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Statement

Setting (pp. 487--488). $L_k$ ($k\ge2$) is the lattice of integer points $(x_1,\ldots,x_k)$. A point is visible when its coordinates have no common divisor greater than $1$, and nonvisible otherwise; the origin counts as nonvisible. Visible points are drawn as circles and nonvisible points as crosses. A pattern $P_k$ assigns to each of the $w^k$ points with $1\le x_\lambda\le w$ a circle, a cross, or neither. $P_k$ is realized in $L_k$ when some $(u_1,\ldots,u_k)\in L_k$ makes $(u_1+x_1,\ldots,u_k+x_k)$ visible for every circle $(x_1,\ldots,x_k)$ of $P_k$ and nonvisible for every cross.

Definition (p. 490). For a positive integer $m$, a complete square modulo $m$ is a set of $m^2$ points of $L_2$ forming a complete system of residues modulo $m$: every $(x,y)$ with $0\le x<m$, $0\le y<m$ is congruent modulo $m$, coordinate by coordinate, to exactly one point of the set.

Theorem 1 (p. 490, quoted). "A given pattern $P_2$ can be realized in $L_2$ if and only if the set $C$ of circles in $P_2$ fails to contain a complete square modulo $p$ for every prime $p$."

So the condition concerns the circles alone; the crosses never obstruct a realization (p. 489). A pattern of four circles on a $2\times2$ block cannot be realized, since one of its points has both coordinates even (p. 489).

Corollary 1 (p. 492, quoted). "Every pattern $P_2$ consisting only of crosses can be realized." The condition of Theorem 1 holds vacuously.

The paper also remarks (p. 492) that the construction shows a realizable pattern occurs in $L_2$ infinitely often.

Proof pointer

Pp. 490--492. Necessity: if $C$ contains a complete square modulo $p$, then for every translate some circle lands on a point with both coordinates divisible by $p$. Sufficiency, by the Chinese Remainder Theorem in three steps: for each prime $p\le w$, choose $(u,v)$ modulo $p$ so that no translated circle is $\equiv(0,0)$, using a residue class that $C$ misses (congruences (5)); give each cross $(i,j)$ its own prime $Q(i,j)>w$ and put $(u,v)\equiv(-i,-j)$ modulo it (congruences (6)); then fix $u>0$ and require $v\equiv0$ modulo every prime $q>w$ other than the $Q(i,j)$ dividing one of $u+1,\ldots,u+w$ (congruences (7)), so that no such $q$ divides $v+y$ for $1\le y\le w$.

Read depth

Claims checked: the definitions, Theorem 1, Corollary 1 and the remarks on pp. 489 and 492 were read clause by clause on the page images of the print, and the proof on pp. 490--492 was followed. Nothing here is independently reviewed.

Dependencies

None in the corpus. The proof uses only the Chinese Remainder Theorem.

Source. Fritz Herzog and B. M. Stewart, Patterns of visible and nonvisible lattice points, Amer. Math. Monthly 78 (1971), no. 5, 487--496, doi:10.2307/2317753; the edition read is named on the source card.

Bears on

None of the problem pages directly. Theorem 1 is the criterion from which Corollary 2 follows; that page states the paper's relation to Problem 1212.