Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 130 3 ii Solution Created 2026-10-03 Updated 2026-10-05
First justify the finite bound used in the hint. Fix . If no finite forced a monochromatic positive solution of , consider the rooted tree whose level consists of the solution-free finite colorings of , with restriction as the predecessor map. Every level is nonempty and every vertex has at most children. The König infinity lemma gives an infinite branch, hence a finite coloring of all positive integers with no such solution, contradicting the partition regular matrix hypothesis. This is the compactness bound for partition regularity.
Choose such a and let , the least common multiple. For a given finite coloring of the positive integers, pull it back to byEach is a positive integer in . The defining property of gives of one color under , with . Set . Their colors under agree, andThus reciprocal partition regularity follows. This construction is an involution on the divisors of ; it permits repeated coordinates and never requires a reciprocal of a positive integer to itself be integral without the common scaling factor .
Reciprocal partition regularity 2026-10-05
If a rational matrix is a partition regular matrix, every finite coloring of the positive integers admits monochromatic such thatChoose a compactness bound for partition regularity for the number of colors, put , and pull back the coloring by on . A monochromatic positive solution of gives ; then . The least common multiple guarantees that every is a positive integer.