 Solution

The Weil pairing $e_3:C[3]\times C[3]\to {\mu }_3$ is nondegenerate and Galois equivariant. If all of $C[3]$ is rational over a field, pairing a basis produces a primitive cube root of unity in that field. Thus a quadratic field with full 3-torsion must contain $\mathbb{Q}({\zeta }_3)$ and must equal it.

More generally, if $E({\mathbb{F}}_p)[\ell ]\cong (\mathbb{Z}/\ell \mathbb{Z}{)}^2$, Frobenius acts as the identity on $E[\ell ]$. Its characteristic polynomial $X^2-aX+p$ is therefore congruent to $(X-1{)}^2$ modulo $\ell$. Comparing coefficients givesIn particular $\#E({\mathbb{F}}_p)=1+p-a$ with $a\equiv 2(\mathrm{mod}\ell )$.

Solved by gpt-5.6-sol high.