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Statement

Setting (p. 357). $S$ is a finite subset of $\mathbb{R}^n$, and an $S$-walk is a sequence $\{z_i\}$ with $z_{i+1}-z_i\in S$ for all $i$. Theorem 2 opens Section III, "Three dimensional case".

Theorem 2 (p. 360, quoted). "If $S$ is a set of vectors which do not all lie in the same plane, then there exists an infinite $S$-walk in which no $5^{11}+1$ vectors are collinear."

The proof (p. 360) states that it suffices to treat $S=\{i,j,k\}$, the three orthonormal unit vectors, and builds one explicit walk $W$ for that set. So the walk has at most $5^{11}$ points on any line, while by Theorem 1 every infinite walk with lattice steps in the plane has, for each $K$, $K$ collinear points.

After the proof (pp. 362--363). The paper says the bound could be sharpened considerably by the same method, sketches how, and writes of the true maximum number of collinear points in $W$ that it "undoubtedly is three" (p. 363). Lidbetter later found six collinear points of $W$ and proved that $W$ has no 189 collinear points; see the Lidbetter card.

Remark 3 (p. 363, quoted). "Theorem 2 also holds in the case where $S\subset R^2$, provided that there are three elements $e_1$, $e_2$, and $e_3$ of $S$, such that $e_1\times e_2$, $e_2\times e_3$, and $e_3\times e_1$ are linearly independent over the rationals. In other words, the condition that the elements of $S$ be lattice points is necessary for Theorem 1."

The question left open (p. 363, quoted). "The above theorems leave unanswered the question of whether it is possible to have an infinite $S$-walk with no three collinear points for some $S\subset Z^n$ (in particular, can $n=3$?)."

Source. Joseph L. Gerver and L. Thomas Ramsey, On certain sequences of lattice points, Pacific J. Math. 83 (1979), no. 2, 357--363, doi:10.2140/pjm.1979.83.357, as identified on the source card. Theorem 2 is stated on p. 360 and proved on pp. 360--362, with Figures 1 and 2 on p. 361.

Read depth. Claims checked: the statement, Remark 3 and the closing question were read clause by clause on the printed pages. The proof was read but not checked step by step; in particular the configuration claims the paper illustrates by Figures 1 and 2 were not rederived. Nothing here is independently reviewed.

Proof pointer

Pp. 360--362, for $S=\{i,j,k\}$. The walk's step sequence is built in blocks: $A_0=(i)$, and $A_{n+1}$ concatenates seven copies of $A_n$, three unchanged and four transformed by a permutation of the unit vectors, alone or with reversal of order, so that $A_n$ has $7^n$ terms and begins $A_{n+1}$; the infinite walk $W$ is the sequence of partial sums of the limiting step sequence. Projected onto the plane perpendicular to $i+j+k$, each block $z_{7^n\nu},\dots,z_{7^n(\nu+1)}$ of $7^n+1$ consecutive points of $W$ lies in a trapezoid with $60^\circ$ base angles and base proportional to $4^n$, and the seven trapezoids of one order fit together inside a trapezoid of the next order (Figure 1). For indices whose difference lies between $7^n$ and $7^{n+1}$, the coordinate sum of $z_p-z_q$, which is proportional to its component along $i+j+k$, has absolute value $|p-q|$, while the perpendicular component is bounded above and below by multiples of $4^n$. Collinear points have equal ratios of the two components, which confines all index differences among them to within eleven orders, giving at most $7^{11}$ collinear points; since a line meets at most five of the seven sub-trapezoids of a trapezoid, the count drops to $5^{11}$.

Bears on

• Problem 193: the problem asks whether every infinite walk in $\mathbb{Z}^3$ with steps from a finite set must contain three collinear points. Theorem 2 gives, for $S=\{i,j,k\}$, an infinite walk with at most $5^{11}$ points on any line; it does not exclude three collinear points, and the paper's closing question on p. 363 leaves exactly that case open. The problem's negative answer is Cambie and Kalviainen's Theorem 1, a different walk.