Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 130 / 2

The Hales-Jewett theorem states that for positive integers $m,k$ there is $N$ such that every $k$-coloring of the words $[m{]}^N$ contains a monochromatic combinatorial line.

Here is the standard focused-line proof. Induct on the alphabet size $m$, the case $m=1$ being immediate. Assume the result for $m-1$ and every number of colors. For $1\le s\le k$, prove inductively that some dimension has the following alternative: either there is a monochromatic combinatorial line, or there are $s$ color-focused lines, meaning that the lines without their common focus are monochromatic in $s$ distinct colors. For $s=1$, restrict to words on $[m-1]$ and use the induction hypothesis on $m$.

For the step from $s-1$ to $s$, let $n$ work for $s-1$ and view a longer word as $(a,b)\in [m{]}^n\times [m{]}^M$. Color each $b\in [m-1{]}^M$ by the complete patternwhich uses at most $k^{m^n}$ colors. Taking $M=HJ(m-1,k^{m^n})$ gives a line on which this entire pattern is constant. Append its missing $m$-letter endpoint. Applying the $s-1$ alternative to the induced coloring of the first block and joining the active coordinate sets produces either a monochromatic line or $s$ lines with one common focus and distinct colors. At $s=k$, the focus has one of the $k$ colors, so it completes the line carrying that color. This proves the theorem.

To deduce the Van der Waerden theorem, let $N=HJ(m,k)$ and color a word $(x_1,\dots ,x_N)\in [m{]}^N$ by the color of $x_1+\cdots +x_N$. On a combinatorial line with active set $I$, these sums area monochromatic arithmetic progression of length $m$.

Solved by gpt-5.6-sol high.

The Strengthened Van der Waerden theorem says that every finite coloring contains, for each prescribed $m$, a monochromatic setWe prove the finite form by induction on the number $k$ of colors. The case $k=1$ is immediate. Let $n$ work for $m$ and $k-1$ colors, and apply the ordinary Van der Waerden theorem to obtain a monochromatic progressionIf one of $d,2d,\dots ,nd$ has the progression's color, say $rd$, thenworks. Otherwise $d,2d,\dots ,nd$ use at most $k-1$ colors. By the induction hypothesis their indices contain $b,b+r,\dots ,b+(m-1)r$ together with $r$ in one color. Multiplying by $d$ yieldswhich is the required progression together with its common difference.

Solved by gpt-5.6-sol high.

Restrict the coloring to the diagonal by setting $d(n)=c(n,n)$. We give the direct focusing proof of a monochromatic three-term arithmetic progression. For each $1\le s\le k$, induction constructs a finite interval in which either there is a monochromatic three-term progression or there are $s$ color-focused two-term progressions. The case $s=1$ is the pigeonhole principle. For the induction step, take sufficiently many equal blocks that two have identical color patterns. Translate the $s-1$ focused pairs in the first block to the second and join corresponding points. These give $s-1$ focused pairs at a translated focus; the pair formed by the old and translated focuses supplies the last one. If its color repeated one of the previous colors, the associated pair and the focus would already form a monochromatic three-term progression. Thus the alternative holds.

At $s=k$, the common focus has one of the $k$ colors and completes the pair of that color, so there are $a,r>0$ with $d(a)=d(a+r)=d(a+2r)$. Thereforeis the required monochromatic two-dimensional progression. This proves the result without invoking the Van der Waerden theorem as a black box.

Solved by gpt-5.6-sol high.

The finite-witness version of part c can be iterated by the standard product argument: color one finite witness by the entire color pattern it induces on a second witness, find a diagonal triple with one pattern, and then find a diagonal triple inside that common pattern. This is the special four-point case of the Gallai theorem for an integer lattice. Apply that theorem toA monochromatic homothetic copy $(a,b)+rS$ is exactlyas required.

Solved by gpt-5.6-sol high.