What is the upper bound of order of $ E(F_{599})$ Can you please help me to slove this problem:
Let E be the elliptic curve $y^{2}=x^{3}+1$ over the finite field $F_{599}$.
Using Hasse's theorem find What is the upper bound of order of $ E(F_{599})$  ?
Thanks a lot.
Upd: What I have done?
I thought that I should just to put $q=599$ into Hasse's theorem. 
http://en.wikipedia.org/wiki/Hasse%27s_theorem_on_elliptic_curves
But then, why in$F_{23}$ case I should get that the upper bound for $N$, the order of  $ E(F_{23})$ is 16?? Why $N$ is it not equal to 15?
$|15-(23+1)|\leq 2\sqrt{23}$
 A: Let $E$ be an elliptic curve defined over a finite field $\mathbb{F}_q$, where $q$ is some power of a prime $p$. The Hasse bound says that 
$$-2\sqrt{q} \leq \# E(\mathbb{F}_q) - (q+1) \leq 2\sqrt{q},$$
or equivalently,
$$q+1-2\sqrt{q} \leq \# E(\mathbb{F}_q) \leq q+1+2\sqrt{q}.$$
In other words, the number of points on $E$ defined over $\mathbb{F}_q$ is approximately $q+1$, and the error (in absolute value) in this estimation is less or equal to $2\sqrt{q}$.
Some examples. Let $E:y^2=x^3+1$.


*

*Let $\mathbb{F}_q = \mathbb{F}_{13}$, a finite field with $13$ elements. Then, $\# E(\mathbb{F}_{13})=12$, so the actual error in approximating $\# E(\mathbb{F}_{13})$ by $q+1=13+1=14$ in this case is $2$. The Hasse bound predicts that the error is less or equal to $2\sqrt{13}=7.21110\ldots.$.

*Let $\mathbb{F}_q = \mathbb{F}_{23}$, a finite field with $23$ elements. Then, $\# E(\mathbb{F}_{23})=24=23+1$, so the actual error in approximating $\# E(\mathbb{F}_{23})$ by $q+1=23+1$ in this case is zero. The Hasse bound predicts that the error is less or equal to $2\sqrt{23}=9.59166\ldots.$.

*Let $\mathbb{F}_q = \mathbb{F}_{599}$, a finite field with $599$ elements. Then, $\# E(\mathbb{F}_{599})=600=599+1$, so the actual error in approximating $\# E(\mathbb{F}_{599})$ by $q+1=599+1$ in this case is zero. The Hasse bound predicts that the error is less or equal to $2\sqrt{599}=48.948953\ldots.$.

*More generally, let $q=p\geq 5$ be a prime congruent to $2\bmod 3$ (such as $p=23$ or $p=599$; notice that $13\equiv 1 \bmod 3$ so it does not fit here), and let $\mathbb{F}_p$ be a finite field with $p$ elements. Then, $E/\mathbb{F}_p$ is supersingular, and therefore $\# E(\mathbb{F}_{p})=p+1$, so the actual error in approximating $\# E(\mathbb{F}_{p})$ by $p+1$ in this case is zero. The Hasse bound predicts that the error is less or equal to $2\sqrt{p}$. 
