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If a rational matrix is a partition regular matrix, then for each fixed number of colors $k$ some finite integer interval $[T]$ already forces a monochromatic positive solution. Otherwise the solution-free finite colorings of successive intervals form a finitely branching tree with every level nonempty. The König infinity lemma gives an infinite branch, contradicting partition regularity. More generally, this argument applies to any family of configurations each using finitely many positive integers.

For a partition regular matrix $A$ and any property $P$ of its positive solution vectors, either every finite coloring admits a monochromatic solution with $P$, or every finite coloring admits one without $P$. If the first alternative fails, take a witnessing coloring and pair it with any other coloring. Partition regularity of the product coloring forces a solution avoiding $P$ in the other coloring. The alternatives need not be exclusive.

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:

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

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.

Rado's criterion. Let $A$ be a matrix over the rational numbers, with columns $a_1,\dots ,a_q$. A partition regular matrix is one for which every finite coloring of the positive integers gives a monochromatic positive vector $x\in {\mathbb{N}}^q$ with $Ax=0$; repeated coordinates are allowed. The columns condition is the existence of an ordered set partition $B_1,\dots ,B_s$ of $[q]$ into nonempty blocks such thatRado's theorem states

Necessity. Clear denominators so the columns are integer vectors. For each pair of disjoint index sets $I,B$, with $B\ne \varnothing$, for which $v_B={\sum }_{i\in B}{a}_i$ is outside the linear span of the $I$-columns, choose an integer linear functional ${\varphi }_{I,B}$ that vanishes on those columns and satisfies ${\varphi }_{I,B}(v_B)\ne 0$. Such a functional exists by extending a basis of the span to one also containing $v_B$, defining a functional on that basis, and clearing its rational denominators.

There are only finitely many such pairs. Choose a prime number $p$ dividing none of these nonzero integer evaluations. If there are no such pairs, any prime will do. Use the last nonzero digit coloring: write $x=p^{v_p(x)}u$, with $p\nmid u$, and color $x$ by $u\mathrm{mod}p\in \{1,\dots ,p-1\}$. Since $A$ is a partition regular matrix, this coloring supplies a monochromatic positive solution, so all its unit parts have a common nonzero residue $d$.

Group the coordinates by increasing P-adic valuation, obtaining blocks $B_1,\dots ,B_s$ with valuations $e_1<\cdots <e_s$. Suppose for some $j$ the sum $v_{B_j}$ lay outside the linear span of the earlier columns, with earlier index set $I$. Apply ${\varphi }_{I,B_j}$ to $Ax=0$. Earlier terms vanish exactly. Divide the remaining integer equation by $p^{e_j}$ and reduce modulo $p$. Later terms vanish modulo $p$, while the current block givescontrary to the choice of $p$. Thus every block sum lies in the earlier linear span, and for the first block this means it is zero. This proves the columns condition by the finite separating-functional proof of the columns condition.

Sufficiency. Suppose the columns condition holds. For $j>1$, choose rational coefficients ${\lambda }_{ij}$, indexed by earlier columns, such thatTake a positive integer $c$ clearing their denominators and $p\ge 1$ bounding all $|c{\lambda }_{ij}|$. We use the allowed monochromatic m-p-c set theorem with $s$ levels. Its M-p-c set is the full setwhere the generators make every displayed expression positive. Positivity is part of this convention, rather than a rule that discards nonpositive expressions. For $i\in B_h$, setAll these numbers lie in the monochromatic set $D$. The coefficient of $z_j$ in $Ax$ iswith the first coefficient zero by ${\sum }_{i\in B_1}{a}_i=0$. Hence $Ax=0$. This Rado solution inside an m-p-c set proves sufficiency.

The equality-or-inequality dichotomy. If every finite coloring gives a monochromatic solution with $x_1=x_2$, the first alternative holds. Otherwise fix a finite coloring $\chi$ admitting no such solution. Given any other finite coloring $\psi$, use the refinement of a finite coloring $n\mapsto (\chi (n),\psi (n))$. Since $A$ is a partition regular matrix, it supplies a solution monochromatic for both colorings. Its first two coordinates cannot be equal, by the defining property of $\chi$, so it is the required unequal solution for $\psi$. Thus one of the two universal alternatives always holds. This is the partition regularity dichotomy under an extra constraint; the alternatives are not asserted to be mutually exclusive.

