= Solution
For
$$
E:y^2=x^3+ax+b,
$$
the <chord-and-tangent group law> gives, for distinct nonopposite points $P=(x_1,y_1)$ and $Q=(x_2,y_2)$,
$$
\lambda=\frac{y_2-y_1}{x_2-x_1},
\qquad
x(P+Q)=\lambda^2-x_1-x_2,
\qquad
y(P+Q)=-y_1+\lambda(x_1-x(P+Q)).
$$
For doubling, replace the slope by
$$
\lambda=\frac{3x_1^2+a}{2y_1}.
$$
The point at infinity is the identity and $-(x,y)=(x,-y)$.
Let the $x$-coordinates of $P,Q,P+Q,P-Q$ be $x_1,x_2,x_3,x_4$. Applying the addition formula with $Q$ and $-Q$ gives
$$
x_3+x_4
=\frac{(y_2-y_1)^2+(y_2+y_1)^2}{(x_2-x_1)^2}-2(x_1+x_2).
$$
Substituting $y_i^2=x_i^3+ax_i+b$ and simplifying yields
$$
x_3+x_4
=\frac{2(x_1x_2+a)(x_1+x_2)+4b}{(x_1-x_2)^2}.
$$
Multiplying the two addition formulas and eliminating $y_1y_2$ in the same way gives
$$
x_3x_4
=\frac{(x_1x_2-a)^2-4b(x_1+x_2)}{(x_1-x_2)^2}.
$$
These identities are the algebraic source of two <parallelogram laws>. Applied to pullbacks of the pole divisor of $x$, they imply
$$
\deg(\varphi+\psi)+\deg(\varphi-\psi)
=2\deg\varphi+2\deg\psi
$$
for <isogenies of elliptic curves>. Together with $\deg(n\varphi)=n^2\deg\varphi$, this makes the degree a <quadratic form>. Applied to the <Absolute logarithmic Weil height> of the four $x$-coordinates, with bounded terms removed by passage to the limit defining the <canonical height of an elliptic curve>, they similarly give
$$
\widehat h(P+Q)+\widehat h(P-Q)
=2\widehat h(P)+2\widehat h(Q),
\qquad
\widehat h(nP)=n^2\widehat h(P).
$$
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