# Why does doubling move from product-is-square to each-is-square?

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?

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 $E$ 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) \\ &\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 $E$, 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 $$P \in mE(K) \;\Longleftrightarrow\; \text{f(P) and g(P) are m'th powers in K^*.}$$

• Note in particular that for $m=2$ this works for all curves of the form $y^2 = x (x-a) (x-b)$, not just the "congruent number" curves $y^2 = x(x-d)(x+d)$. – Noam D. Elkies Aug 13 '16 at 5:14

A geometric view to complement Joe Silverman's cohomological answer:

Suppose more generally that $E$ is the "product-is-square" elliptic curve $$y^2 = (x-e_1)(x-e_2)(x-e_3)$$ for some pairwise distinct $e_i$, and let $C$ be the "each-is-square" curve $$\begin{cases} y_1^2 = x-e_1, \cr y_2^2 = x-e_2, \cr y_3^2 = x-e_3, \end{cases}$$ Then $E$ is a double cover of the $x$-line, and $C$ is a $\{\pm1\}^3$ cover with $E$ as an intermediate cover with Galois group isomorphic with $({\bf Z} / 2{\bf Z})^2$; explicitly the $4:1$ map $C \to E$ is $$(x,y_1,y_2,y_3) \mapsto (x, y_1 y_2 y_3).$$ By Riemann-Hurwitz, $E$, like $C$, has genus $1$, so this $({\bf Z} / 2{\bf Z})^2$ cover $C \to E$ is unramified. Hence if we choose for the origin of $C$ one of the preimages of the origin of $E$, the cover must be the doubling map.

• Isn't it simpler just to show the multiplication by 2 map? – joro Aug 13 '16 at 5:37
• Yes, but the OP already knows that calculation and asked for a more conceptual explanation. – Noam D. Elkies Aug 13 '16 at 5:40
• I don't understand: you write that the cover "is" the doubling map, but for what identification of $C$ with $E$? – RP_ Aug 13 '16 at 15:54
• The "magical" ingredient here is that there's only one unramified $({\bf Z} / 2{\bf Z})^2$ cover of $E$ (because $\pi_1 \cong {\bf Z}^2$ has a unique quotient isomorphic with $({\bf Z} / 2{\bf Z})^2$), so $C$ is automatically identified with $E$ once we choose an origin of the group law. – Noam D. Elkies Aug 13 '16 at 16:31
• It is perhaps worth pointing out that, in order for your whole argument to work, it is very important that $C(\mathbb{Q}) \neq \emptyset$ (I mean, $E$ has many $(\mathbb{Z}/2\mathbb{Z})^2$-covers without rational points). – Daniel Loughran Aug 14 '16 at 13:35