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Suppose nonzero integer coefficients have a nonempty zero-sum subset $I$, and fix $i_0\in I$. Put $s=|a_{i_0}|$ and $C={\sum }_{j\notin I}{a}_j$. A Brauer progression theorem configuration containing a sufficiently long monochromatic progression of step $d$ and the point $sd$ gives a positive solution: take every outside variable equal to $sd$, and the inside variables from the progression with offsets ${\lambda }_i$, where ${\lambda }_{i_0}=-\mathrm{sgn}(a_{i_0})C$ and the other offsets are zero. Shifting all inside offsets by the same integer keeps their weighted sum unchanged, because ${\sum }_{i\in I}{a}_i=0$. This removes negative offsets while preserving the equation.

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.

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.

Van der Waerden theorem says that for every pair of positive integers $r,k$ there is $W(r,k)$ such that every finite colouring of $[W(r,k)]=\{1,\dots ,W(r,k)\}$ using at most $r$ colours contains a monochromatic arithmetic progression of length $k$ and positive common difference. Consequently every finite colouring of $\mathbb{N}$ contains arbitrarily long monochromatic arithmetic progressions.

Here is a colour-focusing proof of Van der Waerden theorem. Induct on $k$, simultaneously for all $r$. For $k=1$, take $W(r,1)=1$. Suppose $W(R,k)$ exists for every number of colours $R$, and fix $r$. We will construct finite bounds $F_t$, $1\le t\le r$, with the following property: every 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, all focused at the same point $f\in [F_t]$. Specifically, their terms are“Focused” means that adjoining $f$ completes the corresponding length-$(k+1)$ arithmetic progression if $f$ has its colour.

For $t=1$, put $F_1=2W(r,k)$. Find a monochromatic length-$k$ arithmetic progression in the first $W(r,k)$ points. Its next point is at most $2W(r,k)$ and supplies the focus. For $k=1$, give the selected singleton common difference $1$.

Suppose $F_t=N$ has been constructed. SetDivide $[2NM]$ into $2M$ consecutive blocks of length $N$. Colour the first $M$ block indices by their full colour profiles, that is, the ordered list of their $N$ colours. There are at most $r^N$ profiles, so the induction hypothesis on $k$ gives equally profiled blocks with indicesWhen $k=1$, choose $D=1$. Otherwise $D>0$ and $j_0+kD\le 2M$, since the last selected block is at most $M$ and $D\le M$.

Assume the original finite colouring contains no monochromatic length-$(k+1)$ arithmetic progression. Applying the property of $F_t$ to the common profile gives $t$ focused length-$k$ arithmetic progressions at a position $f$, with common differences $d_i$ and distinct colours $q_i$. The colour $q$ of position $f$ differs from every $q_i$: otherwise the profile already contains a length-$(k+1)$ monochromatic arithmetic progression, and so does any selected block.

Put $j_{\ast }=j_0+kD$ and $F=N(j_{\ast }-1)+f$, a point in the original interval. For $1\le h\le k$, the pointlies in a selected block at profile position $f-h{d}_i$, so has colour $q_i$. Similarly $F-hND$ has colour $q$. Thus $t+1$ length-$k$ arithmetic progressions of distinct colours focus at $F$. This proves the step in $t$. At $t=r$, the focus necessarily has the colour of one of the $r$ progressions, giving a monochromatic length-$(k+1)$ arithmetic progression. Hence $W(r,k+1)$ exists, completing the mathematical induction and the proof of Van der Waerden theorem.

For the partition regularity deduction we first prove a strengthened progression lemma, the Brauer progression theorem: for positive $r,s,L$ with $L\ge 2$, there is $B(r,s,L)$ such that every finite colouring of $[B(r,s,L)]$ using at most $r$ colours contains a monochromatic setInduct on $r$. For $r=1$, take $B(1,s,L)=\max (s,L)$ and $a=d=1$. Given the result for $r-1$, putVan der Waerden theorem supplies a colour-$q$ arithmetic progression $a+j{d}_0$, $0\le j<K$, inside $[W]$. If some $sj{d}_0$, $1\le j\le N$, also has colour $q$, take $d=j{d}_0$ and use the $L$ terms $a,a+j{d}_0,\dots ,a+(L-1)j{d}_0$. They are in the progression because $(L-1)j\le K-1$.

Otherwise the integers $sj{d}_0$, $1\le j\le N$, use at most $r-1$ colours. Colour $j\in [N]$ by the colour of $sj{d}_0$ and apply the induction hypothesis. It supplies a monochromatic set $\{se,b,b+e,\dots ,b+(L-1)e\}$. Multiplying by $s{d}_0$ gives the required configuration with $a=s{d}_{0}b$ and $d=s{d}_{0}e$: its distinguished member is $s^2{d}_{0}e=sd$. All of these points are at most $sN{d}_0\le sNW$, so the finite bound is valid. This proves the Brauer progression theorem using only the just-proved Van der Waerden theorem.

