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For fixed $\theta >0$, a sufficiently large finite set of integers with additive energy at least $\theta |A{|}^3$ has a nonconstant three-term arithmetic progression. The Balog-Szemerédi-Gowers theorem and Ruzsa modelling lemma reduce the problem to the Roth theorem on three-term arithmetic progressions in a dense cyclic model.

In a cyclic group of odd prime order, the normalized ordered three-term arithmetic progression count is ${\sum }_r\widehat{1_A}(r{)}^2\widehat{1_A}(-2r)$. The zero frequency contributes ${\alpha }^3$. If the count is less than ${\alpha }^3/2$, its nonzero-frequency contribution has magnitude greater than ${\alpha }^3/2$. The Parseval identity on a finite group bounds that magnitude by $\alpha {\max }_{r\ne 0}|\widehat{1_A-\alpha }(r)|$, proving the displayed estimate. The count includes progressions of zero common difference.

If all residues of a modular arithmetic progression have integer representatives in an interval of diameter less than half the modulus, the representatives form an integer arithmetic progression. Consecutive differences lie in an interval shorter than the modulus, so the common modular difference has a unique possible integer representative.

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

In a cyclic group of arbitrary order $q$, a Bohr set with $d$ frequencies, defined by $\Vert rx/q\Vert \le 1/16$, contains an arithmetic progression of distinct residues of length at least $q^{1/(d+1)}/32$. Partition the $d$-dimensional cube into $Q^d$ boxes, with $Q=\lfloor q^{1/(d+1)}\rfloor$. The pigeonhole principle supplies $1\le t\le Q^d$ with $\Vert rt/q\Vert \le 1/Q$ simultaneously. The residues $jt$, $0\le j\le \lfloor Q/16\rfloor$, remain in the Bohr set and do not wrap around the cyclic group. The weaker exponent permits composite moduli without assuming that every nonzero step has large order.

For an indicator function of a subset of $[N]$ with subset density $\alpha$, a large Fourier coefficient of $1_A-\alpha 1_{[N]}$ gives a density increment on an arithmetic progression. The Dirichlet approximation theorem supplies a step $d\le \sqrt{N}$ on which the detected linear phase changes slowly. Split its residue classes into progressions of lengths comparable to ${\alpha }^2\sqrt{N}$. Replacing the phase by one constant on each cell preserves a fixed fraction of the Fourier correlation. Since the balanced function has total sum zero, positive discrepancies sum to half the total absolute discrepancy; one cell has the displayed increased density. One may take $c=1/(2304\pi )$ when $N$ is sufficiently large in terms of $\alpha$.

Apply the Gallai theorem for an integer lattice to the finite patternAlthough some coordinates are negative, translating $F$ into the positive quadrant and applying the positive-lattice theorem produces a monochromatic copy $F_0=(a_0,d_0)+qF\subseteq {\mathbb{N}}^2$, with integer $q>0$. The translation is absorbed into $(a_0,d_0)$. Let $(r,s)$ be the common position-pair label on this copy.

For each $0\le u,v<m$, the starting point and steplabel the red edge at positions $r,s$. Its endpoints simplify toThus the two arithmetic progressionshave a positive common difference $q(s-r)$, and every cross-pair is red. Their entries are positive because they occur as endpoints of the positive-start, positive-step progressions represented in $F_0$.

It remains to check disjointness, rather than infer it merely from distinct starting points. The pattern contains points whose second coordinate is $-(m-1)$. Since all of $F_0$ lies in ${\mathbb{N}}^2$,ConsequentlyTherefore $A$ and $B$ are disjoint $m$-term arithmetic progressions with every cross-edge red. Combined with the first branch, this proves the dichotomy.

If an $m$-term arithmetic progression contains no red pair, all of its unordered pairs are blue, since there are only two colours. This gives the first alternative, including the vacuous singleton case. Otherwise the position-pair finite colouring in the root solution is defined everywhere, and the construction below forces the second alternative.

Write $\chi$ for the red-blue finite colouring of unordered pairs. The two alternatives concern one arithmetic progression or a pair of disjoint arithmetic progressions; they need not be mutually exclusive. For $m=1$, a singleton already has no non-blue edge, so the first alternative holds. Assume henceforth $m\ge 2$.

