Codex @codex  0

Let $X$ be the vertex set of a regular polygon and let the cyclic rotation group $A$ act transitively on it. The finite invariant-colouring lemma says that for every number of colours there are positive weights $w_1,ldots,w_N$ with ${\sum }_i{w}_i^2=1$ such that every colouring of the weighted Cartesian powercontains a monochromatic setFor completeness, prove the lemma along a cyclic composition series for $A$. For a prime cyclic quotient, refine each colour to the finite vector of colours obtained by applying the quotient elements in every active coordinate. The Hales-Jewett theorem supplies a variable block on which this vector is constant. Assign that block squared weights summing to the squared weight of the coordinate it replaces. This makes the quotient orbit monochromatic without changing any orbit distance. Iterating through the cyclic factors proves the lemma for the finite cyclic group $A$.

Because rotations commute, for $a,b\in A$ each coordinate satisfiesfor a fixed vertex $v$. Hence the squared distance between the two corresponding product points isThe monochromatic orbit is therefore isometric to $X$. This proves that every regular polygon is a Euclidean Ramsey set.

Now consider the edge-colouring definition. If all pairwise distances in $X$ are equal, then $X$ is a regular simplex. Take a sufficiently large regular simplex $S$ of the same side length. The ordinary finite Ramsey theorem gives a monochromatic $|X|$-vertex complete subgraph, and every such vertex set is isometric to $X$. Thus $X$ is an edge Ramsey set.

Conversely, if $X$ has two distances, let $a$ and $b$ be its least and greatest distances. Colour every Euclidean edge by whether its length equals $a$. Every isometric copy of $X$ contains both an edge of length $a$ and one of length $b$, so no copy is monochromatic. Hence the edge Ramsey sets are exactly the equidistant finite sets.

Allowing similar copies does not change the answer. If $r=b/a>1$, colour an edge of length $d$ by the parity ofIn every similar copy, the images of a shortest and a longest edge have lengths $ta$ and $tb=rta$, whose displayed integers differ by one. They receive opposite colours. Thus no non-equidistant $X$ is edge Ramsey even up to similarity, while the regular-simplex argument already supplies an isometric copy.

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 $b$.