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Let $p$ be a prime number(maybe generic large odd prime). Let $k$ be integer with $(k,p)=1$. The equation in question is $$X^2Y+Y^2+X=kXY.$$ In homogeneous form, it is alternating $$X^2Y+Y^2Z+Z^2X=kXYZ.$$ The elliptic curve always contain three points (0,0,1), (1,0,0) and (0,0,1).

My question: Is there any formula relating the number of points in $\mathbb Z/p\mathbb Z$ to something else ? $$\#\{(X,Y)\in (\mathbb Z/p\mathbb Z)^2 : X^2Y+Y^2Z+Z^2X=kXYZ\}= \;?$$ I am expecting the right-hand side be something related to certain characters either cubic or related to cubic fields (involving $k$ for sure).

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  • $\begingroup$ I imagine the first thing to do is put it in Weierstrass form, have you done that? $\endgroup$
    – Kevin Casto
    Commented Mar 14, 2017 at 0:36
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    $\begingroup$ A Weierstrass form is obtained as follows: If one fixes $x=x_0$, then the points on the curve are the solutions to $y$ in the equation $x_0^2y+y^2+x_0=kx_0y$. Rewrite this as $y^2+(x_0^2-kx_0)y+x_0=0$; this is a quadratic with discriminant $(x_0^2-kx_0)^2-4x_0 = x_0^4-2kx_0^3+k^2x_0^2-4x_0$. This then means the curve is birational to $\eta^2 =x^4-2kx^3+k^2x^2-4x$, and so, by a standard change of variables to $v^2=-4u^3+k^2u^2-2ku+1$, and so further to $w^2=z^3+k^2z^2+8kz+16$. $\endgroup$
    – R.P.
    Commented Mar 14, 2017 at 0:51
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    $\begingroup$ There's a nice formula for $k=0$ and, less immediately, for $k=-6$ (same formula, in fact). I think these are the only cases where the curve has CM, so one does not expect simple formulas for other $k$. $\endgroup$
    – Noam D. Elkies
    Commented Mar 14, 2017 at 4:36

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It's not clear to me why you expect $\#E(\mathbb F_p)$ to be related to cubic characters or to $\mathbb Q(\sqrt[3]{k})$. As far as I can tell, your curves are not cubic twists of a single curve. The easiest way to check this, as Kevin Casto suggests, is to covert to Weierstrass form and compute the $j$-invariant. (Added: Using Rene's calculation, $j(E_k)=\frac{k^{12} - 72 k^9 + 1728 k^6 - 13824 k^3}{k^3 - 27}$.) Thus you don't have a twist family. On the other hand, your question "Is there any formula relating the number of points in $\mathbb Z/p\mathbb Z$ to something else?" has an affirmative answer, but probably not one you want. Namely if you fit $\#E_k(\mathbb F_p)$ appropriately into an $L$-series, then the Mellin transform will be a modular form (courtesy of Wiles et al.), so the number of points is related to the coefficients of the Fourier expansion of a modular form.

For $k=1$, let $N_p=\#E_1(\mathbb F_p)$. Then the first few values are $$N_{5}=9, N_{7}=9, N_{11}=6, N_{13}=13, N_{17}=21, N_{19}=18, N_{23}=24, N_{29}=24, N_{31}=36, N_{37}=45, N_{41}=42, N_{43}=45, N_{47}=45.$$ This sequence is not in the OEIS.

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    $\begingroup$ The $k=1$ curve has conductor $26$; it might make sense to add to OEIS the $N_p$ sequences for all isogeny classes of conductor up to say $100$. $\endgroup$
    – Noam D. Elkies
    Commented Mar 14, 2017 at 4:34
  • $\begingroup$ @NoamD.Elkies Sure, feel free to do so if you get a chance. I've submitted a few sequences to OEIS, but never tried to submit a long list. $\endgroup$
    – Joe Silverman
    Commented Mar 14, 2017 at 13:46
  • $\begingroup$ Then again maybe the natural thing to upload is the coefficients of the modular form, from which we soon recover $N_p = p + 1 - a_p$. $\endgroup$
    – Noam D. Elkies
    Commented Mar 14, 2017 at 14:23
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The number of projective solutions is $p+1$ plus the Legendre symbol $\left( \frac{k}{p}\right)$ plus a finite symbol hypergeometric sum involving $(k/3)^3$.

To show this, take Rene's Weierstrass form

$ w^2=z^3+k^2z^2+8kz+16$

Multiply z by $k^4$ and $w$ by $k^6$.

$ w^2=z^3+k^6z^2+8k^9z+16k^{12}$

We obtain a polynomial in $t=k^3$ with discriminant $256 t^8 (t-27)$ so it has three singular points $0,1,\infty$. Hence it has singularities only at $t=0,t=27, t=\infty$. Because the original curve has good reduction at $k=0$, the new curve has potentially good reduction, which because the discriminant has a zero of order $8$ means monodromy of order $3$. At $t=27$, there is semistable reduction, hence unipotent local monodromy. At $t=\infty$ there was semistable reduction for the original curve (easiest to tell from the original form $XYZ$) so the cubic pullback must also have semistable reduction, so the local monodromy is minus a unipotent.

