Weighted Zero-Sum Problem With Quadratic Residues

Given a ring \(R\) and a subset \(A \subseteq R\), the
\(A\)-weighted Davenport constant is the least integer \(\mathsf{D}_A(R)\) such
that any sequence of terms from \(R\) of length \(\mathsf{D}_A(R)\) has a
nontrivial subsequence \(g_1 \cdots g_\ell\), where the \(g_i\) are terms of the
subsequence, such that \(0 = a_1 g_1 + \cdots + a_\ell g_\ell\) for some
\(a_i \in A\).

Let \(R = \mathbb{Z}/n\mathbb{Z}\) for an integer \(n \ge 2\), and let
\(U_n^2 = \{u^2 : u \in U_n\}\) be the set of all squares of invertible units.
It is proved that the weighted Davenport constant
\(\mathsf{D}_{U_n^2}(\mathbb{Z}/n\mathbb{Z})\) is equal to \(2\Omega(n) + 1\)
when \(\gcd(n, 10) = 1\) or \(\gcd(n, 6) = 1\), extending a recent result of
Chintamani and Moriya [CM] and another of Adhikari, David and Jiménez Urroz
[ADJ]. Indeed, we show that
\[
\mathsf{D}_{U_n^2}(\mathbb{Z}/n\mathbb{Z}) = 2\Omega(n) + \min\{v_5(n), v_3(n)\} + 1
\]
for odd \(n\) with either \(v_3(n) = 0\) or \(v_5(n) = 0\). As part of the
proof, we show how certain sequences of terms from an abelian group can be
used to create a pairwise balanced design with \(\lambda = 1\).