$$\Large\textbf{Problem}$$ Let $E$ be an elliptic curve defined by $y^2 = x^3 + ax$ where $a \in \mathbb{Z}$ is fourth-power free. Then \begin{aligned} E(\mathbb{Q})^{\text{tor}} = \left\{ \begin{array}{l l} \mathbb{Z}_2 \times \mathbb{Z}_2 & \text{if $-a$ is a square}\\ \mathbb{Z}_4 & \text{if $a = 4$}\\ \mathbb{Z}_2 & \text{if $a \neq 4$ and $-a$ is not a square} \end{array} \right. \end{aligned}
Assume that $a \geq 2$ is a prime. Prove that $E$ as above does not have $p$-torsion for any prime $p > 3$ i.e. that for any $p > 3$, $E(\mathbb{Q})$ does have any points of order $p$.
$$\Large\textbf{Attempts and Ideas}$$ We need to show that $E$ does not have $p$-torsion for any prime $p > 3$. Since $a \geq 2$ is prime, $(0,0)$ is the point of order $2$ in $E(\mathbb{Q})$ and in $E(\mathbb{Q})^{\text{tor}}$. By the discriminant of the elliptic curve formula, $\mathrm{Discr}_E = -4a^3$. Since $a$ is prime, by Nagell-Lutz theorem, if $(\alpha, \beta) \in E(\mathbb{Q})^{\text{tor}}$ and $\beta^2 \mid -4a^3$, then $\beta \in \{\pm 1, \pm 2, \pm a\}$. Either $a = 2$ or $a = 3$ or $a$ is any other odd prime.
If $a = 2$, then there are no rational points of the form $(\alpha,\beta)$, where $\beta \in \{\pm 1,\pm 2,\pm 4\}$. So $E(\mathbb{Q})^{\text{tor}}$ contains $\infty$ and $(0,0)$, the point of order $2$.
If $a = 3$, then $\beta \in \{\pm 1, \pm 2, \pm 3\}$. The torsion points are $(1,2)$ and $(1,-2)$. But since $a = 3 \neq 4$ and $-a$ is not a square, $E(\mathbb{Q})^{\text{tor}} = \mathbb{Z}_2$. So $(1,2)$ and $(1,-2)$ are points of order $4$.
If $a$ is any other odd prime, then by hypothesis there are no other torsion points in $E(\mathbb{Q})^{\text{tor}}$. So this group contains $\infty$ and $(0,0)$.
I am stuck for this part since I am not quite sure about my approach. Any comments or thoughts?
This is Proposition 6.1 in Chapter X of "The Arithmetic of Elliptic Curves" by J. H. Silverman, so you can look it up there.