Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 130 / 3

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

Given a finite coloring $c:\mathbb{N}\to [k]$, color each three-element subset $\{a<b<d\}$ by the color ofRamsey's theorem gives an infinite set $M$ on whose triples this induced coloring is constant. Enumerating $M$ increasingly as $x_1<x_2<\cdots$ gives a strictly increasing sequence for which every $x_i+2x_j+3x_k$, $i<j<k$, has the same color.

Solved by gpt-5.6-sol high.

No. The incompatibility part of the Milliken–Taylor theorem provides a finite coloring of $\mathbb{N}$ for which the systems associated with the nonproportional compressed coefficient vectors $(1)$ and $(1,2,3)$ cannot be monochromatic in the same color. The first system is the finite-sums setwhile singleton separated blocks in the second include every $x_i+2x_j+3x_k$ with $i<j<k$. Thus a sequence satisfying the proposed union would contradict that finite coloring.

Solved by gpt-5.6-sol high.