The cardinal of the Mordell-Weil group is prime for certain elliptic curves over $\mathbb{F}_{q}$ for certain $q$. Let $p\in\{2,3\}$ and $r\in\mathbb{Z}_{\geq 2}$. I would like to find if there exists an elliptic curve defined over $\mathbb{F}_{p}$ such that $|E(\mathbb{F}_{p^{r}})|$ is a prime number. If $p>3$ the answer is no as a consequence of Hasse's bound, but I am not able to prove/find a counterexample if $p\leq 3$.
Any hint would be appreciated.
 A: Because $E(\Bbb{F}_{p^n})$ is a subgroup of $E(\Bbb{F}_{p^m})$ whenever $n\mid m$ we have "necessary" conditions: $r$ has to be a prime, and $|E(\Bbb{F}_p)|=1$. A rare exception might be when $E(\Bbb{F}_p)=E(\Bbb{F}_{p^r})$, but this is difficult to arrange (may be impossible - I haven't tried).
Let's try $p=2$. If we have $|E(\Bbb{F}_p)|=1$, then the zeros $\omega_1$ and $\omega_2=\overline{\omega_1}$ of the $L$-function of $E$ satisfy the system
$$
\left\{\begin{array}{lcl}
\omega_1+\omega_2&=&2,\\
\omega_1\omega_2&=&2.
\end{array}
\right.
$$
The solution of this system is $\omega_{1,2}=1\pm i.$
Given such a curve, the Hasse-Davenport relations imply
$$
N_r=|E(\Bbb{F}_{p^r})|=2^r+1-\omega_1^r-\omega_2^r.
$$
A bit of calculation then reveals that this is a prime number for
at least $r=2,3,5,7,11,19$. With $r=13$ (resp. $r=17$) we get $N_r=53\cdot157$
(resp. $N_{17}=137\cdot 953$).
So we need to find such a curve defined over $\Bbb{F}_2$. The equation
$$
y^2+y=x^3+x+1
$$
looks like it will work. The r.h.s. is $=1$ for all $x\in\Bbb{F}_2$ and the l.h.s. is $=0$ for all $y\in\Bbb{F}_2$, so only the point at infinity is there.
With this curve it is actually easy to check by hand that $N_2=5$. If $x\in\Bbb{F}_2$ the r.h.s. is $=1$ and we have two matching choices for $y$, namely the primitive cubic roots of unity $\in\Bbb{F}_4\setminus\Bbb{F}_2$. OTOH if $x\in\Bbb{F}_4\setminus\Bbb{F}_2$ then the r.h.s. $\notin\Bbb{F}_2$ but the l.h.s. $\in\Bbb{F}_2$ for all $y\in\Bbb{F}_4$. Hence we only have four affine points: $(0,\alpha), (0,\alpha+1), (1,\alpha), (1,\alpha+1)$.