A row is a partition regular matrix over the positive integers if every finite colouring of $\mathbb{N}$ admits a positive vector $x$ whose coordinates have one colour and whose product with the row is zero. Coordinates may repeat. Since the coefficients are rational numbers, multiplying them by a common positive denominator leaves the equation and every zero subset sum unchanged, so write the row as nonzero integer coefficients $b_1,\dots ,b_n$.

To prove necessity, suppose no nonempty subset of the $b_i$ sums to zero. Choose a prime $p>{\sum }_i|b_i|$ and use the last nonzero digit coloringwhere $v_p$ is the P-adic valuation. If positive $x_i$ all had the same colour $u$ and satisfied ${\sum }_i{b}_i{x}_i=0$, let $m={\min }_i{v}_p(x_i)$ and $I=\{i:v_p(x_i)=m\}$. After division by $p^m$ and reduction modulo $p$, the equation becomesAs $u\ne 0$, this says $p\mid {\sum }_{i\in I}{b}_i$. The latter sum is nonzero and has absolute value at most ${\sum }_i|b_i|<p$, a contradiction. This explicitly proves the necessary obstruction without invoking a general partition regularity criterion.

Conversely, suppose $\varnothing \ne I\subseteq \{1,\dots ,n\}$ and ${\sum }_{i\in I}{b}_i=0$. Select an index $j\in I$, and putFor any finite colouring, the Brauer progression theorem gives one-coloured $\{sd,a,a+d,\dots ,a+(L-1)d\}$. The indices $u$ and $u+q$ both lie between $0$ and $L-1$. Define positive, same-coloured coordinates byThenThis also covers $I$ equal to the whole index set and $q=0$. Reversing the clearing of denominators proves the exact criterion:

Use $[m]=\{1,\dots ,m\}$. A combinatorial line in $[m{]}^n$ has one nonempty active set of coordinates, all carrying the same variable letter, with every other coordinate fixed. The Hales-Jewett theorem states that for every finite alphabet size $m$ and number of colours $k$, some $n$ makes every $k$-finite colouring of $[m{]}^n$ contain a monochromatic combinatorial line.

Here is a complete proof through the alphabet insensitivity lemma. A $d$-dimensional combinatorial subspace is obtained from $d$ pairwise disjoint nonempty coordinate blocks, one independently variable letter on each block, and fixed letters elsewhere. For two distinct letters $a,b$, a colouring restricted to such a combinatorial subspace is $(a,b)$-insensitive if switching $a$ and $b$ in any of its parameter coordinates leaves the colour unchanged, with all other letters and coordinates arbitrary.

We first prove that any desired dimension $d$ admits such an insensitive combinatorial subspace, using only the pigeonhole principle. Define finite bounds recursively bySplit the ambient coordinates into a prefix of length $L$ and a suffix of length $N$. For $0\le j\le L$, let the prefix $p_j$ consist of $j$ copies of $b$ followed by $L-j$ copies of $a$. Give it the entire colour profileThere are at most $k^{m^N}=L$ such profiles, but $L+1$ prefixes. Hence two prefixes $p_s,p_t$, $s<t$, have the same profile. Replace their differing coordinates $s+1,\dots ,t$ by one variable letter, obtaining a prefix word $v(x)$ with $v(a)=p_s$ and $v(b)=p_t$. Then $c(v(a)u)=c(v(b)u)$ for every suffix $u$.

Colour the suffixes by the vectorwhich has at most $k^m$ colours. By the dimension-$d-1$ induction hypothesis, there is a $(d-1)$-dimensional suffix combinatorial subspace on which this vector colouring is $(a,b)$-insensitive. Combining it with the variable block of $v$ gives dimension $d$. Switching the first parameter between $a,b$ preserves colour by the equal prefix profiles; switching any other parameter preserves every entry of $c^{\prime }$ by induction. This proves the alphabet insensitivity lemma, including $d=1$, where the suffix is empty.

Now induct on $m$ to prove the Hales-Jewett theorem. For $m=1$, any one-variable word gives a one-point monochromatic combinatorial line. Suppose the theorem holds for $m-1$, and let $d$ be a sufficient dimension for $k$ colours on $[m-1{]}^d$. The lemma supplies a $d$-dimensional combinatorial subspace insensitive to $(m-1,m)$. Restrict its parameter colouring to $[m-1{]}^d$. By the alphabet induction hypothesis it contains a monochromatic combinatorial line. Its $m$th point has the same colour as its $(m-1)$st point: switch the active parameter coordinates from $m-1$ to $m$, one at a time. Therefore the full line over $[m]$ is monochromatic. Its actual active coordinate set is the nonempty union of the active parameter blocks. This completes both inductions.

