Some properties of the distribution of the numbers of points on elliptic curves over a finite prime field

TL;DR

This paper investigates the distribution of the number of points on elliptic curves over finite prime fields, providing explicit evaluations of sums involving the trace of Frobenius for various parameters and congruence classes.

Contribution

It offers new explicit formulas for sums of Frobenius traces on elliptic curves over finite fields, using elementary exponential sum techniques and quadratic form results.

Findings

Derived formulas for sums of Frobenius traces over finite fields.

Evaluated sums for specific congruence classes of primes.

Provided an explicit cubic sum formula for primes congruent to 5 mod 6.

Abstract

Let $p \geq 5$ be a prime and for $a, b \in \mathbb{F}_{p}$, let $E_{a,b}$ denote the elliptic curve over $\mathbb{F}_{p}$ with equation $y^2=x^3+a\,x + b$. As usual define the trace of Frobenius $a_{p,\,a,\,b}$ by \begin{equation*} \#E_{a,b}(\mathbb{F}_{p}) = p+1 -a_{p,\,a,\,b}. \end{equation*} We use elementary facts about exponential sums and known results about binary quadratic forms over finite fields to evaluate the sums $\sum_{t\in\mathbb{F}_{p}} a_{p,\, t,\, b}$, $\sum _{t \in \mathbb{F}_{p}} a_{p,\,a,\, t}$, $ \sum_{t=0}^{p-1}a_{p,\,t,\,b}^{2}$, $ \sum_{t=0}^{p-1}a_{p,\,a,\,t}^{2}$ and $ \sum_{t=0}^{p-1}a_{p,\,t,\,b}^{3}$ for primes $p$ in various congruence classes. As an example of our results, we prove the following: Let $p \equiv 5$ $($mod 6$)$ be prime and let $b \in \mathbb{F}_{p}^{*}$. Then \begin{equation*} \sum_{t=0}^{p-1}a_{p,\,t,\,b}^{3}= -p\left((p-2)\left(\frac{-2}{p}\right) +2p\right)\left(\frac{b}{p}\right). \end{equation*}

Tables

Table 1. Table 1. f k ​ ( p ) subscript 𝑓 𝑘 𝑝 f_{k}(p) for small primes p ≡ 5 𝑝 5 p\equiv 5 ( ( ( mod 6 ) ) ) and small odd k 𝑘 k .

$p╲k$	5	7	9	11
5	-275	-2315	-20195	-179195
11	-10901	-358061	-12030821	-411625181
17	-36737	-1582913	-68613377	-3016710593
23	8257	2763745	304822657	27903893665
29	-35699	-396299	184745341	35260018501
41	-654401	-88683041	-12260782721	-1716248660321

Equations

$$\#E_{a,b}({\mathbb{F}}_p)=p+1-a_{p,a,b}.$$

$$\sum _{t=0}^{p-1}{a}_{p,t,b}^3=-p\left((p-2)\left(\frac{-2}{p}\right)+2p\right)\left(\frac{b}{p}\right).$$

$$\#E_{a,b}({\mathbb{F}}_p)=p+1-a_p.$$

$$a_p=-\sum _{x\in {\mathbb{F}}_p}\left(\frac{x^3+ax+b}{p}\right),$$

$$-2\sqrt{p}\le a_p\le 2\sqrt{p}.$$

$$f(k)=\#\{(a,b)\in {\mathbb{F}}_p^{\ast }\times {\mathbb{F}}_p^{\ast }:a_{p,a,b}=k\}.$$

$$\frac{p-1}{2}\bigg{|}f(k).$$

$$f_1(k)=\#\{(a,b)\in {\mathbb{F}}_p\times {\mathbb{F}}_p\setminus \{(0,0)\}:a_{p,a,b}=k\}.$$

$$f_1(k)=f_1(-k).$$

$$a_{p,a,b}=-a_{p,v^{2}a,v^{3}b}.$$

$$j=\frac{2^8{3}^3{a}^3}{4a^3+27b^2},$$

$$j(t):=\frac{2^8{3}^{3}a(t{)}^3}{4a(t{)}^3+27b(t{)}^2},$$

$$\sum _{t\in {\mathbb{F}}_p}{a}_{p,a(t),b(t)}^2=p^2+O(p^{3/2}).$$

$$S_R(p)=\sum _{a,b=0}^{p-1}{\left[\sum _{x=0}^{p-1}\left(\frac{x^3-ax-b}{p}\right)\right]}^{2R}$$

