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It's hard to do a Google search on this problem.

If I was using Maple correctly, there are no other positive solutions with n at most 10000.

I know some of these Diophantine questions succumb to known methods, and others are extremely difficult to answer.

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@Qiaochu, Do you mean 'Yep, it's the only answer' or 'Yep, it succumbs to known methods' or 'Yep, it's extremely difficult to answer'? :) – David Roberts Feb 22 '11 at 23:40
Write as $(8n+1)^2=(4m)^3-48$, which is Mordell's equation for $k=-48$. A quick google comes up with the following, There's only two solutions for $4m\le10^{10}$. One solution is $m=1$ (hence, $n=0$). The other must be the one you state. – George Lowther Feb 22 '11 at 23:43
A google search gives the following by Keith Conrad describing the methods. In this case, you'd factorize in $\mathbb{Z}[\zeta_3]$. – George Lowther Feb 23 '11 at 0:21
@George, that's $(8n+4)^2=(4m)^3-48$. The problem is discussed on pages 208-209 of Mordell, Diophantine Equations. The solution is ascribed to W Ljunggren, Einige Bemerkungen uber die Darstellung ganzer Zahlen durch binare kubische Formen mit positiver Diskriminante, Acta Math 75 (1942) 1-21. – Gerry Myerson Feb 23 '11 at 0:40
I don't really understand this fact: For solving an equation of the form $aX^{2}+bX+C = M^{3}$ why do we need to translate this problem into algebraic number theoritical methods. Can't we have a purely elementary solution. :( – S.C. Oct 2 '11 at 12:41
up vote 13 down vote accepted

sage: E = EllipticCurve([0,0,1,0,-1])

sage: E

Elliptic Curve defined by y^2 + y = x^3 - 1 over Rational Field

sage: E.integral_points()

[(1 : 0 : 1), (7 : 18 : 1)]

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In other words: There is a lot of research available on integral points on elliptic curves and the resulting algorithm is implemented in sage and magma. See for instance chapter XIII of Smart's "Tha Algorithmic Resolution of Diophantine Equations". The sage documentation of this function refers to Petho A., Zimmer H.G., Gebel J. and Herrmann E., Computing all S-integral points on elliptic curves Math. Proc. Camb. Phil. Soc. (1999), 127, 383-402. – Chris Wuthrich Feb 23 '11 at 9:23

Let $\omega$ be a third root of unity, then $\mathbb{Z}[\omega]$ is a PID.

We have $m^3 = n^2 + n + 1 = (n-\omega)(n-\omega^2)$.

$\gcd(n-\omega,n-\omega^2) = \gcd(n-\omega,\omega-\omega^2) \mid (1-\omega)$, and $(1-\omega)$ is the ramified prime lying over $3$ in $\mathbb{Z}[\omega]$, so from unique factorization of $m^3$ we get that either $(n-\omega)$ and $(n-\omega^2)$ are both roots of unity times cubes, or one is a root of unity times $(1-\omega)$ times a cube and the other is a root of unity times $3$ times a cube. In the second case, $m$ is a multiple of $3$, but then $n^2 + n + 1 \equiv 0 \mod 9$, which is impossible.

If $(n-\omega)$ and $(n-\omega^2)$ are cubes, say $a^3$ and $\bar{a}^3$, then their difference $\omega^2-\omega$ is $a^3-\bar{a}^3 = (a-\bar{a})(a^2+a\bar{a}+\bar{a}^2)$. Thus $a-\bar{a}$ is either a root of unity or a root of unity times $(1-\omega)$, and it must be the latter since $a-\bar{a}$ is pure imaginary. Thus $\Im a \le \Im (\omega-\omega^2) = \sqrt{3}$. The same argument applied to $\omega a$ shows that $\Im \omega a \le \sqrt{3}$, and similarly for other roots of unity times $a$, so $a$ is in a hexagon around the origin that is contained in a circle of radius $2$ around the origin, i.e. $|a| \le 2$, so $m = |a|^2 \le 4$. which doesn't give us any solutions.

Finally we have the case that one of $(n-\omega), (n-\omega^2)$ is of the form $\omega a^3$. Then we have $\pm(\omega^2-\omega) = \omega a^3 - \omega^2 \bar{a}^3$. Write $a = x+y\omega$. Then $\omega a^3 - \omega^2 \bar{a}^3 = (\omega-\omega^2)(x^3+y^3-3x^2y)$, so we have $x^3+y^3-3x^2y = \pm 1$, which is a Thue equation. One solution is $x = -1, y = 2$, leading to the solution $n = 18, m = 7$.

