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Introduce a variable $z_I$ for every nonempty $I\subseteq [k]$ and the homogeneous equations $z_I-{\sum }_{i\in I}{z}_{\{i\}}=0$ for $|I|\ge 2$. In their matrix $A$, partition the columns by the displayed blocks. For each $j$, the vector with coordinate $1$ exactly when $j\in I$ lies in the kernel of a linear map of $A$. Its nonzero coordinates occur in $B_j$ and earlier blocks only; every coordinate in $B_j$ is one. Hence the first block's column sum is zero and every subsequent block sum is in the linear span of earlier columns. The columns condition holds, so Rado's theorem forces a positive monochromatic solution, and all nonempty finite sums of its singleton coordinates have the same colour.

After clearing denominators in a matrix, for every disjoint pair of index sets $I,B$ with $B$ nonempty and ${\sum }_{i\in B}{a}_i$ outside the linear span of the $I$-columns, choose an integer linear functional vanishing on those columns but not on the indicated sum. Choose a prime number dividing none of the finitely many nonzero evaluations. The last nonzero digit coloring for this prime, applied to a monochromatic positive solution, groups coordinates into blocks by P-adic valuation. Applying the appropriate functional and reducing modulo the prime shows that each block sum lies in the linear span of all earlier columns. This proves necessity in Rado's theorem without an asymptotic limiting argument.

For a prime number $p$, write a positive integer as $x=p^{v_p(x)}u$ with $p\nmid u$. Its last nonzero base-$p$ digit is $u\mathrm{mod}p$, an element of $\{1,\dots ,p-1\}$. Assigning this digit as the color gives the last nonzero digit coloring used in the necessity proof of Rado's theorem.

The necessity direction in Rado's theorem colors an integer using initial data from a P-adic valuation. Applying partition regularity and grouping the coordinates of a monochromatic solution by valuation yields the blocks in the columns property; reduction modulo successively higher powers of $p$ supplies the required linear dependences.

For a rational $\ell \times n$ matrix $A$, write its columns as $c_1,\dots ,c_n$. It is partition regular if every finite colouring of $\mathbb{N}$ admits positive integers $x_1,\dots ,x_n$, all of one colour, with $A\mathbf{x}=0$. Repeated coordinates are allowed. The columns condition is an ordered partitioninto nonempty blocks such thatRado's theorem states that a rational matrix is partition regular if and only if it satisfies the columns condition. We prove both implications.

For necessity, clear denominators so the columns are integral. For every disjoint pair of index sets $I,B$ with $B\ne \varnothing$ andchoose a rational linear functional annihilating all columns in $I$ but not $v_B$. Such a functional exists by extending a basis of the indicated linear span to include $v_B$, prescribing zero on the former and one on the latter. Clearing its denominators produces an integer-coefficient functional $f_{I,B}$ with $f_{I,B}(v_B)$ a nonzero integer. There are only finitely many pairs $I,B$. Choose a prime number $p$ dividing none of these nonzero evaluations; if there are no bad pairs, choose any prime.

Use the last nonzero digit coloring for this prime. Namely, write $x=p^{{\nu }_p(x)}u$ with $p\nmid u$ and colour $x$ by $u\mathrm{mod}p\in \{1,\dots ,p-1\}$. By partition regularity, obtain a positive monochromatic solution. Let the distinct P-adic valuations of its coordinates beand put $B_j=\{i:{\nu }_p(x_i)=v_j\}$. All unit parts $u_i$ have a common nonzero residue $t$ modulo $p$.

For a block $B_j$, let $I=B_1\cup \cdots \cup B_{j-1}$. If its column sum were outside the earlier linear span, apply $f_{I,B_j}$ to the equation ${\sum }_i{c}_i{x}_i=0$. Earlier-block contributions are exactly zero. Divide the remaining equation by $p^{v_j}$ and reduce modulo $p$. Higher-valuation terms vanish, leavingThis contradicts the choice of $p$ and $t\not\equiv 0$. Consequently every block sum lies in the earlier linear span. For $j=1$ that span is $\{0\}$, so its sum is zero. This proves the columns condition by the finite separating-functional proof of the columns condition.

For sufficiency, suppose the blocks satisfy the columns condition. Choose rational coefficients ${\alpha }_{ij}$, for $i\in B_1\cup \cdots \cup B_{j-1}$, such thatTake a positive integer $c$ clearing all their denominators and an integer $p\ge 1$ at least as large as every $|c{\alpha }_{ij}|$. An M-p-c set with generators $z_1,\dots ,z_s$ consists of all valueswith every such value positive. The generators are therefore chosen with $c{z}_h>p{\sum }_{j>h}{z}_j$. Positivity is part of the configuration, not achieved by deleting inconvenient values.

