Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 130 / 1

Ramsey's theorem for $r$-sets says that every finite coloring of $[\mathbb{N}{]}^{(r)}$ has an infinite monochromatic set. We prove it by mathematical induction on $r$. The case $r=1$ is the infinite pigeonhole principle. Suppose the result holds for $r-1$, and let $c:[\mathbb{N}{]}^{(r)}\to [k]$. Choose $a_1$, then use the induction hypothesis on the coloring $F\mapsto c(F\cup \{a_1\})$ to obtain an infinite set $M_1$ on which this color is constant, say $d_1$. Inductively chooseso that $c(F\cup \{a_j\})=d_j$ for every $F\in [M_j{]}^{(r-1)}$. Some color $d$ occurs for infinitely many $d_j$. If $j_1<j_2<\cdots$ are the corresponding indices, every $r$-set from $\{a_{j_1},a_{j_2},\dots \}$ has color $d$: take its least-indexed element $a_{j_s}$, after which its other $r-1$ elements lie in $M_{j_s}$. This proves the theorem.

Solved by gpt-5.6-sol high.

Suppose the Finite Ramsey theorem failed for fixed positive integers $r,k,m$. For every $n$ choose a $k$-coloring $c_n:[n{]}^{(r)}\to [k]$ with no monochromatic $m$-set. There are only finitely many colorings of $[r{]}^{(r)}$, so an infinite subsequence of the $c_n$ agrees there. Pass to a further infinite subsequence agreeing on $[r+1{]}^{(r)}$, and continue. The diagonal argument produces compatible colorings $d_N:[N{]}^{(r)}\to [k]$ such that $d_{N+1}{|}_{[N{]}^{(r)}}=d_N$ and no $d_N$ has a monochromatic $m$-set.

Define $d(F)=d_N(F)$ whenever $N\ge \max F$. Compatibility makes this a well-defined finite coloring of $[\mathbb{N}{]}^{(r)}$. By Ramsey's theorem it has an infinite monochromatic set, whose first $m$ elements contradict the defining property of a sufficiently large $d_N$. This compactness argument proves the finite statement.

Solved by gpt-5.6-sol high.

Let $c:\mathbb{N}\to [k]$ be the given finite coloring. Color each $t$-element subset $\{y_1,\dots ,y_t\}$ of $\mathbb{N}$ byBy Ramsey's theorem there is an infinite set $Y$ whose $t$-element subsets all receive the same induced color. Enumerate it increasingly as $x_1<x_2<\cdots$. Then every sum $x_{i_1}^2+\cdots +x_{i_t}^2$ with $i_1<\cdots <i_t$ has that color. The argument works for every positive integer $t$; primality is not needed for this part.

Solved by gpt-5.6-sol high.

No. It is enough to take the prime number $t=2$. By the monochromatic sums-and-products obstruction, there is a finite coloring $\chi$ of $\mathbb{N}$ for which no infinite set has all its pairwise sums and pairwise products in one color. Refine $\chi$ by also recording the parity of the 2-adic valuation.

If a sequence $(x_i)$ made both requested families monochromatic, put $y_i=x_i^2$. If the set of distinct $y_i$ were infinite, an injective subsequence would make all pairwise sums $y_i+y_j$ and products $y_i{y}_j$ monochromatic under $\chi$, a contradiction. Otherwise some $y=x_i^2$ occurs infinitely often. Two occurrences give the sum $2y$ and the product $y^2$, butbecause $v_2(y)$ is even. The refining colors differ, another contradiction.

Solved by gpt-5.6-sol high.