I am trying to find all solutions to
(1) $y^3 = x^2 + x + 1$, where $x,y$ are integers $> 1$
I have attempted to do this 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 approach.
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 $W = \frac{(-1 + \sqrt{-3})}{2}$. 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 approach are
- Is it valid?
- How could it be simplified?
Solution:
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 reasoning, 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 approach is valid because I don't know how Mathematica found solutions to expressions such as $a^3 + b^3 - 3ab^2=-1$. 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.
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 $P=(0,1)$, and the rest of the rational points are all the multiples of $P$:
\begin{align*} & \vdots \\ -3P &=\left(-\frac{703}{216},\frac{73}{36}\right)\\ -2P &=\left(18 , 7 \right)\\ -P &=\left(-1,1\right)\\ 0 & = \infty\\ P &=\left(0,1\right)\\ 2P &=\left(-19 , 7\right)\\ 3P &=\left(\frac{487}{216},\frac{73}{36}\right)\\ &\vdots \end{align*} Now one can use the rest of the theory of elliptic curves to show that the only integral points are $(18,7)$, $(-1,1)$, $(0,1)$, and $(-19,7)$. So the only solution $(x,y)$ with $x,y>1$ is $(18,7)$ as you found.