I am interested in the following problem:
Prove that points of finite order on the curve $ y^2=x^3+Dx$ where $D$ is non zero integer is described as follows: $$\{P\in C (\mathbb Q): P \text{ has finite order}\}\cong\begin{cases}\mathbb Z/4\mathbb Z &\text{if $D=4d^4$ for some $d$}\\ \mathbb Z/2\mathbb Z \bigoplus \mathbb Z/2\mathbb Z &\text{if $D=-d^4$ for some $d$}\\ \mathbb Z/2\mathbb Z &\text{otherwise} \end{cases}$$
I came across the above problem in exercise 3 from the book Rational Points on Elliptic Curves by Joseph H. Silverman. We of course need to use Nagell–Lutz theorem, however, I am not sure how to proceed.
Any help will be highly appreciated.
The question in the OP is Exercise III.3.7 (c)** at page 1041-05 in [Silverman, Tate: Rational points on elliptic curves]. Since it comes with a double asterisk, we will deal equivalently with an other exercise.
This is Exercise IV.4.9 at page 142 in [Silverman, Tate: Rational points on elliptic curves], it deals with curves over $\Bbb Q$ $$ E(b)\ :\ y^2 = x^3 + bx\ , $$ where $b$ is a non-zero integer which is fourth power free.
Note: If $b$ is (initially) rational with a denominator $>1$, multiply with the fourth power of the denominator to get an integer instead. Then eliminate the primes at powers $\ge 4$ in the prime factor decomposition using the observation, that the curve $ y^2=x^3+bu^4x $ can be written equivalently after division with $u^6$ as $(y/u^3)=(x/u^2)^3+b(x/u^2)$ and the substitution $X=x/u^2$, $Y=y/u^3$ leads to an equation of type of $E(b)$ in $(X,Y)$.
Note: In loc. cit. the letter used for the coefficient in $x$ is $b$, not $D$ as in the OP and in Exercise III.3.7 (c)**. (So we have $D$ still available for the discriminant $D=-4b^3$.)
For the convenience of the reader, let us recall the Exercise 4.8 preceding the Exercise 4.9:
The Exercise 4.8 comes with small steps. In (a) we note that the cyclic group $F^\times$ has $(p-1)$ elements, which is $2$ modulo four. Then both maps $u\to u^4$ and $v\to v^2$ have as image the subgroup $G=(F^\times)^2$ of odd order $(p-1)/2$ and index two in $F^\times$. Note that squaring is an isomorphism of $G$. We then work with the new variable $U=u^2$, and it remains to count for every element (in our case $-4b$) in $F$ the representations as difference $v^2-U^2$ of two squares. It is enough to consider the elements $\pm 1$. We may exchange $u\leftrightarrow v$, wo we have to count the number of solutions for $v^2-U^2=1$. With the same argument, all equations of the shape $v^2-U^2=-4b$ have the same number of solutions, $b\ne 0$. We count separately the case $b=0$, there are $1+2(p-1)=2p-1$ realizations, $v=0$ is one of them, for the other values of $v$ we have to take $U=\pm v$. So each of the equations $v^2-U^2=-4b$ has $$\frac{p^2-(2p-1)}{p-1}=(p-1) $$ solutions. This gives (a). For (b) we simply check: $$ \begin{aligned} y^2 - x^3 &= \frac 18(u^2+v)^2\Big[\ 2u^2-(u^2+v)\ \Big] = \frac 18(u^2+v)^2(u^2-v) = \frac 18(u^2+v)(u^4-v^2) \\ &= \frac 18(u^2+v)4b=bx\ . \end{aligned} $$ For (c), we observe that for each point $(x,y)$ with $x\ne 0$ we can uniquely associate $u=y/x$, which determines $v$. It remains to consider $x=0$ and the infinity point.
Now we come back to 4.9 and are in position to apply the theorem (Reduction mod $p$ Theorem, page 123) saying roughly that for all but finitely many primes $p$ the composition $\Phi\to E(\Bbb Q)\to E(\Bbb F_p)$ is an injection, thus the order of $\Phi$ divides $E(\Bbb F_p)$. Using IV.4.8, $|\Phi|$ divides $(p+1)$ for all but finitely many values of $p$ which are $3$ modulo four. For prime density reasons $|\Phi|$ divides $4$.
(This density argument is the canon tool used to move the double asterisk to an other front.)
The Exercise IV.4.9 reduces now to the study of all possibilities for the group $\Phi$ of order either $2$ or $4$. Since the point $(0,0)$ is a two-torsion point.
The situation of a group $\Phi$ of the shape $\Bbb Z/2\oplus \Bbb Z/2$ is possible if there is an other point of order two. Such a point has the $y$-component zero, so the equation $x^3+bx=0$ must have further solutions, so $b$ is minus a square. This is the case (b2).
The situation $\Phi=\Bbb Z/4$ is possible, iff there is a point $T(s,t)$ with $2T=(0,0)$. Point doubling formulas show that after considering the slope $m=(3s^2+b)/(2t)$ of the curve in $T$ we have $0=m^2-2s$ and $0=m(s-0)-t$. Together with the equation of the curve $t^2=s(s^2+b)$ we eliminate $m$, get $$ t^2=\frac s2(3s^2+b) = 2s^3=s(s^2+b)\ . $$ So $b=s^2$ is a square, and from $t^2=2s^3$ we have $s=2d^2$. This leads to $b=s^2=4d^4$, which is case (b1).
In any other situation $|\Phi|\ne 4$, so $|\Phi|$ remains $2$, and $\Phi$ is generated by $(0,0)$, the situation (b3).