Let $E\_d$ be the group of rational points on the elliptic curve $$y^2=(x-d)x(x+d)$$


It is well-known – and easy to verify by elementary algebra – that $(x,y)\in E\_d$ belongs to $2E\_d$ precisely when all three of $x-d$, $x$, $x+d$ are squares.


This seems too neat to be a mere coincidence.


Is there a natural generalisation of this fact? Or a more conceptual way to understand why it ought to be true?



Why does doubling move from product-is-square to each-is-square?
It looks as if you're working over $\mathbb Q$. And you are using an elliptic curve in which all of the 2-torsion is rational, so $E[2]$ is isomorphic to (say) $\boldsymbol\mu\_2^2$ as a Galois module, where $\boldsymbol\mu\_2=\{\pm1\}$ is the group of square roots of 1. Okay, now consider the injection (this comes from basic Kummer theory of elliptic curves)
\begin{align\*}
E(\mathbb Q)/2E(\mathbb Q) &\hookrightarrow
H^1(G\_{\overline{\mathbb Q}/\mathbb Q},E[2]) \\
&\cong H^1(G\_{\overline{\mathbb Q}/\mathbb Q},\boldsymbol\mu\_2^2) \\
&\cong H^1(G\_{\overline{\mathbb Q}/\mathbb Q},\boldsymbol\mu\_2)^2 \\
&\cong \mathbb Q^\*/(\mathbb Q^\*)^2\times \mathbb Q^\*/(\mathbb Q^\*)^2.
\end{align\*}
(The final isomorphism is standard Kummer theory.) So this at least tells you that whether $P$ is in $2E(\mathbb Q)$ should depend on whether certain values are squares. If you trace through all of the maps, which requires choosing a basis for $E[2]$, you'll find (for one of the choices) that it is given by
$$ P \longmapsto \bigl(x(P),x(P)-d\bigr), $$
so $P$ is in $2E(\mathbb Q)$ if and only if both $x(P)$ and $x(P)-d$ are squares.
Of course, from the equation of $E$, any two of $x(P)$, $x(P)-d$ and $x(P)+d$ being a square forces the third one to also be square. Finally, if $x(P)=0$ or $x(P)=d$, one finds during the analysis that one needs to use a different formula. For example, since $x$ and $(x-d)(x+d)$ differ multiplicatively by a square, one has
$$ 0 \longmapsto (-d^2,-d) = (-1,-d) \in \mathbb Q^\*/(\mathbb Q^\*)^2\times \mathbb Q^\*/(\mathbb Q^\*)^2.
$$


To answer your second question, if $E[m]\subset E(K)$, then one similarly gets an injection
$$ E(K)/mE(K) \hookrightarrow K^\*/(K^\*)^m\times K^\*/(K^\*)^m, $$
so (more or less) there are rational functions $f$ and $g$ in $K(E)$ such that for $P\in E(K)$ we have
$$ P \in mE(K) \;\Longleftrightarrow\; 
\text{$f(P)$ and $g(P)$ are $m$'th powers in $K^\*$.}
$$


