation $E(u,v) = 0$ can be simplified to that of 
an elliptic curve in Weierstrass form:
$$\mathcal{E} : y^2 = x^3 - 864x - 5616$$
Throwing following commands to Magma
Q<x> := PolynomialRing(Rationals());
E00  := EllipticCurve(x^3-864*x-5616);
Generators(E00);

We find elliptic curve $\mathcal{E}$ has two generators
$$\begin{array}{lll}
P: (x,y) = (48,252) &\rightarrow (u,v) = (0,-1) &\rightarrow (a,b,c) = (0,-1,1)\\
Q: (x,y) = (-8,28) &\rightarrow (u,v) = (\frac23,\frac13) &\rightarrow (a,b,c) = (2,1,3)
\end{array}
$$
These leads to solutions we already know. 
Given any elliptic curve, the rational points on it forms an abelian group
with respect to some operation we simply refer to as "addition"
One can generate more rational points on $\mathcal{E}$ by "adding" copies of $P$ and $Q$ together.
After a few experiment, I notice adding $P$ to another point on $\mathcal{E}$ just trigger a permutation of $(a,b,c)$ in corresponding solution. This gives
us nothing new.
The generator $Q$ is more interesting. Let $nQ$ be a shorthand for $\underbrace{Q + Q + \cdots + Q}_{n \text{ copies}}$.
With help of a CAS, I get
$$\begin{array}{rll}
2Q &: (x,y) = (160,1988) 
&\rightarrow (u,v) = \left(\frac{19}{21},-\frac{52}{21}\right)\\
3Q &: (x,y) = \left(-\frac{143}{9},\frac{1729}{27}\right)
&\rightarrow (u,v) = \left(\frac{5275}{3258},\frac{1817}{3258}\right)\\
4Q &: (x,y) = \left(\frac{225616}{5041},\frac{76232996}{357911}\right)
&\rightarrow (u,v) = \left(-\frac{124904}{3096807},-\frac{2847511}{3096807}\right)\\
  & \vdots
\end{array}
$$
By collecting those $nQ$ whose $u,v > 0$, we obtain
following non-trivial positive integer solutions of $a,b,c$.
$$\begin{align}
3Q \rightarrow (a,b,c) =& (5275,1817,3258)\\
5Q \rightarrow (a,b,c) =& (15051171563,10840875082,4904676969)\\
7Q \rightarrow (a,b,c) =& (46875396961726681714,\\
& \;81821352777652044467,\\
& \;29381282043563909553)
\end{align}
$$
There are more solutions from $9Q, 11Q, 13Q,\ldots$. It seems all $nQ$ with odd $n$ leads to positive integer solutions in $(a,b,c)$. However, I don't have a proof of that.