What is known about the numbers $M_p = \left\vert C(\mathbb{F}_p )\right\vert$?

Solution 1:

The numbers you actually want to look at turn out to be

$$a_p = p + 1 - |C(\mathbb{F}_p)|.$$

There is a lot to say about these numbers. The Hasse-Weil bound implies that $|a_p| \le 2 \sqrt{p}$. The Weil conjectures explain why this bound holds: it's because (for all but finitely many $p$) there is a pair of conjugate algebraic numbers $\alpha_p, \overline{\alpha}_p$ of absolute value $\sqrt{p}$ such that

$$|C(\mathbb{F}_{p^n})| = p^n + 1 - \alpha_p^n - \overline{\alpha}_p^n$$

for all $n$, which gives

$$a_p = \alpha_p + \overline{\alpha}_p.$$

The Sato-Tate conjecture describes the asymptotic distribution of these numbers as $p$ varies. Finally, one way to state the modularity theorem (which was not a theorem when Silverman-Tate was published!) in this case is that there is a modular form

$$f(q) = \sum_{n \ge 0} b_n q^n$$

which must be a "normalized cuspidal Hecke eigenform of weight $2$ and level $N$," where $N$ is a positive integer called the conductor of $C$, such that $b_0 = 0$ ("cuspidal"), $b_1 = 1$ ("normalized"), and

$$a_p = b_p$$

for all primes $p$ not dividing $N$. So essentially the best you can do as far as finding a formula for the $a_p$ goes is to find a concise description of this modular form.

Sage informs me that this curve has conductor $36$ and that there is a unique normalized cusp form of weight $2$ and level $36$, which is necessarily the Hecke eigenform we want. This implies that $y^2 = x^3 + 1$ must in fact be (isogenous to?) the modular curve $X_0(36)$. The $q$-expansion of this modular form begins

$$q - 4 q^7 + 2 q^{13} + 8 q^{19} - 5 q^{25} + \dots$$

and according to this paper that I found, it is $\eta (6 \tau)^4$ where $\eta$ is the Dedekind eta function. The $q$-expansion of this modular form, in full, is therefore

$$f(q) = q \prod_{n=1}^{\infty} (1 - q^{6n})^4.$$

See also the pentagonal number theorem, which leads to an expression for $a_p = b_p$ in terms of a signed sum over the set of ways $\frac{p - 1}{6}$ can be written as a sum of $4$ pentagonal numbers.

Edit: Here is a proof that $|C(\mathbb{F}_p)|$ is divisible by $12$ when $p \equiv 1 \bmod 3$. As Greg Martin says in the comments, this is equivalent to showing that $C(\mathbb{F}_p)$ has full 2-torsion and a 3-torsion point. The 2-torsion points on an elliptic curve in Weierstrass normal form are precisely those of the form $(x_0, 0)$, together with the point at infinity, and hence the number of 2-torsion points is the number of roots of $x^3 + 1$, plus one.

Now, if $p \equiv 1 \bmod 6$, then $\mathbb{F}_p^{\times}$ contains an element of order $6$, and hence $\mathbb{F}_p$ has all sixth roots of unity. Since the roots of $x^3 + 1$ are sixth roots of unity, it follows that $x^3 + 1$ has $3$ roots over $\mathbb{F}_p$ in this case, and hence there are $4$ 2-torsion points, as desired.

It remains to show that there is also a nontrivial $3$-torsion point. Recall that a point is $3$-torsion iff its tangent line intersects the curve with multiplicity $3$. In fact, the points $(0,\pm 1)$ both have this property: their tangent lines take the form $y = \pm 1, x = t$, and substituting this in gives $t^3 = 0$. So there are always at least $3$ 3-torsion points, as desired.

Solution 2:

If $p$ is $\equiv 1 \bmod 3,$ then we may write $p = a^2 + 3 b^2$, where $a$ is pinned down by requiring $a \equiv 1 \bmod 3$ (and I won't bother to pin down $b$). Then, writing $a_p = 1 + p - | C(\mathbb F_p)|$, as usual, one has that $$a_p = 2a,$$ or equivalently, $$| C(\mathbb F_p)| = 1 + p - 2 a .$$

In terms of Qiaochu's answer, the relevant modular form is a CM modular forms, and so can be expressed in terms of a Hecke character for the CM field (which is $\mathbb Q(\zeta_3)$ in this case).

More directly, the curve $C$ is a CM elliptic curve, with CM by $\mathbb Z[\zeta_3]$ (the element $\zeta_3$ acts by $(x,y) \mapsto (\zeta_3 x, y)$), and so it corresponds to a Hecke character of $\mathbb Q(\zeta_3)$. Explicitly computing this Hecke character gives the above result.