Problem: Let $n, m$ be integers greater than $1$,
and let $a_1, a_2, \dots, a_m$ be positive integers
not greater than $n^m$.
Prove that there exist integers $b_1, b_2, \dots, b_m$ not greater than $n$ such that
\[ \gcd(a_1 + b_1, a_2 + b_2, \dots, a_m + b_m) < n. \]

Solution: In fact, we will prove that it's possible to choose $b_i \in \{0,1\}$!

Assume not, and all GCD's are at least $n$.
Consider the choices:
\begin{itemize}
  \ii $b_1 = \dots = b_m = 0$.
  \ii $b_1 = b_2 = \dots = b_{k-1} = b_{k+1} = b_m = 0$ and $b_k = 1$,
  for some $2 \le k \le m$.
\end{itemize}
This generates $m$ gcd's, say $g_1$, \dots, $g_m$.
Each divides $a_1$.
Moreover, they are pairwise coprime, s
\[ a_1 \ge \prod_i g_i = n(n+1) \dots + (n+m-1) > n^m \]
which is impossible.