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On the product of an elliptic curve with itself, let $D_{+}$ and $D_{-}$ be the diagonal and the graph of negation. The principal divisor of $x(P)-x(Q)$ is $D_{+}+D_{-}-2(\{O\}\times E)-2(E\times \{O\})$: equality of the coordinates means $Q=\pm P$, while $x$ has a double pole at $O$. Pulling back along $(\varphi ,\psi )$ and taking degrees proves the parallelogram law for the degree of an isogeny when $\varphi ,\psi$ are nonzero and $\varphi \ne \pm \psi$. The exceptional cases follow from $\mathrm{deg}[2]=4$. The same argument uses the quotient by negation in characteristic two. Thus degree, with degree zero assigned to the zero map, is a positive quadratic form on $\mathrm{Hom}(E_1,E_2)$.

For the real l-infinity sequence space, every coordinate of an extreme point must be an endpoint of $[-1,1]$; any slack coordinate permits opposite perturbations. For the real absolutely summable sequence space, a unit vector with two nonzero coordinates permits a transfer of mass between them, so only signed coordinate vectors are extreme. For the real l2 sequence space, the parallelogram law makes every unit vector extreme. In all three cases interior points of the unit ball are nonextreme.

Expand using the bilinearity and symmetry of the real inner product:Addition proves the parallelogram law,For the maximum norm on ${\mathbb{R}}^2$, take $x=(1,0)$ and $y=(0,1)$. All four vectors $x,y,x+y,x-y$ have maximum norm one. The left side of the parallelogram law is then $2$, whereas the right side is $4$. The maximum norm cannot come from an inner product.

An isogeny of elliptic curves over $k$ is a nonconstant morphism $\varphi :E_1\to E_2$ defined over $k$ and preserving the identity. Such a morphism is a group homomorphism: apply the Mumford rigidity lemma to $\varphi (P+Q)-\varphi (P)-\varphi (Q)$, whose restriction to either coordinate axis is zero. A nonconstant morphism of smooth projective curves is finite and surjective, so its degree of an isogeny is the positive integerDegrees multiply under composition. A separable isogeny has degree equal to the number of geometric points in its kernel of an isogeny. In positive characteristic of a field, inseparable isogenies also occur, for example the Frobenius isogeny; their total degree includes inseparable multiplicity, so counting geometric kernel points alone is insufficient. Every isogeny of degree $d$ has a dual isogeny $\hat{\varphi }$ satisfying $\hat{\varphi }\varphi =[d]$ and $\varphi \hat{\varphi }=[d]$.

Pointwise addition and negation make the homomorphism group of elliptic curves $H={\mathrm{Hom}}_k(E_1,E_2)$ an abelian group. Its zero is the constant identity map; every other element is an isogeny, because its nonconstant image is the whole target curve. Define $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We prove the positive definiteness of degree on elliptic-curve homomorphisms, including arbitrary characteristic.

Let $L={\mathcal{O}}_{E_2}(O)$, a line bundle associated to a divisor of degree one. On $E_2\times E_2$ write $m(P,Q)=P+Q$, $d(P,Q)=P-Q$ and let ${\pi }_i$ be the projections. The divisor proof of the degree parallelogram law givesHere is a direct justification. The degree-two Weierstrass coordinate $x$ has a double pole at $O$, and $x(P)=x(Q)$ precisely when $Q=P$ or $Q=-P$. Consequentlywhere $D_{+}$ is the diagonal and $D_{-}$ the graph of negation. Since $D_{+}=d^{-1}(O)$ and $D_{-}=m^{-1}(O)$, this principal divisor gives the displayed line bundle identity. This remains valid in characteristic two: the general Weierstrass equation of an elliptic curve of a smooth algebraic curve has a nontrivial generic involution $(x,y)\mapsto (x,-y-a_{1}x-a_3)$, so the two graphs still occur. This identity is also the symmetric-line-bundle form of the Theorem of the square.

Pull back the bundle identity along $(\varphi ,\psi ):E_1\to E_2\times E_2$ and take degrees of line bundles. For a nonzero map, $\mathrm{deg}({\varphi }^{\ast }L)=q(\varphi )$; for a zero map its pullback is trivial and has degree zero. Thus the degree parallelogram law isUsing bundles rather than pulling back the rational function itself is essential when $\varphi =\pm \psi$, where that function can vanish identically. Negation preserves degree, so $q(-\varphi )=q(\varphi )$. Applying the parallelogram identity to $n\varphi$ and $\varphi$ gives the quadratic degree recursion for elliptic multiplication, and induction from $q(0)=0$ yields $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$.