$$S_1(p)$$

$$S_2(p)$$

$$S_3(p)$$

$$S_4(p)$$

$$S_5(p)$$

$$\sum _{t=0}^{p-1}{a}_{p,t,b}^2=p\left(p-1-\left(\frac{-1}{p}\right)\right).$$

$$\sum _{t=0}^{p-1}{a}_{p,a,t}^2=p\left(p-1-\left(\frac{-3}{p}\right)-\left(\frac{-3a}{p}\right)\right).$$

$$\sum _{t=0}^{p-1}{a}_{p,t,b}^3=-p\left((p-2)\left(\frac{-2}{p}\right)+2p\right)\left(\frac{b}{p}\right).$$

$$\sum _{t=0}^{p-1}e\left(\frac{jt}{p}\right)=\{\begin{array}{ll}p,& p|j,\\ 0,& (j,p)=1.\end{array}$$

$$G_p=\{\begin{array}{ll}\sqrt{p},& p\equiv 1(\mathrm{mod}4),\\ i\sqrt{p},& p\equiv 3(\mathrm{mod}4).\end{array}$$

$$\left(\frac{z}{p}\right)=\frac{1}{G_P}\sum _{d=1}^{p-1}\left(\frac{d}{p}\right)e\left(\frac{dz}{p}\right).$$

$$\sum _{d\in {\mathbb{F}}_p\setminus \mathbb{H}}\left(\frac{d}{p}\right)=-\sum _{d\in \mathbb{H}}\left(\frac{d}{p}\right).$$

$$\sum _{t\in {\mathbb{F}}_p}{a}_{p,t,b}=-\sum _{x\in {\mathbb{F}}_p}\sum _{d=1}^{p-1}\frac{1}{G_P}\left(\frac{d}{p}\right)e\left(\frac{d(x^3+b)}{p}\right)\sum _{t\in {\mathbb{F}}_p}e\left(\frac{tdx}{p}\right)$$

$$\sum _{t\in {\mathbb{F}}_p}{a}_{p,t,b}=-\sum _{d=1}^{p-1}\frac{p}{G_P}\left(\frac{d}{p}\right)e\left(\frac{db}{p}\right)=-p\left(\frac{b}{p}\right).$$

$$\sum _{t\in {\mathbb{F}}_p}{a}_{p,a,t}=-\sum _{x\in {\mathbb{F}}_p}\sum _{d=1}^{p-1}\frac{1}{G_p}\left(\frac{d}{p}\right)e\left(\frac{d(x^3+ax)}{p}\right)\sum _{t\in {\mathbb{F}}_p}e\left(\frac{dt}{p}\right)=0.$$

$$\{a_{13,1,t}:0\le t\le 12\}=\{-6,-4,2,-1,0,5,1,1,5,0,-1,2,-4\}.$$

$$\sum _{t=0}^{p-1}{a}_{p,t,b}^2=p\left(p-1-\left(\frac{-1}{p}\right)\right).$$

$$\begin{gathered} \sum _{t\in {\mathbb{F}}_p}{a}_{p,t,b}^2=\frac{1}{G_p^2}\sum _{d_1,d_2=1}^{p-1}\left(\frac{d_1{d}_2}{p}\right)\sum _{x_1,x_2\in {\mathbb{F}}_p}e\left(\frac{d_1(x_1^3+b)+d_2(x_2^3+b)}{p}\right) \\ \times \sum _{t\in {\mathbb{F}}_p}e\left(\frac{t(d_1{x}_1+d_2{x}_2)}{p}\right). \end{gathered}$$

Full text

[TABLE]

We use elementary facts about exponential sums and known results about binary quadratic forms over finite fields to evaluate the sums $\sum_{t\in\mathbb{F}_{p}}a_{p,\,t,\,b}$, $\sum_{t\in\mathbb{F}_{p}}a_{p,\,a,\,t}$, $\sum_{t=0}^{p-1}a_{p,\,t,\,b}^{2}$, $\sum_{t=0}^{p-1}a_{p,\,a,\,t}^{2}$ and $\sum_{t=0}^{p-1}a_{p,\,t,\,b}^{3}$ for primes $p$ in various congruence classes.

