Given an integer $n\ge 3$, let $u_1,...,u_n$ be pairwise coprime integers $\ge 2$, $𝒟$ a family of nonempty proper subsets of $\{1,...,n\}$ with “enough” elements, and $\epsilon$ a function $𝒟\to \{\pm 1\}$. Does there exist at least one prime $q$ such that $q$ divides ${\prod }_{i\in I}{u}_i-\epsilon \left(I\right)$ for some $I\in 𝒟$, 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 $\epsilon$ and $𝒟$ are subjected to certain restrictions.We use the result to prove that, if ${\epsilon }_0\in \{\pm 1\}$ and $A$ is a set of three or more primes that contains all prime divisors of any number of the form ${\prod }_{p\in B}p-{\epsilon }_0$ for...

Let $f\left(X\right)\in \mathbb{Q}\left[X\right]$ be a monic polynomial which is a power of a polynomial $g\left(X\right)\in \mathbb{Q}\left[X\right]$ of degree $\mu \ge 2$ and having simple real roots. For given positive integers $d_1,d_2,\ell ,m$ with $\ell \<m$ and gcd$(\ell ,m)=1$ with $\mu \le m+1$ whenever $m\<2$, we show that the equation$$f\left(x\right)f(x+d_1)\cdots f(x+(\ell k-1)d_1)=f\left(y\right)f(y+d_2)\cdots f(y+(mk-1)d_2)$$with $f(x+j{d}_1)\ne 0$ for $0\le j\<\ell k$ has only finitely many solutions in integers $x,y$ and $k\ge 1$ except in the case$$m=\mu =2,\ell =k=d_2=1,f\left(X\right)=g\left(X\right),x=f\left(y\right)+y.$$