Prime Power Residue and Linear Coverings of Vector Space over ${\mathbb{F}}_q$

Abstract

Let $q$ be an odd prime and $B=\{b_j{\}}_{j=1}^l$ be a finite set of nonzero integers that does not contain a perfect $q^{th}$ power. We show that $B$ has a $q^{th}$ power modulo every prime $p\ne q$ and not dividing ${\prod }_{b\in B}b$ if and only if $B$ corrresponds to a linear hyperplane covering of ${\mathbb{F}}_q^k$. Here, $k$ is the number of distinct prime factors of the $q$-free part of elements of $B$. Consequently: $(i)$ a set $B\subset \mathbb{Z}\setminus \{0\}$ with cardinality less than $q+1$ cannot have a $q^{th}$ power modulo almost every prime unless it contains a perfect $q^{th}$ power and $(ii)$ For every set $B=\{b_j{\}}_{j=1}^l\subset \mathbb{Z}\setminus \{0\}$ and for every $(c_j{)}_{j=1}^l\in ({\mathbb{F}}_q\setminus \{0\}{)}^l$ the set $B$ contains a $q^{th}$ power modulo every prime $p\ne q$ and not dividing ${\prod }_{j=1}^l$ if and only if the set $\{b_j^{c_j}{\}}_{j=1}^l$ does so.

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• Almost No Finite Subset of Integers Contains a $q^{th}$ Power Modulo Almost Every Prime