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Consider the elliptic curve $$E'\colon y^2 = g(x) = x^3 + \frac{1}{432} .$$ One can check that the discriminant of $E'$ is $-1/432$, that $E'$ has complex multiplication, and that $(-1/432)^{1/3} = (-1/6 \sqrt[3]{2})$ is a root of $g(x)$. Hence there exists a CM elliptic curve over $\mathbb{Q}$ such that the cube root of the discriminant is a root of the defining polynomial.

My question is of the same flavor of this with added assumptions, in particular, we want $$E \colon y^2 = f(x) = x^3 + Ax + B$$ to be a non-CM elliptic curve over $\mathbb{Q}$ in Weierstrass form. Let $\Delta_E$ denote the discriminant of $E$, and for the time being, we want $\Delta_E$ to be a rational square. I want to know whether there exists such an elliptic curve over $\mathbb{Q}$ such that $f(\Delta_E^{1/3}) = 0$. The condition on the discriminant tells us that there are 3 real, distinct roots $f(x)$, which only seems to complicate matters. Any help and/or references would greatly be appreciated.

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My guess is that the answer to your question is "no", i.e., there is no elliptic curve without CM with the specified property on the discriminant, and I was very close to prove it... Hopefully you can finish it up somehow. However, there is another CM curve with the same property, namely $E: y^2=x^3-1/16x$!

Let $E: y^2=f(x)=x^3+Ax+B$, for some $A,B\in\mathbb{Q}$ and discriminant $\Delta_E$, be such that $\Delta$ is a perfect square, and $f(\Delta_E^{1/3})=0$. Notice that this means $$0 = \Delta_E +A\Delta_E^{1/3}+B.$$ If $A=0$, then $E$ has CM (with $j$-invariant equal to $0$), so let us assume that $A\neq 0$. Therefore, $\Delta_E^{1/3}=(-B-\Delta_E)/A\in\mathbb{Q}$. It follows that $\Delta_E$ is a cube in addition to a square, so $\Delta_E=\delta^6$ for some $\delta\in\mathbb{Q}$. Thus, $$0 = \delta^6+A\delta^2+B$$ and so $B=-(\delta^6+A\delta^2)$. Moreover, $$\delta^6=\Delta_E=-16(4A^3+27B^2) = -16(4A^3+27(\delta^6+A\delta^2)^2).$$ This condition defines a curve $C:\delta^6+16(4A^3+27(\delta^6+A\delta^2)^2)=0$ which happens to be of genus $1$. There are $4$ points that are easy to find: $$[d,A,z]=[0,0,1],[0,1,0],[1/2,-1/16,1],[-1/2,-1/16,1].$$ The first two correspond to $A=0$ and a solution with $z=0$, so we discard those two. The third and fourth correspond to $E: y^2=x^3-1/16x$, which is a curve with CM (with $j=1728$). Now, $C$ is of genus $1$ and using $P=[0,1,0]$ as the origin point, we can find a (surprisingly simple) Weierstrass equation $$E': y^2=x^3+729$$ and a map $\phi: C\to E'$ given by $$[243d^{10}z + 486d^6Az^4 + 9/16d^4z^7 + 135d^2A^2z^7,-2187d^{11} - 4374d^7Az^3 - 81/16d^5z^6 - 1215d^3A^2z^6, -A^2z^9].$$

It turns out $E'$ has rank $0$ and torsion $\mathbb{Z}/6\mathbb{Z}$, generated by $R=(18,-81)$. The point $[1/2,-1/16,1]$ of $C$ maps to $R$ via $\phi$, and $[-1/2,-1/16,1]$ of $C$ maps to $-R$ via $\phi$. The map $\phi$ is not defined at $[0,1,0]$ or $[0,0,1]$. So the question is... are there points on $C$ that map to $\mathcal{O},2R,3R$, or $4R$? If so, there could be additional solutions to the question.

I leave here the Magma code I used:

S := AffineSpace(Rationals(),2);

f:=-16*(4*A^3+27*(d^6+A*d^2)^2)-d^6;

f;

C := Curve(S,f);

Genus(C);

P:=ProjectiveSpace(Rationals(),2);

C:=Curve(P,-432*d^12 - 864*d^8*A*z^3 - d^6*z^6 - 432*d^4*A^2*z^6 - 64*A^3*z^9);

PointSearch(C,100);

E,phi:=EllipticCurve(C,C![0,1,0]);

E;

Rank(E);

TorsionSubgroup(E);

Generators(E);

P:=phi(C![1/2,-1/16,1]);

2*P;

a,b:=HasPreimage(2*P,phi);

a,b;

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