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I have an elliptic curve $$ c_1y^2 + a_1xy + a_3 = c_2x^3 + a_2x^2 + a_4x + a_6 $$ with integers $a_1,a_2,a_3,a_4,a_6,c_1,c_2$ and I would like to find all integer solutions of this elliptic curve. I have a system (Sage or MAGMA) which can find all integer solutions of an elliptic curve $$ y'^2 + a_1x'y' + a_3 = x'^3 + a_2x'^2 + a_4x' + a_6 $$ which corresponds to the special case $c_1=c_2=1.$ Since I'm working over $\mathbb{Q}$ there is no trouble with the characteristic, and I can find a change of variables $x'=\alpha x+\beta,\ y'=\gamma y+\delta$ to obtain this form. But I'm looking for integral solutions to $(x,y)$ rather than $(x',y')$. Can I still solve this problem with the (black-box) solver?

Alternatively, are there solvers that take $a_1,a_2,a_3,a_4,a_6,c_1,c_2$ and find all integer solutions?


Example

Suppose I'm trying to find integer solutions of $3y^2 - 1 = x^3 + x.$ Let $x'=x/3$ and $y'=y/3,$ then the new curve is $y'^2 - 1/27 = x'^3 + x'/9$. But the integer solution $(x,y)=(1,1)$ in the original corresponds to the rational solution $(x',y')=(1/3,1/3)$ in the second, and hence would be missed if looking only for the integer solutions to the second equation.

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  • $\begingroup$ I don't see the problem, if you can obtain $x',y'$ and you have the change of coordinates from $x,y$, just apply the inverse change of coordinates to get $x,y$. Am I missing something? $\endgroup$
    – Ferra
    May 26, 2016 at 9:35
  • $\begingroup$ @Ferra: Yes, you are. I can find solutions where $x'$ and $y'$ are integers, but I want solutions where $x$ and $y$ are integers, and the two are not equivalent (and may even be disjoint, say if $\alpha$ and $\gamma$ are integers and $\beta$ and $\delta$ are non-integer rationals). I've added an example into the question above. $\endgroup$
    – Charles
    May 26, 2016 at 13:52
  • $\begingroup$ I see your point, but usually if you use sage or magma you can get all points over $\mathbb Q$ and not only over $\mathbb Z$. Thus you can get the points with integral coefficients of the other curve just by checking whether $(x-\beta)/\alpha$ and $(y'-\delta)/\gamma$ are integers. Otherwise I see no (generic) way to get all points with integral coefficients starting from integral points on the other curve... $\endgroup$
    – Ferra
    May 26, 2016 at 14:56
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    $\begingroup$ @Ferra: But very often there are infinitely many rational solutions, so it's not possible to check them individually. $\endgroup$
    – Charles
    May 26, 2016 at 14:58

1 Answer 1

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Suppose $(X,Y)$ is an integral point on the elliptic curve $$c_1y^2 + a_1xy + a_3 = c_2x^3 + a_2x^2 + a_4x + a_6.$$ Then plugging in $X$ and $Y$ and multiplying through by $c_1^3c_2^2$ shows that $$(c_1^2c_2Y)^2+a_1(c_1c_2X)(c_1^2c_2Y)+a_3=(c_1c_2X)^3+a_2c_1(c_1c_2X)^2+a_4c_1^2c_2(c_1c_2X)+a_6c_1^3c_2^2,$$ so the point $(c_1c_2X,c_1^2c_2Y)$ is an integral point on the elliptic curve $$y^2+a_1xy+a_3=x^3+a_2c_1x^2+a_4c_1^2c_2x+a_6c_1^3c_2^2.$$ Now use your black box to find integral points on this elliptic curve, and then for each integral point check whether the $x$- and $y$-coordinates are divisible by $c_1c_2$ and $c_1^2c_2$, respectively, to determine whether it corresponds to an integral point on the original elliptic curve.

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    $\begingroup$ Thank you! Would you like a bounty? $\endgroup$
    – Charles
    Sep 19, 2020 at 1:54
  • $\begingroup$ @Charles That's very generous of you. I don't need more reputation, but I don't mind more either. It's up to you. $\endgroup$ Sep 19, 2020 at 10:31

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