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Number theory problem:

Find the all positive integer solutions to: $$x^3-x^2+x=3y^3$$

Here are my attempts:

$$x(x^2-x+1)=3y^3$$

$$x(x+1)(x^2-x+1)=3y^3(x+1)$$

$$x(x^3+1)=3(x+1)y^3$$

$$x^4+x^2=3(x+1)y^3$$

I can not see how can I proceed.

User
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    Seems that $(-1,-1)$ and $(0,0)$ are the only solutions, but I have no proof for this. But brute force shows that for another solution , we must have $|x|>10^7$ – Peter Oct 27 '22 at 10:04
  • @Peter Yes, Wolfram Alpha failed. You wrote an algorithm for this result on computer, right, $x>10^7$? – User Oct 27 '22 at 10:08
  • I programmed it with PARI/GP and checked whether $\frac{x^3-x^2+x}{3}$ is a perfect power (perfect powers other than $-1,0$ and $1$ did not occur ar all) to get a larger search limit. – Peter Oct 27 '22 at 10:10
  • @Peter Thanks for clarification. But note that, we are looking for only positive integer solutions. – User Oct 27 '22 at 10:13
  • Yes, I now noticed that. I checked however also the negative $x$ and the solution set seems complete (in particular , there is probably no solution in positive integers) – Peter Oct 27 '22 at 10:14
  • @Peter Yes I agree, because Wolfram Alpha also failed. – User Oct 27 '22 at 10:16
  • Only Suzi can solve these questions. Where is she? – Bob Dobbs Oct 27 '22 at 11:31
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    @BobDobbs The sister of Alexa ? :) – Peter Oct 27 '22 at 11:42
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    @User In your last line, the $x^4 + x^2$ should be $x^4 + x$ instead. – John Omielan Oct 27 '22 at 12:36
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    Unless I made a mistake this curve is rationally equivalent to the curve $v^2+v=u^3-7$. According to the database it is a rank zero curve, i.e. there are only finitely many rational points on it. – Jyrki Lahtonen Oct 27 '22 at 12:52
  • The steps in the transformation are simple. First homogenize to $X^3-X^2Z+XZ^2=3Y^3$. Then pick the affine chart $X=1$ and multiply by $9$ to get $9-9(Z/X)+9(Z/X)^2=27(Y/X)^3$. Finally write it in terms of $u=3Y/X$ and $v=3Z/X-2$. – Jyrki Lahtonen Oct 27 '22 at 12:56
  • So the rational points are all torsion and can be listed as $(u,v)=\infty, (3,4)$ and $(3,-5)$. These correspond to $[X:Y:Z]=[0:0:1]$, $[1:1:2]$ and $[1:1:-1]$ respectively. Therefore it should be the case that all the rational solutions are $(x,y)\in{(0,0),(1/2,1/2),(-1,-1)}$. – Jyrki Lahtonen Oct 27 '22 at 13:20
  • Mind you, the database says that the curve is also equivalent to the Fermat curve $X^3+Y^3=Z^3$. This means that proving the above list of rational solution to be complete is equivalent to proving Fermat's last for the exponent $3$. – Jyrki Lahtonen Oct 27 '22 at 13:21

1 Answers1

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As hinted in the LMFDB database (see my comments under the question), this elliptic curve is birationally equivalent to the Fermat cubic. In addition to the transformation described in the comments, I needed to bring it into the short Weierstrass form $r^2=s^3-432$, and then use the transformation given here. A bit of cleaning up gave me the following.

By expanding we see that $$ (x+1)^3+(2x-1)^3=9x^3-9x^2+9x. $$ This means that multiplying the given equation by nine we can rewrite it in the form $$ (x+1)^3+(2x-1)^3=(3y)^3. $$ By the well-studied case $n=3$ of Fermat's last this equation is possible for rational $x,y$ only if one of the three numbers $x+1$, $2x-1$ or $3y$ vanishes. This quickly leads to the following:

All the rational solutions are $(0,0)$, $(-1,-1)$ and $(1/2,1/2)$. Hence there are no solutions with positive integers.

Jyrki Lahtonen
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    See here for an on-site elementary proof of the case $n=3$ of Fermat's last. – Jyrki Lahtonen Oct 27 '22 at 14:22
  • Very nice solution. If you have established that the curve is birationally equivalent to Fermat's cubic then the rational points found are necessarily of torsion in your group. A Mazur result that is a classic of elliptic curve theory states (1978) that the torsion subgroup is isomorphic to one of the following groups:

    $\mathbb Z/n\mathbb Z$ with $1\le n\le 10$ or $n=12$

    $\mathbb Z/2\mathbb Z \text { x }\mathbb Z/2n\mathbb Z$ with $1\le n\le 4$.

    It would be nice in your curve to know what is the order of the two non-trivial torsion points $P$ you have found (suite)

     – Ataulfo Oct 28 '22 at 09:30
  • For this you have to calculate $nP$ till you find $nP=0$ which is not straightforward at all (even graphically for only $2P$ it is not clear because of the shape of the curve $x^3-x^2+x=3y^3$). (Obviously we are refered to the elliptic curve defined over $\mathbb Q$) – Ataulfo Oct 28 '22 at 09:31
  • @Piquito The LMFDB entry says that that torsion subgroup is cyclic of order three. Makes sense as the Fermat curve has three rational points also. – Jyrki Lahtonen Oct 28 '22 at 09:38
  • Sure but I said this for the sum to do with the functions defining the sum in your curve. Of course that you cannot find other points because they does not exist. Regards. – Ataulfo Oct 28 '22 at 13:35
  • Very Impressive the work done in your link, it is gigantic. Could you tell me how you place the treated curve in it? – Ataulfo Oct 28 '22 at 13:50