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Several monochromatic arithmetic progressions are color-focused when they have different colors and extend by one further term to the same point. Such focused families give an elementary induction proof of the length-three case of the Van der Waerden theorem.

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.

For a single homogeneous equationwith nonzero integer coefficients, Rado theorem says that it is a partition regular equation exactly whenfor some nonempty $I\subseteq \{1,\dots ,r\}$.

For necessity, suppose no nonempty coefficient sum vanishes. Choose a prime $p$ dividing none of the finitely many nonzero numbers ${\sum }_{i\in I}{c}_i$. Colour $x\in \mathbb{N}$ by the first nonzero base-$p$ digitIf $x_1,\dots ,x_r$ had one colour, divide the equation by the least power of $p$ occurring among them and reduce modulo $p$. The terms of minimum valuation givefor a nonzero $u$, contradicting the choice of $p$.

For sufficiency, reorder so that $c_1+\cdots +c_t=0$. The standard focusing lemma derived from the Van der Waerden theorem says the following: given a finite colouring, a finite monochromatic solution of the first $j$ blocks of the columns condition can be chosen together with a common difference $d$ so that every bounded translate of every chosen entry by a multiple of $d$ retains its colour. To prove the lemma, refine the colour of $a$ to the finite vector $(\chi (a),\chi (2a),\dots ,\chi (Ra))$, apply van der Waerden to a sufficiently long progression in this refined colouring, and take a common multiple of the finitely many resulting coefficients as $d$.

Start with the zero-sum block $\{1,ldots,t\}$, for which equal variables already solve its contribution. Add each remaining coefficient as a singleton block. In one dimension its block sum is a rational multiple of any fixed nonzero earlier coefficient, so the focusing lemma chooses a bounded translate that cancels this new contribution while preserving the common colour. Induction over the remaining indices gives a monochromatic solution of the full equation. This proves the single-equation form of Rado's theorem.

Now let $a_1,ldots,a_n,b$ be positive and put $A=a_1+\cdots +a_n$. If $b=At$, then $x_1=\cdots =x_n=t$ is a monochromatic solution in every colouring. Conversely, every solution has $x_i\le b/a_i$. Give each integer at most $b/{\min }_i{a}_i$ its own colour and colour all larger integers with one extra colour. A monochromatic solution must have all $x_i=t$, so $b=At$. Thus the inhomogeneous equation is partition regular exactly when $A$ divides $b$.

Solved by gpt-5.6-sol high.