On Hilbert cubes in certain sets
Given \(k\geq 1\), a set \(H\subset \mathbb{N}\) is called a cube of size \(k\) if there exist \(a>0\) and \(x_1,\dots,x_k\) such that \(H=\{a+\sum_{i=1}^k\varepsilon_ix_i:\varepsilon=0\text{ or }1\}\). If the set of \(x\)'s is infinite the cube is called infinite. If \(\mathcal{A}\subset \mathbb{N}\) and \(n\in\mathbb{N}\) then \(F_{\mathcal{A}}(n)\) denotes the size of the largest Hilbert cube contained in \(\mathcal{A}\cap\{1,\dots,n\}\). Finally, let \(\mathcal{S}\), \(\mathcal{P}\), and \(\mathcal{P}_k\) denote the set of all squares, the set of all primes, and the set of positive integers composed of the primes not exceeding \(k\), respectively. The authors prove that (1) for \(n>n_0\) we have \(F_{\mathcal{S}}(n)<48(\log n)^{1/3}\), (2) for \(\varepsilon>0\), \(n>n_0(\varepsilon)\) we have \((1.1-\varepsilon)\log\log n<F_{\mathcal{P}}(n)<(16+\varepsilon)\log n\), (3) the sets \(\mathcal{P}\), \(\mathcal{P}_k\), \((k\in\mathbb{N},k\geq 2)\) do not contain an infinite cube. To prove (1) the authors are led to study the modular analogue of the problem for a fixed prime \(p\). If \(f(p)\) denotes the size of the largest cube with elements taken modulo \(p\) such that each element is a square modulo \(p\), then for \(\varepsilon>0\), \(p>p_0(\varepsilon)\) we have \((1.1-\varepsilon)\log\log p<f(p)<12 p^{1/4}\).
scientific article; zbMATH DE number 1371467