Use the permitted monochromatic m-p-c set theorem to obtain one such monochromatic set. For each $i\in B_h$, putEvery $x_i$ is a positive element of that set, because its non-leading coefficients are integers of magnitude at most $p$. In the sum ${\sum }_i{c}_i{x}_i$, the coefficient of $z_1$ is $c{\sum }_{i\in B_1}{c}_i=0$, and the coefficient of each $z_j$ with $j>1$ isThus $A\mathbf{x}=0$. This is an explicit Rado solution inside an m-p-c set and proves sufficiency.

For the Finite sums theorem, defineTo deduce its monochromaticity from Rado's theorem, introduce variables $z_I$ for all nonempty $I\subseteq [k]$ and equationsLet $A_k$ be their coefficient matrix, and partition its columns by $B_j=\{I:\min I=j\}$. For a fixed $j$, the vector with $I$-coordinate $1$ when $j\in I$, and $0$ otherwise, solves this homogeneous system: in every row both sides count whether $j$ belongs to its indexing set. HenceThe first block sum is zero, and each later block sum lies in the linear span of earlier columns. This proves the columns partition for finite-sums systems. The positive monochromatic solution supplied by Rado's theorem has $x_i=z_{\{i\}}>0$, and its other coordinates are exactly all the nonempty finite sums. Therefore $FS(x_1,\dots ,x_k)$ is monochromatic. For $k=1$ a singleton is immediate. Equivalently, the full positive M-p-c set with $m=k,p=c=1$ directly contains every such sum, by taking its leading index to be the least index in $I$.

For a rational matrix $A$ with columns $a_1,\dots ,a_n$, partition regularity means that every finite colouring of the positive integers admits a positive monochromatic vector $x$ with $Ax=0$. Repeated coordinates are allowed. The columns condition is an ordered partition $\{1,\dots ,n\}=B_1\bigsqcup \cdots \bigsqcup B_s$ into nonempty blocks such thatRado's theorem says that $A$ is partition regular if and only if it satisfies the columns condition. We prove both directions.

For necessity, clear denominators so the columns are integer vectors. For every pair of disjoint index sets $I,B$ with $B\ne \varnothing$ for which ${\sum }_{i\in B}{a}_i$ is not in the rational linear span of the I-columns, choose an integer linear functional ${\ell }_{I,B}$ that vanishes on those I-columns but not on that sum. Such a functional exists by finite-dimensional linear algebra; clear its rational denominators. There are only finitely many pairs, so choose a prime number $p$ dividing none of the nonzero integersIf there are no offending pairs, any prime number suffices.

Colour a positive integer $x=p^{v_p(x)}u$, $p\nmid u$, by $u\mathrm{mod}p$. This is the last nonzero digit coloring, using $p-1$ colours. Partition regularity gives a positive monochromatic solution. Group its coordinate indices by their distinct P-adic valuations $e_1<\cdots <e_s$, writing $B_j=\{i:v_p(x_i)=e_j\}$. All units $x_i/p^{e_j}$ have the same nonzero residue $\alpha$ modulo $p$.

For each $j$, let $I=B_1\cup \cdots \cup B_{j-1}$. If its block sum failed the required span condition, apply the chosen ${\ell }_{I,B_j}$ to ${\sum }_i{a}_i{x}_i=0$. The earlier terms vanish exactly, not just modulo a power of $p$. Divide by $p^{e_j}$; terms with later valuations are still divisible by $p$. Reduction modulo $p$ then yieldscontrary to the choice of $p$. Thus every required span relation holds, and with $I=\varnothing$ the first block sum is zero. This is the finite separating-functional proof of the columns condition; it avoids an unjustified passage from congruences to rational equality.

For sufficiency, assume the columns condition. Choose rational coefficients ${\lambda }_{ij}$, for $j\ge 2$ and $i\in B_1\cup \cdots \cup B_{j-1}$, such thatChoose a positive integer $c$ clearing every denominator, and a positive integer $p\ge \max |c{\lambda }_{ij}|$, taking $p=1$ if the list is empty. By the allowed monochromatic m-p-c set theorem, there are positive generators $z_1,\dots ,z_s$ such that all numbersare positive and have one colour. This is a full m-p-c set; positivity is part of its definition, not a rule discarding negative expressions. For $i\in B_h$, putThese coordinates all belong to that monochromatic m-p-c set. Collecting coefficients of $z_j$ givesThus the Rado solution inside an m-p-c set proves partition regularity and completes the theorem.

