TITLE: Prime numbers satisfying a congruence relation
QUESTION [7 upvotes]: Let $p_1, p_2, \ldots, p_n$ prime numbers such that for each $i, 1 \le i \le n$, 
$$\prod\limits_{j \neq i} p_j  \equiv 1 \pmod{p_i}.$$ For example, $2,$ $3$ and $5$ satisfies these conditions. Then, it is true that one of $p_1, p_2, \ldots, p_n$ must be $2$?

REPLY [2 votes]: It is not what you are asking but you can make some conjecture on the density of solution, assuming independence and uniform distribution of congruences. Based on this you can do numerical experiments and compare with the truth.
Let $m=\prod_{j=1}^i p_j$ squarefree, if for some $q$ the set $\{ p_1,\ldots, p_i,q\}$ satisfies your condition then $q$ is the unique $a\in [1,m]$ such that $a \frac{m}{p_j}\equiv 1\bmod p_j$. Such a $a$ always exists, what is rare is that $a$ is prime and $m\equiv 1\bmod a$.
Thus we can estimate the probability that $m$ works as $\approx\frac1{\log m}\times \frac{\log m}m$ where $\frac1{\log m}$ is probability that $a$ is prime and $\frac{\log m}m\approx \Bbb{E}[\frac1a]$ is the probability that $m\equiv 1\bmod a$.
And hence the number of solutions with $m\le M$ is expected to be $$\approx \sum_{m\le M} \frac{|\mu(m)|}{m}\approx C \log M$$