As an example of our results, we prove the following: Let $p\equiv 5$ $($mod 6$)$ be prime and let $b\in\mathbb{F}_{p}^{*}$. Then

[TABLE]

1. Introduction

Let $p\geq 5$ be a prime and let $\mathbb{F}_{p}$ be the finite field of $p$ elements. For $a,b\in\mathbb{F}_{p}$, let $E_{a,b}$ denote the elliptic curve over $\mathbb{F}_{p}$ with equation $y^{2}=x^{3}+a\,x+b$. Denote by $E_{a,\,b}(\mathbb{F}_{p})$ the set of $\mathbb{F}_{p}$– rational points on the curve $E_{a,\,b}$ and define the trace of Frobenius, $a_{p}$, by the equation

[TABLE]

A simple counting argument makes it clear that

[TABLE]

where $\left(\frac{.}{p}\right)$ denotes the Legendre symbol. We recall some of the arithmetic properties of the distribution of $a_{p}$. The following theorem is due to Hasse [4]:

Theorem 1**.**

The integer $a_{p}$ satisfies

[TABLE]

Since we wish to look at how $a_{p}$ varies as the coefficients $a$ and $b$ of the elliptic curve vary, it is convenient for our purposes to write $a_{p}$ for the elliptic curve $E_{a,b}$ as $a_{p,\,a,\,b}$. The following result is well known (an easy consequence of the remarks on page 36 of [3], for example).

Proposition 1**.**

Let the function $f:\mathbb{Z}\to\mathbb{N}_{0}$ be defined by setting

[TABLE]

Then for each integer $k$,

[TABLE]

The following result can be found in [2] (page 57).

Proposition 2**.**

Define the function $f_{1}:\mathbb{Z}\to\mathbb{N}_{0}$ by setting

[TABLE]

Then for each integer $k$,

[TABLE]

The following result is also known ([3], page 37, for example).

Proposition 3**.**

Let $v$ be a quadratic non-residue modulo $p$. Then

[TABLE]

To better understand the distribution of the $a_{p,\,a,\,b}$ it makes sense to study the moments. The j -invariant of the elliptic curve $E_{a,b}$ is defined by

[TABLE]

provided $4a^{3}+27b^{2}\not=0$. Michel showed in [7] that if $\{E_{a(t),\,b(t)}:t\in\mathbb{F}_{p}\}$ is a one-parameter family of elliptic curves with $a(t)$ and $b(t)$ polynomials in $t$ such that

[TABLE]

is non-constant, then

[TABLE]

In [2] Birch defined

[TABLE]

for integral $R\geq 1$, and proved

Theorem 2**.**

111 In [2], Birch incorrectly omitted the factor of $p-1$ in his statement of Theorem 2.

For $p\geq 5$,

[TABLE]

where $\tau(p)$ is Ramanujan’s $\tau$-function.

Theorem 2 evaluates sums of the form $\sum_{a,\,b=0}^{p-1}a_{p,\,a,\,b}^{2R}$ in terms of $p$ and these results were derived by Birch as consequences of the Selberg trace formula .

In this present paper we instead use elementary facts about exponential sums and known results about binary quadratic forms over finite fields to evaluate the sums $\sum_{t\in\mathbb{F}_{p}}a_{p,\,t,\,b}$, $\sum_{t\in\mathbb{F}_{p}}a_{p,\,a,\,t}$, $\sum_{t=0}^{p-1}a_{p,\,t,\,b}^{2}$, $\sum_{t=0}^{p-1}a_{p,\,a,\,t}^{2}$ and $\sum_{t=0}^{p-1}a_{p,\,t,\,b}^{3}$, for primes $p$ in particular congruence classes. In particular, we prove the following theorems.

Theorem 3**.**

Let $p\geq 5$ be a prime, and $a,\,b\in\mathbb{F}_{p}$. Then

(i) $\sum_{t\in\mathbb{F}_{p}}a_{p,\,t,\,b}=-p\left(\frac{b}{p}\right)$,

(ii)$\sum_{t\in\mathbb{F}_{p}}a_{p,\,a,\,t}=0.$**

This result is elementary but we prove it for the sake of completeness.

Theorem 4**.**

Let $p\equiv 5$ $($mod 6$)$ be prime and let $b\in\mathbb{F}_{p}^{*}$. Then

[TABLE]

Theorem 5**.**

Let $p\geq 5$ be prime and let $a\in\mathbb{F}_{p}^{*}$. Then

[TABLE]

Theorem 4 and Theorem 5 could be deduced from Theorem 2, but we believe it is of interest to give elementary proofs that do not use the Selberg trace formula.

