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The Diophantine equation is $n^2 + n + 1 = m^3$ my attempt to solve it shows there is no solutions to this equation, but in fact there are four. I could not find my mistake so I hope someone could point it out please.


Factoring in the Eisenstein integers $\mathbb{Z}[\omega]$ with $\omega^2 + \omega + 1 = 0$: $n^2 + n + 1 = (n - \omega)(n + 1 + \omega)$.

If $d$ is a common divisor then $d$ also divides their difference $1 + 2 \omega$ which is a prime.

If the greatest common divisor is $1$ then both numbers are cubes, but that is impossible because $(a + b \omega)^3 = (a^3 - 3 a b^2 + b^3) + 3 (a^2 b - a b^2)\omega$ and the $\omega$ term of the factors are not multiples of $3$.

If the greatest common divisor is $1 + 2 \omega$ then $(n - \omega)(n + 1 + \omega) = -3 \frac{n - \omega}{1+2 \omega}\frac{n + 1 + \omega}{1 + 2 \omega}$ so either $n - \omega$ or $n + 1 + \omega$ is $1 + 2 \omega$ times a cube, but that would imply $2 a^3 - 3 a^2 b - 3 a b^2 + 2 b^3 = (a-2b)(a+b)(2a-b) = \pm 1$ which is impossible.


Edit: Does he make the same mistake here http://mathforum.org/library/drmath/view/68612.html ?

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2 Answers 2

up vote 5 down vote accepted

You could have easily detected the mistake yourself if you had just plugged in one of the solutions and seen where your proof asserts something incorrect. E.g. let's take the obvious solution $n=0$, $m=1$. Your factors are then $-\omega$ and $1+\omega$. Are they cubes? No, they are only cubes up to a unit. To say that two numbers are coprime means that their greatest common divisor is a unit. Clearly, a unit divides any number, so you can never do better than that. See also this discussion if you want to know how to take that into account.

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So if (a,b) = 1 and ab = x^3 then a = u p^3 and b = u^-1 q^3? –  quanta Apr 2 '11 at 14:26
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@quanta: Exactly. Provided of course that the class number of the ring you are working in is 1, or at least coprime to 3. Otherwise, you have to work with the ideals $(a)$ and $(b)$. –  Alex B. Apr 2 '11 at 14:31

From $$\mbox{cube}= -3 \frac{n - \omega}{1+2 \omega}\frac{n + 1 + \omega}{1 + 2 \omega}$$ doesn't the inference "so either $n−\omega\ $ or $n+1+\omega\ $ is $1+2\omega\ $ times a cube" also need $-3$ to be prime? This is not true in the Eisenstein integers.

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What I was thinking here is that, since (n-w)/(1+2w) and (n+1+w)/(1+2w) are relatively prime.. one of them is relatively prime to 1+2w (the common factor we divided out) and so it must be a cube. Although I should be saying a unit times a cube.. I am not 100% sure that's what I said is true! –  quanta Apr 2 '11 at 15:27
    
You are right, I didn't even read that far. One can fix that, but of course here, also, everything is only up to units. –  Alex B. Apr 2 '11 at 15:27
    
@Alex Thanks for your comment, I honestly wasn't sure. As an amateur of number theory, I easy fall into the same sort of traps that quanta did. –  Byron Schmuland Apr 2 '11 at 15:41

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