Numerical evidence suggests a conjecture that the number of
points of certain elliptic curve over $\mathbb{F}\_p$ is
either $p$ or $p+2$ for $p$ of certain form.


Let $p$ be prime of the form $p=27a^2+27a+7$ and $(a/b)$ denote
the Kronecker symbol.


For integer $k$ nonzero modulo $p$ define $E\_k / \mathbb{F}\_p : y^2=x^3+2k^3$.


$E\_k$ is the quadratic twist of $y^2=x^3+2$.


**Conjecture 1**: $\#E(\mathbb{F}\_p)=p+1 + (2k^3 / p)$.


In other words for $p$ of the given form $\#E(\mathbb{F}\_p)$ is either
$p$ or $p+2$.



> 
> Is Conjecture 1 true?
> 
> 
> 


Example sage session with 200 bit $p$



```
sage: a=2^100+8;p=27*a^2+27*a+7;k=3;Kp=GF(p);E=EllipticCurve([Kp(0),Kp(2*k^3)])
sage: o=E.order();o2=p+1+kronecker(2*k^3,p);o==o2
True

```


Conjecture: The number of points modulo $p$ of certain elliptic curve is $p$ or $p+2$ for $p$ of form $p=27a^2+27a+7$
The conjecture is true. To prove it, we can restrict to $k=1$ as explained in the comments. Let $\chi$ denote a cubic Dirichlet character modulo $p$; this exists as $p\equiv 1\pmod{3}$, and it is unique up to complex conjugation. Let $\psi$ denote the (unique) quadratic Dirichlet character modulo $p$. 


The number of points of the *affine* elliptic curve $E$ modulo $p$ equals
$$\sum\_{y\bmod p}\left(1+\chi(y^2-2)+\overline{\chi}(y^2-2)\right)=p+\sum\_{y\bmod p}\chi(y^2-2)+\overline{\sum\_{y\bmod p}\chi(y^2-2)}.$$
The $y$-sum on the right hand side equals
$$\sum\_{y\bmod p}\chi(y^2-2)=\sum\_{z\bmod p}\chi(z-2)\bigl(1+\psi(z)\bigr)=\sum\_{z\bmod p}\chi(z-2)\psi(z).$$
Here $\chi(z-2)=\chi(2-z)$, because $\chi(-1)=\chi((-1)^3)=\chi^3(-1)=1$. Therefore,
$$\sum\_{y\bmod p}\chi(y^2-2)=\sum\_{z\bmod p}\chi(2-z)\psi(z)=\chi(2)\psi(2)J(\chi,\psi),$$
where $J(\chi,\psi)$ is the corresponding [Jacobi sum](https://en.wikipedia.org/wiki/Jacobi_sum).


To evaluate the Jacobi sum $J(\chi,\psi)$, we rely on Chapter 6 of Rose: A course in number theory (2nd ed., Oxford University Press, 1994). By (the last part of) Exercise 10 at the end of this chapter,
$$J(\chi,\psi)=\chi(4)J(\chi,\chi).$$
Now we observe that $4p=1+27(2a+1)^2$, and this is the only way to write $4p$ as $u^2+27v^2$ with $u\equiv 1\pmod{3}$ and $v$ positive, cf. Theorem 2.5 and the subsequent comments. Therefore, by Lemma 2.6,
$$J(\chi,\chi)=(3a+2)+(6a+3)e^{\pm 2\pi i/3}=\frac{1\pm(6a+3)i\sqrt{3}}{2},$$
the choice of the $\pm$ sign depending on which cubic character we denoted by $\chi$.


Putting everything together,
$$\sum\_{y\bmod p}\chi(y^2-2)=\chi(2)\psi(2)J(\chi,\psi)=\chi(8)\psi(2)\frac{1\pm(6a+3)i\sqrt{3}}{2}.$$
Here, $\chi(8)=\chi(2^3)=\chi^3(2)=1$, and so
$$\sum\_{y\bmod p}\chi(y^2-2)+\overline{\sum\_{y\bmod p}\chi(y^2-2)}=\psi(2)\sum\_{\pm}\frac{1\pm(6a+3)i\sqrt{3}}{2}=\psi(2).$$


To summarize, the number of points of the *affine* elliptic curve $E$ modulo $p$ equals $p+\psi(2)$.