Both restrictions can fail. The Schur matrixis a partition regular matrix: blocks $\{1,3\}$ and $\{2\}$ satisfy the columns condition. Color $n$ by $v_2(n)\mathrm{mod}2$, using the 2-adic valuation. An equal-coordinate solution would have $x_3=2x_1$, and $v_2(2x_1)=v_2(x_1)+1$, so its coordinates cannot be monochromatic. This exhibits failure of the equality restriction.

Conversely, $A=(1-1)$ satisfies the columns condition in one block. Every solution has $x_1=x_2$, so even the constant one-color finite coloring admits no unequal-coordinate solution. Consequently

A rational matrix is a partition regular matrix when every finite coloring of the positive integers admits a monochromatic positive vector in its kernel. Its columns have the columns property if their indices can be partitioned into ordered nonempty blocks $B_1,\dots ,B_s$ such that the columns in $B_1$ sum to zero and, for $j>1$, the sum over $B_j$ lies in the rational linear span of the columns in the earlier blocks. Rado's theorem states that a rational matrix is partition regular if and only if its columns have this property.

For one equation, clear denominators and writeThe one-row columns property is equivalent to the existence of a nonempty $I\subseteq [n]$ with

First suppose such an $I$ exists. Choose $i_0\in I$, put $c=|c_{i_0}|$, $C={\sum }_{j\notin I}{c}_j$, and choose $p\ge |C|$. The monochromatic m-p-c set theorem, whose finite induction proof uses the Van der Waerden theorem, gives positive $z_1,z_2$ for which all numbersare positive and have one color. SetThe middle coefficient is the integer $-C\mathrm{sgn}(c_{i_0})$, of absolute value at most $p$, so all the $x_i$ belong to the monochromatic set. Their $z_1$ contribution vanishes because the coefficients over $I$ sum to zero, and their $z_2$ contribution isThus the equation is partition regular.

Conversely, suppose no nonempty subset of the coefficients sums to zero. Choose a prime number $p$ that divides none of the finitely many nonzero subset sums. Color each positive integer by its last nonzero digit coloring in base $p$. If a monochromatic solution existed, let $v$ be the smallest P-adic valuation among its coordinates and let $I$ index the coordinates of valuation $v$. After division by $p^v$ and reduction modulo $p$, all $x_i$ with $i\in I$ have the same nonzero last digit $r$, while the other terms vanish. The equation would givecontrary to the choice of $p$. This proves the Rado theorem for one equation.

Let $A$ be a rational matrix with columns $a_1,\dots ,a_n$. It is a partition regular matrix when every finite coloring of the positive integers has a monochromatic vector $x$ with $Ax=0$. It has the columns property when the column indices have an ordered partition $B_1,\dots ,B_r$ such thatand, for every $s>1$,Rado's theorem states that $A$ is partition regular if and only if it has the columns property.

First suppose $A$ is partition regular, and clear its denominators. For a prime number $p$, apply the last nonzero digit coloring in base $p$: if $x=p^{v_p(x)}u$ with $p\nmid u$, its color is $u\mathrm{mod}p$. Choose a monochromatic solution and group its coordinates into blocks $B_1,\dots ,B_r$ of equal $p$-adic valuation, in increasing order of valuation. Only finitely many ordered partitions are possible, so one partition occurs for infinitely many primes.

For any such prime, divide ${\sum }_i{x}_i{a}_i=0$ by the lowest power of $p$ and reduce modulo $p$. All coordinates in $B_1$ have the same nonzero leading digit $d$, while later blocks vanish, soThe integer vector ${\sum }_{i\in B_1}{a}_i$ is divisible by infinitely many primes and therefore equals zero.

Now fix $s>1$, and let $t$ be the common valuation on $B_s$. Reduction modulo $p^{t+1}$ givesIf the first sum of columns were outside the rational linear span of the earlier columns, an integer linear functional would vanish on every earlier column but not on that sum. Applying it to the congruence would say that infinitely many primes divide one fixed nonzero integer, a contradiction. Thus the displayed partition has the columns property.

