On Réalis’s solution of the “cubic Markov equation” I am interested in the Diophantine equation
$$X^3+Y^3+Z^3=3XYZ$$
and its solutions. In Nouv. Corr. Math. 1879 (page 7), Réalis claims that, other than the trivial solution $x=y=z$, the complete solution in integers is given by
$$
x=(a-b)^3+(a-c)^3, \quad y=(b-c)^3+(b-a)^3, \quad z=(c-a)^3+(c-b)^3.
$$
It’s easy to show that these values suffice — by simply substituting and doing the algebra — but the result seems just a little too convenient for my tastes…
Does anyone know how to prove [or disprove] the claim? Are there any known solutions that don’t fit Réalis’s parameterization?
 A: This is quite wrong. We have
$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x + y \omega + z \omega^2)(x + y \omega^2 + z \omega)$$
where $\omega = e^{ \frac{2 \pi i}{3} }$ is a primitive third root of unity. This identity comes from recognizing the LHS as the determinant of the circulant matrix $\left[ \begin{array}{ccc} x & y & z \\ z & x & y \\ y & z & x \end{array} \right]$ and using a standard fact about the eigenvalues of circulant matrices. Setting this product equal to zero, we need to solve
$$x + y + z = 0 \text{ or } x + y \omega + z \omega^2 = 0 \text{ or } x + y \omega^2 + z \omega = 0.$$
The first equation is very easy and we get a parameterized family of solutions $\boxed{ x = t, y = s, z = - t - s }$, which amounts to the identity
$$t^3 + s^3 - (t + s)^3 = - 3 ts(t + s)$$
which is not hard to check. The given parametric solution only allows values of $x, y, z$ which are a sum of two cubes so it doesn't include all of these solutions.
For the second two equations, we have $1 + \omega + \omega^2 = 0$ so $\omega^2 = -1 - \omega$. This means the first equation is equivalent to
$$x + y \omega - z(1 + \omega) = (x - z) + (y - z) \omega = 0$$
so $\boxed{ x = y = z }$, and the second equation also gives this. We can also argue without using complex numbers by instead writing
$$\begin{eqnarray*} (x + y \omega + z \omega^2)(x + y \omega^2 + z \omega) &=& x^2 + y^2 + z^2 - xy - yz - zx \\
 &=& \frac{(x - y)^2 + (y - z)^2 + (z - x)^2}{2} \end{eqnarray*}$$
which is clearly $\ge 0$ with equality case $x = y = z$.