By Katz's hypergeometric rigidity that means the $H^1$ of this family of curves is geometrically isomorphic to the $\mathcal H\begin{pmatrix} 1 ,1 \\ \chi_3 ,\overline{\chi}_3 \end{pmatrix}$ hypergeometric sheaf, so the trace function is a constant times the hypergeometric trace function

$\sum_{a,b \in \mathbb F_p | ab =t/27 } \chi_3\left(\frac{a}{a-1}\right) \overline{\chi}_3\left(\frac{b}{b-1}\right) = \sum_{a,b \in \mathbb F_p | ab =27/t } \chi_3\left(a-1\right) \overline{\chi}_3\left(b-1\right)$

where $\chi_3$ is a character of order $3$

so the count of points on your original curve should have the form

$$p+1 + c_p \sum_{a,b \in \mathbb F_p | ab =27/t} \chi_3\left(a-1\right) \overline{\chi}_3\left(b-1\right) $$

where $c_p$ is a constant (I think one can check a root of unity) depending on $p$.

Probably there is a direct and elementary proof of this by manipulating the sum, which should also give the constant.

If we consider the map from this curve to the curve of $A:B:C:D$ in $\mathbb P^3$ satisfying $A+B+C=kD$ and $ABC=D^3$ given by $A=X^2Y, B= Y^2Z,C=Z^2X, D=ABC$ we see that it has an inverse - note that $X/Z = A/D$, $Y/X= B/D$, so our inverse can take $(A:B:C:D)$ to $(AD, AB, D^2)$. So it's sufficient to count the projective solutions to this equation (unless there are some singularities and the inverse doesn't exactly exist - I don't want to calculate this right now but it would be easy to).

For an additive character $\psi$ of $\mathbb F_p$, the exponential sum

$\sum_{A,B,C,D | ABC = D^3} \psi(A+B+C-kD)$

is $p$ times the number of projective solutions of the equation minus the number of projective solutions of $ABC=D^3$ (which is $(p-1)^2$ where $D$ is nonzero plus $3p$ where $D$ is zero, or $(p^2+p-1)$) plus $1$.

Hence the count of points is $p+1$ plus $\frac{1}{p} \sum_{A,B,C,D | ABC = D^3} \psi(A+B+C-kD)$

Up to some other terms which I believe magically cancel, this is $\frac{1}{p(p-1) } \sum_{A,B,C,D ,\chi} \psi(A+B+C-kD)\chi(A) \chi(B) \chi(C) \chi^{-3}(D) $ which can be written in terms of Gauss sums as $\frac{1}{p(p-1) } \sum_\chi G(\chi,\psi)^3 G(\chi^{-3},\psi) \chi^3(-k)$.

If $p=1$ mod $3$ one can make cubic characters appear using a Hasse-Davenport relation $G(\chi^{-3},\psi) = - G(\chi^{-1},\psi) G(\chi_3 \chi^{-1},\psi) G(\overline{\chi}_3 \chi^{-1},\psi)\chi^{-3}(3)/(G(1,\psi) G(\chi_3,\psi) G(\overline{\chi}_3,\psi)$

using $G(\chi,\psi) G(\chi^{-1},\psi) = p \chi(-1)$ and multiplicativity of $\chi$ we obtain

$$\frac{1}{(p-1)(G(1,\psi) G(\chi_3,\psi) G(\overline{\chi}_3,\psi) } \sum_\chi G(\chi,\psi)^2 G(\chi_3 \chi^{-1},\psi) G(\overline{\chi}_3 \chi^{-1},\psi) \chi(k^3/27)$$

where the inner sum is another formulation of the hypergeometric sum $\mathcal H\begin{pmatrix} 1 ,1 \\ \chi_3 ,\overline{\chi}_3 \end{pmatrix}$ applied to $\chi^3/27$.

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    $\begingroup$ How can there be a $\chi_3$ if $p$ is allowed to be $-1 \bmod 3$? (Presumably once one has the correct formula it can be proved by more elementary and accessible means than invoking "Katz's hypergeometric rigidity" and a "hypergeometri[c] sheaf" . . .) $\endgroup$
    – Noam D. Elkies
    Commented Mar 14, 2017 at 4:29
  • $\begingroup$ @NoamD.Elkies In this case, I think if you sum over $\alpha \in \mathbb F_{p^2}$ with norm $t/27$ of $\chi_3(\alpha/(\alpha-1)) \chi_2(\alpha)$, where $\chi_3$ and $\chi_2$ are the quadratic characters over $\mathbb F_{p^2}$, one gets the right answer. One can unify these formulas by viewing the sum as over the algebra $\mathbb F_p[\zeta]/(\zeta^2+\zeta+1)$, which is the field $\mathbb F_{p^2}$ if $p=-1 \mod 3$ and is $\mathbb F_p \times \mathbb F_p$ if $p = 1 \mod 3$. $\endgroup$
    – Will Sawin
    Commented Mar 14, 2017 at 4:38
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    $\begingroup$ Isn't $p+1+\sum_z \chi_2(z^3+k^3z^2+8k^3z+16k^3)$ just as good a formula? What sort of information can one gain from your formula? $\endgroup$
    – Chris Wuthrich
    Commented Mar 14, 2017 at 9:16
  • $\begingroup$ @ChrisWuthrich Not very much, but some. There are some theorems about hypergeometric functions that can be applied to e.g. calculate the distribution of $a_p$. While this can make it helpful to transform other explicit sums into hypergeometric functions, there are also theorems about elliptic curves that accomplish the same task. However if for some reason you want to study the Fourier transform of this function than there are theorems about hypergeometric sheaves that could be helpful. $\endgroup$
    – Will Sawin
    Commented Mar 14, 2017 at 14:53

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