To obtain the Finite sums theorem from Rado's theorem, introduce a positive variable $y_F$ for every nonempty $F\subseteq \{1,\dots ,k\}$ and imposePartition the columns of this system into $B_j=\{F:\min F=j\}$. The $B_1$-column sum is zero: in a row indexed by $F$ containing $1$, the contributions of $y_F$ and $y_{\{1\}}$ cancel, and other rows receive neither. For $j>1$, the $B_j$-column sum iswhere $a_F$ is the column of $y_F$. Those columns lie in earlier blocks. Hence the columns condition holds. A monochromatic positive solution gives $x_i=y_{\{i\}}$ and $y_F={\sum }_{i\in F}{x}_i$, provingFor $k=1$ the system has no rows and the conclusion is immediate. The construction is also visible directly in the sufficiency proof: a monochromatic $(k,1,1)$-set contains every nonempty sum of its generators, classified by the smallest chosen index.

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

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.

A rational matrix $A=(a_1\cdots a_n)$ is partition regular over the positive integers if every finite colouring of $\mathbb{N}$ admits a vector $x\in {\mathbb{N}}^n$ whose coordinates all have the same colour and satisfy $Ax=0$. Equal coordinates are allowed. Rado's theorem says that this holds exactly when the columns can be partitioned into nonempty blocks $I_0,\dots ,I_t$ withThis is the columns condition. Clearing an overall denominator lets us prove necessity for an integer matrix.

Suppose the columns condition fails. There are only finitely many ordered partitions of the finite column index set. For each ordered partition, choose a failing block: its sum $b_j$ is outside the rational span of preceding columns, with that span interpreted as zero for the first block. Finite-dimensional linear algebra gives a rational linear functional vanishing on that span but not on $b_j$. For completeness, take a basis of the span, append $b_j$, extend to a basis of the column space, and prescribe values zero on the first basis vectors and one on $b_j$. Clearing the functional's denominators gives an integer row functional $\ell$ with $\ell (b_j)\ne 0$.

Choose one prime $p$ larger than the absolute values of all these finitely many nonzero integers $\ell (b_j)$. Colour each positive integer by its first nonzero base-$p$ digit:Assume a monochromatic solution $x$ exists, and partition its coordinates into blocks of equal P-adic valuation, ordered by increasing valuations $v_0<\cdots <v_t$. Their first nonzero digit is a common $d\ne 0(\mathrm{mod}p)$. For this ordered partition take its chosen failing block $I_j$ and functional $\ell$. Apply $\ell$ to ${\sum }_i{a}_i{x}_i=0$. All preceding-block terms disappear exactly. Divide the remaining integer equation by $p^{v_j}$ and reduce modulo $p$. Later blocks disappear, while the current block givesThis contradicts the prime's choice. Thus the colouring has no monochromatic solution, proving necessity. This is the finite separating-functional proof of the columns condition; it needs neither a limiting argument nor a bound on the valuations themselves.

For sufficiency, suppose the blocks satisfy the columns condition. Choose rational coefficients $q_{ji}$, for $i$ in preceding blocks, such thatLet $c$ be a positive integer clearing their denominators, and choose an integer $P\ge 1$ bounding all $|c{q}_{ji}|$. Use the allowed monochromatic m-p-c set theorem with $t+1$ generators. We use the triangular convention in which the resulting positive integer set contains every numberin one colour. This is an m-p-c set, with the generator indices reversed if the alternative lower-triangular convention is used. The permitted theorem applies to either convention, since it applies to every finite number of generators. Its set lies in $\mathbb{N}$, so all of the displayed combinations are positive.

For $i\in I_j$, defineEach coordinate belongs to that monochromatic set. Summing the column contributions and collecting coefficients of $u_k$ givesFor $k=0$ the inner sum is empty and $b_0=0$. This proves sufficiency by a Rado solution inside an m-p-c set, and completes the theorem.

For the requested application, chooseThe matrix, in the coordinate order $(x,y,z,w)$, has columnsThe block $I_0=\{x,z,w\}$ has zero sum. The remaining column satisfies $a_y=-a_z/3+2a_w$, so it is in the span of the first block. By Rado's theorem, positive monochromatic $x,y,z,w$ satisfy the system. In particular,One can see the positivity directly in the same triangular construction: a monochromatic set containing $3u+\lambda v$ for $|\lambda |\le 6$, together with $3v$, givesAll four are positive because they belong to that positive m-p-c set. This is a partition-regular system forcing an ordered Schur relation.

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 $i$.