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I'm studying elliptic curves and have stumbled upon this problem:

Let p be a prime number such that 3 does not divide p − 1. Let E be an elliptic curve defined like this:

$E = \{ (x,y ) \in \mathbb{F}_{p}^2 | \quad Y^{2} = X^{3} + 7 \} .$

The Goal is to compute $| E(\mathbb{F}_{p})|$.

I've seen Hasse's bound: $|E(\Bbb F_p)| \geq p+1-2\sqrt p > 1, \quad\forall p \geq 5$

And that the number of points is $N=1+\sum_{x\in\Bbb{F}_p}\left(1+\left(\frac{x^3+ax+b}p\right)\right).$

(I thought that for $p > 3$, $p$ can be written as $p = 3*k + 2$ for some $k \in \mathbb{Z}$ since p-1 is not divisible by 3. However i don't see how i could use this.)

Has anyone got an idea how to compute the number of points?

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You made all the relevant observations, so let me outline an answer in the form of exercises for you.

Let $p$ be an odd prime with $p \equiv 2 \pmod{3}$.

Exercise 1: show that the map $\varphi: \mathbf F_p^\times \rightarrow \mathbf F_p^\times$ given by $\varphi(x) = x^3$ is an automorphism.

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Exercise 2: use exercise 1 to show that $$\sum_{x \in \mathbf F_p}\left(\frac{x^3 + 7}{p}\right) = \sum_{x \in \mathbf F_p}\left(\frac{x}{p}\right) = 0 $$

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Exercise 3: conclude that $|E(\mathbf F_p)| = p+1$.

Can you solve it now?

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