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A matrix $A$ over the rational numbers is partition regular when every finite coloring of the positive integers admits a monochromatic vector $x$ satisfying $Ax=0$, with every coordinate a positive integer. Repeated coordinates are permitted.

For the Ehrenfeucht-Mostowski theorem, let $M$ be an infinite first-order structure and $I$ any linear order. Choose a full Skolem expansion $M^{\ast }$: add witness functions for existential formulas, repeat in the expanded languages, and take their union. We will construct a model of its theory containing distinct generators $c_i$, $i\in I$, forming an order-indiscernible sequence.

Adjoin the elementary diagram of $M^{\ast }$, the inequalities $c_i\ne c_j$ for $i\ne j$, and for every expanded-language formula $\varphi (x_1,\dots ,x_r)$ and every two increasing $r$-tuples of indices, the sentencesEvery finite subcollection is satisfiable in $M^{\ast }$. To prove this, choose a countably infinite sequence of distinct elements of $M$, and for each of the finitely many relevant arities color its increasing index tuples by the finite vector of truth values of the formulas of that arity. Repeated applications of the Ramsey's theorem give an infinite subsequence homogeneous for all these finitely many finite colorings. Interpret the finitely many mentioned $c_i$ by its elements in their index order. All indiscernibility and distinctness constraints hold, and the finite diagram fragment holds because the ambient structure is still $M^{\ast }$.

By compactness theorem, a model $N^{\ast }$ of the full enlarged theory exists. Its Skolem hull $H$ generated by the $c_i$ is elementary: whenever an existential formula with parameters in $H$ holds in $N^{\ast }$, its chosen witness function returns a value in $H$, so the Tarski-Vaught test applies. Its reduct is a model of $\mathrm{Th}(M)$. If the index order has size $\kappa$, the hull has size at most $\kappa +|L|+{\aleph }_0$, and it contains $|I|$ distinct generators.

Every order automorphism $h$ of $I$ induces the automorphismTo check well-definedness, write any two terms being compared over the same increasing list of their generators. Their equality is an expanded-language formula; indiscernibility of the order-indiscernible sequence preserves its truth under $h$. The same argument preserves each relation, and the term construction preserves the functions. Applying $h^{-1}$ gives the inverse. The extension is unique in the specified Skolem language since every hull element is a term in the generators; no uniqueness among all automorphisms of the reduct is being asserted. This proves the Ehrenfeucht-Mostowski theorem, including its usual automorphism consequence.

For simple typed set theory with atoms, use sorts $0,1,\dots$, predicates $S_i$ for sets, and typed membership from sort $i$ to sort $i+1$. Atoms have no members, extensionality compares sets, and every well-typed formula defines a set of the next sort. Typical ambiguity is the schema $\varphi \leftrightarrow {\varphi }^{+}$ for every closed typed sentence, where the superscript raises all types by one.

Here are concrete full typed models before imposing the ambiguity schema. Choose increasing natural numbers $n_{-1}<n_0<n_1<\cdots$ and setThe auxiliary $D_{-1}$ merely interprets the set predicate at the lowest sort. Define typed membership byThen the other objects of $D_{i+1}$ are atoms, regardless of their old membership. Every subset of $D_i$ has its representative in $S_{i+1}$, giving comprehension even for formulas with higher-sort quantifiers and parameters. Extensionality holds for the set representatives and atoms have no typed members. The rank gaps leave infinitely many atoms at each sort, and all sorts are infinite.

Consider finitely many ambiguity instances with unraised sentences ${\varphi }_1,\dots ,{\varphi }_t$, and choose a fixed window length $r$ sufficient to include every sort used and its auxiliary predecessor. Color every increasing $r$-tuple of natural numbers by the vector of truth values of these unraised sentences in its finite rank-window interpretation. There are only $2^t$ colors. The Ramsey's theorem gives an infinite homogeneous set. Choose $r+1$ successive points from it. The first $r$ points and last $r$ points have the same color, while the latter interpretation of ${\varphi }_j$ is exactly the former model's interpretation of ${\varphi }_j^{+}$. The auxiliary predecessor shifts too, so this remains true for sentences mentioning the lowest set predicate. Extend the chosen rank sequence arbitrarily to interpret the remaining sorts. This yields a model of typed set theory satisfying all the specified ambiguity instances.