To deduce the Van der Waerden theorem, choose the dimension $n$ for alphabet size $m$ and pull back a colouring of a sufficiently long integer interval alongAll images lie in $[N]$ for $N=m{\sum }_{j=1}^n{m}^{j-1}$. A combinatorial line with active coordinates $D$ maps towhere $a$ is the contribution of the fixed coordinates. Thus every finite colouring of the positive integers contains monochromatic arithmetic progressions of every finite length. The argument also provides the finite-interval version. Length one is immediate.

The Strengthened Van der Waerden theorem requires the common difference itself to have the progression's colour. We prove a useful stronger, scaled version, the Brauer progression theorem: for every $k,t,\ell \ge 1$, a finite bound $B(k,t,\ell )$ guarantees a monochromatic setThe desired strengthening is $t=1$. For $\ell =1$ choose $a=t,d=1$; for one colour any interval covering the displayed configuration suffices. For the induction on $k$, assume $M=B(k-1,t,\ell )$ and $\ell \ge 2$. Put $L=(\ell -1)M+1$, let $W$ be a Van der Waerden theorem bound for $k$ colours and length $L$, and take $B(k,t,\ell )=tMW$. Inside $[W]$, obtain a one-colour progression $a+j\delta$ for $0\le j<L$, of colour $c$.

All numbers $tj\delta$, $1\le j\le M$, lie in the larger interval because $\delta \le W$. If one has colour $c$, then the subsequence $a,a+j\delta ,\dots ,a+(\ell -1)j\delta$, together with $tj\delta$, is the desired configuration with $d=j\delta$. Otherwise the colouring $j\mapsto c(t\delta j)$ of $[M]$ uses at most $k-1$ colours. Apply the induction hypothesis to obtain a one-colour configuration $\{te,b,b+e,\dots ,b+(\ell -1)e\}$ there. Multiplication by $t\delta$ gives common difference $d=t\delta e$ and distinguished point $td=t^2\delta e$, all of the same original colour. This proves the scaled theorem using only the Van der Waerden theorem.

The fixed-active-size strengthening is false. It already fails for $m=k=2$. For any proposed positive $n,d$, if $d>n$ there is no eligible line. Otherwise colour a word byThe two points of a line with exactly $d$ active coordinates have counts $q$ and $q+d$ of the letter $2$. Their quotient floors differ by exactly one, so their colours differ. This fixed-active-size obstruction for combinatorial lines is valid for every proposed $n,d$. The colouring may depend on $d$, since the assertion chooses $n,d$ before demanding the conclusion for every colouring.

Clear a common denominator first, so that all coefficients are nonzero integers. This changes neither the solutions nor partition regularity. We prove the one-equation criterion directly; no form of Rado's theorem is used as an assumption.

For necessity, choose a prime number $p>{\sum }_i|a_i|$. Colour a positive integer by its first nonzero base-$p$ digit: writing $x=p^{v_p(x)}u$ with $p\nmid u$, use the colour $u(\mathrm{mod}p)\in \{1,\dots ,p-1\}$. Suppose the equation has a monochromatic solution. Let $v$ be the smallest P-adic valuation of its entries and put $I=\{i:v_p(x_i)=v\}$, a nonempty set. Divide the equation by $p^v$ and reduce modulo $p$. All entries indexed by $I$ contribute the same nonzero digit $u$, and all other terms vanish. HenceIt follows that $p$ divides ${\sum }_{i\in I}{a}_i$. The absolute value of this sum is less than $p$, so $\boxed{\sum _{i\in I}{a}_i=0}$.

For sufficiency, take a nonempty zero-sum index set $I$ and put $S={\sum }_{i\notin I}{a}_i$. If $S=0$, the full coefficient sum is zero and the constant vector $(1,\dots ,1)$ is already a monochromatic solution in every colouring. Otherwise choose $j\in I$, put $t=|a_j|$, and set $b_j=-\mathrm{sgn}(a_j)S$, $b_i=0$ for $i\in I\setminus \{j\}$. ThenAdd the same sufficiently large nonnegative integer $H$ to all the $b_i$, obtaining $c_i=b_i+H\ge 0$. The sum is unchanged because ${\sum }_{i\in I}{a}_i=0$. Let $K={\max }_{i\in I}{c}_i$.

Apply the scaled Brauer progression theorem, proved above from the permitted Van der Waerden theorem, to obtain one-colour numbers $td$ and $a,a+d,\dots ,a+Kd$. DefineAll entries are positive and monochromatic, andTherefore the row is partition regular exactly when some nonempty coefficient subset sums to zero. Repetition of entries is allowed by the definition of partition regularity.

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.