Torsion on $y^2=x^3+d$

Question

A question that I am stuck on is: prove that the $\mathbb{Q}$-torsion subgroup of the elliptic curve $y^2=x^3+d$ has order dividing 6. Any hints on how to start would be nice. I tried saying something about the reduced curve, but the lack of information about $d$ was a problem. I guess it amount to trying to say something about the Jacobi symbol $\big( \frac{x^3+d}{p}\big)$ for $p\nmid 6d$, but I can't see it.

Answer 1

You can see it in this way: for all but finitely many primes $p$, there is an injective morphism of groups $E(\mathbb Q)_{tors}\to E(\mathbb F_p)$. Now take $p\equiv -1\bmod 6$. Then there are no elements of order $3$ in $\mathbb F_p^*$ and this tells you that $x\to x^3+d$ is a bijection of $\mathbb F_p$. Therefore for such $p$'s you must have $|E(\mathbb F_p)|=p+1$. Now since $m=|E(\mathbb Q)_{tors}|\mid p+1$ for all these $p$'s, you must have that all but finitely many primes $\equiv -1\bmod 6$ are $\equiv -1\bmod m$, and this is possible iff $m\mid 6$ because of Dirichlet's theorem on primes in arithmetic progression.