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For positive $r,k$, let $W(r,k)$ bound the interval needed to force a monochromatic length-$k$ arithmetic progression in every finite colouring using at most $r$ colours. The colour-focusing proof of Van der Waerden theorem inducts on $k$, simultaneously for every number of colours, and then creates progressions of distinct colours with a common next point.

Assume $W(r,k)$ exists for every $r$. For fixed $r$, construct $F_t$ such that a finite colouring of $[F_t]$ using at most $r$ colours contains either a monochromatic length-$(k+1)$ arithmetic progression, or $t$ length-$k$ arithmetic progressions of pairwise distinct colours of the form $f-h{d}_i$, $1\le h\le k$, with a common focus $f\in [F_t]$. Take $F_1=2W(r,k)$: a length-$k$ arithmetic progression in its first half has its next point inside the whole interval. For $k=1$, give a singleton step $1$.

Given $N=F_t$, put $M=W(r^N,k)$ and $F_{t+1}=2NM$. Regard the first $M$ length-$N$ blocks as coloured by their full colour profiles. Some block indices $j_0,j_0+D,\dots ,j_0+(k-1)D$ share a profile. If there is no monochromatic length-$(k+1)$ arithmetic progression, this profile supplies $t$ focused progressions with focus position $f$, steps $d_i$, and distinct colours $q_i$. The colour $q$ of position $f$ differs from every $q_i$. In the later block $j_{\ast }=j_0+kD\le 2M$, put $F=N(j_{\ast }-1)+f$. For each $i$, the points $F-h(ND+d_i)$ have colour $q_i$: they lie in the selected block $j_{\ast }-hD$ at position $f-h{d}_i$. The points $F-hND$ have colour $q$. Thus $t+1$ distinct colours focus at $F$. For $t=r$, its own colour completes one of the $r$ focused progressions. The base case $W(r,1)=1$ completes the mathematical induction.

The common difference of an arithmetic progression is the fixed difference between consecutive terms.

For any fixed density strictly below one, random subsets of an integer interval can have that density yet avoid arithmetic progressions longer than a constant times the logarithm of the interval length. A union bound over possible progressions bounds their occurrence, while concentration of the random cardinality supplies a dense realization. Passing to a subset gives an exact prescribed cardinality.

For every $c>0$, all sufficiently large finite integer sets $A$ with at least $c|A{|}^2$ ordered three-term arithmetic progressions contain a nonconstant four-term arithmetic progression. Cauchy-Schwarz inequality gives $E(A)\ge c^2|A{|}^3$. The Balog-Szemerédi-Gowers theorem supplies a large small-doubling subset. A bounded-modulus Freiman 2-isomorphism gives a positive-density cyclic model; apply the assumed four-term case of Szemerédi's theorem to representatives, then lift the two consecutive second-difference relations.

Suppose the support of a probability distribution consists of all consecutive integers from $a$ to $b$, including both endpoints, with $a\le b$. Its cardinality is $n=b-a+1$, and a discrete uniform distribution on it satisfiesFor a short derivation, set $K=X-a$, whose possible values are $0,\dots ,n-1$. The finite sums giveand henceAdding $a$ shifts the expected value by $a$ and leaves the variance unchanged. For completeness, the first sum follows by pairing $k$ with $n-1-k$. For the second, sum $(k+1{)}^3-k^3=3k^2+3k+1$ from $k=0$ to $n-1$: the left side is $n^3$, so inserting the first sum gives the stated result.

The edge case $a=b$ has $n=1$: $X=a$ with probability one, expected value $a$, and variance zero. On an arithmetic progression $\{a,a+h,\dots ,a+(n-1)h\}$ with $h>0$, the same argument instead gives expected value $a+h(n-1)/2$ and variance $h^2(n^2-1)/12$. Equal spacing is the extra hypothesis behind these formulas; the discrete uniform distribution on $\{1,2,4,8\}$ requires the general sums above.

The doubling constant of a nonempty finite set $S\subseteq G$ in an abelian group $G$ isIt measures the cardinality of the sumset relative to the original set. For the integer examples, $K(A)=7/4$ and $K(B)=10/4=5/2$. A nonconstant arithmetic progression with $n$ members has doubling constant $(2n-1)/n$, which stays below two as $n$ grows. This is the sense in which its sumset has small growth.

In any abelian group, $K(S)\ge 1$: fixing $s_0\in S$ puts a translated copy $s_0+S$ inside $S+S$. The stronger integer bound gives $K(S)\ge 2-1/|S|$ only in that setting. A finite subgroup $H$ has $K(H)=1$, as does every coset $x+H$, since its sumset is $2x+H$. In particular $H=\{0,2,4\}\subseteq \mathbb{Z}/6\mathbb{Z}$ has doubling constant one.

