No-three-in-line problem on a torus: periodicity

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Problem: Let $\tau_{m,n}$ be the maximum size of a subset of the discrete torus

with no three points on a toric line, where a line is the image modulo $(m,n)$ of a primitive integer line in $\mathbb Z^2$. Let $\sigma_z(n)=\tau_{z,n}$. The conjecture is, with the paper’s convention $\mathbb N=\{1,2,\dots\}$,

If one uses $\mathbb N=\{0,1,2,\dots\}$, the literal $k=0$ case is false, since $\sigma_6(3)=4$. The surrounding table and convention indicate $k\ge1$.

Result: The conjecture is true.

Let $N=3\cdot 2^k$, $k\ge1$. By the Chinese remainder theorem,

For $\varepsilon\in\{0,1\}$, consider the slice of points whose first coordinate has parity $\varepsilon$. Fix any affine line $\ell\subset\mathbb F_3^2$. The set of all torus points with first-coordinate parity $\varepsilon$ and $(x\bmod 3,y\bmod 3)\in\ell$ is itself a toric line in $T_{6,N}$. Hence any no-three-in-line set contains at most two points in each such set.

Therefore, for each parity $\varepsilon$, the multiset of residues $(x\bmod3,y\bmod3)$ arising from selected points in that parity slice has at most two points on every affine line of $\mathbb F_3^2$. Such a multiset has size at most $4$: if a point has multiplicity $2$, no other point may occur; otherwise it is an ordinary cap in $\mathbb F_3^2$, and every cap in the affine plane of order $3$ has size at most $4$. Thus each parity slice contributes at most $4$ points, so

It remains to show $8$ points are attainable. On $T_{6,6}$, the set

has no three collinear. Reducing a line in $T_{6,N}$ modulo $6$ gives a line in $T_{6,6}$, so the same configuration embeds into $T_{6,N}$ for every $N=3\cdot2^k$. Hence

Combining both bounds gives

for all $k\ge1$.

Citation: No prior proof of this conjecture was found in the cited paper; the argument above resolves it directly.