Thus every finite subset of the typed axioms together with all ambiguity instances has a model. Applying multisorted compactness theorem givesThe final model satisfies first-order typed comprehension, with quantification over its own sorts and sets as in Henkin semantics.

For $a<b$, give the pair $\{a,b\}$ the color determined by $v_2(b-a)\mathrm{mod}2$, where $v_2$ is the 2-adic valuation: use red for even valuation and blue for odd valuation. This finite coloring works because the pairs $\{a,a+d\}$ and $\{a,a+2d\}$ have valuations $v_2(d)$ and $v_2(d)+1$. They therefore have opposite colors. In particular, no three-term arithmetic progression has all its pairs monochromatic.

For the second assertion, $m=1$ is immediate: every singleton is a blue complete arithmetic progression, since it has no pairs to check. Assume $m\ge 2$. If some $m$-term arithmetic progression has all its pairs blue, we are done. Otherwise, every such arithmetic progression contains a red pair. For each $(a,d)\in {\mathbb{N}}^2$, choose, say lexicographically, indices $0\le r<s<m$ for which $\{a+rd,a+sd\}$ is red, and color $(a,d)$ by this chosen pair $(r,s)$. This is a finite coloring with at most $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ colors.

We need a lattice pattern that works whichever pair becomes the common color. Fix, before applying the Gallai theorem for an integer lattice, the finite setTranslate $F$ into the positive quadrant and apply the Gallai theorem for an integer lattice to that translated pattern. Absorbing the translation into the base point gives integers $a_0,d_0$ and $q>0$ such that $(a_0,d_0)+qF\subseteq {\mathbb{N}}^2$ is monochromatic. Write $(r,s)$ for its common color. Since $(0,0)\in F$, the base point itself has positive coordinates.

For each $0\le i,j<m$, the parameter pointbelongs to that monochromatic copy and has chosen red pair $(r,s)$. Its red endpoints simplify toConsequently, puttinggives two $m$-term arithmetic progressions of positive common difference $q(s-r)$, with every cross-pair red. These arithmetic progressions really are disjoint: the parameter point corresponding to $i=m-1,j=0$ has positive second coordinate, so $d_0>q(m-1)$. HenceThis proves the arithmetic-progression bipartite Ramsey dichotomy, including the required disjointness.

Here is the matrix form of Rado's theorem. A matrix $A$ with entries in the rational numbers and columns $a_1,\dots ,a_n$ is a partition regular matrix if every finite coloring of the positive integers admits a vector $x\in {\mathbb{N}}^n$ whose coordinates are monochromatic and for which $Ax=0$. Its columns condition asks for an ordered partition into nonempty blocks $B_1,\dots ,B_s$, withRado's theorem states that these two conditions are equivalent. Coordinates need not be distinct; positivity is required for every coordinate.

First prove necessity using the finite separating-functional proof of the columns condition. Clearing denominators does not change either the kernel or the columns condition, so assume that $A$ has integer entries. For each pair of disjoint index sets $I,B\subseteq \{1,\dots ,n\}$, with $B$ nonempty, for which $b_B={\sum }_{i\in B}{a}_i$ is outside the rational linear span of the $I$-columns, choose an integer linear functional ${\ell }_{I,B}$ such thatSuch a linear functional exists: extend a basis over the rational numbers of the indicated linear span by $b_B$, prescribe values zero on the former basis and one on $b_B$, extend to a basis of the ambient vector space over the rational numbers, and then clear the functional's denominators. There are only finitely many pairs $(I,B)$. Choose a prime number $p$ larger than the absolute values of all their nonzero evaluations; if there are none, choose any prime number.

