This a repost of a question which was asked at MathStackExchange, but got no answer so far, so I am trying here.
Let $n\ge 1$ and let $Q= \{q_1,\dotsc, q_n\}$ be a set of $n$ odd primes, all different and such that $Q \neq \{3\}$.
Show that there is no such set $Q$ such that for every $q_j$ it holds that $2+\prod_{k \neq j} q_k $ is divisible by $q_j$
(with the convention that an empty product is $1$).
Clearly this is true when $n=1$ since in this case, $q_1 \neq 3$.
This is also true for $n=2$ since if $q_1 < q_2$ and $q_1$ divides $q_2+2$, it is not possible that $q_2$ divides $q_1+2$.
For $n=3$, this is shown as follows: suppose that there exist three odd primes $p< q < r$ such that $3\le p \lt q$ and $r\ge 7$ such that $p$ divides $q\cdot r +2$, $q$ divides $p\cdot r +2$ and $r$ divides $p\cdot q +2$, then $p\cdot q$ divides $p\cdot q \cdot r^2 +4 +2r(p+q)$, then there exists an integer $K$ such that $K\cdot p\cdot q=1+r\frac{p+q}{2}$. But $ \frac{1}{p\cdot q}+ \frac{r}{2q}+ \frac{r}{2p} \le \frac{1}{15}+ \frac{r}{10}+ \frac{r}{6}= \frac{1+4r}{15}$. Then $ 1 +r\frac{p+q}{2} \le \frac{1+4r}{15} p\cdot q$ , then $K \le \frac{1+4r}{15}$. On the other hand, $r$ divides $p\cdot q +2$ then $r$ divides $K\cdot p\cdot q+ 2K= 2K+1 + r\frac{p+q}{2}$, then $r$ divides $2K+1 \le \frac{17+8r}{15}$ then $7r \le 17 $ which contradicts $r\ge7$.
It can be shown that a similar proof (showing a contradiction on the size of the largest member of $Q$) works as long as $n\le 8$. The case $n=9$ does not bring a contradiction but a finite limit ($958$) to the size of the largest member of $Q$, then the case $n=9$ can be settled by checking a large but finite number of cases. Hence, should such a set $Q$ exist, it has to contain at least $10$ different prime numbers. Similarly, this method shows that if $3$ is not allowed, $Q$ must contain at least $28$ different prime numbers.
For the context, I have found that a proof of the above statement for any $n$ would imply that the conjecture raised in Trying to prove a power sum divisibility conjecture is true:
For positive integers $a$ and $n$, $2a + 3 \nmid 1^n + 2^n + \dotsb + (a - 1)^n + a^n$.
I would also accept as an answer a disproof of the above claim or the exhibition of such a set $Q$.