Theorem 6**.**

Let $p\equiv 5$ $($mod 6$)$ be prime and let $b\in\mathbb{F}_{p}^{*}$. Then

[TABLE]

2. Proof of the Theorems

We introduce some standard notation. Define $e(j/p):=\exp(2\pi ij/p)$, so that

[TABLE]

Define

[TABLE]

Lemma 1**.**

Let $\left(\frac{.}{p}\right)$ denote the Legendre symbol, modulo $p$. Then

[TABLE]

Proof.

See [1], Theorem 1.1.5 and Theorem 1.5.2. ∎

We will occasionally use the fact that if $\mathbb{H}$ is a subset of $\mathbb{F}_{p}$,

[TABLE]

We will also occasionally make use of some implications of the Law of Quadratic Reciprocity (see [5], page 53, for example).

Theorem 7**.**

*Let $p$ and $q$ be odd primes. Then

$($a)$$\left(\frac{-1}{p}\right)=(-1)^{(p-1)/2}.

$($b$)$ $\left(\frac{2}{p}\right)=(-1)^{(p^{2}-1)/8}$.

$($c$)$ $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{((p-1)/2)((q-1)/2)}$.*

We now prove Theorems 3, 4, 5 and 6,

Theorem 3. Let $p\geq 5$ be a prime, and $a,\,b\in\mathbb{F}_{p}$. Then

$($i$)$ $\sum_{t\in\mathbb{F}_{p}}a_{p,\,t,\,b}=-p\left(\frac{b}{p}\right)$,

$($ii$)$ $\sum_{t\in\mathbb{F}_{p}}a_{p,\,a,\,t}=0.$*

Proof. (i) From (1.1) and (2.3), it follows that

[TABLE]

The inner sum over $t$ is zero unless $x=0$, in which case it equals to $p$. The left side therefore can be simplified to give

[TABLE]

The last equality follows from (2.3).

$($ii$)$: From (1.1) and (2.3), it follows that

[TABLE]

The inner sum over $t$ is equal to [math], by (2.1), since $1\leq d\leq p-1$.

$\Box$

The result at (ii) follows also, in the case of primes $p\equiv 3$ (mod 4), from the fact that $a_{p,\,a,\,t}=-a_{p,\,a,\,p-t}$. However, this is not the case for primes $p\equiv 1$ (mod 4). For example,

[TABLE]

The results in Theorem 3 are almost certainly known, although we have not been able to find a reference.

Theorem 4. Let $p\equiv 5$ $($mod 6$)$ be prime and let $b\in\mathbb{F}_{p}^{*}$. Then

[TABLE]

Proof. From (1.1) and (2.3) it follows that

[TABLE]

The inner sum over $t$ is zero, unless $x_{1}\equiv-d_{1}^{-1}d_{2}x_{2}(\text{mod }p)$, in which case it equals $p$. Thus

[TABLE]

Since the map $x\to x^{3}$ is one-to-one on $F_{p}$, when $p\equiv 5(\text{mod }6)$, the $x_{2}^{3}$ in the inner sum can be replaced by $x_{2}$. Thus the inner sum is zero unless $d_{2}^{2}-d_{1}^{2}\equiv 0(\text{mod }p)$, in which case it equals $p$. It follows that

[TABLE]

We have once again used (2.3) to compute the sums, noting that the sums above start with $d_{1}=1$. The result now follows since $p/G_{p}^{2}\times(-1|p)=1$ for all primes $p\geq 3$.

$\Box$

Remarks: (1) It is clear that the results will remain true if $a(t)=t$ is replaced by any function $a(t)$ which is one-to-one on $F_{p}$.

(2) It is more difficult to determine the values taken by $\sum_{t\in\mathbb{F}_{p}}a_{p,t,b}^{2}$ for primes $p\equiv 1(\text{mod }6)$. This is principally because the map $x\to x^{3}$ is not one-to-one on $F_{p}$ for these primes (so that (2.1) cannot be used so easily to simplify the summation), but also because the answer depends on which coset $b$ belongs to in $\mathbb{F}_{p}^{*}/\mathbb{F}_{p}^{*3}$.