Color $x>0$ by its last nonzero digit coloring value, namelywhere $v_p$ is the P-adic valuation. By partition regularity, choose a positive solution $Ax=0$ all of whose coordinates have the same nonzero digit $u$. Group its indices into nonempty blocks $B_1,\dots ,B_s$ according to increasing distinct P-adic valuations ${\nu }_1<\cdots <{\nu }_s$.

Suppose that a block sum $b_j$ is outside the rational linear span of the earlier columns, and put $I=B_1\cup \cdots \cup B_{j-1}$, taking $I=\varnothing$ for $j=1$. Apply the selected linear functional ${\ell }_{I,B_j}$ to $Ax=0$. The earlier terms vanish exactly. Divide the remaining integer equality by $p^{{\nu }_j}$ and reduce modulo $p$. Later blocks vanish modulo $p$, while each coordinate in $B_j$ contributes its common digit $u$. We obtainwhich is impossible by the choice of $p$ and because $u\not\equiv 0(\mathrm{mod}p)$. Thus each block sum lies in the earlier linear span. For the first block, this span is zero, so $b_1=0$. This proves the columns condition.

For sufficiency, assume the columns condition and choose rational coefficients ${\lambda }_{ij}$, for $i\in B_1\cup \cdots \cup B_{j-1}$, such thatChoose a positive integer $c$ clearing all these coefficients' denominators and a positive integer $p$ with $p\ge |c{\lambda }_{ij}|$ for every coefficient. Use the permitted monochromatic m-p-c set theorem, in the suffix convention for an m-p-c set, to obtain positive generators $z_1,\dots ,z_s$ such that the full setis positive and monochromatic. The positivity of the full m-p-c set is important: we do not discard expressions that happen to be negative. For $i\in B_h$, setAll these coordinates belong to $D$, so they are positive and monochromatic. On collecting coefficients of each $z_j$, their matrix product isThis explicit Rado solution inside an m-p-c set proves sufficiency and completes the proof of Rado's theorem.

To deduce the Finite sums theorem directly from the columns condition, introduce one positive variable $y_F$ for each nonempty $F\subseteq \{1,\dots ,k\}$ and imposeFor $k=1$ there are no equations and any positive integer works. For $k\ge 2$, write $a_F$ for the column belonging to $y_F$. Partition the columns into blocks $B_j=\{F:\min F=j\}$ in the order $j=1,\dots ,k$. Consider the row indexed by a set $H$ with $|H|\ge 2$. The sum of the columns in $B_j$ has row valueFor $j=1$ this is always zero. For $j>1$ it equals one exactly for those $H$ with $\min H<j$ and $j\in H$. Each such $a_H$ is the negative coordinate vector in row $H$, and it belongs to an earlier block. Thereforewhich verifies the columns condition, rather than assuming the desired Finite sums theorem. Apply Rado's theorem to this system. Its monochromatic positive solution satisfies $y_F={\sum }_{i\in F}{x}_i$ with $x_i=y_{\{i\}}$, and hence

Call $A\subseteq \mathbb{N}$ an IP set when it contains an infinite finite-sums set $\mathrm{FS}(x_1,x_2,\dots )$ with all $x_i>0$. We first need the relative consequence of the Hindman theorem: a finite union can be an IP set only if one of its constituents is an IP set. This does not follow by simply applying the theorem to a coloring of all integers, because its monochromatic finite-sums set might lie outside the given set.

Here is a reduction that proves the partition regularity of IP sets. Suppose $\mathrm{FS}(x_i)\subseteq A$ and $A$ is finitely colored. Index the generators by $i\ge 0$. For $n>0$, let $S(n)$ be the set of positions of the nonzero digits in its binary expansion, and color $n$ by the color of ${\sum }_{i\in S(n)}{x}_i$. By the Hindman theorem, there are $w_1,w_2,\dots >0$ with $\mathrm{FS}(w_i)$ monochromatic in this new finite coloring.

