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For $b>1$ and an integer $q\ge 2$, assign $x\ge 1$ the color $\lfloor {\log }_{b}x\rfloor \mathrm{mod}q$. The floor function partitions the domain into half-open geometric bins. If $1<{\log }_b(y/x)<q-1$, their bin indices differ by an integer in $\{1,\dots ,q-1\}$, so their colors differ. For example, $b=3/2$ and $q=3$ separates every ratio in $[1.9,2]$. Composing this finite coloring with another logarithm can obstruct monochromatic multiplicative configurations whose logarithms have such ratios.

Let ${\sigma }_N=\sqrt{Np(1-p)}$ and use the floor function to chooseInterpret an empty integer interval as an empty sum. These bounds give exactly the event $a<(K_N-Np)/{\sigma }_N\le b$, where $K_N$ has the binomial distribution of parameters $N,p$. Since $0<p<1$, the central limit theorem for a sum of independent Bernoulli distribution variables gives convergence in distribution of that standardized count to the standard normal distribution. Its cumulative distribution function is continuous at both endpoints, soThe TeX's additional malformed inequality involving $p/k_a(N)$ and $p/k_b(N)$ is absent from the PDF and is not a condition on the requested integers.

Use the cyclic logarithmic coloringHere the floor function places $x$ in the unique half-open real interval $[(3/2{)}^j,(3/2{)}^{j+1})$. If $y/x\in [1.9,2]$, thenbecause $3/2<1.9\le 2<(3/2{)}^2$. For any real $t$ and $1<u<2$, $\lfloor t+u\rfloor -\lfloor t\rfloor$ is either $1$ or $2$. Applying this to the logarithms of $x$ and $y$ shows that their bin indices differ by $1$ or $2$, and hence have different residues in modular arithmetic modulo $3$. This proves the required separation, including both endpoints of the prescribed ratio interval. The half-open bins remove any ambiguity at their boundaries.

Fix a number $k$ of colors. For a two-point set whose points are distance $d$ apart, take the vertices of a regular simplex with $k+1$ vertices and side length $d$. The pigeonhole principle gives two vertices of one color, and they form the required congruent copy. Thus every two-point set, equivalently every line segment, is a Euclidean Ramsey set.

For an equilateral triangle of side length $d$, take a regular simplex with $2k+1$ vertices and side length $d$. The pigeonhole principle gives three vertices of one color, and every three vertices of a regular simplex form an equilateral triangle. Hence every equilateral triangle is Euclidean Ramsey.

To prove the product theorem for Euclidean Ramsey sets, let $S_X$ be a finite Ramsey witness for $X$ under $k$ colors. There are at most $k^{|S_X|}$ possible color patterns on $S_X$. Choose a finite Ramsey witness $S_Y$ for $Y$ under that many colors. Given a $k$-coloring of $S_X\times S_Y$, color each $y\in S_Y$ by the complete patternThere is a copy $Y^{\prime }\cong Y$ on which this pattern is constant. The common pattern on $S_X$ contains a monochromatic copy $X^{\prime }\cong X$. Every point of $X^{\prime }\times Y^{\prime }$ then has the same original color, and the orthogonal product is congruent to $X\times Y$.

A rectangle is the Cartesian product of two line segments, so it is Euclidean Ramsey. Three suitable vertices of a rectangle form a right triangle; any subset of a monochromatic set is monochromatic. Thus every right triangle is Euclidean Ramsey.

It remains to show that the collinear set $\{0,1,2\}$ behaves differently. In every ${\mathbb{R}}^m$, use the finite coloringA congruent copy has the form $a-v,a,a+v$ with $\Vert v\Vert =1$ for the Euclidean norm. The parallelogram law givesPut $n_{+}=\lfloor 2\Vert a+v{\Vert }^2\rfloor$, $n_{-}=\lfloor 2\Vert a-v{\Vert }^2\rfloor$, and $n_0=\lfloor 2\Vert a{\Vert }^2\rfloor$. The errors introduced by the three floor functions show thatIf all three points had one color, the integer in the middle would be divisible by ten, which is impossible. This proves the three-term unit arithmetic progression is not Euclidean Ramsey assertion.

Write the Chebyshev theta function asTaking natural logarithms in part a givesand henceBecause $\sqrt{n}\log n/n\to 0$, the right-hand side is at least $n/2$ for every sufficiently large integer $n$.

Now let $x$ be sufficiently large and set $n=\lfloor x/2\rfloor$ using the floor function. Then $2n\le x$ and $n\ge x/3$, so the monotonicity of $\vartheta$ gives