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I have the following elliptic curve: $$ E: \quad Y^{2} = X^{3} + 1 ~ \text{over} ~ \mathbb{F}_{q}, ~ \text{where} ~ q \equiv 1 ~ (\text{mod} ~ 3). $$ I want to know the number of points on this curve. Any ideas on how I could approach the problem?

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    $\begingroup$ Read Ireland & Rosen's A Classical Introduction to Modern Number Theory, esp. section 3 of Chapter 18. They cover the case of prime numbers $p$, but if you understand general finite fields then you can extend their treatment of fields ${\mathbf F}_p$ for primes $p \equiv 1 \bmod 3$ to general finite fields with order $q \equiv 1 \bmod 3$. $\endgroup$
    – KCd
    Jun 21 '14 at 20:46
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When $q \equiv 1 \pmod 3$ and $q>3$ , there is a unique way (up to the sign of $b$) to write $q = a^2+3b^2$ with $a \equiv 1 \pmod 3$.

I think the explicit class field theory for $\Bbb Q(\sqrt{-3})$ can make a parallel between the action of $Frob_q$ on the points of finite order of the elliptic curve and the multiplication by $a+b\sqrt{-3}$ (or its conjugate) on $(\Bbb Q/\Bbb Z)^2$. Then the number of points is the number of fixpoints of $Frob_q$ which is $det(Frob_q-1) = Norm(a-1+b\sqrt{-3}) = q+1-2a$.

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