How many duplication formulas exist for the Mordell curve family $Y^2-X^3=c$? For the Mordell equation
$$
Y^2-X^3 = c,
$$
Bachet gave a famous duplication formula which translates one rational solution $(x_1,y_1)$ into a second rational solution $(x_2,y_2)$. Réalis gave a second such formula.
My question is: How many others are there?
 A: The projective curve $E$ given by the equation $Y^2Z = X^3 + cZ^3$
is a smooth curve of genus $1$ over $R = \mathbb Q[c, c^{-1}]$. You are essentially asking for a description of $$\text{End}_{R}({E}).$$
The curve $E$ has the rational point
$$\mathcal O := [0:1:0] \in E(R)$$
which we can use as a base-point, so that $E$ is an elliptic curve over $R$. I believe that the only nonconstant endomorphisms of the curve $E/R$ are the multiplication by $n$ maps $[n]: E \to E$. 
Any nonconstant endomorphism $\sigma$ is of the form $\sigma = t \circ \sigma'$ where $t$ is a translation $P \mapsto P+\sigma(\mathcal O)$ and $\sigma'$ is an isogeny (an endomorphism preserving the point $\mathcal O$). 
Therefore, to describe $\text{End}_{R}({E})$, we should describe:


*

*The Mordell-Weil group $E(R)$;

*The endomorphism ring of $E/R$ as an elliptic curve.


I claim that $E(R) = \{\mathcal O\}$, assuming that the following is true:

For infinitely many $c \in \mathbb Q^\times$, the specialization $E_c/\mathbb Q$ has trivial Mordell-Weil group.

In fact, one certainly expects that for a positive proportion of $c \in \mathbb Q^\times$, $c$ ordered by height, the specialization $E_c$ has rank $0$. This is likely true, and the data is convincing, but I'm not sure if this is known at the moment. 
Let me explain how $E(R) = \{\mathcal O\}$ follows from this conjecture. Suppose that $P = [f : g : h] \in E(R)$, with $f,g,h \in R$. Suppose that $P \neq \mathcal O$, so that $h \neq 0$. Then for almost all $c_0$, $h(c_0) \neq 0 $. Choose such a $c_0$ for which $E_{c_0}$ has trivial Mordell-Weil group. Then $[f(c_0), g(c_0), h(c_0)] \in E_{c_0}(\mathbb Q)$ is a rational point with nonzero $Z$-coordinate, hence a nontrivial point on $E_{c_0}$; this contradicts the choice of $c_0$. Hence $E$ has trivial Mordell-Weil group. 
I claim also that $\text{Hom}(E, E) = \mathbb Z$, and that this follows from the following fact:

For almost all $c \in \mathbb Q^\times$, the specialization $E_c/\mathbb Q$ has endomorphism ring $\mathbb Z.$

This is obviously true as finitely many $j$-invariants are excluded, and the $j$-invariant of $E_c$ is a polynomial in $c$. The proof of $\text{Hom}(E, E) = \mathbb Z$ using this fact is similar to the one above. 
Thus, the only nonconstant endomorphisms of the curve $E/R$ are the multiplication by $n$ maps $[n]: E \to E$. 