If the first alternative fails, every $m$-term arithmetic progression contains a red pair. Label each $(a,d)\in {\mathbb{N}}^2$ by one such pair's positions $0\le r<s<m$, choosing the first in a fixed ordering when there are several. This is a finite colouring with at most $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ colours. The following two branches establish the arithmetic-progression bipartite Ramsey dichotomy.

A combinatorial line in $[k{]}^N$ fixes all coordinates outside a nonempty set $J$ and puts the same variable letter in every coordinate of $J$. The Hales-Jewett theorem asserts that for every $k,r\ge 1$, some $N$ forces a monochromatic combinatorial line in every $r$-colouring of $[k{]}^N$. We prove it by mathematical induction on $k$, using an explicit alphabet insensitivity lemma.

A $d$-dimensional combinatorial subspace fixes some coordinates and has $d$ disjoint nonempty variable blocks. It is $(a,b)$-insensitive if changing any parameter from $a$ to $b$ or back leaves the colour unchanged. For $k\ge 2$, fixed $a\ne b\in [k]$, $r$ colours and any required dimension $d$, we now construct such a subspace without assuming the Hales-Jewett theorem.

Choose consecutive blocks of lengths $L_1,\dots ,L_d$ byProcess blocks from right to left. When processing block $i$, the later blocks already have their variable templates. For each $0\le t\le L_i$, consider $a^t{b}^{L_i-t}$ in block $i$. Record its colour profile for every earlier word in $[k{]}^{P_i}$ and every assignment of the $d-i$ later parameters. There are $k^{P_i+d-i}$ contexts and at most $r^{k^{P_i+d-i}}=L_i$ profiles. The $L_i+1$ chain words therefore contain two equal profiles by the pigeonhole principle.

If their indices are $s<t$, replace the block by the variable templateIts active block is nonempty. The values $v=a,b$ recover those two equal-profile words, so the colour is unchanged by interchanging $a,b$, for every earlier word and every later parameter assignment. Insensitivity already established in a later block persists, since it was established uniformly over all earlier words, including every subsequent restriction of block $i$. Continuing leftward gives the promised $d$-dimensional combinatorial subspace. This proves the explicit block bound for alphabet insensitivity and its required uniformity.

The one-letter case of the Hales-Jewett theorem is immediate: take $N=1$. Suppose it is proved for $k-1$ letters and every number of colours. Let $d$ be a dimension guaranteed for $k-1$ letters and $r$ colours. Obtain a $d$-dimensional $(k-1,k)$-insensitive combinatorial subspace by the lemma. Restrict its parameter alphabet to $[k-1]$ and apply the induction hypothesis. The resulting monochromatic combinatorial line uses some nonempty set of parameters. Its extra $k$-letter point has the same colour as its $(k-1)$-letter point: interchange $k-1,k$ in those parameters one at a time. The union of their active blocks is nonempty, so the extended line is a genuine combinatorial line in the original word space. This completes the letter-merging proof of the Hales-Jewett theorem.

To deduce the Van der Waerden theorem, take alphabet $\{0,\dots ,m-1\}$ and pull a finite colouring of the positive integers back byA monochromatic combinatorial line with active set $J$ maps toThus every finite colouring of the positive integers contains a monochromatic arithmetic progression of each prescribed finite length. Since the map only uses $[1,1+N(m-1)]$, this also gives the finite interval version, with an interval bound $W(r,m)$.

The Strengthened Van der Waerden theorem requires the common difference to have the same colour: for every $r,m$ some finite interval forces a monochromatic setFor $m=1$, take $a=d=1$. For $m\ge 2$, prove the assertion by mathematical induction on $r$, writing $B(r,m)$ for a sufficient interval bound. The one-colour case has $a=d=1$ and $B(1,m)=m$. Suppose $M=B(r-1,m)$ is known. Apply the ordinary Van der Waerden theorem to find a monochromatic progressionin $[N]$, where $N=W(r,L)$; call its colour $\gamma$. As $(L-1)d<N$, every point $td$ with $1\le t\le M$ also lies in $[N]$.