Edit: Mathematica claims that the only solutions to $x^3+y^3-3x^2y = 1$ are $(x,y) = (-2, -3), (-1, -1), (-1, 2), (0, 1), (1, 0), (3, 1)$. Mathematica's documentation says it computes an explicit bound on the size of a solution to a Thue equation based on the Baker-Wustholz theorem in order to solve it, and in this case it seems like the bound was small enough.

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zeb: Nice, but a disappointing finish. We started off by looking for integer points on an elliptic curve, so it seems to have gone in a bit of a circle. – George Lowther Feb 23 '11 at 0:43
Mordell, following Ljunggren, solves $x^3-3xy^2-y^3=1$ on pages 208-209 of Diophantine Equations, as noted in my other comment. He has to go to the degree 6 field ${\bf Q}(\sqrt\xi)$, where $\xi$ is a root of $z^3-3z+1=0$, find the 4 fundamental units, and apply 2-adic methods. – Gerry Myerson Feb 23 '11 at 0:47
George - I realized this almost immediately after posting my answer. Well, at least the new equation seems simpler to me (are Thue equations easier to deal with than general elliptic curves?) – zeb Feb 23 '11 at 0:58
@zeb: Write $p(z)=z^3-3z+1$, so the equation becomes $p(y/x)=\pm x^{-3}$, so the theory of Diophantine approximation tells you that there are only finitely many solutions. That is $y/x$ approximates the roots of $p$. Although, as Gerry mentions, this is already solved by Mordell. – George Lowther Feb 23 '11 at 1:12

This is an old question, and has already been well-answered, but what I've got to say is slightly too long for a comment...

The equation $x^2+x+1 = y^3$ is of interest to finite geometers because $x^2+x+1$ is the number of points (and lines) in a finite projective plane of order $x$.

People have mentioned Ljunggren's name in comments above. The paper that's relevant is this:

Ljunggren, Wilhelm Einige Bemerkungen über die Darstellung ganzer Zahlen durch binäre kubische Formen mit positiver Diskriminante. (German) Acta Math. 75, (1943). 1–21.

I heartily recommend the Mathscinet review of that article, which says (amongst other things)...

... that Nagell [Norsk Mat. Forenings Skr. (I) no. 2 (1921)] proved that the equation

(1) $x^2+x+1=y^n$

has only trivial solutions unless $n$ is a power of $3$...

... And that Ljunggren then proved that (1) has only two nontrivial solutions, namely (18,7) and (-19, 7), for n=3.

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I realise this is a massive revival/refresh of this question, but I just found a completely elementary solution of this problem, outlined in this MSE question and my own answer.

Does anyone know if there are other completely elementary solutions?

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It is unclear to me why y-1 divides x. Are you assuming y-1 is a power of two times a prime? Gerhard "Would Like Very Elementary Proof" Paseman, 2016.03.09. – Gerhard Paseman Mar 9 at 17:47
(3.3)-(3.2) yields $y-1=a-d$. Since (4) implies $$ x(3y-2) = x\left(3(y-1)+1\right) \equiv 0 \pmod{a-d}, $$ we conclude $(y-1) \mid x$. Right? Note that even without working out $c=1$, this conclusion holds. – Kieren MacMillan Mar 9 at 17:51
(I'll be submitting this result for publication; the proof outlined on MSE will be made rigorous with those sorts of details.) – Kieren MacMillan Mar 9 at 18:00
I am still not seeing c=1. I do see (a-d) dividing both x and y-1 however. If you can show me c=1, then I will work through the rest of it. If you (figure out my address and) send me an email, there is the chance I can review your MS before you submit it. We can discuss that issue over email. Gerhard "Seeing C's Is Not Unseemly" Paseman, 2016.03.09. – Gerhard Paseman Mar 9 at 18:29
I just reviewed the rest. If you can prove c =1 ( or work the argument to success with (y-1)q=cx with small enough c ), I will offer to review your manuscript before submission. Gerhard "For The Advancement Of Science" Paseman, 2016.03.09. – Gerhard Paseman Mar 9 at 19:04

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