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I want to solve the diophantine equation $y^2+y=x^3-1$.

I thought about writing it as $(y-\omega)(y-\overline{\omega})=x^3$, where $\omega=(-1+i\sqrt{3})/2$. If we prove that $y-\omega$ and $y-\overline{\omega}$ are relatively prime in $\mathbb{Z}[\omega]$, then it follows that there is $u\in\{\pm 1,\pm \omega,\pm\omega^2\}$ and $a,b\in\mathbb{Z}$ such that $$y-\omega=u(a+b\omega)^3=u(a^3+b^3-3ab^2+3ab(a-b)\omega).$$ I think we may be able to solve it by considering all the possible $a,b$ for a given $u$.

We two cases:

  • Case 1: $u=\pm 1$. This case is impossible since we would have $1=\pm 3ab(a-b)$ as an equality of integers.
  • Case 2: $u=\pm \omega$. Here we have that $a^3+b^3-3a^2b=\pm 1$ and $y=\mp 3ab(a-b)$. I don't know what to do.

To sum up, I would like to know how to prove that $y-\omega$ and $y-\overline{\omega}$ are relatively prime in $\mathbb{Z}[\omega]$ and how to deal with case 2.

Gabriel
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    Should case 3 be $u=\pm\overline{\omega}$ instead? And there is no need to use $\pm$ here since $\pm1$ are perfect cubes. – Batominovski Nov 13 '19 at 21:50
  • @WETutorialSchool, I actually misremembered the units of $\mathbb{Z}[\omega]$. They are ${\pm 1,\pm \omega, \pm \omega^2}$. I'll edit the post. – Gabriel Nov 13 '19 at 21:53
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    I would also absorb case 3 into case 2, since $y-\omega=-(-y-1-\overline{\omega})$. So practically, you only have one case left: $$y-\omega=\omega(a+b\omega)^3.$$ – Batominovski Nov 13 '19 at 21:55
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    Note that $(y-\omega)-(y-\overline{\omega})=-\sqrt{3}i$ with norm $3$. Hence if $y-\omega$ and $y-\overline{\omega}$ are not coprime, then $3$ divides $y^2+y+1$. Thus, $3$ divides $x$, making $3^3$ a factor of $x^3=y^2+y+1$. You can show that $y^2+y+1$ is not divisible by $9$ for all integers $y$. – Batominovski Nov 13 '19 at 22:01
  • @WETutorialSchool could you explain why if $y-\omega$ and $y-\overline{\omega}$ are not coprime, then $3$ divides their product? – Gabriel Nov 13 '19 at 22:19
  • Does it help to know that one solution is $y=-1, x=1$ – tomi Nov 13 '19 at 22:27
  • @tomi, Wolfram-Alpha gives $(1,-1)$, $(1,0)$, $(7,18), and $(7,-19)$ as all the solutions. But I don't know how I could use this. – Gabriel Nov 13 '19 at 22:34
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    If $y-\omega$ and $y-\overline{\omega}$ are not coprime, then both of them are divisible by $\sqrt{3}i$. That is, $y-\omega=\sqrt{3}i\alpha$ and $y-\overline{\omega}=-\sqrt{3}i\overline{\alpha}$ for some $\alpha\in \Bbb{Z}[\omega]$. Therefore, $$y^2+y+1=(y-\omega)(y-\overline{\omega})=3(\alpha\overline{\alpha}).$$ – Batominovski Nov 14 '19 at 04:02
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    LMFDB entry of this elliptic curve confirms the list of integer points given by @Gabriel. A lot of other information is also there. An elementary argument for the list of integer points may be possible. – Jyrki Lahtonen Nov 14 '19 at 12:48
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    I've answered the same question here. Using Algebraic Number Theory leads me to a Thue equation so that is still not enough, though I'm not sure if I missed something. However both this question and the resulting Thue equation are published results. – Yong Hao Ng Nov 15 '19 at 02:34
  • In particular a norm argument like what @WETutorialSchool has mentioned resolves the relative prime issue and your Case 2 is the remaining Thue Equation in problem. – Yong Hao Ng Nov 15 '19 at 02:49

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