Showing that it is not possible that for every $q_j$ it holds that $2+\prod_{k \neq j} q_k $ is divisible by $q_j$.

Solution 1:

Your second question can be proven to be true when one uses Dirichlet's theorem on primes in arithmetic progression. Let $q_{1}, q_{2}, ..., q_{n-1}$ be any distinct odd primes. Choosing any prime $p>q_{n-1}$ that satisfies the $n-1$ congruences

$j)\text{ }\text{ }\text{ }2 + p\prod_{k =1, k \neq j}^{n-1}q_{k} \equiv 0 \mod(q_{j})$

for $j = 1,...,n-1$ and setting $p = q_{n}$ solves the lemma. Note that here we used the Chinese remainder theorem; the above system of modular equations has a unique solution modulo $q_{1}q_{2}...q_{n-1}$ that is necessarily coprime to each $q_{j}$. Later we applied Dirichlet's theorem which tells us that any arithmetic sequence $an+b$ with $(a,b) = 1$ contains infinitely many primes.