Help with proof of $y^3 = x^2 + x + 1$

I have attempted a proof using...I think they are called 'quadratic integers'. It would be great if someone could verify the steps and suggest simplifications to this approach. I am also wondering whether my use of Mathematica invalidates this proof.

I am trying to find all solutions to

(1) $y^3 = x^2 + x + 1$, where $x,y$ are integers $> 1$

My exploration is based on a proof I read that $x^3 + y^3 = z^3$ has no non-trivial integer solutions. This proof uses the ring Z[W] where $W = (-1 + \sqrt{-3})/2$. I don't understand most of this proof, or what a ring is, but I get the general idea. The questions I have about my attempted proof are

1) Is it valid? 2) How could it be simplified?

Proof:

Let $w = (-1 + \sqrt{-3})/2$. (Somehow, this can be considered an "integer" even though it doesn't look anything like one!)

Now $x^3 - 1 = (x-1)(x-w)(x-w^2)$ so that, $(x^3 - 1)/(x-1) = x^2 + x + 1 = (x-w)(x-w^2)$. Hence

$y^3 = x^2 + x + 1 = (x-w)(x-w^2).$

Since $x-w, x-w^2$ are coprime up to units (so I have read) both are "cubes". Letting $u$ be one of the 6 units in Z[w], we can say

$x-w = u(a+bw)^3 = u(c + dw)$ where

$c = a^3 + b^3 - 3ab^2, d = 3ab(a-b)$

Unfortunately, the wretched units complicate matters. There are 6 units hence 6 cases, as follows:

1) $1(c+dw) = c + dw$

2) $-1(c+dw) = -c + -dw$

3) $w(c+dw) = -d + (c-d)w$

4) $-w(c+dw) = d + (d-c)w$

5) $-w^2(c+dw) = c-d + cw$

6) $w^2(c+dw) = d-c + -cw$

Fortunately, the first two cases can be eliminated. For example, if $u = 1$ then $x-w = c+dw$ so that $d = -1 = 3ab(a-b).$ But this is not possible for integers $a,b$. The same reasoning applies to $u = -1$.

For the rest I rely on a program called Mathematica, which perhaps invalidates my proof, as you will see.

We attack case 5. Here

$x = c-d = a^3 + b^3 - 3a^2b$, and $c = a^3 + b^3 - 3ab^2 = -1.$

According to Mathematica the only integer solutions to $c = -1$ are $(a,b) = (3,2), (1,1), (0,-1), (-1,0), (-1,-3), (-2,1).$

Plugging these into $x = c-d$ we find that no value of x that is greater than 1. So case 5 is eliminated, as is 6 by similar reasoning.

Examining case 4 we see that $d-c = -(a^3 + b^3 - 3a^2*b) = -1$ with solutions $(-2,-3), (-1,-1), (-1,2), (0,1), (1,0), (3,1).$

Plugging these values into $x = d = 3ab(a-b)$ yields only one significant value, namely $x = 18$ (e.g. (a,b)=(3,1) . The same result is given by case 4. Hence the only solution to (1) is $7^3 = 18^2 + 18 + 1$

However, I'm unsure this proof is valid because I don't know how Mathematica found solutions to expressions such as $a^3 + b^3 - 3ab^2$. These seem more difficult than the original question of $y^3 = x^2 + x + 1$, although I note that Mathematica could not solve the latter.

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I don't know why... but I must say I genuinely liked your approach. What exactly compelled you to do that initial treatment with the w? –  frogeyedpeas Jun 17 '13 at 15:14

I have not verified your work. This is just to say that $y^3=x^2+x+1$ is the equation of an elliptic curve (with $j$-invariant equal to $0$). Using Sage or Magma, you can find that this elliptic curve has rank $1$ and trivial torsion subgroup. The group of rational points is generated by the point $P=(1,0)$, and the rest of the rational points are all the multiples of $P$:
\begin{align*} & \vdots \\ -3P &=\left(\frac{73}{36} , -\frac{703}{216}\right)\\ -2P &=\left(7 , 18 \right)\\ -P &=\left(1,-1\right)\\ 0 & = \infty\\ P &=\left(1,0\right)\\ 2P &=\left(7 , -19\right)\\ 3P &=\left(\frac{73}{36} , \frac{487}{216}\right)\\ &\vdots \end{align*} Now one can use the rest of the theory of elliptic curves to show that the only integral points are $(7,18)$, $(1,-1)$, $(1,0)$, and $(7,-19)$. So the only solution $(x,y)$ with $x,y>1$ is $(7,18)$ as you found.