Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 10 / 2

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.

The two versions of partition regularity are equivalent. If the row is partition regular on all positive integers, pull a colouring of the even positive integers back by $n\mapsto 2n$. A monochromatic solution $\mathbf{y}$ for this pulled-back colouring gives the even solution $\mathbf{x}=2\mathbf{y}$ in the original colouring, since the equation is homogeneous.

Conversely, restrict any colouring of all positive integers to its even members. A monochromatic solution supplied by regularity on the even integers is already a valid solution in the full domain. This argument is an instance of scaling invariance of homogeneous partition regularity and does not require a new coefficient criterion.

Again use integer coefficients after clearing denominators and put $S={\sum }_i{a}_i$. If $S=0$, choose any odd positive integer $u$ and take all $x_i=u$. This gives a monochromatic solution in any colouring.

For necessity, suppose $S\ne 0$ and choose $K$ with $2^K>|S|$. Colour the odd positive integers by their residue classes modulo $2^K$. In any monochromatic candidate, $x_i\equiv u(\mathrm{mod}{2}^K)$ for a common odd $u$. The equation would imply $uS\equiv 0(\mathrm{mod}{2}^K)$. Since $u$ is coprime to $2^K$, this forces $2^K\mid S$, impossible for nonzero $S$ with $|S|<2^K$. Thus $\boxed{\text{regular over odd positive integers}⟺\sum _i{a}_i=0}$. This is one-equation partition regularity over odd integers; its obstruction concerns the full coefficient sum rather than just a zero-sum subset.