Before proving the next theorem, it is necessary to recall a result about quadratic forms over finite fields. Let $q$ be a power of an odd prime and let $\eta$ denote the quadratic character on $\mathbb{F}_{q}^{*}$ (so that if $q=p$, an odd prime, then $\eta(c)=\left(c/p\right)$, the Legendre symbol). The function $v$ is defined on $\mathbb{F}_{q}$ by

[TABLE]

Suppose

[TABLE]

is a quadratic form over $\mathbb{F}_{q}$, with associated matrix $A=(a_{ij})$ and let $\triangle$ denote the determinant of $A$ ($f$ is non-degenerate if $\triangle\not=0$).

Theorem 8**.**

Let $f$ be a non-degenerate quadratic form over $\mathbb{F}_{q}$, $q$ odd, in an even number $n$ of indeterminates. Then for $b\in\mathbb{F}_{q}$ the number of solutions of the equation $f(x_{1},\dots,x_{n})=b$ in $\mathbb{F}_{q}^{\,n}$ is

[TABLE]

Proof.

See [6], pp 282–293. ∎

Theorem 5. Let $p\geq 5$ be prime and let $a\in\mathbb{F}_{p}^{*}$. Then

[TABLE]

Proof. Once again (1.1) and (2.3) give that

[TABLE]

The inner sum over $t$ is zero, unless $d_{1}\equiv-d_{2}(\text{mod }p)$, in which case it equals $p$. Thus

[TABLE]

We have used the fact that $p/G_{p}^{2}\times(-1|p)=1$ for all primes $p\geq 3$. The inner sum over $d_{1}$ equals $-1$, unless one of the factors $x_{1}-x_{2}$, $x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}+a$ equals [math], in which case the sum is $p-1$. The equation $x_{1}=x_{2}$ has $p$ solutions and, by (2.6) with $q=p$, $n=2$, $f(x_{1},x_{2})=x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}$ and $A=\left(\begin{smallmatrix}1&(p+1)/2\\ (p+1)/2&1\end{smallmatrix}\right)$, the equation $x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}=-a$ has

[TABLE]

solutions. However, we need to be careful to avoid double counting and to examine when $x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}=-a$ has a solution with $x_{1}=x_{2}$. The equation $3x_{1}^{2}=-a$ will have two solutions if $\left(\frac{-3a}{p}\right)=1$ and none if $\left(\frac{-3a}{p}\right)=-1$. Hence the number of solutions to the equation $3x_{1}^{2}=-a$ is $\left(\frac{-3a}{p}\right)+1$. Thus the number of solutions to $(x_{1}-x_{2})(x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}+a)=0$ is

[TABLE]

Thus

[TABLE]

The right side now simplifies to give the result.

$\Box$

Before proving Theorem 6, we need some preliminary lemmas.

Lemma 2**.**

Let $p\equiv 5$ $($mod 6$)$ be prime. Then

[TABLE]

Proof.

If $z=0$, the left side of (2.8) becomes

[TABLE]

The second equality follows since $\{y^{3}:y\in\mathbb{F}_{p}\}=\{y:y\in\mathbb{F}_{p}\}$ for the primes $p$ being considered, the third equality follows from (2.1) and the last equality follows from (2.4).

If $z\not=0$, then the left side of (2.8) equals

[TABLE]

Now replace $y$ by $yz$ and then $z^{3}$ by $z$ (justified by the same argument as above) and finally $e$ by $ef$ to get this last sum equals

[TABLE]

We wish to extend the last sum to include $z=0$. If we set $z=0$ on the right side of the last equation (and denote the resulting sum by ”r.s.”) and sum over $d$ and $y$ we get that

[TABLE]

Replace $f$ by $f(e+1)^{-1}$ in the second sum above and then

[TABLE]

It follows that the left side of (2.9) equals

[TABLE]

and thus that the left side of (2.8) equals

[TABLE]

The second equality in (2.10) follows upon replacing $y$ by $ye^{-1}$ and then $e$ by $e^{-1}$. ∎

Lemma 3**.**

Let $p\equiv 5$ $($mod 6$)$ be prime. Then

[TABLE]

where

[TABLE]

Proof.