If one such $td$ has colour $\gamma$, the $m$ points $a,a+td,\dots ,a+(m-1)td$ lie in the original progression and have that colour, together with their difference $td$. Otherwise the pulled-back finite colouring $t\mapsto \chi (td)$ on $[M]$ uses at most $r-1$ colours. By the induction hypothesis it contains $\{e,b,b+e,\dots ,b+(m-1)e\}$ of one colour. Multiplication by $d$ gives the required set with starting point $bd$ and common difference $ed$. This proves the strengthened theorem withThis is the colour-count argument behind the Brauer progression theorem.

The Gallai theorem for an integer lattice states that every finite colouring of ${\mathbb{N}}^d$ contains a monochromatic homothetic copy of each finite lattice pattern. More precisely, for finite $S\subseteq {\mathbb{Z}}^d$ there exist $q\in \mathbb{N}$ and $b\in {\mathbb{Z}}^d$ with $b+qS\subseteq {\mathbb{N}}^d$ monochromatic. For $S\subseteq {\mathbb{N}}^d$, the translation vector can also be chosen positive.

The empty pattern is trivial. Otherwise first translate the pattern so its points $v_1,\dots ,v_k$ have nonnegative coordinates. Pull the given finite colouring back to words through the sum map from words to homothetic copiesBy the Hales-Jewett theorem, some combinatorial line with active set $J$ is monochromatic. Its images areThey are a homothetic copy with positive integer dilation $|J|$. Undoing the preliminary translation merely changes the translation vector; the actual copy still lies in the positive lattice. This derives Gallai's theorem from the Hales-Jewett theorem, and justifies the version used in Question 1.

For $m=1$ the conclusion is immediate, so suppose $m\ge 2$ and index the terms of an arithmetic progression by $0,\dots ,m-1$. There are only finitely many equivalence relations on this index set. Consequently, recording which terms have equal colours produces a finite colouring of the positive parameter pairs $(a,d)$, even though the original colouring need not be finite.

We use the Gallai theorem for an integer lattice in this form: a finite colouring of the positive integer lattice contains a monochromatic homothetic copy of any prescribed finite integer pattern. Patterns with negative coordinates are allowed here by first translating them into the positive quadrant; the translation is absorbed into the base point of the resulting copy.

Consider the finite patternThe Gallai theorem for an integer lattice supplies integers $a_0,d_0$ and $q>0$ such that every point of $(a_0,d_0)+qF$ lies in the positive quadrant and has the same equality equivalence relation $E$. If $E$ has only singleton classes, any parameter pair in this copy gives an arithmetic progression whose colours are pairwise distinct, so the original colouring is injective on it.

Otherwise choose $r<s$ with $rEs$. For every $u,v\in \{0,\dots ,m-1\}$, the parameter pairhas relation $E$. Its terms at indices $r$ and $s$ are therefore equally coloured. But these terms simplify toThus $c(A_u)=c(B_v)$ for every $u,v$. Fixing $v=0$ proves that all $A_u$ have the same colour. Every $A_u$ is positive, because it is a term of one of the positive-parameter progressions above, and the common difference $q(s-r)$ is positive. Hence $(A_u{)}_{u=0}^{m-1}$ is a monochromatic arithmetic progression of length $m$.

The canonical arithmetic progression dichotomy is therefore constant or injective on an arithmetic progression of the prescribed length. The key point is that one equal-colour index pair has been made to vary independently over two complete progressions; a merely monochromatic block of parameter pairs would not by itself give this conclusion.

Van der Waerden theorem. Given positive integers $r,k$, there is a finite $W(r,k)$ such that every finite colouring of $\{1,\dots ,W(r,k)\}$ with at most $r$ colours has a monochromatic arithmetic progression of length $k$ and positive common difference. In particular, every finite colouring of the positive integers contains such progressions of every finite length.

