# Where is the mistake in this incorrect proof in Eisenstein integers?

The Diophantine equation is $n^2 + n + 1 = m^3$ my attempt to solve it shows there is no solutions to this equation, but in fact there are four. I could not find my mistake so I hope someone could point it out please.

Factoring in the Eisenstein integers $\mathbb{Z}[\omega]$ with $\omega^2 + \omega + 1 = 0$: $n^2 + n + 1 = (n - \omega)(n + 1 + \omega)$.

If $d$ is a common divisor then $d$ also divides their difference $1 + 2 \omega$ which is a prime.

If the greatest common divisor is $1$ then both numbers are cubes, but that is impossible because $(a + b \omega)^3 = (a^3 - 3 a b^2 + b^3) + 3 (a^2 b - a b^2)\omega$ and the $\omega$ term of the factors are not multiples of $3$.

If the greatest common divisor is $1 + 2 \omega$ then $(n - \omega)(n + 1 + \omega) = -3 \frac{n - \omega}{1+2 \omega}\frac{n + 1 + \omega}{1 + 2 \omega}$ so either $n - \omega$ or $n + 1 + \omega$ is $1 + 2 \omega$ times a cube, but that would imply $2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 = (a-2b)(a+b)(2a-b) = \pm 1$ which is impossible.

Edit: Does he make the same mistake here http://mathforum.org/library/drmath/view/68612.html ?

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You could have easily detected the mistake yourself if you had just plugged in one of the solutions and seen where your proof asserts something incorrect. E.g. let's take the obvious solution $n=0$, $m=1$. Your factors are then $-\omega$ and $1+\omega$. Are they cubes? No, they are only cubes up to a unit. To say that two numbers are coprime means that their greatest common divisor is a unit. Clearly, a unit divides any number, so you can never do better than that. See also this discussion if you want to know how to take that into account.
@quanta: Exactly. Provided of course that the class number of the ring you are working in is 1, or at least coprime to 3. Otherwise, you have to work with the ideals $(a)$ and $(b)$. – Alex B. Apr 2 '11 at 14:31
From $$\mbox{cube}= -3 \frac{n - \omega}{1+2 \omega}\frac{n + 1 + \omega}{1 + 2 \omega}$$ doesn't the inference "so either $n−\omega\$ or $n+1+\omega\$ is $1+2\omega\$ times a cube" also need $-3$ to be prime? This is not true in the Eisenstein integers.