  Given an integer $n \ge 3$, let $u_1, \ldots, u_n$ be pairwise coprime
integers $\ge 2$, $\mathcal D$ a family of nonempty proper subsets of $\{1,
\ldots, n\}$ with "enough" elements, and $\varepsilon$ a function $ \mathcal D
\to \{\pm 1\}$. Does there exist at least one prime $q$ such that $q$ divides
$\prod_{i \in I} u_i - \varepsilon(I)$ for some $I \in \mathcal D$, but it does
not divide $u_1 \cdots u_n$? We answer this question in the positive when the
$u_i$ are prime powers and $\varepsilon$ and $\mathcal D$ are subjected to
certain restrictions. We use the result to prove that, if $\varepsilon_0 \in
\{\pm 1\}$ and $A$ is a set of three or more primes that contains all prime
divisors of any number of the form $\prod_{p \in B} p - \varepsilon_0$ for
which $B$ is a finite nonempty proper subset of $A$, then $A$ contains all the
primes.