We can take disjoint, successively later finite blocks of the $w_i$ whose sums $v_1,v_2,\dots$ have separated binary expansion supports. Start with $v_1=w_1$. Having chosen finitely many blocks, take $K$ larger than every binary position used in their sums. On a fresh tail of the $w_i$, consider $2^K+1$ partial sums, including the zero partial sum. Two have the same residue class modulo $2^K$, by the pigeonhole principle. Their positive difference is a consecutive block sum $v_{j+1}$ divisible by $2^K$. Discard all indices up to that block's end and repeat. Thus every bit of $v_{j+1}$ is above every bit in the preceding block sums.

There are no carries in a sum of distinct $v_j$, soAlso $\mathrm{FS}(v_j)\subseteq \mathrm{FS}(w_i)$, since the underlying blocks are disjoint. Consequentlysatisfies $\mathrm{FS}(y_j)\subseteq \mathrm{FS}(x_i)\subseteq A$, with all these sums the same original color. This proves the relative assertion. In particular, if $A_1\cup \cdots \cup A_r$ contains an infinite finite-sums set, color its points by the first constituent containing them and apply this argument; one $A_j$ is an IP set.

Let $\mathcal{I}$ be the family of sets that are not IP sets. It is closed under subsets and, by the preceding argument, under finite unions. It contains every finite set, since positive successive partial sums of an infinite sequence are unbounded. It does not contain $\mathbb{N}$. Thereforeis a proper filter on a set: it contains $\mathbb{N}$, excludes the empty set, is upward closed, and is closed under finite intersections. Its members are exactly the IP-star sets, the sets meeting every IP set. Indeed, failure to meet an IP set is equivalent to having an IP complement. This identifies the sets that must belong to the required ultrafilter.

Apply the ultrafilter lemma to extend the IP-star filter $\mathcal{F}$ to an ultrafilter $\mathcal{U}$. For completeness, Zorn lemma applies to its proper filter extensions ordered by inclusion, because the union of a chain is still a proper filter on a set. A maximal extension decides every set $A$: if $A$ cannot be adjoined while keeping the filter proper, some filter member is disjoint from $A$, forcing $A^c$ into the filter. This is the defining ultrafilter property.

If $A\in \mathcal{U}$ were not an IP set, then $A^c\in \mathcal{F}\subseteq \mathcal{U}$, contradicting properness. Hence every member of $\mathcal{U}$ contains an infinite finite-sums set. Conversely, any ultrafilter with finite-sums members must contain every IP-star set, since it cannot contain such a set's non-IP complement. Thus this construction captures precisely the forced filter, without making an unnecessary claim about idempotence.

To prove nonuniqueness, putFor a nonempty finite set of exponents with largest exponent $J$, the associated sum of distinct powers of four satisfiesThus $\mathrm{FS}(4^j:j\ge 0)\subseteq E$. Doubling the inequality givesso $\mathrm{FS}(2\cdot 4^j:j\ge 0)\subseteq E^c$. These alternating dyadic intervals contain disjoint IP sets.

Every member of $\mathcal{F}$ meets $E$ and also $E^c$, because it meets every IP set. A finite intersection of members of $\mathcal{F}$ is again a member, so both $\mathcal{F}\cup \{E\}$ and $\mathcal{F}\cup \{E^c\}$ have the finite intersection property. Extend the proper filters they generate to ultrafilters ${\mathcal{U}}_0$ and ${\mathcal{U}}_1$. Each extends $\mathcal{F}$, so the previous argument makes each an ultrafilter with finite-sums members. But $E\in {\mathcal{U}}_0$ and $E^c\in {\mathcal{U}}_1$, so

Pull the given finite coloring $\chi :\mathbb{N}\to [k]$ back to a finite coloring of unordered pairs by puttingThe ordering $a<b$ makes this a well-defined function on the edges of a complete graph. By Ramsey's theorem, some infinite set $X=\{x_1<x_2<\cdots \}$ has all its pairs of one color $r$ under $\tilde{\chi }$. ConsequentlyPossible coincidences between these products cause no difficulty, since the original finite coloring assigns each positive integer a single color.

