<s>[INST] Cubic diophantine equation with a prime $x^3 + y^3 + z^3 - 3xyz = p$. Question: Find all triple positive integers $(x, y, z)$ so that 
$$x^3 + y^3 + z^3 - 3xyz = p,$$
where $p$ is a prime number greater than $3$.

I have tried the following: The equation factors as
$$(x + y + z) (x^2 + y^2 + z^2-xy-yz-zx) = p.$$
Since $x + y + z> 1$, we must have $x + y + z = p$ and 
$$x^2 + y^2 + z^2-xy-yz - zx = 1.$$
The last equation is equivalent to 
$$(x-y)^2 + (y-z)^2 + (z-x)^2 = 2.$$
Without loss of generality you can assume that $x ≥ y ≥ z$, we have $xy ≥ 1$ and $xz ≥ 2$, implying 
$$(xy)^2 + (yz)^2 + (zx)^2 ≥ 6> 2.$$
Who can help me and correct me, thank you.
 [/INST] Strangely enough, the solution is finite.
for the equation:
$$X^3+Y^3+Z^3-3XYZ=q=ab$$
If it is possible to decompose the coefficient as follows:  $4b=k^2+3t^2$
Then the solutions are of the form:
$$X=\frac{1}{6}(2a-3t\pm{k})$$
$$Y=\frac{1}{6}(2a+3t\pm{k})$$
$$Z=\frac{1}{3}(a\mp{k})$$
Thought the solution is determined by the equation Pell, but when calculating the sign was a mistake. There's no difference, but the amount should be. Therefore, the number of solutions of course.
I may be wrong, though. We still need to check other options.
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