Let $D\in\mathbb{N}$ with $D\equiv 2\pmod{3}$. Describe the torsion group over $\mathbb{Q}$ of the elliptic curve \begin{equation}Y^2=X^3+DX.\end{equation}
Idea: Firstly we recall that for an elliptic curve $y^2=x^3+Ax+B$ with $A,B$ integers, the discriminant $\Delta:=4A^3+27B^2$. In our case we have \begin{equation}\Delta=4D^3.\end{equation} Therefore any prime $p\ne 2$, $p\nmid D^3$ will be a prime of good reduction. This latter condition implies we are looking for primes $p>2, p\nmid D$. Since $D\equiv 2\pmod 3$ we know that $3\nmid D$ so $3$ is a prime of good reduction. So we reduce the curve modulo $3$.
The overall goal here is to list the points on the curve $\tilde{\mathcal{E}}(\mathbb{F}_3)$; we know that the order of $\mathcal{E}_{\textrm{tors}}(\mathbb{Q})$ divides this number, so we can get an upper bound on the number of elements and we can work from there.
The problem here is without knowing anything more about the value of $D$ it is hard to say whether a point will be on the curve. So I tried looking at specific values of $D$. Firstly take the obvious choice $D=2$. Then our curve is $Y^2=X^3+2X$; reducing modulo $3$ we have the points $(0,0)$ and the point at infinity $\textbf{o}$; so the order of $\mathcal{E}_{\textrm{tors}}(\mathbb{Q})$ is $1$ or $2$. Noting that $(0,0)$ lies on the curve and is of order $2$ we deduce that $\mathcal{E}_{\textrm{tors}}(\mathbb{Q})=C_2$ since we found a point of order $2$ on the curve.
If we do the same for $D=5$ we get the same result, but it isn't easy to see how to actually generalise this for all $D$.