# Normalizing an elliptic curve to find integer solutions

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.

• 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? May 26, 2016 at 9:35
• @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. May 26, 2016 at 13:52
• 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... May 26, 2016 at 14:56
• @Ferra: But very often there are infinitely many rational solutions, so it's not possible to check them individually. May 26, 2016 at 14:58

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.