Write $\mathbb{N}=\{1,2,\dots \}$ and let $\chi$ be the given finite coloring of the edges of the complete graph on the positive integers. Construct an increasing sequence $(v_i)$ and nested infinite sets $A_0\supset A_1\supset \cdots$ as follows. Set $A_0=\mathbb{N}$. Having chosen $A_{i-1}$, take $v_i=\min A_{i-1}$, partition $A_{i-1}\setminus \{v_i\}$ by the color of $\{v_i,w\}$, and use the infinite pigeonhole principle to choose an infinite set $A_i$ on which this color is constant, say $r_i$. Every element of $A_i$ exceeds $v_i$.

The infinite pigeonhole principle applied again to $(r_i)$ gives an infinite set of indices $I$ and a color $r$ such that $r_i=r$ for all $i\in I$. For $i<j$ in $I$, nesting gives $v_j\in A_i$, so $\chi (\{v_i,v_j\})=r$. Thus the infinite set $X=\{v_i:i\in I\}$ is a monochromatic complete graph:This proves the required infinite case of Ramsey's theorem; finiteness of the color set is used at both applications of the infinite pigeonhole principle.

Let $\rho$ be the cyclic logarithmic coloring from part (iii). Define a finite coloring of the positive integers byThe inner logarithm is at least $1$ whenever $n\ge 2$, so this is defined everywhere. Suppose an increasing infinite sequence $(x_i)$ made every product with distinct indices monochromatic. Fix an index $i$ with $x_i\ge 2$ and write $a={\log }_2{x}_i>0$. The sequence is unbounded, so choose $j>i$ with $b={\log }_2{x}_j\ge 28a$. Both ordered products are among the supposedly monochromatic values, but their logarithms satisfyIndeed, the first inequality is equivalent to $b\ge 28a$, and the second uses $a>0$. Part (iii) therefore gives $\chi (x_i{x}_j^2)\ne \chi (x_j{x}_i^2)$, a contradiction. Thus the assertion with all distinct ordered indices is false. The obstruction needs an infinite unbounded sequence; it does not assert that every pair of different positive integers gives different colors.

A run of a word is a maximal consecutive block of equal letters. For every $N\ge 1$, use the following finite coloring of $[3{]}^N$, with at most nine colors:We show that this run-count obstruction to interval-active lines excludes every monochromatic combinatorial line with an interval active set $[a,b]$. If $a=1$, its three words already have different first coordinates, so their colors differ. Suppose $a>1$ and denote the letter immediately to the left by $p$. All contributions to $R(w)$ away from the two possible boundaries of $[a,b]$ are independent of its active letter $x$; no internal active boundary contributes a change.

If $b=N$, the only variable contribution is $1_{\{p\ne x\}}$, which is $0$ at $x=p$ and $1$ at another letter. If $b<N$, write $q$ for the letter immediately to the right. The variable contribution isWhen $p=q$, its values are $0$ and $2$. When $p\ne q$, it equals $1$ at $x=p$ or $x=q$, and equals $2$ at the third letter. In every case these values are distinct modulo $3$. Hence no dimension works for alphabet size three and nine colors, which disproves the proposed universal strengthening. For the PDF's illustrative line, choosing either adjacent letter merges one active run with a neighboring run, while choosing any other letter merges neither; this is exactly the boundary effect measured above. Recording the first letter also handles active intervals meeting the beginning, including the entire coordinate set.

