Elliptic curve $Y^2=X^3+X$

Question

Let $p$ be prime s.t. $p=a^2+4b^2, a\equiv 1\,mod\,4$.

Then, for $f=X^3+X$, $a_E(p)=2a$?

(Here, $a_E(p):=p+1-\#\{(x,y)\in\{0,...p-1\}^2|x^3+x\equiv y^2\,mod\,p\}$)

Answer 1

For the direct reference i will cite a related result from [Koblitz, Introduction to elliptic curves and modular forms]. It is a Theorem from II, §2, The zeta function of $E_n$, pages 59, 60, where $E_n$ is the "congruent number problem" elliptic curve: $$ E_n\ :\ y^2 = x^3-n^2x\ . $$ The results needed to prove the result in the sequel are using the Gauss and Jacobi sums, defined for a multiplicative character $\chi:\Bbb F_q^\times\to\Bbb C^\times$ by $$ \begin{aligned} g(\chi) &= \sum_x\chi(x)\psi(x)\ ,\\ J(\chi_1,\chi_2) &= \sum_x\chi_1(x)\chi_2(1-x)\ . \end{aligned} $$ (Above $\psi$ is an additive character, the sums are over elements in the field.)

It is related, because the given curve $Y^2=X^3+X$ and the curve $E_2$ of equation $y^2=x^3-4x$ are isogenous. Now let us cite / state:

Let $E_n$ be the curve $y^2=x^3-n^2x$, considered over the field $\Bbb F_p$ with $p$ elements. Here, the prime $p$ does not divide $2n$. Then $$ \begin{aligned} Z(E_n/\Bbb F_p, T) &=\frac{1-2aT+pT^2}{(1-T)(1-pT)}\\ &=\frac{(1-\alpha T)(1-\beta T)}{(1-T)(1-pT)}\ , \end{aligned} $$ where $\beta=\bar\alpha$, $a$ is the real part of $\alpha$ (or of $\beta$). Moreover:

• $\alpha,\beta$ are $\pm i\sqrt p$ in case of an odd prime congruent to $3$ modulo four,
• in the case $p$ is an odd prime congruent to $1$ modulo four, then $\alpha\in\Bbb Z[i]$ is characterized by having the norm $p$, and by being congruent with $\left(\frac np\right)$ modulo $2+2i$.

This is a generalization of the posted result.

The result for $p\equiv 3$ modulo four is simple to obtain, the polynomial $f$ given by $x^3+x$, respectively $x^3-n^2x$ being odd, so we can group points $\ne 0$ in $\Bbb F_p$ in pairs $\{x,-x\}$, and exactly one of $f(x)$, $f(-x)=-f(x)$ is a squaer.

The result for $p\equiv 1$ modulo four and $n=2$ leads to the posted question. (The $2$ is a square, so $\left(\frac2p\right)$ is $=1$.)

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Comment:

Here are some numerical examples, supported by sage code for the first primes:

sage: E = EllipticCurve( QQ, [1, 0] )
sage: E
Elliptic Curve defined by y^2 = x^3 + x over Rational Field
sage: for p in primes(200):
....:     if p%4 != 1:    continue
....:     ap = E.an(p)
....:     a = ZZ(ap/2); b = sqrt(p-a^2)
....:     print "p = %3s :: a_E(p) = %+3s = 2*(%+3d) :: p = %3s = (%+3d)^2 + %2s^2" % (p, ap, a, p, a, b)
....: 
p =   5 :: a_E(p) =   2 = 2*( +1) :: p =   5 = ( +1)^2 +  2^2
p =  13 :: a_E(p) =  -6 = 2*( -3) :: p =  13 = ( -3)^2 +  2^2
p =  17 :: a_E(p) =   2 = 2*( +1) :: p =  17 = ( +1)^2 +  4^2
p =  29 :: a_E(p) =  10 = 2*( +5) :: p =  29 = ( +5)^2 +  2^2
p =  37 :: a_E(p) =   2 = 2*( +1) :: p =  37 = ( +1)^2 +  6^2
p =  41 :: a_E(p) =  10 = 2*( +5) :: p =  41 = ( +5)^2 +  4^2
p =  53 :: a_E(p) = -14 = 2*( -7) :: p =  53 = ( -7)^2 +  2^2
p =  61 :: a_E(p) =  10 = 2*( +5) :: p =  61 = ( +5)^2 +  6^2
p =  73 :: a_E(p) =  -6 = 2*( -3) :: p =  73 = ( -3)^2 +  8^2
p =  89 :: a_E(p) =  10 = 2*( +5) :: p =  89 = ( +5)^2 +  8^2
p =  97 :: a_E(p) =  18 = 2*( +9) :: p =  97 = ( +9)^2 +  4^2
p = 101 :: a_E(p) =   2 = 2*( +1) :: p = 101 = ( +1)^2 + 10^2
p = 109 :: a_E(p) =  -6 = 2*( -3) :: p = 109 = ( -3)^2 + 10^2
p = 113 :: a_E(p) = -14 = 2*( -7) :: p = 113 = ( -7)^2 +  8^2
p = 137 :: a_E(p) = -22 = 2*(-11) :: p = 137 = (-11)^2 +  4^2
p = 149 :: a_E(p) = -14 = 2*( -7) :: p = 149 = ( -7)^2 + 10^2
p = 157 :: a_E(p) = -22 = 2*(-11) :: p = 157 = (-11)^2 +  6^2
p = 173 :: a_E(p) =  26 = 2*(+13) :: p = 173 = (+13)^2 +  2^2
p = 181 :: a_E(p) =  18 = 2*( +9) :: p = 181 = ( +9)^2 + 10^2
p = 193 :: a_E(p) = -14 = 2*( -7) :: p = 193 = ( -7)^2 + 12^2
p = 197 :: a_E(p) =   2 = 2*( +1) :: p = 197 = ( +1)^2 + 14^2

These explicit numerical examples are here, hoping that the questions gets more support.