Conversely, suppose $A$ has the columns property. For $s>1$, choose rational numbers $q_{is}$ such thatLet $d_{is}$ equal $1$ for $i\in B_s$, equal $-q_{is}$ in the earlier blocks, and equal zero in the later blocks. For $s=1$, let $d_{i1}$ be the indicator of $B_1$. Thenfor every $s$. Choose a positive integer $c$ clearing all denominators and a positive integer $p$ with $|c{d}_{is}|\le p$ whenever $s$ is later than the block containing $i$.

By the monochromatic m-p-c set theorem, the coloring contains a monochromatic $(r,p,c)$-set with generators $z_1,\dots ,z_r$. DefineIf $i\in B_t$, thenso every $y_i$ lies in that one monochromatic M-p-c set. Moreover,Thus $A$ is partition regular, proving Rado's theorem.

For the equation $4x+3y-6z=0$, a monochromatic solution in the last nonzero base-$p$ digit coloring would, at the least $p$-adic valuation among $x,y,z$, force a nonempty subset sum of $4,3,-6$ to vanish modulo $p$. The seven possible sums areNone is divisible by $5$, so base $5$ gives no monochromatic solution. The smaller primes do admit solutions: $(x,y,z)=(3,2,3)$ works for base $2$, whose coloring has one color, and $(3,4,4)$ is monochromatic in base $3$. Hence the smallest prime is

For $x+2y-7z=0$, use the last nonzero base-$11$ digit itself. The nonempty subset sums of $1,2,-7$ arenone zero modulo $11$. This is the required $10$-coloring.

For a $5$-coloring, identify each nonzero residue $d$ modulo $11$ with $-d$, giving the five colorsIf a monochromatic solution existed, reduction at the least $11$-adic valuation would give a signed nonempty subset sum of $1,2,-7$ equal to zero modulo $11$. Singles have absolute residues $1,2,7$; pairs have absolute residues $1,3,5,6,8,9$; and triples have absolute residues $4,6,8,10$. None is zero modulo $11$, which proves that this $5$-coloring works.

Let $c_1,\dots ,c_n$ be the columns of the rational partition regular matrix $A$, and clear denominators so that they are integer vectors. We use the P-adic columns lemma. For a large prime number $p$, color each positive integer by a sufficiently long initial block of the unit part of its P-adic valuation, together with its valuation modulo the block length. Partition regularity supplies a monochromatic $x=(x_1,\dots ,x_n)$ withGroup the indices according to the successive $p$-adic orders of the $x_i$. At the lowest order, division by the common power of $p$ and reduction modulo the chosen large power showsComparing the next nonzero blocks of base-$p$ digits shows successively thatFor completeness, these congruences may be made exact by taking the digit block longer than every determinant and coordinate formed from the fixed columns: a nonzero such integer cannot be divisible by the resulting power of $p$. There are only finitely many ordered partitions of $[n]$, so passing through arbitrarily long blocks leaves one partition satisfying all the displayed identities. This is precisely the columns property.

Rado's theorem states that a rational matrix is partition regular if and only if it has the columns property. By hypothesis each $A_i$ is partition regular, so choose a columns partition for each one. Form the block-diagonal matrixTaking at stage $j$ the union of the $j$th blocks from the individual partitions, with empty blocks added after a partition ends, gives the columns property for $B$: each row block sees exactly the corresponding dependence for its $A_i$. Hence $B$ is partition regular by Rado's theorem. A monochromatic vectorin the kernel of $B$ satisfies $A_i{x}_i=0$ for every $i$, and all entries of all the $x_i$ have the same color.

Rado's theorem states that a rational matrix is partition regular if and only if it has the columns property.

If a rational matrix $A$ is a partition regular matrix, every finite coloring of the positive integers admits monochromatic $y_1,\dots ,y_n$ such thatChoose a compactness bound for partition regularity $T$ for the number of colors, put $S=\mathrm{lcm}(1,\dots ,T)$, and pull back the coloring by $t\mapsto S/t$ on $[T]$. A monochromatic positive solution $x$ of $Ax=0$ gives $y_i=S/x_i$; then $(1/y_i)=x_i/S$. The least common multiple guarantees that every $y_i$ is a positive integer.