---

Statement

Theorem 3 (p. 3), quoted: "Let $p$ be a sufficiently large prime number. Suppose that $\mathbf g$ is an invertible affine transformation of $\Pi$ such that $\mathbf g-I$ is also invertible. Then for any two–coloring of the plane $\Pi$ and any $a\neq0$ there is a monochromatic triple $\{x,y,z\}$ such that $y=x+s$, $s\in\mathcal S_a$ and $z=x+\mathbf g(s)$."

Here $\Pi=\mathbb F_p\times\mathbb F_p$, $I$ is the identity map, and for $j\neq0$ the sphere is $\mathcal S_j=\{x\in\Pi:\|x\|:=x_1^2+x_2^2=j\}$ (p. 2). No measurability condition arises, since $\Pi$ is finite.

Source. I. D. Shkredov, On some problems of Euclidean Ramsey theory, arXiv:1507.02727v2 (22 July 2015), Theorem 3, p. 3; Lemma 2, pp. 2--3; Corollary 5, p. 5. The copy read is identified in the source digest.

Read depth. Claims checked: the statement and Lemma 2 were read clause by clause on the page images; the proof (pp. 3--4) was read for structure only. Nothing here is independently reviewed.

Proof pointer

Pp. 3--4. Write each color as its density plus a balanced function of mean zero and expand the count of triples $x$, $x+s$, $x+\mathbf g(s)$ with $s\in\mathcal S_a$. The main term is the density cubed times $|\mathcal S|p^2$. The three two-function terms are bounded by $2\sqrt p\,|A|$ through Parseval and the Fourier bounds of Lemma 2, using the invertibility of $\mathbf g$ and of $\mathbf g-I$; the cubic terms of the two colors cancel. Summing over both colors gives a positive count; the paper's last step holds "provided by $p>1000$, say" (p. 4). Not checked here.

Dependencies

• Lemma 2 (pp. 2--3): $|\mathcal S_j|=p+2\theta\sqrt p$ with $|\theta|\leqslant1$, and for all $r\neq0$ the Fourier transforms of $\mathcal S_j$ and of $\mathbf g(\mathcal S_j)$, for any invertible $\mathbf g$, are at most $2\sqrt p$ in absolute value. The paper proves the last bound by Gauss sums and Weil's bound for Kloosterman sums (its [11]) and refers to Iosevich and Koh (its [6], Lemma 2) for the others.

Used by

• Corollary 4 (p. 4).
• Corollary 5 (p. 5), not paged separately: for every sufficiently large prime $p$, every two-coloring of $\Pi$ and every $a,b\neq0$ with $a/b$ a quadratic residue have a monochromatic collinear triple $\{x,y,z\}$ with $\|y-x\|=a$ and $\|z-y\|=b$, where $\|\cdot\|$ is the quadratic form above.

Bears on

• Problem 173: only as an analog over the finite plane $\mathbb F_p\times\mathbb F_p$; it is not a statement about colorings of $\mathbb R^2$.