The cardinality of elliptic curves over finite field Given an elliptic curve over $\mathbb Q$ as $y^2=f(x)$ where $f(x)$ is a cubic polynomial.
In some places I read that if $p$ is a prime of good reduction then we have that $E(\mathbb F_p)=p+1$. Is this really true? If yes, why? Is there a (if possible simple) explanation? If it's not true in general, then what are the conditions?
I only know that we have $E(\mathbb F_p)\leq2p+1$.
Thanks in advance!
 A: That is false in general. For example if $E\colon y^2=x^3+x+1$, then $E(\mathbb F_5)=9$.
It is true that if $p>2$ is a prime of good supersingular reduction, then the trace of the Frobenius at $p$ is $0$, so that $E(\mathbb F_p)=p+1$.
A: Expanding on Ferra's answer. This is a true statement for $p > 3$ a prime of supersingular reduction, i.e. one for which the endomorphism ring is an order in a quaternion algebra. This is because we have that if $\phi_p$ is frobenius then we know that: 
$ [|{E(\mathbb{F}_p)}|] = [\text{deg}(1 - \phi_p)] = (1 - \phi_p)(1 - \widehat{\phi}_p) = [1] + [\text{deg}(\phi_p)] - [\text{Tr}(\phi_p)]$
And 
$\widehat{\phi_p} = [\text{Tr}(\phi_p)] - \phi_p$ is inseparable because our elliptic curve is supersingular, thus we know that $p | \text{Tr}(\phi_p)$ as the only inseparable isogenies of the form $[n], n \in \mathbb{Z}$ are those where $n = 0$ mod $p$. But by Hasse's estimate we have that $|\text{Tr}(\phi_p)| \leq 2\sqrt{p}$ and thus we have $np \leq 2\sqrt{p}$ and so for $p \neq 2,3$ we have $n = 0$ and thus $[|{E(\mathbb{F}_p)}|] = [1] + [\text{deg}(\phi_p)] = [1 + p]$ and so we are done.
