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Van der Waerden theorem. Given positive integers $r,k$, there is a finite $W(r,k)$ such that every finite colouring of $\{1,\dots ,W(r,k)\}$ with at most $r$ colours has a monochromatic arithmetic progression of length $k$ and positive common difference. In particular, every finite colouring of the positive integers contains such progressions of every finite length.

Here is a colour-focusing proof of Van der Waerden theorem. We induct on $k$, simultaneously for all numbers of colours. The assertion for $k=1$ is immediate. Suppose $W(r,k)$ exists for every $r$, and fix $r$. We construct bounds $F_t$, for $1\le t\le r$, with this property: every colouring of $[F_t]$ either has a monochromatic progression of length $k+1$, or has $t$ progressions of length $k$, of distinct colours, which share a next point $f\in [F_t]$. Write those progressions aswith positive steps $d_i$ and all displayed points in $[F_t]$.

For $t=1$, put $L=W(r,k)$ and $F_1=2L$. A monochromatic progression of length $k$ in $[L]$ has its next point in $[2L]$, since its step is at most $L$; for $k=1$ choose step $1$. This establishes the initial focusing property.

Suppose $F_t=N$ has been constructed. Put $M=W(r^N,k)$ and $F_{t+1}=2NM$. Divide the first $NM$ integers into $M$ consecutive blocks of size $N$. Colour a block by its full colour profile, an $N$-tuple having at most $r^N$ possible values. The induction hypothesis on $k$ gives block indiceswith the same colour profile. For $k=1$ use any one block and $D=1$. Suppose that the whole interval has no monochromatic progression of length $k+1$. Its repeated profile has $t$ focused length-$k$ progressions with positions $f-h{d}_i$ and distinct colours $q_i$. The colour $q$ of position $f$ differs from every $q_i$, since equality would complete a length-$(k+1)$ progression in the block.

Set $j_{\ast }=j_0+kD$ and $F=N(j_{\ast }-1)+f$. The index $j_{\ast }$ is at most $2M$, so this focus is in the full interval. For $1\le h\le k$, the pointis in one of the chosen blocks and has colour $q_i$. Also $F-hND$ is at position $f$ in that block and has colour $q$. We have produced $t+1$ focused length-$k$ progressions of distinct colours. This proves the recursive focusing property. At $t=r$, the focus itself must have one of the $r$ colours, so one of the progressions extends to length $k+1$. Thus $W(r,k+1)=F_r$ is a valid bound, completing the mathematical induction and the requested proof.

To obtain the equation criterion, we first prove the small strengthening of Van der Waerden theorem that we need. For any positive integers $r,s,L$, there is a finite bound forcing a monochromatic configurationThis is the Brauer progression theorem, and the following argument proves it from the theorem just established. Induct on $r$. One colour is immediate. Given a bound $M$ for $r-1$ colours with the same $s,L$, let $T=W(r,(L-1)M+1)$ and colour an interval at least as long as $\max (T,sMT)$. Find a monochromatic progression $a,a+d,\dots ,a+(L-1)Md$ in $[T]$, of colour $q$; take $d=1$ if this progression has one term. If some $jsd$, $1\le j\le M$, has colour $q$, its first $L$ suitably spaced terms $a,a+jd,\dots ,a+(L-1)jd$ and the point $s(jd)$ give the configuration. Otherwise the colouring of the indices $1,\dots ,M$ obtained from $j\mapsto c(jsd)$ uses at most $r-1$ colours. The induction hypothesis supplies indices $se,b,b+e,\dots ,b+(L-1)e$ of one colour. Multiplication by $sd$ gives the desired configuration with step $esd$. This completes the strengthening without assuming any Rado theorem.

Multiply the rational coefficients by a common positive denominator, so that all $a_i$ are nonzero integers; this does not alter the equation or partition regularity. By definition, the row is partition regular when every finite colouring admits positive monochromatic $x_1,\dots ,x_n$ satisfying ${\sum }_i{a}_i{x}_i=0$. The variables need not be distinct.

For necessity, choose a prime number $p>{\sum }_i|a_i|$. Write each positive $x$ uniquely as $p^{v_p(x)}u$, with $p\nmid u$, and colour $x$ by $u\mathrm{mod}p$. This is a finite colouring with $p-1$ colours. If a monochromatic solution exists, let $v={\min }_i{v}_p(x_i)$ and $I=\{i:v_p(x_i)=v\}$. Dividing the equation by $p^v$ and reducing modulo $p$ giveswhere $u_0\ne 0(\mathrm{mod}p)$ is the common colour. Thus ${\sum }_{i\in I}{a}_i\equiv 0(\mathrm{mod}p)$. Its absolute value is strictly less than $p$, so ${\sum }_{i\in I}{a}_i=0$. The index set $I$ is nonempty. This proves the leading-residue obstruction to partition regularity.

Conversely, suppose ${\sum }_{i\in I}{a}_i=0$ for a nonempty $I$. If $I$ is all indices, taking every variable equal to any one positive integer already gives a monochromatic solution. Otherwise fix $i_0\in I$, put $s=|a_{i_0}|$, $C={\sum }_{j\notin I}{a}_j$, and define integer offsetsLet $u={\min }_{i\in I}{\lambda }_i$ and choose $L=1+{\max }_{i\in I}{\lambda }_i-u$. The proved Brauer progression theorem gives a monochromatic configuration consisting of $sd$ and a length-$L$ progression starting at $b$. SetAll variables are positive and of one colour. Their weighted sum isThis explicitly gives a Brauer configuration for a one-row zero-sum equation. We concludeBoth directions have been proved independently of the general Rado theorem.