Why is 1331 the only cube of the form $x^2 + x − 1$? The Wikipedia (https://en.wikipedia.org/wiki/1000_(number)#1300_to_1399) mentions that 1331 is the only cube of the form $x^2 + x − 1$, for $x = 36$. What is the proof?
 A: According to SAGE, there are 6 integral points on the curve.
sage: EllipticCurve([0,0,1,0,1]).integral_points(both_signs=True)
[(-1 : -1 : 1),(-1 : 0 : 1),(1 : -2 : 1),(1 : 1 : 1),(11 : -37 : 1),(11 : 36 : 1)]

A: Here is the more elementary way that one would have solved this before LLL. I will still use the computer for some calculations, but technically one could do all this by hand. Worse, I won't even do all of the steps, but I hope the part I show explains the method, for those that may be interested. Maybe someone wants to complete it or finds a shorter way.
First, setting $K=\mathbb{Q}(\theta)$ with $\theta^2+\theta-1$, a.k.a the Golden ratio, we will rewrite our equation as $$N_{K/\mathbb{Q}}(y-\theta) = x^3$$ to be solved in $X$ and $y$ in $\mathbb{Z}$. The field $K$ has class number 1 and the units are generated by $\theta$. From this, using a little argument about the $\sqrt{5}$-adic valuation, one can show that, there must be a $i\in \{0,1,2\}$ such that $$y-\theta = \theta^i\cdot \alpha^3$$ for some $\alpha\in\mathbb{Z}[\theta]$, the ring of integers of $K$. Write $\alpha = a+b\,\theta$ for $a,b\in\mathbb{Z}$ and split into the three cases according to $i$.
Case $i=0$ : Spanning out $\alpha^3$ and equating the coefficients in front of $\theta$, we get $$\begin{align*} y &= a^3+3ab^2-b^3 \\ -1&=b\cdot(3a^2-3ab+2b^2)\end{align*}$$ Now the second line tells us that $b=\pm 1$, but neither of the choices allows us to find $a$ as an integer, so that case does not occur.
Case $i=1$ : This time the equations are $$\begin{align*} y &= 3a^2b-3ab^2+2b^3\\-1 &=a^3-3a^2b+6ab^2-3b^3\end{align*}$$ One can note here aside that the second line is a model of the curve $3$-isogenous to the original curve over $\mathbb{Q}$. This second (Thue) equation, we rewrite as $$-1 = N_{L/\mathbb{Q}}(a-b\xi)$$ where $L=\mathbb{Q}(\xi)$ and $\xi^3-3\xi^2+6\xi-3=0$. This $L$ has also trivial class number and the units are generated by $1-\xi$ of norm $1$. We are now looking for units of norm $-1$ in $L$ with no $\xi^2$ term. Once more we split into three cases $j\in\{0,1,2\}$. We want to solve $$a+b\xi = - (1-\xi)^j\cdot(1-\xi)^{3k} = -(1-\xi)^j\cdot(-2+3\xi)^k.$$
Subcase $j=0$ : We may rewrite it as $$a-b\xi = - (1+3(\xi-1))^k = -\bigl(1+3(\xi-1)k+9(\xi-1)^2k(k-1)/2 + 27\cdots \bigr)$$ and consider this as a power series in $k$ over $\mathbb{Q}_3(\xi)$, which is a totally ramified extension of $\mathbb{Q}_3$. We can compare the power series coefficient in front of $\xi^2$, to get an equation of the form $$0=9 k (k-1)\cdot \bigl(1/2 -9/24 (k-2)(k-3)+\cdots\bigr).$$ By Strassmann's theorem, the power series has at most two solutions in $\mathbb{Q}_3$. But it is easy to spot that there are two such solutions, namely $(a=-1,b=0,k=0)$ and $(a=2,b=3,k=1)$. These correspond to $(x,y)=(-1,0)$ and $(11,36)$ repsectively.
Subcase $j=1$ : leads to a single solution $a-b\xi=-1+\xi$ which corresponds to $(x,y)=(1,-2)$.
Subcase $j=2$ : produces a unit power series and has therefore no solutions.
Case $i=2$ : I believe one just gets the $-P$ for all the $P$ found in the case $i=1$. But I admit that I have not checked.
Finally, I should say that for questions on integral points on elliptic curve I often turn to Smart's "The algorithmic Resolution of Diophantine Equations" that explains things well. Like the $3$-adic method used in the subcases in chapter III.