Upon changing the order of summation slightly, we get that

[TABLE]

If $y=0$, the inner double sum over $d$ and $z$ is zero, unless $f=\pm 1$, if which case it equals $p(p-1)$ and the right side of (2.11) equals

[TABLE]

By similar reasoning, if $y=-1$, the right side of (2.11) also equals

[TABLE]

Thus

[TABLE]

where the last equality follows upon replacing $f$ by $f(y+1)^{-1}$ and $e$ by $ey^{-1}$. The inner sum over $d$ and $z$ is zero unless

[TABLE]

in which case the inner sum is $p(p-1)$. We distinguish the cases $e^{2}=f^{2}$ and $e^{2}\not=f^{2}$. If $e^{2}=f^{2}$, then necessarily $e^{2}=f^{2}=1$ and the sum on the right side of (2.12) becomes

[TABLE]

If $e^{2}\not=f^{2}$ then

[TABLE]

and since $y\not=0,-1$, we exclude $f^{2}=1$ and $e^{2}=1$. After substituting for $y$ in the sum in the final expression in (2.12), we find that

[TABLE]

where

[TABLE]

∎

Lemma 4**.**

Let $p\equiv 5$ $($mod 6$)$ be prime and let $S^{**}$ be as defined in Lemma 3. Then

[TABLE]

Proof.

Clearly we can remove the restrictions $f\not=e$, $f\not=1$ and $e\not=-1$ freely. If we set $f=-e$, we have that

[TABLE]

The last equality follows from (2.4). Thus

[TABLE]

If $f$ is set equal to 0 in the sum above we get

[TABLE]

If $f$ is set equal to -1 in this sum we get

[TABLE]

Thus

[TABLE]

If we set $e=0$ in this latest sum we get

[TABLE]

If we set $e=1$ in this sum we get

[TABLE]

Thus

[TABLE]

∎

Lemma 5**.**

Let $p\equiv 5($ mod $6)$ be prime. Then

[TABLE]

Proof.

If $f$ is replaced by $f+1$ and then $e$ is replaced by $e+f$, the value of the double sum above does not change. Thus

[TABLE]

We evaluate the inner sum using (2.3).

[TABLE]

The next-to-last equality follows since the sum over $f$ in the previous expression is 0, unless $d_{1}=-d_{2}$, in which case this sum is $p$. The sum over $d_{2}$ equals $p-1$ if $e=p-1$ and equals $-1$ otherwise. Hence the sum at (2.16) equals

[TABLE]

the last equality following from the remark after (2.7). ∎

Corollary 1**.**

Let $S^{*}$ and $S^{**}$ be as defined in Lemma 3. Then

[TABLE]

Proof.

Lemmas 4 and 5 and the fact that $(-3|p)=-1$ if $p\equiv 5(\,mod\,6)$ give (i). Lemma 3 and part (i) give (ii). ∎

Theorem 6. Let $p\equiv 5$ $($mod 6$)$ be prime and let $b\in\mathbb{F}_{p}^{*}$. Then

[TABLE]

Proof. Let $g$ be a generator of $\mathbb{F}_{p}^{*}$. It is a simple matter to show, using (1.1), that

[TABLE]

Thus the statement at (2.17) is equivalent to the statement

[TABLE]

Let $S$ denote the left side of (2.18). From (1.1) and (2.3) it follows that

[TABLE]

The inner sum is zero, unless $dx+ey+fz=0$ in $\mathbb{F}_{p}$, in which case it equals $p$. Upon letting $x=-d^{-1}(ey+fz)$, replacing $e$ by $de$ and $f$ by $fe$, we get that

[TABLE]

which was what needed to be shown, by (2.18). The second equality above follows from Lemma 2. Above $S^{*}$ is as defined in Lemma 3 and in the next-to-last equality we used Corollary 1, part (ii). In the last equality we used once again the fact that $p/G_{p}^{2}(-1|p)=1$.

$\Box$

3. Concluding Remarks

Let $p\equiv 5$ $($mod 6$)$ be prime, $b\in\mathbb{F}_{p}^{*}$ and $k$ be an odd positive integer. Define

[TABLE]

(It is not difficult to show that the right side is independent of $b\in\mathbb{F}_{p}^{*}$)

By Theorem 6

[TABLE]

We have not been able to determine $f_{k}(p)$ for $k\geq 5$ (We do not consider even $k$, since a formula for each even $k$ can be derived from Birch’s work in [2]). We conclude with a table of values of $f_{k}(p)$ and small primes $p\equiv 5$ $($mod 6$)$, with the hope of encouraging others to work on this problem.