= Not every $x$ belongs to the elliptic curve over a non quadratically closed field
One major difference between the <elliptic curve over a finite field> or the <elliptic curve over the rational numbers> the <elliptic curve over the real numbers> is that not every possible $x$ generates a member of the curve.
This is because on the <equation Definition of the elliptic curves> we see that given an $x$, we calculate $x^3 + ax + b$, which always produces an element $y^2$.
But then we are not necessarily able to find an $y$ for the $y^2$, because not all <field (mathematics)>[fields] are not <quadratically closed fields>.
For example: with $a = 1$ and $b = 1$, taking $x = 1$ gives:
$$
y^2 = 1^3 + 1 \times 1 + 1 = 3
$$
and therefore there is no $y \in \Q$ that satisfies the equation. So $x = 1$ is not on the curve if we consider this <elliptic curve over the rational numbers>.
That $x$ would also not belong to <Elliptic curve over the finite field> $\F_4$, because doing everything $\mod 4$ we have:
$$
\begin{aligned}
0*0 &= 0 & &\mod 4 \\
1*1 &= 1 & &\mod 4 \\
2*2 &= 4 &= 0 &\mod 4 \\
3*3 &= 9 &= 1 &\mod 4 \\
\end{aligned}
$$
Therefore, there is no element $y \in \F_4$ such that $y \times y = 2$ or $y \times y = 3$, i.e. $2$ and $3$ don't have a <multiplicative inverse>.
For the <real numbers>, it would work however, because the <real numbers> are a <quadratically closed field>, and $\sqrt{3} \in \R$.
For this reason, it is not necessarily trivial to determine the <number of elements of an elliptic curve>.