Rado's theorem

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

The positive-variable system $3x+y=3z$, $x-2y=w$ is partition regular. Its columns for $x,z,w$ sum to zero, and the remaining $y$ column lies in their rational span, satisfying the columns condition. Thus every finite colouring has monochromatic positive $x,y,z,w$ with $2y<x$. A triangular m-p-c set gives the explicit solution $x=3u$, $y=3v$, $z=3u+v$, $w=3u-6v$, with positivity guaranteed by membership in that set.

For nonzero rational coefficients $a_1,\dots ,a_n$, the equationis partition regular exactly when some nonempty subset of the coefficients has sum zero. This is the one-row form of the columns property.

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.

For nonzero integer coefficients $a_i$, choose a prime number $p>{\sum }_i|a_i|$ and colour each positive integer by its nonzero leading residue after removing its power of $p$. In a monochromatic zero-sum solution, divide by the least occurring power of $p$. Reduction modulo $p$ then makes the sum of the coefficients attached to the least-valuation variables vanish modulo $p$. Its absolute value is less than $p$, so that sum vanishes as an integer. This gives the necessity direction of the one-equation partition regularity criterion without using the general Rado theorem.

A rational homogeneous equation ${\sum }_i{a}_i{x}_i=0$ has a monochromatic positive odd solution in every finite colouring of the odd integers exactly when its full coefficient sum is zero. Sufficiency uses a constant odd vector. For necessity, clear denominators and, if the total $S$ is nonzero, colour by odd residue classes modulo $2^K>|S|$. A common odd residue is invertible, so a solution would require $2^K\mid S$, impossible. This is stronger than the zero-sum-subset criterion for unrestricted partition regularity.