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Forthe chord-and-tangent group law gives, for distinct nonopposite points $P=(x_1,y_1)$ and $Q=(x_2,y_2)$,For doubling, replace the slope byThe 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 $-Q$ givesSubstituting $y_i^2=x_i^3+a{x}_i+b$ and simplifying yieldsMultiplying the two addition formulas and eliminating $y_1{y}_2$ in the same way gives

These identities are the algebraic source of two parallelogram laws. Applied to pullbacks of the pole divisor of $x$, they implyfor isogenies of elliptic curves. Together with $\mathrm{deg}(n\phi )=n^2\mathrm{deg}\phi$, 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

Fix a number $k$ of colors. For a two-point set whose points are distance $d$ apart, take the vertices of a regular simplex with $k+1$ vertices and side length $d$. The pigeonhole principle gives two vertices of one color, and they form the required congruent copy. Thus every two-point set, equivalently every line segment, is a Euclidean Ramsey set.

For an equilateral triangle of side length $d$, take a regular simplex with $2k+1$ vertices and side length $d$. The pigeonhole principle gives three vertices of one color, and every three vertices of a regular simplex form an equilateral triangle. Hence every equilateral triangle is Euclidean Ramsey.

To prove the product theorem for Euclidean Ramsey sets, let $S_X$ be a finite Ramsey witness for $X$ under $k$ colors. There are at most $k^{|S_X|}$ possible color patterns on $S_X$. Choose a finite Ramsey witness $S_Y$ for $Y$ under that many colors. Given a $k$-coloring of $S_X\times S_Y$, color each $y\in S_Y$ by the complete patternThere is a copy $Y^{\prime }\cong Y$ on which this pattern is constant. The common pattern on $S_X$ contains a monochromatic copy $X^{\prime }\cong X$. Every point of $X^{\prime }\times Y^{\prime }$ then has the same original color, and the orthogonal product is congruent to $X\times Y$.

A rectangle is the Cartesian product of two line segments, so it is Euclidean Ramsey. Three suitable vertices of a rectangle form a right triangle; any subset of a monochromatic set is monochromatic. Thus every right triangle is Euclidean Ramsey.

It remains to show that the collinear set $\{0,1,2\}$ behaves differently. In every ${\mathbb{R}}^m$, use the finite coloringA congruent copy has the form $a-v,a,a+v$ with $\Vert v\Vert =1$ for the Euclidean norm. The parallelogram law givesPut $n_{+}=\lfloor 2\Vert a+v{\Vert }^2\rfloor$, $n_{-}=\lfloor 2\Vert a-v{\Vert }^2\rfloor$, and $n_0=\lfloor 2\Vert a{\Vert }^2\rfloor$. The errors introduced by the three floor functions show thatIf all three points had one color, the integer in the middle would be divisible by ten, which is impossible. This proves the three-term unit arithmetic progression is not Euclidean Ramsey assertion.

Set $h(O_E)=0$ andfor $P\ne O_E$. The canonical height of an elliptic curve isThe duplication formula is a rational function of degree four in $x$, so the rational-map height estimate in part (a) givesTherefore successive terms of $4^{-r}h([2^r]P)$ differ by at most $C_E{4}^{-r-1}$, and the limit is well defined.

Shifting the limit immediately gives $\hat{h}([2]P)=4\hat{h}(P)$. The addition formula likewise givesApply this to $[2^r]P,[2^r]Q$, divide by $2\cdot 4^r$, and pass to the limit to obtain the exact parallelogram lawTaking $Q=P$ starts an induction on $|n|$ that yieldsfor every integer $n$.

The collinear set $\{0,1,2\}$ is not a Euclidean Ramsey set. In every dimension, coloring $x$ by $\lfloor 2\Vert x{\Vert }^2\rfloor$ modulo ten prevents a monochromatic congruent copy; the parallelogram law would force the corresponding three integer parts to have a second difference strictly between two and six, which cannot be divisible by ten.