$$\Large\textbf{Problem}$$ Let $E:y^2 = x^3 + 1$ be an elliptic curve. For each prime $5 \leq p \leq 13$, describe the group $E(\mathbb{F}_p)$, the Mordell-Weil group.
$$\Large\textbf{Attempts and Ideas}$$ I am having lots of thoughts about this problem, so I jot them down.
Since $E(\mathbb{F}_p)$ is a nonempty group, the identity $\infty$ (this is how my professor defines the identity in Mordell-Weil group) must be in $E(\mathbb{F}_p)$.
For any prime $p$ such that $5 \leq p \leq 13$, we see that there is a point of order $2$ in the group $E(\mathbb{F}_p)$ since there exists $x \in \mathbb{F}_p$ such that $0 = x^3 + 1$ holds.
It remains to determine the points of finite order $n \geq 2$. If $E:y^2 = x^3 + ax^2 + bx + c$, then
$$\mathrm{Discr}_E = -4a^3b + a^2b^2 + 18ab - 4b^3 - 27c^2$$
Since $a = b = 0$ and $c = 1$, $\mathrm{Discr}_E = -27$. The strong form of Nagell-Lutz Theorem tells us that if $\beta^2 | -27$ and $(\alpha, \beta) \in E(\mathbb{F}_p)$, then $\beta \in \{\pm 1, \pm 3\}$. Write
$$\beta^2 = \alpha^3 + 1$$
We need to check if there are points of finite order. If $\beta = \pm1$, then $ \alpha$ does not exist since it does not belong to $\mathbb{F}_p$ for any prime $5 \leq p \leq 13$ (at this point $\alpha = 0 \not\in \mathbb{F}_p$). If $\beta = \pm 3$, then $\alpha^3 = 8$, which implies $\alpha = 2$.
Overall, $E(\mathbb{F}_p)$ consists of $\infty$, the point of order $2$ and two points of order $4$ (I believe that this is isomorphic to $\mathbb{Z}_4$).
$$\Large\textbf{Notes}$$
This problem is similar to "The Group of points of Elliptic curve $y^2 = x^3 + 1$ over $\mathbb{F}_5$", but different since the given problem asks me to describe the group for prime $5 \leq p \leq 13$ (hopefully not a duplicate!).
Other than that, leave comments/answers/thoughts about the details I posted here.
Take for instance $p=5$. The curve $E: y^2=x^3+1$ has good reduction at $5$, so $E/\mathbb{F}_5$ is an elliptic curve, in particular, $E(\mathbb{F}_p)$ is a finite abelian group. You can easily find all the points of $E$ in $\mathbb{P}^2(\mathbb{F}_5)$, and verify that there are $6$ points (counting the point at infinity). Since $E(\mathbb{F}_p)$ is finite abelian of order $6$, it must be isomorphic to $\mathbb{Z}/6\mathbb{Z}$. Indeed, you can also verify that $P=(2,3)$ is a generator, with $$2P=(0,1),\ 3P=(4,0),\ 4P=(0,4),\ 5P=(2,2),$$ and $6P=\infty$. You can work out the structure of $E(\mathbb{F}_p)$, for any $p>3$ in a similar way. For example, you can verify that $E(\mathbb{F}_7)\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/6\mathbb{Z}$, or that $E(\mathbb{F}_{17})\cong \mathbb{Z}/18\mathbb{Z}$.