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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 $I$.