  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 $p \neq q$ 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 $\big(c_{j}\big)_{j=1}^{l}
\in\Big(\mathbb{F}_{q}\setminus\{0\}\Big)^{l}$ the set $B$ contains a $q^{th}$
power modulo every prime $p \neq q$ and not dividing $\prod_{j=1}^{l}$ if and
only if the set $\{b_{j}^{c_{j}}\}_{j=1}^{l}$ does so.