Define the bilinear degree pairing for elliptic-curve homomorphisms byIt is symmetric and odd in each argument. Two applications of the parallelogram law giveAdding proves $B(x+z,y)=B(x,y)+B(z,y)$, and symmetry proves additivity in the other argument. Also $B(\varphi ,\varphi )=q(\varphi )$. Therefore $q$ is an integer-valued quadratic form. The integral polarization is $2B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$; $B$ itself may be half-integral.

Finally $q(\varphi )>0$ for every nonzero $\varphi$, since an isogeny has positive finite degree. If $n\varphi =0$ with $n\ne 0$, the identity $q(n\varphi )=n^{2}q(\varphi )$ forces $\varphi =0$, so $H$ is a torsion-free abelian group. On any finite-dimensional rational span in $H\otimes \mathbb{Q}$, the pairing has a rational matrix and is positive on every nonzero rational vector. By density it is nonnegative on real vectors; if that rational matrix were singular, its kernel would contain a nonzero rational vector, contradicting strict positivity. Thus its real extension is a positive-definite quadratic form. This establishes the required positivity as well as the quadratic identity. For example, on the subgroup of multiplication maps $[n]$, $q([n])=n^2$; on the endomorphism ring of an elliptic curve with complex multiplication it is the imaginary-quadratic field norm, explaining the lattice geometry behind isogeny degrees.

A point $x$ of a set $S$ is an extreme point if $x=(1-t)y+tz$, with $y,z\in S$ and $0<t<1$, forces $y=z=x$. For a convex set it is enough to test midpoint decompositions: a nontrivial segment containing $x$ in its interior contains a smaller segment with midpoint $x$. The nonempty convex set ${\mathbb{R}}^2$ has no extreme points, since every $x$ is the midpoint of $x+e_1$ and $x-e_1$.

For the real l-infinity sequence space, a vector in the closed unit ball is extreme precisely when every coordinate is $1$ or $-1$. If $|x_j|<1$ for some $j$, choose $0<\delta \le 1-|x_j|$. The distinct vectors $x\pm \delta e_j$ remain in the unit ball and have midpoint $x$, so $x$ is not extreme. Conversely suppose all coordinates have modulus one and $x=(y+z)/2$, with $\Vert y{\Vert }_{\infty },\Vert z{\Vert }_{\infty }\le 1$. In each real interval $[-1,1]$, an endpoint can be a midpoint only if both entries equal that endpoint. Thus $y_j=z_j=x_j$ for every $j$ and $y=z=x$. Therefore

For the real absolutely summable sequence space, any point with $\Vert x{\Vert }_1<1$ is nonextreme: choose a nonzero coordinate perturbation of norm less than $1-\Vert x{\Vert }_1$. If $\Vert x{\Vert }_1=1$ and two coordinates $x_j,x_k$ are nonzero, take $0<\delta <\min (|x_j|,|x_k|)$ and putBoth perturbations preserve the signs of those two coordinates. One adds $\delta$ to one absolute value and subtracts it from the other, so $\Vert x+h{\Vert }_1=\Vert x-h{\Vert }_1=1$. Hence such an $x$ is not extreme. The only remaining candidates are $\pm e_j$. If $e_j=(y+z)/2$ with $\Vert y{\Vert }_1,\Vert z{\Vert }_1\le 1$, its $j$th coordinate forces $y_j=z_j=1$. That consumes the entire norm allowance, forcing all other coordinates to vanish. The argument for $-e_j$ is identical. Thus

For the real l2 sequence space, interior points of the unit ball again permit sufficiently small opposite perturbations and are not extreme. If $\Vert x{\Vert }_2=1$ and $x=(y+z)/2$, write $y=x+h$, $z=x-h$. The parallelogram law givesThe left side is at most one, so $h=0$ and $y=z=x$. ConsequentlyThese are the extreme points of real sequence-space unit balls: signs in ${\ell }^{\infty }$, coordinate spikes in ${\ell }^1$, and the whole sphere in ${\ell }^2$. The restriction to real scalars matters for the first description; the complex supremum ball has arbitrary unit-modulus coordinates instead.

Let $M$ be the linear subspace of $L^2(P)$ consisting of classes admitting $\mathcal{G}$-measurable representatives. It is closed. Indeed, if $Y_n\in M$ converges in $L^2$ to $Y$, choose a subsequence converging almost surely: take $\Vert Y_{n_j}-Y{\Vert }_2^2\le 2^{-3j}$ and apply Markov inequality and the Borel-Cantelli lemma to errors exceeding $2^{-j}$. The pointwise limit of its $\mathcal{G}$-measurable representatives, set to zero wherever the finite limit does not exist, is $\mathcal{G}$-measurable and equals $Y$ almost surely. This also works when $\mathcal{G}$ is not complete.

