Problem 1631

If $n = 2$, the statement holds for any primes $p, q, r$. We show that the result fails for $n \ge 3$. Suppose $n = 3$ and $p < q < r$ are some primes. Let $a_1 = 1$ and $a_2 = r+1$. Consider the numbers $r+jq$, $0 \le j < p$. One of these numbers is divisible by $p$, say $r+1$ or $r+2$. Take $a_3 = r+1$ or $a_3 = r+2$. Then $a_3 - a_1$ is divisible by $p$; $a_3 - a_2$ is divisible by $q$; and $a_2 - a_1$ is divisible by $r$. None of $p, q, r$ divide all the differences. For $n > 3$, consider $a_1, a_2, a_3$ as above; and $a_j = pqr(j-3) + 1$ for $j \ge 4$. We may easily check that each of the differences $a_j - a_k$ is divisible by at least one of $p, q, r$, but there is no single prime among $p, q, r$ which divides all the differences.