# 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?

• let me swap your variables and try to solve y^2 + y = x^3 + 1. That looks like an elliptic curve, and in fact it's this one: lmfdb.org/EllipticCurve/Q/225/c/2 In fact what you write is not quite true, since 0^2 + 0 - 1 = (-1)^3, and 1^2 + 1 - 1 = 1^3. But the solution you've got is sort of the interesting one. Jun 9, 2021 at 18:42
• @NoahSchweber Why? There is non-trivial mathematics here, and the OP knows her mathematics Jun 9, 2021 at 20:05
• To correct Thomas Browning's comment above, the curve is actually rank $1$, so there are infinitely many rational solutions, stating $(61/25, 434/125)$, $(-71/144, 973/1728)$, $(-779/1849, -125945/79507)$ ... Jun 10, 2021 at 0:46
• According to LMFDB, $y^2+y=x^3+1$ has integral solutions as follows. $(x,y)=(-1, 0) , (-1, -1) , (1, 1) , (1, -2) , (11, 36) , (11, -37).$ Hence only positive integral solutions are $(x,y)=(1, 1),(11, 36).$ Using generator $(x,y)=(1,1)$, we can get positive rational solutions such as $(61/25, 434/125),(408241/45796, 256170307/9800344),(479399/2739025, 2812459393/4533086375)$. Jun 10, 2021 at 3:47
• ... The Dipohantine equation $u^2-v^3=k$ is known as *Mordell's equation". It has been extensively studied; in particular, Mordell has shown that for any integer $k$ the equation has only a finite number of solutions. For some values of $k$, Mordell's equation can be solved using an elementary argument, not appealing to elliptic curves. I do not know whether the case $k=80$ admits an elementary solution, just search the web / arXiv.
– Seva
Jun 11, 2021 at 8:14

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.

• Regarding the $i=2$ case: Given a solution to $y-\theta = \theta^2 \alpha^3$, rewrite it as $y - \theta = \theta^{-1} \beta^3$ for $\beta = \theta \alpha$. Then apply the Galois automorphism of $K$ to get $y-(1-\theta) = - \theta \beta^3$ or $(1-y) - \theta = \theta \beta^3$. So, if $y$ solves the $i=2$ case then $1-y$ solves the $i=1$ case (and vice versa). And thanks for this answer! Jun 11, 2021 at 12:29

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)]

• The question is “What is the proof?” — so “Because the computer says so” seems a slightly unsatisfying answer, on its own. It would be greatly improved by some information about how SAGE computes this, since that’d provide at least a recipe for the proof. In 5 minutes of Googling I wasn’t easily able to work out what algorithm is used — the best I could find was the SAGE documentation and a couple of old MO answers: mathoverflow.net/a/46778/2273, mathoverflow.net/a/42044/2273 Jun 10, 2021 at 15:13
• @PeterLeFanuLumsdaine: I agree. It is more than nothing though, and the algorithms of SAGE are pretty reliable. Jun 10, 2021 at 15:50
• Oh yes, I completely agree this answer is an excellent starting point! Jun 10, 2021 at 15:54
• Jun 10, 2021 at 16:05