Here is a proof of the orthogonal projection construction using only completeness. Put $d={\inf }_{W\in M}\Vert X-W{\Vert }_2$ and take $W_n$ approaching this infimum. The parallelogram law givesCompleteness and closedness give a minimizer $Y\in M$. For every $Z\in M$, expand $\Vert X-Y-tZ{\Vert }_2^2$ for real $t$. Its minimum at zero impliesThis $Y$ is unique almost surely, because the difference of two possible choices belongs to $M$ and is orthogonal to itself. It is the conditional expectation $E(X\mid \mathcal{G})$, as also follows by testing indicators of events in $\mathcal{G}$.

For the jointly Gaussian pair, if $u>0$, setThe residual $R=X-Y$ has mean zero and covariance $\mathrm{Cov}(R,G)=v-(v/u)u=0$. Since $(R,G)$ is jointly Gaussian, its characteristic function factors when this covariance is zero, so $R$ and $G$ are independent. Hence $E(RZ)=E(R)E(Z)=0$ for every square-integrable function $Z$ of $G$; independence first proves this for bounded functions, and truncation plus Cauchy-Schwarz inequality gives the general case. This proves the Gaussian conditional expectation formula, including a zero residual variance. If $u=0$, $G=\mu$ almost surely and the covariance inequality gives $v=0$; $Y=\nu$ is the answer.

The closest point theorem in a Hilbert space says that, for every nonempty closed convex set $C\subset H$ and $x\in H$, there is exactly one $p\in C$ minimizing $\Vert x-p\Vert$.

Put $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The midpoint belongs to $C$ because it is a convex set. The parallelogram law givesThus $(y_n)$ is a Cauchy sequence. Completeness of the Hilbert space and closedness of $C$ give a limit $p\in C$, with $\Vert x-p\Vert =d$. Applying the same identity to two minimizers gives their squared distance at most zero, proving uniqueness.

The resulting projection is characterized byIndeed, differentiate $\Vert x-p-t(z-p){\Vert }^2$ at $t=0^{+}$; the minimum there gives the inequality. Conversely, expanding $\Vert x-z{\Vert }^2$ proves minimality from this inequality. For a closed linear subspace, both signs of each direction are allowed, so $x-p$ is orthogonal to that subspace: this recovers the orthogonal projection.

Work first where all three coordinates are finite and the joining line is not vertical. By the chord-and-tangent group law, the third intersection of $y=\lambda x+\mu$ with the elliptic curve is $-(P_1+P_2)$, whose first coordinate is also $x_3$. Substitution gives the monic polynomialIts roots are $x_1,x_2,x_3$, counted with intersection multiplicity. Comparing its coefficients with the elementary symmetric polynomials provesThis symmetric-coordinate identity for elliptic-curve addition is an identity of rational functions. It includes tangent cases by specialization. Literally assigning finite $s_i$ when a point is $O$ would be meaningless; such cases must be interpreted on the projective curve. A nonsingular equation in the printed short form has characteristic different from two.

Here is how the identity yields the quadratic form for the degree of an isogeny. With $u=x(P)$, $v=x(Q)$, write $s=u+v$ and $q=uv$. As an equation for $t=x(P+Q)$, it becomesBoth $x(P+Q)$ and $x(P-Q)$ satisfy it, because replacing $Q$ by $-Q$ does not change $v$. Thus its two branches are the sum and difference maps. Their common denominator is the square of $u-v$. The following divisor on an algebraic curve calculation keeps track of cancellations and exceptional points rigorously, instead of assuming degrees of displayed numerators always add.

On $E_2\times E_2$, equality of the two first coordinates means $Q=P$ or $Q=-P$. Each coordinate has a double pole at $O$. Consequently the principal divisor identity iswhere $D_{+}$ is the diagonal and $D_{-}$ is the graph of negation. Equivalently, the two zero divisors are the inverse images of $O$ under difference and sum, exactly the two branches identified above. Their generic multiplicities are one; the divisor identity also accounts for their intersections at 2-torsion points.