Here is a colour-focusing proof of Van der Waerden theorem. We induct on $k$, simultaneously for all numbers of colours. The assertion for $k=1$ is immediate. Suppose $W(r,k)$ exists for every $r$, and fix $r$. We construct bounds $F_t$, for $1\le t\le r$, with this property: every colouring of $[F_t]$ either has a monochromatic progression of length $k+1$, or has $t$ progressions of length $k$, of distinct colours, which share a next point $f\in [F_t]$. Write those progressions aswith positive steps $d_i$ and all displayed points in $[F_t]$.

For $t=1$, put $L=W(r,k)$ and $F_1=2L$. A monochromatic progression of length $k$ in $[L]$ has its next point in $[2L]$, since its step is at most $L$; for $k=1$ choose step $1$. This establishes the initial focusing property.

Suppose $F_t=N$ has been constructed. Put $M=W(r^N,k)$ and $F_{t+1}=2NM$. Divide the first $NM$ integers into $M$ consecutive blocks of size $N$. Colour a block by its full colour profile, an $N$-tuple having at most $r^N$ possible values. The induction hypothesis on $k$ gives block indiceswith the same colour profile. For $k=1$ use any one block and $D=1$. Suppose that the whole interval has no monochromatic progression of length $k+1$. Its repeated profile has $t$ focused length-$k$ progressions with positions $f-h{d}_i$ and distinct colours $q_i$. The colour $q$ of position $f$ differs from every $q_i$, since equality would complete a length-$(k+1)$ progression in the block.

Set $j_{\ast }=j_0+kD$ and $F=N(j_{\ast }-1)+f$. The index $j_{\ast }$ is at most $2M$, so this focus is in the full interval. For $1\le h\le k$, the pointis in one of the chosen blocks and has colour $q_i$. Also $F-hND$ is at position $f$ in that block and has colour $q$. We have produced $t+1$ focused length-$k$ progressions of distinct colours. This proves the recursive focusing property. At $t=r$, the focus itself must have one of the $r$ colours, so one of the progressions extends to length $k+1$. Thus $W(r,k+1)=F_r$ is a valid bound, completing the mathematical induction and the requested proof.

To obtain the equation criterion, we first prove the small strengthening of Van der Waerden theorem that we need. For any positive integers $r,s,L$, there is a finite bound forcing a monochromatic configurationThis is the Brauer progression theorem, and the following argument proves it from the theorem just established. Induct on $r$. One colour is immediate. Given a bound $M$ for $r-1$ colours with the same $s,L$, let $T=W(r,(L-1)M+1)$ and colour an interval at least as long as $\max (T,sMT)$. Find a monochromatic progression $a,a+d,\dots ,a+(L-1)Md$ in $[T]$, of colour $q$; take $d=1$ if this progression has one term. If some $jsd$, $1\le j\le M$, has colour $q$, its first $L$ suitably spaced terms $a,a+jd,\dots ,a+(L-1)jd$ and the point $s(jd)$ give the configuration. Otherwise the colouring of the indices $1,\dots ,M$ obtained from $j\mapsto c(jsd)$ uses at most $r-1$ colours. The induction hypothesis supplies indices $se,b,b+e,\dots ,b+(L-1)e$ of one colour. Multiplication by $sd$ gives the desired configuration with step $esd$. This completes the strengthening without assuming any Rado theorem.

Multiply the rational coefficients by a common positive denominator, so that all $a_i$ are nonzero integers; this does not alter the equation or partition regularity. By definition, the row is partition regular when every finite colouring admits positive monochromatic $x_1,\dots ,x_n$ satisfying ${\sum }_i{a}_i{x}_i=0$. The variables need not be distinct.

For necessity, choose a prime number $p>{\sum }_i|a_i|$. Write each positive $x$ uniquely as $p^{v_p(x)}u$, with $p\nmid u$, and colour $x$ by $u\mathrm{mod}p$. This is a finite colouring with $p-1$ colours. If a monochromatic solution exists, let $v={\min }_i{v}_p(x_i)$ and $I=\{i:v_p(x_i)=v\}$. Dividing the equation by $p^v$ and reducing modulo $p$ giveswhere $u_0\ne 0(\mathrm{mod}p)$ is the common colour. Thus ${\sum }_{i\in I}{a}_i\equiv 0(\mathrm{mod}p)$. Its absolute value is strictly less than $p$, so ${\sum }_{i\in I}{a}_i=0$. The index set $I$ is nonempty. This proves the leading-residue obstruction to partition regularity.