Small doubling constant is useful because it constrains repeated growth, as in the Plünnecke-Ruzsa inequality. It signals structure such as arithmetic progressions and subgroups; precise descriptions depend on the ambient abelian group and are the subject of the Freiman-Ruzsa theorem. The additive energy bound below gives a simpler immediate consequence: $E(S)\ge |S{|}^3/K(S)$.

Colour each interval $[2^j,2^{j+1})\cap \mathbb{N}$ by the parity of $j$. Every infinite arithmetic progression of positive common difference d meets every sufficiently large such interval, since the interval has length greater than d. Both colours occur infinitely often. Thus neither colour class contains an infinite arithmetic progression, despite the finite-length conclusion of the Van der Waerden theorem.

There is no nonconstant four-term arithmetic progression of rational square numbers. If $r^2,s^2,1,t^2$ formed one, then $(-2s^2,2rst)$ would lie on $y^2=x(x+1)(x+4)$; its explicitly known Mordell-Weil group forces $s^2=1$ and the common difference to vanish.

For $p\ge 5$ and every $\alpha >0$, a sufficiently large subset of ${\mathbb{F}}_p^n$ of density at least $\alpha$ contains a nontrivial four-term arithmetic progression.

Let $I=[n]$ for odd $n\ge 3$, embed it in the cyclic group $G=\mathbb{Z}/(2n+1)\mathbb{Z}$, and let $A\subseteq I$ have density of a finite subset $\alpha$. If $A$ has no nonconstant three-term arithmetic progression and $n\ge 4{\alpha }^{-2}$, its balanced indicator function $f=1_A-\alpha 1_I$ satisfiesIndeed, the normalized trilinear arithmetic progression count has the Fourier analysis on a finite abelian group formula $\Lambda (h_1,h_2,h_3)={\sum }_r{\hat{h}}_1(r){\hat{h}}_2(-2r){\hat{h}}_3(r)$. Its values on $1_A$ and $\alpha 1_I$ are $\alpha n/|G{|}^2$ and ${\alpha }^3(n^2+1)/(2|G{|}^2)$. Telescoping their difference into three terms containing $f$, the Parseval identity and Cauchy-Schwarz inequality bound its magnitude by $3\max |\hat{f}|\alpha n/|G|$. The difference is at least ${\alpha }^3{n}^2/(4|G{|}^2)$, giving the claim. The zero Fourier coefficient vanishes by the definition of $\alpha$.

Use four parts indexed by $0,1,2,3$ in a cyclic group $G$, and include the triple missing part $i$ when $L_i\in A$. A transversal three-uniform tetrahedron then has the four values $S_1-i{S}_0$, where $S_0={\sum }_j{x}_j$ and $S_1={\sum }_{j}j{x}_j$, so it encodes a four-term arithmetic progression. Constant progressions give $|A||G{|}^2$ edge-disjoint tetrahedra. This is the bridge from the tetrahedron removal lemma to the Szemerédi theorem.

For a probability measure-preserving system, every measurable $B$ of positive measure and every integer $k\ge 2$ admit an integer $n\ge 1$ with $\mu ({\bigcap }_{j=0}^{k-1}{T}^{-jn}B)>0$. The sets are preimages; neither invertibility nor an ergodic transformation is required. The Furstenberg correspondence principle converts this simultaneous return into arithmetic progressions and yields the Szemerédi theorem. The case $k=2$ follows from the Poincaré recurrence theorem.

For functions bounded by one on a cyclic group of prime order greater than three, three applications of Cauchy-Schwarz bound the four-term arithmetic progression average by the smallest Gowers U3 norm of its factors. Zero extension yields the corresponding interval bound with an absolute factor. This makes Gowers U3 norm a measure of uniformity relevant to four-term arithmetic progression counts.

A geometric progression is a sequence obtained by repeatedly multiplying by one fixed common ratio $r$:Here $a$ and $r$ may be real numbers or complex numbers. The recurrence is meaningful even when $a=0$ or $r=0$; expressing $r$ as a ratio of successive terms requires a nonzero denominator. As a polynomial convention, $r^0=1$, including at $r=0$.

For example, quantities $100,80,64,51.2,\dots$ retain four fifths of their preceding value at every step. The decrement changes, but the retained fraction does not. With the same initial value and common ratio $2$, the terms are $100,200,400,800,\dots$. This multiplicative structure is useful whenever the same proportional rule repeats: it turns many updates into the single exponentiation $r^n$. An arithmetic progression instead repeats an additive increment.

For nonzero $a$, the modulus of $a_n$ tends to zero when $|r|<1$, stays fixed when $|r|=1$, and grows without bound when $|r|>1$. For a negative real common ratio, successive nonzero terms also alternate in sign. Thus the terms $1,-1/2,1/4,-1/8,\dots$ decay in size while crossing zero at every step. Adding the terms gives a geometric series; the individual terms and their accumulated partial sums are different sequences.