For nonzero isogenies of elliptic curves $\varphi ,\psi :E_1\to E_2$ with $\varphi \ne \pm \psi$, pull this identity back by $(\varphi ,\psi )$. Taking degrees of the resulting principal divisor on $E_1$ gives the divisor proof of the degree parallelogram law:If one map is zero, this is immediate. If $\varphi =\pm \psi$, it follows from $\mathrm{deg}([2]\circ \varphi )=4\mathrm{deg}\varphi$. This also proves the parallelogram law for a general Weierstrass equation of an elliptic curve in characteristic two: the same divisor identity uses the quotient by negation, even though the particular symmetric-coordinate formula above is unavailable.

Put $q(\varphi )=\mathrm{deg}\varphi$ and $q(0)=0$. Applying the parallelogram law to $n\varphi ,\varphi$ gives the recurrence $q((n+1)\varphi )+q((n-1)\varphi )=2q(n\varphi )+2q(\varphi )$, and hence $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$. Its polarization $B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$ is symmetric and integer-valued. For fixed $\psi$, the parallelogram law makes $B(\cdot ,\psi )$ odd and gives $B(x+y,\psi )+B(x-y,\psi )=2B(x,\psi )$. Interchanging $x,y$ and adding proves additivity, since integers have no two-torsion. Thus $B$ has bilinearity, and $q$ is a positive quadratic form on $\mathrm{Hom}(E_1,E_2)$, with $q(\varphi )>0$ whenever $\varphi \ne 0$.

Galois cohomology and the finite division quotient. Let $K$ be a number field, $G_K=\mathrm{Gal}(\overline{K}/K)$ with its profinite topology, and $M$ a discrete continuous $G_K$-module. Its Galois cohomology is continuous group cohomology. In degrees zero and one,Thus a crossed homomorphism measures the failure of a choice to be Galois-invariant, and changing the choice changes the cocycle by a coboundary. A short exact sequence of such modules gives a long exact sequence in group cohomology. The connecting map sends an invariant element to the cocycle obtained by lifting it and comparing its Galois translates.

For an elliptic curve $E/K$ and $n\ge 2$, multiplication by $n$ is surjective on $E(\overline{K})$ and has kernel $E[n]$. The Kummer exact sequence of an elliptic curve therefore givesExplicitly, choose $nQ=P$ and set $\kappa (P)(\sigma )=\sigma Q-Q$. Replacing $Q$ changes this by a coboundary; replacing $P$ by $P+nR$, $R\in E(K)$, does not change it. If its class is zero, some torsion translate of $Q$ is Galois-invariant, so $P\in nE(K)$. This proves injectivity of the Kummer map of an elliptic curve.

The whole group $H^1(K,E[n])$ need not be finite. The crucial restriction is ramification. Let $S$ contain the finite primes of bad reduction of an elliptic curve and those dividing $n$. At $v\notin S$, good reduction and prime-to-residue-characteristic multiplication imply that every division point of a local point is in an unramified extension, by the formal-group lifting argument. Thus every global Kummer class restricts trivially to inertia there.

Choose a finite Galois extension $L/K$ containing $E[n]$ and the $n$th roots of unity, and enlarge $S$ to include its ramified primes. Over $L$, the torsion module is constant and can be identified with ${\mu }_n^2$. By Kummer theory,For classes unramified outside the primes $S_L$ above $S$, valuations outside $S_L$ must be divisible by $n$. Their restriction images therefore lie in $L(S_L,n{)}^2$, the square of an S-unramified power class group. Its finiteness can be seen without assuming the Mordell-Weil theorem: the finiteness of S-unramified Kummer classes follows fromThe S-unit group is finitely generated by the Dirichlet unit theorem and its valuations at $S_L$. The localized ideal class group is a quotient of the ordinary ideal class group of $L$, which is finite by the finiteness of the ideal class group. The Minkowski bound for ideal classes supplies an integral ideal of bounded norm in each class, and only finitely many integral ideals have bounded norm.

The kernel of restriction from $H^1(K,E[n])$ to $H^1(L,E[n])$ is also finite: a cocycle trivial after restriction can, after subtracting a coboundary, be made to factor through the finite group $\mathrm{Gal}(L/K)$. There are only finitely many maps from that group to the finite module $E[n]$. This is the finite-extension kernel of a Kummer map mechanism. Consequently the possible Kummer classes form a finite set, provingThis is the Weak Mordell-Weil theorem. Imposing membership in the local Kummer image at every place refines the finite unramified collection to the n-Selmer group. Its relation with the obstruction to a globally rational point iswhere the last group is the $n$-torsion of the Tate–Shafarevich group. Thus local solubility gives a finite computable upper bound but can leave a genuine global obstruction.

