# Torsion group of $y^2 = x^3 + B$ has order dividing $6$

Let $$B$$ be a positive integer. Show that the torsion subgroup of the elliptic curve $$y^2 = x^3 + B$$ has order dividing $$6$$ (Feel free to assume Dirichlet's theorem, that for any coprime $$a$$ and $$d$$ there are infinitely many primes $$p$$ with $$p\equiv a \pmod d$$, but not any stronger version of it.)

The discriminant is $$\Delta = 27B^2$$ so we can use reduction mod $$p$$ for any prime $$p\geq 5$$, $$p\nmid B$$. Using primitive roots one can easily show that if $$p\equiv 2 \pmod 3$$ then every remainder mod $$p$$ is a cube and hence that $$y^2 = x^3 + B$$ has exactly $$p+1$$ points modulo $$p$$ (including the point at infinity). So the order of the torsion group must divide $$p+1$$ for all $$p\equiv 2 \pmod 3$$. Then what?

Any help appreciated!

Then what?

Then you have an integer $$n$$ which divides $$p + 1$$ for almost all (i.e. all except finitely many) prime numbers $$p \equiv -1 \mod 6$$.

Suppose that $$n$$ does not divide $$6$$. Let $$m = \operatorname{lcm}(6, n)$$. I claim that there is an integer $$d$$ which satisfies the following properties:

• $$d \equiv -1 \mod 6$$;
• $$d \not\equiv -1 \mod m$$;
• $$d$$ is coprime to $$m$$.

The existence of such a $$d$$ is a simple exercise in elementary number theory.

Now by Dirichlet's theorem, the arithmetic sequence $$\{mk + d: k \geq 1\}$$ contains infinitely many prime numbers.

In particular, since these prime numbers are all $$\equiv -1\mod 6$$, there must be at least one prime number $$p$$ of the form $$mk + d$$ such that $$n$$ divides $$p + 1$$.

But this contradicts our construction, as $$p = mk + d \equiv d \not\equiv -1 \mod m$$.