Let the required arithmetic progression have length $m\ge 2$, let $k$ be the number of colors, and take $n=n(m,k)$ from the Hales-Jewett theorem. Given any finite coloring $\chi$ of $[mn]$, color a word over an alphabet $w\in [m{]}^n$ by $\chi (w_1+\cdots +w_n)$. For a monochromatic combinatorial line, let $A\ne \varnothing$ be its active coordinate set and let $C$ be the sum of its fixed coordinates. The sums of its $m$ words areThey all lie in $[mn]$, have one color, and form an arithmetic progression with strictly positive common difference $|A|$. Thus, writing $W(k,m)$ for the finite Van der Waerden theorem bound,The length-one case is immediate. Restricting a finite coloring of the positive integers to this finite integer interval gives the infinite-domain formulation of the Van der Waerden theorem as well.

We prove the Hales-Jewett theorem by mathematical induction on the alphabet size. A combinatorial line has a nonempty active coordinate set, with one common letter in all those coordinates. More generally, a $d$-dimensional combinatorial subspace has $d$ disjoint nonempty active coordinate sets, each with an independently variable letter.

First prove the alphabet insensitivity lemma, rather than assume it. Fix distinct letters $a,b\in [m]$, $k$ colors, and dimension $d$. Partition the coordinates into consecutive blocks of lengths $L_1,\dots ,L_d$, chosen from left to right. Put $S_0=0$ and chooseConstruct the variable blocks from right to left. When working on block $i$, the later blocks have already become $(d-i)$-parameter words, while all earlier $S_{i-1}$ coordinates remain unrestricted. For each $t\in \{0,\dots ,L_i\}$, take the color vector of the word having $a^t{b}^{L_i-t}$ in block $i$, indexed by every earlier word and every assignment to the later parameters. There are $m^{S_{i-1}+d-i}$ entries and hence at most $L_i$ different vectors. The pigeonhole principle gives $t<s$ with identical vectors. Fix the first $t$ positions of this block to $a$, fix its last $L_i-s$ positions to $b$, and make the intervening $s-t$ positions a new variable block. Substituting $a$ or $b$ there gives exactly the two equal color vectors.

This equality holds for every earlier word and every later parameter assignment. Restricting the earlier coordinates in subsequent steps therefore preserves it. Likewise, the insensitivity already proved in a later block survives every restriction of earlier coordinates. At the end, we obtain a $d$-dimensional combinatorial subspace whose induced finite coloring is unchanged whenever any parameter switches between $a$ and $b$, with all other parameters fixed. Multiple switches preserve the color by successive single switches. This is the alphabet insensitivity lemma in full.

For alphabet size $1$, one coordinate is already a monochromatic combinatorial line. Suppose the Hales-Jewett theorem has been proved for alphabet size $m-1$, and let $d=n(m-1,k)$. Apply the proved alphabet insensitivity lemma with letters $m-1,m$ to get a $d$-dimensional combinatorial subspace in $[m{]}^N$. Restrict its parameter words to $[m-1{]}^d$. The induction hypothesis gives a monochromatic combinatorial line in these parameters. The union of its active parameter blocks is nonempty, so its image is a combinatorial line in the original coordinates. Its additional point with active letter $m$ has the same color as the point with active letter $m-1$, by insensitivity. Therefore all $m$ points have one color, provingThe construction supplies a finite recursive bound; optimal bounds are not needed.

First justify the finite bound used in the hint. Fix $k$. If no finite $T$ forced a monochromatic positive solution of ${\sum }_i{a}_i{x}_i=0$, consider the rooted tree whose level $t$ consists of the solution-free finite colorings of $[t]$, with restriction as the predecessor map. Every level is nonempty and every vertex has at most $k$ children. The König infinity lemma gives an infinite branch, hence a finite coloring of all positive integers with no such solution, contradicting the partition regular matrix hypothesis. This is the compactness bound for partition regularity.