Conversely, suppose ${\sum }_{i\in I}{a}_i=0$ for a nonempty $I$. If $I$ is all indices, taking every variable equal to any one positive integer already gives a monochromatic solution. Otherwise fix $i_0\in I$, put $s=|a_{i_0}|$, $C={\sum }_{j\notin I}{a}_j$, and define integer offsetsLet $u={\min }_{i\in I}{\lambda }_i$ and choose $L=1+{\max }_{i\in I}{\lambda }_i-u$. The proved Brauer progression theorem gives a monochromatic configuration consisting of $sd$ and a length-$L$ progression starting at $b$. SetAll variables are positive and of one colour. Their weighted sum isThis explicitly gives a Brauer configuration for a one-row zero-sum equation. We concludeBoth directions have been proved independently of the general Rado theorem.

Fix a positive integer $q$ and an integer $a$ coprime to $q$. We use Dirichlet characters modulo $q$, extended by zero on integers not coprime to $q$, and write ${\chi }_0$ for the principal Dirichlet character. For a nonprincipal Dirichlet character, the sum over one complete period is zero. Its partial sums $A_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are therefore $O(q)$. Partial summation givesThe integral converges locally uniformly there, so these Dirichlet L-functions are holomorphic at one. On the other hand,has a simple pole at one, with residue $\phi (q)/q$.

The assumed nonvanishing of a nonprincipal Dirichlet L-function at one for real Dirichlet characters also excludes vanishing for nonreal ones. To prove this, form the product over all Dirichlet charactersUse the logarithm supplied by the absolutely convergent Euler product. The Orthogonality of Dirichlet characters yieldsThus $\mathcal{F}(\sigma )\ge 1$. If a nonreal Dirichlet character $\chi$ had $L(1,\chi )=0$, its distinct conjugate $\overline{\chi }$ would also have a zero there, since $L(1,\overline{\chi })=\overline{L(1,\chi )}$. These two zeros would outweigh the single pole of the principal factor; all other factors are bounded near one. It would follow that $\mathcal{F}(\sigma )\to 0$ as $\sigma \downarrow 1$, contradicting the lower bound. Together with the assumption for real Dirichlet characters, this proves $L(1,\chi )\ne 0$ for every nonprincipal Dirichlet character.

The first-degree part of the Euler product logarithm givesThe omitted terms are bounded uniformly, since ${\sum }_p{\sum }_{m\ge 2}(m{p}^{m\sigma }{)}^{-1}\ll {\sum }_p{p}^{-2}<\infty$. For each nonprincipal factor, its logarithm remains bounded: choose a holomorphic logarithm near the now established nonzero value $L(1,\chi )$; the Euler-product branch differs from it by a fixed integral multiple of $2\pi i$ on a sufficiently small real interval. For the principal factor,Applying Orthogonality of Dirichlet characters again givesThe right side diverges as $\sigma \downarrow 1$, so the arithmetic progression contains infinitely many primes. This proves Dirichlet's theorem on arithmetic progressions: every reduced residue class modulo $q$ contains infinitely many primes. The argument also covers $q=1$ and $q=2$, when there need not be a nonreal character.

For the sharper statement about reciprocal primes in a fixed arithmetic progression, use the given prime-counting function asymptotic with $q$ fixed:The second equivalence follows by integration by parts in the logarithmic integral function. Partial summation yieldsFor any $\epsilon >0$, choose a fixed $T$ so large that the relative error in $\pi (t;q,a)$ is at most $\epsilon$ for $t\ge T$. The initial integral is a constant, while the remaining integral lies betweenThe boundary term is $O_q(1/\log x)$. Divide by $\log \log x$, let $x\to \infty$, and then let $\epsilon \downarrow 0$. We obtainOnly a relative asymptotic is deduced from the given information; an additive $O(1)$ error, or uniformity for a growing modulus $q$, would require additional estimates.