Heights and finite generation. For the other essay, normalize the absolute values on a field $K$ so that the product formula holds. The Absolute logarithmic Weil height of $[x_0:x_1]\in {\mathbb{P}}^1(K)$ isAt the infinite places the local degrees are one or two and the usual real or complex modulus is used. The product formula makes this independent of scaling; it is also unchanged by extending $K$. Over $\mathbb{Q}$, using coprime integral coordinates gives $h([u:v])=\log \max (|u|,|v|)$, the naive height on the projective line. The Northcott theorem says that points of bounded height and bounded field degree form a finite set; in particular this holds over the fixed field $K$. One way to see the finiteness is the height-Mahler measure formula: for an algebraic number of degree $d$, the coefficients of its primitive minimal integer polynomial are bounded by binomial factors times $\exp (dh)$. Bounds on $d$ and $h$ therefore leave only finitely many integer polynomials and hence only finitely many algebraic numbers.

Set $h_E(P)=h(x(P))$ for $P\ne O$, and $h_E(O)=0$. Since $x:E\to {\mathbb{P}}^1$ has degree two, bounded $h_E$ gives only finitely many points of $E(K)$. The height growth under a morphism of the projective line states that a degree-$d$ morphism gives $h(f(t))=dh(t)+O(1)$. The upper bound follows from its homogeneous coordinate polynomials; the lower bound follows from their having no common zero, using a resultant identity at each place. Applied to the degree-four duplication map, it gives a uniform constant $C$ withConsequently $4^{-j}{h}_E(2^{j}P)$ is Cauchy: the absolute difference of consecutive terms is at most $C{4}^{-(j+1)}$. Define the canonical height of an elliptic curve byIt is nonnegative and satisfies $\hat{h}(2P)=4\hat{h}(P)$. The factor one-half is the conventional normalization for the divisor $(O)$, since the first-coordinate height belongs to $2(O)$.

The addition-divisor identity also gives the bounded-error height identityuniformly in $P,Q$. This is the height form of the identity for the sum and difference pullbacks of the line bundle associated to $2(O)$; it includes points where the affine addition formulas have a zero denominator. Apply it to $2^{j}P,2^{j}Q$, divide by $2\cdot 4^j$, and take the limit. Thus $\hat{h}$ satisfies the parallelogram law, and its polarization is a bilinear form, the height pairing. Nonnegativity implies the Cauchy-Schwarz inequality for that pairing, by evaluating its nonnegative quadratic polynomial on $mP+nQ$ for all integers $m,n$ and approximating real ratios by rational numbers. In particularThese properties do not presuppose finite generation. Bounded canonical height gives a finite set by the bounded difference and the Northcott theorem. Torsion points have canonical height zero. Conversely, if $\hat{h}(P)=0$, all $2^{j}P$ have bounded naive height, so two coincide; their difference makes $P$ torsion. Thus the height detects precisely the free part, once finite generation is proved.

Use the Weak Mordell-Weil theorem with $n=2$ and choose finitely many coset representatives $R_i$. Write $P=2Q+R_i$, and put $H={\max }_i\hat{h}(R_i)$. The height inequality yieldsIteration enters the finite set $\{P:\hat{h}(P)\le H+1\}$, since after $j$ steps the height is at most $H+2^{-j}\max (\hat{h}(P)-H,0)$. Unwinding $P=2Q+R_i$ expresses every point using that finite set and the $R_i$. This proves the canonical-height proof of Mordell-Weil finite generation and henceThe Fundamental theorem of finitely generated abelian groups supplies this decomposition. The Mordell-Weil theorem combines a finite quotient from arithmetic Galois cohomology with a contracting height descent; either ingredient alone would not establish finite generation.

The addition law and part (a) give the approximate parallelogram identityReplace $P,Q$ by $2^{n}P,2^{n}Q$, divide by $4^n$, and let $n\to \infty$. The error disappears, leaving the exact parallelogram lawTogether with $\hat{h}(2P)=4\hat{h}(P)$, induction on $|m|$ yieldsfor every integer $m$.

Let $x,y\in {\ell }^2$ satisfy $\Vert x{\Vert }_2=\Vert y{\Vert }_2=1$. The parallelogram law givesIf $\Vert x-y{\Vert }_2\ge \epsilon$, thenand thereforeThis depends only on $\epsilon$, so the l2 sequence space is a uniformly convex Banach space, as asserted by the uniform convexity of l2.

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.

The l2 sequence space is a uniformly convex Banach space. If $\Vert x{\Vert }_2=\Vert y{\Vert }_2=1$, the parallelogram law givesThus $\Vert x-y{\Vert }_2\ge \epsilon$ permits the modulus $\delta (\epsilon )=1-\sqrt{1-{\epsilon }^2/4}$.