Integer Tiles with Unsupported Cyclotomic Divisors at Three Primes

Jeff Pickhardt

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Integer Tiles with Unsupported Cyclotomic Divisors at Three Primes

Abstract

Let A(X) = a∈A Xa denote the mask polynomial of a finite subset of a cyclic group. Kiss, Łaba, Marshall, and Somlai asked whether a set satisfying the Coven–Meyerowitz conditions (T1) and (T2) can have an unsupported cyclotomic divisor. They proved that this is impossible when the modulus has at most two distinct prime divisors and constructed an example with four distinct prime divisors. We close the intervening three-prime case. We construct an explicit set A⊂Z810000, |A|= 27000, which tiles Z810000, satisfies both (T1) and (T2), and has the unsupported divisor Φ30. Its prime-power cyclotomic divisors are exactly Φ2α , Φ3α , Φ5α , α∈{2,3,4}, whereas Φ2, Φ3, and Φ5 do not divide A(X). The construction is not tied to {2,3,5}. For every triple of distinct primes p < q < r, and every split point c satisfying two divisibility conditions that are always solvable, the same recipe produces such a tile in Z(pqr)4 (Theorem 11.1); the set above is the case (p,q,r) = (2,3,5). The answer to the three-prime question is therefore positive at every modulus of that form, matching the generality of the four-prime family it sharpens, and it has been verified by machine for (2,3,5), (2,3,7), (2,5,7) and the all-odd triple (3,5,7). The construction therefore sharpens the previous four-prime example and shows that unsupported-divisor exclusion cannot prove the Coven–Meyerowitz conjecture in the three-prime regime, even when the set under consideration is itself a genuine tile satisfying (T2). The exhibited tiling complement also satisfies (T2), so this construction is not a counterexample to the Coven–Meyerowitz conjecture and does not resolve either direction of Fuglede’s conjecture in dimension one. We further classify all tiling complements of A. Every one of them is a product in the first two coordinates, and consequently meets each residue class modulo 30 exactly once; hence every complement of A satisfies (T1) and (T2). The example therefore cannot be strengthened into a Coven–Meyerowitz counterexample by any choice of complement. This is a complement-sensitive obstruction: one-factor cyclotomic data permits the unsupported divisor, while the joint structure of the tiling forces (T2) on the other factor.