Choose such a $T$ and let $S=\mathrm{lcm}(1,\dots ,T)$, the least common multiple. For a given finite coloring $\chi$ of the positive integers, pull it back to $[T]$ byEach $S/t$ is a positive integer in $[S]$. The defining property of $T$ gives $x_1,\dots ,x_n\in [T]$ of one color under $\tilde{\chi }$, with ${\sum }_i{a}_i{x}_i=0$. Set $y_i=S/x_i$. Their colors under $\chi$ agree, andThus reciprocal partition regularity follows. This construction is an involution on the divisors of $S$; it permits repeated coordinates and never requires a reciprocal of a positive integer to itself be integral without the common scaling factor $S$.

Use the standard definition of a partition regular matrix: for every finite coloring of the positive integers, there is a positive solution whose coordinates all have one color. Coordinates may repeat. Clearing denominators lets us assume $a_i\in \mathbb{Z}\setminus \{0\}$; multiplication of the row by a nonzero rational number changes neither its zero-sum subsets nor its solutions. We prove both directions of the Rado theorem for one equation directly, without invoking any form of Rado's theorem.

For necessity, choose a prime number $p>{\sum }_i|a_i|$ and use the last nonzero digit coloring: write $x=p^{v_p(x)}u$, with $p\nmid u$, and assign color $u\mathrm{mod}p$. Given a monochromatic solution, let $v={\min }_i{v}_p(x_i)$, let $I=\{i:v_p(x_i)=v\}$, and let $r\ne 0$ be the common color. Divide ${\sum }_i{a}_i{x}_i=0$ by $p^v$ and reduce modulo $p$. Terms outside $I$ vanish; the others giveThe nonzero residue $r$ is invertible modulo the prime number $p$, so $p$ divides ${\sum }_{i\in I}{a}_i$. This sum has absolute value less than $p$, forcing it to be zero. The set $I$ is nonempty by its definition.

For sufficiency we first derive the needed Brauer progression theorem from the permitted Van der Waerden theorem. Fix positive integers $s,\ell$. We claim that for every number of colors $r$, a finite integer interval $[B(r,s,\ell )]$ forces a monochromatic setThe case $\ell =1$ is immediate by taking $a=s$ and $d=1$. For $r=1$ and $\ell \ge 2$, take $a=d=1$ in an interval of length at least $\max (s,\ell )$. Suppose the claim holds for $r-1$ and write $M=B(r-1,s,\ell )$. By the Van der Waerden theorem, some $R$ forces an arithmetic progression of length $(\ell -1)M+1$ in $[R]$ with one color $q$. Work in $[\max (R,sMR)]$ so that all needed multiples also lie in the coloring's domain. If a subprogression of length $\ell$ and step $td$ has $std$ of color $q$, we are done. Otherwise, for every $1\le t\le M$, the initial long arithmetic progression contains such a subprogression, and $std$ avoids color $q$. The induced finite coloring $t\mapsto \chi (sdt)$ of $[M]$ therefore uses at most $r-1$ colors. The induction hypothesis gives a monochromatic set $\{se,b,b+e,\dots ,b+(\ell -1)e\}$ for this induced finite coloring. Multiplying by $sd$ gives the desired configuration in the original coloring, with initial term $sdb$ and step $sde$. This proves the claim for $\ell \ge 2$; the only use below has $\ell \ge 3$.

Now suppose ${\sum }_{i\in I}{a}_i=0$ for a nonempty $I$. Choose $i_0\in I$, put $B={\sum }_{i\notin I}{a}_i$, and put $s=|a_{i_0}|$. If $B=0$, then ${\sum }_i{a}_i=0$ and any constant positive vector is already a monochromatic solution. Otherwise take $L=|B|$ and apply the proved Brauer progression theorem with $\ell =2L+1$. All of $sd$ and $a+jd$ for $0\le j\le 2L$ have one color. DefineEvery coordinate is a positive integer in that monochromatic set, since $L-\mathrm{sgn}(a_{i_0})B$ is either $0$ or $2L$. Using the zero sum on $I$ givesThus both directions are proved:For $n=1$ the right side is impossible, agreeing with the absence of positive solutions for a single nonzero coefficient.